TITLE: Rachinsky quintets QUESTION [45 upvotes]: This 1895 painting of Nikolai Bogdanov-Belsky shows mental calculations in the public school of Sergei Rachinsky. Boys in a Russian village school try to calculate $(10^2+11^2+12^2+13^2+14^2)/365$ in their heads. One of the methods of solution is based on the equality $10^2+11^2+12^2=13^2+14^2$. Now this Rachinsky equality can be considered as a generalization of the well-known Pythagorean triple (3,4,5), $3^2+4^2=5^2$, and in analogy with the Pythagorean triples one can define Rachinsky quintets as a set of five positive integers $(a,b,c,d,e)$ such that $a^2+b^2+c^2=d^2+e^2$. It is known that all primitive Pythagorean triples $(a,b,c)$ such that $a^2+b^2=c^2$ are generated by Euclid's formula $a=m^2-n^2$, $b=2mn$, $c=m^2+n^2$, where $m$ and $n$ are positive integers such that $m>n$, $m$ and $n$ are coprime, and $m \not\equiv n \bmod 2$. Can one establish an analogous result for Rachinsky quintets? REPLY [2 votes]: The following recipe (algorithm) generates all solutions. It may be viewed as a parametrization in a general(ized) sense. W.l.o.g. we may assume that $c$ is odd. Then $$ \left(\frac{x-y}2\right)^2 + \left(\frac{u-v}2\right)^2 + c^2 \ =\ \left(\frac{x+y}2\right)^2 + \left(\frac{u+v}2\right)^2 $$ where three conditions hold: $\ x\equiv y\equiv 1\ \mbox{mod}\ 2$ $\ u\equiv v\equiv 0\ \mbox{mod}\ 2$ $\ u\cdot v = c^2-x\cdot y$ i.e. we may take arbitrary $x$ and $y$ as in condition 1, and then one decomposes $c^2-x\cdot y$ (see condition 3), where $\ u\ v\ $ are as in condition 2; of course $\ 4\,|\,c^2-x\cdot y\ $ (and the expressions under the squares are integers).<|endoftext|> TITLE: Which groups are the unitary group of a $C^*$-algebra QUESTION [18 upvotes]: Which groups are the unitary group of a $C^*$-algebra? Does anyone know anything in this direction? REPLY [2 votes]: Here is a partial answer: for any given group $G$ there exists a C*-algebra $A$ whose unitary group contains $G$ as a subgroup. Namely, take $A$ to be the (full) group C*-algebra $C^*(G)$ of $G$. This follows from the fact that the functor $U \colon \mathbf{Cstar} \to \mathbf{Grp}$ taking unitary groups is right adjoint to the functor $C^* \colon \mathbf{Grp} \to \mathbf{Cstar}$ taking (full) group C*-algebras. Moreover, both functors are faithful and reflect isomorphisms. See Nassopoulos.<|endoftext|> TITLE: Submission of papers to ArXiv or similar QUESTION [5 upvotes]: This is an extension of this question and this question on MathStackExchange. I have developed a formula for almost primes which is far more accurate asymptotically than Landau's well known $$\pi_k(n) \sim \left( \frac{n}{\log n} \right) \frac{(\log\log n)^{k-1}}{(k - 1)!}$$ and I have a paper that I would like to submit to arXiv.org, whilst it is in the process of being reviewed (which I realise can take a very long time). However, the submission guidelines say that it must be endorsed by someone currently affiliated with ArXiv. I have heard that it is possible to publish on vixra.org, but that it is not really advisable. I don't know whether this is an appropriate question to ask, but users on MathStackExchange have suggested that I ask here whether anyone would be willing to endorse a publication of the paper on ArXiv. I am not affiliated with any university - I just have a (an over?!) healthy interest in the subject! Unfortunately, I do not know of anyone in the field, and am a little wary of sending the paper to people I do not know. REPLY [12 votes]: I don't know anything about almost primes, but I think you're mistaken not to send your paper to people in the field. If you're afraid of someone stealing your idea, I wouldn't worry --- having already submitted it to a journal establishes your priority (besides the fact that having your idea stolen seems very unlikely in this situation). A much more likely outcome is that your paper won't be read. However, if you do the research to find people who would be interested in this exact problem --- which, if you have internet, should be easy --- you increase the chance of being read. Being polite and asking for suggestions could also be more likely to produce a helpful response. Most professional mathematicians are not interested in stealing your ideas, but are very interested in encouraging others who share similar interests. Approach others with this in mind.<|endoftext|> TITLE: The example of mechanical system that has a Mobius strip as their configuration space QUESTION [9 upvotes]: Can you give examples for mechanical system that has a Mobius strip as their configuration space? REPLY [15 votes]: Imagine a rod of length d confined inside a spherical shell of diameter d and free to rotate within it. Provided the two ends of the rod are indistinguishable and the rod is cylindrically symmetric, the configuration space of the rod is the real projective plane (for an application of this sort of picture in nature, see this answer). If you glue a block to the inside surface of the sphere, this confines the rod further and now the configuration space is the Mobius strip. REPLY [14 votes]: Here is an example in which the Mobius strip is also physically visible. Namely, the old eight-track tape system, on which I listened to the Cars and Led Zeppelin as a teenager, has an endless tape with one twist, giving the basic tape-position configuration space the nature of a Mobius strip. (The configuration of the head position keeps track of an additional four positions across the tape.)<|endoftext|> TITLE: Is "stackiness" transitive? (and a couple other basic questions about stacks) QUESTION [6 upvotes]: Say, $B$ is a category fibered in groupoids over some category $C$, and $A$ is a category fibered in groupoids over $B$. Suppose $A$ is a stack (over whatever site) over $B$, and $B$ is a stack over $C$, is $A$ also a stack over $C$? "Conversely", if $A$ is a stack over $C$ and $B$ is a stack over $C$, then must $A$ be a stack over $B$? Also, do properties like being algebraic, having a coarse moduli space, etc, depend on what you consider the base category to be? (Ie, if $A$ as a stack over $B$ is algebraic/has a CMS, then must $A$ as a stack over $C$ be algebraic/have a CMS as well?) REPLY [10 votes]: OK, I think the first and second question are not sufficiently precise. I am going to assume you mean: C is a site and B is endowed with the topology inherited from C. In this case the Stacks project contains a lemma stating that the answer to your second question is "yes". See Lemma Tag 06NX. Strangely, the statement that the composition A --> B --> C turns A into a fibred category over C if A --> B and B --> C define fibred categories (and similarly for categories fibred in groupoids and for stacks with topologies as above) seems to be missing from the Stacks project. I believe these are true, easier to prove than the lemma above, and they should be added there (now done). An example of a statement of the kind you seem to be asking for in your final paragraph, in the setting of algebraic stacks, is Lemma Tag 05XY. Enjoy!<|endoftext|> TITLE: The Arnold – Serre debate QUESTION [66 upvotes]: I have read (but I cannot now find where) that V. I. Arnold & J.-P. Serre had a public debate on the value of Bourbaki. Does anyone have more details, or remember or know what was said? REPLY [9 votes]: In retrospect it does not seem that Arnold's 2+5=5+2 comment was that effective particularly in view of the information provided by Joël. On the other hand, there is one aspect of this matter that did not come out sufficiently in the earlier answers. Namely, Arnold is a fabulous pedagogue. To give a quick example, his book Mathematical methods of classical mechanics has benefited thousands, if not millions, of readers. Many people share his doubts about the effectiveness of the Bourbaki method. Verbal excesses of course cannot be justified but there is a genuine issue there nonetheless.<|endoftext|> TITLE: Can I express any odd number with a power of two minus a prime? QUESTION [15 upvotes]: I have been running a computer program trying to see if I can represent any odd number in the form of $$2^a - b$$ With b as a prime number. I have seen an earlier proof about Cohen and Selfridge regarding odd numbers that are nether a sum or a difference of a power of two and a prime, and I was curious to see if anyone has found an odd number that couldn't be represented using the above formula. REPLY [14 votes]: Erdos's argument essentially gives that there is an infinite arithmetic progression of odd numbers containing no element of the form power of two minus a prime. Define \begin{array}{cccc} i & a_i & m_i & p_i \\ 1 & 1 & 2 & 3 \\ 2 & 2 & 4 & 5 \\ 3 & 4 & 8 & 17 \\ 4 & 8 & 16 & 257 \\ 5 & 16 & 32 & 65537 \\ 6 & 0 & 64 & 641 \\ 7 & 32 & 64 & 6700417 \\ \end{array} Set $A_0=\{x:x\equiv 1 \pmod{8}\}$, $A_i=\{x:x\equiv 2^{a_i} \pmod{p_i}\}$ ($1\leq i\leq 7$), $B=A_0\cap A_1\cap\cdots\cap A_7$. $B$ is an AP consisting of odd numbers. Assume that $x\in B$ and $x=2^n-q$ for some odd prime $q$. For some $1\leq i\leq 7$, we have $n\equiv a_i \pmod{m_i}$, therefore $x\equiv 2^{a_i}-q \pmod{p_i}$. As $x\in A_i$, we also have $x\equiv 2^{a_i} \pmod{p_i}$, consequently $q=p_i$. But this is impossible as $2^n\equiv 0 \pmod{8}$ for $n\geq 3$, $p_i\equiv 1,3,5 \pmod{8}$ and as $x\in A_0$, $x\equiv 7 \pmod{8}$.<|endoftext|> TITLE: $\mu$-invariant and Pontryagin dual of Selmer group of elliptic curves 1 QUESTION [5 upvotes]: 1) What are the examples of elliptic curves over $\mathbb{Q}$ with good reduction and $\mu$-invariant $\geq 2$ at $p = 3$ and how to find them $?$ 2) Let $\Lambda = \mathbb{Z}_{p}[[T]] $ and $ K=\mathbb{Q}_{\infty} $ be the cyclotomic $ \mathbb{Z}_{p} $-extension of $ \mathbb{Q} $. Then the Pontrjagin dual $ X_{E}(\mathbb{Q}_{\infty}) $ of $ Sel_{E}(\mathbb{Q}_{\infty})_{p} $ is a finitely generated torsion $ \Lambda $-module and one has a pseudo-isomorphism $$ X_{E}(\mathbb{Q}_{\infty}) \sim (\bigoplus_{i=1}^{s}\Lambda/(f_{i}(T)^{a_{i}}))\bigoplus(\bigoplus_{j=1}^{t}\Lambda/(p^{\mu_j})) $$ Are there any known methods for computing these decompositions for elliptic curves over $\mathbb{Q}$ $?$ REPLY [8 votes]: 1) 91b3 is an example with $\mu=2$ at $p=3$. Recall that the $\mu$-invariant (with respect to a prime $p$) only changes when there is an isogeny of degree $p$. More precisely it changes just be the quotient of the real Néron periods. Hence we are looking here for curves with cyclic isogenies of degree 9 defined over $\mathbb{Q}$ whose kernel is in $E(\mathbb{R})$. I guess one could parametrise them. 2) You have to assume that the reduction of $E$ is ordinary if you want the Selmer group to be torsion. Using the (know proven) main conjecture, one can get the characteristic series of $X_E$ to any precision (using sage, magma, ... all based on Pollacks earlier work for his tables). In practice this allows one to determine the decomposition very often. First note that there is (conjecturally) a curve in the isogeny class for which the $\mu$-invariant is zero. Better work with this one as you then know that $\mu=0$ by computing sufficiently many coefficients. Now the only trouble could come from repeated factors. Of course very often the characteristic series is a square-free polynomial. Conjecturally again the entries $a_i$ should all be zero, but it may be hard to prove that. Usually however, some information about the curve in the first few layers of the $\mathbb{Z}_p$-extension will suffice. REPLY [2 votes]: Regarding 1), Robert Pollack's tables - http://math.bu.edu/people/rpollack/Data/data.html - will be probably useful.<|endoftext|> TITLE: A question about the first Cohen model QUESTION [5 upvotes]: Consider the first Cohen model, i.e. let $M$ be a countable transitive model of ZFC + $V=L$, let $\mathbb P$ be the poset consisting of finite partial functions from $\omega\times\omega$ to $2$, let $M[G]$ the generic extension adding a sequence $\{a_n\}_{n\in\omega}$ of generic subsets of $\omega$, and let $M\subset N\subset M[G]$ the symmetric model consisting of the sets $\sigma_G$, where $\sigma\in SM^\mathbb P$ is a hereditarily symmetric name, where $\sigma$ is symmetric if there exists a finite set $x\subset\omega$ such that each permutation $f$ of $\omega$ fixing $x$ induces an automorphism of $\mathbb P$ which induces a map $M^\mathbb P\longrightarrow M^\mathbb P$ that fixes $\sigma$. Now, the ultrafilter $G$ does not belong to $N$, and so, the map $SM^\mathbb P\longrightarrow N$ given by $\sigma\mapsto \sigma_G$ is not definable in $N$. However, for each finite $x\subset \omega$, we can consider the set (class in $M$) $SM^\mathbb P_x$ of all $\sigma\in SM^\mathbb P$ fixed by all permutations fixing $x$, as well as $N_x=\{\sigma_G\mid \sigma\in SM^\mathbb P_x\}$. I guess that the map $SM^\mathbb P_x\longrightarrow N_x$ given by $\sigma\mapsto \sigma_G$ should be definable in $N$ from the set $A=\{a_n\mid n\in\omega\}\in N$, the map $f_x:x\longrightarrow N$ given by $f_x(n)=a_n$ (notice that $f_x\in N$) and maybe other parameters in $M$. My question is if this is true, and the motivation is that I am trying to simplify the exposition of Halpern and Lévy "The Boolean Prime Ideal Theorem does not Imply the Axiom of Choice" by showing that the Cohen model is a model of the theory SP described there, but avoiding the formal languaje used to formulate it. I see that all the axioms have a clear translation in standard terms of forcing but the last one, which asserts more or less what I am trying to prove. Anyway, do you know if something like this is already done? REPLY [5 votes]: While not an answer to your question directly, here is some information which you might consider useful. (See edit for that part.) In Jech The Axiom of Choice he points out that the proof of Halpern-Levy is much more difficult than the proof of Halpern that $\sf BPI$ holds in the Mostowski permutation model. He points out that it seems impossible to escape the technically difficult Halpern-Lauchli lemma. Chapter 5 includes a proof that in Cohen's first model the order principle holds, and a similar construction is given. You might want to look at that. Flegner's book Models of ZF Set Theory also has a section devoted for this proof. However the book itself was written when forcing was still being developed, and it's often not as readable as other modern sources, making it less ideal source for learning actual proofs. I don't recall any other major books about the axiom of choice where the proof is presented. But one perhaps can give a whole other proof using the following information. We say that $M\models\mathsf{SVC}(s)$ if for every $x\in M$ there is (in $M$) an ordinal $\alpha$ such that $M$ knows about a surjection from $s\times\alpha$ onto $x$. Every model of $\sf AC$ is a model of $\sf SVC(\{\varnothing\})$, but symmetric extensions are always models of $\sf SVC$ for some set. In the paper where the axiom was suggested by Andreas Blass the following theorem is given: Suppose that $M$ satisfies $\mathsf{SVC}(s)$, then $M$ satisfies $\sf BPI$ if and only if there exists an ultrafilter $U$ on $s^{<\omega}$ such that for every $x\in s$, $\{p\in s^{<\omega}\mid s\in\operatorname{range}(p)\}\in U$. It might be worth pursuing this approach, if one just wants to show the consistency of $\sf\lnot AC+BPI$. Note that if $D$ is any class of hereditarily symmetric names such that for some finite $E\subseteq\omega$ every permutation fixing all the elements in $E$ fixing all the names in $D$, then the following class name is also symmetric: $$f=\{(1,\langle(\dot x)^\check{},\dot x\rangle^\bullet)\mid \dot x\in D\}$$ (Where the $\bullet$ indicates some canonical encoding of the name for the ordered pair from the names appearing in it) To see this, note that if $\pi$ is a function fixing $E$ (and thus all the names in $\dot D$) then we have $$\begin{align} \pi f &=\{(\pi1,\pi\langle(\dot x)^\check{},\dot x\rangle^\bullet)\mid \dot x\in D\}\\ &=\{(1,\langle\pi(\dot x)^\check{},\pi\dot x\rangle^\bullet)\mid \dot x\in D\}\\ &=\{(1,\langle(\dot x)^\check{},\dot x\rangle^\bullet)\mid \dot x\in D\}\\ &=f \end{align}$$ Now we have that $f_\sigma(\dot x)=\dot x_\sigma$, for every $\dot x\in D$.<|endoftext|> TITLE: Sums of primes that are themselves prime QUESTION [19 upvotes]: I'm not a math expert so this may be a trivial question; if $p_i$ is the $i$-th prime, let: $$S(n) = \sum_{i=1}^n p_i$$ be the sum of the first $n$ primes and $$P(n) = | \{1 \leq i \leq n \mid S(i) \mbox{ is prime} \} | $$ be the number of the sums $S(i), 1 \leq i\ \leq n$ that are prime. Do we have $$\lim_{n \to \infty} \frac{P(n)}{n} = 0\;?$$ Where can I find a proof? EDIT: I generated the sequence $P(n)$ and found it on OEIS: Numbers n such that n is prime and is equal to the sum of the first k primes for some k., but there is not a lot of information there. REPLY [3 votes]: I wonder if for questions of this type, the following probabilistic paradigm is useful: Instead of adding all prime numbers, let us "flip a coin" and include the next prime number with a fixed probability $0 TITLE: Different definitions of the rank of a module QUESTION [13 upvotes]: I have seen different definitions of a rank of a module $M$ over a commutative ring $R$. Here in nLab, for quite general modules, the rank is defined locally at $p\in \mathrm{Spec}(R)$ as the dimension over the residue field $\kappa(p)$ of the $\kappa(p)$ vector space $M_p/pM_p$. When $M$ is of finite type, this is the local minimum of generators of $M_p$ over $R_p$. The usual definition of the rank of a module when this module is projective of finite type is locally the rank of the free module $M_p$ over $R_p$. (see mathoverflow.net/questions/29993 for instance). This function rank is locally constant, and therefore when $R$ is a domain, is the dimension of the vector space $M\otimes K$ over $K=R_0$. When $R$ is a domain, Matsumura defines the rank of a quite general module $M$ by the maximum number of independent elements, that is the dimension of the vector space $M\otimes K$ over $K=R_0$. Clearly, all these definitions coincides when $M$ is projective of finite type. When $R$ is a domain, and $M$ only of finite type, I see no reason why they should outside of the ideal $0$. I mean, in that case, to be projective for $M$ is the same as to be locally free. But if $M$ is not projective, the local minimum of generators of $M_p$ (definition 1) does not necessarily coincides with the global maximum of independent elements (definition 3), right? Did I misunderstand something ? If I didn't, which definition should be used? And in the case of $M$ is only of finite type? EDIT: In case the ring $R$ is noetherian and $M$ is of finite type, we have the following generalization of Matsumura's definition, if $Q$ is the total ring of fraction of $R$, with the equivalence of all the following statements: $M$ is of rank $r$. $M\otimes Q$ is a free rank $r$ module over $Q$. The maximum number of independent elements of $M$ is $r$. There exists a submodule $N$ of $M$ such that $M/N$ is torsion. For all prime $p$ associated to $R$, $M_p$ is free of rank $r$ over $R_p$. I like this last definition because this is the most global one when $R$ is not a domain (provided the module actually have a finite rank ...), and it is additive on exact sequence when two of the terms actually have finite rank. REPLY [12 votes]: For a finitely generated module $M$ over a commutative ring $R$, the first definition gives a rank function $r_M: \operatorname{Spec} R \rightarrow \mathbb{N}$, whereas the third definition gives $r_M((0))$. When $M$ is projective, $r_M$ is locally constant, so when $\operatorname{Spec} R$ is connected -- so when $R$ is a domain -- $r_M$ is constant and may be identified with its value at $(0)$. I think you are asking for reassurance that for finitely generated non-projective modules over a domain, the rank function need not be constant. That's certainly true: take $\mathbb{Z}/p\mathbb{Z}$ over $\mathbb{Z}$ (or even over $\mathbb{Z}_p$): then the rank function is $1$ at $(p)$ and otherwise $0$, so the first definition really is not the same as the third. Is this a problem? I don't think so. Asking for terminology in mathematics to be globally consistent seems to be asking for too much: even more basic and central terms like "ring" and "manifold" do not have completely consistent definitions across all the mathematical literature: rather, they overlap enough to carry a common idea. That is certainly the case here. Similarly, asking which of 1) and 3) is "right" doesn't seem so fruitful. It is true that the first definition records more information than the third definition. But it's just terminology, and it is often useful to have a term which records exactly the information in the third definition: e.g. the "rank of an abelian group" is a very standard and useful notion, and usually often it means 3). (I am a number theorist, and in number theoretic contexts this definition of rank is quite standard. Apparently it is less so elsewhere...) For a finitely generated module over a PID, of course the rank in the sense of 3) is exactly what you need in addition to the torsion subgroup in order to reconstruct the module. In this context the rank function 1) gives some information about the torsion but not complete information -- e.g. $\mathbb{Z}/p\mathbb{Z}$ and $\mathbb{Z}/p^2\mathbb{Z}$ have the same rank function -- so the rank function as in 1) does not seem especially natural or useful. I am (almost) sure there are other contexts where it would be natural and useful to think in terms of the rank function as in 1). Added: Following quid's comments I checked out Fuchs's text on infinite abelian groups, and indeed the rank of an arbitrary abelian group is defined in a way so as to take $p$-primary torsion into account. (I simply didn't know this was true.) Thus the rank defined there would be the sum over all the values of the rank function in the sense of 1). All this seems to indicate, even more than my answer, that different notions of rank proliferate. Added Later: Quid also suggests consideration of the quantity $\operatorname{mg}(M)$, the minimal number of generators of an $R$-module $M$: this is a cardinal invariant which is (clearly) finite if and only if $M$ is finitely generated. Let $R$ be a Dedekind domain with fraction field $K$. For a maximal ideal $\mathfrak{p}$ of $R$ and a finitely generated $R$-module $M$, let $M[\mathfrak{p}^{\infty}]$ be the submodule of $M$ consisting of elements annihilated by some power of $\mathfrak{p}$. Then $M[\mathfrak{p}^{\infty}]$ is a finitely generated torsion module over the DVR $R_{\mathfrak{p}}$ and is thus a direct sum of $\operatorname{mg}(M[\mathfrak{p}^{\infty}])$ copies of $R_{\mathfrak{p}}/\mathfrak{p}^k R_{\mathfrak{p}}$. Let us define $tr(M,\mathfrak{p})$ to be this number of copies. (When $M$ is infinitely generated I believe one should also count copies of $K_{\mathfrak{p}}/R_{\mathfrak{p}}$ in a certain sense in order to recover Fuchs's $p$-primary torsion rank in the $R = \mathbb{Z}$ case. Let me omit this for now.) Now define $R(M) = r_M((0)) + \sum_{\mathfrak{p} \in \operatorname{MaxSpec} R} tr(M,\mathfrak{p})$ When $R = \mathbb{Z}$ then $R(M) = \operatorname{mg}(M)$ is the "total rank" in Fuchs's sense. More generally $R(M) = \operatorname{mg}(M)$ when $R$ is a PID. However, I wanted to point out that in general the function $\operatorname{mg}$ behaves rather badly. This is discussed in $\S$ 6.5.3 of my commutative algebra notes. In particular, when $M$ is finitely generated projective, $\operatorname{mg}(M) \geq R(M) (= r_M((0))$ but can be larger. However, it is much more restricted than what I knew about before reading the comments on this question. In particular, it follows from the Forster-Swan Theorem that when $M$ is projective of rank $n$ then $\operatorname{mg}(M) \in \{n,n+1\}$. (In an earlier version of this answer I knew only that $\operatorname{mg}(M) \leq 2n-1$ and "guessed" that it could be that large over suitable Dedekind domains. Not a terrible guess, perhaps, but not the most educated one either...)<|endoftext|> TITLE: Mathematical study of Mpemba effect? QUESTION [11 upvotes]: It has been known since the days of Aristotle and Descartes that under certain circumstances warm water freezes faster than cold water. This effect is now commonly known as the Mpemba effect, named after a student who rediscoverd the effect in the sixties. Several theories have been proposed to explain the effect, but so far none of them seem to be generally accepted, see, e.g., this discussion on http://physics.stackexchange.com. In 2012 the Royal Society of Chemistry offered £1000 to the person or team producing the best and most creative explanation of the phenomenon, see http://www.rsc.org/mpemba-competition/. One problem is that many factors might play a role. The theories that try to explain the effect involve, for example, evaporation, convection, gas dissolved in the water, or interactions on molecular level, and it is difficult to design experiments that allow to isolate these factors. Are there any mathematical studies (exact solutions for special cases, numerical analysis, simulations, etc.) based on the equations proposed to describe or explain the Mpemba effect? Do they allow to isolate different influences and to compare them with experiments, e.g. by simulating heat flow with convection and/or evaporation. Does anybody here know of any such work? Or does anybody have a reference on simulations of similarly complex thermodynamical systems like a heat flow with convection and/or evaporation? PS: I discovered two papers by a group of Chinese chemical physicists, seeO:H-O Bond Anomalous Relaxation Resolving Mpemba Paradox and Mpemba Paradox Revisited -- Numerical Reinforcement. The second uses a finite element method to solve a one-dimensional model. I am not an expert in numerical analysis, but I believe modern mathematics should be able to go further than this. PPS: I changed the formulation of the second paragraph, following Theo Johnson-Freyd's remark. REPLY [5 votes]: Since this question is still open, I take the liberty of pointing to a recent survey of the status of the Mpemba effect, Pathological Water Science -- Four Examples and What They Have in Common, which draws the following conclusion: If confounding factors (such as evaporation, dissolved gases, mixing by convective currents, inefficient thermal contacts) are removed, hot water may indeed freeze earlier than cold water, but not because it cools off more quickly: the hot water remains warmer than the cold water during the cooling process. Nucleation sites in the container may lower the freezing temperature to anywhere between $0$ and $-45^\circ$ C and it may happen that the container filled with the cold water needs to be supercooled to a much lower temperature before the water freezes than the container filled with the hot water. The freezing temperature is a reproducible but unpredictable property of the container. If it would be possible to have two containers with identical nucleation sites, then cold water would freeze earlier than hot water.<|endoftext|> TITLE: Intermediate submodels without Boolean algebras QUESTION [11 upvotes]: My question is motivated by the following well-known fact regarding intermediate submodels of generic extensions. I would like to know if it can be proven using posets without the need for Boolean algebras. Fact: Let $\mathbb{R}$, $\mathbb{S}$ be partial orders in $V$. Let $J$ be $\mathbb{R}$-generic over $V$ and let $K \in V[J]$ be $\mathbb{S}$-generic over V. Then there is some partial order $\mathbb{T} \in V[K]$ and $H$ that is $\mathbb{T}$-generic over $V[K]$ such that $$V[K][H]=V[J].$$ Remark: As I understand it, a standard way to prove this can be sketched as follows: By Jech03-Lem15.4, there is some Boolean algebra $\mathbb{B}$ which is a complete subalgebra of $RO(\mathbb{R})$ and such that $V[K]=V[\mathbb{B} \cap J']$, where $J'$ is the obvious counterpart of $J$ in $RO(\mathbb{R})$. Let $\mathbb{T} = RO(\mathbb{R})/(\mathbb{B} \cap J')$. (There are a number of ways of doing this but they all come to the same thing. I'm more comfortable treating the Boolean algebras as posets.) Let $H = J'/(\mathbb{B} \cap J')$. Then it can be shown that we may define $J'$ from $K$ and $H$ and vice versa; so, $$V[J]=V[J']=V[K][H].$$ Alternatively instead of steps 2. and 3., we might define $\mathbb{T}$ as the $\mathbb{S}$-name, $RO(\mathbb{R})/\mathbb{B}$, and show that $\mathbb{S} \times \mathbb{T}$ is forcing equivalent (or even isomorphic) to $\mathbb{R}$. My question: Can the fact be demonstrated without recourse to Boolean algebras; i.e., with posets alone? Clearly steps, (2) and (3) above (or something like them) will work, but step (1) seems problematic. If we want to follow the plan above, we want to find some $\mathbb{P}$ which is completely embedded in $\mathbb{R}$ and such that $V[K]=V[\mathbb{P} \cap J]$; thus, avoiding a detour through Boolean algebras. However, Kunen provides a counter-example in his new Kunen11-ExV.4.50. Since the book is relatively new, I'll write it in: "Exercise V.4.50 Assume in $M$ that $\mathbb{Q} = Fn(\omega,\kappa)$ with $\kappa > 2^\omega$, and let $\mathbb{H}$ be $\mathbb{Q}$-generic over $M$. Then there is a real number $A \in M[H]$ such that: $A$ is random over $M$; $A \subseteq I \in M$, where $I$ is the set of rationals; but there is no poset $\mathbb{P}$ such that $\mathbb{P} \subseteq_c \mathbb{Q}$ and $M[A]=M[H \cap \mathbb{P}]$." This suggests that the path through 1. is blocked. Is there another way through? In Kanamori, Proposition 10.10, we have the statement of similar fact. Kanamori seems to suggest that it can be established with partial orders. A More Precise Problem: Can the fact above be established when V is a countable transitive model of $ZF-P$? Remark: Without powerset we cannot guarantee the existence of $RO(\mathbb{R})$ so the obvious implementation of the strategy above is blocked. REPLY [4 votes]: One possible way to prove the theorem without using the Boolean valued techniques is to use Bokovsky's theorem given in Characterization of generic extensions of models of set theory. I may mention there are several proofs of this fact without refering to Boolean algebras, see for example Schindler's paper The long extender algebra or the paper On the set-generic multiverse by Friedman-Fuchino-Sakai. Note that the theorem is an immediate corollary of Bukovsky's theorem.<|endoftext|> TITLE: Why don't the algebraic and geometric adjoints of the Lefschetz operator agree? QUESTION [14 upvotes]: One of the standard conjectures in algebraic geometry is that an operator $\Lambda$ on the cohomology algebra of a projective variety is algebraic. To my lying eyes it looks like there are two definitions of the operator $\Lambda$ out there and that the two don't agree. On the algebraic side, let $X$ be a projective variety of dimension $n$, equipped with the first Chern class $w$ of an ample line bundle $A$. We then get a Lefschetz operator on cohomology by setting $L u = u \wedge w$ for a class $u \in H^*(X)$. Because of the Lefschetz theorems we can define another operator $\Lambda_a u = (L^{n-k+2})^{-1} \circ L \circ L^{n-k}$ on $k$-classes $u$ when $k \leq n$ (when $k > n$ there's a similar definition that doesn't matter here). By definition this operator satisfies $\Lambda_a L = \operatorname{id}$. On the geometric side, let $X$ be a compact Kahler manifold of dimension $n$, equipped with a Kahler metric $\omega$. (If the cohomology class of $\omega$ is entire then it is the curvature form of an ample line bundle on $X$ and everything is algebraic.) As before we get a Lefschetz operator on forms and cohomology by setting $L_g u = u \wedge \omega$. The metric $\omega$ also defines a Hodge star operator $*$ that operates on forms, which defines an inner product on forms, and the adjoint of $L$ with respect to that inner product is $\Lambda_g$. The Kahler identities say (amongst other things) that $[L,\Lambda_g]\,u = (k-n) u$ for a $k$-form $u$. These operators descend to the cohomological level, either by representing cohomology classes by harmonic forms or by working out purely cohomological definitions of the operators from primitive decompositions of cohomology classes (that the two agree necessarily passes through harmonic representatives). Now, to me these definitions do not agree. For one, the commutation identity $[L,\Lambda_g]\,u = (k-n) u$ on the geometric side is incompatible with $\Lambda_a L = \operatorname{id}$ on the algebraic side. I don't think this difference is just a matter of overcoming not having all of Hodge theory in arbitrary characteristic; if we have primitive decompositions of classes we can define the $*$ and $\Lambda$ operators just fine on cohomology without reference to differential forms, which I would think is what we'd want to do if we wanted to emulate Hodge theory on Kahler manifolds in the algebraic world. Apparently the people who defined these things (Grothendieck and co. I imagine) didn't agree. Why not? Why is one of the standard conjectures stated in terms of an operator that doesn't seem to be used in the classical world? REPLY [5 votes]: In addition to Kleiman's "Algebraic cycles and the Weil conjectures" mentioned by abx, I can highly recommend reading the first article in the Motives proceedings (Jannsen/Kleiman/Serre) by Kleiman: "The standard conjectures". It introduces several $\Lambda$ and $\star$ operators, and shows that when one is algebraic, then so are the others. Further, it shows implications to weaker conjectures, such as algebraicity of the Künneth projectors.<|endoftext|> TITLE: Universal graphs on higher cardinals QUESTION [5 upvotes]: The Rado graph contains every finite graph as induced subgraph, and its also holds for countable graphs. So it is an universal graph of size $\aleph_0$, which contains all graphs of size $\aleph_0$ as induced subgraph. Question: Is it also true for higher cardinals? Does there exist a graph of size $\kappa$, which contains all graphs of size $\kappa$ as induced subgraph? If not, what is the condition for $\kappa$? REPLY [8 votes]: An old result of Shelah states that if one adds $\lambda > \kappa^{++}$ Cohen subsets to a cardinal satisfying $\kappa^{<\kappa}$, then in the resulting universe the universality number for graphs on $\kappa^+$ is $\lambda$. It is possible to have a universal graph of size $\aleph_1$ even when CH fails (Mekler and Shelah independently).<|endoftext|> TITLE: Harvey Friedman's strict reverse mathematics vs. Cook-Nguyen's V$^0$ QUESTION [12 upvotes]: Harvey Friedman posted several manuscripts [1] proposing a program for "strict" reverse mathematics, in the sense that the base theory should be mathematically natural and coding-free. In them he describes a number of two-sorted systems (with sorts for integers and finite sets or sequences of integers) weaker than the base theory RCA$_0$ of reverse mathematics. I wonder if it is known whether any of Friedman's systems are essentially equivalent to the two-sorted system called V$^0$ in Cook-Nguyen [2] (which if I understand correctly was introduced by Zambella [3]). [1] http://people.math.osu.edu/friedman.8/pdf/InevLogStr082907.pdf among others [2] Stephen Cook, Phuong Nguyen. Logical foundations of proof complexity, 2010. [3] Domenico Zambella. Notes on polynomially bounded arithmetic, JSL, Sep. 1996. REPLY [8 votes]: If I understand the notation correctly, Theorem 8.28 in [1] shows that FSTZ is a notational variant of $V^0$ (and the same holds for all the theories Friedman shows to be definitionally equivalent to FSTZ in later sections, but see below for a caveat). Specifically, FSTZ is biinterpretable with $V^0$ as follows. (I will use Zambella’s axiomatization for simplicity.) The natural interpretation of $V^0$ in FSTZ is given by using nonnegative integers for the number sort, and sets of nonnegative integers for the second sort (with set equality—present in Zambella’s language, but not in Friedman’s language—interpreted by extensionality). The translation of $\Sigma^b_0$-comprehension is provable by Theorem 8.28, the axiom $A\ne\varnothing\to\exists x\in A\,(A\exp(n,i)$. This contradicts the maximality of $\exp(n,i)$ unless $i=r-1$, in which case $\exp(n,m)\le\exp(n,r-1)<\exp(n,r)$ as $\exp(n,m)\in A$. I think that Lemma 11.7 holds regardless: if we take $M$ a countable model of FSTZ, we can find an order isomorphism $f$ of (the first-order part of) $M$ and its initial segment $I$ such that $n^m$ exists for all $m\in I$ and $n\in M$ (using the fact that the order reduct of $M$ is recursively saturated), and then we can define $\exp(n,m)=n^{f(m)}$. However, this does not give an interpretation, let alone definitional equivalence.<|endoftext|> TITLE: Reference and best bounds of $\sum_{n\leq x}\frac{\mu(n)}{n}$ QUESTION [7 upvotes]: Could someone please provide information about the best possible known bounds of the sum $$A(x)=\sum_{n\leq x}\frac{\mu(n)}{n}?$$ Unconditionally, $A(x)=O(e^{-c\sqrt{\log x}})$ is known to me. Does there exist any better bound conditionally or unconditionally? I am expecting a result like $A(x)=O(\frac1{\sqrt x})$ if Riemann Hypothesis is assumed. Is this true? What I could get is up on truth of RH $A(x)=O(x^{-1/2+\epsilon})$, which is not very hard to prove. Any reference regarding conditional (RH) bound of $A(x)$ will be highly appreciated. Thanks. REPLY [12 votes]: As Alexey has pointed out the problem can be reduced, via summation by parts, to understanding the asymptotic of Mertens' sum $$M(x) := \sum_{n\leq x} \mu(n).$$ Conditional on the Riemann hypothesis, the best bound to date on Mertens' sum, due to Soundararajan in 2009, is $$M(x) \ll x^{1/2} e^{(\log(x))^{1/2} (\log\log(x))^{14}}.$$ In particular, this implies $$\sum_{n\leq x} \frac{\mu(n)}{n} \ll \frac{e^{(\log(x))^{1/2} (\log\log(x))^{14}}}{x^{1/2}}.$$<|endoftext|> TITLE: Floors of powers of reals, how much do the first few determine the next? QUESTION [17 upvotes]: Call an integer sequence $\mathbf{x}=\left( x_1,x_2,\cdots \right)$ feasible if it is $f(r)=\left(\lfloor r \rfloor, \lfloor r^2 \rfloor, \lfloor r^3 \rfloor, \ldots, \lfloor r^n \rfloor, \ldots \right)$. for some real $r \gt 1$. Q: if the members of $f(r)$ all have the same parity, must $r$ be an integer? I believe(d) so (I was wrong see below.) The motivation of the question is not so much about parity as about how much an initial segment of a feasible sequence limits the next term. An affirmative answer to the question would imply that for every non-integer $r \gt 1$, there must be some $k$ so that $x_1,x_2,\cdots,x_k$ uniquely determines $x_{k+1}$. Given only the value $x_k$ we know the approximate value of $r$ and I claim that (provided $k$ is fairly large compared to $r$) this restricts the possible values of $x_{k+1}$ to an interval of length roughly $r.$. In more detail: Given only $x_k$ we know that $\sqrt[k]{x_k} \le r \lt \sqrt[k]{x_k+1}$ so $$(x_k)^{1+1/k}\le x_{k+1} \lt (x_k+1)^{1+1/k} \tag{ * }.$$ Suppose that $k$ is a good bit larger than $r$ and let $s=\sqrt[k]{x_k}.$ Then $\sqrt[k]{x_k+1} \approx s+\frac{1}{ks^{k-1}}$ and the bounds $( * )$are approximately $$s^{k+1} \le x_{k+1} \lt s^{k+1}+s(1+1/k)$$. Of course $x_{k+1}$ must be an integer. Also, we can say $$\max_{j \le k} \sqrt[j]{x_j}\le r \lt \min_{j \le k}\sqrt[j]{x_j+1}.$$ later Thanks for the good answers. I now realize that the answer to my question turns out to be No, $r$ need not even be algebraic. No matter how many initial terms are given, Even if one term is forced by the previous ones, choices can then be made, one term at a time, which always allow several choices for the next term. This might require $x_1 \ge 3$ or a similar condition. I think that one can maintain freedom of choice as long as one refrains from making the greatest possible choice (unless forced to do so, in which case the next choice will not be forced.) I will describe things in general terms here and add a graphic illustration as an answer: In addition to $(*)$ above, showing that $x_k$ alone allows about $s(1+1/k)$ possible values for $x_{k+1}$ one has $$\lfloor s^{k-1} \rfloor \le x_{k-1} \le s^{k-1}+\frac{1-1/k}s.$$ Which means that there might be two possible values for $x_{k-1}$, but not more. Here is an example with a forced term followed by a sketch of how to find such. Typically a feasible a sequence $[235,x_2,x_3,x_4,x_5]$ will allow about $235+\frac{235}{5}=282$ possible values for $x_6$ and over $75000$ possible values for both $x_6,x_7.$ However the sequence $[235, 55677, 13137758, 3100000161, 731479476931,x_6,x_7]$ has only $5$ completions: one must have $x_6=172600708777398$ and $| x_7-40727054702138152| \le 2.$ In this case $x_4 \approx 3100000000$ was the third thing I tried and I was fortunate to find out that $3100000162^{\frac54}=731479476931.00004144$. Thus the restriction $$\sqrt[5]{731479476931} \le r \lt \sqrt[4]{3100000162}$$ forced by the sequence is very strong. REPLY [3 votes]: Here is an illustration of what things look like given that $x_1=\lfloor r \rfloor=3.$ I think that a larger value than $3$ would work even better. Then we know $3^k \le x_k \le 4^k-1$. So there are $781$ possible values for $x_5$ and, by my calculations, $991$ possibilities for the sequence $(x_1,x_2,x_3,x_4,x_5)$. This means that most of the time $x_5$ alone determines all the previous terms. In fact there are $196$ choices which can be arrived at in exactly two ways and only $7$ which can be arrived at in three ways. One is $x_5=569$ which can be due to any one of $$ [3,12,44,159,569],\ [3,12,44,160,569], \ [3,12,45,160,569]$$ One can imagine the successive terms $x_k$ as describing an upward path in a tree with a node labelled $x_k$ located at the point $(\sqrt[k]{x_k},k)$. There may be other nodes labelled $x_k$ at that level. For example there are three nodes labelled $569$ at level $5$. They are at $(x,5)$ for $x=\sqrt[5]{569},\sqrt[4]{160},\sqrt[3]{45}$ Here are selected portions of that graph. The first three levels and above that portions of the next two levels above $46$ and $47$ in the third level. If the tree was extended to more levels eventually each node would have $3$ or $4$ continuations to the next level. If one takes an upward path which moves always to the right (never directly vertically) as it goes up, then it will be consistent to keep doing this in any way desired. SO there is freedom to choose the parity. If one avoids both the leftmost and rightmost upward edge leaving each node then each term will determine all the earlier ones.<|endoftext|> TITLE: Examples of finite groups with "good" bijection(s) between conjugacy classes and irreducible representations? QUESTION [33 upvotes]: For symmetric group conjugacy classes and irreducible representation both are parametrized by Young diagramms, so there is a kind of "good" bijection between the two sets. For general finite groups see MO discussion. Question: What are the finite groups where some "good" bijection(s) between conjugacy classes and irreducible representations are known ? "Good" bijection is an informal "definition", nevertheless I hope example of S_n and other examples listed below, may convince that the question makes sense. I think that it is far too optimistic to have one unique bijection for general group, but it seems to me that for certain classes of groups there can be some set of "good" bijections. Let me briefly discuss below some properties which "good" bijection may satisfy, and may be discuss details in the next question. Some examples: 1) symmetric group S_n 2) Z/2Z is naturally isomprhic to its dual, as well as $Z/2Z \oplus Z/2Z$ see e.g. MO "fantastic properties of Z/2Z" 3) Generally for abelian finite groups: among all set-theoretic bijections $G \to \hat G$, some are distinguished that they are group isomorphisms. So we have non unique, but a class of "good" bijections. 4) For GL(2,F_q) Paul Garret writes: "conjugacy classes match in an ad hoc fashion with specic representations". (See here table at page 11). 5) G. Kuperberg describes relation of the McKay correspondence and that kind of bijection for A_5 (or its central extension), see here. 6) If I understand correctly here at MO D. Jordan mentions that bijection exists for Coxeter groups. (I would be thankful for detailed reference). 7) Dihedral groups $D_{2n}$ - see answer by Glasby below 8) Finite Heisenberg group with $p^{2n+1}$ elements, also known as extraspecial group - see answer by Glasby below 9) Quternionic group $Q_{8}$ - this actually can be seen as a particular example of the item above. Or note that it is $Z/2Z$ central extension of $Z/2Z \oplus Z/2Z$, and $ Z/2Z \oplus Z/2Z$ has natural bijection as mentioned in item 2 above, and it is easy to extend it to $Q_8$. 10) It seems that for Drinfeld double of a finite group (and probably more generally for "modular categories") there is known some analog of natural bijection. There is such remark at page 5 of Drinfeld Doubles for Finite Subgroups of SU(2) and SU(3) Lie Groups. R. COQUEREAUX, Jean-Bernard ZUBER: In other words, there is not only an equal number of classes and irreps in a double, there is also a canonical bijection between them. There can be several properties which "good" bijection may satisfy, at least for some "good" groups 1) Respect the action of $Out(G)$. Actually the two sets are not isomorphic in general as $Out(G)$-sets (see MO21606), however there are many cases where they are isomorphic see G. Robinson's MO-answer. 2) Reality/Rationality constraints. Again in general there is no correspondence see MO, but there are some cases where corresponding properties of classes and irreps agree - see J. Schmidt's answer on that question. Two properties below are even more problematic 3) It might be that product on conjugacy classes have something to do with tensor product of representations (at least for abelian groups we may require these two fully agree). 4) If to think about kind of "orbit method" ideology, and think that conjugacy class is in some sense perversed coadjoint orbit, we may hope that structure of conjugacy representation, should somehow respect the "good" bijection. For example for $S_n$ we proved that irrep corresponding to Young diagram "d" lives inside the conjugacy subrepresentation realized as functions on the conjugacy class parametrized exactly by "d". (See MO153561, MO153991 for some discussion of conjugacy (adjoint) representation). 5) For algebraic groups over finite fields conjugacy classes and irreps sometimes naturally divided into families (e.g. conjugacy classes are often parametrized by equations $ F_{t_i}(x_k) = 0 $ - changing "t" we get different conjugacy classes in the same "family"). So we may hope that good bijection respects the families. (It is works fine for Heisenberg group, but for UT(4,p) I have met some problems). REPLY [9 votes]: I think it is one of the wunderful beauties of the representation theory of finite groups of Lie type $G(\mathbb{F_{p^n}})$ such as $GL_2(\mathbb{F_p})$ mentioned above, that irreducible representations are in some canonical (but not perfect) correspondence to conjugacy classes in the group with the dual root system $G^\vee$ ($\sim$ Langlands dual, this is usually isomorphic or a central extension, but $B_n,C_n$ are interchanged). In these cases, what you are asking is an essential part of Langlands philosophy for a finite field. Most of what I shall write below I think attributes to Lusztig - it workes very roughly as follows: We all known Jordan decomposition: A matrix $x$ can be decomposed into a unipotent and semisimple matrix $x_sx_u$. Since unipotent matrices are an additive group and the diagonal matrices a multiplicative group, over the field $\mathbb{F_{p^n}}$ the elements $x_s$ have order $p^k$ and the elements $x_u$ have order prime to $p$. ...Analogously hold (more complicated!): Every character $\chi$ of $G(\mathbb{F_{p^n}})$ can be decomposed in a unipotent and a semisimple character $\chi_u, \chi_s$ (such that the dimensions multiply to $\dim(\chi)$), which are described as follows: The semisimple characters can be characterized by having a dimension prime to $p$. They are in a quite transparent way in bijection to semisimple classes $[s]$ (i.e. of order prime to $p$) in the dual group $G^\vee(\mathbb{F_{p^n}})$....basically by looking what they do on the torus. The unipotent characters are much more complicated to deal with. In fact I do not even know (which doesnt mean there isn't) a definition independent of the theory of flag varienty cohmology and Deligne-Lusztig characters; typically on distinguishes between cuspidal unipotent representations and between unipotent representations induced from cuspidals of smaller parabolics. ...Nevertheless: $\chi_u$ is a unipotent character of the centralizer of $[s]$, so it is a representation of an algebraic subgroup of the dual $G^\vee(\mathbb{F_{p^n}})$. It was a difficult conjecture (I think proven?) that these are now again in bijection to special unipotent classes in again the dual (of the dual). Example: For $G=PSL_2$ the dual is $G^\vee=SL_2$. The unipotent representations can be constructed from the set of Deligne-Lusztig characters and one fixed Irreps of the Weyl group. This is moreover true for $PSL_n$, so we have $n!$ unipotent representations. The two extremal cases are $\chi_u$ the trivial representation and $\chi_u$ the Steinberg representation of dimension $p$ (resp. $p^{n(n-1)/2}$). In the more general case we recover the picture known from $\mathbb{S}_n$: Unipotent $SL_n$-Irreps $\chi_u$ biject from $\mathbb{S}_n$-Irreps biject to $n$-partitions biject to $\mathbb{S}_n$ conjugacy classes, but also to Jordan blocks of unipotent matrices (with eigenvalue $1$) in $SL_n$.<|endoftext|> TITLE: Latest stand of core model theory? QUESTION [34 upvotes]: What is the "strongest" core model to this day? In particular, how far are we from a core model for supercompact cardinals? There are rumors of some notes from a workshop in 2004: http://www.math.cmu.edu/~eschimme/AIM/LongDescription.html But I couldn't find any more details with respect to a new core model. Also, could someone recommend the clearest and most rigorous exposition of the "strongest" core model so far? REPLY [13 votes]: The following may not be an answer to your question, but I think it is related. I have taken it from the introduction of a joint work I am doing with James Cummings and Sy Friedman: An important development in large cardinal theory is the construction of inner models $M$ all of whose sets are definable from ordinals and which serve as good approximations to the entire universe $V$. The former means that $M$ is contained in $HOD$, the universe of hereditarily ordinal definable sets, and the latter can be interpreted in a number of ways. One such interpretation is that the cardinal structure of $M$ is ``close'' to that of $V$ in the sense that $\alpha^+$ of $M$ equals $\alpha^+$ of $V$ for many cardinals $\alpha.$ This is for example the case if $V$ does not contain $0^{\sharp}$ and $M$ equals $L$, or if $V$ does not contain an inner model with a Woodin cardinal and $M$ is the core model $K$ for a Woodin cardinal. The following theorem shows that we can't hope to approximate the cardinals of $V$ by those of (inner models of) $HOD$ in general: Theorem Suppose $GCH$ holds and $\kappa$ is a $\kappa^{+4}-$supercompact cardinal. Then there is a generic extension $V^*$ of $V$ in which $\kappa$ remains inaccessible and for all infinite cardinals $\alpha <\kappa,$ $(\alpha^{+})^{HOD}<\alpha^{+}.$ In particular $W=V_{\kappa}^{V^{*}}$ is a model of $ZFC$ in which for all infinite cardinals $\alpha, (\alpha^{+})^{HOD}<\alpha^{+}.$<|endoftext|> TITLE: Right triangle with edge lengths equal to regular unit polygon edge lengths QUESTION [14 upvotes]: This question came up naturally recently from a blog post of John Baez. There is an observation of Euclid that edges of a pentagon, hexagon, and decagon inscribed in a unit circle form the edges of a right triangle: Euclid used this to construct the icosahedron (see Baez's description for more details). I observed that this follows from a dual description in terms of the dual pythagorean theorem (see also Greg Egan's description with more details). There are 4 right triangles associated to the 5 Platonic solids with the property that the edge lengths are equal to the edges of regular polygons inscribed in a unit circle: Question: Are there any other right triangles with edge lengths equal to the lengths of edges of regular polygons inscribed in a unit circle? Note that a regular polygon is allowed to be immersed, that is a star (or polygram as Sam Nead points out in a comment). Generalizing the algebraic analysis in Baez's post, this may be formulated in terms of a diophantine equation involving cyclotomic numbers, but I'll leave this formulation up to the reader. As observed in the blog post, there is a dual description stating that there are 3 regular polygons with edge lengths 1, such that the circumradii of these polygons form the altitudes of a right triangle (which satisfy the dual pythagorean theorem). Addendum: As Will Sawin points out in the comments, there is also the Galois conjugate hexagon, pentagram, decagram. In fact, one obtains examples associated to all of the Kepler-Poinsot polyhedra. The pentagram, hexagon, decagram is associated to the great dodecahedron and great icosahedron. The pentagon, hexagon, decagon is associated to the small stellated dodecahedron, and the triangle, decagon, decagram to the great stellated dodecahedron, if I've figured correctly. Postscript: It appears that a transformation of this question was answered by Gordan in 1877. Gordan showed that the only solutions to $1+\cos(\phi_1)+\cos(\phi_2)+\cos(\phi_3)=0$ with $0<\phi_i<\pi$ and $\phi$ a rational multiple of $\pi$ are permutations of $(\frac23\pi,\frac23\pi,\frac12\pi)$ and $(\frac23\pi,\frac25\pi,\frac45\pi)$ (see. p. 109 of Coxeter). This equation is equivalent to the equation in Sawin's answer, $\zeta_1+\zeta_1^{-1}+\zeta_2+\zeta_2^{-1}-\zeta_3-\zeta_3^{-1}=2$ where $\zeta_1=e^{\phi_1 i}, \zeta_2=e^{\phi_2 i}, \zeta_3=e^{(\pi-\phi_3)i}$ (the degenerate solutions in Gordan's equation are eliminated by assuming $0<\phi<\pi$). The polygon going through the midpoints of edges is similar to the Petrie polygon, which is the projection of the sequence of edges, and lies in the Coxeter plane. The duality in orthogonal triangles associated to each of the regular polyhedra then follows from the symmetry of these equations, but doesn't seem to be observed directly in Coxeter's book. In fact, if $h$ is the degree of the polygon, then we see a further symmetry of the solutions by letting $\frac{1}{r}+\frac{1}{h}=\frac12$. We see that $\{p,q,r\}$ satisfies equation 6.71 on p.108 of Coxeter: $$\cos^2(\pi/p)+\cos^2(\pi/q)+\cos^2(\pi/r)=1.$$ The solutions are $\{3,3,4\}$ and $\{3,5,\frac52\}$. On p. 109 (see also regular polyhedron), one sees that permuting the two solutions to this equation, one obtains 3 and 6 polyhedra respectively, indicating a further symmetry not apparent from the right triangle equation. REPLY [7 votes]: These are the equivalent images for the Kepler-Poinsot polyhedra: One way to see why one of the polygons (the "green" one) is invariant under taking the dual of the polyhedron is to note that if the faces of the polyhedron are $n$-gons and the vertex figures are $v$-gons, the "yellow" polygon (whose vertices are all the midpoints of all the edges incident on some vertex of the polyhedron) will be a $v$-gon, with each edge subtending an angle of $\frac{2\pi}{v}$, while the "blue" polygon centred on the vertex will be a $\frac{2n}{n-2}$-gon, where each edge subtends an angle of $\frac{(n-2)\pi}{n}=\pi-\frac{2\pi}{n}$ (i.e. the interior angle between the edges of an $n$-gon). The difference of the squares of the sines of half these angles is: $$\sin\left(\frac{\pi}{v}\right)^2 - \sin\left(\frac{(n-2)\pi}{2n}\right)^2 = -\cos\left(\pi \left(\frac{1}{n}+\frac{1}{v}\right)\right) \cos\left(\pi \left(\frac{1}{n}-\frac{1}{v}\right)\right)$$ where the RHS is now clearly invariant under an exchange of $n$ and $v$. And of course this quantity is the squared sine of half the angle subtended by each edge of the "green" polygon. And for completeness, here are some images of the construction in the degenerate case, where the polyhedron becomes a polygon and the "yellow" polygon becomes a digon:<|endoftext|> TITLE: Higher dimensional generalization of the Hardy-Littlewood conjecture? QUESTION [7 upvotes]: The famous Hardy-Littlewood conjecture on prime-tuples states that if $\{h_1, \cdots, h_k\} = \mathcal{H}$ is an admissible set, that is, for every prime $p$ the set $\mathcal{H}$ does not contain a complete residue system modulo $p$, then there exist infinitely many positive integers $n$ such that the tuple $$\displaystyle (n+h_1, n+h_2, \cdots, n+h_k)$$ consists of only prime numbers. That is, each entry is prime. The study of this conjecture is intimately connected to the study of bounded gaps between primes. Indeed, the confirmation that there exist infinitely often bounded gaps between primes is a confirmation of a weaker form of the above conjecture, namely that for fixed positive integer $k$ and a integer $l \geq 1$ (where we would like $l$ to be as small as possible the product $$\displaystyle (n+h_1)\cdots(n+h_k)$$ contains at most $k+l$ many prime factors (counting multiplicity) between them, and for $l$ sufficiently small this in particular implies that there are at least two factors $n+h_i, n+h_j$ are prime. My question concerns a higher dimensional analogue of the Hardy-Littlewood conjecture, namely looking at linear forms of more than one variable. Suppose that we are given the linear polynomials $L_j(x_1, \cdots, x_m) = a_1^{(j)} x_1 + \cdots + a_m^{(j)} x_m + h_j$, $1 \leq j \leq k$, such that $a_i^{(j)} \in \mathbb{N}$ for each $i,j$, and the content of each $L_j$ is 1. Then does one expect that for infinitely many $\textbf{x} = (x_1, \cdots, x_m) \in \mathbb{N}^m$ that the tuple $$\displaystyle (L_1(\textbf{x}), \cdots, L_k(\textbf{x}))$$ is a prime tuple (that is, every entry is prime)? I suspect that this would be much easier to prove than the Hardy-Littlewood conjecture due to the many degrees of freedom available when $m \geq 2$. If so, has anyone related results already? REPLY [8 votes]: This is sometimes called Dickson's conjecture (questions of this form were raised in his 1904 paper, available here). Tao and Green's paper Linear Equations in the Primes give unconditional results along these lines, when combined with the resolution of the Gowers' norm inverse conjecture by Green, Tao and Ziegler, and the Mobius nilsequence conjecture by Green and Tao.<|endoftext|> TITLE: How do you call the problem of approximating a continuous distribution with a simple discrete distribution? QUESTION [5 upvotes]: The following problem came up on the Mathematica forum as "Generating a list of integers that roughly satisfy a distribution": Given $n$, find $n$ integers (possibly with duplicates) whose distribution approximates a given normal distribution. For example, given a normal with mean 25 and standard deviation 7, the following multiset could be a decent candidate : {11,15,17,18,20,21,22,23,24,25,25,26,27,28,29,30,32,33,35,39} I understand that the term "approximates" is not properly defined and there may be several ways of fixing that. For example, one could specify that random samples taken from the discrete distribution should "look like" (again, need to specify what "looks like means") they are drawn from the continuous distribution (conditioning the normal distribution on integrality or after rounding). Are there references on that problem? REPLY [2 votes]: The literature on order statistics might suggest approximating $X$ with $\{E[X_{k:n}]:1\le k\le n\}$. For your $X$ with $n=20$, this gives {11.9, 15.1, 17.1, 18.6, 19.8, 20.9, 21.9, 22.8, 23.7, 24.6, 25.4, 26.3, 27.2, 28.1, 29.1, 30.2, 31.4, 32.9, 34.9, 38.1} So in a sample of size 20 from $X$, 11.9 is the expectation for the smallest element, and 15.1 is the expectation for the second-smallest element. Then rounding gives {12, 15, 17, 19, 20, 21, 22, 23, 24, 25, 25, 26, 27, 28, 29, 30, 31, 33, 35, 38}<|endoftext|> TITLE: Using Eichler-Selberg trace formula to compute dimension of modular forms? QUESTION [11 upvotes]: Is it possible to use Eichler-Selberg trace formula to compute the dimension of modular forms of weight $k$ for $SL(2,\mathbb Z)$? This was computed by classical methods such as Riemann-Roch. REPLY [4 votes]: An adelic reference for $GL(2, \mathbb{Z})$ is also "Traces of Hecke operators" by Knightly and Li. You choose matrix coefficent at the infinite place and the characteristic function $GL_2(\mathbb{Z}_p)$ times the center, and plug it into the Arthur trace formula. They do that almost for weight $k \geq 3$, but they work with matrix coefficient. This gives the dimension formula. This is likely to be more fruitful for generalization to higher rank or deeper level, where not always a classical treatment is available.<|endoftext|> TITLE: Singularities of moduli spaces of curves QUESTION [9 upvotes]: Let $\overline{M}_{g,n}$ be the moduli space of $n$-pointed genus $g$ Deligne-Mumford stable curves. This is a normal projective scheme. Then $$codim_{\overline{M}_{g,n}}Sing(\overline{M}_{g,n})\geq 2.$$ For instance for $g = 1, n = 2$ wa have that $\overline{M}_{1,2}$ is a rational surface with four singular points. Does there exist any value of $g$ and $n$ for which $Sing(\overline{M}_{g,n})$ is in codimension at least $3$ ? REPLY [14 votes]: No. In fact more is true: the locus of all $n$-pointed curves of genus $g-1$ with a single elliptic tail $E$, such that $\mathrm{Aut}(E)=\mathbf Z/6$, has codimension two in $\overline M_{g,n}$ and consists of noncanonical singularities. This was famously determined by Harris and Mumford in their paper on the Kodaira dimension of the moduli space of curves (they work with $n=0$ but this makes no difference).<|endoftext|> TITLE: Records in $Z$-numbers and a relaxation QUESTION [5 upvotes]: A Z-number is a (non-zero) real number $x$ such that the fractional parts $$\left\lbrace x \left(\frac 3 2\right)^ n \right\rbrace $$ are less than $\frac12$ for all natural numbers $n$. It is not known whether $Z$-numbers exist. First, I am interested in finite records, i.e. large $k$ such that for explicit $x > 1$ the fractional part is $ < \frac12$ for $n = 1 \ldots k$. So far got $k=12$ for $x \approx 2.81365$ and don't think $ x < 2$ is possible in this case. Second, consider the following relaxation. Let $f$ be a strictly increasing function $\mathbb{N} \to \mathbb{N} $ Is it possible: $$\left\lbrace x \left(\frac 3 2\right)^ {f(n)} \right\rbrace < \frac12 $$ for all natural $n$? REPLY [4 votes]: Suppose the fractional parts of $x, \frac{3}{2} x, (\frac{3}{2})^2 x, ... (\frac{3}{2})^kx$ are under $\frac{1}{2}$, but the fractional part of $(\frac{3}{2})^{k+1} x$ is between $\frac{1}{2}$ and $1$. Then consider $y = x + 2^k$. For $n \le k, \lbrace (\frac{3}{2})^n y\rbrace = \lbrace (\frac{3}{2})^n x \rbrace$ while $\lbrace (\frac{3}{2})^{k+1} y \rbrace = \lbrace (\frac{3}{2})^{k+1} x \rbrace - \frac{1}{2} \lt \frac{1}{2}.$ So, if we start with an arbitrary $x$, then one of the $2^{k}$ numbers $x, x+1, ... x+2^{k}-1$ will have fractional parts under $\frac{1}{2}$ when you multiply by $\frac{3}{2}, ..., (\frac{3}{2})^k$. Here is some Mathematica code which implements this: Clear[j]; j[1] = 1; j[n_] := j[n] = If[FractionalPart[j[n - 1] (3/2)^n] < 1/2, j[n - 1], j[n - 1] + 2^(n - 1)] For the choice $j[1]=1$, this computes the integer from $1$ to $2^n$ which has the right fractional parts when multiplied by $\frac{3}{2}, ..., (\frac{3}{2})^n$. $j[4]=1, j[5]=17, j[20] = 386737, j[100] = 719590229933913224019274229425$. Yes, we can take $f(n)=3n$. Lemma: For any integer $a$, $[(\frac{3}{2})^3 a, (\frac{3}{2})^3(a+\frac{1}{2})]$ contains at least one interval $[b,b+\frac{1}{2}]$ for some integer $b$. This follows because $\frac{1}{2}(\frac{3}{2})^3 \gt 1+\frac{1}{2}$. Start with an interval $[a_1,a_1+\frac{1}{2}]$. Inductively pick $a_{n+1}$ so that $[a_{n+1},a_{n+1}+\frac{1}{2}] \subset (\frac{3}{2})^3 [a_n,a_n+\frac{1}{2}].$ Then the intervals $(\frac{3}{2})^{-3n}[a_n,a_n+\frac{1}{2}]$ are nested hence their intersection contains an element $x$ so that $\lbrace x (\frac{3}{2})^{3n} \rbrace \lt \frac{1}{2}$ for all $n$. You can choose $x\ne 0$.<|endoftext|> TITLE: Generalising the adherence operator and its closure properties with regard to regular (rational) languages QUESTION [7 upvotes]: Let $X$ be an alphabet and denote by $X^{\omega}$ the set of all infinite sequences (i.e. words) in $X$. A subset $L \subseteq X^{\omega}$ is called $\omega$-regular if it is acceptable by some Büchi-Automaton, equivalently if it is of the form $$ L = \bigcup_i^n U_i V_i^{\omega} $$ for regular languages $U_i, V_i$ in the usual sense, also $L$ is regular iff it is acceptable by a Müller-Automata (see Wikipedia). Let $\xi \in X^{\omega}$ be some infinite word, denote by $A(\xi)$ the set of its prefixes, these are finite words and so $A(\xi) \subseteq X^*$, likewise define $F(\xi)$ to be the set of factors of $\xi$. Now define the following language operator (called adherence) on $X^{\omega}$: $$ \mbox{Adh}(L) := \{ \xi \in X^{\omega} : A(\xi) \subseteq A(L) \}. $$ It is $\xi \in \mbox{Adh}(L)$ iff every prefix of $\xi$ is the prefix of some word from $L$. Now if $L$ is $\omega$-regular, $A(\xi)$ is $\omega$ regular too, if given a Büchi-Automata for $L$ just declare every state on an acceptance path (i.e. a path having an infinite number of final states) a final state too, the automata with the usual acceptance condition for finite words accepts $A(\xi)$. Now $\mbox{Adh}(L)$ is accepted by this automaton according to the Büchi-condition. I want to generalise this, define the operator $$ \mbox{Adh}_F(L) = \{ \xi \in X^{\omega} : F(\xi) \subseteq F(L). $$ It is $\xi \in \mbox{Adh}_F(L)$ iff every factor of $\xi$ is a factor of some word from $L$. Now I want to know If $L$ is $\omega$-regular, is $\mbox{Adh}_F(L)$ also $\omega$-regular? I conjecture that in general not to be true, because to test an infinite word $\xi$ for this condition, if the automaton would read the $n$-te symbol, it need to trace back to to all the position $1,2,\ldots, n$ of $\xi$ and test from them if the factor starting at this position and ending at the $n$-th position is contained in $F(L)$, but in general a finite automata can not save the last $n$-th positions for abitrary $n$. On the other side I am not able to represent some known non-regular languages as $\mbox{Adh}_F(L)$, and furthermore for a regular set $L$ the set $F(L)$ is regular too (given an automata for $L$ on every path which is final, i.e. leads to a final state, put an $\varepsilon$-transition from the start state to that state). So maybe I have overlooked some property of factor sets which make $\mbox{Adh}_F$ regular... REPLY [3 votes]: Then answer is yes. Since $F(L)$ is a regular language, it suffices to prove the following result: If $K$ is a regular language, then $R(K) = \{u \in A^\omega \mid F(u) \subseteq K \}$ is $\omega$-regular. Since regular languages are closed under complement, it suffices to show that $R(K^c)$ is $\omega$-regular. Observing that $R(K^c) = \{u \in A^\omega \mid F(u) \subseteq K^c \}$, we get $$ (R(K^c))^c = \{u \in A^\omega \mid F(u) \cap K \not= \emptyset \} = A^*KA^\omega $$ Thus $(R(K^c))^c$ is $\omega$-regular, and since $\omega$-regular languages are closed under complement, $R(K^c)$ is $\omega$-regular.<|endoftext|> TITLE: A question about extensions of Markov semigroups QUESTION [5 upvotes]: I'm cross-posting this question from MSE. It's the first time I do this so I'm unsure of etiquette regarding how to cross-post, if this irritates anyone please vote this down and I'll delete the post. Also if any reply appears to the MSE post I'll update this one. Suppose that $\{T(t)\}_{t\geq 0}$ is a Markov semigroup on the space of continuous bounded functions defined on $\mathbb{R}^n$ and that $\mu$ is an invariant measure of the semigroup. Then, under some technical assumptions, $\{T(t)\}_{t\geq0}$ can be extended to a strongly continuous semigroup on $L^p(\mathbb{R}^n,\mu)$ for every $p\geq1$. Can this be generalised to Markov semigroups on more general function spaces (in particular, spaces of functions that do not take values necessarily in $\mathbb{R}^n$)? Ideally, I'm looking for a result that covers both the case of functions that are defined on $\mathbb{R}^n$ and those that are defined on $\mathbb{N}^n$. Any references to texts containing these type of results would also be great. Edit: Typo, should have been $p\geq 1$ and not $p\geq 0$. REPLY [3 votes]: I think this holds in quite some generality by the following simple argument. Let $S$ be a Polish space, let's say. If $T(t)$ is a Markov-Feller semigroup on $C_b(S)$ with kernel $p_t(x,dy)$, then note that for $f \in C_b(S) \cap L^p(\mu)$, $1 \leq p < \infty$, \begin{equation} ||T(t)f||_{L^p(\mu)}^p = \int_S \left| \int_S p_t(x, d y) f(y) \right|^p \mu(dx) \leq \int_S \left(\int_S p_t(x,dy) |f(y)|^p\right) \mu(dx) = ||f||_{L^p(\mu)}^p, \end{equation} which shows that $(T(t))$ is contracting with respect to the $L^p(\mu)$-norm. Approximation of $f \in L^p(\mu)$ by continuous functions should then give you the extension you are looking for. Note, $L^p(\mu)$ for $p < 1$ is not a Banach space so talking about strongly continuous semigroups does not seem to make sense. Update: simplified argument.<|endoftext|> TITLE: Where to publish math that's original but easy but requires theorems not every undergrad knows? QUESTION [10 upvotes]: I'm struggling to find a home for a certain paper of mine. The problem is that the mathematical content of it is extremely easy: a couple definitions and a theorem with a five-line proof, invoking the Recursion Theorem and the Smn Theorem. It's not really appropriate for a logic journal, because it's too easy (are there any logic journal equivalents to "Mathematics Magazine")? But every easy math journal I send it to rejects it saying that the Recursion Theorem and Smn Theorem are too obscure for mathematicians! Due to its easiness, it might even be more appropriate as a "Problem", in a place like AMM's monthly list of problems, except it definitely requires Recursion Theorem and Smn theorem. Are there any places where such a problem would be welcome? REPLY [8 votes]: You should look beyond the standard outlets. The so-called "micropublishing revolution" is slowly emerging in mathematics and science in general. Original research ideas and results that aren't necessarily suitable for journal publication regularly find their way here on MathOverflow and in a variety of alternative outlets such as blog posts, social networks and whatnot. The idea is not new, Doron Zeilberger's 39th Opinion was posted in 1999 and Michael Nielsen has already written several books and essays on the topic. Though these new outlets don't have the notoriety of journals and other traditional outlets, they all achieve the basic goal of disseminating research ideas. (A five-line proof doesn't need any peer review or any other "added value" that journals provide.) Look around until you see something that fits your result. Don't worry about community acceptance and similar illusions, the people who are interested in your result will eventually find it and use it. News travels much farther and faster than it used to!<|endoftext|> TITLE: Is a composite of (co)monadic adjunctions (co)monadic? QUESTION [5 upvotes]: I think this is probably elementary, but some searching (and asking on the chatroom) hasn't turned up a result. Could anyone point me to a reference for (or counterexample to) the following statement? Given categories $\mathcal{C}, \mathcal{D}, \mathcal{E}$ and comonadic adjunctions $\mathcal{C} \rightleftarrows \mathcal{D}, \mathcal{D} \rightleftarrows \mathcal{E}$, the composite adjunction is comonadic. REPLY [3 votes]: Here is a positive result: Let $\mathcal{C} \to \mathcal{D}$ and $\mathcal{D} \to \mathcal{E}$ be two monadic functors. Assume that the underlying functor of the first monad $\mathcal{D} \to \mathcal{D}$ preserves reflexive coequalizers (this happens quite often!). Then, the composition $\mathcal{C} \to \mathcal{D} \to \mathcal{E}$ is monadic, too. This is sketched here. (Does anybody know a canonical reference for this result?) There you can also find another counterexample by Jim Dolan.<|endoftext|> TITLE: Are infinite dimensional Lie algebras related to unique Lie groups? QUESTION [12 upvotes]: For every finite dimensional Lie algebra $g$, there is a unique simply-connected Lie group $G$ whose Lie algebra is $g$. Is this true in the infinite dimensional case? REPLY [21 votes]: No. Van Est and Korthagen (1964) gave perhaps the first example of what they called a "non-enlargible Lie algebra", having no corresponding Lie group. Needless to say, this kind of question largely depends on the precise definitions adopted for infinite-dimensional Lie groups and Lie algebras. A canonical reference is Milnor's Remarks on infinite-dimensional Lie groups, which can be found on the internet and contains also some positive results: The author outlines a theory of Lie groups modelled on arbitrary complete locally convex topological vector spaces (CLCTVSs).This category includes everything one would want to call a Lie group, with possibly a few exceptions such as diffeomorphism groups of noncompact manifolds. (...) Every Lie group as defined here has an associated (topological) Lie algebra. For finite-dimensional groups the converse is true, but this is not so in general (even for Banach Lie algebras). The passage from Lie algebra to Lie group depends on the notion of regularity. (...) All known Lie groups are regular. A connected, simply connected, regular Lie group is uniquely determined (up to isomorphism) by its Lie algebra. For a more recent survey I would recommend Neeb's Towards a Lie theory of locally convex groups.<|endoftext|> TITLE: Understanding Bruhat's notion of Schwartz function QUESTION [9 upvotes]: I am trying to understand Bruhat's generalized Schwartz functions over (Hausforff) locally compact Abelian groups [1], following this paper [2] by Osborne. There, the Schwartz-Bruhat space $\mathscr{S}(G)$ of complex functions over a (Hausdorff) locally compact Abelian group $G$ is defined based on the notion of differential operators on groups. I have trouble understanding how differential operators are defined in groups that are discrete; see questions 1 and 2 below. For simplicity I only look at groups of the form For simplicity (and as in the previous post ) I will always assume that $G$ is a group of the form $$G=\mathbb{R}^a\times\mathbb{T}^b\times\mathbb{Z}^c\times F$$, where $\mathbb{T}^m$ is an $m$-dimensional torus and $F$ an finite Abelian group given in the form $$F=\mathbb{Z}_{d_1}\times\cdots\times \mathbb{Z}_{d_e}.$$ For groups of the form $G$, the space $\mathscr{S}(G)$ is defined as follows: a function $f$ is in $\mathscr{S}(G)$ if $f$ is infinitely differentiable, and if $P(\partial)f\in L^2(G)$ for every polynomial (in the $\mathbb{R}^a\times \mathbb{Z}^c$ variables) differential operator $P(\partial)$ on $G$. Question 1 What is a ``polynomial differential operator $P(\partial)$'' over a group $G$? I am vaguely aware that groups of the form $G$ are the abelian Lie groups that are compactly-generated (cf. [3]). Therefore, I assume there must exist a robust-notion of differentiability of functions over these groups. Still I do not know how differential operators on function spaces over discrete groups such as $\mathbb{Z}^c$ or $F$ should be defined. Question 2 Why does $P(\partial)$ have to be a polynomial differential operator only in the $\mathbb{R}^a\times \mathbb{Z}^c$ variables. I do not understand why $\mathbb{T}^b$ and $F$ are not mentioned. Does it perhaps something to do with compactness? Related posts. I have opened another post asking questions about an alternative equivalent way to define the Schwartz-Bruhat space via functions of rapid decay. Also, my first question seems to be related to this question. REPLY [11 votes]: I strongly recommend you to read the François Bruhat paper, that Osborne cites. For an arbitrary locally compact (not necessarily abelian) group $G$ Bruhat defines smooth function $\varphi:G\to{\mathbb C}$ as the one that can be locally represented as a composition $$ \varphi=\psi\circ\pi $$ where $\pi:G\to H$ is a continuous homomorphism into a Lie group $H$, and $\psi:H\to{\mathbb C}$ is a "usual" smooth function on $H$ (considered as a smooth manifold). (Actually, this is the definition for the so-called LP-groups, and the general definition becomes obvious after that.) Bruhat describes the most important properties of the space ${\mathcal E}(G)$ of smooth functions on $G$ (it is not obvious, for example, that ${\mathcal E}(G)$ is always dense in ${\mathcal C}(G)$). As far as I remember, he gives also a definition of a differential operator on $G$, but I am not sure, anyway you can also take a look at my paper, where differential operators on $G$ are defined as linear maps $$ D:{\mathcal E}(G)\to {\mathcal E}(G) $$ which preserve the support of functions: $$ \text{supp}(D\varphi)\subseteq \text{supp}\ \varphi,\qquad \varphi\in {\mathcal E}(G). $$ ($\text{supp}\ \varphi$ is the set of points in $G$ where the germ of $\varphi$ is non-zero). One can prove (see again my paper) that every such operator has unique decomposition $$ D=\sum_{\alpha\in{\mathbb N}_I}\xi_\alpha\cdot\partial e_\alpha, $$ where $I$ is the set of indices (in general, infinite) for a given basis $\{e_i;\ i\in I\}$ in the tangent space $T_1(G)$ (=Lie algebra of $G$), ${\mathbb N}_I$ the set of multi-indices over $I$, $\partial e_\alpha$ are the corresponding translation-invariant differential operators on $G$, and $\xi_\alpha$ are some smooth functions on $G$ (defined by $D$). $D$ is called a polynomial differential operator, if the coefficients $\xi_\alpha$ are polynomials, i.e. usual polynomials of finite sets of real characters $\chi:G\to{\mathbb R}$ (see Hewitt-Ross; obviously, each polynomial on ${\mathbb R}^a\times {\mathbb T}^b\times {\mathbb Z}^c \times F$ vanishes on ${\mathbb T}^b$ and $F$). Using these notions one can define the Schwartz-Bruhat space ${\mathcal S}(H)$ on an abelian Lie group $H$ as consisting of smooth functions $\psi:H\to{\mathbb C}$ with compactly generated support and with the property $$ \sup_{x\in H}|D\psi(x)|<\infty, $$ for every polynomial differential operator $D$ on $H$. And if $G$ is an arbitrary locally compact abelian group (not necessarily a Lie group) then the space ${\mathcal S}(G)$ is defined as the direct limit (=union) of the spaces ${\mathcal S}(H)$ where $H$ runs through all Lie quotient groups of $G$: $$ {\mathcal S}(G)=\bigcup_H {\mathcal S}(H). $$ I believe this is equivalent to Osborne's definitions. By $K^m$ he means, I think, the $m$-th power of a set $K\subseteq G$, defined by induction $$ K^1=K,\qquad K^{m+1}=K^m\cdot K, $$ where $A\cdot B=\{a\cdot b;\ a\in A,\ b\in B\}$ (this is for your new post).<|endoftext|> TITLE: Maps from the moduli space of abelian surfaces with level stucture to curves QUESTION [5 upvotes]: Let $A_2(N)$ denote the moduli space of principally polarized abelian surfaces with level $N$ stucture. The absolute Igusa invariants $i_1$ $i_2$ and $i_3$ give three different maps from $A_2(1)$ to $\mathbb P^1$. Now what I want to do is construct a family of rational maps $f_N : A_2(N) \to C(N)$ where all the $C(N)$ are curves, and such that the genus of $C(N)$ eventually goes to $\infty$. Now I don't know wether this is possible at all so my first question is: Are there $N$ for which there is a (dominant rational) map $A_2(N) \to C$ where $C$ is a curve of nonzero genus? And my second question is: Is $\mathbb C(i_k)$ algebraically closed in $\mathbb C(A_2(N))$? REPLY [6 votes]: The answer to the first question is negative. It follows from an old result of Matsushima that $b_1(\mathcal{A}_g(n))=0$ for $g\geq 2$ (Annals of Math. 75 (1962), 312-330). If there is a dominant map $\mathcal{A}_g(n)\rightarrow C$, $H^1(C,\mathbb{Q})$ injects into $H^1(\mathcal{A}_g(n),\mathbb{Q})$, which implies $g(C)=0$.<|endoftext|> TITLE: Faltings-Riemann-Roch Theorem QUESTION [10 upvotes]: I found the famous Faltings book ``Lectures on arithmetic Riemann-Roch theorem". In the book, very analytic techniques such as Garding inequality or heat kernel are explained. I have no idea where such analytic tools must come in to prove algebraic theorem. Because we want to calculate the Euler-Poincare number for the cohomology of vector bundle $E$ that is defined in a purely algebraic manner. So Question 1: Why do we need to consider such analytic aspects to formulate Riemann-Roch theorem for arithmetic surfaces over Spec $\mathbb{Z}$? I also found in Lang's book that to formulate Arithmetic Riemann-Roch theorem for arithmetic surface over Spec $\mathbb{Z}$, one might have to take care of Arakelov metric for arithmetic surface over infinite place $R$. Question 2: What does it mean to ``choose'' a nice metric such as Arakelov-metric for base-changed arithmetic surface over real place? Does it change the structure of the given arithmetic surface? Or is it incomplete to formulate Faltings-Riemann Roch Theorem that we merely consider the algebraic defining equation of arithmetic surface over Spec $\mathbb{Z}$? REPLY [2 votes]: I also have to admit I'm a non-expert, and that this is probably more accurately a long comment. I should also add that this just builds off of Dietrich's answer which probably has more direct motivation. Question 1: Why do we need to consider such analytic aspects to formulate Riemann-Roch theorem for arithmetic surfaces over Spec Z? In some sense, Arakelov geometry doesn't study $\operatorname{spec}\mathbb{Z}$ directly but studies the 'mythological' proper model of $\mathbb{Q}$, notated as $\widehat{\operatorname{spec}\mathbb{Z}}$ and varieties/schemes over it. The 'valuative criterion for properness' suggests that we need to add points for the missing completions for $\mathbb{Z}$, but this is exactly adding points for the archimedian places. The analytic aspects merely reflect this. At this point the analogies start stacking up, and the picture is not 100% clear to me. Let me take a shot at the second question then. Question 2: What does it mean to ``choose'' a nice metric such as Aeakelov-metric for base-changed arithmetic surface over real place? You should consider a choice of metric on the associated analytic space(or object) as something like fpqc-descent-data for the arithmetic variety. I'm not sure how Faltings does this, if he does at all, but I recently took a course on Arakelov geometry and at least for a short while, to denote an arithmetic surface we'd write $\overline{X}$ for a pair $(X,\mu)$ where $\mu$ is a Kähler form on $X(\mathbb{C})$. My point in bringing up notation, which is typically somewhat boring to discuss, is that one needs to keep in mind that you're studying $\overline{X}$ directly, not necessarily $X$. [N.B. I say for a short while because we only briefly went over what Arakelov's work, the course was very pointed towards arithmetic intersection theory and basically everything was quasi-projective, giving an obvious choice of a Kähler form] Does it change the structure of the given arithmetic surface? Or is it incomplete to formulate Faltings-Riemann Roch Theorem that we merely consider the algebraic defining equation of arithmetic surface over Spec Z? As I said, you should consider it descent-data, and different metrics are something like different compactifications of the same 'affine piece'. We dealt with this issue more with vector bundles, and again we'd write $\overline{E}$ for a pair $(E,h)$ where $E$ was a vector bundle on the scheme and $h$ was a (nice) metric on $E(\mathbb{C})$. A very concrete example to show how these metrics lead to different structures is via Picard groups. Classically, on $\mathbb{P}^1$ we have denumerably-many non-isomorphic line bundles, but restricted to $\mathbb{A}^1$ they all become isomorphic. Compare this to the Arakelov situation: on $\operatorname{spec}\mathbb{Z}$ every line bundle is trivial but on $\widehat{\operatorname{spec}\mathbb{Z}}$ the line bundles are parametrized by a choice of (conjugation invariant) hermitian metric $h$ on $\mathbb{C}$. Such a metric is determined by the real number $h(1)$, and so we can say $\operatorname{Pic}(\widehat{\operatorname{spec}\mathbb{Z}}) \cong \mathbb{R}$, though we'd normally write this with the hat 'homotopied' leftward: $\widehat{\operatorname{Pic}}(\operatorname{spec}\mathbb{Z})$. This even looks like the classical situation, where $\operatorname{Pic}(\mathbb{A}^1) \cong 0$ and $\operatorname{Pic}(\mathbb{P^1}) \cong \mathbb{Z}$ The point here is that: yes! It is not enough to formulate the analogue of RR or GRR just considering the algebraic part. We have to consider the algebraic part along with some analytic data to be able to ask the correct question. Note: There are formalizations which make symbols like $\widehat{\operatorname{spec}\mathbb{Z}}$ mathematically meaningful and well-defined, for example Durov's definitions in his colossal thesis New Approach to Arakelov Geometry (PDF warning!). I'd even go so far as to suggest reading chapter zero of his thesis, it's non-technical and I found it helpful.<|endoftext|> TITLE: Has this generalization of a determinant (assigning multiplicities to the rows) been studied? QUESTION [7 upvotes]: I'm working on some questions in tropical geometry, and my problem led me to create the following generalization of a determinant: Let $A$ be an $m \times n$ matrix with $m \le n$, and positive integer multiplicities $m_i$ assigned to the rows with $\sum_{i=1}^m m_i =n$ so $A$ is "square with multiplicity". I consider labeled partitions $\mathcal I= \{I_1, \dots, I_m\}$ of $\{1, \dots, n\}$ with $|I_i| = m_i$. Then the "determinant with multiplicity" is $$ \sum_{\mathcal I} \operatorname{sgn}(\mathcal I)\prod_{i=1}^m \prod_{k\in I_i} A_{ik}. $$ Notice that when $m=n$ (so all $m_i=1$) then this is just the usual determinant. I actually haven't thought about what $\operatorname{sgn}(\mathcal I)$ should be in general, because I created that definition by lifting from a tropical definition-- I call a matrix "tropically non-singular with multiplicity" if $$ \max_{\mathcal I} \left\{\sum_{i} \sum_{k \in I_i} A_{ik}\right\}, $$ is achieved exactly once. I expect "tropically non-singular with multiplicity" to have some nice properties generalizing those of the usual (Strumfels) tropical non-singularity. Some of those proofs use a lift to Puiseaux series and then the properties of the usual determinant, so if I want to imitate those proofs, I would need to understand the properties of the "determinant with multiplicity". So my question is: does anybody recognize the definitions above as something that has been studied before? If so, can you point me to a reference? EDIT Some more motivation: As a function, the second display above is the same as the tropical determinant/permanent of the actual square matrix formed by repeating the $i$th row $m_i$ times. I prefer my version because it eliminates repeated terms that "come only from" the repeated rows, that is, a matrix with a repeated row is not automatically considered to be singular. I guess this is funny classically, but I think I have a tropical geometric interpretation that makes sense. REPLY [8 votes]: Summary There is a clear connection to the permanent, which, if I understand correctly, yields the same tropicalization, and a more speculative connection to the determinant. The reason is that OP's formula can be rewritten as an immanant-like expression evaluated on an auxiliary matrix $\widehat{A}$ obtained by repeating the rows of $A$ according to the multiplicities $m_i.$ For arguably the most natural choice of the coefficients in the expansion, where they are all equal $1$, one obtains the permanent of $\widehat{A}$ (up to a constant multiple). On the other hand, one can also choose the coefficients for an arbitrary $m\leq n$ in a way that reproduces the determinant when $m=n.$ I don't know whether this leads to a meaningful generalization of the determinant. Analysis Modulo the choice of the coefficients, the procedure outlined in the question amounts to the following. Replace the rectangular $m\times n$ matrix $A$ with the square $n\times n$ matrix $\widehat{A}$ by repeating the $i$th row of $A$ $m_i$ times. Consider the row expansion of the "immanant" $\sum_{\sigma\in S_n}k_{\sigma}\prod_{i=1}^n \widehat{A}_{i,\sigma(i)},$ where $k_\sigma$ are some coefficients that need to be specified ("the signs"). The partition $\mathcal{I}=\{I_1,I_2,\ldots,I_m\}$ of the set $\{1,2,\ldots,n\}$ of the column indices from the original formula is $\large\{\sigma$(first $m_1$ indices), $\sigma$(next $m_2$ indices), $\ldots$, $\sigma$(last $m_m$ indices)$\large\}$ and the coefficient $\operatorname{sgn}(\mathcal{I})=\sum k_\sigma,$ with the sum taken over all permutations $\sigma$ corresponding to that partition $\mathcal{I}.$ In the case that $k_\sigma=1$ for all $\sigma\in S_n,$ we get the permanent of $\widehat{A},$ which is equal to the constant $(\prod_{i=1}^{m}m_i!)$ times (OP's expression for all coefficients equal to $1$). Although this is non-standard in the context of permanents, the constant can be eliminated by setting $k_{\sigma}=0$ for permutations $\sigma$ that "scramble" some of the parts $I_k$, cf the next paragraph. One can also consider the following "determinant" choice of $k_\sigma.$ It is defined to be $\operatorname{sgn}(\sigma)$ if the descent set of $\sigma$ is contained in $\{m_1,m_1+m_2,\ldots,m_1+\ldots+m_{m-1}\}$ and $0$ otherwise. In other words, for each partition $\mathcal{I}$ consider the element $\sigma_{\mathcal{I}}\in S_n$ which maps the first $m_1$ indices to $I_1$ preserving the order, the next $m_2$ indices to $I_2$ preserving the order, and so on, and set $\operatorname{sgn}(\mathcal{I})=\operatorname{sgn}(\sigma_{\mathcal{I}}).$ The resulting expression is always non-trivial and for $m=n,$ it yields $\det(A).$ On the other hand, if $m TITLE: Faster multiplication with a restricted set of multiplicands? QUESTION [12 upvotes]: Let $A$ be a set of $k>1$ distinct elements from a semigroup. We wish to compute the product $$ p=b_1 b_2 \cdots b_n$$ where each $b_i\in A$. Clearly $n-1$ multiplications suffice to compute $p$; can we do it with fewer? Let $m=m(b_1,...,b_n)$ be the minimum number of multiplications required to compute $p$. My specific questions are: Can we compute $m$ in polynomial time? Can we answer question #1 constructively, i.e. actually figure out which $m$ terms can be multiplied together to compute $p$? If not, can we do it with some approximation guarantee (like "at worst we use $2m$ terms"?) The interesting case (for my application, anyway) occurs when $n\gg k$ (so there are many common subexpressions) but the semigroup is noncommutative (so you cannot simply rearrange terms and use repeated doubling). Let me outline one simple (but insufficient) approach. Suppose that we begin by computing all products of (not necessarily distinct) pairs from $A$. Then we can compute $p$ with $k^2+(\lceil n/2 \rceil -1)$ multiplications by multiplying the terms two at a time. If $k^2\ll n$, then we have essentially halved the number of multiplications. Analogous reasoning can be extended to provide roughly an $\lfloor \log_k(n)-\log_k(\log_k(n)) \rfloor -1$ improvement in the number of multiplications. However, for highly repetitive strings we could do much better still. This problem arose in the context of parallel inference algorithms for hidden Markov models; the noncommutative semigroup in question consists of non-negative matrices over the reals. REPLY [3 votes]: This question has already been studied in the literature under the term "word chains". Doing a search on that term will turn up some relevant papers, such as http://dl.acm.org/citation.cfm?id=33329 http://www.sciencedirect.com/science/article/pii/0020019089901671 http://dl.acm.org/citation.cfm?id=79917 Basically one cannot get a much better compression (in general) than the one you propose. Caution: there is a different, unrelated problem also called "word chain" in the literature. Addendum: Yu. V. Merekin has also worked on this problem, but to my knowledge it has only appeared in Russian. http://www.mathnet.ru/php/person.phtml?personid=17890&option_lang=eng Addendum 2: And finally, there is a complexity result here: https://www.cs.virginia.edu/~shelat/papers/GrammarIEEE.pdf although I never managed to understand the proof.<|endoftext|> TITLE: Irreducibility of a class of polynomials QUESTION [10 upvotes]: This question is directly inspired by this question. Consider polynomials of the form $$p(x) = \prod_{i=1}^n(x-i)^2 - d.$$ For which values of $n$ and $d$ is $p(x)$ irreducible? There is a theorem of Polya, cited in the original question, which states that: If for $n$ integral values of $x,$ a polynomial $q(x)$ of degree $n$ has values which are different from $0,$ are smaller in absolute value than $$\frac{\lceil n/2\rceil!}{2^{\lceil n/2\rceil}},$$ then $q(x)$ is irreducible over $\mathbb{Q}.$ This seems to not quite apply to the given type of polynomial (though it would, if there were no squares). In this paper, there is a reference to a paper of Brauer, Brauer, and Hopf(!) from 1926 which addresses the question (of Schur) on irreducibility of polynomials of the form $g(f(x)),$ but it seems that their $g$ always has constant coefficient $1,$ so this is not directly applicable. Surely, technology has advanced since those days, and even the Galois groups can be computed (I assume the Galois group is the full symmetric group for most choices of $n$ and $d.$) What makes the whole thing strange is that the original question (with $n=2013$ and $d=2014$) was supposedly an entrance exam question at Peking university. So this is either child abuse, or there is some simple trick. REPLY [6 votes]: I prove here that $\prod_{i=1}^n (x-a_i)^2 + d$ is irreducible over $\mathbf{Q}$ whenever the $a_i$ are distinct integers, $n\ge 6$, and $d$ is squarefree, $d>1$, and $d\not\equiv 3\pmod{4}$. The hypotheses on $d$ ensure that the ring of algebraic integers in $\mathbf{Q}(\sqrt{-d})$ is $\mathbf{Z}[\sqrt{-d}]$, and that the only units in this ring are $\pm 1$. Write $f(x):=\prod_{i=1}^n (x-a_i)$, and let $\alpha$ be a root of $f(x)^2+d$, so that $\beta:=f(\alpha)$ satisfies $\beta^2=-d$. Suppose that $f(x)^2+d$ is reducible over $\mathbf{Q}$, so that $[\mathbf{Q}(\alpha):\mathbf{Q}]<\deg(f(x)^2+d)=2n$. Since $$[\mathbf{Q}(\alpha):\mathbf{Q}]=[\mathbf{Q}(\alpha):\mathbf{Q}(\beta)]\cdot [\mathbf{Q}(\beta):\mathbf{Q}] = 2[\mathbf{Q}(\alpha):\mathbf{Q}(\beta)],$$ it follows that $[\mathbf{Q}(\alpha):\mathbf{Q}(\beta)] TITLE: Computing Bredon Cohomology of Z/2-spheres? QUESTION [5 upvotes]: Can anyone suggest me how to calculate explicitly the Bredon Cohomology of the sphere(at least for 2-dimensional case) with antipodal Z/2-action with constant coefficient system associated to integers? REPLY [11 votes]: EDIT: per request. We can be pretty explicit in this case: the $n$-sphere with antipodal action has a $\Bbb Z/2$-equivariant cell structure whose $k$-skeleton is the $k$-sphere. As a result, we get a cellular chain complex of coefficient systems: $$ 0 \leftarrow C_0 \leftarrow C_2 \leftarrow \cdots \leftarrow C_n $$ where $C_k$ is the coefficient system associated to $(S^k,S^{k-1})$. According to Bredon's definitions, we calculate the value of the coefficient system on the trivial orbit $*$ as $H_k((S^k)^{\Bbb Z/2},(S^{k-1})^{\Bbb Z/2})$, which is trivial; we calculate the value on the orbit $\Bbb Z/2$ as $H_k(S^k,S^{k-1}) = \Bbb Z \oplus \Bbb Z$ where the generator $g$ of $\Bbb Z/2$-action swaps the two factors. Therefore, in each degree the coefficient group is the same coefficient system $C$. More, one can calculate the cellular chain maps using the same homology calculation (one knows that it has to recover the homology of $S^n$) and find that it is a sequence of maps $$ 0 \leftarrow C \stackrel{1-g}{\leftarrow} C \stackrel{1+g}{\leftarrow} \cdots $$ Now we apply Hom out to any coefficient system $M$. You can show $Hom(C,M)$ is $M(\Bbb Z/2)$. The resulting chain complex is $$ 0 \to M(\Bbb Z/2) \stackrel{1-g}{\to} M(\Bbb Z/2) \stackrel{1+g}{\to} M(\Bbb Z/2) \stackrel{1-g}{\to} \cdots $$ Substituting $M = \Bbb Z$, we get the cochain complex computing the cohomology of $\Bbb{RP}^n$. This is part of a more general result here when the action is free and properly discontinuous, and the group $\Bbb Z/2$ acts trivially on the value $\Bbb Z$ of the coefficient system on $\Bbb Z/2$. In these cases, the Bredon cohomology coincides with cohomology of the orbit space. Therefore, we're getting the cohomology of $(S^n) / \Bbb Z/2 = \Bbb{RP}^n$ with coefficients in $\Bbb Z$ (hence $\Bbb Z$ in degree 0, $\Bbb Z/2$ in even positive degrees, and 0 otherwise).<|endoftext|> TITLE: Deviation bound for the maximum of the norm of Wiener process QUESTION [6 upvotes]: Let $W(t)$ be an $n$-dimensional Wiener process. Denote by $\chi_n^2$ a chi-squared random variable with $n$ degrees of freedom. I have recently found the following inequality given without proof: $$ {\mathbf P} \left\{ \max_{t\in[0,2]} \|W(t)\|> x\right\}\leqslant 2 {\mathbf P} \bigl\{\chi_n^2>x^2/2\bigr\}. $$ Why is this bound true? What other deviation inequalities for the Bessel processes are known? REPLY [3 votes]: A general bound that's sometimes useful is given by Fernique's theorem. It's a general fact about Gaussian measures on Banach spaces, but in this case it gives the following: there are constants $C, \epsilon > 0$ (depending on the dimension $n$) such that $${\mathbf P} \left\{ \max_{t\in[0,2]} \|W(t)\|> x\right\}\leqslant C e^{-\epsilon x^2}.$$ You can find a proof in these lecture notes of mine (see Section 4.2).<|endoftext|> TITLE: Which sigma-ideals in a sigma-algebra are ideals of null sets? QUESTION [13 upvotes]: My question is motivated, to be somewhat vague, by an attempt to see how much a measure space is defined by the set of null sets. In other words, assume we are not given a concrete measure on a space but are just told which sets are null - what can we say about the measure space? Note that, for example, membership in $L^\infty$ depends only on the questions of which sets are null. Concretely, let $ \left (X, \mathcal{B} \right ) $ be a measurable space. A $\sigma$-ideal $\mathcal{I}$ is a subset of $\mathcal {B}$ which is closed under countable unions, contains the null set, and such that whenever $I \in \mathcal{I}$ and $B \in \mathcal{B}$, $I \cap B \in \mathcal{I}$. The standard example of $\sigma$-ideal is the family of null sets in a measure space $\left (X, \mathcal{B}, \mu \right) $. Question: Given an arbitrary $\sigma$-ideal in a $\sigma$-algebra, when is it the family of null sets for some measure on the space? I have not been able to find an answer to this question that does not amount to a tautology, for example, assuming that there exists a positive-definite function on $\mathcal{B}$ which essentially satisfies the axioms of a measure. My approach to the question so far has been to attempt to define an analogue of $L^\infty$ for this case: for a measurable $f:X \to \mathbb{C}$, define the essential supremum $ \left \|f \right \|_\infty$ to be the infimum over all positive numbers $M$ such that $|f(x)| \le M$ outside of some set in $\mathcal{I}$, and let $L^\infty (X, \mathcal{B}, \mathcal{I})$ be the space of all functions with finite essential supremum. It is easy to check that $L^\infty (X, \mathcal{B}, \mathcal{I})$ is always a Banach space, and by defining the $*$ operation to be conjugation as usual, it in fact becomes a $C^*$-algebra. If we knew that it is a Von-Neumann algebra (actually, a $W^*$-algebra, meaning it has a predual), it would be easy to conclude that $\mathcal{I}$ is the family of null sets for a measure, since then our space would be isomorphic to $L^\infty (Y, \mathcal{S}, \mu)$ and this isomorphism would necessarily take characteristic functions to characteristic functions. From here, I can think of two ways to proceed: Define also an analogue of $L^1$ in this case: since Radon-Nikodym allows us to identify $L^1$ with the space of all finite measures which are absolutely continuous wrt to a given measure, we define $L^1(X,\mathcal{B},\mathcal{I})$ to be the set of all measures which are zero on all sets in $\mathcal{I}$. The first problem with this is that in general, $(L^1)* \neq L^\infty$. This can be remedied by making a very reasonable assumption on $\mathcal{I}$, which I can say more about if needed, but even then, I don't see how to proceed. Find a representation of $L^\infty(X,\mathcal{B},\mathcal{I})$ on a separable Hilbert space. However, I don't know how to construct a suitable Hilbert space without some sort of tautological assumption on the $\sigma$-algebra of the style I have written above. Any reference or ideas would be greatly appreciated. REPLY [5 votes]: Your question has already been excellently answered from two points of view: (a) looking at the quotient $\sigma$-algebra (measurable sets modulo null sets): when is it a measure algebra? [Joseph Van Name, citing Jech] (b) looking, as you suggested, at the $C^*$-algebra of bounded measurable functions modulo the almost everywhere null functions: when is it a von Neumann algebra? [Dmitri Pavlov, citing I. Segal] Case (a) is more specialized than case (b), which in turn is not the most general nontrivial case. After two sections with definitions and basic facts (about measure spaces and $*$-operator algebras), the final section (the core of this answer) considers the missing cases (so someone might want to skip to the final section). 1: Four levels of measure spaces In terms of measure spaces $(X,\mu)$ (where the $\sigma$-algebra is the domain of definition of $\mu$ and the $\sigma$-ideal corresponds to $\mu$-null sets) the four cases are, in increasing level of generality: $\bullet$ (a) finite or equivalently $\sigma$-finite $\mu$ (using a countable partition of $X$ in sets $X_i$ with $\mu$-finite measure, rescaling $\mu$ by a suitable factor $a_i$ on each $X_i$ gives a convergent sum of the $a_i\mi(X_i)$ hence a finite measure). $\bullet$ (b') strictly localizable case. The measure space is direct sum of finite measure spaces, i.e. there is a (possibly uncountable) partition of $X$ in sets $X_i$ of finite $\mu$-measure such that: $\mu(Y)$ is defined iff each $\mu(Y\cap X_i)$ is defined, and then $\mu(Y)$ is the sum of the $\mu(Y\cap X_i)$. [Typical examples are Radon measures, the only measures considered by Bourbaki.] $\bullet$ (b) localizable case. The boolean algebra of measurable sets modulo null sets is complete (not only $\sigma$-complete); equivalently: the vector lattice of real measurable functions modulo null functions is conditionally Dedekind complete (not only $\sigma$-complete); equivalently: the $C^*$-algebra of complex measurable functions modulo null functions is a $W^*$-algebra (it has a predual as Banach space, equivalently: it has a faithful representation in a Hilbert space as von Neumann algebra [i.e. as a self-adjoint set of bounded linear operators which is the commutant of its commutant]). The equivalent conditions above only depend upon the "abstract" quotient algebra (measurable sets modulo null sets); on the contrary, the concept of strict localizability depends upon the "concrete" realization as algebra of sets. This is the only difference between the two concepts: localizability is about "measure algebras" (no $X$ is present), strict localizability is about "measure spaces" (a structured $X$). Assuming that $\mu$ is complete (all subsets of a $\mu$-null set are in the domain of $\mu$) and "semi-finite" (there are no sets of infinite measure with only null sets as subsets of finite measure), the following conditions are equivalent to localizability: Radon - Nykodym (finite measures absolutely continuous with respect to $\mu$ are represented by elements of $L^1$); $L^\infty$ is the Banach dual of $L^1$ by means of the natural duality $(f,g)\mapsto\int_X fgd\mu$. Under the same two assumptions, the existence of a lifting (a subalgebra of the bounded measurable functions that has exactly one representant for each element of $L^\infty$, or equivalently the analogue for $L^\infty$ as vector lattice, or equivalently the analogue for the Boolean algebra of measurable sets) is equivalent to strong localizability and so it is a strictly stronger condition. See Fremlin's book (free at Fremlin's home page) for more equivalent conditions for (strict) localizability, and examples to show distinctions of the two concepts (and strangeness of the incomplete or non semi-finite cases). $\bullet$ (c) as "general" case, one could take an arbitrary $\mu$. Then each $\sigma$-ideal in a $\sigma$-algebra gives a $\mu$. null on the $\sigma$-ideal and infinite outside (as noted by Stefan Waldmann). Avoiding such uninteresting cases, the most general interesting case the semi-finite measures; each $\mu$ produces an associated semi-finite measure $\mu'$ (the largest semi-finite $\mu'\leq\mu$): $\mu'(Y)$ is the sup of the finite $\mu(Y')$ with $Y'\subseteq Y$. Example: the semi-finite measure associated to a $\mu$ that takes only $0,\infty$ as values is the identically zero measure; hence, in general, the associated semi-finite measure gives a different boolean algebra of measurable sets modulo null sets and a different $L^\infty$. Strictly localizable (in particular, $\sigma$-finite) measures are semi-finite. Completing a measure (by declaring measurable each symmetric difference of a set in the domain of $\mu$ and a subset of a $\mu$-null set, and then extending $\mu$ in the only possible way) does not change (up to canonical isomorphisms) the (metric) boolean algebra of measurable sets modulo null sets, nor the space of step functions modulo null functions and its completions (the spaces $L^p$), nor the semi-finiteness (localizability, $\sigma$-finiteness, finiteness) status of the measure. As noted by Dmitri Pavlov, almost all notrivial theorems require localizability, and in fact are often equivalent to it. So localizability is in a sense the most central among the possible conditions on a measure space. There are a few exceptions. First: theorems requiring conditions stronger than localizability. Some cases are easily equivalent to finiteness (the constants are integrable, uniform convergence for integrable functions implies convergence of the integrals and more generally $L^q$ is contained in $L^p$ when $0\leq p TITLE: Proportion of square-free integers $n$ with $\gcd(n,\varphi(n))$ a prime QUESTION [13 upvotes]: A classical theorem of Erdos shows that the density of numbers n with $\gcd(n,\varphi(n))=1$ is zero. Actually, something stronger is proven: the number of integers $n\leq N$ with $\gcd(n,\varphi(n))=1$ is $(1+o(1))\frac{e^{-\gamma}N}{\log(\log(\log(N)))}$. Now I was wondering what can be said on the number (say $t(N)$) of square-free integers $n\leq N$ with $\gcd(n,\varphi(n))$ prime. To be honest, I am not interested on a full asymptotic description of this function of $N$. I am just happy to know whether $t(N)/N$ tends to zero. REPLY [10 votes]: Peter Mueller has given a full answer to your question already. But let me say a word about what's going on "behind the scenes." The key observation, which goes back at least to Erdos but is possibly older, is that for any fixed integer $M$, the set of $n$ with $\phi(n)$ divisible by $M$ has asymptotic density $1$. This is because all $n$ outside of a set of density zero have a prime factor $\equiv 1\pmod{M}$, which in turn traces back to the fact that the sum of the reciprocals of the primes $\equiv 1\pmod{M}$ diverges. Anyway, let's apply this with $M:= \prod_{p \leq z}p$ for some large $z$. (Here $p$ always denotes a prime.) Then $M \mid \phi(n)$ for almost all $n$, and so $\gcd(n,M) \mid \gcd(n,\phi(n))$ for almost all $n$. So if $\gcd(n,\phi(n))$ is prime for such an $n$, then $\gcd(n,M)=1$ or $\gcd(n,M)=p$. Now what are the odds that $\gcd(n,M)=1$ or $p$? The proportion of $n$ with $\gcd(n,M)=1$ is precisely $\phi(M)/M = \prod_{p \leq z}(1-1/p)$, and this is $\ll \frac{1}{\log{z}}$ by Mertens' theorem. What are the odds that $\gcd(n,M)=p$? Well, in this case, $p$ better be a prime $\leq z$, and then $\gcd(n,M)=p$ if and only if $p \mid n$ and $\gcd(n/p, M/p)=1$. So for each $p \leq z$, the odds are $$ \frac{1}{p} \frac{\phi(M/p)}{M/p} = \frac{1}{p} \prod_{\substack{q \leq z \\ q \neq p}}(1-1/q) \ll \frac{1}{p \log{z}};$$ summing over $p \leq z$ gives $$ \ll \frac{\log\log{z}}{\log{z}}. $$ Putting everything together, we see that the upper density of $n$ with $\gcd(n,\phi(n))$ prime is $\ll (\log\log{z})/\log{z}$ for each fixed large $z$. Since $z$ can be taken arbitrarily large, this density must in fact be $0$. Hence, $t(N)/N\to 0$. This is the same strategy Erdos used to prove the result of his that you quoted, except that one doesn't fix $z$ but instead takes $z$ as a function of $N$ (roughly $\log\log{N}$). One has to be much more careful about error terms if one does this though.<|endoftext|> TITLE: What does "variété à coins" translate to in English? QUESTION [7 upvotes]: I am attempting to translate Borel's "Cohomologie de $\text{SL}_{n}$ et valeurs de fonctions zeta aux points entiers" paper into English. Since I know no French, this is a rather crude process heavily involving Google Translate (and also a little common sense). However, I am unable to interpret the phrase "variété à coins". I am led to believe that this translates to something like "wedge variety" or "corner variety". Neither of these terms are familiar to me and a quick Google search turned up nothing, which leads me to my question... Would someone be kind enough to provide me with an English translation of the phrase "variété à coins"? For completeness, I include the sentence where the phrase is to be found:- "Mais en fait la construction précédente fournit une autre démonstration de l’injectivité de $j^{*}_{\Gamma}$; qui, à l’encontre de celle de [4], ne fait pas intervenir la compactification de $X/\Gamma$ en une variété à coins (cf. 5.6)." The reference for the paper is:- Borel, Armand. Cohomologie de $\text{SL}_{n}$ et valeurs de fonctions zeta aux points entiers. Ann. Sc. Norm. Super. Pisa 4 (1977) 613-636. REPLY [9 votes]: The translation is "manifold with corners".<|endoftext|> TITLE: A problem on the boundedness of maximal operator by using linearization method QUESTION [5 upvotes]: We know that the maximal operator is bounded on $L^{p}(\mathbb{R}^{n})$ where $n\geq 1$ and $10$ is a fixed constant and $f(x)$ is positive. This operator has been given by the form in Lacey's book: \begin{align*} Tf(x)=\sum_{I\in \mathfrak{D}}\frac{\chi_{E(I)}}{|I|}\int_{I}f(y)dy \end{align*} where $\mathfrak{D}$ denotes the set of dyadic intervals on $\mathbb{R}$ and for each $I\in \mathfrak{D}$ associate $E(I)\subseteq I$ so that the sets $\{E(I): I\in \mathfrak{D}\}$ are disjoint subsets. Lacey points out that $Mf(x)\leq 2Tf(x)$ in his book.(I could not verify this even if taking $C$ instead of $2$.) To Bound the maximal operator, we calculate \begin{align*} TT^{\star}f(x)=\sum_{I\in \mathfrak{D}}\sum_{J\in \mathfrak{D}}\chi_{E(I)}\frac{<\chi_{I},\chi_{J}>}{|I|}\frac{<\chi_{E(J)},f>}{|J|}\leq Tf(x)+T^{\star}f(x). \end{align*} Therefore, from the inequalities \begin{align*} \left \| Tf \right \|^{2}_{2}\leq \left \| T \right \|^{2}\left \| f \right \|^{2}_{2}\leq \left \| TT^{\star} \right \|\left \| f \right \|^{2}_{2}\leq 2\left \| T \right \|\left \| f \right \|^{2}_{2} \end{align*} we have $\left \| T \right \|\leq 2$ and then $\left \| Mf \right \|_{2}\leq C\left \| Tf \right \|_{2}\leq 2C\left \| f \right \|_{2}$. However, for the key step, the description of the choice of $E(I)$ in Lacey's book seems not too clear (maybe to me) . Hence, $\mathbf{my \ question}$ is: How to choose a desirable set $\{E(I): I\in \mathfrak{D}\}$ such that the pointwise inequality $Mf(x)\leq CT(x)$ is valid？ REPLY [4 votes]: I have just obtained an answer to this question but there would have some modifications that differ from Lacey's description. The detail is as follows (forgive me to pose it here): First, we should proof a Lemma (it is just as an exercise in Lacey's book, p.4) : $\textbf{Lemma.}$ $\mbox{}$For any interval $I$, there is an interval $J\in \mathfrak{D}\cup \mathfrak{D'}$ with $I\subseteq J$ and $|J|\leq 8|I|$, where $\mathfrak{D'}$ is another choice of dyadic intervals defined by $\mathfrak{D'}=\left \{ \left [j2^{k},(j+1)2^{k}\right)+(-1)^{k}\frac{2^{k}}{3}: j,k\in \mathbb{Z}\right \}$. Proof of the Lemma: $\mbox{}$ Suppose the length of $I$ satisfies $2^{k}\leq |I|<2^{k+1}$ with $k\in \mathbb{Z}$. We should only separate it into two cases. Firstly, $I\subseteq J$ where $J\in\mathfrak{D}$ and $|J|=2^{k+1} $. Then it is easy to get $|J|\leq 8|I|$. Secondly, on the other hand, there must exists only one point $j_{0}2^{k+1}$ contained in $I$. We denote $J_{1}=\left[j_{0}2^{k+1},(j_{0}+1)2^{k+1}\right)$ and $J_{s}=J_{1}+(s-1)2^{k+1}$, $s\in \mathbb{N}$, in which $J_{1}$ is the right interval which has a nonempty intersection with $I$. Note that there exists only one interval $J'\in \mathfrak{D}$ such that $J'\supseteq J_{1}$ with $|J'|=2^{k+3}$. Then, if $J'\bigcap J_{4}=\varnothing$, we choose $J=J'\in \mathfrak{D}$ and the $J$ satisfies the Lemma. If $J'\bigcap J_{4}\neq \varnothing$ (hence $J_{4}\subseteq J'$), from the two inequalities $2^{k+1}<\frac{2^{k+3}}{3}$ and $\frac{2^{k+3}}{3}<2^{k+3}-2^{k+1}$, we have $J'-\frac{2^{k+3}}{3}\supseteq I$. Suppose that the interval $J''\in \mathfrak{D}$ is the left one which is adjacent to $J'$ with legth $|J''|=2^{k+3}$. Also we have $J''+\frac{2^{k+3}}{3}\supseteq I$. Hence, there must exists a $J\in \mathfrak{D'}$ with length $|J|=2^{k+3}$ such that $J\supseteq I$ and $|J|\leq 8|I|$. Therefore we complete the proof of the Lemma. Now we can construct the set $\left \{E(I)\right \}$ and the operator $T$: For any fixed $x\in \mathbb{R}$, suppose that $Mf(x)<\infty$, there exists an interval $Q\ni x$ such that $Mf(x)\leq \frac{2}{|Q|} \int_{Q}f(y)dy$, (recall that $f\geq 0$). By Lemma, there exists an $I_{x}\in \mathfrak{D}\bigcup \mathfrak{D'}$ such that $I_{x}\supseteq Q$ and $|I_{x}|\leq 8|Q|$. Hence $Mf(x)\leq \frac{16}{|I_{x}|} \int_{I_{x}}f(y)dy$. We relabel $\left \{ I_{x}:x\in\mathbb{R} \right \}$ as $\left \{I_{j}:j\in\mathbb{N} \right \}$. For each $j$, we set $$\tilde{E}(I_{j})=\left \{ x\in I_{j}:Mf(x)\leq \frac{16}{|I_{j}|}\int_{I_{j}}f(y)dy \right \}$$ and let $$E(I_{j})=\tilde{E}(I_{j})\backslash \bigcup_{k=1}^{j-1}\tilde{E}(I_{k})$$ for $j>1$, and $E(I_{1})=\tilde{E}(I_{1})$. We can easily get $\bigcup_{k=1}^{\infty}E(I_{k})=\mathbb{R}$ and $E(I_{j_{1}})\bigcap E(I_{j_{2}})= \varnothing$, if $j_{1}\neq j_{2}$. Analogously, denote $$Tf(x)=\sum_{j=1}^{\infty}\frac{\chi_{E(I_{j})}(x)}{|I_{j}|}\int_{I_{j}}f(y)dy.$$ The most important thing is that the similar inequality $$Tf(x)\leq Mf(x)\leq 16Tf(x)$$ is now still valid for $x\in \mathbb{R}$. $\textbf{Supplement}$ $\mbox{}$ The operator $T$ defined in the above is a little different from the one in Lacey's book. However, the rest proof for the boundedness of maximal operator $M$ on $L^{2}(\mathbb{R})$ is still followed by the method of Lacey's. I have modified some mistakes in the proof of the Lemma. Now the proof will become more clear.<|endoftext|> TITLE: What is the order of the isotopy group of the Brieskorn homology 3-sphere? QUESTION [8 upvotes]: Let $\Sigma(p,q,r)$ be the Brieskorn homology 3-sphere with $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}<1$ (so not the 3-sphere or the Poincare sphere). The fundamental group is given by $$ \pi_1(\Sigma(p,q,r)=\langle a,b,c\, |\, a^p=b^q=c^r=abc \rangle $$, the centrally extended triangle group for $(p,q,r)$. The isotopy group $\pi_0(Diff(\Sigma(p,q,r)))$ must be finite as well as the outer automorphisms of the fundamental group $Out(\pi_1(\Sigma(p,q,r))$. Furthermore the map $$\pi_0(Diff(\Sigma(p,q,r)))\to Out(\pi_1(\Sigma(p,q,r))$$ is injective ( I found this result in D. McCullough, Virtually geometrically finite mapping class groups of 3-manifolds, J. Differential Geom. 33 (1991), no. 1, 1–65.) Now to my questions: Are there new results to show that this map is an isomorphism? Or, what is the order of the isotopy group? In particular, I'm interested in the example $\Sigma(2,5,7)$ (bounding a contractable smooth 4-manifold). I do not find any result about the outer automorphism of tre centrally extended triangle group. What is known? REPLY [10 votes]: In fact, much more is known: $Diff(\Sigma(p,q,r))\simeq Isom(\Sigma(p,q,r))$ when $\frac1p+\frac1q+\frac1r <1$ by a result of McCullough and Soma. The metric is a homogeneous metric on $\Sigma(p,q,r)$ modeled on the homogeneous space $\widetilde{SL_2(\mathbb{R})}$. In this case, the isometry group preserves the Seifert fibering, so descends to isometries of the quotient 2-orbifold $S^2(p,q,r)$, which is made from two hyperbolic triangles with angles $\pi/p,\pi/q,\pi/r$. There is a reflection isometry fixing $(p,q,r)$, which lifts to an action on the fiber by orientation reversal. The quotient of $\Sigma(p,q,r)$ by such an isometry is an orbifold whose underlying singular locus is a Montesinos link. If $p=q\neq r$, there is also an orientation preserving isometry given by an involution which exchanges the two orbifold points and the two isometric triangles. This isometry lifts to $\Sigma(p,q,r)$ preserving the fibers. If $p=q=r$, then there is a dihedral group action, together with the reflection. Thus, the identity component of $Isom(\Sigma(p,q,r))\cong S^1$, obtained by rotating along the fibers. Also, there is an extension by a finite group of isometries of $S^2(p,q,r)$. In the case of $\Sigma(2,5,7)$, this is just an extension by $\mathbb{Z}/2\mathbb{Z} \cong \pi_0(Diff(\Sigma(2,5,7)))$.<|endoftext|> TITLE: Constructing a vector field with given zeros on a torus QUESTION [6 upvotes]: By the Hopf-Poincaré theorem, the sum of the indices of the zeros of a vector field on the d-dimensional torus must equal zero. Given an even number of points $x_i$ on a d-dimensional torus, and numbers $n_i=\pm 1$ such that $\sum_i n_i=0$, does there always exist a vector field on the torus which has a zero of index $n_i$ at $x_i$ (and no other zeros)? If so, how would one go about constructing such a vector field? REPLY [6 votes]: Yes, this can be done with the Pontryagin-Thom construction. This answer is essentially the same as Pietro Majer's, by the way, but perhaps a bit more "hands on" and also more sketchy. The first step is to note that vector fields on the $d$-torus with zeros at points $x_i$ can be constructed from maps from the $d$-torus punctured at each $x_i$ to the $(d-1)$-dimensional sphere $S^{d-1}$. So how do we actually construct a map from this punctured $d$-torus to $S^{d-1}$ such that the degree of the map restricted to a sphere around the puncture $x_i$ is precisely $n_i$? Let us begin by noting that $\sum_i n_i=0$ implies that we can draw disjoint oriented curves in the punctured $d$-torus each beginning and ending at one of the punctures such that the net number of outgoing arrows at each $x_i$ is precisely $n_i$. (These oriented curves correspond roughly to the paths that the zero-antizeros make under the diffeomorphisms described by Pietro Majer). Here and below I will include some pictures I had lying around which eventually constructs a 2d vector field (actually a line field) in a square patch of the plane with three zeros, one of index -2, and two of index +1. In this first picture I have simply drawn in red curves as the oriented curves that I have just described. The punctures / eventual zeros are the blue points. The idea of the Pontryagin-Thom construction is to identify these oriented curves as the inverse image of the north pole on $S^{d-1}$. Let us now take tubular neighborhoods of these curves which are also all disjoint. A cross-section of each tube is a $(d-1)$-disk with a distinguished normal (pointing along the orientation of the curve) which we map to $S^{d-1}$ by a stereographic projection-like map so that the center of the disk maps to the north pole, the boundary of the disk maps to the south pole, and the orientation of the disk is such that the distinguished normal would correspond to the outward pointing normal of the standard $S^{d-1}$ embedded in $R^d$. In this picture I show the construction of the map in the tubular neighborhoods. I have depicted $\mathbb{RP}^1$ (as the set of orientations of double-headed arrows in the plane) on the right instead of $S^1$ but as the two spaces are homeomorphic the idea is the same. The red curves map to the north pole, in the case of this $\mathbb{RP}^1$, the vertical direction. The cross-section of the tubular neighborhood in this case is just an interval, and we map the open interval to the circle minus the south pole (or as depicted, $\mathbb{RP}^1$ minus the horizontal direction). This is the meaning of the pink and maroon shaded areas -- they map to the pink and maroon shaded parts of $\mathbb{RP}^1$. We've now constructed a map from the union of the tubular neighborhoods of the oriented curves to $S^{d-1}$. To complete the construction, note that we can map the rest of the punctured $d$-torus to the south pole of $S^{d-1}$ since the boundaries of all the tubular neighborhoods were also mapped there. This picture shows the resulting line field (to get a vector field, simply take the angles of the lines and multiply them all by 2). Observe that in the parts of the picture which were pink, the line field rotates from vertical to horizontal by turning clockwise, and in the maroon parts, the line field rotates from vertical to horizontal by turning counter-clockwise. I leave it to you to check the indices at each of the blue points. I've sketched the construction of a continuous map from the punctured $d$-torus to $S^{d-1}$ - this can be made smooth with partitions of unity. (I believe there are additional restrictions on analytic vector fields, e.g. the fact that meromorphic functions on elliptic curves must have at least two poles) See Milnor's "Topology from the differentiable viewpoint" for more details on the Pontryagin-Thom construction. REPLY [2 votes]: In two dimensions, an explicit construction can be given: First find a doubly periodic meromorphic function f with given zeros and poles (you can construct f explicitly in terms of the p function), and then take your vectorfield to be $f(z)/(1+|f(z)|^2)$.<|endoftext|> TITLE: Containment of $c_0$ QUESTION [12 upvotes]: I have the following question. I guess it's quite simple for experts. Unfortunately, I could not come up with an answer yet. Let $X$ be a Banach space which contains no copy of $c_0$. Does it impply that $X''$ (the bidual of $X$) contains no copy of $c_0$? REPLY [16 votes]: No. For complicated and important examples, consider any $\mathcal{L}_\infty$ space that does not contain a subspace isomorphic to $c_0$. The first such examples were constructed by Bourgain and Delbaen in the early 1980s. Some had their duals isomorphic to $\ell_1$. The constructions there were subsequently used by Argyros and Haydon to produce a space $X$ s.t. $X^*$ is isomorphic to $\ell_1$ but every operator on $X$ is of the form $\lambda I +K$ with $K$ compact. Since then much more has been done. For a simple example, consider $(\sum_{n=1}^\infty \ell_\infty^n)_1$. This space obviously does not contain a copy of $c_0$, and it is just an exercise to to point this prove that its dual contains a norm one complemented subspace isometric to $\ell_1$. Charles Stegall was the first to point this out. I think the right reference is lemma 1 in: Stegall, C. Banach spaces whose duals contain l1(Γ) with applications to the study of dual L1(μ) spaces. Trans. Amer. Math. Soc. 176 (1973), 463–477.<|endoftext|> TITLE: Group cochains invariant under the action of the symmetric group QUESTION [17 upvotes]: Let $G$ be a finite group and $A$ an abelian group. Recall the cochain groups $$ C^k = \{f: G^k \to A\} $$ and the coboundary map $$ \delta : C^k \to C^{k+1} $$ $$ (\delta f)(g_1, \ldots, g_{k+1}) = f(g_2, \ldots, g_{k+1}) + \cdots + (-1)^i f(g_1, \ldots, g_i g_{i+1}, \ldots, g_{k+1}) + \cdots + f(g_1, \ldots, g_k) $$ for the group cohomology of $G$ with coefficients in $A$. To each $k$-tuple $(g_1,\ldots, g_k)$ we can associate a labeling of the (oriented) edges of the standard $k$-simplex as follows. The oriented edge from vertex $i$ to vertex $j$, with $i TITLE: Decomposing the conjugacy representation of Sym$(n)$ for small $n$ QUESTION [5 upvotes]: I am trying to compute the decomposition of the conjugacy representation of some small symmetric groups. Perhaps someone has undertaken a similar calculation. My own calculations are quite slow, even for $n = 8$. I suppose my question is how large could one expect to scale this computation, if one desires an explicit decomposition. I am aware there are estimates of the parameters known for large $n$ (e.g. see http://www.maths.qmul.ac.uk/~twm/ConjRep.pdf). REPLY [2 votes]: Just some remarks which might be of interest. 1) Functions on every conjugacy class are subrepresentation (highly reducible typically). For S_n both conjugacy class and irreps are parametrized by Young diagrams. Lemma: functions on conjugacy class contain corresponding irrep. ("Corresponding" means conjugacy class and irrep parametrized by same diagram). We come to it some time ago, but it might be known to experts. (See some motivation for this question below). 2) Quite an amazing claim was discovered by A.Frumkin and R. Adin: "The conjugacy character of $S_n$ tends to be regular". They prove that in certain sense: the characters of regular and conjugacy representations are almost the same for large "n". More precise quantitative results in this direction were obtained in: Y. Roichman, Decomposition of the conjugacy representation of the symmetric groups, Israel J. Math. 97 (1997), 305–316. and in recent paper (already quoted by OP): DECOMPOSITION OF THE CONJUGACY REPRESENTATION FOR SYMMETRIC GROUPS AND SUBGROUP GROWTH Thomas W. Muller and Jan-Christoph Schlage-Puchta Here is some speculative motivation for interest in such a question: Conjugacy classes are known to be in bijection with irreps for all groups. Can we construct at least for some groups such bijection in some "good" way ? It sounds like orbit method. (Where functions on orbit more or less give desired irrep). However in orbit method one takes COadjoint orbits, not the adjoint ones, while conjugacy classes can be seen as exponentials of adjoint orbits. So we need kind of metric to identify adjoint and coadjoint orbits - for reductive groups over C we have Killing form. $S_n$ is reductive group $GL(F_1)$ - "field with one element". So may be some "Killing form" also exists ? The lemma above fits to this picture. Well, it is extremely speculative. There is orbit method for finite groups, but it works for nilpotent groups, not for groups like $GL(F_q)$, and, of course, $F_1$ is speculation by itself.<|endoftext|> TITLE: Reflexive (hyperbolic) graphs QUESTION [5 upvotes]: Is there an effective description of the graphs such that exactly one eigenvalue (of the conventional adjacency matrix) is $>2$ whereas all others are $\le2$? By "effective" I mean something that would let one iterate through all such graphs with reasonably few (say, a few dozens) vertices, or, better yet, to prove theorems about such graphs :) (Ideally, a list similar to that of Dynkin diagrams.) The question is motivated by this one: "spectrum of an adjacency matrix" and by my own research. It seems wild, but I'm not an expert. REPLY [2 votes]: I think this question is so hard, since we do not have any control on other eigenvalues, specially on the minimum of them. As an evidence (and maybe useful for your work), recently S. M. Cioab‎$‎\breve{a}‎$‎, W. H. Haemers, J. Vermette and W. Wong‎ in the paper with name: "The graphs with all but two eigenvalues equal to ‎‎$‎\mp‎$‎‎ 1", characterized all graphs that have two eigenvalues $r>1$ and $s<-1$, and all other eigenvalues are $1$ and $-1$. The way of their proof is very interesting and special. So, if $r>2$ and all other eigenvalues is less than or equal $1$, we have very difficult task to describe these graphs. Do you have any evidences that hopeful you for solving this problem in general?<|endoftext|> TITLE: Hilbert scheme of points on a surface QUESTION [5 upvotes]: Let $X$ be a complex surface and $X^{[n]}$ be the Hilbert scheme of finite analytic subspaces $Z$ for which $dimH^0(Z,\mathcal{O}_Z)=n$. I have trouble understanding $X^{[n]}$. That's what i've worked out: If we take $n$ distinct points $p_1,p_2,\cdots,p_n$ there is nothing to see ($I_Z=\{f\in\mathcal{O}_X|f(p_1)=\cdots f(p_n)=0\}$) If two points coincide ($p_1=p_2$) then i think the subspace $Z$ will have a defining ideal of the form $I_Z=\{f\in \mathcal{O}_Z|f(p_1)=0,df_{p_1}v=0,f(p_3)=0,\cdots,f(p_n)=0\}$ for a certain $v\in T_{p_1}X$. Then if $X^{(n)}$ is the symmetric n-product of $X$ and $h:X^{[n]}\rightarrow X^{(n)}$ is the Hilbert-Chow morphism it is that $h^{-1}((p_1,p_1,p_3,\cdots,p_n))\simeq \mathbb{P}^1$ i.e. varying the direction of $v\in T_{p_1}(X)$ we obtain different points in $h^{-1}((p_1,p_1,p_3,\cdots,p_n))$. But now if the three points coincide how is $I_Z$ defined (i can not understand it)? what is the fiber of $h$ above $(p_1,p_1,p_1,p_4,\cdots,p_n)$? I suppose it's "bigger" than $\mathbb{P}^1$.. How is the case of $i$ coinciding points ($3 TITLE: Are primes of density 0 in $a\cdot b^n+c$? QUESTION [7 upvotes]: Hooley proves in Applications of Sieves to the Theory of Numbers that there are only $o(x)$ numbers $n\le x$ such that $n\cdot2^n+1$ is a (Cullen) prime. The proof generalizes to forms $n\cdot2^{n+a}+b$. It seems evident that the same result would hold without the $n$ out front, that is, for $a,b,c$ with $a>0$ and $b>1$ there are only $o(x)$ primes of the form $a\cdot b^n+c$ with $n\le x.$ Has this been proved? This is essentially the same as a question asked on math.se which was never answered. REPLY [13 votes]: Unfortunately, proving such a result for a sequence like $2^n-3$ (say) seems very difficult. The reason one can handle $n\cdot2^n+1$ by sieve methods is because that sequence is equidistributed modulo each odd number $m$. (This equidistribution statement is maybe due to Rieger; I think Hooley uses a weaker result.) In the same tract, Hooley actually discusses the very question you ask about, maybe in a special case like $2^n-3$. But he is only able to get the $o(x)$ result under the assumption of GRH and another auxiliary ad-hoc assumption.<|endoftext|> TITLE: Counterexamples in universal algebra QUESTION [36 upvotes]: Universal algebra - roughly - is the study, construed broadly, of classes of algebraic structures (in a given language) defined by equations. Of course, it is really much more than that, but that's the best one-sentence definition I can give. I'm very new to universal algebra. So far I've found it incredibly interesting, mainly because it looks at things I was already interested in from a new (to me) perspective, and that's always good; but I don't at all have a firm command of even the basics. For example, the recent question Relatively free algebras in a variety generated by a single algebra made me realize that I'd naively accepted a very false statement: until I thought about it, I'd sort of taken for granted that A is always relatively free in Var(A). I'm sure this isn't the only false belief I have about universal algebra, and I'm sure I'll hold more in the future; and I'm also sure I'm not alone in this. So my question is: What are some notable counterexamples (to reasonable hypotheses a student in universal algebra might have) in universal algebra? I'm specifically interested in universal algebra because, well, it's fairly universal; it seems reasonable that a counterexample in universal algebra would be of interest to algebraists of many different stripes, and hopefully many outside algebra as well. At the same time, universal algebra is distinct enough that counterexamples in universal algebra would hopefully have their own flavor not found necessarily in questions like "counterexamples in group theory," "counterexamples in ring theory," etc. In that vein, I'd especially appreciate counterexamples about topics firmly within universal algebra - say, congruence lattices, or Mal'cev conditions - which nonetheless have "something to say" to other areas of mathematics. REPLY [2 votes]: For a while it was conjectured that modularity was the "weakest" congruence identity in the following sense: if there was some non-trivial identity satisfied by the congruence lattice of every algebra in a variety, then the variety would have to be congruence modular. This was refuted by Polin in 1977.<|endoftext|> TITLE: A group-theoretic perspective on Frankl's union closed problem QUESTION [37 upvotes]: Here is a group theoretic phrasing of a special case of the union closed conjecture: Question: Given a finite group $G$, is there an element of prime power order which is contained in at most half the subgroups of $G$? Motivation: Frankl's union closed sets conjecture has an equivalent phrasing in terms of lattices. It says that in every finite lattice there is a join irreducible element which is less than or equal to at most half the elements in the lattice. Finite lattices are always isomorphic to intervals of subgroups $[H,G]$ for groups $H,G$ (i.e. the lattice of subgroups $H\subseteq K \subseteq G$, with the subgroup relation). It is not known whether it suffices to take $H$ and $G$ to be finite. I wonder if anything about Frankl's conjecture is known for the case when $H$ is the trivial group. And that is precisely what is asked above. Notice that the elements of prime power order are in correspondence with the join irreducibles of the lattice of subgroups of $G$. Is the answer to the question above known? Is this known for special classes of groups? REPLY [14 votes]: In a somewhat different direction from Alireza: the conjecture is true for a large family of groups, including all abelian groups and many supersolvable groups. Let me start with the abelian case. Pick an element $g$ of highest possible prime-power order in an abelian group $G$. Then $\langle g \rangle$ has a complement: that is, there is a subgroup $K$ such that $K \langle g \rangle = G$ and $K \cap \langle g \rangle = 0$. In particular, $K \cong G / \langle g \rangle $ by the Isomorphism Theorems, and for any subgroup $X$ with $\langle g \rangle \subseteq X$ there is a corresponding subgroup $X \cap K$ which does not contain $\langle g \rangle$. In fact, the same argument applies to any $G$ and prime-power order element $g$ if 1) $g$ generates a normal subgroup, and 2) we can find a complement $K$ to $\langle g \rangle$ in G. In this situation, $[\langle g \rangle, G] \cong [1,K]$. (Edit: deleted discussion of supersolvable groups, which is irrelevant in light of update below.) UPDATE: The conjecture is true for all finite solvable groups. Proof: Let $G$ be a solvable group. Then $G$ has a normal subgroup $N$ of prime index, and some element of prime-power order $g \notin N$. Since $N$ is maximal in $G$, we have $\langle g,N \rangle = G$, and since $N$ is normal we have $\langle g,N \rangle = \langle g \rangle N$. Then by Dedekind's identity, we get that $\langle g \rangle (H \cap N) = H \cap G = H$ for any $H$ containing $\langle g \rangle$. The last tells us that the map from the interval $[\langle g \rangle,G] \rightarrow [1,N]$ given by $H \mapsto H \cap N$ is an injection. Since $N$ doesn't contain $g$, we get the conjectured statement. $\square$ Indeed, the above works whenever $G$ has a maximal normal subgroup of prime index. (E.g., for symmetric groups.) (Thanks to John Shareshian for several useful comments and discussion.) UPDATE 2: By combining my argument for solvable groups with Alireza's argument for (certain) finite simple groups, we can prove the conjecture for all groups having a quotient satisfying a certain condition. This very likely holds for all finite groups, but as I'll explain, a little bit of work remains on the group theory side. Theorem 1: If $G$ is a finite group with a normal subgroup $N$ such that $G/N$ is generated by (at most) two elements of prime power order, then $G$ satisfies the Frankl condition. Proof: A routine argument shows that we can pick $g,h \in G$ of prime power order such that $g$ and $h$ project in the quotient to generators of $G/N$. (If $G/N$ is cyclic, assume $g=h$.) Let $K = \langle g, h \rangle$, and notice that $KN = \langle N, g, h\rangle = G$. Then wlog there are at least as many subgroups containing $h$ as containing $g$ (else switch $h$ and $g$). This gives an injection $\varphi_1$ from the set of subgroups containing $g$ but not $h$ to the set of subgroups containing $h$ but not $g$. Moreover, if $H$ is a subgroup containing both $h$ and $g$, then $H \supseteq K$. It then follows from the Dedekind identity that $K (H \cap N) = H \cap KN = H$, hence the map $\varphi_2$ sending $H \mapsto H \cap N$ is an injection from $[K,G] \rightarrow [1,N]$. Since the intervals $[\langle h \rangle, G]$ and $[1,N]$ are disjoint, combining the two maps gives an injection from $[\langle g \rangle, G]$ to its complement, and we see the Frankl condition to hold. $\square$ Remark 1: The solvable case (above) is the case when $G/N$ is cyclic -- here, $K = \langle g,h \rangle = K$. The case where $N=1$ solved by Alireza has $K = G$, in which case we map $G$ to $1$ and all other subgroups containing $g$ to a subgroup containing $h$. Remark 2: The proof has a particularly pleasing explanation if we can take $g$ and $h$ to be conjugate. In this case, the intervals $[\langle g \rangle, G]$ and $[\langle h \rangle, G]$ are isomorphic lattices, with the isomorphism given by conjugation of subgroups. As we shall see below, this stronger conjugacy condition often does seem to occur. Now, if $N$ is a maximal normal subgroup, then $G/N$ is simple, so to verify the conjecture for all groups it suffices to verify that every finite simple group is generated by two elements of prime-power order. I managed to put together the following from results in the literature: Theorem 2: All but possibly finite many of the finite simple groups are generated by two elements of prime power order. The two elements may be taken to be conjugate. Any exceptions are classical groups. The conjugacy part will come from the following lemma: Lemma: If a finite simple group $G$ is generated by $x$ and $g$, where $x$ has order 2, then $G = \langle g, g^x \rangle$. Proof (of Lemma): Suppose not, and let $H = \langle g, g^x \rangle$. Then $H$ is easily seen to permute with $\langle x \rangle$, hence $H$ has index 2, contradicting simplicity of $G$. $\square$ We say a group is $(2,q)$-generated if it is generated by an element of order 2, together with an element of order $q$. To prove Theorem 2, it suffices to show that (modulo possible exceptions) every finite simple group is (2,$q$)-generated for some prime-power $q$. We go through the list of groups from the Classification Theorem. Many finite simple groups are known to be (2,3)-generated, and this condition has been well-studied. ((2,3)-generated groups can be represented as quotients of $PSL_2(\mathbb{Z})$, explaining the high level of interest in this condition.) A. Liebeck and Shalev in "Classical groups, probabilistic methods, and the (2,3)-generation problem" showed that, excluding the groups $PSp_4(q)$, all but finitely many of the classical groups are (2,3)-generated. B. Di Martino and Cazzola in "$(2,3)$-generation of $PSp(4,q),\ q=p^n,\ p\neq 2,3$" showed that $PSp_4(q)$ is (2,3)-generated, except in characteristic 2 or 3. C. Malle, and Lübeck and Malle in a series of papers culminating in "(2,3)-generation of exceptional groups" show that all exceptional groups of Lie type are (2,3)-generated, except for the Suzuki groups and $G(2)' \cong PSU_2(9)$. (Nick Gill also referenced this paper in his comment to Alireza's answer.) D. Woldar in "On Hurwitz generation and genus actions of sporadic groups" showed the sporadic groups to be (2,3)-generated, except for $M_{11}$, $M_{22}$, $M_{23}$, and $McL$ E. GA Miller proved in 1901 that the alternating groups $A_n$ are (2,3)-generated for $n \neq 6,7,8$. $A_6, A_7, A_8$ are easily verified to all be (2,5)-generated. Liebeck and Shalev in Proposition 6.4 of the same paper referenced above show that all but finitely many of the groups $PSp_4(2^n)$ and $PSp_4(3^n)$ are (2,5)-generated. Suzuki, in his original paper on the topic, showed that the Suzuki groups are (2,4)-generated. For the remaining sporadics, $M_{11}$ and $M_{22}$ are (2,4)-generated, while $M_{23}$ and $McL$ are (2,5)-generated. (The best place I found to read about this is in a series of papers on symmetric genus of groups. These papers are by various authors, frequently including Conder and Woldar.) Theorem 2 follows from the list above, together with the Classification of Finite Simple Groups. There is very likely a nicer approach to some of the simple group stuff than what I do above -- I'm far from an expert. (Maybe one of the real experts around here can remove the possible exceptions from the statement!!) I found the slides linked here of Maxim Vsemirnov on (2,3)-generated groups very helpful for understanding the exceptions to (2,3)-generation. In particular, he has a list of groups that are not (2,3)-generated on page 15 (which he conjectures to be complete). I've verified with GAP that all are (2,$q$)-generated for some prime $q$, except for the PSp's and $\Omega_8^+(2), P\Omega_8^+(3)$. Corollary: The conjecture holds for any $G$ with a maximal normal subgroup $N$ such that $G/N$ is not one of the exceptional (small) classical groups in Theorem 2.<|endoftext|> TITLE: Is there a similar theorem in the partially hyperbolic case? QUESTION [6 upvotes]: Theorem 5.10.3 from Introduction to dynamical systems, by Brin & Stuck: Let $f:M\rightarrow M$ be an Anosov diffeomorphism. Then the following are equivalent: $NW(f)=M$, every unstable manifold is dense in $M$, every stable manifold is dense in $M$ $f$ is topologically transitive, $f$ is topologically mixing. I want to know weather it is Ok to replace "Anosov diffeomorphism" with "Partially hyperbolic diffeomorphism" in the above theorem? You can find the definitions of hyperbolicity and partial hyperbolicity here and here REPLY [3 votes]: I think this runs into trouble at condition 1. For example, let $f : M \to M$ be an Anosov diffeomorphism that satisfies these conditions. The map $g: M \times S^1 \to M \times S^1$ defined by $g = f \times identity$ is partially hyperbolic (center direction is along the $S^1$ fibers) and the non-wandering set is the whole product manifold. But none of conditions 2-5 hold, since for any open set $U \subset S^1$ we have $g(M \times U) = M \times U$<|endoftext|> TITLE: On the generating functions for Euler characteristic of Hilbert schemes of points QUESTION [5 upvotes]: Let $d>1$ be an integer. If $n\geq 0$ is an integer we have a notion of $d$-dimensional partitions of $n$; the number of these, denoted $p_d(n)$, is the number of ways we can stack $n$ ($d$-dimensional) boxes in a corner of a $d$-dimensional "room". No closed formula is known for $p_d$, for any $d>1$. As far as I know, the generating function $\mathcal P_d$ for $p_d$ is known for $d=2,3$, but for no higher $d$'s: \begin{align} \mathcal P_2=\sum_{n\geq 0}p_2(n)t^n&=\prod_{k\geq 1}(1-t^k)^{-1},\notag\\ \mathcal P_3=\sum_{n\geq 0}p_3(n)t^n&=\prod_{k\geq 1}(1-t^k)^{-k}.\notag \end{align} However, it seems to me that to find $p_d(n)$ is to find the number of "higher dimensional Young Tableaux", and these correspond to monomial ideals in $\mathbb C^d$. So it should be true that $$p_d(n)=\chi(\textrm{Hilb}^n(\mathbb C^d)_0),$$ the topological Euler characteristic of the punctual Hilbert scheme. It is also true that, if $S$ is a smooth projective surface and $Y$ is a smooth projective threefold, then \begin{align} \sum_{n\geq 0}\chi(\textrm{Hilb}^nS)t^n&=\mathcal P_2^{\chi(S)}\,\,\,\,\,\,\,\textrm{(Göttsche's formula)}\notag\\ \sum_{n\geq 0}\chi(\textrm{Hilb}^nY)t^n&=\mathcal P_3^{\chi(Y)} \,\,\,\,\,\,\,\textrm{(Cheah's formula)}\notag \end{align} Question: do we have such formulas for any $d$? in other words, do we have $$\sum_{n\geq 0}\chi(\textrm{Hilb}^nX)t^n=\mathcal P_d^{\chi(X)}$$ for any smooth projective $X$ of dimension $d$? REPLY [2 votes]: Yes. Write $\mathcal P_d= 1 + p_d$, so $\mathcal P_d^{\chi(X)}= \sum_{k=0}^{\infty} \left( \begin{array}{c} \chi( X) \\ k \end{array}\right) p_d^k$. I will show that $\left( \begin{array}{c} \chi( X) \\ k \end{array}\right) p_d^k$ is the generating function for the stratum of $Hilb^n X$ consisting of subschemas that are supported on $k$ distinct points. This subscheme is a fibration over the variety $\left( \begin{array}{c} X \\ k \end{array}\right)$, the variety of all sets of $k$ distinct points in $X$. We can easily check that the Euler characteristic of $\left( \begin{array}{c} X \\ k \end{array}\right)$ is $\left( \begin{array}{c} \chi(X) \\ k \end{array}\right)$. The Euler characteristic of a fibration is the Euler characteristic of the base times the Euler characteristic of the fiber. So we must show that the Euler characteristic of the fiber is $p_d^k$. But this is clear - it's just the Hilbert scheme of subschemes supported exactly at $k$ distinct fixed points, which is just a $k$-fold product of the hilbert scheme of nonempty subschemes supported at a single point, which is $p_d$.<|endoftext|> TITLE: Topological classification of Morse-Smale flows QUESTION [5 upvotes]: Does anyone know of papers that mention the classification of non-singular Morse-Smale (NMS) flows up to topological equivalency? I am particularly interested in the flows on manifolds of dimension 3. But all I can find so far is the paper by Bin Yu on the classification of the NMS flows defined on the three-sphere. REPLY [5 votes]: Generally, if you hope a very clean list (like topological classification of surfaces) to completely classify NMS flows on 3-manifolds (even in three sphere) up to topological equivalence. It seems hopeless. One reason is that heteroclinic trajectories connecting saddle orbits will lead a complete list quite wild. If you reduce the requirement, there are several directions: if you only need to classify up to indexed links of periodic orbits, Wada give a nice algorithm to deal with the question in three sphere; if you don't need a clean list, there exists an attempt to consider complete classification. But the list seems as complicated as NMS flows themselves; if we don't care "heteroclinic trajectories connecting saddle orbits" (Certainly this is very small subset of all NMS flows), we can do some discussions, for instance, see here. More intersting discussions of NMS flows on three manifolds always are connected with the other topics. The following directions are what I know: Similar the connection of gradient flows and handle decompositions, Asimov connect NMS flows with round handle decompositions. And J. Morgan (1979) use it to nearly detect which 3-manifolds admitting NMS flows. And Yano considered for a given 3-manifold, which homotopy classes of nonsingular vector fields admit MS flows. People use NMS flows to represent the homotopy classes of nonsingular vector fields, see here. there are intersting connection between the periodic orbits of some NMS flows and integrable Hamiltonian systems, see here. People use some parts of NMS flows to construct more complicated system, for instance, Anosov flows on 3-manifolds, see here and here.<|endoftext|> TITLE: Are there principal $G$-bundles whose holonomy group is $G$? QUESTION [11 upvotes]: While studying universal constructions on principal bundles, I've stuck on a quite a basic question, namely: Given a Lie group $G$, does there exist a principal $G$-bundle $\pi \colon P \to B$, for some base $B$, that admits a connection $\theta$ whose holonomy group is the full group $G$? I suppose the answer is yes, but I have no idea of how to prove it. REPLY [16 votes]: As I mentioned in my comment, when $G$ is connected, you can do this with $B=\mathbb{R}^2$ and $P$ being the trivial bundle $P = G\times\mathbb{R}^2$. Here is one construction: Let $\frak{g}$ be the Lie algebra of $G$ and let $\gamma:TG\to\frak{g}$ be the canonical left invariant, $\frak{g}$-valued $1$-form on $G$. Let $(r,\theta)$ be polar coordinates on $\mathbb{R}^2$, let $f = f(\theta):S^1\to\frak{g}$ be a curve such that there are values $\theta_1,\ldots,\theta_m$ such that the elements $f(\theta_i)$ for $1\le i\le m$ form a basis for $\frak{g}$ and such that $f'$ vanishes in an $\epsilon$-interval about each $\theta_i$ for some $\epsilon > 0$, and let $\rho=\rho(r)$ be a smooth, even function of $r$ that vanishes identically in a neighborhood of $r=0$ but has $\rho'(r)\equiv1$ for $r\ge1$. Now consider the connection form $$ \alpha= \gamma + \mathrm{Ad}(g^{-1})\bigl(\rho(r)f(\theta)d\theta\bigr), $$ where $g:P\to G$ is the projection onto the first factor. (When $G$ is a matrix group, this is just $\alpha = g^{-1}dg +g^{-1}\bigl(\rho(r)f(\theta)d\theta\bigr)g$.) Then I claim that the holonomy group of this connection is all of $G$. Since $G$ is connected, it suffices to show that the Lie algebra of the holonomy group is all of $\frak{g}$ because, by Borel and Lichnerowicz, we know that the holonomy is a Lie subgroup of $G$. To do this, note that the given trivialization is is $\alpha$-parallel along each radial line through the origin $(r=0)$, since, along such a line, $d\theta=0$, implying that $\alpha = g^{-1}dg$ along such a line. On the other hand, because the curvature is $$ d\alpha + \tfrac12[\alpha,\alpha] = \mathrm{Ad}(g^{-1})\bigl(\rho'(r)f(\theta)\ dr\wedge d\theta\bigr), $$ one sees that, if one takes parallel translation starting at $\bigl(e,(0,0)\bigr)\in G\times\mathbb{R}^2=P$ above the curve that goes out along the radial path from $r=0$ to $r=1$ at angle $\theta_i$, goes counterclockwise around the box $[1,1{+}t]\times[\theta_i,\theta_i{+}\epsilon]$, and then returns to the origin ($r=0$) back along the initial radial path, then this will wind up at the point $\bigl(\exp(tf(\theta_i)),(0,0)\bigr)\in G\times\mathbb{R}^2=P$. (Moreover, $\alpha$-parallel translation of the curvature of $\alpha$ at $(1,\theta_i)$ back to the origin along the radial segment will yield a curvature that takes the value $f(\theta_i)$ at $\bigl(e,(0,0)\bigr)\in G\times\mathbb{R}^2=P$, so the Lie algebra of the holonomy group is all of $\frak{g}$.) More directly, concatenating $m$ of these 'lassoes' about the points $(1,\theta_i)$ with appropriate values of $t$ will yield an $\alpha$-horizontal curve connecting $\bigl(e,(0,0)\bigr)\in P$ to the point $$ \bigl(\exp(t_1f(\theta_i))\cdots\exp(t_mf(\theta_m)),(0,0)\bigr)\in P. $$ Thus, one can reach an open set in the fiber over $(0,0)$ starting from $\bigl(e,(0,0)\bigr)\in P$ and traveling along $\alpha$-horizontal curves. Thus, since $G$ is connected, the holonomy of $\alpha$ is all of $G$.<|endoftext|> TITLE: Analogy between topology and algebraic geometry QUESTION [16 upvotes]: In topos theory, there are many generalizations of topological concepts. For example, open, closed, proper and etale morphisms between toposes. However, there are also such analogous concepts in algebraic geometry. My question is that do these concepts actually coincide? I mean, for example, a proper morphism of schemes actually induces a proper geometric morphism between some toposes induced by schemes (like etale topos)? Although I mensioned about only morphisms, I want to know such analogous concepts in topology and algebraic geometry which are coincide at the level of toposes. REPLY [3 votes]: I found this question while I was also wondering about proper maps of schemes versus proper geometric morphisms of their small étale toposes. I will add a partial result for future reference. I'll consider toposes of sheaves of sets (no ringed toposes). A topos $\mathcal{E}$ is tidy or strongly compact iff the global sections functor $\Gamma : \mathcal{E} \to \mathbf{Sets}$ preserved filtered colimits, and it is proper or compact iff $\Gamma$ preserves directed unions of subterminal objects. All tidy toposes are proper. It is shown in SGA4, Exposé VI, Exemple 1.22 that the small étale topos $X_\mathrm{\acute{e}t}$ is tidy if and only if $X$ is quasi-compact quasi-separated (and similarly for the small Zariski topos). So if $X$ is a proper scheme, then $X_\mathrm{\acute{e}t}$ is a tidy (in particular, proper) topos. But if $X$ is an affine scheme, then the associated topos $X_\mathrm{\acute{e}t}$ is proper as well.<|endoftext|> TITLE: Generalization of Euler four square formula? QUESTION [5 upvotes]: It is well known that the number of distinct decompositions of a positive natural number N into sum of four squares of integers is equal to 8*sum(d : d|N and not(4|d)) In other words, this is the number of distinct integer solutions of the quadratic form x^2+y^2+z^2+w^2=N How do I count solutions for more general quadratic forms? For my purposes it suffice to limit the question to diagonal forms a*x^2+b*y^2+c*z^2+d*w^2=N In fact if it helps to simplify the question. The specific instance of interest right now is x^2+y^2+z^2+2*w^2=2^L , where L is an integer. Thanks, Alex-- REPLY [6 votes]: The representation numbers for your specific quaternary quadratic form are determined in Theorem 5.1 of this paper by Alaca^2, Lemire, and Williams: http://people.math.carleton.ca/~williams/papers/pdf/330.pdf In particular, that theorem confirms the conjecture Noam made in the comments. For related results, see also this paper http://people.math.carleton.ca/~williams/papers/pdf/310.pdf and the references therein.<|endoftext|> TITLE: Is it always possible to "encircle" exactly $n$ points in an infinite subset of $\mathbb{R}^d$ without limit points? QUESTION [9 upvotes]: Let $d$ be a positive integer, and let $\mathbb{R}^d$ be endowed with the Euclidean metric. Given an infinite set $S \subset \mathbb{R}^d$ without limit points and a positive integer $n$, is there always a point $p \in \mathbb{R}^d$ and a radius $r \in \mathbb{R}^+$ such that the $r$-neighbourhood of $p$ contains exactly $n$ points in $S$? If the answer is yes, what can be said in general about in which metric spaces this holds? REPLY [13 votes]: Yes, this is possible. Pick $p$ generic, so that $$\not \exists\,\,\, x,y\in S\quad\text{s.t.}\quad d(p,x)=d(p,y).$$ Then slowly increase the radius until the $r$-neighborhood of $p$ contains exactly $n$ points of $S$.<|endoftext|> TITLE: Differentials in the Lyndon-Hoschild-Serre Sequence for p=0 QUESTION [6 upvotes]: I'm interested in whether there is a simple description of the differentials in the first column of the LHS spectral sequence (the column with $E_2^{0,q}=H^0(BK,H^q(BG))=H^q(BG)^K$ for a short exact sequence $$ 1 \to G \to H \to K \to 1. $$ I believe these should be much simpler to understand than the general differentials, since when the above sequence splits these differentials are all zero (unless I'm mistaken), which is certainly not the case for $p\geq 1$. I would be satisfied with some formulas for small $q$ (say $\leq 6$ or so) such as $d_3(x) =$ the contraction of $x$ with the extension class $\omega$. REPLY [7 votes]: Johannes Huebschmann has written several papers on this topic. Here are three references: Automorphisms of group extensions and differentials in the Lyndon-Hochschild-Serre spectral sequence. J. Algebra 72 (1981), no. 2, 296–334. Group extensions, crossed pairs and an eight term exact sequence. J. Reine Angew. Math. 321 (1981), 150–172. Sur les premières différentielles de la suite spectrale cohomologique d'une extension de groupes. (French. English summary) C. R. Acad. Sci. Paris Sér. A-B 285 (1977), no. 15, A929–A931. In these papers you will find descriptions of the differentials $d_2^{0,2}:E_2^{0,2}\to E_2^{2,1}$, $d_2^{0,1}:E_2^{0,1}\to E_2^{2,0}$ and $d_3^{0,2}:E_3^{0,2}\to E_3^{3,0}$, among others.<|endoftext|> TITLE: Radon-Nikodym derivatives as limits of ratios QUESTION [8 upvotes]: Let $\mu_1$ and $\mu_2$ be measures with $\mu_1 \ll \mu_2$. Suppose we can characterize (a version of) their Radon-Nikodym derivative this way: $$\frac{d\mu_1}{d\mu_2}(x) = \lim_{n \to \infty} \frac{\mu_1(B_n)}{\mu_2(B_n)}$$ where $\bigcap_{n \in \mathbb{N}} B_n = \{x\}$. Q. Under what conditions for $\mu_1$, $\mu_2$ and $(B_n)_{n \in \mathbb{N}}$ does this hold a.e.? For example, it appears to hold if $\mu_1$ and $\mu_2$ are defined on the Borel sets of $\mathbb{R}$ and have densities w.r.t. Lebesgue measure, and $(B_n)_{n \in \mathbb{N}}$ is a sequence of intervals converging on $\{x\}$. I'm interested in the most general known conditions under which Radon-Nikodym derivatives can be defined analogously to how derivatives are defined in differential calculus. EDIT: In real life, I've just been pointed to a fairly general theorem: Bogachev, Measure Theory, vol. 1, Theorem 5.8.8. Restated: Theorem. Let $\mu_1$ and $\mu_2$ be nonnegative measures on the Borel sets of $\mathbb{R}^n$ that are finite on all balls, such that $\mu_1 \ll \mu_2$. Let $(B_n)_{n \in \mathbb{N}}$ be a descending sequence of balls, each with center $x$, with intersection $\{x\}$. Then the above equation holds $\mu_2$-a.e. I suspect this can be extended to sequences of measurable sets with a uniformity condition, as in Rudin's generalization of Lebesgue differentiation (Real and Complex Analysis, Theorem 7.10). REPLY [5 votes]: This question has been thoroughly investigated in the literature a while back. For example see most of the book Hayes, C. A.; Pauc, C. Y., Derivation and martingales, Ergebnisse der Mathematik und ihrer Grenzgebiete. 49. Berlin-Heidelberg-New York: Springer-Verlag. VII, 203 p. (1970). ZBL0192.40604., or chapter 7 in the book Edgar, G. A.; Sucheston, Louis, Stopping times and directed processes., Encyclopedia of Mathematics and Its Applications 47. Cambridge: Cambridge University Press (ISBN 978-0-521-13508-5/pbk). xii, 428 p. (2010). ZBL1189.60074. Three results which stand out are that it is enough to assume any one of the following conditions: 1 the underlying space is the real line ($\mu_1,\mu_2$ are arbitrary measures finite on compacts), and $B_n$ are intervals which contain $x$ and whose length goes to zero: see J.L. Doob's book Doob, Joseph L., Measure theory, Graduate Texts in Mathematics. 143. New York: Springer-Verlag. xii, 210 p. (1994). ZBL0791.28001. 2 the sets $B_n$ have some uniformity in their shape and they contain $x$ in their interior (as you described in the book by Rudin) 3 $f:=\frac{d\mu_1}{d\mu_2}$ is not just in $L^1(\mu_2)$ but rather satisfies stronger integrability properties. For example in $\mathbb{R}^n$ if $\mu_2$ is the Lebesgue measure it is enough to ask that $|f|\log^+(|f|)^{n-1}\in L^1(\mu_2)$; this latter result appears in both books cited above, and also as Theorem 2.2.1 in the book Khoshnevisan, Davar, Multiparameter processes. An introduction to random fields, Springer Monographs in Mathematics. New York, NY: Springer. xix, 584 p. (2002). ZBL1005.60005.<|endoftext|> TITLE: twisted poincare duality QUESTION [5 upvotes]: Let $M$ be a closed smooth oriented manifold of dimension $n$. Suppose that $\pi:L\longrightarrow M$ is a line bundle with a flat connection $\nabla$. Consider the space of $L$-valued differential forms: $$\Omega^{*}(L):=\Omega^{*}(M)\otimes L$$ The flat connection induce a differential operator on $\Omega^{*}(L)$, therefore we can define the $L$-valued de Rham cohomology (or twisted cohomology) $H^{*}(M;L)$. So is there a Poincar$\acute{e}$ duality for such twisted cohomology by a flat line bundle? REPLY [4 votes]: Yes, there is a perfect pairing $$ H^p(M;L)\otimes H^q(M;L^*)\to\mathbb{R}, $$ $p+q=n$.<|endoftext|> TITLE: conjugacy classes in anisotropic semisimple groups QUESTION [5 upvotes]: You have an anistropic semisimple algebraic group $G$ defined over a non-archimedean local field $k$. When can you say that the $k$-rational part of the conjugacy class of a $k$-rational point is compact? REPLY [3 votes]: Since the literature is rather complicated (and scattered) when $k$ fails to be perfect, it's worthwhile to add to what Aakumadula says. Over a perfect field, say in characteristic 0, the answer to your questions is straightforward based on older work of Borel and Tits. As I commented, Cor. 9.4 in their foundational paper on reductive groups here shows that (over a local field of characteristic 0) $G_k$ is compact in the ultrametric topology iff $G$ is reductive and $k$-anisotropic. Moreover, their Cor. 8.5 shows in this anisotropic situation that $G_k$ consists of semisimple elements. (As in their $\S1$ and the earlier Chevalley seminar, it's true quite generally that in a reductive group the semisimple classes are precisely the closed ones.) There are many more relevant papers by Tits, sometimes in collaboration with Borel or Bruhat, in which he explores the structure and classification of reductive groups over local fields. An interesting early paper in a Brussels conference volume includes some remarks on the terms $k$-isotropic and $k$-anisotropic, which arise in part from the study of quadratic forms and orthogonal groups but tend to contradict existing notions of "isotropic". In any case, these terms have persisted. This paper is probably hard to track down (and unfortunately there is no collection of his papers): Groupes semi-simples isotropes. 1962 Colloq. Th´eorie des Groupes Alg´ebriques (Bruxelles, 1962) pp. 137–147. Librairie Universitaire, Louvain; Gauthier-Villars, Paris One result stated here by Tits shows how delicate the cas of imperfect fields is: here a simple algebraic group $G$ defined over $k$ is $k$-anisotropic if and only if $G_k$ contains no "good" unipotent element other than the identity (in a certain sense of "good" explored further in a paper with Borel in Invent. Math. 12 (1971), which can be found online at GDZ. For a more modern treatment of structure theory of reductive groups over arbitrary (especially local) fields, you should probably look at the monograph Pseudo-reductive Groups by Conrad-Gabber-Prasad (Cambridge Univ. Press, 2010).<|endoftext|> TITLE: Chern classes of the sheaf of LOG differentials QUESTION [6 upvotes]: Let $\Omega_X^1(\log D)$ be the (locally free) of logarithmic differentials on a smooth projective variety $X$ with respect to a simple normal crossing divisor $D$. What are the Chern classes of $\Omega_X^1(\log D)$ in terms of the Chern classes of $\Omega_X^1$ and $D$? I think, the first one is $$c_1(\Omega_X^1(\log D))= c_1(\Omega_X^1)+D.$$ REPLY [3 votes]: Let $D_1,\ldots,D_m$ be the irreducible components of $D$. The reasoning in Prp 2.3 in MR1240599 gives an exact sequence $$0\longrightarrow \Omega_X^1\longrightarrow \Omega_X^1(\log D)\longrightarrow\bigoplus_{i=1}^m{\mathcal O}_{D_i}\longrightarrow 0.$$ Thus, the total Chern class is given by $$c\big(\Omega_X^1(\log D)\big)=c(\Omega_X^1)\Big/\prod_{i=1}^m\big(1-[D_i]\big).$$ Mohammad's formula for the first Chern class is correct.<|endoftext|> TITLE: Is 0' of PA degree relative to a non-low set? QUESTION [6 upvotes]: Definitions: A set $X$ is of PA degree relative to a set $Y$ if every infinite $Y$-computable binary tree has an infinite $X$-computable path. A set $X$ is low if $X'$ is computable from $\emptyset'$. Easy facts: By the relativized Low basis theorem and using the fact that a low relative to a low is still low, $\emptyset'$ is of PA degree relative to every low set. Of course, if $\emptyset'$ is of PA degree relative to a set $X$, then $X$ is $\emptyset'$-computable. Question: Are there non-low set $Y$ such that $0'$ is of PA degree relative to $Y$ ? REPLY [6 votes]: No, by the Arslanov completeness criterion $0'$ is only DNC (Diagonally non-computable) relative to low sets. And PA implies DNC.<|endoftext|> TITLE: Equivalence of Hadamard Graph and Hadamard Matrix QUESTION [5 upvotes]: I'm reading Distance Regular Graphs by Brouwer, Cohen, and Neumaier. In section 1.8, they explained Hadamard graphs. Conversion from a Hadamard Matrix into a Hadamard Graph An $n$-Hadamard graph $G$ is a graph on $4n$ vertices defined in terms of a Hadamard matrix of order $n$ $H_n = h_{ij}$ as follows: Define $4n$ symbols $r_i^+$, $r_i^-$, $c_i^+$, and $c_i^-$, where $r$ stands for row and $c$ stands for column and take these as the vertices of the graph. Then add two types of edges between row vertices and column vertices based on the sign of $h_{ij}$: \begin{equation*} \text{parallel edges $(r_i^+, c_j^+)$ and $(r_i^-, c_j^-)$ if $h_{ij} = +1$} \\ \text{crossing edges $(r_i^+, c_j^-)$ and $(r_i^-, c_j^+)$ if $h_{ij} = -1$} \end{equation*} Then the graph $G$ will be a bipartite graph where the set of vertices is partitioned into row vertex set of $2n$ vertices and column vertex set of $2n$ vertices. And there will be $2n^2$ edges. Equivalence between Graph and Matrix In theorem 1.8.1 in their book, they showed that $G$ is a distance-regular graph with an intersection array $\{n,n-1,\frac{n}{2},1;1,\frac{n}{2},n-1,n\}$ if and only if the matrix $H$ is Hadamard matrix of order $n$. My Question Their proof of this theorem seems rather brief and I have hard time in understanding the equivalence. Especially, I don't understand what role two orthogonal rows or columns in the matrix $H_n$ play in the graph $G$ so that distance-regularity is achieved. The book cited three papers for the proof, but I cannot find any of them in the Internet. Can anyone explain what's the idea or intuition behind the proof of equivalence? REPLY [6 votes]: First, the book by Brouwer, Cohen and Neumaier is known for a degree of terseness; any one who uses it will have struggled with it at some point. You describe a construction that, from an $n\times n$ Hadamard matrix, produces a bipartite graph on $4n$ vertices that is regular of degree $n$. This graph has diameter four; moreover for each vertex $u$ there is a unique vertex $u'$ at distance four from $u$. It follows that the the $4n$ vertices may be partitioned into $2n$ pairs. So the way to proceed is to prove that these comments about pairs of vertices at distance four are correct and then, using this, prove that the parameters $b_i$ and $c_i$ are well-defined. This reduces to showing that if $v$ is at distance two from a vertex $u$, then $v$ has exactly $n/2$ neighbours in common with $u$. (This is where the orthogonality of the rows of $H$ enters.)<|endoftext|> TITLE: Inequality in information theory QUESTION [8 upvotes]: I am reading the paper "chain independence and common information" (http://ttic.uchicago.edu/~yury/papers/independ.pdf). In this paper, an inequality is used several times (without proof) which looks interesting to me. I would appreciate if anybody can help me prove it. The inequality is as follows: For random variables $X$, $Y$, $Z$ we have $$H(Z)\leq H(Z|X) + H(Z|Y) + I(X;Y)$$ where $H(\cdot)$ is the entropy and $I(\cdot;\cdot)$ is mutual information. As far as I see in the paper, there is no restriction on the structure of random variables $X,Y$ and $Z$, so they are arbitrary. (Look at the first page or second page of the paper in the proof of Theorem I) REPLY [3 votes]: I'd prefer to keep the symmetry of $X$ and $Y$ in the argument. Then $$ H(Z|X) + H(Z|Y) + I(X,Y) \\ = H(X\vee Z) - H(X) + H(Y\vee Z) - H(Y) + H(X) +H(Y) -H(X\vee Y) \\ = H(X\vee Z) + H(Y\vee Z) - H(X\vee Y) \\ = 2 H(Z) + H(X|Z) + H(Y|Z) - H(X\vee Y) \\ \ge 2H(Z) + H(X\vee Y|Z) - H(X\vee Y) \\ = H(Z) + H(X\vee Y\vee Z) - H(X\vee Y) \\ \ge H(Z) $$ Several lines here are totally obvious, but I preferred to keep all the details. In what concerns the notation, I have actually replaced the random variables from the original formulation with the associated partitions of the base probability space (so that the partition $X\vee Y$ corresponds to the joint distribution of $X$ and $Y$, etc.). Actually the language of partitions is much more appropriate for dealing with problems of this kind. Of course, this argument (as well as the original question) only makes sense if all the partitions ($\equiv$ random variables) involved are discrete and have finite entropy.<|endoftext|> TITLE: A binomial determinant fomula QUESTION [11 upvotes]: Is there an existing or elementary proof of the determinant identity $ \det_{1\le i,j\le n}\left( \binom{i}{2j}+ \binom{-i}{2j}\right)=1 $? REPLY [17 votes]: This is true, and in fact you can show a slightly more general fact: $$\det_{1\le i,j\le n}\left( \binom{x_i}{2j}+ \binom{-x_i}{2j}\right)=\prod_{i=1}^n x_i^2 \prod_{i TITLE: A lagrangian version of the Withney theorem QUESTION [9 upvotes]: Let $M$ be a smooth n dimensional manifold. Is there an smooth embedding $f:M \to \mathbb{R}^{2n}$ whose image is a Lagrangian submanifold of $\mathbb{R}^{2n}$? REPLY [12 votes]: No: being a Lagrangian submanifold of $\mathbb{R}^{2n}$ imposes strong conditions on $M$. If $M$ is embedded in $\mathbb{R}^{2n}$, the bundle $T_M\oplus N$ (normal bundle) is trivial; if $M$ is Lagrangian, the symplectic form induces an isomorphism $N\cong T_M^*$. Thus $T_M\oplus T_M$ is trivial; this implies for instance that the Pontryagin classes of $M$ are trivial in $H^*(M,\mathbb{Q})$. So e.g. $\mathbb{CP}^2$ cannot be embedded as a Lagrangian submanifold of $\mathbb{R}^{8}$.<|endoftext|> TITLE: What is known about equiconsistency of PFA and existence of supercompact cardinals? QUESTION [10 upvotes]: Question: What is the last status of known partial results and new approaches on the problem of equiconsistency of PFA and existence of supercompact cardinals? REPLY [22 votes]: We do not know that $\mathsf{PFA}$ and supercompactness are equiconsistent. In a sense, we are far from knowing, but in another sense, most set theorists who have worked on the problem are almost certain. The problem is that the tools we have for deriving lower bounds in consistency strength are not strong enough to reach supercompactness (or even $\kappa^+$-strong compactness). As far as the current tools can reach, we know that $\mathsf{PFA}$ is at least as strong as that. Typically, lower bounds are established not by using the full power of $\mathsf{PFA}$ but instead by verifying that some combinatorial consequence of $\mathsf{PFA}$ already requires at least that lower bound. For years, the only combinatorial consequence of $\mathsf{PFA}$ that we could extract such strength from was the failure of square principles, first established by Todorcevic, in Stevo Todorcevic. A note on the proper forcing axiom, in Axiomatic Set Theory (Boulder, Colorado, 1983), Contemporary Mathematics, vol. 31, American Mathematical Society, Providence, RI, 1984, 209-218. MR0763902 (86f:03089). Using this, Sargsyan proved in On the strength of PFA. I that if $\mathsf{PFA}$ holds, then there is an inner model containing all the reals and satisfying $\mathsf{AD}_{\mathbb R}+$"$\Theta$ is regular". This is stronger than the existence of a proper class of Woodin cardinals and a proper class of strong cardinals. Sargsyan's techniques already give us (a bit) more than this, and the bounds should increase as descriptive inner model theory reaches stronger determinacy models. This is the current best bound we have. Recently, alternatives to the approach via square have also been established. Neeman (see also his work with Schimmerling) proved in Itay Neeman. Hierarchies of forcing axioms II, J. of Symbolic Logic, vol. 73 (2008), pp. 522–542. MR2414463 (2009d:03123), that the strength of $\mathsf{PFA}$ restricted to $\mathfrak c$-linked posets is that of a $\Sigma^2_1$-indescribable cardinal, and the restriction to $\mathfrak c^+$-linked posets should be what he calls a $\Sigma^2_1$-indescribable $1$-gap $[\kappa,\kappa^+]$. There is a natural hierarchy associated with the notion of $\Sigma^2_1$-indescribability, that ends up in supercompactness, so this gives us a new route to attempt establishing lower bounds. Already at the level described in the paper, though, the fine structural models required for the results are beyond what we can reach currently. Another alternative was suggested in Matteo Viale, and Christoph Weiß. On the consistency strength of the proper forcing axioms, Advances in Mathematics, vol. 228 (2011), no. 5, 2672-2687. MR2838054 (2012m:03131). Their results show that the usual iteration techniques for forcing $\mathsf{PFA}$ require at least a strongly compact cardinal, and if the forcing used is proper, then a supercompact is indeed needed. This is not the same as saying that we indeed need supercompactness in strength, but it goes to explain the "almost certainty" I mentioned above. Two remarks need to be added, both related to the strength of failures of square. I. There are many combinatorial variants of square principles, of varying strength, for instance the family of principles $\square_\kappa,\square_\kappa^2,\dots,\square_\kappa^{<\omega},\square_\kappa^\omega,\dots,\square_\kappa^\kappa=\square_\kappa^*$, and also $\square(\lambda)$ and its variants. I will not discuss here all known results; they are due to a number of authors, including (in rough historical order) Todorcevic, Magidor, Cummings, and Strullu. (The list is grossly incomplete, and I apologize for this.) We have that $\mathsf{PFA}$ implies not just the negation of $\square_\kappa$ for $\kappa\ge\omega_1$ but, in fact, of $\square_\kappa^{\omega_1}$. It actually implies the failure of $\square(\lambda,\omega_1)$ for $\lambda\ge\omega_2$. The stronger principle $\mathsf{MM}$ reaches farther. For instance, $\mathsf{PFA}$ is consistent with $\square_{\kappa,\omega_2}$ holding for all $\kappa\ge\omega_2$, while $\mathsf{MM}$ implies $\square_\kappa^*$ fails whenever $\kappa$ has cofinality $\omega$, etc. A number of set theorists (notably, Magidor) have taken this as partial evidence that in fact strong compactness and supercompactness have different consistency strength, with strong compactness sufficing for the failures of square visible to $\mathsf{PFA}$, and supercompactness being responsible for the additional failures coming from $\mathsf{MM}$. Together with $\mathsf{PFA}$ and $\mathsf{MM}$, in the last 30 years or so, a variety of reflection principles have been identified, some incompatible with $\mathsf{MA}$ (such as Todorcevic's Rado conjecture), some being a consequence of $\mathsf{PFA}$ (say, Moore's Mapping reflection principle), and some being consequences of $\mathsf{MM}$ (the Strong reflection principle, for example). In spite of their obvious differences and mutual incomparability, all these principles imply failures of square, and so they have significant consistency strength (the bounds discussed by Sargsyan). Unless all these principles turn out to be equiconsistent with supercompactness, a finer analysis than we can currently anticipate will be needed to separate their strength. The reason for the last remark is that the way strength is derived via failures of square can be understood as a proof by contradiction: If the principle under consideration does not imply the existence of inner models with certain large cardinals, then we can use this smallness assumption to build appropriate local core models $K$ (which are fine structural models and therefore satisfy square principles at suitable cardinals). We can then argue that the failure of square implies that these models must compute successors incorrectly. The point is that $\square_\kappa$ is upwards absolute between models where $\kappa$ and $\kappa^+$ remain cardinals. This is a violation of weak covering (one of the key properties of $K$), which implies that $K$ cannot actually exist, meaning that the large cardinals we were after were indeed reached during the attempted construction of $K$. An entirely different approach seems needed if (as Magidor suspects) these reflection principles end up having different consistency strength. In particular, weak versions of square are equivalent to the strongest known versions in fine structural models, so weak covering or its relatives cannot be enough to distinguish lower bounds between these principles. II. Recently, we have begun to understand how single failures of square principles are significantly weaker than consecutive failures. In particular, the strength of $\lnot\square(\kappa)+\lnot\square_\kappa$ has been investigated. This started with work of Schimmerling and continued in Ronald Jensen, Ernest Schimmerling, Ralf Schindler, and John Steel. Stacking mice, J. Symbolic Logic, vol. 74 (2009), no. 1, 315–335. MR2499432 (2010d:03087). Independently of the study of square principles, the "stacking" technique described in this paper has been significant in modern investigations in inner model theory. The strength reached from the failure of both $\square_\kappa$ and $\square(\kappa)$ for $\kappa\ge\omega_3$ is comparable to that reached through Sargsyan's method -- the goals and techniques are different: Sargsyan looks at the failure of $\square_\kappa$ for a singular strong limit cardinal $\kappa$, and is directly developing the core model induction, while the techniques in the stacking paper are purely inner model theoretic and pursue showing the existence of non-domestic mice. See here for Square principles in $\mathbb P_{max}$ extensions, work involving me, Larson, Sargsyan, Schindler, Steel, and Zeman. We look at the failure of $\square(\omega_2)$ and $\square_{\omega_2}$, in ($\mathsf{ZFC}$) forcing extensions of models of determinacy. Looking at strengthenings of this failure led to the isolation of the notion of $\Pi^2_1$ subcompactness as playing a key role. That this is indeed the appropriate large cardinal has been essentially confirmed in Equiconsistencies at subcompact cardinals, by Neeman and Steel. The goal here has not been to identify the strength of $\mathsf{PFA}$ or $\mathsf{MM}$, but rather of some of their fragments, below $\mathsf{MM}(\mathfrak c^+)$. We expect that a good understanding at this level will be key for an eventual understanding of the strength of $\mathsf{PFA}$ via an appropriate stratification (in this respect, the goal is similar to that pursued in the hierarchies paper by Neeman mentioned above).<|endoftext|> TITLE: What are simplicial topological spaces intuitively? QUESTION [18 upvotes]: (This is a repost of a question from MSE. I hope there is more to say.) I tend to imagine simplicial objects in a category as some kind of "topological objects", with a notion of homotopy. Simplicial sets are like topological spaces, simplicial groups are like topological groups, and so on. This rough idea has some precise formulations as mentioned here. However, as it was pointed to me in an answer to MSE question, this heuristic could not be expected to work when weak equivalences are defined not by equivalences of geometric realizations. Simplicial spaces are useful and quite common: for instance, the nerve of a topological category is a simplicial space. Applying this to the simplicial construction of $\mathrm{B}G$, where $G$ is a Lie group, we see that $\mathrm{B}G$ is a simplicial manifold, hence we can do Chern-Weil theory on it. Segal's $\Gamma$-categories use simplicial spaces essentially. Well, actually bisimplicial sets, since all they need is to form nerves of various categories, but that's more or less the same thing in the context of this question. Singular simplicial set functor $\mathrm{Top} \to \mathrm{sSet}$ can be readily modified to produce a simplicial space by using a compact-open topology on the space of mappings. Unfortunately here the additional data is redundant, so there are no lessons to extract from the most familiar example. I can follow simple arguments involving simplicial spaces, since they are more or less the same as arguments about simplicial sets, which I learned to be (somewhat) happy about. However, this understanding is purely formal. So my question is: how to imagine simplicial spaces? Is there an informal interpretation in terms of topological spaces with additional structure (probably not)? The answer to the same question on MSE by Zhen Lin seems to indicate there is no interpretation beyond diagrams of spaces. I suppose it's absolutely true for trisimplicial (or even higher) sets, but bisimplicial sets look feasible for a direct intuitive description. REPLY [25 votes]: If $\mathcal{A}$ is an abelian category, then the Dold-Kan correspondence supplies an equivalence between the category of simplicial objects of $\mathcal{A}$ and the category of nonnegatively graded chain complexes in $\mathcal{A}$. One can therefore think of simplicial objects as a generalization of chain complexes to non-abelian settings. In homological algebra, chain complexes often arise by choosing ``resolutions'' of objects $X \in \mathcal{A}$: that is, chain complexes $$ \cdots \rightarrow P_2 \rightarrow P_1 \rightarrow P_0$$ with homology $$H_{i}( P_{\ast} ) = \begin{cases} X & \text{ if } i = 0 \\ 0 & \text{ otherwise }. \end{cases}$$ Among these, a special role is played by projective resolutions: that is, resolutions where each $P_n$ is a projective object of $\mathcal{A}$. If $X_{\ast}$ is a simplicial space, it might be helpful to think of $X_{\ast}$ as a resolution of the geometric realization $| X_{\ast} |$. It plays the role of a ``projective resolution'' if $X_{\ast}$ is degreewise discrete: that is, if it is a simplicial set.<|endoftext|> TITLE: colorings of ${\mathbb Z}^d$ with constraints QUESTION [8 upvotes]: For a lattice $\mathbb Z^d$, denote by lattice line any line that contains two (and thus infinitely many) lattice points. For $2\le k2$. For a prime $p$, take for $D$ the cube $\{0,1,\dots,p-1\}^d$ with a point $x\in D$ getting color $d-\#x$, where $\#x$ is the number of coordinates equal to zero. It is easy to check that this yields a $C_d(d+1,2)$ for each prime $p$; these colorings will be called "simplex constructions". The special case $d=2$ yields $C_2(3,2)$'s which are again non-isomorphic to the above ones, with e.g. $D= \pmatrix{ 0 & 1 & 1\\ 1 & 2 & 2 \\ 1 & 2 & 2 }$ for $p=3$. A $C_d(2^d,2)$ is defined by a $2^d$-hypercube $D$ in which each color is used once (and it is not hard to see that all $C_d(2^d,2)$ are of this form). We may call this one the "rainbow construction" and note that it generalizes immediately to a $C_d(k^d,k)$. For this one, $k$ does not need to be a prime! Questions: For which triples $(d,n,k)$ do such colorings $C_d(n,k)$ exist? Which periods (i.e. side lengths of $D$) can occur for a given triple? Are there other examples where $D$ is not a hypercube? It seems like there are no $C_d(n,2)$ for $n>d+1$ unless $n=2^d$. Is there an easy way to prove that? Now for $k>2$, apart from the rainbow construction, it is much harder to find general constructions. I wonder if it is possible to tweak certain combinatorical designs in order to obtain a suitable $D$ or, for dimension $d=2$, to build one from an appropriate set of latin squares. Or is there a construction for $k=3$ based on the above simplex constructions? As $k$ grows, searching for small $D$'s by hand feels a bit like solving a sudoku. While a $C_2(6,3)$ is still quite easy to find (e.g. $D= \pmatrix{ 0 & 1 & 2 & 0\\ 1 & 2 & 1 &3 \\ 0 & 1 & 2 & 0\\ 4& 3&4&5 }$), I have the impression that no $C_2(n,3)$ exists for $n>6$, except of course $n=9=3^2$. The following question seems well motivated: For given naturals $d,k\ge2$, denote by $m(d,k)$ the biggest $n2$, but it's fine for $d=2$. You can make a slightly more general construction for a $C_2(2k-1,k)$ with period $p^{k-1}$: Given a lattice point $(x,y)$, let $n_x$ be the number of times $p$ divides $x$, up to a maximum of $k-1$, and similarly for $n_y$. Then color the point $(x,y)$ with the number $n_x + n_y$. Since we're free to add $1$ to both $n$ and $k$, this means we can construct every $C_2(n,k)$ for $k \leq n < 2k$. By adding these to the rainbow constructions which give us $C_2(k^2,k)$, we can also fill chunks of the range $2k \leq n \leq k^2$, in particular, something like the range $k \leq n < C k^2$ for some $C < 1$. There are no $C_2(n,k)$ for $n > k^2$, so we have the asymptotic answer.<|endoftext|> TITLE: Characteristic polynomials of trees and E8 QUESTION [9 upvotes]: In thinking about constructing manifolds via surgery or plumbing, the following combinatorial problem comes up: If T is a tree with adjacency matrix A and I is the identity matrix of the same order, when is |det(A-I)| = 1 possible? The fact that this is true for the E8 tree allows us to construct manifolds following Milnor et al. I've thought about this problem a bit and can't see an obvious reduction to similar problems, involving root systems, etc. Can it be shown that no other examples exist? What simple argument am I missing? A correction after the comments I got below - I originally put |det(A-I)|=1 but actually meant |det(A+2I)|=1 or equivalently |det(A-2I)|=1 since trees are bipartite and their eigenvalues are symmetric around 0. Chris's observation is interesting as well... I now have to go back and do some rechecking of small trees. ::::grin:::: REPLY [4 votes]: Using sage, and up to 16 vertices included, one finds only two trees satisfying $|det(A+2I)|=1$, namely $E_8$ and $E_{10}$. EDIT: I also confirm that there are 18 such trees with 18 vertices. Here they are in the graph6 string notation (suitable for input in sage): ['Q??GOGA?O??C?AGAA?C???@d?O_', 'Q??GOGA?O@?A?A?@G?C??G?gO@O', 'Q???GGA?O@??_?O?C???kC@OGG_', 'Q???GGA?O??C_AO?C???oC@OG?o', 'Q??GOGA?O??C?C?AG@C??G@COB?', 'Q???G???OG?COCGAO??F?C?GGCG', 'Q????CA??O?_O?G??AO@aCPA?G_', 'Q????CA?O?O?O?G??D_GaCA?GAG', 'Q???GGA?O@???@_?G??OSG@OOO_', 'Q???GGA??A?A?GGCO?A??A?cOK_', 'Q???GGA??A?A?GOCO??ODGA?OCG', 'Q????G??_??@OGGOO??L?CE?GC_', 'Q????C???Q?_O?G??AO@aCHAA?_', 'Q????C???_@G@GO??AG@_G_E?G_', 'Q????_G@?C???B_?G?H?`?P??`G', 'Q???G??@?C?GOGG?O??BACCGGK?', 'Q????GA????@?DG?A@?OBOKA@@?', 'Q????GAA???@?C_?G?G_HGQ??b?'] For example, you can type Graph('Q????GAA???@?C_?G?G_HGQ??b?') to get the last one.<|endoftext|> TITLE: What did Shimura say about $y^2 + y = x^3 - x$? QUESTION [6 upvotes]: From the introduction of Ribet-Stein: Shimura showed that if we start with the elliptic curve $E$ defined by the equation $y^2 +y = x^3 −x^2$ then for “most” $n$ the image of $\rho$ is all of $\mathrm{GL}_2(\mathbf{Z}/n\mathbf{Z})$. Here $\rho$ is the representation of $\mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$ on the $n$-torsion of $E$. What is the original paper where Shimura shows this? (Is there an online copy?) REPLY [10 votes]: Goro Shimura, A reciprocity law in non-solvable extensions. J. Reine Angew. Math. 221 1966 209--220.<|endoftext|> TITLE: Can a partition free family in $2^{[n]}$ always be enlarged to one of size $2^{n-1}$? QUESTION [5 upvotes]: Let $\left[ n \right]=\{{1,2,\cdots,n\}}$ and call a family $\mathcal{F} \subset 2^{\left[n\right]}$ partition-free if it does not contain any partition of $\left[n\right]$. A recent question asked for the maximal size of such a set. The answer given is $2^{n-1}$ since having more than that allows a two member partition. Q: can every partition-free family be enlarged to one of this size? There is an interesting comparison to Intersecting Families (ones which contain no two disjoint members.) Here again the maximal size is $2^{n-1}$ because we never can have both a set and its complement. Any intersecting family can be enlarged to one of this size (a nice exercise). Thus there are many examples built from the lines of projective planes, weighted voting schemes etc. Also, any maximal intersecting family can be changed into any other by repeatedly switching a minimal member with its complement. It is almost true that an intersecting family $\mathcal{F}$ is partition free. If $\left[n\right] \in \mathcal{F}$ (as will be the case for a maximal intersecting family) then we must replace $\left[n\right]$ with $\emptyset$ and then we will have a partition free-set. There are certainly many other maximal partition-free families: From $\left[ 9 \right]$ start with all the sets with more than half the elements. Replace the whole set by $\emptyset$ and any set of size $6$ which does contain the element $1$ by its complement. REPLY [14 votes]: Yes. Suppose $\mathcal{F}$ is partition-free of smaller size. Then there is some $A$ for which $\mathcal{F}$ contains neither $A$ nor its complement. It must be possible to add either $A$ or its complement to $\mathcal{F}$ to get a larger partition-free family, as otherwise $\mathcal{F}$ would already have to contain partitions of both $A$ and its complement, and therefore a partition of $[n]$.<|endoftext|> TITLE: Where are Georg Cantor's Original Manuscripts? QUESTION [20 upvotes]: Georg Cantor is famous for introducing transfinite numbers and set theory. A main part of his mathematical point of view about this new type of "numbers" and this new "realm of mathematics" cannot be found in his official publications. Maybe one can find it in his unpublished manuscripts, papers and letters. Question 1: Where are Georg Cantor's original manuscripts? Question 2: Were some of these manuscripts lost during World War II and subsequent years? Question 3: Are some of these manuscripts in museums or personal collections now? REPLY [21 votes]: According to I. Grattan-Guinness, in Materials for the History of Mathematics in the Institut Mittag-Leffler, much of Cantor's archive disappeared from his house in Halle in 1945, at the end of the war. An important collection remains in the Institut Mittag-Leffler in Sweden, listed in Towards a Biography of Georg Cantor. They are not online, but you might contact their library for help. It is in this collection that one "lost" manuscript of Cantor resurfaced in 1970. REPLY [20 votes]: Cantor's estate is now at the university library in Göttingen; you can find a list (PDF, in German) here. As far as I can see, these have not been digitized, but I didn't look very hard. Due to the obvious interest to historians of mathematics, his writings have of course been edited; the list above contains references, mostly to Gesammelte Abhandlungen mathematischen und philosophischen Inhalts, edited by Ernst Zermelo, 1932 (link is to Springer's 2013 reprint), Briefe, edited by Herbert Meschkowski and Winfried Nilson, Springer 1991. In particular, the exchange with Dedekind (which both works contain) might be of interest to you.<|endoftext|> TITLE: Applications of Atiyah-Singer using pseudodifferential operators QUESTION [21 upvotes]: Though the Atiyah-Singer index theorem holds for pseudodifferential operators, all the applications of the index theorem I know of only need it for Dirac-type operators. I know that pseudodifferential operators play a major role in the K-theoretic proof of the index theorem, but it seems to me that they are of no use for any applications of it. What are examples of applications of the Atiyah-Singer index theorem, which essentially use the computation of the index of a pseudodifferential operator which is not a Dirac-type operator? Or phrased it another way: what are the benefits of knowing that the Atiyah-Singer index theorem holds for pseudodifferential operators and not only for Dirac-type operators? Why do I care: for Dirac-type operators we can prove Atiyah-Singer using the heat kernel method, whereas this is in general not possible for pseudodifferential operators (so for them we have to use other proofs). So I was asking myself whether there is any need to have these other proofs that work also for pseudodifferential operators besides the fact that these other proofs further our understanding of the index theorem. REPLY [13 votes]: Index theory is fundamentally about a homomorphism $$K_n(M) \to \mathbb{Z}$$ from the top degree K-homology of $M$ (even dimensional) to the integers called the analytic index map. It is called this because every graded self-adjoint elliptic (pseudo)differential operator on $M$ determines a class in $K_n(M)$ and the analytic index map sends the class of an elliptic operator to its index. The content of the index theorem is that the analytic index map agrees with another map (the topological index) from K-homology to the integers defined via algebraic topology (essentially the Poincare-Thom construction). Baum and Douglas constructed a geometric model of the K-homology of a CW-complex $X$ in which a cycle consists of a spin$^c$ manifold equipped with its spin$^c$-Dirac operator and a reference map from the manifold into $X$. They proved that the abelian group generated by these cycles (with relations that I will not specify) really is the K-homology of $X$, i.e. the homology theory determined by the Bott spectrum. The classical index theorem for spin$^c$-Dirac operators is really about calculating the topological index of the fundamental class of a spin$^c$-manifold. So since the Fredholm index is well-defined on K-homology classes (the existence of the analytic index map) and K-homology classes can be represented by Dirac operators (Baum-Douglas), this means that any index problem can be reduced to an index problem involving Dirac operators. There are index theorems involving operators which are not obviously related to Dirac type operators - Toeplitz operators come to mind (this is related to Sonke Hansen's answer), for instance - but even for those operators the index theory is controlled by Dirac operators. Now some comments about the role of pseudodifferential operators in all this. Pseudodifferential operators typically enter into index theory for one of three reasons: To prove that a certain operator is Fredholm. To construct homotopies. To emphasize the stability of the index. First, to prove that an operator is Fredholm it is typically easiest to show that it is invertible up to compact operators. One of the goals of pseudodifferential operator theory is to construct pseudo-inverses for differential operators in this sense, and many people like to prove that differential operators are Fredholm by explicitly constructing pseudo-inverses. This is more or less the line that Atiyah-Singer took in their original papers. Second, to do calculations in index theory one often must construct homotopies between different operators (exploiting the homotopy invariance of the Fredholm index), and even if you only care about differential operators these homotopies usually must pass through pseudodifferential operators. Finally, the K-homology picture of index theory emphasizes how insensitive the Fredholm index is to the details of a differential operator. When constructing the K-homology class of an elliptic operator (using the Baum-Douglas model or otherwise) one notices that the construction depends only on the asymptotic properties of the principal symbol; for example, if the principal symbols of two different operators agree outside a compact set at each point then the K-homology classes will be the same. Thus the index theorem for pseudodifferential operators comes almost for free, and I think this is why index theorems are often formulated for pseudodifferential operators (rather than to handle a specific set of examples).<|endoftext|> TITLE: Probability two matching runs of coin tosses QUESTION [7 upvotes]: If you toss a coin $2\ell-1$ times you get a sequence of outcomes, say, $HTHTHTH$ for $\ell = 4$. I am trying to work out the probability that there are at least two runs (in other words contiguous subsequences of outcomes) of length $\ell$ that are identical. In this case $HTHT$ occurs twice, starting at the first and third positions. The problem I am having is how to deal with the overlap between runs. Is there a simple (or not so simple) observation that makes this problem solvable? I worked it out for some small examples. l=2. Prob is 2/8 l=3. Prob is 8/32 l=4. Prob is 28/128 l=5. Prob is 86/512 l=6. Prob is 250/2048 l=7. Prob is 680/8192 l=8. Prob is 1792/32768 l=9. Prob is 4562/131072 l=10.Prob is 11344/524288 l=11.Prob is 27614/2097152 The sequence $2,8,28,86,250$ is http://oeis.org/A118047 but after that it diverges. If it turns out that an exact solution is not feasible, I would be very interested in any large $\ell$ approximations. Is there a direct argument which tells us that the probabilities in this problem are bounded above by the case where the runs of length $\ell$ are all independent? REPLY [6 votes]: Inspecting the bound of Bjorn Kjoss-Hanssen's answer with the numerical results in the question, it seems that the upper bound is about twice as large as the true value. I give a refined upper bound that saves this factor of $2$, and which I expect is close to the correct answer. Numerically it seems that $\binom{\ell}{2}2^{-\ell-1}$ is very close to the right answer; the upper bound I give gets the (presumably) right coefficient for the $\ell^2$ term but not the $\ell$ term. Suppose the string of length $2\ell -1$ is given by values $a(n)$ with $1\le n\le 2\ell -1$. Suppose that such a string of length $2\ell-1$ has two strings of length $r$ that match, where $r\ge \ell$ and is the maximal length of strings that match. Then there exist $1\le i< j$ such that $a(i)=a(j)$, $a(i+1)=a(j+1)$, $\ldots$, $a(i+r-1)=a(j+r-1)$ (note that this means $j\le 2\ell-r$), and also we have $a(i-1)\neq a(j-1)$ (this condition doesn't exist if $i=1$) and $a(i+r) \neq a(j+r)$ (this condition doesn't exist when $j+r=2\ell$). If $1 TITLE: compact-open topology on $B(H)$ QUESTION [18 upvotes]: In topology, it is common to use the compact-open topology on the set of continuous maps between two given topological spaces. Let now $H$ be a Hilbert space and $B(H)$ the set of continuous linear maps from $H$ to itself. In functional analysis, there are many topologies that people like to use on $B(H)$. Does any one of them agree with the compact-open topology? If yes, which one? REPLY [2 votes]: As an introductory remark, there has been a considerable amount of work done on topologies on spaces of operators in Hilbert spaces, in particular with regard to their relevance to the theory of von Neumann algebras. In my opinion there are two essential criteria which one should apply: the topology should be complete and its dual should be a natural space of operators. The compact open topology fails the second one---its dual is the space of finite rank operators. (Incidentally, the most immediate natural candidate, the norm topology, also passes the first one but fails the second---in this case, the dusl is too large). Further candidates (which have already been mentioned)---the weak, strong, ultraweak and ultrastrong topologes---also fail this test. A suitable family of natural topologies which pass both tests with flying colours (whereby the dual spaces are the space of nuclear operators) was introduced 40 years ago in the Comptes Rendues paper "Topologies dans l'espace des operateurs sur les espaces de Hilbert" (1973). These are the finest locally convex topologies which agree with the topology of compact convergence on the unit ball. (There are four, since we can consider convergence with resepct to the weak or strong topologies and also their symmetric versions, i.e., ones for which the operation of taking adjoints is continuous). Some of their basic properties can be found in the above article. This shows that the answer to your question is pretty well always no and goes on to answer the (implicit?) continuation---what is the "correct" topology on $B(H)$?<|endoftext|> TITLE: Decidability of differential equations QUESTION [6 upvotes]: Is there anything well-known about the algorithmic decidability of the satisfiability of an ODE $\dot{x}=f(x)$, $x: [0,1]\to R^n$ with an initial condition $x(0)=x_0$, given that $f(x)$ belongs to some specified class of functions -- for example, component-wise polynomial? REPLY [3 votes]: In the the paper Boundedness of the Domain of Definition is Undecidable for Polynomial ODEs by Daniel S. Graça et al. (preprint version) they prove the undecidability of deciding the boundedness of the maximal domain of solution of a polynomial ODE.<|endoftext|> TITLE: $\mathrm{Bessel}^3$ Integral QUESTION [13 upvotes]: I'm trying to calculate the following integral: $\int_0^\infty \mathrm{BesselJ}[l_0,k_0r] \cdot \mathrm{BesselJ}[l_1,k_1r] \cdot \mathrm{BesselJ}[l_0-l_1,kr] \cdot r\,dr$ ($\mathrm{BesselJ}[n,x]$ is the Bessel function of the first kind of order $n$) I assume that $l_0,l_1$ are integers, and that $k_0,k_1,k>0$. The result has nice implications in quantum mechanics - to explain selection rules. I was able to prove that the integral is zero when $k,k_0,k_1$ cannot be the lengths of a triangle. So there is a non-zero result when $|k_0-k_1| < k < k_0+k_1$ However, I don't know how to calculate the integral when it is not zero. Can anyone help? I tried using Mathematica to get a numerical answer (which is fine for me), but I do not think I can count on it. Mathematica is giving a non-zero result for the regime where the integral should be zero. REPLY [18 votes]: In fact, precisely your integral has been computed in closed form, in: Annie Gervois and Henri Navelet, Some integrals involving three Bessel functions when their arguments satisfy the triangle inequalities, J. Math. Phys. 25 (1984), no. 11, 3350–3356. Their result is $$ \int_0^\infty J_m(ar)J_n(br)J_{m+n}(cr)r\,dr = \begin{cases} 0&\text{if }c^2 < (a-b)^2\\ \dfrac{\cos(mB-nA)}{\pi ab\sin C}&\text{if }c^2 = a^2+b^2-2ab\cos C\\ 0&\text{if }c^2 > (a+b)^2 \end{cases} $$ where in the second case, $A, B, C$ are the angles opposite to sides $a, b, c$ of the resulting triangle. (They give formulas valid for real $m, n$, which reduce to the above when $m, n\in\mathbf Z$ as you assume.) To put this in your notation, take $(m,n,a,b,c)=(l_0,-l_1,k_0,k_1,k)$ and use that $A+B+C=\pi$ and $J_n=(-1)^nJ_{-n}$. We get $$ \int_0^\infty J_{l_0}(k_0r)J_{l_1}(k_1r)J_{l_0-l_1}(kr)r\,dr = \begin{cases} 0&\text{if }k < |k_0-k_1|\\ \dfrac{\cos([l_0-l_1]K_1-l_1K)}{\pi k_0k_1\sin K}&\text{if }|k_0-k_1| |k_0+k_1| \end{cases} $$ where $K_0,K_1,K$ are the angles opposite to $k_0,k_1,k$. The denominator is $2\pi$ times the triangle's area.<|endoftext|> TITLE: The intersection of $n$ cylinders in $3$-dimensional space QUESTION [10 upvotes]: A standard question in vector calculus is to calculate the volume of the shape carved out by the intersection of $2$ or $3$ perpendicular cylinders of radius $1$ in three dimensional space. Such shapes are known as Steinmetz Solids, and for two and three cylinders we have $$V_2=4\cdot \frac{4}{3},$$ $$ V_3=6(2-\sqrt{2})\cdot \frac{4}{3}.$$ Here, and throughout, when we say that the cylinders intersect, we mean that the centers of the cylinders (which are lines) all intersect at a single point. Moreton Moore wrote an article where he calculates the volume of the intersection of $4$ and $6$ cylinders which arise from identifying faces of the octahedron and dodecahedron respectively. In this case, we have $$V_4=\frac{4}{3}\cdot9(2\sqrt{2}-\sqrt{6}),$$ $$V_6=\frac{4}{3} 4\left(3+2\sqrt{3}-4\sqrt{2}\right).$$ Define $$K_n=\inf\{V: V \text{ is the volume of the intersection of }n\text{ cylinders of radius }1\}.$$ Since we are assuming that the centers of the cylinders share a common point, the sphere will be contained in any configuration of cylinders. The maximum volume is infinite, and the limiting volume will be the sphere, that is $$K_\infty=\pi\frac{4}{3}.$$ I believe that for reasons of symmetry, for $n=2,3,4,6$ we have $K_n=V_n$. Naturally, this leads to the question: What is the optimal configuration of cylinders for each $n$, and what is the resulting volume? While this may be a difficult problem, I am interested in the more approachable problem of evaluating the rate of convergence to the sphere. Can we identify how fast $K_n$ approaches $K_\infty$? Question: Let $$f(n)=\frac{K_n}{4/3}-\pi.$$ Then $f(n)$ is monotonically decreasing to $0$, and $f(n)$ measures how fast $K_n$ approaches the volume of the sphere. At what rate is $f(n)$ decreasing to $0$? Can we find the order of magnitude of $f(n)$? REPLY [4 votes]: This is a heuristic, suggesting $f(n)=O( 1/n^2)$. Consider the sphere of radius $r$. Each cylinder intersects this sphere in a great circle. The great circles divide the sphere into a number of regions. By induction we can see that this is at most $n(n-1)+2$. So if the sphere is relatively evenly divided into regions, each region must have diameter asymptotic to $1/n$. The intersection of the $n$ cylinders is a ball, plus a little piece jutting out in each of these regions. The little piece is approximately a cone of radius $O(1/n)$ and slope $O(1/n)$, so it has height $O(1/n^2)$. Since the area of the base is $O(1/n^2)$, it has volume $O(1/n^4)$. So the sum of all the pieces has volume $O(1/n^2)$. The fact that the cone is on a spherical base and not a flat base cuts its thickness by $1/2$,so is responsible for only a constant factor. So assuming that: i. It's possible to choose the great circles to be relatively evenly distributed ii. Weirdly-shaped regions don't cause any problems iii. There's not some other error in these heuristic arguments: the asymptotic is $f(n) = O(1/n^2)$. Edit: We can make the upper bound part of this more rigorous using the probabilistic method. Place all the cylinders randomly. Fix a point on the sphere. Consider a ray out from that point - how far does that ray travel in the intersection of the cylinders? It is at most $d^2/2+$-higher order terms, where $d$ is the distance to the nearest great circle corresponding to a cylinder. (In fact, if we express the distance as an angle $\theta$, it is exactly $\sec \theta - 1$.) Instead of finding the nearest great circle corresponding to a cylinder, we can just look in a random direction and find the nearest great circle corresponding to a cylinder in that direction. So we look along some great circle passing through our point. Each great circle corresponding to a cylinder intersects this one at two random antipodal points. So we need to find the distribution of the distance to the nearest point among $n$ randomly chosen points on the unit circle. This converges rapidly to an exponential distribution of mean $1/n$. Because this distribution decays so rapidly, only the lowest-order term for the radius or the local contribution to volume matters when computing the expected value of the volume. The lowest-order term is proportional to the radius, which is proportional to $d^2$, whose expected value is $1/n^2$. Averaging over all the points, the volume is $O(1/n^2)$ Edit 2: We can make the lower bound rigorous using a simple estimate. The extra radius at a given point is $\sec \theta-1$, where $\theta$ is the distance to the nearest great circle corresponding to a cylinder. For any distance $\epsilon$, the points of distance no more than $\epsilon$ from a single great circle form a thin tube of area at most $ 4 \pi \epsilon$. So the total area of points that are within $\epsilon$ of any great circle form an area of at most $ 4 \pi n \epsilon$. The area of points that are not within $\epsilon$ of any great circle has area at least $4 \pi (1- n \epsilon)$. So if we set $\epsilon=2n/3$, say, then one-third the points of the sphere will extend outwards by at least $\sec \epsilon -1 \ geq \epsilon^2/2 = 2/9n^2$. This gives a lower bound of the form $O(1/n^2)$. REPLY [2 votes]: Not an answer; just an illustration for $n=3$:            (Image by A.J. Hildebrand, Lingyi Kong, Abby Turner, Ananya Uppal, from this link.) REPLY [2 votes]: The rate of convergence of $f(n)$ to zero seems about the same as the rate of the best approximation of the ball's volume by the volume of a circumscribed centrally-symmetric polyhedron with $2n$ facets, which should be known already. (The problem of approximating the ball by circumscribed polyhedra has been studied extensively.) When $n$ is large, the volume of the intersection of $n$ cylinders, each tangent to two opposite faces of the best approximating $2n$-faced circumscribed polyhedron is very close to the volume of the polyhedron. For example, look at the illustration posted by Joe: already for $n=3$, the intersection of the three cylinders looks a bit like the circumscribed cube. The faces are curved, but for large $n$, as all faces become very small, the curved faces become relatively less curved and extremely close to the flat ones.<|endoftext|> TITLE: Cyclic subgroups of GL(n,q) QUESTION [8 upvotes]: Let $q$ be a prime power. It is well known that all Singer subgroups (subgroups of order $q^n-1$) in $GL(n,q)$ are conjugate. My question is: If $H$ is a cyclic subgroup of order $m$ in $GL(n,q)$, $m\mid (q^n-1)$ and $m$ is quite large (for example, $m= (q^n-1)/2$ for $q$ odd), is it necessary that $H$ must be contained in some Singer subgroup of $GL(n,q)$? REPLY [11 votes]: Yes. Put the generator in rational canonical form. Because its order is prime to $q$, it is semisimple, so this puts it in block diagonal form where each block is a companion matrix of an irreducible. It is contained in a Singer subgroup if and only if it has a single $n \times n$ block. Suppose it is not in a Singer subgroup. Then the order of the matrix is the lcm of the order of each block. Since each $k \times k$ block is in a Singer subgroup, its order divides $q^{k}-1$. The lcm of all these copies of $q^{k}-1$ is at most their product divided by $q-1$, since $q-1$ divides them all, and there are at least two blocks. Since the $k$s add up to $n$, the product is less than $q^n-1$. So if it is not Singer, the order must be less than $q^n-1/ (q-1)$. So as long as $m \geq (q^n-1)/(q-1)$, there is a contradiction, and it must be Singer.<|endoftext|> TITLE: Reference for the triple covering of A_6 QUESTION [19 upvotes]: I would like to ask for a reference (book, paper ...) for the following nice construction, which I have found as an exercise in some notes of a course by R. Borcherds. For $n=6$ or $7$ (and only in that case), the alternating group $\mathfrak{A}_n$ has a nontrivial central extension by $\mathbb{Z}/3$. The construction in the case $n=6$ is as follows. Define an oval in $\mathbb{P}^2(\mathbb{F}_4)$ as a set of 6 points which are in general position (i.e. no 3 of them are collinear). It is easy to show that any projective frame of $\mathbb{P}^2(\mathbb{F}_4)$ is contained in a unique oval, and from this that the stabilizer of an oval in $PGL_3(\mathbb{F}_4)$ is isomorphic to $\mathfrak{A}_6$. The inverse image of this $\mathfrak{A}_6$ in $SL_3(\mathbb{F}_4)$ provides the nontrivial extension by $\mathbb{Z}/3$. Does anyone know a reference for this? and/or an analogous construction for $\mathfrak{A}_7$? REPLY [17 votes]: The oval construction for $3\cdot A_6$ can be found on p.110 of Symmetric Generation of Groups With Applications to Many of the Sporadic Finite Simple Groups by Robert Curtis. An e-version is here. In fact Curtis does better: he constructs $3\cdot A_6\cdot 2^2$. If you read on to p.112 and beyond you will also find a construction of $3\cdot A_7$, as you requested. (It seems like the e-version is freely available on the web. But if you can't get it, contact me and I'll email it to you.)<|endoftext|> TITLE: A family of skew-symmetric matrices corresponding to cycles in graphs QUESTION [8 upvotes]: When investigating loops in Markov chains I ran into the following observation. A cycle in a graph $G$ with $n$ vertices may be represented by a matrix $\Gamma \in \mathbb R^{n \times n}$ having the following properties: (i) $\Gamma = -\Gamma^T$. (ii) $\Gamma \mathbb 1 = 0$, i.e. $\sum_{j = 1}^n \Gamma(i,j) = 0$ for all $i =1, \dots, n$. (iii) $\Gamma(i,j) = 0$ whenever there is no edge between vertices $i$ and $j$. (iv) $\Gamma \neq 0$. The simplest example of such a matrix is $\Gamma = \begin{pmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{pmatrix}$. The representation mentioned above is made more precise by the following proposition. Proposition Let $G$ be a graph over $n$ vertices and let $\Gamma$ satisfy the conditions (i)-(iv) above. Then there exists a cycle, i.e. a non-empty sub-graph $C = (V_C, E_C)$ with $V_C = \{ x_0, x_1, \dots, x_k\}$ and $E_C = \{x_0 x_1, x_1 x_2, \dots, x_k x_0 \}$, such that for every $x_i x_j \in E_C$, $\Gamma(x_i, x_j) > 0$. Conversely, if there exists a non-empty cycle $C = (V_C, E_C)$, then there exists a matrix $\Gamma$ satisfying (i)-(iv), and $\Gamma(i,j) > 0$ for $(i,j)\in E_C$. Proof: "$\Rightarrow$": Since $\Gamma \neq 0$ and $\Gamma$ is skew-symmetric, there exists a pair $(i_0,i_1)$, $i_1 \neq i_0$, such that $\Gamma(i_0, i_1) > 0$. Since $\Gamma$ is skew-symmetric, $\Gamma(i_1, i_0) < 0$, and because rows sum to zero, there must be a positive element on row $i_1$. Suppose this is at position $(i_1, i_2)$. Again $i_2 \neq i_1$. We may repeat this procedure until we encounter a node $i_k$ that we already obtained (which will surely happen within $n-1$ steps). If this vertex is $i_0$, we are done. If this vertex is $i_l = i_k$ for some $0 < l < k - 1$ (note $l=k$ is impossible by skew-symmetry), we obtain a cycle $\{x_{i_l}, x_{i_{l+1}}, \dots, x_k\}$ with the required properties by removing vertices $i_0, \dots, i_{l-1}$. "$\Leftarrow$": Let the entries of $\Gamma(i,j) = 1$ and $\Gamma(j,i) = -1$ whenever there is a edge between $i$ and $j$ in the directed cycle $x_0, x_1, \dots, x_k$, and $\Gamma(i,j) = 0$ otherwise. For any $i$, $\sum_{j=1}^n \Gamma(i,j) = \sharp \{\mbox{directed edges out of $i$}\} - \sharp \{\mbox{directed edges into $i$}\} = 0$, so that (ii) is satisfied. The other conditions (i), (iii), (iv) are clearly satisfied. $\square$ This result seems quite basic to me, but I have trouble finding references in the literature. What would a matrix satisfying (i)-(iv) be called? I would like to understand the structure of the set of matrices satisfying (i)-(iv) for particular adjacency structures of graphs. Also results on the spectra of such matrices might prove helpful. Basically anything related to matrices satisfying (i)-(iv) would be of interest to me. Note: in the (applied) literature on Markov chains I have found one other reference to these matrices. This is Sun, Gomez, Schmidhuber, Improving the asymptotic performance of Markov chains by inserting vortices, 2010. They prove a couple of interesting results related to such matrices but do not provide a reference to any literature on this topic. The notion of skew-adjacency matrices (see e.g. Cavers et al, Linear Algebra and its Applications, 2012) seems related but is different. (In particular, a matrix satisfying (i)-(iv) and containing only 0 and +/- 1 entries is a skew adjacency matrix for the cycle it represents, but in general not for the original graph containing such a cycle.) It would help a lot if any of you could provide a reference for this kind of theory. Many thanks in advance. REPLY [2 votes]: The object you're interested in is an element of the kernel of the combinatorial divergence operator whose definition you give in (ii). But there's quite a nice theory of these kind of things that may be in the spirit of what you (were, I see this is from 2014) after. In what follows, rather than have your matrix be all zeroes and $\pm1$, I'm going to generalize and allow it to have any real numbers. Let $X$ be a finite set of vertices and consider the following three spaces for $k \in \{0,1,2\}$: $$ \Omega^k(X) = \big\{\phi : X^{k+1} \to \mathbb{R} \; | \; \phi(x_{\sigma(0)}, \ldots, x_{\sigma(k)}) = \textrm{sign}(\sigma) \phi(x_0,\ldots, x_k)\big \}$$ where $\sigma$ is a permutation of $\{0, \ldots, k\}$. The space $\Omega^0$ is the space of real valued functions from your vertex set and so isomorphic to $\mathbb{R}^{|X|}$. The space $\Omega^1$ is the set of edge flows on the complete graph on $|X|$ vertices (where if we simply don't want an edge, we give it a zero flow weight). The permutation condition implies these are skew symmetric $|X| \times |X|$ matrices. The space $\Omega^2$ is the space of all triangular flows on the graph (an element of it is an alternating 3-tensor, or, a skew symmetric $|X| \times |X| \times |X|$ hypermatrix). We may also define some natural linear maps between these spaces that are natural combinatorial analogues of the classic vector calculus operations. The gradient is a map: $$ \textrm{grad}: \Omega^0 \to \Omega^1 $$ via $\textrm{grad}(a)_{ij} = a_j - a_i$. Similarly, the curl is a map: $$ \textrm{curl}: \Omega^1 \to \Omega^2 $$ via $\textrm{curl}(A)_{ijk} = A_{ij} + A_{jk} + A_{ki}$. Note that $\Omega^0$ inherits an inner product from $\mathbb{R} ^{|X|}$. Define a nice simple inner product structure on $\Omega^1$ via: $$ \langle A, B \rangle_1 = \sum_{\{i,j\} \in E} A_{ij} B_{ij} $$ where $E$ is just the edge set of your graph. Similarly for $\Omega^2$: $$ \langle\mathcal{A}, \mathcal{B}\rangle_2 = \sum_{\{ijk\} \in T} \mathcal{A}_{ijk} \mathcal{B}_{ijk}. $$ where $T$ is the set of triangles. Now using these we can construct the adjoints of $\textrm{grad}$ and $\textrm{curl}$: $$ \textrm{grad}^* = -\textrm{div} = -\sum_{j \in X} A_{ij} $$ $$ \textrm{curl}^* = \sum_{k \in X} \mathcal{A}_{ijk}. $$ Then we have the two sequences: $$ \Omega^0(X) \xrightarrow{\text{grad}} \Omega^1(X) \xrightarrow{\text{curl}} \Omega^2(X) $$ and $$ \Omega^0(X) \xleftarrow{\text{ grad}^*} \Omega^1(X) \xleftarrow{\text{ curl}^*} \Omega^2(X). $$ Theorem (Combinatorial Hodge/Helmholtz Decomposition): $\Omega^1$ admits an orthogonal decomposition as: $$ \Omega^1 = \textrm{im}(\textrm{grad}) \oplus \underbrace{\textrm{ker}(\textrm{curl}) \cap \textrm{ker}(\textrm{div}) \oplus \textrm{im}(\textrm{curl}^*)}_{= \textrm{ker}(div)}. $$ Elements in the first component are precisely those flows that admit a potential function; flows in $\textrm{ker}(\textrm{div})$ are precisely those in your example: they are flows with everywhere zero divergence, and hence are sums of cycles and occupy a linear subspace orthogonal to those flows that admit a potential function.<|endoftext|> TITLE: How feasible is it to prove Kazhdan's property (T) by a computer? QUESTION [90 upvotes]: Recently, I have proved that Kazhdan's property (T) is theoretically provable by computers (arXiv:1312.5431, explained below), but I'm quite lame with computers and have no idea what they actually can do. So, my question is how feasible is it to prove property (T) of a given group, say $\mathrm{Out}(F_{r>3})$ (a famous open problem), by solving the equation below by a computer? Even the case of $\mathrm{SL}_{r>2}({\mathbb Z})$ where property (T) is known is unclear. A group $\Gamma$, generated by a finite subset $S$ and with its non-normalized Laplacian denoted by $$\Delta=\sum_{x\in S} (1-x)^*(1-x)=\sum_{x\in S} (2-x-x^{-1})\in{\mathbb Z}[\Gamma],$$ has property (T) iff the equation in ${\mathbb Z}[\Gamma]$, $$ m \Delta^2 = n \Delta + \sum_{i=1}^k l_i \xi_i^*\xi_i $$ has a solution in $k,m,n,l_i\in{\mathbb Z}_{>0}$ and $\xi_i\in{\mathbb Z}[\Gamma]$. REPLY [38 votes]: I think it is appropriate to let MO users know (the OP himself knows it well) that this question was recently solved: it is feasible to provide a computer based proof for property (T) using the Ozawa Equation and this method was applied successfully to the groups mentioned in the question. In particular, the famous open problem alluded to above is now solved. The computation is done via semidefinite programming, as suggested by David Speyer in his answer. It is now known that the group $\mathrm{Aut}(F_n)$ has property (T) iff $n\geq 4$. For $n\geq 5$ this is shown in Kaluba-Kielak-Nowak. The case $n=5$ was treated earlier in Kaluba-Nowak-Ozawa and the case $n=4$ was recently settled in Nitsche. Some earlier positive results were obtained in Netzer-Thom and Fujiwara-Kabaya. $\mathrm{Aut}(F_2)$ and $\mathrm{Aut}(F_3)$ are known not to have (T). This is all quite remarkable.<|endoftext|> TITLE: Doubly primitive groups with simple socle QUESTION [6 upvotes]: The classification of doubly transitive groups with simple socle is known. A good account of such classification can be found for example in this paper: Cameron, Peter J. Finite permutation groups and finite simple groups. Bull. London Math. Soc. 13 (1981), no. 1, 1--22. MR0599634 (83m:20008) Another reference is Chapter 7 (Section 7.7) of Dixon, John D.; Mortimer, Brian. Permutation groups. Graduate Texts in Mathematics, 163. Springer-Verlag, New York, 1996. xii+346 pp. ISBN: 0-387-94599-7 MR1409812 (98m:20003) I am particularly interested in the following permutation groups of the classification: $PSL(d,q)$ with $d>2$, $Sp(2d,2)$ with $d>2$, $PSU(3,q)$ with $q>2$, $Sz(q)$ (Suzuki groups) with $q^{2n+1}>2$ and $Ree(q)$ (Ree groups) with $q^{3n+1}>3$. Let $G$ be any of these groups and suppose that $G$ acts $2$-transitively on a set $\Omega$. My question is the following: Which of these permutation groups are doubly primitive on $\Omega$? (Doubly primitive means that the one-point stabilizer $G_\alpha$ is primitive on $\Omega\setminus\{\alpha\}$.) Motivation: I need to prove that the centers of the one-point stabilizers of the groups I mentioned are trivial. This can be done case-by-case. (Unfortunately, I cannot find a general argument.) I realized that the calculations of the centers could be avoided in the cases where $G$ is doubly primitive on $\Omega$. (This happens for example when $G=Sp(2d,2)$, see this MO Question.) REPLY [7 votes]: None of these groups are $2$-primitive except for ${\rm Sp}(2d,2)$. For ${\rm PSL}(d,q)$ with $d>2$, the $2$-point stabilizer fixes two projective points, say $\langle v_1 \rangle$ and $\langle v_2 \rangle$, so it also fixes other points, such as $\langle v_1+v_2 \rangle$. These fixed points, other than $\langle v_1 \rangle$, form a block of imprimitivity for the action of $G_{\langle v_1 \rangle}$ on $\Omega \setminus \langle v_1 \rangle$. In ${\rm PSU}(3,q)$, the point stabilizer $G_1$ has the structure $N \rtimes H$, where $N$ is a (nonabelian) group of order $q^3$ (it's a Sylow $p$-subgroup of $G$)and $H = G_{1,2}$ (the $2$-point stabilizer) is cyclic of order $(q^2-1)/(q+1,3)$. But $H$ does not act irreducibly on $N$, and fixes an elementary abelian sugbroup of order $q$, so $H$ is not maximal in $G_1$, and hence $G$ is not $2$-primitive. The situation for ${\rm Sz}(q)$ and $R(q)$ is very similar. In the first case, $G_1 = N\rtimes H$ with $|N|=q^2$, $|H|=q-1$ and in the second case $|N|=q^3$, $|H|=q-1$. In both cases, $H$ fixes a subgroup of order $q$ in $N$ so is not maximal in $G_1$. The $2$-primitivity of the two $2$-transitive actions of ${\rm Sp}(2d,2)$ was explained in the answers to the previous question. But you are correct in saying that the centres of the point stabilizers are trivial for all of these groups. The structures of these stabilizers are all very well understood. For ${\rm PSL}(d,q)$, we have $G_1 = N \rtimes H$, where $N$ is elementary abelian of order $q^{d-1}$, and $H$ is a subgroup of ${\rm GL}(d-1,q)$ that contains ${\rm SL}(d-1,q)$, and acts faithfully and naturally on $N$. So the action has no fixed points, and hence $G_1$ has trivial centre. In the other cases, $G_1$ is as described above. The unique minimal normal subgroup of $G_1$ is the centre of $N$ and has order $q$. Again $H$ acts without fixed points on this subgroup.<|endoftext|> TITLE: Stiefel-Whitney classes of virtual vector bundles QUESTION [6 upvotes]: Let $E=\xi-\eta$ be a virtual vector bundle over a compact base $B$, which we may assume is a CW complex. A quick and dirty way to define the total Stiefel-Whitney class $w(E)\in H^\ast(B;\mathbb{Z}/2\mathbb{Z})$ would be to say that, morally, $E\oplus\eta\cong \xi$, and so we can use the Whitney product formula $w(E\oplus\eta)=w(E)w(\eta)$ to get $$w(E) = w(\xi)w(\eta)^{-1},$$ the element $w(\eta)\in H^\ast(B;\mathbb{Z}/2\mathbb{Z})$ being invertible since $B$ is compact. I am interested in a more righteous definition, along the lines of Whitney's original definition for honest bundles in terms of obstructions to finding linearly independent sections over skeleta. I think that such a definition exists (maybe in terms of sectioning a certain Hom-bundle) but I wasn't able to find a reference. Hence I ask: Can the Stiefel-Whitney classes of a virtual vector bundle be defined obstruction theoretically, and if so, is there a good reference in the literature describing this construction? REPLY [5 votes]: Given $\eta$, there is a trivial bundle $N$ such that $\eta\subset N$. Then $-\eta\sim N/\eta$, and the classes of $-\eta$ can be defined as the loci where a certain number of constant sections of $N$ stops being independent modulo $\eta$. This can certainly be restated more formally in terms of obstructions.<|endoftext|> TITLE: Minimal generating sets of groups QUESTION [14 upvotes]: I am not exactly a group theorist, so this may be well-known. Let $G$ be a finitely generated group such that the cardinality of minimal generating sets of $G$ is bounded above. Does it follow that $G$ is finite? This is true if $G$ is abelian, but I have no idea about the general case. REPLY [3 votes]: This is a variant of Derek Holt's answer. Note that if $X$ is a finite group, and $S = \{x_{1},x_{2},\ldots,x_{r} \}$ is an irredundant generating set for $X$ ( that is, if any $x_{i}$ is omitted, the remaining elements do not generate $X$), then $\langle x_{1} \rangle < \langle x_{1},x_{2} \rangle < \ldots \langle x_{1},\ldots x_{r-1} \rangle < X$ is a strictly increasing sequence of subgroups of $X.$ It follows that $r \leq \log_{2}(|X|),$ in fact $r \leq \log_{p}(|X|),$ where $p$ is the smallest prime divisor of $|X|.$ Hence if $G$ is an infinite group, all of whose proper subgroups are finite, of order at most $n,$ then any irredundant generating set for $G$ has at most $1 + \log_{2}(n)$ elements.For if the chosen generating set has $s$ elements, then any $s-1$ elements irredundantly generate a proper subgroup, say $H$, of $G$, and $H$ has order at most $n.$ This bound can obviously be improved with finer knowledge of the possible prime divisors of orders of proper subgroups of $G.$<|endoftext|> TITLE: Whitney stratification and affine grassmanian QUESTION [7 upvotes]: Let $G$ a simply connected group over $\mathbb{C}$ and $Gr:=G(\mathbb{C}((t)))/G(\mathbb{C}[[t]])$ the affine grassmannian. By Cartan decomposition we have a partition of stratas indexed by $\lambda\in X_{*}(T)^{+}$. Let $Gr_{\lambda}$ such a strata and $\overline{Gr}_{\lambda}$ the closure in $Gr$. Then, $\overline{Gr}_{\lambda}=\coprod\limits_{\mu\leq\lambda}Gr_{\mu}$, is it a Whithney stratification? REPLY [5 votes]: The point is following: $\overline{Gr^\lambda}$ is a finite dimensional variety acted upon by the pro-algebraic group $G(\mathbb{C}[[t]])$. This action factors through the action of some finite dimensional algebraic group called $G(\mathcal{J}^l)$. The strata you describe are just the orbits of this group action and there are finitely many of them. (See explanation below) It is well known, that if a complex algebraic group $G$ acts on a complex variety $X$ algebraically with only finitely many orbits, then the $G$-orbits form a Whitney stratification of $X$, see Algebraic Stratifications of $G$-varieties for details. Remark: Let $\mathcal{J}^l:= \mathbb{C}[[t]]/t^l \mathbb{C}[[t]]$ for $l\in \mathbb{N}$. Then $G(\mathcal{J}^l)$ is a finite dimensional algebraic group. Let $G(\mathcal{O}^l)$ be the kernel of the canonical homomorphism $G(\mathbb{C}[[t]])\to G(\mathcal{J}^l)$. If you choose an embedding $G\hookrightarrow GL_n(\mathbb{C})$ you can think of $G(\mathcal{O}^l)$ as the subgroup of $G(\mathbb{C}[[t]])$ where all entries on the diagonal are in the form of $1+t^l\cdot f$ with $f\in \mathbb{C}[[t]]$ and all other entries are in the form of $t^l f$. Anyway it is straight forward to show that for any $\lambda$ there exists an $l\gg 0$ such that $G(\mathcal{O}^l)$ operates trivially on $\overline{Gr^\lambda}$, in other words the $G(\mathbb{C}[[t]])$ action factors through $G(\mathcal{J}^l)$<|endoftext|> TITLE: Can we prove that the ring of formal power series over a noetherian ring is noetherian without axiom of choice? QUESTION [10 upvotes]: Let $A$ be a commutative ring with an identity. Suppose that every non-empty set of ideals of $A$ has a maximal element. Let $A[[x]]$ be the formal power series ring over $A$. Can we prove that every non-empty set of ideals of $A[[x]]$ has a maximal element without Axiom of Choice? Remark The same question was asked in MSE. REPLY [5 votes]: I believe that the answer is yes. Let $A$ be a commutative ring with $1$, such that any non-empty set of ideals has a maximal element. For each ideal $I\leq A[[x]]$, and each $n\in \mathbb{N}$, define $I^{(n)}$ to be the set $$\{0\}\cup\{a\in A\ :\ a\text{ is the leading non-zero coefficient, in degree $n$, of some }f\in A[[x]]\}.$$ This is an ideal of $A$. Also define $I^{(\omega)}=\bigcup_{n\in \mathbb{N}}I^{(n)}$, which is also an ideal of $A$. Now suppose $\{I_{\alpha}\}_{\alpha}$ is a set of ideals in $A[[x]]$. The set of ideals $\{I_{\alpha}^{(\omega)}\}_{\alpha}$ has a maximal element, say $I^{(\omega)}_{\beta}$. Further, it isn't hard to prove that the form of the noetherian hypothesis we are assuming implies finite generation of ideals (provable in ZF). Thus, we may fix a generating set $a_1,a_2,\ldots, a_k$ for $I^{(\omega)}_{\beta}$. This means that $I^{(\omega)}_{\beta}=I^{(n)}_{\beta}$ for some $n\in \mathbb{N}$, which we also fix. (For instance, we can take the integer $n$ to be the lowest degree where each of $a_1,a_2,\ldots, a_k$ occurs as a leading coefficient.) Now, restrict the set of $\alpha$ to only those for which $I_{\alpha}^{(n)}=I_{\beta}^{(n)}$. (Note that if $I_{\alpha}^{(n)}\neq I_{\beta}^{(n)}$, then $I_{\alpha}$ does not contain $I_{\beta}$. So this restriction costs us nothing.) Next, we may as well also assume $\beta$ is chosen such that $I_{\beta}^{(n-1)}$ is a maximal element of $\{I_{\alpha}^{(n-1)}\}_{\alpha}$. We then may assume $I_{\beta}^{(n-2)}$ is maximal in $\{I_{\alpha}^{(n-2)}\}_{\alpha}$, after restricting $\alpha$ again. Recursively repeating this process (only finitely many times), the resulting $\beta$ is such that $I_{\beta}$ is maximal in $\{I_{\alpha}\}_{\alpha}$.<|endoftext|> TITLE: Intersections of complex submanifolds in $\mathbb{C}^N$ QUESTION [5 upvotes]: This is an exercise from Gromov's Partial differential relations. (page 5) Let $V$ and $V'$ be two closed complex submanifolds in $\mathbb{C}^N$ of complimentory dimension. Prove that $V$ and $V'$ intersect if the following sets are compact for all k. $V_k = \{(v,v') \subset V\times V'|dist(v,v') \leq k$}. I was looking for a differential geometry approach to solving this problem along the lines of Theorem 2 of Frankel(1961) but anything would do. REPLY [9 votes]: By compactness, there is a pair of points at the minimal distance. On the other hand, any positive distance can be made smaller (e.g., cf. the proof of Lefschetz theorem in Milnor's "Morse theory": the critical points of the distance function to a complex manifold always have positive index).<|endoftext|> TITLE: Etymology of cuspidal representations QUESTION [16 upvotes]: In the literature on representation theory of $GL_2(\Bbb F_p)$ and $GL_2(\Bbb Q_p)$, the irreducible representations with trivial Jacquet module are often called "cuspidal" or "supercuspidal". Why are these representations called cuspidal and who started calling them that? Also, is there any difference between cuspidal and supercuspidal? I imagine this is related to modular forms but haven't been able to find any explanation in the literature. These representations seem mysterious to me and maybe understanding their etymology will give me a better sense of how they fit into the bigger picture. REPLY [10 votes]: To reinforce what's already been said and add some references, I'd emphasize first that Harish-Chandra's "philosophy of cusp forms" was indeed a driving force in the study of representations over both finite and local fields. Borel organized the 1965 AMS summer institute at Boulder around these ideas and the emerging Langlands program. Certainly the term "cusp" arose in the early work on arithmetic groups and fundamental domains, leading indirectly to the idea of "discrete series" of representations. This starts with $\mathrm{SL}_2(\mathbb{R})$ and was developed very generally by Harish-Chandra for real semisimple Lie groups. He and many others also opened the way to study of such groups over local fields as well as over finite fields. His conference talk Eisenstein series over finite fields (in the volume Functional Analysis and Related Fields, Springer, 1970) reflects his interest in carrying classical ideas over to such fields. Concerning the character theory of finite groups of Lie type, there is of course a long history going back to work of Frobenius and Schur, in which it became clear that parabolic induction alone would not produce all characters. In his influential 1955 paper on characters of finite general linear groups, J.A. Green was able to use combinatorial methods to fill in the missing characters. But his student Bhama Srinivasan had more difficulty with the case of $\mathrm{Sp}_4(\mathbb{F}_q)$, which helped to shift attitudes toward the systematic study of "discrete series" or "cuspidal characters" in terms of Harish-Chandra's philosophy. Here one associates series of characters to the various types of finite tori, ranging from split to anisotropic ("compact"). The latter lead to cuspidal characters, and the 1976 paper by Deligne-Lusztig made possible the use of sophisticated geometric methods for their study. (Lusztig himself had been influenced by Green in his earlier construction of some discrete series characters satisfying "cusp conditions" for general linear groups, in his paper Ann. of Math. Studies 81 (1974).) Along the way there was also a special year at IAS in 1968-69, organized by Borel and influenced by Harish-Chandra, Springer, and others. In particular, Springer had two relevant series of lectures written up in Lecture Notes in Mathematics 131 (Springer, 1970), including Cusp forms for finite groups. It's difficult to disentangle all the overlapping contributions here, but the terminology (codified in Roger Carter's 1985 book) is appropriate even if indirect. As noted already, terminology for the groups over local fields developed along similar philosophical lines, but with the addition of terms like "supercuspidal" which doesn't come up over finite fields.<|endoftext|> TITLE: Did ancient mathematicians know Euler's characteristic for convex polyhedra? QUESTION [32 upvotes]: The formula $V-E+F=2$ is so simple that I can't believe that it was really Euler (or perhaps Descartes) who first observed it (I mean the formula itself in some generality, not necessarily a valid proof). To have a concrete question: Is there any reference to this formula in ancient mathematics? REPLY [13 votes]: Once I was lecturing to high school students about this theorem. My proof began with the words: "suppose that the net is drawn on a surface of a rubber ball...". One student asked: "Did rubber exist at the time of Euler?" I think the moral of this story is that the very statement of the question was foreign to ancient Greek mathematics. Did they have a definition of ARBITRARY polytope? Or even an arbitrary convex polytope? I doubt it. (See this related post). Descartes proof was earlier than that of Euler, and his statement was in terms of solid angles (which is equivalent to the Gauss-Bonnet theorem, but stated in terms of elementary geometry). I think it was Euler who recognized for the first time that we have a topological fact here. And this was a great insight. Like the problem of the same Euler about the Konigsberg bridges. It is not difficult, but before Euler these things apparently were not considered part of mathematics. That's why these questions were not raised in antiquity. EDIT. Many things were not discovered for the simple reason that no one was looking for these things. Let me give a more recent example. Could Fermat, Descartes, Huygens, Euler, etc. discover linear programming and simplex method? There is no doubt they could. After all, this is elementary mathematics which can be easily taught in high school. But look what had really happen. Fourier had a paper on systems of linear inequalities (where he invented what is called Fourier-Motzkin elimination nowadays). And in the end of XIX century (!) the editor of Complete works of Fourier had to APOLOGIZE for including this paper: he wrote that "every work of such a grand master has to be included", where "every" evidently means "even such trivial nonsense":-) And linear programming became a part of mainstream mathematics only during WWII, when it was really needed.<|endoftext|> TITLE: Simultaneous geometric separator QUESTION [5 upvotes]: A geometric separator is a line that separates a given set of shapes to two subsets of approximately the same size (up to a constant), while intersecting only a small number of shapes. When a geometric separator exists, it is very useful because it allows us to solve difficult computational geometry problems in a divide-and-conquer manner. Example: Given a set of $n$ disjoint axis-parallel squares in the plane, there is a rectangle such that at most $2n/3$ squares are inside it, at most $2n/3$ are outside it, and at most $O(\sqrt{n})$ are intersected by it. Smith and Wormald (1998) prove this theorem, as well as thousands of generalizations with various applications. However, there is one generalization I haven't found yet, and I really want to know whether it is true: Given $m$ sets, each with $n$ disjoint axis-parallel squares of various sizes, is there a rectangle such that, at most $2mn/3$ squares are inside it, at most $2mn/3$ are outside it, and at most $O(\sqrt{n})$ squares of each collection are intersected by it? Note that, because the shapes in each of the $m$ sets are disjoint, the union of all sets is $m$-thick. Therefore, by theorem 39 in the original paper, it is possible to partition the collections such that the total number of intersected squares is $O(m\sqrt{n})$. Therefore, when $m=O(1)$, the existence of a simultaneous separator is guaranteed. On the other hand, when $m \to \infty$, a simultaneous separator may not exist - see the comment by fedja. The interesting case is when $m=\Omega(\sqrt{n})$. In this case, the existing theorem allows $\Omega(n)$ intersected squares in a single collection. My question is, basically, whether it is possible to bound the number of intersected squares per collection. REPLY [3 votes]: Expanding fedja's comment, and assuming that the separating rectangle is axis-parallel: The answer is "no" even if $m=O(n)$, and more generally, whenever $m$ is not bounded. Let $f:\Bbb N \rightarrow \Bbb N$ be an arbitrary unbounded function. Let $m_1 = \frac{99}{100} \cdot f(n)$. Arrange $n$ unit squares into an $\sqrt{n} \times \sqrt{n}$ grid, forming a square $Q$, and repeat this collection $m_1$ times. The boundary of every axis-parallel separating rectangle contains a horizontal or a vertical segment of length $\sqrt{n}/2$ inside $Q$. Tile $Q$ with $nf(n)/400$ squares of side length $20/\sqrt{f(n)}$, and partition them into $m_2=f(n)/400$ collections, each collection forming roughly an $\sqrt{n} \times 400\sqrt{n}/f(n)$ rectangle. Repeat this tiling once more but rotate the partition by $\pi/2$. In total, we have at most $m_1 + 2m_2 < f(n)$ collections, and every separating axis-parallel rectangle intersects some collection in $\sqrt{nf(n)}/20$ squares. For arbitrarily rotated separating rectangles, the construction can be modified: we tile $Q$ by larger squares, of side length $cf(n)^{-1/4}$, partition the tiling into $O(f(n)^{1/2})$ collections, and repeat this $O(f(n)^{1/2})$ times, each time "rotating" the partition by a small angle of size $O(f(n)^{-1/2})$.<|endoftext|> TITLE: Best ranking in tournament: polynomial time algorithm? QUESTION [6 upvotes]: This question was posed by my colleague Torbjörn Lundh in his paper Which Ball is the Roundest? A Suggested Tournament Stability Index, Journal of Quantitative Analysis in Sports 2(3), 2006. We have discussed it a number of times without finding a solution. The context is a tournament, where $n$ players or teams meet pairwise in games. We assume for simplicity that each game results in one team winning and the other losing (no ties). Given the results of all $n \choose 2$ games, the task is to produce a ranking of the teams, that is, to order them from best to worst. This sort of problem goes back at least to the 19th century, and arises when a number of subjects are asked to rank for instance the taste of different wines. It's hard to directly produce an ordering of say ten glasses of wine based on tasting them. It's much easier to repeatedly make pairwise comparisons, is wine A better or worse than wine B? But the final answers need not be consistent, and we arrive at the problem of interpreting a series of such pairwise comparisons. One possibility is the method used in sports tournaments, where teams are ranked according to the number of won games (with various tiebreak rules when the scores are equal). But we might also be interested in the ranking that minimizes the number of inconsistencies, that is, minimizes the number of games that are won by the lower ranked team. Is there a polynomial time algorithm that finds, from a table of results of all $n \choose 2$ games, a ranking with minimum number of inconsistencies? It's easy to see that a ranking that minimizes the number of inconsistencies might have to place a team with fewer victories above a team with more victories. This is because a ranking minimizing the number of inconsistencies can have no inconsistencies between teams adjacent in the ranking (if there were, we could improve by just switching them). So suppose in a tournament that team A loses to team B and wins all other games, team B loses two other games and all other teams win fewer games. Then it can't be optimal in the sense of minimizing the number of inconsistencies to rank A first and B second. Another example is shown in Torbjörn's paper. There are various superficially similar problems that are known to be NP-complete (Torbjörn mentions in his paper the directed optimal linear arrangement and the quadratic assignment problem), and therefore it would be quite surprising if there were an efficient algorithm for finding a ranking that minimizes the number of inconsistencies. But we haven't been able to encode any known NP-complete problem in terms of a tournament requiring a ranking. REPLY [3 votes]: The problem you are looking for is called the Feedback Arc Set problem for tournaments (FAST). It was proved to be NP-hard by Alon: http://www.tau.ac.il/~nogaa/PDFS/paley.pdf and separately Charbit, Thomassé and Yeo: http://www.liafa.jussieu.fr/~charbit/Recherche/publis/minfastour.ps<|endoftext|> TITLE: Population dynamics for fish arriving via a Poisson process and living for a time given by some (not necessarily symmetric) general distribution QUESTION [5 upvotes]: Imagine we have a hypothetical population of fish in a pond. The fish cannot reproduce, but are introduced by a Poisson process (with some known and fixed rate parameter independent of the total number of fish), and survive for a time given by some general distribution $G$. Starting from a state where there are $N$ fish in the pond, what do we need to know about $G$ in order to calculate the fish population dynamics over time (the rate of increase or decrease over time)? Consider that if $G$ is bimodal (early age / child mortality often does this to lifetime probability distributions), the mean lifetime expectation may not be very helpful. Is the median lifetime more useful in this regard? Or is the answer always going to be that "it depends"? Edit: I suppose there are two parts to this question, the second one conditional on the first: (1) Is there some simple or minimal metric that we can measure about $G$ (mean, median, etc.) that serves the purpose of allowing us to determine fish population dynamics over time? (2) If not, and given perfect information about the probability distribution $G$ (or at least a good enough approximation from experimental / simulation data), we must necessarily have enough information to determine the fish population dynamics, but how do we do this in practice? For a fun example, let's imagine the lifetime probability distribution $G$ looks like the positive component of some damped harmonic oscillator (after appropriate scaling, etc.): http://tinyurl.com/o5o8p9l (WolframAlpha query "Plot[(Re[Sqrt[e^(-t)*Sin[2*pi*t]]])^2, {t, 0, 3.5}]" --- the full URL failed to hyperlink properly due to the presence of special characters). One could also imagine those peaks having arbitrary or random heights. This kind of scenario could occur if the primary mechanism for fish mortality was fishing, and a set of catch or kill rules were put in place based on a fish's age (maybe it has to be a prime number, etc). I'd argue that the mean or median value for random variates sampled from $G$ might not be too useful in estimating the fish population dynamics (though I could certainly be wrong). Are there more "robust" metrics? REPLY [2 votes]: To know the total number of fishes in the pond at time $t$, one needs to know the lifetimes of the $N$ initial fishes. The number $N_t$ of fishes in the pond at time $t$ not in the pond at time $0$ is $$ N_t=\int_0^tB_{t-s}(s)\,\mathrm ds, $$ where, for every $t$ and $u$, $B_t(u)\mathrm d t$ denotes the number of fishes introduced in the pond in the time interval $(t,t+\mathrm dt)$ with an initial lifetime at least $u$. Thus, $E(B_t(u))=\lambda P(Z\geqslant u)$, where $Z$ has distribution $G$, and, for every $t$, $$ E(N_t)=\lambda\, E(\min(Z,t)), $$ which converges to the stationary value $E(N_\infty)=\lambda\,E(Z)$, as was to be expected. Likewise, the number $N_t(u)$ of fishes in the pond at time $t$ not in the pond at time $0$ and whose residual lifetime at time $t$ is at least $u$ is $$ N_t(u)=\int_0^tB_{t-s}(s+u)\,\mathrm ds, $$ hence $$ E(N_t(u))=\lambda\, E(\min((Z-u)^+,t)), $$ which converges to the stationary value $$ E(N_\infty(u))=\lambda\,E((Z-u)^+). $$ In other words, the distribution of the residual lifetime $R$ of a typical fish in the pond at time $t$ when $t$ and $\lambda$ are large is characterized by the identities $$ P(R\geqslant u)=\frac{E((Z-u)^+)}{E(Z)}, $$ hence the density $f_R$ of $R$ is such that, for every $u\geqslant0$, $$ f_R(u)=\frac{P(Z\geqslant u)}{E(Z)}, $$ and the moments of $R$ are given by $$ E(R^n)=\frac{E(Z^{n+1})}{(n+1)E(Z)}. $$ In the other direction, every density $f_R$ is nonincreasing and finite everywhere, and the distribution of $Z$ can de deduced from $f_R$ thanks to the identities $$ P(Z\geqslant u)=\frac{f_R(u)}{f_R(0)}. $$<|endoftext|> TITLE: When does the choice of the generic matter? QUESTION [26 upvotes]: It is a somewhat curious phenomenon that, in forcing arguments, one usually doesn't care about any particular properties of the generic filter being used (this isn't strictly true; there are cases where we force below some sort of master condition, for example, but this basically amounts to asking my question for the cone below the condition). This position might possibly be preferred by mathematicians who interpret talking about generic filters as "semantic sugar" for purely syntactic arguments about Boolean truth values, but if we are prepared to talk about generics as real objects, it seems weird that we don't pay very much attention to the structure of the objects which generate our extensions. But perhaps we needn't always care about the particular generic. Let $M$ be a transitive model of set theory. Call a notion of forcing $\mathbb{P}\in M$ forcing agnostic (over $M$) if for any two $M$-generic filters $G,H\subseteq \mathbb{P}$ the two extensions $M[G]$ and $M[H]$ are elementarily equivalent. There is an obvious reformulation of forcing agnosticism: $\mathbb{P}$ is forcing agnostic iff the Boolean value, with respect to the Boolean algebra associated to $\mathbb{P}$, of any sentence (without parameters in the forcing language) is either 0 or 1. This immediately implies that any almost homogeneous forcing is forcing agnostic; in fact, any two extensions by almost homogeneous forcing are elementarily equivalent in the language augmented with constants for elements of the ground model. Is there a characterization of forcing agnostic posets? Is this a purely structural property of the poset or does the ambient model matter, i.e. can a poset be forcing agnostic over some models but not over others? I would also welcome any examples of forcing arguments where some care is needed in choosing the generic (in addition to ensuring a particular condition gets in). REPLY [7 votes]: Here is another kind of example where the choice of the generic filter $g$ used to build a generic extension $M[g]$ of a transitive model $M$ is important. Suppose that we have some set $x$ that we want to be an element of the generic extension $M[g]$. If $x$ is not already in the ground model $M$, then a mutual genericity argument shows that for any condition $p$ there are lots of generic filters $g$ containing $p$ with $x \notin M[g]$. So if we want $x \in M[g]$ then we must choose $g$ rather carefully. For example, if $M$ is a countable mouse with a Woodin cardinal and $x$ is a real, then by Woodin's "genericity iteration" theorem (or a variant due to Neeman) there is an iteration $i: M \to M^*$ of $M$ and an $M^*$-generic filter $g$ such that $x \in M^*[g]$. This theorem is used all the time in inner model theory, and there is no way to state it using only the forcing language of $M$ or of $M^*$.<|endoftext|> TITLE: Has Witten's perturbation on de Rham complex been studied on other elliptic complexes? QUESTION [17 upvotes]: In his famous work, Supersymmetry and Morse theory, Witten perturbs de Rham complex by perturbing the exterior derivative $$d_h=e^{-ht}de^{ht}.$$ And he proves Morse inequality using some spectral information of the Laplacians of the complex. Has similar perturbation on the the other elliptic complexes been studied? Complexes like Dolbeault complex or signature complex, Or maybe in general, twisted spin complexes. Can one expect new information from such a perturbed complexes? REPLY [11 votes]: Hormander's approach to solving the $\bar \partial$ problem is basically this, and his paper is from 1965, predating Witten's work by a couple of decades! By varying the "weight" function $h$, you can get families of estimates on the solution of $\bar \partial$ problem. Check out Hormander's 1965 ACTA paper for more details. Really a fabulous paper. He does not make explicit that he is "perturbing the $\bar \partial$-complex", but that is exactly what he is doing. I am sure Witten was reading Hormander. It is also possible to perturb this complex in more dramatic ways. My advisor (Jeff McNeal) has done a lot of this work. You might be interested in his survey paper "$L^2$ Estimates on Twisted Cauchy Riemann Complexes". My thesis work is basically applying such twisted complexes (which can sometimes give estimates which are unavailable without the twisting) to prove some new results on approximation by holomorphic functions in $\mathbb{C}^n$.<|endoftext|> TITLE: Why is adopting Russell's Axiom of Reducibility as strong as eliminating the Ramified Hierarchy? QUESTION [5 upvotes]: In order to respond to concerns of impredicativity, Bertrand Russell developed a system of ramified second-order logic, which is like regular second-order logic except the comprehension schema is divided into levels. The comprehension schema for level $0$ sets does not allow any formulas with second-order quantifiers. For any natural number $n$, the comprehension schema for level $n+1$ sets allows quantification over sets of level $n$ and below. This ensures that no set is defined using quantification over itself. The resulting system, however, proved too weak to do much mathematics. (Although it turns out it can do more than Russell assumed; see here.) So Russell adopted his controversial axiom of reducibility, actually an axiom schema which for each natural number $n$, states that for any set $X$ of level $n$, there exists a set $Y$ of level $0$ such that $X$ and $Y$ contain the same elements. Now, it is commonly asserted that the axiom of reducibility is equivalent to simply eliminating the ramified hierarchy and just working in standard second-order logic. But I don't see why. At first glance, it seems to me that saying that every set, period, is coextensional with a level $0$ set, is a stronger statement than saying that every set of any given level is coextensional with a level $0$ set. Aren't there sets that aren't coextensional with sets of any level? Any help would be greatly appreciated. Thank You in Advance. EDIT: Here's another way to phrase my question. Consider the following two possible axiom schemata: For any formula $\phi(x)$ in the language of second-order arithmetic, there exists a set of level $0$ whose elements are the ones that satisfy $\phi(x)$ For any formula $\phi(x)$ with only graded quantifiers (i.e. quantifiers of the form "for all sets of level..." or "there exists a set of level..."), there exists a set of level $0$ whose elements are the ones that satisfy $\phi(x)$. Is the first axiom schema stronger than the second, or are they equivalent? REPLY [2 votes]: Using the language of second-order arithmetic with both graded and ungraded quantifiers, axiom schema 1 proves more theorems than axiom schema 2. The proof is that 2 does not imply 1, because we can take a non-trivial model for 2, and expand it by having all subsets of $N$ as ungraded subsets.<|endoftext|> TITLE: Local positivity of solutions to linear differential inequalities (Chaplygin's theorem) QUESTION [7 upvotes]: According to the entry "Differential inequality" of the Encyclopedia of Mathematics http://www.encyclopediaofmath.org/index.php/Differential_inequality the following result is due to Chaplygin (1919). Let $y: [0, a)\to \mathbb{R}$ satisfy the differential inequality $$ y^{(m)}+\alpha_1(x) y^{(m-1)}+\cdots+ \alpha_{m}(x) y> f $$ and let $z: [0, a)\to \mathbb{R}$ be a solution of the same equation with the equality sign $$ z^{(m)}+\alpha_1(x) z^{(m-1)}+\cdots+ \alpha_{m}(x) z= f $$ both with the same initial condition at $x=0$. Then there is some $b>0$, $b0$, there is some neighborhood of $(0,0)$ (in the product topology) over which $K^{(n-1)}$ is positive, and hence a triangle $0\le \xi \le x < b\le c/2$ over which $K^{(n-1)}$ is positive. But on the diagonal $K^{(i)}$, $i=1,\cdots, n-2$, vanishes so upon integration on $x$ we obtain that $K^{(i)}$, is positive on $0\le \xi < x < b$ for every $i=1,\cdots, n-1$. From the assumption we have that $Ly\ge 0$, where $Ly$ is continuous. The uniqueness of the solution to the differential equation $L y= f$ implies the easily verifiable formula $$ y(x)=\int_0^x K(x,\xi) Ly(\xi) \, d \xi , $$ which under differentiation gives more generally $$ y^{(i)}(x)=\int_0^x K^{(i)}(x,\xi) Ly(\xi)\, d \xi , \qquad i=0,1,\cdots, n-1, $$ thus $y^{(i)}\ge 0$ on $[0,b]$, and the equality $y=0$ on $[0,b]$ is possible only if $Ly=0$ on $[0,b]$. $\square$<|endoftext|> TITLE: When are Galois representations with open image attached to elliptic curves? QUESTION [7 upvotes]: Let $K$ be a number field with absolute Galois group $G_K$. Let $\rho:G_K \rightarrow GL_2(\hat{\mathbb{Z}})$ be a Galois representation such that the image of $\rho$ is open in $GL_2(\hat{\mathbb{Z}})$, and the determinant of $\rho$ is cyclotomic i.e. $\det \circ \rho = \chi$ where $\chi$ is the cyclotomic character. Then under what conditions does $\rho$ come from the Galois representation on the Tate module of an elliptic curve? REPLY [3 votes]: You have to assume that all the local Galois representations are compatible. Even then the representation can come from an automorphic form (as pointed out in the comments). In the case when $K=\mathbb{Q}$, David Zywina has computed all the possible indices of the image of $\rho$ (the result is a little more elaborate) in "On the possible images of the mod l representations associated to elliptic curves over $\mathbb{Q}$".<|endoftext|> TITLE: Can a unit square be cut into rectangles that tile a rectangle with irrational sides? QUESTION [24 upvotes]: For arbitrary positive integers $m$ and $n$, if we dissect a unit square into an $m\times n$ rectangular grid of $1/m\times 1/n$ rectangles, we can reassemble these $mn$ rectangles into an $n/m\times m/n$ rectangle, which is of square-rational proportion $m^2/n^2$. Is there any essentially different way to rectangularly dissect a unit square to a rectangle----that is, to a $1/r\times r$ rectangle, of proportion $r^2$, where $r$ is irrational? Notes: The usual rules for dissection problems apply (finitely many pieces, with no gaps or overlaps of positive planar measure, and no discarded or additional pieces). This question was posted, unanswered, on MathStackExchange for a month. For what it's worth, my failed attempts to find such a dissection lead me to think it impossible. REPLY [8 votes]: Suppose that the unit square has been cut into $n$ rectangles: $S=\cup_{i=1}^nR_i,$ where each rectangle $R_i$ has dimensions $s_{2i-1}\times s_{2i}$. Also let $r$ be an irrational number and let $R^*$ be a $1/r \times r$ rectangle. We will define an additive "weight" (which can take negative values) on certain plane sets such that $w(S)=1$ but $w(R^*)=0$. Let $L=\{{1,r,1/r,s_1,s_2,\dots,s_{2n}\}}.$ Clearly any rectangle which can be assembled using the $R_i$ has sides which can be expressed as $\sum_{i=1}^{2n}c_is_i,$ where each $c_i \in \{{0,1\}}$. As usual, let $\operatorname{span}L$ be the set of all linear combinations $\sum_{j=1}^{2n}c_js_j+c'+c''r+c'''\!\!\ 1/r$ with the coefficients $c \in \mathbb{Q}.$ There is a subset $B \subseteq L$ which is a basis in that $\operatorname{span}B=\operatorname{span}L$ and each $r \in \operatorname{span}L$ can be uniquely expressed as a rational linear combination of elements of $B$. Furthermore, we can arrange to have $1 \in B$ and $r \in B.$ Also we may have $1/r \in B$ unless $r$ is the root of an integer quadratic equation $ar^2+br+c=0,$ in which case $1/r=\frac{-a}{c}r -\frac{b}{c}.$ Define the function $f:\operatorname{span}L \to \mathbb{Q}$ by taking $f(x)$ to be the coefficient of $1$ in the unique rational linear combination from $B$ which equals $x$. Then $f$ is a linear transformation: $f(x_1+x_2)=f(x_1)+f(x_2).$ Also, $f(x)=x$ for $x$ rational, and $f(r)=0$ for our special chosen irrational number. Now, as in the other proofs, use $f$ to give a weight $w_f(R)=f(u)f(v)$ to any rectangle $R$ with sides $u \times v$ with $u,\!v \in \operatorname{span}L,$ and extend the weight additively to any plane set which is a (finite) union of such rectangles with disjoint interiors. As before, $w_f(S)=1 \ne 0=w_f(R^*)$. Hence no dissection of $S$ can be used to assemble $R^*.$<|endoftext|> TITLE: Distribution of subset-sums QUESTION [11 upvotes]: Let $A$ be a set of $n$ integers uniformly distributed in $\{0,\dots,N-1\}$. Let $S$ be the set of subset-sums modulo $N$ of $A$. Let $f_{n,N}(k)$ be the probability that $|S|=k$. Is there an expression for $f_{n,N}(k)$? In particular, I am interested in the case where $N=2^n$. REPLY [3 votes]: You might be interested in this paper: Li, Jiyou; Wan, Daqing, Counting subset sums of finite Abelian groups, J. Comb. Theory, Ser. A 119, No. 1, 170-182 (2012). ZBL1229.05289. Abstract: In this paper, we obtain an explicit formula for the number of zero-sum $k$-element subsets in any finite abelian group.<|endoftext|> TITLE: There's something strange about $\sqrt{\big(j(\tau)-1728\big)d}$ QUESTION [52 upvotes]: Given discriminant $d$ and j-function $j(\tau)$, I was looking at, $$F(\tau) = \sqrt{\big(j(\tau)-1728\big)d}$$ which appears in Ramanujan-type pi formulas. Let $C_d$ be the odd prime factors of the constant term of the minimal polynomial for $F(\sqrt{-d})$. Then for prime $d>3$, $$\begin{aligned} C_{5} &= 5, 11, 19.\\ C_{7} &= 3, 7, 19.\\ C_{11} &=7, 11, 19, 43.\\ C_{13} &=3, 13, 43.\\ C_{17} &=17, 19, 43, 59, \color{red}{67}.\\ C_{19} &=3, 19, \color{red}{67}.\\ C_{23} &=3, 7, 11, 19, 23, 43, \color{red}{67}, 83.\\ C_{29} &=7, 23, 29, \color{red}{67}, 107.\\ C_{31} &=3, 11, 23, 31, 43.\\ C_{37} &=3, 7, 11, 37, \color{red}{67}, 139.\\ C_{41} &=23, 31, 41, 43, 83, 139, \color{blue}{163}.\\ C_{43} &=3, 7, 19, 43, \color{blue}{163}.\\ C_{47} &=3, 11, 19, 31, 43, 47, \color{red}{67}, 107, 139, \color{blue}{163}, 179.\\ C_{53} &=7, 11, 43, 53, 131, \color{blue}{163}, 211.\\ C_{59} &=3, 5, 11, 23, 31, 43, 47, 59, \color{red}{67}, 211, 227.\\ C_{61} &=3, 19, 47, 61, \color{blue}{163}.\\ C_{67} &=3, 7, 11, 31, 43, \color{red}{67}.\\ C_{71} &=5, 7, 11, 23, 47, 59, \color{red}{67}, 71, \color{blue}{163}, 283.\\ \vdots\\ C_{163} &=3, 7, 11, 19, 59, \color{red}{67}, 127, \color{blue}{163}, 211, 571, 643.\\ C_{167} &=3, 43, \color{red}{67}, 103, 131, 139, 151, \color{blue}{163}, 167, 227, 307,\dots 659.\\ \end{aligned}$$ and so on. Notice that the d with $C_d$ divisible by $163$ are the first few primes of Euler's prime-generating polynomial, $$P_1(n) = n^2+n+41 = 41, 43, 47, 53, 61, 71, 83, 97,\dots$$ and the lesser known, $$P_2(n) = 4n^2+163 = 163, 167, 179, 199,\dots$$ Similarly, the d with $C_d$ divisible by $67$ intersect with, $$Q_1(n) = n^2+n+17 = 17, 19, 23, 29, 37, 47, 59, 73, 89,\dots$$ and, $$Q_2(n) = 4n^2+67 = 67, 71, 83, 103,\dots$$ Q: Does anybody know the reason for this "numerology"? REPLY [65 votes]: "Numerology" such as you've observed is explained in the paper Gross, B.H., and Zagier, D.: On singular moduli, J. reine angew. Math. 355 (1985), 191$-$220. MR772491 (86j:11041) which gives more generally the factorizations of the constant terms of the minimal polynomials of $j(\tau) - j(\tau')$ where $\tau,\tau'$ are quadratic imaginaries not equivalent under ${\rm PSL}_2({\bf Z})$. Your $j(\tau) - 1728$ is the special case $\tau' = i$. Before seeking patterns in the appearance of factors such as $67$ and $163$, one might wonder why all the constant terms, which are roughly exponential in $\sqrt d$, factor into such small prime factors in the first place. The reason is that these are the primes $p$ for which the elliptic curve $E: y^2 = x^3 - x$, which has $j$-invariant $1728$, is also the reduction mod $p$ of a curve of invariant $j(\sqrt{-d}\,)$, and thus has an action of ${\bf Z}[\sqrt{-d}\,]$. Since $E$ already has an action of ${\bf Z}[i]$ [with $i$ acting by $(x,y) \mapsto (-x,iy)$], this makes $E$ supersingular. The condition that the endomorphism ring of a supersingular curve $E \bmod p$ accommodate both ${\bf Z}[i]$ and ${\bf Z}[\sqrt{-d}\,]$ comes down to the representability of $4d$ by the quadratic form $a^2+pb^2$. In particular $p < 4d$, which explains why all the prime factors are small. Your $Q_1$ and $Q_2$ are obtained by setting $b=1$ and $b=2$, but eventually higher $b$ arise too, e.g. you'll find $p=67$ among the factors of $C_{151}$ (for which the minimal polynomial has degree $7$) because $151 = \frac14 (1^2 + 67 \cdot 3^2)$, even though $151$ is not represented by either $n^2+n+17$ or $n^2+67$.<|endoftext|> TITLE: Can a Suslin line be 2-entangled? QUESTION [15 upvotes]: A Suslin line is a linear order $L$ which is dense with no endpoints, complete, and ccc but not separable. I'm wondering what kind of order-preserving maps there are from $L$ into $L$. Specifically, Question: Can there be a Suslin line $L$ such that for every one-to-one, monotonic function $f$ from an uncountable subset of $L$ into $L$, the set of $x\in\mathrm{dom}(f)$ with $f(x)\neq x$ is countable? The motivation for this question is the following. A linear order $L$ is $n$-entangled if for every uncountable set $A$ of pairwise-disjoint $n$-tuples in $L$, and for every $s : n\to 2$, there are $a,b\in A$ such that $a_i < b_i$ if and only if $s(i) = 0$, for all $i < n$. One can show that a linear order $L$ is $2$-entangled if and only if it is rigid, in the sense that every one-to-one, monotonic function on an uncountable subset of $L$ is equal to the identity on a co-countable subset of its domain. It's not difficult to show that a weakening of the Open Coloring Axiom implies there are no $2$-entangled sets of reals. However, OCA is consistent with the existence of a Suslin line. So an answer to the above question would provide evidence for an answer to the following. Question: Is OCA consistent with the existence of a $2$-entangled linear order? Edit: I forgot to mention why entangledness is relevant. A $3$-entangled linear order is necessarily separable, so $2$-entangledness is the most you might possibly get out of a Suslin line. REPLY [3 votes]: Regarding the second question: Definition: $A \subset \mathbb{R}$ of power $\aleph_1$ is called an increasing set if in every uncountable set of pairwise disjoint finite sequences from A there are two sequences $$ and $$ having the same length such that for all i $a_i TITLE: What is the connection between direct/inverse image of set maps and direct/inverse image functors of sheaves? QUESTION [5 upvotes]: Given a function of sets $f:X\to Y$, one defines the direct image and inverse image maps: $$ f_*:\mathcal{P}(X)\to\mathcal{P}(Y) $$ $$ f^{-1}:\mathcal{P}(Y)\to\mathcal{P}(X) $$ In the usual way. On the other hand, given a continuous map of topological spaces $f:X\to Y$, and a nice category $\mathcal{C}$, one defines the direct image and inverse image functors between the corresponding categories of $\mathcal{C}$-valued sheaves: $$ \bar{f}_*:Sh_{\mathcal{C}}(X)\to Sh_{\mathcal{C}}(Y) $$ $$ \bar{f}^{-1}:Sh_{\mathcal{C}}(Y)\to Sh_{\mathcal{C}}(X) $$ (the bar over $f$ is just to distinguish between the notations). Now, the terimnology suggests some connection between the notions, so I tried to work it out myself. It turns out that there is a rather obvious connection, but it conflicts with the terminology! Namely, we can view $\mathcal{P}(X)$ and $\mathcal{P}(Y)$ as poset categories with respect to inclusion, and this turns $f_*$ and $f^{-1}$ into functors (as they preserve inclusions). This seems like the obvious first step, but we can already see that something is wrong, since the pair $(f_*,f^{-1})$ is andjoint, but in the wrong order! Namely, $f_*$ is the left adjoint and $f^{-1}$ is the right (translation: $f(A)\subseteq B \iff A\subseteq f^{-1}(B)$ and not the other way around) as opposed to the situation with sheaves. Now, we can go further and interpreate the category $\mathcal{P}(X)$ as a category of sheaves on some space. Indeed, it is isomorphic (!) to the category of "truth-valued" sheaves on the discrete space on $X$. Under this isomorphism, the functor $f^{-1}$ indeed corresponds to $\bar{f}^{-1}$ (note that no colimit is needed in the definition of $\bar{f}^{-1}$ since the space is discrete). But as we can already expect, $f_*$ does not correspond to $\bar{f}_*$. In fact, since $f^{-1}$ preserves union as well as intersections, and hence cocontinuous as well as continuous, it has a right adjoint as well: $$ f_!:\mathcal{P}(X)\to\mathcal{P}(Y) $$ that can be defined by $$ f_!(A)=\{y\in Y \mid f^{-1}(y)\subseteq A\} $$ and in fact it is this functor that corresponds to $\bar{f}_*$ and not $f_*$. So, finally, my question is this: Is there some other connection between the two notions that makes the analogy work, or is it just unfortunate coincidence in terminology? REPLY [2 votes]: Here is a setting in which the question can be made more precise. Consider a continuous map $f: X \to Y$ of topological spaces which is also open. Let us denote by $O(X)$ and $O(Y)$ the posets of open subsets of $X$ and $Y$ respectively. Since $f$ is both continuous and open the image of an open is open and the inverse image of an open is open. As in your description, we obtain an adjunction between the posets $O(X)$ and $O(Y)$. Now consider the categories $Sh(X)$ and $Sh(Y)$ of sheaves of sets on $X$ and $Y$ respectively. We have the inverse image functor $f^*: Sh(Y) \to Sh(X)$ and direct image functor $f_*: Sh(X) \to Sh(Y)$. Now each open $U \in O(X)$ represents a set valued sheaf $rU \in Sh(X)$ and similarly for $Y$. Hence the following well-defined version of your question naturally arises: 1) For $V \in O(Y)$, is it true that $f^*(rV)$ is the sheaf represented by $f^{-1}(V)$? 2) For $U \in O(X)$, is it true that $f_*(rU)$ is the sheaf represented by $f(U)$? It is easy to verify that the answer to the first question is yes and the answer to the second question is no. However, in this particular case the functor $f^*$ admits a left adjoint $f_!: Sh(X) \to Sh(Y)$ as well, and (2) will become true if we replace $f_*$ with $f_!$. In particular, it seems that there is a certain clash of terminology here, which appears to be accidental. You might ask why didn't they call the functor $f_!$ "direct image" instead of $f_*$. One possible reason that comes to mind is that $f_*$ always exists while $f_!$ does not in general. In any case, there doesn't seem to be any hidden connection beyond this.<|endoftext|> TITLE: Epidemic threshold QUESTION [5 upvotes]: Need some help / ideas to proceed. Stuck for a while on this. In the literature of epidemic theory, it is found that the epidemic threshold is $1/\lambda_{\max}(A)$ where $\lambda_{\max}(A)$ is the largest eigenvalue of the adjacency matrix $A$, thus implying that the outbreak is basically dependent on the spectral radius of the underlying contact graph. Say, now, there's a new kind of virus that spreads by a different mechanism (i.e., no longer purely based on the existence of a link between two nodes). For instance, say each time it infects a specific type of people (nodes) (e.g., male/female, young/old, Asian/European/American/African etc.) and then it mutates and infect different kind of people (nodes) next time instance. For this, I come up with a matrix, say $X$, that records the probability of the virus infecting each node (i.e., $x_{ij}$ = probability of node $j$ being infected if $i$ is the infectant). It is a non-symmetric square hollow matrix with positive values for all off-diagonal elements in the matrix. With this infection probability matrix $X$ in mind, I'm struggling on how to derive the epidemic threshold. Is it still $1/\lambda_{\max}(X)$ by analogy of the conventional epidemic theory? Any idea, suggestions, hints or pointers will be appreciated. Secondly, if I have two of such matrices $X_1$ and $X_2$ essentially describing the behaviour of two viruses, how do I compare them using their corresponding probability matrix, $X$? Thank you in advance. REPLY [2 votes]: The problem is addressed in Peng, Chengbin, Xiaogang Jin, and Meixia Shi. "Epidemic threshold and immunization on generalized networks." Physica A: Statistical Mechanics and its Applications 389.3 (2010): 549-560. Link An epidemic will become extinct if and only if the spectral radius of matrix $X$ is smaller than 1.<|endoftext|> TITLE: Strong measure zero sets and selection principles QUESTION [8 upvotes]: A set of reals $X$ is strong measure zero if for any sequence of positive real numbers $ ( \epsilon_n ) _{n \in \omega }$ there is a sequence of open intervals $ ( a_n ) _{n \in \omega }$ which covers $X$ and such that each $ a_i $ has length less than $ \epsilon _i $. Does anyone know where can I find a proof that the selection principle $\mathcal S_1(\mathcal O,\mathcal O)$ implies Borel property of strong measure zero? I have seen this in The Combinatorics of open covers, but the reference there is to an article which is not in English. Thank you, REPLY [2 votes]: To help with @Student's last comment, I post here Baillif's answer. The proof is very short, actually: given the sequence $\epsilon_i$, let $U_i$ be the cover of $X$ by open intervals of length $\epsilon_i$. Then $S_1(O,O)$ ensures that there is a cover of $X$ consisting of (at most) one member of each $U_i$. A more general statement is proved in Theorem 9 in "Finite powers of strong measure zero sets" by M. Scheepers. – Mathieu Baillif<|endoftext|> TITLE: Families of pairwise incomparable subsets of the integers QUESTION [12 upvotes]: Certain maximal objects whose existence follows from Zorn's Lemma have received some set-theoretic attention. Examples are maximal independent families and maximal almost disjoint families. There is quite a bit of information out there about the two cardinal invariants $\mathfrak i$ and $\mathfrak a$, the minimal sizes of maximal independent families of subsets of $\omega$ and of maximal almost disjoint families of subsets of $\omega$. However, I have not seen anything concerning maximal incomparable families: Let $\mathcal A$ be a family of subsets of $\omega$ that are both infinite and coinfinite. We call $\mathcal A$ incomparable if the elements of $\mathcal A$ are pairwise incomparable with respect to the relation $\subseteq^*$ where $A\subseteq^*B$ if $A\setminus B$ is finite. Note that both independent families and almost disjoint families are incomparable. Question: What is the minimal size of a maximal incomparable family of subsets of $\omega$? I am guessing that this cardinal invariant is equal to $2^{\aleph_0}$ or equal to $\mathfrak i$, but mostly for psychological/social reasons: It should have been looked at before, unless it is equal to one of the known invariants. REPLY [6 votes]: Michael Hrusak has recently answered my question: The cardinal invariant that I am looking for is indeed $2^{\aleph_0}$. We are looking for the minimal size of a maximal infinite family of pairwise incomparable elements of the Boolean algebra $\mathcal P(\omega)/fin$. We can ask the same question for any infinite Boolean algebra $B$. For a Boolean algebra $B$ and $b\in B$ let $B\upharpoonright b=\{a\in B:a\leq b\}$. Let $d(B)$ be the minimal size of a dense subset of $B$. The following is true for any infinite Boolean algebra $B$: The minimal size of a maximal infinite family of pairwise incomparable elements of $B$ is at least $\min\{d(B\upharpoonright b):b\in B\setminus\{0\}\}$. Note that in the case of $B=\mathcal P(\omega)/fin$ we have $d(B\upharpoonright b)=2^{\aleph_0}$ for all non-zero $b\in B$, hence this fact solves the original problem. To see this fact, let $A$ be an infinite set of pairwise incomparable elements of $B$ of size $<\min\{d(B\upharpoonright b):b\in B\setminus\{0\}\}$ and let $a\in A$. Since $A$ is infinite, both $a$ and $-a$ are different from $0$. Let $\langle A\rangle$ denote the Boolean algebra generated by $A$. Since $A$ is infinite, $\langle A\rangle$ is of the same size as $A$. By the size of $A$, there are $x\in B\upharpoonright -a$ and $y\in B\upharpoonright a$, both non-zero, such that no non-zero element of $\langle A\rangle$ is below $x$ or $y$. Let $a'=x\vee(a-y)$. An easy computation shows that $a'$ is incomparable with all elements of $A$, showing that $A$ was not maximal.<|endoftext|> TITLE: When are those subgroups of $\mathrm{SL}(2, \mathbb{C})$ discrete? QUESTION [10 upvotes]: Let $A = \pmatrix{1 & 0 \\ \alpha & 1} $ and $ B = \pmatrix{1 & 1 \\ 0 & 1}$, where $\alpha \in \mathbb{C}$ is a complex parameter. Now consider the family of representations $r_{\alpha}$ of the free group on two generators $F_2 = \langle a,b\rangle$ in $\mathrm{SL}(2, \mathbb{C})$ setting $r_{\alpha}(a) = A$ and $r_{\alpha}(b) = B$. One can see that when $\alpha$ is transcendental over $\mathbb{Q}$, the representation $r_{\alpha}$ is faithful (see T. Church & A. Pixton "Separating twists and the Magnus representation of the Torelli group" Lemma 5.1). The question I am interested in is the following : when is (or is not) $r_{\alpha}(F_2)$ a discrete subgroup of $\mathrm{SL}(2, \mathbb{C})$ ? I suppose this is a difficult question of dynamics, I am curious if anyone has ever studied similar questions. REPLY [14 votes]: The free discrete subgroups consist of the closure of the Riley slice of Schottky space (notice that one may assume $Re(\alpha)\geq 0, Im(\alpha)\geq 0$, since $\alpha \mapsto -\alpha, \overline{\alpha}$ preserves discreteness). Here's a picture of the first quadrant of the Riley slice: The exterior of the black fractal represents free discrete groups that are generalized Schottky groups. The colored regions correspond to the combinatorics of the Ford domain. These have been investigated in detail by Akiyoshi, Sakuma, Wada, and Yamashita (see also their monograph where the picture was taken from). The black curve consists of geometrically finite groups with a cusp or degenerate groups. By the density conjecture, it is known that all free two-parabolic generator groups lie in the boundary of the Riley slice. Moreover, it is known that there is a group realizing each ending lamination parameterized by $\mathbb{R/Z}-\mathbb{Q/Z}$. In the exterior of the Riley slice, there are many more non-free discrete groups, such as those corresponding to 2-bridge links. In fact, all of the torsion-free discrete groups correspond to 2-bridge links. Note that for 2-bridge links, although they are known to be generated by 2 parabolics, the precise value of $\alpha$ which gives the parabolics has not been determined. One may see an example of this for the twist knots, for which $\alpha$ has been worked out by Hoste and Shanahan. For the discrete groups with torsion, there are certain other orbifolds closely related to the 2-bridge links; see some incomplete notes of mine. Conjecturally, all of the discrete groups are obtained by extending the "pleating locus" for each rational number $\in \mathbb{Q/Z}$ (representing a simple closed curve on the 4-punctured torus) through the cusp, and into the complement of Schottky space. The points where these elements are primitive elliptics should correspond to discrete groups (with some caveats); these pleating rays are displayed in the picture. In fact, the discrete groups in the complement of the Riley slice closure form a discrete set, with limit points at the boundary of the Riley slice.<|endoftext|> TITLE: Submitting a companion paper with detailed proofs ? QUESTION [36 upvotes]: In papers there are often sketchy arguments in proofs that I find hard to understand. Filling in the gaps is laborious and time-consuming. According to the post www.mathoverflow.net/questions/40729/does-a-referee-have-to-check-carefully-the-proof, referees seem to be faced with this problem as well. Presently I'm preparing my first paper for submission to a journal. Of course I approve the usual conventions and write more or less sketchy proofs myself in order to keep the paper short (around 15 pages). On the other hand, I checked the proofs carefully. So I could support the publishing process by this idea: Submit two versions of the paper: a short one, designated for publishing a long one, assigned for the referee with proofs given in full detail Is this a good idea that simplifies the referee's life (and, maybe, helps getting the paper accepted) or is it, in contrast, maybe even a no-go ? I appreciate your opinions very, very much. Thanks in advance. N.b. I intentionally ask the question on MO (and not on academia.stackexchange.com) because I think checking proofs is particular to mathematics and doesn't occur this way in most other fields of science. REPLY [12 votes]: Here is an illuminating quote from Terence Tao's journal submission guidelines; I could not have said it better: "Give appropriate amounts of detail. A paper should dwell at length on the most important, innovative, and crucial components of the paper, and be brief on the routine, expected, and standard components of the paper. In particular, a paper should identify which of its components are the most interesting. Note that this means interesting to experts in the field, and not just interesting to yourself; for instance, if you have just learnt how to prove a standard lemma which is well known to the experts and already in the literature, this does not mean that you should provide the standard proof of this standard lemma, unless this serves some greater purpose in the paper (e.g. by motivating a less standard lemma). Conversely, some computations which you are very familiar with, but are not widely known in the field, should be expounded on detail, even if these details are “obvious” to you due to your extensive work in this area."<|endoftext|> TITLE: Reference for a generalization of Γ-spaces to monoidal model categories QUESTION [10 upvotes]: Γ-spaces were introduced by Segal in 1969 as models for what can be now described as commutative ∞-monoids and ∞-groups in cartesian symmetric monoidal ∞-categories, e.g., E_∞-spaces and connective spectra. The word “cartesian” means that the tensor product coincides with the categorical product, i.e., A⊗B := A×B. In Lurie's Higher Algebra one finds Definition 2.1.3.1, which, combined with Definition 2.0.0.7 and Example 2.1.2.18, gives a model for commutative ∞-monoids in any symmetric monoidal quasicategory. In Theorem 4.1.4.4 Lurie proves that in the case of left proper tractable symmetric monoidal model categories that satisfy the monoid axiom his notion of a commutative monoid is equivalent to the traditional one. Using Example 4.1.3.6 or Proposition 4.1.3.10, which construct a symmetric monoidal quasicategory from a symmetric monoidal (simplicial) model category, one can try to unfold Lurie's definition and see what it means for the case of a symmetric monoidal quasicategory coming from a symmetric monoidal model category C. Very roughly speaking, we obtain the data of a functor X from the category of finite (unpointed) sets to C and for any finite family p of finite sets we have a weak equivalence (also known as a Segal map) X_p: X_{∐p} → ⨂_i X_{p_i}, satisfying the obvious associativity conditions. The above description is simplified quite a bit compared to what actually results from expanding Lurie's definition, but should suffice as the first approximation. Needless to say, this description closely matches that of a Γ-object in the case of a cartesian monoidal model category. Recall that a Γ-object is, roughly speaking, a functor Y from finite pointed sets to some model category C such that for any finite family p of finite pointed sets the canonical Segal map Y_p: Y_{∐p} → ∏_i Y_{p_i} induced by the canonical maps ∐p → p_i (the ith component is the identity map and all other components are trivial) is a weak equivalence. It is easy to see how one can construct an object X satisfying the conditions above from a Γ-object Y. The functor X is constructed from the functor Y by adding disjoint basepoints to all finite (unpointed) sets. Likewise, the Segal maps are obtained from the corresponding Segal maps of Y by adding basepoints to the elements of p. The main difference between these two descriptions is that in the cartesian case all Segal maps can be canonically recovered from the maps ∐p → p_i of pointed finite sets, whereas in the general case they must be supplied as an additional data. (This also explains the choice of unpointed (as opposed to pointed) finite sets in the above description.) Two earlier papers by Leinster (http://arXiv.org/abs/math/9912084v2, http://arXiv.org/abs/math/0002180v1) also discuss this definition, in the context of categories with weak equivalences that are closed under monoidal products, though unlike Lurie's book they do not construct an equivalence to the usual definition, and the condition on weak equivalences excludes virtually any known example of a monoidal model category (apart from those in which all objects are cofibrant). Although the above definition of a “monoidal Γ-object” seems to be rather natural, I was unable to locate any references (apart from Lurie's book and Leinster's papers) that mention it or any similar construction in the framework of monoidal model categories. I am specifically interested in statements that show an equivalence between such a model and the traditional definition. (Lurie's Theorem 4.1.4.4 proves such a statement for a significantly weaker model, with many additional homotopy coherences.) Another question is whether one can construct a model structure on monoidal Γ-objects so that its bifibrant objects have cofibrant components and all Segal maps are weak equivalences. Are there any references that discuss models for commutative ∞-monoids similar to Segal's Γ-spaces in the framework of (noncartesian) monoidal model categories? REPLY [5 votes]: The following paper by Tom Leinster http://arxiv.org/abs/math/0002180 defines for any symmetric monoidal model category $M$ and any (symmetric) operad $P$ (in $Set$) the notion of an $\infty$-algebra (which he calls homotopy algebra) over $P$ in $M$, in a way similar to Segal's $\Gamma$-spaces description of $\infty$-commutative monoids in cartesian monoidal model categories. The main point is the following: To any operad $P$ one can associate its prop $Prop(P)$. This is a symmetric monoidal category with objects $\{0,1,...\}$, such that the monoidal product is given on objects by $n\otimes m = n+m$. $Prop(P)$ has the property that for any symmetric monoidal category $M$, the category of algebras over $P$ in $M$ is equivalent to the category of symmetric monoidal functors from $Prop(P)$ to $M$. If $M$ is a symmetric monoidal model category, Leinster defines a homotopy algebra over $P$ in $M$ to be a colax symmetric monoidal functor $A:Prop(P)\to M$ such that the structure maps $$A(I)\to I,\:\:A(n+m)\to A(n)\otimes A(m)$$ are always weak equivalences (see Definition 2.2.1). There is also an obvious generalization to the enriched case. The case of the commutative operad $Com$ is treated in Section 3.2. The prop $Prop(Com)$ is equivalent the category of (unpointed) finite sets and functions, with the monoidal product given by disjoint union. In Proposition 3.1.1 it is shown that if $M$ is a cartesian monoidal category, then there is an isomorphism of categories between the category of colax symmetric monoidal functors from $Prop(Com)$ to $M$ and the category of all functors from $\Gamma^{op}$ to $M$. In Proposition 3.1.2 it is shown that if $M$ is a cartesian monoidal model category then a colax symmetric monoidal functor $Prop(Com)\to M$ has weak equivalences as structure maps (as above) iff the corresponding functor $\Gamma^{op}\to M$ satisfies Segal's condition. Thus, a homotopy commutative monoid in $M$ in the sense of Leinster is precisely a $\Gamma$ object in $M$ satisfying Segal's condition. However, Leinster's definition generalizes and makes sense also if our monoidal model category $M$ is not cartesian. The only problem is that Leinster doesn't prove the equivalence of his definition of a homotopy algebra over an operad $P$ in symmetric monoidal model category $M$, with the more known definition using a cofibrant replacement of $P$.<|endoftext|> TITLE: Erdős-Renyi graph restricted to largest connected component QUESTION [8 upvotes]: Suppose we have an instance of Erdős-Renyi $G(n,p)$ graph with $p = d/n$. Thus the expected node degree is $d$ which we will fix, while letting $n \to \infty$. Then, there will be more than one connected component (CC) for large $n$. Suppose that we pick the largest CC and relabel the nodes randomly from $1,\dots,s$ where $s$ is the size of the component. Is there a way to describe the distribution of these connected graphs as $n \to \infty$? The edges are no longer independent, but can we approximate the whole distribution with another Erdős-Renyi graph with modified parameters (say increased $p$)? REPLY [5 votes]: If $d>1$, the paper Anatomy of the giant component: the strictly supercritical regime by Ding, Lubetzky and Peres might be what you are looking for. The paper also has references for the cases $d<1$ and $d=1$.<|endoftext|> TITLE: Do exist infinitely differentiable, compactly supported non zero solutions of the free Schrodinger equation? QUESTION [5 upvotes]: I would like to get an answer for the following problem (and possibly be pointed to the relevant literature): given the one dimensional free Schrodinger equation $ i \, f_t + f_{xx}/2 = 0$ for the function $f$ of real variables $t$ and $x$, it is asked whether any solution exists, defined for all $x$ and in a neighborhood of $t=0$, meeting the following conditions (restated following Terry Tao answer): $f(x,t)$ is infinitely differentiable in $x$ and $t$, and $f(x,t)$ is compactly supported in $x$ for every $t$, i.e., do exist any compactly supported $C^\infty$ initial condition $f_0(x) \equiv f(x, 0)$ that evolves into a compactly supported solution in a neighborhood of $t=0$? I expect an answer in the no, according to the following intuition: $f(x,t)$ can be obtained by convolution of $f_0(x)$, in the variable $x$, with the kernel (propagator) $K(x,t)=\left(2 \pi\, t\right)^{-1/2} \exp( i\, x^2/(2 t)- i \pi/4)$. Inspection of easily obtainable closed form solutions for compaclty supported and regular (albeit not $C^\infty$) $f_0$, as well as asymptotic analysis for $ t \to 0$ suggest that $f$ is bound to oscillate rapidly, whilst tailing off to the $x$ infinity, for any $t\ne 0$. In Fourier transform, the compact support of $f_0(x)$ should prevent the entire funcion $f_0(k)$ from falling at infinity rapidly enough to temper the exponential term in the transformed solution $f(k, t) = \exp(-i\,t\,k^2/2) f_0(k)$, thus contradicting for $t\ne 0$ the Schwartz condition $|f(k,t)| < A_n (1+|k|)^{-n} \exp(-B | \mbox{Im}( k)|)$ met by $f(k,t)$ in case $f(x,t)$ has compact support (in general, references on the precise characterization of the Fourier transform of the space of $C^\infty$ compactly supported functions, in terms of both upper and lower bounds on their asymptotic behavior in the complex plane, would be appreciated). If so, and as a motivation for asking, any $C^\infty$ non-zero solution $f(x,t)$ that is compactly supported at $t=0$ has non-zero values that reach the $x$ infinity for any $t>0$ – a fairly curious and counterintuitive conclusion (to me) both mathematically, since a step by step integration in the variable $t$ should fail for any compactly supported inital condition $f_0$, and physically, since a free particle - as idealized by the Schrodinger equation - should be impossible to confine, in principle, for a finite time interval within a finite region of space (in practice of course, the far reaching tails of the wavefunction would quickly fade beyond detecdability). Thanks in advance for your answers REPLY [8 votes]: I think the question you stated is not the one you intended, since the Schrodinger evolution is indeed a smooth flow in the Schwartz class, as can be seen by Fourier analysis, and so smooth compactly supported data $f_0$ always leads to a smooth global solution which is also Schwartz in space. My guess is that you are asking instead to exclude the possibility of a nontrivial solution which is compactly supported at more than one time (not just at the initial time $t=0$). This was recently established (in quite great generality) in a series of papers by Escuriaza, Kenig, Ponce and Vega: see e.g. the survey http://arxiv.org/abs/1110.4873 . The methods are related to those used to establish unique continuation for elliptic operators, e.g. Carleman type estimates. There is also a close connection with the Hardy uncertainty principle, which ties in with your intuition that the relationship between the decay of a function and of its Fourier transform should be relevant.<|endoftext|> TITLE: Finding a basis for the (linear combinations) span of a matrix group, efficiently? QUESTION [5 upvotes]: I have an algorithm whose bottleneck is the following task: Let $\mathbb{F}$ be a finite field. Given a set of $k$ invertible matrices $g_1,\dots,g_k\in GL_n(\mathbb{F})$, let $G=\langle g_1,\dots,g_k \rangle\le GL_n(\mathbb{F})$ be the group they generate, and assume that the characteristic of $\mathbb{F}$ does not divide $|G|$. Let $\operatorname{span}G\subseteq M_n(\mathbb{F})$ be the vector space spanned by linear combinations of the elements of $G$, or equivalently, the matrix algebra genrated by $g_1,\dots,g_k$. Problem: Find, efficiently, a basis for the vector space $\operatorname{span}G$. The naive algorithm has running time $O(kn^6)$. I am looking for something that is at most $O(kn^{2\omega})$, with $\omega$ being the linear algebra constant ($\omega\approx\log_27\approx 2.81$). One possible direction may be as follows: (1) Find a basis (i.e., a conjugating matrix) such that the group decomposes into a direct sum of irreducible representations (irreps). (2) If the irreps are absolutely irreducible, we can take the standard basis for each, and we are done. Can (1) be achieved in time $O(kn^{2\omega})$ (or faster)? Is (2) correct in general? That is, in the above setting, must irreps be absolutely irreducible? And if not, is there anything more efficient than $O(kn^6)$ for irreps? (This question is related to this and that questions.) (Note: The probability that random $n^2$ elements of $G$ span may be negligible in general. Perhaps $O(n^2\log n)$ would do but I do not know that, either.) REPLY [2 votes]: The MeatAxe algorithm (which is available in GAP and Magma) can be used to solve (1). It is very fast in practice and I think it computes a decompostion into irreducibles in time $O(n^4)$ in the coprime case. In the case of non absolutely irreducible representations, it is not difficult to rewrite the representation over the larger field, which would make it possible to write down a basis of the space spanned by the original matrices.<|endoftext|> TITLE: Almost uniquely generated groups QUESTION [6 upvotes]: This is inspired by this question. Does there exist an infinite finitely generated group having (a) a unique (b) finitely many inclusion-minimal generating set(s) up to automorphisms? ♫ 2.$ $ Which finite groups have a unique inclusion-minimal generating set up to automorphisms? REPLY [3 votes]: Here is an honors thesis studying exactly this property for finite groups, which the author calls UMP, and all conclusions in the various answers are in the thesis. More examples are also given. A related paper by the same author is here<|endoftext|> TITLE: Probability a random Toeplitz matrix is singular QUESTION [10 upvotes]: Consider Toeplitz matrices where the entries in the first row and column (which define the whole matrix) are independently chosen to be either $1$ or $0$ with probability $1/2$. Define $p_n$ to be the probability that such a uniformly chosen $n$ by $n$ Toeplitz matrix is singular (over $\mathbb{R}$). Is it known that $$\lim_{n\to \infty} p_n = 0 ?$$ The equivalent question for random Bernoulli matrices was resolved by Komlós (1963). I see Probability of random (0,1) Toeplitz matrix being invertible where the exact value of $p_n$ was asked for (and with no answer to the part related to my question). For $n = 1,\dots,11$ the number of singular matrices is $1,4,13,50,153,522,1648,5173,15047,43892, 123417$. Sadly this is not in OEIS. The following picture shows the probability that a random $0-1$ Toeplitz matrix is singular for increasing values of $n$. Feb 6 2014: Fixed some of the numbers of singular matrices where the determinant was very close to zero. At the risk of asking a really trivial question, is it even obvious how to show that the probability is non-increasing? REPLY [2 votes]: The determinant follows the base-time-height formula, i.e., it is the product of distances from projecting a row to the linear space spanned by its previous rows. This can be understood from the Gram-Schmidt orthogonalization. So if the matrix is singular, then one of the distance should be zero. Following this line, one can cast the attempts into the framework established by Terry Tao and Van Vu. For your reference, a link for this reasoning in Terry's blog is "https://terrytao.wordpress.com/2010/03/05/254a-notes-7-the-least-singular-value/". However, the difference now is that because of the Toeplitz construction, the distances as random variables are dependent. Two bad consequences of this are that concentration inequalities for independent case can no longer be used, and that conditioning on the normal vector of a random hyperplance does not induce independence anymore. This is the key difference. However, if this is uccessfully dealt with, one can substantially advance Tao and Vu's framework and reasoning to dependent case, just as the Green function comparison technique of L. Erdos and HZ Yau has been extended to the dependent case (e.g., covariance/correlation matrix). A second way can be to see if universality of smallest eigenvalue holds for your Toeplitz matrix and the matrix of Komlós (1963). If it does, then you are done. However, universality on the edge is usually different from that in the bulk, and your case is intrinsically different due to dependence among the entries of the matrix. (I am no expert in this.)<|endoftext|> TITLE: Minimal normally generating subsets of minimal generating sets QUESTION [10 upvotes]: Let $G$ be a finitely generated group. The weight $w(G)$ of $G$ is defined to be the minimum number of elements of $G$ whose normal closure in $G$ is the whole of $G$ (this is sometimes also called the normal rank). Obviously, $d(G^{\operatorname{ab}})\leq w(G) \leq d(G)$, where $d(G)$ is the rank of $G$. A minimal generating set for $G$ does not necessarily contain a minimal normal generating set. The question is: does there always exist such a minimal generating set, i.e. one that realises the rank and contains a subset realising the weight? If not, which conditions on $G$ would guarantee the existence of such a generating set? REPLY [3 votes]: The answer is positive for finitely generated soluble groups. (And also for finitely generated simple groups, but it's easy!) Claim. Let $G$ be a finitely generated soluble group. Then $G$ has a generating set with $d(G)$ elements which contains a subset with $w(G)$ elements whose normal closure is $G$. Proof. Let $W(G)$ be the intersection of the maximal normal subgroups of $G$. It follows from the definition of $W(G)$ that a vector in $G^n$ with $n \ge w(G)$ normally generates $G$ if and only if its image under the natural map $G \rightarrow \overline{G} \Doteq G/W(G)$ normally generates $\overline{G}$. Let $S$ be a generating vector of $G$ of length $d(G)$. By [1, Folgerung 2.10 and Satz 6.4], the group $\overline{G}$ is Abelian and we have $w(G) = w(G_{ab}) = w(\overline{G})$. Thus we can find a Nielsen transformation $\psi \in \text{Aut}(F_{d(G)})$ such that the last $d(G) - w(G)$ components of $S \cdot \psi$ lie in $W(G)$ [2, Theorem 1.1]. By the previous remark, this means that the first $w(G)$ components of $S \cdot \psi$ normally generate $G$. [1] R. Baer, "Der reduzierte Rang einer Gruppe", 1964. [2] D. Oancea, "A note on Nielsen equivalence in finitely generated abelian groups", 2011.<|endoftext|> TITLE: Kummer's quartic surface and the Dirac operator QUESTION [10 upvotes]: (source) The picture shows 1936 photograph by Man Ray of a Kummer surface plaster model in the collection of the Institute Henri Poincaré (from Mathematics, Art and Science of the Pseudosphere by Kenneth Brecher). I was surprised to learn that this fancy surface has something to do with the quantum theory of relativistic electrons. In a 1928 paper A Symmetrical Treatment of the Wave Equation, Arthur Eddington, disappointed that Dirac’s equation did not appear in tensor form, tried to reformulate it in a tensor language and introduced so called E-numbers. Eddington's E-number is a linear combination (real or complex) of the imaginary unit i and fifteen basic elements $E_{\mu\nu}$, $\mu<\nu$, $\mu,\nu=0,1,2,3,4,5$, with the following properties (no summation is assumed over repeated indices) $$E_{\mu\nu}=-E_{\nu\mu},\;\;E_{\mu\nu}*E_{\mu\nu}=-1,$$ $$E_{\mu\nu}*E_{\mu\sigma}=-E_{\mu\sigma}*E_{\mu\nu}=E_{\nu\sigma},\;\; E_{\mu\nu}*E_{\sigma\tau}=E_{\sigma\tau}*E_{\mu\nu}=\pm i E_{\lambda\rho},$$ where $\mu,\nu,\sigma,\tau,\lambda,\rho$ are all different from each other and in the last relation the positive or negative sign is taken according as $(\mu,\nu,\sigma,\tau,\lambda,\rho)$ is an even or odd permutation of $(0,1,2,3,4,5)$. Soon (in 1932) it was shown by Oscar Zariski that the (projective) geometry behind this algebraic system was that of Kummer's quartic surface. In fact Eddington discovered Majorana spinors and the modern account of this connection can be found in the papers Some remarks on the algebra of Eddington's E-numbers by Nikos Salingaros and The Kummer Configuration and the Geometry of Majorana Spinors by Gary W. Gibbons. It is known (originally due to Majorana) that the quantum mechanics of photons can also be based on a Dirac-like equation (see also Photon Wave Function by Iwo Bialynicki-Birula. Is it possible to extend Eddington's considerations in this case too? And what projective geometry structure will be behind it? I would be grateful for references about the modern mathematical meaning(s) of Eddington's construction. REPLY [3 votes]: This question is interesting. I have done a little work in this direction, looking at brachistochrones over the Dirac matrices, like a geodesic idea. Here's a link to a paper that might be relevant: "Time Dependent Biqubits", Peter Morrison (Wayback Machine link) (current link). I agree with Prof. Eddington in that there are definitely links between tensors or complex matrices and the Dirac equation. My work looked at whether it was possible to formulate the Dirac equation in terms of fundamental non-commutative matrix calculus instead of the standard approach that uses differential operators to define the algebra. It is, but by the same caveat, I would say it is a a little more involved; however, the simplicity of representation is attractive. Thank you for posting this idea. It's good to know I'm not alone in looking at the properties of these types of curious surfaces etc. Incidentally, in looking at geodesics on a 4-dimensional submanifold of a 16-dimensional space, the electron rest mass came out neatly as a constant of motion, consistent with what we observe.<|endoftext|> TITLE: What are your favorite concrete examples of limits or colimits that you would compute during lunch? QUESTION [30 upvotes]: (The title was initially "What are your favorite concrete examples that you would compute on the table during lunch to convince a working mathematician that the notions of limits and colimits are not as dreadful as they might appear?" but it was too long.) This is a pretty basic question which may be more appropriate for another website. However, the examples I am looking for should appeal to a working mathematician. That is why I am asking here. Tomorrow I shall have lunch with a mathematician whom notions of limits and colimits make nervous in general. He feels that calculating small examples may help him overcome his fear. More precisely, what he is asking for are examples of various small diagrams (including fancy ones, for instance with loops) in familiar categories (topological spaces, abelian groups, &c.) whose limit or colimit we could calculate together over lunch, in the hope that he would get a better understanding of what are limits and colimits when they are taken over diagrams other than those giving pullbacks or pushouts (of which he already has a feeling). Do you know of some particular instances of diagrams, in categories familiar to the working mathematician, whose calculation of the limit or colimit seems particularly illuminating? Or at least which could help a nervous mathematician overcome his fear of general limits and colimits? I could come up with ad hoc examples, but perhaps there is better than that? EDIT: Sorry for the tardy edit. I went to sleep after asking the question and just woke up thinking "I should have made it Community Wiki, added a big-list tag and provided more details". Thanks for the answers so far. (By the way, I myself am unsure as to what extent this question is appropriate for MO, but if the three votes to close could be shortly explained I would nevertheless appreciate it.) The examples should appeal to a mathematician working in geometry and topology. For some reasons he really would like to make concrete computations. It seems that he has been faced with the following situation (to which I myself have never been faced; that is why I am asking here, in the hope that someone else already has been in the same situation as his): he is given a fancy (not the most usual) small diagram in Top or Ab or whatnot and wants to compute the limit or the colimit. Somehow he is afraid of this. I feel that my question is somewhat too broad and unprecise, but this is the best I have come up with given what I was asked myself. REPLY [5 votes]: Group actions! Treat $G$ as a category $\mathcal{G}$. Then a $G$-action is a functor $\mathcal{G}\to \mathcal{C}$. Its colimit is the orbit space and its limit is the fixed points.<|endoftext|> TITLE: Curvature as infinitesimal holonomy QUESTION [13 upvotes]: Let $P \to M$ be a principal $G$-bundle, assume as much regularity as you want (compact $G$ or compact base manifold, ect). Via parallel transport, a connection $A$ on $P$ gives rise to the holonomy map $$ Hol_p: L_m M \to G $$ where $p \in P$ is a chosen reference point projecting on $m \in M$ and $L_mM$ denotes the loop group based at $m$ (i.e piecewise smooth, closed curves starting at $m$). Question: What is the exact relationship between the curvature $F_A$ of $A$ and the derivative of $Hol_p$ at the constant loop (the infinitesimal holonomy). In the abelian case, I can answer this question as follows: Let $\gamma_s$ be a family of loops which represent a tangential vector $X \in T_\gamma L_mM$, that is $\gamma_0 = \gamma$ and $\frac{d}{ds}{\big|_0} \gamma_s = X$. Then the derivative of $Hol_p$ in the direction of $X$ evaluates to $$ \frac{d}{ds}{\big|_0} Hol_p (\gamma_s) = \frac{d}{ds}{\big|_0} \exp (\int_{\gamma_s} A) = (\exp)'_0 \int_0^1 \frac{d}{ds}{\big|_0} A(\dot \gamma_s(t)) dt = \int_0^1 dA(\frac{d}{ds}{\big|_0} \gamma_s(t), \dot \gamma_0(t)) dt, $$ That is, in this case we get a close expression for the derivative of $Hol_p$ at an arbitary loop $\gamma$ in terms of the curvature $F_A = dA$. Furthermore, the Ambrose-Singer theorem follows from this expression. I was hoping to get a similar result also for the non-abelian case. Sidequestion: conjugacy classes of (topological) homomorphisms $h$ between the loop gorup and $G$ uniquely determine a equivalence class of principal bundles with curvature by a result of Kobayashi (1954). Which additional properties on $h$ have to be impsoed to characterize all (equivalence classes of) connections on a fixed principal bundle $P$. REPLY [5 votes]: I have no idea about the sidequestion. For the main question, there is a general answer applicable for any parallel transport, not only for holonomy. Let $\gamma_{s}$ a family of smooth paths such that $\gamma_{s}(0)=p,\gamma_{s}(1)=q$ for every $s$ and fixed points $p$, $q$, and let $P_{s}(t’,t)$ the parallel transport from $\gamma_{s}(t)$ to $\gamma_{s}(t’)$ along $\gamma_{s}$. Here I try to prove the formula $$\begin{equation} \frac{d}{ds}P_s(1,0)= \int_{0}^{1}P_{s}(1,t)F(\partial_{t}\gamma_{s}(t),\partial_{s}\gamma_{s}(t))P_{s}(t,0)dt \end{equation}, $$ where $F$ is the curvature. (There might appear a negative sign depending on convention. I use a convention where $F=dA+A\wedge A, \nabla=d+A, F(w,v)=F_{\nu\mu}w^{\nu}v^{\mu}$.) Especially if $\gamma_{0}$ is the constant loop, then $\partial_{t}\gamma_{0}(t)$ is always zero, so the derivative map of $Hol_{p}$ at the constant loop is just the zero map. The proof goes as if we are discussing a connection of vector bundles. Let $H_{t}(s’,s)$ the parallel transport from $\gamma_{s}(t)$ to $\gamma_{s’}(t)$ along $\gamma_{\bullet}(t)$ and let $S(s,s’;t)=P_{s}(1,t)H_{t}(s,s’)P_{s’}(t,0)$. Then, $$ P_{s’}(1,0)-P_{s}(1,0)=S(s,s’;1)-S(s,s’;0) =\int_{0}^{1}\partial_{t}S(s,s’;t)dt =\int_{0}^{1}\lim_{\epsilon\rightarrow 0} P_s(1,t+\epsilon)\frac{H_{t+\epsilon}(s,s’)P_{s’}(t+\epsilon,t)-P_{s}(t+\epsilon,t)H_{t}(s,s’)}{\epsilon}P_{s’}(t,0)dt, $$ so $$ \partial_{s}P_{s}(1,0)=\int_{0}^{1}\lim_{(\epsilon,\delta)\rightarrow 0}P_{s}(1,t+\epsilon)\frac{H_{t+\epsilon}(s,s+\delta)P_{s+\delta}(t+\epsilon,t)-P_{s}(t+\epsilon,t)H_{t}(s,s+\delta)}{\epsilon\delta}P_{s+\delta}(t,0)dt. $$ From the parallelogram discussion follows the identity $$ \lim_{(\epsilon,\delta)\rightarrow 0} \frac{H_{t+\epsilon}(s,s+\delta)P_{s+\delta}(t+\epsilon,t)-P_{s}(t+\epsilon,t)H_{t}(s,s+\delta)}{\epsilon\delta}=F(\partial_{t}\gamma_{s}(t),\partial_{s}\gamma_{s}(t)), $$ and thus also does the formula I’m proving. I’d like to comment that integrating both sides of the formula reads a relationship between $P_{s’}(1,0)-P_{s}(1,0)$ and the curvature, if we focus on connections of vector bundles. (If you want to discuss a $G$-principal bundle $Q$, you should remember that each $P_{s}(1,0)$ can be canonically identified with an element of $(Q_{q}\times Q_{p}\times G)/\sim$, where $(v,u,g)\sim(vy,ux,y^{-1}gx)$. Then the formula I have shown describes the path $P_{\bullet}(1,0)$ in this manifold in terms of the curvature.)<|endoftext|> TITLE: Preservation of some stationary sets by sufficiently closed forcing QUESTION [8 upvotes]: The following statement can be proven using elementary submodels and sufficiently generic conditions: "If $S \subseteq cof(<\kappa) \cap \kappa^+$ is stationary, and $\kappa^{<\kappa} =\kappa$, then the stationarity of $S$ is preserved by $\kappa$-closed forcing." If we just assume $\kappa$ is regular, do we need the cardinal arithmetic? REPLY [7 votes]: No: your cardinal arithmetic assumption can be dropped when $\kappa$ is regular. This follows from $I[\lambda]$ analysis. $S \subseteq \lambda \cap \mathrm{cof}(\kappa)$ (where $\kappa < \lambda$ are regular) is said to be $\textit{in $I[\lambda]$}$ if there is a sequence of sets $\langle a_i \mid i < \lambda \rangle$ and a club $C\subseteq \lambda$ such that every $\delta \in S \cap C$ is approachable w.r.t $\vec{a}$. $\delta \in S$ is said to be approachable w.r.t. $\vec{a}$ when there is an unbounded subset $A \subseteq \delta$ of order-type $\kappa$ such that $\{ A \cap \alpha \mid \alpha < \delta \} \subseteq \{ a_i \mid i < \delta \}$. Shelah proved the following: $S \subseteq \lambda \cap \mathrm{cof}(\kappa)$ is indestructible by $\kappa^+$-closed forcings if and only if, for every large regular $\theta >> \lambda$ and every $x \in H(\theta)$, there are an elementary submodel $M \ni x$ and $\delta \in S$, and an unbounded $A \subseteq \delta$ such that: $\delta= M\cap \lambda$, $\mathrm{otp}(A)= \kappa$, $\{ A \cap \alpha \mid \alpha < \delta \} \subseteq M$. If $\kappa$ are regular, $\kappa^+ \cap \mathrm{cof}(< \kappa) \in I[\kappa^+]$. By 1, it is easy to check that if a stationary set $S \subseteq \kappa^+ \cap \mathrm{cof}(\nu)$($\nu< \kappa$: regular) is in $I[\kappa^+]$ then $S$ is indestructible by $\nu^+$-closed forcings, and hence is indestructible by $\kappa$-closed forcings. By Shelah's result 2, $I[\kappa^+] \restriction \mathrm{cof}(< \kappa)$ is improper, so every stationary set $S \subseteq \kappa^+ \cap \mathrm{cof}(<\kappa)$ is preserved by $\kappa$-closed forcings, without any cardinal arithmetic. For proofs, see Cummings' article ``Notes on Singular Cardinal Combinatorics.''<|endoftext|> TITLE: n-cocycles of finite abelian groups from cohomology group QUESTION [7 upvotes]: Question: Given a generic finite abelian group $G=\mathbb{Z}_{N^{(1)}} \times \cdots \times \mathbb{Z}_{N^{(k)}}$. (1) What is the explicit forms of its cohomology group (see my definition) in a generic $n$: $$ H^n(G,R/\mathbb{Z})=H^n(G,U(1)) =? $$ (2) what are their explicit $n$-cocycles? $$ \omega_{}^{}(A_1,A_2,\dots, A_n): (G)^n \to U(1) $$ Please be very explicit in $n=4$ for priority (and the secondary are $n=5$ and $n=2$) of interests. But the generic formulas for any $n$ are the best. Thank you. References are very welcome. ps. below I provide the answer I know for $n=3$. Please feel free providing partial answers for other $n$. REPLY [7 votes]: 3 years later. My recent paper (https://arxiv.org/abs/1703.03266) answers this question. More precisely, let $\mathbb{k}$ be an algebraically closed field of characteristic zero. By $\mathbb{k}^*$ we denote the multiplicative group $\mathbb{k}-\{0\}$. Let $G=\mathbb{Z}_{m_{1}}\times\cdots \times\mathbb{Z}_{m_{n}}$ where $m_i|m_{i+1}$ for $1\le i\le n-1$ and $(B_{\bullet},\partial_{\bullet})$ be its normalized bar resolution. Applying $\text{Hom}_{\mathbb{Z}G}(-,\mathbb{k}^*)$ one gets a complex $(B^*_{\bullet},\partial^*_{\bullet})$. Let $\alpha:=(\alpha_{11},\dots,\alpha_{1n},\dots,\alpha_{k1},\dots,\alpha_{kn})$ where $0\leq\alpha_{ij} TITLE: A hard integral identity on MathSE QUESTION [82 upvotes]: The following identity on MathSE $$\int_0^{1}\arctan\left(\frac{\mathrm{arctanh}\ x-\arctan{x}}{\pi+\mathrm{arctanh}\ x-\arctan{x}}\right)\frac{dx}{x}=\frac{\pi}{8}\log\frac{\pi^2}{8}$$ seems to be very difficult to prove. Question: I worked on this identity for several days without any success. Is there any clue how to prove this integral identity? REPLY [88 votes]: I have proved this equality by means of Cauchy’s Theorem applied to an adequate function. Since my solution is too long to post it here, I posted it in arXiv: Juan Arias de Reyna, Computation of a Definite Integral, arXiv:1402.3830. The function $$G(z)=\frac{\log(1+(1+i)\,f(z)\,)}z$$ where $$f(x)=\frac{\operatorname{arctanh}(x)-\arctan(x)}{\pi}$$ extended analytically.<|endoftext|> TITLE: Even unimodular lattices with root system $32 A_1$ QUESTION [5 upvotes]: I'm studying Venkov's proof of the classification of even unimodular rank 24 lattices, and it prompted the following question. For an even unimodular lattice $L$, let $R(L)= \{ x \in L : (x,x) =2\}$ be its root system. Question. Can one "roughly" classify, up to isomorphism, all rank $32$ even unimodular lattices with root system $32 A_1$? Or are there too many of these? (Note: there are at least $10^8$ even unimodular lattices of rank 32.) Remark. There are no even unimodular lattices of rank 8 and 16 with root system $8 A_1$ and $16A_1$, respectively. There is a unique rank $24$ even unimodular lattice with root system $24 A_1$ by the work of Niemeier (and Venkov). Any comments are appreciated! REPLY [2 votes]: Another construction (basically, different terminology, but the same outcome) is the maximal (of rank 16) isotropic subgroups in the discriminant group (in the sense of Nikulin) of $32A_1$, which is $32\langle\frac12\rangle$, not containing a sum of four generators. This is a finite problem, but I wouldn't do that manually.<|endoftext|> TITLE: History and motivation for Tannaka, Krein, Grothendieck, Deligne et al. works on Tannaka-Krein theory? QUESTION [17 upvotes]: I am trying to wrap my mind around Tannaka-Krein duality and it seems quite mysterious for me, as well, as its history. So let me ask: Question: What was the motivation and historical context for works of major contributors to the "Tannaka-Krein theory" (in a broad sense)? Just to name a few names: Tannaka, Krein, Saavedra, Deligne, Milne, Lurie, (it seems Grothendieck should also be in this list(?)). Let me explain some points in the history which seems to me puzzling: Mark Krein was a famous Soviet mathematician, but he was an expert in analysis, it seems it is the only paper by him devoted to algebra (See discussion below). How did he come to it? Why did he not continue? Similar question about Tadao Tannaka. "His interest in mathematics lied mainly in algebraic number theory", And it seems similar to Krein, it is the only work by devoted to group theory. (See his publication list). P. Deligne seems to have devoted quite much efforts on "Tannakian formalism" and more generally on tensor categories. What was his motivation? He is a leading algebraic geometer. So probably the subject should be quite important in algebraic geometry? What is its importance? Wikipedia article starts with a sentence: "...natural extension to the non-Abelian case is the Grothendieck duality theory." What is the role of Grothendieck in this history ? And what is "Grothendieck duality theory" - wikipedia links to something not related. Important work was done by Saavedra. It seems not so much is known about him, his motivation, his other works. J. Lurie seems to develop the theory further (see e.g. MO question Tannakian formalism). What is the motivation? List of references (it seems original articles by Tannaka and Krein are not available electronically) Tadao Tannaka, Über den Dualitätssatz der nichtkommutativen topologischen Gruppen, Tohoku Math. J. 45 (1938), n. 1, 1–12 (project euclid has only Tohoku new series!) M.G. Krein, A principle of duality for bicompact groups and quadratic block algebras, Doklady AN SSSR 69 (1949), 725–728. in Russian: М. Г. Крейн, Принцип двойственности для бикомпактной группы и квадратной блок-алгебры, Докл. АНСССР, 69:6 (1949), 725–728. N. Saavedra Rivano, Cat´egories tannakienns, Lecture Notes in Math., vol. 265, Springer-Verlag, Berlin–New York, 1972. Deligne, P., and Milne, J.S., Tannakian Categories, in Hodge Cycles, Motives, and Shimura Varieties, LNM 900, 1982, pp. 101-228". ( http://www.jmilne.org/math/xnotes/tc.html ) Some remarks about Mark Krein. Part of his publication list is here, strangely enough the paper on "Tanaka-Krein duality" is not contained in this list. I have found an article devoted to overview of his works related to group theory: L. I. Vainerman. On M. G. Krein's works in the theory of representations and harmonic analysis on topological groups Ukrainian Mathematical Journal 46 (1994), no. 3, 204-218. It seems he had several papers dating from 1940-1949 which were related to "Tannaka-Krein theory". He started as student of Nikolai Chebotaryov, who is famous for Chebotarev density theorem, but actually was also working on Lie groups: famous results Ado theorem and Jacobson-Morozov theorem were obtained by his students Igor Ado and Morozov, who worked in Kazan city Russia. But it is not clear whether Krein was influenced by Chebotarev in this respect, since they meet around 1924 in Odessa city, and the paper was written in 1949, when Chebotarev already passed and long before he moved from Odessa to Kazan city, while Krein stayed in Odessa. Anatoly Vershik in his paper devoted to 100-anniversary of M. Krein suggests that it might be that "success of Gelfand's theory of commutative normed rings" influenced Krein. REPLY [19 votes]: (Edited to correct mistakes signaled in comments below). I don't know much about the first steps on the theory, Krein and Tannaka. I can just say their works answer a question that seems very natural now, and that I think was natural even then. Since the beginning of the 20th century, representations of groups had been studied, used in many part of mathematics (from Number Theory, think of Artin's L-function to mathematical physics) and more and more emphasized as an invaluable tool to study the group themselves. It was therefore natural to see if a group (compact say) was determined by its representations. But then, I want to insist on the fundamental role played by Grothendieck in the development of the theory. This role comes in two steps. First Grothendieck developed a pretty complete end extremely elegant theory for a different but analog problem: the problem of determining a group (profinite say) by its category of sets on which it operates continuously. It is what is called "Grothendieck Galois Theory", for Grothendieck did that in the intention of reformulating and generalizing Galois theory, in a way that would contain his theory of the etale fundamental groups of schemes. What Grothendieck did, roughly, was to define an abstract notion of Galois Category. Those categories admit special functors to the category of Finite Sets, called Fibre Functors. Grothendieck proved that those functors are all equivalent and that a Galois category is equivalent to the category of finite sets with G-action, where G is the group of automorphism of a fibre functor. He then goes on in establishing an equivalence of categories between profinite groups and Galois categories, with a dictionary translating the most important properties of objects and morphisms on each side. This was done in about 1960, and you can still read it in the remarkable original reference, SGA I. Already at this time, according to his memoir Recoltes et Semailles, Grothendieck was aware of Krein and Tannaka's work, and interested in the common generalization of it and his own to what would become Tannakian category, that is the study of categories that "look like" categories of representations over a field $k$ of a group, As he had many other things on his plate, he didn't work on it immediately, but after a little while gave it to do to a student of him, Saavedra. As Grothendieck was aware, the theory is much more difficult than the theory of Galois categories. Saavedra seems to have struggled a lot with this material, as would have probably done 99.9% of us. He finally defended in 1972, two years after Gothendieck left IHES, and at a time he was occupied by other, in part non-mathematical subject of interest. Saavedra defined a notion of Tannakian category (as a rigid $k$-linear tensor category with a fibre functor to the category of $k'$-vector space, $k'$ being a finite extension of $k$) but he forgot one important condition (then $End(1)=k$) and some of the important theorems he states are false without this condition. After that, mathematics continued its development and Tannakian categories began to sprout up like mushrooms (e.g. motives (69, more or less forgotten until the end of the 70's), the dreamt-of Tannakian category of automorphic representations of Langlands (79), to name two extremely important in number theory). Then Milne and Deligne discovered in 1981 the mistake mentioned above in Saavedra's thesis, gave a corrected definition of Tannakian'a category, and were able to prove the desired theorems in the so-called neutral case, when $k'=k$ (I believe with arguments essentially present in Saavedra). Later with serious efforts, Deligne proved those theorems in the general case. Modern theory have added many layers of abstraction on that.<|endoftext|> TITLE: Origin of the term "weight" in representation theory QUESTION [8 upvotes]: In representation theory, there are the related concepts of weights and roots. Since both are kinds of generalised eigenvalues, and eigenvalues are roots of e.g. the characteristic polynomial, the word "root" makes sense to me (at least, the question is reduced to why zeros of polynomials / equations are called "roots".) But I wondered: Who used the term weight (or poids, or Gewicht, or ...) for the first time? And for what (if any) specific reason? This site does not know the word "weight" in this meaning. (But see the entry "radix" about roots (of equations).) The only thing I could find on the internet is this (unanswered) stackexchange question. REPLY [5 votes]: Robert Bryant's comment motivates me to mention the "weighty" historical monograph Emergence of the Theory of Lie Groups (Springer, 2000) written by Thomas Hawkins. As usual with terminology such as "weight", the history reaches back into nineteenth century's invariant theory (Cayley, G. Kowalewski) but becomes most relevant to modern Lie theory in the work of Elie Cartan about a century ago. The early part of Chapter 8 in Hawkins' book is most pertinent but not easy reading. Though Cartan's use of the term poids (weight, Gewicht) was not the earliest mathematical occurrence, it does seem to have been the first use in connection with what we now call weights of representations. There is also a long history involving the term "root" (and its offshoot "secondary root"), going back to antiquity, but here the work of Killing anticipates Cartan's more definitive treatment of semisimple Lie groups and what we now call their Lie algebras. The history is not at all easy to untangle, but I think Hawkins was thorough in his study of the development of ideas along with terminology. Terminology in this particular subject should not be taken too seriously, I think, and sometimes the names given to things are either misleading or inappropriate (including concepts named after people). Still, we are stuck with the language, which is almost impossible to change.<|endoftext|> TITLE: How was Christoffel a 'whimsical eccentric'? QUESTION [5 upvotes]: I've seen several citations of a letter from Weierstrass, talking about his dispute with Kronecker, in which he refers to Christoffel as a 'whimsical eccentric' (presumably the German original is even more flavorful). In what respect was he whimsical or eccentric? One reference is Barrow, 'Pi in the Sky,' page 200, quoting the letter from Weierstrass. Was Christoffel generally nutty, or was Weierstrass just citing his support for Kronecker? The Google translation of the German helpfully supplied by Suvrit is: When a strange fellow as Christoffel says that in 20-30 years the current function theory is carried to the grave and the whole analysis to be worked out in the theory of forms, so you answered that with a shrug. Subquestion: what does he mean here by 'die Theorie der Formen' here? REPLY [13 votes]: (a) "Kauz" is literally a "brown (or tawny) owl," colloquially a "queer (or strange, familiarly a 'rum') fellow" (Muret-Sanders German-English Dictionary); certainly a more pungent expression than "whimsical eccentric," but not as dismissive as "crank" or "crackpot"; a rather precise English colloquial equivalent is "an odd duck." (b) To illustrate that this is an apt characterization of Christoffel, cf. the opening sentence of a paper of his of 1888, "Lehrsätze über arithmetische Eigenschaften der Irrationalzahlen" (see his Gesammelte mathematische Abhandlungen, vol. 2, p. 216): "Da die Irrationalzahlen demnächst abgeschafft werden sollen, scheint es mir nicht unbillig, doch noch einmal zu erwägen, ob das, was bis auf heutigen Tag über diesen Zahlen ans Tageslicht gekommen ist, wirklich von solcher Art is, dass ihnen nunmehr die Existenzberechtigung aberkannt werden muss." ("Since the irrational numbers are about to be got rid of, it seems to me not unreasonable to consider yet once again whether what has come to the light of day up to now about these numbers is really of such a kind that from now on the right to exist must be denied them.") (c) Although Christoffel doesn't name names in this statement, it seems rather clear that his reference is to Kronecker. (d) To underscore his character as an odd duck: the aim of this paper is to present what Christoffel regards as the first ever account--the first ever discovery--of "genuinely arithmetical" properties of the irrational numbers; and by doing this, to establish that they have the right, not merely to exist, but to be regarded as genuine numbers. (e) From this I conclude further that the remark Weierstrass ascribes to Christoffel was most probably made by the latter sarcastically (although Weiserstrass may not have recognized this); the "forms" C. refers to had, I should think, nothing to do with the Christoffel symbols in Riemannian geometry, but with the algebraic forms that play a central role in Kronecker's theory of algebraic numbers: Christoffel does not think Kronecker's algebraic methods will succeed in eliminating the real numbers.<|endoftext|> TITLE: higher dimensional analogues of the Manin-Drinfeld theorem QUESTION [10 upvotes]: The Manin-Drinfeld theorem asserts that a divisor on the compact modular curve $X_0(N)$ which is supported on the cusps is torsion. Equivalently, if $Y_0(N)$ is the open modular curve, the mixed Hodge structure $H^1(Y_0(N), \mathbb{Q})$ splits. Are there some kind of generalizations of this theorem to higher-dimensional Shimura varieties and their toroidal compactifications? REPLY [5 votes]: This is perhaps more of a comment than an answer. In various texts by Günter Harder he uses the terminology that the Manin-Drinfeld principle holds for the Shimura variety $X$, the coefficient system $\mathbb V$ and the cohomology theory $H(-)$ (usually $\ell$-adic cohomology or mixed Hodge theory) if there is a direct sum decomposition $$ H(X,\mathbb V) = H_!(X,\mathbb V) \oplus H_{Eis}(X,\mathbb V)$$ into inner and Eisenstein cohomology. The former is the image of compactly supported cohomology in ordinary cohomology, the latter is the cokernel. However I don't actually know any general result ensuring that the Manin-Drinfeld principle holds - but this may very well be due to my own ignorance. But you could start looking through his book (Eisensteinkohomologie und die Konstruktion gemischter Motive) as well as the various manuscripts on his webpage. You will at least find plenty of stuff of the form "if the Manin-Drinfeld principle holds, then we also know that..."<|endoftext|> TITLE: Is every commutative group structure underlying at least one (unitary, commutative) ring structure QUESTION [5 upvotes]: From the theorem of classification of finitely generated abelian groups, we can see that every finitely generated commutative group can be considered as the additive structure underlying (at least) one unitary and commutative ring. My question is about possible generalization of this result. Question: Is it true that with every abelian group G it is possible to build (at least) one ring(resp: unitary ring, commutative ring) whose additive structure is the group structure on G ? Gérard Lang REPLY [2 votes]: No, there are many abelian groups such that only zero multiplication can be defined over them. Even if you can define a non-zero multiplication on a group, the ring obtained may have no unit, also it may not be associative, and so on. A detailed discussion of this subject can be find in L. Fuchs: (Infinite abelian groups). Two former Ph.D students of our department have also some papers in the subject. You can find their works searching the names: "A. Najafizadeh, F. Karimi, A. M. Aghdam", and key words "torsion free groups, additive group of rings". One more interesting problem is to determine which subgroups of an abelina group $A$ can be realized as ideals in some ring with additive group $A$. You can find some good results concerning this problem.<|endoftext|> TITLE: Does an injective group homomorphism between countable abelian groups that splits over every finitely generated subgroup, necessarily split? QUESTION [7 upvotes]: Let $G_1$ and $G_2$ be countable abelian groups, and let $\iota\colon G_1\to G_2$ be an injective group homomorphism (so that we may regard $G_1$ as a subgroup of $G_2$). Suppose that for every finitely generated subgroup $K$ of $G_2$, there is a map $\pi_K\colon K\to G_1$ such that $\iota \circ \pi_K\circ \iota= \mbox{id}_{\iota(G_1)\cap K}$ (or if you want to omit $\iota$ from the notation, this would be $\pi_K(g)=g$ whenever $g\in G_1\cap K \subseteq G_2$. In other words, for every finitely generated subgroup of $G_2$, one can find a splitting. Does it follow that there is a "global" splitting $\pi\colon G_2\to G_1$ for $\iota$? I don't want to assume that the "partial" splittings $\pi_K$ have any sort of coherence (by this I mean that I don't want to assume that whenever $K\subseteq K'$, the splitting $\pi_{K'}$ can be chosen so that $\pi_{K'}$ restricted to $K$ is just $\pi_K$). I also don't want to assume that $G_1$ is finitely generated. Of course the conclusion would be that $G_1$ is a direct summand in $G_2$. REPLY [4 votes]: Here is another concrete example constructed as in Yves Cornulier's comment. Let $G_2$ be the group defined by the (abelian group) presentation $G_2=\langle\ x,y_i\ (i \ge 0) \mid y_i^2 = y_{i-1}x\ (i \ge 1)\ \rangle$ and $G_1 = \langle x \rangle$. Any finitely generated subgroup is contained in $\langle y_i,x\rangle$ for some $i$, which is free abelian of rank $2$, and any subgroup of that splits over its intersection with $\langle x \rangle$. But a complement of $G_1$ in $G_2$ would be generated by elements $y_i' = y_ix^{k_i}$ for some integers $k_i$ with $y_i'^2 = y'_{i-1}$ for all $i > 1$. But $y_i'^2 = y_{i-1}x^{2k_i+1}$, so $k_{i-1} = 2k_i+1$ for all $i>1$, which is clearly not possible.<|endoftext|> TITLE: Liftability of mod p Hilbert modular forms of parellel weight QUESTION [5 upvotes]: If $f$ is a mod $p$ Katz modular form of weight $k$ for $k \geq 2$, there is a classical modular form $g$ such that reduction of $g$ is $f$. Is there a similar property holds for Hilbert modular forms of higher weights? Namely, let $f$ be a mod $p$ Hilbert modular form of parallel weight $k$ for some big $k$ (in the sense of Katz). Then, does there exist a classical Hilbert modular form $g$ of same weight that gives $f$ by reduction? If so, is there any bound for $k$? REPLY [5 votes]: Let $\overline{\mathcal{M}}(N,\mathcal{O})$ be the minimal compactification of the Hilbert modular scheme over $\mathbb{Z}_p$ for a totally real field $F$ and $\omega$ be the determinant of the co-normal sheaf ($p\nmid N$). It is known that $\omega$ is ample. We have an exact sequence of sheaf such that the first map is multiplying by $p$. $0\rightarrow \omega^{k} \rightarrow \omega^{k} \rightarrow \omega^{k}/p\rightarrow 0$ Since $\overline{\mathcal{M}}(N,\mathcal{O})$ is projective an $\omega$ is ample, then for a large $k$, $H^{1}(\overline{\mathcal{M}}(N,\mathcal{O}), \omega^{k})$ is trivial. Moreover, the support of $\omega^{k}/p$ is the special fibre of $\overline{\mathcal{M}}(N,\mathcal{O})$, Hence by applying the global section to the above exact sequence, we find that $M_k(N,\mathbb{Z}_p) \otimes \mathbb{F}_p$ surjects to $M_k(N,\mathbb{F}_p)$. You can jump on the weight by multiplying by the Hasse modular form. In the case where we have a modular curve $X_1(N)$, we can use the same argument combined with Riemann-Roch to show that $H^1(X_1(N),\omega^{k})$ is trivial for a large $k$.<|endoftext|> TITLE: Origin of symbols used for half-sum of positive roots in Lie theory? QUESTION [6 upvotes]: The Weyl character formula is a central result in the finite dimensional representation theory of semisimple Lie groups, algebraic groups, Lie algebras. Related questions on MO include these here and here. But there is another, mainly notational, question which I'm still curious about: What is the origin of symbols such as $\rho$ and $\delta$ used to denote the half-sum of positive roots in Lie theory, or equivalently the sum of fundamental dominant weights, relative to a given simple system of roots? I haven't made an exhaustive search of the early literature, but here is what I've noticed: 1) Apparently Weyl himself (Math. Z., 1925-26) avoided using a symbol for this purpose, but instead wrote his character formula in terms of weight coordinates. He denoted what we would call the fundamental dominant weights by $\varphi_1, \dots, \varphi_n$. (Brauer's 1937 note in C.R. Acad. Sci. Paris on tensor product decompositions follows Weyl, also avoiding a specific symbol. ) Weyl lectured on Lie algebras (previously called 'infinitesimal groups") and Lie groups at IAS in 1933-35, with notes written up by Jacobson and Brauer respectively during their visits. (I don't recall what's in the mimeographed notes, but I recall seeing a copy years ago in the IAS library. Are those notes available online somewhere?) 2) A number of people used the notation g, including Freudenthal ((nag. Math., 1954-56), then Kostant (Trans. Amer. Math. Soc., 1959), Possibly the German word Gewicht for "weight'" influenced Freudenthal's choice of $g$. 3) In work influenced by the formulas of Weyl and Kostant, both Steinberg and Cartier used instead the notation $\phi$ in their short notes published in Bull. Amer. Math. Soc., 1961. 4) Jacobson wrote essentially the first textbook treatment of Lie algebras (1962), where (perhaps influenced by Weyl's lectures?) he chose the symbol $\delta$ for the half-sum of positive roots. This convention was followed in the 1969 lecture notes Topics in Lie Algebras by Samelson and the 1974 text by Varadarjan (both published in later editions by Springer). 5) As far as I know, the first use of $\rho$ occurs in Serre's 1965 Algiers lectures, written up in a short but highly influential set of lecture notes and published in typescript in 1966 for the W.A. Benjamin Lecture Notes series with the title Algebres de Lie semi-simples; see page VII-17. (Around 1987 an English translation was published by Springer.) Of course, Serre was then active in the Bourbaki group. Their Chap. IV-VI of the treatise Groupes et algebres de Lie appeared in 1968 but must have been in development for some years, since references were made to Chap. VI in the 1965 Borel-Tits IHES paper on reductive groups (with tentative numbering later changed in the published chapters). Borel was also active in Bourbaki, but he and Tits did little then with representation theory and went their own way on root system notation. In my 1972 Springer GTM I still used $\delta$, probably because I had picked up some of the theory from Jacobson's book while at Cornell in 1962. After buying a copy of Serre's lectures in 1967, I based my supplementary lectures on structure and classification of semisimple Lie algebras at a summer school in Maine in 1968 on his approach but didn't treat representation theory. These lectures increased my interest in writing a textbook, which I got more inspired to do by the 1971 BGG paper. (In 1969 I did buy a copy of Bourbaki but was already used to writing $\delta$ and using other root system notation.) ADDED: Notation is essential in mathematics but often problematic. For me the term "origin" combines the interrelated notions of "first use" and "rationale". To correct what I wrote above, another look at Harish-Chandra's early work shows that he actually used the symbol $\rho$ for the half-sum of positive roots as early as 1951. on page 69 of his important paper here. Like many other authors, he was not totally consistent in his choices; he also used $\rho$ for "representation" in the same paper. Concerning one comment by Francois, I did consider asking some of the (retired) people involved in this history, but didn't want to bother them with a relatively minor question. I'm also not sure how reliable anyone's recollection (including mine) would be after half a century or more. I regret now that I never raised this question with people like Jacobson and Serre long ago when I had informal contact with them. One other remark is that the really odd notation $\phi$ used in the two 1961 notes (based in part on Kostant's 1959 paper) in Bull. Amer. Math. Soc. by Steinberg and Cartier might possibly have been a printer's misreading of their handwritten symbols $\varrho$ or such. To complicate matters, one reviewer substitutes Kostant's symbol $g$ for $\phi$. REPLY [3 votes]: Re 5): Cartier already uses $\rho$ in Séminaire «Sophus Lie» (1955): exposé 19, p. 1, and in Weyl's character and dimension formulas: exposé 21, pp. 7-9.<|endoftext|> TITLE: The time to drift a binary string from one state to another via deterministic selection of two possible random bit mutation procedures QUESTION [7 upvotes]: I have some length $L$ binary string consisting of an ordered array of bits: $(b_1, b_2, ..., b_{L})$, where all bit values $b_i$ are originally set to zero. I'd like a particular set of $n$ bits to hold the value $1$. For example, if $L = 5$, starting from $00000$ we might want to reach the state $01101$ where a specific set of $n = 3$ bits have the value $1$. However, I can only do two things: [1] Choose a bit in the string at position $i$ where $b_i = 0$, with uniform random probability over all bits in the string with this value, and set $b_i = 1$. To be clear, here, if there are nine bits with the value $0$ and one bit with the value $1$, a particular bit with the value $0$ would be selected for a $0 \to 1$ transition with probability $\frac{1}{9}$. [2] Choose a bit in the string at position $i$ where $b_i = 1$, with uniform random probability over all bits in the string with this value, and set $b_i = 0$. What distribution do we have for the length of time until we reach the state where at least $k \leq L$ of the ordered array of bits have the desired $0$ and $1$ value ($k = L$ implying that only the aforementioned $n$ positions have the value $1$)? What if we make it increasingly likely for us to select the previously flipped bit in [2], for example, if the immediately preceding $0 \to 1$ transition had a probability $r$ times greater of being reversed in [2] than any other bit with the value $1$ being flipped to the value $0$ (implying that the immediately previously flipped bit has a $1 \to 0$ transition probability during [2] of $Prob(1 \to 0) = \frac{r}{(q+r)}$ if there are $q$ OTHER $1$ bits that can be flipped to $0$ with probability $\frac{1}{(q+r)})$? As $r \to \infty$, one would imagine that it would be possible to quickly drift any binary string from some initial state to another desired state. Alternative Question Formulation: There's another fun way to ask this question. Given that the set of single bit flip transitions for a length $L$ binary string can be represented with a graph $G$ having the topology of an $L$-dimensional hypercube, draw this hypercube and replace each edge with a pair of directed RED and BLUE edges (that can only be traversed in a single direction). RED directed edges imply a $0 \to 1$ bit flip, and BLUE directed edges imply a $1 \to 0$ bit flip. Starting one vertex $v_i$, our task is to reach another vertex $v_j$ in the fewest number of hops. However, we can only decide to select a RED or BLUE edge (pointing in the correct direction to allow traversal) at any given point in time, and then we need to pick one of the edges of the desired color randomly with uniform probability. In terms of the "bias" in the original posting, we can say that we're $r$ times more likely to move back to the previously occupied vertex than any other vertex if we previously traversed a RED directed edge and now decide to traverse a BLUE directed EDGE. Update - I wrote "...What distribution do we have for the length of time until we reach the state where at least $k \leq L$ of the ordered array of bits have the desired $0$ and $1$ value..." in the original posting. However, this could unnecessarily complicate things. So please feel free to set $k = L$. Also feel free to assume that the optimal "strategy" here is to immediately switch to process [2] when the wrong bit has its value flipped from $0$ to $1$. Update 2 - Intuitively, letting $m_0$ be the number of $0$ bits and $m_1$ be the number of $1$ bits, we can probably guess that the length of time to reach the target string will probably have some dependency on $\delta = || m_1 - m_0 ||$, where smaller values of $\delta$ (i.e. where $m_0 \approx m_1$) imply longer average times to reach the desired string target structure. This "seems" like it might be true because ${L}\choose{m_0}$ and ${L}\choose{m_1}$ are simultaneously maximized at this limit. Simulations (!!! where we immediately try to flip back "incorrect" $1$ bits by applying procedure [2] for as long as necessary to do so) appear to back up this intuition. For a length $L = 10$ binary string where the target string has $k$ bits with value $1$ (where exactly these bits are in the string should be irrelevant), and performing $10^4$ trials: $k=0$ trivially implies a {mean, median} $= (\mu, \mu_{1/2}) = (0, 0)$ $k=1$ yields a {mean, median} $= (\mu, \mu_{1/2}) = (19.3202, 13)$ $k=2$ yields a {mean, median} $= (\mu, \mu_{1/2}) = (66.2872, 46)$ $k=3$ yields a {mean, median} $= (\mu, \mu_{1/2}) = (152.303, 107)$ $k=4$ yields a {mean, median} $= (\mu, \mu_{1/2}) = (258.273, 180)$ $k=5$ yields a {mean, median} $= (\mu, \mu_{1/2}) = (333.897, 237)$ $k=6$ yields a {mean, median} $= (\mu, \mu_{1/2}) = (321.758, 226)$ $k=7$ yields a {mean, median} $= (\mu, \mu_{1/2}) = (238.742, 167)$ $k=8$ yields a {mean, median} $= (\mu, \mu_{1/2}) = (130.412, 94)$ $k=9$ yields a {mean, median} $= (\mu, \mu_{1/2}) = (50.1086, 37)$ $k=10$ trivially implies a {mean, median} $= (\mu, \mu_{1/2}) = (L, L) = (10, 10)$ However, we also see see that the length of the string $L$ and $\delta = || m_1 - m_0 ||$ do not provide enough information to predict waiting times, specifically that for the same value of $\delta$, $m_1 > m_0$ implies a larger number of necessary bit flips to reach the target string. I suppose this makes sense given that the initial string is populated with only $0$-valued bits. I see no reason why these relationships wouldn't hold with the "previous flipped bit" selection bias discussed at the end of the original question posting, though I could obviously be wrong about this. Let's add some $L = 20$ values for the unbiased case (we're only doing $10^3$ trials here because these trials take a lot longer): $L=20$ and $k=9$ yields a {mean, median} $= (\mu, \mu_{1/2}) = (118,487.382, 82,308)$ $L=20$ and $k=10$ yields a {mean, median} $= (\mu, \mu_{1/2}) = (127,382.07, 90,858)$ As Will Sawin nicely points out, for $k = 11$, we could quickly drift our string to the state where we have all-$1$ bit values, and then achieve roughly the same number of expected bit flips (in the unbiased case!) as if we had $k = 9$. REPLY [3 votes]: Ignoring the specific binary string structure of the problem for a moment, let's generalize it to an arbitrary digraph $G$ with vertices $V$ and edges $E$ divided into two disjoint subsets $E = E_\mathrm R \sqcup E_\mathrm B$ (for "red" and "blue" respectively). Equivalently, let us denote the "red" and "blue" neighbors of each vertex $v \in V$ as $N_\mathrm R(v)$ and $N_\mathrm B(v)$, where $v' \in N_c(v) \iff (v,v') \in E_c$ for $c \in \{\mathrm R, \mathrm B\}$. (Note that a single vertex $v$ can be a red neighbor of $v'$ and a blue neighbor of $v''$ at the same time — the color is really a property of the edges, not of the nodes.) We wish to minimize the expected number of steps needed to reach an arbitrary goal set $A \subset V$ from an arbitrary initial vertex $v_0 \in V$, given that, at each step $k$, we have the choice of picking $v_{k+1}$ uniformly at random from either $N_\mathrm R(v_k)$ or $N_\mathrm B(v_k)$. Let us denote the expected number of steps to reach $A$ from an initial vertex $v$ using strategy $\mathcal S$ by $T_\mathcal S(v)$. (For consistency, I'm assuming that $A$ is in fact reachable from any vertex $v \in V$.) For $v \in A$, we naturally have $T_\mathcal S(v) = 0$, while for $v \notin A$ we have $$ T_\mathcal S(v) = 1 + \frac{P_\mathrm R(\mathcal S,v)}{|N_\mathrm R(v)|} \sum_{v'\in N_\mathrm R(v)} T_\mathcal S(v') + \frac{P_\mathrm B(\mathcal S,v)}{|N_\mathrm B(v)|} \sum_{v'\in N_\mathrm B(v)} T_\mathcal S(v') \tag{1} \label{answer-155160-1} $$ where $P_\mathrm R(\mathcal S,v)$ and $P_\mathrm B(\mathcal S,v) = 1 - P_\mathrm R(\mathcal S,v)$ are the respective probabilities of jumping to a red or a blue neighbor of $v$ under the strategy $\mathcal S$. (Of course, if $N_c(v) = \emptyset$ for either $c \in \{\mathrm R, \mathrm B\}$, we must have $P_c(\mathcal S,v) = 0$ and simply $T_\mathcal S(v)$ $=$ $1$ $+$ $\frac{1}{|N(v)|} \sum_{v'\in N(v)} T_\mathcal S(v')$, where $N(v) = N_\mathrm R(v) \cup N_\mathrm B(v)$.) (Note that I'm only considering "Markovian" strategies, where $P_\mathrm R(\mathcal S,v)$ and $P_\mathrm B(\mathcal S,v)$ are constant for each $\mathcal S,v$ and, in particular, independent of the way we may have ended up at $v$. It's not hard to show that the optimal strategy must indeed be Markovian, but I'll skip that bit for brevity.) The nice thing about equation $\eqref{answer-155160-1}$ is that it's linear! Therefore, we can find the optimal solution to it, subject to the constraints that $P_\mathrm R(\mathcal S,v) \ge 0$ and $P_\mathrm B(\mathcal S,v) \ge 0$ for all $v$, using linear programming. (In fact, from the form of the constraints, we can immediately tell that the optimal solution $\mathcal S^*$ will have $P_\mathrm R(\mathcal S^*,v) \in \{0,1\}$ for all $v$, except possibly for degenerate cases where any value of $P_\mathrm R(\mathcal S^*,v)$ will yield the same $T_{\mathcal S^*}(v)$.) I have not actually tested this method in practice yet, but I see no reason why it shouldn't work. Note that, for large $L$ (where optimizing a system of $2^L$ linear equations may not be practical), we may want to exploit the structure of the problem by collapsing equivalent states: in particular, it should be sufficient to represent each state by two numbers $(n_0, n_1)$ giving respectively the number of mismatched $0$ and $1$ bits in the string. This will reduce the number of distinct states from $2^L$ to $(n+1)(L-n+1)$ $\le$ $(L/2+1)^2$.<|endoftext|> TITLE: Moore decomposition, dual to Postnikov tower QUESTION [7 upvotes]: Let $X$ be a CW complex with given cohomologies $H^n(X; \mathbb{Z}) = G$, $H^m(X; \mathbb{Z}) = H$ and other reduced cohomologies are zero. Which additional algebraic information/structures do I need to identify the homotopy type of $X$? Similar question for three or more nonzero cohomologies. Can the answer be similar to the answer to the dual question about Postnikov invariants? REPLY [6 votes]: This isn't really an answer but it's a bit long for a comment. Do you really mean cohomology rather than homology? The dual of a Postnikov tower is usually considered to be a homology decomposition, building a space up one homology group at a time. A space with only two nonzero cohomology groups could have up to four nonzero homology groups, by the universal coefficient theorem. Incidentally, another reference for basics on homology decompositions in addition to the Baues book in Mark Grant's answer is my algebraic topology book starting on page 464. Two-stage Postnikov systems were studied a lot in the 60s and 70s, partly because of their connections with secondary cohomology operations. Among two-stage homology decompositions are the so-called "two-cell complexes" $X = S^n \cup e^m$ whose classification is the same as computing $\pi_{m-1}S^n$ modulo the effect of composing with reflections of $S^{m-1}$ and $S^n$. In the general case $S^{m-1}$ and $S^n$ are replaced by Moore spaces, so one is considering homotopy groups with coefficients. There should be a significant literature on this, but I'm not very familiar with it. I believe that Joseph Neisendorfer has written quite a bit about this. It seems that things become considerably more complicated for three-stage Postnikov systems, and probably the same is true for homology decompositions.<|endoftext|> TITLE: Isometric embeddings of metric spaces in Hilbert spaces QUESTION [6 upvotes]: There are plenty of isometric embeddings of metric spaces in Banach spaces. Nevertheless, I have been unable to find any significant result on isometric embeddings into Hilbert spaces. My question is: how can one recognize those metric spaces that are isometrically embeddable into Hilbert spaces? Later edit: I have removed two paragraphs from my original question, which created a lot of confusion among those who answered it. I take responsibility for mixing "metric spaces" isometries and "differential geometry" isometries. I apologize. REPLY [3 votes]: The answer was given in the papers of I. Schoenberg and von Neumann, MR1501980, MR1503439, MR0004644.<|endoftext|> TITLE: Is every T0 2nd countable space the quotient of a separable metric space? QUESTION [12 upvotes]: Suppose the space $X$ has a countable basis and $X$ is $T_{0}$. Must there exist a separable metrizable space $Y$ and a quotient map q:$Y \rightarrow X$? (Some surrounding facts: Every metrizable space is 2nd countable iff it's separable. Every 2nd countable space is 1st countable and hence Frechet and hence sequential and hence the quotient of a locally compact metrizable space. ( However in the canonical proof, $Y$ is the disjoint union of a typically very large collection of convergent sequences [Franklin] and usually not separable, even if $X$ itself is a separable metric space). If $X$ is $T_{0}$ and regular and 2nd countable then $X$ is metrizable (Urysohn metrization)). For a non $T_{0}$ counterexample let $X$ have cardinality larger than the real numbers and employ the indiscrete topology.) If the answer is `no' can a counterexample $X$ be $T_1$ or even $T_2$? (Edit: the answer is `yes' and Francois Dorais and Andrej Bauer provide two explicit solutions below and also point out relevant references. The $T_{1}$ case was settled by Paul Strong. As shown below similar tactics settle the $T_{0}$ case. The question is relevant to topological domain theory. For example ``The similarity between our definitions and results and those of Schroder was first observed by Andrej Bauer, who proved that the sequential spaces with admissible representations are exactly the T0 (quotients of countably based) spaces...''. From the paper `Topological and Limit-space Subcategories of Countably Based Equilogical Spaces.' by Menni and Simpson. homepages.inf.ed.ac.uk/als/Research/Sources/subcats.pdf REPLY [7 votes]: Here is an elementary construction of such a quotient. Let $X$ be a $T_0$-space and $B_0, B_1, B_2, \ldots$ a countable base for $X$. For any point $x \in X$ define $N(x) = \{i \in \mathbb{N} \mid x \in B_i\}$, the index set of basic neighborhoods of $x$. Given a sequence $\alpha : \mathbb{N} \to \mathbb{N}$ let $i(\alpha) = \{\alpha(k) \mid k \in \mathbb{N}\}$, the image of the sequence. The Baire space $\mathbb{N}^\mathbb{N}$ is countably based and $0$-dimensional with the ultrametric $$d(\alpha, \beta) = 2^{-\min_k (\alpha_k \neq \beta_k)}.$$ Let $D$ be the subspace $$D = \{\alpha \in \mathbb{N}^\mathbb{N} \mid \exists x \in X . i(\alpha) = N(x)\},$$ which consists of those sequences that enumerate the index set of some point in $X$. Note that the point $x$ in the definition of $D$ is unique for a given $\alpha$, if it exists, because $X$ is $T_0$. Define the map $q : D \to X$ by $$q(\alpha) = \text{"the $x$ such that $i(\alpha) = N(x)$"}.$$ It is a basic exercise in topology to verify that $q$ is a quotient map. Thus, every countably based $T_0$-space is the quotient of a $0$-dimensional countably based ultrametric space. Also note that the map $N : X \to \mathcal{P}(\mathbb{N})$ is an embedding when the codomain is equipped with the Scott topology.<|endoftext|> TITLE: Representations of S_n induced from centralizers of elements QUESTION [6 upvotes]: Does anyone have a reference for a good description of representations of $S_{n}$ obtained by inducing up from $C_{S_{n}}(\pi)$, for some element $\pi$ of $S_{n}$? (I'd prefer an efficient combinatorial description if there is one.) This seems like it should be doable, since I'm fairly sure that centralizers of elements are just going to be products (direct and/or wreath) of cyclic groups and smaller symmetric groups, so those representations should all be understandable combinatorially. Any references that people have would be great. Edit: There are references for inducing from cyclic subgroups, as given in the answer to Decomposition of induced representations in S_n, and also Stembridge here: http://www.ams.org/mathscinet-getitem?mr=1023791, but I'm looking a bit more general than that. REPLY [3 votes]: It is known from work of Roger Richardson and others that $G = S_{n}$ has a ``model", that is, we may choose one element $t_{i}$ from each conjugacy class of elements of order $2$, and a certain sign character $\lambda_{i}$ of each $C_{G}(t_{i}),$ such that each nontrivial irreducible character of $S_{n}$ occurs once and only once as an irreducible constituent of some ${\rm Ind}_{C_{G}(t_{i})}^{G}(\lambda_{i}).$ This is a more precise illustration of the fact that since all complex irreducible representations of the symmetric group are realizable over the real field ( even the rational field), we have $\sum_{ \chi \in {\rm Irr}(G)} \chi(1) = 1 +\sum_{i=1}^{k} [G:C_{G}(t_{i})]$ for the group $G = S_{n},$ (where $k = \lfloor \frac{n}{2} \rfloor $ is the number of conjugacy classes of elements of order $2$ of $S_{n}$) by a general formula obtained using the Frobenius-Schur indicator, which simplifies in the case of $S_{n}$ because we always have $\nu(\chi) = 1.$ Hence the sum of the induced characters at least has the right degree, but it is rather more subtle to obtain the finer result as stated. REPLY [2 votes]: The representations $Ind_H^{G} \rho$, where $H=C_{G}(g)$ (for some $g\in G$) and $\rho$ is an irreducible representation of $C_{G}(g)$, all turn out to be irreducible representations of the quantum double of $G$ (denoted sometimes as $D(G)$). The action of the Hopf algebra $D(G)$ on this irreducible space is well known. For example, when reviewing this part in our paper, we describe the action of the Hopf algebra $D(S_n)$ on its irreducible representations. So if you want to know the action of $S_n$ on the induced representation $Ind_{C_{S_n}(\pi)}^{S_n}$, then you can use Eq. 3.9 by taking the trivial element of the dual group (i.e., replace $h^\ast$ by $\sum_{h\in G} h^\ast$). The rest of section 3 is specific to $S_n$. The centralizers are direct products of wreath products of the type $C_m \wr S_k$ as you mentioned. The irreducible representations of these groups can be obtained using Clifford theory. We also describe a coset factorization of $C_{S_n}(\pi)$ in $S_n$. However, this does not answer how the induced representation decomposes into irreducible representations of $S_n$.<|endoftext|> TITLE: Construction of the Casson invariant QUESTION [11 upvotes]: What is the easiest construction of the Casson invariant? The original construction using representation spaces (as found, for instance, in Akbulut-McCarthy) is very technical since you have to perturb things to make them transverse. One possibility would be to do some kind of Kirby calculus construction. Such a thing can be found in Walker's book and Lescop's book, but they constructed much more general things so their work is also very technical. REPLY [21 votes]: Hoste has a combinatorial formula for the Casson invariant for integral homology $3$-spheres. Let $\Sigma$ be an integral homology $3$-sphere obtained via surgery on a framed link $L = K_1 \cup \dots \cup K_n$ in $S^3$ with framings $1/q_i$. Additionally, we assume $\mathrm{lk}(K_i, K_j) = 0$ for $i \neq j$ (this can always be arranged, see Lemma 12.2 in Saveliev's Lectures on the Topology of $3$-Manifolds. In fact, that lemma says that we can take $q_i = \pm 1$). We can write the Conway polynomial of $L$ in the form $$\nabla_L(z) = z^{n-1}(a_0(L) + a_1(L) z^2 + \cdots + a_m(L) z^{2m}).$$ With the above notations, Hoste's formula is $$\lambda(\Sigma) = \sum_{L' \subset L} \left( \prod_{i : K_i \subset L'} q_i\right) a_1(L').$$ The sum is taken over all sublinks $L'$ of $L$. You can see the details of Hoste's formula in his original paper: Hoste, Jim. A formula for Casson's invariant. Trans. Amer. Math. Soc. 297 (1986), 547-562 Hoste's original paper does not show that $\lambda$ as defined by his formula is actually an invariant of integral homology $3$-spheres, however. The missing ingredient is a Kirby move that preserves the unlinking of all components of $L$ as required for Hoste's formula to be stated. Such a move is constructed by Habiro in the following paper: Habiro, Kazuo. Refined Kirby calculus for integral homology spheres. Geometry & Topology 10 (2006), 1285–1317. Combining the results of the above two papers gives a nice combinatorial construction of the Casson invariant as an invariant of integral homology $3$-spheres.<|endoftext|> TITLE: Cholesky decomposition of a positive semi-definite QUESTION [8 upvotes]: We know that a positive definite matrix has a Cholesky decomposition,but I want to know how a Cholesky decomposition can be done for positive semi-definite matrices?The following sentences come from a paper. "There are two assumptions on the specified correlation matrix R. The first is a general assumption that R is a possible correlation matrix, i.e. that it is a symmetric positive semidefinite matrix with 1’s on the main diagonal. While implementing the algorithm there is no need to check positive semi-definiteness directly, as we do a Cholesky decomposition of the matrix R at the very start. If R is not positive semi-definite, the Cholesky decomposition will fail." Thank you for your answer. REPLY [3 votes]: If you don't mind including some permutations, you can get a variant of Cholesky that still has the rank-revealing property: $$P^T R P = R_1^T R, \quad R_1 = \begin{bmatrix} R_{11} & R_{22} \\ 0 & 0 \end{bmatrix}.$$ This is a matter of simple greedy pivoting. For the algorithm and more details, see Higham's "Cholesky Factorization".<|endoftext|> TITLE: Grothendieck spaces and total subspaces of the dual QUESTION [10 upvotes]: There is probably an embarrassingly simple counter-example to my question but I couldn't figure it out myself. Let me give it a try here. A Banach space $X$ is Grothendieck if weak*-convergent sequences in $X^*$ converge weakly (that is, with respect to the weak topology introduced by functionals in $X^{**}$). Standard examples of such spaces include reflexive spaces and $C(K)$-spaces for $K$ Stonian. I would like to relax this condition a bit, so my question is: Let $X$ be a Banach space such that $X^*$ is weak*-separable and let $T$ be a total subspace of $X^*$. Suppose that each sequence $(f_n)_{n=1}^\infty \subset T$ which converges weak*, converges also weakly. Can we conclude that $X$ is Grothendieck? REPLY [6 votes]: Here is a natural way to build a counterexample. Let $T$ be a weakly sequentially complete space and set $X=T^*$. Consider $T$ as a subspace of $X^*$, so that a sequence in $T$ is weak$^*$ convergent iff it is weakly Cauchy and hence weakly convergent to an element of $T$. You want $T$ non reflexive and $X$ should not be Grothendieck. $T=\ell_1$ satisfies the first condition but not the second. What about $T=(\sum E_n)_1$ for a carefully chosen sequence $(E_n)$ of finite dimensional spaces? You need to choose the sequence so that $X$ is not Grothendieck. One way of doing that is to make sure that $X^*$ has a complemented non reflexive separable subspace. Now if you take a Banach space $Y$ and a sequence $E_1\subset E_2 \subset \dots$ with $\cup E_n$ dense in $Y$ and set $T=(\sum E_n)_1$, then $Y^*$ is isometrically isomorphic to a norm one complemented subspace of $X=T^*$ [J] and hence $Y^{**}$ is isometrically isomorphic to a norm one complemented subspace of $X^*$. So you just need $Y$ to be non reflexive and complemented in its bidual; e.g., $Y$ can be any separable non reflexive quasi-reflexive space. [J] Johnson, William B. A complementably [sic] universal conjugate Banach space and its relation to the approximation problem. Proceedings of the International Symposium on Partial Differential Equations and the Geometry of Normed Linear Spaces (Jerusalem, 1972). Israel J. Math. 13 (1972), 301–310 (1973).<|endoftext|> TITLE: Genericity by names QUESTION [8 upvotes]: If $P$ is a notion of forcing in $M$, then $G$ is a $P$-generic filter over $M$ if $G\subseteq P$ is a filter, and for every $D\in M$ which is a dense subset of $P$, $G\cap D\neq\varnothing$. Equivalently we can replace $D$ being dense by being pre-dense, open dense, or a maximal antichain. Given such a generic filter, and $\dot x$ which is a $P$-name, we can define the interpretation of $\dot x$ by the filter $G$ by recursion, $$\dot x^G=\{\dot y^G\mid\exists p\in G:\langle p,\dot y\rangle\in\dot x\}.$$ Clearly, we don't need genericity in order to interpret names. We can talk about interpretation using arbitrary filters. Question. Given a filter $G$, is there some reasonable condition stating that $G$ is generic if and only if it interprets certain names (e.g. names which are forced to be ordinals) "properly"? REPLY [6 votes]: Your question is fairly open-ended, but here is one way to do it. For any antichain $A\subset P$, consider the name $\dot a_A=\{\langle\check a,a\rangle\mid a\in A\}$, which is a mixture of the elements of $A$ on the antichain $A$. This is the name that is trying to be an element of $A$, with values determined by $A$ itself. Theorem. The following are equivalent, for a filter $G\subset P$. $G$ is $V$-generic. $(\dot a_A)^G\in A$ for each maximal antichain $A\subset P$. $(\dot a_A)^G$ is nonempty for every maximal antichain. Proof. The filter $G$ is generic if and only if it meets every maximal antichain, and in this case, 2 will hold, and clearly 2 implies 3. If 3 holds, then there must be some $a\in A\cap G$, and so $G$ will be $V$-generic. QED Miha suggests a similar idea in the comments, namely, take any nonempty set $x$, and consider the names $\dot x=\{\langle\check y,a\rangle \mid y\in x\text{ and }a\in A\}$, for any maximal antichain $A$. Miha's observation is that although every condition forces $\check x=\dot x$, nevertheless if $G$ is not generic and misses the antichain $A$, then $\dot x^G$ will be empty. This idea arrives at the following, which seems to meet the requirement of your question. Theorem. (Miha Habič, in comments) The following are equivalent: $G$ is $V$-generic. $G$ interprets all names for non-zero ordinals as non-zero ordinals. That is, if $\mathbb{1}\Vdash(\tau$ is a nonzero ordinal), then $\tau^G$ is a nonzero-ordinal. $G$ interprets all names for the number $1$ correctly. Proof. Clearly 1 implies 2 and 3. Conversely, if 2 or 3 holds, let $A$ be any maximal antichain and let $\tau=\{\langle 0,a\rangle \mid a\in A\}$. This is a name forced to be the ordinal $1$, and it is easy to see that $\tau^G$ is either $0$ or $1$, depending on whether $G$ meets $A$. So $1$ holds. QED One can use dense sets instead of antichains, which will avoid a dependence on AC. Lastly, let me add then when you are entertaining the idea of interpreting names via a filter that is not generic, then you should not be using the value as you have defined it, but rather you should be using the Boolean ultrapower instead. In this method, you use the Boolean algebra rather than just a partial order, and define the equivalence on names $\sigma=_G\tau\iff[\![\sigma=\tau]\!]\in G$, and similarly for $\sigma\in_G\tau$. Now, the ultrapower map is $x\mapsto [\check x]_G$, meaning the equivalence class under this equivalence relation. The result is an elementary embedding $$j:V\to\check V_G\subseteq V^{\mathbb{B}}/G,$$whose target is the ground model of the corresponding quotient $V^{\mathbb{B}}/G$. The following theorem appears in the Boolean ultrapower paper: Theorem. The following are equivalent: $G\subset P$ is $V$-generic. The Boolean ultrapower by $G$ is an isomorphism of $V$ with $\check V/G$. This theorem is part of the explanation that "non-generic" is the right generalization of "non-principal" when considering the ultrapower on a complete Boolean algebra instead of merely on a powerset.<|endoftext|> TITLE: Are admissible open subsets of an affinoid space of countable type? QUESTION [6 upvotes]: A rigid analytic space $Y$ over a complete non-archimedean valued field $k$ is said to be of countable type if it has a countable (possibly finite) admissible covering by affinoids over $k$. Suppose that $X$ is an affinoid space over such a $k$ and $U$ is an admissible open subset of $X$ with respect to the strong $G$-topology on $X$ (see Non-Archimedean Analysis by Bosch, Guntzer, Remmert section 9.1.4). In section 4 of 'On one-dimensional separated rigid spaces'. Indag. Math. (N.S.) 6 (1995), no. 4, 439–451 by Q. Liu and M. van der Put there is given an example of such a pair $X$ and $U$ with $U$ not of countable type. However, I believe that example only works if the residue field is uncountable. My question is whether there is such an example if the residue field is itself countable. I suspect that there are examples but would be pleased if not. REPLY [3 votes]: If the field admits a countable dense subfield is countable, the corresponding Berkovich space will be a metrizable topological space. Therefore, every open subset is metrizable, hence paracompact, hence each connected component will be countable at infinity.<|endoftext|> TITLE: classifying $\infty$-toposes for topological/localic groups? QUESTION [9 upvotes]: Let $G$ be a locally compact topological group (or more generally a localic group). Is there an infinity topos which classify principal $G$ bundles ? More precisely, is there an $\infty$-topos $BG$ such that for every localic topos $\mathcal{L}$ the category of geometric morphism from $\mathcal{L}$ to $BG$ is equivalent to the category of $G$ principal bundle over $\mathcal{L}$, where a $G$-principal bundle over $\mathcal{L}$ is a locale $\mathcal{X}$ endowed with a $G$ action and an invariant map $p: \mathcal{X} \rightarrow \mathcal{L}$ such that: 1)$p$ is an open surjection. 2) The canonical map $\mathcal{X} \times G \rightarrow \mathcal{X} \times_{\mathcal{L}} \mathcal{X}$ is an isomorphism. I am especially interested in the cases where $G= \mathbb{U}$ (the group of complex number of module $1$) and $G=\mathbb{R}$. Of course, if $G$ is pro-discrete, then the answer is yes: it suffice to consider the infinity topos associated to the $1$-topos of continuous $G$ set. In the general case, one should look for an infinity topos of spaces endowed with a $G$ action (up to homotopy), but my knowledge of homotopy theory is not enough to see if this trivially work/does not work or if it is a difficult question... REPLY [9 votes]: The projection map $p: \mathbf{R} \rightarrow \ast$ induces a fully faithful embedding of sheaf categories $p^{\ast}: Shv(\ast) \rightarrow Shv( \mathbf{R} )$. This is equally true for sheaves of sets, sheaves of spaces, or other variants. It follows that for any topos (or $\infty$-topos) $\mathcal{X}$, the ($\infty$-)category of geometric morphisms from $\ast$ to $\mathcal{X}$ embeds fully faithfully into the ($\infty$-)category of geometric morphisms from $\mathbf{R}$ into $\mathcal{X}$. As a consequence, there can't be a classifying topos (or $\infty$-topos) for $U(1)$-bundles in the sense you describe (the category of $U(1)$-bundles on a point does not embed fully faithfully in the category of $U(1)$-bundles on $\mathbf{R}$). The same counterexample works if you replace $U(1)$ by any topological group $G$ for which there exists a nonconstant continuous map $\mathbf{R} \rightarrow G$.<|endoftext|> TITLE: Why people usually consider reductive groups in GIT? QUESTION [9 upvotes]: Where do people essentially use the reductive groups in the theory of GIT? Or how does reductive groups simplify the constructions in GIT? I found that the property of completely reducible of reductive groups (in character 0) is fascinating (see Mumford's GIT book page 26-27), but I don't know if it is this property that makes the reductive groups so pervasively in GIT, and how does it used in the theory. REPLY [10 votes]: Since the comments are already getting long, I'll add this in community-wiki format to clarify a few points. I should emphasize that I'm not at all a specialist in GIT but have dealt with neighborhing problems involving algebraic groups. First, there are fundamental differences between characteristic 0 (where Mumford mostly worked) and prime characteristic. In the former case, a linear algebraic group splits as a semidirect product of its unipotent radical and a reductive (Levi) subgroup which is unique up to conjugacy. In the latter case, such a splitting may fail. More important here is the fact that "reductive" is equivalent in characteristic 0 to "linearly reductive": all finite dimensional representations as an algebraic group are completely reducible. This fails badly in prime characteristic. On the other hand, unipotent groups have the nice property that their algebraic actions have only closed orbits. (And their only irreducible representation is the trivial one.) In terms of classical invariant theory, all reductive groups (even in prime characteristic) turn out to have the good property that the associated rings of polynomial invariants are finitely generated. (This was settled in prime characteristic by Haboush's proof using algebfaic geometry of the Mumford Conjecture, that reductive implies "geometrically reductive". There is an algebraic proof in Jantzen's book Representations of Algebraic Groups.) Anyway, Mumford's initial slim volume has grown in its third edition to a larger book, with contributions first by Fogarty and then by Kirwan. But the prefaces Mumford wrote show pretty clearly what he had in mind and why he wanted to work especially with reductive groups. While unipotent groups have their own interest, much classical work on moduli problems involves reductive groups. Their actions involve orbits which need not be closed, as already seen in the adjoint representation (where only the semisimple elements live in closed orbits). So Mumford's ideas about stable and semistable points are delicate, but important to sort out.<|endoftext|> TITLE: Finitely presentable objects in functor categories QUESTION [7 upvotes]: Given a locally finitely presentable category $\mathcal{C}$ it is well-known that every functor category $[\mathcal{A},\mathcal{C}]$ (where $\mathcal{A}$ is a small category) is also locally finitely presentable. Is there a concrete description of the finitely presentable objects in such categories? If this is not possible in general, what about special cases like $\mathcal{A}=\mathcal{Set}_{fin}^{op}$? REPLY [6 votes]: When $\mathcal{A}$ is a finite category, the finitely presentable objects in $[\mathcal{A}, \mathcal{C}]$ are precisely the diagrams that are componentwise finitely presentable: see e.g. Proposition 2.23 here. The general case is harder to describe. There are two steps: First, determine the finitely presentable objects in $[\operatorname{ob} \mathcal{A}, \mathcal{C}]$; these will contain e.g. "finitely supported" families of finitely presentable objects. Then, determine the left adjoint of the restriction functor $[\mathcal{A}, \mathcal{C}] \to [\operatorname{ob} \mathcal{A}, \mathcal{C}]$; by general nonsense, the restriction functor is finitely accessible and monadic, and the finitely presentable objects of $[\mathcal{A}, \mathcal{C}]$ will be the objects in the smallest full subcategory that is closed under finite colimits and that contains the free "algebras" generated by the finitely presentable objects in $[\operatorname{ob} \mathcal{A}, \mathcal{C}]$.<|endoftext|> TITLE: Simple-minded coherence of tricategories QUESTION [6 upvotes]: Recall Mac Lane's version of coherence for monoidal categories, which one can state informally as follows: "Simple-minded" coherence for monoidal categories Let $A$, $A^\prime$ be two bracketings on the monoidal product $A_{1} \otimes \ldots \otimes A_{n}$ (with possibly some additional occurances of the monoidal unit $I$). Let $i_{1}, i_{2}: A \rightarrow A^\prime$ be two isomorphisms built from associator and left/right unitors. Then $i_{1} = i_{2}$. This theorem tells us that there is always canonical way to identify two products of $A_{1},\ldots, A_{n}$ with possibly different bracketings. Here I say "simple-minded" to mean that it can be presented as a statement of the form "some diagram commutes". However, coherence for monoidal categories can also refer to the following result: "Strictifying" coherence for monoidal categories Any monoidal category is equivalent to a strict monoidal category. The strictifying version of coherence is an important theorem on its own right, but it also implies the simple-minded version of coherence with the following argument. Let $i_{1}, i_{2}: A \rightarrow A^\prime$ be two isomorphisms as above in a monoidal category $M$ and let $T: M \rightarrow N$ be an equivalence into a strict monoidal category. Then $T(i_{1}), T(i_{2})$ are both isomorphic to analogous composites of associators and left/right unitors in $N$, with the isomorphism given by constraint cells of $T$. Since $N$ is strict, the analogous composites in $N$ are both equal to the identity and we conclude that $T(i_{1}) = T(i_{2})$. (The argument seems to be well known, although I learned it from these short notes of Tom Leinster.) My question concerns known results about such "simple-minded" coherence for monoidal bicategories (ie. one object tricategories), which I had trouble finding in the literature. Recall that coherence for tricategories as proved by Gordon, Street, Power has the following form: "Strictifying" coherence for tricategories Any tricategory is triequivalent to a Gray-category, ie. to a category enriched over the category of strict 2-categories equipped with the Gray tensor product. In particular, any monoidal bicategory is equivalent to a Gray monoid. This is certainly an important and powerful result. However, what is not clear to me is how to extract from this some "simple-minded" corollaries, ie. statements about some diagrams (say, of 3-cells) commuting in any tricategory. I believe some argument similar in spirit to the one from notes of Tom Leinster should work, but triequivalences (or more generally, homomorphisms of tricategories) are such complicated objects that it is not quite obvious for me how to do this. Is there any general framework for proving that some classes of diagrams commute in every tricategory? Has this been covered in the literature? What are the possible references? For example, is the following naive generalization of Mac Lane coherence true? "Naive" version of coherence for monoidal bicategories Let $A$, $A^\prime$ be two bracketings on the monoidal product $A_{1} \otimes \ldots \otimes A_{n}$ (with possibly some additional occurances of the monoidal unit $I$) in a monoidal bicategory. Let $i_{1}, i_{2}: A \rightarrow A^\prime$ be two 1-cells built from associators, left/right unitors and their pseudoinverses. Moreover, let $j_{1}, j_{2}: i_{1} \sim i_{2}$ be two isomorphisms built from the constraint $2$-cells. Then $j_{1} = j_{2}$. Here I am using the algebraic definition of a monoidal bicategory, ie. the associatiors and unitors of the monoidal structure come with a chosen pseudoinverse and unit/counit modifications that make them into adjoint equivalences. I believe in this these unit/counit 2-cells should also be considered "constraint", but I hope I can be forgiven for leaving this a little vague just to see what might be true and what isn't. I stumbled upon this type of questions while studying possible definitions of a dual pair of objects in a monoidal bicategory. I frequently find it very problematic to prove any uniqueness results due to the relevant computations being difficult. Hence, I am looking for techniques that could simplify working with a general tricategory. REPLY [3 votes]: I am still certainly no expert on tricategories, but I believe I have found the needed result. The question is answered in a paper of Nick Gurski, "An algebraic theory of tricategories" and probably also in his new book ("Coherence in Three-dimensional Category Theory"). I will reference the former, where Corollary 10.2.3. reads Let $X$ be a 2-locally discrete category-enriched 2-graph. Then in the free tricategory on $X$, $FX$, every diagram of 3-cells commutes. Translating it into the easier language of monoidal bicategories we obtain the following. Recall a category-enriched graph $E$ is simply a set of objects $Ob(E)$ together with a category $E(a, b)$ for any pair $a, b \in Ob(E)$. In particular, any bicategory is a category-enriched graph. A category enriched graph is locally discrete if all the categories $E(-, -)$ are discrete, ie. do not contain non-identity morphisms. Let $E$ be a locally discrete category-enriched graph. In the free monoidal bicategory $FE$ on $E$ any diagram of $2$-cells commutes. Free monoidal bicategories (in the above sense) have universal properties with respect to strict homomorphisms so it is easy to deduce that in any monoidal bicategory, any diagram of 2-cells that can be presented as image of some diagram in $FX$ under a strict homomorphism, where $X$ is some locally discrete category-enriched graph, necessarily commutes. In particular the "naive" version of coherence for monoidal bicategories I asked for above is true. The situation is not ideal, however, as one could also want to obtain a similar result for free monoidal bicategories on more general generating data (say, allow generating 1-cells of the type $I \rightarrow A \otimes B$ or $C \otimes D \rightarrow E$). This generalization turns out to be false. This is related to the fact that a one-object monoidal bicategory is "morally the same" as a braided monoidal category (a result due to Gordon, Power, Street), with the braiding given by a clever composition of 2-cells. As braiding are in general not symmetries, some diagrams of constraint 2-cells in monoidal bicategories do not commute in general.<|endoftext|> TITLE: how is the dual connection defined? QUESTION [5 upvotes]: Let $E$ be a vector bundle (i.e. locally free $\mathcal{O}_X$-module) on some smooth algebraic variety $X$ and let $\nabla: E \to E \otimes \Omega^1_X$ be an integrable connection. I have seen that $\nabla$ induces a connection $\nabla^\vee$ on the dual vector bundle $E^\vee=\mathcal{Hom}_{\mathcal{O}_X}(E, \mathcal{O}_X)$. 1) How is $\nabla^\vee$ defined? 2) If $X$ is not complete and $\nabla$ has regular singularities along the boundary, is the same true for $\nabla^\vee$? 3) Are the local systems of horizontal solutions of $\nabla$ and $\nabla^\vee$ dual? REPLY [8 votes]: The dual connection $\nabla^\vee$ on $E^\vee$ is defined by $$\langle \nabla^\vee \phi, s\rangle = d\langle \phi, s\rangle - \langle\phi, \nabla s\rangle,\quad \forall \phi\in E^\vee, s\in E,$$ where $\langle-,-\rangle:E^\vee\otimes E\rightarrow \mathcal{O}$ is the pairing. One can check that the formula is $\mathcal{O}$-linear in terms of $s$ and that $\nabla^\vee$ satisfies the Leibniz rule. A connection on $\overline{X}$ with regular singularities along $D=\overline{X}-X$ is the same as the datum of $\nabla_v: E\rightarrow E$ for all vector fields $v\in T_{\overline{X}}$ tangent to $D$. If the original connection $\nabla$ is defined for such vector fields, so is the dual connection. It does not extend to an ordinary connection: on $\mathbf{A}^1$ the dual of the connection $d+\frac{dz}{z}$ on the trivial line bundle is $d-\frac{dz}{z}$. Yes, this is the case when Riemann-Hilbert is an equivalence, i.e. when $E \cong dR(E)\otimes \mathcal{O}_X$, where $dR(E)$ is the associated local system. Indeed, in this case there are maps between $dR(E^\vee)$ and $dR(E)^\vee$ in both directions (one map is given by the obvious pairing, the other map is given by the $\mathcal{O}$-linear extension of functionals).<|endoftext|> TITLE: Constructible subset of constructible set QUESTION [6 upvotes]: Let $X$ be a topological space. Let $F \subset E \subset X$ be subsets. Assume that $E$ is constructible in $X$ and that $F$ is constructible in $E$. Is it true that $F$ is constructible in $X$? We use the definition of constructible as defined in EGA I (first edition). This is true if $X$ has a basis for the topology consisting of quasi-compact opens, see Lemma Tag 09YJ. In particular, it holds if $X$ is a scheme. Thus our question is a bit frivolous as the important use case is covered. Still: Is there a counter example for general $X$? Is it true if $X$ is quasi-compact? REPLY [3 votes]: If $F$ is open in $E$ (where $E$ carries the subspace topology from $X$), then $F = E \cap U$ where $U$ is open in $X$, hence $F$ is constructible in $X$ (since constructibles are closed under intersection). Similarly, if $F$ is closed in $E$, then $F$ is constructible in $X$. The constructibles in $E$ constitute the smallest Boolean algebra that contains the open and closed sets in $E$. The constructibles in $X$ are a Boolean algebra which, we have just seen, include the open and closed sets in $E$. Hence constructibles in $E$ are included in the the constructibles of $X$, as anticipated.<|endoftext|> TITLE: What properties of knots lead Lord Kelvin to hypothesize that atoms were knots in the ether? QUESTION [14 upvotes]: I've often heard that Lord Kelvin was one of the first people to study knot theory, as he hypothesized that atoms were knots in the ether. I assume that he had some compelling evidence for this fact. What properties of knots did Lord Kelvin consider similar to atomic properties? REPLY [2 votes]: It wasn't just Kelvin---the three premier Scottish physicists of the 19 th century, Maxwell, Thomson (Kelvin) and Tait were all fascinated by knots. Some details can be found in the article by D.S. Silver "Scottish Physics and knot theory's odd origins" which is available online. The fact that this didn't lead to important advances in physics should not blind one to their other seminal contributions and it is arguable that this work led to the development of modern knot theory.<|endoftext|> TITLE: Presenting $\mathbb{Q}[[t]]$ as an explicit colimit of smooth $\mathbb{Q}$-algebras: an explicit example for the Popescu's theorem QUESTION [12 upvotes]: By the seminal Popescu's theorem, $R=\mathbb{Q}[[t]]$ is a filtered colimit of smooth $\mathbb{Q}$-algebras. Could you give me a hint: which $\mathbb{Q}$-algebras can yield such a colimit? My problem is that subalgebras of $R$ seem to be 'usually' singular. Upd. 1. I realized that $R$ actually contains plenty of smooth $\mathbb{Q}$-subalgebras; yet I am not sure that I can prove that $R$ can be presented as their filtered colimit. Is this true; are there any nice references on these matters? Upd. 2. Possibly, it suffices to consider henselizations of freely generated $\mathbb{Q}$-subalgebras of $R$ here. In any case, I would like to understand presentations of regular complete local rings as 'smooth colimits' better. REPLY [5 votes]: One can prove Popescu's theorem directly in this special case (due to the two strong assumptions -- characteristic $0$ and one-dimensionality -- present here). The basic point is that any 'singular' subalgebra $A \subset R$ may be resolved by a proper birational map $X \to \mathrm{Spec}(A)$ by Hironaka, and that the map $\mathrm{Spec}(R) \to \mathrm{Spec}(A)$ lifts to $X$ by the valuative criterion. Write $R$ as a filtered colimit of finitely generated $\mathbb{Q}$-subalgebras $A_i \subset R$, so $\mathrm{Spec}(R) = \lim \mathrm{Spec}(A_i)$. Each $A_i$ is a domain, and the map $\mathrm{Spec}(R) \to \mathrm{Spec}(A_i)$ maps the generic point to the generic point. For each $i$, consider the cofiltered projective system $ C_i := \{X_{i,j} \to \mathrm{Spec}(A_i)\}$ of all proper maps which are isomorphisms over the generic point of the target. Pullback along $\mathrm{Spec}(A_{i'}) \to \mathrm{Spec}(A_i)$ gives a functor $C_i \to C_{i'}$, so we get a two-variable cofiltered projective system $\{X_{i,j}\}$ of schemes. Also, the map $\mathrm{Spec}(R) \to \mathrm{Spec}(A_i)$ admits a unique lift to $\mathrm{Spec}(R) \to X_{i,j}$ for each $j$ by the valuative criterion for properness. Let $B_{i,j}$ be the local ring of $X_{i,j}$ at the image of the closed point of $\mathrm{Spec}(R)$, so there is a natural map $B_{i,j} \to R$ extending the given map $A_i \to R$. Varying over the projective system, we get a filtered system $\{B_{i,j}\}$ of rings with a map to $R$. One checks that $\mathrm{colim} B_{i,j} \simeq R$; the surjectivity is completely trivial, while injectivity comes from the fact that $B_{i,j}$'s are domains birational to $A_i$ for a cofinal set of $j$'s. Moreover, by resolutions, for a fixed $i$, the $B_{i,j}$'s are smooth over $\mathbb{Q}$ for a cofinal set of $j$'s. Thus, we have expressed $R$ as a filtered colimit of smooth $\mathbb{Q}$-algebras.<|endoftext|> TITLE: In $L$, does there exist a definable non-principal ultrafilter on $\mathbb{N}$ QUESTION [7 upvotes]: The axiom of constructibility $V=L$ leads to some very interesting consequences, one of which is that it becomes possible to give explicit constructions of some of the "weird" results of AC. For instance, in $L$, there is a definable well-ordering of the real numbers (since there is a definable well-ordering of the universe). Since AC holds true in $L$, the ultrafilter lemma must be true. Does this mean that a definable non-principal ultrafilter on $\mathbb{N}$ exists in $L$, given by an explicit formula? If so, what is the formula? REPLY [8 votes]: Let me emphasize what Noah de-emphasized (by putting it in parentheses) in his answer: Instead of the "really cheap" idea of the first free ultrafilter in $L$-order, one can use the $L$-order of subsets of $\mathbb N$ to greedily construct an ultrafilter. The $L$-ordering of the subsets of $\mathbb N$ is a $\Delta^1_2$ relation with some additional properties (I believe Moschovakis uses the terminology $\Delta^1_2$-good well-ordering). One can then deduce that the greedily constructed ultrafilter is a $\Delta^1_2$ set of subsets of $\mathbb N$. Since free ultrafilters on $\mathbb N$ never have the Baire property and are never Lebesgue measurable, this shows that the classical results, that analytic and coanalytic sets have the Baire property and are measurable, cannot be extended higher in the projective hierarchy in ZFC (without additional hypotheses like large cardinals or determinacy). Similar greedy constructions based on the $L$-ordering of the reals produce $\Delta^1_2$ examples of lots of other sets of reals that, in the absence of $V=L$, would be obtained by applying the axiom of choice and thus might not be definable at all. Examples include Hamel bases over $\mathbb Q$, Vitali sets, and Bernstein sets.<|endoftext|> TITLE: global dimension QUESTION [5 upvotes]: Suppose we are given finite dimensional semisimple $k$-algebras $A_1,..., A_r$. now we consider the matrix algebra$A=\begin{pmatrix} A_1 & M_{1,2} & \dots & M_{1,r} \\ 0 & A_2 & \dots & M_{2,r} \\ \vdots & 0 & \ddots & \vdots \\ 0 & \dots & 0 & A_r \end{pmatrix}$ with finitely generated bimodules $M_{i,j}$. Now suppose $\mathrm{gldim}(A)$ is finite. Furthermoe suppose we are given a finite group $G$ acting on $A$ such that $\mathrm{char}(k)$ does not divide the order of $G$. Do we have $\mathrm{gldim}(A^G)<\infty$? I guess that first of all one can consider $ \mathrm{gldim}\begin{pmatrix} R & M \\ 0 & S \end{pmatrix}=\mathrm{max}\{\mathrm{pdim}_R M+1, \mathrm{gldim}R\} $, where $R$ and $S$ are finite dimensional semisimple $k$-algebras and $M$ a finitely generated $R-S$ bimodule and then proceed by induction... REPLY [6 votes]: Yes, $A^G$ must have finite global dimension, as it has the same kind of triangular form as $A$. The Jacobson radical of $A$ is $$J(A)=\begin{pmatrix} 0 & M_{1,2} & \dots & M_{1,r} \\ 0 & 0 & \dots & M_{2,r} \\ \vdots & 0 & \ddots & \vdots \\ 0 & \dots & 0 & 0 \end{pmatrix},$$ and $$A/J(A)=A_1\times\dots\times A_r,$$ which can be identified with a subalgebra of $A$ in the obvious way. The primitive central idempotents $e_1,\dots,e_s$ of $A/J(A)$ can be ordered so that $i TITLE: Can any curve be embedded into $\mathbb{P}^3$? QUESTION [13 upvotes]: We know from Hartshorne's "Algebraic geometry" that if a curve $C$ is smooth then there exists a closed immersion of $C$ into $\mathbb{P}^3$. Does this still hold if $C$ is not generically reduced or singular at finitely many points? REPLY [25 votes]: No. If a curve embeds into $\mathbb{P}^3$, its tangent space at every point has dimension $\leq 3$. This is a strong restriction on the possible singularities. For instance, a curve locally isomorphic to the union of the lines $x=y=z=0$, $x=y=t=0$, $x=z=t=0$, $y=z=t=0$ in $\mathbb{A}^4$ cannot be embedded in $\mathbb{P}^3$.<|endoftext|> TITLE: Why study CM abelian varieties? QUESTION [13 upvotes]: I know that abelian varieties of CM type have central importance in algebraic geomtry and number theory. There are many conjectures and concepts related to them like Andre-Oort, Coleman conjecture, Shimura varieties, etc... . But why CM abelian varieties came into consideration in first place? Why did mathematicians begin to study CM abelian varieties? I am in particular intereseted to know the relation to "special functions". REPLY [8 votes]: Why did mathematicians begin to study CM abelian varieties? Because they could. The study of CM abelian varieties arguably starts with Fagnano's work on the length of the lemniscate. In 1799, Gauss linked the length of the lemniscate to the arithmetic-geometric mean and in the entries of Gauss's diary from 1814, including famously (but not restricted to) the last one, he studied some properties of biquadratic reciprocity and their links with the division of the lemniscate which he understood to derive from properties of Punktgitter (lattices of points) in the complex planes, i.e CM elliptic curves. As you see, there is a continuous path from traditional questions in geometry arguably going back to Antiquity (Fagnano's motivation was to rectify the ellipse) to accounts of complex multiplication quite close to the modern version, and the latter was known (at least to Gauss) in the early XIXth century. Legendre, Jacobi, Abel, Galois all knew about CM from one perspective or another. REPLY [8 votes]: (An elaboration of Timo Keller's answer.) One of the main sources of historical interest in CM abelian varieties is the pursuit of Kronecker's Jugendtraum and Hilbert's 12th problem. If $K$ is an imaginary quadratic field, then one can always find an elliptic curve $E$ with CM by the full ring of integers $\mathfrak o_K$ of $K$, and then it turns out that, (i) $K(j(E))$ is the Hilbert class field of $K$, and (ii) all the abelian extensions of $K$ can be constructed by adjoining to $K(j(E))$ values of a certain "modular function" (the Weber function $h$) evaluated at torsion points of $E$. For the precise statement and further elaboration, pick up any book that deals with the CM theory of elliptic curves, e.g., Silverman's second book on elliptic curves. What about more general number fields $K$: can we explicitly construct their abelian extensions in a similar fashion? The answer to this general question is wide open. But, anyhow, one possible approach is to try to think about how to generalize the above picture, in which case we ask: what should replace the field $K$, the elliptic curve $E$, and the special functions $j$ and $h$? Hecke gave the first attempt at answering these questions. In his thesis he takes up the case of a real quadratic $K$ and uses a certain "Hilbert modular function", but doesn't get far. Later on he tries his hand at dealing with certain biquadratic extensions, and constructs unramified abelian extensions of them by adjoining singular values of two-variable Hilbert modular functions. Implicitly, the geometric object taking the place of $E$ here is a certain abelian surface with CM. Next comes Shimura who takes this one step further and looks at a general CM number field $K$. His $E$ and special functions are then CM abelian varieties and certain "Siegel modular functions". He uses this setup to construct certain class fields, except not over $K$, but over a certain other field $K^*$, the so-called "reflex field". (In the case of elliptic curves, we have $K=K^*$, and we get all the class fields of $K$.) So, unfortunately, unlike in the case of elliptic curves and imaginary quadratic fields, the higher-dimensional theory is more intricate doesn't give as complete an answer. For more details, you can start with Shimura's Automorphic functions and number theory (LMN 1968). REPLY [6 votes]: CM elliptic curves and their torsion points are used to explicitly construct ray class fields of imaginary quadratic number fields. E.g., the $j$-invariant generates the Hilbert class field.<|endoftext|> TITLE: Projective embedding in families of curves QUESTION [8 upvotes]: Let $\pi:\mathcal{X} \to B$ be a family (flat, projective, surjective morphism) of projective curves (not necessarily reduced) where $B$ is smooth, irreducible. Suppose that for some closed point $b_0 \in B$, $\pi^{-1}(b_0)=\mathcal{X}_{b_0}$ can be embedded into $\mathbb{P}^n$ for some integer $n$. Does there exist an open neighbourhood $U \subset B$ containing $b_0$ such that for all $b \in U$, $\pi^{-1}(b)$ can be embedded into $\mathbb{P}^n$ for the same integer $n$? REPLY [13 votes]: No, that is not true. I am sure that somebody else has answered this on MO before. I guess the simplest example is a family of genus 6 curves, where some fibers are embeddable as plane quintics. If memory serves, the canonical image of a general genus 6 curve (non-hyperelliptic) is the intersection of a Pfaffian quintic del Pezzo surface and a quadric hypersurface in $\mathbb{P}^5$. So there is an "explicit" irreducible family of genus 6 curves dominating the moduli space $\mathcal{M}_6$. Of course $\mathcal{M}_6$ has dimension $3(6)-3 = 15$. By parameter counts, the dimension of the moduli space of plane quintics is $$\text{dim}H^0(\mathbb{P}^2,\mathcal{O}(5)) - \text{dim}\textbf{GL}_3 = 21 - 9 = 12.$$ Since $12<15$, a typical genus 6 curve is not isomorphic to a plane quintic. (In terms of the canonical embedding, the condition is that the curve is contained in a Veronese surface in $\mathbb{P}^5$.)<|endoftext|> TITLE: Grothendieck group of curves QUESTION [6 upvotes]: The Grothendick group of a smooth curve over $k=\bar{k}$ is $K_0(C)=\mathbb{Z}\oplus\mathrm{Pic}(C)$ and is finitely generated only for $\mathbb{P}^1$. When is $K_0(C)$ finitely generated when $k\neq\bar{k}$? REPLY [6 votes]: The degree map gives a short exact sequence $$ 0 \to J(C) \to Pic(C) \to \mathbb{Z} \to 0,$$ where $J(C)$ denotes the Jacobian of $C$. This is an abelian variety of dimension $g$, where $g$ is the genus of $C$. In particular, $Pic(C)$ is always finitely generated when $C \cong \mathbb{P}^1$. To deal with the case of higher genus, you might as well ask: for which fields $k$ do abelian varieties over $k$ of positive dimension have finitely generated groups of rational points? Clearly this holds for finite fields, and the Mordell-Weil theorem also says that this holds for number fields. More generally, this is known to hold for fields which are finitely generated over their prime field.<|endoftext|> TITLE: An intutive reason why a "distance" metric may be a poor one for a procedure where we attempt to modify a string (mutating 0 OR 1 bits) QUESTION [5 upvotes]: If I'm attempting to mutate one arbitrarily chosen binary string $s_a$, to another arbitrarily chosen binary string $s_b$, in the smallest number of steps (i.e. with the smallest number of mutations) via a procedure where I decide to either: (1) Randomly and with uniform probability choose a "0" bit and flip it to a "1" bit. E.g. if there are three "0" bits in our string $s_a$ at some decision / mutation step, and $10^2$ "1" bits, here a particular "0" bit would be selected for mutation to a "1" bit with probability $\frac{1}{3}$. (2) Randomly and with uniform probability choose a "1" bit and flip it to a "0" bit. Is there a good intuitive reason why I might not want to always choose (1) or (2) based on which option maximizes the probability that the Hamming distance to the target string will be smaller in the next step? Let me write a specific statement about where my thinking is failing: I thought it would be the case that any string within a fixed Hamming distance of the target string would have a (topologically) indistinguishable representation on the $n$-dimensional hypercube where vertices represent all possible states of a length $n$ binary string (i.e. a length $n$ Hamming code - http://en.wikipedia.org/wiki/Hamming_distance). Given that the $n$-dimensional hypercube is a distance transitive graph, all vertices within a fixed distance $k$ of another vertex should be identical up to isomorphism of the $n$-cube (http://en.wikipedia.org/wiki/Distance-transitive_graph). Now, since "passing through" each set of topologically indistinguishable vertices within a Hamming distance $(1,2,...)$ of the target string must be done to reach the target string, what advantage could we have in not maximizing our chance to move one step closer to the target string at each step of the aforementioned decision process? The above is NOT intended to be anything other than an illustration of where my intuition is likely failing. Note: This question is related to The time to drift a binary string from one state to another via deterministic selection of two possible random bit mutation procedures. I also asked this question in a long-winded manner in a (now deleted) question that received no answers, but where Douglas Zare in the comments claimed that a "Hamming distance" strategy for the above decision process failed in simple counterexamples. I haven't been able to see this, it's on me and I thank him for his time. But I'd really like to understand better what's going on, so I'd like to ask the above with the "soft-question" tag. Let me provide some support that my intuition is failing here. Here are simulations for the time to mutate some length $n = 10$ string $s_a = 0000000000$ to another string $s_b$ with $k$ $1$ bits under a strategy where we immediately try to flip back "incorrect" $1$ bits by applying procedure [2] for as long as necessary to do so. Where the target string $s_b$ has $k$ bits with value $1$ (where exactly these bits are in the string should be irrelevant), and performing $10^4$ trials: $k=0$ trivially implies a {mean, median} $= (\mu, \mu_{1/2}) = (0, 0)$ $k=1$ yields a {mean, median} $= (\mu, \mu_{1/2}) = (19.3202, 13)$ $k=2$ yields a {mean, median} $= (\mu, \mu_{1/2}) = (66.2872, 46)$ $k=3$ yields a {mean, median} $= (\mu, \mu_{1/2}) = (152.303, 107)$ $k=4$ yields a {mean, median} $= (\mu, \mu_{1/2}) = (258.273, 180)$ $k=5$ yields a {mean, median} $= (\mu, \mu_{1/2}) = (333.897, 237)$ $k=6$ yields a {mean, median} $= (\mu, \mu_{1/2}) = (321.758, 226)$ $k=7$ yields a {mean, median} $= (\mu, \mu_{1/2}) = (238.742, 167)$ $k=8$ yields a {mean, median} $= (\mu, \mu_{1/2}) = (130.412, 94)$ $k=9$ yields a {mean, median} $= (\mu, \mu_{1/2}) = (50.1086, 37)$ $k=10$ trivially implies a {mean, median} $= (\mu, \mu_{1/2}) = (L, L) = (10, 10)$ Now, under the strategy where we try to pick procedure [1] vs. procedure [2] based on which one maximizes the chance of moving closer to $s_b$ (in terms of Hamming distance) at any given time step, we have: $k=5$ yields a {mean, median} $= (\mu, \mu_{1/2}) = (436.988, 303)$ Which is a lot worse than what we found for the strategy where we immediately try to eradicate incorrect bit flips by mutating using procedure [2] until they are removed. REPLY [4 votes]: I'll appeal to the linked question's answers for computational details. Intuition only: (as requested here) It is "easier" using your procedure to move from $00110^{n-4}$ to $11000^{n-4}$ than it is to go from $0000^{n-3}$ to $1110^{n-3}$, despite the Hamming distance being smaller in the latter case. You have to "win the lottery" three times in the latter case. Therefore, a priori it is unclear that reducing Hamming distance at each step is optimal.<|endoftext|> TITLE: Flag manifolds (=R-spaces): quotients by parabolic subgroups vs. isotropy representation QUESTION [8 upvotes]: Real flag manifolds (also known as R-spaces) can be defined in two ways which I believe are equivalent although some fine print may have escaped me: as a quotient of a semisimple real Lie group $G$ by a parabolic subgroup (subgroup containing a Borel), as an orbit of the isotropy representation (action of the point stabilizer on the tangent space) of the Riemann symmetric space associated to $G$. Where can I find a explicit statement and proof of this equivalence, ideally in book form? Preferably with additional details on the connection between flag manifolds and Riemann symmetric spaces (e.g., on when one is the other, and conversely)? REPLY [9 votes]: A real flag manifold is $M=G/P$ where $G$ is a real semisimple Lie group and $P$ is a parabolic subgroup of $G$. Here parabolic means that the Lie algebra $\mathfrak p$ contains a minimal parabolic subalgebra $\mathfrak p_{min}$ of $\mathfrak g$. In order to define this, let $\mathfrak g=\mathfrak k+\mathfrak s$ be a Cartan decomposition, $\mathfrak a$ is a maximal Abelian subspace of $\mathfrak s$, $\mathfrak g =\mathfrak g_0+\sum_{\lambda\in\Sigma} \mathfrak g_\lambda$ be the root decomposition with respect to $\mathfrak a$. We have $\mathfrak g_0=\mathfrak m+\mathfrak a$ where $\mathfrak m$ is the centralizer of $\mathfrak a$ in $\mathfrak k$. Choose a positive subsystem $\Sigma^+$ and let $\mathfrak n = \sum_{\lambda\in\Sigma^+}\mathfrak g_\lambda$ (nilpotent Lie algebra). We get the Iwasawa decomposition $\mathfrak g =\mathfrak k+\mathfrak a+\mathfrak n$. Finally, $\mathfrak p_{min}=\mathfrak m+\mathfrak a+\mathfrak n$. In general, $\mathfrak p = \mathfrak p_{min} + \sum_{\lambda\in\langle\Phi\rangle}\mathfrak g_{-\lambda}$, where $\Phi$ is a subset of of the system of simple roots of $\Sigma$. It turns out $K$, as a subgroup of $G$, acts by left translations on $G/P$ and this action is transitive. In fact, the isotropy algebra at the basepoint is $\mathfrak k\cap\mathfrak p$, and if you count dimensions, you see that the orbit is open. It is also closed by compactness of $K$, so $K/K\cap P=G/P$. It remains to identify the left hand side with an orbit of the isotropy representation of the symmetric space $G/K$, namely, the action of $K$ on $\mathfrak s$, but this is not hard. I can refer you to the nice book "An Introduction to Lie Groups and the Geometry of Homogeneous Spaces", by Andreas Arvanitoyeorgos (http://www.ams.org/bookstore?fn=20&arg1=stmlseries&ikey=STML-22). Edit: Indeed $\mathfrak k\cap\mathfrak p =\mathfrak m + \sum_{\lambda\in\langle\Phi\rangle}(\mathfrak g_{\lambda}+\mathfrak g_{-\lambda})\cap\mathfrak k$ and you can select a point $x\in\mathfrak a\subset\mathfrak s$ such that $\lambda(x)=0$ if and only if $\lambda\in\langle\Phi\rangle$. The desired $K$-orbit is that through $x$.<|endoftext|> TITLE: Text for Algebraic Number Theory QUESTION [16 upvotes]: I have the privilege of teaching an algebraic number theory course next fall, a rare treat for an algebraic topologist, and have been pondering the choice of text. The students will know some commutative algebra, some homological algebra, and some K-theory. The texts I am now considering are (1) Frohlich and Taylor, Algebraic Number Theory, (2) Janusz, Algebraic Number Fields, and (3) Milne, Algebraic Number Theory (online). I am leaning toward (1): it seems very well written and has sections on cubic, biquadratic and sextic fields as well as quadratic and cyclotomic fields. I do know that when you get into a course you discover drawbacks that aren't apparent when you're reading the text before the course begins. Frohlich and Taylor is designed for a year long course and I only have one semester, so will need to do some selecting. I regret not using any homological techniques, also. I would appreciate suggestions, either about these or other texts. REPLY [7 votes]: My favorite textbook on the subject is "Number fields" by Marcus. Things are done in a very explicit way, there are hundreds of exercises, and I've usually found that the proofs of the classical theorems (finiteness of the class number, structure of the units, ...) are done in a way that I really like.<|endoftext|> TITLE: Which morphisms of ring spectra are of effective descent for modules? QUESTION [7 upvotes]: There is a well understood bifibration of $\infty$-categories over the $\infty$-category of commutative ring spectra whose fiber over a ring $R$ is the category of $R$-module spectra. This is in analogy to the usual Grothendieck fibration over commutative rings in discrete algebra. One then can ask which morphisms of commutative ring spectra are of effective descent for modules. In other words, one can ask whether or not the category of $R$-modules is equivalent to the category of descent data for some morphism $R\to S$ (which is constructed as as the homotopical totalization of the Amitsur complex). Rognes has introduced a notion of faithful morphism of ring spectra which is the following: if $\phi:R\to S$ is a morphism of ring spectra, then it is faithful if for every $R$-module $M$, $M\wedge_RS\simeq \ast$ implies that $M\simeq\ast$. Lurie has also introduced a notion of faithfully flat: the morphism $\phi$ is faithfully flat if $\pi_0(\phi):\pi_0R\to\pi_0S$ is faithfully flat and $\pi_iS\cong \pi_iR\otimes_{\pi_0R}\pi_0 S$. Are either of these sufficient to imply effective decent for modules? What if we say that morphism is faithful in the sense of Rognes but also a Galois extension for some stably dualizable group $G$ or even a Hopf-Galois extension? Is there a classification of effective descent morphisms in this setting? Thanks! REPLY [7 votes]: In DAG7, Theorem 6.1, Lurie proves faithfully flat descent for modules: the functor $A \mapsto \mathrm{Mod}_A$ is a hypercomplete sheaf for the fpqc topology on $E_\infty$-rings. This, in particular, gives effective descent for module spectra along faithfully flat maps.<|endoftext|> TITLE: an elementary question on K-theory QUESTION [8 upvotes]: Sorry for asking such an elementary question. 1) What is Quillen's $K_1$ of a (nice) scheme $X$? If $X=Spec(k)$, I guess one gets $k^\times$, is that correct? What about the case of a curve $C$ over some field k, is it true that $K_1(C)=k(C)^\times$? 2) I would also like to know how the localization sequence of $K$-theory is related to the tame symbol in the case of curves. I guess this is the case when one takes the empty set as a closed subscheme. What about the localization sequence of $C$ a curve and $S$ a bunch of points? I would be happy if someone could help me with these questions. I find K-theory pretty hard so far... REPLY [10 votes]: As is already apparent from Steven Landsburg's answer, the question is not as elementary as one might think at first glance. Already for $K_1$ of curves, there are a couple of as yet unresolved problems. This is mainly due to the fact that the localization sequence relates the K-theory of curves over function fields to the K-theory of their (higher-dimensional) models. Concerning (1): Let me discuss a bit what is known about the structure of $K_1$ of curves. Generalities: Some of the structure of $K_1$ becomes clearer when looking at motives or motivic cohomology. If $C$ is a smooth projective curve over a field $k$, then there is a splitting of the motive (Chow or Voevodsky, does not matter): $$ M_k(C)\cong \mathbb{Z}_k\oplus \operatorname{Jac}(C/k)\oplus\mathbb{Z}_k(1)[2]. $$ Using Bott periodicity of algebraic K-theory, the two summands $\mathbb{Z}$ and $\mathbb{Z}(1)[2]$ each contribute one copy of $K_1(k)$ to $K_1(C)$. The remaining part of $K_1(C)$ comes from the motive $\operatorname{Jac}(C/k)$, which as the notation suggests comes from the Jacobian. There are several ways to define groups related to this Jacobian part of $K_1(C)$. Rationally, we can use the Adams eigenspace decomposition $$ K_1(C)_{\mathbb{Q}}\cong\bigoplus_{q\geq 0} H^{2q-1,q}(C,\mathbb{Q}). $$ Known vanishing results for motivic cohomology, imply that the summands are trivial unless $q=1,2$. One copy of $K_1(k)$ is the weight one part ($q=1$), and the weight two part ($q=2$) is the combination of the other $K_1(k)$ and the stuff coming from the Jacobian motive. There are integral ways of defining the weight two part: we have $H^{3,2}(C,\mathbb{Z})=CH^2(C,1)=\ker(K_1(C)\to K_1(k(C))$, and Bloch defined the group $V(C)=\ker(CH^2(C,1)\to K_1(k))$. This is therefore a group which encodes integrally the stuff in $K_1(C)$ coming from the Jacobian motive. This also is the part of $K_1(C)$ that is not yet understood, and subject to several conjectures. [The situation of smooth affine curves can be dealt with using the localization sequence, reducing to the projective curves. The analoguous thing to $V(C)$ in the case of smooth affine curves is $SK_1(k[C])=\ker(\det:K_1(k[C])\to K_1(k))$, which alternatively is the abelianization of $SL_\infty(k[C])$. That this group can be rather non-trivial is in the answer of Steven Landsburg.] Now I will outline some known results and open problems concerning the "Jacobian part" of $K_1$ of curves, the case distinction being in terms of the base field. Finite fields: Everything is known in the case of smooth curves over finite fields, cf. Theorems VI.6.4 and VI.6.7 of Weibel's $K$-book. The "Jacobian part" is trivial, so that $K_1(C)\cong\mathbb{F}_q^\times\oplus\mathbb{F}_q^\times$ for $C$ a smooth projective curve over $\mathbb{F}_q$. The localization sequence allows to deal with arbitrary smooth curves over $\mathbb{F}_q$. Global fields: For curves over number fields, there are conjectures of Vaserstein and Bloch for curves over number fields. Vaserstein's conjecture asks if $SK_1(C)$ is torsion if $C$ is a smooth affine curve, Bloch's conjecture asks if $V(C)$ is torsion for $C$ smooth projective. Both conjectures are equivalent by a result of Raskind: W. Raskind. On $K_1$ of curves over global fields. Math. Ann. 288 (1990) 179-193. You can find statements of the conjectures, references to the relevant papers and the proof of the equivalence in that paper. These conjectures state that the part of $K_1(C)$ coming from the Jacobian is torsion (and hence trivial over the algebraic closure of a global fields). Though the function field case is not discussed in this paper, probably similar statements would be consequences of Parshin's conjecture. Some further discussion on $K_1$ of elliptic curves over number fields can be found in the answers to this MO-question. As noted there, there is a recent paper of Kondo and Yasuda, in which they provide some computations of $K_1$ of elliptic curves over function fields. I think their computations would imply a complete calculation of $V(E)$ of an elliptic curve $E$ provided the Parshin conjecture is true. Generic fields: Finally, the statements outlined above say that $K_1$ of curves over "small fields" can be understood and is not too big (although this depends on serious conjectures). In general, this would appear to fail rather badly. There is a preprint of Rosenschon and Srinivas in which they show that for $k\subsetneq K$ an extension of algebraically closed fields of characteristic $0$ and $C$ a sufficiently generic smooth projective curve over $K$, the product map from K-theory $k^\times\otimes\operatorname{Pic}(C)\to CH^2(C,1)$ is injective, so that the Jacobian part of $K_1(C)$ is quite huge. Summing up, we are still quite far from actually understanding $K_1$ of curves. Nevertheless, we might hope that the structure of $K_1$ of curves can be considered an elementary question some day in the future. Concerning (2): To also say something about your question (2), the tame symbol appears as a map in a colimit of localization sequences for K-theory. Take the curve $C$, finitely many points as closed subscheme $Z$ and the open complement $U$ - this yields a localization sequence. Taking the limit over larger and larger closed subschemes $Z$ gives a localization sequence connecting the K-theory of the curve $C$, the K-theory of its function field $k(C)$, and the direct sum of K-theories of the residue fields of the closed points of $C$. The tame symbol is then the boundary map $K_2(k(C))\to\bigoplus_{\kappa\in C^{(1)}} \kappa(x)^\times$.<|endoftext|> TITLE: What is the proof-theoretic ordinal of PA + Con(PA), PA + Con(PA + Con(PA)) etc., and why? QUESTION [12 upvotes]: I seem to remember having read that the proof-theoretic ordinal (sup of ordinals the theory can prove well-ordered) of $\mathsf{PA} + \mathsf{Con}(\mathsf{PA})$ is the same as that of $\mathsf{PA}$, namely $\epsilon_0$. However, the one particular source where I remember reading something like this is The Realm of Ordinal Analysis by Michael Rathjen (bottom of p.9), but this is less unambiguous than I thought. Is this statement true? If so, do $\mathsf{PA} + \mathsf{Con}(\mathsf{PA} + \mathsf{Con}(\mathsf{PA}))$ etc. have the same proof-theoretic ordinal as well? What's a good reference for this? REPLY [16 votes]: This is an instance of a general phenomenon: adding true $\Pi_1$ sentences to a reasonable theory doesn't change its proof-theoretic ordinal. $Con(T)$ is $\Pi_1$, so if $T$ is any consistent theory, $PA+Con(T)$ still has proof-theoretic ordinal $\epsilon_0$. This is a standard fact in the area, though I'm not sure it's included in the usual sources on ordinal analysis. An easy way to see this is to notice that the usual cut-elimination proof for $PA$ goes through essentially unchanged $PA+\sigma$ whenever $\sigma$ is $\Pi_1$ and true. In general, $PA+\forall x\psi(x)$ doesn't have any new "computational" information when $\forall x\psi(x)$ is true. One would like to say that $PA+\forall x\psi(x)$ is conservative for "properly $\Pi_2$ formulas"---$\Pi_2$ formulas which aren't just $\Pi_1$ formulas in disguise. One way to make this precise is: suppose $PA+\forall x\psi(x)\vdash\forall x\exists y\phi(x,y)$. There's a corresponding computable function $f$ so that for each $x$, $f(x)$ is least such that $\phi(x,f(x))$ holds. Then $\forall x\exists y\phi(x,y)$ is equivalent to $tot(f)$, the formula $\forall x\exists y f(x)=y$. It need not be the case that $PA\vdash tot(f)$: if $PA\not\vdash\forall x\exists y\phi(x,y)$ and $\phi(x,y)$ is just $\psi(x)$ then $f$ is the function which checks whether $\psi(x)$ holds and outputs $0$ if so and runs forever if not. However there is always a related computable function $f'$ given as follows: on input $x$, if $f(x)$ terminates in $s$ steps and for each $i\leq s$, $\psi(i)$ holds, $f'(x)$ outputs $f(x)$; if there is an $s$ such that $\neg\psi(s)$ and $f(x)$ has not terminated in $s$ steps, $f'(x)$ outputs (say) $0$. Then $PA\vdash tot(f')$, and, since $\forall x\psi(x)$ is true, $\forall x f(x)=f'(x)$. In particular, this means that adding true $\Pi_1$ formulas doesn't prove the totality of any functions which grow more quickly than those provably total in $PA$. This is basically what we'd expect in a well-behaved theory: adding (true) universal formulas shouldn't give new existential information. (A poorly behaved theory could have an axiom like $\sigma\rightarrow\forall x\exists y\phi$, so this isn't true in general.)<|endoftext|> TITLE: 'Cohomologically approximating' a $\mathbb{Q}[[t]]$-scheme by a one over the henselization of $\mathbb{Q}[t]$? QUESTION [7 upvotes]: For certain matters the henselization $R$ of $\mathbb{Q}[t]$ at $0$ is a 'reasonable approximation' for $\mathbb{Q}[[t]]$ (Artin's approximation theorem and so on). Now, I would like to study certain (\'etale) cohomological properties of a finite type $\mathbb{Q}[[t]]$-scheme $S$; I would like to say that there exists an $R$-scheme $S'$ that is 'very similar to $S$' ('from the cohomological point of view'). What are the possible methods for doing so? I guess that I should embedd $S$ into a 'family' and then apply either Artin's approximation (so, this a sort of deformation), or smooth base change, or both. Yet I would certainly be deeply grateful for any hints (and references); in particular, where can I find an argument for presenting $S$ as a 'member of a family'? I can certainly say more on these cohomological issues; yet they are rather specific. They are somewhat related (yet more complicated) to the weight-monodromy conjecture; cf. http://www.jstor.org/stable/40067932 Upd. As we discussed in my previous question, $\mathbb{Q}[[t]]$ is the limit of its finitely generated subalgebras. Yet the corresponding etale pullback functors are not very nice, since $\mathbb{Q}[[t]]$ is not 'pro-smooth' over these subalgebras. So I suspect that 'approximating' $\mathbb{Q}[[t]]$ by $R$ is more convenient for my purposes. REPLY [6 votes]: For the last question, the standard approach is to use the results in EGA IV.8, in particular Theorem 8.8.2: there is a finitely generated $\mathbb{Q}$-subalgebra $A\subset\mathbb{Q}[[t]]$ and a scheme of finite type $S_0$ over $U=\mathrm{Spec}(A)$ such that $S\to \mathrm{Spec}\,\mathbb{Q}[[t]]$ is induced from $U$ by the natural base change $i:\mathrm{Spec}\,\mathbb{Q}[[t]]\to U$. You may impose on $S_0$ various properties of $S$ (see e.g. EGA IV, (8.10.5)). Then, as you say, one may apply Artin's (in fact, Greenberg's) approximation theorem to $i$, to get an $R$-point of $U$ arbitrarily close to $i$, in the $t$-adic sense. By base change you get $S'\to \mathrm{Spec}(R)$. It remains to use the étale cohomology toolbox (constructibility theorem?) to see that $S$ and $S'$ have "similar" cohomologies.<|endoftext|> TITLE: Why the term "monad" in homological algebra? QUESTION [8 upvotes]: Which is the origin and the reason for the choice of the term "monad" in the sense of homological algebra? Does this concept have any relation whatsoever to the "monads" from category theory? REPLY [5 votes]: A projective resolution "splits" into monads in the following way. Each exact complex of projectives $\dots \to P_{n+1} \to P_n \to P_{n-1} \to \dots $ is a gluing of monads $0\to Z_{n} \to P_n \to Z_{n-1} \to 0$, where $Z_{k} $ is the kernel of the map $P_{k} \to P_{k-1} $. So, a monad is an elementary piece of any projective resolution, like Leibniz's monads are the elementary pieces of more complicated entities (as far as I know).<|endoftext|> TITLE: Torsors and the fpqc topology QUESTION [6 upvotes]: Fix a scheme $S$, a group scheme $G/S$ (let us say smooth, maybe even affine with some finiteness conditions if you like), and suppose I have some other $S$-scheme $P$ with a right $G$-action. We want to investigate the question of whether $P$ is a $G$-torsor (in the fppf topology). If I understand correctly, then since $G$ is smooth it is equivalent for $P$ to be a torsor in the etale topology, which is a priori a relatively strong condition. On the other hand, an a priori weaker statement is that it be a torsor in the fpqc topology. My question is: suppose I can find an fpqc (but not finitely presented) cover $\{S_i \rightarrow S\}$ with $P\times_S S_i \rightarrow S_i$ isomorphic to the trivial $G\times_S S_i$-torsor. Are there circumstances under which I may conclude that $P$ is in fact an fppf torsor? One (perhaps silly) example is if I have $S$ equal to the spectrum of some small field $k$ with an interesting Galois group, and I am able to find some huge complicated transcendental extension $K/k$ over which $P$ becomes a trivial torsor. Under what conditions does this imply the existence of a finite separable extension over which $P$ becomes a trivial torsor? REPLY [14 votes]: One way to define an fppf (or etale, or...) torsor is to require that $G \times_S P \rightarrow P \times_S P$ given by $(g, p) \mapsto (gp, p)$ is an isomorphism and that $P$ has a section fppf (resp., etale, etc.) locally. The first condition can be checked even fpqc locally: use fpqc descent for the property 'an isomorphism'; in particular, it holds in your setup. The second condition is satisfied whenever $P \rightarrow S$ is itself fppf: this is one possible choice of an fppf cover over which $P$ acquires a section; this also holds in your setup, since $P$ is smooth. In fact, in your setup $P$ will be a torsor even for the etale topology due to the same argument: smooth morphisms have sections etale locally on the base, see [EGA IV, 17.16.3 (ii)].<|endoftext|> TITLE: Permutation Group Question QUESTION [6 upvotes]: A question about permutation groups: I wonder if someone who is expert in permutation group theory could answer the following question. Let $x \in S_n$ (the symmetric group) be an involution which is the product of k disjoint transpositions. For any permutation $y \in S_n$, let f(y) be the length of a shortest reduced word in $x, y, \mbox{ and } y^{-1}$ that equals the identity. Finally, let F(k,n) be the maximum of f(y) over all (non-involutions) $y \in S_n$ ($n >= 2k$). Question: Is there an easy way to determine F(k,n)? For (very) small k the evidence suggests something like F(k,n) = 12k. For k=1 this is easy to show by a case analysis, based on how the transposition intersects the cycles of y. Even for k=2 I found the problem harder than expected. Perhaps this has been studied before? Of course this could be restated as a question about the maximum girth of Type II, 3-regular Cayley graphs on $S_n$. REPLY [2 votes]: This is not a complete answer but maybe more an extended comment. First, it is not hard to find an upper bound that does not depend on $n$, as you claimed. Note that $x$ has support $2k$ and hence so does $x^y$. This implies that $xx^y$ has support at most $4k$ hence is contained in some copy of $S_{4k}$. Since the word $xx^y$ has length $4$, an upper bound on $F(n,k)$ is $4$ times the maximal order of an element in $S_{4k}$. This is not tight and can easily be improved. For example, we know that $xx^y$ is the product of two involutions, each of which is a product of $k$ disjoint transpositions. Let $g(k)$ be the maximal order of such an element. Then we have that $F(k,n)\leq 4g(k)$. It is easy to see that $g(1)=3$ hence you recover $F(1,n)\leq 12$. If I did not make a mistake, we have $g(2)=5$ (with an example given by $x=(12)(34)$ and $x^y=(13)(25)$ with product a $5$-cycle). This gives $F(2,n)\leq 20$, which shows that your conjecture of $12k$ is wrong. Finally, note that there are explicit upper bounds known on the maximal order of an element of $S_{4k}$, hence you can already get explicit upper bounds on $F(n,k)$ as a function of $k$. These are probably not tight and the best way to improve them would probably to obtain upper bounds on $g(k)$.<|endoftext|> TITLE: subsets of groups which have to be closed no matter what QUESTION [12 upvotes]: One example of a subset of a group $G$ which has to be closed in any topology on $G$ compatible with the group operations is a centraliser. Are there any other interesting examples? REPLY [25 votes]: Subsets of a group that are closed with respect to any Hausdorff group topology are called unconditionally closed. Clearly, all algebraic sets are unconditionally closed, where a subset of a group $G$ is called algebraic if it is an intersection of finite unions of the sets of solutions to some equations with coefficient from $G$. A.A.Markov proved that for countable groups the converse is also true: $$ \hbox{unconditionally closed = algebraic}. $$ For uncountable groups, this is not always the case as follows from works of S. Shelah (under CH) and G. Hesse.<|endoftext|> TITLE: Mapping class group vs automorphism group in cobordism category QUESTION [10 upvotes]: Let $3Cob$ be the category whose objects are closed surfaces and whose morphisms are diffeomorphism classes of cobordisms. By sending a diffeomorphism $\phi$ of a surface $X$ to its associated cobordism cylinder $M_\phi$ we get a homomorphism $$ M : \pi_0 (Diff(X)) \rightarrow Aut_{3Cob}(X) $$ which in general dimensions need not be injective or surjective (see eg. Dan Freed's lecture notes). My understanding is that $M$ is indeed injective and surjective for $3Cob$. Is this true, and what is the easiest way to prove it? As a follow-up question, what is known about the group $Aut_{nCob}(Y)$ for a closed $(n-1)$-manifold $Y$ for general $n$? Can it be expressed in terms of a more "traditional" automorphism group of $Y$? If not, does passing to the $\infty$-category setting give this question a nicer answer? REPLY [5 votes]: Injectivity is not hard to prove: (I'm quoting an email from Bruce.) Let f,g be pseudo-isotopic diffeomorphisms of a surface X. Then they are homotopic. Therefore they are isotopic. (see eg. Farb and Margulit, A primer on mapping class groups, pg 43). Here is how you prove surjectivity; this amounts to proving that an invertible cobordism is a product (respecting all identifications). Let's call your surface F, so we have cobordisms A and B with $A\cup_F B = F \times I$ where $\partial A = F \cup F\times\{0\}$. (I am ignoring the actual identification maps you need to keep track of things in your category, but the argument would work as stated with more notation and care.) Then I claim that both A and B are product cobordisms. To see this, note that the projection $F \times I\to F \times \{0\} $ gives a map $F \to A \to F \times \{0\}$. The degree of this map is $1$, and so a standard covering space argument shows that the induced map $\pi_1(F) \to \pi_1(F \times \{0\})$ is surjective. By the Hopfian property of surface groups, it is an isomorphism. The same argument applies to the maps induced by including F into B. Now van Kampen's theorem implies that the maps from F to A and B are isomorphisms on $\pi_1$, and so it follows that these are h-cobordisms. Since $F\times I$ embeds in $R^3$, it is irreducible (you can avoid this step by quoting the solution of the Poincaré conjecture, but that's overkill) and so work of Stallings (On fibering certain 3-manifolds. 1962 Topology of 3-manifolds and related topics (Proc. The Univ. of Georgia Institute, 1961) pp. 95–100) implies that they are both products. I believe that the proof of Stallings' theorem would actually give you a product respecting all identifications so that the resulting isotopy induces the given automorphism. You can also get this by Waldhausen's general theory of Haken manifolds (On irreducible 3-manifolds which are sufficiently large. Ann. of Math. (2) 87 1968 56–88), but Stallings' proof in this special case is easier. Either proof of the product theorem would essentially proceed by finding a hierarchy for A and B that looks like you took a system of curves and arcs cutting F into a disk, and crossed with I. Since you know that $A\cup_F B = F \times I$, you can probably short-cut this argument by starting with such a hierarchy for $F \times I$, consisting of cylinders and disks. Then you would use standard 3-manifold arguments to say that these cylinders and disks can be assumed to hit F in circles and arcs in the same pattern as they do in $F \times \{0\}$. This (and irreducibility) would imply that the intersection of the cylinders and disks cuts each of A and B up into a ball, from which you get the product structure using Alexander's theorem.<|endoftext|> TITLE: Push-outs of fully faithful (enriched) functors QUESTION [5 upvotes]: I'd like to know a reference where the following property, that I believe to be true, is checked: given a diagram of categories and functors $B\leftarrow A\rightarrow C$ (I'm actually interested in the enriched case, over a closed symmetric monoidal category), if $A\rightarrow C$ is fully faithful then so is the functor to the (strict) push-out $B\rightarrow B\cup_A C$. I should say that I have been able to check this in some specific cases that I needed, but a reference would help me to reduce a paper in four pages ;-) REPLY [7 votes]: A proof of this fact is given in Proposition 3.1 of Alexandru Stanculescu, Constructing model categories with prescribed fibrant objects, arXiv:1208.6005. He attributes this result to an earlier paper from the 80's by Fritsch and Latch.<|endoftext|> TITLE: A question on the morphism between Hilbert schemes QUESTION [5 upvotes]: Let $L_1,L_2$ be two irreducible component of two different Hilbert schemes parametrizing closed subscheme in $\mathbb{P}^n$ and $\mathbb{P}^{n-1}$, respectively. Denote by $\pi_1: \mathcal{X}_1 \to L_1,\pi_2: \mathcal{X}_2 \to L_2$ the corresponding universal families of closed subschemes. Assume that there exists a hyperplane in $\mathbb{P}^n$, say $H$ and an open set $U$ in $L_1$ such that for all $u \in U$, $\pi_1^{-1}(u)$ intersect $H$ transversally and the intersection $\pi_1^{-1}(u).H$ is a fiber of $\pi_2$. Hence gives a set theoretic map from $U$ to $L_2$. The question is whether this induces a morphism of schemes (from $U$ to $L_2$)? If not, is it known under what conditions this could hold true? Note that the intersection $\pi_1^{-1}(u).H$ gives only a subscheme in $\mathbb{P}^n$ but there is a natural identification of this with a subscheme in $\mathbb{P}^{n-1}$ obtained by the fibered product $\mathbb{P}^{n-1} \times_{\mathbb{P}^n} \pi_1^{-1}(u)$, where the morphism from $\mathbb{P}^{n-1}$ to $\mathbb{P}^n$ is a closed immersion induced by the vanishing of $H$. EDIT: Assume if necessary that $U, L_2$ are smooth. REPLY [3 votes]: The Hilbert scheme is universal for flat families of closed subschemes of $\mathbb P^{n-1}$. So if $\mathcal X_1(U) \cap H$ is flat over $U$, then we obtain a map $U \to Hilb( \mathbb P^{n-1})$ by the universaL property. Since all the points land in the irreducible component $L_2$, the map factors through $L_2$. A hypersurface in a flat family is flat as long as its equation is not a zero-divisor in any fiber. Because the intersection is transverse, this condition is satisfied.<|endoftext|> TITLE: Is a fair lottery possible? QUESTION [16 upvotes]: I'm trying to come up with a scheme for a lottery where each individual has roughly the same chance of becoming the winner, regardless of the number of tickets one holds. So no individual should have an incentive to buy a second ticket, but even if they decide to do so, it should not gain them any significant advantage over any of the other players. Ofcourse, the simplest solution would be to just refuse to sell them multiple tickets, but since all players are anonymous, I cannot differentiate between ticketholders and tickets, so that's impossible. A working solution would be to increase the house-edge exponentially (or decrease the jackpot) as soon as a new ticket is sold, so that buying another one lowers the expected profit more than is gained by having an extra ticket. No rational player would want do that, unless they have no previous tickets. But the problem with that is that only a very small amount of tickets can be sold untill the house has 100% chance of winning, so in practice nobody will ever win. Is there a better formula possible? By dynamicly changing parameters like the ticket-price, jackpot, or win-chance with every new ticket? The lottery does not have to be economicly feasible, its no problem if the organizer always makes a loss from it for example. I have a feeling I'm asking for the impossible, but since the proposal above would work fine for very small groups, there should be a formula which works in larger groups too. REPLY [6 votes]: With the various restrictions in place the answer to the question as asked seems to me to be 'no, there is no such system'. It was said there is one prize to be distributed (one person to be selected) and it was also said the tickets could even be free. And, the lottery should be fair, so that different individuals have essentially the same chance to win they holding one or more ticktets regardless. And, of course the key point that one cannot distinguish tickets from individuals. However, the prize can be adjusted in the process. So there are tickets $\{1, \dots, n\}$ distributed, and there is a partition of the set $\{1,\dots, n\}$ into disjoint nonempty sets $I_1, \dots, I_k$ corresponding to the tickets one individual holds. As described the only thing the organizer of the lottery knows is $n$. But, they have no control over the $I_j$. It is impossible to assign winning propabilties $p(i)$ to the different tickets so that $\sum_{i \in I_j}p(i)$, the probability individual $j$ will win, is about equal for all $j$ and each possible choice of $I_j$, except $p(i)= 0$ or $n\le 2$. Now, the above assumes that tickets are distributed and then some 'drawing' is done possibly following some complicated system. If one does not insist on this there are of course ways to assign in a in some sense fair way one prize in a way that it is pointless or even bad to get more than one ticket (even if it is free). The "unique is the winner" was already mentioned, one could also do "first come, first served" (the first ticket sold always wins). Finally, in OP and in some answers something else is discussed, which in my opinion is not really "fair" in the sense of the question. Merely, these would be systems to make it unreasonable to get more than one ticket but still if somebody would they would be in some sense better of than the others in the same game. Yet, yes, I think there one could (in a certain formalization) show that one needs at least exponential decrease of expected win to make this work.<|endoftext|> TITLE: $S$-Tate-Shafarevich groups of elliptic curves QUESTION [13 upvotes]: Let $S$ be a finite set of places of a number field $k$ and let $E$ be an elliptic curve over $k$. Define the ''$S$-Tate-Shafarevich group" of $E$ to be $$Ш(E,S) = \ker\left(H^1(k,E) \to \prod_{v \not \in S}H^1(k_v,E_v)\right).$$ Note that the normal Tate-Shafarevich group is $Ш(E) = Ш(E,\emptyset)$. Recall that the Tate-Shafarevich conjecture states that $Ш(E)$ is finite. My question concerns the corresponding problem for $Ш(E,S)$. Is $Ш(E,S)$ conjecturally finite? If the answer to this is yes, then an obvious next question is whether this is equivalent to finiteness of $Ш(E)$, i.e. Is it known that $Ш(E) \subset Ш(E,S)$ has finite index? REPLY [7 votes]: Global duality (as for instance on page 29 of Rubin's "Euler systems") gives you a description of the cokernel of your inclusion. Let $p$ be a prime. Then the $p$-primary part of the quotient of $Ш(E,S)$ by $Ш(E)$ is dual to the cokernel of the map $$ \mathfrak S_p(E/k) = \varprojlim_n \,\mathrm{Sel}_{p^n}(E/k) \ \to \ \bigoplus_{v\in S} E(k_v)^{*}.$$ Here the source is the compact $p$-adic Selmer group of $E/k$, which is a finitely generated $\mathbb{Z}_p$-module of rank $r$, believed to be the rank of $E(k)$. The target is the sum of the $p$-adic completions of the local points $E(k_v)$. If $v$ is not above $p$, then this group is finite and otherwise it is a finitely generated $\mathbb{Z}_p$-module of rank $[k_v:\mathbb{Q}_p]$. One can make quite precise conjectures as to what the corank of the $p$-primary part of $Ш(E,S)$ should be in terms of the rank of $E(k)$ and $k/\mathbb{Q}$ by the above duality. Roughly speaking some sort of independence of elliptic logarithms tells you that the above map should have image as large as possible. For instance over $k=\mathbb{Q}$, it is clear that the corank of the $p$-primary part of $Ш(E,S)$ will be $0$ is $p$ if is not in $S$ or if $r\geq 1$, and $1$ otherwise. For larger $k$ it is a bit harder to formulate as it will depend on the field of definition of $E$, but generally speaking it tends to be large as soon as a $p$-adic place of large degree is in $S$.<|endoftext|> TITLE: Binomial Identity QUESTION [8 upvotes]: I recently noted that $$\sum_{k=0}^{n/2} \left(-\frac{1}{3}\right)^k\binom{n+k}{k}\binom{2n+1-k}{n+1+k}=3^n$$ Is this a known binomial identity? Any proof or reference? REPLY [3 votes]: This is a response to Joe Alfano's comment. Define the two discrete functions $$F(n,k):=(-1)^k3^{n-k}\binom{n+k}k\binom{2n-k-2}{n-2k} \qquad \text{and} \qquad G(n,k):=(-1)^{k-1}3^{n+1-k}\binom{n+k-1}{k-1}\binom{2n-k-1}{n-2k}.$$ The proof for the identity $\sum_{k\geq0}F(n,k)=0$ follows from $$F(n,k)=G(n,k+1)-G(n,k).\tag1$$ Equation (1) can be verified say up on dividing through by $F(n,k)$ and simplifying. The rest comes from summing (1) over all integers $k$ and where the convention $\binom{a}b=0$ (if $b>a$ or $b<0$) is brought to bear. In this regard, the RHS of (1) telescopes $\sum_kG(n,k+1)-\sum_kG(n,k)=0$. Hence $\sum_{k\in\mathbb{Z}}F(n,k)=\sum_{k=0}^{\lfloor\frac{n}2\rfloor}F(n,k)=0$, as desired. ================= Now, back to the original problem from the OP. Define the functions $$A(n,k):=\frac{(-1)^k}{3^{n+k}}\binom{n+k}k\binom{2n+1-k}{n-2k}, \qquad \text{and} \qquad B(n,k):=\frac{(-1)^{k-1}}{3^{n+k}}\binom{n+k}{k-1}\binom{2n+2-k}{n+1-2k}.$$ Again, one checks $A(n+1,k)-A(n,k)=B(n,k+1)-B(n,k)$ and then sum over the integers $k$. The outcome is $\sum_kA(n+1,k)-\sum_kA(n,k)=0$. That is to say, the sum $\sum_kF(n,k)=\sum_{k=0}^{n/2}\frac{(-1)^k}{3^{n+k}}\binom{n+k}k\binom{2n+1-k}{n-2k}$ is identically a constant. A quick check at $n=0$ reveal that the constant is $1$. The proof ends here.<|endoftext|> TITLE: Shelah's proof that proper forcing preserves P-points QUESTION [5 upvotes]: In Proper and Improper forcing, VI.5; Claim 5.1 part 1 is the following: If $F$ is a P-point in $V$, $P$ is a proper forcing notion and $\Vdash_P `` F$ generates an ultrafilter" Then the ultrafilter generated by $F$ is a P-point in $V^P$. The proof starts like this: let $p\Vdash \{\tilde A_n : n<\omega\}$ be such that $p\Vdash \tilde A_n\in F$. By properness, for some $q\geq p$ and $A_{n,m}\in F$ for $n,m<\omega$ we have $q\Vdash \tilde A_n\in\{A_{n,m} : m <\omega\}$ To finish the proof, he uses that $F$ is a P-point in V, and finds an almost-subset of all the $A_{n,m}$. My question is how is the bolded statement true? It appears (since each $A_{n,m}\in V$) that it would follow that P cannot add any subsets of $\omega$, which is clearly not true. What am I missing? EDIT: Thanks Mohammad Golshani for the comment. I now believe that the bolded statement can be true. But how do we get $q$ (and the collection of $A_{n,m}$)? Another question is once we have $q$, why is $q$ in $G$? Is the collection of such $q$ dense below $p$? REPLY [5 votes]: In more details: Let $N$ be a countable elementary submodel of some sufficiently large $H(\chi)$ containing $p$, $P$, $F$, and the sequence of names $\{\dot A_n:n<\omega\}$. Then the countable set $N\cap F$ works for the collection of $A_{n, m}$, and any $(N,P)$-generic extension of $p$ will work for $q$.<|endoftext|> TITLE: Are superstrong stronger than strongly compact cardinals? (or vice versa) QUESTION [8 upvotes]: In the last part of Kanamori's excellent "The Higher Infinite" there is a small diagram about the strength and consistency strength of some major large cardinal axioms. Below supercompact cardinals there are two then-incomparable cardinals. Superstrong cardinals and strongly compact cardinals, both are stronger than Woodin cardinals Has there been any progress on this problem since then? If yes, is there some reference? If not, can someone give basic outline as to why this problem is difficult? (I am somewhat familiar with the problem of extender models for very large cardinals, so an concise similarities if they exist would suffice). REPLY [6 votes]: In the comments, you ask Are strongly compact cardinals superstrong? The answer is no, not necessarily, and indeed, strongly compact cardinals need not even be strong, nor even a little bit strong. The reason is that it is relatively consistent that the least strongly compact cardinal is the same as the least measurable cardinal, and in this case such a cardinal $\kappa$ will not even be $(\kappa+2)$-strong. This was the first instance of the so-called "identity crises" phenomenon, discovered by Magidor, and similar phenomenon have now been uncovered for many other cardinals (for example, the least weakly compact cardinal can be unfoldable, weakly measurable and nearly $\theta$-supercompact). That is for outright implication, but meanwhile, for consistency strength, it is a different story. As far as we know, strongly compact and supercompact cardinals have the same consistency strength. Here, the relevant observation is: Theorem. If $\kappa$ is $2^\kappa$-supercompact, witnessed by $j:V\to M$, then $\kappa$ is superstrong in $M$. This is the reason that supercompact cardinals are strictly stronger than superstrong cardinals in consistency strength. Proof: The map $j\upharpoonright V_{\kappa+1}:V_{\kappa+1}\to M_{j(\kappa)+1}$ has size $2^\kappa$ and is therefore inside $M$. Using this part of $j$ to define an extender $E$, the derived extender of $j$, the model $M$ can produce an embedding $j_E:M\to N$, which agrees with $j$ on $V_{\kappa+1}$. In particular, this means $M_{j(\kappa)+1}\subset N$, which means that $E$ witnesses that $\kappa$ is superstrong from the perpsective of $M$. QED Perhaps one of the inner model theorists will post an answer explaining how much of the inner model theory one can undertake from a strongly compact cardinal, and I suppose that is really what you are asking.<|endoftext|> TITLE: Are there proofs of Rice Theorem without using the undecidability of some problem? QUESTION [6 upvotes]: Most proofs of Rice theorem seem to be based on the undecidability of the halting problem. They are "reduction-based". Are there "direct" elementary proofs, perhaps based on diagonalization? I think that the answer is "YES there are". However most textbooks I know (such as [Odifreddi, Moret, Jones, Phillips] present reduction-based proofs. REPLY [4 votes]: Here is a proof in a more 'constructive' form that can be directly applied to any real-world programming language. Just to make clear, a property of programs must be a predicate that only depends on the halting and output behaviour, in other words two equivalent programs will have exactly the same properties. I will use $X \equiv Y$ to denote that two programs are equivalent, and so $P$ is a property iff $P$ is a predicate and $P(X)=P(Y)$ for any programs $X \equiv Y$. For any property $P$ of programs such that $P(T)$ and $\neg P(F)$ for some programs $T$,$F$:   If $P$ is decidable:     Let $D$ be a program that decides $P$     Let $M$ be the following program on input $x$:       Create $N$ to be the following program on input $y$:         If $x$ is an invalid program: Return "" [may be unnecessary in some languages]         Return $x(x)(y)$ [using closures or some interpreter written in the language itself]       If D(N): Return F       Return T     Let $R$ be the program that $M(M)$ creates in the variable $N$     Clearly $R \equiv M(M)$     Also $M(M)$ halts because $D$ halts on all inputs     Thus $M(M) \equiv T \vee M(M) \equiv F$     $P(M(M)) \Leftrightarrow P(R) \Leftrightarrow D(R) = True \Leftrightarrow M(M) \equiv F \Leftrightarrow \neg P(M(M))$     Contradiction   Therefore $P$ is not decidable<|endoftext|> TITLE: "Rice (like) Theorem" for primitive recursive functions? QUESTION [8 upvotes]: As primitive recursive (PR) functions seem to be so important (see for instance Kleene normal form Theorem) we may expect that many decision questions related to PR functions are undecidable. (INPUT) A description of a PR function f (say by the usual definition or by a LOOP program), (QUESTION) Does the (mathematical) function f have the property P? perhaps conjecture something like: *Any non trivial (universally or existentially) quantified property of PR functions is undecidable. Note. Example of such a nontrivial, quantified property $P:\;\;\forall n [ (f(n)=0) \vee (f(n)>=5) ]$ Answer to "Is it clear that your example property is undecidable?" – Noah S 13 mins ago Given a TM $M$ with index $e$ define f(n) that outputs 1 if the computation $M(0)$ halted in exactly $n$ steps, and outputs 0 otherwise. There is a PR such function $f$; P(f) holds iff the computation $M(0)$ does not halt. REPLY [5 votes]: It happens that there do exist non-trivial universal properties that are decidable. An example of such a property is expressed in terms of what we could call ``primitive recursive Kolmogorov complexity''. Definition. If $v=(v_0,\ldots,v_{n-1})$ is a sequence of natural numbers then let $K_{pr}(v)$ be the size of a shortest LOOP program computing a function $f$ extending $v$, i.e. satisfying $f(0)=v_0,\ldots,f(n-1)=v_{n-1}$. Unlike the usual notions of Kolmogorov complexity, $K_{pr}$ is computable. However it is not primitive recursive. For a function $f$, let $f|n$ be the finite sequence $(f(0),\ldots,f(n-1))$. Claim. The property $$\forall n, K_{pr}(f|n)\leq n$$ is decidable, given a LOOP program for $f$. Proof. Given a loop program $p$ for $f$, one has $K_{pr}(f|n)\leq |p|$ for all $n$. In order to check the property, one can only look at $n<|p|$. As $K_{pr}$ is computable, the property is decidable. Observe that the property is not decidable if one is only given $f$ as oracle, as no finite prefix of $f$ is sufficient to ensure the property: for each finite sequence $v=(v_0,\ldots,v_{n-1})$ there is $v_n$ such that $K_{pr}(v_0,\ldots,v_n)>n+1$ hence no extension of $(v_0,\ldots,v_n)$ satisfies the property (the property is a closed subset of the Baire space that has empty interior). More generally and for the same reasons, if $h:\mathbb{N}\to\mathbb{N}$ is a computable non-decreasing unbounded function then the property $P_h$ defined by $$f\in P_h\iff\forall n, K_{pr}(f|n)\leq h(n)$$ is decidable given a LOOP program for $f$ but not given $f$ as oracle. If $h(1)$ is sufficiently large then $P_h$ is non-empty as it contains all the functions computed by LOOP programs of size $\leq h(1)$. An analog of Rice and Rice-Shapiro theorem. So far, we know some basic properties that are decidable: extending a finite sequence $v$ (decidable given $f$), the property $P_h$ of having $h$-compressible prefixes (decidable given a LOOP program). Now, there is an analog of Rice and Rice-Shapiro theorems, stating that they form a ``subbasis'' (as in topology) of the semi-decidable properties: every semi-decidable property can be obtained as a union of finite intersections of these simple properties. Theorem. Let $P$ be a property of primitive recursive function. The following are equivalent: $f\in P$ is semi-decidable given a LOOP program for $f$, $P$ is a computable disjunction of properties of the form $$ f\text{ extends $v$ and }\forall n, K_{pr}(f|n)\leq h(n). $$ The result is more general as it applies to any class of total computable functions that can be computably enumerated (for instance the polynomial-time computable functions, the provably total computable functions, etc.). All this can be found in the paper http://arxiv.org/abs/1503.05025.<|endoftext|> TITLE: Separating unit disks by lines QUESTION [7 upvotes]: Given $n\ge 2$. For a real $d>2$, consider a constellation $C$ of $2n$ disks of radius $1$ in the plane such that $h(C)$, the minimal distance between any two of their centers, is equal to $d$. Let $d_n$ be the smallest $d$ such that any such constellation is separable, i.e. that there is a line which has $n$ disks on either side of it. We'll call such lines separating lines. They may touch some disks but not intersect them. (It is easy to see that such a $d$ and thus $d_n$ exists, by using a different scaling: Choose $2n$ points as centers, start with very small disks and increase the common radius of the disks as much as possible without destroying the separability. Then rescale to radius 1.) Denote by $\mathcal C_n$ the set of all separable constellations of $2n$ unit disks. Let's call a $C\in\mathcal C_n$ tight if each separating line touches at least three of the disks. Obviously, an extremal $C\in\mathcal C_n$ (i.e. such that $h(C)=d_n$) must be tight, and so we have the equivalent definition $d_n=\max \{h(C): C\in\mathcal C_n \text{ is tight}\}$. A tight constellation must have a somewhat high degree of symmetry, presumably a $s$-fold rotational symmetry for some $s\ge3$. If there is a disk at the center (and thus $s$ odd), $C$ seems to be much "better" (i.e. closer to extremal) than if not. A trivial lower bound for $n\ge 4$ is $d_n\ge8\cos\frac\pi{4n-2}$, attained for a constellation $C$ that has one disk at the center and the others forming a regular $(2n-1)$-gon around it. For $n=2,3$, the same construction is extremal and yields $d_n=2\csc\frac\pi{4n-2}$ (note that for $n\le 3$, $h(C)$ is the circumradius of the $(2n-1)$-gon, whereas for $n\ge4$ it's its side). The image shows the $n=3$ case with $h(C)\approx 6.472$ and the $n=8$ case with $h(C)\approx 7.956$. I'd conjecture that if $2n-1$ is prime, this is best possible. So: Is it true that $\liminf\limits_{n\to\infty}\ d_n=8$? Now there are other tight constellations, which may be better for certain values of $n$, e.g. for $n=8$ the following, where 2 of the 10 separating lines are also drawn: I haven't calculated this one exactly, but $h(C)$ appears to be slightly bigger than $8$. Note that for each $n$ there is an increasing sequence $d_n2$, the search for extremal constellations is much more tricky than for $k=2$, but I think there are good reasons that again, those must have an $s$-fold symmetry. By this type of construction, it is in fact not hard to show that $\liminf\limits_{n\to\infty}\ d_{4,n}\ge 16$ (and conjecturing that for $4n-1$ prime this is again the best possible constellation would mean that equality holds). For even $k=6,8,...$, $\liminf\limits_{n\to\infty}\ d_{k,n}$ does not seem to increase further, it rather seems to stick at $16$. Very strange. May there be then a universal bound for $d_{k,n}$? Note that similar constructions for odd $k$ yield much smaller $h(C)$'s, as no separating line touches the central disk, which is where the situation is most tight. For fixed $k$, what can be said about $\limsup\limits_{n\to\infty}\ d_{k,n}$? REPLY [5 votes]: An earlier version of this answer used a uniformly random point process but that isn't quite good enough (it will lead to a small number of close-together pairs). Here's a patched version using a more complicated random process. Choose $r=c_1\frac{n}{\log n}$ and $d=c_2\frac{\sqrt{n}}{\log n}$ for constants $c_1$ and $c_2$ to be determined later. We will choose $2n$ points from a disk of radius $r$, centered at the origin, by repeating the following process $n$ times: find a point $p$ within the disk that is at distance at least $d$ from all other points already chosen and from the origin, and add both $p$ and $-p$ to the set. Any bisector of this set will have to pass near the origin, so to show that the point set is inseparable we need only show that (with high probability) each strip of unit width through the origin is covered by at least one point. We can find a family of $O(r)$ strips of constant width such that each unit-width strip contains one of the members of this family. Each of the strips in the family has an area that is at least a $\frac{\log n}{n}$ fraction of the area of the whole disk. By adjusting $c_1$ and $c_2$ appropriately we can ensure that the area of the $r$-disk covered by $d$-disks around chosen points is less than half the area of the whole $r$-disk, and therefore that at each step one of these strips has probability proportional to $\frac{\log n}{n}$ of being hit. Therefore, the probability that it is hit at least once throughout the whole process is $1-1/n^{c_3}$ where $c_3$ can be made arbitrarily large by adjusting $c_1$ and $c_2$. By making $c_3$ larger than one, we ensure that with high probability all the strips are hit and therefore that the point set is inseparable. Thus, $d_n$ can grow at least proportionally to $\frac{\sqrt n}{\log n}$, much larger than a constant.<|endoftext|> TITLE: Resource for learning quantum mechanics from the viewpoint of representation theory QUESTION [11 upvotes]: Quantum mechanics is deeply connected with representation theory. Therefore, I'm looking for a textbook or article which presents quantum mechanics in a representation theoretic manner. Could anyone provide me a reference to such a resource? I don't care if it's a math or physics book as long as it makes use of the language of representation theory mathematicians are familiar with. REPLY [2 votes]: There is an upcoming book of Peter Woit, available for download if you click on the link. From the Preface: These are the course notes prepared for a class taught at Columbia during the 2012-13 academic year. The intent was to cover the basics of quantum mechanics, from a point of view emphasizing the role of unitary representations of Lie groups in the foundations of the subject. [..] I just noticed that these lecture notes were also mentioned in the comments.<|endoftext|> TITLE: Very Large Cardinal Axioms and Continuum Hypothesis QUESTION [10 upvotes]: Are very large cardinal axioms like $I_0$, $I_1$, $I_2$ consistent with $CH$ and $GCH$? REPLY [2 votes]: I just want to make one simple observation, which I think got lost in the history of the subject. Unfortunately, I do not speak German and so it will be nice if someone who knows German actually reads the sources (and corrects me if I am wrong). I believe Cantor's original conjecture was that any infinite subset of reals either has the same size as the natural numbers or the same size as the real numbers. Under AC this becomes the most cited version of CH, namely that $2^\omega=\omega_1$. However, I believe it is a mistake to interpret Cantor's conjecture under ZFC. To me, one gets the feeling that Cantor's original CH is true by simply playing with actual sets, he himself showed that every closed set satisfies his original CH. And so the proper interpretation of Cantor's original CH under ZFC should be that every definable subset of reals is either countable or has the same size as the reals. Of course, "definable" is open to interpretation as well, but large cardinals did settle a lot of these. For example, under large cardinals, there is no counterexample to Cantor's original CH among projective sets, sets in $L(\mathbb R)$, etc. In this sense, Goedel was not wrong at all when he suggested that large cardinals may help settle the continuum hypothesis. In fact, far from it. The large cardinals freeze so much of the definable part of the universe that it is astonishing how right Goedel was. For example, Woodin showed that under large cardinals the theory of Chang model cannot be changed, he also showed that after collapsing the power set of the supercompact, the theory of $L(\Gamma_{uB},\mathbb R)$ cannot be changed, all of these is what Goedel anticipated and it is, at its roots, motivated by his beliefs.<|endoftext|> TITLE: Why aren't fields called "bodies" instead? QUESTION [44 upvotes]: The discrepancy regarding the names of commutative division algebras in German and English has always startled me. In English they are called fields, whereas their original German name is Körper (hence the $K$), a word which usually means "body" in everyday language. Nearly all other English names of algebraic objects are direct translations of the corresponding German terms (or maybe also the other way round), so I'm wondering which historic development lead to this divergence in naming conventions. By looking up the Wikipedia article on fields in the respective languages, I found out that most languages seem to have adopted the German name. For example a field is called corps commutatif in French, cuerpo in Spanish, Σώμα in Greek, corpus in Latin, Ciało in Polish, kropp in Swedish and Norwegian. However there also seem to be some languages where the name for fields translates to "field", such as Russian (Поле) and Italian (campo). Dutch seems to assume a special role, since the term used in the Netherlands seems to be lichaam, whereas the Belgians call it veld. Does anyone know how this strange division in naming conventions did evolve? REPLY [16 votes]: It might be worth noting that the word "Feld" is also (sometimes) used in German, at least in the compound "Galoisfeld" (for "Galois field"). And, this is not some modern-day re-translation from English but classical usage (of Witt for example), see the discussion in comments of https://mathoverflow.net/a/18638/9072 for references. In my opinion this provides additional support for anon's answer that quite at the start there were two different names and just in the one language the one 'won' whereas in the other one the other 'won' the 'competition'. Regarding other theories and in general, it might be worth noting that "Körper" does not only mean body, like in human-body, but is also used to designate certain structures/organizations/groups, like "Justizkörper" (comprising judges, attorneys, and so on). And, in that meaning "Körper" fits a lot better with "Gruppe" (group) and also with "Ring" (also used as name for organizations, like in English) and also "Verband" (again used to designate organizations, the math meaning being lattice, in the order sense, providing another example where it is not a translation). See https://mathoverflow.net/a/117354/9072 for some discussion. So that "Körper" is really not all that sensual in that context but quite corporate (which ultimately derives from the same source, I think).<|endoftext|> TITLE: Algebraic equations for modular parameterizations QUESTION [7 upvotes]: I was wondering if there some place where for some small $N$ I can find explicit modular parameterizations in an algebraic way. One type of model for $X_0(N)$ is just given by a single algebraic relation $\phi(j,j')$ between $j(\tau)$ and $j(N\tau)$. So given an elliptic curve $E$ of conductor $N$ I would really like to find $X, Y \in \mathbb Q(j)[j']/\phi$ such that $\mathbb Q(X,Y) \cong \mathbb Q(E)$. Explicit parameterizations by other models of $X_0(N)$, for example using the weber f function instead of j, would also work for me. And in the light of the remark of David Loeffler below maybe even better. As long as there is a know expression for the $j$-invariant on that model. I'm especially interested in the case where $N = 121$. REPLY [2 votes]: There's a similar question on MO at: Where can I find a comprehensive list of equations for small genus modular curves? You might also check out Y. Yang, "Defining equations of modular curves," Advances in Mathematics 204 (2006), 481-508. Yang surveys the problem and then gives a method for finding defining equations with small coefficients by using explicit properties of some generalized Dedekind $\eta$ functions. He also gives lists of equations for $X_0(N)$ for $N \leq 50$. (So not $N=121$, but the techniques should work.) In his examples the coefficients are usually no more than $2$ digits.<|endoftext|> TITLE: PFA: A New Godel's Program & A New Large Cardinal Ladder (Updated) QUESTION [6 upvotes]: We know $PFA$ implies $2^{\aleph_0}=\aleph_2$. Q1. What does $PFA$ say about other values of continuum function? Does proper forcing axiom carry any further information about values of continuum function in other cardinals greater than $\aleph_{0}$? In the other words, is there any other known non-trivial result in the form "$ZFC+PFA\vdash 2^{\aleph_{\alpha}}=\aleph_{\beta}$" for some ordinals $\alpha , \beta$, or all other situations are consistent with $PFA$ like Easton's theorem in presence of some large cardinal? In the direction of my previous question on Godel's program for deciding $CH$ and $GCH$ using adding large cardinal axioms to $ZFC$ and based on the impact of $PFA$ on $CH$ which is compatible with Godel's conjecture on refuting $CH$ and $GCH$ using large cardinal axioms. One can consider existence of another large cardinal tree in the shadow of the current standard tree of large cardinal assumptions including $PFA$ in a rank near supercompacts. (The equiconsistency of PFA and supercompacts is not proved yet but we assume it for straightforwardness of the below diagram.) Maybe Godel's conjecture and program are true if we replace the standard tree of large cardinals with another tree with equiconsistent steps. Please note the following imaginary large cardinal ladder which includes two parallel trees of "large cardinal" assumptions: Q2. Can we complete the middle part of the following ladder by some axioms like $A$, $B$, $C$, ..., $X$ which are equiconsistent with usual large cardinal axioms and also refute $CH$, $GCH$ and $V=L$ in direction of Godel's program? Update. Using above approach to large cardinal axioms (inspired by success of $PFA$ in deciding $CH$ and its possible equiconsistency with existence of supercompact cardinals) we can restate Godel's program in deciding independent statements of set theory as follows: For every independent statement $\sigma$ from $ZFC$ there is a (standard) large cardinal axiom $I$ and a statement $J$ (a non-standard large cardinal axiom) such that: (a) Consistency of $ZFC+I$ implies consistency of $ZFC+J$. (b) $ZFC+J$ decides $\sigma$ (i.e. $ZFC+J\vdash \sigma$ or $ZFC+J\vdash \neg\sigma$) In the other words one can decide any independent statement of mathematics using some standard/non-standard large cardinal axiom. Q3. Is the above thesis true? I would like to thank Asaf Karagila for his comment in my previous question that was the main guide for this question. REPLY [2 votes]: Question 2 admits a trivial answer, because any assertion implies $0\neq 1$, provably so in ZFC, and so for $A$ you can just use "there is an inaccessible cardinal". Similarly, the existence of $0^\sharp$ already implies $V\neq L$, and so you can use $B=$ "$0^\sharp$ exists". And for C, you can use "there is a measurable cardinal and $\neg$CH"; this is equiconsistent with a measurable cardinal, and it implies $\neg$CH. For $X$, you can just use the Reinhardt cardinal again, since any statement implies $0\neq 1$ in ZFC. Similarly, question 3 also has such a kind of unsatisfactory simple answer. Suppose that $\sigma$ is any statement at all, and $I$ is any large cardinal axiom. Since any particular model of ZFC+I will either satisfy $\sigma$ or satisfy $\neg\sigma$, we can deduce from Con(ZFC+I) that one of Con(ZFC$+I+\sigma$) or Con(ZFC$+I+\neg\sigma$) holds. Let $J$ be either $I+\sigma$ or $I+\neg\sigma$ accordingly, so we get property (a) as a material implication, and property (b) holds because $J$ settles $\sigma$ explicitly.<|endoftext|> TITLE: Does O'Nan-Scott depend on CFSG? QUESTION [16 upvotes]: My question is in the title. Some context: there are two versions of the O'Nan-Scott theorem. The first, weaker version, is due to O'Nan and Scott (independently) and gives the structure of the maximal subgroups of a finite symmetric group: A maximal subgroup of $S_n$ is one of: $S_k\times S_{n-k}$, the stabiliser of a $k$-set (that is, intransitive), $S_a \mathrm{wr} S_b$ with $n=ab$, the stabiliser of a partition into b parts of size a (that is, imprimitive), or primitive (that is, preserves no nontrivial partition) and of one of the following types: $\mathrm{AGL}(d,p)$ with $n=p^d$, $S_\ell \mathrm{wr} S_k$ with $n=\ell^k$, the stabiliser of a product structure,, a group of diagonal type, or an almost simple group. Question 1: Am I right in thinking that this theorem does not depend on the Classification of Finite Simple Groups (CSFG)? (Note, of course, that you need CFSG if you want to explicitly list the almost simple groups at (4). But such an explicit list is not a part of the stated theorem.) The second, stronger, version of the O'Nan-Scott theorem is due to Aschbacher, O'Nan and Scott, and gives the structure of all finite primitive permutation groups. I'll state an abbreviated version. A primitive subgroup of $S_n$ is of affine type, of wreath type in a product action, of diagonal type, of twisted wreath type, or almost simple. All the proofs of this result that I have seen depend on the Schreier Conjecture, which asserts that the outer automorphism group of a finite simple group is solvable. This Conjecture is known to be true but only as a consequence of CFSG. So.... Question 2: Is it true that all known proofs of the Aschbacher-O'Nan-Scott theorem depend on CFSG? REPLY [14 votes]: According to the book "Permutation groups" by Peter Cameron, your first version, concerning maximal subgroups of $S_n$ was announced in 1979 before the announcement of the completion of CFSG, so it did not depend on CFSG at that point. There is a complication however, because the earliest versions of O'Nan-Scott erroneously omitted the twisted wreath product case, which was pointed out later by Aschbacher. But the twisted wreath products are not maximal in $S_n$ - they are contained in the product type $S_l \wr S_k$ maximals, so I don't think that affects the maximality theorem. According to Chapter 4 of Dixon and Mortimer's book on Permutation Groups, whihch deals with the O'Nan-Scott Theorem, the only place where the Schreier conjecture is used (Theorem 4.7B of Dixon and Mortimer) is in the analysis of the primitive groups with a regular nonabelian normal subgroup (which is exactly the twisted wreath product case), and I believe that no proof of that result is known that does not use the Schreier Conjecture. So if you want to avoid it, then you just have to leave that case without further information. I heard Chris Parker from Birmingham give a talk for students on the O'Nan-Scott Theorem a few years ago, and I remember he did address the question of use of CFSG, and I believe it was exactly as I have described here, but you could write to him to check! Addendum: Just to emphasize the dependence on the Schreier Conjecture, if that conjecture had been false, then it is conceivable that there could be two simple groups $S,T$ with $T \le {\rm Aut}(G)$ but $T \not\le {\rm Inn}(G)$, in which $T$ leaves no nontrivial proper subgroup of $S$ invariant, in which case the semidirect product $S \rtimes T$ acting on the cosets of $T$ would be a primitive permutation group having $S$ as regular normal subgroup.<|endoftext|> TITLE: Summary of Lie-Algebra integration tactics QUESTION [10 upvotes]: If this is in the scope of MO, I would like to gather here the known tactics of Lie algebra integration, since it appear surprisingly hard to find such a compendium, library or any other kind of collections on this topic. A Lie algebra integration should be a tactic (a functor or maybe something less formal) that assigns a Lie group to any given Lie algebra and preferable a Lie group morphism, to any Lie algebra morphism. In the most general scenario this would be applicable to infinite Lie algebras and over arbitrary fields, but as I said,less general tactics are ok. === 1.) A well known integration method, is of course the integration by path. This technique is functorial (over at least finite dimensional, real Lie algebras AFAIK) and assigns a simply connected Lie group to any given finite dim. Lie algebra. An introduction can be found in the paper "Integrability of Lie brackets" of Crainic and Fernandes and the references therin. REPLY [7 votes]: The method of integration that I like most is via the theory of central extensions. To start with note that if $\mathfrak{g}$ is a subalgebra of $\mathfrak{gl}_n$, then the subgroup $\langle\exp(\mathfrak{g})\rangle$ of $GL_n$ that is generated by $\exp(\mathfrak{g})$ integrates $\mathfrak{g}$. Thus we are done with integrating $\mathfrak{g}$ if we find a faithful representation of $\mathfrak{g}$ (which exists by Ado's Theorem). The problem with the above procedure is that the faithful representation of $\mathfrak{g}$ is not very canonical. However, there is a very canonical representation, the adjoint representation $\operatorname{ad}\colon\mathfrak{g}\to \mathfrak{der(\mathfrak{g}})\leq \mathfrak{gl}(\mathfrak{g})$. This is not faithful, the kernel is the center $\mathfrak{z}(\mathfrak{g})$ of $\mathfrak{g}$ and thus we have a canonical central extension $$\mathfrak{z}(\mathfrak{g})\to\mathfrak{g} \to \mathfrak{g}_{\operatorname{ad}}:=\operatorname{ad}(\mathfrak{g}). $$ We now set out to integrate this central extension. If we set $G_{\operatorname{ad}}:=\langle\exp(\mathfrak{g_{\operatorname{ad}}})\rangle\leq\operatorname{Aut}(\mathfrak{g})$, then this is a connected Lie group integrating $\mathfrak{g}_{\operatorname{ad}}$, and so is its universal covering $\widetilde{G}_{\operatorname{ad}}$. Since $\pi_2(\widetilde{G}_{\operatorname{ad}})=\pi_1(\widetilde{G}_{\operatorname{ad}})=0$ (this is the case for any simply connected finite-dimensional Lie group and was used in different disguise by Cartan in one of his first proofs of Lie's Third Theorem) there now exists a unique central extension $$\mathfrak{z}(\mathfrak{g})\to \widehat{G}\to \widetilde{G}_{\operatorname{ad}}$$ that integrates the above central extension of Lie algebras. The latter follows from the exact sequence$$\operatorname{Hom}(\pi_1(K),Z)\to \operatorname{Ext}(K,Z)\xrightarrow{L}\operatorname{Ext}(\mathfrak{k},\mathfrak{z})\to \operatorname{Hom}(\pi_2(K),Z)\oplus\operatorname{Hom}(\pi_1(K),...) $$ (where $L$ is the Lie differentiation and $...$ is some abelian group not of big importance for this discussion) that can be found in the paper of Neeb that was already cited by Peter Michor (and also in Theorem 7.12 of Neeb's "Central extensions of infinite-dimensional Lie groups" Ann. Inst. Fourier (Grenoble) 52 (2002)(5):1365–1442 (MR1935553)). The nice thing about the latter approach is that it is functorial at each step it also works in infinite dimensions it can (in infinite-dimensions) be used to show that certain Lie algebras do not integrate it can be used in the non-integrable case to show how non-integrable Lie algebras still integrate to etale Lie 2-groups I could elaborate on any of the above points, but this is perhaps a bit off topic from the original question.<|endoftext|> TITLE: a $L^1$ convergence for backward martingale QUESTION [7 upvotes]: I have a question which may be naive, but I can not find the related result in the classical reference such as "Foundations of Modern Probability" and "Probability"(Billingsley). So if someone knows the result please let me know, many thanks! Let $\{M_n\}_{n\geq 0}$ be a backward martingale, i.e. $$E[M_n|\mathcal{F}_{n+1}]=M_{n+1},~ \forall n\geq 0$$ where $\mathcal{F}_n:=\sigma(M_k,~ k\geq n)$. Now suppose $$\lim_{n\to\infty}M_n=M,~ a.s.$$ My question is whether we have $$\lim_{n\to\infty}E[|M_n-M|]=0$$ Clearly by Fatou Lemma and Jensen's inequality $M\in L^1$. If we suppose $M_0\in L^p$ with $p>1$ then it is just an application of Doob's inequality, but when $p=1$ I do not know how to prove it. Thanks a lot for your help! REPLY [4 votes]: The following is contained in Durrett's book Probability theory and examples. First remark: the limit always exists. Consider indeed $U_n$ the number of upcrossings of $[a,b]$ by $M_0,\dots,M_n$. Then $(b-a)\mathbb E(U_n)\leqslant \mathbb E(M_0-a)^+$. Using towering property of conditional expectation, we obtain $\mathbb E(M_0\mid\mathcal F_n)=M_n$. If $M_0\in\mathbb L^1$, then the family $\{\mathbb E(M_0\mid \mathcal G),\mathcal G\subset\mathcal F \}$ is uniformly integrable and we conclude convergence in $\mathbb L^1$.<|endoftext|> TITLE: Resolution of the indeterminacy locus of a rational map away from an irreducible component QUESTION [5 upvotes]: Suppose I have a rational map between two smooth, complex, projective varieties (or a meromorphic map between two compact, complex, manifolds) $X$ and $Y$. Ordinarily one eliminates the indeterminacy locus by resolving the singularities of the graph. Assume the indeterminacy locus has co-dimension $2$ irreducible components and fix one of these, say $Z$. My question is, can one still resolve the indeterminacy away from this irreducible component? More precisely, is there a sequence of blowups along smooth centres, so that on the resolution I obtain a rational (meromorphic) map into $Y$ whose indeterminacy locus is exactly the proper (strict) transform of the component $Z$? For my purposes I believe it would be enough to know that the indeterminacy locus is the total transform of $Z$ (the full pre-image). REPLY [2 votes]: If $Y$ is projective, at least the weaker question (indeterminacy locus is preimage of $Z$) has, I guess, a positive answer: In this case, the rational/meromorphic map, call it $f$, is given by a linear system $|V|$ where $V \subset \operatorname{H}^0(X,L)$ and $L$ is some line bundle on $X$. Let $\mathfrak b$ denote the base ideal of $V$, i.e. the image of the evaluation map $V \otimes L^{-1} \to \mathcal O_X$. Possibly after replacing $V$ and $L$, we can assume that the base locus $\operatorname{Bs}(V) = \operatorname{supp}(\mathcal{O}_X / \mathfrak b)$ has codimension at least $2$ in $X$. Under this assumption $\operatorname{Bs}(V)$ is precisely the indeterminacy locus of $f$. Since $Z$ is an irreducible component of $\operatorname{supp}(\mathcal O_X / \mathfrak b)$, there exist ideal sheaves $\mathcal J, \mathcal K \subset \mathcal O_X$ such that $$\mathfrak b = \mathcal J \cap \mathcal K, \qquad Z = \operatorname{supp}(\mathcal O_X / \mathcal J) \qquad \text{and} \qquad Z \not\subset \operatorname{supp}(\mathcal O_X / \mathcal K).$$ If $X$ is not assumed to be projective, use Siu's "Noether-Lasker Decomposition of Coherent Analytic Subsheaves". Otherwise you can also use the decomposition given by EGA $IV_2$ 3.2.6 (see also this MO question Primary decomposition for non-affine schemes). By Hironaka, there exists a principalization of $\mathcal K$, i.e. a morphism $g \colon W \to X$ that is a composition of smooth blow-ups, such that the inverse ideal sheaf $g^{-1} \mathcal K$ is invertible, i.e. there exists a divisor $E \subset W$ such that $g^{-1}\mathcal K = \mathcal I_E$. Furthermore $g$ is an isomorphism over the complement of $\operatorname{supp}(\mathcal O_X / \mathcal K)$, in particular generically over $Z$. A good reference for this is "A Simplified Proof of Desingularization and Applications" by Bravo, Encinas and Villamayor U.: Principalization is Theorem 2.5 and after Theorem 8.3 the authors remark that it is also true for compact complex spaces. Since $g^{-1}\mathfrak b \subset g^{-1}\mathcal K = \mathcal I_E$, the linear subspace $g^*V \subset \operatorname{H}^0(W,g^*L)$ is contained in the image of $\operatorname{H}^0(W,g^*L - E) \to \operatorname{H}^0(W,g^*L)$. Let $V_W \subset \operatorname{H}^0(W,g^*L - E)$ denote the preimage of $g^*V$ under this injection. Then the base ideal of $V_W$ is $$\mathfrak b_W = \mathcal I_E^{-1} \cdot g^{-1}(\mathcal J \cap \mathcal K).$$ Clearly $\mathfrak b_W|_{g^{-1}(X \setminus Z)} = \mathcal O_{g^{-1}(X \setminus Z)}$, hence $\operatorname{Bs}(V_W) = \operatorname{supp}(\mathcal O_W / \mathfrak b_W) \subset g^{-1}(Z)$.<|endoftext|> TITLE: Can a non-commutative C*-algebra be a minimal operator space? QUESTION [12 upvotes]: By an operator space structure on a Banach space $X$ I mean a sequence of norms on spaces $M_n \otimes X$ that satisfies Ruan's axioms. Among such admissible sequences there is always the smallest one (if we impose some normalisation, say $\|e_{11} \otimes x\|=\|x\|$), the one obtained by embedding $X$ in some $C(K)$-space; it is sometimes referred to as "commutative" operator space structure. It is then reasonable to ask: can a non-commutative $C^{\ast}$-algebra (which has a privileged operator space structure induced by $\ast$-homomorphic embedding into $B(\mathcal{H})$) be completely isomorphic (or isometric) to a minimal operator space? I suspect that the answer is no (at least in the completely isometric case) but I can't come up with a proof. REPLY [9 votes]: I don't think so. If A is a noncommutative C*-algebra with the min operator space structure then it's double dual $A^{**}$ will also have the min operator space structure (this fact can be found in any operator space text) and be noncommutative. Then $A^{**}$ will have two noncommuting projections. One can then use ideas from Takesaki Vol 1, Chapter V.1 (the last topic in the section) to build a 2x2 matrix algebra inside of $A^{**}.$ This would force the natural operator space structure on 2x2 matrices to be minimal, which it isn't.<|endoftext|> TITLE: Examples of complete distributive lattices that are not Heyting algebras QUESTION [6 upvotes]: Here is a short question with a possibly simple and short answer: I need an example of a complete distributive lattice that is not a Heyting algebra which should be an infinite complete lattice that does not satisfy the infinite distributivity law (in the finite world all lattices are complete). I haven't seen this matter addressed in literature yet (not even as an exercise) but it turns out that the wikipedia page on Heyting algebras (http://en.wikipedia.org/wiki/Heyting_algebra) might be giving the wrong idea in the second paragraph: "Every Boolean algebra is a Heyting algebra when a=>b is defined as usual as \neg a v b, as is every complete distributive lattice[clarification needed] when a=>b is taken to be the supremum of the set of all c for which a ^ c \leq b." Thank you in advance. By the way, the great book of D. Scott et al, "Continuous lattices and domains" might be giving an answer in O-45, page 39. But I didn't get there yet .. REPLY [5 votes]: The situation mentioned in Andreas Blass's answer happens for most complete Heyting algebras that one encounters. In other words, the dual of a frame is usually not a frame even though it is a distributive lattice. For example, the lattice of closed subsets of any non-discrete $T_{1}$-space is not a Heyting algebra. A bounded distributive lattice $L$ is said to be subfit if whenever $a,b\in L$ and $a\vee c=1\Rightarrow b\vee c=1$ for all $c$, then $a\leq b$. For example, if $(X,\mathcal{T})$ is a $T_{1}$-space, then the frame $\mathcal{T}$ is subfit. Therefore subfitness is a very weak separation axiom that works particularly well in point-free topology. Therefore, most frames that people study in point-free topology are subfit. Let $L$ be a complete lattice. Then we say that an element $a\in L$ is linear if whenever $I$ is a set and $b_{i}\in L$ for $i\in I$, then $$a\wedge\bigvee_{i\in I}b_{i}=\bigvee_{i\in I}(a\wedge b_{i}).$$ We say that an element $a\in L$ is colinear if whenever $I$ is a set and $b_{i}\in L$ for $i\in I$, then $$a\vee\bigwedge_{i\in I}b_{i}=\bigwedge_{i\in I}(a\vee b_{i}).$$ $\mathbf{Lemma}$ Every complemented element in a complete distributive lattice is linear and colinear. $\mathbf{Proof}$ Since the notions of linearity and colinearity are dual, we only need to show that each element on $L$ is linear. Suppose that $L$ is a complete distributive lattice and $a,c\in L$ are complemented elements. Take note that $$a\vee(a\wedge\bigvee_{i\in I}b_{i})=a=a\vee\bigvee_{i\in I}(a\wedge b_{i}).$$ Furthermore, $$c\vee(a\wedge\bigvee_{i\in I}b_{i})=(c\vee a)\wedge(c\vee\bigvee_{i\in I}b_{i})=(c\vee\bigvee_{i\in I}b_{i})$$ $$=\bigvee_{i\in I}(b_{i}\vee c)=\bigvee_{i\in I}((b_{i}\vee c)\wedge(a\vee c))=\bigvee_{i\in I}(b_{i}\wedge a)\vee c=(\bigvee_{i\in I}(b_{i}\wedge a))\vee c.$$ Therefore. $$a\wedge\bigvee_{i\in I}b_{i}=(a\wedge c)\vee(a\wedge\bigvee_{i\in I}b_{i})= (a\vee(a\wedge\bigvee_{i\in I}b_{i}))\wedge(c\vee(a\wedge\bigvee_{i\in I}b_{i}))$$ $$=(a\vee\bigvee_{i\in I}(a\wedge b_{i}))\wedge(c\vee\bigvee_{i\in I}(a\wedge b_{i})) =(a\wedge c)\vee\bigvee_{i\in I}(a\wedge b_{i})=\bigvee_{i\in I}(a\wedge b_{i}).$$ $\mathbf{QED}$ $\mathbf{Theorem}$ Let $L$ be a complete distributive fit lattice. Then an element $a\in L$ is complemented if and only if it is colinear. $\mathbf{Proof}$ $\rightarrow$. This direction has been proven already in the above lemma. $\leftarrow$. Suppose $a\in L$ is colinear. Let $c=\bigwedge\{x\in L|a\vee x=1\}$. Then $a\vee c=1$ as well by colinearity. Now suppose that $a\wedge c\neq 0$. Then by subfitness, there is some $x\neq 1$ with $1=x\vee(a\wedge c)=(x\vee a)\wedge(x\vee c)$. Therefore $x\vee a=1$, so $x\geq c$, hence $x\geq x\vee c=1>x$, a contradiction. We therefore conclude that $a\wedge c=0$, so $a$ and $c$ are complements. $\mathbf{QED}$ In particular, if $L$ is a subfit frame and $L^{*}$ is the lattice with the same underlying set as $L$ but with the reverse ordering, then $L^{*}$ is a frame if and only if $L$ is a complete Boolean algebra. Also, if $X$ is a $T_{1}$-space and $\mathcal{C}$ is the lattice of closed sets of $X$, then $\mathcal{C}$ is a frame if and only if $X$ is discrete. All these facts about frames can be found in Picado and Pultr's new book [1]. Picado, Jorge, and Aleš Pultr. Frames and Locales: Topology without Points. Basel: Birkhäuser, 2012.<|endoftext|> TITLE: Does there exist an uncountable separable metric space $X$ such that every subset of $X$ is a Borel set? QUESTION [15 upvotes]: Is it consistent with ZFC that there exists an uncountable separable metric space $X$ such that every subset of $X$ is a Borel set? If the continuum hypothesis holds, or more generally $2^{\aleph_{0}}<2^{\aleph_{1}}$ , then each uncountable separable metric space contains non-Borel sets since there are only $2^{\aleph_{0}}$ Borel sets. Therefore, in order for there to be an uncountable separable metric space $X$ where every subset of $X$ is Borel, we would need to at least have $2^{\aleph_{0}}=2^{\aleph_{1}}$. What other conditions are necessary in order for there to exist an uncountable separable metric space $X$ where every subset of $X$ is Borel? REPLY [18 votes]: Under Martin's Axiom plus the negation of CH, every set $X$ of reals of size $<\mathfrak c$ is a Q-set, which means that every subset of $X$ is an $F_\sigma$-set with respect to the subspace topology that $X$ inherits from the real line. For more about such sets, see Arnie Miller's chapter, "Special subsets of the real line" in the Handbook of Set-Theoretic Topology.<|endoftext|> TITLE: Fixed points and their continuity QUESTION [6 upvotes]: Let $f : I^2 \to I$ be a continuous map, where $I := [0,1]$ is the unit interval. It is a basic fact that for each $y\in I$, the function $x \mapsto f(x,y)$ admits a fixed point. I want to ask whether one can always choose those fixed points as a continuous function of $y$. Question: Does there always exist a continuous path $\gamma : I \to I$ such that $f(\gamma(y),y) = \gamma(y)$ for every $y\in I$? REPLY [8 votes]: Not necessarily. Let $f(x,0)=0$, $\,f(x,1/2)=x$, $\,f(x,1)=1$, and define $f(x,y)$ for $01/2$, so $g$ cannot be continuous at $y=1/2$.<|endoftext|> TITLE: Determining Tightness/Overtwistedness of Contact Structure using Lift of Structure to the Universal Cover QUESTION [6 upvotes]: All: I would appreciate any ideas, refs., etc. on the following: Let $M^3$ be a contact 3-manifold, and let $X$ be its universal cover. Then the contact structure, say $\eta$ on $M^3$ lifts to a contact structure $\eta'$ on $X$. Just hoping to get some information about the relationship between the tightness/overtwistedness and other contact properties of $(X, \eta')$ and the contact properties of $(M^3, \eta)$ . Can we, e.g., directly conclude from the tightness/overtwistedness of $(x, \eta')$ the tightness/overtwistedness of $(M^3 \eta)$?; are there any other known relationships between the contact properties of the two spaces? Maybe some properties of Legendrian knots and their twisting? Thanks. REPLY [6 votes]: Overtwisted disks lift to overtwisted disks, so if $(M^3, \eta)$ is overtwisted then so is any cover. The reverse is not true. There are tight contact structures with finite covers that are overtwisted. If the universal cover is tight, the contact structure is called universally tight. There may be simpler references, but for examples of universally tight and virtually overtwisted contact structures, see Ko Honda, On the Classification of Tight Contact Structures II. J. Differential Geometry Vol 55, Number 1 (2000), 83-143.<|endoftext|> TITLE: A preprint of Sela concerning the work of Kharlampovich-Miyasnikov QUESTION [25 upvotes]: Yesterday, Z. Sela published a preprint in arXiv which claims that the solution of Olga Kharlampovich and Alexi Miyasnikov for the Tarski problem on decidablity of the first order theories of free groups and torsion free hyperbolic groups contains mistakes and so, that problem which was announced to be solved in 2006, is still an open problem. At this time, I am interested to know, which important theorems of Group theory, Model theory and Algebraic geometry over groups discovered in the period of 2006-2013 applied result of Kharlampovich-Miyasnikov. Edit: An answer of Kharlampovich and Miyasnikov for the preprint of Sela is just published in arXiv. They explained briefly that there was no serious mistakes in their work, and many errors discovered by Sela are already have been corrected. See this link: http://arxiv.org/abs/1402.0482 REPLY [21 votes]: Sela does not have an objective view of Kharlampovich, Myasnikov's work. We will post a paper dismissing his statements about "fatal mistakes". It just takes time. There are some typos and inessential errors that were fixed in later works. Sela himself has many such mistakes. The objects in the two works (Sela's and our work) are similar but not identical. They are not amenable to a crude direct interpretation; some of our statements would not be true if interpreted via a ``computer translation'' into his language (and vice versa). One example is Sela's wrong Theorem 7 from his paper 6 on Diophantine Geometry. This theorem describes groups elementarily equivalent to a non-abelian free groups. Sela claims that our Theorem 41 is wrong too. But our theorem is stated using our concept of regular NTQ groups and is correct. This shows that regular NTQ groups are not completely identical to hyperbolic $\omega$-residually free towers. Many of his critical comments resulted from such an exact translation of our concepts into his language. Additional misrepresentations result from not remembering that some statement was made two pages before (such as that we only consider fundamental sequences satisfying first and second restrictions). The decidability of the elementary theory of a free group is used in the proof of the decidability of the theory of a torsion free hyperbolic group (our recent preprint in the arxiv) and to make quantifier elimination algorithmic. One can use it to approach the theory of a free product of groups with decidable elementary theories (Malcev's problem). One can also use it to deal with Right Angled Artin groups. The algorithm to find the abelian JSJ decomposition of a limit group was constructed in our paper "Effective JSJ decompositions" that appeared before, it is used in the proof of the decidability of the theory. Actually many Algebraic Geometry over free groups questions are solved algorithmically (see references on pages 508-514), finding irreducible components of finite systems of equations, analogs of elimination and parametrization theorems in classical Algebraic Geometry etc Olga Kharlampovich<|endoftext|> TITLE: Summation of a series QUESTION [7 upvotes]: I would like to sum the series $$ \sum_{n=0}^\infty \frac{1}{(1+a^2 (n+1/2)^2) ^{3/2}} . $$ It arose when trying to perform a calculation on superconductivity. In particular I am interested in its diverging behaviour for $a\to 0$. I have tried turning it into a contour integral by defining $$ f(z)=tan(\pi z)(1+a^2 z^2) ^{-3/2} $$ The integral along the real axis gives (modulo factors of $\pi$ and 2) the desired sum. The integral over a great cicle in the upper half plane should vanish, but there is a branch cut to take care of. When I choose it to lie from $\frac i a$ to $+i \infty$ my integral along this branch cut from $i \infty$ to the pole diverges and I can get no meaningful answer. Maybe I am using the wrong technique for evaluating this sum. Mathematica also refuses to solve it. Any suggestion on this problem would be much appreciated. REPLY [6 votes]: This series is a good candidate for the Poisson summation formula. Let $f(x) = \frac{1}{(1+a^2 x^2)^{3/2}}$ and $f_{\frac{1}{2}}(x) = f(x+1/2)$. Define the sum $J = \sum_{n\in \mathbb{Z}} f_{\frac{1}{2}}(n)$. It is related to the desired sum $I = \sum_{n=0}^{\infty} f_{\frac{1}{2}}(n)$ by the formula $I=J/2$, because $f_{\frac{1}{2}}(-n-1) = f_{\frac{1}{2}}(n)$. By the Poisson summation formula, we have the equality $$ J = \sum_{n\in\mathbb{Z}} f_{\frac{1}{2}}(n) = \sum_{p\in\mathbb{Z}} F_{\frac{1}{2}}(2\pi p) , $$ where $F(k) = \int_{-\infty}^{\infty} dx\, e^{-ikx} f(x)$ and $F_\frac{1}{2}(k) = \int_{\infty}^{\infty} dx\, e^{-ikx} f_{\frac{1}{2}}(x) = e^{ik/2} F(k)$. The idea is that for each larger value of $p$, the terms of the sum in Fourier space will be of larger and larger subleading order in $a$, so that only the first few terms of that serious would be needed to get a good asymptotic estimate. Now, we can use some integration by parts and complex contour integration methods to evaluate the actual Fourier transform $F(k)$. \begin{align} F(k) &= \int_{-\infty}^{\infty} dx\, e^{-ikx} \frac{1}{(1+a^2 x^2)^{3/2}} \\ &= \int_{-\infty}^{\infty} dx\, e^{-ikx} \frac{d}{dx}\frac{x}{(1+a^2 x^2)^{1/2}} \\ &= ik \int_{-\infty}^{\infty} dx\, e^{-ikx} \frac{x}{(1+a^2 x^2)^{1/2}} \\ \end{align} The integration by parts helps make the subsequent deformed contour integrals finite. For $k>0$ the integral can be deformed into the lower half complex plane, while for $k<0$ it can be deformed into the upper half. In each case, the new contour will run along both sides of a branch cut, extending from either $x=i/a$ or $x=-i/a$ outward to infinity. Since $f(x)$ is real and symmetric, so is $F(k)$. So it is enough to consider only $k>0$ and the lower branch cut. Note that the imaginary part of the square root will be negative along the branch of that contour running to infinity. Letting $x=-i(z+1)/a$ and summing the contributions to the contour from either side of the branch cut we get \begin{align} F(k) &= (ik) (-2i/a) e^{-k/a} \int_0^\infty dz\, e^{-kz/a} \frac{-i(z+1)/a}{-i\sqrt{(z+1)^2-1}} \\ &= \frac{2k}{a^2} e^{-k/a} \int_0^\infty dz\, \frac{e^{-kz/a}}{\sqrt{z}} \frac{1+z}{\sqrt{2+z}} \\ &\sim \frac{2k}{a^2} e^{-k/a} \int_0^\infty dz\, \frac{e^{-kz/a}}{\sqrt{z}} \frac{1}{\sqrt{2}} \left(1+ \frac{3}{4}z - \frac{5}{32} z^2 + \cdots \right) \\ &\sim \frac{\sqrt{2\pi}}{a} e^{-k/a} \frac{k^{1/2}}{a^{1/2}} \left(1 + \frac{3}{8}\frac{a}{k} - \frac{15}{128}\frac{a^2}{k^2} + \cdots \right), \end{align} where the last two steps constitute an asymptotic series in $a/k$ obtained by expanding the integrand in a non-everywhere uniformly converging power series. Note that the singularity at $z=0$ is of integrable type. It would have been non-integrable without the integration by parts done before deforming the contour. Note also that the above formulas don't work for $k=0$. Fortunately, we can obtain that value directly, $$ F(0) = \int_{-\infty}^{\infty} dx\, \frac{1}{(1+a^2x^2)^{3/2}} = \int_{-\infty}^{\infty} dx\, \frac{d}{dx} \frac{x}{(1+a^2x^2)^{1/2}} = \frac{2}{a} . $$ So, finally, an $a\sim 0$ asymptotic formula for the desired sum is \begin{align} I &= \frac{1}{2}\left(F(0) + 2\cos(\pi)F(2\pi) + \cdots \right) \\ &\sim \frac{1}{a} - \frac{2\pi}{a^{3/2}} e^{-2\pi/a} + \cdots , \end{align} where more terms could be computed as needed.<|endoftext|> TITLE: Mean time to get $k$ heads for a coin with growing bias QUESTION [7 upvotes]: For integers $n,k\geq 1$ we repeatedly toss a coin and count the number of heads that occur. The probabilty of getting a head is $min(t/n,1)$ where $t$ is the current discrete time step. I am trying to work out the asymptotics of the mean time to get $k$ heads. My claim is the following. Claim: Let $X$ be a random variable which represents the number of coin tosses at the $k$th head. If $k$ is $o(n)$ and $n \rightarrow \infty$, we have: $$\mathbb{E}(X) \sim \sqrt{2n}\frac{\Gamma(k+\frac{1}{2})}{(k-1)!}$$ My "proof" uses a limit approximation of a nonhomogeneous Bernoulli process by a nonhomogeneous Poisson process. We then compute the expected number of tosses by transforming the nonhomogeneous Poission process to a homogeneous process by an application of the inverse transform method. Let $T_k$ be the time taken to see $k$ events in a Poisson process with increasing rate $t/n$, $t >0$. Define $$\Lambda(t) \overset{\text{def}}{=}\;\; \int_0^t \frac{x}{n} dx = \frac{t^2}{2n}$$ with inverse $$\Lambda^{-1}(y) = \sqrt{2 y n}.$$ The key observation is that $T_k$ has the same distribution as $\Lambda^{-1}(S_k)$, where $S_k$ is the time taken to see $k$ events in a Poisson process with rate $1$. We want \begin{align*} \mathbb{E}(T_k) &= \mathbb{E}(\sqrt{2n S_k}) \\ &= \sqrt{2n} \int_0^\infty \frac{x^{1/2} x^{k-1} e^{-x}}{(k-1)!} \mathrm{d}x\\ &= \frac{\Gamma(k+\frac{1}{2})}{(k-1)!}. \end{align*} The problem is that when $k$ grows with $n$, as is permitted in the claim, I don't see how to give a formal justification for this line of reasoning. In particular, how can you justify the limit approximation? Any help gratefully received. As Douglas Zare points out, when $k$ grows with $n$ (and is also $o(n)$), the claim is equivalent to $$\mathbb{E}(X) \sim \sqrt{2nk}.$$ If my particular approach can't be justified, is there another way to get the same result which has a surer footing? REPLY [3 votes]: Let me just consider the case when $k$ is growing, but still $o(n)$ (in fact, we can even let $k$ go up to $(1/2-\epsilon)n$). I will show that the expected number of coin tosses is about $\sqrt{2nk}$ as conjectured. In fact the argument shows more, getting bounds for the probability of getting $k$ heads after exactly $y$ tosses. Let $P(k,y)$ denote the probability that we get $k$ heads exactly after $y$ tosses (so the $y$-th toss is heads, and there are $k-1$ heads up till then). If $y\le n$, then note that $P(k,y)$ equals $y/n$ times the coefficient of $x^{k-1}$ in $\prod_{i\le y-1} (x(i/n)+(1-i/n)) =Q_y(x)$ say. Now $Q_y(x)$ is a polynomial with non-negative coefficients, and therefore the coefficient of $x^{k-1}$ is bounded by $Q_y(r)r^{-(k-1)}$ for any positive real number $r$. Thus we have that $$ P(k,y) \le \frac{y}{n} \min_{r>0} Q_y(r) r^{-(k-1)}. $$ Now note that $$ Q_y(r) = \prod_{i\le y-1} (1+ (r-1)i/n) \le \exp\Big(\sum_{i=1}^{y-1} \frac{(r-1)i}{n} \Big) = \exp\Big( \frac{(r-1)y(y-1)}{2n} \Big). $$ A little calculus shows that $(r-1) y(y-1)/(2n) - (k-1)\log r$ attains its minimum when $r=2n(k-1)/(y(y-1))$. Taking this value for $r$, we obtain with $t=y(y-1)/(2n(k-1))$ $$ P(k,y) \le \frac{y}{n} \exp\Big( -(k-1)(t-1-\log t) \Big). $$ Now note that $t-1-\log t$ is approximately $(t-1)^2/2$ if $t$ is close to $1$; and that if $|t-1|>\delta$ then $(t-1-\log t)$ is at least $C\delta |t-1|$ for an absolute positive constant $C$. Thus if $|y(y-1)/(2n(k-1))-1| > \delta$ then from the above estimates it follows that $$ P(k,y) = O\Big(\frac{y}{n} \exp\Big(-C\delta (k-1)\Big|\frac{y(y-1)}{2n(k-1)}-1\Big|\Big) \Big). $$ In other words, the probability is tiny (exponentially small in $k$) unless $y(y-1)$ is very close to $2n(k-1)$; that is, unless $y$ is about $\sqrt{2nk}$. From this it is easy to see that the expected value is $\sim \sqrt{2nk}$ as conjectured. If one argues carefully using the saddle point method, one could obtain asymptotics for $P(k,y)$; as the argument above indicates, the likely values of $y$ are sharply concentrated around $\sqrt{2nk}$. Added: In the range where $y=o(n^{2/3})$ and $k=o(\sqrt{y})$, directly from the definition one can compute that $$ P(k,y) \sim \frac{y}{n} e^{-y^2/(2n)} \frac{(y^2/(2n))^{k-1}}{(k-1)!}. $$ So in the range $k=o(n^{1/3})$ one can compute the distribution directly (getting Gaussian fluctuations) -- this will also give the right result for bounded $k$. For larger $k$ one can use the previous argument of bounding the probability away from the peak (or work harder and get asymptotics using the saddle point method).<|endoftext|> TITLE: Is Kolmogorov complexity (KC) relevant for proof theory? QUESTION [6 upvotes]: Note. The title was modified. Previous title was "Every theorem t has a proof no more complex than~|t|. Is this right?" The question ("Is Kolmogorov complexity (KC) relevant for proof theory?") arises because every theorem has a proof with low Kolmogorov complexity". To be more specific, Every theorem $t$ has a proof $p$ with $K(p)\leq K(t)+c\leq|t|+c'$, where $c$ and $c'$ are constants. Proof: It is assumed that there is a proof checking algorithm $C(x)$ that outputs TRUE if $x$ is a correct proof, FALSE otherwise. Then there exists a fixed (depending on the formal system) Turing machine $M$ with Input: $t$ a string (that may be a theorem) - $M$ enumerates all the proofs until (and if) $t$ is proved. - When $t$ is proved, print the proof $p$ and halt. Thus, if $t$ is a theorem, $M(t)$ prints a proof of $t$. Using the universal TM, we get $ K(p) \leq K(t) + c \leq |t| + c' $ where $c$ and $c'$ are constants. Thus, and apart from a constant, no theorem needs a proof with Kolmogorov complexity greater than $|t|$. To me this seems a bit strange. In other words, the number of bits of "inspiration" (or "non-deterministic bits", or "oracle bits") needed to prove any theorem $t$ is at most $|t|+c$. If this is true, the very lengthy proofs that sometimes are needed to prove some simple to state theorems are necessarily very regular / structured / compressible (synonymous). REPLY [2 votes]: After making the above comment I did a quick Google search and found a paper entitled "Kolmogorov Complexity and Computational Complexity" by Lance Fortnow. The paper is about resource-bounded variants of Kolmogorov complexity. After giving the standard definition and some consequences he writes the following. Of course this definition of Kolmogorov complexity makes any complexity theorist cringe: what good is a small program [for] $x$ if it takes the life of the universe to produce $x$? Perhaps he could give you more precise pointers to the literature. [Edit] - Your main observation has been made before, by Henry Cohn, on MathOverflow. See the last paragraph of his answer here.<|endoftext|> TITLE: Reasons for the use of Nisnevich topology in motivic homotopy theory QUESTION [30 upvotes]: The objects of interest in motivic homotopy theory are "spaces"-which are simplicial sheaves of sets on the big Nisnevich site $Sm/k$ of smooth schemes of finite type over a field $k$. I understand that we need to enlarge the category of smooth schemes over $k$ to spaces, since $Sm/k$ is not closed under limits and colimits. Why does one use Nisnevich topology? If one considers the Zariski site over $Sm/k$, is the category of simplicial sheaves over it not complete/co-complete? Can someone please give an example demonstrating this? Is this the only reason for using Nisnevich topology (apart from the fact that it retains useful properties of both Zariski and etale topologies)? REPLY [26 votes]: Here are some comments about the use of topologies in motivic homotopy theory. This is based on the discussion in Morel-Voevodsky's "A^1-homotopy theory of schemes" p.94-95 (MV below), I only add some background and references. I also comment on the differences between the development of the unstable and stable theories. I am not an expert, so please comment/edit out innacurate statements. First some model category generalities. Various model structures on categories of simplicial (pre)sheaves on any Grothendieck site $(\mathcal{C},\tau)$ can be constructed. For a survey see: http://nlab.mathforge.org/nlab/show/model+structure+on+simplicial+presheaves and the references therein. Many of those are Quillen equivalent and can be thought as different presentations of the $(\infty,1)$-category of $(\infty,1)$ $\tau$-sheaves on $\mathcal{C}$. As usual in model category theory, they are suited for deriving different functors (i.e. some natural functors will be Quillen with respect to some structures but not others). The resulting model categories are all left proper, simplicial and combinatorial, so by a theorem of Smith (see http://nlab.mathforge.org/nlab/show/Bousfield+localization+of+model+categories#Existence )) left Bousfield localizations at a set of maps exist. In particular if $\mathcal{C}$ is a small category of schemes over a fixed $S$ which is stable by $S$-fiber products and contains $\mathbb{A}^1_S$, one can localize at the set of maps $X\times_S \mathbb{A}^1_S\rightarrow X$. The Morel-Voevodsky category $\mathcal{H}(S)$ can be obtained by this procedure with $\mathcal{C}=Sm/S$, $\tau=Nis$ and using as starting point the so-called injective model structure on simplicial sheaves. It should be clear from the above that many variants are possible, some giving alternative models for $\mathcal{H}(S)$ (using projective model structures, simplicial presheaves instead of sheaves, or even more exotic choices like cubical presheaves, etc.) and some giving different categories (using another topology: Zariski, étale, using the category of all schemes instead of smooth ones, etc.). The passage to the stable theory via model categories of spectra is also formal (in the sense that it can be done in great generality and with variants) but rather subtle. See e.g. Riou "Catégorie homotopique stable d'un site suspendu avec intevalle" or Ayoub's thesis. The question now becomes: in which respects are $\mathcal{H}(S)$ and $\mathcal{SH}(S)$ nicer than the alternatives ? And how much does this depend on the choice of the Nisnevich topology ? Here are some possible answers: 1) Characterisation of $\tau$-local simplicial (pre)sheaves. For a general site (and in particular for the étale site over a general scheme), this is a complicated condition which can be expressed only in terms of a descent condition for hypercovers (see Dugger-Hollander-Isaksen, "Hypercovers and simplicial presheaves"). For the Zariski and the Nisnevich site, one can show a "Brown-Gersten" property: $\tau$-locality can be rephrased as some squares of simplicial sets associated to $\tau$-distinguished squares being homotopy cartesian. See MV proposition 1.16 for the case of the Nisnevich topology. Aside: this argument has been abstracted by Voevodsky in "Homotopy theory of simplicial sheaves in completely decomposable topologies" and used to compare $\mathcal{H}(k)$ for $k$ a field admitting resolution of singularities with a similar category defined with $\mathcal{C}=Sch/k$ and $\tau=cdh$. In the setting of triangulated categories of mixed motives, this argument can be pushed in various directions to exploit various forms of resolution of singularities, see Cisinski-Deglise, "Triangulated categories of mixed motives", section 3.3. The Brown-Gersten property in turn plays an important role in MV. For instance, it implies that the property of being $\tau$-local is stable by filtered colimits. It is also used in MV to construct an explicit A^1-localisation functor and to study the functoriality of $\mathcal{H}(S)$. On the other hand I am not sure if the use of the Brown-Gersten property for all this is unavoidable: at least in the context of $\mathcal{SH}(S)$, there are arguments which do not use this (see Riou's paper mentioned above or Ayoub's thesis) and which consequently give some results for the étale topology. On the other hand, in following works of Morel on $\mathcal{H}(k)$, there are really substantial applications of the Brown-Gersten condition, see "A^1-algebraic topology over a field" Chap. 8 and App. A. 2) Homotopy purity (or localisation). The proof of this major theorem in MV requires a topology at least as fine as the Nisnevich topology and to work with $\mathcal{C}=Sm$. The idea is that to reduce the theorem to the case of the closed immersion $Z\rightarrow A^n_Z$ (where one can write explicit $A^1$-homotopies), one uses the local structure of smooth pairs in the étale topology (cf EGAIV 17.12.2) and the fact that étale morphisms to Henselian local schemes (i.e. points of the Nisnevich site) which have a section on the closed point have a section. 3) Compactness properties. The cohomological dimension of the small Nisnevich site on a noetherian scheme is bounded by the Krull dimension, see MV 1.8 and Thomason-Trobaugh E.6.c. This is very different from the case of the étale topology. This has important consequences for the stable theory. It implies in particular that $\mathcal{SH}(S)$ is compactly generated. See e.g. the last paragraphs of Ayoub's thesis. I do not know if there are analoguous statements to be made in the unstable case. 4) Nisnevich descent for motivic cohomology and algebraic K-theory. The descent properties of algebraic K-theory have been studied long before motivic homotopy theory. The definitive, pre-$A^1$-homotopy result (Nisnevich descent for algebraic K-theory of regular schemes) is in Thomason-Trobaugh, and this is used as an input in the proof of representability of algebraic K-theory in MV p.139. On the other hand, algebraic K-theory with integral coefficients does not satisfy étale descent and hence cannot be represented in the étale $A^1$-homotopy category.<|endoftext|> TITLE: Topological rigidity for negatively curved manifolds? QUESTION [10 upvotes]: I was wondering if two compact oriented manifold carrying a Riemannian metric with negative sectional curvature, whose fundamental groups are isomorphic, are necessarily diffeomorphic (or homeomorphic) ? In the hyperbolic case, the ingredients in the proof of Mostow rigidity strongly rely on the hyperbolic structure, and it makes me think that such a result, if true, must have a proof essentially different. One the other hand, the topological differences between hyperbolic manifolds and only negatively curved manifold (which are known to differ when the dimension is greater than $4$) makes it hard to think of a counterexample for the non-specialist that I am. REPLY [17 votes]: I'm assuming that your intention is that negatively curved means having negative sectional curvature. Your question, with regard to uniqueness up to homeomorphism, is a special case of the Borel Conjecture (topological uniqueness of aspherical manifolds) which is still unsolved despite much progress. In the special case of negative curvature, there is a monumental series of works of Farrell and Jones suggesting the correctness of the Borel conjecture. I would look on Mathscinet for these papers, including several detailed surveys. The smooth version of the Borel conjecture does not hold; one can change the smooth structure of a negatively curved manifold, and still have a negatively curved metric. See [Farrell, F. T.; Jones, L. E. Negatively curved manifolds with exotic smooth structures. J. Amer. Math. Soc. 2 (1989), no. 4, 899–908.]<|endoftext|> TITLE: reference for Lindelof Hypothesis implying finitely many zeros off critical line? QUESTION [5 upvotes]: Can anyone give me a reference for the following theorem on the Riemann zeta function? If the Lindelof Hypothesis is true (that is $\zeta(\sigma+it)=O(t^\epsilon)$ as $t\rightarrow\infty$), then there are only finitely many zeros of $\zeta(s)$ off the critical line $Re(s)=\frac{1}{2}$. I've heard that this is the case (and I think I read it on the internet, but can't find it again), but I'm looking for an actual proof. Many thanks for any help with this! REPLY [13 votes]: The Lindelof hypothesis (LH) does not seem to give such precise information about the zeros of $\zeta(s)$. Here are three known implications of Lindelof on the zeros, but they will be seen to fall far short of showing finitely many exceptions to RH. Thus, no result of the form stated in the question exists in the literature. Backlund showed that LH implies that for large $T$ and any $\epsilon >0$ there are at most $o(\log T)$ zeros of $\zeta(s)$ with real part bigger than $1/2+\epsilon$ and imaginary part between $T$ and $T+1$. (For comparison there are about constant times $\log T$ zeros of $\zeta(s)$ with imaginary part between $T$ and $T+1$.) LH implies the density hypothesis: For any $\sigma>1/2$, the number of zeros of $\zeta(s)$ with real part $\ge \sigma$ and imaginary part between $0$ and $T$ is denoted by $N(\sigma,T)$. Then LH implies the bound $N(\sigma,T) = O(T^{2(1-\sigma)+\epsilon})$. A theorem of Halasz and Turan: LH implies that $N(3/4+\epsilon,T)= O(T^{\epsilon})$. These results may be found in the books of Titchmarsh, Ivic or Edwards on zeta.<|endoftext|> TITLE: What are Kirby diagrams of candidate exotic 4-manifolds? QUESTION [10 upvotes]: It is an open problem whether there exist smooth manifolds homeomorphic, but not diffeomorphic to the standard $S^4$. The same is true for the 4-torus and several other manifolds. Handle decompositions of 4-manifolds can be written down as Kirby diagrams: Dotted circles represent 1-handles, undotted, numbered links represent 2-handles. Is there a reference that lists candidate exotic manifolds, expressed in Kirby diagrams? Edit: I am mainly interested in diagrams for manifolds that are not known, but conjectured to be exotic. REPLY [15 votes]: There is no comprehensive list in the format you ask about; you will probably want to look at the original papers. Searching for "exotic" and "4-manifold" on Mathscinet gives > 100 responses, and probably there are other phrases you could use for such a search. There are a number of examples listed in the book of Gompf and Stipsicz, but the subject has progressed quite a bit since that book was published. The use of handle pictures in describing exotic 4-manifolds is not quite what you are suggesting, though. Typically, an exotic manifold would be described by some other construction (log transforms, fiber sums, knot surgery, rational blowdown, etc.) which enables you to compute a Seiberg-Witten or other gauge-theory invariant. The hard part is typically doing the right sequence of such operations in order to get the right topological type; even computing the fundamental group may be hard. It is usually possible, although often requires considerable effort, to draw a handle picture afterwards. This might be used to uncover some interesting property of the exotic manifold. But I don't know of any examples (at least any simply-connected ones) where one starts with an interesting handle diagram and then deduces exoticity. The point is that one still doesn't know how to effectively compute gauge-theory invariants from the handle picture.<|endoftext|> TITLE: Condition for a local ring whose maximal ideal is principal to be Noetherian QUESTION [7 upvotes]: The statement "a local ring whose maximal ideal is principal is Noetherian" is (I think) false. The ring of germs about $0$ of $C^\infty$ functions on the real line seems to be a counterexample since $e^{-1/x^2}\in \left(x^n\right)$ for all $n\geq 1$. If I add to the hypothesis that the ring is a domain, then (I think) the statement is true. I'm trying to figure out if this must be true (I suspect not). Is there a nice example of a local Noetherian ring whose maximal ideal is principal that is not a domain? Is there a better, weaker condition to add to the hypothesis so that sufficiency holds? In other words, "if R is a local ring whose maximal ideal is principal, then R is Noetherian if and only if R is [what is the best thing to put here]?" Here local rings are assumed to be commutative with unity. REPLY [10 votes]: Here is an elementary argument. Let $R$ be a local ring with a principal maximal ideal $M=mR$. Since $1+M\subseteq R^*$, we have $ma\mid a$ only if $a=0$. Moreover, every nonunit is divisible by $m$, hence if there is a nonzero $a\in\bigcap_nM^n$, we can construct an increasing sequence of ideals $a_0R\subsetneq a_1R\subsetneq a_2R\subsetneq\cdots$ where $a_0=a$ and $ma_{n+1}=a_n$, in particular $R$ is not noetherian. On the other hand, if $\bigcap_nM^n=0$, every nonzero element can be written as $um^n$ for some $u\in R^*$ and $n\in\omega$, so all nonzero ideals are of the form $M^n$. This makes $R$ a discrete valuation ring unless $m$ is nilpotent. Thus, Fact: If $R$ is a local ring with a principal maximal ideal $M$, the following are equivalent. $R$ is noetherian. $\bigcap_{n\in\omega}M^n=0$. $R$ is a DVR, or there is $n\in\omega$ such that all ideals of $R$ are $R,M,M^2,\dots,M^n=0$.<|endoftext|> TITLE: Is there an analog of Sperner's lemma for the Hopf invariant? QUESTION [36 upvotes]: Recall that Sperner's lemma is essentially a combinatorial version of the topological statement "A map from $S^n$ to $S^n$ with degree one cannot be nullhomotopic." My question is, does there exist an analogous combinatorial lemma corresponding to the statement "A map from $S^3$ to $S^2$ with Hopf invariant one cannot be nullhomotopic"? REPLY [6 votes]: Sperner's Lemma. Let me start by talking through the implications for topology of Sperner's Lemma. The point is a map $S^1 \rightarrow S^1$ is null-homotopic exactly if it extends to a map $D^1 \rightarrow S^1$. Then the 3 coloring of Sperner's Lemma gives a map from the boundary of the triangulated triangle, to the boundary of a triangle (with the vertices corresponding to the three colors). The condition on the coloring amounts to requiring the map to be degree 1. Since a map from a simplicial disc to a triangle maps only to its boundary if and only if no 2-simplex maps to the entire triangle, the conclusion follows. Null-homotopic maps. Now, as Vidit Nanda has noted, a map from $S^3 \rightarrow S^2$ is similarly null-homotopic if and only if it extends to a map $D^4 \rightarrow S^2$. Thus, to show a simplicial map $f$ from the boundary of a triangulated 4-simplex $\Delta$ to the boundary of a tetrahedron is not null-homotopic, it suffices to show that some 3-simplex $\sigma$ in $\Delta$ is "rainbow" -- that is, that the 4 vertices of $\sigma$ map to the 4 vertices of the tetrahedron. (Rainbow being a good word if you think of the 4 vertices as being named Red, Blue, Green, and Yellow.) The Hopf invariant. At this point, I can answer Jim's question: Yes, there exists a combinatorial analogue of Sperner's Lemma showing that a map with Hopf invariant 1 is not null-homotopic. The definition of simplicial cohomology is purely combinatorial, and the above allows us to translate "null-homotopic" into combinatorics. There surely exists a (possibly complicated) encoding of "Hopf invariant 1" into statements about colors. Towards a concrete answer According to P Hilton's "An introduction to homotopy theory" Chapter 6, we can define the Hopf invariant for $f:S^3 -> S^2$ as follows: take a point $x$ of $S^2$. The inverse image of $x$ under $f$ is a (homology 1-) cycle $C$, and so there is a homology 2-chain $D$ which has $C$ as its boundary. Since $f$ collapses the boundary of $D$ (and leaves the rest alone), we get that $f(D)$ is a homology 2-cycle in $S^2$. If $f(D)$ is the fundamental cycle, then we say $f$ has Hopf invariant 1. Thus, a map has Hopf invariant 1 if we can find $D$ and $C$ as above such that the induced map from $D/C \rightarrow S^2$ has degree 1. Hilton's formulation of the Hopf invariant however does seem to be somewhat more tractable to turning into a combinatorial condition than the standard one from e.g. Wikipedia -- Hilton's is easier to work with simplicially! Incidentally, Hilton seems to say that his formulation is closer than the standard to Hopf's original statements. Let me try to translate Hopf-invariant-one into a coloring condition: A theorem: Let $f$ be a 4-coloring of the vertices of a triangulated 5-simplex $\Delta$, such that there is a subcomplex $D \subseteq \partial \Delta$ which is a triangulated polygon (i.e., 2-disc) satisfying the following coloring properties: i) The boundary of $D$ is monochromatic (say, Red), ii) No interior vertices of $D$ are colored Red, and iii) The subcomplex of $D$ induced by the interior vertices of $D$ satisfies the Sperner condition on Green, Blue, and Yellow -- that is, there are boundary vertices $v_1, v_2, v_3$ colored Green, Blue, and Yellow, such that the outside path from $v_1$ to $v_2$ has all vertices Green or Blue, etc. Then $\Delta$ has a rainbow (Red-Green-Blue-Yellow) simplex. An illustration of the conditions on $D$ follows: Proof: If there is a rainbow simplex on $\partial \Delta$, then the result is trivial. Otherwise, $f$ induces a map $\partial \Delta \rightarrow S^2$, and the conditions above directly imply that $D/\partial D$ (with the induced coloring) satisfies the conditions of Sperner's Lemma for the boundary of the 2-simplex. We get that the map $f \vert_D$ induces a degree 1 map from $D / \partial D \cong S^2 \rightarrow S^2$, and thus that $f \vert_{\partial \Delta}$ has Hopf invariant 1. Thus, $f \vert_{\partial \Delta}$ is not null-homotopic, and the theorem follows from the discussion above of null-homotopic maps. $\square$ Loose ends The above theorem is a start towards what @JimBelk asked for, but there are several things that are unsatisfactory about it: It is not clear to me that the homology chain $D$ in the definition of the Hopf invariant need be a disc. This would be the main obstacle to proving from the above theorem that a map with Hopf invariant 1 is not null-homotopic. (If I understand Hopf's map correctly though, one can indeed take the homology chain to be a disc here. Since this is the only example in these dimensions, requiring $D$ to be a disc might be within the boundaries of good taste.) It would be nice to have a direct combinatorial proof of the above theorem. The main obstacle is seeing how the interior of the simplicial $D^4$ yields an interior of $D / \partial D$. (If you can understand some 3-dimensional interior of $D / \partial D$, then I guess it'll follow from Sperner.)<|endoftext|> TITLE: Background to understand Gromov's green book QUESTION [7 upvotes]: I have a decent background in differential geometry. I have read John Lee's introduction to smooth manifolds and doCarmo's Riemannian Geometry. I was trying to read Misha Gromov's Metric structures for Riemannian and non Riemannian spaces but am finding it extremely difficult to follow. Any suggestions for books/papers as prerequisites ? REPLY [5 votes]: Slightly more advanced books on Riemannian geometry can help, in particular Petersen, and Gallot-Hulin-Lafontaine. Gromov draws a lot of his examples from Riemannian geometry ! Reading Burago-Burago-Ivanov "Introduction to metric geometry" is a also a good idea. It will give the necessary understanding of intrinsic metric spaces. With this a good part of the book should be understandable.<|endoftext|> TITLE: Weak versions of Bertrand's postulate QUESTION [6 upvotes]: We are interested in the following statement: For each $n>1$ and $x>2$ there is at least one prime $p$ satisfying $x2$. Thanks for your help. REPLY [3 votes]: Pal Erdös' version can be found in "proofs from the book". This was rather elementary.<|endoftext|> TITLE: "Introduction to mathematical logic" book from a formalist perspective QUESTION [8 upvotes]: I'm looking for books that introduce the reader to mathematical logic assuming the perspective of a formalist. I've found that many books are more or less written for the platonist - like Kunen's Foundations of Mathematics, where he even implicitly says on pp. 191 that his book, if I understood it right, is primarily written for platonists, but also explains how a formalist would understand his book (in a whopping 3 pages compared to a couple of sentences for platonist view!). I'm looking for a book that doesn't wait until page 191 to explain this to me, but constantly conveys the formalist viewpoint. It is important that the books clearly explains the distinction between theory and metatheory and where different theorems of the metatheory live in (e.g. the soundness theorem can be perceived to be a theorem of ZFC since the relevant parts of the metatheory can be coded in ZFC). I looked at every book from the thread Ask for recommendations for textbook on mathematical logic and none was what I was looking for. Closest to my needs came Kunen - who at least mentions formalism and how his book should be read according to this perspective. This contrasts with other logic books who don't mention anything, and Cori and Lascar's book - for their excellent introduction concerning the vicious circle in what mathematical logic studies - and Goldrei's book on logic, which is not on the list. To give an explanation for this, perhaps, unusual request: I find it that I understand mathematical theories best when the setting in which the theory "lives" in is clearly outlined so that working in that theory is just formal manipulations of symbols - of course I can attach meaning and intuition to these manipulations, but there has to be a "fixed" setting to work in. From what I've read this aligns understanding aligns best with the perspective of formalism. But sadly mathematical logic is always somewhat vague and in basic core always seems to be somewhat obscure (Kunen says in the above mentioned book for example on page 190 that we cannot say exactly what metatheory is. Now I accept that we can't begin with formal setting based on nothing, because there has to be an informal description of the most basic formal elements of our setting, but I would hope that there are books that explain in more detail that in a single paragraph what metatheory really is. Additionally the lecturer at a course I'm taking also believes in some absolute mathematical objects - I assume he is a platonist - since he frequently says things like "no, now we're not talking about a formalized version of the natural numbers, we're talking about the real natural numbers", which totally annoys me because for me, there are no real natural numbers). REPLY [7 votes]: I'd like to elaborate a bit on the suggestion, mentioned by some others, that your complaint is not really about platonism versus formalism, but about the common practice of being not completely clear about the metatheory. Given that the distinction between theory and metatheory is so important, why don't the books make it completely clear at all times which metatheory is in force? To some extent this can be blamed on poor exposition, but I'd like to suggest that there are good reasons behind this practice, which are important to understand. As an example, consider the irrationality of √2. The proof of this theorem is almost always presented in textbooks without any explicit statement about what formal system the proof is supposed to be taking place in. So how is the reader supposed to tell whether the proof is correct? Wouldn't it be better to state at the beginning that the proof is supposed to be carried out on the basis of 1st order Peano arithmetic, or whatever? Well, this could be done, but there are a couple of reasons why this is not typically done. The most important is that almost everyone finds it easier to understand the proof that √2 is irrational if it is presented in the usual manner. Even if they want to verify a statement such as "`√2 is irrational' is a theorem of PA", they find it easier to do this by first getting an intuitive understanding of the proof, and then making a "second pass" through the proof to verify that each step in the argument can be mimicked by a formal deduction in PA, rather than taking a purely formal string such as ∀a∀b¬(Sa⋅Sa=SS0⋅Sb⋅Sb) and mechanically checking each step of a formal proof of it. A second reason is that it is often the case that we want to verify that "√2 is irrational" is a theorem not only of PA, but of various other formal systems. Again, while one possible approach is to go through the entire formal verification process every time one wants to switch to another formal system, it is far more practical if the reader can understand the "content" of the argument and verify for himself or herself that all the necessary steps can be carried out on the basis of whatever formal system is currently of interest. The flexibility is useful. My guess is that despite your stated views, you count yourself among the vast majority of people who are able to read and verify the correctness of the proof that √2 is irrational without having to have it completely formalized first. After all, almost everyone who studies mathematics encounters this argument before learning the details about formal systems. My guess is that the trouble begins when the theorems in question get a lot more complicated. For example, consider the proof of Goedel's 1st incompleteness theorem. This is a lot more complicated than the proof that √2 is irrational, and there might be parts that you have difficulty with. This is the point where you might worry about the validity of what's being asserted, and might wish that the metatheory were clearly defined so that you could fall back on formally verifying that A follows from B on the basis of the metatheory. I'd like to suggest, though, that almost always, it's not the lack of clarity about the metatheory that's the real problem; it's that the reasoning is just not being explained clearly enough for you. With rare exceptions, if the reasoning is correct, it's going to be correct in any "reasonable" choice of metatheory. At this point you might say, "I understand everything you're saying, but still, the textbooks and professors sometimes talk about sets and integers as if they're real things, and I just don't believe that they're real! So I can't follow what they're arguing, and I have a sneaking suspicion that they're using philosophically unjustified assumptions about the reality of integers or sets in their supposedly mathematically rigorous proofs." Don't worry, this isn't what is happening. Any time that someone refers to the "actual" integers or the "standard model" or something like that, you can always make sense of it by taking your metatheory to be some set theory such as ZFC, inside which there is a unique set of natural numbers. People don't always say ZFC explicitly, for the same reason as before: It doesn't have to be ZFC; various other set theories would work just fine, and it's useful to be flexible. In the end, I think that rather than look for a book that presents things from a formalist point of view, you are better off developing the skill of reading a proof and recognizing what assumptions are needed to carry it through. Eventually you need to develop this skill anyway. For example, for the proof of Goedel's 2nd incompleteness theorem, you need to be able to go through the proof of the 1st incompleteness theorem and verify that it can be formalized in (for example) PA. If you have this skill then you don't need to be explicitly told which metatheory is in force. And probably the best way to develop the skill is to do some formalization exercises. Looking for such exercises (perhaps in the context of modern proof assistants such as Coq or Mizar) might be more fruitful than looking for a formalist textbook.<|endoftext|> TITLE: Given a polygon with holes, find a maximum distance pair in two subsets QUESTION [5 upvotes]: I am curious about the following problem: Given a polygon with holes and two convex subsets, $S$ and $T$, find points $s \in S, t \in T$ such that the shortest path between the two points has maximal length w.r.t. all point pairs in $S$ and $T$. Note that both $s$ and $t$ might be interior points of the corresponding regions. If $S$ would only be a point, i.e. the start point is fixed, we could build a Shortest Path Map (continuous Dijkstra) and could thereby find the farthest point. Does anybody have any idea how to handle the case when $S$ is not just a point? REPLY [5 votes]: This* question is addressed in a 2010 paper, Sang Won Bae, Matias Korman, Yoshio Okamoto. "The Geodesic Diameter of Polygonal Domains." (arXiv link) They describe an algorithm that achieves a worst-case complexity of $O(n^{7.73})$ for a polygon of $n$ vertices, or $O(n^7 (\log n + h))$ when expressed also as a function of the number of holes $h$. The reason the problem is so complicated is that the diameter-achieving points might lie in the interior of the polygon (i.e., not on its boundary). In that case, there are at least five distinct shortest paths, a result previously established for the geodesic diameter of convex polyhedra. *The OP's question was changed slightly as I was preparing this answer, but I believe the added restriction to $S$ and $T$ does not make the problem any easier (in general). Fig.1 from the Bae-Korman-Okamoto paper:<|endoftext|> TITLE: Rationality in Families QUESTION [7 upvotes]: I am looking for examples of flat, projective families $\pi:X\to B$ of schemes over $\mathbb C$ such that the general fiber $X_b$ is rational (i.e. birational to $\mathbb P^n$), while the special fiber $X_0$ is irrational. It is not known whether such a family exists with $\pi$ smooth. Instead, I would like to see examples where The general fiber is smooth. The special fiber is singular, irreducible. Hassett-Pirutka-Tschinkel have produced smooth examples of the opposite phenomenon in https://arxiv.org/pdf/1603.09262.pdf. REPLY [10 votes]: That's quite easy: take a family of cubic surfaces where the special fiber is a cone (for instance $X$ given by $X^3+Y^3+Z^3+uT^3=0$ in $\mathbb{A}^1\times \mathbb{P}^3$). Of course a cone over an elliptic curve is not rational.<|endoftext|> TITLE: Rosenthal like inequality for weak $\mathbb L^p$-norms QUESTION [8 upvotes]: Let $p$ be a real number greater than $1$. It is well known (see Hall and Heyde's Martingale limit theory and its applications, Theorem 2.10) that there exists a constant $C_p$ such that if $(X_i)_{i=1}^n$ is a real valued martingale difference with respect to the filtration $(\mathcal F_i)_{i=1}^n$ (that is, $(S_j:=\sum_{i=1}^jX_i)_{j=1}^n$ is a martingale with respect to this filtration), then $$\frac 1{C_p}\mathbb E\left(\sum_{i=1}^nX_i^2\right)^{p/2}\leqslant \mathbb E\left|\sum_{i=1}^nX_i\right|^p\leqslant C_p\mathbb E\left(\sum_{i=1}^nX_i^2\right)^{p/2}.$$ Hence the $\mathbb L^p$ norm of the partial sum is controlled by those of the quadratic variation. Now define for a real valued random variable $X$: $$\lVert X\rVert_{p,\infty}:=\left(\sup_{t\geqslant 0}t^p\mu\{|X|\geqslant t\}\right)^{1/p}.$$ This is equivalent to a norm (namely $N(X):=\sup_{\mu(A)>0}\mu(A)^{-1+1/p}\int_A|X|\mathrm d\mu$). I would like to know whether there is a similar inequality to Rosenthal's one, that is, a control of $N(S_n)$ in terms of those of $N\left(\sqrt{\sum_{i=1}^nX_i^2}\right)$ plus maybe an other term. This seems to be a natural question which has probably been investigated, but I didn't manage to find a reference. There are weak-$\mathbb L^p$ versions of Rosenthal's inequality for independent random variables, but I would like to see a reference dealing with an extension to martingale differences, namely: Let $(\Omega,\mathcal F,\mu)$ be a probability space $p\gt 2$. Is there a constant $C_p$ such that if $n$ is an integer and $(X_j)_{1\leqslant j\leqslant n}$ is a martingale difference with respect to the filtration $(\mathcal F_j)_{1\leqslant j\leqslant n}$ with $\lVert X_j\rVert_{p,\infty}\lt\infty$, then $$C_p^{-1}\left\lVert \sqrt{\sum_{j=1}^nX_j^2}\right\rVert_{p,\infty}\leqslant \left\lVert \sum_{j=1}^nX_j\right\rVert_{p,\infty}\leqslant C_p\left\lVert \sqrt{\sum_{j=1}^nX_j^2}\right\rVert_{p,\infty}~? $$ REPLY [3 votes]: The result you want is mentioned in Remark 6 in Johnson, W. B.(1-TXAM); Schechtman, G.(IL-WEIZ) Martingale inequalities in rearrangement invariant function spaces. Israel J. Math. 64 (1988), no. 3, 267–275 (1989).<|endoftext|> TITLE: Are all complex zeros of $\zeta(s) \pm \zeta(-s)$ on the line with $\Re(s)=0$? QUESTION [7 upvotes]: My conjecture is that all zeros in the strip $-1 \le \Re(s) \le 1$ of $\zeta(s) \pm \zeta(-s)$ are on the line $\Re(s)=0$. I did find three complex zeros for $\pm =+$ (i.e. 12 in total) and two complex zeros for $\pm = -$ (i.e. 8 in total) that lie outside this strip, however their imaginary part seems capped at $|\Im(s)| < 6$ and also rapidly converges to the real zeros that in their turn converge to $s=2k, k \in \mathbb{Z}, |k| > 9$. The root-finding graph below for $\pm = -$ illustrates the point: I could not find a function on the web that directly links $\zeta(s)$ to $\zeta(-s)$, but via this formula (derived here), $$\displaystyle \frac{\xi(s)}{\xi(-s)} = \prod_{n=1}^\infty \left(1- \frac{1}{(\rho_n+s)} \right) \left(1- \frac{1}{(1-\rho_n +s)} \right)$$ with $\xi(s) = \frac12 s(s-1) \pi^{-\frac{s}{2}} \Gamma(\frac{s}{2}) \zeta(s)$ being the Riemann Xi-function and $\rho_n$ the n-th nontrivial zero of $\zeta(s)$, I derived that $\zeta(s) \pm \zeta(-s)=0$ is equivalent to: $$\displaystyle -\frac{(s+1) \, 2 \pi^{s+1}}{s(s-1) \sin \left(\frac{\pi \, s}{2} \right) \Gamma \left(\frac{s}{2}\right)^2} \prod_{n=1}^\infty \left(1- \frac{1}{(\rho_n+s)} \right) \left(1- \frac{1}{(1-\rho_n +s)} \right) = \pm 1 $$ This function suggests that the non-trivial zeros could play a role in the conjecture (and maybe the RH even implies it), however grateful for any thoughts on how I might proceed in proving (or falsifying) this. Many thanks. ADDITIONAL QUESTION: The Haseo Ki paper in Joro's answer below, proves that all but finitely many complex zeros of $\zeta(0+s) \pm \zeta(0-s)$ are on the line $\Re(s)=0$. I now wondered if this implies that the same must be true for: $$\zeta^{k}(0+s) \pm \zeta^{k}(0-s)$$ with $k=k^{th}$ derivative of $\zeta(s)$. I tested it for $k=1..12$ and up to $\Im(s)=99$ and in the strip $-1 \le \Re(s) \le 1$. I again only find zeros lying on $\Re(s)=0$ (note there are always a finite few zeros outside this strip for lower $\Im(s)$). Is this a consequence of Haseo Ki's proof? REPLY [10 votes]: Finiteness of counterexamples follows from the paper On the zeros of sums of the Riemann zeta function, Haseo Ki From p.1: For fixed real number $\sigma_0$ define $$ H(\sigma_0,s) = \zeta(\sigma_0 + s) + \zeta(\sigma_0 -s) \text { or } \zeta(\sigma_0 + s) - \zeta(\sigma_0 -s) $$ Your question is about $\sigma_0 = 0$. From p. 3, Theorem 1: 2) If $\sigma_0 \le 0$, then all but finitely many complex zeros of $H(\sigma_0,s)$ are on $\Re(s)=0$. This is unconditional. A generalization: 1) If $\sigma_0 \le \frac12$, then all but finitely many complex zeros of $H(\sigma_0,s)$ are on $\Re(s)=0$ provided that $\zeta(s)$ has only finitely many complex zeros in $\Re(s) < \sigma_0$. Related Theorem 3, p.4. (1) Let $\sigma_0 < \frac12$ . Then, all zeros of $H(\sigma_0,s)$ in $|\Im(s)| \ge 100$ are on $\Re s = 0$ provided that $\zeta(s)$ in $\Re(s) < \sigma_0$ and $\Im(s) \ne 0$ has no zeros. (2) The Riemann hypothesis holds if and only if for any $\sigma_0 <\frac12$ , all zeros of $H(\sigma_0,s)$ in $|\Im(s)| \ge 100$ are on $\Re(s) = 0$.<|endoftext|> TITLE: Can one explain Tannaka-Krein duality for a finite-group to ... a computer ? (How to make input for reconstruction to be finite datum?) QUESTION [30 upvotes]: Consider a finite group. Tannaka-Krein duality allows to reconstruct the group from the category of its representations and additional structures on it (tensor structure + fiber functor). Somehow trying to wrap my mind into details sometimes a feeling arise that it is really "abstract nonsense" which hides something under the carpet. So the question is about how to get more down-to-earth level of understanding. Finite group - is something very explicit - it is easy to "explain" to a computer. On the other hand the Tannaka-Krein reconstruction requires as the input the "category+fiber functor+monoidal structure" - it seems to me not very explicit thing - it is "difficult" to explain to a computer, more precisely this input datum seems to be infinite. That does not seems to be satisfactory. So my question: Question: Can one describe the input for Tannaka-Krein reconstruction theorem (for finite group) as some FINITE datum ? If yes, we can put this datum to a computer and reconstruct the group. If it is impossible that would be quite strange. REPLY [39 votes]: $\DeclareMathOperator\Rep{Rep}\DeclareMathOperator\Vect{Vect}\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Mod{Mod}\DeclareMathOperator\GL{GL}\DeclareMathOperator\Hom{Hom}$The infinitude of the input data is deceptive. For example, an infinite finitely presented group may appear to be infinite input data but it's fully specified by a finite alphabet and a finite set of words in it. The same sort of thing happens here; everything is "finitely presented" in a suitable sense. First, let's recall the statement of Tannaka-Krein duality here so we can see exactly how the group arises. Let $G$ be a finite group and let $\Rep(G)$ be its category of representations (say over $\mathbb{C}$). $\Rep(G)$ is equipped with a monoidal structure in the form of the tensor product of representations and with a forgetful functor $F : \Rep(G) \to \Vect$. We are interested in studying natural automorphisms $\eta : F \to F$ compatible with the tensor product in the sense that the obvious diagram (which can be found on page 8 of Commelin - Tannaka duality for finite groups for completeness) commutes. Every element of $G$ gives an element of this group of natural automorphisms, which we'll denote $\Aut^{\otimes}(F)$. Theorem: The natural map $G \to \Aut^{\otimes}(F)$ is an isomorphism. So our reconstruction begins first with the group $\Aut(F)$ of natural automorphisms of $F$, then cuts out a subgroup satisfying some extra compatibility conditions. Let's start with figuring out what data we need to describe $\Aut(F)$. Explicitly, an element of $\Aut(F)$ is a family $\eta_V : F(V) \to F(V)$ of automorphisms of the underlying vector spaces of all representations of $G$ which is compatible with all morphisms between representations. There is a universal such representation generated by a single element, namely $\mathbb{C}[G]$ regarded as a representation via left multiplication, and compatibility with all morphisms $\mathbb{C}[G] \to V$ implies that any such natural transformation $\eta$ is completely determined by what it does to $\mathbb{C}[G]$. On $\mathbb{C}[G]$ any such natural transformation needs to be compatible with right multiplication, and so must be left multiplication by some invertible element of $\mathbb{C}[G]$. Conversely any such element gives an element of $\Aut(F)$. Hence $$\Aut(F) \cong \mathbb{C}[G]^{\times} \cong \prod_i \GL(V_i)$$ where the $V_i$ are the irreducible representations of $G$. (Secretly we are using the Yoneda lemma but I wanted to be completely explicit.) This reflects the fact that the data of the category and the fiber functor is equivalent to the data of the number and dimensions of the irreducible representations respectively. So we can use these as our initial data: Data 1, 2: The number $n$ of irreducible representations and their dimensions $d_i$. This step of the reconstruction reflects a more general fact, namely that if $R$ is a $k$-algebra, $\Mod(R)$ the category of left $R$-modules, and $F : \Mod(R) \to \Vect$ the forgetful functor, then the natural endomorphisms of $F$ are canonically isomorphic to $R$ (as a $k$-algebra), which again follows from the Yoneda lemma. The tricky part is how to describe the influence of the tensor product. Explicitly, let $\eta : F \to F$ be an element of $\Aut(F)$ again. Then there are two ways to use $\eta$ to write down an automorphism of the underlying vector space of a tensor product $F(V \otimes W)$: On the one hand, there are two maps $\eta_V : F(V) \to F(V)$ and $\eta_W : F(W) \to F(W)$, and we can tensor them to get a map $\eta_V \otimes \eta_W : F(V) \otimes F(W) \to F(V) \otimes F(W)$. Since $F$ is a tensor functor this gives a map $F(V \otimes W) \to F(V \otimes W)$. On the other hand, there is a map $\eta_{V \otimes W} : F(V \otimes W) \to F(V \otimes W)$. $\eta$ lies in $\Aut^{\otimes}(F)$ if and only if these are equal. More explicitly, if $\eta$ is left multiplication by some element $\sum c_g g \in \mathbb{C}[G]$, then the first map is $$\left( \sum c_g \rho_V(g) \right) \otimes \left( \sum c_g \rho_W(g) \right)$$ while the second map is $$\sum c_g \rho_V(g) \otimes \rho_W(g)$$ and setting $V = W = \mathbb{C}[G]$ and comparing coefficients shows that these can't be equal unless $\sum c_g g = g$ for some $g$. So how do we describe this restriction to a computer? The key point is that since everything in sight is compatible with direct sums it suffices to restrict our attention to the irreducible representations $V_i$, so we only need to compare the maps $\eta_{V_i} \otimes \eta_{V_j}$ and $\eta_{V_i \otimes V_j}$ for all $i, j$. This means that we need to describe How two maps $F(V_i) \to F(V_i), F(V_j) \to F(V_j)$ tensor to give a map $F(V_i \otimes V_j) \to F(V_i \otimes V_j)$ for all $i, j$, and How knowing $\eta_{V_i}$ for all $i$ determines $\eta_{V_i \otimes V_j}$ for all $i, j$. This requires that we know first of all the decompositions $$V_i \otimes V_j \cong \bigoplus_k m_{ijk} V_k$$ of the tensor products of the irreducibles into irreducibles. So this is our third piece of data: Data 3: The multiplicities $m_{ijk}$. These multiplicities are equivalent to the data of the character table of $G$, and in particular it is possible to compute Data 1, 2 from this data, so in some sense Data 1, 2 are redundant. But we already know that the character table is not enough. The multiplicities only tell us about $\eta_{V_i \otimes V_j}$, but not about $\eta_{V_i} \otimes \eta_{V_j}$. To get our last piece of data, let's think about how a computer would represent $\eta$. Specifying linear transformations $\eta_{V_i} \in \GL(V_i)$ requires writing down a basis of each $V_i$. These bases give rise to two different bases of the tensor products $V_i \otimes V_j$: on the one hand the tensor product basis, and on the other hand the basis coming from the decomposition into irreducibles $\bigoplus_k m_{ijk} V_k$. So the final piece of data we need is the identification between these: Data 4: For every pair $i, j$, the $(d_i \times d_j)$ by $(d_i \times d_j)$ transition matrix from the tensor product basis to the decomposition basis of $V_i \otimes V_j$. (This is essentially the data of the identifications $F(V \otimes W) \cong F(V) \otimes F(W)$ making $F$ a tensor functor: unlike in the case of maps between monoids, being a tensor functor is a structure, not a property, because these maps are required to satisfy coherence conditions.) This is the data we need to write $\eta_{V_i} \otimes \eta_{V_j}$ and $\eta_{V_i \otimes V_j}$ as matrices with respect to the same basis. These transition matrices should be definable over a splitting field of $G$ at worst, so this really is finite data. Edit, 12/26/21: Wow, sorry everyone, there's actually a small gap here that needs to be fixed. The "decomposition basis" I described above is not well-defined because the decomposition of $V_i \otimes V_j$ into irreducibles isn't unique when the multiplicities are greater than $1$! Instead we need to proceed as follows: the truly fully canonical decomposition into irreducibles takes the form $$V_i \otimes V_j \cong \bigoplus_k \text{Hom}_G(V_k, V_i \otimes V_j) \otimes V_k$$ where the tensor product on the RHS is just the ordinary tensor product of vector spaces. To get a decomposition basis as claimed above we need to make additional choices, namely a choice of basis for each homspace $\Hom_G(V_k, V_i \otimes V_j)$. I don't think this choice really ends up mattering but I think it does still need to be made? Maybe a slightly nicer approach could avoid it.<|endoftext|> TITLE: Irreducible representations of compact groups QUESTION [5 upvotes]: Let G be a compact group (or even profinite - Galois group). Let $V$ be a vector space over the field ${\mathbb F}_p$ with $p$ elements, $p$ a finite prime, such that $V$ is a contable product of ${\mathbb F}_p$ with the product topology. Let s be an irreducible continuous representation of G on $V$. Must s be finite dimensional? If not, what conditions can we add to ensure this? For example: G is profinite/abelian/solvable/finitely generated... REPLY [5 votes]: Yes, $V$ is finite-dimensional. More generally, whenever a compact group $G$ act continuously on $V\neq 0$, then there is a finite-codimensional invariant closed subspace $W\neq V$. To see this, let $G$ act on the Pontryagin dual $\hat{V}$. Then this action is continuous and $\hat{V}$ is discrete. Let $v$ be a nonzero element of $\hat{V}$: then its stabilizer $G_v$ is an open subgroup of $G$, hence has finite index. So $Gv$ is finite, and hence generates a nonzero invariant finite-dimensional subspace of $\hat{V}$. By duality, it corresponds to a closed invariant subspace $W\neq V$. This even shows that in full generality, every compact $\mathbf{F}_p[G]$-module is profinite as a $\mathbf{F}_p[G]$-module. Edit: I answer your two questions in the comments: 1) yes, the argument also works when $V$ is an arbitrary profinite abelian group and $G$ a compact group: $V$ is then profinite as $G$-module (same argument, using that $\hat{V}$ is locally finite). 2) is it true that there always exists an irreducible subrepresentation?: no: indeed pick $R=\mathbf{F}_p[[t]]$, $V=R$ (additive group) and $G=R^\times=R\smallsetminus tR$ (multiplicative group). Since $R^\times$ generates $R$ as an $R$-algebra, a $G$-submodule of $V=R$ is the same as an ideal of $R$. The ideals of $R$ are $(0)$ and the $(t^n)$, so $(0)$ is the only finite one. Hence there is no irreducible submodule.<|endoftext|> TITLE: good reference for the Hitchin fibration QUESTION [7 upvotes]: can you please recommend me a good reference to learn about the Hitchin fibration in the language of algebraic geometry? REPLY [4 votes]: At the moment I am not sure what to recommend if you want short introduction. But if you something like a longer text giving various perspective is Okay for you I would suggest: Ron Donagi, Eyal Markman Spectral curves, algebraically completely integrable Hamiltonian systems, and moduli of bundles, http://arxiv.org/abs/alg-geom/9507017. As a kind of remark, according to my experience looking from solely algebraic geometry point of view is somehow not quite good to get complete understanding. I am not aware of any "natural" explanation why Hitchin's hamiltonians Poisson commute (i.e. fibration would be Lagragian) given in terms of algebraic geometry. Original Hitchin's approach is via differential geometry and Beilinson-Drinfeld point of view is via representation theory.<|endoftext|> TITLE: Rank and hyperbolic volume QUESTION [11 upvotes]: Suppose $M$ is a hyperbolic $3$-manifold whose fundamental group has rank $r.$ What is the best (lower) bound on the volume of $M?$ Similar question for rank of $H_1.$ There is a bunch of papers of Culler and Shalen on related subjects, but they seem to care about "small" manifolds, whereas this question is more on the asymptotic dependence. REPLY [10 votes]: In Counting arithmetic lattices and surfaces, Mikhail Belolipetsky, Tsachik Gelander, Alex Lubotzky and Aner Shalev prove the following Theorem. Let $H$ be a connected simple Lie group of real rank one. Then there is an effective computable constant $C=C(H)$ such that for any lattice $\Gamma < H$ we have $r(\Gamma)\le C\cdot \mathrm{vol}(\Gamma\backslash H)$, where $r(\Gamma)$ is the minimal number of generators of $\Gamma$. Applying this to $SO(1,3)$ gives $\mathrm{vol}(M) \ge 1/C\cdot r(\pi_1(M))$ in your case.<|endoftext|> TITLE: An equation involving perfect numbers QUESTION [5 upvotes]: Let $s,x_1,x_2,\cdots, x_s$ be natural numbers not neccesarily distinct. I am interested in solving the equation $$(x_1+x_2+\cdots +x_s)^s=2^s(x_1\cdot x_2\cdots x_s)^2$$ Some Notes: I have found two solutions $(x_1,\cdots ,x_s)$ 1) We can see that equality is satisfied if $x_s= 2^{p-1}(2^p-1)$ is an even perfect number and all the other $x_i$ are the proper divisors of $x_s$. (That is why i started to investigate the equation) Of course we can disregard the restriction that $x_s$ is perfect and show easily that we also get a solution if: $$x_1=1,x_2=2^1,...,x_n=2^{n-1},x_{n+1}=2^n-1,x_{n+2}=2^1(2^n-1),...$$ and $x_s=2^{n-1}(2^n-1)$.(The number $n$ is not necessarily prime ) 2) A trivial one: $$x_1=x_2=\cdots =x_s=\frac{s}{2}$$ . Is it possible to find other solutions or to prove that there are only 2 solutions, those mentioned above? Thanks in advance. REPLY [8 votes]: I think the computer found other solutions for $s=4$: [x1,x2,x3,x4] [ 2 , 27 , 150 , 1 ] [ 2 , 3 , 6 , 121 ] [ 27 , 150 , 2 , 1 ] [ 50 , 6 , 3 , 1 ]<|endoftext|> TITLE: automorphisms of fat points QUESTION [12 upvotes]: Let $k$ be an algebraically closed field. I am looking for an easily quotable description of automorphism groups of $\mathrm{Spec} k[x]/(x^n)$. I could compute explicit matrix representations for several small $n$, but a more intrinsic description would be desirable. REPLY [13 votes]: Smooth points are all alike; every fat point is happy in its own way.                                                                                                                   "Anna Karenina", Leo Tolstoy Let $k$ be an algebraically closed field. Denote $A_n:=k[x]/x^{n+2}$, $G_n:=\text{Aut}A_n$, and let $N_n\subset G_n$ stand for the unipotent radical of $G_n$. Since there is no problem in finding a faithful matrix representation of $G_n$, I guess, Dima asks about a coordinate-free description of $G_n$. We will mainly deal with $N_n$ because $G_n=N_n\rtimes T$ is a semidirect product, where $\text{G}_m\simeq T\subset G_n$ has no canonical choice (though all tori are conjugated in $G_n$). Answer. We assume for simplicity $k$ of characteristic $0$. For sufficiently big $n$, $N_{n+1}$ is a central extension (with that very centre) of $N_n$. This extension is determined by a certain orbit of the group $\text{Out}N_n$ naturally acting on the projective plane ${\mathbb P}_kH^2(N_n,k)$. The orbits are: a dense one, two $1$-dimensional orbits $L'$ and $H'$ dense in the projective lines $L$ and $H$, and a couple of fixed points $f_0,f_1\in L$, where $f_0$ is the intersection $L\cap H$. Our choice determining the extension $0\to k\to N_{n+1}\to N_n\to1$ is $H'$. The action of $T$ can be easily defined on the way. The interesting stuff appears at the end, when the point becomes really fat. Boring stuff. Every element $g\in G_n$ is determined by the polynomial $f_g:=gx$, where $f_g(x)=f_0x+f_1x^2+\dots+f_nx^{n+1}$ with $f_i\in k$ and $f_0\ne0$. The composition in $G_n$ is expressed in these terms as $f_{g_1g_2}(x)\equiv f_{g_2}\big(f_{g_1}(x)\big)\mod x^{n+2}$. Note that $g\in N_n$ iff $f_0=1$ and $g\in T$ iff $f_1=\dots=f_n=0$. Sometimes, it is more convenient to use other coordinates: $f_g(x)=x\Big(h_g(x)+\displaystyle\sum_{i=0}^n(p_1x)^i\Big)$, where $h_g(x)=\displaystyle\sum_{i=2}^np_ix^i$. So, $p_1=f_1$ and $p_i=f_i-f_1^i$ for $i\ge2$. The Lie algebra $L_n$ of $G_n$ is formed by all the derivations from $A_n$ to $A_n$, i.e., by all the derivations from $k[x]$ to $k[x]$ that preserve $\text{Ideal}(x^{n+2})$. The formula $df(x)=f'(x)dx$ correctly defines a derivation $d\in L_n$ by its value $dx\in A_nx$. Taking $d_ix=x^{i+1}$ for $i=0,1,\dots,n$, we obtain a basis of $L_n$ subject to $[d_i,d_j]=(j-i)d_{i+j}$ for $i+j\le n$ and $[d_i,d_j]=0$ for $i+j>n$. The Lie algebra $M_n$ of $N_n$ is simply spanned by $d_1,\dots,d_n$. 0. $N_0=1$ and $G_0=\text{G}_m$. 1. $N_1=\text{G}_a$ and $G_1=N_1\rtimes\text{G}_m$ is the semidirect product with respect to $\text{G}_m=\text{Aut}\text{G}_a$. 2. $N_2=\text{G}_a\oplus\text{G}_a$ (the corresponding coordinates are, for instance, $p_1,p_2$) and $G_1=N_1\rtimes\text{G}_m$, where the action of $c_0\in\text{G}_m$ is given by the rule $c_0\cdot(p_1,p_2):=(c_0p_1,c_0^2p_2)$. Preliminaries. The homomorphism $\pi:A_n\to A_{n-1}$ induces the surjective homomorphisms $\pi:G_n\to G_{n-1}$ and $\pi:N_n\to N_{n-1}$ whose kernel $\text{G}_a\simeq C_n\subset N_n$, formed by all $c\in N_n$ with $f_c=x(1+fx^n)$, $f\in k$, lies in the centre of $N_n$ and coincides with this centre unless $n=2$. A few more well-known and easy facts. Under the assumption that there is a rational section of $\pi$ (which is valid in our case of $X_h=N_{n+1}$ and $X=N_n$), a central extension $0\to k\to X_h\overset\pi\to X\to1$ of algebraic groups is given by an (arbitrary) element $h\in H^2(X,k)$ of rational $2$-cohomologies of $X$ (the action of $X$ on $k=\text{G}_a$ is trivial). The groups $\text{Aut}X$ and $\text{G}_m$ act on $X$ and on $k$, hence, on $H^2(X,k)$. Assuming that $k$ coincides with the centre of $X_{h_i}$ for both $i=1,2$, the groups $X_{h_1}$ and $X_{h_2}$ are isomorphic iff $h_1$ and $h_2$ are in a same orbit of $\text{Out}X\times\text{G}_m$ (the inner automorphisms of $X$ act trivially on $H^2(X,k)$). In our case, the central extension is nontrivial. Denote by $0\ne[h_n]\in H^2(N_n,k)$ an element providing the extension $\pi:N_{n+1}\to N_n$. By the above, the group $N_{n+1}$ is in fact given by the choice of the orbit of $[h_n]$ with respect to the action of $\text{Out}N_n$ on ${\mathbb P}_kH^2(N_n,k)$. Our task is therefore to indicate this orbit and to explain how $T\simeq\text{G}_m$ acts on $N_{n+1}$ (already knowing how does $T$ act on $N_n$). The homomorphism $\pi$ induces a $k$-linear map $\pi^*:H^2(N_{n-1},k)\to H^2(N_n,k)$ and, unless $n=2$, a homomorphism $\pi:\text{Out}N_n\to\text{Out}N_{n-1}$. Another elementary remark: an element $\alpha\in\text{Out}N_{n-1}$ belongs to the image $\pi\text{Out}N_n$ (i.e., is liftable) iff $[h_{n-1}]\in{\mathbb P}_kH^2(N_{n-1},k)$ is a fixed point of $\alpha$. Using the Chevalley-Eilengebrg complex of $M_n$, one can calculate $H^2(M_n,k)\simeq H^2(N_n,k)$ for $n\ge2$ (the action of $M_n$ on $k$ is trivial), getting the following result. For any $n\ge4$, we have the cocycle $a:=d_2^*\wedge d_3^*$; for any $n\ge6$, we have the cocycle $b:=d_2^*\wedge d_5^*-3d_3^*\wedge d_4^*$; for any $n\ge2$, we have the cocycle $h_n:=\displaystyle\sum_{1\le i TITLE: Image of J splitting QUESTION [6 upvotes]: The image of J space is defined for $p$ odd as the homotopy fibre $J_{(p)}$ of the self map $$\psi^k - 1: BU_{(p)} \to BU_{(p)}.$$ Here, $\psi^k$ is an Adams operation, and $k \in \mathbb{N}$ descends to a topological generator of the $p$-adic units. For $p=2$, $J_{(2)}$ is defined as the homotopy fibre of a lift of an orthogonal form of this map, $\psi^3-1: BO_{(2)} \to BSpin_{(2)}$. It is known that $J_{(p)}$ splits off of the $p$-local sphere, $Q S^0_{(p)}$. A map $e: Q S^0_{(p)} \to J_{(p)}$ is given by the unit of the ring spectrum whose zeroth space is $J_{(p)}$. A one-sided inverse to $e$ can be constructed from the solution of the Adams conjecture. My question is: who is responsible for the proof of this result? Mahowald's "The order of the image of the J-homomorphism" constructs the map $e$ and proves the resulting surjection $\pi_* S^0_{(p)} \to \pi_* J_{(p)}$, but does not phrase the result in terms of a splitting of the infinite loop spaces. Also, he works only at $p=2$. The result is stated as Corollary 4.6 in May-Quinn-Ray-Tornehave's "$E_\infty$ ring spaces and $E_\infty$ ring spectra," but since that book appeared 7 years after Quillen's proof of the Adams conjecture and Mahowald's paper, I wonder if the proof appeared somewhere else in between these publications. (Not so) secretly, I'm hoping that Peter will jump in and fill us in on the history of this question. REPLY [9 votes]: my friend, I have an email! But I can offer the history. First, although the $E_{\infty}$ book was published in 1977, it is a shotgun marriage of a bunch of earlier preprints that were rejected for publication. Despite the ongoing development of infinite loop space theory at the time, most algebraic topologists didn't care. You are referring to Corollary 4.2 (not 4.6) in Chapter VIII, which is joint with Tornehave. As I wrote on p. 203, specifically referring to the relevant Section 4, ``These results were first proven, quite differently, by the second author'', that is, by Tornehave. He was already notorious for not publishing anything, and that is borne out by Math Reviews. I had a preprint from him, via his adviser (and my student) Ib Madsen. I wrote Tornehave to ask permission to publish jointly and then reworked the material. Remark 4.6 (that is the 4.6 you probably had in mind) describes his original proof. Detail: the splitting is of the identity component $SF$ of the $p$-local sphere infinite loop space $QS^0$. Incidentally, you refer to Mark's work at the prime $2$, and it is worth emphasizing that although $SF$ splits as a space at $p=2$, it does not split as an infinite loop space or even a $1$-fold loop space. There is a still open conjecture by Haynes Miller and Stewart Priddy about what does happen; see ``On $G$ and the stable Adams conjecture". Geometric applications of homotopy theory (Proc. Conf., Evanston, Ill., 1977), II, pp. 331–348, Springer Lecture Notes in Math. Vol 658, 1978.<|endoftext|> TITLE: Example of a space for which $V \cong Hom(V,V)$ QUESTION [18 upvotes]: Let $V$ be a topological linear space, and let $\operatorname{Hom}(V,V)$ be the space of continuous linear maps from $V$ back to $V$, equipped with a suitable topology. Is there a non-trivial example which satisfies the equivalence $V \cong \operatorname{Hom}(V,V)$? REPLY [2 votes]: A simple and natural example of such a space is that of the row-finite matrices. These play a central row in the theory of summation of divergent series since many of the classical methods are implemented by such matrices (prominent example---Cesaro summation and its higher degree variants). The above space has a natural locally convex structure and it enjoys the requested property. EDIT after Tom's comment. Am typing on my iPad and so can't use many math symbols. The standard reference on infinite matrices and summability is "Infinite matrices and sequence spaces" by R. Cooke (now a bit dated). On nuclear spaces, spaces of operators and tensor products it is Grothendieck's thesis. If you are interested in such topics, I recommend its study, a potentially life-changing experience. The basic spaces involved are the nuclear Fréchet space $\omega$ of all sequences and its dual $\phi$, the nuclear Silva space of all finite sequences. The row finite matrices map the former into itself and it is easy to see that all such continuous linear mappings arise in this way. The rest is just formal manipulation of tensor products and the corresponding operator spaces, one further ingredient being the fact that $\omega$ is isomorphic to $\omega \otimes \omega$ (but, notabene, not in any natural way). Now if $E$ is a nuclear Fréchet space which is isomorphic to $E\times E$ and there are many such and we put $V=E\otimes E'$, then $$L(V,V)=V'\otimes V=(E\otimes E')\otimes(E\otimes E')'=(E\otimes E')\otimes(E'\otimes E'')=(E\otimes E')\otimes(E'\otimes E)=(E\otimes E) \otimes(E'\otimes E')=E\otimes E'=V$$ (equality denote "is isomorphic to", $\otimes$ any tensor product since everything in sight is nuclear).<|endoftext|> TITLE: is the Hodge conjecture birationally invariant? QUESTION [19 upvotes]: Let $X$ and $Y$ be two birational smooth projective varieties over the complex numbers. Assume $X$ satisfies the Hodge conjecture. Is it known that the Hodge conjecture holds for $Y$? REPLY [28 votes]: Of course, abx is completely correct in saying that the truth of the Hodge conjecture is not a birational invariant. That said, something slightly weaker is true: if $X$ and $Y$ are $K$-equivalent, then the Hodge conjecture is true for $X$ if and only if it is true for $Y$. Here we say two smooth projective varieties $X, Y$ are $K$-equivalent if there exists a third smooth projective variety $Z$ and birational morphisms $f: Z\to X, g: Z\to Y$, such that $f^*\omega_X\simeq g^*\omega_Y$. For example, birational Calabi-Yau varieties satisfy this property. The theory of motivic integration then implies that $[X]=[Y]$ in the Grothendieck group of varieties, $K_0(\text{Var})$. But now, this paper of Donu Arapura and Su-Jeong Kang shows that the truth of the Hodge conjecture for $X$ depends only on its class in $K_0(\text{Var})$. So the bottom line is: no, the Hodge conjecture is not a birational invariant in general. But it is for Calabi-Yau varieties, and it is a "$K$-equivalence invariant."<|endoftext|> TITLE: Ergodicity of composition with a rotation QUESTION [8 upvotes]: Let $T$ be an arbitrary Lebesgue measure-preserving automorphism of the unit interval $I$. Let $R_{\alpha}$ denote rotation by $\alpha$, i.e. $R_{\alpha}(x)=x+\alpha \pmod{1}$ for $x \in I$ and $\alpha \in \mathbb{R}$. Is it true that the composition $R_{\alpha} \circ T$ is ergodic for (Lebesgue) almost every $\alpha \in \mathbb{R}$? REPLY [9 votes]: No, for trivial reasons. Take as your $T$ the map $x\mapsto-x\mod 1$. Than the map $x\mapsto\alpha-x\mod 1$ is a reflection around $\alpha/2$, and is never ergodic.<|endoftext|> TITLE: Can Galois conjugates of lattices in SL(2,R) be discrete? QUESTION [7 upvotes]: Let $\Gamma$ be a lattice in $SL(2,\mathbb{R})$. Suppose that the trace field of $\Gamma$ is a totally real number field of degree $d$. This gives $d$ homomorphisms $\rho_i:\Gamma\to SL(2,\mathbb{R})$ (here $i=1,\ldots, d$), which act as Galois conjugation on traces. Here $\rho_1$ denotes the identity homomorphism. Question 1: Can any of the $\rho(\Gamma_i)$ be discrete when $i\neq 1$? Question 2: Can the group $\{(\rho_2(g), \rho_3(g), \ldots, \rho_d(g):g\in \Gamma\}$ ever be discrete in $SL(2,\mathbb{R})^{d-1}$? A positive answer to the question 1 of course also gives a positive answer to question 2. I am most interested in the case when $\Gamma$ is not cocompact, and I would even be happy to know the answer when $\Gamma$ is a triangle group. REPLY [7 votes]: You probably want to consider the invariant trace field $k$ instead of trace field $K$. The trace field $K$ is a multi-quadratic extension of $k$, and $Gal(K/k)$ acts trivially on the $PSL(2,\mathbb{R})$ representations of $\Gamma$ (it acts non-trivially on $\Gamma < SL(2,\mathbb{R})$, but with the same image in $PSL(2,\mathbb{R})$). For example, the lattice associated to the modular curve $X_0^+(N)$ will have trace field $\mathbb{Q}(\sqrt{N})$. This is an extension of the lattice $\Gamma_0(N)$ (which has integral trace) by the Fricke involution. The Galois automorphism sends the involution to its negative in $SL_2(\mathbb{R})$, but has the same image in $PSL(2,\mathbb{R})$, and therefore the same action on $\mathbb{H}^2$. The answer to your question is yes for punctured torus groups. This follows from results of Brian Bowditch. A representation of the punctured torus group in $SL(2,\mathbb{R})$ corresponds to a non-zero real solution of the Markoff equation. In Proposition 4.11, it is shown that conversely a non-zero solution gives a discrete fuchsian representation. So if the trace field is totally real, all non-trivial Galois conjugates will be discrete fuchsian groups.<|endoftext|> TITLE: Reference request for an early theorem of Gromov QUESTION [5 upvotes]: In his talk Misha Gromov- How does he do it, Jeff Cheeger mentions a theorem of Gromov proved sometime in the early 70's. Theorem: Every manifold admitting a sequence of metrics such that the diameter and curvature go to zero is finitely covered by a nilmanifold. Further, he says that the proof is not easy even today; something usually associated with Gromov! It would be great to read the original paper, and even better if some book contains it. Please help me find them. Thanks. REPLY [7 votes]: The book "Gromov's almost flat manifolds" by Buser and Karcher is a nice exposition of Gromov's result.<|endoftext|> TITLE: riemann mapping theorem for skew-fields of quaternions and beyond QUESTION [7 upvotes]: Is there any known generalization of the riemann mapping theorem over skew-fields of quaternions and beyond or at least a conjectured formulation of it? In a less focused way, how far does the main results of complex analysis and fourier analysis carry over to quaternions. REPLY [5 votes]: Let $\mathbb{H}$ denote the field of quaternions and $\mathbb{S}$ denote the unit sphere of imaginary quaternions. We have a Riemann mapping theorem for axially symmetric sets inside the field of quaternions $\mathbb{H}$. The set $\Omega$ is axially symmetric if the set $\{ x+Jy~|~J\in\mathbb{S} \},~J\in\mathbb{S}$ are contained in $\Omega$ (we have rotation invariance around the axis). Let $\Omega$ be an axially symmetric open set in $\mathbb{H}$. Extend $\Omega$ to $\mathbb{C}_I$ (the restricted complex plane that can be identified with $\mathbb{C}$) for $I\in\mathbb{S}$. Then if $\mathbb{B}$ is the open unit ball, then we have a bijective mapping of a quaternionic slice function $f$ that bijectively maps from $\Omega$ to $\mathbb{B}$. Check out the following paper by Sorin Gal and Irene Sabadini: http://arxiv.org/pdf/1406.5516.pdf<|endoftext|> TITLE: Euclidian norm of Gaussian vectors QUESTION [15 upvotes]: Let $X \sim \mathcal{N}(0, \Sigma)$ be a Gaussian vector in dimension $N$. I am interested by the probability density of the random variable $\lVert X \lVert_2$. If $\Sigma = {I}_N$, we recognize the $\chi$-law. We especially know that the probability density is given by $$p(x) \propto x^{{N} -1} \mathrm{e}^{-\frac{x^2}{2}} 1_{x\geq 0} .$$ In the general case, we can decompose the matrix $\Sigma = P^t D P$ with $P$ orthogonal and $D=D(\lambda_1,\cdots,\lambda_N)$ diagonal and $\lVert X \lVert_2 \sim \lVert \mathcal{N} (0,D)\lVert_2$. What can we say about the probability density of $\lVert X \lVert_2$ in this general case? Thanks by advance. REPLY [12 votes]: After you diagonalize the covariance matrix, you have $||X||^2=\sum_{k=1}^n X_i^2$, right? And $X_1^2,\ldots,X_n^2$ are independent $\chi(1)$-distributed r.v. (up to a scaling with $\frac{1}{\sqrt{\lambda_i}})$, i.e., they are gamma-distributed, but with different scale parameter. So you can apply the results of this paper: P. G. Moschopoulos (1985) The distribution of the sum of independent gamma random variables, Annals of the Institute of Statistical Mathematics, 37, 541-544. See also A. M. Mathai (1982) Storage capacity of a dam with gamma type inputs, Annals of the Institute of Statistical Mathematics, 34, 591-597.<|endoftext|> TITLE: Is there a largest prime p such that J_0(p) completely splits into elliptic curves QUESTION [7 upvotes]: The question in the title is related to a more general question. Namely does there exist an integer $N$ such that for all curves $C/\mathbb C$ of genus $> N$ one has that not all simple isogeny factors of $J(C)$ are elliptic curves. Now since over $\mathbb Q$ the isogeny factors of $J_0(p)$ correspond to galois orbits of newforms it seems to me that the question wether there is a largest prime $p$ such that $J_0(p)$ splits completely into elliptic curves over $\mathbb Q$ should be easier and I wonder wether the answer is already known. I suspect that the answer is yes and that $p = 37$ is the largest prime such that $J_0(p)$ completely splits into elliptic curves. Indeed using Cremona's database of elliptic curves I verified that 37 is the biggest prime below 300000 such that $J_0(p)$ completely splits into elliptic curves. The reason I only ask it for prime levels $p$ is that I already showed that if $p = 37$ is indeed the largest prime such that $J_0(p)$ completely splits into elliptic curves, then $N = 1200$ is the largest composite number such that $J_0(N)$ completely splits into elliptic curves. REPLY [9 votes]: In fact we know the list of all $N$, whether even or odd, for which $J_0(N)$ is isogenous to a product of elliptic curves. See Takuya Yamauchi, On $\mathbb Q$-simple factors of Jacobian varieties of modular curves, Yokohama Math. J. 53 (2007), no. 2, 149-160. An alternative proof (which also corrects a minor error in Yamauchi's list) is given in section 5 of the recent paper Noam D. Elkies, Everett W. Howe, and Christophe Ritzenthaler: Genus bounds for curves with fixed Frobenius eigenvalues, Proc. Amer. Math. Soc. 142 (2014), 71-84. arXiv: 1006.0822. See page 82 of the online version or page 12 of the arXiv preprint.<|endoftext|> TITLE: Eigenvalues of generalized Vandermonde matrices QUESTION [5 upvotes]: Given a strictly increasing sequence $00$) shows that the coefficients are always alternating? If this observation is true, then there is perhaps a (more or less) naturally associated symmetric matrix lurking behind. REPLY [8 votes]: Expanding on my comment, the phenomenon that you are observing is explained by noting that: The kernel $x^y$ is strictly totally positive (STP), i.e., for any choices of $0 < x_1 < \cdots < x_n$ and $y_1 < y_2 < \cdots < y_n$ the matrix $K_{ij} = x_i^{y_j}$ is (strictly) totally positive, i.e., all minors are positive. Since it is STP, its eigenvalues are also positive---for a proof see Spectral Properties of Totally Positive Kernels and Matrices, in Total Positivity and its Applications, M. Gasca, C. A. Micchelli (eds.), pp. 477-511, Kluwer Academic Publishers, 1996 For a proof of the STP property of this kernel, notice that it is essentially a special case of the $e^{xy}$ kernel (since $x_i > 0$, we can write it as an exponent). The STP property of $e^{xy}$ is well-studied, and can be found for example in Pinkus's recent book (Chapter 4): Totally Positive Matrices.<|endoftext|> TITLE: Is the following consequence of the Lang conjecture known? QUESTION [6 upvotes]: This came up in a discussion with a colleague of mine, who studies PDEs. He was asking for a function $f \colon \mathbb{N} \rightarrow \mathbb{N}$ such that, for all but finitely many $n$, the equation $$ x^4 + y^4 = n $$ has at most $f(n)$ solutions in integers $x,y$. My own feeling was that there should be a constant function $f$ that does the trick. Indeed, if the "weak Lang conjecture" (see below) is true, then even the number of rational solutions to the above equation is bounded uniformly over all $n$. Assuming the weak Lang conjecture then, we can take $f$ to be a constant function. My question is whether we can prove this statement without assuming any unknown conjecture: Does there exist $N \in \mathbb{N}$ such that, for all $n \in \mathbb{N}$, the equation $x^4+y^4=n$ has at most $N$ solutions in integers $x,y$? The statement of the weak Lang conjecture is as follows: if $X$ is a variety of general type over a number field $K$, then the set $X(K)$ of rational points of $X$ is not Zariski dense. In their article "Uniformity of rational points" (JAMS, 1997), Caporaso, Harris, and Mazur prove that this conjecture implies the existence of constants $B(K,g)$ such that every smooth genus $g$ curve $X$ over a number field $K$ has at most $B(K,g)$ rational points (see Theorem 1.1 in their paper; note that they probably assume $X$ irreducible as well, although they do not explicitly state this). REPLY [6 votes]: Here are two papers that do not allow to conclude positively to your question. The reason is that one does not know how to bound uniformly the Mordell-Weil rank of abelian varieties of given dimension over a given number fields, even in well behaved families. 1. A general bound depending on the size of the coefficients In his paper ``Borne polynomiale pour le nombre de points rationnels des courbes'', Journal de théorie des nombre de Bordeaux 23 no. 1 (2011), p. 251-255, Gaël Rémond has given a general explicit bound for the number of solutions of polynomial equations in two variables with coefficients in a number field, assuming the equation has finitely many solutions, of course. By Faltings's theorem, this is the case of your curve, so his bound applies and says that there are at most $n^{2^{3^{16}}}$ solutions. NB. The exponent of $n$ is equal to $1.721783764\, 10^{12958354}$ and this bound is both unpractical and certainly non-optimal. NB. Rémond bounds the Mordell-Weil rank in terms of the size of the equation. 2. A bound for twists of a given curve The paper A uniform bound for rational points on twists of a given curve, J. London Math. Soc. (2) 47 (1993) 385-394, by Joseph Silverman shows a partial uniformity of the number of rational points among twists of a given curve. However, the bound depends on the Mordell-Weil rank of the specific curve, so is not really uniform. He also gives the example of Catalan curves of the form $ax^m+bx^n=1$, where $m$ and $n$ are fixed and $a,b$ varies among non-zero rational numbers.<|endoftext|> TITLE: Examples of "nice" properties of algebraic extensions of $\mathbb{Q}$ QUESTION [5 upvotes]: I am writing a short survey of some "nice'' properties of algebraic extensions of $\mathbb{Q}$. Let's say a property (P) is nice if every finite extension of $\mathbb{Q}$ satisfies (P), and if $K \subseteq L \subseteq \overline{\mathbb{Q}}$ and $L$ satisfies (P), then $K$ satisfies (P). For the property to be interesting, there should be some example of an infinite extension of $\mathbb{Q}$ that does NOT satisfy it. Edit: and there should be an example of an infinite extension that DOES satisfy it! Here are some examples, described in terms of an algebraic extension $K/\mathbb{Q}$: -$K^\times/K^\times_{tors}$ is free abelian. (Or replace $\mathbb{G}_m$ with something else (e.g. abelian variety) and ask the same question). -The ring of integers in $K$ is a Noetherian (and hence a Dedekind domain), or more generally a Lasker ring. These properties are discussed at length in chapter 12 of Ribenboim's book, The Theory of Classical Valuations. -The Northcott and Bogomolov properties (defined in Bombieri and Zannier's paper A note on heights in certain infinite extensions of $\mathbb{Q}$). These properties have analogous statements in terms of heights on abelian varieties, etc. -Conditions on the completions of $K$ at various places, e.g. $K_v$ is a local field for some place $v$, for all $v$, for all $v$ above a fixed rational prime... The case where $K_v$ is a local field with uniformly bounded local degrees is described in detail by Checcoli, Fields of algebraic numbers with bounded local degrees and their properties. If you can think of other properties that fit in this list, please give one per answer, ideally with an explanation of why it is "nice", if it's not obvious. REPLY [9 votes]: For $E$ a fixed elliptic curve (or more generally, abelian variety) over $\mathbb Q$, the condition "$E(K)$ is a finitely generated abelian group" is an answer to your question. It is satisfied by any finite extension $K$ of $\mathbb Q$, by the theorem of Mordell-Weil. If $K \subset L$ and the condition is satisfied by $L$, then it is also satisfied by $K$ since $E(K)$ is a subgroup of $E(L)$ and a subgroup of a finitely generated abelian group is still finitely generated. The condition is definitely not satisfied for $K=\overline{ \mathbb Q}$. But a theorem of Ribet (combined with results of Kato, Rohrlich and Greenberg) states that this condition is satisfied for $K$ the maximal abelian extension of $\mathbb Q$ unramified outside a fixed set of primes. For other infinite extensions of $\mathbb Q$, the situation is very complex (certainly not fully understood even conjecturally), but there are a good deal of deep conjectures, and a good lot of them proved, concerning extensions of number fields whose Galois group is an analytic pro-$p$-group: this is the so-called "Iwasawa theory for elliptic curve". Of course, the abelian Iwasawa theory, where we deal with abelian such extension, that is of Galois group $\mathbb Z_p^n$, is more advanced, but by no means complete. For a thorough discussion of that theory, see Greenberg's Introduction to Iwasawa Theory for Elliptic Curves.<|endoftext|> TITLE: Robinson Arithmetic and Composite Numbers QUESTION [6 upvotes]: Define a number $n$ to be composite if it can be written as $a\cdot b$ for some $a,b$ where $a,b\neq 1$. Define $p$ to be prime if $p=a\cdot b$ implies $a=1$ or $b=1$. The theorem that every composite number has a prime factor seems to require a bit more induction than just everything is either 0 or a successor. So what is a model of Robinson Arithmetic where that theorem fails? REPLY [11 votes]: The nonnegative part of a discrete ordered ring always satisfies Robinson Arithmetic, so many examples can be found there. For a specific one, take the ring $R$ of formal Puiseux polynominals of the form $$a = a_0 + a_1 T^{q_1} + \cdots + a_k T^{q_k}$$ where $0 < q_1 < \cdots \lt q_k$ are rationals and $a_0,a_1,\ldots, a_k$ are integers. The order is given by $a \lt b$ if the most significant term of $b-a$ is positive (i.e. we think of $T$ as an infinitely large number). In this ring $T$ is composite (e.g. $T = T^{2/3}\cdot T^{1/3}$) but it has no prime factor.<|endoftext|> TITLE: Area covered by Brownian motion of 2D disc QUESTION [8 upvotes]: I would like to know the expected value for the area covered by a disc of radius $R$ whose center undergoes Brownian motion (diffusion). Specifically, let $\mathbf{X}_t$ represent a two-dimensional Brownian motion, and define the covered set $C_t = \bigcup_t B(\mathbf{X}_t,R)$ to be the union of discs centered at every point along the realized trajectory. Define $A_t$ to be the area of the set $C_t$. What is the expected value $\mathbb{E}[A_t]$ as a function of $t$? I am specifically interested in this for the mathematical modeling of a chemical reaction and would be happy with an approximate solution that works in the short-to-moderate time regime, where $|x_t|$ is at most (say) $5R$. I think that this time regime is too short to use the lattice random walk result for the number of distinct sites ($t / \log t$ from Dvoretzky and Erdös). Thank you. REPLY [4 votes]: Your process $C_t$ is better known by the amusing name of Wiener sausage, and its expected area $E[A_t]$ has been extensively studied (though more so for long times than short). I found a 2010 paper by Yuji Hamana which gives small-time asymptotics for the area of the $d$-dimensional Wiener sausage; for $d=2$ it looks like we have, as $t \to 0$, $$E[A_t] \sim \pi R^2 + \sqrt{8 \pi t} R + o(\sqrt{t}).$$ There's also a 2013 paper by Kôhei Uchiyama that obtains a similar result. MR2761916 Hamana, Yuji. On the expected volume of the Wiener sausage. J. Math. Soc. Japan 62 (2010), no. 4, 1113–1136. Open access at Project Euclid. MR2988115 Uchiyama, Kôhei. The expected area of the Wiener sausage swept by a disc. Stochastic Process. Appl. 123 (2013), no. 1, 191–211.<|endoftext|> TITLE: Wanted: a "Coq for the working mathematician" QUESTION [80 upvotes]: Sorry for a possibly off-topic question -- there are four StackExchange subs each of which could be construed as the proper place for this question, and I've just picked the one I'm most familiar with. I'm trying to obtain a working knowledge of the Coq programming language sufficient to be able to encode and (have it) verify constructive proofs in combinatorics and abstract algebra. The proofs, or at least a portion of them high enough to make me happy, are not really probing the boundaries of constructive mathematics; they require no univalence, no coinduction, and probably not even the full strength of inductive types other than $\mathbb N$ (think Peano arithmetic). Obviously my first idea was to scour the internet for tutorials, but I found the kind of introduction that I want to be surprisingly scarce if not unavailable. So I'm wondering if someone knows such a source or one is currently being written (if so, how can one contribute?). Here is an outline of what I am looking for: The focus should be on using Coq for verifying proofs in constructive mathematics, or in even more conservative subsets of it. It should not be on playing around with additional axioms (univalence, classical logic) or other systems (temporal logic etc.); nor should it be on practical software verification. It also shouldn't be a logics textbook with Coq used as illustration for the concepts. I'm abstractly interested in each of the things just mentioned, but what I really am looking for is a tutorial written for mathematicians in general rather than for logicians or HoTTists. It should provide hands-on examples of proofs which aren't totally toylike or artificial. (I imagine things like "the conjugate of the conjugate of a partition is the old partition" or "if you sort each row of a matrix in increasing order and then do the same with each columns, then the rows are still increasing" or "$-1$ is a quadratic residue modulo the odd prime $p$ iff $p \equiv 1 \mod 4$".) Ideally it should give some guidance on how to set up a Coq project (i. e., more than one .v file) and how to write literate Coq/TeX. The text should provide the reader with a survival kit of syntax and basic tactics as early on as possible that theoretically allows formalizing any constructive proof in elementary combinatorics and number theory with enough patience. It should then go on with more advanced tactics that make this less of a pain, possibly even those from ssreflect (if that package is sufficiently stable). The texts I'm currently trying to follow are Coq'Art and Software Foundations, but it seems they don't really fit the above description very much (and the only English version of Coq'Art is from 2004, which isn't very recent in computer science reckoning). EDIT: Slightly off-topic, I'm also wondering what it would take to release Coq bundled with its most important user-made contributions like ssreflect, making the learning curve at least a little bit less steep by removing the PITA of compiling an ocaml ecosystem from source... EDIT 2: Florent Hivert's Coq-Combi project seems to be the thing I was trying to build, and I suspect that reading its sources will be a good step towards learning Coq at least for me. On the other hand, Pierre-Yves Strub offers a bundle with Coq and SSReflect for Windows, which solves another problem I was having (the caveat being that the versions are not the newest). It looks like Coq is becoming usable :) EDIT3: I feel that, with the progress done by now (particularly by Florent Hivert and others on Coq-Combi), I could learn Coq and get sufficiently experienced in it within half a year of not having to worry about publications, teaching and pretending to keep track of current developments. I am wondering if this is mainly me, or everyone in combinatorics is a half-year away from being able to teach their work to their computer. Meanwhile, I would like to share a talk by Neil Strickland I have just discovered, which is a far better rant about the current state of affairs than I could ever write. REPLY [5 votes]: A relatively new project, Lean, has created a lot of activity lately, see Lean and Mathlib with its lots of learning material, driven in part by successful attempts by Kevin Buzzard to teach Lean to maths undergraduates.<|endoftext|> TITLE: Who proved "sets in every generic are already in the ground model?" QUESTION [11 upvotes]: Suppose $\mathbb{P}$ is a notion of forcing in the ground model $V$, and $X$ is a set which is in $V[G]$ for every $\mathbb{P}$-generic filter $G$. Then $X\in V$ already, by a fairly simple (if tedious) argument. I'm working on a paper in which this fact is mentioned, and I would like to cite it properly; however, I can't seem to find any citation for it, and I don't recall when I learned the argument. So my question is: What is a citation for "if $X$ is in every forcing extension by some poset $\mathbb{P}$, then $X$ is already in the ground model?" REPLY [5 votes]: Here is the computability-theoretic version: Suppose $G=G_0\oplus G_1$ is 1-generic and suppose $X$ is computable from both: so $X=\Phi_1^{G_0}=\Phi_2^{G_1}$. Then also $X=\Phi^{G_0}=\Phi^{G_1}$ where (Posner's trick) we define $\Phi$ by: noneffectively pick a number $n_0$ with $G_0(n_0)\ne G_1(n_0)$ and use this to decide whether to run $\Phi_1$ or $\Phi_2$. The set $S := \{\sigma = \sigma_0\oplus\sigma_1: \Phi^{\sigma_0}(n)\downarrow \ne \Phi^{\sigma_1}(n)\,\exists n\}$ is $\Sigma^0_1$ and by assumption $G$ does not meet it, i.e., $\sigma\in S$ implies $\sigma$ is not a prefix of $G$. Then by 1-genericity there is a prefix of $G$, $\tau$, such that no extension of $\tau$ lands in $S$. Then $X$ is computable. Indeed, given $n$ we may search for any extension of $\tau$ making $\Phi^{\tau_0}(n)$ converge; then $X(n)=\Phi^{\tau_0}(n)$.<|endoftext|> TITLE: What other monoidal structures exist on the category of sets? QUESTION [36 upvotes]: I know of the following monoidal structures over $\mathbf{Set}$ (taken from here): The Cartesian product $\otimes=\times$ (categorical product) The disjoint union $\otimes=+$ (categorial coproduct) $A\otimes B = A\times S\times B + A + B$ for some given set $S$ $A\otimes B = (A\times S\times B\times S)^* \times (A + A\times S \times B) + (B\times S\times A\times S)^* \times (B + B\times S \times A)$ where $*$ is the Kleene star (free monoid construction) If $A=\emptyset$ then $A\otimes B=B$, if $B=\emptyset$ then $A\otimes B=A$, otherwise $A\otimes B=I$ I want to know if there are any more that are known or, conversely, if there are any demonstrations that additional examples don't exist. Also, what if we restrict to, say, braided or symmetric monoidal categories, or traced monoidal categories (or some example), are there any results/examples regarding the existence/non-existence of monoidal structures over $\mathbf{Set}$ in these cases? REPLY [3 votes]: Here is another possible source of tensor products on $\mathbf{Set}$: transferring tensor products from categories $\mathbf{Set}$ is coreflective in. Unfortunately I have no good example. Suppose $(\mathbf{C},\otimes,I)$ is a monoidal category with a coreflection from $\mathbf{Set}$; that is, there are functors $F\colon \mathbf{Set} \to \mathbf{C}$ and $G \colon \mathbf{C} \to \mathbf{Set}$ such that $F$ is left adjoint to $G$ with isomorphic unit $\eta_X \colon X \to GF(X)$. Moreover, suppose that $FG \colon \mathbf{C} \to \mathbf{C}$ is monoidal. Then $X \otimes Y := G(FX \otimes FY)$ is a monoidal product on $\mathbf{Set}$. (See e.g. Theorem 5 here.) For example, these conditions are automatically fulfilled if $\mathbf{C}$ is the category of topological spaces and $G$ is the forgetful functor. However, a quick browse gives results classifying the tensor products on $\mathbf{Top}$ (*), so this is probably not a good source of examples. Are there others? (*) That are canonical - meaning they just give the tensor product $\times$ on $\mathbf{Set}$ - and closed; but we are precisely looking for noncanonical ones...<|endoftext|> TITLE: Expected maximum inner product QUESTION [7 upvotes]: If you sample $n$ vectors each with $m$ entries, with each entry chosen from the set $\{-1, 1\}$, how can you calculate the expected maximum absolute value of the inner product between all pairs of vectors? That is let us call the vectors $v_i$ and let $X_{n,m} = \max_{i \ne j} |\langle v_i,v_j \rangle|$. I would like $\mathbb{E}(X_{n,m})$. We can assume both $n$ and $m$ are large. We can also assume that $n^c \leq m \leq n^{d}$ for constant $c,d > 0$. REPLY [9 votes]: Depending on how sharp an answer you want, concentration inequalities might suffice. For one pair of (random) vectors, the inner product $X$ is distributed the same way as the sum of $m$ independent $\pm 1$-valued random variables. The simplest Chernoff inequality tells us that $$ \mathbb P(|X| > a) < 2e^{-a^2/2m}. $$ So $|X|$ is exponentially unlikely to take values greater than anything just a tiny bit larger than $\sqrt m$. Since $m$ and $n$ are related polynomially, summing over the $n^2$ pairs of vectors doesn't make the failure probabilty substantially greater, so with very high probability $|X_{m,n}|$ will be less than $m^{1/2+o(1)}$.<|endoftext|> TITLE: Is Grothendieck classification of tensor norms and Kuratowski's 14 sets theorem somehow related? QUESTION [19 upvotes]: It is known that there are only 14 reasonable tensor norms in $Ban$. On the other hand it is well known fact for topologists that one can obtain only 14 different sets from a given set applying closure and complementation operations. It seems like a pure coincidence but... both constructions starts with the single object (a set/ the injective tensor norm), then apply only predescribed operations (closure and complement/ transpose, dual and right projective associate) and get only 14 different objects in the end. Is this a simple coincidence or there is something deep behind the scenes? Maybe someone knows other examples? REPLY [9 votes]: I asked David Sherman this question and this was his response: In the Kuratowski setup, if you omit complementation, you get 7 elements. Adding complementation gives you a disjoint upside-down copy. In the Grothendieck setup, if you omit duality, you get 8 elements (two five-element chains with common inf and sup, like an O). Adding duality gives you an upside-down copy, but it is not disjoint -- two elements agree between the copies. '' So, it seems that the answer is NO.<|endoftext|> TITLE: Irreducible polynomial QUESTION [14 upvotes]: Let n be a positive integer greater than 1 and $p_n(x)$ be a polynomial of degree $n$ such that $p(0)= 2, p(1)=3, p(2)=5$, and in general $p(k)$ is the $(k+1)$th prime, for $0\leq k \leq n$. Is $p_n(x)$ irreducible in $\mathbb Q [x]$? REPLY [7 votes]: "Most" polynomials are irreducible. More precisely, in this case, the heights of the coefficients of $p_n(x)$ are (at least) exponential in $n$. Among all polynomials of degree $n$ whose coefficients have height on the order of $C^n$ for some fixed $C>1$, only a tiny proportion will be reducible. So once you've checked that $p_n(x)$ is irreducible for the first few $n$, say $n\le10$, it then becomes likely that the rest of them will be irreducible. But I doubt that this has anything to do with the primes. If you took some other "random looking" sequence of integers $a_k$ that grows appropriately and set $p_n(k)=a_k$ for $0\le k\le n$, you should get similar properties. For example, something weird like $a_k=\lfloor \pi (k+1) \log(k+2)\rfloor$.<|endoftext|> TITLE: Explicit calculation of Weil Deligne representations QUESTION [7 upvotes]: According to Grothendieck monodromy theorem, l-adic galois representations of a local field corresponds to Weil-Deligne representations. However, given a galois representation, it is usually difficult to find the Weil-Deligne representations. My questions are: (1) Is it possible to describe explicitly the Weil-Deligne representation associated to the Tate module of an elliptic curve over a local field? How about Tate curve (corresponding to a prime element) for example? (2) In the situation as in (1), is it possible to calculate the L function and epsilon factor associated to the Weil-Deligne representation? (3) If these are possible, how can one do for Tate curves? Please give me any advice! REPLY [5 votes]: Yes, it is possible. For all this explained clearly and in detail, see David Rohrlich's paper "Elliptic curves and the Weil-Deligne group" along with the accompanying "Student's supplement to "Elliptic curves and the Weil-Deligne group" if needed.<|endoftext|> TITLE: Factor subset of finite group QUESTION [5 upvotes]: Let $G$ be a group of order $n$ and $d$ a positive divisor of $n$. Is it true that there exists a subset $A$ of $G$ with $d$ elements and a subset $B$ such that $G=AB$ and $|AB|=|A||B|$ (equivalently, the product $AB$ is direct)? Note: If $G$ has the property that there exists a subgroup of $G$ with order $d$ or $n/d$, then we can show that the answer is positive (see A question about finite groups (a weak version of the converse of Lagrange theorem)). Now, is the answer positive for ${\rm PSL}(2,8)$, ${\rm PSL}(2,11)$, ${\rm PSL}(2,13)$, ${\rm SL}(2,11)$, ${\rm SL}(2,13)$, ${\rm PSL}(2,17)$, ${\rm A}_7$, ${\rm PSL}(2,19)$, ${\rm A}_5 \times {\rm A}_5$? About $PSL_2(8)$ (in the last following comment, by Russ Woodroofe). If it has subgroups of indexes $6$, $14$ and $21$ with the property $P(*,*)$, then we can say that $PSL_2(8)$ has $P(*,*)$. Note that $PSL_2(8)$ has subgroups of indexes $6$ (that is solvable and so $P(*,*)$ is true for it), $14$ (that is also solvable) and $21$ (with order $84$). Now, is $P(*,*)$ valid for the last subgroup? (if yes, then $PSL_2(8)$ has $P(*,*)$). REPLY [6 votes]: You can handle several of the groups on the list from your edit with a modification of the argument of Marty Isaacs. For the rest, you can find guidance as to what a counterexample would look like. Say that $G$ satisfies $P(a,b)$ if $ab = |G|$ and there are subsets $A$ and $B$ of cardinalities $a$ and $b$ such that $G = AB$. Write $P(*,*)$ when $G$ satisfies $P(a,b)$ for any $a,b$ with $ab = |G|$. The following has exactly the same proof as the solvable argument. Lemma: If $H$ is a subgroup of $G$ satisfying $P(a,b)$, and $[G:H] = t$, then $G$ satisfies $P(a,bt)$. Then if $G$ has a maximal subgroup of prime index satisfying $P(*,*)$, then $G$ also satisfies $P(*,*)$. That handles $A_5$, $PSL_2(7)$, and $A_5 \times A_5$. Moreover, if $G$ has a maximal subgroup of index $t$, then $G$ satisfies $P(a,b)$ for every pair $a,b$ with t dividing $b$ (and of course $ab=|G|$). This lets us handle $A_6$, which has maximal subgroups of index 6, 10, and 15. If $ab = 60$ then either $a$ or $b$ is divisible by one of these. $A_7$ then follows, since $A_6$ is a maximal subgroup of $A_7$ having (prime) index 7. Similar (but slightly more involved) arguments to those for $A_6$ show that $A_8$ satisfies $P(*,*)$. The first open case is $PSL_2(8)$, which has order $504 = 2^3 \cdot 3^2 \cdot 7$, and maximal subgroups with indices 9, 28, and 36. Since every subgroup satisfies $P(*,*)$, the possible places where $PSL_2(8)$ could fail $P(*,*)$ are $P(12, 42)$ and $P(21, 24)$. By the way, GAP will calculate the relevant data for these arguments for you pretty easily. The following commands will do it (replace $G$ with your favorite group): G:=PSL(2,7);; Print(Size(G), " = ", FactorsInt(Size(G)), "\n"); Set(List(MaximalSubgroups(G), x->Index(G, x)));<|endoftext|> TITLE: Can an ultraproduct be infinite countable? QUESTION [8 upvotes]: In exercise 4 page 456 of Hodges "Model Theory" it is required to show that if an ultrafilter $\mathcal{U}$ is not $\omega_1$-complete, then every ultraproduct $\prod_I A_i/ \mathcal{U}$ has cardinality $< \omega$ or $\geq 2^\omega$. Does this suggest there can be ultraproducts $\prod_I A_i/ \mathcal{U}$ which have cardinality $\omega$ if we assume that the ultrafilter is (non-principal) $\omega_1$-complete? Is this possible? REPLY [12 votes]: Yes, but the existence of such ultrafilters is a large cardinal hypothesis; it is equivalent to the existence of a measurable cardinal. If each $A_i$ is countably infinite and $\cal U$ is countably complete ($\omega_1$-complete), then the ultraproduct $\Pi_i A_i/{\cal U}$ will be countable. To see this, fix enumerations of each $A_i$, and observe that any element $x\in\Pi_iA_i/{\cal U}$ is represented by some function $\vec x=\langle x_i\rangle_i$, where $x_i$ is the $n_i^{th}$ element of $A_i$. But by countable completeness, the value of $n_i$ must be constant on a set in $\cal U$, and so the function $n\mapsto [g_n]_{\cal U}$ where $g_n(i)$ is the $n^{th}$ element of $A_i$ is a bijection of $\omega$ with the ultraproduct. So it is countably infinite, as desired. Meanwhile, it is important to note that every countably complete nonprincipal ultrafilter $\cal U$ on any set is actually $\kappa$-complete for some measurable cardinal $\kappa$, and so the hypothesis carries large cardinal strength. A stronger result is possible, once you realize that any countably complete ultrafilter is actually $\kappa$-complete for a much larger cardinal. If $\cal U$ is a $\kappa$-complete ultrafilter on a set $I$ and the structures $A_i$ are uniformly bounded in size below $\kappa$, in a language of size less than $\kappa$, then the ultrapower $\Pi_i A_i/{\cal U}$ is isomorphic to one of the $A_i$. The reason is that we may assume $\kappa$ is a measurable cardinal, and in particular, it is also inaccessible. And so there are fewer than $\kappa$ many structures in that language of size at most $\delta$, for any fixed $\delta<\kappa$. It follows by $\kappa$-completeness that the measure $\cal U$ must concentrate on a set of indices $i$ for which the $A_i$ are all isomorphic to each other. And it follows by the reasoning in the first argument that this common $A_i$ is also isomorphic to the ultrapower $\Pi_i A_i/{\cal U}$. The result mentioned in the previous paragraph is the size $\kappa$ analogue of the fact for ultrafilters $\cal U$ on $\omega$ that the ultrapower $\Pi_i A_i/{\cal U}$ of structures $A_i$ having uniformly bounded finite size in a finite language is isomorphic to one of them, namely, to the one whose isomorphism type occurs on a set in ${\cal U}$.<|endoftext|> TITLE: How many unit cylinders can touch a unit ball? QUESTION [64 upvotes]: What is the maximum number $k$ of unit radius, infinitely long cylinders with mutually disjoint interiors that can touch a unit ball? By a cylinder I mean a set congruent to the Cartesian product of a line and a circular disk. The illustrations, copied from 2, show various configurations of six cylinders, perhaps all possible - up to isometry. The question is about 25 years old; the answer is conjectured to be 6, but the conjecture is still unconfirmed. Heppes and Szabó 1 proved in 1991 that $k\le 8$; nine years later Brass and Wenk 2 improved this result to $k\le 7$. Also, one would like to know: What does the configuration space of six unit cylinders touching a unit ball look like? In particular: Is the configuration space connected? Even more specifically, referring to the configurations shown below: Within the configuration space, is a continuous transition possible from the configuration in Figures 1 and 2 to the configuration in Figure 3? The last question may be possible to verify by a natural candidate, but the necessary computations seem too tedious to me... 1 Heppes, Aladar and Szabó, Laszló. "On the number of cylinders touching a ball." Geom. Dedicata 40 (1991), no. 1, 111–116; MR1130481. 2 Brass, Peter and Wenk, Carola. "On the number of cylinders touching a ball." Geom. Dedicata 81 (2000), no. 1-3, 281–284; MR1772209. REPLY [6 votes]: For a javascript rendered 3D model of the path of configurations reported by Ogievetsky and Shlosman in arXiv:1805.09833 see this page on my website. The configurations in this path are composed of two "three-pole teepees" pointed in antipodal directions. This is also true of the Figure 3 configuration in the question. But a crucial difference is that in Ogievetsky and Shlosman's configurations, the two teepees are both right-handed. In Figure 3, the two teepees have different handedness. You can set up the calculation of OS in the case of opposite-handed teepees. Just switch $\delta$ to $-\delta$ for $D$, $E$, and $F$ in their Equation (5). Now solving for $d_{AB}=d_{AD}=d_{BD}$ is even easier. From $d_{AD}=d_{BD}$ I get $\varkappa=0$, and from $d_{AB}=d_{AD}$, I get $$s+\frac{3(s-1)}{t+4-3s}+\frac{s-s^2}{s+t}=0\text,$$ where $s=(\sin\phi)^2$ and $t=(\tan\delta)^2$. There are two real solutions for $t$ when $0\le s\le \tfrac13$, and then the solutions become complex for $\tfrac13 TITLE: Is any continuous group homomorphism from R to C* an exponential map? QUESTION [5 upvotes]: Consider $\mathbb{R}$ to be an additive topological group, and $\mathbb{C}^{\ast}$ to be a multiplicative topological group. Is the following statement true? If so, then how can one prove it? Statement: any continuous group homomorphism $\mathbb{R} \to \mathbb{C}^{\ast}$ is always of the form $exp(\lambda x)$, $\lambda \in \mathbb{C}^{\ast}$. REPLY [7 votes]: Yes. First, decompose $\mathbf{C}^\times$ as $\mathbf{R}_{>0}\times\mathbf{T}$, where $\mathbf{T}=\{z\in\mathbf{C}^\times:\vert z\vert=1\}$. We have $\mathbf{R}_{>0}\cong\mathbf{R}$ via the logarithm, and any continuous homomorphism $\mathbf{R}\rightarrow\mathbf{R}$ is necessarily $\mathbf{R}$-linear, and hence given by $x\mapsto sx$ for some (unique) $s\in\mathbf{R}$. The set of continuous homomorphisms $\mathbf{R}\rightarrow\mathbf{T}$ is the Pontryagin dual of $\mathbf{R}$, which is isomorphic to $\mathbf{R}$ via $s\mapsto(x\mapsto\exp(ixs))$. Combining these facts, one finds that the continuous homomorphisms $\mathbf{R}\rightarrow\mathbf{C}^\times$ are of the form $x\mapsto \exp(zx)$ for $z\in\mathbf{C}$. REPLY [4 votes]: It is certainly true that the only $differentiable$ homomorphisms are exponential maps. To see this let $f$ be a differentiable homomorphism, and note that $f(0) = 1$. Write $f'(0) = a$, so $\lim_{t \to 0} (f(t) - 1)/t = a$. Now we compute $f'(x)$ for arbitrary $x$. We have $$ f'(x) = \lim_{t \to 0} \frac{f(x+t) - f(x)}{t}, $$ and $f(x+t) = f(x)f(t)$ since $f$ is a homomorphism. This yields $$ f'(x) = \lim_{t \to 0} f(x)\frac{f(t) - 1}{t} = a f(x). $$ The unique solution to this differential equation with initial condition $f(0) = 1$ is $f(x) = \exp(ax)$. Added later:${\ \ }$ I now have a complete proof using only first-year calculus techniques. This does not rely on my previous argument. Define $F(x) = \int_0^x f(t) dt$, and note that $F(x) \ne 0$ for $x \ne 0$ and $F'(x) = f(x)$ by the fundamental theorem of calculus. For a fixed constant $y$, make the change of variables $u = t + y$. This yields $$ F(x) = \int_y^{x+y} f(u-y) du = \frac1{f(y)}\int_y^{x+y} f(u)du = \frac1{f(y)} (F(x+y) - F(y)). $$ This yields $$ f(y)F(x) + F(y) = F(x+y) = f(x)F(y) + F(x), $$ where the second equality follows by interchanging $x$ and $y$. Then $F(x)(f(y) - 1) = F(y)(f(x) - 1)$ for all $x,y$. We can assume $f$ is not the constant $1$, so choose $a$ with $f(a) \ne 1$ and write $k = F(a)/(f(a) - 1)$, so $k \ne 0$ . Then $F(x) = k(f(x) - 1)$, and we have $f(x)= F(x)/k + 1$. Since $F$ is differentiable, so is $f$, and differentiation yields $f'(x) = F'(x)/k = f(x)/k$. Then $f(x) = \exp(x/k)$, and we are done.<|endoftext|> TITLE: Stable moduli interpretation of $\mathbb{R}\mathrm{P}^\infty_{-1}$ QUESTION [12 upvotes]: I attended a talk recently which closed with the following tantalizing facts: there is a naturally occurring map of spectra $$K(\mathbb{S}) \to \Sigma \mathbb{C}\mathrm{P}^\infty_{-1},$$ which can be rotated to a sequence $$\mathbb{C}\mathrm{P}^\infty_{-1} \to ? \to K(\mathbb{S}).$$ The right-hand spectrum has an interpretation as a sort of stable moduli of finite CW-complexes, and the left-hand spectrum has an interpretation (due to Michael Weiss and someone whose name I've forgotten --- very likely Ib Madsen) as a sort of moduli of Riemann surfaces, stabilized as the genus grows large. The suggestion, then, was that we find some other interesting moduli to fill in the question-mark part of the sequence. The spectrum $\mathbb{C}\mathrm{P}^\infty_{-1}$ is not so familiar to me. On the other hand, structure encoded in the spectrum $\mathbb{R}\mathrm{P}^\infty_{-1}$ is visible all over homotopy theory (two big e.g.s: the vector fields on spheres problem / Hopf invariant one theorem, and the stabilized EHP spectral sequence / Lin's theorem / the root invariant). Hearing this description of stunted complex projective space has me curious: Is a stable moduli theoretic description of the homotopy type $\mathbb{R}\mathrm{P}^\infty_{-1}$ known? (Disclaimer: I'm almost totally ignorant of this area of topology. I'm happy to correct this to a sane question if there's something askew.) REPLY [6 votes]: Let $T$ denote the tautological bundle over $BO(d)$ or $BSO(d)$. Suppose you have a bundle $E\to B$ of smooth closed $d$-manifolds, with vertical tangent bundle $V$. Using a Gysin map and a classifying map we get maps $\Sigma^\infty B_+\to E^{-V}\to BO(d)^{-T}$ of Thom spectra, and thus a map $B\to\Omega^\infty BO(d)^{-T}$ of spaces. With orientation conditions we can change the target to $\Omega^\infty BSO(d)^{-T}$. When $d=2$ we have $SO(2)=S^1$ and so the target is $\Omega^\infty \mathbb{C}P^\infty_{-1}$. Madsen and Weiss proved that when $E\to B$ is the universal bundle of Riemann surfaces of high genus, the map $B\to\Omega^\infty\mathbb{C}P^\infty_{-1}$ is homologically highly connected (but the fundamental groups are different unless you do some kind of plus construction or group completion to the source). There are various generalisations due to various combinations of Madsen, Weiss, Galatius and Randall-Williams, I don't have a good knowledge of the precise state of play. Anyway, for any not-necessarily-orientable bundle $E\to B$ of smooth closed 1-manifolds you get a map $B\to\Omega^\infty\mathbb{R}P^\infty_{-1}$. I don't know if there is a universal example where this map is an equivalence.<|endoftext|> TITLE: A Question About the Elliott-Natsume-Nest Proof of Bott Periodicity QUESTION [6 upvotes]: In Wegge-Olsen’s book K-Theory and C$ ^{*} $-Algebras, there is an outline of a proof of Bott Periodicity (the proof is due to George Elliott, Toshikazu Natsume and Ryszard Nest). The first step of the proof outline goes as follows: Suppose that there are natural transformations $ \Phi^{0} $ from the $ K_{0} S $-functor to the $ K_{1} $-functor and $ \Phi^{1} $ from the $ K_{1} S $-functor to the $ K_{0} $-functor, where $ S $ denotes the suspension functor. In other words, for every C$ ^{*} $-algebra $ \mathscr{A} $ (not necessarily unital), there are abelian-group homomorphisms $$ \Phi^{0}_{\mathscr{A}}: {K_{0}}(S(\mathscr{A})) \to {K_{1}}(\mathscr{A}) \quad \text{and} \quad \Phi^{1}_{\mathscr{A}}: {K_{1}}(S(\mathscr{A})) \to {K_{0}}(\mathscr{A}) $$ such that for every C$ ^{*} $-algebraic homomorphism $ \alpha: \mathscr{A} \to \mathscr{B} $, the following diagrams commute: $$ \require{AMScd} \begin{CD} {K_{0}}(S(\mathscr{A})) @>{{K_{0} S}(\alpha)}>> {K_{0}}(S(\mathscr{B})) \\ @V{\Phi^{0}_{\mathscr{A}}}VV @VV{\Phi^{0}_{\mathscr{B}}}V \\ {K_{1}}(\mathscr{A}) @>>{{K_{1}}(\alpha)}> {K_{1}}(\mathscr{B}) \end{CD} \quad \quad \quad \begin{CD} {K_{1}}(S(\mathscr{A})) @>{{K_{1} S}(\alpha)}>> {K_{1}}(S(\mathscr{B})) \\ @V{\Phi^{1}_{\mathscr{A}}}VV @VV{\Phi^{1}_{\mathscr{B}}}V \\ {K_{0}}(\mathscr{A}) @>>{{K_{0}}(\alpha)}> {K_{0}}(\mathscr{B}) \end{CD} $$ Suppose further that $ \Phi^{0}_{\mathbb{C}} $ and $ \Phi^{1}_{\mathbb{C}} $ are isomorphisms. Then prove that $ \Phi^{0}_{\mathscr{A}} $ and $ \Phi^{1}_{\mathscr{A}} $ are isomorphisms for every C$ ^{*} $-algebra $ \mathscr{A} $. Wegge-Olsen’s hint is to consider the commutative diagrams above for the homomorphism \begin{align*} \alpha: \mathbb{C} & \to \mathscr{A} \otimes \mathbb{K}; \\ 1 & \mapsto p \otimes e_{11}, \end{align*} where $ p $ is a fixed choice of a projection in $ \mathscr{A} $ and $ e_{11} $ is a rank-$ 1 $ projection in $ \mathbb{K} \stackrel{\text{def}}{=} {M_{\infty}}(\mathbb{C}) $. My difficulty: I do not see how \begin{align*} {K_{0}}(\alpha): & {K_{0}}(\mathbb{C}) \to {K_{0}}(\mathscr{A} \otimes \mathbb{K}), \\ {K_{1}}(\alpha): & {K_{1}}(\mathbb{C}) \to {K_{1}}(\mathscr{A} \otimes \mathbb{K}), \\ {K_{0} S}(\alpha): & {K_{0}}(S(\mathbb{C})) \to {K_{0}}(S(\mathscr{A} \otimes \mathbb{K})) \quad \text{and} \\ {K_{1} S}(\alpha): & {K_{1}}(S(\mathbb{C})) \to {K_{1}}(S(\mathscr{A} \otimes \mathbb{K})) \end{align*} are isomorphisms. If this hurdle can be overcome, then $ \Phi^{0}_{\mathscr{A}} $ and $ \Phi^{1}_{\mathscr{A}} $ are clearly isomorphisms. Thank you for any assistance! REPLY [3 votes]: A vague guess: Consider something by putting your two Diagramms together (for the map $\alpha:\mathbf{C} \rightarrow A$) so that they fuse to one diagram with three lines. In the first diagram you replace $A$ by $S(A)$ so that double Suspension $SS(A)$ is involved. The first column is then an isomorphism $K_0(SS(\mathbf{C}) \cong K_0(\mathbf{C})$. Chase in the Diagramm, to see that the second column is also something (a isomorphism (??!)) for the subgroup generated by $[p]$. Then vary over all $p$. Maybe something like this.. or a starting Point to think further.<|endoftext|> TITLE: What is "special" maximal compact subgroup of algebraig group over local field? QUESTION [5 upvotes]: Learning the theory of Langlands correspondence, I met the notion of "special" maximal compact subgroup of a (reductive) algebraic group over a local field. Here, I think the word "compact" is used in analytic meaning. The textbook I used did not explain the definition of "special". My questions: (1) What is the definition of "special" maximal compact subgroup? (2) Is there any concrete example of maximal compact subgroup which is NOT "special"? Please give me any advice. REPLY [5 votes]: A sort-of explanation, at least to connect to some little examples: a subgroup of a reductive $p$-adic group $G$ is "special" or "good" if it fixes a "special/good" vertex in the affine building. A vertex is special/good when it is special/good in every apartment in which it lies, in the building (in the unique maximal apartment system). Each apartment is an affine Coxeter complex attached to Coxeter group $(G,S)$ with generators $S$. A vertex in an affine Coxeter complex is special when its stabilizer surjects to the "linear parts" quotient of the affine Coxeter group by its translations. That is, the fixer of that point surjects to the spherical Weyl group of $G$. Examples: all maximal compacts in $SL_n(\mathbb Q_p)$ are special/good, because all vertices look the same. E.g., $SL_3(\mathbb Q_p)$'s apartments are simplicial complexes that look like the equilateral-triangle tesselation of $\mathbb R^2$. But for $Sp_4(\mathbb Q_p)$ (four-by-four matrices) there are two different types of vertices: the apartments look like planes tesselated by squares with diagonals added touching vertices $(m,n)$ with $m,n$ of the same parity (is one way to describe it in words...). Thus, half the vertices will have the vertical and horizontal $1$-simplices and four diagonals connected to them, while the other half of the vertices have only the four vertical and horizontal simplices connected to them. The former are special, the latter are not special.<|endoftext|> TITLE: history of calculus of several variables QUESTION [13 upvotes]: Everybody knows that Leibniz and Newton (or Newton and Leibniz, if you wish) invented calculus, i.e. they developed the notion of differentiability for a function of one real variable. But who had for the first time the right idea about generalizing their notions to functions of several variables? I am talking in particular about total differentiability. REPLY [10 votes]: Functions of several variables in the modern sense were never systematically considered in the 17th century. Instead the idea of partial differentiation arose in connection with families of curves, through the idea of differentiating with respect to the parameter (as one does for example when finding the envelope of a family of curves). For a synopsis of how this idea arose in the 1697 correspondence between Leibniz and Johann Bernoulli, see my review of Engelsman's Families of Curves and the Origins of Partial Differentiation (1984), an excellent study. Newton never discussed partial derivatives in any systematic way. REPLY [9 votes]: To Carlo Beenakker's references about total differentials (which came first), one might add a word about Alexis Fontaine, inventor of the notation (notion?) of partial derivative: Fontaine, Alexis Le calcul intégral. Première méthode (19 Nov. 1738), in Mémoires donnés à l'Académie royale des sciences, non imprimés dans leur temps. Imprimerie Royale, Paris, 1764. Page 25: Let $\mu$ be a function of several quantities $p, x, y, z,$ &c. To denote the coefficient of $dx$ in the difference of $\mu$, I'll write $\frac{d\mu}{dx}$; to denote that of $dy$, I'll write $\frac{d\mu}{dy}$, &c. By the same reason (...) I'll write $\frac{dd\mu}{dxdy}$ or $\frac{dd\mu}{dydx}$; for, as we shall soon demonstrate, these two expressions are one and the same thing N.B.: Although not published until 1764, Fontaine’s memoir was discussed in correspondence of Euler with Clairaut (17 Sep, 19 Oct, 26 Dec 1740) and D. Bernoulli (7 Mar & 16 May 1739). Euler, Leonhard Principia motus fluidorum (31 Aug 1752). Novi Comm. Acad. Sci. Petrop. 6 (1761) 271-311. Page 276: This system of notation was first used by the illustrious Fontaine; as it appreciably simplifies calculation, I will also adopt it here. Kline, Morris Mathematical thought from ancient to modern times. Oxford University Press, 1972. Page 425: it was Alexis Fontaine des Bertins (1705-71), Euler, Clairaut, and d'Alembert who created the theory of partial derivatives. Greenberg, John L. Alexis Fontaine's "fluxio-differential method'' and the origins of the calculus of several variables. Ann. of Sci. 38 (1981), no. 3, 251–290. Page 253: In the years 1738-1741 Fontaine produced an exposition of the calculus of several variables which we recognize to have been remarkably general for those years. (...) Fontaine prefaced his exposition by four elementary but fundamental results in the calculus of several variables: $$ \frac{d}{dx}\int F\,dy=\int\frac{dF}{dx}dy $$ (the so-called Leibnizian 'differentiation under the integral sign'); $$ \frac{d^2F}{dxdy}=\frac{d^2F}{dydx} $$ (the equality of mixed second-order partial differential coefficients); $$ nF=\frac{dF}{dx}x+\frac{dF}{dy}y+\frac{dF}{dz}z+\dots $$ if $F=F(x,y,z,\dots)$ is a homogeneous expression of degree $n$ in a finite number of variables $x, y, z, \dots$; and $$ \alpha\frac{d\pi}{dx}-\pi\frac{d\alpha}{dx}+\frac{d\alpha}{dp}-\frac{d\pi}{dy}=0, $$ which is the 'equation of condition' for the integrability of a total differential $dx+\alpha dy+\pi dp$ in three variables $x,y$ and $p$. [Note: A few pages later Fontaine writes the general case just as we now would: the differential $Ldp+Mdx+Ndy$ becomes total when multiplied by an integrating factor only if $$ L\Bigl(\frac{dM}{dy}-\frac{dN}{dx}\Bigr)+ M\Bigl(\frac{dN}{dp}-\frac{dL}{dy}\Bigr)+ N\Bigl(\frac{dL}{dx}-\frac{dM}{dp}\Bigr)=0.] $$ Greenberg, John L. Alexis Fontaine's integration of ordinary differential equations and the origins of the calculus of several variables. Ann. of Sci. 39 (1982), no. 1, 1–36. Page 10: Fontaine was among those who pioneered the use of (...) total differentials, while the introduction of whole symbol notations $\frac{dF}{dx}, \frac{dF}{dy}, \frac{d^2F}{dx^2}, \frac{d^2F}{dxdy}, \frac{d^3F}{dy^3},\dots$ for differential coefficients (...) was his innovation. As a means of defining partial differential coefficients, the total differential was the earliest development in a recognizable calculus of several variables. Greenberg, John L. The Origins of Partial Differentiation. Ann. of Sci. 42 (1985), no. 4, 421-429. Page 428: Euler did not invent the 'whole symbol' notation $$ \left(\frac{dy}{dx}\right) $$ for partial differential coefficients. The notation appeared in print for the first time in Clairaut's published Paris Academy of Sciences mémoires for 1739 and 1740 on the problems that Fontaine had raised. By 1752, Euler knew that Fontaine had first introduced the notation.<|endoftext|> TITLE: An infinite, amenable, finitely presentable group with no non-trivial finite quotients QUESTION [15 upvotes]: My question is a simple one: is there a group with the properties in the title? In the absence of the 'finitely presentable' hypothesis, an example is provided by Juschenko--Monod's construction of a finitely generated, infinite, simple, amenable group. On the other hand, all the standard examples that I know of finitely presentable groups without finite quotients seem to be non-amenable. There are a number of participants on MO who have constructed interesting examples of amenable groups. My guess is that the answer to this question is unknown, but if so I'd also be interested in informed conjectures. Is it at all probable that such a group doesn't exist? REPLY [4 votes]: It's a classical open question. The finitely generated case was also open until the 2012 examples of Juschenko-Monod (for which finite generation and non-existence of finite quotients was established by Matui 2006). Matui also checked that these examples are not finitely presented. Such groups can certainly not be elementary amenable, since it is an elementary exercise to show that an infinite f.g. elementary amenable group always admits an infinite virtually abelian quotient and hence admits finite quotients of unbounded cardinal.<|endoftext|> TITLE: Eigenvectors of a particular transition matrix QUESTION [14 upvotes]: I am considering a Markov chain with $n$ states with a particularly nice structure. The transition matrix is as follows: \begin{equation}\mathbf{P}=\begin{pmatrix} 0 & 0& \dots&0 & 0 &1\\ 0 & 0& \dots&0 & \frac{1}{2}&\frac{1}{2}\\ \vdots& & & & & \vdots \\ 0 &\frac{1}{n-1}& \dots&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}\\ \frac{1}{n} &\frac{1}{n} &\dots&\frac{1}{n} &\frac{1}{n} &\frac{1}{n} \end{pmatrix} \end{equation} I already deduced that the eigenvalues of the matrix are $\lambda_i=(-1)^{i-1}\frac{1}{i}$ for $1\leq i\leq n$. However, I feel that there should also exist closed-form expressions for the eigenvectors of this matrix. Any help in proving or disproving this feeling is appreciated. Thanks. REPLY [12 votes]: EDIT2. (5th Nov, 2014). Based on Darij's comments, am editing the answer to improve its clarity. The answer below shows how to get both eigenvalues and eigenvectors (my original answer was just for eigenvectors). Eigenvalues The key idea is to consider $P^{-1}$. Some (Markovian) guessing leads us to the following subdiagonal matrix: \begin{equation*} L_n := \begin{pmatrix} 0 &&&\\ -1 & 0 &&\\ 0& -2 & 0 &&\\ &&\ddots&\ddots\\ \dots& && 1-n & 0 \end{pmatrix}. \end{equation*} Compute now the matrix exponential $V=\exp(L_n)$, which is a lower-triangular matrix with the lower-triangle given by the binomial coefficients: \begin{equation*} V_{ij} = (-1)^{(i-j)}\binom{i-1}{j-1}. \end{equation*} A quick experiment shows that (something that can be verified by a moderately tedious induction): \begin{equation*} VP^{-1}V^{-1} = V\begin{pmatrix} &&&1-n & n\\ &&2-n & n-1&\\ &\dots&\dots&\\ -1& 2 &&&\\ 1&&&& \end{pmatrix}V^{-1} = \begin{pmatrix} 1 & * & \cdots &*\\ & -2 & * &* \\ &&\ddots&*\\ &&&(-1)^{n+1}n \end{pmatrix}, \end{equation*} which is upper triangular, so we can immediately read off the eigenvalues of $P^{-1}$ (and hence $P$). Eigenvectors Although $V$ does not diagonalize $P^{-1}$, we observe that it turns $P^{-2}$ into the bidiagonal matrix: \begin{equation*} B := VP^{-2}V^{-1} = \begin{pmatrix} 1 & -2(n-1)\\ & 4 & -3(n-2)\\ &&9 & -4(n-3)&\\ &&&\ddots & \ddots\\ \\ &&&&(n-1)^2 & -n(1)\\ &&&&& n^2 \end{pmatrix}. \end{equation*} If we succeed in diagonalizing $B$ in closed-form, then we are done. Suppose, $SBS^{-1}=\Lambda$, then we obtain ($\Lambda$ can be read off of the diagonal): \begin{equation*} S^{-1}\Lambda S = VP^{-2}V^{-1} \implies SVP^{-2}V^{-1}S^{-1} = \Lambda = \text{Diag}([i^2]_{i=1}^n), \end{equation*} which shows that $SV$ diagonalizes $P^{-2}$, completing the answer. Technical Lemma It remains to find $S$. This reduces to the system of equations: \begin{equation*} SB = \Lambda S\quad\leftrightarrow\quad B^TS^T=S^T\Lambda. \end{equation*} I prefer to solve the latter formulation. Let us write $U := S^T$. Consider the $j$th eigenvalue $\lambda_j=j^2$; denote the corresponding column of $U$ by $u$. Then, the equation to solve is \begin{equation*} \begin{split} B^Tu = j^2u,\qquad \implies & u_1 = j^2u_1\\ -(k+1)(n-k)u_k + (k+1)^2u_{k+1} = j^2u_{k+1}. \end{split} \end{equation*} Examining these equations, we see that $U$ is a lower-triangular matrix, with $1$s on its diagonal (which comes from the equation with $k+1=j$). The subsequent values are nonzero. Solving the recurrences for a few different choices of $j$, we can guess the general solution with some help from Mathematica: \begin{equation*} u_k = \frac{(j+1)_{k-j}(n-k+1)_{k-j}}{(2j+1)_{k-j}(k-j)!},\qquad k \ge j. \end{equation*}<|endoftext|> TITLE: Commutative algebras whose bidual is commutative QUESTION [11 upvotes]: Let $k$ be a commutative ring and $A$ a commutative $k$-algebra. Call $D(A) := \mathrm{Hom}_k(A,k)$ the dual of $A$ as a $k$-module, and $DD(A) := \mathrm{Hom}_k(D(A),k)$ the dual of the latter. Let $\Phi\colon A\to DD(A)$ be the canonical map $a \mapsto (u\mapsto u(a))$. Define a multiplication on $DD(A)$ as follows: if $\xi,\eta \in DD(A)$, define $\xi\bullet\eta$ to be $D(A) \ni u \mapsto \eta(y\mapsto \xi(x\mapsto u(xy))) \in k$. This is clearly $k$-bilinear, and furthermore $\Phi(a) \bullet \eta = \eta \bullet \Phi(a)$ is $u \mapsto \eta(y \mapsto u(ay))$ (for $a \in A$ and $\eta \in DD(A)$); in particular, $\Phi(a)\bullet\Phi(b) = \Phi(ab)$. Clearly this is "the correct" multiplication on $DD(A)$. I'm sure the following will come as a surprise to others as it did to me: this product is not necessarily commutative. For a counterexample, consider $A = k[t]$ the ring of polynomials over a finite field $k$: then $D(A) = k^{\mathbb{N}}$ as a $k$-vector space, and $DD(A)$ contains at least the elements $\Lambda_{\mathscr{F}}\colon u \mapsto \lim_{\mathscr{F}} u$ where $\mathscr{F}$ is an ultrafilter on $\mathbb{N}$ and their linear combinations (apparently these don't exhaust $DD(A)$: see here; but this doesn't matter); now one can easily check that if $\mathscr{F}$ and $\mathscr{G}$ are ultrafilters on $\mathbb{N}$ then $\Lambda_{\mathscr{F}} \bullet \Lambda_{\mathscr{G}} = \Lambda_{\mathscr{F}+\mathscr{G}}$ where $\mathscr{F}+\mathscr{G} = \{U \subseteq \mathbb{N} : \{j \in \mathbb{N} : U-j \in \mathscr{F}\} \in \mathscr{G}\}$ is the standard addition on $\beta\mathbb{N}$ defined here (§3.2 "Addition on the Stone–Čech compactification of the naturals") and which is not commutative (see, e.g., Hindman & Strauss, Algebra in the Stone-Čech Compactification (1998), §4.2). So here's my question: What nice conditions on the $k$-algebra $A$ guarantee that $DD(A)$ is commutative? I'm pretty sure that $A$ being finite (i.e., of finite type as a $k$-module) is sufficient, but even this I don't have an appropriate reference for (e.g.: in Vasconcelos, Arithmetic of Blowup Algebras (1994), prop. 1.1.15, the author does not bother to define the multiplication on $DD(A)$). Contrariwise, does someone have a counterexample to $DD(A)$ being commutative that does not require ultrafilters or some use of the axiom of choice? Edit: I believe the following gives a positive answer ($DD(A)$ is commutative) when $k$ is a noetherian integral domain and $A$ is a finite $k$-algebra. Indeed, when $k$ is a noetherian integral domain with fraction field $F$, if $M$ is a $k$-module of finite type, then we can write a presentation $k^s \to k^r \to M \to 0$ (with $r,s$ natural numbers), and by comparing the obvious $0 \to D_k(M) \otimes_k F \to F^r \to F^s$ and $0 \to D_F(M \otimes_k F) \to F^r \to F^s$ (where $D_k(M) := \mathrm{Hom}_k(M,k)$ as a $k$-module), we see that the natural map $D_k(M) \otimes_k F \to D_F(M \otimes_k F)$ is an isomorphism — and also, $D_k(M)$ is a $k$-submodule of this. Dualizing twice (and using the fact that $D_k(M)$ is a $k$-module of finite type, being a submodule of $k^r$), we see that $D_k D_k(M) \otimes_k F$ is isomorphic to $D_F D_F(M\otimes_k F) = M\otimes_k F$ (finite dimensional vector space over a field!), and $D_k D_k(M)$ is a $k$-submodule of it. Now if $M = A$ is a finite $k$-algebra, one can check that the multiplication on $D_k D_k (A)$ is indeed the one obtained by restricting the multiplication on $D_F D_F(A\otimes_k F) = A\otimes_k F$ to it: but this multiplication is commutative. So $D_k D_k (A)$ is a commutative $k$-algebra (indeed, a subalgebra of $A\otimes_k F$). This is the case I was interested in, but I'm leaving the question open, since maybe someone can say something more general or more interesting about the question. Edit 2: I'm told that the product I defined is known (at least in the context of normed algebras) as the Arens multiplication [note: the EoM article contains a number of typos / missing symbols: beware; here is another text defining the Arens product]. So my question could be rephrased as: "In the context of pure algebra, when is a commutative algebra Arens-regular?" REPLY [3 votes]: Not an answer, but: It shouldn't be surprising that this product isn't commutative: your definition has no obvious symmetry between $\xi$ and $\eta$, and so to me it's quite unclear that it's "the correct" multiplication on the double dual. Furthermore an attempt at a more symmetric construction fails. Namely, let's think about what happens when we dualize a multiplication $A \otimes A \to A$. We get a diagram $$A^{\ast} \to (A \otimes A)^{\ast} \leftarrow A^{\ast} \otimes A^{\ast}$$ where the second arrow is not an isomorphism in general and hence is pointing in the wrong direction to give us a comultiplication on $A^{\ast}$. Dualizing a second time gives us a diagram $$A^{\ast \ast} \leftarrow (A \otimes A)^{\ast \ast} \to (A^{\ast} \otimes A^{\ast})^{\ast} \leftarrow A^{\ast \ast} \otimes A^{\ast \ast}$$ where again the middle arrow is not an isomorphism in general and hence is pointing in the wrong direction to give us a multiplication on $A^{\ast \ast}$.<|endoftext|> TITLE: Comparing two numbers given their factorization QUESTION [11 upvotes]: I'm not an expert, but given the integer factorization of two numbers $a,b$: $$a = p_{i_1}^{a_1}...p_{i_n}^{a_n}, \quad b = p_{j_1}^{b_1}...p_{j_m}^{b_m}$$ What is the time and space compexity of checking if $a > b$ ? Suppose that the factors and the exponents are given in binary and the whole input over alphabet $\{0,1,\wedge,;,*\}$ looks like: $$p_{i_1}\wedge{a_1}*...*p_{i_n}\wedge{a_n};p_{j_1}\wedge{b_1}*...*p_{j_m}\wedge{b_m}$$ What algorithms can be used? REPLY [9 votes]: According to this paper from 2013 by Etessami, Stewart, and Yannakakis, the time complexity of the [a priori] harder problem where the $p_k$ are not required to be prime is still open. In that paper the authors show that certain conjectures in number theory would imply that the simple algorithm ("compute enough bits of the logs") runs in polynomial time. The authors also observe that unconditional on any conjectures the algorithm runs in polynomial time for fixed $m$ and $n$ by Baker's Theorem. Of course, it is possible that the version of the problem with $p_k$ prime is strictly easier than the more general one.<|endoftext|> TITLE: Existence of maximal totally ramified $p$-extension of a local field QUESTION [5 upvotes]: This relates to this question: Existence of maximal totally ramified extensions of an arbitrary CDVF Let $K$ be a local field with finite residue field of characteristic $p>0$. Does there exist a maximal totally ramified $p$-extension of $K$? In other words, if $K^{tame}$ is the maximal tamely ramified extension of $K$, does the short exact sequence $$ 1 \longrightarrow Gal(K^{sep}/K^{tame}) \longrightarrow Gal(K^{sep}/K) \longrightarrow Gal(K^{tame}/K) \longrightarrow 1 $$ split? My guess would be that it doesn't, but maybe there is some trick I am missing here. REPLY [3 votes]: The answer seems to be yes. I think this was proved by Kuhlmann, Pank, and Roquette, "Immediate and purely wild extensions of valued fields", Manuscripta math. 55 (1986), 39-67. A short proof is given in Efrat's book on valuation theory, p. 203.<|endoftext|> TITLE: Jordan-Hölder theorem for planar algebras? QUESTION [5 upvotes]: First recall the Jordan-Hölder theorem for groups: Theorem (Jordan-Hölder): Let $G$ be a group, and let $$ G=G_1 \supset G_2 \supset \dots \supset G_r = \{ e \} $$ be a normal tower such that each group $G_i /G_{i+1}$ is simple, and $G_i \neq G_{i+1}$ for $0 TITLE: Uniqueness of composition series for profinite groups QUESTION [6 upvotes]: (It is possible that an answer to this question can be found in the literature, but I couldn't find anything after searching for about an hour.) Let $G$ be a compact, totally disconnected, second countable group. (Equivalently, a profinite group: it is the inverse limit of finite groups.) It makes sense to talk about normal series: $\{e\} \lhd \cdots\cdots G_2\lhd G_1\lhd G$ where $G_n$ is a closed subrgoup of $G$, $G_n$ is normal in $G_{n-1}$ and $G_{n-1}/G_n$ is finite and $\bigcap_{n\in\mathbb{N}}G_n=\{e\}$. This is the same as having an inverse limit expresion $$G\to \cdots \to H_2\to H_1\to \{e\}$$ where each $H_n$ is finite. Call the normal series a composition series if each $G_{n-1}/G_n$ is simple. (In the inverse limit picture, this is the same as the kernel of each map $H_n\to H_{n-1}$ being simple.) Are the simple factors appearing in a composition series unique up to permutation? The result is false without compactness, but I do not know if it is true with it. REPLY [2 votes]: Yes, the composition factors are unique up to permutation, and this can be derived from the Jordan-Hölder theorem for finite groups. If $G$ is second countable, it has a composition series: one can obtain this by starting with some countable basis of identity neighbourhoods (which may be taken to consist of open normal subgroups of $G$, since every open neighbourhood of the identity contains an open normal subgroup), taking intersections to form a descending chain of open normal subgroups with trivial intersection, and then refining the series. So let us assume that we have a composition series $(G_n)$. The equivalence class of $(G_n)$ is defined by the number of times, say $f(S)$, each isomorphism type $S$ of finite simple group appears as a factor $G_{n-1}/G_n$. Let $S$ be a finite simple group. Then there are two possibilities: either $f(S)$ is a non-negative integer, or $f(S) = \aleph_0$. If $f(S)$ is finite, then there is a finite discrete quotient $G/G_n$ of $G$ such that some (hence any) composition series for $G/H$ has $f(S)$ simple factors isomorphic to $S$. Moreover, every finite discrete quotient of $G$ has at most $f(S)$ copies of $S$ in any composition series: This is clear for $G/G_n$ for all $n$ assuming Jordan-Hölder for finite groups, and then we use compactness to show that any finite discrete quotient $G/H$ of $G$ is a quotient of $G/G_n$. If $f(S)$ is infinite, then $G$ has finite quotients with composition series that have arbitrarily many factors isomorphic to $S$. Thus $f(S)$ is an invariant of $G$ as a topological group, independent of the choice of composition series $(G_n)$, so all composition series are equivalent. One can define transfinite descending composition series for arbitrary profinite groups. Here I am not sure about uniqueness: the difficulty is in counting the cardinality of times a composition factor appears (assuming it appears infinitely many times).<|endoftext|> TITLE: A question on non noetherian ring QUESTION [6 upvotes]: Let $R$ be a reduced commutative non noetherian ring of dimension $d$ and $a$ a non zero divisor. Can I say that Krull dimension of $R/(a)$ is at most $d - 1$? REPLY [5 votes]: Since $a$ is a non-zerodivisor of $R$, it does not lie in any of the minimal primes of $R$. Therefore any chain of primes in $R$ that all contain $(a)$ has no minimal prime in the chain and can therefore be extended to a larger chain of primes in $R$, namely, by including a prime properly contained in the smallest prime in the chain. So $\dim R/(a) < \dim R$.<|endoftext|> TITLE: What are the automorphisms of a perfectoid Tate algebra? QUESTION [8 upvotes]: Let $K$ be a complete nonarchimedean field. The classical Tate algebra $K\langle T \rangle$ has lots of automorphisms, e.g., any substitution $T\mapsto a_1T+a_2T^2+\cdots$, where $a_1\in \mathcal{O}_K^\times$, $a_n\in\mathfrak{m}_K$ for $n\geq 2$, and $a_n\to 0$ as $n\to \infty$. This question is about perfectoid spaces. Let $K$ be a perfectoid field (in particular $K$ is a non-discrete valued field). Let $A$ be the ``perfectoid Tate algebra'' $K\langle T^{1/p^\infty}\rangle$. This is obtained by inverting $p$ on the $p$-adic completion of $\mathcal{O}_K[T^{1/p^\infty}]=\bigcup_{n\geq 1} \mathcal{O}_K[T^{1/p^n}]$. If the characteristic of $K$ is $p$, then $K$ is perfect. In that case, the automorphisms of $K\langle T\rangle$ from the first paragraph extend uniquely to $A$. Other automorphisms arise by composing these with $p$th powers and roots. My question is, are there any other automorphisms? For instance, suppose $K$ has characteristic $p$ and $\varpi\in K$ has positive valuation. Then the substitution $$ T\mapsto T+\varpi T^{1/p}+\varpi^2 T^{1/p^2}+\cdots $$ is certainly an endomorphism of $A$, but I cannot decide if it is an automorphism. REPLY [9 votes]: Recall the following elementary fact in commutative algebra: Fact: Say $f:R \to S$ is a map of commutative rings. Assume there exists a $p \in R$ that is a non-zero divisor in both $R$ and $S$, and that $R$ and $S$ are $p$-adically complete. Then $f$ is an isomorphism if and only if $f_1$ is so; here $f_n$ is reduction modulo $p^n$ of $f$. [ Proof: Indeed, it is elementary, using the torsion freeness, to go from $f_1$ being an isomorphism, to $f_n$ being an isomorphism for all $n$, and then one takes a limit. ] This immediately implies that the map mentioned in the question is an isomorphism (as it is defined on the integral perfectoid Tate algebra, and coincides with the identity modulo $\varpi$). In particular, there indeed are more automorphisms than the ones coming from Frobenius or the classical Tate algebra.<|endoftext|> TITLE: Does the generator of a 1-parameter group of Banach space isometries know which elements are entire? QUESTION [11 upvotes]: Let $X$ be a complex Banach space. Let $(\sigma_t)_{t \in \mathbb{R}}$ be a 1-parameter group of linear isometries of $X$ which is strongly continuous i.e. $t \mapsto \sigma_t(x)$ is continuous for each $x \in X$. An element $x \in X$ is said to be entire (for the given flow $\sigma$) if $t \mapsto \sigma_t(x)$ admits an entire extension: that is, extends to a complex differentiable map $\mathbb{C} \to X$. There is also the equivalent "weak" definition which uses the dual $X^*$ to frame things in terms of complex-valued maps. By a routine smoothing argument, the entire elements form a dense subspace of $X$. Now, associated to $\sigma$, is its generator $D$. This is the closed operator $D : X \to X$ whose domain is all the $x$ for which $t \mapsto \sigma_t(x)$ is differentiable at $t=0$ (which then implies $t \mapsto \sigma_t(x)$ is a $C^1$ map) given on its domain by $$D(x) = \frac{d}{dt} \sigma_t(x) \bigg|_{t =0 } .$$ Inductively, one gets that the domain of $D^k$ is all the $x$ for which $t \mapsto \sigma_t(x)$ is $C^k$ and, on this domain, $$D^k(x) = \frac{d^k}{dt^k} \sigma_t(x) \bigg|_{t = 0}.$$ Now if $x$ is entire then, from general theory of power series, the entire extension of $t \mapsto \sigma_t(x)$ is given by the norm-convergent series $$ \sum_{n=0}^\infty \frac{1}{n!}z^n D^n(x).$$ My question is about the converse: Question: If $x \in X$ is such that the series $\sum_{n=0}^\infty \frac{1}{n!}z^n D^n(x)$ has infinite radius of convergence (hence defines an entire mapping $\mathbb{C} \to X$), is it then true that $\sigma_t (x) = \sum_{n=0}^\infty \frac{1}{n!}z^n D^n(x)$ for all $t \in \mathbb{R}$ so that $x$ is actually entire? Let me explain why I don't think this is immediate. To my eye, the natural line of reasoning would be to look at the derivatives of $t \mapsto \sigma_t(x)$ and $t \mapsto \sum_{n=0}^\infty \frac{1}{n!} t^n D^n(x)$. A bit of calculation shows that both solve the same initial value problem: \begin{align*} \frac{d}{dt} f(t) = D(f(t)) && f(0) = x. \end{align*} The trouble is, however, that $D$ is not a bounded operator. So there is not, to my knowledge, a good uniqueness theorem for this initial value problem. Ultimately, my goal here is to understand how to recover $\sigma$ from $D$ (if indeed this is possible at this level of generality). I had figured a natural approach would be to try to recover the dense subspace of entire elements since, there, we have an explicit formula for the flow. REPLY [3 votes]: As it turns out, the basic idea is so simple that I had really might as well add a brief summary. As before, $(\sigma_t)$ is a strongly-continuous, isometric flow on a Banach space $X$ and $D$ is its infinitesimal generator. As I mentioned above, on the face of it the domain of $D$ is just the $x$ for which $\frac{d}{dt} \sigma_t(x) |_{t =0}$ exists, but it follows easily from the group law that, if $x \in \operatorname{dom}(D)$, then $t \mapsto \sigma_t(x)$ is $C^1$ and $\frac{d}{dt} \sigma_t(x) = \sigma_t(D(x))$ is satisfied. Claim: If $f : \mathbb{R} \to \operatorname{dom}(D)$ is a $C^1$ solution to the initial value problem \begin{align*} \frac{d}{dt} f(t) = D(f(t)) && f(0) = x_0 \in \operatorname{dom}(D), \end{align*} then $f(t) = \sigma_t(x_0)$ for all $t \in \mathbb{R}$. Proof: Since $$ \frac{d}{dt} \sigma_{-t}( f(t)) = - \sigma_t(D(f(t))) + \sigma_t(\frac{d}{dt}f(t))=0,$$ we conclude $\sigma_{-t}(f(t))$ is constant, hence $\sigma_{-t}(f(t)) = \sigma_0(f(0)) = x_0$ for all $t \in \mathbb{R}$. Applying $\sigma_t$ on both sides gives $f(t) = \sigma_t(x_0)$, as desired. The above claim shows that $\sigma_t(x)$ is determined by $D$ when $x \in \operatorname{dom}(D)$. In particular, this is true when $x$ is entire, which answers my question. Moreover, since the domain of $D$ is norm-dense in $X$, this implies the flow can be recovered from its infinitesimal generator. Note the proof of the claim is really no different from the standard proof in elementary, single-variable calculus that, if $\frac{dx}{dt} = \delta x(t)$ and $x(0) = x_0$, then $x(t)=x_0e^{ \delta t}$. Finally, in the Engel and Nagel book which the other answers mention, the IVP in the claim is replaced by the integral equation $$ f(t) = x_0 + D \left( \int_0^t f(s) \ ds \right).$$ Continuous functions $f : \mathbb{R} \to X$ satisfying this equation are called "mild solutions". In the same way, one can check that $t \mapsto \sigma_t(x_0)$ is the unique mild solution. An advantage to this appraoch is that, because of the extra smoothing afforded by integrating before applying $D$, this actually works for every $x_0 \in X$, and not just for $x_0 \in \operatorname{dom}(D)$ as with the IVP.<|endoftext|> TITLE: Suslin trees in ccc ideals QUESTION [9 upvotes]: A countably complete ideal $I$ on a set $Z$ ideal is c.c.c. when there is no uncountable family of pairwise disjoint $I$-positive subsets of $Z$. If such an ideal exists, then there exists a weakly Mahlo cardinal $\kappa$ and a $\kappa$-complete, c.c.c. ideal $J$ on $\kappa$. An example of such an ideal is the collection of measure zero subsets of a real-valued measurable cardinal. The existence of such an ideal, which is nowhere maximal ("ultra"), requires a large continuum. The standard proof of this is as follows. Build a tree of subsets of $\kappa$, with root $\kappa$. Suppose by induction that each level of the tree up to level $\alpha$ is a partition of $\kappa$. For every $J$-positive member of the partition, partition this into two disjoint $J$-positive sets. For members in $J$, don't split them. This gives you level $\alpha+1$ and a finer partition. At limit stages, take intersections. By the c.c.c., we must have all members of the partition in $J$ at level $\omega_1$, and no cofinal branches. There are $2^{<\omega_1} = 2^\omega$ many measure zero sets at that stage, so we have $\kappa$ equals a union of continuum many sets from $J$, so $2^\omega \geq \kappa$. My questions is: Does this process of building the tree terminate before $\omega_1$? If it does not, then by the c.c.c., some subtree of what we've constructed is a normal Suslin tree. So in some cases, we know the process does halt at a countable stage: (a) If we start with a measurable $\kappa$ and force MA$_\kappa$ with the Solovay-Tennenbaum forcing, we get no Suslin trees, and a c.c.c. ideal on $\kappa$. (b) By a result of Laver, we can start with a model of MA$_{\omega_2}$ and a measurable above, and then force a real-valued measurable cardinal by random real forcing, keeping SH. So the obvious question is, is there any way to get a c.c.c., $\kappa$-complete ideal $J$ on $\kappa$, where we can find a Suslin tree embedded in the algebra of $J$-positive sets? Can this be done by adding measurably many Cohen reals? REPLY [9 votes]: Such trees must have countable height because every $x \in \kappa$ has to leave the tree at some level below $\omega_1$ so that by ccc-ness of the ideal the tree dies at a countable level.<|endoftext|> TITLE: Why can you deform singularities in two dimensions but not in higher dimensions? QUESTION [8 upvotes]: I've been trying to read this paper to understand deformations of surface quotient singularities. I'm particularly interested in when one can deform certain cyclic quotient singularities into other cyclic quotient singularities. I've heard that in higher dimensions, isolated quotient singularities are rigid. More precisely, if $X$ is an isolated quotient singularity by a finite group and $\dim X \geq 3$ then the only deformation of $X$ is the trivial one. What is some explanation or intuitive reason for why you can deform quotient surface singularities but you can't deform higher dimensional quotient singularities? REPLY [11 votes]: The rigidity of quotient singularities in dimension greater or equal than $3$ was established by Schlessinger in his paper Rigidity of quotient singularities, Invent. Math. 14 (1971). Roughly speaking, he proved that if $(X, \,x)$ is a local scheme with an isolated singularity at $x$ and $\dim X \geq 3$, then deforming $X$ is equivalent to deform the punctured spectrum $U=X-x$. More precisely, the key result in the paper is the following Proposition. Let $X$ be a geometric local scheme such that $X$ has an isolated singularity at his closed point $x$. If $\textrm{depth}_xX \geq 3$ then there is an isomorphism $$T_X^1 \cong H^1(U, \, \Theta_X),$$ where $U:=X - x$. More precisely, under these assumptions the formal deformation theories of $X$ and its punctured spectrum $U$ are equivalent. Now, assume that $X=Y/G,$ where $Y$ is smooth and $G$ is a finite group of automorphisms acting with an isolated fixed point $y \in Y$. Let $p \colon Y \to X$ the quotient map and $x=p(y)$. Then $\textrm{depth}_xX \geq \textrm{depth}_yY$ and, if $\textrm{depth}_yY \geq 2$, one has $\Theta_Y=(p_*\Theta_X)^G.$ From this, if $\textrm{depth}_yY \geq 3$, it is not too difficult to show that $H^1(U, \, \Theta_X)=0$, hence using the Proposition one obtains $T_X^1=0,$ that is $X$ is rigid. You can read Schlessinger's paper for further details.<|endoftext|> TITLE: How can classifying irreducible representations be a "wild" problem? QUESTION [26 upvotes]: Let $q$ be a prime power and $U_n(\mathbb{F}_q)$ be the group of unitriangular $n\times n$-matrices. I've read and heard in several places (see e.g. this mathoverflow question) that classifying irreducible complex characters of $U_n(\mathbb{F}_q)$ is a wild problem. This seems to be in contrast to the fact that $U_n(\mathbb{F}_q)$ is a finite group, and hence there are only finitely many irreducible represenations, i.e. classifying them for a fixed $n$ and $q$ is a finite problem. So my first question is: $1.$ In what sense is classifying irreducible characters for $U_n(\mathbb{F}_q)$ a wild problem? The only reference I found so far is the following paper by Poljak not accessible to me: [Poljak:On the Characters of Irreducible Complex Representations of the Group of Triangular Matrices over a Prime Finite Field. (Ukrainian) Dopovīdī Akad. Nauk Ukraïn. RSR, 1966, 434-436. So my other questions are: $2.$ What is the rough idea to prove this? (A bit more precise than "the problem is translated to the problem of classifying indecomposable representations of a wild quiver). $3.$ Are there other instances of classifying irreducible representations being a wild problem in this particular sense? REPLY [6 votes]: Well, to understand how this problem is wild it may be useful to contrast it with the situation of finite reductive groups where we do have a classification statement. The first part of this post considers the reductive case and then after the cut I give some information about $\mathrm{U}_n(q)$. Hopefully some of this is clarifying or useful for your question. Assume $\mathbf{G}$ is a connected reductive algebraic group over an algebraic closure $\mathbb{K} = \overline{\mathbb{F}_p}$ of the finite field of prime order $p>0$. Furthermore, let us assume that $F : \mathbf{G} \to \mathbf{G}$ is a (generalised) Frobenius endomorphism of $\mathbf{G}$ admitting an $\mathbb{F}_q$-rational structure $G = \mathbf{G}^F$, which is a finite reductive group. Now, we want to "sensibly" parameterise the irreducible characters of $G$. In general this means that the parameterisation should be given, as far as possible, in terms of data from the algebraic group $\mathbf{G}$ and should not depend on $q$. However, a priori it is not clear at all that such a classification can be achieved. It is one of the great feats of finite group representation theory that Lusztig managed to obtain such an elegant classification theory. We roughly recap this below. Assume $\mathbf{T} \leqslant \mathbf{G}$ is an $F$-stable maximal torus of $\mathbf{G}$. To every irreducible character $\theta \in \mathrm{Irr}(T)$ ($T = \mathbf{T}^F$) Deligne and Lusztig have defined, using $\ell$-adic cohomology, a virtual character $R_{\mathbf{T}}^{\mathbf{G}}(\theta)$ of the group $G$. We call this a Deligne--Lusztig character of $G$. The following curcial fact is known about these virtual characters: The character of the regular representation of $G$ is a sum of Deligne--Lusztig characters. In particular, every $\chi \in \mathrm{Irr}(G)$ occurs in some $R_{\mathbf{T}}^{\mathbf{G}}(\theta)$. This is somewhat surprising as this is not true of every class function on $G$ (unless $\mathbf{G} = \mathrm{GL}_n(\mathbb{K})$)! With this one has a chance to at least bunch the irreducible characters together but to do this we need more information about the Deligne--Lusztig characters. Let $\mathbf{G}^{\star}$ be a dual group of $\mathbf{G}$ and let $F^{\star} : \mathbf{G}^{\star} \to \mathbf{G}^{\star}$ be a (generalised) Frobenius endomorphism such that $G^{\star} = {\mathbf{G}^{\star}}^{F^{\star}}$ is a dual group of $G$. We will denote by $\nabla(\mathbf{G},F)$ the set of all pairs $(\mathbf{T},\theta)$ where $\mathbf{T} \leqslant \mathbf{G}$ is an $F$-stable maximal torus and $\theta \in \mathrm{Irr}(\mathbf{T}^F)$. The group $G$ acts naturally on this set by conjugation and we denote the orbits of this action by $\nabla(\mathbf{G},F)/G$. Conversely we denote by $\nabla^{\star}(\mathbf{G},F)$ the set of all pairs $(\mathbf{T}^{\star},s)$ where $\mathbf{T}^{\star} \leqslant \mathbf{G}^{\star}$ is an $F^{\star}$-stable maximal torus and $s \in T^{\star} = {\mathbf{T}^{\star}}^{F^{\star}}$ is a semisimple element. The group $G^{\star}$ also acts naturally on $\nabla^{\star}(\mathbf{G},F)$ by conjugation and we denote by $\nabla^{\star}(\mathbf{G},F)/G^{\star}$ the orbits of this action. With this notation we have the following result: We have a natural bijection $\nabla(\mathbf{G},F)/G \to \nabla^{\star}(\mathbf{G},F)/G^{\star}$ which satisfies $(\mathbf{T},1_T) \mapsto (\mathbf{T}^{\star},1)$. If $(\mathbf{T},\theta)$, $(\mathbf{T}',\theta') \in \nabla(\mathbf{G},F)$ are in the same $G$-orbit then we have $R_{\mathbf{T}}^{\mathbf{G}}(\theta) = R_{\mathbf{T}'}^{\mathbf{G}}(\theta')$. In particular, if $(\mathbf{T},\theta)$ corresponds to $(\mathbf{T}^{\star},s)$ under the above bijection then we may simply write $R_{\mathbf{T}^{\star}}^{\mathbf{G}}(s)$ for $R_{\mathbf{T}}^{\mathbf{G}}(\theta)$. With this in hand we may state one of the most important theorems concerning Deligne--Lusztig characters. Assume $(\mathbf{T}^{\star},s)$ and $({\mathbf{T}^{\star}}',s')$ are not in the same $G^{\star}$-orbit then $R_{\mathbf{T}^{\star}}^{\mathbf{G}}(s)$ and $R_{{\mathbf{T}^{\star}}'}^{\mathbf{G}}(s')$ have no irreducible constituent in common. Now, for any semisimple element $s \in G^{\star}$ we denote by $\mathcal{E}(G,s)$ the set of all irreducible characters $\chi \in \mathrm{Irr}(G)$ such that $\chi$ is a constituent of $R_{\mathbf{T}^{\star}}^{\mathbf{G}}(s)$ for some $F^{\star}$-stable maximal torus $\mathbf{T}^{\star}$ containing $s$. The set $\mathcal{E}(G,s)$ is called a Lusztig series of $G$ and by the above we have a disjoint union $$\mathrm{Irr}(G) = \bigsqcup_{(s)} \mathcal{E}(G,s)$$ where the union runs over all $G^{\star}$-conjugacy classes of semisimple elements of $G^{\star}$. Let us now assume that the centre $Z(\mathbf{G})$ of $\mathbf{G}$ is connected (similar statements hold when $Z(\mathbf{G})$ is not connected but this is more complicated to state). One of the most amazing parts of Lusztig's classification result is that, under this assumption, there exists a bijection $$\mathcal{E}(G,s) \to \mathcal{E}(C_{G^{\star}}(s),1)$$ where $C_{G^{\star}}(s)$ is the centraliser of $s$ in $G^{\star}$. Note that, as $Z(\mathbf{G})$ is connected we have $C_{\mathbf{G}^{\star}}(s)$ is a connected reductive algebraic group. As $F^{\star}(s) = s$ we have $F^{\star}$ induces a (generalised) Frobenius endomorphism of $C_{\mathbf{G}^{\star}}(s)$ and so $C_{G^{\star}}(s) = C_{\mathbf{G}^{\star}}(s)^{F^{\star}}$ is a finite reductive group. The crucial part of Lusztig's classification result is given by the following result: Assume $\mathbf{H}$ is a connected reductive algebraic group and $F : \mathbf{H} \to \mathbf{H}$ is a (generalised) Frobenius endomorphism of $\mathbf{H}$. Let $\mathbf{T}_0 \leqslant\mathbf{B}_0\leqslant \mathbf{H}$ be an $F$-stable maximal torus and Borel subgroup of $\mathbf{H}$. This data determines a Coxeter system $(\mathbf{W},\mathbb{S})$ where $\mathbf{W} = N_{\mathbf{H}}(\mathbf{T}_0)/\mathbf{T}_0$ and $\mathbb{S}$ is a set of Coxeter generators determined by $\mathbf{B}_0$. By our choices $F$ induces an automorphism $\gamma : \mathbf{W} \to \mathbf{W}$ which stabilises $\mathbb{S}$ (i.e. it is an automorphism of the Coxeter system $(\mathbf{W},\mathbb{S})$). Lusztig has then shown that there is a bijection $$\mathcal{E}(H,1) \to X(\mathbf{W},\gamma)$$ where $X(\mathbf{W},\gamma)$ is a set whose definition depends only on $\mathbf{W}$ and $\gamma$. This is somehow the truly amazing thing. That these irreducible characters $\mathcal{E}(H,1)$ can be parameterised "independently of $q$". This gives us the classification that we desire. In fact, it is really only the classification of the semisimple conjugacy classes of $G^{\star}$ that really depends on $q$. Note that this mimics the classification of the conjugacy classes of $G$. In general a conjugacy class of $G$ is a product $(s)\mathcal{O}$ where $(s)$ is a semisimple conjugacy class of $G$ and $\mathcal{O}$ is a unipotent conjugacy class of $C_G^{\circ}(s) = C_{\mathbf{G}}^{\circ}(s)^F$. The unipotent conjugacy conjugacy classes are parameterised "independently of $q$" and depend only upon the action of $F$ on the root system of $C_{\mathbf{G}}(s)$. Let us now consider the case where $\mathbf{U}_n$ is the group of unitriangular matrices in $\mathrm{GL}_n(\mathbb{K})$ and $F : \mathbf{U}_n \to \mathbf{U}_n$ is the Frobenius endomorphism $F(x_{ij}) = (x_{ij}^q)$ so that $\mathbf{U}_n^F = \mathrm{U}_n(q)$. One may now take the strategy for finite reductive groups and apply it to our situation here. In particular we would like to do the following: Break $\mathrm{Irr}(\mathrm{U}_n(q))$ into series $\mathcal{F}_i$ such that the irreducible characters in each $\mathcal{F}_i$ can be parameterised "independently of $q$". The theory of supercharacters gives a way to determine such series $\mathcal{F}_i$ (see "Supercharacters and Superclasses for Algebra Groups" by Diaconis and Isaacs). In particular, these are given by the transitive closure of the condition that two irreducible characters occur as constituents of a supercharacter. These series have a nice combinatorial description but in contrast to the case of finite reductive groups once cannot parameterise $\mathcal{F}_i$ independently of $q$. In this sense the problem is wild. See this paper by Marberg for a recap on the supercharacter theory of $\mathrm{U}_n(q)$ http://arxiv.org/pdf/1005.4150v4.pdf Another approach to the character theory of $\mathrm{U}_n(q)$ is given by the theory of character sheaves on $\mathbf{U}$. This theory (conjectured by Lusztig) was developed by Boyarchenko and Drinfeld. In Theorem 1.13 of http://arxiv.org/pdf/1006.2476v3.pdf Boyarchenko gives a classification of the irreducible characters of $\mathrm{U}_n(q)$ in terms of minimal idempotents of the $\mathbf{U}$-equivariant bounded derived category of $\overline{\mathbb{Q}_{\ell}}$-constructible sheaves on $\mathbf{U}$. This may also give a way to see why the classification of such characters is wild by showing that the classification of such minimal idempotents is also a wild problem. One also has a way to break the irreducible characters up into series based on the theory of $L$-packets. This approach is similar to the Lusztig series encountered in the case of finite reductive groups.<|endoftext|> TITLE: Hamiltonicity criteria for sparse graphs QUESTION [6 upvotes]: Given a sparse graph, how can one go about proving that it is Hamiltonian? (Assuming it actually is, of course). There are three main classes of criteria for Hamiltonicity that I am aware of: Dirac-type conditions ($\delta \geq \frac{n}{2}$, i.e. high minimum degree). Spectral conditions. Erdos-Chvatal-type conditions ($\kappa \geq \alpha$, i.e. connectivity greater than independence number). However, none of these approaches is able to settle even the Hamiltonicity of $C_{n}$, the $n$-cycle! Apparently, the reason is that these approaches work best for dense graphs. Is there an alternative criterion that works well for sparse graphs? REPLY [5 votes]: Partial answer. According to Eppstein It is known that it is NP-complete to test whether a Hamiltonian cycle exists in a 3-regular graph, even if it is planar (Garey, Johnson, and Tarjan, SIAM J. Comput. 1976) or bipartite (Akiyama, Nishizeki, and Saito, J. Inform. Proc. 1980) or to test whether a Hamiltonian cycle exists in a 4-regular graph, even when it is the graph formed by an arrangement of Jordan curves (Iwamoto and Toussaint, IPL 1994). So in general you can't expect complete answer to your question. Another reason for Hamiltonicity are forbidden subgraphs, which force Hamiltonicity (possibly with a finite number of exceptions). Not sure if this works for sparse graphs, perhaps search the web. Found this: Forbidden subgraphs, hamiltonicity and closure in claw-free graphs There are theorems of the form: Every 2-connected $XY$-free graph is hamiltonian. Where $X,Y$ are explicit small graphs. For certain type of graphs polynomial time algorithms exist for HC, so if the they work for sparse graphs you can find a cycle, e.g. this paper<|endoftext|> TITLE: Equivariant homotopy, simplicially QUESTION [9 upvotes]: It is a classic result of Kan that the homotopy categories (with appropriate model structures) of simplicial sets and of topological spaces (in fact, one could only care about CW-complexes) are equivalent. One could ask if there is a similar result for G-equivariant homotopy category of spaces, where G is a topological group (usually a Lie group) and a morphism of G-spaces is a map, strictly commuting with G-action: $$f(gx) = g\cdot f(x)$$ Since the G-equivariant category admits a very nice theory of cell decompositions and in fact is equivalent to a homotopy category of space presheaves on the category of G-orbits (Elmendorf's theorem), one could guess that such a result should be true. However, I would like to see it not on the level of $\mathrm{SSet}$-presheaves on orbits, but rather directly in the standard definitions. I see 2 problems. Firstly, while cellular and simplicial descriptions of spaces are very similar, they are not equivalent in any sense I could see (besides giving the same homotopy category). The second problem is that if we pass from topological spaces to simplicial sets, we should also change a topological group into a simplicial group. While there is an obvious way to go back and forth (singular complex $\mathrm{Sing}: \mathrm{Top} \to \mathrm{SSet}$ and geometric realization $|\cdot|: \mathrm{SSet} \to \mathrm{Top}$), it is very unobvious that this gives the correct correspondence, since $X \to |\mathrm{Sing}X|$ is just a weak homotopy equivalence, and the equivariant category doesn't respect homotopy equivalences in general, both for spaces and groups. A contractible space can have different non-equivalent G-structures (e.g. free and trivial), and homotopy equivalent (as spaces) groups can give different equivariant categories. So the question is, can we generalize Kan's theorem to the equivariant setting? If we do, then how does this correspondence look like and how far is it from the intuition? For example, are there finite-dimensional simplicial groups G, such that their G-homotopy categories are not equivalent to equivariant categories of Lie groups? Can we associate with a space $X$ some simplicial group $G_X$, such that the $G_X$-equivariant category is the same as fibrations over $X$? REPLY [3 votes]: Perhaps you should look at the series of papers by Dwyer and Kan, starting with W. G. Dwyer and D. M. Kan, Singular functors and realization functors, Nederl. Akad. Wetensch. Indag. Math., 87, (1984), 147 – 153. I do not think you will find all the answers that you require there, as your reluctance to see things in terms of simplicial presheaves is a hinderance from their point of view, but their discussion (and in later papers by others) may help. In fact I seem to remember that Dwyer and Kan's viewpoint (taken up by Dror later on) implies that there is a variant of the singular complex / geometric realisation adjointness that does the job. The standard simplicial models are more or less replaced by the $G/H\times \Delta^n$.<|endoftext|> TITLE: Fields of characteristic zero via ultraproducts QUESTION [7 upvotes]: Is every noncountable field of characteristic zero the ultraproduct (using a non principal ultrafilter over the set of prime numbers) of fields of positive characteristic? REPLY [12 votes]: No. Ultraproducts over nonprincipal ultrafilters on a countable index set are always $\aleph_1$-saturated. This rules out many fields just on cardinality basis (the cardinality of the ultraproduct must satisfy $\kappa=\kappa^\omega$), but even if the field has cardinality $2^\omega$, it does not have to be $\aleph_1$-saturated. Even more importantly, it is not true that every field of characteristic $0$ is elementarily equivalent to an ultraproduct of fields of nonzero characteristic. In other words, there exist first-order sentences that are satisfiable in a field of characteristic $0$, but not in any field of positive characteristic. One such sentence is “every sum of two squares is a square, and $-1$ is not a square”.<|endoftext|> TITLE: Iterated semi-direct products QUESTION [11 upvotes]: Let $G$ be a finite group. Suppose that we can write $G= A \rtimes B$ and also $A = C \rtimes D$. Further suppose that C is normal in $G$ (not just in $A$). Then can we write $G = C \rtimes E$ where $E=G/C$? Of course, if $|C|$ and $[G:C]$ are relatively prime, then this follows from the Schur-Zassenhaus theorem. We can of course write $G= A \rtimes B = (C \rtimes D) \rtimes B$ and we would like to have some kind of "associativity" of semi-direct product so that $G= C \rtimes (D \rtimes B)$. The question is whether the semi-direct product "$D \rtimes B$" actually makes sense (that $D$ and $B$ act on $C$ is clear). I am mainly interested in whether this works in the case that we have the following extra hypotheses (i) $G$ is soluble, (ii) $C$ is a $p$-group, (iii) $D$ is cyclic of order prime to $p$, and (iv) $B$ is cyclic (the question is only interesting if $p$ divides $|B|$ due to (iii) and Schur-Zassenhaus). We could also strengthen (i) to (v) $C$ is an elementary abelian $p$-group. I don't know whether any of these extra hypotheses are helpful or not. One can also phrase the problem in terms of splitting of short exact sequences. One has a commutative diagram $ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}$ $$ \begin{array}{c} & & 1 & & 1 & & \\ & & \da{} & & \da{} & & \\ & & C & \ra{=} & C\\ & & \da{} & & \da{} & & \\ 1 & \ra{} & A & \ra{} & G & \ra{} & B & \ra{} & 1 \\ & & \da{} & & \da{\pi} & & \da{=} \\ 1 & \ra{} & D & \ra{} & E & \ra{} & B & \ra{} & 1 \\ & & \da{} & & \da{} & & \\ & & 1 & & 1 & & \end{array} $$ in which all rows and columns are exact. We are assuming that all the short exact sequences split (including the last row), apart from the one containing $\pi$; we want to show that this one also splits. It is easy to see what the section $\varepsilon: E \rightarrow G$ should be in terms of the other sections, but unfortunately I can't seem to show that it is actually a homomorphism. I have the feeling that there is either an easy solution or any easy counterexample satisfying the extra hypotheses above, so I am missing something either way. (I did try some small examples, which seemed to work.) REPLY [3 votes]: Here's a short answer, assuming only that $C$ is an abelian $p$-group and $\vert D \vert$ is relatively prime to $p$. Remember that $G \cong H\rtimes K$ iff normal subgroup $H$ has a complement in $G$ isomorphic to $K \cong G/H$. A theorem of Gaschütz says that a normal abelian $p$-group $C$ has a complement in $G$ if and only if $C$ has a complement in a Sylow $p$-subgroup $P$ containing $C$. Since $( \vert D \vert, p)=1$, we have that $C$ is a Sylow $p$-subgroup of $A$. If $P$ is a Sylow $p$-subgroup of $B$, then by counting we get that $CP$ is a Sylow $p$-subgroup of $G$. Then $P$ is a complement to $C$ in $CP$, hence $C$ has a complement in $G$, as desired. $\square$<|endoftext|> TITLE: Is irreducibility sufficient for uniqueness of invariant distribution for a Feller semigroup? QUESTION [6 upvotes]: Let $(T_t)$ be a strongly continuous semigroup of positive operators on $C(K)$, where $K$ is a compact space. Assume also that $T_t1 =1 $ for every $t\geq 0$. (This is also called a Feller semigroup.) Since $K$ is compact we know that there exists a probability measure $\mu$ on $K$ satisfying $\mu T^*_t = \mu $ for every $t\geq 0$ (i.e. $\mu$ is invariant). My question is: to show that $\mu$ is the unique invariant probability distribution, is it sufficient to show that $(T_t)$ is irreducible? Recall that a semigroup is by definition irreducible if the resolvent $R_\lambda=(\lambda-L)^{-1}$ ($L$ is the generator of $(T_t)$) maps for sufficiently large $\lambda$ nonnegative nonzero functions into strictly positive functions. I thought this should be true by applying some version of the Krein-Rutman theorem, but did not find a suitable reference. The closest I found is Proposition 3.5. on p. 185 of this book (link at Springer site), from which, if I understand well, I can just conclude that $\text{dim (ker } L) = 1$, but not $\text{dim (ker } L^*) = 1$. REPLY [5 votes]: Take $K = \{0,1\}^{\mathbf{Z}^2}$ and take for $T_t$ the Glauber dynamic for the Ising model below the critical temperature. Then $T_t$ is Feller and irreducible, but it has two distinct ergodic invariant measures. If however you know that $T_t$ is strong Feller (or asymptotically strong Feller) then irreducibility does imply uniqueness of the invariant measure.<|endoftext|> TITLE: Basics on anabelian geometry and Grothendieck's section conjecture QUESTION [6 upvotes]: Even I can find similar questions and some answers on that questions, most of them are not quite unsatisfactory to me. Maybe this is a very stupid question, but there is no other place that I can ask this. I want to undersatnd Grothendieck's section conjecture and its recent results and also I want to study anabelian geometry. I just finished to read Hartshorne's algebraic geometry book. Is there any good start point for those things which contains some basics? If my background is not enough, please let me konw a kind of 'road-map' for anabelian geometry and section conjecture. REPLY [2 votes]: For the section conjecture you might look at Jakob Stix's great book Evidence for the section conjecture in the theory of arithmetic fundamental groups Habilitationsschrift, School of Mathematics and Computer Science at the Ruprecht-Karls-Universität Heidelberg, January 2011, x+190 pages, to appear as Rational Points and Arithmetic of Fundamental Groups. Evidence for the Section Conjecture Springer Lecture Notes in Mathematics 2054, xx+pp.249, Springer, 2013. "The first exposition of foundational material on the arithmetic of fundamental groups with respect to the Section Conjecture of anabelian Geometry: from the history of the subject to the state of the art of the conjecture. Numerous approaches to the Section Conjecture are discussed with open questions to stimulate future research. Assuming the basics, the more advanced chapters are self contained and can be read independently."<|endoftext|> TITLE: Riemann surfaces of $w^3 = (z-a)(z-b)(z-c)$ QUESTION [8 upvotes]: I've been playing around with Riemann surfaces of cubics, and it seems to me that all coverings of the Riemann sphere from equations of the form $w^3 = q(z)$, where $q(z)$ is a cubic with three distinct roots, must be isomorphic. Is this correct? (Argument given below.) Mainly I want to know if there are any good references on this specific kind of cover. Argument: we have a critical point of multiplicity 3 at each of the roots of $q(z)$. Monodromy around a small counterclockwise circuit about any of these points multiplies $w$ by the same cube root of unity ($\neq 1$). So monodromy around a circuit enclosing all three roots of $q(z)$ leaves $w$ unchanged, so no branch points over $\infty$. Now we can move the three roots of $q(z)$ to any other positions using a Möbius transformation, so by the previous paragraph, we should be able to establish an analytic isomorphism between the surfaces via continuation. REPLY [11 votes]: Yes, you can view this equation as a family of curves (cyclic covers of $\mathbb{P}^{1}$) , as you said we can move each three points to $0,1,\infty$ and hence your family in fact reduces to a point in the moduli space of curves (or $A_{g}$). Your curve will have genus $1$ and is smooth so is an elliptic curve. You can find alot about cyclic covers of $\mathbb{P}^{1}$ in the book "cyclic coverings, calabi-Yau manifolds and complex multiplication" by J.C.Rohde. The main stream of this book as the name suggests is something else, but the generalities on cyclic covers is treated quite nicely here.<|endoftext|> TITLE: Is $Lex(\mathcal B,\mathsf{Set}_*)$ an $\mathbb F_1$-linear category? QUESTION [11 upvotes]: Following Anton Deitmar, let $\mathcal B$ be an "$\mathbb F_1$-linear category" (Deitmar uses the term "Belian"); i.e., $\mathcal B$ is balanced, pointed, contains finite products, kernels, and cokernels, and has the property that every morphism with zero cokernel is an epimorphism. Let $\mathsf{Set}_*$ denote the category of pointed sets, and let $Lex(\mathcal B,\mathsf{Set}_*)$ denote the category of left-exact functors $F: \mathcal B\to\mathsf{Set}_*$ (that is, $F$ preserves finite limits). I want to show that $Lex(\mathcal B,\mathsf{Set}_*)$ is also $\mathbb F_1$-linear if $\mathcal B$ is small; in particular, the properties that still concern me are that $Lex(\mathcal B,\mathsf{Set}_*)$ is balanced, and that every morphism with $0$ cokernel is an epimorphism. I know that $Lex(\mathcal B,\mathsf{Set}_*)$ is a reflective subcategory of $Fun(\mathcal B,\mathsf{Set}_*)$ and that finite products, kernels, and cokernels exist in $Lex(\mathcal B,\mathsf{Set}_*)$ (it's also clear that $Lex(\mathcal B,\mathsf{Set}_*)$ is pointed), and I know that $Fun(\mathcal B,\mathsf{Set}_*)$ is $\mathbb F_1$-linear. My hope was originally to prove that if a natural transformation $\eta : F\to G$ is mono/epi in $Lex(\mathcal B,\mathsf{Set}_*)$, then it is mono/epi in the full functor category, which would show that $Lex(\mathcal B,\mathsf{Set}_*)$ is balanced (and I had a similar plan to show that morphisms with $0$ cokernel were epi), although a proof of this is turning out to be quite elusive (and I'm no longer convinced that one exists). I was able to show that if a natural transformation is a monomorphism in $Lex(\mathcal B,\mathsf{Set}_*)$, it is a monomorphism in the full functor category - epimorphisms are the root of my problems. Then my question is really a few related questions (hopefully they're related enough to warrant posting all of them here): Are epimorphisms in $Lex(\mathcal B,\mathsf{Set}_*)$ also epimorphisms in $Fun(\mathcal B,\mathsf{Set}_*)$? If not, does anyone know of an alternate proof (or a resource where I could find one) that a bimorphism in $Lex(\mathcal B,\mathsf{Set}_*)$ is also an isomorphism? Is there any reasonable condition one can place on $\mathcal B$ to force epis in $Lex(\mathcal B,\mathsf{Set}_*)$ to remain epis in $Fun(\mathcal B,\mathsf{Set}_*)$ or to force bimorphisms in $Lex(\mathcal B,\mathsf{Set}_*)$ to be isomorphisms? (I doubt that this has an answer in the affirmative, as my experience indicates that many of the properties of a functor category depend on the codomain rather than the domain.) REPLY [6 votes]: Here is a counterexample (modulo a small statement I'm not sure how to prove). Building on the comments, let $B$ be (a skeleton of) the opposite of the category of at most countable groups. This satisfies all of the desired conditions. There is a functor $\text{Grp} \to \text{Lex}(B, \text{Set}_{\ast})$ sending a group $G$ to the representable functor $\text{Hom}(-, G)$ (which lands in $\text{Set}_{\ast}$ since $\text{Grp}$ has a zero object) which is full and faithful; you can see this by evaluating on the subcategory given by the free groups, which gives a right adjoint $\text{Lex}(B, \text{Set}_{\ast}) \to \text{Grp}$ (this is the statement I'm not sure how to prove). In particular, $\text{Grp}$ is a coreflective subcategory, so colimits in $\text{Grp}$ agree with those in $\text{Lex}(B, \text{Set}_{\ast})$. But $\text{Grp}$ has morphisms with zero cokernel which are not epimorphisms: take any inclusion $f : H \to G$ of a proper subgroup $H$ into a group $G$ whose normal closure is all of $G$ (for example, take $G$ to be simple and $H$ to be proper and nontrivial). Hence the same is true of $\text{Lex}(B, \text{Set}_{\ast})$. For the purposes of imitating the proof of Freyd-Mitchell I think you should be looking at $\text{Lex}(B, \text{Set}_{\ast})^{op}$ as I mentioned in the comments.<|endoftext|> TITLE: Function extensionality: does it make a difference? why would one keep it out of the axioms? QUESTION [51 upvotes]: Yesterday I was shocked to discover that function extensionality (the statement that if two functions $f$ and $g$ on the same domain satisfy $f\left(x\right) = g\left(x\right)$ for all $x$ in the domain, then $f = g$) is not an axiom in the standard constructive logic of Coq. Of course, one can add it as an axiom in one's files, but it is not obviously available in any pre-defined tactics. I am left wondering... 1. What is the thinking behind considering function extensionality as foreign to the calculus of constructions? I thought that from a computational perspective, a function really is there to be applied to things, and cannot carry any more information than what it does to them (and its type). For a moment I suspected that Coq avoids function extensionality in order to allow applying results to models like arbitrary enriched categories whose internal homomorphisms carry some more information than plain morphisms. But this is not the case: Coq (since version 8.4) has a implementation of eta-expansion (saying that any function $f$ equals the function sending every $x$ in its domain to $f\left(x\right)$, provided the types are right). In an enriched category, this would pour any additional structure of an internal Hom down the drain. (I must say the eta-expansion in Coq feels rather weird, too -- it is triggered by the reflexivity tactic. I expected it to be a tactic on its own...) Having eta-expansion but no extensionality is seriously confusing: one can have $f\left(x\right) = g\left(x\right)$ for all $x$, and yet one cannot rewrite the $f\left(x\right)$ in "the function sending every $x$ to $f\left(x\right)$" as a $g\left(x\right)$. And there I thought the bound variable in a lambda term would be like the bound variable in a forall quantification? 2. On a more practical note (and more on-topic in MathOverflow), how much do the axiom of function extensionality and the (weaker) axiom of eta-expansion contribute to the strength of the logic? (At this point I have to admit that I don't really know the definition of the logic involved, so I'll just say I'm talking about the logic of Coq with no additional axioms assumed; it has so far been agreeing with my intuitive understanding of constructivism, until extensionality came along.) If I can define two terms $a$ and $b$ of type $\mathrm{nat}$ (natural numbers) and use function extensionality (resp. eta-reduction, both ways) to prove their equality, can I also do it without? If I can prove a (more complicated) statement using function extensionality, is there a way to transform it into a (possibly clumsier) statement which can be proven without function extensionality and which can be transformed into the former statement using some straightforward applications of function extensionality? I'm sorry for logical naivety. The way I am posing the question, I fear it would qualify as soft; nevertheless I am pretty sure that there is some precise statements to be made here (or maybe just a reference to a textbook to be given). REPLY [58 votes]: I am going to draw heavily from Github discussion on HoTT book pull request 617. There are different kinds of equality. Let us say that equality is "intensional" if it distinguishes objects based on how they are defined, and "extensional" if it distinguishes objects based on their "extension" or "observable behavior". In Frege's terminology, intensional equality compares sense and extensional equality compares reference. To use Russell's example, intensionally the morning star and the evening star are clearly not the same (because their definitions are different), but they are extensionally the same because they both denote the same object (planet Venus). A more mathematical example is comparison of $x + y$ and $y + x$. These are extensionally equal, but intensionally differ because (the usual) definition of $+$ treats its arguments asymmetrically. It should be clear that two functions may be extensionally equal (have same behavior) even though they differ intensionally (have different definitions). It is possible for two kinds of equality to coexist. Thus in type theory there are two equalities. The intensional one is called "judgmental" or "definitional equality" $\equiv$ and the extensional one is known as "propositional equality" $=$. Mathematicians are aware of $=$ as a "real thing" while they think of $\equiv$ as "formal manipulation of symbolic expressions" or some such. We may control the two kinds of equality and the relationship between them with additional axioms. For instance, the reflection rule collapses $\equiv$ and $=$ by allowing us to conclude $a \equiv b$ from $a = b$ (the other direction is automatic). There are also varying degrees of extensionality of $=$. Without any extra axioms, $=$ is already somewhat extensional. For instance, we can prove commutativity of $+$ on natural numbers by induction in the form $x + y = y + x$, but we cannot prove $x + y \equiv y + x$. Function extensionality is an axiom which describes what constitutes an "observation" on functions: by saying that two functions are equal when they give equal values we are in essence saying that only values matter (but not for example the "running time" or some other aspect of evaluation). Another axiom which makes $=$ "more extensional" is the Univalence axiom. It is hard to do mathematics without function extensionality, but type theorists have their reasons for not including it as an axiom by default. But before I explain the reason, let me mention that there is a standard workaround. We may introduce user-defined equalities on types by equipping types with equivalence relations. This is what Bishop did in his constructive mathematics, and this is what we do in Coq when we use setoids. With such user-defined equalities we of course recover function extensionality by construction. However, setoids are often annoying to work with, and they drag in technical problems which we would prefer to avoid. Incidentally, the setoid model shows that function extensionality does not increase the power of type theory (it is a model validating function extensionality built in type theory without function extensionality). So why don't type theorist adopt function extensionality? If we want to have a type theory with nice properties, and a useful proof assistant, then $\equiv$ should be "easy" to deal with. Technically speaking, we would like a strongly normalizing $\equiv$. By assuming function extensionality we throw into type theory a new constant funext without explaining how it interacts with the process of strong normalization, and things break. Type theorists would say that we failed to explain the computational meaning of funext. Consequently, Coq does not adopt function extensionality because that would lead to a lot of problems. Coq would not be able to handle $\equiv$ automagically anymore, and the whole system would just have worse behavior. Type theorists of course recognize that having a good computational explanation of function extensionality, and more generally of the univalence problem, would be extremely desirable. This is why the HoTTest open problem is to give a computational interpretation of the univalence axiom. Once this is accomplished, we ought to have at our disposal type systems and proof assistants which are much more natural from a mathematician's point of view. Until then, you can always assume funext as an axiom and work around the resulting complications. To see how this can be done, have a loot at the Funext axiom in the HoTT library. [This P.S. is outdated after the question was edited.] P.S. The title of your question points to a common leap of reasoning from "not accepting function extensionality" to "denying function extensionality". While there are models in which function extensionality has counter-examples, one should be aware of the difference between "not accept" and "deny". (I am complaining because this sort of leap is often made about the law of excluded middle, and there it leads to absurdity.)<|endoftext|> TITLE: Is there a generalization of linear algebra that allows fractional ranks? QUESTION [6 upvotes]: The rank is the number of linearly independent rows/cols of a matrix. Generally, we think of linear independence as a binary property. But we could imagine an alternative definition that allows for numbers in the range [0,1]. Then, we could have fractional ranks. I'm curious if there's any use to such generalizations of rank/independence, or if anyone has even thought about it before? REPLY [11 votes]: von Neumann thought about this; the keyword is continuous geometry.<|endoftext|> TITLE: Is there a finitely generated residually finite group with solvable word problem that does not embed in a finitely presented residually finite group? QUESTION [15 upvotes]: The famous Higman embedding theorem says that every recursively presented group embeds in a finitely presented group. This is a convenient tool to construct finitely presented groups with bizarre properties from recursively presented ones, which are usually easier to construct. One cannot hope for an exact analogue of Higman's theorem in the setting of residually finite groups because finitely presented residually finite groups have solvable word problem and hence their finitely generated subgroups do as well. But Meskin constructed finitely generated recursively presented groups with undecidable word problem. I know of no other obstruction to embedding a finitely generated residually finite group into a finitely presented one, so I ask the following question. Does there exist a finitely generated residually finite group with decidable word problem that cannot be embedded in a finitely presented residually finite group? REPLY [5 votes]: A preprint by E. Rauzy appeared today on the arXiv, and gives a negative answer to this question. In other words (if the proof is correct), there exists a f.g. residually finite group with decidable word problem which does not embed in a f.p. residually finite group.<|endoftext|> TITLE: Picard group of Drinfeld upper half space QUESTION [10 upvotes]: Let $K$ be a $p$-adic field and $\Omega^{(n)}_K$ the $n$-dimensional Drinfeld upper half space over $K$ (which is a rigid analytic space over $K$). Is the Picard group of $\Omega^{(n)}_K$ known? More generally, I would like to know the Picard group of $\Omega^{(n)}_K$ base changed to any finite extension $L$ of $K$ (which is not the same as $\Omega^{(n)}_L$). I'd appreciate any references where this question, or similar questions, are considered. REPLY [2 votes]: In this paper of Junger, all these Picard groups are shown to be zero. I think this was known for a long time for $n=1$, see for example the book Rigid Analytic Geometry and its Applications by Fresnel and van der Put.<|endoftext|> TITLE: Perfectly centered break of a perfectly aligned pool ball rack QUESTION [95 upvotes]: Imagine the beginning of a game of pool, you have 16 balls, 15 of them in a triangle <| and 1 of them being the cue ball off to the left of that triangle. Imagine that the rack (the 15 balls in a triangle) has every ball equally spaced apart and all balls touching all other appropriate balls. All balls are perfectly round. Now, imagine that the cue ball was hit along a friction free surface on the center axis for this triangle O-------<| and hits the far left ball of the rack dead center on this axis. How would the rack react? I would imagine this would be an extension of newtons cradle and only the 5 balls on the far end would move at all. But in what way would they move? Thanks REPLY [18 votes]: This is more of a comment inspired by Jim Belk's answer than an answer to the question in itself. It is also more physics than mathematics. However, I hope I can help readers see that this system is related to some very interesting science. I wanted to say a little bit about the dynamics of the energy transfer in the billiard break. Very naïvely, you might think of a triangular lattice of interacting particles (the billiard rack) as acting like a discrete model for a crystalline solid. Solids support sound waves (just knock on your table and listen), and so again naïvely we might expect that the billiard break occurs via some strong compressional sound waves propagating through the system, akin to those vibrations induced when knocking on wood. However, this naïve intuition is wrong! A system of billiards all just barely touching, like Newton's cradle, is an example of a sonic vacuum (see e.g. this expository physics article). The term was coined by Vitali Nesterenko to describe mechanical systems where the speed of sound vanishes and the tiniest perturbation must lead to supersonic shocks -- behavior very far from that of ordinary sound. The usual formula for the speed of sound in say a chain of identical masses joined by identical springs is $v_s=\sqrt{k/m}a$ where $k$ is the spring constant, $m$ is the mass, and $a$ is the length of the springs. In a system interacting via a nonlinear force law like Hertz's, the effective spring constant is $df/d\delta|_{\delta=0}=0$ (where $\delta$ is the overlap between particles) which actually vanishes! Thus the speed of sound is zero. This is a very special situation which occurs because the particles are all just barely touching. The consequence of all this is that energy cannot propagate via ordinary sound, and what appears instead is a shock wave, satisfying a nonlinear PDE. In Jim Belk's pictures, this is what we see as the "shock front" colored yellow traveling from the point of impact through the billiards. Nesterenko wrote down a continuum limit for systems of Hertzian particles (the following is specialized to particles arranged on a triangular lattice): $\frac{\partial^2\xi}{\partial t^2}=c^2\frac{\partial^2}{\partial x^2}\left[\xi^{3/2}+\frac{2R^2}{5}\xi^{1/4}\frac{\partial^2}{\partial x^2}(\xi^{5/4})\right]$ (If one wants to study shock waves in systems of particles interacting with other force laws, then some of the exponents here change). Also, none of the above discussion really depends on the "perfectly centered" initial condition. I think the paper I cite below may give you some idea of what can happen as we vary the initial angle. It might be interesting to compare the solutions of this equation with suitable boundary conditions to simulations along the line of Jim Belk's for a very large pool ball rack (large enough at least so that one can follow the shock wave for an appreciable length of time). The derivation of the above equation is apparently in Nesterenko's book "Dynamics of Heterogeneous Materials", though I learned it from this very relevant paper coauthored by some friends of mine: Transmission and reflection of strongly nonlinear solitary waves at granular interfaces, by A. M. Tichler, L. R. Gomez, N. Upadhyaya, X. Campman, V. F. Nesterenko, V. Vitelli. In this paper they study the propagation of shocks through interfaces between packings of billiard balls of two different masses. I think you'll agree it's quite close to the billiard break. Here's figure 2 of Tichler et al's paper.<|endoftext|> TITLE: Reference Request: Compact manifolds with boundary have the homotopy type of a CW-complex QUESTION [13 upvotes]: Let $M$ be a compact manifold (possibly non-smooth) manifold with boundary $\partial M$. Is the inclusion $\partial M\hookrightarrow M$ homotopy equivalent to the inclusion of a subcomplex into a CW-complex, i.e. is there a CW-complex $X$ with a subcomplex $Y$ and homotopy equivalences $g\colon X\rightarrow M$ and $h\colon Y\rightarrow \partial M$, such that the obvious diagram commutes up to homotopy? I think this must be true, but I can't find a reference. I would be also interested, if someone has one for the smooth case. REPLY [4 votes]: Kirby and Siebenmann proved that any compact topological manifold has the homotopy type of a finite complex (announced in Kirby, R. C.; Siebenmann, L. C. On the triangulation of manifolds and the Hauptvermutung. Bull. Amer. Math. Soc. 75 1969 742–749, with proofs in their book "Foundational essays on Topological Manifolds").<|endoftext|> TITLE: origin of spectral sequences in algebraic topology QUESTION [29 upvotes]: I have the following somewhat vague question. I am not sure if it is appropriate for this forum, please feel free to close (or migrate to stackexchange). I have been "brought up" as an algebraic geometer. Spectral sequences are obviously ubiqutous and useful in this subject. The conclusion I have drawn from my exposure to spectral sequences there is that they express how you can compute ordinary (non-derived) invariants of "derived objects". An alternative way of saying this that whenever you have spectral sequence you should identify it as being either a grothendieck spectral sequence (you are really doing a derived a composition), or a hyper(co)homoloy spectral sequence (you legitimately want to know ordinary invariants of derived objects). I have now begun studying (classical) stable homotopy theory, and there seems to be a bewildering set of spectral sequences. I cannot explain any of them in the above terms, but many of them "feel like they are close to being of the above type". Let me give some examples. Consider the spectral sequence of a homotopy limit: $\lim^* \pi_* E_\bullet \Rightarrow \pi_* \operatorname{holim} E_\bullet$ (I'm writing $*$ for all indices to avoid getting into details.) If you pretend that there is a nice functor $D\pi: SH \to DAb$ taking a spectrum to a chain complex with homology groups the homotopy groups of the spectrum ($h_* D\pi = \pi_*$) and which commutes with homotopy limits, then the above "is just the hyperhomology spectral sequence". Unfortunately I'm fairly sure $D\pi$ cannot exist. If we keep up the pretense for a bit, we could try to say $Map(E, F) = RHom(D\pi E, D\pi F)$ (this is getting real silly now, since $D\pi$ is fully faithful and essentially surjective), and then the Atiyah-Hirzebruch spectral sequence also becomes "just a hyperhomology spectral sequence". It seems similarly imaginable that the Atiyah-Hirzebruch spectral sequence is an incarnation of the Leray-Serre spectral sequence (for the inclusion $i: * \to X$), although I am less sure how to even put this in symbols. I could go on; by dreaming up $D\pi$ (or a related gadget) to have various (eventually contradictory) properties many spectral sequences can be "interpreted" in this way. But enough woffling. Now to my real question. Is there a way in which sense can be made of these ideas? For example by replacing $DAb$ by a more complicated abelian category? Alternatively, is there a better organising principle for spectral sequences in algebraic topology? Notes If X is a topological abelian group (spectrum), then $D\pi X = N_\bullet Sing(X)$ (normalised chain complex of the singular simplicial abelian group of X) has some of the properties dreamed up above. Since the spectral sequences apply to topological abelian groups and their maps and in this case reduce to the hyperhomology spectral sequences I have shown, this perhaps explains why the topologists' spectral sequences feel familiar. Thanks, Tom REPLY [2 votes]: Here is how I look at the situation after thinking hard about the answers. I will say things in the language of triangulated categories and homotopy limits being somewhat imprecise. I suppose some hares would say that this is really about $(\infty,1)$-categories; unfortunately I don't know enough about these gadgets. We are in the following situation: we have triangulated categories $D_1$ and $D_2$ together with a triangle functor $\alpha: D_1 \to D_2$ (additive, preserving distinguished triangles, etc). We would like to "explain" or "describe" the effect of $\alpha$ on objets. Clearly this necessitates first a way of describing objects in the first place. The most conventional way I can think of is to use homological functors $\pi_i: D_i \to A_i$ into abelian categories (i.e. functors turning distinguished triangles into long exact sequences). I will write $\pi_{i*}X$ for the graded object $\pi_i(\Sigma^*X),$ where $\Sigma$ is the shift functor. Ideally, we would like a way to compute $\pi_{2*} \alpha X$ in terms of $\pi_{1*}X$ and perhaps some extra data. It is sort of clear that this cannot work in full generality, because $\pi_1$ can throw away arbitrary amounts of information. So we need a more "internal" way of describing objects. One way to do this is using filtrations. That is, we associate to $X$ a sequence $\dots \to X_{k-1} \to X_k \to X_{k+1} \to \dots$ such that $X = \operatorname{holim} X_\bullet$ or $X = \operatorname{hocolim} X_\bullet.$ For this to be useful we need the "subquotients" $\operatorname{cone}(X_k \to X_{k+1})$ to be "nice", e.g. live in a subcategory we understand well. The "internal" and "external" approach are not unrelated; for example a t-structure on $D_1$ has as heart an abelian subcategory, every object acquires a filtration and a "cofiltration" (I think), and the "subquotients" live in the abelian subcategory, effectively giving us a compatible $\pi_1.$ So the strategy to describe $\alpha(X)$ is now this: Find the (co)filtration $X_\bullet$ and relate $\alpha(\operatorname{ho(co)lim}X_\bullet)$ to $\operatorname{ho(co)lim}\alpha(X_\bullet).$ Relate the subquotients of $X_\bullet$ to the subquotients of $\alpha X_\bullet.$ Relate the subquotients of $\alpha X_\bullet$ to $\pi_2\operatorname{ho(co)lim} \alpha(X_\bullet).$ Step one works if $\alpha$ commutes with filtered homotopy limits or colimits (then take the filtration or cofiltration as appropriate). I don't know good conditions for this, but it seems to be common. Certainly (?) homotopy (co)limits commute with homotopy (co)limits, so we can get the spectral sequence of a homotopy (co)limit in this language. Also mapping spaces commute appropriately, so we can get the AH-SS. Step two really is the input to this entire game. Note that $\alpha$ commutes with finite homotopy (co)limits (i.e. cones) so this is reasonable. Step three is the spectral sequence of a filtered homotopy type. There is always a spectral sequence, but it may or may not converge to the groups of the homotopy (co)limit (depending on exactness properties of homotopy (co)limits on $A_2$). I think many spectral sequences can be looked at in this way (certainly Grothendieck, AH, homotopy limit/colimit; probably also Adams). To expand on Ben Wieland's example of the AH-SS, in this case $D_1 = D_2 = SH,$ $\pi_1 = \pi_2 = \pi$ is the stable homotopy groups, corresponding to the natural t structure, with heart abelian groups, $X$ is a fixed spectrum, $\alpha(E) = Map(X, E).$ Then $\alpha$ sends homotopy limits to homotopy limits, so we should use the filtration corresponding to $\pi,$ writing $E$ as the homotopy limit of its Postnikov filtration. The subquotients are Eilenberg-Maclane spectra corresponding to the homotopy groups of $E,$ and so we get a spectral sequence $$ E_2^{*,*} = \pi_*\alpha(\text{subquotients of }E_*) = \pi_* Map(X, H\pi_*E) = H^*(X, \pi_*E) \Rightarrow \pi_*\alpha(E) = H^*(X, E)$$ which is what we wanted.<|endoftext|> TITLE: Solve $f(x)=\int_{x-1}^{x+1} f(t) \mathrm{d}t$ QUESTION [5 upvotes]: Solve $f(x)=\int_{x-1}^{x+1} f(t) \mathrm{d}t$. When $f$ is a function, it looks like the only solution is $f(x)=0$. But what if we allow distributions, such as the Dirac delta? REPLY [25 votes]: By differentiating we obtain $$f'(x)=f(x+1)-f(x-1)$$ This type of equations was addressed in the MO question On equation f(z+1)-f(z)=f'(z) Let $\lambda$ be any (complex) root of the equation $$\lambda=e^\lambda-e^{-\lambda},$$ which is equivalent to $z=2\sin z$, as Noam wrote. to this $\lambda$ a solution $f(x)=e^{\lambda x}$ is associated. So you have infinitely many exponential solutions. Any linear combination is also a solution. Then, depending on your assumptions of $f$ you can consider various limits of those linear combinations in the appropriate topology for your functions/distributions class. Edit. For a complete theory of this kind of equations see "Fonctions moyenne-periodiques", a theory created by Delsart and Schwartz in 1940-s. This is generalized in the modern theory of "equations of convolution-type", see, for example Hormander, Linear Partial differential operators. In this theory one considers equations $u\star f=0$, where $u$ is a distribution. In your case, $u=\delta-\chi,$ where $\chi$ is the characteristic function of $[-1,1]$. The method of solution is an appropriate version of Fourier--Laplace transform, depending on your class of functions/distributions. Ordinary Fourier transform of $u$ is $U(\lambda)=1-2\sin\lambda/\lambda$, whose roots give you the exponential solutions. The question whether we obtain all solutions in this way is addressed by the Malgrange approximation theorem (Theorem 16.4.1 in Hormander).<|endoftext|> TITLE: On the number of polynomials that divide $x^{q-1}-1$ in some subspaces of $\mathbb F_q[x]$ QUESTION [9 upvotes]: Let $\mathbb F_q$ be a finite field (where $q$ is in general a power of a prime), and let $e, k$ be positive integers with $k \leq e < q-1$. Let $f_0(x), \ldots, f_k(x) \in \mathbb F_q[x]$ be polynomials such that $$e= \deg f_0 > \deg f_1 > \cdots > \deg f_k,$$ and let us define $W:= \langle f_0, f_1, \ldots, f_k \rangle$ the subspace of dimension $k+1$ of $\mathbb F_q[x]_{\leq e}$ seen as vector space over $\mathbb F_q$. Furthermore let $T_e$ be the set defined by $$T_e:=\left\{p(x) \in \mathbb F_q[x] \;|\; \deg p =e \;, lc(p)=1, \; p(x)|x^{q-1}-1 \right\}. $$ I'd like to find $$ M_{e,k,q}:=\max | W \cap T_e|, $$ where the maximum is over all the subspaces of the form described above. Three simple cases are: When $e=k$ then $M_{e,e,q}= \binom{q-1}{e}$. When $e=q-2$ then $M_{q-2, k, q} =k+1$. When $k=1$ then $M_{e,1,q}=q-e$. When $k=0$ then $M_{e,0,q}=1$. Maybe it could be useful the fact that, for $e TITLE: Adams' theorems on the Hopf-Whitehead J-homomorphism QUESTION [17 upvotes]: The J-homomorphism is a well-known and classical map $\pi_n (O(k)) \to \pi_{n+k} (S^k)$, or after stabilizing with respect to $k$, a map $J_n:\pi_n (O) \to \pi_{n}^{st}$, from the stable homotopy of orthogonal groups to stable homomotopy of spheres. The main results on $J_n$ were proven by Adams in a classic series of four papers. The results are: the image of $J_{4r-1}$ (in which case the source group is $\mathbb{Z}$) is cyclic of order $den (\frac{B_r}{4r})$ (denominator of Bernoulli numbers). For $n\equiv 0,1 \pmod 8$, $J_n$ is injective (the source is $\mathbb{Z}/2$). Another theorem by Adams is that the unit map $\pi_{n}^{st} \to \pi_n (BO)= \pi_{n-1} (O)$ hits all $\mathbb{Z}/2$-groups in the image. These results play an important role in differential topology (for example in the classification of exotic spheres), which is why from time to time, I struggle to understand these results. But I am rather foreign to stable homotopy theory and I am scared away by this battle with homological algebra and stable homotopy theory and never manage to get the main points from Adams' papers. However, for the case $n=4r-1$, there is a version of the proof without leaving the mathematical terrain I am used to navigate in. There are two parts: proving that the image of $J$ has \it{at least} the order $den (\frac{B_r}{4r})$ requires an invariant an a device for its computation. The invariant is the $e$-invariant $e: \pi_{4n-1}^{st} \to \mathbb{Q}/\mathbb{Z}$ and the computation of $e \circ J$ is done using characteristic classes. All this is well-explained in Hatcher's book project on $K$-theory, with a hands-on definition of the $e$-invariant. Proving that the image of $J$ has \it{at most} the order $den (\frac{B_r}{4r})$ requires a construction of a nullhomotopy. It follows from the Adams conjecture and some Bernoulli numerology. Besides the first proofs of the Adams conjecture by Quillen and Sullivan, there exist two proofs which I understand (by Becker-Gottlieb and a simplification of it by E. Brown, which I wrote up some years ago). Here are my questions: Is there an argument for the injectivity of $J_{n}$, $n \equiv 0,1 \pmod 8$, which is similarly direct as the argument in Hatcher's book? The $J$-homomorphism gives a map of spectra $\Sigma^{-1} KO \to S$. What is the composition with the unit map $S \to KO$, and what is a low level explanation for it? EDIT: this is too naive, see Neil's comment below. Is there a low-level description for the image of the unit map $S \to KO$ on homotopy groups? Or do have to learn it the hard way? REPLY [18 votes]: This answer probably won't be coming from the perspective that you want, since it'll use even more stable homotopy theory than Adams did. But I think it's pretty clear in its own way. First let me make a small correction to Neil's comment. Actually, from the J-homomorphism construction you only get a map of spectra from the connective cover of $\Sigma^{-1}ko$ to $gl_1(S^0)$. So you have to kill that bottom Z in ko. But actually the better thing to do is to "find" the same Z below $gl_1(S^0)$ instead, by changing the target to $pic(S^0)$, the classifying spectrum for invertible spectra. So the J-homomorphism should be thought of as a map of spectra $ko\rightarrow pic(S^0)$. In these terms it has an easy intuitive description too: $ko$ classifies vector bundles, $pic(S^0)$ classifies stable spheres, and the J-homomorphism sends a vector bundle to its one-point compactification. Now you see why it's the multiplicative structure on $S^0$ that's relevant, because this operation sends direct sum of vector bundles to smash product of spheres. This isn't just me being pedantic, by the way. It's good to understand exactly what structure the J-homomorphism carries, and it's also very handy to have that extra Z around. One measure of this is the following: Claim: Let $J: ko \rightarrow pic(S^0)$ be any map of spectra which sends the unit class in $\pi_0 ko$ to the class of the 1-sphere in $\pi_0 pic(S^0)$. Then the image of J on homotopy groups satisfies the ``lower bound'' estimates established by Adams for the J-homomorphism. Unless I'm misreading your question it's these lower bound estimates that you're interested in. So what I'm saying is that for those purposes, all you need to know about the J-homomorphism, besides the natural structure it carries, is that it ``sends 1 to 1'' on that bottom Z we attached (or rather, never bothered to forget). The way we'll prove the above claim is to give a more modern version of Adams' e-invariant argument. This will be based on Rezk's study of logarithmic operations coming from the Bousfield-Kuhn functor. The motivation is that, as Neil says, the spectrum $pic(S^0)$ (or $gl_1(S^0)$) is kind of mysterious. That's too bad for us, since apparently we need to understand it to prove the above claim. However, the Bousfield-Kuhn functor tells you that for any prime p, the "part of a spectrum which mod p K-theory understands", meaning the K/p-localization of the spectrum, only depends on the n^{th} space of the spectrum, for any n you like. This is some amazing consequence of periodicity. The upshot is that the K/p-localization doesn't care that $pic(S^0)$ has this exotic infinite loop structure, and you get a natural map $$pic(S^0)\rightarrow L_{K/p}S^{1}$$ which exhibits the latter as the K/p-localization of the former. Now, a priori this map is somewhat mysterious on the level of homotopy groups. But actually Rezk made a fantastic study of it in his paper on logarithmic cohomology operations (which the above is essentially an instance of.) One consequence of Rezk's work is a calculation of the effect of the above map on $\pi_0$. It is as follows. First, recall that $\pi_{-1} L_{K/p}S$ identifies with $Hom(Z_p^\times,Z_p)$ (noncanonically, this is just $Z_p$, the $p$-adic integers.) Then, the above map $log:pic(S^0)\rightarrow L_{K/p}S^{1}$ sends the class of the $1$-sphere to the homomorphism $Z_p^\times\rightarrow Z_p$ given by $$x\mapsto \frac{1}{2p}log(x^{p-1}).$$ (I might have a sign wrong, but that doesn't matter for our purposes.) Here $log$ stands for the $p$-adic logarithm. It's an interesting thing that the above expression makes sense and is primitive. That is, the set of all $(p-1)^{st}$ powers of $p$-adic units is exactly the domain of convergence of $log$, and, also, $2p$ is the GCD of all the values of $log$. We deduce from this that the composition $log\circ J:ko\rightarrow L_{K/p}S^{1}$ sends the unit class in $\pi_0ko$ to that same homomorphism. Now the point is that $K/p$-local homotopy is computationally friendly. Also, $ko$ is nearly $K/p$-local: its $K/p$-localiztion is the $p$-adic $KO$, via the connective cover map $ko\rightarrow KO$. Thus it's not hard to calculate that a homotopy class of maps $ko\rightarrow L_{K/p}S^{1}$ is completely determined by its effect on $\pi_0$. So actually the above minimal information tells us exactly what $log\circ J$ is. Then if we look on higher homotopy groups, we see that the image of $log\circ J$ has size given by Adams' upper bound. This implies the claim. By the way, I think the cleanest way to perform the above calculations is to fix the generator $g = exp(\frac{2p}{p-1})$ for the group $Z_p^\times/\pm$ of p-adic Adams operations acting on p-adic $KO$. Then you get to write down the fiber sequence $$L_{K/p}S \rightarrow KO\overset{1-\psi^g}{\rightarrow} KO,$$ and the result is that $log\circ J$ identifies with $ko\rightarrow KO\rightarrow L_{K/p}S^{1}$, where the last map is the boundary map in the above sequence. (Again my signs might be off, sorry for that.)<|endoftext|> TITLE: "Fraïssé limits" without amalgamation QUESTION [13 upvotes]: All structures are countable with countable signature. Given a structure $\mathcal{A}$, the age of $\mathcal{A}$, $Age(\mathcal{A})$, is the set of structures isomorphic to finitely-generated substructures of $\mathcal{A}$ (see https://en.wikipedia.org/wiki/Age_(model_theory)). (This isn't really a set, but we can restrict attention to those structures contained in $H_{\omega_{17}}$, say.) $Age(\mathcal{A})$ automatically satisfies the Joint Embedding and Hereditariness properties, but may lack the Amalgamation property: an example of this is the undirected graph $\mathcal{G}$ with integers as vertices, and an edge between ever pair $(z, z+1)$. Given a collection $\mathbb{K}$ of finite structures closed under isomorphism with the Joint Embedding, Hereditariness, and Amalgamation properties, we can form the Fraïssé limit of $\mathbb{K}$, the unique homogeneous structure $F(\mathbb{K})$ with $Age(F(\mathbb{K}))=\mathbb{K}$. Specifically, let $\mathbb{P}(\mathbb{K})=\mathbb{P}$ be the poset whose elements are finite sequences $\emptyset\prec \mathcal{A}_0\prec . . . \prec\mathcal{A}_n$ of structures in $\mathbb{K}$, ordered by reverse extension; then forcing with $\mathbb{P}$ produces the Fraïssé limit. Formally, there is a countable collection $\mathcal{D}$ of dense sets in $\mathbb{P}$ such that the direct limit of the terms of any $\mathcal{D}$-generic filter through $\mathbb{P}$ is isomorphic to the Fraïssé limit. Now if $\mathbb{K}$ lacks the Amalgamation property, there will be no homogeneous structure with age $\mathbb{K}$. The poset $\mathbb{P}(\mathbb{K})$, however, still makes sense, and we can consider the collection $Fil(\mathbb{K})$ of all structures formed by (taking direct limits of terms of) maximal filters through $\mathbb{P}(\mathbb{K})$. Moreover, we can compare structures in $Fil(\mathbb{K})$ in terms of relative genericity. For $D\subseteq \mathbb{P}(\mathbb{K})$ dense, let $\mathcal{O}_D$ be the set of structures which can be built by filters hitting $D$. These sets $\mathcal{O}_D$ then generate a topology $\tau_\mathbb{K}$ on $Fil(\mathbb{K})$. This space is terrible, but seems reasonably natural if what we're interested in is how generic elements of $Fil(\mathbb{K})$ are. As it turns out, elements of $Fil(\mathbb{K})$ can be extremely generic, even though the Fraïssé limit as such may not exist. Specifically, let $Gen(\mathbb{K})$ be the class of isomorphism classes of structures $\mathcal{A}$ such that in every open $\mathcal{O}_D$ there is a copy $\mathcal{B}\cong\mathcal{A}$. Then, for example, consider the graph $\mathcal{G}$ from the first paragraph and let $\mathbb{K}=Age(\mathcal{G})$. $Fil(\mathbb{K})$ is precisely the class of acyclic infinite graphs in which each vertex has degree at most 2. However, the graph $\mathcal{G}$ is special among these graphs, since there is a countable collection $\mathcal{D}$ of dense subsets of $\mathbb{P}(\mathbb{K})$ such that any $\mathcal{D}$-generic filter corresponds to $\mathcal{G}$. This means $Gen(\mathbb{K})=\{\mathcal{G}\}$, and so in some sense $\mathcal{G}$ is the closest thing to a Fraisse limit of $Age(\mathcal{G})$ as we might reasonably expect to exist. In general, if $Gen(Age(\mathcal{A}))$ has precisely one element, denote that element by $G(\mathcal{A})$, and call it the generalized Fraïssé limit of $Age(\mathbb{A})$. My questions are as follows. First, the obvious: What are some sources for learning about this construction? In particular, what's it actually called? Second, When is $Gen(Age(\mathcal{A}))$ nonempty? When does it have exactly one element? Finally, the fact that $G(\mathcal{G})=\mathcal{G}$ motivates me to ask: Suppose $Gen(Age(\mathcal{A}))$ has exactly one element. What can we say about $G(\mathcal{A})$ in relation to $\mathcal{A}$? For example, what structures $\mathcal{A}$ do we have $G(\mathcal{A})=\mathcal{A}$? A couple quick observations: any homogeneous structure $\mathcal{A}$ will certainly have $G(\mathcal{A})=\mathcal{A}$. On the other hand, consider the graph $\mathcal{G}_\omega$ consisting of the disjoint union of infinitely many copies of $\mathcal{G}$; then $G(\mathcal{G}_\omega)=\{\mathcal{G}\}$, so not every structure has this property even if it has singleton $Gen$. Finally, note that we will always have $G(G(\mathcal{A}))=G(\mathcal{A})$ if $G(\mathcal{A})$ exists. REPLY [5 votes]: Kubiś wrote a lot on this subject. This. This. I also have a small contribution on this subject. There is also an even more abstract point of view, started by Rosický in '80. The general motto is that the amalgamation property has a key role in the construction of the Fraïssé limit, any weakening of it corresponds to a weakening of the universal property limit.<|endoftext|> TITLE: history of quaternion algebras QUESTION [23 upvotes]: Who is responsible for the generalization of Hamilton's quaternions to other types of quaternion algebras, and when did this occur? In particular, Hamilton's quaternions are the 4-dimensional algebra generated over $\mathbb{R}$ by the elements $1, i, j, ij$, where $i^2=j^2=-1$, and $ij=-ji$, invented in the 1840s. In a generalized quaternion algebra, we allow $\mathbb{R}$ to be replaced by any field $K$ not having characteristic 2, then we choose $a,b\in K-\{0\}$, and let $i^2=a$, $j^2=b$. The primary reference folks give for the latter version is Marie-France Vigneras' book Arithmetique Des Algebres De Quaternions, published in 1980. I would love to see the references in that book, but unfortunately it's out of print, and the (choppy) English translation I find online leaves them out. If one follows the evolution of Hamilton's idea, one finds that Graves' invention of the octaves (later octonians) is the most directly inspired successor (but not a generalization). There is also a lesser known construction by Macfarlane where he investigates what happens when $a=b=1$ (using $a,b$ as above), dating back to 1891, but it seems that did not catch on. Perhaps such structures were studied before they were called "quaternion algebras," since this definition also characterizes all possible central-simple 4-dimensional algebras over a field (char$\neq2$). If this is the case, I wonder who was studying them before they were given that name, and I wonder who was responsible for connecting it back to Hamilton's terminology. I'm asking not only out of a curiosity about the history, but also because I want to see what were the motivations and intentions behind the generalization when it was first introduced, as compared to current applications. REPLY [32 votes]: In the early 1900s, Dickson introduced what he called generalized quaternion algebras over any field $K$ of characteristic not 2. These are exactly what we'd call quaternion algebras over $K$. His definition was in terms of a basis with rules for their products, and he gave a criterion for these to be division rings. In particular, these were the first noncommutative division rings besides the quaternions of Hamilton, aside from subrings of Hamilton's quaternions. Three of Dickson's works where he introduces these algebras are (1) Linear Algebras, Trans. AMS ${\bf 13}$ (1912) 59-73. (2) Linear Associative Algebras and Abelian Equations, Trans. AMS ${\bf 15}$ (1914), 31-46. (3) Algebras and Their Arithmetics, Univ. of Chicago Press, 1923. In (1) he gives the defining equations for a generalized quaternion algebra and the norm criterion for it to be a division algebra (pp. 65-66), though without using the label "generalized quaternion algebra." He writes in a footnote that this work goes back to 1906. In (2) he constructs cyclic algebras, without using that name, and calls the special case of dimension 4 a generalized quaternion algebra. (Archaic terminology alert: Dickson refers to equations defining cyclic Galois extensions as uniserial abelian equations.) In (3) he defines cyclic algebras again without using that name (p. 65), writes them as $D$, refers to them as algebras of type $D$ (p. 68), and remarks in a footnote on p. 66 that Wedderburn calls them Dickson algebras. Near the end of the book he looks at generalized quaternion algebras over the rationals with $\mathbf Q(i)$ as a maximal subfield, and as a particular example he uses the Hamilton quaternions over $\mathbf Q$ to describe all rational and integral solutions of certain quadratic Diophantine equations in several variables. Think about how a sum of four squares factors over the quaternions to imagine how arithmetic properties of quaternions could be useful to analyze a Diophantine equation involving a sum of four squares; this is similar in spirit to the way the Gaussian integers are useful in studying a Diophantine equation involving a sum of two squares. The term "arithmetics" in the title of the book is, as far as I can tell, Dickson's label for what we'd call maximal orders, so the book would be called today "Algebras and Their Maximal Orders." The motivation for Dickson's interest in quaternion algebras was the earlier development of integral Hamilton quaternions, due first to Lipschitz (all integral coefficients, which is clunky in the same way that $\mathbf Z[\sqrt{-3}]$ is compared to $\mathbf Z[(1+\sqrt{-3})/2]$) and then to Hurwitz (all integral or all half-integral coefficients). To study quadratic Diophantine equations in several variables going beyond a sum of four squares, Dickson was led to extend the original definition of quaternions. His main interest was developing the right theory of generalized integral quaternions, rather than just a theory over a field. A further reference in this direction is Dickson's paper (4) On the theory of numbers and generalized quaternions, Amer. J. Math ${\bf 46}$ (1924), 1-16. Concerning the connection to central simple algebras, Dickson was the first to show any division algebra that is 4-dimensional over its center is cyclic, at least outside characteristic 2 since he didn't have a good definition in characteristic 2. Taking into account Wedderburn's theorem that every finite-dimensional central simple algebra over a field is a matrix algebra over a division algebra, and that matrix algebras are cyclic, Dickson had shown that every 4-dimensional CSA over a field not having characteristic 2 is an algebra of "his" type. (Wedderburn proved the analogous theorem for dimension 9.)<|endoftext|> TITLE: Linear independence of the square roots over Q QUESTION [7 upvotes]: Does there exist a real number $a$ such that the numbers $\sqrt{n^2 + a^2}$ (for all natural $n$) are linearly independent over the field of rational numbers? It is evident that $a$ cannot be rational. Is it possible to prove independence for $a=\pi$? REPLY [15 votes]: Suppose that $S_c:=\{\sqrt{n^2+c^2}:n\in\mathbb{N}\}$ is linearly dependent over $\mathbb{Q}$. This means that there is a finite list of rational numbers $a_1,\ldots,a_r$ so that $$ \sum_{n=1}^r a_n\sqrt{n^2+c^2} = 0.$$ Hence the set of $c$ values such that $S_c$ is $\mathbb{Q}$-linearly dependent is smaller than the set of finite sequences of rational numbers. Since the latter set is countable, so is the set of such $c$. Hence the set $S_c$ is $\mathbb{Q}$-linearly independent for almost all $c\in\mathbb{R}$. Of course, this Cantor argument is useless for proving anything about any particular $c$, such as $c=\pi$. REPLY [5 votes]: If $\pi$ is transcendental, the $\sqrt{n^2+\pi^2}$ are linearly independant over ${\mathbb Q}$: take a linear combination and notice that $\sqrt{n^2+\pi^2}$ is the only member of the family that is not smooth at $\pi=in$ (this proof would also show independance over ${\mathbb Q}(\pi)$). If one wants to be more precise, one can argue in this way. Let $C$ be the algebraic curve over ${\mathbb Q}$ defined by equations $Y_n^2=X^2+n^2$ (for $n$ in a finite set $I$), then the point $\pi$, $\sqrt{\pi^2+n^2}$, $n\in I$ is a generic point of this curve as $\pi$ is transcendental. So, if a function vanishes at $\pi$, it is identically $0$. One can also use down to earth arguments by taking the product of the $\sum \pm a_n\sqrt{X^2+n^2}$ to get a polynomial in $X$ with rational coefficients: if this polynomial is $0$ at $\pi$ then it is identically $0$ and one of the factors is identically $0$, etc.<|endoftext|> TITLE: A generalized K- theory via generalized idempotents QUESTION [6 upvotes]: Edit After the answer by Neil Strickland, I add the word "a ring" in this new version. In the literature, there is a concept of generalized idempotent: an n- idempotent is an element $a$ of a Banach algebra or a ring with $a^{n}=a$. Can the 3 equivalent relations, Murray-Von Neumann, similarity and homotopy on 2-idempotents be generalized to n-idempotents,for arbitrary $n>2$? Does this processes gives us a useful and new type of K theory? We know that "Vector bundles" are the topological analogy of 2-idempotents. Now what is a topological analogy for generalized idempotents? REPLY [14 votes]: Let $E_n(A)$ be the set of $n$-idempotents in $A$, and let $u_1,\dotsc,u_n$ be the elements of $E_n(\mathbb{C})$. Let $E'_n(A)$ be the set of $n$-tuples $e_1,\dotsc,e_n\in E_2(A)$ with $e_ie_j=0$ for $i\neq j$, and $\sum_ie_i=1$. Define $f\colon E'_n(A)\to E_n(A)$ by $f(e_1,\dotsc,e_n)=\sum_iu_ie_i$. Then it is not hard to see that $f$ is bijective. Thus, $E_n(A)$ does not really tell you anything that is not already determined by $E_2(A)$.<|endoftext|> TITLE: Are rounded rectangle billiard dynamics ergodic? QUESTION [15 upvotes]: Bunimovich proved that the billiard-ball dynamics in the Bunimovich stadium is ergodic.             (Image from Microwave_billiards_and_quantum_chaos.) Q. Is it known that the billiard-ball dynamics in a rounded rectangle is ergodic?      Here the corners are quarter circles; so can be viewed as a stadium with vertical segment inserts. Perhaps Bunimovich's proof also covers rounded rectangles? If anyone knows, I'd appreciate a pointer—Thanks! REPLY [4 votes]: The relevant result is stated in Bunimovich, "Conditions of stochasticity of two-dimensional billiards" Chaos 1, 187-193 (1991): If the circles to which the arcs belong are completely contained in the billiard, the dynamics is ergodic. So Yes, if the rounded arcs have equal radius. If not, it is easy to construct rectangles for which the above condition fails. Conjecture: There is an example with non-equal radii which is not ergodic.<|endoftext|> TITLE: Is $G/T$ a projective variety? QUESTION [8 upvotes]: Let $G$ be a semisimple Lie group and $T$ be its maximal torus. Can we say that $G/T$ is a projective variety?. Is there any proof or counterexample for it? REPLY [10 votes]: Try $G=SL(2,\mathbb{R})$; the $T$ is the stabilizer of a metric, so $G/T$ is the set of metrics on $\mathbb{R}^2$ with unit volume, certainly not projective, because they are just positive definite symmetric $2 \times 2$ matrices with determinant 1, an affine hypersurface. Projective varieties are compact. REPLY [7 votes]: Suppose that $G$ is compact, connected, and semisimple. Let $T\subseteq G$ be a maximal torus. Take the complexification $G_{\mathbb{C}}$ of $G$, and choose a Borel subgroup $B\subseteq G_{\mathbb{C}}$ containing $T$. Using the Iwasawa decomposition $G_{\mathbb{C}}=GB$, we see that $G$ acts transitively on $G_{\mathbb{C}}/B$. Also, the stabilizer of the identity coset is $T$, giving us a $G$-equivariant diffeomorphism $G/T\cong G_{\mathbb{C}}/B$. The thing to note is that $G_{\mathbb{C}}/B$ naturally carries the structure of a complex projective variety. Hence, $G/T$ inherits the projective variety structure for which the isomorphism $G/T\cong G_{\mathbb{C}}/B$ is an isomorphism of projective varieties. Note, however, that the projective variety structure of $G/T$ depends on the choice of $B$ containing $T$, or equivalently, the choice of positive roots for the adjoint representation of $T$ on $\mathfrak{g}_{\mathbb{C}}$.<|endoftext|> TITLE: What's a good reference for the following obstruction theory yoga? QUESTION [5 upvotes]: Fix a colored operad, which I will leave implicit, and a field $\mathbb K$ of characteristic $0$. By algebra in this post I will mean a dg algebra over $\mathbb K$ for the given colored operad. I will let $Y$ denote any algebra, and I will let $X$ denote an algebra such that if you forget the differential, then $X$ is free on a well-ordered set of generators, and such that for each generator, its differential is a composition of strictly earlier (for the well-ordering) generators. Here are some facts I know how to prove, but my proofs are in places long-winded. They are not in any way due to me — they are "classical results", where "classical" is defined as "something the author learned in graduate school". My question is: what is a good reference that proves these results? One may try to build an algebra homomorphism $\eta: X \to Y$ inductively. When trying to $\eta(x)$ for a generator $x$, the only condition is that $\partial \eta(x) = \eta(\partial x)$, the latter having already been defined. The induction can continue iff $\eta(\partial x)$ is exact (it is already closed), and this is measured by the class of $\eta(\partial x)$ in $\mathrm H_{\deg x - 1}(Y)$. Here $x$ has homological degree $x$, and I'm using homological conventions, so that $\deg(\partial x) = \deg x - 1$. I'm also being a bit sloppy with notation — really I mean $\mathrm H_\bullet($the part of $Y$ with the appropriate colors for $\eta(x)$ to be valued there$)$. Suppose that $\eta(\partial x)$ is exact. Different choices for $\eta(x)$ might affect whether later steps of the induction succeed. Changing $\eta(x)$ by something exact will not affect the success or failure of later steps. So the "true" set of choice for $\eta(x)$ is a torsor for $\mathrm H_{\deg x}(Y)$. Recall Sullivan's simplicial dg commutative algebra $\Omega(\Delta_\bullet)$, whose $k$-simplices are the dgca $\mathbb K[t_0,\dots,t_k,\partial t_0,\dots,\partial t_k] / (\sum t_i = 1, \sum \partial t_i = 0)$. By definition, the space $\hom_\bullet(X,Y)$ of homomorphisms $X \to Y$ is the simplicial set whose $k$-simplices are $\hom_{\text{algebras}}(X,Y \otimes \Omega(\Delta_k))$. This simplicial set satisfies the Kan horn filling condition. Suppose we have chosen a homomorphism $\eta : X \to Y$. What is the homotopy type of the connected component of $\eta$ in $\hom_\bullet(X,Y)$? For $k\geq 1$, the $k$th homotopy group $\pi_k(\hom_\bullet(X,Y);\eta)$ is an extension of abelian groups, one for each generator $x$ of $X$, such that the $x$th group describes $\pi_k($space of choices for $\eta(x))$. The $x$th group in the extension is precisely $\mathrm{H}_{\deg x + k}(Y)$. REPLY [3 votes]: Theo, there is a nice, abstractly developed, obstruction theory in Baues's "Combinatorial Foundation of Homology and Homotopy": http://books.google.es/books/about/Combinatorial_Foundation_of_Homology_and.html?id=JejY53ixOloC&redir_esc=y Your context can be fitted into the general framework of this book with little effort. This would give you a much more structured obstruction theory than what you suggest in 1 and 2. Computations are possible, and Baues himself has successfully applied it in many papers (and other books). In 3, that's going to be a Kan complex whenever $X$ is cofibrant, your conditions may not be enough, but probably close. As for 4, and to some extent for a different answer to 1 and 2, you can take a look at Bousfield's "Homotopy spectral sequences and obstructions" http://link.springer.com/article/10.1007%2FBF02765886 This is a long paper, and Bousfield gives usually very short conceptual arguments, so it's indeed very dense, not easy to understand, and very difficult for explicit computations, but it is an invaluable source of knowledge.<|endoftext|> TITLE: Decidability of a matrix product being the identity QUESTION [8 upvotes]: Given a finite set $S$ of $n\times n$ integer matrices, it is known that for $k\geq 3$ it is undecidable whether some product of them (allowing repetitions) is the zero matrix (called the mortality problem). It is also known (http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.102.448&rep=rep1&type=pdf) that if $k=2$, then it is decidable whether some product of the matrices is the identity matrix. Is it known whether for $k$ sufficiently large, it is undecidable whether some product of matrices in $S$ is the identity matrix? This has some connection with a question raised by Kontsevich. REPLY [3 votes]: There is a new algorithm (in manuscript) that shows that Identity Problem for SL(2,Z) is in NP, meaning that it is also NP-complete. Also point to point reachability for SL(2,Z) is also shown recently to be decidable, see proceeding of MFCS 2016 http://drops.dagstuhl.de/opus/frontdoor.php?source_opus=6492 Which question was raised by Kontsevich and what is the connection with identity matrix? Could you please comment.<|endoftext|> TITLE: Reverse Math of High Sets? QUESTION [5 upvotes]: Is there a standard principle in reverse math that is known to be equivalent (over $RCA_0$) to the existence of a set of high (Turing) degree? I'm interested in the general case, but would be happy to learn of such a principle for $\omega$-models. I haven't been able to find much discussion on this topic... but then, I don't have much experience in reverse math yet. If the answer is obvious (say, $ACA_0$), please forgive me! To clarify: the specific principle I'm interested in is the statement that "for all $X$, there exists some $Y\ge_T X$ with $Y'=X''$", appropriately rephrased to avoid the explicit use of the jump operator. REPLY [8 votes]: In the paper "On a conjecture of Dobrinen and Simpson regarding almost everywhere domination", Binns, Lerman, Solomon and I constructed $\omega$-models of this "high" principle which demonstrate it does not imply WKL, WWKL, or $G_\delta$-regularity.<|endoftext|> TITLE: Equivariant classifying spaces from classifying spaces QUESTION [27 upvotes]: Given compact Lie groups $G$ and $\Pi$, there is a notion of "$G$-equivariant principal $\Pi$-bundle", and a corresponding notion of classifying space, often denoted $B_G\Pi$, so that $G$-equivariant maps $X\to B_G\Pi$ (up to equivariant homotopy) correspond to $G$-equivariant $\Pi$-bundles on $X$ (up to equivalence), for nice $X$. There is a natural map of $G$-spaces $$ f\colon B_G\Pi \to \mathrm{Map}(EG, B\Pi), $$ which in some sense encodes the fact that a $G$-equivariant $\Pi$-bundle $P\to X$ gives rise to a principal $\Pi$-bundle of the form $P\times_G EG\to X\times_G EG$. It's natural to ask whether $f$ is a $G$-equivariant weak homotopy equivalence. This amounts to showing that for any closed subgroup $H\leq G$, the induced map on fixed points $$ f^H\colon (B_G\Pi)^H \to \mathrm{Map}(EG,B\Pi)^H\approx\mathrm{Map}(BH, B\Pi)$$ is a weak equivalence. In general, this is not the case, but it is known to be true for arbitrary compact $G$ when either $\Pi$ is finite, or $\Pi$ is compact abelian. The reference for both of these is: JP May, Some remarks on equivariant bundles and classifying spaces. Astérisque No. 191 (1990). Case 1 follows from covering space theory, while case 2 needs a bit of work and is proved in Lashof, May, Segal, Equivariant bundles with abelian structural group. Contemp. Math. 19, (1983). The obvious guess is that there is a common generalization, where 1. or 2. is replaced by $\Pi$ is an extension of a finite group by a compact torus (or more cleanly, $\Pi$ is a compact Lie group with contractible simply connected cover). Is this true? Has someone proved it? REPLY [11 votes]: Added Aug 2016: I've written this up, available at https://arxiv.org/abs/1608.02999 $\def\Hom{\mathrm{Hom}} \def\Map{\mathrm{Map}} \def\ad{\mathrm{ad}}$ I think this is true. I'll sketch a possible proof here. I haven't carefully checked everything, and there are things that need checking. Feel free to do that. First, we can assume $H=G$: we want to show that $(B_G\Pi)^G\to \mathrm{Map}(BG,B\Pi)$ is an equivalence if $G$ is compact Lie and $\Pi$ is compact Lie and a 1-type. We might as well consider the induced map on homotopy fibers over the maps to $B\Pi$ (induced by evaluating at the basepoint of $BG$.) That is, we want to show $$ \Hom(G,\Pi) \to \Map_*(BG,B\Pi) $$ is a weak equivalence. Here $\Hom(G,\Pi)$ is topologized as a subspace of $\Map(G,\Pi)$. We know that this is homeomorphic to $\coprod_{[\phi]} \Pi/C_\Pi(\phi(G))$, a coproduct over conjugacy classes of homomorphisms $G\to \Pi$ (see Nearby homomorphisms from compact Lie groups are conjugate). Given this, it is already clear we get an equivalence when $G$ is compact and connected (reduce to the case where $\Pi$ is a torus). It's not so easy to see why this is so for general $G$: although we can "compute" both sides, the accounting is different and hard to match up. Here's an attempt at a general proof, based on the ideas which work when $\Pi$ is a torus (which involve the idea of continuous cochains as in Graeme Segal, "Cohomology of topological groups"). It should fit into some already-known technology (cohomology of topological groups with coefficients in a topological 2-group?), but I don't want to bother to figure out what or how. Consider data consisting of a group $\Pi$ (a compact Lie 1-type as above), with connected component $\Pi_0$ (which is abelian), a vector space $V$, a group homomorphism $\exp\colon V\to \Pi$, an action $\ad\colon \Pi/\Pi_0\to \mathrm{Aut} V$, such that $\exp(\ad(\pi)v)=\pi \exp[v] \pi^{-1}$. (It's a kind of crossed module.) Given this, define $E(G, (\Pi,V))$ to be the space of pairs $(f,v)$ where $f\colon G\to \Pi$ and $v\colon G\times G\to V$ are continuous maps, satisfying $f(g_1)f(g_2)=\exp[ v(g_1,g_2)] f(g_1g_2)$, $v(g_1,g_2)+v(g_1g_2,g_3)= \ad(f(g_1))v(g_2,g_3) + v(g_1,g_2g_3)$. (I might want to additionally require a normalization: $f(e)=e$. Or not.) The examples I have in mind are $E:= E(G, (\Pi, T_e\Pi))$ and $E^0:= E(G, (\Pi,0))$. The claims are as follows. $E$ is weakly equivalent to $\Map_*(BG,B\Pi)$. To compute the mapping space, you need to climb the cosimplicial space $[k] \mapsto \Map_*(G^k, B\Pi)$. Because $B\Pi$ is a 2-type, you don't need to climb very far. The idea is that if you do this, and you keep in mind facts such as: $\Pi$ is equivalent to $\Omega B\Pi$, and the fibration $(v,\pi)\mapsto (\pi, \exp[v]\pi) \colon V\times\Pi\to \Pi\times \Pi$ is equivalent to the free path fibration $\Map([0,1],\Pi)\to \Pi\times \Pi$, you see that you get an equivalence. (I came up with the definition of $E$ exactly by doing this.) That's kind of sketchy. More concretely: $\Map_*(BG,B\Pi)$ can be identified with the space of maps between pointed simplicial spaces, from $G^\bullet$ to $S_\bullet:=\bigl([n]\mapsto \Map_*(\Delta^n/\mathrm{Sk}_0\Delta^n, B\Pi)\bigr)$. The space $E$ is also a space of maps between such, from $G^\bullet$ to $N_\bullet$, where $N_\bullet$ is a simplicial space built from the crossed module $(\Pi,V)$ (the nerve of the crossed module, as in https://mathoverflow.net/q/86486 ) with $N_n \approx \Pi^n\times V^{\binom{n}{2}}$. It's not to hard to show that $N_\bullet$ and $S_\bullet$ are weakly equivalent Reedy fibrant simplicial spaces; they both receive a map from $\Pi^\bullet$, which exhibits the equivalence. (But note: showing that $N_\bullet$ is Reedy fibrant relies crucially on the fact that $\exp$ is a covering map.) $E^0$ is homeomorphic to $\Hom(G,\Pi)$. Yup. The inclusion $E^0\subseteq E$ is a weak equivalence. To see this, let $C^1:=\Map(G,V)$, as a topological group under pointwise addition. There is an action $C^1\curvearrowright E$, by $u\cdot (f,v)=(f',v')$ where $f'(g) := \exp[u(g)] f(g)$, $v'(g_1,g_2) := u(g_1)-u(g_1g_2) + \ad(f(g_1))u(g_2) + v(g_1,g_2)$. It's useful to note that for any $(f,v)\in E$, the resulting map $G\xrightarrow{f} \Pi\to \Pi/\Pi_0$ is a homomorphism. Thus we write $E=\coprod E_\gamma$ for $\gamma\in \Hom(G,\Pi/\Pi_0)$, and $C^1$ acts on each $E_\gamma$. Consider $(f,0)\in E_\gamma^0= E_\gamma\cap E^0$. Note that $u\cdot (f,0)$ has the form $(f',0)$ for some $f'$ if and only if $u\in Z^1_\gamma$, where this is the set of $u\colon G\to V$ such that $u(g_1)-u(g_1g_2) + \ad\gamma(g_1) u(g_2)=0$. So the action passes to an injective map $C^1\times_{Z^1_\gamma} E^0_\gamma\to E_\gamma$. In fact, it should be a homeomorphism. To see that it's surjective, fix $(f,v)\in E_\gamma$; we need to solve for $u\in C^1$ such that $u(g_1)-u(g_1g_2)+\ad\gamma(g_1)u(g_2)=v(g_1,g_2)$. This amounts to the vanishing of $H^2$ in the complex $C^\bullet_\gamma$ of continuous cochains: $C^t_\gamma:=\Map(G^{t}, V_\gamma)$ (where the differential uses the action $\ad\gamma\colon G\to\mathrm{Aut}(V)$). The vanishing is because $G$ is compact, so we can "average" over Haar measure to turn a non-equivariant contracting homotopy on $D^\bullet_\gamma=\Map(G^{\bullet+1}, V_\gamma)$ into a contracting homotopy on $C^\bullet_\gamma = (D^\bullet_\gamma)^G$. Given this, since both $C^1$ and $Z^1_\gamma$ are contractible groups, (in fact, $Z^1_\gamma=V/V^{\gamma(G)}$ by $H^1=0$), we should have that $C^1\times_{Z^1_\gamma} E^0_\gamma$ is weakly equivalent to $E^0_\gamma$. Note: in the case that $\Pi$ is abelian, we simply get a homeomorphism $C^1\times E^0\approx E$.<|endoftext|> TITLE: Hitting time for two out of three random walk particles QUESTION [10 upvotes]: I'm imagining a simple random walk on $\mathbb{Z}$ with three independent particles (maybe add laziness so they don't jump over each other). Suppose the particles are initially placed at, say, $-10$, $0$ and $10$. The distance between any two particles evolves like a random walk, so the expected hitting time for those two specific particles is infinite. Now, if we require only that any two of the three particles hit, is the expected hitting time still infinite? REPLY [3 votes]: Expanding the comment above to a complete proof of Noam Elkies' formula: Let $\tau$ be the first meeting time for independent simple random walks $X_t, Y_t$ and $Z_t$ started at the even points $x TITLE: Conductor CM abelian variety QUESTION [5 upvotes]: This is probably well known but I am not an expert in the subject. Given an abelian variety $A$ of dimension $g$ with CM by $O_K$ where $K$ is a CM field of degree $2g$, let $N_A$ be the norm of the conductor of $A$ as defined here http://en.wikipedia.org/wiki/Conductor_of_an_abelian_variety My question is : can one compare $N_A$ to $d_K$, the absolute discriminant of $K$? More precisely, can one expect $\log(N_A) < c \log (d_K)$ where $c$ depends on $g$ only. Many thanks in advance! REPLY [5 votes]: The short answer is no, since if we take the quadratic twist of $A$ by $\sqrt{D}$, then the conductor of $A$ will more-or-less acquire divisibility by the primes dividing $D$. So we can make the conductor arbitrarily large without changing the CM field $K$. For example, if $D$ is a square-free odd integer, then the conductor of $E_D:y^2=x^3+Dx$ is divisible by $D^2$, but the CM field $\mathbb{Q}(i)$ is independent of $D$. On the other hand, you might find it useful to know that for any prime $\mathfrak{p}$, if $A$ has potential good reduction at $\mathfrak{p}$ (which is the case for CM abelian varieties), then the power of $\mathfrak{p}$ dividing $N_A$ is bounded by a constant that depends only on $g$ and $[K:\mathbb{Q}]$ (independent of $A$ and $\mathfrak{p}$). For example, for an elliptic curve $E/\mathbb{Q}$ with potential good reduction at $p$, we have $\operatorname{ord_p(N_E)}\le 2$ for $p\ge5$, $\operatorname{ord_p(N_E)}\le 5$ for $p=3$, and $\operatorname{ord_p(N_E)}\le 8$ for $p=2$.<|endoftext|> TITLE: Quantities whose generating functions are symmetric QUESTION [6 upvotes]: This is inspired by an old Putnam problem from 2005, and a solution given by Professor Greg Martin (a Professor of Mathematics at the University of British Columbia, also a user on MO). The question is Question (Putnam 2005): For non-negative integers $m,n$, let $f(m,n)$ denote the number of $n$-tuples $(x_1, \cdots, x_n)$ of integers such that $|x_1| + \cdots + |x_n| \leq m$. Show that $f(m,n) = f(n,m)$. Greg's proof essentially boiled down to showing that the generating function $$\displaystyle G(x,y) = \sum_{m,n \geq 0} f(m,n) x^m y^n$$ is symmetric in $x,y$. Update: Since it seems that MAA took down the Putnam directory and the old solutions are no longer easily accessible, I shall give the proof here. The credit goes entirely to Professor Martin. We write $$\displaystyle G(x,y) = \sum_{n \geq 0} \sum_{m \geq 0} f(m,n)x^m y^n$$ $$\displaystyle = \sum_{n \geq 0} \sum_{m \geq 0} x^m y^n \sum_{\substack{k_1, \cdots, k_n \in \mathbb{Z} \\ |k_1| + \cdots + |k_n| \leq m}} 1$$ $$\displaystyle = \sum_{n \geq 0} y^n \sum_{k_1, \cdots, k_n \in \mathbb{Z}} \sum_{m \geq |k_1| + \cdots + |k_n|} x^m$$ $$\displaystyle = \sum_{n \geq 0} y^n \sum_{k_1, \cdots, k_n \in \mathbb{Z}} \frac{x^{|k_1| + \cdots + |k_n|}}{1 - x}$$ $$\displaystyle = \frac{1}{1-x}\sum_{n \geq 0} y^n \left(\sum_{k \in \mathbb{Z}} x^{|k|}\right)^n$$ $$\displaystyle = \frac{1}{1-x} \sum_{n \geq 0} y^n \left(\frac{1+x}{1-x}\right)^n$$ $$\displaystyle = \frac{1}{1-x} \frac{1}{1 - y(1+x)/(1-x)}$$ $$\displaystyle = \frac{1}{1-x-y-xy}.$$ This seemed a fascinating approach to me back then (2005 was the first time I wrote the Putnam, and Greg was our Putnam coach at UBC), and even more so today when I looked back at it given that some of my work involves some clever generating function arguments (based on the answers given to me by Richard Stanley on a question I posted here). So the question I pose is: Are there any other interesting quantities $f(n_1, \cdots, n_k)$ involving parameters $n_1, \cdots, n_k$ with $k \geq 2$ say that are symmetric in the parameters, and the proof comes from showing that the generating function $$\displaystyle \sum_{n_1, \cdots, n_k} f(n_1, \cdots, n_k)x_1^{n_1} \cdots x_k^{n_k}$$ is symmetric? REPLY [2 votes]: The following isn't an example of showing directly that the coefficients are symmetric but, establishing that the generating function is symmetric is instrumental in the calculation of an auxiliary quantity involved. I think an amazing example of this kind of proof is Stanley's proof for the number of reduced decompositions of the long permutation. Specifically because a bijective proof of this fact (such as Edelman and Greene's) is considerably harder to establish. Moreover, this proof uses what seems like a minimal amount of knowledge about reduced decompositions (again compared to the bijective proof). What follows is a very rough outline of the proof. See "On the Number of Reduced Decompositions of Elements of Coxeter Groups - Stanley" for details. Setup: Let $S_n$ be the symmetric group and $s_i:=(i,i+1)$ be an adjacent transposition. For any permutation $w$, write $w=s_{a_1}s_{a_2}\cdots s_{a_p}$, where $p$ is equal to the length of $w$ (the number of inversions in $w$). Call this sequence $(a_1,\cdots,a_p)$ a reduced decomposition of $w$, and denote by $\mbox{Red}(w)$ the set of all reduced decompositions of $w$. Let $w_0:=(n,n-1,\cdots,1)$ be the long permutation (the one with the most inversions in $S_n$, so $p=\binom{n}{2}$). For example, in $S_3$, if $w=321$, then $$\mbox{Red}(w)=\{(1,2,1),(2,1,2)\},$$ so $|\mbox{Red}(w)|=2$. Let $\lambda:=\lambda(n)$ be a tableau shape $(n-1,n-2\cdots,1)$ (a staircase tableau) and $f_\lambda$ to be the number of Standard Young tableau of shape $\lambda$. Theorem (Stanley): $|\mbox{Red}(w_0)|=f_\lambda$ Sketch of Proof: The key idea is to define the so called Stanley Symmetric function (the umbrella term for these is Gessel quasisymmetric functions). For any $w\in S_n$, again let $p$ be the length of $w$. Then define $$F_w(x_1,\cdots,x_p)=\sum_{(a_1,\cdots,a_p)\in \mbox{Red}(w)}\sum_{\substack{1\leq k_1\leq k_2\cdots\leq k_p\\ k_i TITLE: How to characterize Dirac's gamma matrices in differential geometry? QUESTION [6 upvotes]: I want to understand what is the interpretation of Dirac gamma matrices in differential geometry. Basically, I am considering the Dirac matrices as 3-indexed tensors, which means a tensor with 1 Lorentz (obeying the [1, -1, -1, -1] metric) index and 2 spinor indices (obeying the flat metric [1, 1, 1, 1]). The problem is that such a tensor seems to be ill-defined in book definitions, I was unable to identify it as a tensor product of elements of the tangent space and cotangent space of some manifold. Separating the Lorentz and spinor indices, it is clear that the Lorentz index stands for a 1-rank tensor (i.e. vector or one-form, depending on raise/lower index) of the spacetime manifold, while the spinor indices stand for a 2-rank tensor of a flat manifold (or a Clifford algebra?). I also know that the Lorentz and spinor indices are two representations of the Lorentz group, but still, how to characterize that in terms of manifolds and differential geometry? REPLY [2 votes]: When you have a vector space $V$ with a symmetric non-degenerate bi-linear form $g$ you can ask for the existence of an algebra generated by the elements of $V$ which obey the following relation. $$vu+uv=-2g(u,v)$$ Such a property for the algebra multiplication is needed in the famous paper of Dirac, you can find it here, in which he is trying to find a differential operator which is the square root of the Laplacian on the flat Lorentzian space. The relation is the result of a very easy manipulation of the symbol of the operator (which we know is of order zero) an the fact that it is the root of the Laplacian. what he founds, by manipulating the Pauli matrices, are matrices assigned only for the "orthonormal" basis $e_i$. Which is enough to create all algebra. because it shows that in 4 dimensional case this algebra can be realized as an sub-algebra of $M_4(C)$ (by dimension argument it actually is exactly $M_4(C)$. Finding this matrices not only gives the algebra but also gives an irreducible representation of the Clifford algebra. The space of $C^4$ on which $M_4(C)$ acts is the space of spinors. (Note that he was able to find the root of the Laplacian as a differential operator by paying the price of moving into higher rank space). So Dirac gamma matrices are representations for a basis of Lorentz vector space which satisfies the Clifford multiplication relation. Note that there are other basis for clifford algebra like Weyl and Majorana basis (see wikipeia page) To generalize this on general manifold, one has to do this process on cotangent space of every single point of the manifold. This is doable for any manifold CL(T*M), However, it is not always possible to have a vector bundle (which is called spinor bundle if it exists) on which the Clifford algebra acts irreducibly. It turns out that the existence of the spinor bundle has a topological obstruction. The second Whitney class should vanish and in this case the manifold is called spin manifold. Then one can go ahead and define the Dirac operator.<|endoftext|> TITLE: How many distinct homeomorphism classes of lens spaces are there with a fixed p? QUESTION [6 upvotes]: This question is about the topological classification of lens spaces. Fix $p$ a positive integer, not necessarily a prime. From Brody, The topological classification of lens spaces, Annals of Math. (2) 71 pp 163-184, the lens spaces $L(p,q_1)$ and $L(p,q_2)$ are homeomorphic if and only if either $q_1\equiv \pm q_2$ (mod $p$) or $q_1q_2\equiv \pm 1$ (mod $p$). Now the question is, for fixed $p$, how many homeomorphism classes of lens spaces are there with this fixed $p$. In particular, is there a formula $f(p)$ that gives the number of homemorphism classes as a function of $p$. Example: When $p=7$, then $q$ can be chosen from the set $\{1,2,3,4,5,6\}$ (since any such $q$ is coprime to 7) and we find that $L(7,1)\approx L(7,6)$ (where $\approx$ denotes homeomorphism) and $L(7,2)\approx L(7,3)\approx L(7,4)\approx L(7,5)$ so that there are two distinct homeomorphism classes of lens spaces with $p=7$. Thus $f(7)=2$. Example: When $p=8$, then $q$ can be chosen from the set $\{1,3,5,7\}$ (since $p$ and $q$ need to be coprime) and then $L(8,1)\approx L(8,7)$ and $L(8,3)\approx L(8,5)$ so in this case there are again two homeomorphism classes of lens spaces with $p=8$. Thus $f(8)=2$. Remark: For any $p$, the permissible $q$'s can be found and then Brody's formula can be applied to each such pair of $q$'s to determine which are homeomorphic to each other. Thus $f(p)$ can be computed by hand for any $p$. My question is, is there a formula for $f(p)$? REPLY [24 votes]: This is an interesting question which can be solved by applying Burnside's lemma (as suggested by Qiaochu Yuan in a comment) and some knowledge of the group of units modulo $n$. The relevant facts about the groups of units mostly derive from the Chinese remainder theorem, quadratic reciprocity, and the knowledge that the group of units modulo a power of an odd prime is cyclic. This is all rather elementary group theory and number theory, but I found it to be a very instructional exercise. I have written the details below, partly for my own benefit.$\newcommand{\totient}{\varphi}$$\newcommand{\congruent}{\equiv}$$\newcommand{\To}{\longrightarrow}$$\newcommand{\ZZ}{\mathbb{Z}}$$\newcommand{\ZZmod}[1]{\ZZ/ #1 \ZZ}$$\newcommand{\set}[1]{\lbrace #1 \rbrace}$$\newcommand{\units}[1]{U(#1)}$$\newcommand{\size}[1]{\mathopen{}\mathclose{\left\lvert #1 \right\rvert}}$ I will first describe the final answer. Let $n > 2$ be an integer. The number $\psi(n)$ of distinct homeomorphism classes of $3$-dimensional lens spaces whose fundamental group has order $n$ is $$ \psi(n) = \frac14 \bigl( \totient(n) + 2^{l(n) + \epsilon(n)} \bigr) $$ where $\totient$ is Euler's totient function. Moreover, $\epsilon(n)$ is $1$ if $-1$ is a quadratic residue modulo $n$, and zero otherwise. As will be shown below, $\epsilon(n) = 1$ if and only if $4$ does not divide $n$ and each of the odd prime divisors of $n$ is congruent to $1$ modulo $4$. Finally, describing $l$ requires the prime factorization of $n$: $$ n = {p_1}^{\alpha_1} \cdots {p_k}^{\alpha_k} $$ where the numbers $p_i$ are distinct primes, and the $\alpha_i$ are positive integers. Assume further that $p_1$ is the smallest prime divisor of $n$. Then $l(n)$ is given by $$ l(n) = \begin{cases} k & \text{if }\ p_1 \neq 2 \\ k + \min\set{\alpha_1 - 2 , 1} & \text{if }\ p_1 = 2 \end{cases} $$ Notation. To maintain a clear distinction between rings and groups, $\ZZmod{n}$ will denote the ring of integers modulo $n$, where $n$ is a positive integer. On the other hand, $C_n$ will denote a cyclic group with $n$ elements. The group of units in the ring $\ZZmod{n}$ is denoted by $\units{\ZZmod{n}}$. I will generally use multiplicative notation for the groups appearing in this answer. Applying Burnside's lemma. We follow here the suggestion of Qiaochu Yuan in a comment below the question. Fix an integer $n > 2$. Consider the group $G = C_2 \times C_2$, where $C_2 = \set{-1,1}$ with generator $-1$. Make $G$ act on $\units{\ZZmod{n}}$ in the following way: the generator $(-1,1)$ of the first copy of $C_2$ acts by taking a number to its negative (the additive inverse in $\ZZmod{n}$), and the generator $(1,-1)$ of the second copy of $C_2$ takes a number to its multiplicative inverse. From Brody's classification mentioned in the question, the set of homeomorphism classes of lens spaces with fundamental group of order $n$ is in bijection with the set of orbits for the action of $G$ on $\units{\ZZmod{n}}$. Let $\psi(n)$ denote the size of the quotient set $\units{\ZZmod{n}}/G$. The so-called Burnside's lemma implies that $$ 4\, \psi(n) = \size{G} \cdot \psi(n) = \sum_{g\in G} \size{\units{\ZZmod{n}}^g} \tag{1a} $$ where $\units{\ZZmod{n}}^g$ denotes the set of fixed points of the action of $g$ on $\units{\ZZmod{n}}$, and $\size{\bullet}$ returns the size of a set. The fixed point sets corresponding to the first copy of $C_2$ are quickly determined. The unit $(1,1)\in G$ fixes everything in $\units{\ZZmod{n}}$ and thus contributes $$ \size{\units{\ZZmod{n}}^{(1,1)}} = \size{\units{\ZZmod{n}}} = \totient(n) $$ The fixed points for the generator $(-1,1)\in G$ of the first copy of $C_2$ are the units $x \in \units{\ZZmod{n}}$ such that $2x = 0$ in $\ZZmod{n}$. The existence of such an invertible $x$ entails that $2 \congruent 0 \mod n$. Since $n>2$, $$ \size{\units{\ZZmod{n}}^{(-1,1)}} = 0 $$ Square roots modulo $n$. The fixed points for the remaining elements of $G$ are related to the number of square roots modulo $n$. For each $a\in\units{\ZZmod{n}}$, define the set of square roots of $a$: $$ R(n,a) = \set{ x\in\units{\ZZmod{n}} \mid x^2 = a } $$ In particular, the set $R(n,a)$ is non-empty if and only if $a$ is a quadratic residue modulo $n$. Now, the fixed point set for the elements $(1,-1)$ and $(-1,-1)$ in $G$ consist precisely of the units which square to $1$ and $-1$, respectively: $$ \begin{align*} \units{\ZZmod{n}}^{(1,-1)} &= R(n,1) \\ \units{\ZZmod{n}}^{(-1,-1)} &= R(n,-1) \end{align*} $$ Replacing everything back into (1a): $$ 4\, \psi(n) = \totient(n) + \size{R(n,1)} + \size{R(n,-1)} \tag{1b} $$ More generally, for any abelian group $A$ and an element $a\in A$, define the set: $$ R(A,a) = \set{ x \in A \mid x^2 = a } $$ Then the sets of square roots modulo $n$ are an instance of this general definition: $R(n,a) = R\bigl(\units{\ZZmod{n}},a\bigr)$. Consider the squaring homomorphism $\sigma : A\to A$ on an abelian group $A$ which sends an element $x \in A$ to its square $x^2$ — we are using multiplicative terminology here. Then $R(A,a)$ is the fibre of $\sigma$ over $a \in A$: $R(A,a) = \sigma^{-1}(a)$. In particular, if $1 \in A$ denotes the identity element of $A$, $R(A,1)$ is the kernel of the squaring homomorphism, and is therefore an abelian group. The fibres of a group homomorphism are always torsors over the kernel, so $R(A,a)$ is a torsor over the group $R(A,1)$ for any $a\in A$. Hence, $\size{R(A,a)} = \size{R(A,1)}$ whenever $R(A,a)$ is non-empty. We conclude from equation (1b): $$ 4\, \psi(n) = \totient(n) + \size{R(n,1)}(1 + \epsilon(n)) \tag{1c} $$ where $\epsilon(n) = 1$ if $R(n,-1)\neq\emptyset$ and $\epsilon(n) = 0$ if $R(n,-1)=\emptyset$. In other words, $\epsilon(n) = 1$ if and only if $-1$ is a quadratic residue modulo $n$. Square roots and the Chinese remainder theorem. As before, set the prime factorization of $n$ to $n = {p_1}^{\alpha_1} \cdots {p_k}^{\alpha_k}$, where the numbers $p_i$ are distinct primes, and the $\alpha_i$ are positive integers. According to the Chinese remainder theorem, there is an isomorphism of rings $$ \ZZmod{n} = (\ZZmod{{p_1}^{\alpha_1}}) \times \cdots \times (\ZZmod{{p_k}^{\alpha_k}}) $$ determined by the canonical projections from $\ZZmod{n}$ onto each of the factors in the target. Consequently, we deduce an isomorphism between the groups of units: $$ \units{\ZZmod{n}} = \units{\ZZmod{{p_1}^{\alpha_1}}} \times \cdots \times \units{\ZZmod{{p_k}^{\alpha_k}}} $$ Since $R\bigl(A\times B,(a,b)\bigr) = R(A,a) \times R(B,b)$, taking the set of square roots of $a\in\units{\ZZmod{n}}$ on each side of the previous expression produces: $$ R(n,a) = R({p_1}^{\alpha_1},a) \times \cdots \times R({p_k}^{\alpha_k},a) \tag{2} $$ In each term above, $a$ really stands for the projection of $a$ in the corresponding group. Determining when $-1$ is a quadratic residue. Recall that $a$ is a quadratic residue modulo $m$ if and only if $R(m,a) \neq \emptyset$. As a consequence of the isomorphism (2), $a \in \units{\ZZmod{n}}$ is a quadratic residue modulo $n$ if and only if $a$ is a quadratic residue modulo ${p_i}^{\alpha_i}$ for every $i\in\set{1,\ldots,k}$. The following claim reduces the case of a prime power modulus to the case of a prime modulus. Proposition: Let $p$ be an odd prime, and $\alpha$ a positive integer. A unit $a \in \units{\ZZmod{p^\alpha}}$ is a quadratic residue modulo $p^\alpha$ if and only $a$ is a quadratic residue modulo $p$. Proof: Consider the projection map $\pi : \ZZmod{p^\alpha} \to \ZZmod{p}$, which is a ring map. Then $\pi^{-1}\bigl(\units{\ZZmod{p}}\bigr) = \units{\ZZmod{p^\alpha}}$: an integer $m$ becomes invertible in $\ZZmod{p^\alpha}$ if and only if it is coprime to $p^\alpha$, which happens precisely when $m$ is not divisible by $p$ and thus invertible in $\ZZmod{p}$. Hence, we obtain a short exact sequence of groups: $$ 1 \To \pi^{-1}(1) \To \units{\ZZmod{p^\alpha}} \To \units{\ZZmod{p}} \To 1 \tag{ES} $$ The group $\pi^{-1}(1) = 1 + p(\ZZmod{p^\alpha})$ has order $p^{\alpha-1}$, which is odd. Therefore, the squaring homomorphism on $\pi^{-1}(1)$ is necessarily a self-bijection, and so every element in $\pi^{-1}(1)$ is a square. Chasing around the preceding short exact sequence, it follows that when $a \in \units{\ZZmod{p^\alpha}}$ is a quadratic residue modulo $p$, it is also a quadratic residue modulo $p^\alpha$.  ■ The first supplement to the law of quadratic reciprocity now implies the following result. Corollary: If $p$ is an odd prime, and $\alpha$ is a positive integer, then $-1$ is a quadratic residue modulo $p^\alpha$ if and only $p \congruent 1 \mod 4$. It remains to deal with the case of powers of the prime two. Observe that $-1$ is a quadratic residue modulo $2$. However, $-1$ is not a quadratic residue modulo $4$, and thus cannot be a quadratic residue modulo $2^\alpha$, for any $\alpha > 1$. The following proposition collects what we have learned in this section. Proposition: Given an integer $n > 1$, $-1$ is a quadratic residue modulo $n$ — i.e. $\epsilon(n) = 1$ — if and only if: each odd prime divisor of $n$ is congruent with $1$ modulo $4$, and $4$ does not divide $n$. Counting square roots of unity. To finish the proof, we need to find the number of square roots of the unit in the ring $\ZZmod{n}$, that is, the size of $R(n,1)$. By (2), this is reduced to calculating the size of $R({p_i}^{\alpha_i},1)$ for each $i$. We will use the fact that $R(C_m,1) \cong C_1$ (where $1 \in C_m$ is the identity element) is trivial if $m$ is odd, and $R(C_m,1) \cong C_2$ if $m$ is even. Claim: For $p$ an odd prime, and $\alpha$ a positive integer, $R(p^\alpha,1) \cong C_2$. Proof: This follows from the fact that $\units{\ZZmod{p^\alpha}}$ is a cyclic group of even order equal to $p^{\alpha-1}(p-1)$ (stated on the relevant wikipedia page). Alternatively, it may be deduced from two results stated in the proof of the first proposition from the previous section: the short exact sequence (ES), and the observation that the squaring homomorphism is a bijection on the kernel of that exact sequence.  ■ Claim: For $\alpha > 2$, $R(2^\alpha,1) \cong C_2 \times C_2$. Moreover, $R(4,1) \cong C_2$ and $R(2,1) \cong C_1$. In particular, $\size{R(2^\alpha,1)} = 2^{\min\set{\alpha-1,2}}$ for each positive integer $\alpha$. Proof: This is a consequence of the following facts (see the wikipedia page): $\units{\ZZmod{2}} \cong C_1$, so that $R(2,1) \cong C_1$; $\units{\ZZmod{4}} \cong C_2$, so that $R(4,1) \cong C_2$ for $\alpha > 2$, $\units{\ZZmod{2^\alpha}} \cong C_2 \times C_{2^{\alpha-2}}$, so that $R(2^\alpha,1) \cong C_2 \times C_2$. For reference, section 4.5 of Helmut Hasse's book Number theory proves the results, identifying the groups $\units{\ZZmod{p^\alpha}}$, which were used in this proof and in the previous one.  ■ Using the isomorphism (2), these two claims prove that $R(n,1)$ is the product of a certain number of cyclic groups of order $2$. We calculate: $$ \size{R(n,1)} = \prod_{i=1}^k \size{R({p_i}^{\alpha_i},a)} = 2^{l(n)} $$ Here, $l(n) = l({p_1}^{\alpha_1} \cdots {p_k}^{\alpha_k})$ is as defined at the beginning: $$ l(n) = \begin{cases} k & \text{if }\ p_1 \neq 2 \\ k + \min\set{\alpha_1 - 2 , 1} & \text{if }\ p_1 = 2 \end{cases} $$ where we assume that $p_1$ is the smallest prime number which divides $n$. Finally, equation (1c) implies $$ 4\, \psi(n) = \totient(n) + 2^{l(n)}(1+\epsilon(n)) = \totient(n) + 2^{l(n) + \epsilon(n)} $$ since $\epsilon(n)$ only takes the values $0$ and $1$.<|endoftext|> TITLE: Why does the gamma function use the symbol $\Gamma(\,)$? QUESTION [6 upvotes]: I am aware of some of the history of the gamma function $\Gamma(z)$, partly through a 2009(!) MO question "Who invented the gamma function?"—Euler, Bernoulli, etc. My question does not seem to be answered in that discussion, or in other historical accountings I can easily locate: Q. Why was the symbol $\Gamma$ chosen for the generalized factorial? Were $\alpha(z)$ and $\beta(z)$ already "taken" and so $\Gamma$ was a natural successor? Or was the choice due to the shape of the uppercase $\Gamma$? Or some other reason? Or lost to history? REPLY [7 votes]: The first use of $\Gamma$ in this sense is due to Legendre (1811). It is unknown why he choose that notation but some speculations are recorded at http://jeff560.tripod.com/functions.html They range from the Gamma being an inverted L (from Legendre) and variants, to the in my opinion more integresting that the logarithm of $\Gamma$ is related to the Euler-Mascheroni constant, which is mentioned in that work of Legendre and was (then) denoted $C$ (which makes $\Gamma$ as a Greek capital C natural) or also $\gamma$ (that usage was present before the work of Legendre too).<|endoftext|> TITLE: Steenrod operations in algebraic geometry QUESTION [6 upvotes]: What are some applications of Steenrod operations (or similar constructions) in algebraic geometry? I am dimly aware of the the use of these Voevodsky's work on motivic cohomology, and would be happy if someone explained that in a comprehensible way. However, the primary motivation for asking the question is to see if such operations on the coherent (or de Rham) cohomology on algebraic varieties (in positive/mixed characteristic) have concrete geometric consequences. REPLY [5 votes]: Not sure if this counts, but a much older application of $Sq^1$ (appearing as a Bockstein) and some of its cousins appears in Serre's paper on Witt vector cohomology. He constructed a family of operations, in characteristic $p$, $\beta_r: H^1(\mathcal{O}_X) \rightarrow H^2(\mathcal{O}_X)$ called the higher Bocksteins and it's reasonable to think of the first of these as a $Sq^1$. A concrete application was given by Mumford who showed that a surface over a field of positive characteristic has smooth Picard scheme if and only if all of the Bocksteins on $H^1(\mathcal{O}_X)$ vanish. He later used this fact (amidst many others) in his and Bombieri's classification of algebraic surfaces in positive characteristic.<|endoftext|> TITLE: Are deformations of a scheme some kind of a "derived gerbe" under the cotangent complex? QUESTION [15 upvotes]: (This is probably a very naive question. My understanding of the cotangent complex is quite vague.) Let me first recall the picture for deformations of a smooth morphism: If $f:X_0\to S_0$ is a smooth morphism of schemes and $S_0\to S$ is a closed immersion defined by a square zero ideal $I$, the stack (in the Zariski topology) over $X_0$ associating to an open $U_0\subseteq X_0$ the category of smooth lifts $U$ of $U_0$ over $S$ is naturally a gerbe under the sheaf $$\mathcal{G}_{X_0/S}:=\mathcal{H}om (\Omega^1_{X_0/S_0}, f^* I).$$ This just means that deformations locally exist and are locally unique, and that the sheaf of automorphisms of a local deformation is canonically isomorphic to $\mathcal{G}_{X_0/S}|_{U_0}$. It is a formal consequence of this that (a) there is an obstruction class $o(X_0/S) \in H^2(X_0, \mathcal{G}_{X_0/S})$ to deforming $X_0$ over $S$, (b) if it vanishes then such deformations are a torsor under $H^1(X_0, \mathcal{G}_{X_0/S})$, and (c) the group of automorphisms of a deformation is $H^0(X_0, \mathcal{G}_{X_0/S})$. The above three conclusions hold generally for any morphism $f$ if we set $\mathcal{G}_{X_0/S} = R\mathcal{H}om (L_{X_0/S_0}, f^* I)$ where $L_{X_0/S_0}$ is the cotangent complex. Therefore it seems natural to ask whether, as in the smooth case, local deformations form a some sort of "gerbe" under $\mathcal{G}_{X_0/S}$. Of course the problem is that $\mathcal{G}_{X_0/S}$ is not a sheaf of groups but an object of the derived category. So it would have to be some "derived gerbe". Question: Is there a notion of a "derived gerbe" (under an object $\mathcal{G}$ of the derived category of sheaves of abelian groups), such that local deformations of $X_0/S$ as above form a "derived gerbe" under $R\mathcal{H}om (L_{X_0/S_0}, f^* I)$, and such that the analogues (a)-(c) above hold (for any $\mathcal{G}$)? REPLY [4 votes]: A related question is answered in https://arxiv.org/abs/1101.4069. edit: My paper https://arxiv.org/abs/1712.01384 takes a very similar approach. It deforms modules instead of algebras, so it's simpler but less relevant. I see no reason to think of this as a gerbe but as a torsor for a group stack. I believe this perspective is due to Grothendieck's book "Categories cofibrees additives et complexe contagent relatif" and a solid modern treatment through a Gromov-Witten lens is established in https://arxiv.org/abs/1111.4200<|endoftext|> TITLE: To whom is the internal characterization of $Q$-groups due? QUESTION [5 upvotes]: A group is said to be a $Q$-group if the character of any complex representation is rational valued. A well-known internal characterization of $Q$-groups is the following: $G$ is a $Q$-group if and only if elements which generate the same cyclic subgroup of $G$ lie in the same conjugacy class. See for example: Corollary 2 to theorem 29 in Serre's Linear Representations of Finite Groups as suggested in https://mathoverflow.net/a/10652/9672, or Kletzing's Structure and Representations of $Q$-groups. I would be grateful if someone could tell me where this theorem first appears (to whom is it due). REPLY [9 votes]: It's hard to pin down who first made this remark because it is a corollary of the very basic fact that if $\sigma$ is an automorphism of the cyclotomic field $Q_{|G|}$, then there exists an integer $m$ coprime to $|G|$ such that for every character $\chi$ of $G$, we have $\chi(x)^\sigma = \chi(x^m)$. The earliest explicit reference I can find is in Huppert's book Endliche Gruppen I in 1967. It appears there as Satz V 13.7 (b), on page 537.<|endoftext|> TITLE: Relation between Milnor ring and middle dimensional homology of hypersurface QUESTION [8 upvotes]: I have suspected that the following is well-known: If $P$ is a homogeneous polynomial of degree $d$ in $n$ variables (for example, Fermat quintic $x_1^5 + \cdots + x_5^5$). The Milnor ring is ${\mathbb C}[x_1, \ldots, x_n]$ modulo the Jacobian ideal of $P$. On the other hand, $P$ defines a hypersurface $S$ in ${\mathbb P}^{n-1}$. The middle dimensional cohomology of $S$ (or maybe the primitive part) should correspond to the ${\mathbb Z}_d$-invariant subspace of the Milnor ring. For example, for the Fermat quintic, the Milnor ring has dimension 1024 and the $h_3$ of the Calabi-Yau 3-fold defined by $P$ is 204 which is equal to the dimension of the ${\mathbb Z}_5$-invariant subspace of the Milnor ring. But I don't know if there is any elementary literature which discusses this correspondence. And I would like to see some geometric treatment using vanishing cycles/Lefschetz pencils/thimbles to see this correspondence. For example, for Fermat quintic, there are 1024 Lefschetz thimbles if we perturb $P$ by adding a small linear term; but why only the ${\mathbb Z}_5$-invariant part survives at "infinity"? REPLY [5 votes]: This is implied by Griffiths' celebrated presentation of the Hodge filtration of a hypersurface in terms of the Jacobian ring. He gives an isomorphism of the primitive part of $H^{p,q}$ of a smooth hypersurface of degree $d$ in $\mathbb{P}^{n+1}$ with the degree $(p+1)d-n-2$ part of the Jacobian ring. You can find an excellent presentation in Voisin's book Hodge Theory and Complex Algebraic Geometry.<|endoftext|> TITLE: Original manuscript of Archimedes' cattle problem QUESTION [16 upvotes]: Wikipedia states that [Archimedes' cattle problem] was discovered by Gotthold Ephraim Lessing in a Greek manuscript containing a poem of forty-four lines, in the Herzog August Library in Wolfenbüttel, Germany in 1773. I have accepted that statement as fact for a long time, but I have never seen the Greek original of that poem, nor a scan/photograph/transcription of it, or even an inventory reference for it in the Herzog August Library, anywhere, nor can I find it on the web. This makes me wonder whether that "history" is even true at all, or whether Lessing or someone else might have made it up in order to make an 18th century problem look more interesting. I'd appreciate any reference to the Greek original, or more background showing that (as I now suspect) there is no (ancient) Greek original after all. REPLY [11 votes]: Here is a link to the full document at the Hochschule für angewandte Wissenschaften in Augsburg: http://www.hs-augsburg.de/~harsch/graeca/Chronologia/S_ante03/Archimedes/arc_bous.html REPLY [10 votes]: Although I cannot quell your doubts, at least the problem statement/poem in Greek can be found here: ARCHIMEDIS OPERA OMNIA CUM COMMENTARIIS EUTOCII B. G. Teubner, Leibzig, Volume 2 (1881), Pages 448 & 450. (Webpage link) Here is the beginning:       The full poem, more scholarly references, translations, etc., may be found at this link, maintained by Chris Rorres.<|endoftext|> TITLE: Class Forcing and Genericity: Predense sets vs Dense classes QUESTION [11 upvotes]: In short my question is: why do we use definable classes in the definition of genericity for class forcing, instead of predense sets. To elaborate, in Sy's book and indeed other sources on the subject of class forcing, $G$ is generic if it meets every definable, dense class of $M$, the ground model; a natural extension of the usual situation. Of course classes can get a bit tricky so one would naturally wonder whether one can use predense sets in their place: intuitively the answer should be "no," since otherwise we'd probably go ahead and do so. Indeed, Stanley outright says that this "definable genericity" is stronger than "internal genericity" [1]. Thus my question is really twofold: What is the problem with the standard argument that genericity defined for predense sets is the same: Let $\mathbb P$ be a definable class forcing. Let $D$ be a predense set of $M$, let $\tilde D$ be the class of extensions of $D$, which is definable ($\tilde D=\{p\in\mathbb P\mid \exists q\in D(p\leq q)\}$) and dense (as usual: if $p\in\mathbb P$ then there is a $q\in D$ compatible with $p$, hence an $r \in \tilde D$ extending $q$.) So by definable genericity, there is $r\in G\cap\tilde D$, which must come from extending a $q\in D$, which is also in $G$ since it's a filter. I assume I'm doing something naughty with those proper classes, but I don't see where. So, with a little more thought it's obvious that it's the other direction that's the issue, i.e., if $G$ is internally generic, when is it definably generic. Thus I would like to know about this with respect to the pretameness condition that Sy defines: $\mathbb P$ is pretame iff given a $p$, any $M$-definable sequence of dense classes can be refined to a sequence (in $M$) of predense sets below some $q\leq p$. so: Does class forcing work out OK if we use predense sets? Can the extension satisfy ZFC(-) (in the language with a predicate for the generic) If not in general, are there useful conditions which make it work? What are some examples of cases where the difference matters? I'm actually most interested in the situation of Prikry forcing over a model with an $M$-ultrafilter. [1] Stanley, M.C., 2003, Outer Models and Genericity. JSL. REPLY [12 votes]: Let me point out that meeting all pre-dense sets is not generally the same as meeting all dense classes. Consider the forcing $\mathbb{P}$ that adds a generic function from $\text{Ord}$ to $V$. (One can use this forcing to force global choice — see Victoria Gitman's account.) That is, conditions are functions $p:\alpha\to V$ for some ordinal $\alpha$, and the order is extension of functions. If a filter $G\subset\mathbb{P}$ meets all dense classes, then it is easy to see that $\cup G$ is a surjective function total function from $\text{Ord}$ to $V$: for any set $x$, it is dense that the conditions have $x$ in their range, and for any ordinal $\alpha$, it is dense that the conditions are defined at $\alpha$. But that argument breaks down completely if one tries to use only predense sets: the reason is that the forcing $\mathbb{P}$ has no pre-dense sets at all! (except for sets that contain the empty function) For any set $B$ of nontrivial conditions, there is a condition $p$ that is incompatible with every element of $B$. So it turns out that every filter vacuously meets all pre-dense sets, and the forcing technology would not be doing what we want. You might reply that this example is because we are using the partial order, rather than the Boolean algebra. But that reply has problems, because in a case like this, there is no way to formalize the Boolean completion without going to meta-classes, as the antichains are generally proper classes here. So we cannot so easily go to the Boolean completion of the forcing. One can transform $\mathbb{P}$ into forcing that also adds sets, for example, by using the generic filter to determine what happens next, for example, to code $G$ into the GCH pattern.<|endoftext|> TITLE: Uncountable atomless subalgebras of the Boolean algebra of all Jordan measurable sets in [0,1] QUESTION [5 upvotes]: Definition: Suppose $\mathcal A$ is the Boolean algebra of all Jordan measurable sets in $I=[0,1]$ (i.e $\mathcal A=\{A\subseteq I: \mu(\partial(A))=0\}$, where $\mu$ is the Lebesgue measure and $\partial$ means the topological boundary). Definition: "Interval algebra" suppose $\mathcal B$ on $I=[0,1]$ is the algebra generated by the sets of the form $([a_{0},b_{0})\cup[a_{1},b_{1})\cup...\cup[a_{n},b_{n}))\cap I$ where $0\leq a_{0}< b_{0} TITLE: Is the representation category of quantum groups at root of unity visibly unitary? QUESTION [18 upvotes]: Let $\mathfrak g$ be a simple Lie algebra. By taking the specialization at $q^\ell=1$ of a certain integral version¹ of the quantum group $U_q(\mathfrak g)$, and by considering a certain quotient category² of the category of tilting modules³ over that Hopf algebra, one obtains a fusion category. Moreover, by using the universal $R$-matrix and the so-called charmed element, one can endow that fusion category with the structure of a modular tensor category. Can this approach be used to further equip this category with the structure of a unitary modular tensor category? In other words, is it possible to equip the objects of this category with some kind of extra structure (something like an inner product) such that the adjoint $f^*:W\to V$ of a morphism $f:V\to W$ makes sense, and such that the braiding and twist are unitary? ¹: That integral version is usually denoted $U_q^{res}(\mathfrak g)$. It is generated by the usual elements $K_i$, $E_i$, $F_i$, along with the divided powers $E_i^{[r]}:=\frac{E_i^r}{[r]!}$ and $F_i^{[r]}:=\frac{F_i^r}{[r]!}$. ²: This quotient category has the same objects as the original category. The hom-spaces are modded out by the subspace of negligible morphisms, where a morphisms $f:V\to W$ is called negligible if $tr_q(fg)=0$ for any $g:W\to V$. ³: A module is tilting if it admits a filtration whose associated graded pieces are Weyl modules, and also admits a filtration whose associated graded pieces are duals of Weyl modules. Here, the Weyl modules, are the ones that "look like" irreps of $\mathfrak g$. REPLY [12 votes]: The answer is "Sometimes." Following the notation of Rowell's "From Quantum Groups to Unitary Modular Tensor Categories", let $\mathcal C(\mathfrak g, l, q)$ be the category corresponding to $U_q(\mathfrak g)$ such that $q=e^{\pi \imath/l}$.Denote by $m$ is the ratio of the square lengths of a long root to a short root. As mentioned there, this only gets one as far as being a ribbon fusion category (also called a pre-modular category). The questions of unitarity and modularity for $\mathcal C(\mathfrak g, l,q)$ are addressed in section 4. Specifically $\mathcal C(\mathfrak g, l,m)$ is unitary if $m\vert l$. This is attributed to both Wenzl and Xu. Modularity in the case that $m\vert l$ is attributed to Kirillov (among others). Numerous cases where $m\not\vert l$ are treated in Rowell. As a related but different matter, assume one has a modular tensor category $\mathcal C$ and wants to know if it can be made unitary. Then by theorems 3.2 (Every braiding of a unitary fusion category is unitary) and 3.5 (Every braiding of a unitary fusion category admits a unique unitary ribbon structure) of Galindo's "On Braided and Ribbon Unitary Fusion Categories", it suffices to determine whether or not $\mathcal C$ admits the structure of a unitary fusion category.<|endoftext|> TITLE: Is this obfuscation scheme unbreakable? QUESTION [8 upvotes]: I've just come across this popular article about a breakthrough (which can be purchased here), published in Foundations of Computer Science (FOCS), 2013 IEEE 54th Annual Symposium by a team of cryptographers (Sanjam Garg, Craig Gentry, Shai Halevi, Mariana Raykova, Amit Sahai and Brent Waters) showing that the workings of a computer programme can be obfuscated. The author writes : This obfuscation scheme is unbreakable, the team showed, provided that a certain newfangled problem about lattices is as hard to solve as the team thinks it is. Would someone care to share with us the problem about lattices that is alluded to here ? REPLY [9 votes]: "Is this obfuscation scheme unbreakable?" "Well.. no." said people a couple of years later. On GGHRSW13 specifically: Cryptanalyses of Candidate Branching Program Obfuscators See also (concurrent, similar flavor): Cryptanalysis of Indistinguishability Obfuscations of Circuits over GGH13 Zeroizing Attacks on Indistinguishability Obfuscation over CLT13 Some interesting candidates for secure obfuscation do remain. See, for instance: a) Secure Obfuscation in a Weak Multilinear Map Model b) Preventing CLT Zeroizing Attacks on Obfuscation<|endoftext|> TITLE: Are Fourier series of length 2 'asymmetric enough' to generate all crossing patterns? - A reformulation of the Fourier-(1,1,2) knot question QUESTION [7 upvotes]: Given $N$ pairs of distinct real numbers $t_i, t'_i \in [0,1]$, $i = 1,\ldots,N$, we ask if there is a function $f(x) = \cos(2\pi mx+\alpha) + \gamma\cdot \cos(2\pi nx+\beta)$, with $m, n \in \mathbb{N}$, $\alpha, \beta, \gamma \in \mathbb{R}$, so that for all $i$: $f(t_i) > f(t'_i)$? If yes, this proves that all knots are Fourier-(1,1,2) knots, that is, possess a parametrization with Fourier series of length 1,1,2 in coordinates $x,y,z$. Because: a) the parametrization in the x-y plane is rich enough to generate all knots (see Lamm, 1998) and b) by interchanging $t_i$ and $t'_i$ every crossing pattern can be achieved. As references see the articles http://arxiv.org/abs/q-alg/9711013 (Kauffman, 'Fourier knots', 1997) http://arxiv.org/abs/1210.4543 (Lamm, 'Fourier knots', 1998) http://arxiv.org/abs/0707.4210 (Boocher et. al, 'Sampling Lissajous and Fourier knots', 2007) http://arxiv.org/abs/0708.3590 (Hoste, 'Torus knots are Fourier-(1,1,2) knots', 2007) If the answer is no, we ask more generally if functions with a bounded number of cosine terms suffice (e.g. with bound 3). We remark that Fourier-(1,1,1) knots are Lissajous knots. These are too symmetric to yield all knots (see also Tying knots with reflecting lightrays). Edit1 (12Feb14). We give an example with 4 pairs of distinct numbers for which a single cosine function $\cos(2\pi mx+\alpha)$ does not suffice: Choose $t_1, t'_1$ and $t_2, t'_2$ arbitrarily (but distinct) in $[0,1]$. Let $t_3 = t_1 + 0.5$, $t'_3 = t'_1 + 0.5$ and $t_4 = t'_2 + 0.5$, $t'_4 = t_2 + 0.5$ (note the interchanged ' in $t_2$). We then have: if $f(t_1) > f(t'_1)$ then for odd $m$: $f(t_3) < f(t'_3)$ and for even $m$: $f(t_4) < f(t'_4)$. This shows how the symmetries $\cos(x+2\pi) = \cos(x)$ and $\cos(x+\pi) = -\cos(x)$ prevent a solution in a Fourier series of length 1. The same argument applies for a series of length two (or more) if the frequencies are all odd or all even. Edit2 (09July15). The article http://arxiv.org/abs/1507.00880 (Marc Soret, Marina Ville) solves the Fourier-(1,1,2) knot problem using Kronecker's theorem! REPLY [3 votes]: If all $t_i$, $t'_i$ are rational numbers with common denominator $d$, then it is enough to examine frequencies $m, n \le d$. After David's answer I decided that it could be worthwhile to look for counterexamples for small $d$ and used an optimization routine to minimize the sum over $\max(0,f(t'_i)-f(t_i))$ for each combination of $m, n$. Counterexamples seem to exist for 10 or more pairs. I found the following for $d=20$ and $d=23$: [3,13,2,12,16,18,8,9,10,19;1,17,14,7,4,15,20,1,6,5]/20 [7,9,5,4,14,16,10,20,6,15;17,19,8,12,3,1,18,2,11,13]/20 [9,8,12,19,1,16,18,10,5,17;7,13,4,20,6,21,2,14,11,15]/23 This should be read as $t_1=3/20$, $t'_1=1/20, \ldots$ for the first example. Open questions: a) Show that the above examples are indeed counterexamples for Fourier series of length 2. b) Can you find counterexamples with less than 10 pairs? c) Study the question for Fourier series of length 3. I would like to add that in a numerical study a small tolerance should be introduced: e.g. sum over $\max(0,f(t'_i)-f(t_i)+10^{-6})$ to avoid functions which are zero at $t_i$, $t'_i$. These occur because of the identity $\cos(x)-\cos(y)=-2 \sin(\frac{x+y}{2})\sin(\frac{x-y}{2})$ for frequencies with $m+n=d$ and $\gamma=-1$. Update (6Mar14): I found potential counterexamples with 8, 7 and 6 pairs: [10,16,11,3,9,20,15,6;12,4,19,5,1,2,13,8]/20, [12,5,10,14,4,11,7;8,1,3,6,2,9,13]/14, [10,11,3,2,1,7;8,9,4,5,6,12]/12.<|endoftext|> TITLE: A palindromic polynomial and its derivative have the same number of zeros outside the unit circle. Reference? QUESTION [5 upvotes]: I am trying to find the original reference for a lemma attributed to Cohn (as in Schur-Cohn method): Let $A(z)$ be a palindromic or skew-palindromic polynomial, and denote its derivative by $A'(z)$. Then $A(z)$ and $A'(z)$ have the same number of zeros outside the unit circle. The lemma can be found at http://www2.ece.ohio-state.edu/~randy/publications/RLM_journal/J11.pdf, statement on p. 105, proof on p. 116, but I don't see a reference there. I presume that if I could find the historical papers on the Schur-Cohn method it would not be hard to find, but I am not having much luck with that so far. (BTW, I am including the signal-analysis tag because this result seems to appear in the signal processing literature). REPLY [6 votes]: MR0058748 (15,419a) Ancochea, Germán, Zeros of self-inversive polynomials, Proc. Amer. Math. Soc. 4, (1953) 900–902. 41.1X This is a simple and elegant proof of a theorem of A. Cohn [Math. Z. 14, 110–148 (1922)]. A polynomial $g(z)$ with its zeros symmetric with respect to the unit circle $C$ has the same number of zeros outside $C$ as its derivative $g′(z)$. The proof uses only Rouché's theorem and the continuity of the zeros of a polynomial as functions of its coefficients. Reviewed by F. F. Bonsall The Cohn reference is given in full as A. Cohn, Über die Anzahl der Wurzeln einer algebraischen Gleichung in einer Kreise, Math. Zeit. vol. 14 (1922) pp. 110-148. MR1544543 From another review, it seems there is also a proof in Marden's book, Geometry of polynomials, second edition, pp. 198–206.<|endoftext|> TITLE: Genus of a graph QUESTION [6 upvotes]: Let $G$ is a simple undirected graph. Suppose $G$ has two subgraphs $G_1$ and $G_2$, such that $E(G_1)\cap E(G_2) =\emptyset$ ($E(G_i)$, stand for the set of edges of $G_i$). Then is it true that genus of $G$ is greater than or equal to the sum of genera of $G_1$ and $G_2$? REPLY [14 votes]: No. The two subgraphs can share the surface more efficiently than that. Take a graph $G$ with genus $g\ge 1$ and duplicate each edge. If you don't like double edges, subdivide them with new vertices. Then you can divide the new graph into two edge-disjoint subgraphs homeomorphic to $G$, therefore each having genus $g$, yet you can still draw the whole graph on the same surface.<|endoftext|> TITLE: Status of local Langlands conjecture over positive characteristic QUESTION [7 upvotes]: I am interested to know what the status of Local Langlands Conjectures in positive characteristic is? By a positive characteristic local field, I mean a field of the form $\mathbb{F}_q((t))$. A nice summary of what is known for characteristic zero local fields is given in the first paragraph of http://arxiv.org/pdf/1204.0132v2.pdf I'm looking for a similar description in the positive characteristic case. REPLY [8 votes]: I am not an expert, but what I think I know on the subject is: For $GL_n$, the local Langlands conjecture has long been known by work of Laumon, Rapoport and Stuhler (Inventiones Math, 1993). Of course, in this case, even the global correspondence is now known, due to the Fields-Medal-winning work of Laurent Lafforgue. For general reductive groups, the local Langlands correspondence is not known at this date but there is movement right now. Vincent Lafforgue (Laurent's younger brother) has recently released a paper proving the direction "automorphic --> Galois" of the global correspondence. In this paper, he announces a work in preparation of himself and Genestier aiming at establishing the local Langlands correspondence for reductive groups. So when this paper is released, the answer to your question may well be "solved!".<|endoftext|> TITLE: Complete (possibly official) list of "What is..." articles from the Notices of the AMS QUESTION [17 upvotes]: Does it exist online and where can one find it? (For example, these two sources are not official; is the longer one complete?) REPLY [3 votes]: It seems that this is an official (and complete) list https://www.ams.org/cgi-bin/notices/amsnotices.pl?article_id=whatis&article_type=gallery&gallery_type=whatis<|endoftext|> TITLE: Why is proving $C^{\infty}$ regularity of sub Riemannian geodesics so hard? QUESTION [9 upvotes]: In Montgomery's A Tour of Subriemannian Geometries, Their Geodesics and Applications, problem 10.1 in Chapter 10 asks "Is every minimizing geodesic smooth ?". Can someone explain what are the major difficulties faced in proving the same ? Suggestions for suitable references are also welcome. Thanks. REPLY [2 votes]: According to the MathSciNet, the first general contribution to the solution of the problem is given in the following paper: E. Hakavuori, E. Le Donne, Non-minimality of corners in subriemannian geometry. Invent. Math. 206 (2016), no. 3, 693–704. (MathSciNet review). The paper mentioned by Richard Montgomery is: E. Le Donne, G. P. Leonardi, R. Monti, D. Vittone, Extremal curves in nilpotent Lie groups. Geom. Funct. Anal. 23 (2013), no. 4, 1371–1401. (MathSciNet review).<|endoftext|> TITLE: Monograph or rich survey on infinite-dimensional Riemann manifolds QUESTION [10 upvotes]: I'm working with the space of smooth curves $\mathcal{C}$ in a smooth manifold $M$, having (different, pre-determined) fixed endpoints. I'd like to endow it with a Riemann structure (I already have a few options in mind). (Since $M$ is not necessarily simply-connected, I guess that no nice topology could make $\mathcal{C}$ connected.) However, I know nothing about infinite-dimensional manifolds, hence my question: could anybody please point me to some monograph or rich survey article on the subject? I've read a brief presentation by Andrew Stacey from which I've understood that infinite-dimensional Riemann manifolds can be strong or weak, the weak ones can in particular be weak co-Riemannian, and that there might also be problems regarding the local-triviality of the tangent bundle. I guess that the results that I'm looking for are the ones that best apply to spaces of smooth curves with fixed endpoints (which rules out the theory of loop spaces). (One of the reasons I'm trying to do this is in order for me to compactify this manifold and still get "something with a distance". Maybe having this end in mind could narrow down the focus area of your answers.) REPLY [6 votes]: Firstly, I would suggest taking a look at the work of Peter Michor, who is actually a frequent contributor to MO. Regarding Riemannian metrics on spaces of curves, you might be interested in this nice recent paper. Also, for more on infinite-dimensional Riemannian geometry including some metric aspects, take a look at the work of Brian Clarke, Boris Khesin, Jonatan Lenells, Gerard Misiolek, Stephen Preston and collaborators, such as this and this paper.<|endoftext|> TITLE: A combinatorial problem concerned with logic circuits QUESTION [8 upvotes]: Consider a logic circuit with two-bit gates only. The length of each gate is the number of bit lines that the gate crosses. How hard is to compute the maximum length for a given circuit? Notice that two circuits are, say, isomorphic, if they differ only up to a permutation of the bit lines. Notice that there are clearly $n!$ ways to arrange $n$ bit lines. REPLY [10 votes]: If I understand your question correctly, you're trying to find a permutation of the bit lines so the maximum gate "length" is as small as possible. This is called the bandwidth problem: Given a graph $G = (V,E)$ find a permutation $\pi : V \rightarrow [1 \ldots n]$ such that $$\max_{(i,j) \in E} |\pi(i) - \pi(j)|$$ is minimized. The wikipedia article has more information on the complexity (it's NP-hard and APX-hard) and algorithms. The bandwidth problem is interesting also because it was used to introduce the idea of volume-respecting embeddings.<|endoftext|> TITLE: Exponential of a specific hypergeometric series QUESTION [5 upvotes]: This is motivated by this question. Let $f$ be the hypergeometric series $ f(x) = 2 x \, _{4}F_3([1, 1, 4/3, 5/3], [2, 2, 2], 27 x) $ which is explictly given by $ f(x) = \sum_{n \geq 1} \frac{(3n-1)!}{n!^3} x^n $ and appears in some studies of Mahler measures. Is is true that $\exp(f)$ has only positive integer coefficients ? The first few coefficients are $1, 2, 17, 218, 3404, 59644$. If true, is there a combinatorial meaning for these coefficients ? In the same vein, what can one say on the coefficients of $\exp(-f)$ which start $1, -2, -13, -158, -2431, -42454, -802790$ ? REPLY [2 votes]: Here's another approach. It resembles Ira's solutions, although less combinatorial. Maybe it will be of use. Let $a_n := \frac{1}{3} \binom{3n}{n,n,n}$. ($a_1=1$ and $a_n \in \mathbb{Z}$, as $a_n = \binom{3n-1}{n-1,n,n}$.) Note that $f(x) = \sum_{n \ge 1} \frac{a_n}{n} x^n$. Hence, the function we're dealing with is $\zeta_{\{a_n\}}(x) := \exp( \sum_{n \ge 1} \frac{a_n}{n} x^n)$, which we'll refer to as "the zeta function of $\{a_n\}$". Lemma 1: Consider a sequence $\{ a_n \}_{n\ge1}$ of rational numbers, with $a_1=1$. $\zeta_{\{a_n\}}(x) \in \mathbb{Z}[[x]]$ iff $\forall n \in \mathbb{N}: n\mid \sum_{d\mid n} \mu(n/d) a_d$. This lemma reduces our problem to proving a congruence: $\forall n \in \mathbb{N}: 3n\mid \sum_{d\mid n} \mu(n/d) \binom{3d}{d,d,d}$. Lemma 2: The condition $\forall n \in \mathbb{N}: n\mid \sum_{d\mid n} \mu(n/d) a_d$ is equivalent to $a_{np^k} \equiv a_{np^{k-1}} \mod {p^k}$ for any prime $p$ and $n,k \in \mathbb{N}$. So we need to prove the congruence $\binom{3np}{np,np,np} \equiv \binom{3n}{n,n,n} \mod {3np}\mathbb{Z}_{p}$ (note how I turned to $p$-adics, as it simplifies computations and presentation). This congruence is a simple consequence of either: A combinatorial argument, reminiscent of the combinatorial proof of (a weak version of) Wolstenholme's theorem. Properties of the $p$-adic Gamma function. The congruence is equivalent to $\frac{\Gamma_{p}(3np)}{\Gamma_{p}^3(np)} \equiv 1 \mod {3np \mathbb{Z}_{p}}$. Kazandzidis congruences give something much stronger (although a small modification is necessary for the even prime and the second prime): $$\frac{\Gamma_{p}(3np)}{\Gamma_{p}^3(np)} \equiv 1 \mod {n^3p^3 \mathbb{Z}_{p}}$$ Proofs of both lemmas are not hard. The first lemma follows by writing $\zeta_{a_n}(x)$ as $\prod_{n\ge1} (1-x^n)^{-\frac{b_n}{n}}$, where taking the logarithmic derivative shows $b_n = \sum_{d\mid n} \mu(n/d) a_d$. This gives a proof of one direction. For the other direction, notice that $$[x^n]\prod_{i=1}^{\infty} (1-x^i)^{-\frac{b_i}{i}} = [x^n]\prod_{i=1}^{n-1} (1-x^i)^{-\frac{b_i}{i}} + b_n$$ and use an induction argument. The second lemma is just a "coupling" argument (note how $\mu(ap) = -\mu(a)$ for $p\nmid a$).<|endoftext|> TITLE: Connections on a Lie Group QUESTION [5 upvotes]: A Lie group $G$ can be considered as a reductive homogeneous space in at least two different ways; $G/\{e\}$ and $G\times G/G^*$. In the first case, the canonical connection associated with the reductive decomposition has zero curvature and non-zero torsion (if $G$ isn't abelian). This connection coincides with the Cartan (-)-connection. In the second case, the canonical connection has zero torsion and generally nonzero curvature. Can we express a Lie group as a reductive homogeneous space for which the canonical connection has both nonzero curvature and torsion? We could take as our connection a combination of the two connections described above. However, how do we know that this will be a canonical connection with respect to some reductive decomposition? REPLY [2 votes]: (this is also related to The (-)-Connection on a Lie Group, Metric Connections on a Lie Group) If we want to speak about bi-invariant metric connections,then we cannot drop the assumption of compactness. Actually, a compact connected Lie group $G$ with a bi-invariant Riemannian metric $\rho$ can be viewed as a Riemannian symmetric space of the form $((G\times G)/{\rm diag}(G), \rho)$. If this Lie group is in addition simple, then it can be considered as a compact, isotropy irreducible, Riemmanian symmetric space, the so-called of Type II in Helgason's book. The classical Cartan-Schouten theorem is about a compact simple Lie group $G$, and it states that the unique flat bi-invariant metric connections on $G$ are the so-called +1 and -1 connections, say $\nabla^{\pm 1}$. They have non-zero (skew)torsion $T^{\pm 1}(X, Y)=\pm [X, Y]$. Moreover, $\nabla^{\pm 1}T^{\pm 1}=0$. In general, one can construct a 1-dimensional family $\{\nabla^{t} : t\in R\}$ of bi-invariant metric canonical connections on $G$, which joins the Levi-Civita connection and the $\pm 1$-connections. This family occurs by a reductive decomposition $\frak{g}\oplus\frak{g}=\Delta_{\frak{g}}\oplus{\frak{m}}_{t}$, which generalizes (and includes) the classical Cartan decomposition of $G$ (the latter induces the L-C connection on $G$). For example, see Section 4/page 8 of the following paper http://arxiv.org/pdf/1111.5044.pdf Notice that by the term canonical, we usually mean these (bi-invariant) connections on $G\cong (G\times G)/{\rm diag}(G)$, say $\nabla$, for which the $\nabla$-parallel tensor fields are exactly the $(G\times G)$-invariant tensor fields. This 1-parameter family of bi-invariant metric connections on $G$, has non-trivial (parallel) skew-torsion (except the trivial case of the Levi-Civita connection) and only the values $\pm 1$ give rise to flat metric connections. For all the other values of the parameter $t$, the associated curvature is non-zero. An easy way to compute the curvature and the torsion (or its covariant derivative) is by using the correspondence between bi-invariant affine connections on $G$ and bilinear maps $\lambda : \frak{g}\times\frak{g}\to\frak{g}$ which are ${\rm Ad}(G)$-equivariant, i.e. $\lambda({\rm Ad}(g)X, {\rm Ad}(g)Y)={\rm Ad}(g)\lambda(X, Y)$ for any $X, Y\in\frak{g}$ and $g\in G$. Then $$R(X, Y)=[\Lambda(X), \Lambda(Y)]-\Lambda([X, Y])$$ $$T(X, Y)=\Lambda(X)Y-\Lambda(Y)X-[X, Y],$$ where $\Lambda :\frak{g}\to{\rm End}(\frak{g})$ is the equivariant endomorphism associated to $\lambda$, i.e. $\Lambda(X)Y=\lambda(X, Y)$. It is easy to see that $\lambda$ induces a bi-invariant metric connection on $G$, if and only if $\Lambda(X)\in\frak{so}(\frak{g})$ for any $X\in\frak{g}$, i.e. $$\langle \Lambda(X)Y, Z\rangle+\langle Y, \Lambda(X)Z\rangle =0 \quad \forall \ X, Y, Z\in\frak{g}.$$ For example, the 1-parameter family of bi-invariant canonical metric connections on $G$ is induced by the bilinear map $\lambda(X, Y)=((1-t)/2)[X, Y]$ (up to scalar and sign), but it depends how we consider the reductive decomposition $\frak{g}\oplus\frak{g}=\Delta_{\frak{g}}\oplus{\frak{m}}_{t}$.<|endoftext|> TITLE: Almost Hadamard matrices QUESTION [11 upvotes]: As well-known, a Hadamard matrix is a square matrix with all coefficients $\pm 1$ and pairwise orthogonal rows or columns. Such matrices exist conjecturally in every dimension divisible by $4$. Call a matrix with an odd number $n$ of columns an "almost Hadamard matrix" if all its coefficients are $\pm 1$ and if all scalar products between distinct rows are $\pm 1$. We want to maximize the number of rows in almost Hadamard matrices: If $n\equiv 3\pmod 4$ this is easy: erase the last column of a Hadamard matrix of size $n+1$ (provided such a matrix exists). This yields a matrix with $n+1$ rows which is best possible. If $n\equiv 1\pmod 4$ the number of rows cannot exceed $n$ (argument: up to replacing rows by their opposites, we can suppose that all rows have an even number of coefficients $-1$. All scalar products between rows are now equivalent to $n$ modulo $4$ and a rank computation of the corresponding Gram matrix gives the result). A solution with $n-1$ rows in the case $n\equiv 1\pmod 4$ is obtained by adding an arbitrary last column (with coefficients $\pm 1$) to a Hadamard matrix of size $n-1$. For $n=5$ (and of course for $n=1$) there is a solution with $n$ rows but I am unaware of the existence of solutions with $n$ rows for any $n\equiv 1\pmod 4$ greater than $5$. Are there any? Is there any literature on such almost Hadamard matrices (perhaps under a different terminology, they are for example related to systems of equiangular lines)? REPLY [5 votes]: You can find most useful information on this page, which however has not been updated since 2012. The initial question is different, but the partial solution they have found is relevant here. Quote: The Hadamard maximal determinant problem asks when a matrix of a given order with entries -1 and +1 has the largest possible determinant. If the determinant of such an extremal matrix $M=M_n$ for $n\equiv1\pmod4$ attains Barba's bound $$\sqrt{ (2n-1) (n-1)^{n-1}},$$ which obviously needs $(2n-1)$ to be a perfect square, then $$M^TM=MM^T=J_n+(n-1)I_n$$ where $J_n$ is the all-$1$-matrix. So this matrix is an "almost Hadamard" (a.H.) matrix. (Moreover, all scalar products are $+1$.) Examples are given for $n=5,13,25,41,61,113$. If $2n-1=(2q+1)^2$ and $q$ is a prime power (note that the examples correspond to $q=1,2,3,4,5,7$), there is an explicit construction according to the paper A. E. Brouwer, An infinite series of symmetric designs, Math. Centrum Amsterdam Report ZW 202/83 (1983). If $(2n-1)$ is not a perfect square at all, we cannot conclude inversely that no square a.H. matrix exists, though I have a (kind of) heuristic argument against its existence: What we can conclude is that $M^TM$ must have a determinant smaller than Barba's bound because it must be a square. This means that the set of column vectors cannot be split in two sets $V$ and $W$ such that $v^T_iv_j=w_i^Tw_j=+1$ and $v^T_iw_j=-1$. (Nor, similarly, for the rows.) Before explaining why, here is an illustration of what I mean with those two sets. Take for instance the matrix $M$ given for $n=13$ in Gottfried Helms' comment. After some row and column permutations we have $$MM^T=-J_{13}+2J_5\oplus J_8+12I_{13}$$ and $$M^TM=-J_{13}+2J_3\oplus J_{10}+12I_{13},$$ written in compact form: $MM^T=$ n 1 1 1 1 - - - - - - - - 1 n 1 1 1 - - - - - - - - 1 1 n 1 1 - - - - - - - - 1 1 1 n 1 - - - - - - - - 1 1 1 1 n - - - - - - - - - - - - - n 1 1 1 1 1 1 1 - - - - - 1 n 1 1 1 1 1 1 - - - - - 1 1 n 1 1 1 1 1 - - - - - 1 1 1 n 1 1 1 1 - - - - - 1 1 1 1 n 1 1 1 - - - - - 1 1 1 1 1 n 1 1 - - - - - 1 1 1 1 1 1 n 1 - - - - - 1 1 1 1 1 1 1 n and similarly for $M^TM$. That is, the rows of $M$ come in two sets of sizes $5$ and $8$, while the columns of $M$ form two sets of sizes $3$ and $10$. Now the determinant of such a block matrix is always $(2n-1) (n-1)^{n-1}$, independently of the block sizes. So if $(2n-1)$ is not a square, the rows and columns cannot be partitioned in such a way, which means that an $n\times n$ a.H. matrix cannot have this kind of "symmetries". And heuristically, the existence of such a matrix seems to me about as improbable as the existence of a projective plane of non prime-power order... BUT based on the data of the above site, e.g. the $n=93$ example, I would conjecture that $n-1$ almost orthogonal vectors always exist.<|endoftext|> TITLE: Kernel of the differential in de Rham complex in positive characteristic QUESTION [5 upvotes]: Roughly, I'd like to ask how does the first terms in de Rham complex behaves for singular varieties. Let $Y$ be a potentially singular integral scheme over a perfect field $k$ of characteristic $p$ and $F : Y \to Y$ be its Frobenius morphism. The first mapping in de Rham complex of $Y$ (after application of $F_*$) is just the differential: $$d : F_*\mathcal{O}_Y \to F_*\Omega^1_{Y/k}.$$ Its kernel clearly contains the image of the Frobenius morphism $F : \mathcal{O}_Y \to F_*\mathcal{O}_Y$. What are the sufficient assumptions for which the kernel $\textrm{Ker} (d)$ is in fact given by $F$? For now, I can prove it for normal $Y$ with a torsion-free differential module $\Omega^1_Y$. The proof, more or less, boils down to localization at the generic point (which is injective by torsion-freeness) and then using normality ($a^p = f$ for $a \in k(Y)$ and $f \in \mathcal{O}_Y$ implies $a \in \mathcal{O}_Y$). Is there any well-known, explicit class of varieties for which the kernel is bigger? Maybe, there is a large class of normal singularities for which $\Omega^1_Y$ is torsion-free? Do you know any reference for this kind of problems? REPLY [2 votes]: Does not hold for a cusp in char 2 (for example $F_2[t^2, t^3]$), but does hold for a node in all characteristics.<|endoftext|> TITLE: Does this prime-gaps pattern occur infinitely often? QUESTION [6 upvotes]: Let $p_n$ be the $n$-th prime. For each integer $k \ge 0$, do there exist an infinite number of $k+3$ consecutive primes $(p_n, p_{n+1}, \ldots, p_{n+2+k})$ so that (1) The gap between the 1st and 2nd, and between the 2nd and last, are equal: $p_{n+1}-p_n = p_{n+2+k}-p_{n+1}$. (2) There are $k$ primes between the 2nd and last, i.e., between $p_{n+1}$ and $p_{n+2+k}$. For $k=0$, the answer is Yes by the recent breakthroughs on prime gaps. Here are some examples:       One could whimsically imagine "skipping" a flat stone on the primes, where the first bounce covers the gap between the 1st and 2nd primes, followed by $k+1$ smaller bounces that together cover the same gap before sinking on the last prime. REPLY [15 votes]: This would follow from the $k$-tuple conjecture in the following way. Choose an admissible tuple $d_1, \ldots, d_{k+2}$, such that $d_2-d_1=d_{k+2}-d_2$. If $n\in[d_1, d_{k+2}]$ is an integer, such that $\{d_1, \ldots, d_{k+2}\}\cup\{n\}$ is admissible, pick a prime number $p$ such that for all $i$ we have that $n-d_i$ is not divisible by $p$. Pick integers $e_i>d_{k+2}$, such that $\{d_1,\ldots, d_{k+2}, e_1, \ldots, e_\ell\}$ is admissible, and the set $\{d_1,\ldots, d_{k+2}, e_1, \ldots, e_\ell\}$ covers all residue classes modulo $p$ with the exception of $n\pmod{p}$. Repeat until you arrive at an admissible set $D$, such that $D\cap[d_1, \ldots, d_{k+2}]=\{d_1, \ldots, d_{k+2}\}$, and for all $n\in[d_1, d_{k+2}]\setminus\{d_1, \ldots, d_{k+2}\}$ we have that $D\cup\{n\}$ is not admissible. Apply the $k$-tuple conjecture to the set $D$. If $x+d$ is prime for all $d\in D$, and $x$ is sufficiently large, then $x+d_1, \ldots, x+d_{k+2}$ are consecutive prime numbers.<|endoftext|> TITLE: When does a dominant map generically admit a section? QUESTION [5 upvotes]: Let $f:X\to Y$ be a dominant map of algebraic varieties. Say that $f$ generically admits sections if there exists a Zariski (not etale) open set $U \subseteq Y$ over which $f$ has a section $\sigma:Y\to X$. For example, $f:\mathbb C\to \mathbb C$ taking $z\mapsto z^2$ does not generically admit sections. Is there a characterization of such maps in terms of more standard notions? I don't mind assuming characteristic $0$ so then we can use the generic smoothness theorem to assume $f$ is a smooth morphism (through shrinking $X$ and $Y$). Quasiprojective is fine too. Feel free to add tags as seems appropriate. REPLY [4 votes]: So, as you allow, let's assume characteristic zero and perhaps even working over $\mathbb C$? It seems to me that Graber-Harris-Mazur-Starr gives you at least a necessary condition. Jason might correct me if this is wrong, or add something in the other direction. Anyway, GHMS introduces the notion of a pseudosection which is a subvariety $Z\subseteq X$ that dominates the base with a rationally connected general fiber and they prove (essentially) that the existence of a pseudosection is equivalent to $f$ admitting a section after a base change to the normalization of any curve in the base. As mentioned in the comments, over a smooth curve the existence of a section follows from the general fiber being rationally connected by an earlier result of Graber-Harris-Starr. So, I would proceed as follows: Suppose there exists a section over $U$, a Zariski open subset of $Y$. Restrict $f$ to $U$ and observe that the section of $f\big|_U$ implies that any further base change to a smooth curve admits a section. Then by GHMS this $f\big|_U$ admits a pseudosection. Take the closure of this pseudosection in the original $X$. This closure will be a pseudosection of the original $f$. In other words, if $f$ admits a generic section as you require, then it admits a pseudosection. This shows for example why your example does not admit a generic section (not that you need this condition, I'm just saying that it does imply that). So, the remaining question is whether having a pseudosection is enough for admitting a generic section. Clearly it is enough over a curve by [GHS]. I don't think it is enough in general and I think it is possible that the exact condition that would be needed here is not known. Again, I defer to Jason to possibly comment on this.<|endoftext|> TITLE: Which smooth compactly supported functions are convolutions? QUESTION [22 upvotes]: If $f,g$ are smooth functions with support in the interval $[-r,r]$ for some $r>0$, then their convolution $f*g$ is smooth with support in $[-2r,2r]$. My question is about the converse: Given smooth $h$ with support in $[-2r,2r]$, can I always write it as $h=f*g$ with $f,g$ as above? (By Fourier transform, one can formulate this problem also as a decomposition of entire functions of exponential type $2r$ into a product of entire functions of exponential type $r$ with additional restrictions on the real line.) REPLY [14 votes]: After some more searching I found the solution in the literature. In the paper L. Ehrenpreis, "Solution of some problems of division. IV", Amer. J. Math. 82 (1960), 522-588 Ehrenpreis posed the question if any $h\in C_c^\infty({\mathbb R}^n)$ can be represented as a convolution $f*g$ of two functions $f,g\in C_c^\infty({\mathbb R}^n)$, this question is therefore known as the "Ehrenpreis factorization problem". For $n\geq2$ the answer is no (shown 1978-1980 by several authors, cited in the paper below), but for $n=1$ such a factorization is always possible. This has been proven much later in R. S. Yulmukhametov, "Solution of the Ehrenpreis factorization problem", Sb. Math. 190 (1999) 597, doi:10.1070/SM1999v190n04ABEH000400 via the complex analysis approach, i.e. by factoring the entire Fourier transform of $h$ (and in particular its zeros) in an appropriate manner. In his Theorem 10, Yulmukhametov also answers the sharpened version, including the support conditions supp $h\subset[-2r,2r]$, supp $f$, supp $g\subset[-r,r]$, affirmatively. This is precisely the question posted here.<|endoftext|> TITLE: computing second cohomology $H^2(O_a,\mathbb{Z})\cong \mathbb{Z}^n$ of a generic coadjoint orbit QUESTION [5 upvotes]: Let $G$ be a compact , connected and simply connected Lie group and $a\in\mathfrak{g}^*$ (dual of Lie algebra of Lie group $G$). Then let $O_a$ be a generic coadjoint orbit then can we say $H^2(O_a,\mathbb{Z})\cong \mathbb{Z}^n$ for some $n$. Is there any counter example? REPLY [2 votes]: The cohomology of regular coadjoint orbits of a compact Lie group can be computed using Morse theory. This approach is due to Bott and Samelson (1955). Let $A \in \mathfrak{g}^{*}$ be a regular element, $\langle\cdot,\cdot\rangle$ the Cartan-Killing form, and consider the coadjoint orbit $\mathcal{O}_{B}$ of a regular element $B \in \mathfrak{g}^{*}$, then $\langle A,\cdot\rangle: \mathcal{O}_{B}\to\mathbb{R}$ is a Morse function whose critical set is the Weyl orbit of $B$, and the index of a critical point $w(B)$ turns out to be twice the number of root hyperplanes crossed by traversing the line from $A$ to $w(B)$. This gives a cell decomposition concentrated in even degrees, so the cohomology vanishes in odd degrees and in particular $H^{2}(\mathcal{O}_{B},\mathbb{Z})$ has the form you suggest.<|endoftext|> TITLE: Higher Weierstrass points on curves of genus 3 QUESTION [6 upvotes]: So this question is directly related to a comment made by David Mumford in his Lecture 1 given at U. Michigan in 1974 entitled: What is a curve and how explicitly can we describe them ? Mumford claims that if you take a (non-hyper elliptic) smooth projective curve $C$ over $\mathbb{C}$ of genus 3 and embed it in $\mathbf{P}^2$ via its canonical map (denoting the image of the curve again by $C$), then there are exactly $108$ points $x\in C$ for which there is a conic passing through $x$ with contact order (with respect to $C$) equal to $6$. Q1: How does one prove that you have only finitely many such conics touching $C$ and have contact order $6$? This seems to suggest, that for most points $P\in C$, the best contact order of a conic passing through $P$ is $5$. Q2: In general if $C\subseteq \mathbf{P}^2$ is a fixed embedded smooth projective curve and $x\in C$ is a point, then for a fixed degree $d$, how does one compute the maximum contact order at $x$ among all smooth projective curves $D$ of degree $d$ in $\mathbf{P}^2$ passing through $x$ (is this computable)? REPLY [6 votes]: Will is right of course: there exists a curve of degree $d$ with a contact of order $k$ with $C$ at $P$ iff $h^0(C,\mathcal{O}_C(dH-kP))>0$, where $H$ is the divisor of a line. So: Q1) $2H-6P$ has degree 2, so the condition is that it lies on the canonical theta divisor $\Theta $ of $J^2C$. So we are looking at the intersection of $\Theta $ with $6_*C$, the image of $C$ under multiplication by 6 in $JC$ (I am implicitely choosing some base point here to identify $J^2C$ with $JC$). Now $6_*$ acts as $6^p$ on $H_p(JC)$, so the (co)homology class of $6_*C$ is $6^2[C]=6^2\frac{\Theta ^2}{2} $. Thus $(\Theta .6_*C)=6^2\frac{\Theta ^3}{2}=6^2.3=108$. Q2) The same ideas will give you the general answer. Let $e:=\deg(C)$, $g:=g(C)$. We look at $dH-kP$. If $kde-(g-1)$, there is none in general; there can be some, of course, for some particular pairs $(C,x)$. The interesting case is $k=de-(g-1)$; then we find $N$ contact points, with $N=(\Theta . k_*C)=k^2g$ (of course these points must be counted with multiplicity).<|endoftext|> TITLE: Did the notion of "angle" originate with Thales? QUESTION [14 upvotes]: Thales (circa 600BC—roughly 50 years before Pythagoras, 200 years before Plato, and 300 years before Euclid) certainly knew and reasoned with the concept of a planar angle. Are there earlier historical references to angles in a mathematical sense? Or is the origin of the notion lost in the mists of antiquity? REPLY [5 votes]: The division of a circle into 360 degrees may well have originated from ancient Babylonian mathematics (1800 BC), as evidenced by a clay tablet from Shush. Here is a quote from David Wallis, History of angle measurement: See also A History of Pi (pages 21-22).<|endoftext|> TITLE: Applications of homotopy purity theorem of Morel-Voevodsky QUESTION [16 upvotes]: One of the most important theorems in motivic homotopy theory is the homotopy purity theorem of Morel-Voevodsky which says that the motivic Thom space of the normal bundle $\mathcal N_{Z/X}$ of a closed immersion $Z \hookrightarrow X$ is $\mathbb A^1$-weak equivalent to $X/X-Z$. What are the main applications of this in motivic homotopy? Why is this theorem very crucial? REPLY [9 votes]: Your question comes because of the vague use of the term purity. That's ok, moreover because Morel and Voevodsky didn't explain in that paper why that result deserved the term purity. That result that you mention is a final step on a sequence of arguments which gives purity, but it is not purity itself. Find the link between them at the end. What is purity: The term purity comes from the purity isomorphism. As Mayer-Vietoris, the purity isomorphism is a property you expect on any cohomology. Let's be more precise: whatever context you are if $H^\bullet$ is a cohomology and $Z\hookrightarrow X$ is a "good" closed immersion (smooth and sometimes even proper-compact) one should have an isomorphism $$ \bar i_*\colon H^{*-2d,*-d}(Z) \buildrel{\sim}\over{\to} H^{*,*}_Z(X) \qquad \mbox{Purity isomorphism.} $$ In the formula $d$ is the codimension of $Z$ in $X$. Denote $U$ the open complement of $Z$. Together with the local long exact sequence and forgetting support the purity isomorphism gives the so called Gysin long exact sequence $$ \cdots\to H^{*-1,*}(U)\to H^{*-2d,*-d}(Z)\buildrel{i_*}\over{\to}H^{*,*}(X)\to H^{*,*}(U)\to\cdots $$ The term "Gysin" comes from the fact that $i_*$ is called the "Gysin morphism". And now back to your question: Why is purity important?: There go my two main reasons: 1.- Purity is a standard property of cohomology. Take any book of any type of cohomology, you will find it. No purity? Then no cohomology. It night be good, fun or pretty, but it is not cohomology what one is doing. It is as simple as that. If Morel and Voevodsky want to show that they found the adequate homotopy category of schemes (which is expected to describe cohomologies) then they have to prove purity. 2.- Purity is the "nontrival" basic property of cohomology. Whatever context you are Mayer-Vietoris, inverse image, ect... are usually direct from definition. Purity is always a theorem and requires hypothesis. Surprisingly for me, the Local long exact sequence in motivic homotopy is not trivial, and you will see a lot fuzz around it too (even more because it is preliminary to purity). As I said purity is not important for applications, it is a requirement to start speaking about cohomology. But if you want a concrete application, an important one, there it goes: nowadays higher Riemann-Roch relies essentially on purity. Link between MV result and purity: People in Algebraic Geometry nowadays usually rely on the so called "Thom isomorphism" for establishing the purity. By construction of the Thom space you have $$ H(\mathrm{Th}(N_{Z/X}))\simeq H(Z). $$ With the isomorphism proved by Morel and Voevodsky you have: $$ H(Z)\simeq H(\mathrm{Th}(N_{Z/X}))=\mathrm{Hom}_{\mathbf{SH}(X)}(\mathrm{Th}(N_{Z/X}),E)\buildrel{\mathrm{MV}}\over{\simeq} \mathrm{Hom}_{\mathbf{SH}(X)}(X/X-U,E)= H_Z(X) $$ Where $\mathbf{SH}(X)$ is the stable homotopy category and $E$ is the ring spectrum giving your cohomology. (If you dont want to use $\mathbf{SH}$ you can put the $\mathbb{Z}\times BGL$ instead of $E$ and $\mathrm{Hom}$ in the homotopy category and obtain it for $K$-theory).<|endoftext|> TITLE: Are these two notions of "dualizable" spectra equivalent? QUESTION [11 upvotes]: A spectrum $X$ is dualizable if the natural map $$Map(X,\mathbb S) \wedge X \rightarrow Map(X,X)$$ is an equivalence of spectra. This is equivalent to having evaluation and coevaluation maps in the stable homotopy category $$ X \wedge DX \rightarrow \mathbb S $$ $$ \mathbb S \rightarrow DX \wedge X $$ for which the usual composites $$ X \rightarrow X \wedge DX \wedge X \rightarrow X $$ $$ DX \rightarrow DX \wedge X \wedge DX \rightarrow DX $$ give the identity in the homotopy category (cf. Lewis-May-Steinberger III.1.2). It is also well-known that this implies that $X \rightarrow D(DX)$ is an equivalence of spectra (LMS III.1.3(i)), but my question is Is the converse true? Does $X \overset\sim\rightarrow D(DX)$ imply that $X$ is dualizable? I understand that everything I have said holds in an arbitrary closed symmetric monoidal category, but I am willing to consider arguments that only work for spectra (or $R$-module spectra). REPLY [7 votes]: Let $R$ be bounded below, bounded above and nontrivial, and work in $R$-modules. Let $X = \bigvee_{n\in\mathbb{Z}} \Sigma^n R$. Then $X \overset{\simeq}\to D(DX)$ is an equivalence, but $X$ is not dualizable. (Edit: Increased generality, based on a comment by Nardin.)<|endoftext|> TITLE: Which large cardinals are upward reflecting? QUESTION [7 upvotes]: Let the first order formulas $p(x)$ and $wi(x)$ assert "$x$ is a large cardinal of type $p$" and "$x$ is weakly inaccessible" respectively. The large cardinal type $p$ is upward reflecting if $ZFC\vdash \forall \kappa~~(p(\kappa)\longrightarrow\exists \lambda>\kappa~~wi(\lambda))$. In the other words existence of a large cardinal of type $p$ like $\kappa$ implies existence of a large cardinal (at least weakly inaccessible) above $\kappa$. Which large cardinals are upward reflecting? In particular, are Shelah cardinals upward reflecting? REPLY [3 votes]: As you are mainly interested in Shelah cardinals, let me add a few more remarks about them. Given a Shelah cardinal $\kappa$, let $wt(\kappa),$ the witnessing number of $\kappa,$ be the least $\lambda$ such that for each $f:\kappa\to\kappa,$ there is an extender $E$ in $V_\lambda,$ witnessing the Shelahness of $\kappa$ with respect to $f$. Then we can show that: 1) for all $\xi TITLE: Lebesgue entropy zero and positive topological entropy QUESTION [8 upvotes]: I am looking for examples of volume preserving $C^{\infty}$ diffeomorphisms $f$ of a surface, which have positive topological entropy ($h(f) > 0$), but that the Lebesgue measure entropy (metric entropy) is equal to zero. It was pointed out to me that such horrible things exist. These examples should be very weird in nature as positive topological entropy implies the existence of a homoclinic periodic point (by a theorem of A. Katok), which implies the existence of some horseshoe for a power of $f$. I am also looking for conditions on a diffeomorphism $f$ of a surface or a measure $\mu$ invariant under $f$ (For example $f$ being real analytic or $\mu$ being SRB and $h_{\mu}(f) >0$) that would imply it has positive lebesgue entropy. Any reference on these questions on anything somewhat related to this would be greatly appreciated. Thanks!! REPLY [3 votes]: I do not have an direct answer but since I fear the answer might not be known and you also ask for related stuff let me mention a couple of results. In the $C^1$ case, Mañe-Bochi's result asserts that far from Anosov systems, this is the ``typical'' phenomena. However, $C^1$-generic sets of diffeomorphisms may contain no smooth ones. In the $C^\infty$ case, one could imagine that it is possible to start with some diffeomorphism which has positive entropy (a linear Anosov in $T^2$ for example) and opening elliptic islands in the periodic orbits until they fill up a full lebesgue measure subset, but I am not sure if this can be done (nor if the existence of such an example is known). I guess a good place to look for this type of phenomena is looking at the problem of coexistence (for example, in section 1.1.2 of that paper a way of constructing the above map is proposed but it claims not to be known if the measure of the complement of the islands is positive).<|endoftext|> TITLE: Explicit Kodaira-Spencer map of hyperelliptic curves QUESTION [6 upvotes]: Let $g\geq 2$, and $$\mathcal T=\{(t_1,\cdots,t_{2g+2})~|~t_i\neq t_j,\forall i\neq j\}.$$ For any $t=(t_1,\cdots,t_{2g+2})\in \mathcal T$, let $$Y_t=\left\{y^2=\prod_{i=1}^{2g+2}(x-t_i)\right\}.$$ Thus we get a family of hyperelliptic curves $\mathcal Y \to \mathcal T$. Let $t^0=(t_1^0,\cdots,t_{2g+2}^0) \in \mathcal T$ be a fixed point (we assume $t_i^0\neq 0,\forall i$). Then we have the Kodaira-Spencer map $$\Theta_{\mathcal T,t^0} \longrightarrow H^1(Y_{t^0}, \Theta_{Y_{t^0}}).$$ Take dual, we get a map (also called Kodiara-Spencer map): $$\kappa:H^0(Y_{t^0}, \omega^{\otimes 2}_{Y_{t^0}}) \longrightarrow \Omega^1_{\mathcal T,t^0}.$$ Now one can take $$\frac{x^i}{y^2}(dx)^2, 0\leq i \leq 2(g-1);~ \frac{x^j}{y}(dx)^2, 0\leq j \leq g-3,$$ as a basis of $H^0(Y_{t^0}, \omega^{\otimes 2}_{Y_{t^0}})$. And take $$dt_1, \cdots, dt_{2g+2},$$ as a basis of $\Omega^1_{\mathcal T,t^0}$. Then can we write down the Kodaira-Spencer map explicitly under the above two basis? REPLY [3 votes]: Caution: I was thinking over the calculation that led to my proposed answer below, and I realized that I had neglected a term that I haven't actually justified as being zero, so now I'm less sure that these formulae are correct. There may be another term in the formula for $\kappa(\phi_i)$ and I don't see how to rule it out. As a result, I am withdrawing my answer until I have had time to determine whether the extra term really does vanish. My apologies to those who trusted my claim and up-voted it! Unfortunately, I am not allowed to delete an 'accepted answer', so I'm putting this caution at the top so that readers will know not to regard it as correct (yet). I haven't checked all of the details, but I think that the following is correct and gives you a basis for an answer: We can get rid of the action of the Möbius group of linear fractional transformations, by, instead, looking at the manifold $$ \mathcal{S} = \{(s_0,\ldots,s_{2g-2})\ |\ s_i\not=0,1\ \text{and}\ s_i\not=s_j\ \text{when}\ i\not=j \ \}\subset \mathbb{C}^{2g-1} $$ and letting $s\in\mathcal{S}$ correspond to the genus $g$ hyperelliptic curve $C_s$ defined by $$ y^2 = x(x-1)(x-s_0)(x-s_1)\cdots(x-s_{2g-2}). $$ The hyperelliptic involution is $\iota(x,y) = (x,-y)$. A basis for the $\iota$-even holomorphic quadratic differentials on $C_s$ (a space of dimension $2g-1$ when $g>1$) is given by $$ \frac{x^i\ dx^2}{x(x-1)(x-s_0)(x-s_1)\cdots(x-s_{2g-2})}\ \text{when}\ 0\le i\le 2g-2, $$ and a basis for the $\iota$-odd holomorphic quadratic differentials on $C_s$ (a space of dimension $g-2$ when $g>1$) is given by $$ \psi_j = \frac{x^i\ dx^2}{y}\qquad\text{when}\ 0\le j\le g-3. $$ It turns out that a more convenient basis for the $\iota$-even holomorphic quadratic differentials is $$ \phi_i = \frac{dx^2}{x(x-1)(x-s_i)}\qquad \text{when}\ 0\le i\le 2g-2. $$ Then, it seems, in this case, that the (dual) Kodaira-Spencer map is given by $$ \kappa(\psi_j) = 0\qquad \text{when}\ 0\le j\le g-3 $$ while there is a (nonzero) constant $c$ (which might depend on $g$, but I don't think so) such that $$ \kappa(\phi_i) = \frac{c\ ds_i}{s_i(s_i-1)}\qquad \text{when}\ 0\le i\le 2g-2. $$ (This formula even works when $g=1$, by the way.) I think you can get your desired formula by making this equivariant with respect to the action of the Möbius group $\mathrm{PSL}(2,\mathbb{C})$, but I suspect that the general formula in this case won't be particularly nice.<|endoftext|> TITLE: Reference for homeomorphism between "analytic" compactification of $M_{g,n}$ and Deligne-Mumford compactification QUESTION [5 upvotes]: There are several natural ways to endow the compactification of the space of marked Riemann surfaces $M_{g,n}$ ($2g+n\geq 3$), with a topology, which is defined using "differential geometric or analytic" notions, e.g. the topology inherited from the "augmented Teichmüller space". As far as I understand, most of these natural topologies coincide, since the marked Riemann surfaces (or the associated hyperbolic surfaces) are (sufficiently) rigid. On the other hand there is the classical Deligne-Mumford compactification of $M_{g,n}$, the space of genus g nodal curves with n marked points, which inherits a natural topology as a projective variety. What is a good source which shows that the compactification of $M_{g,n}$, endowed with any one of those "analytic" topologies, is homeomorphic to the Deligne-Mumford compactification? Sometimes this result is attributed to: Harvey, William; Chabauty spaces of discrete groups. Discontinuous groups and Riemann surfaces (Proc. Conf., Univ. Maryland, College Park, Md., 1973), pp. 239–246. Ann. of Math. Studies, No. 79, Princeton Univ. Press, Princeton, N.J., 1974. But in my understanding this result is not really claimed or proved there, or if it is, then only in a quite implicit way. Can anybody point me to some more explicit reference? Thanks! REPLY [3 votes]: See John Hubbard and Sarah Koch: "An analytic construction of the Deligne-Mumford compactification of the moduli space of curves".<|endoftext|> TITLE: Connectedness of sub-varieties via Hilbert polynomials QUESTION [5 upvotes]: Let $X$ be a sub-variety of $\mathbb CP^n$ and let $p_X(k)$ be its Hilbert polynomial. It is well known that some basic invariants of $X$ (such as its dimension) can be read from $p_X(k)$. I am interested to know to which extent one can read connectedness of $X$ from $p_X(k)$. Here are two precise questions. ($n$ is fixed). Question 1. Is there a classification of polynomials, for which it is known that every subvariety in $\mathbb CP^n$ with given Hilbert polynomial is connected? If there is no such a classification, I would like at least to know a large list of such polynomials. Question 2. Suppose that $P$ is the Hilbert polynomial of a complete intersection on $\mathbb CP^n$ of positive dimension. Is it true that any subscheme of $\mathbb CP^n$ with Hilbert polynomial $P$ is connected? REPLY [4 votes]: Let me give a negative answer to Question 2, which shows that the answer to Question 1 should be almost always negative (perhaps with very few exceptions). The Hilbert polynomial of a complete intersection of type $(2,2)$ in $\mathbb{P}^3$ is $4k$. But this is also the Hilbert polynomial of the union of a plane quartic curve (embedded in $\mathbb{P}^3$) and two points lying anywhere in $\mathbb{P}^3$.<|endoftext|> TITLE: Gödel's Constructible Universe in Infinitary Logics (A Possible Approach to HOD Problem) QUESTION [22 upvotes]: Gödel's constructible universe ($L$) is defined using definable power set operator in first order logic ($\mathcal{L}_{\omega ,\omega}$). One can produce such a universe in infinitary logics in the same way using corresponding notions of formulas and definability. Obviously $L$ becomes larger when the logic has more expression power. For each cardinal $\kappa$ define $L_{\kappa}$ to be Gödel's constructible universe in the infinitary logic $\mathcal{L}_{\kappa,\kappa}$ and $L_{\infty}$ is Gödel's constructible universe in $\mathcal{L}_{\infty,\infty}$. (1) Is $L_{\kappa}$ a model of $ZFC$ for each cardinal $\kappa$? What about $ZFC+GCH$? (2) What is $L_{\infty}$? (3) Is there a (possibly large) cardinal $\kappa$ such that $L_{\kappa}$ is Dodd-Jensen core model, $L[U]$, $HOD$, etc? (4) What are the consistency strengths of the existence of non-trivial elementary embeddings from $\langle L_{\kappa},\in\rangle$ to itself for different $\kappa$s in the sense of infinitary logic $\mathcal{L}_{\kappa,\kappa}$? Note that by Prof. Hamkins' answer for $L_{\infty}$ finally it reach Kunen's inconsistency but what about a given cardinal $\kappa$? Are all these consistency strengths for different cardinals bounded by some large cardinal axiom and there is a gape between consistency strength of the existence of a non-trivial elementary embedding from $\langle L_{\infty},\in\rangle$ to itself and consistency strengths of the existence of such elementary embeddings for $L_{\kappa}$s? (5) If there is a cardinal $\kappa$ such that $L_{\kappa}=HOD$, is it possible to determine consistency strength of the existence of a non-trivial (first-order) elementary embedding from $\langle HOD,\in\rangle$ to itself by analyzing the growth speed of the consistency strength of existence of such embeddings for $\langle L_{\lambda}, \in\rangle$s in $\lambda <\kappa$? (6) What is $L_{\kappa}$ for the least strongly compact cardinal $\kappa$? REPLY [18 votes]: Theorem. $L_\infty$ is the entire set-theoretic universe $V$. Proof. I claim that every set will arise in the construction process, because eventually it will become explicitly definable by a formula. In infinitary logic, there are far more than only countably many formulas, and one can cook up a formula to define a specific set, by using the formulas that define its elements. What I claim specifically is that for every set $a$ there is a ${\cal L}_{\infty,\infty}$ formula $\psi_a(x)$, such that in any transitive set $M$ with $a\subset M$ we have $a=\{ x\mid M\models\psi_a(x)\}$. Suppose that this is true for each $a\in A$. Consider the formula $$\psi_A(u)=\bigvee_{a\in A}(u=\{x\mid \psi_a(x)\}).$$ In any transitive $M$ with $A\subset M$, it follows that $\psi_A(u)$ will hold if and only if $u=a$ for some $a\in A$. Thus, $A=\{u\mid M\models \psi_A(u)\}$, and so $A$ is also definable. Thus, by $\in$-induction, we've verified that every set is definable, and so every set in $V$ eventually arises in your universe $L_\infty$. QED This argument is similar to the fact that if one undertakes the constructible universe using second-order logic, rather than infinitary logic, the result is $\text{HOD}$. Theorem.(Myhill & Scott) The constructible universe in second-order logic is the same as HOD. $$L_{SO}=\text{HOD}.$$ Proof. Any set that appears in $L_{SO}$ is ordinal definable, and so $L_{SO}\subset \text{HOD}$. Conversely, if $A$ is a set of ordinals in $\text{HOD}$, then it is ordinal definable in some $V_\alpha$, and so once $L_{SO}$ has constructed up to some stage $\theta$ of size at least $|V_\alpha|$, then in second order logic we can define $A$ as a subset by saying "there is a relation on $\theta$ which makes it isomorphic to $\langle V_\alpha,\in\rangle$, such that the formula is true for the corresponding ordinals. That is, the second-order logic allows us to summon a copy of $V_\alpha$ and run the definitions inside it. QED For a reference, see Myhill, Scott, Ordinal definability. 1971 Axiomatic Set Theory (Proc. Sympos. Pure Math., Vol. XIII, Part I, Univ. California, Los Angeles, Calif., (1967) pp. 271–278 Amer. Math. Soc., Providence, R.I. . Your models $L_\kappa$ arise in Chang's paper in the very same journal issue: C.C. Chang, Sets constructible using $L_{\kappa,\kappa}$, 1971 Axiomatic Set Theory (Proc. Sympos. Pure Math., Vol. XIII, Part I, Univ. California, Los Angeles, Calif., 1967) pp. 1–8 Amer. Math. Soc., Providence, R.I. MR Abstract. Let $C_\alpha^\omega$ be the αth level in the Gödel ramified hierarchy of constructible sets. The author studies the hierarchy $C_α^κ$ that results on using $L_{κκ}$ definability (in place of $L_{ωω}$ definability) in generating this hierarchy, where κ is a regular cardinal. The class $C_κ=⋃_αC_α^κ$ is a model of ZF; in fact, it is the smallest transitive model containing all the ordinals closed under <κ-termed sequences (Theorems I and II). It need not be a model of the axiom of choice (Theorem IV). Other results show what happens to the GCH and Scott's result on measurable cardinals in this model. These results depend on a generalization of Gödel's collapsing lemma (Theorem V). Finally, the author shows how to improve results about indiscernibles in $C_ω$ by using infinitary formulas. In particular, he shows that if there is a Ramsey cardinal then $C_{ω_1ω}$ is an $L_{ω_1ω}$-elementary substructure of $C_ω$, thus improving Silver's result. The case $\kappa=\omega_1$ is known as the Chang model.<|endoftext|> TITLE: Are there known expressions for total variation distance between $N(0,\sigma_1^2)$ and $N(0,\sigma^2)$ QUESTION [9 upvotes]: Are known expressions for total variation distance between $N(0,\sigma^2)$ and $N(0,\sigma^2+\epsilon)$ for small $\epsilon$? The only thing I seem to find is things are expression about the mean but not if we change variance slightly. REPLY [4 votes]: There's also a softer argument based on properties of the heat kernel, which applies in higher dimensions as well, in Lemma 4.9 of this paper of Klartag. It shows that in $n$ dimensions, the total variation distance between centered Gaussian distributions with covariances $\alpha I_n$ and $\beta I_n$ is at most $$ C \sqrt{n} \left|\frac{\beta}{\alpha} - 1\right|, $$ where $C > 0$ is an absolute constant (which can be made explicit if you want).<|endoftext|> TITLE: Delay Differential Equations Numerical methods QUESTION [5 upvotes]: I have a general question about delay differential equations. I know that even simple ones hardly have analytic solutions and mine clearly doesn't have any as it is a system of non-linear delay differential equations. I am looking for ways of solving them numerically. I know a few algorithms to solve ODEs numerically. Using a basic algorithm (say Euler's method), the equation is : $$f_{n+1} = f_n + hf'_n$$ where $h$ is the step size. For DDE, is it possible to use $f_{n-a}$ instead of $f_n$ for some $a$ corresponding to the delay? How does it work for more complicated methods such as the Runge-Kutta family or the PECE methods? If it works, has there been any study on the stability of the solvers? REPLY [7 votes]: There is an excellent monograph on the topic by Bellen and Zennaro. You can find quite sophisticated FORTRAN code on the homepage of Nicola Guglielmi. There is also a good analysis of numerical methods for stiff delay equations on the corresponding scholarpedia article.<|endoftext|> TITLE: Distribution of $n$-th roots modulo a smooth number QUESTION [11 upvotes]: Let $n \ge 2$. Let $p_1,\dots,p_m$ be distinct primes $\equiv 1 \pmod{n}$. Let $N=p_1 p_2 \cdots p_m$. If $\gcd(a,N)=1$ and the equation $x^n \equiv a \pmod{N}$ has a solution then it has $n^m$ solutions. This question concerns how these $n$-th roots are distributed as $a$ varies, particularly with $n$ fixed and $m$ large. Question. Do there exist $\alpha$ and $\beta$ with $0 \le \alpha \le \beta <1$ and $\beta-\alpha$ small, such that for every $a$ coprime to $N$ that is an $n$-th power modulo $N$, there is $b$ satisfying $\alpha N \le b \le \beta N$ and $b^n \equiv a \pmod{N}$? Can we take $\beta-\alpha=O(n^{-m})$? REPLY [3 votes]: This is a partial answer. The idea is from my recent preprint http://arxiv.org/abs/1509.03768 Additionally assuming that $p_1, \cdots, p_m$ are least among the congruence $1$ mod $n$, By repeated application of Linnik's theorem, there is an absolute constant $L$ such that $N< n^{L^m}$. We use the following result on exponential sums over $\mathbb{Z}_N^{*}$ by J. Bourgain: Let $N\geq 1$. For any $\epsilon>0$, there exist a constant $\delta_0=\delta_0(\epsilon)>0$ such that for any subgroup $H$ of $\mathbb{Z}_N^{*}$ with $|H|>N^{\epsilon}$, $$ \max_{(m,N)=1}\left|\sum_{a\in H}e^{2\pi i m\frac {a}N}\right|0$ such that $$ \left||\{ 0 TITLE: Climbing quickly up $L$ QUESTION [5 upvotes]: This question is motivated by Joel David Hamkins' answer to Gödel's Constructible Universe in Infinitary Logics (A Possible Approach to HOD Problem), in which he shows that, if we replace first-order definability by $\mathcal{L}_{\infty\infty}$-definability in the definition of $L$, we wind up building all of $V$. This relies on using $\mathcal{L}_{\infty\infty}^V$; clearly if we take $$ \overline{L}_{\alpha+1}=\{A\subseteq\overline{L}_\alpha: \exists \varphi\in \overline{L}_\alpha\cap\mathcal{L}_{\infty\infty}(A=\varphi^{\overline{L}_\alpha})\},$$ we just wind up with $L$ itself at the end. Moreover, we also climb up $L$ at about the same speed as in the usual construction, since if I have an infinitary formula in $L_\alpha$, the subset of $L_\alpha$ it defines will show up in $L_{\alpha+17}$ (say). A similar observation, however, seems to fail for second-order logic: if we interpret the second-order quantifiers as ranging over $L$, so that $$\hat{L}_{\alpha+1}=\{X\subseteq\hat{L}_\alpha:\exists \varphi\in \mathcal{L}_{II}(L\models \text{"}\varphi^{\hat{L}_\alpha}=A\text{"})\},$$ then the only obvious upper bound on the $\hat{L}$-hierarchy is $\hat{L}_{\alpha}\subseteq L_\beta\implies \hat{L}_{\alpha+1}\subseteq L_{\vert\beta\vert^+}$ (by condensation); and this is really a trivial observation. So in principle the $\hat{L}$-hierarchy could grow quite fast. My question is: How fast does the $\hat{L}$-hierarchy grow? REPLY [6 votes]: In one sense I think you almost have the answer anyway. Assume $V=L$. For any cardinal $\beta$ let $C_\beta = \{ \gamma\in (\beta,\beta^+)| \,L_\gamma \prec L_{\beta^+}\}$ ($\prec $ meaning First Order elementarily). Let $\hat L_0 =L_\omega=HF$. Then $\hat L_1 = L_{\gamma_1}$ where $\gamma_1$ is the minimal element of $C_\omega$. Let $\gamma_2$ be the next element of $C_\omega$. Then $\hat L_2 = L_{\gamma_2}$. This is because Second Order definability over $L_\beta$ is essentially equivalent to First Order definability over $L_{\beta^+}$; one deduces that S.O. definability over $\hat L_\tau \cup \{\hat L_\tau\}$ yields everything in $L_\delta$ where $\delta$ is least in $C_\beta$ greater than $On\cap \hat L_\tau$. In short: if, setting $\gamma_0=\omega \, , \langle\gamma_\tau\,|\, \tau\in On\rangle$ enumerates $Card \cup \bigcup_{\beta\in Card} C_\beta$ in ascending order, $\hat L_\tau = L_{\gamma_\tau}$.<|endoftext|> TITLE: Chebotarev density theorem for $k$-almost primes QUESTION [15 upvotes]: Consider a finite Galois extension $L$ of $\mathbb Q$, of Galois group $G$. Let $k \geq 1$ be a fixed integer. Let $D$ be a subset of $G^k$ invariant by conjugation and by the natural action of the symmetric group $S_k$ on $G^k$. Let $A=A(L,D)$ be the set of all integers $n$ which are products of $k$ primes $n=p_1\dots p_k$ (variant: $k$ distinct primes) such that $(Frob_{p_1},\dots,Frob_{p_k}) \in D.$ (Here $Frob_p$ is a Frobenius at $p$ in $G$, assuming $p$ is unramified in $L$. if one of the $p_i$ is ramified, by convention say that $n$ is not in $A$.) Let $A(x)$ be the number of integers $n0$. Let $N_k=N(k;a_1,\ldots,a_r)$ denote the set of integers that are products of $k$ primes with exactly $a_j$ of these primes chosen from the set $P_j$. Thus $a_1+\ldots+a_r=k$ and let's assume that all $a_j \ge 1$. I claim that for fixed $k$ and as $x\to \infty$ $$ \sum_{\substack {{n\le x}\\ {n\in N_k}}} 1 \sim k\prod_{j=1}^{r} \frac{\alpha_j^{a_j}}{(a_j)!} \frac{x}{(\log x)} (\log \log x)^{k-1} $$ The argument that follows is standard. The case $k=1$ follows from our assumption, and now suppose that $k\ge 2$. Put $z=x^{1/\log \log x}$ and note that by partial summation $$ \sum_{p\in P_j, p\le z} \frac{1}{p} \sim \alpha_j \log \log z \sim \alpha_j \log \log x $$ while $$ \sum_{p \in P_j, z< p \le x} \frac{1}{p} \sim \alpha_j \log \log \log x. $$ Now suppose $n=p_1\cdots p_k$ is an element of $N(k;a_1,\ldots,a_r)$. We distinguish two cases: when the largest prime factor of $n$ is bigger than $z$ but all the other prime factors are smaller than $z$, and when the two largest prime factors of $n$ are both larger than $z$. (Of course there is also the case when all prime factors of $n$ are smaller than $z$, but this is clearly negligible.) The first case will be the main term and the second case will be a slightly smaller error. Let's look at the first case. Suppose the largest prime factor (say $p$) of $n$ lies in the set $P_j$, so that the product of the smaller primes, say $m$, lies in $N(k-1;a_1,\ldots,a_{j}-1,\ldots,a_r)$. These terms give, upon summing over the last prime $p$, (note that as $m\le z^{k-1}$ we have $\log x \sim \log (x/m)$) $$ \sim \sum_{\substack{{m\in N(k-1;\ldots)} \\ {p|m\implies p\le z}} } \frac{\alpha_j x}{m \log (x/m)} \sim \frac{\alpha_j x}{\log x} \prod_{t=1, t\neq j}^{r} \frac{1}{a_t!} \Big(\sum_{p\le z, p\in P_t} \frac 1p\Big)^{a_t} \frac{1}{(a_j-1)!} \Big(\sum_{p\le z, p\in P_j}\frac 1p\Big)^{a_j-1}. $$ This is $$ \sim a_j \prod_{t=1}^{r} \frac{\alpha_t^{a_t}}{a_t!} \frac{x}{\log x} (\log \log x)^{k-1} $$ and summing it over $j$ from $1$ to $r$ gives the claimed asymptotic. Now for the second case. The sum over the largest prime $p$ gives a quantity bounded by (let $m$ be the product of the remaining primes and note that $m\le x^{(k-1)/k}$) $$ \frac{x/m}{\log (x/m)} \ll \frac{x}{m \log x}. $$ Now sum this over the possible values of $m$, keeping in mind that $m$ has $k-1$ prime factors, and the largest prime factor of $m$ is assumed to lie in $[z,x]$. Thus these terms are bounded by $$ \ll \frac{x}{\log x} \Big(\sum_{p\le x} \frac{1}{p} \Big)^{k-2} \Big(\sum_{z< p\le x} \frac{1}{p}\Big) \ll \frac{x}{(\log x)} (\log \log x)^{k-2} (\log \log \log x). $$ Application to the Chebotarev question. Suppose $C_1$, $\ldots$, $C_r$ are distinct conjugacy classes in $G$. Suppose each element of the set $D$ (which is a $k$-tuple) consists of $a_1$ entries from $C_1$, $a_2$ entries from $C_2$, $\ldots$, $a_r$ entries from $C_r$. A little combinatorics shows that the size of the set $D$ (assumed to be conjugation and permutation invariant) is $$ |D| = k!\prod_{j=1}^{r} \frac{|C_j|^{a_j}}{a_j!}. $$ In our work above take $P_j$ to be the primes $p$ for which Frobenius lies in the class $C_j$. Then by Chebotarev our assumed asymptotic for the primes in $P_j$ holds with $\alpha_j=|C_j|/|G|$. It follows that the number of integers in the original question is $$ \sim \frac{|D|}{|G|^k} \frac{x}{\log x} \frac{(\log \log x)^{k-1}}{(k-1)!}. $$ This answers the Chebotarev question (assuming I got my combinatorics right) at least when $k$ is fixed. When $k$ grows, one has to do more work, but the Selberg-Delange method should work in this case with some effort; note here that one needs to use the structure of sets of Chebotarev primes, rather than just an arbitrary subset of the primes as above.<|endoftext|> TITLE: Injectivity of matrix "fingerprint" QUESTION [7 upvotes]: Consider $S$, the set of all $n\times m$ real matrices with specified row sums $(r_1,...,r_n)$, column sums $(c_1,...,c_m)$, and strictly positive entries. For any matrix $A$, define $$ D_A(i,j)=\frac{A_{i,j}A_{i+1,j+1}}{A_{i+1,j}A_{i,j+1}} $$ Let $T:S\rightarrow (R^+)^{(n-1)\times (m-1)}$ be the map where the $(i,j)$-th entry of $T(A)$ is $D_A(i,j)$. Question: Given $T(A)$, can we recover $A\in S$ (i.e. is $T$ injective)? Motivation: This question arose in the context of a MathOverflow question about a matrix version of the log-likelihood ratio. Comment #1: For any $1\leq i \frac{x_{pq}x_{rs}}{x_{ps}x_{rq}}$$ which shows $T(X)\neq T(X')$.<|endoftext|> TITLE: 3-cocycle representatives for the dihedral group $D_{2n}$? QUESTION [12 upvotes]: I am looking for a reference for a complete list of 3-cocycle representatives for $H^3(D_{2n},\mathbb{C}^\times)$, where $$ D_{2n}=\langle a, b\mid a^2=b^2=(ab)^n=e\rangle $$ is the dihedral group of order $2n$. Here's what I believe so far, but I do not know much group cohomology. First, I believe there are isomorphisms $$ H^3(D_{2n},\mathbb{C}^\times)\cong H^4(D_{2n},\mathbb{Z})\cong H_3(D_{2n},\mathbb{Z}). $$ I calculated these groups in GAP for $2\leq n\leq 10$, and I got the following formula: $$ H_3(D_{2n},\mathbb{Z}) \cong \begin{cases} \mathbb{Z}/2 \oplus \mathbb{Z}/2 \oplus \mathbb{Z}/n & \text{if $n$ is even}\\ \mathbb{Z}/2 \oplus \mathbb{Z}/n & \text{if $n$ is odd.} \end{cases} $$ Are these formulas correct? Is there a reference for this, or does it follow by some spectral sequence calculation? Why is there an extra factor of $\mathbb{Z}/2$ when $n$ is even? Second, I believe I have a formula for 3-cocycles representing the $\mathbb{Z}/n$ factor, which is the factor I most care about. To give this formula, I use the following notation. First, represent the elements of $D_{2n}$ by the alternating words in $a,b$ with length at most $2n-1$ which always start with $a$. For such a word $w$, let $|w|$ be its length. If $\zeta_n=\exp(2\pi i / n)$, then for $k=0,\dots, n-1$, I believe the maps $$ \lambda_k(x,y,z) = \begin{cases} (\zeta_n^k)^{(-1)^{|x|} \lfloor (|x|+1)/2\rfloor } & \text{if $|y|$ is even and $|y|+|z|\geq 2n$}\\ (\zeta_n^k)^{(-1)^{|x|+1} \lfloor (|x|+1)/2\rfloor } & \text{if $|y|$ is odd and $|z|>|y|$}\\ 1 & \text{else} \end{cases} $$ give a complete list of normalized 3-cocycle representatives for the $\mathbb{Z}/n$ factor of $H^3(D_{2n},\mathbb{C}^\times)$. Certainly the $\lambda_k$ act like $\mathbb{Z}/n$ under multiplication, but is there an easy way to see they are not cohomologous? Is there an easier way to express these formulas, perhaps by choosing a different presentation of $D_{2n}$? Finally, the formulas above yield the trivial 3-cocycle when restricted to the two copies of $\mathbb{Z}/2$ generated by $a$ and $b$. To see this, just note $\lambda_k$ is normalized, $\lambda_k(a,a,a) = 1$, and if $w=(ab)^{n-1}a$, which is equal to $b$, then $\lambda_k(w,w,w) = 1$. (In both cases, $|y|$ is odd and $|y|=|z|$.) I believe any 3-cocycle representatives for the other cohomology classes should not restrict trivially to at least one of these copies of $\mathbb{Z}/2$. Partial answers would be greatly appreciated as well! REPLY [5 votes]: Pardon that this answer is very partial and incomplete, hopefully it provides some comments that may be useful to you. The cohomology groups can be computed (you have the correct result), following this Reference by Propitius: The explicit 3-cocycles for the odd Dihedral group is known and given in its Eq.(3.28): $$ {\omega\left( (A,a),(B,b),(C,c) \right) \;= }\\ \exp \left( \frac{2\pi i p}{(2N+1)^2} \{ (-)^{B+C} a \left( (-)^Cb+c-[(-)^Cb+c] \right) +\frac{(2N+1)^2}{2}ABC\} \right) . \;\;\; $$ Here is the notational conventions: The two generators $X$ and $R$ of the odd dihedral group $D_{2N+1}$ are subject to the conditions $$ R^{2N+1} \; = \; e, \qquad X^2 \; = \; e, \qquad XR \; = \; R^{-1} X, $$ with $e$ the unit element of $D_{2N+1}$. Label the elements of $D_{2N+1}$ by the 2-tuples $$ (A,a) := X^A R^a \qquad \qquad \mbox{with $A \in 0,1$ and $a \in -N, -N+1,\ldots,N$.} $$ the multiplication law defines as $$ (A,a) \cdot (B,b) = ([A+B] \text{mod}2,[(-1)^B a + b]\text{mod}(2N+1))\, , $$ The $\text{mod}(2N+1)$ produces values between $-N, -N+1,\ldots,N$.<|endoftext|> TITLE: Number of solutions of linear homogenous Diophantine equation inside a box QUESTION [9 upvotes]: Let $a_1, ..., a_d$ be positive reals and consider the linear Diophantine equation $$ \sum_i a_in_i = 0. $$ I am interested in estimating the number of integer solutions of this equation inside a box $[-N_1, N_1] \times ... \times [-N_d, N_d]$ and also the minimal basis for the corresponding integer lattice in terms of $a_1,...,a_d$. More specifically, it seems that if the number of solutions is of order $N^{d-1}$ then $(a_1, ...., a_d)$ should be proportional to some integer vector of bounded norm. Presumably, smth much more precise is known. I believe that such kind of questions are well studied so any good reading on this topic would be highly appreciated. REPLY [3 votes]: Siegel's Lemma is a famous solution to this type of problem with a system of linear forms, giving an estimate for the basis of solutions.<|endoftext|> TITLE: Surreal numbers, ultrapowers of $\Bbb R$, ordinal-valued functions and the slow-growing hierarchy QUESTION [11 upvotes]: Philip Ehrlich's paper “The Absolute Arithmetic Continuum and the Unification of All Numbers Great and Small”, The Bulletin of Symbolic Logic 18 (1) 2012, pp. 1-45. claims as a theorem that, in NBG, $\mathbf{No}$ is the unique class-sized field compatible with Alling's hyperreal axioms. I was inspired to experiment a bit with ultrapowers of $\Bbb R$ to see what sorts of connections might be lying between the surreals and hyperreals, and came across an interesting construction that I'd like to know if has been well-studied before. One can consider the usual hyperreals $^*\Bbb R$ as equivalence classes of infinite sequences $(n_1,n_2,n_3,n_4,...)$ of real numbers, where the equivalence class is given by a non-principal ultrafilter on $\Bbb N$. Each real embeds into the hyperreals via the diagonal embedding, so that we have $r \to (r,r,r,r,...)$. Addition and multiplication are done pointwise. Notationally, we will then define $\omega$ as the sequence corresponding to $a(n) = n$, aka $(1,2,3,4,5,...)$, noting that I'm leave out 0 in the set of naturals. We can use the transfer principle to extend any real function $f(r)$ to a hyperreal function $^*f(^*r) = (f(r_1), f(r_2), f(r_3), ...)$, where the $r_n$ are the entries in the tuple for $^*r$. So by the above, we can immediately get the following: $\omega+1 = (2,3,4,5,...)$ $\omega \cdot 2 = (2,4,6,8,...)$ $\omega^2 = (1,4,9,16,...)$ $\omega^\omega = (1,4,27,256...)$ Since this gives us all finite expressions of polynomials and power towers, this immediately leads to an order-preserving embedding of all ordinals less than $\epsilon_0 = \omega^{\omega^{\omega^{...}}}$ into $^*\Bbb R$, and if we consider the ordinals as being a semiring under commutative addition and multiplication, this embedding is a semiring homomorphism. We can go further: $\omega - 1 = (0,1,2,3,...)$ $\omega/2 = (1/2,1,3/2,2,...)$ $\sqrt{\omega} = (1,\sqrt{2},\sqrt{3},\sqrt{4},...)$ $\log{\omega} = (0,\log{2},\log{3},\log{4},)$, $-\omega = (-1,-2,-3,...)$ $\sin(\omega) = (\sin(1), \sin(2), \sin(3), \sin(4)...)$ $1/\omega = (1,1/2,1/3,1/4,...)$ $\text{round} \left(\frac{\omega}{2} \right) = (1,1,2,2,3,3,...)$ If we define an infinite summation of reals to be its sequence of partial sums, and likewise with infinite products, we get some more nice structure: $\sum_{1}^\infty 1 = \omega$, which is a countably infinite sum of 1's $\sum \Bbb N = 1+2+3+4+... = (1,3,6,10,15,...) = \frac{\omega(\omega+1)}{2}$, with the terms being summed in ascending order $\prod \Bbb N = 1\cdot2\cdot3\cdot4\cdot... = (1,2,6,24,...) = \omega!$, with the terms being multiplied in ascending order Most of these can be identified with surreal numbers, though $\omega!$ is definitely a new one. Also interesting is that whether or not $\text{round} \left(\frac{\omega}{2} \right) = \frac{\omega}{2}$ is dependent on whether the odd or even indices are in the ultrafilter, so the question of whether $\omega$ is "odd" or "even" is undecidable. Likewise, since it isn't really clear where $\sin(\omega)$ lies on the real line, this also seems very likely to be undecidable. If we want to be bold and define ordinal tetration, we can embed a few more ordinals as well. Let's identify $^\omega \omega$ with $\epsilon_0$. Then we get: $\omega^{\omega^{\omega^{...}}} =$$^{\omega} \omega = (1,$$^{2}2,$$^{3}3,$$^{4}4,...) = (1,2^2,3^{3^{3}},4^{4^{4^{4}}}...)$, corresponding to $a(n) =$ $ ^{n}n$ All of the ordinals so far, including this one, lead to the same functions given by the slow-growing hierarchy. That is, for every ordinal $\alpha$ that we've encountered so far, its embedding into the ordinals (as constructed by the definition of $\omega$ and the transfer principle) is more or less the same as the sequence in the slow-growing hierarchy $g_\alpha(n)$, with a few insignificant details like how I didn't include zero in the naturals. If there are canonical fundamental sequences for ordinals greater than $\omega^{\omega^{\omega^{\omega^{...}}}}$, then it seems plausible that they could also be recovered in this way too. So we've recovered all of this beautiful non-Archimedean structure just by defining $\omega = (1,2,3,...)$ and applying the transfer principle. My question, which is admittedly "big picture," is: has this sort of construction been studied before, either in non-standard analysis, or in surreal analysis, or in the study of growth rates of sequences and functions, or anywhere? What is going on, and why do these things sync up so nicely, and is there more information about this sort of connection between these different topics? It seems to touch on a lot of different things, but I'm having trouble finding a launching off point to see the big picture of all of this. Any more information would be much appreciated. REPLY [3 votes]: Alec, I'm afraid your construction does not work. You claim you obtain the surreals by forming the Dedekindean Completion of what you call the surrationals. However, since No--the surreals-- is not Dedekindean Complete, your claim is false. In fact, the Dedekindean Completion of No has "cardinality" $2^{\aleph_{On}}$, which is not even well-defined in NBG (with Global Choice). No is a (fully) saturated model for the theory of real-closed ordered fields and it has long been known that no such model is Dedekindean complete. On the other hand, there are real-closed fields that are $\kappa$-saturated but not (fully) saturated that are Dedekindean Complete. See, for example, H. J. Keisler and J. Schmerl Making the hyperreal line both saturated and complete. J. Symbolic Logic 56 (1991), no. 3, 1016–1025. REPLY [2 votes]: (beginning of post deleted due to a missed error, thanks to Philip Ehrlich) It is possible to extend the usual hyperoperation sequence $\mathcal{H}_\omega$ on $\mathbb{N}$ to a recursive sequence of operations $\mathcal{H}$ on $O_n$, with the following result: $$\mathcal{H}_{_\Omega}(\alpha,\beta)=\begin{cases} \mathcal{S}\alpha, & \text{if} \ \Omega=0. \\ \alpha, & \text{if} \ \Omega=1 \ \text{and} \ \beta=0. \\ \mathcal{S}\alpha,& \text{if} \ \Omega=1 \ \text{and} \ \beta=1. \\ 0, & \text{if} \ \Omega=2 \ \text{and} \ \beta=0. \\ \alpha, & \text{if} \ \Omega=2 \ \text{and} \ \beta=1. \\ 1, & \text{if} \ \Omega\geq3 \ \text{and} \ \beta=0. \\ \alpha, & \text{if} \ \Omega\geq3 \ \text{and} \ \beta=1. \\ \mathcal{H}_{_{\Omega-1}}\big(\alpha,\mathcal{H}_{_\Omega}(\alpha,\beta-1)\big), & \text{if} \ \Omega=\mathcal{S}\bigcup\Omega \ \text{and} \ 1<\beta=\mathcal{S}\bigcup\beta. \\ \bigcup_{\delta<\beta}\mathcal{H}_{_\Omega}(\alpha,\delta), & \text{if} \ \Omega=\mathcal{S}\bigcup\Omega \ \text{and} \ \ 1<\beta=\bigcup\beta . \\ \bigcup_{\rho<\Omega}\mathcal{H}_{_\rho}(\alpha,\beta), & \text{if} \ 0\neq\Omega=\bigcup\Omega \ \text{and} \ 1<\beta=\mathcal{S}\bigcup\beta. \\ \bigcup_{\rho<\Omega}\bigcup_{\delta<\beta}\mathcal{H}_{_\rho}(\alpha,\delta), & \text{if} \ 0\neq\Omega=\bigcup\Omega \ \text{and} \ 1<\beta=\bigcup\beta. \\ \end{cases}$$ This 'transfinite hyperoperation sequence' is a very 'large' object in the sense that each hyperoperation in the sequence is a proper class, but it is well defined under the appropriate reflection principle or under ETR and satisfies nice relations like $\mathcal{H}_4(\omega,\omega)=^\omega\omega=\omega^{\omega^{\omega^{\dots}}}$ in your question, and $\mathcal{H}_5(\omega,\omega)=^{^{^{\dots}\omega}\omega}\omega$, so on and so forth. I hope this assists with understanding what is happening with the hyperoperation-related piece of the questions you've mentioned, as I find them fascinating as well!<|endoftext|> TITLE: Classification of automorphism groups of groups of order $p^4$ QUESTION [10 upvotes]: For the purpose of classifying another algebraic structure which is parametrised by the choice of a group and of an automorphism I need the classification up to isomorphism of automorphism groups of p-groups of order $p^4$. I duly searched the web for a while and all the group theory manuals I could lay my hands on but I didn't find anything, not a hint. Could someone provide a reference? I already posted the same question on maths stack exchange and Professor Holt guessed that there is no publication on the subject. I repost here to verify his guess thoroughly. A good reference for the classification of groups of order $p^4$ would be useful too, since I found the following thesis On p-groups of low power order but a different reference to help me compare the possible approaches to the problem would be of great help! REPLY [4 votes]: There is a description of the automorphism groups of groups of order $p^4$ in the Thesis by B. Girnat: "Klassifikation der Gruppen bis zur Ordnung $p^5$." Staatsexamensarbeit, TU Braunschweig, 2003, whose advisor was B. Eick. The description is given in terms of the action on a set of generators of the respective group. If I may add a personal note, I have been searching for this information for a long time, and I found it incredible that such a basic and old topic had no treatment whatsoever published in any form on the web. But I was assuming that anything interesting can be found in English (either written or translated into it). This assumption has been proven wrong, and I state this here as a warning for the newbies like me!<|endoftext|> TITLE: A question about tiling Hilbert Space QUESTION [10 upvotes]: Let H be an infinite dimensional and separable Hilbert Space. Let e be a positive real number-which can be arbitrarily small. Does there exist a denumerably infinite set S of pairwise disjoint and pairwise congruent subsets of H such that (1) the union of S is H and (2) each set which is an element of S has a diameter not greater than e? Although the sets belonging to S must be bounded they do not have to be open or closed in H. Note that if H were a finite dimensional Euclidean space, my question would have an affirmative answer-since the sets belonging to S could be multi-dimensional semi-closed intervals.........The motivation for asking this question is to obtain a clearer understanding of the problems which arise when one attempts to define a measure for infinite dimensional spaces that is (to the greatest extent possible) a generalization or extension of n-dimensional Lebesgue measure. REPLY [10 votes]: Yes, there are such tilings, but I don't know of any "nice" ones. The example below makes use of the axiom of choice, so you can guess how horrible it is. :) Take any dense countable subgroup $J$ of $H$ - say, the subgroup of vectors with rational coordinates, such that finitely many of those are nonzero. Now for every coset $x + J$ there is a representative of norm $\le \varepsilon$. Choose such a representative from every coset, and let's call $A$ the set of these. Now $A$ is bounded by construction, $A + j, j \in J$ are disjoint, and their union is the whole $H$ (since every coset contains exactly one representative in $A$).<|endoftext|> TITLE: Does every commutative variety of algebras have a cogenerator? QUESTION [16 upvotes]: By a commutative variety $\mathcal{V}$ I mean a classical variety of algebras for some $(\Sigma,E)$, such that each pair of operations in $\Sigma$ commutes. Equivalently (i) every interpretation of every operation defines an algebra homomorphism, or (ii) the hom-sets have pointwise algebraic structure, or (iii) $\mathcal{V}$ forms a symmetric closed monoidal category with its tensor product (the latter being definable as a bi-functor in any variety). By a cogenerator I mean some $K \in \mathcal{V}$ such that any two distinct morphisms $\alpha,\beta : A \to B$ have a respective `predicate' $h : B \to K$ satisfying $h \circ \alpha \neq h \circ \beta$. I have two questions: Do commutative varieties necessarily have a cogenerator? Do locally finite commutative varieties necessarily have a finite cogenerator? I include some positive examples. If $(\Sigma,E)$ at least contains a binary operation and a respective unit, then $\mathcal{V}$ is essentially the modules for some commutative semiring. Examples are abelian groups with cogenerator $\mathbb{Q}/\mathbb{Z}$, vector spaces with cogenerator $k$, join-semilattices with zero with cogenerator $2$. Other examples with a two element cogenerator include sets, pointed sets, semilattices without zero, and the variety defined by a single operation and equation $u(x) = u(y)$ (pointed sets with an additional initial object). Actions of a commutative monoid have a cogenerator, since they form a topos $[M,\mathsf{Set}]$. If the monoid is finite, there is a finite cogenerator. Edit Let me add that: I do not know if commutative monoids have a cogenerator, or whether there is one which works for the finitely-generated = finitely-presentable $\mathbb{N}$-modules. The commutative inverse monoids define a commutative variety. They extend commutative monoids with a single involutive unary operation, which (i) preserves the monoid structure, (ii) satisfies $x = x \cdot u(x) \cdot x$. They have a cogenerator $2 \times \mathbb{Q}/\mathbb{Z}$ where $2$ is the two-chain. REPLY [12 votes]: The answer is no. Let $A$ be the algebra with universe $\{0,1\}$ and fundamental operations $f(x,y,z)=x+y+z \pmod{2}$ and $g(x)=x+1\pmod{2}$. Then $f$ and $g$ commute with each other and with themselves, so the variety generated by $A$ is commutative. This variety has a weird property: on every $B\in \mathcal V(A)$ the operation $g$ interprets either as a fixed point free involution (type 1) or as the identity function (type 2). Moreover, every group of exponent 2 can be modified slightly to make it an algebra of type i in this variety for i = 1 OR 2. Now, suppose that $K$ is a cogenerator for $\mathcal V(A)$. Necessarily $|K|>1$. $K$ is not of type 1. Assume otherwise. Let $B\in\mathcal V(A)$ be the 2-element type 2 algebra. There are maps $\alpha,\beta\colon B\to B$ where $\alpha = id$ and $\beta$ is a constant function. These maps cannot be separated by a map $h\colon B\to K$, since there is no homomorphism from a type 2 algebra to a type 1 algebra. $K$ is not of type 2. Assume otherwise. If $B\in\mathcal V(A)$ is of type 1, then $id, g\colon B\to B$ cannot be separated by any map $h\colon B\to K$ since you cannot separate elements of the same $g$-orbit of a type 1 algebra by a homomorphism into a type 2 algebra.<|endoftext|> TITLE: How hard is it to determine if a weighted graph can be isometrically embedded in R^3? QUESTION [5 upvotes]: Consider a graph $G$ with nonnegative edge weights. Question: In $\mathbb{R}^3$, how hard is it to assign coordinates to vertices such that the Euclidean length of each edge is equal to its weight? Question: Does it get any easier if $G$ is the 1-skeleton of a simplicial surface? (A similar question was already answered here, but an answer was given only for the special case of complete graphs.) REPLY [7 votes]: There are several possible ways to interpret your question. Let me mention three, all very different: (1) For general graphs, say you want to decide if there is such realization at all, and if yes find it approximately. This is called Graph Realization Problem and it is well studied both theoretically and practically (e.g. it is easily NP-hard). For general references see So's thesis, and for connection to graph colorings our paper below. A. So, A Semidenite Programming Approach to the Graph Realization Problem, Ph.D. thesis, Stanford, 2007. I. Pak and D. Vilenchik, Constructing uniquely realizable graphs, Discrete & Computational Geometry, vol. 50 (2013), 1051-1071. (2) For triangulated surfaces, say you want to approximately compute the convex realization (the decision problem is easy by the Alexandrov theorem). This is done by a recent analysis of the Bobenko-Izmestiev algorithm mentioned in the Igor Rivin's answer: D. Kane, G. Price, E. Demaine, A Pseudopolynomial Algorithm for Alexandrov’s Theorem, Algorithms and Data Structures, Lecture Notes in CS, Vol. 5664, 2009, pp 435-446. (3) For triangulated surfaces, say you want to compute the exact vertex coordinates. First, note that even in $\Bbb R^2$ the equilateral triangle cannot be realized using rational coordinates. What happens is that you are solving a large system of quadratic equations. The vertex coordinates are then solutions of equations of large degree with coefficients being polynomials in squared edge-lengths. In our paper with Fedorchuk we use a Bezout's theorem type argument to show that the upper bound on these degrees is $2^m$, where $m=3n-6$ is the number of edges of the triangulated surfaces with $n$ vertices. Note that for even the most trivial triangulated surfaces such as the cyclic polytope (think of a "snake of tetrahedra") there is a natural exponential lower bound $2^{n-4}$, see the paper. M. Fedorchuk and I. Pak, Rigidity and polynomial invariants of convex polytopes, Duke Math. J., vol. 129 (2005), 371-404.<|endoftext|> TITLE: Can the Veronese variety of degree k have k+1 linearly dependent points? QUESTION [5 upvotes]: For $k>2, d>2$, is it possible for the Veronese variety $v_k(\mathbb P^{d-1})$ to contain $k+1$ points that are not linearly independent? For $d = 2$, an answer in the negative is given by the Vandermonde determinant. Is a similar statement true in general? REPLY [9 votes]: Distinct points $x_1,\dots,x_{k+1}$ are always sent to linearly independent points in $v_k(\mathbb P^{d-1})$. Let $f_i$ be a linear form that vanishes at $x_i$ but not at $x_{k+1}$. Let $f$ be the product of all the $f_i$. Since $f$ is a form of degree $k$, it gives a linear form on the $k$th symmetric power which vanishes at the linear span of all $v_k(x_i)$ but not $v_k(x_{k+1})$.<|endoftext|> TITLE: Can $\mathbb{R}$ be partitioned into dedekind-finite sets? QUESTION [9 upvotes]: Assuming $ZF$ itself is consistent, it is consistent that there are sets $D$ which are infinite but cannot be placed in bijection with any of their proper subsets; such sets are called "strictly Dedekind-finite." Consistently, there is even a Dedekind-finite set of reals. My question is, is it consistent to be able to partition $\mathbb{R}$ into strictly Dedekind-finite sets? The simplest way to produce strictly Dedekind-finite sets of reals is to use Cohen forcing and take a symmetric submodel of the resulting forcing extension. We can also create $\kappa$-many disjoint strictly Dedekind-finite sets in a similar fashion, for any $\kappa$, without collapsing $\kappa$ (of course, the continuum is bumped above $\kappa$). However, as far as I can see, there is no simple way to adapt this to provide a partition of $\mathbb{R}$ into such sets. One simple thing I've been able to figure out: suppose $\kappa$ is a (well-ordered) cardinal, $\{D_i: i\in I\}$ is a partition of $\mathbb{R}$ into strictly Dedekind-finite sets, and $\kappa^+$ injects into $\mathbb{R}$. Then there cannot be an injection $I\rightarrow\kappa$. However, letting $\Psi$ be the least ordinal not injectible into $\mathbb{R}$, it is not even clear to me that we cannot partition $\mathbb{R}$ into $\Psi$-many strictly Dedekind-finite pieces; and of course this says nothing about the case when the index set $I$ itself is non-well-orderable. REPLY [15 votes]: YES WE CAN! Suppose that there is an infinite Dedekind-finite set of real numbers $A$ (e.g. Cohen's first model). Simple cardinal arithmetic shows that, $$|\Bbb R|\leq|\Bbb R\times A|\leq|\Bbb{R\times R}|=|\Bbb R|.$$ Clearly $\Bbb R\times A$ can be partitioned into infinite Dedekind-finite sets, simply consider $\{\{r\}\times A\mid r\in\Bbb R\}$. Now use a bijection of $\Bbb R\times A$ with $\Bbb R$ to transport this partition to a partition of $\Bbb R$ without changing the cardinality of its parts.<|endoftext|> TITLE: Weak forms of the Axiom of Choice QUESTION [11 upvotes]: Let $n\geq 2$ be a natural number and consider the following: $AC(n)$: For each family $\{X_i\}_{i \in I}$ of $n$-element sets the product $\prod_{i\in I}X_i$ is non-empty. Is it known that for which values of $m$ and $n$, $AC(m)$ and $AC(n)$ are equivalent !? REPLY [3 votes]: Isn't the point that this is really all about finite groups? Given a group $G$, the $G$-axiom of choice says that for any collection of sets with a simply transitive action of $G$, we may choose from each set a coset of a proper subgroup. Then $AC(n)$ is equivalent to the axiom of choice for all groups $G$ with a fixed-point-free action on $n$ objects. The axiom of choice for a subgroup or a quotient group of a group implies the axiom of choice for that group. I believe that all implications between different finite axioms of choice follow from these facts.<|endoftext|> TITLE: How come Cartan did not notice the close relationship between symmetric spaces and isoparametric hypersurfaces? QUESTION [18 upvotes]: Elie Cartan made fundamental contributions to the theory of Lie groups and their geometrical applications. Among those, we can list the introduction of the remarkable family of Riemannian symmetric spaces, and his four papers that revolutionized the theory of isoparametric hypersurfaces. How come Cartan did not notice the close relationship between symmetric spaces and isoparametric hypersurfaces? Starting in 1926, Cartan developed his theory of symmetric spaces and published papers between 1927 and 1935. He first introduced them as Riemannian manifolds with parallel curvature tensor, under the name "espaces $\mathcal E$". Then he noticed that an equivalent, more geometric definition is to require that the geodesic symmetric around any point is an isometry and, around 1929, changed their name to "espaces symetriques'". The second definition implies that a symmetric space is a homogeneous space $G/K$ and there is a decompositon $\mathfrak g=\mathfrak k+\mathfrak p$ into the eigenspaces of an involution induced by conjugation by the geodesic symmetry at the basepoint. The adjoint action of $K$ on $\mathfrak p$ is equivalent to the linear isotropy representation of the symmetric space. The rank of $G/K$ is the dimension of a maximal flat, and coincides with the codimension of the principal orbits of this representation. Isoparametric hypersurfaces in space forms are hypersurfaces with the simplest local invariants, namely, they have constant principal curvatures. They existed before Cartan, but between 1937 and 1940 he published four papers that completely revolutionized the field. Among other things, he showed that isoparametric hypersurfaces in spheres is a much more interesting subject than in Euclidean or hyperbolic spaces. Denote by $g$ the (constant) number of principal curvatures. The initial cases are not very interesting; in a sphere $S^{n+1}$, an isoparametric hypersurface with $g=1$ is an umbilic sphere, and with $g=2$ is the standard product of two spheres. Cartan showed that in case $g=3$ there are exactly four examples, of dimension $n=3d$ where $d=1$, $2$, $4$ or $8$ is the uniform multiplicity of the principal curvatures, each related to an embedding of a projective plane over one the normed division algebras $\mathbb R$, $\mathbb C$, $\mathbb H$, $\mathbb O$. He notes that those examples are all homogeneous and determines their isometry groups; in particular, he is pleased with the appearance of the exceptional group $F_4$ the case $n=24$, "(...) the first appearance of the simple $52$-dimensional group in a geometric problem (...)"; this group had already appeared in his classification of symmetric spaces. Later Cartan discusses the case $g=4$ and shows there are only two examples where the multiplicities of principal curvatures are all equal, namely, one in $S^5$ and one in $S^9$. Cartan ends his third paper on the subject (Sur quelques familles remarquables d'hypersurfaces, C. R. Congres Math. Liege (1939), 30-41. Also in: Oeuvres Completes, Partie I11, Vol. 2, 1481-1492.) with three questions, one of which asking whether there exist isoparametric hypersurfaces in spheres with $g>3$ such that multiplicities of principal curvatures are unequal. In 1971, Hsiang and Lawson published a paper (Minimal submanifolds of low cohomogeneity, J. Differential Geom. 5 (1971), 1-38.) including a classification of (maximal) subgroups of $SO(n+2)$ whose principal orbits have cohomogeneity $1$ in $S^{n+1}$, and remarked that they precisely coincide with the linear isotropy representations of symmetric spaces of rank two. In 1972, Takagi and Takahashi (On the principle curvatures of homogeneous hypersurfaces in a sphere, Differential Geometry, in Honor of K. Yano, Kinokuniya, Tokyo (1972), 469--481.) remarked that Hsiang-Lawson's result yields a classification of homogeneous isoparametric hypersurfaces in spheres and computed their invariants; in particular, they found examples with $g=4$ and unequal multiplicities. The relation is that the principal orbits of the linear isotropy representations of symmetric spaces rank two (resp. arbitrary rank) yield beautiful examples of isoparametric hypersurfaces (resp. submanifolds) in spheres. This relation is relatively easy to explain today. Cartan was the master of both subjects in the late 1930's. Is there anything interesting that can be said about the situation of differential geometry and Lie group theory at that time that prevented him to grasp this connection? REPLY [2 votes]: I think this question is a little subject to opinion, so hopefully it would be considered okay to share mine. At that time there were a lot of different approaches for how to fit Lie groups into a broader algebraic context. The end of the 1800s saw much debate over things like this (a nice book on this by Crowe: https://www.researchgate.net/publication/244957729_A_History_of_Vector_Analysis). I would say it took at least 40 years to clean things up into a more streamlined system. And some elements of it still merit looking back at the original sources! (I've gotten some interesting results of my own by updating ideas from the early 1900s into modern methodology.) Sometimes just comparing one person's notation and terminology with another's brings out enlightening relationships that, in hindsight, seem obvious. But carrying out those comparisons is non-trivial and often takes a long time to occur.<|endoftext|> TITLE: Iteration of a 2D map involving absolute value: phase transition? QUESTION [7 upvotes]: I was looking at this map: $f(x,y) \mapsto (|x-y|,x)$, starting from some point with coordinates $(x,y) \in [0,1]^2$, and iterating: $(x,y),\, f(x,y), \, f^2(x,y), \,f^3(x,y), \ldots$. It displays behavior surprising to me, with something akin to a phase transition when $x \approx y$, when it then burrows toward $(0,0)$ in a methodical fashion:       Two questions: Q1. Is there some sense in which there are two distinct regimes of behavior to this map? Q2. What tools are available to study maps like this? Incidentally, I started with $f(x,y) \mapsto (x-y,x)$, which results in the hexagon $$ (x,y), \, (x-y,x), \, (-y,x-y), \, (-x,-y), \, (-x+y,-x), \, (y,-x+y) \;. $$ REPLY [13 votes]: Let $(x_n,y_n)$ denote the coordinates of the $n$-th iterate of the map. Put $t=y/x$ and $t_n=y_n/x_n$. Note that $t_{n+1}=1/|t_n-1|$. We claim that the sequence $(x_n,y_n)$ tends to $(0,0)$ if and only if $t$ is irrational. Note that if $t$ is irrational, then so are all the $t_n$; similarly if $t$ is rational then so are all the $t_n$ (but this sequence may include $\infty$ which we consider rational here). Suppose first that $t$ is rational. Write $t_n = a_n/b_n$ in reduced terms. Then we may see that $\max(a_n,b_n)$ is a non-increasing sequence of natural numbers, and must decrease at some point if $\max(a_n,b_n) \ge 2$. Thus eventually we will arrive at a situation where $\max(a_n,b_n)=1$. This means that we are in the cases where either $x_n$ or $y_n$ equals zero, or when $x_n=y_n$. In all these cases the map cycles through three possibilities forever. Now suppose that $t$ is irrational. Note that $\max(x_n,y_n)$ is a non-increasing sequence for all $n$; let's call this quantity the size of the point $(x_n,y_n)$. So it suffices to show that every once in a while the size of a point has become substantially smaller than a previous value. Let's consider some point $(x_n,y_n)$. If $t_n <1$ then note that at the next step we'll have $t_{n+1}>1$. So we may suppose that $t_n >1$. Now suppose first that $t_n>2$. Note that $(x_{n+1},y_{n+1}) = (x_n(t_n-1),x_n)$, $(x_{n+2},y_{n+2})= (x_n(t_n-2), x_n(t_n-1))$, and $(x_{n+3},y_{n+3})=(x_{n},x_n(t_n-2))$. So in three steps the size has diminished from $t_n x_n$ to $x_n \max(1,t_n-2)$. Repeating this $[t_n/2]$ times, we see that the size diminishes by a factor of $t_n$. Now suppose $2> t_n > 1.1$. Then after one step the size has diminished by a factor of $1/1.1$. Finally suppose that $1.1 > t_n>1$. After one step we get a value of $t_{n+1}$ which is bigger than $10$. We can now apply our previous argument to that case. Thus in all cases, after some number of steps the size of a point will be smaller by a factor $< 1/1.1$, and thus the points must tend to $(0,0)$. Note that if $t_n$ is large, then the size of iterates diminishes linearly for a while; whereas if $t_n$ is small then the size diminishes geometrically for a while. This may explain the ``phase transitions" seen in the pictures.<|endoftext|> TITLE: When do iterated conditional expectations converge? QUESTION [8 upvotes]: Take a probability space $(\Omega,\mathcal{F},\mathbf{P})$ and random variable $X$ satisfying $\mathbf{E}[|X|]<\infty$. Define the iterated expectations of X as follows: $X_0 = X$, and, inductively, $X_k = \mathbf{E}[X_{k-1} \, | \, \mathcal{G}_k]$, where $\mathcal{G}_k \subseteq \mathcal{F}$ is some $\sigma$-algebra. Assume that all the $\mathcal{G}_k$ come from some finite set of $\sigma$-algebras. In words, we are repeatedly taking conditional expectations with respect to various information. Can we conclude that the sequence $(X_k)_k$ converges $\mathbf{P}$-almost surely? In my comment on the question I include some of what I know about this, but I suspect that this is easy if one looks at it right, or covered in a standard reference. REPLY [5 votes]: Here's what's known, courtesy of Omer Tamuz. Here's what's known, with references below. Amemia and Ando (1965) prove weak convergence in $L^2$. This also covers the finite-$\Omega$ case. Convergence in norm in $L^2$ seems to still be open. It is known to converge if your sequence of algebras is periodic, as is shown by Delyon and Delyon (1999). If you more generally allow projections to a closed subspaces of $L^2$, then there is a recent example by Kopecká, E. & Paszkiewicz (2017) where no convergence in norm occurs. But not every projection to a subspace corresponds to conditional expectation according to some sub-sigma-algebra. Amemiya and T. Ando, Convergence of random products of contractions in Hilbert space, Acta Sci. Math. (Szeged) 26 (1965), 239–244. B. Delyon and F. Delyon, Generalization of von Neumann’s spectral sets and integral representation of operators, Bull. Soc. Math. Fr. 127 (1999), 25–41. Kopecká, E. & Paszkiewicz, A, Strange products of projections, Israel Journal of Mathematics, April 2017, Volume 219, Issue 1, pp 271–286<|endoftext|> TITLE: Consistency of P1 on Kunen QUESTION [8 upvotes]: It's the first time I'm posting here so I don't know if I really should put this question here... I tried to post it on math.stackexchange, but a friend told me I would get better results by posting it here. So I decided to give it a try. The original posting is here. https://math.stackexchange.com/questions/668929/consistency-of-p-1-on-kunen [I have deleted it, it had no answers] I'm finally finished studying Kunen's "Introduction to Independence proofs" but I'm having trouble on understanding something in the last theorem of the book (!!!). I think that in order to show that $2^{\omega_1}\leq \kappa$ in $M[G]$ I must show that $|\mathbb P_\xi| < \kappa$ if $\xi<\kappa$ by induction (as in the proof of MA), but I can't. Can someone give me some advice? In $M$, assume CH, and assume that $\kappa>\omega_1$, $\kappa$ is regular and $2^{<\kappa}=\kappa$; then there is a $\mathbb P \in M$ such that $\mathbb P$ is $\omega_1$-closed and $\omega_2$-c.c. in $M$, and whenever $G$ is $\mathbb P$-generic over $M$, $2^{\omega_1}=\kappa$ and $P_1$ hold in $M[G]$. $P_1$ is the statement that whenever $\mathscr A \subset \mathscr P(\omega_1)$, $|\mathscr A|<2^{\omega_1}$ and $\forall F \subset \mathscr A(|F|<\omega_1 \rightarrow |\omega_1 \setminus \bigcup F|=\omega_1$ there exists $d \subset \omega_1$ such that $|d|=\omega_1$ and $\forall x \in \mathscr A(|x \cap d|<\omega_1)$. What I have done so far: Let $f \in M$ such that $f: \kappa \rightarrow \kappa \times \kappa$ is onto and $\forall \xi, \eta, \gamma<\kappa(f(\xi)=\langle \eta, \gamma\rangle \rightarrow \eta \leq \xi)$. We shall build within $M$ an iterated forcing construction of the form $\langle \langle \mathbb P_\xi: \xi \leq \kappa\rangle, \langle \pi_\xi: \xi < \kappa\rangle\rangle$ such that for all $\xi$, $\pi_\xi$ is the standard $\mathbb P_\xi$-name for $\mathbb Q_\tau$ for a $\tau$ such that $|\tau|\leq \kappa$ and $1\Vdash(\tau \subset \mathscr P(\check {\omega_1}))$. This will imply that for each $\xi\leq\kappa$, $|\mathbb P_\xi|\leq \kappa$. By induction: Step 0 is trivial. For the induction step, $\pi_\xi=\{\langle op(\check s, \sigma), 1\rangle: s \subset \omega_1 \wedge |s|\leq \omega_1 \wedge 1 \Vdash(\sigma \subset \tau \wedge |\sigma|<\omega_1) \wedge \sigma$ is a $\mathbb P_\xi$ nice name for a subset of $\tau\}$. By CH within $M$, we have $\omega_1^\omega=\omega_1$ possibilities for $s$. Within M, $\kappa^{\omega_2}=\kappa$. By the previous theorems of the book, all the $\mathbb P_\xi$ will have $\omega_2$-c.c., therefore we will have $\kappa^{\omega_2}=\kappa$ antichains and since $|\tau|\leq \kappa$ we will have we will have at most $\kappa^\kappa$ nice names for subsets of $\tau$. Well, here is where I am stuck. I'm not able to conclude that $|\pi_\xi| \leq \kappa$. IF I had $|\tau|<\kappa$ I would be able to conclude that, since $\kappa^\lambda=\kappa$ if $\lambda<\kappa$. Supposing $|\mathbb P_\xi|<\kappa$ whenever $\xi<\kappa$ makes no difference at this point. So I tried to workaround supposing that $|\tau|<\kappa$. Then the argument works and the limit step follows easy. So let's try to continue the construction. Suppose we have already constructed $\mathbb P_\xi$ as above. Let $\langle \sigma_\gamma^\xi: \gamma < \kappa\rangle$ enumerate all $\mathbb P_\xi$-names $\sigma$ such that for some $\lambda<\kappa$, $\sigma$ is a nice name for a subset of $(\lambda \times \omega_1\check)$ and let $\tau^\xi_\gamma$ be such that $$1 \Vdash \tau^\xi_\gamma=\{x \subset \omega_1: \exists \mu < \check \kappa(x=\{v:\langle \mu, \nu\rangle \in \sigma^\xi_\gamma)\}$$ Now I need to find a $\tau^\xi_\gamma$ such that $|\tau^\xi_\gamma|<\kappa$. Therefore I can't use the maximal principle. I tried something like $$\tau^\xi_\gamma=\{\langle \sigma, p\rangle: \sigma \text{ is a nice name for a subset of }\check \omega_1 \wedge p \Vdash \exists \mu < \check \kappa(\sigma=\{v: \langle\mu, \nu\rangle \in \sigma^\xi_\gamma)\}$$ So, for this having chance to work, I have to assume in my previous induction that $|\mathbb P_\xi|<\kappa$. But I can't guarantee that there are less than $\kappa$ such nice names. So I'm really stuck again. REPLY [4 votes]: Note that in fact it suffices to consider those $\tau$'s which have size $<\kappa.$ The reason is simply as follows: Suppose $M[G]$ is the final extension and we are going to show that $P_1$ holds in it. So we consider some $A\subset P(\omega_1)$ of size $<\kappa.$ As the forcing has $\omega_2-$c.c., $A$ has a name of size $<\kappa,$ and in fact by our book-keeping function it is some $\tau$ which is appeared during the iteration. So from the beginning we can just consider those $\tau$'s which have size $<\kappa,$ and if some $\tau$ has size $\kappa,$ we can do the trivial iteration in that step.<|endoftext|> TITLE: Non-vanishing $\mathrm{lim}^1$-term for the cohomology of a CW-complex QUESTION [7 upvotes]: Let $h$ be an additive cohomology theory. If we want to compute $h^*(X)$ for an infinite CW-complex $X$, a standard method is to use the Milnor sequence $$ 0 \to \mathrm{lim}^1_k h^{n-1}(X^{(k)}) \to h^n(X) \to \mathrm{lim}_k h^n(X^{(k)}) \to 0, $$ where $X^{(k)}$ is the $k$-skeleton of $X$. If $h$ is singular cohomology, then the $\mathrm{lim}^1$-term always vanishes. What are examples of CW-complexes $X$ and cohomology theories $h$, where this $\mathrm{lim}^1$-term does not vanish? REPLY [13 votes]: For our CW-complex I'm going to take $X = \Bbb{CP}^\infty$ (as a based space), whose skeleta are $\Bbb{CP}^n$. The cohomology theory will be more difficult to construct. For any $k \geq 1$, let $E_k$ be the spectrum which is the homotopy fiber of the map $$ Sq^{2^k} \cdots Sq^8 Sq^4 Sq^2: \Sigma^2 H\Bbb{Z}/2 \to \Sigma^{2^{k+1}} H\Bbb{Z}/2 $$ so that, for any space $Y$, we have a long exact sequence in cohomology with at least the terms $$ \cdots \to H^1(Y) \to H^{2^{k+1}-1}(Y) \to (E_k)^0(Y) \to H^2(Y) \stackrel{Sq^{2^k} \cdots Sq^2}{\longrightarrow} H^{2^{k+1}}(Y) \to \cdots $$ (The cohomology operation is actually taking the $(2^k)$'th power of the elements in degree 2.) In particular, we find that: $$ (E_k)^0(\Bbb{CP}^n) = \begin{cases} \Bbb{Z}/2 &\text{if }n < 2^{k}\\ 0 &\text{if }n \geq 2^{k}\\ \end{cases} $$ Now I'll put all these cohomology theories together by taking $E = \bigvee E_k$, so that $E^0(Y) = \bigoplus (E_k)^0(Y)$ for $Y$ a finite CW-complex. We then find that $$ E^0(\Bbb{CP}^n) = \bigoplus_{n < 2^k} \Bbb{Z}/2. $$ This is a decreasing family of abelian groups with a nonzero $lim^1$-term. To prove that, you can use the short exact sequence of inverse systems $$ 0 \to \bigoplus_{n < 2^k} \Bbb{Z}/2 \to \bigoplus_{k \geq 1} \Bbb{Z}/2 \to \bigoplus_{n \geq 2^k} \Bbb{Z}/2 \to 0, $$ which gives a nontrivial $lim^1$-sequence $$ 0 \to 0 \to \bigoplus_{k \geq 1} \Bbb{Z}/2 \to \prod_{k \geq 1} \Bbb{Z}/2 \to lim^1 \to 0. $$ This illustrates the basic issue: we can explicitly build our cohomology theory so that no cohomology classes survive to the whole of $\Bbb{CP}^\infty$, but it can take arbitrarily long to figure it out. You can imagine how to build more general examples than this, but this one is convenient to calculate.<|endoftext|> TITLE: Candidates for non-sofic groups QUESTION [11 upvotes]: What are the "simplest" examples of countable groups that are not known to be sofic? REPLY [10 votes]: The simplest candidate I know of is Higman's group $\langle a,b,c,d\mid a^b=a^2, b^c=b^2, c^d=c^2, d^a=d^2\rangle$ (where, as usual, $a^b$ means $b^{-1}ab$). Terry Tao wrote a nice blog post about it here. Residually finite and amenable groups are both known to be sofic. As Tao explains, Higman's group has no finite quotients (hence is 'highly non-residually finite') and a non-abelian free subgroup (hence is certainly non-amenable).<|endoftext|> TITLE: The biggest class of schemes which the reduction principle holds QUESTION [5 upvotes]: In A. Bondal, M. van den Bergh's paper, Generators and representability of functors in commutative and noncommutative geometry , the "reduction principle" of quasi-compact, quasi-separated schemes is shown. It is stated as follows. Assume $X = U_1 \cup U_2 $ with $U_1, U_2$ open and put $U_{12} = U_1 \cap U_2$. Let $P$ be a property satisfied by some schemes such that (1) $P$ is true for affine schemes. (2) If $P$ holds for $U_1, U_2, U_{12}$ as above, then it holds for $X$. Then $P$ holds for all quasi-compact quasi-separated schemes. I understood this statement. However, the next Remark3.3.2 says It is easy to see that the class of quasi-compact quasi-separated schemes is the biggest class of schemes to which the reduction principle is applicable (for all properties $P$). I'm not sure why this characterization of quasi-compact, quasi-separated schemes holds. Is there an obvious property $P$ which satisfies above properties and holds only for quasi-compact, quasi-separated schemes? REPLY [8 votes]: Yes, such a property $P$ exists, namely "being qcqs" (i.e., quasi-compact and quasi-separated). It is trivial that qcqs answers your question, if it indeed satisfies your conditions on $P$. Let's check the conditions: $P$ is true for affine schemes. http://stacks.math.columbia.edu/tag/01S7 If $P$ holds for $U_{1}$, $U_{2}$, $U_{12}$ as above, then it holds for $X$. The qc part is (easy) topology. For qs one can use http://stacks.math.columbia.edu/tag/01KO, and see that #3 is satisfied for $X$. Indeed, take $S = \mathrm{Spec}(\mathbb{Z})$, and cover it with the trivial covering $\mathrm{Spec}(\mathbb{Z})$. The rest of the lemma says that we have to cover $X$ with affine opens $V_{j}$: well, we may (and do) choose the $V_{j}$ to all be in $U_{1}$ or $U_{2}$. All we need to check is that $V_{j} \cap V_{j'}$ is covered by a finite number of affine open subsets of $X$. Well, take any such cover $W_{k}$. Suppose $V_{j} \subset U_{i}$, and $V_{j'} \subset U_{i'}$. We have $$ V_{j} \cap V_{j'} = (V_{j} \cap U_{i}) \cap (U_{i'} \cap V_{j'}) = V_{j} \cap (U_{i} \cap U_{i'}) \cap V_{j'}. $$ On the right hand side everything is qcqs (either because affine, or by assumption) [edit:] and all intersections take place in qcqs schemes (namely $U_{i}$ and $U_{i'}$) [/edit]. Hence so is the intersection, and therefore $V_{j} \cap V_{j'}$ is qcqs (in particular qc). This allows us to take a finite subcover of $W_{k}$. Edit: The last paragraph was not very well-phrased. The point is that a priori $V_{j} \cap V_{j'}$ need not be qcqs (even though both $V_{j}$ and $V_{j'}$ are qcqs). After all, the intersection takes place in $X$, and we do not know that $X$ is qcqs. (Indeed, it is what we are trying to prove.) However, we know that $V_{j} \subset U_{i}$, and $V_{j'} \subset U_{i'}$, and this makes it possible to rewrite the intersection into intersections taking place in qcqs schemes. And of those, we know that the intersection is qcqs: item (6) of http://stacks.math.columbia.edu/tag/01KU.<|endoftext|> TITLE: Minimal number of intersection of curves in $\mathbb P^2$ QUESTION [9 upvotes]: Let $C_1$ and $C_2$ be two smooth curves of degrees $m$ and $n$ in $\mathbb CP^2$. By Bezout's theorem the maximal number of their intersections is $mn$. I wonder if the minimal possible number is known for all pairs $(m,n)$? In particular for which pairs $(m,n)$ there can be exactly one point of intersection? REPLY [16 votes]: For all pairs. Suppose $n\geq m$. Take for $C_1$ a curve with an inflection point of order $m$, say $F=0$ with $F(X,Y,Z)=ZY^{m-1}+X^m+Z^m$. Then take $C_2$ defined by $G(X,Y,Z)F(X,Y,Z)+Z^n=0$, where $G$ is general of degree $n-m$. Then $C_1\cap C_2$ is reduced to the point $p:=(0,1,0) $. To make sure that $C_2$ is smooth, we apply Bertini's theorem to the linear system of curves given by $G(X,Y,Z)F(X,Y,Z)+\lambda Z^n=0$. The only base point is $p$, and if $G(p)\neq 0$ the curve is smooth at $p$, so we are done.<|endoftext|> TITLE: Banach-Mazur distance to complex $\ell^1$ of a space containing real $\ell^1$ QUESTION [5 upvotes]: Consider a complex Banach space $X$ with a real subspace isometric to $\ell^1_{\mathbb R}$. What is the best constant $c$ such that $X$ contains a complex subspace $c$-isometric to $\ell^1_{\mathbb C}$? I guess this is a very classical question, but I could not find an answer. In this question, $\ell^1_{\mathbb R}$ (respectively $\ell^1_{\mathbb C}$) is the real (resp. complex) Banach space of summable sequences of real (resp. complex) numbers. And two Banach spaces $Y,Z$ are called $c$-isometric if there is a linear isomorphism $T:Y\to Z$ such that $\|T\| \|T^{-1}\| \leq c$. The motivation is that I want to understand the complex Banach spaces of trivial (Rademacher) type. A famous theorem of Maurey and Pisier, usually stated for real Banach spaces, shows that $\ell^1_{\mathbb R}$ is finitely representable in any (real) space of trivial type $\mathbb R$. My question is essentially equivalent to "what happens when the field of scalars is $\mathbb C$?". REPLY [6 votes]: Schechtman and I discussed your question this morning and have these comments. You can get $1+\epsilon$. The usual argument for improving the constant works in the complex case as well as the real case; i.e., if a complex Banach space contains a subspace complex isomorphic to $\ell_1$, then for all $\epsilon >0$ it contains a subspace $1+\epsilon$-complex isomorphic to $\ell_1$. To see that a complex Banach space that contains a real subspace isomorphic to $\ell_1$ also contains a complex subspace that is complex isomorphic to $\ell_1$, apply Dor's extension to the complex case of Rosenthal's $\ell_1$ theorem. The full theorem says that a (real or complex) Banach space contains an isomorphic copy of $\ell_1$ iff the space contains a bounded sequence that has no weakly Cauchy subsequence. For other values of $p$ the situation is different. By the Odell-Schlumprecht distortion theorem, for every $M$ there is a an equivalent renorming of $\ell_p$, $1 TITLE: Uniform-in-p classification* of p-groups of order p^n for each fixed n? QUESTION [8 upvotes]: To what extent is there/can there be a description that is uniform in p (for p sufficiently large) of the p-groups of order $p^n$, for each fixed n? Note 1: I used the word "description" rather than "classification" because I understand that classifying p-groups is notoriously open and difficult. The word "description" is meant to be headed in the general direction of "classification," but somewhat more generous and general. Note 2: I'm asking more about the possibility of such a uniform-in-$p$ description, and what it could/must/can't look like, rather than just asking about what kind of classification results are currently known or the difficulty of such a classification. Note 3 (added Feb 10, 2014 2:40pm EST): I changed it from "independent in $p$" to "uniform in $p$" because I think that better captures my intention. I don't mean completely independent of (large) p. As in example (2) and Derek Holt's answer, if there is some description of, say, groups of order $p^{12}$ that depends on, say, the residue class of $p$ mod 20 (a fixed number), and on the number of solutions mod $p$ of a fixed (set of) equation(s) over $\mathbb{Z}$, then I'd be happy with that. The hope is sort of that there should be a finite description, uniform in $p$; that one doesn't need a completely new description for infinitely many values of $p$. Motivating Examples 1) One of the examples which motivated this question: there is a classification of p-groups of order $p^4$ (going back to Burnside, I believe) for p=2, p=3, and then - crucially for this question - all $p \geq 5$. 2) Another infinite family of partial examples (partial because it only concerns some and not all groups of order $p^n$ for some fixed $n$): isomorphism classes of p-groups of class 2, exponent p, with $G/[G,G]$ of rank $t$ and $[G,G]$ of rank $z$ are in bijective correspondence with orbits of $z$-dimensional spaces of antisymmetric $t \times t$ matrices under change of basis on $G/[G,G]$. I understand that any description of the orbits of the latter action may vary with p (see this related question). But I think I nonetheless want to count this description as "independent of p", since the change-of-basis group and the action can all be defined over $\mathbb{Z}$ and then just taken mod p to get the picture for a given p (if I understand correctly). 3) Added Feb 10, 2:40pm EST: Derek Holt pointed out examples of du Sautoy and Vaughan-Lee suggesting that a classification of groups of order $p^{10}$ should be incredibly difficult. Although I find the difficulty interesting, in this question I'm more interested in the logical/model-theoretic/algebro-geometric/what-have-you possibility of the existence of such a description. For the particular examples in Derek Holt's answer, they give a classification of some groups of order $p^{10}$ that depend only on the value of $p$ mod 12 and on the number of solutions mod $p$ of the pair of integer equations $x^4 + 6x^2 - 3 = y^2 - x^3 + x = 0$. Although this number can vary a lot with $p$, this is still a single, finite description that is uniform in $p$ (and a very interesting example!). REPLY [12 votes]: I believe that it is extremely unlikely that there could be such a description. In fact, groups of order $p^n$ for all primes $p$ have been fully classified for $n \le 7$. See here for summary information. Note, in particular, that the number of isomorphism classes increases with $p$ when $n \ge 5$. The Higman PORC conjecture (polynomial on residue classes) is that, for each $n$, this number is a polynomial function of $p$ and of $p \pmod k$ for a finite collection of values of $k$. But a family of examples constructed by Vaughan-Lee and du Sautoy of order $p^{10}$ (see here for the full paper, which is $76$ pages long), while not actually disproving the conjecture, suggests that it is very unlikely indeed that it is true. Just looking quickly at that construction, which depends on the geometry of elliptic curves, will probably help you to understand just how complicated a uniform description of all groups of order $p^{10}$ would be. REPLY [9 votes]: For each $n$, the groups of order $p^n$ with $p\geq n$ are regular (a notion defined by Philip Hall). There are many equivalent ways to define regular $p$-groups; one is that if $a$ and $b$ are any elements of the group, then $$(ab)^{p^r} = a^{p^r}b^{p^r}c_1^{p^r}\cdots c_t^{p^r},$$ where $c_i$ are elements of the commutator subgroup of $\langle a,b\rangle$; in essence, a group is regular if the operation of taking $p$th powers interacts "well" with taking commutators. Hall showed that in a regular $p$-group, one can define "type invariants" which are similar to the invariant factors for finite abelian groups. Though they do not completely determine the groups the way the invariant factors do for abelian groups, they are usually a very good first reduction towards the analysis. The key observation and why I'm mentioning this in response to your query is that if $p\geq n$, then a group of order $p^n$ is necessarily regular (more generally, if the group is of class $c$ and $p>c$, then the group is regular; in particular, since a group of order $p^n$ is of class at most $n-1$, the observation just made follows); in fact, Hall says in his original paper (Philip Hall, A contribution to the theory of groups of prime power order, Proc. London Math. Soc. (series 2) 36, no. 1 (1934), pp. 29-95 ), if we fix $n$, then "most" groups of order $p^n$ are regular (since only those with $p\lt n$ may fail to be regular). This leads, classically, to a separation of $p$-groups into those of "small class" (when the class is smaller than $p$), and "the rest". This also means that when classifying groups of order $p^n$, the analysis usually breaks into two different cases: when the group is regular (which includes all $p\geq n$), and when the group is irregular; the latter case leads to a case-by-case analysis for small primes. This occurs in the classification of groups of order $p^3$, where we deal with odd primes on the one hand, and then deal with the groups of order $8$ separately. Likewise Burnside's work on group of order $p^4$. The same again occurs in the classification of groups of order $p^6$ (James, Rodney; The groups of order $p^6$ ($p$ an odd prime), Mathematics of Computation 34 no. 150 (1980), pp. 613-637), and $p^7$ (E. O'Brien, M.R. Vaughan-Lee, The groups with order $p^7$ for odd prime $p$, J. Algebra 292 (2005), 243-258); the latter deals with $p\geq 7$ uniformly, and then separately and directly with the groups of order $3^7$ and $5^7$. (The groups of order $2^7$ are dealt with in a completely separate work with different methods, $2$ being the oddest prime of all as usual...) The latest work uses Lie rings and algebras as a starting point. There are algorithms that are known to produce and check isomorphism types (see the paper by O'Brien and Vaughan-Lee for references). The older classification ($p^6$) was based on Philip Hall's notion of isoclinism as a first, rough classification scheme. In both cases, though, the classification tends to look like a long list of group presentations. It was hoped, via Higman's PORC conjecture (I see this has just been noted by Derek Holt in his response) that at least a uniform count might exist for all sufficiently large primes; some more recent evidence suggests it might not be quite that simple.<|endoftext|> TITLE: Computational approach deciding whether a set of Wang Tile could tile the space up to some size QUESTION [5 upvotes]: As an applied person, I'm facing one practical problem deciding whether a set of Wang tile could tile the plane periodically or aperiodically. Although both problems seem undecidable, but I'm on a more practical aspect. Say, if the program accidentally ("or systematically") find some "periodic structure", then it stops and tells me there exists periodic pattern. If during running, it enumerates all the use of tile and finds that it simply cannot tile the plane, then it tell me this set of tiles cannot tile the plane. Even if the program didn't stop, then after running some steps, it returns me a few most ordered patterns that that could "possibly tile the plane". For practical purpose, I simply assume if the tessellation are up to some size (maybe 1000*1000) then I say "it could tile the plane practically". So my most interested question is: is there any established programs or algorithms that "try" to help me analyze a set of tile even if it might not halt ("but I could define some imposed halting condition"). For context why I am interested in this problem, here's the links: coloring in lattice Reference for Wang Tile Periodic Tiling of Wang tile Cross-posted to: https://cs.stackexchange.com/questions/21502/computational-approach-deciding-whether-a-set-of-wang-tile-could-tile-the-space REPLY [4 votes]: There are a few theory papers tackling issue involved in Wang Tile: http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.216.7285&rep=rep1&type=pdf Also recent development considering finite cases of Wang tile is considered here. http://front.math.ucdavis.edu/1305.2796 http://front.math.ucdavis.edu/1212.3380 There are not yet exact answers to this problem as far as I know, but hope this helps....<|endoftext|> TITLE: Can we decompose a polynomial into difference of convex polynomials? QUESTION [6 upvotes]: Given a multivariate polynomial $p(x_1, ..., x_n)$ on $\mathbb{R}^n$, can we always decompose it into the difference of two convex polynomials? i.e., is there a pair of convex polynomials $f$ and $g$, such that: $\forall (x_1, ..., x_n) \in \mathbb{R}^n$, $p(x_1, ..., x_n) = f(x_1, ..., x_n) - g(x_1, ..., x_n)$? If given a bounded region $\mathbb{N}$ which is a convex subset of $\mathbb{R}^n$, can we then decompose polynomial $p(x_1, ..., x_n)$ such that in $\forall (x_1, ..., x_n) \in \mathbb{N}$ , $p(x_1, ..., x_n) = f(x_1, ..., x_n) - g(x_1, ..., x_n)$ where $f$ and $g$ are convex polynomials. If we can (no matter under the unbounded or bounded case), is there any constructive method to find such a pair of $f$ and $g$ for a given $p$? REPLY [6 votes]: Polynomials of degree $n$ which are convex in the given domain form a cone (subset which is closed under "+" and multiplication on nonnegative numbers) in the finite-dimensional space of polynomials of degree $n$. Lemma For a given cone $C$ in a vector space $V$ the following statements are equivalent (a) $C$ contains an inner point (b) $\forall v \in V \exists s, t \in C$ such that $s-t=v$ (c) linear span of $C$ is $V$ Proof (b)<=>(c) obvious (a)=>(c) obvious (c)=>(a) pick a basis $(e_1, ..., e_n)$ in $C$ and take $e_1 + ... + e_n$, its inner For the bounded region, theorem follows obviously, because slight deformation of strictly convex polynomial is convex in the bounded domain. For the unbounded region it can be done as follows: solve it for the homogeneous polynomials (where you can restrict to the unit sphere and use that it is compact) and then take the homogeneous convex polynomial which is inner and add the standard quadratic form. REPLY [2 votes]: For a bounded convex region as the domain, the answer is yes. Let $ Hess(p) $ be the hessian matrix of p (the matrix of the second derivatives by each pair of variables). This is an $ n\times n $ symmetric matrix each of whose elements is a real polynomial. The elements of this matrix are bounded functions on N, say the absolute value of each element is at most m everywhere in N. Define $$ f(x_1, \dots, x_n) := nm(x_1^2 + \dots + x_n^2) $$ Then f is convex, and $ Hess(f) = 2nmE $ where E is the identity matrix of size n. Now $ g := f - p $ is a polynomial, and $ Hess(g) = Hess(f) - Hess(p) $ at any point in N is a symmetric real matrix whose diagonal elements are at least $ 2nm - m $ but other elements have absolute value at most m so it is postive definite. This implies that g is convex in N. I don't know if a similar argument could work in the unbounded case. Edit: mentioned the matrix being symmetric, which I used implicitly.<|endoftext|> TITLE: alternative to Kontsevich formality QUESTION [7 upvotes]: Has anyone considered an alternative approach to Kontsevich formality in which the DGCA of poly-vector fields is deformed to an $L_\infty$-algebra? REPLY [2 votes]: You might want to have a look at $\S2.2$ of this paper by Boris Shoikhet, where Boris constructs an exotic $L_\infty$-structure on poly-vector fields on a (possibly infinite dimensional) affine space, which deforms the usual Schouten-Nijenhuis graded Lie structure. This exotic $L_\infty$-structure has been used in a wise way in another paper by Merkulov and Willwacher in order to split Kontsevich's quantization procedure into two independant steps (only one of them depending on the choice of an associator). Furthermore, this exotic $L_\infty$-structure has been shown (by Willwacher) to be the unique non-trivial $L_\infty$-deformation of the Schouten-Nijenhuis graded Lie structure that is universal (in the sense that it doesn't depend on the dimension $n$ of the affine space one is working with, including $n=+\infty$). For poly-vector fields on a finite dimensional affine space this exotic $L_\infty$-structure seems to be quasi-isomorphic to the Schouten-Nijenhuis one (Merkulov-Willwacher write in their paper that this is a folklore conjecture; but it seems to be proven in Shoikhet's paper, see the Corollary of the Main Theorem in $\S2.3$).<|endoftext|> TITLE: Tannaka–Krein duality QUESTION [6 upvotes]: First I would like to stress that maybe I don't have a necessary background from the theory of Lie groups. I met the topic of Tannaka–Krein duality while reading the book of Gracia–Bondia, Varilly and Figureoa "Elements of noncommutative geometry". This theory resembles the Gelfand–Najmark theory for commutative $C^*$-algebras but there are some points which are different. I will follow the aproach from this book. The setting is the following: they start with the Hopf algebra $H$ over $\mathbb{R}$ and consider the set of all algebra homomorphisms $\mathcal{G}(H)$. This set could be equipped with the so called convolution product and it turns out to be a group. One can define the topology of pointwise convergence on $\mathcal{G}(H)$ and $\mathcal{G}(H)$ becomes a topological group. Then a suitable functional $J$ is introduced: as far as I understood, this is a prototype of a Haar integral. It must satisfy certain axioms: a commutative Hopf algebra together with this functional is called a commutative skewgroup. With this extra structure on $H$ it it shown that $\mathcal{G}(H)$ becomes a compact group. Then comes a theorem concerning Hopf algebras: starting from the commutative skewgroup $H$ one can form a compact group $\mathcal{G}(H)$ and then form an algebra of representative functions $\mathcal{R}(\mathcal{G}(H))$. The theorem states that: $H$ is isomorphic to $\mathcal{R}(\mathcal{G}(H))$ The isomorphism is the same as in Gelfand–Najmark and some steps in the proof are similar. Using Stone–Weierstrass theorem one shows that the image (call it $V$) of the canonical "to be" isomorphism is dense in $\mathcal{R}(\mathcal{G}(H))$ and one needs to check that it is closed. Then the authors claim that it is enough to prove that $V$ is a $\mathcal{G}(H)$-module (meaning that there is an action, say from the left, of $\mathcal{G}(H)$ on $V$ such that $g \triangleright ( \cdot)$ is linear) and they refer to the book of Brocker and Dieck "Representations of compact Lie group". I've checked this theorem and have two questions: Q1. It is about compact Lie groups. In Gracia–Bondia's book there is nothing about the Lie group structure on $\mathcal{G}(H)$. How to fix this problem? Q2. This theorem provides equivalent conditions for a $G$-submodule of $\mathcal{R}(G)$ to be closed. I don't see why these conditions are (if really they are!) automatically satisfied in out situation. Is this really the case? Further, in Gracia–Bondia's book the authors formulate a second theorem: If $G$ is a compact Lie group then $\delta \colon x \mapsto \delta_x$ (where $\delta_x(f)=f(x)$) is an isomorphism between compact Lie groups $G$ and $\mathcal{G}(\mathcal{R}(G))$. I've checked the proof and as far as I understood the steps the situations is the following: a) the fact that it is an group homomorphism uses only algebraic structures b) injectivity follows from Peter-Weyl thm. and continuity is immediate. The problem is with surjectivity: let us denote $\mathcal{G}:=\mathcal{G}(\mathcal{R}(G))$ and $\hat{}:\mathcal{R}(G) \to \mathcal{R}(\mathcal{G})$ is an isomorphism from the previous theorem. One shows that $\delta^*:F \mapsto F \circ \delta$ is right inverse of $\hat{}$ and therefore is also an isomorphism. Since $\mathcal{R}(G)$ and $\mathcal{R}(\mathcal{G})$ are dense in $C(G)$ and $C(\mathcal{G})$ resp. it follows that $\delta^*$ extends to the isomomorhism between $C(G)$ and $C(\mathcal{G})$ and thus $\delta$ is surjective. A continuous bijection beetween compact spaces is a homeomorphism, and a continuous homomorphism of Lie groups is smooth so we are done. But here is my third question: Q3. Where in the proof we used the fact that $G$ is a Lie group? Do we really need this asumption? The proof uses the previous theorem but in the formulation of the previous theorem there is nothing about Lie structure. It seems to me that this Lie structure is somehow hidden. So, to summarize let me ask: Q4. Which of this two theorems has to be reformulated (if any) and how should it look like? Are the proofs which I've sketched here correct? And the final question: Q5. why it doesn't work for $\mathbb{C}$? Where is a key point of the argument? REPLY [4 votes]: On Q1 and Q3: The Lie structure is not needed, and that's why you don't see it. Pedro Lauridsen Ribeiro also referred to this in the comments. You also said “a continuous homomorphism of Lie groups is smooth”, which implies that if two Lie groups are isomorphic as topological groups, then they are isomorphic as Lie groups. On Q4: The theorems are correct, and don't need to be reformulated. Your sketches also seem to be correct. On Q5: It does not work for $\mathbb{C}$, because that is not a compact Lie group. You used compactness, for example in saying that a “continuous bijection beetween compact spaces is a homeomorphism”. [On a sidenote: this whole Tannaka–Krein duality business works in a lot of different settings. For a baby case, let me shamelessly plug my bachelor's thesis. It gives the proof for finite groups, and also computes what goes wrong in the case you take $\mathbb{Z}$. This might help you understand why compactness is needed.]<|endoftext|> TITLE: Why is the standard flop a flop? QUESTION [11 upvotes]: I have seen at least two ways to define flops (and similarly flips). We start with $Y \to X$, a surjective birational morphism, contracting a locus of codimension at least 2, such that $K_Y$ is relatively trivial. 1) We start with $\pi\colon Y \to X$ and take Proj(\oplus \pi_*O(mH)) for $H$ relatively ample or anti-ample. 2) It is the unique $W \to X$ satisfying the same assumptions as $Y \to X$ but such that $W$ is not isomorphic to $Y$. Why are these two definitions equivalent? (for flips one has instead $K_Y$ is relatively anti-ample and $K_W$ ample) There is also a special class of flops, sometimes called standard. 3) Let $C$ be a $(-1,-1)$ curve in $Y$. Blow it up. The exceptional locus is $P^1 \times P^1$. Contract in the other direction and obtain $W$. Why is this equivalent to other two definitions? EDIT: I've edited the question following SK's answer. I guess what I really would like to understand is the line in SK's answer which says: "The fact that there are only these two possible maps follows from the fact that one of them is Proj(⊕π∗Y(−mH)) and the other is Proj(⊕π∗Y(mH)) and one proves that any such small morphsim has to be one of them." How do you prove that? REPLY [9 votes]: It looks like you are mixing up flips and flops. 1) looks more like a flip, though not really the usual definition. 2) is indeed a flop, but this is also not the usual definition. For certainty look up the definition in Kollár-Mori 98. (Also, flips go from $K$ being anti-ample to ample, not the other way around. This is actually important!!) The essence of flips and flops is the following: As you say start with a small projective birational morphism $\pi:Y\to X$ and let $-H$ be a $\pi$-ample line bundle on $Y$. Then the $H$-flip of $\pi$ is another small projective birational morphism (isomorphic over the same open set as $\pi$) $\pi^+:Y^+\to X$ such that if $H^+$ denotes the strict transform of $H$ on $Y^+$, then $H^+$ is $\pi^+$-ample. Now traditionally a $K_Y$-flip is called a flip and an $H$-flip for some $H$ when $K_Y$ is $\pi$-trivial is called a flop. The fact that there are only these two possible maps follows from the fact that one of them is $\mathrm{Proj}(\oplus \pi_*\mathscr O_Y(-mH))$ and the other is $\mathrm{Proj}(\oplus \pi_*\mathscr O_Y(mH))$. Actually, it is probably even better to say that if $H^+$ is $\pi^+$-ample, then $X^+$ is isomorphic to $\mathrm{Proj}(\oplus \pi_*\mathscr O_Y(mH))$ regardless any other condition. This proves that the $H$-flip is unique. (However, this does not mean that these actually obviously exist, because you need these rings to be finitely generated). To get that there are no other small morphisms to compete for being the flip/flop of $\pi$ one needs to assume that the relative Picard number of $\pi$ is $1$. In practice, this is usually a given from the way these small morphisms arise. With that assumption, if $\pi$ is $K$-trivial, then no matter what $H$ one picks, the corresponding $H$-flip (i.e., a flop of $\pi$) will be the same. For number 3, the point is that that $C$ is contractible and so is the curve that $\mathbb P^1\times \mathbb P^1$ contracts to on $W$ and those contractions form a flop, not the blow up, that's just a way to construct it. I suppose it is called standard, because this construction was known before flips and flops were invented and no one really cared about contracting $C$, because it is much nicer to blow it up. However, this method doesn't always work.<|endoftext|> TITLE: Riemann's quote cited by Lakatos: what is the context? QUESTION [15 upvotes]: "If only I had the theorems! Then I should find the proofs easily enough." This quote is generally attributed to Bernhard Riemann. In particular, on page 9 in Proofs and refutations by Imre Lakatos, where it is referred to page 487 in Die mathematische Methode. Logisch erkenntnistheoretische Untersuchungen im Gebiete der Mathematik, Mechanik und Physik. Berlin: Springer, 1924, by Otto H\"older. I read German but I do not have access to the book by H\"older. Does anyone know the original quote in German? The context? Where does it come from? What is in the book by H\"older? REPLY [10 votes]: Wenn ich nur erst die Sätze habe! Die Beweise werde ich schon finden. as quid said. .<|endoftext|> TITLE: Cohomology of the tangent sheaf of $\mathbb{P}(1,2,3)$ QUESTION [7 upvotes]: Using the exact sequence $$0\mapsto\mathcal{O}_{\mathbb{P}^{2}}\rightarrow\mathcal{O}_{\mathbb{P}^{2}}(1)^{\oplus 3}\rightarrow T_{\mathbb{P}^{2}}\mapsto 0$$ it is easy to compute $H^{1}(\mathbb{P}^{2},T_{\mathbb{P}^{2}}) = H^{2}(\mathbb{P}^{2},T_{\mathbb{P}^{2}}) = 0$ while $h^{0}(\mathbb{P}^{2},T_{\mathbb{P}^{2}}) = 8$. On the singular variety $\mathbb{P}(1,2,3)$ by $T_{\mathbb{P}(1,2,3)}$ I mean $\mathcal{H}om(\Omega_{\mathbb{P}(1,2,3)},\mathcal{O}_{\mathbb{P}(1,2,3)})$. Is there an analogous way (or a completely different way) of computing the cohomology groups of $T_{\mathbb{P}(1,2,3)} = \mathcal{H}om(\Omega_{\mathbb{P}(1,2,3)},\mathcal{O}_{\mathbb{P}(1,2,3)})$ ? If it helps $\mathbb{P}(1,2,3)$ can be embedded in $\mathbb{P}^{6}$ as a singular Del Pezzo surface of degree six. REPLY [3 votes]: A sheaf D of differentials on any weighted projective space WPS have been constructed by I.Dolgachev (1982). He computed cohomology of the D(n)' s and generalized the Bott theorem to WPS' s. ( Before him, C.Delorme computed cohomology of the sheaves O(n) and studied duality for WPS' s (1975)).The ref. is : I.Dolgachev, Weighted projective varieties,in "Group Actions and Vector Fields" , Lect. N. Math. 956, Springer-Verlag, 1982,pp. 34-72. (ref. for C. Delorme is included).<|endoftext|> TITLE: Is there a known solution to $f(x) = (1-x)f(x^2)$? QUESTION [16 upvotes]: The functional equation $f(x) = (1-x)f(x^2)$ (with $f(0)=1$) has a simple solution that can be expressed as a rapidly converging infinite product $$f(x) = \prod_{n=0}^\infty (1 - x^{2^n}) = (1-x)(1-x^2)(1-x^4)(1-x^8)\cdots$$ and as a more slowly converging power series $$f(x) = \sum_{n=0}^\infty (-1)^{H(n)}x^n = 1 - x - x^2 + x^3 -\cdots,$$ where $H(n)$ is the Hamming weight of $n$ (i.e., the number of 1's in the binary representation of $n$). Is anything known about this function $f$? Does it have a simple description in terms of more familiar functions? REPLY [3 votes]: I don't know, whether the following counts as "is anything known", but I found it interesting. Consider the separation according to the Hamming weight: $$ \begin{array}{} a_0(x) &= 1 &\\ a_1(x) &= & x^1+x^2 + x^4 + x^8 + ... \\ a_2(x) &= & x^3+x^5 + x^6 + x^9 + ... \\ a_3(x) &= & x^7+x^{11} + x^{13} + x^{14} + ... \\ ... & = ... \end{array}$$ which are all convergent for $|x| \lt 1$, then $$ s_0(x) = a_0(x) +a_1(x)+a_2(x)+ ... \\ = \sum_{k=0}^\infty a_k(x) \\ = \sum_{k=0}^\infty x^k \\ = { 1 \over 1-x } \tag2$$ and $$ s_1(x) = a_0(x) -a_1(x)+a_2(x)- ... \\ = \sum_{k=0}^\infty (-1)^k a_k(x) \\ = \sum_{k=0}^\infty (-1)^{H(k)} x^k \\ = f(x) \tag3$$ I found it then interesting, that at $x=1/2$ the evaluation $a_1(1/2)$ is known to be a transcendental number. But what about the other $a_k(1/2)$ ? If I recall it correctly they are all transcendental numbers, but don't have it at hand how this has been shown; I think they might be rational multiples of each other): just (another) infinite set of transcendental numbers adding up to a rational one. [update] If I recall correctly the $a_k(1/2)$ can be generated as rational compositions of $a_1(1/2),a_1(1/2^2),a_1(1/2^3),...$ (or very similar) applying the Newton-method of converting a sequence of powersums into symmetric polynomials. For instance, $$a_1(1/2) \approx 0.816421509022 \\ a_2(1/2) \approx 0.175061285686 \\$$ and also $$ [(a_1(1/2)^2-(a_1(1/2) -1/2) )/2,a_2(1/2)] \\ [0.175061285686, 0.175061285686] $$ using Pari/GP, reflecting the second power $(x^1 + x^2 + x^4+x^8+...)^2= 2(x^3 + x^5+x^6 + ...) + (x^2+x^4+x^8+...) \\ = 2 a_2(x)+ a_1(x^2) = 2 a_2(x)+ (a_1(x)-x) $<|endoftext|> TITLE: Equivariant derived category and invariant divisor QUESTION [5 upvotes]: I'm looking for a reference of the following (folklore?) result. Let $X$ be a smooth projective variety equipped with a $G=\mathbb{Z}/2\mathbb{Z}$ action (we consider the simplest case, everything can be generalized to an arbitrary finite group). Suppose that the invariant subset is a smooth divisor $Z$. Put $Y=X/G$ and let $p:X\to Y$ and $i:Z\to X$ denote the projection and inclusion maps correspondingly. Then on has a semiorthogonal decomposition $D^G(X)=\left$, where $D(Y)$ stands for the bounded derived category of coherent sheaves, $D^G(X)$ denotes the equivariant counterpart and $i_*D(Z)$ is equipped with a non-trivial $G$-action. (I might be confusing the order of components.) REPLY [2 votes]: I know you wanted a reference, but let me try to sketch a proof anyway. In general, suppose we have an adjoint pair of functors between triangulated categories (or better, cocomplete pretriangulated dg categories or similar(?)). $$ R: \mathcal C \leftrightarrows \mathcal D: L$$ where $RL \simeq id_D$ (so $L$ is fully faithful). Claim (well known, I guess): we have a semiorthogonal decomposition $ (ker(R), im(L))$. Proof: First, it is clear that $ker(R)$ is contained in the right orthogonal to $im(L)$ by adjunction. On the other hand, if $M$ is contained in the right orthogonal to $ker(R)$, then consider the map $f:LR(M) \to M$ given by the counit of the adjunction. The map $R(f)$ is an isomorphism as $RL\simeq id$, thus the cone of $f$ is in the kernel of $R$. As $M$ is right orthogonal to $ker(R)$, the cone is zero, so $f$ must be an isomorphism. Thus $M$ is contained in the image of $L$. $\square$ Using the notation from the question, we take $\mathcal C=D(X)^G$, $\mathcal D = D(Y)$, $L=p^\ast$ and $R$ is the functor of invariants. Your question is then reduced to: Claim: $i_\ast D(Z) = ker(R)$. Proof: Everything is local, so we may assume $X=Spec(A)$ is affine, then $Y=Spec(A^G)$, and $Z=Spec(A/I)$, where $I$ is the ideal generated by $g(f)-f$, where $f\in A$, and $g$ is the involution. The functor $L= A\otimes_{A^G} (-)$, and $R = (-)^G$. It is clear that $i_\ast D(Z)$ is contained in the kernel of invariants. To see the other containment, we must prove that if $M$ is in the kernel of invariants then it is annihilated by the ideal $I$. Note that $g(m)=-m$ for any $m\in M$. Consider $$ (g(f) - f)(g(m)-m) = 2(g(f)-f)m$$ for $f\in A$ and nonzero $m\in M$. This is a $G$-invariant element of $M$, so must be zero, thus $m$ is annihilated by $g(f)-f$ as required.<|endoftext|> TITLE: What are global sections of the determinant bundle on the Beilinson-Drinfeld Grassmannian? QUESTION [10 upvotes]: Let $X$ be a smooth proper algebraic curve over $\mathbb{C}$, and let $G$ be a reductive group over $\mathbb{C}$. Let $Gr_{X,n}$ be the Beilinson-Drinfeld Grassmannian (for n points in $X$), which classifies triples $\left( (x_i) \in X^n, \mathcal{P}, \beta \right)$, where $(x_i)$ is an $n$-tuple of points in $X$, $\mathcal{P}$ is a $G$-bundle on $X$, and $\beta$ is a trivialization of $\mathcal{P}$ away from the set $\{x_i\}$. We have a natural map $Gr_{X,n} \rightarrow \text{Bun}_G(X)$ given by remembering only $\mathcal{P}$. Here $\text{Bun}_G(X)$ is the moduli stack of $G$-bundles on $X$. This stack carries a natural line bundle $\mathcal{L}$ called the determinant line bundle. My question is what are the global sections of $\mathcal{L}$ restricted to $Gr_{X,n}$? If instead of the Beilinson-Drinfeld Grassmannian, I look at the ordinary affine Grassmannian (which classifies $G$-bundles on a curve trivialized away from a fixed point), the answer to this question is the following: global sections of $\mathcal{L}$ are given by $V(\Lambda_0)^*$, the dual of the basic representation of the affine Kac-Moody group associated to $G$. I want to know whether there is an answer to my question in terms of representation theory of the affine Lie algebra. Let me say why I ask this question. I am interested in the following situation: we can consider the subspace of $Gr_{X,2}$ where the second point $x_2$ is fixed. This space maps to $X$ by remembering the first point, and this map is known to be flat. If we look at the fiber of this map over any point except $x_2$, we get a product of two copies of the affine Grassmannian. If the first point coincides with $x_2$, we get a single copy of the affine Grassmannian. Because of flatness of the map to $X$ and a higher cohomology vanishing statement, on the level of sections of the determinant bundle, we expect to find that $V(\Lambda_0)^* \otimes V(\Lambda_0)^* \cong V(\Lambda_0)^*$ (c.f. Theorem 1.2.2 in the following paper (http://arxiv.org/abs/0710.5247) by Xinwen Zhu where this construction is carried out geometrically for Demazure modules, where everything is finite dimensional). Is there any technical problem in using the techniques of this paper to conclude that $V(\Lambda_0)^* \otimes V(\Lambda_0)^* \cong V(\Lambda_0)^*$ (perhaps with some completion), and if so, can this map be algebraically understood? I was hoping that understanding global sections on the full Beilinson-Drinfeld Grassmannian would be a step towards understanding this. REPLY [9 votes]: In a recent paper https://arxiv.org/abs/2003.12930 the space of global sections on Schubert subvarietis of Beilinson-Drinfeld Grassmanian was computed. It turns out to be global Demazure module over the current algebra $\mathfrak{g}[t]$, introduced in https://arxiv.org/abs/1912.07988. So the (dual) sections on the whole BD Grassmanian should be a direct limit of global Demazure modules. I think it would be very interesting to describe it in a more explicit way.<|endoftext|> TITLE: Not quite adjoint functors QUESTION [5 upvotes]: What are standard and/or natural examples of pairs of functors $F:C\leftrightarrows D:G$ and unnatural bijections $\hom_D(Fx,y)\to\hom_C(x,Gy)$ for all $x$ and $y$? Can one do this so that the bijections are natural in $x$ and not in $y$? REPLY [12 votes]: Let $C$ and $D$ both be the category of finite-dimensional (say real) vector spaces and invertible linear maps between them, let $F$ be the identity, and let $G$ take a vector space to its dual. ($G$ is not functorial on all linear maps, but it is on the invertible ones.) Then $G y \cong y$ unnaturally, so $\hom(F x, y) = \hom(x, y) \cong \hom(x, G y)$ naturally in $x$ but unnaturally in $y$.<|endoftext|> TITLE: Is this knot invariant already treated somewhere in the literature? QUESTION [8 upvotes]: Fix a knot type $K \subset S^3$, and consider the set $$Y_K = \{ \mbox{Diagrams of }K \} / \mbox{planar isotopy}.$$ We can turn $Y_K$ into a metric space by considering the distance induced by Reidemeister moves: $d(D_0,D_1) = $ minimum length of a sequence of Reidemeister moves (each move is possibly followed by a planar isotopy), and $D_0,D_1$ are two diagrams of $K$. We can build a graph $\mathcal{G}_K$ as follows: the vertices are given by the points of $Y_K$, while the edges are given by the pairs $(D_0,D_1)$ such that $d(D_0,D_1)=1$. This construction parallels Hirasawa and Yoshiaki's in "The Gordian Complex of Knots" for the Gordian metric space of knots, but unlike the Gordian complex, each vertex of $\mathcal{G}_K$ has only finitely many edges (the number of edges at a vertex $D$ corresponds to the possible inequivalent diagrams - up to planar isotopy - that can be reached from $D$ by a single Reidemeister move). For each knot $K$ we can thus define a function $$\varphi_K : Y_K \longrightarrow \mathbb{N}$$ called the simplicity, as $\varphi_K (D) = \#\{$edges of $\mathcal{G}_K$ that contain $D$ in their boundary$\}$. As an example the "standard" diagram of a trefoil trefoil-1 has a lower simplicity than trefoil-2 (you need to take into account that the former is 3-symmetric). Taking the minimum of the simplicity over all diagrams of a knot yields an invariant $\mathcal{S}(K)$. Is the invariant $\mathcal{S}(K)$, or any possible application/connection, known? REPLY [2 votes]: I think the minimal degree of "Reidemeister complex" has some meaning. I know one paper about this topic. A distance for diagrams of a knot<|endoftext|> TITLE: Big mono-chromatic subgraphs of vertex 2-colourings QUESTION [11 upvotes]: I'm not a graph theorist, but the following quantity came up in my work and I'm curious if it has been studied. Given a connected finite graph $\Gamma = (V,E)$ define: $$ c(\Gamma) = \min_{f : V \rightarrow \{0,1\}} \max_{G \subseteq \Gamma} |V(G)| $$ where the minimum runs over all vertex 2-colourings and the maximum runs over all connected mono-chromatic subgraphs $G \subseteq \Gamma$. Are there non-trivial lower bounds for $c(\Gamma)$ for general graphs $\Gamma$? What I'm looking for is some easier to compute quantity which will lower bound $c(\Gamma)$. For example, if we take a cyclic graph of odd order we get $c(C_{2k+1}) = 2$. Complete graphs give us: $c(K_n) \geq n/2$. It has been pointed out to me that $c(\Gamma)$ can be thought of as measure of how far from bi-partite $\Gamma$ since we have that $c(\Gamma) = 1$ for all bi-partite $\Gamma$. Edit: Consider the adjacency graph of an $n \times n$ Hex board, which write $H_n$. It's well known that when the board is full (its vertices have been 2-coloured) some player has won. This forces a path between non-adjacent edges of the grid. Thus we have $c(H_n) = n$. Note that $H_n$ has $|V| = n^2$ and maximal degree six. Is it well understood why $H_n$ has such nice lower bounds on $c(H_n)$? I've read Gale's famous article on Hex and Brouwer, but the larged mono-chromatic subgraph comes out like a rabbit from a hat. REPLY [6 votes]: This is a funny coincidence :-) I posted a manuscript on arXiv 2 days ago, where we show some results about the complexity of computing what you call $c(\Gamma)$. See the final section of https://arxiv.org/abs/1402.2475 We show that it is NP-hard to approximate $c(\Gamma)$ within a constant multiplicative factor, even in very specific families of graphs, such as 2-degenerate triangle-free planar graphs, or 2-degenerate graphs of girth at least 8, or graphs with girth larger than any given constant (the girth is the length of a shortest cycle). Even if you are not interested by the complexity results, the proofs give you ways to construct graphs $G$ within these classes for which $c(G)$ is unbounded from above (showing that the lower bounds you are looking for are not constant for these classes). Are you interested in a particular class of graphs, besides triangular grids?<|endoftext|> TITLE: Realisation of the noncommutative torus as a universal $ C^{*} $-algebra QUESTION [17 upvotes]: One of the most basic examples in noncommutative geometry is the so-called noncommutative torus, denoted here by $ \mathbb{T}_{\theta} $. As far as I know, there are several equivalent constructions of it: as the $ C^{*} $-algebra of a foliation; as a crossed-product $ C^{*} $-algebra; as a universal $ C^{*} $-algebra. I’m interested in the last construction. It is defined (here I follow M. Khalkhali’s book Basic Noncommutative Geometry) as a universal unital $ C^{*} $-algebra generated by two unitaries $ u $ and $ v $ with relation $ u v = \lambda v u $, where $ \lambda = e^{2 \pi i \theta} $. The author describes the concrete realisation of $ \mathbb{T}_{\theta} $ as follows. He defines two unitary operators $ U,V: {L^{2}}(\mathbb{S}^{1}) \to {L^{2}}(\mathbb{S}^{1}) $ by the formulas $$ Uf(x) = e^{2 \pi i x} \cdot f(x), \quad Vf(x) = f(x + \theta) $$ (where we think of $ \mathbb{S}^{1} $ as $ \mathbb{R} / \mathbb{Z} $ to keep additive notation) and forms the $C^{*} $-algebra generated by these two unitaries. Then he then omits the proof that this $ C^{*} $-algebra is indeed universal. I’ve asked one person who is more familiar than me with noncommutative geometry, and this person said that this is folklore and that he doesn’t know where I could find it in the literature. Hence, I would like to know if there is some standard procedure to handle such examples or if each example needs a particular method? REPLY [13 votes]: According to what I have seen in the literature so far, the standard procedure consists of two main steps: Prove the existence of a universal $ C^{*} $-algebra $ A_{\theta} $ generated by two unitaries $ u $ and $ v $ that satisfy $$ u v = e^{2 \pi i \theta} v u. $$ Note: We are assuming that $ \theta $ is irrational. Prove that $ A_{\theta} $ is simple, and conclude that the concrete realization given above is indeed universal. To accomplish Step 1, there are several methods. I understand that you have read Davidson’s book, so let me describe an approach different from his that is more algebraic in nature. Let $ \mathcal{P} $ denote the free associative unital $ \mathbb{C} $-algebra in four indeterminates, $ u $, $ u^{*} $, $ v $ and $ v^{*} $, where the identity element of $ \mathcal{P} $ is denoted by $ \mathbf{1} $. Let $ \mathcal{I}_{\theta} $ denote the (not a priori proper) two-sided ideal $$ \left\langle u u^{*} - \mathbf{1}, u^{*} u - \mathbf{1}, v v^{*} - \mathbf{1}, v^{*} v - \mathbf{1}, u v - e^{2 \pi i \theta} v u \right\rangle. $$ Form the quotient $ \mathbb{C} $-algebra $ \mathcal{A}_{\theta} \stackrel{\text{df}}{=} \mathcal{P} / \mathcal{I}_{\theta} $. Let $ \dot{\mathbf{1}} $, $ \dot{u} $, $ \dot{u}^{*} $, $ \dot{v} $ and $ \dot{v}^{*} $ denote the images of $ \mathbf{1} $, $ u $, $ u^{*} $, $ v $ and $ v^{*} $ in $ \mathcal{A}_{\theta} $ respectively. Then the following monomials are distinct in $ \mathcal{A}_{\theta} $ and define a Hamel basis for it: $$ \dot{\mathbf{1}}, \quad \dot{u}^{m} \dot{v}^{n}, \quad \dot{u}^{m} (\dot{v}^{*})^{n}, \quad (\dot{u}^{*})^{m} \dot{v}^{n}, \quad (\dot{u}^{*})^{m} (\dot{v}^{*})^{n}; \qquad (m,n) \in \mathbb{N}_{0}^{2} \setminus \{ (0,0) \}. $$ As such, $ \mathcal{I}_{\theta} $ is a proper ideal of $ \mathcal{P} $, and so $ \mathcal{A}_{\theta} $ is a unital $ \mathbb{C} $-algebra. Comment: Wegge-Olsen claims that Point (5) is obvious, but I beg to differ as its validity depends on a non-trivial algebraic result called the Diamond Lemma for Ring Theory. Davidson appears to avoid all forms of algebraic machinery by resorting to the GNS Construction. However, as a staunch believer in the Principle of Conservation of Difficulty, I think that Davidson is simply transferring all technical issues from the Diamond Lemma to the GNS Construction, which, as most operator algebraists would agree, is a highly non-trivial result in the representation theory of $ C^{*} $-algebras. Define a $ C^{*} $-representation of $ \mathcal{A}_{\theta} $ to be a triple $ (A,s,t) $, where: $ A $ is a unital $ C^{*} $-algebra. $ s $ and $ t $ are unitary elements of $ A $ satisfying $ s t = e^{2 \pi i \theta} t s $. Given a $ C^{*} $-representation $ (A,s,t) $ of $ \mathcal{A}_{\theta} $, there exists a unique unital $ \mathbb{C} $-algebra homomorphism from $ \mathcal{P} $ to $ A $ defined by $$ u \mapsto s, \quad u^{*} \mapsto s^{*}, \quad v \mapsto t \quad \text{and} \quad v^{*} \mapsto t^{*}. $$ Then as the homomorphism kills $ \mathcal{I}_{\theta} $, we obtain a unital $ \mathbb{C} $-algebra homomorphism $ \pi_{A,s,t}: \mathcal{A}_{\theta} \to A $, once again unique, that satisfies: $ {\pi_{A,s,t}}(\dot{u}) = s $ and $ {\pi_{A,s,t}}(\dot{u}^{*}) = s^{*} $. $ {\pi_{A,s,t}}(\dot{v}) = t $ and $ {\pi_{A,s,t}}(\dot{v}^{*}) = t^{*} $. $ {\pi_{A,s,t}}(\dot{u}) ~ {\pi_{A,s,t}}(\dot{v}) = e^{2 \pi i \theta} ~ {\pi_{A,s,t}}(\dot{v}) ~ {\pi_{A,s,t}}(\dot{u}) $. Question: Do $ C^{*} $-representations of $ \mathcal{A}_{\theta} $ exist? Answer: Yes! The concrete realization $ (\mathscr{B}({L^{2}}(\mathbb{T})),U,V) $ described by the OP is one. Amusingly, the very existence of this concrete realization shows that $ \mathcal{I}_{\theta} $ is a proper two-sided ideal of $ \mathcal{P} $, but I think we can safely say that this is not a demonstration of the fact from first principles. Define a mapping $ \| \cdot \|_{0}: \mathcal{A}_{\theta} \to [0,\infty] $ by $$ \| a \|_{0} \stackrel{\text{df}}{=} \sup (\{ \| {\pi_{A,s,t}}(a) \|_{A} \in \mathbb{R}_{\geq 0} \mid \text{$ (A,s,t) $ is a $ C^{*} $-representation of $ \mathcal{A}_{\theta} $} \}) $$ for each $ a \in \mathcal{A}_{\theta} $. Proof sketch that $ \| \cdot \|_{0} $ is a $ \mathbb{C} $-algebra semi-norm: Thanks to the existence of a $ C^{*} $-representation of $ \mathcal{A}_{\theta} $, we have $$ \left\| \dot{\mathbf{1}} \right\|_{0} = \| \dot{u} \|_{0} = \| \dot{u}^{*} \|_{0} = \| \dot{v} \|_{0} = \| \dot{v}^{*} \|_{0} = 1. $$ As $ \dot{\mathbf{1}} $, $ \dot{u} $, $ \dot{u}^{*} $, $ \dot{v} $ and $ \dot{v}^{*} $ generate $ \mathcal{A}_{\theta} $, it follows that $ \| a \|_{0} < \infty $ for each $ a \in \mathcal{A}_{\theta} $. Knowing now that $ \| \cdot \|_{0}: \mathcal{A}_{\theta} \to [0,\infty) $, it is easily shown to satisfy the axioms of a $ \mathbb{C} $-algebra semi-norm. Notice that $ \| \cdot \|_{0} $ also satisfies the $ C^{*} $-identity. Let $ \mathcal{N} \stackrel{\text{df}}{=} \{ a \in \mathcal{A}_{\theta} \mid \| a \|_{0} = 0 \} $. Then $ \mathcal{N} $ is a $ \mathbb{C} $-subalgebra of $ \mathcal{A}_{\theta} $. Form the quotient $ \mathbb{C} $-algebra $ \mathcal{A}_{\theta} / \mathcal{N} $ to get a pre-$ C^{*} $-algebra, denoting the quotient norm by $ \| \cdot \| $. Complete $ \mathcal{A}_{\theta} / \mathcal{N} $ with respect to $ \| \cdot \| $ to obtain the irrational rotation $ C^{*} $-algebra $ A_{\theta} $. Clearly, $ A_{\theta} $ is unital and is generated by the pair $ ([\dot{u}]_{\mathcal{N}},[\dot{v}]_{\mathcal{N}}) $ of unitary elements. Furthermore, $$ [\dot{u}]_{\mathcal{N}} [\dot{v}]_{\mathcal{N}} = e^{2 \pi i \theta} [\dot{v}]_{\mathcal{N}} [\dot{u}]_{\mathcal{N}}. $$ This completes the construction. Claim: $ (A_{\theta},[\dot{u}]_{\mathcal{N}},[\dot{v}]_{\mathcal{N}}) $ is a universal $ C^{*} $-representation of $ \mathcal{A}_{\theta} $. Proof of Claim Let $ (A,s,t) $ be a $ C^{*} $-representation of $ \mathcal{A}_{\theta} $. Then by the definition of $ \| \cdot \|_{0} $, $$ \forall a \in \mathcal{A}_{\theta}: \quad \| {\pi_{A,s,t}}(a) \|_{A} \leq \| a \|_{0} = \| [a]_{\mathcal{N}} \|. $$ We thus have a unique unital $ * $-homomorphism $ \pi_{A,s,t}^{\mathcal{N}}: \mathcal{A}_{\theta} / \mathcal{N} \to A $ satisfying $$ {\pi_{A,s,t}^{\mathcal{N}}}([\dot{u}]_{\mathcal{N}}) = s \quad \text{and} \quad {\pi_{A,s,t}^{\mathcal{N}}}([\dot{v}]_{\mathcal{N}}) = t, $$ and we can extend this, using continuity, to a unique unital $ * $-homomorphism $ \Pi_{A,s,t}: A_{\theta} \to A $. In other words, $ \Pi_{A,s,t} $ is the only unital $ * $-homomorphism from $ A_{\theta} $ to $ A $ that maps $ [\dot{u}]_{\mathcal{N}} $ to $ s $ and $ [\dot{v}]_{\mathcal{N}} $ to $ t $. Now, suppose that there is another unital $ C^{*} $-algebra $ B $ generated by two unitaries $ u' $ and $ v' $ satisfying $$ u' v' = e^{2 \pi i \theta} v' u' $$ such that for any $ C^{*} $-representation $ (A,s,t) $, there exists a unique $ * $-homomorphism $ \Phi_{A,s,t}: B \to A $ that maps $ u' $ to $ s $ and $ v' $ to $ t $. The following statements are then true: $ \Pi_{B,u',v'}: A_{\theta} \to B $ is the unique $ * $-homomorphism that maps $ [\dot{u}]_{\mathcal{N}} $ to $ u' $ and $ [\dot{v}]_{\mathcal{N}} $ to $ v' $. $ \Phi_{A_{\theta},[\dot{u}]_{\mathcal{N}},[\dot{v}]_{\mathcal{N}}}: B \to A_{\theta} $ is the unique $ * $-homomorphism that maps $ u' $ to $ [\dot{u}]_{\mathcal{N}} $ and $ v' $ to $ [\dot{v}]_{\mathcal{N}} $. $ \Phi_{A_{\theta},[\dot{u}]_{\mathcal{N}},[\dot{v}]_{\mathcal{N}}} \circ \Pi_{B,u',v'}: A_{\theta} \to A_{\theta} $ equals $ \text{id}_{A_{\theta}} $ on a dense subset of $ A_{\theta} $. $ \Pi_{B,u',v'} \circ \Phi_{A_{\theta},[\dot{u}]_{\mathcal{N}},[\dot{v}]_{\mathcal{N}}}: B \to B $ equals $ \text{id}_{B} $ on a dense subset of $ B $. Therefore, $ \Phi_{A_{\theta},[\dot{u}]_{\mathcal{N}},[\dot{v}]_{\mathcal{N}}} \circ \Pi_{B,u',v'} = \text{id}_{A_{\theta}} $ and $ \Pi_{B,u',v'} \circ \Phi_{A_{\theta},[\dot{u}]_{\mathcal{N}},[\dot{v}]_{\mathcal{N}}} = \text{id}_{B} $, and so $ A_{\theta} $ is $ * $-isomorphic to $ B $ via the unique $ * $-isomorphism that sends $ [\dot{u}]_{\mathcal{N}} $ to $ u' $ and $ [\dot{v}]_{\mathcal{N}} $ to $ v' $. This concludes the proof that $ (A_{\theta},[\dot{u}]_{\mathcal{N}},[\dot{v}]_{\mathcal{N}}) $ is indeed a universal $ C^{*} $-representation of $ \mathcal{A}_{\theta} $. $ \quad \blacksquare $ The proof that $ \Pi_{\mathscr{B}({L^{2}}(\mathbb{T})),U,V}: A_{\theta} \to {C^{*}}(U,V) $ is a $ * $-isomorphism follows from the non-trivial fact that $ A_{\theta} $ is a simple $ C^{*} $-algebra (i.e., it has no non-trivial closed proper two-sided ideals). The main idea behind the proof is to use the so-called trace function on $ A_{\theta} $. This trace function does wonders for us. Firstly, it shows that $ A_{\theta} $ contains a non-trivial projection element. Secondly, it shows that $ \mathcal{A}_{\theta} $ is faithfully represented as a $ \mathbb{C} $-algebra in $ A_{\theta} $, i.e., $ \mathcal{N} = \{ 0_{\mathcal{A}_{\theta}} \} $. Observe that in defining the $ C^{*} $-semi-norm $ \| \cdot \|_{0} $, there was no guarantee that each non-zero element of $ \mathcal{A}_{\theta} $ would not be sent by $ \| \cdot \|_{0} $ to $ 0 $. Playing around with the trace function shows that this is indeed the case.<|endoftext|> TITLE: irrationality of the p-adic exponential QUESTION [7 upvotes]: I would like to illustrate my lecture on p-adic numbers with some elementary results. I proved that the series $e^p=\sum_{n\ge0}\frac{p^n}{n!}$ converges in $\mathbb Q_p$ for every prime $p$. Now I would like to teach that $e^p$ is irrational in $\mathbb Q_p$ by elementary methods (this is true by Mahler result on transcendence of $e^x$ in $p$-adic fields). Do you know such an elementary proof. I can't find one in the literature... Good catch! I meant $\frac{p^n}{n!}$ instead of $\frac1n$. I edited it. REPLY [2 votes]: Maybe I harp on this too much, but I think the $p$-adic exponential is not nearly as interesting as the logarithm. It’s defined on a much larger subgroup than the exponential, it’s a homomorphism (unlike the complex logarithm), and its roots are the $p$-power roots of unity. As a function on $1+D$, where $D$ is the open unit disk in $\Bbb C_p$, it’s onto $\Bbb C_p$, so that even though for $z\in\Bbb C_p$ you won’t usually be able to define $e^z$, you can find $\alpha\in\Bbb C_p^*$ such that $\log\alpha=z$.<|endoftext|> TITLE: Lights out game QUESTION [5 upvotes]: I would like to ask about the game Lights Out for a square nxn. In http://mathworld.wolfram.com/LightsOutPuzzle.html there is a list of the number of solutions to the game, and the number of solutions modulo symmetries of the square. Somewhere I have seen a formula for the number of solutions for general n, written as a generating function. However I cannot locate it now. Does anyone know where it is? And is there also a formula for the number of solutions modulo symmetries for general n? Thanks REPLY [3 votes]: http://oeis.org/A075462 The number of solutions to the all-ones lights out problem on an $n\times n$ square, where nonidentical reflected and rotated solutions are considered distinct. (The link gives the Mathematica code to generate the sequence.) Similarly, http://oeis.org/A075463 gives the number of rotation-reflection inequivalent solutions.<|endoftext|> TITLE: Under which constraints are there only finite numbers of irreducible eta product identities? QUESTION [7 upvotes]: For the Dedekind eta function, defined as usual by $\eta(q) = q^{\frac1{24}} \prod\limits_{n=1}^{\infty} (1-q^{n})$, let for brevity $e_k:=\eta(q^k)$. An eta product identity (or eta identity for short) is then defined as a homogenous polynomial in the $e_k$ with integer coefficients such that its Taylor series vanishes identically. There are already more than 6300 of them in Michael Somos' collection. Asking bluntly like that how many eta identities exist, the answer is of course infinitely many, as they form a (graded) vector space.So we may try, as does Somos, to restrict to "irreducible" ones in an appropriate sense. Now it turns out that it is a bit tricky to come up with a good definition of irreducibility for an eta identity. Some requirements are obvious: it should be irreducible as a polynomial, including that not only the $\gcd$ of the coefficients (up to sign) should be $1$, but also the $\gcd$ of the eta exponents (i.e. the indices of the $e_k$'s) involved it should not be expressible as the sum of two shorter eta identities if there are several linear independant eta identities with given level, degree and rank, we should only count $k$ basis elements where $k$ is the dimension of the vector space generated by them, all others would be considered "reducible in terms of this basis". A (minor) problem can occur if we add two identities whose monomials are not all disjoint in a way that some monomials cancel out. In particular, we can take any pair of 3-term identities, say $a+b+c$ and $d+e+f$, then $d(a+b+c)-a(d+e+f)=bd+cd-ae-af$ is a 4-term identity. Should we call it reducible? Some 4-term identities in Somos' collection can be obtained like that, but many can't (a priori). There are also pairs of 4-term identities of the form $a+b+c+d$ and $a+b+kc+e$ with $k\in\mathbb Q$, so their difference yields a 3-term identity, but as it is made up of two longer identities, there would be no reason anyway to call that one reducible. So it seems best to stick to the four above criteria for an irreducible eta identity, and not to worry about some 4-term identities that would fall through. There are still infinitely many irreducible eta identities, because e.g. all Schläfli type modular equations can be written in terms of eta products, so (irreducible) eta identities exist at least for all levels divisible by 4. Now, many other types of modular equations, like e.g. theta functions that can be expressed as eta quotients, tend to come in finite quantities, and I think that we can impose feasible constraints on eta identities such that there are only finitely many fulfilling those constraints. So which constraints might that be? Some questions below are stronger than others. For a given level $N$, are there only finitely many irreducible eta identities? (Note that if length and degree are limited, that follows from the existence of the Sturm bound) (more or less equivalent to 1.) For a given level $N$, is there a maximal length and/or a maximal degree of an irreducible eta identity? It seems like for a given level, the identities of maximal degree tend to have more symmetries, more precisely, they are most often self-dual in the sense defined here, or sometimes, even stronger, each term is self-dual. (I am aware that the collection is not necessarily exhaustive even for rather small levels, so this may be a wrong impression). Are there only finitely many 3-term identities (all levels combined)? If so, then it makes also sense to ask: for a given length $k>3$, are there only finitely many irreducible $k$-term identities? Are there irreducible eta identities for each level $N$ that is not prime? Concerning the last question above: it seems like for odd $N$'s, there are relatively few identities, and the bigger the minimal prime factor, the more intricate they are. E.g. for level $13^2$, according to Somos the "smallest example has 50 terms of degree 14 and rank 182", and for level $11^2$, the smallest example has 156 terms of degree 60 and rank 660. But it does exist! On the other hand, if $N$ has only small prime factors, there are lots of identities. For $N=12$ alone, the collection lists over 1000, the vast majority of them with only four terms. EDIT: Subsequently to Jeremy's answer, Michael Somos has added a section with the (unique) identities of level $p^2$ for primes $p\le 19$. For the record: some of their parameters (summarized from the comments) are $$\matrix{p&|°ree&rank& \#\ terms\cr\hline 2&|&24&48&3 \cr 3&|&12&36&4\cr5&|&6&30&6\cr 7&|&8&56&11\cr 11&|&60&660&156\cr 13&|&14&182&50 \cr 17&|&72&1224&421 \cr 19&|&60&1140&436 }$$ REPLY [4 votes]: I feel like you had a little bit of trouble coming up with a good notion of "irreducible". A lot of the things you say suggest you want identities that "cannot be built from simpler ones." Here's a way of thinking about it that may be helpful. Given a level $N$, consider the polynomial ring $R$ over $\mathbb{C}$ on $d(N)$ generators $u_{d}$. The map $\phi_{N} : R \to \oplus_{k} M_{k}(\Gamma_{0}(24N))$ that maps $u_{d}$ to $e_{24d}$ is a ring homomorphism. The kernel of $\phi$, is precisely the space of eta identities. It seems that you want to be asking for "generators" of this space. The kernel of $\phi_{N}$ is a prime ideal of $R$, and since every ideal of $R$ is finitely generated (by the Hilbert basis theorem), there will be "finitely many irreducible eta identities of level $N$," corresponding to the finitely many generators of $\ker \phi_{N}$, and hence an affirmative answer to question 1 and hence 2. (Of course the notion of irreducible in this context means "it cannot be built from the other identities that we've picked" and so labelling a given eta identity as irreducible or not is somewhat difficult to do - in the same way that choosing a single "best" generating set of an ideal is hard to do.) Question 3 is quite a bit more subtle. For a given level $N$, an eta identity can be divided by one of the terms to give rise to an identity of the form $$ a_{1} + a_{2} + \cdots + a_{r} = 1 $$ where the $a_{i}$ are eta quotients that are weight zero modular functions. These modular functions have no zeroes or poles except at the cusps of $X_{0}(N)$, and there will only be finitely many such, by appealing to results about $S$-unit equations in global function fields (see the paper by R.C. Mason, "Norm form equations I" in J. Number Theory (1986), for example). However, this only applies for a fixed level $N$ - if the level is allowed to vary, there may be many more options. The same thing applies for 4. Every elliptic curve is modular, so if there are infinitely many elliptic curves $y^{2} + ay = x^{3}$ where the modular parametrization has $x$ and $y$ given by eta quotients (and this is a big if), then the answer is yes. The issue with question 5 is the requirement that each identity not come from a relation of lower level. Given a positive integer $N$, if we take $\ker \phi_{d}$ for each divisor $d$ of $N$ and multiply each $k$ value by $N/d$ we get a subset of $\ker \phi_{N}$. Is that subset the whole ideal? I suspect in many cases the answer is yes - the field of functions on $X_{0}(N)$ is indeed generated by $j(z)$ and $j(Nz)$, and so it is plausible that there might not always be an irreducible identity of level $N$. In general, it seems that the image of $\phi$ is a ring of Krull dimension $2$ (one dimension coming from the weight, one dimension coming from the dimension of $X_{0}(N)$). This would imply that $\ker \phi$ is a prime ideal of height $d(N) - 2$ and is hence non-trivial unless $N = 1$ or $N$ is prime. In the case that $N = p^{2}$, there will be an irreducible eta identity, and moreover such an identity will be unique, because a height $1$ prime ideal in a polynomial ring is principal.<|endoftext|> TITLE: Virtual fibering conjecture for cusped hyperbolic manifolds QUESTION [8 upvotes]: I am interested in understanding if the Virtual Fibering Theorem holds in the non-compact case. Agol proved that every closed hyperbolic $3$-manifold has a finite index cover which fibers over the circle. I could not find the same result stated for finite volume, possibly cusped, hyperbolic $3$-manifolds. If such a manifold has a virtually special fundamental group then a result of Agol implies the virtual fibering property. In order to prove that the fundamental group of a closed hyperbolic $3$-manifold is virtually special, one relies on the following: Kahn-Markovic's result on the existence of immersed almost geodesic surfaces in closed hyperbolic $3$-manifolds, which allows to build a CAT(0) cube complex on which the fundamental group acts properly discontinuously and cocompactly. Agol's result which states that a hyperbolic group acting properly discontinuously and cocompactly on a CAT(0) cube complex is virtually special. The first result is stated only for closed hyperbolic $3$-manifolds. The second one holds for (word)-hyperbolic groups, but the fundamental group of cusped hyperbolic $3$-manifolds is hyperbolic relative to the cusp subgroups. I would be interested in knowing if anyone has managed to get around these issues. REPLY [12 votes]: It does hold - Wise proved that finite-volume non-compact hyperbolic 3-manifolds are virtually special, hence virtually RFRS and so virtually fibred by one of Agol's results. Details and references are contained in this survey.<|endoftext|> TITLE: A Converse to the Gauss Bonnet Theorem QUESTION [5 upvotes]: Let $S$ be a compact surface in $\mathbb{R}^{3}$ with the gauss normal map $N:S\to \mathbb{S}^{2}$. Assme that $\phi;\mathbb{S}^{2}\to S$ is a diffeomorphism. Put $F=N\circ \phi$ and represent $F:\mathbb{S}^{2}\to \mathbb{S}^{2}$ in the form $F=(f,g,h)$. then as a consequence of the Gauss Bonnet theorem we have \begin{equation}\iint_{\mathbb{S^{2}}} fdgdh= 4/3 \; \pi\end{equation} (See page 14 of the(printed version) of the book, Non commutative geometry by Alain Connes My question is that ;is the converse of the above statment true? That is: Assume that $F:\mathbb{S}^{2}\to \mathbb{S}^{2}$ is a smooth map with $F=(f,g,h)$ such that the above integral equality hold. Is there a compact surface $S$ with Gauss normal map $N$ and a diffeomorphism $\phi: \mathbb{S}^{2} \to S$ such that $F=N\circ \phi$? REPLY [11 votes]: First of all, the identity holds for any degree 1 map $F:\mathbb S^2\to\mathbb S^2$. Moreover, for any $F=(f,g,h):\mathbb S^2\to\mathbb S^2$, $$ \int_{\mathbb S^2} f\,dgdh = \frac43\pi \deg F. $$ This follows from Stokes' formula. So basically the identity means that the Gauss map of a surface diffeomorphic to the sphere has degree 1. This is by the way not always true, you have to choose $N$ according to the orientation (that you use to define the degree), otherwise you get the value $-\frac43\pi$. In other words, $\phi$ must be orientation-preserving. Back to the question, the answer is trivially yes if $F$ is a diffeomorphism, by Alex Degtyarev's comment. In general, the answer is no because certain types of singularities can not occur. For example, consider $F$ that maps some circle $C\subset\mathbb S^2$ to the north pole but such that the $F$-image of an open disc $D\subset\mathbb S^2$ bounded by $C$ avoids the north and south poles. It is easy to construct such a map of degree 1. Suppose that the desired $S$ and $\phi$ exist and consider the disc $\phi(D)$ in $S$. On its boundary $\phi(C)$ the normal is vertical, hence $\phi(C)$ is contained in some horizontal plane $H$. Then there is a point in the interior of $\phi(D)$ where the tangent plane is horizontal (look at a point furthest from $H$). The normal at this point is vertical contrary to the fact that $F(D)$ avoids the north and south poles.<|endoftext|> TITLE: regular singularities and comparison isomorphism QUESTION [8 upvotes]: Let $X$ be a smooth variety over some field $k \subset \mathbb{C}$. By a theorem of Grothendieck, one has a canonical isomorphism of complex vector spaces $$ \mathbb{H}^j(X, \Omega_{X/k}^\bullet) \otimes_k \mathbb{C} \stackrel{\sim}{\longrightarrow} \mathbb{H}^j(X(\mathbb{C}), \Omega^\bullet_{X^{an}/\mathbb{C}}) $$ between algebraic and analytic de Rham cohomology. The situation with coefficients is more subtle. If $\nabla: E \to E \otimes_k \Omega^1_U$ is some integrable connection on a locally free sheaf $E$ on $X$, there is still a canonical morphism $$ \mathbb{H}^j(X, (E \otimes \Omega_{X/k}^\bullet, \nabla)) \otimes_k \mathbb{C} \longrightarrow \mathbb{H}^j(X(\mathbb{C}), (E^{an} \otimes \Omega^\bullet_{X^{an}/\mathbb{C}}, \nabla^{an})) \quad \quad (\ast) $$ between the algebraic (hyper)cohomology of the complex induced by the connection and the corresponding analytic one. Deligne proves that this is an isomorphism when $\nabla$ has regular singularities. Question: is this theorem and "if and only if", that is, if the map (*) is an isomorphism, is it true that $\nabla$ has regular singularities? If this is the case, a reference would be very much appreciated. REPLY [2 votes]: This should probably be a comment, but it got too long... Your claim seems to be true if the space $X$ is simply connected. I think it should fail in general, but I am having a hard time coming up with a counterexample. If $X$ is simply connected, then $(E^{an}, \nabla^{an})$ is isomorphic to the trivial connection. Suppose that $H^0_{dR}(E,\nabla) \xrightarrow{\sim} H^0_{dR}(E^{an} ,\nabla^{an}) = \mathbb C^r$, where $r$ is the rank of $E$. Note that $H^0_{dR}(E,\nabla) = Hom_{Conn(X)}((\mathcal O_X,d), (E,\nabla))$, and thus we have a map $f: \mathcal (O_X, d)^{\oplus r} \to (E,\nabla)$ such that the analytification $f^{an}$ is an isomorphism. Hence $f$ is an isomorphism as the analytification functor is conservative. In particular $(E,\nabla) \simeq (\mathcal O_X,d)^{\oplus r}$ is regular. Thus, in the simply connected case I showed that if analytification induces an isomorphism on $H^0_{dR}$ then the connection is regular. This stronger statement fails for $X=\mathbb C^\times$. For example, take $E=\mathcal O_X . z^\alpha e^{z}$, where $\alpha \in \mathbb C - \mathbb Z$ (i.e. the trivial line bundle with connection $d + (\alpha/z + 1)dz$). The connection $E$ is non-regular and has no algebraic or analytic flat sections. On the other hand, $H^1_{dR}(E) = \mathbb C$, while $H^1_{dR}(E^{an})=0$ (I think!), so $E$ does not provide a counterexample to your question. I don't really know much about irregular connections, so I would be curious to see a complete answer to this.<|endoftext|> TITLE: Are there two computable binary trees such that each has a branch not computing any branch through the other? QUESTION [9 upvotes]: It is a well-known elementary classical result in computability theory that there are computable infinite binary trees $T\subset 2^{<\omega}$ having no computable infinite branch. (One can build such a tree $T$ as follows: fix a computably inseparable pair of c.e. sets $A$ and $B$, and allow a binary string $t$ into the tree when the pattern of yes/no answers is consistent with being a separation of $A$ and $B$ in as much as these sets are revealed by stage $|t|$. In this way, incorrect guesses about the separation are eventually killed off, stuck in a finite dead part of the tree, and there can be no computable infinite branch because the sets have no computable separation.) What I have need of is a strengthening of that classical result to the following: Question. Are there two computable trees $T,S\subset 2^{\lt\omega}$ such that $T$ has an infinite branch not computing any infinite branch through $S$ and $S$ has a branch not computing any branch through $T$? That is, I want computable trees $S$ and $T$ such that there are infinite binary sequences $s$ and $t$, which are branches through $S$ and $T$, respectively, such that no infinite binary sequence that is Turing computable from $s$ is a branch through $T$ and no infinite binary sequence that is Turing computable from $t$ is a branch through $S$. (So in particular, neither $S$ nor $T$ can have any computable infinite branches.) REPLY [2 votes]: Here is another example. Let $P_1$ be a nonempty $\Pi^0_1$-class which only contains Martin-Lof random reals, and $P_2$ be a nonempty $\Pi^0_1$-class in which no path is either of DNR degree or recursive. The existence of $P_2$ can be found in Rod and Denis's book (Thm 12.4.6, page 584). Now it is obvious no real in $P_2$ computing a real in $P_1$. Let $x$ be a hif random real, then there must be some $y\in P_1$ Turing equivalent to $x$. However, any nonrecursive real Turing below $x$ must be DNR. So no real in $P_2$ can be computed by $y$.<|endoftext|> TITLE: Intermediate submodels which do not satisfy AC QUESTION [7 upvotes]: The following is known: Theorem. Suppose $V[G]$ is a generic extension of $V$ by a set forcing, and let $N$ be a model of $ZFC$ with $V\subseteq N\subseteq V[G].$ Then $N$ is a generic extension of $V$ by a set forcing, in particular $N=V[A],$ for some set of ordinals. It seems that the above theorem is not true if $N$ does not satisfy $AC$. In fact the following abstract is given in a talk by James Cummings (see http://settheory.mathtalks.org/cmu-math-logic-seminar-tues-11-september/): If $c$ is Cohen-generic over $L$, then there is a transitive class model $M$ of $ZF$ intermediate between $L$ and $L[c]$ which is not of the form $L(A)$ for any $A.$ Does anyone know a proof of this fact? REPLY [5 votes]: To my knowledge there is no written proof of this fact. I have all the available notes, which include a very very scattered description of $V_{\omega+1}$ and $V_{\omega+2}$ of this model $M$, and a single lemma which is used to proceed through successor of singular cardinals. I am working on rebuilding this model in a cleaner method, and I am relatively close to finishing (some technical constructions are needed to finish the outline, but the idea itself is completely finished). With luck I might actually finish this soon, and I could write a reasonable outline announcement (and then a full detailed accounts of the construction). I should probably add that I asked all the people involved in the construction of this model, The Bristol model, and what I was told by everyone of them is that it started as some general idea to play with, and by the time they realized someone should be writing things down they already did a lot of the work, so it didn't survive into the notes. Update (April 25, 2017): It's on arXiv now.<|endoftext|> TITLE: Smallest non-zero eigenvalue of a (0,1) matrix QUESTION [20 upvotes]: What's the smallest absolute value possible of a non-zero eigenvalue of an $n$ by $n$ square matrix whose entries are either $0$ or $1$ (all operations are over $\mathbb{R}$)? I would be interested in estimates or bounds as I imagine an exact answer is tricky. I asked this question previously at https://math.stackexchange.com/questions/666493/smallest-non-zero-eigenvalue-of-a-0-1-matrix . REPLY [20 votes]: The correct asymptotic behavour is $n^{-n/2(1+o(1))}$. This is proved in: N. Alon and V. H. Vu, Anti-Hadamard matrices, coin weighing, threshold gates and indecomposable hypergraphs, J. Combinatorial Theory, Ser. A 79 (1997), 133-160.<|endoftext|> TITLE: Universal Central Extension of pi(X), X a compact Riemann surface of genus>1 QUESTION [6 upvotes]: Does a universal central extension exist for the fundamental group of a Compact Riemann Surface of genus1? Please give a detailed explanation.I am unable to justify the statements in Atiyah-Bott Phil Trans Roy Soc 1982 p559 (apparently it exists and some quotient operations are performed) REPLY [10 votes]: Well, the OP is right: Atiyah and Bott indeed use the term "universal central extension" for the fundamental group of a compact surface $\Sigma_g$ of genus $g\geq 1$ -- which is incorrect since that group is not perfect. What they mean is that there is in this case a canonical central extension by $\mathbb{Z}$: indeed these extensions are parameterized by $H^2(\pi _1(\Sigma _g),\mathbb{Z})$; since $\Sigma _g$ is a $K(\Pi ,1)$, this is canonically isomorphic to $H^2(\Sigma _g, \mathbb{Z})=\mathbb{Z}$, and they consider the extension associated to the generator of this group. They describe explicitly this extension $$0\rightarrow \mathbb{Z}\rightarrow \tilde{\pi }_g\rightarrow \pi _1(\Sigma _g)\rightarrow 1 $$as follows: $\tilde{\pi }_g $ is the largest quotient of the free group $F_{2g}$, with generators $a_1,\ldots ,a_g,b_1,\ldots ,b_g$, such that the element $[a_1,b_1]\ldots [a_g,b_g]$ is central.<|endoftext|> TITLE: The intersection of a circle and a rank 3 subgroup of the plane QUESTION [11 upvotes]: Let $A$ be a rank 3 subgroup of the Euclidean plane, i.e. $A = \mathbb{Z} v_1 + \mathbb{Z} v_2 + \mathbb{Z} v_3$, where $v_1, v_2, v_3 \in \mathbb{R}^2$ are three $\mathbb{Q}$-linearly independent vectors. Let $S^1 = \{v \in \mathbb{R}^2 ; \, |v|=1\}$ denote the unit circle in the plane. Is it possible that $|A \cap S^1| = \infty$? Remark: I could find rank 4 subgroups intersecting $S^1$ in infinitely many points. The idea was to take a complex algebraic integer $\alpha$ such that $|\alpha| = 1$ and $\alpha$ is not a root of unity. Then the additive group generated by $1, \alpha, \alpha^2, \dots$ has finite rank. REPLY [4 votes]: I am grateful to Edgardo whose answer supplied me with the Siegel theorem on integral points (I did not know it), and I am happy to see that my approach coincides with Edgardo's in many points. The answer is negative: there is no abelian subgroup $A\subset{\mathbb R}^2$ of rank $3$ such that its intersection $I_A:={\mathbb S}^1\cap A$ with the unit circle ${\mathbb S}^1:=\big\{v\in{\mathbb R}^2\mid|v|=1\big\}$ is infinite. In a few words: we show that such $I_A$ is a part of the intersection of two quadrics in ${\mathbb P}_{\mathbb C}^3$ defined over ${\mathbb Q}$, observe that this intersection in interesting cases is generic, hence, is an elliptic curve, and, finally, apply the Siegel theorem http://en.wikipedia.org/wiki/Siegel_theorem to an affine part of this curve. Let $L_{\mathbb Q}:={\mathbb Q}A$ denote the ${\mathbb Q}$-linear space spanned by $A$. Besides the intersection $I_A$, we will also deal with the intersection $I_{\mathbb Q}:={\mathbb S}^1\cap L_{\mathbb Q}$. One can choose another ${\mathbb Q}$-basis $B'$ in $L_{\mathbb Q}$ and generate another $A'$. It is immediate that $A\subset\frac1nA'$ and $A'\subset\frac1nA$ for a suitable $n\in{\mathbb N}$. We conclude that $I_A$ is infinite for some $A$ iff, for any ${\mathbb Q}$-basis $B$ in $L_{\mathbb Q}$, there are infinitely many elements in $I_{\mathbb Q}$ whose denominators in terms of $B$ are limited. Assuming that some $I_A$ is infinite, we can choose a ${\mathbb Q}$-basis of $L_{\mathbb Q}$ inside $I_{\mathbb Q}$. Indeed, it is easy to pick a couple of ${\mathbb R}$-linearly independent $b_1,b_2\in I_{\mathbb Q}$. They generate a discrete ${\mathbb Z}$-lattice $A_0\subset{\mathbb R}^2$; so, $\frac1nA_0$ is also discrete for any $n\in{\mathbb N}$. Consequently, the intersection ${\mathbb S}^1\cap\frac1nA_0$ is finite. Including $b_1,b_2$ in some basis of $L_{\mathbb Q}$, we can see that $I_{\mathbb Q}\subset{\mathbb Q}A_0$ would imply $I_A\subset\frac1nA_0$ for a suitable $n\in{\mathbb N}$. A contradiction. For a similar reason, any $2$-dimensional ${\mathbb Q}$-linear subspace in $L_{\mathbb Q}$ contains finitely many rational points from $I_{\mathbb Q}$ with limited denominators. Let $b_1,b_2,b_3\in I_{\mathbb Q}$ be a ${\mathbb Q}$-basis in $L_{\mathbb Q}$. Its Gram matrix has the form $G:=\left[\begin{smallmatrix}1&c_3&c_2\\c_3&1&c_1\\c_2&c_1&1\end{smallmatrix} \right]$ with $-1 TITLE: Probability distribution or the distance between two points in $n$-dimensional Euclidean space after a random perturbation of one point QUESTION [5 upvotes]: Take two points, $p_0$ and $p_k$, in $n$-dimensional Euclidean space, where $d(p_0,p_k)$ is the distance between the points. Now, draw an $n$-sphere of radius $r$ centered on $p_0$ and uniformly select a new point, $p_1$, in the volume of the sphere, and a new point $p_2$ on the surface of the sphere. Let $h_1 = d(p_0,p_k) - d(p_1,p_k)$, and $h_2 = d(p_0,p_k) - d(p_2,p_k)$, represent the difference in the distance from $p_0$ to $p_k$ if $p_0$ is moved to $p_1$ or $p_2$, respectively. To clarify, if we blow up another $n$-sphere, $S$, about $p_k$ until $p_0$ "touches" its contour, $h_1$ / $h_2$ will be positive if $p_1$ / $p_2$ is inside $S$, zero if $p_1$ / $p_2$ are on the contour of $S$, and negative if $p_1$ / $p_2$ fall outside $S$. What probability distribution and expectation do we have for $h_1$ and $h_2$? Update [2/15/2014] :: Thanks to Bjørn Kjos-Hanssen's efforts, we have a nice exact expression for the $n = 2$ case for $h_1$ (which implies that we have expectation --- $\int_{1-r}^{1+r} \space R \times f(R) \space dR$). At this point, I think its probably wise to restrict the focus or scope of this question to the $n = 2$ case. Can a PDF for $h_2$ (where we select points along the contour of the circle) be derived in a similar manner? REPLY [3 votes]: Let us consider the case $n=2$. Assume $p_k$ is at the origin and $p_0$ is at the point $(1,0)$ on the $x$-axis, and that $r<1$. Distribution of $h_1$ By subtracting a constant it suffices to find the distribution of $d:=d(p_1,0)$. The density for $d$, $f_d(R)$, is proportional to $R\Theta$ where $\Theta$ is length of the interval of angles $\theta$ for which the point $(R\cos\theta,R\sin\theta)$ is within $r$ of $(1,0)$. Now, calculation shows that $$ |(R\cos\theta,R\sin\theta)-(1,0)| < r $$ is equivalent to $$ |\theta| < \arccos\left(\frac{R^2+1-r^2}{2R}\right) $$ so $f_d(R)$ is proportional to $$ 2R\arccos\left(\frac{R^2+1-r^2}{2R}\right),\quad R\in [1-r,1+r]. $$ and $f_d(R)=0$ for $R\not\in [1-r,1+r]$. For geometric reasons (and one could also check it analytically) the integral of the given expression over the given interval is $\pi r^2$. So $$ f(R)=\frac{2R\arccos\left(\frac{R^2+1-r^2}{2R}\right)}{\pi r^2},\quad 1-r\le R\le 1+r. $$ Distribution of $h_2$ The squared distance from $(1,0)+r(\cos\theta,\sin\theta)$ to the origin is $$ D^2 : = (r\cos\theta+1)^2+(r\sin\theta)^2 = r^2 + 1 + 2r\cos\theta$$ and the probability of picking a point $R$ or more away from the origin is then $1/\pi$ times $\theta_R>0$ where $\theta=\theta_R$ makes $D^2=R^2$. Namely $$ \theta_R = \arccos\left(\frac{R^2-r^2-1}{2r}\right) $$ Then the density is $$ f(R)= \frac{-1}{\pi} \frac{d}{dR} \arccos\left(\frac{R^2-r^2-1}{2r}\right); $$ $$ f(R)=\frac1{\pi}\frac{1}{\sqrt{1-u(R)^2}} \frac{R}{r},\quad 1-r\le R\le 1+r, $$ where $u(R)=\frac{R^2-r^2-1}{2r}$.<|endoftext|> TITLE: Heisenberg subalgebras of affine Lie algebras QUESTION [5 upvotes]: It seems to be "well-known" that (infinite-dimensional) Heisenberg subalgebras of an affine Lie algebra $\hat{\mathfrak{g}}$ corresponding to a finite-dimensional simple Lie algebra $\mathfrak{g}$ of type $A$, $D$, or $E$ are parametrized (up to conjugacy) by the set of conjugacy classes in the Weyl group of $\mathfrak{g}$. For instance, this statement can be found in the paper "112 constructions of the basic representation of the loop group of $E_8$" by Kac and Peterson (MR). However, no proof is given in this paper. (A Heisenberg subalgebra is constructed for each element of the Weyl group, but the assertion that this leads to a parametrization of all of them is not proved.) Does anyone know of a reference (with proof) of this classification of Heisenberg subalgebras of affine Lie algebras? REPLY [4 votes]: Up to the central extension (which doesn't affect the classification) this is a special case of the classification of Cartan subgroups of a reductive group over a field. Since they all split after an etale extension, where they are uniquely conjugate up to Weyl group, they are given by Galois $H^1$ of the field with coefficients in the Weyl group. In our case the Galois group is $\widehat {\mathbf Z}$, so we get just elements of $W$ up to conjugacy. Geometrically this means such a subgroup is conjugate to standard one after passing to a branched cover of the disc (etale cover of punctured disc), and so is classified by the monodromy of the associated cameral cover ($W$-torsor over the punctured disc of all ways to conjugate to a standard Cartan).<|endoftext|> TITLE: Torsion of the Picard group for surfaces isogenous to a product QUESTION [5 upvotes]: We say that a complex surface $S$ is isogenous to an (unmixed) product if there exists a finite group $G$, acting faithfully on two smooth projective curves $C_1$ and $C_2$ and freely on their product (with the diagonal action), so that $S$ is isomorphic to $(C_1 \times C_2)/G$. If $S =(C_1 \times C_2)/G$ is any surface isogenous to a product, the two natural projections of $C_1 \times C_2$ induces two fibrations $f_1 \colon S \to C_1/G$ and $f_2 \colon S \to C_2/G$, whose smooth fibres $F_1$ and $F_2$ are isomorphic to $C_2$ and $C_1$, respectively. Moreover, since the action of $G$ on the product is free, all the singular fibres of $f_1$ and $f_2$ are multiple of smooth curves. Let us consider now the group $\textrm{Tors(Pic}^0 S)$. Inside this group naturally lives the subgroup $H$ generated by $f_1^* \textrm{Tors(Pic}^0 C_1/G)$ and $f_2^* \textrm{Tors(Pic}^0 C_2/G)$. Question. Are there conditions on $C_1$, $C_2$, $G$ and the actions ensuring that $H=\textrm{Tors(Pic}^0 S)?$ In other words, when the torsion part of the Picard group of $S$ is determined by the torsion parts of the bases $C_1/G$ and $C_2/G$ of the two natural fibrations? REPLY [3 votes]: I think the answer to the question as it is stated is negative: if I do understand correctly the question I can prove that, if $G$ is abelian and both curves $C_i/G$ are rational, this is never true. Indeed, since ${\mathbb P}^1$ has no torsion, $H$ is trivial, so it is enough if I construct a non trivial torsion bundle on $S$. Let $p_1,\ldots,p_r$ be the critical values of $f_1$. These are the points whose fibre are not reduced, say of multiplicity $m_i>1$; as you know very well, since $C_1/G$ is rational and $ G $ is abelian, $r\geq 2$ and each $m_i$ is a divisor of the l.c.m. of the other $m_j$. In particular, up to renumbering the points, I can suppose that $d:=\gcd(m_1,m_2)>1$. Set $n_i:=\frac{m_i}{d}$, call respectively $F'$ and $F''$ the reduced fibres over $p_i$ and $p_2$, so that $m_1F'$ and $m_2F''$ are fibres of $f_1$. Then the divisor $D:=n_1F'-n_2F''$ is a nontrivial d-torsion divisor. Indeed $dD=m_1F'-m_2F''=f_1^*p_1-f_1^*p_2$ is principal, and by the standard results on the normal bundles of a multiple fibre taken with the reduced structure $$lD_{|F'}=l(n_1F'-n_2F'')_{|F'}=ln_1F'_{|F'}$$ is trivial iff $ln_1$ is a multiple of $m_1$, that is iff $d$ divides $l$. Let me add some comments. Even if we remove my assumptions, my construction, for each pair of $m_i$ not relatively prime on the same side, produces a torsion bundle, and I do not see any reason for it to be in $H$ even when $H$ is very big. Still these torsion bundle are determined by the fibrations, so I would modify your question as follows Question: Under which assumptions is the group generated by $H$ and the bundles constructed above the whole torsion group of $S$? That could be true in a wider range of cases. One could try to prove it in the regular+abelian case by using a theorem of Armstrong as it was used here for computing some fundamental groups.<|endoftext|> TITLE: Morita equivalence for operator algebras and tensor products, question about proof QUESTION [7 upvotes]: This is a bit of a dumb question I know, but I was reading "Morita equivalence for C*-algebras and W*-algebras" by Rieffel, in this section about Morita equivalences and how they relate to forming tensor products of $W^*$-algebras (page 90) I came across this proof that I just can't seem to grasp and was seeking clarification about a few things. On page 90 there's this proposition: 8.5. Proposition. Let $M$, $M_1$, $N$, $N_1$ be $W^*$-algebras. Let $X$ be a normal $N$-rigged $M$-module and $Y$ a normal $N_1$-rigged $M_1$-module. Then the algebraic tensor product $X \otimes Y$ over the complex numbers , with $N \otimes N_1$-valued inner product defined by $\langle x \otimes y,x' \otimes y' \rangle_{N \otimes N_1} = \langle x,x' \rangle_{N} \otimes \langle y,y' \rangle_{N_1}$ and completed for the usual norm, is a normal $N \otimes N_1$-rigged $M \otimes M_1$-bimodule. If $X$ and $Y$ are in fact equivalence bimodules and if an $M \otimes M_1$-valued inned product is defined by $ \langle x \otimes y , x' \otimes y' \rangle_{M \otimes M_1} = \langle x,x' \rangle_{M} \otimes \langle y,y' \rangle_{M_1}$, then $X \otimes Y$ (completed) becomes and equivalence module. The proof focuses on proving that the inner products are positive and the "rest of the proof of this proposition is carried out by routine calculations", which, as it were, I'm stuck with, pardon my lack of familiarity with the whole subject but I just have to ask: What is this "usual norm" they're referring to? I've seen a tensor product of $W^*$-algebras be referred to as the completion of the algebraic tensor product $X \otimes Y$ with respect to certain norm (don't know if that's Sakai's definition they refer to in the paper). How does the range of $\langle - , - \rangle_{M \otimes M_1}$ and $\langle - , - \rangle_{N \otimes N_1}$ span weakly dense subsets on $M \otimes M_1$ and $N \otimes N_1$ respectively? REPLY [6 votes]: To answer your first question, I think the "usual norm" is the one described on page 63 of Rieffel's paper, i.e. $\lVert z \rVert = \lVert \langle z,z\rangle\rVert^{1/2}$ for $z \in X \otimes Y$. Note that the inner product takes values in a $W^*$-algebra. Depending on your point of view, a $W^*$-algebra is either the same as a von Neumann algebra (i.e. comes along as a weakly closed *-subalgebra of $B(H)$ for some Hilbert space $H$), then the norm on the right hand side of the last equation is the operator norm of $B(H)$. Or $W^*$-algebra refers to the abstract concept of a $C^*$-algebra, which has a predual. In this case the norm is the norm as a $C^*$-algebra. By choosing a faithful representation, you can switch from the latter point of view to the first, so it doesn't really make a difference. In your case the inner product takes values in the spatial tensor product of $W^*$-algebras $N \otimes N_1$. I will suppose that $N \subset B(H)$ and $N_1 \subset B(K)$ for Hilbert spaces $H$ and $K$ as weakly closed $*$-subalgebras. Then the spatial tensor product is given by forming the tensor product $H \otimes K$ of the two Hilbert spaces. Note that $B(H) \otimes_{alg} B(K) \subset B(H \otimes K)$. In particular, $N \otimes_{alg} N_1 \subset B(H \otimes K)$ and $N \otimes N_1$ is the weak closure of $N \otimes_{alg} N_1$, which agrees with the double commutant $(N \otimes_{alg} N_1)''$. By definition the algebraic tensor product is weakly dense in $N \otimes N_1$. The commutant of any subset $S \subset B(H)$ is weakly closed, therefore we have $(N \otimes_{alg} N_1)' = (N \otimes N_1)'$. Now look at the tensor product on $X \otimes Y$ defined by $$ \langle x \otimes y, x' \otimes y' \rangle_{N \otimes N_1} = \langle x, x'\rangle_N \otimes \langle y, y' \rangle_{N_1} $$ and let $\widetilde{N} = \{ \langle x,x' \rangle_N | x, x' \in X \}$ and $\widetilde{N}_1 = \{ \langle y,y' \rangle_{N_1} | y, y' \in Y \}$. We know that $\widetilde{N} \subset N$ and $\widetilde{N}_1 \subset N_1$ are weakly dense by assumption. Anything that commutes with $\widetilde{N}$, therefore also has to commute with $N$ and likewise for $N_1$. Therefore $$ (\widetilde{N} \otimes_{alg} \widetilde{N}_1)' = (N \otimes_{alg} N_1)' = (N \otimes N_1)' $$ which should answer your second question. I hope I didn't mess it up somewhere.<|endoftext|> TITLE: Does there exist a terminal surjective discrete fibration out of $C$? QUESTION [6 upvotes]: Let $DF$ denote the category whose objects are categories and whose morphisms $F\colon R\to S$ are the discrete fibrations. This category has applications to the real-world problem of structuring data. You can think of any discrete fibration $R\to S$ as providing a schematic structure $S$ for more raw data $R$. This question is about finding a best such structure for given data. For any category $R$, one can define the coslice category $$DF_{R/}$$ of discrete fibrations out of $R$. Given a discrete fibration $R\to S$, call $S$ the base space. I want to find a minimal base space $R^{min}$ for a given $R$. To get at that idea, I'll ask for some kind of universal object. I once asked on the categories mailing list whether $DF_{R/}$ has a terminal object in general. As was explained to me by Mark Weber and Thorston Palm, it does not. Basically, if $R=\emptyset$ then we're asking for a terminal object in $DF$, and by cardinality arguments, this does not exist. But in fact I was asking the wrong question. If I want minimal models, I actually want my discrete fibrations to be surjective. The counter-examples provided to me by Weber and Palm fail to cause a problem in that context. So here's the question. Question: Define $DFS$ to be the subcategory of $DF$ in which the morphisms are discrete fibrations $F\colon R\to S$, required to be surjective on objects. Then does $DFS_{R/}$ have a terminal object $R^{min}$ for a given category $R$? Edit provenance: An earlier version of this question got very little attention, so I've edited it to explain an intended application and to clarify the question a bit. The latter edit involved replacing "discrete opfibration" (and notation $DopF$, $DopFS$) with "discrete fibration" (and notation $DF$, $DFS$) throughout, because it looked cleaner. REPLY [3 votes]: In general, the answer is "No": the category $DFS_{C/}$ of surjective discrete fibrations under $C$ need not have a terminal object. This is due to the following: Lemma. Let $C$ be the codiscrete category with $n$ objects. For any group $G$ with $n$ elements, there is a surjective discrete fibration $C \to \mathsf{B}G$ (where $\mathsf{B}G$ is the delooping of $G$). Proof. The regular action of $G$ on itself by right multiplication gives rise to a functor $\mathsf{B}G^{op} \to \mathsf{Set}$ whose category of elements is isomorphic to $C$. $\square$ When $n$ is composite, there are at least two non-isomorphic groups $G \not\cong H$ with $n$ elements, and these will give rise to non-isomorphic discrete fibrations $p \colon C \to \mathsf{B} G$ and $q \colon C \to \mathsf{B} H$. Further, any maps out of $p$ and $q$ in $DFS_{C/}$ will necessarily be isomorphisms (as noted by David's reply to Noam). So if there were a terminal discrete fibration $t \colon C \to S$, we must have $\mathsf{B}G \cong S \cong \mathsf{B}H$, which contradicts our assumption that $G \not\cong H$. But $p$ and $q$ above are minimal in some sense: all maps out of them are isomorphisms. So while $DFS_{C/}$ has no terminal objects, it does have multiple minimal ones. (Still, any attempt to classify minimal objects in $DFS_{C/}$ for all (finite) categories $C$ must include the classification of all (finite) groups!) However, if we are willing to restrict the categories $C$ that we want to consider, then we have: Theorem. Let $C$ be such that for all $x \in C$, the slice $C/x$ has no non-trivial automorphisms. Then $DFS_{C/}$ has a terminal object. I've written the proof in this draft. The idea is to construct a category $S$ with one object for each isomorphism class $[C/x]$ of slice categories in $C$, and then show that we have a discrete fibration $C \to S$.<|endoftext|> TITLE: Deconvolution of sum of two random variables QUESTION [7 upvotes]: Let $Z = X + c \cdot Y$ where $X$ and $Y$ are independent random variables drawn form the same distribution given by the pdf $g()$ and $0 < c < 1$ I have observations of $Z_i$'s and thus can approximate the discrete pdf $f()$ which is the distribution of Z. Thus: $f(x) = (g \ast g^\prime)(x) = \sum_{d \in D} g(d) g((x-d)c)$ where $g^\prime(x) = g(xc)$ How to calculate $g()$ based on $f()$? REPLY [3 votes]: I found this paper Rates of convergence for constrained deconvolution problem which describes how to estimate the distribution of $Z$ from observations of a random process of the form $Z = \alpha X + \beta Y$. The trick is to use the characteristic function of Z and it really works out quite nicely.<|endoftext|> TITLE: Why are the power symmetric functions sums of hook Schur functions only? QUESTION [21 upvotes]: One interesting fact in symmetric function theory is that the power symmetric function $p_n$ can be written as an alternating sum of hook Schur functions $s_{\lambda}$: $$ p_n = \sum_{k+\ell = n} (-1)^\ell s_{k, 1^\ell}. $$ A priori, all that is known is that $p_n$ can be expressed as a sum of Schur functions $s_{\lambda}$, with the sum ranging over all partitions of $n$, not just those of hook shape. The fact that the coefficients of all non-hook shapes are zero is quite interesting and most likely says something about representations of $S_n$. However, this property does not extend to other generalizations of the Schur functions. In particular, when $p_n$ is expanded in the Jack basis $J_{\lambda}$ of symmetric functions with coefficients in $\mathbb{Q}(\alpha)$, such a vanishing phenomenon no longer occurs. In particular, the coefficient of $J_{2,2}$ in the expansion of $p_4$ is non-zero (though, of course, it vanishes when $\alpha = 1$). Is there a representation-theoretic or geometric explanation for why the expansion of power functions in the Schur basis has such a nice form? And is there any sort of generalization that could determine which polynomials in the $p_n$ would have a similar property in one of the generalized bases, specifically the Jack basis? Though it may be too much to ask for, I would love to have a combinatorially explicit, algebraically independent set of polynomials in the $p_i$ that generate the ring of symmetric functions in $\mathbb{Q}(\alpha)$ with the property that each generator is a sum of hook Jack symmetric functions only. REPLY [17 votes]: The coefficient of $s_{\lambda}$ in $p_{\mu}$ is (up to normalizing factors which I won't get right) the trace of the permutation of conjugacy class $\mu$ acting on the representation $Sp(\lambda)$ of $S_n$. You are asking why the $n$-cycle acts with trace $0$ on every irrep except for the hooks (which are $\bigwedge^k \mathbb{Q}^{n-1}$, for various values of $k$.) Actually, a more general statement is true: If $\alpha/\beta$ is a skew shape of size $n$, then $\langle p_n, s_{\alpha/\beta} \rangle$ is $0$ unless $\alpha / \beta$ is a ribbon. For a combinatorial proof, see Section 7.17 in Enumerative Combinatorics. For a more representation-theoretic proof, see Proposition 8.2 in Vershik-Okounkov. In summary: We want to evaluate the central element $$z:= \sum_{\mbox{$w$ an $n$ cycle}} w$$ in $\mathbb{Q}[S_n]$ acting on $Sp(\lambda)$. Define $x_k = (1k)+(2k)+\cdots + (k-1 k)$. Observe that the $x_i$ commute, and that $z = x_2 x_3 \cdots x_n$. Since the $x_i$ commute we should expect them to be simultaneously diagonalizable, and indeed they are. V&O construct a basis $v_T$ for $Sp(\lambda)$, indexed by standard Young tableaux of shape $T$, which is a simultaneous eigenbasis for all $x_i$. In particular, $x_i v_T = 0$ if and only the element $i$ is on the main diagonal of the tableau $T$. (I draw tableaux in the English style, like this, so for me the main diagonal runs from the upper left to the lower right.) For $\lambda$ any non hook shape, there is a box one step down and to the right of the upper left corner. For any $T$, that box contains some $i>1$. Then $x_i v_T=0$ and hence $z v_T=0$. So $z$ acts by $0$ on $Sp(\lambda)$ for any nonhook shape $\lambda$. I thought about it a bit more, and here is what I think is the most direct proof. Define $z_n$ to be the element $\sum_{\mbox{$w$ an $n$ cycle}} w$ in the group algebra of $S_n$. I shall prove the following statement: Lemma If $V$ is a representation of $S_n$ where $z_n$ acts by $0$, then $z_{n+1}$ acts by $0$ on $\mathrm{Ind}_{S_n}^{S_{n+1}} V$. Proof of Lemma: We use the description of induction as a tensor product: $$\mathrm{Ind}_{S_n}^{S_{n+1}} V \cong \mathbb{Q}[S_{n+1}] \otimes_{\mathbb{Q}[S_n]} V,$$ the fact that $z_{n+1}$ is central, and the identity $z_{n+1} = x_{n+1} z_n$. (The notation $x_{n+1}$ was defined in the previous proof.) Therefore, $$z_{n+1} \cdot (w \otimes \vec{v}) = (z_{n+1} w) \otimes \vec{v} = (w z_{n+1}) \otimes \vec{v} = (w x_{n+1} z_n) \otimes \vec{v} = (w x_{n+1}) \otimes (z_n \vec{v})=0$$ for any $w \in S_{n+1}$ and $\vec{v} \in V$, as desired. $\square$ Direct computation shows that $z_4$ acts by $0$ on $Sp(2,2)$. So, inductively, $z_n$ acts by $0$ on $\mathrm{Ind}_{S_4}^{S_n} Sp(2,2)$. Therefore, $z_n$ acts by $0$ on any irreducible summand of $\mathrm{Ind}_{S_4}^{S_n} Sp(2,2)$. For $\lambda$ any non-hook, $Sp(\lambda)$ occurs as a summand of $\mathrm{Ind}_{S_4}^{S_n} Sp(2,2)$ so $z_n$ acts by $0$ on $Sp(\lambda)$. As discussed in the previous proof, this proves the claim.<|endoftext|> TITLE: Can Hausdorff dimension make sets into a Tropical Semiring? QUESTION [5 upvotes]: If $X$ is a metric space, we construct Hausdorff $d$ measure from the outer measure \begin{equation} H^d(U) = \lim_{\delta \to 0}\inf\left\{\sum_{i=1}^\infty \left(\text{diam}(E_i)\right)^d : \bigcup_{i=1}^\infty E_i \supseteq U, E_i \text{ Borel}, \text{diam}(E_i)<\delta\right\} \end{equation} We define the Hausdorff dimension of a (Borel?) set $U\subseteq X$ as $\text{dim}_X(U) = \inf\left\{d : H^d(U) = 0\right\}$. Some basic properties include that if $U_1,U_2 \subseteq X$ then \begin{equation} \text{dim}_X(U_1\cup U_2) = \max\{\text{dim}_X(U_1),\text{dim}_X(U_2)\} \end{equation} and if $X,Y$ are metric spaces, $X\times Y$ has the product topology, and $U_1\subseteq X$, $U_2\subseteq Y$, then \begin{equation} \text{dim}_{X\times Y}(U_1\times U_2) \ge \text{dim}_X(U_1) + \text{dim}_Y(U_2) \end{equation} This inequality is tight. Question 1: Is there a natural setting where the inequality reduces to equality for "nice" sets? Question 2: Since this looks like a max-plus algebra, is there a way to make some kind of tropical semiring of sets using Hausdorff dimension? REPLY [7 votes]: Actually, the inequality for the Hausdorff dimension of product sets goes the other way (fixed in the OP now): $$ \dim_{X\times Y}(U_1\times U_2) \ge \dim_X(U_1)+\dim_Y(U_2). $$ And you probably need to assume something about the metric space, locally compact should do. This inequality is a consequence of Frostman's Lemma (for compact metric spaces it is due to Howroyd), which says that $\dim_X(U)$ is the supremum of all $s$ such that there exists a Borel probability measure on $U$ satisfying $\mu(B(x,r)) \le C\, r^s$. If $\mu_1,\mu_2$ are such measures for $U_1, U_2$ with exponents $s_1, s_2$, then $\mu_1\times\mu_2$ is a measure on $U_1\times U_2$ satisfying $$ \mu_1\times \mu_2(B(x,r)) \le C\, r^{s_1+s_2}, $$ giving the inequality above. Indeed there are examples of sets $U_1$ of Hausdorff dimension $0$ in $\mathbb{R}$ (for example) such that $\dim(U_1\times U_1)=1$. However, there is a very natural condition under which equality holds: this will happen whenever either $U_1$ or $U_2$ have equal Hausdorff and packing dimension. Packing dimension is much less well known than Hausdorff dimension, however in my opinion this shouldn't be the case as it plays a dual role (it is defined in terms of packings rather than coverings) and, as this question illustrates, it is often necessary even if you only care about Hausdorff dimension! The packing dimension $\dim_P U$ of a set $U\subset X$ (for any metric space $X$; I do not write the dependence on $X$ explicitly) is defined as follows. First we define pre-packing $s$-dimensional measure $P_s$ as follows: $$ P_0^s(U) = \sup\left\{ \sum_i r_i^s: B(x_i,r_i) \text{ is a packing of } U\right\}. $$ $P_0^s$ is not countably (or finitely) subadditive in general, so $s$-dimensional packing measure is defined as $$ P^s(U) = \inf\left\{ \sum_i P_0^s(U_i):U\subset \bigcup_i U_i \right\}. $$ Then, as for Hausdorff dimension, the packing dimension is defined as $$ \dim_P(U) = \inf\{ s: P^s(U)=0 \} = \sup\{ s: P^s(U)=+\infty\}. $$ As mentioned above, if $\dim(U_1)=\dim_P(U_1)$, then one does have $$ \dim(U_1\times U_2) = \dim(U_1)+\dim(U_2). $$ This is not completely trivial but follows from the definitions. All of this can be found e.g. in Mattila's book "Geometry of sets and measures in Euclidean space" Now, most (though by no means all) sets one encounters in practice do have equal Hausdorff and packing dimensions. These include, for example, self-similar sets (even allowing overlapping), sets invariant under conformal dynamical systems, many random sets such as Brownian paths and graphs, more general Cantor sets defined inductively if the number and location of the pieces is not "too wild" and so on. So this is a large and natural class on which the hypotheses in the OP are met. Moreover, packing dimension satisfies the opposite inequality for product sets: $$ \dim_P(U_1\times U_2)\le \dim_P(U_1)+\dim_P(U_2). $$ This means that if $\mathcal{U}$ is the class of sets $U$ in a (compact, say) metric space $X$ such that $\dim(U) = \dim_P(U)$, then $\mathcal{U}$ is closed under products. A sufficient (but far from necessary) condition for equality of Hausdorff and packing dimensions of $U$ is the existence of a measure $\mu$ on $U$ such that (for some constant $C>1$) $$ C^{-1} \, r^s \le \mu(B(x,r)) \le C \, r^s \text{ for all }x\in U. $$ Such sets $U$ are called Ahlfors-regular and as you could expect, the common value for the dimension is $s$. I'm not familiar enough with tropical geometry but I suppose that if we consider the space of all finite cartesian products of sets/metric spaces with equal Hausdorff and packing dimension, then Hausdorff dimension would be some sort of tropical morphism from this space to the reals.<|endoftext|> TITLE: Is Turing degree actually useful in real life? QUESTION [10 upvotes]: In theoretical computer science, we classify problems according to their Turing degree. Is there any practical application of this? Edit: Given that we cannot explicitly and mechanically understand the non-computable, what traces of non-computability (nonzero Turing degree) might we encounter in everyday life on Earth? REPLY [6 votes]: Focusing on applications to mathematics - as opposed to real life :P - let me mention two points I find interesting: First, recursiveness shows up occasionally un-asked-for in classification theorems. My personal favorite example of this is Higman's Embedding Theorem, which shows that the finitely generated groups which can be embedded in a finitely presentable group are precisely the finitely generated recursively presentable groups. What makes this result interesting to me is that there is no obvious way to relativize it, leaving open the following (to me, very interesting) question: for which classes $\mathcal{D}$ of degrees is there a structural characterization of the finitely generated $\mathcal{D}$-presentable groups? (E.g., the $\Delta^0_2$-presentable groups.) Calling this an "application" is certainly a stretch, but this is one sort of area that might become interesting in the future, and I'm an optimist. Second, there is recent work by Nabutovsky and Weinberger (see e.g. http://arxiv.org/pdf/math/9711225.pdf; also http://press.princeton.edu/titles/7903.html) using computability theory to study the structure of some complicated moduli spaces. Although the more heavy-duty results use computability theory in their statements, the simpler results involve no computability theory but have proofs relying on the fact that there is no algorithm for determining which "nice" smooth homology spheres are actualy spheres. I think in general, using computability theory to study moduli spaces could be an incredibly interesting future direction of research.<|endoftext|> TITLE: Are there more "types" of derivatives than "symmetric" or "alternating"? QUESTION [11 upvotes]: The $kth$ derivative of a function $f:\mathbb{R}^n \to \mathbb{R}$ can be thought of as a symmetric $k$-tensor. (Well, almost. It is not invariant under coordinate changes, and should really be thought of as part of a jet bundle. But we have nice coordinates on $\mathbb{R}^n$, so lets roll with it) The exterior derivative maps alternating $k$-forms to alternating $k+1$-forms. These two "types" of derivative are really the only ones I have met. Is there a unified notion of "derivative" which contains both of these concepts? Assuming there is, are there other examples besides these two? REPLY [13 votes]: This is a big subject. There are many, many different notions of derivative in different contexts, all of which derive in some way from jet bundles. This question was considered in great generality in the foundational works on so-called 'differential invariants' by Lie, Cartan, and their followers and was reconsidered in modern times by many people. Some of the basic ideas in modern form can be found in R. Palais' Natural operations on differential forms and there is a more comprehensive modern treatment in the book Natural operations in differential geometry by Kolar, Michor, and Slovak. Basically, there are methods for defining and computing the invariants of a given geometric structure under the action of some specified (pseudo-)group of transformations, and they yield, as special cases, the exterior derivative of exterior forms, the covariant derivatives of functions with respect to an affine connection, the Schwarzian derivative of a function with respect to linear fractional transformations, and a host of others. Even the Levi-Civita connection of a Riemannian metric can be regarded as the natural first derivative of the metric, which gives a simple example of a (nonlinear) derivative that is not a tensor. Using Cartan's method of equivalence, for example, one can define the invariants to all orders of most $G$-structures and test them for equivalence under various pseudo-groups.<|endoftext|> TITLE: Discriminants of Clifford algebras QUESTION [5 upvotes]: I have a Clifford algebra defined over a field of characteristic not equal to $2$. Is there a formula for its discriminant in terms of the corresponding symmetric bilinear form (or in terms of its quadratic form)? In particular, let $k$ be a field with $\text{char} \, k \neq 2$, let $n$ be an even positive integer, and let $$ W_n = k\left\langle x_1, x_2, \ldots, x_n \right\rangle / (x_i x_j + x_j x_i - 1 \text{ for } i \neq j). $$ As long as $n$ is even, then the center of this algebra is $C=k[x_1^2, \ldots, x_n^2]$, and $W_n$ is free and finitely generated over $C$. Therefore I can define the discriminant of $W_n$ over $C$ as the determinant of the appropriate matrix of traces; the discriminant will be an element of $C$. I think that the answer should be as follows: let $M$ be the matrix with $(i,j)$ entry equal to $x_i x_j + x_j x_i$ (so $M$ has entries $2x_i^2$ down the diagonal, 1 off the diagonal). Then the discriminant should be $(\det M)^{2^{n-1}}$, up to a multiple by an element of $k$. Can anyone provide a reference? REPLY [2 votes]: It was proved in Theorem 3.7 of "Discriminant Formulas and Applications" by Chan, Young and Zhang.<|endoftext|> TITLE: Forcing for Arbitrary First Order Theories QUESTION [6 upvotes]: Forcing is a relative model construction method for models of $ZF$ as a particular first order theory using models of another first order theory (forcing companion) that in this case is the theory of partial orders ($PO$). Is it possible to develope forcing for other first order theories? In fact I am asking for possible relative model constructions of a first order theory without finite models like $T$ (instead of $ZF$) using models of a "forcing companion" theory like $S$ (instead of $PO$). This model construction method produces countable models of $T$ with special properties using a given model of $T$ and a given model of $S$. Obviously if we define a uniform "forcing companion theory" $S$ for an arbitrary first order theory $T$ then in many cases when countable models of $T$ are not too board up to isomorphism the "forcing theory of $T$" is just a trivial method and produces no interesting models of $T$. But in this question my focus is on first order theories with wild countable and countable models like $ZF$ and $PA$. Every references in this frame work is welcome. REPLY [6 votes]: One of the most robust ways to understand forcing is via the method of Boolean ultrapowers, and this is a purely model-theoretic construction that makes sense to undertake with any first-order theory whatsoever. One may form the Boolean ultrapower of any graph, group, ring, field, partial order, and indeed of any structure in any first-order language whatsoever. The general construction of the Boolean ultrapower has classical roots (due in set theory Vopenka, developed also by Solovay, Scott and others including a very nice presentation by Bell, but also as a purely model-theoretic construction by Mansfield and others), and a general introductory account can be found in my paper Well-founded Boolean ultrapowers as large cardinal embeddings, written jointly with Dan Seabold. One may consider the general class of $\mathbb{B}$-valued models in a given first-order language. Specifically, a $\mathbb{B}$-valued structure in a first-order language (not necessarily set-theoretic) consists of a set of objects $M$, called names, and an assignment $[\![\tau=\sigma]\!]\in\mathbb{B}$ giving the Boolean value that any two names are equal, as well as the Boolean value $[\![R(\vec \sigma)]\!]\in\mathbb{B}$ that a given relation holds at a tuple of names, such that the laws of equality hold with respect to these assignments (and one can also handle function symbols and constants). The basic fact is that the concept of a Boolean-valued model is not particularly connected with set theory, and makes sense for models in any first-order language. For any such $\mathbb{B}$-valued structure, whether it is group, ring, field, partial order or model of set theory, one may collapse it to a classial structure by taking the quotient by an arbitrary ultrafilter on $\mathbb{B}$. Specifically, if $U\subset\mathbb{B}$ is an ultrafilter (no need for any genericity), then one defines $\sigma=_U\tau$ for names just in case $[\![\sigma=\tau]\!]\in U$. This is an equivalence relation, indeed a congruence, and one defines the structure on the resulting quotient structure $R([\sigma]_U)$ just in case $[\![R(\sigma)]\!]\in U$, which is well-defined because the equality axioms had Boolean value one. In this way, any $\mathbb{B}$-valued structure is transformed into a classical $2$-valued structure by the quotient. Thus, we have a generalization of the classical ultrapower construction from ultrapowers on a power set algebra to arbitrary ultrapowers on a complete Boolean algebra, and this is known as the Boolean ultrapower. Specifically, if you have a first-order structure $\cal M=\langle M,\ldots\rangle$ and a complete Boolean algebra $\mathbb{B}$, then consider the set of spanning functions $f:D\to M$, where $D$ is any open dense set in $\mathbb{B}$. Define $[\![R(f)]\!]=\bigvee\{b\in\mathbb{B}\mid R(f(b))\}$, and this produces a Boolean-valued model. If $U\subset\mathbb{B}$ is an ultrafilter in $\mathbb{B}$, define $f=_Ug$ for two spanning functions $f:D\to M$, $g:E\to m$, if $\bigvee\{b\in\mathbb{B}\mid f(b)=g(b)\}\in U$, and let ${\cal M}^{\downarrow\mathbb{B}}/U$. This is an equivalence relation, and we may consider the set of spanning functions modulo this relation, denoted $M^{\downarrow\mathbb{B}}/U$. For any relation $R$ in the language, we define the interpretation of $R$ on this structure by $R([f]_U)$ holds if and only if $\bigvee\{b\mid {\cal M}\models R(f(b))\}\in U$. One may similarly handle constants and functions as explained in the paper. If $U\subset\mathbb{B}$ is an ultrafilter, then the corresponding Boolean ultrapower map is the map $x\mapsto[c_x]_U$, where $c_x$ is the constant map on $\mathbb{B}$ with value $x$. This is a generalization of the ordinary ultrapower construction from ultrafilters on power set algebras to ultafilters on arbitrary complete Boolean algebras. The connection with forcing is that for any complete Boolean algebra $\mathbb{B}$, we may construct the $\mathbb{B}$-valued model $V^{\mathbb{B}}$, whose objects are the $\mathbb{B}$-names with Boolean-valued truth defined in the usual forcing manner. If $U\subset\mathbb{B}$ is any ultrafilter (not necessarily generic in any sense), then one gets an elementary embedding $j:V\to \check V_U\subset \check V_U[[\dot G]_U]\cong V^{\mathbb{B}}/U$, which is precisely the Boolean ultrapower map of $V$ into the ground model of the Boolean extension $V^{\mathbb{B}}/U$ quotiented by the ultrafilter $U$.<|endoftext|> TITLE: Root number of a quadratic twist of an elliptic curve QUESTION [5 upvotes]: Could someone provide a reference for the following fact which is stated without proof in section 4.3 of Alice Silverberg's survey "Open Questions in Arithmetic Algebraic Geometry"? Let E be an elliptic curve over the rationals and let $D$ be a squarefree integer. Let $\psi_D$ denote the Dirichlet character associated to the field $\mathbb Q(\sqrt D)$, and let $N_E$ denote the conductor of $E$. If $4D$ is coprime to $N_E$, then the root numbers of $E$ and its twist $E^{(D)}$ are related by the formula $$w_{E^{(D)}}=\psi_D(-N_E)w_E.$$ REPLY [6 votes]: I will prove a stronger Claim: Let $A$ be a $g$-dimensional abelian variety over $\mathbb{Q}$, let $N_A$ be its conductor, let $D$ be a squarefree integer, let $\psi_D$ be the Dirichlet character associated to $\mathbb{Q}(\sqrt{D})$, and let $A^{(D)}$ be the twist of $A$ by $\mathbb{Q}(\sqrt{D})/\mathbb{Q}$. If $N_A$ is coprime to the discriminant of $\mathbb{Q}(\sqrt{D})/\mathbb{Q}$, then the root numbers $w(A)$ and $w(A^{(D)})$ are related by $$w(A^{(D)})w(A) = \psi_D((-1)^g N_A).$$ Proof. The LHS equals $w(A_{\mathbb{Q}(\sqrt{D})})$; this is "standard", but let me know if you want references (note though that the corresponding formula for local root numbers fails but the local correction terms multiply to $1$ over all places). Expressing as a product of local root numbers, we get that the LHS is $$\prod_{v\mid N_A \infty} w(A_{\mathbb{Q}(\sqrt{D})_v})$$ ($v$ is a place of $\mathbb{Q}(\sqrt{D})$). The RHS, on the other hand, equals $$\psi_D((-1)^g)\prod_{p\mid N_A} \psi_D(p)^{f_p(A)},$$ where $f_p(A)$ is the conductor exponent of $A$ at $p$. Let us compare the two products term by term: $\psi_D(-1) = 1$ iff $\infty$ splits. Since the local root number of a $g$-dimensional abelian variety at an infinite place is $(-1)^g$, we see that $\psi_D((-1)^g)$ matches with the product of the factors contributed by $v\mid \infty$ in the first product. If $p$ splits, then $\psi_D(p) = 1$, whereas the places $v\mid p$ in the first product contribute two identical $\pm 1$ terms, which therefore multiply to $1$. If $p$ is inert, then $\psi_D(p)^{f_p(A)} = (-1)^{f_p(A)}$. If $v$ is the unique place above $p$, then, since $\mathbb{Q}(\sqrt{D})/\mathbb{Q}$ is unramified at $p$, one also has $w(A_{\mathbb{Q}(\sqrt{D})_v}) = (-1)^{f_p(A)}$; for the latter equality, let me shamelessly refer you to Thm. 1.12 of http://arxiv.org/abs/1402.2939. Ramified $p$ give no contribution to either product: afterall, we have assumed that $A$ has good reduction at all such $p$. In conclusion, the products are equal and we're done.<|endoftext|> TITLE: Is there a physical intuition for Darboux's theorem? QUESTION [12 upvotes]: We know that there is a physical interpretation for symplectic manifolds (briefly, the fact that a sympletic form assigns to any Hamiltoninan a vector field which describes the motion of particles). My question is if there is a physical meaning to the fact that all symplectic forms locally look alike, i.e. they are standard locally,(for instance, for the cotangent bundle of a manifold, what does it mean that any non-standard symplectic form locally looks like the standard one- which basically imposes the Hamilton's equations on the hamiltonian H)? REPLY [4 votes]: In Riemannian or semi-Riemannian geometry, the reason that our spaces don't all look alike in a neighborhood of a point is that parallelism fails, and the extent to which parallelism fails can be described as a curvature, which is invariant. For example, if two geodesics both start from a point and later on intersect again at some other point, this is a sign of curvature. It's invariant because intersection is invariant. It doesn't matter what coordinates you pick -- an intersection is an intersection. In general relativity, geodesics are the trajectories of test particles, and they can intersect in ways that imply curvature. In the phase space for a Hamiltonian system, the integral curves never intersect, which is physically because preparing a physical system in a definite state is supposed to determine its future time evolution according to the laws of physics. (As in general relativity, the curves can be the trajectories of particles through phase space, but a point in phase space doesn't just tell you the particle's position, it tells you its entire physical state.) Since the integral curves can't intersect, if someone draws a set of integral curves for you, you can always bend and stretch the picture so that the integral curves look like parallel lines -- in some finite neighborhood.<|endoftext|> TITLE: Number of vectors so that no two subset sums are equal QUESTION [18 upvotes]: Consider all $10$-tuple vectors each element of which is either $1$ or $0$. It is very easy to select a set $v_1,\dots,v_{10}= S$ of $10$ such vectors so that no two distinct subsets of vectors $S_1 \subset S$ and $S_2 \subset S$ have the same sum. Here $\sum_{v \in S_i} v$ assumes simple element-wise vector addition where element addition takes place over $\mathbb{R}$. For example, if we take the vectors that are the columns of the identity matrix as $S$ this will do. What is the maximum number of vectors one can choose that has this property? I previously asked this question on MSE . An explicit construction of $17$ vectors was given by Oleg567 using computer search and an upper bound of $45$ was given by jpvee simply using the observation that $\sum_{k=1}^{17} {46 \choose k} > (17+1)^{10}$ implies that $46$ vectors is impossible. Lower bound improved to $18$ by Oleg567. Upper bound still stuck at $45$ although it seems implausible the true value is far from the current lower bound. Upper bound of $36$ given by Seva. Conjecture Feb 24, 2014. I conjecture the optimal solution size is $\lfloor \frac{1}{2} (n+1) \log_2(n+1) \rfloor$. For $n=2\dots 15$ this is $2, 4, 5, 7, 9, 12, 14, 16, 19, 21, 24, 26, 29, 32$. New lower bound of $19$ by Brendan McKay. New upper bound of $30$ by Brendan McKay. REPLY [5 votes]: Inspired by Seva's probabilistic method, I will show how to improve the upper bound to 30. Imagine we have a 0-1 matrix $A$ of 31 rows and 10 columns. I will show that there are two different subsets of the rows that have the same sum. Define the 10-dimensional random variable $X=(X_1,\ldots,X_{10})$ whose value is the sum of a random subset of the rows. This is the same random variable defined by Seva. Seva proceeded by showing that each $X_j$ is concentrated on a few values. I will improve the bound by considering the components in pairs. Write $X$ as $(Y_{12},Y_{34},Y_{56},Y_{78},Y_{9,10})$, where $Y_{12}=(X_1,X_2)$, $Y_{34}=(X_3,X_4)$, and so forth. The distribution of $Y_{12}$ depends only on the parameters $w_{01},w_{10},w_{11}$, which are respectively the number of times that 01, 10, 11 occur in the first two columns of $A$. (And so 00 occurs $31-w_{01}-w_{10}-w_{11}$ times.) The probability generating function for $Y_{12}$ is $$ F_{12}(x_1,x_2) = 2^{-w_{01}-w_{10}-w_{11}}(1+x_1)^{w_{10}} (1+x_2)^{w_{01}} (1+x_1x_2)^{w_{11}}. $$ (The coefficient of $x_1^ax_2^b$ is the probability that $Y_{12}=(a,b)$.) By trying all possible $w_{01},w_{10},w_{11}$, we find that in each case there is some set $K_{12}$ of 55 values such that $$\textrm{Prob}( Y_{12}\notin K_{12}) \le p = \frac{300387}{2097152} \approx 0.1432.$$ This calculation is easy for a computer: just expand $F_{12}$ and sum the largest 55 coefficients. One worst case is $w_{01}=w_{10}=10, w_{11}=11$. By symmetry, there are also sets $K_{34},\ldots,K_{9,10}$ of size 55 containing at least the fraction $1-p$ of $Y_{34},\ldots,Y_{9,10}$, respectively. By the union bound, at least the fraction $1-5p$ of $Y$ lies in $$K = K_{12}\times \cdots\times K_{9,10}.$$ However, $(1-5p)2^{31} > |K| = 55^5$, Therefore, there are two values the same. It should be possible to do better by grouping the columns even more. I think a non-trivial but plausible computation could handle the 10 columns in two groups of 5.<|endoftext|> TITLE: Convergence rate of the central limit theorem near the center of the distribution QUESTION [9 upvotes]: I'm looking for fast convergence rates for the central limit theorem - when we are not near the tails of the distribution. Specifically, from the general convergence rates stated in the Berry–Esseen theorem http://en.wikipedia.org/wiki/Berry%E2%80%93Esseen_theorem we know that, under certain conditions, the cumulative probability distribution of the scaled mean of a random sample $F_n(x)$ converges to the cumulative normal distribution $\Phi(x)$ with a convergence rate of $n^{-1/2}$, where n is the sample size. However, as stated in http://en.wikipedia.org/wiki/Central_limit_theorem it is well known that "As an approximation for a finite number of observations, it [the central limit theorem] provides a reasonable approximation only when close to the peak of the normal distribution; it requires a very large number of observations to stretch into the tails." Therefore, my question is: If we are given additional assumption that $|x| TITLE: Analogy between the nodal cubic curve $y^2=x^3+x^2$ and the ring $\mathbb{Z}[\sqrt{-3}]$? QUESTION [19 upvotes]: I'm trying to motivate a bit of algebraic geometry in an abstract algebra course (while simultaneously trying to learn a bit of algebraic geometry), and I thought that it might be nice to present an example of the analogy between Riemann surfaces and algebraic integers (since one might say that the entire subject originates from this analogy --- see Dedekind and Weber, 1880). I'm also hoping to motivate the concept of "integral closure" and to make the case that "unique factorization" is spiritually the same as "smooth". Anyway, I thought that perhaps the best examples from each side of the analogy would be $y^2=x^2(x+1)$ and $\mathbb{Z}[\sqrt{-3}]$. The coordinate ring of $y^2=x^2(x+1)$ is $\mathbb{C}[t^2-1,t^3-t]$, which is not integrally closed in $\mathbb{C}(t)$. The integral closure is $\mathbb{C}[t]$. Geometrically $y^2=x^2(x+1)$ is $\mathbb{C}P^1$ with two points identified and the integral closure separates the two points to obtain $\mathbb{C}P^1$. Similarly, $\mathbb{Z}[\sqrt{-3}]$ is not a UFD (read: smooth) and since UFD implies integrally closed, one might naively hope that the integral closure of $\mathbb{Z}[\sqrt{-3}]$ is a UFD. Indeed it is. The integral closure of $\mathbb{Z}[\sqrt{-3}]$ is the ring of Eisenstein integers $\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]=\mathbb{Z}[\omega]$, where $\omega=e^{2\pi i/6}$. Then one can use the very nice geometry of $\mathbb{Z}[\omega]$ to prove that it is Euclidean, hence UFD. One can view the inclusion map $\mathbb{Z}[\sqrt{-3}]\hookrightarrow \mathbb{Z}[\omega]$ as a projection of spectra $\mathrm{Spec}\,\mathbb{Z}[\omega]\twoheadrightarrow \mathrm{Spec}\,\mathbb{Z}[\sqrt{-3}]$. I'm hoping that this projection "looks like" identifying two points of $\mathbb{C}P^1$. (I realize that in general the integral closure of $\mathbb{Z}[\alpha]$ is not UFD, but only has unique factorization of ideals. Pedagogically, though, that seems like too many complications in one example.) Can anyone help me flesh out the analogy? Since it's an abstract algebra class (not an algebraic geometry class) I would like to keep things in naive terms. Maybe you can think of a more appropriate pair of examples? Thanks. REPLY [16 votes]: I think this will be a needlessly confusing example. In algebraic geometry over an algebraically closed field, there are two basic examples of nonnormal curves: the node and the cusp. Explicit equations are $y^2=x^2+x^3$ and $y^2 = x^3$. respectively. If $X$ has a node, and $\tilde{X}$ is its normalization, then $X$ is obtained by identifying two different points of $\tilde{X}$. An analogous example in number theory is $\mathbb{Z}[\sqrt{-7}]$, with normalization $\mathbb{Z}[(1+\sqrt{-7})/2]$. There are two maps $\mathrm{Spec} \ \mathbb{F}_2 \to \mathrm{Spec} \ \mathbb{Z}[(1+\sqrt{-7})/2]$ and $\mathrm{Spec} \ \mathbb{Z}[\sqrt{-7}]$ is the equalizer of these maps. This is what I would compare to $y^2 = x^2 + x^3$. If $X$ has a cusp, and $\tilde{X}$ is its normalization, then $\tilde{X} \to X$ is bijective on points, but elements in the coordinate ring of $\tilde{X}$, if they vanish at the cusp, must vanish to order $>1$. An analogous number theory example is $\mathbb{Z}[\sqrt{8}]$, with normalization $\mathbb{Z}[\sqrt{2}]$. If you squint, the ring $\mathbb{Z}[y]/(y^2 - 2^3)$ even looks like $k[x,y]/(y^2-x^3)$. The example of $\mathbb{Z}[\sqrt{-3}]$ exhibits a phenomenon which you simply don't see over an algebraically closed field: The normalization map is bijective on the set of closed points, but there is a residue field extension. In other words, the two maps from $\mathrm{Spec} \ \mathbb{F}_4$ which Qiaochu mentions have the same image, but differ by the automorphism of $\mathbb{F}_4$. This phenomenon does happen with curves over a non-algebraically closed field: Consider $x^2+y^2=x^3$ over $\mathbb{R}$. But you probably don't want to discuss that in an intro course. I would just use one or both of the examples in the previous paragraphs, and only talk about $\mathbb{Z}[\omega]$ if the students made me. REPLY [5 votes]: As a "public service," so-to-speak, here is a plot of $y^2 = x^3 + x^2$:<|endoftext|> TITLE: Conjecture on NP-completeness of tesselation of Wang Tile up to finite size QUESTION [9 upvotes]: Motivated by these following questions on tessellation: coloring in lattice Reference for Wang Tile Computational approach deciding whether a set of Wang Tile could tile the space up to some size Could we actually show that the problem of trying to tile a finite size square with a given set of Wang tile to be NP-complete? REPLY [6 votes]: The problem is also trivially NP-complete without an anchor tile. The trick is that we can easily force any tile to become an anchor tile. First, take a set of tiles that have a unique way of tiling an $n\times n$ square, e.g., by putting (respectively matching) $(i,j)$'s on their sides where $1\le i,j\le n$. Then take the "direct product" of this tile set with, e.g., the tile set described by Joel except for the position where you want your anchor tile, there keep only that. This gives a suitable tile set. ps. I used to give this as an exercise for undergrads and some of them always solved it.<|endoftext|> TITLE: Does the shortest path between two braids pass through string links? QUESTION [9 upvotes]: One of the fundamental facts underlying the application of braid theory to knot theory is that braids inject into string links. This means that braids $B_1$ and $B_2$, considered inside a cube $I^3$, are related by ambient isotopy if and only if they are related by height-preserving ambient isotopy. This is a non-trivial fact whose proofs are all somewhat complicated (Stallings Theorem/ Magnus expansion/ embedding fibrations). Given that there is no theoretical advantage to injecting braids into string links (distinct braids stay distinct), I wonder whether there is a computational advantage in doing so. Explicitly, given diagrams for $B_1$ and for $B_2$, is the minimum number of Reidemeister moves between them always realized for Reidemeister moves between braids? Or might the `shortest path between two braids' pass through string links? Question: Is there an example of a pair of equivalent braid diagrams, considered as tangle diagrams, such that the minimum number of Reidemeister moves between them is increased if we allow only braid-like Reidemeister moves (i.e. if the result of each Reidemeister move must also be a braid)? REPLY [7 votes]: I do not have the reputation to comment, but I'll delete this after you've read it. There is an idea, mostly from virtual knot theory, called parity (I recommend this reference: arxiv:1211.0403, and I won't bother copying the definition). Fiedler and Stoimenow's "New Knot and Link Invariants" paper from the Knots in Hellas proceedings' book gives you an example of a parity for classical braid (they call it type), and some other exist. I think that a parity projection proof might work to show that in fact no such shortcut can exist. You just need to define the appropriate parity. EDIT: I know about parity, but can't see right now how it would lead to a proof- anything more you can say would be great! The standard idea of a parity-projection proof is that you find a parity which singles out crossings that do not satisfy the property you want and then an arbitrary sequence of Reidemeister moves between two even diagrams. The Reidemeister moves which create odd diagrams are ignored, usually by turning them into the corresponding virtual move. Then, by the axioms of parity, the resulting sequence of diagrams, which each differ by a single extended Reidemeister mover all satisfy the desired property. In the case that you want, using virtual crossings is a bad idea because an isolated crossing would be odd, and that would restrict the next moves. (Think of doing a RM2, then a RM on a arc between the two new crossings, then a RM3. If the RM1 is changed to a virtual RM1, the RM3 is no longer allowed without going to unwelded string links, and classical braids do not embed inside those.) So, my idea would be to use a projection that smooths the crossing according to the orientation of the string link. We still have to define an appropriate parity. Now, this is the part where I am not sure the math actually works, since this construction does not respect the parity axioms, one still has to check that this projection isn't too destructive. I will call this the monotonicity parity, denoted $m(c)$. Let $c$ be a crossing is a string link $L$, and $L_c^+$, $L_c^-$ be the components of $L$ smoothed at $c$ (again, with orientation) that contain the arcs that were smoothed. The, $m(c)=0$ if and only if both $L_c\pm$ are long components. It is quite obvious that this parity is invariant under Reidemeister moves away from the crossing smoothed and planar isotopies, and not too complicated to see that two crossings that can cancel in a RM2 have the same parity and an isolated crossing is odd. The hard part is RM3. The parities of the crossings involved in one is invariant under the move, but what about the requirement that the number of odd crossings in a RM3-like configuration not be odd? It isn't satisfied, precisely by the example sequence I gave above, given that you had a braid after the RM2. This is were everything goes wrong in the general case, but possibly not in the case you care about, since the odd crossings created by moves from a braid diagram are unknots which aren't linked with the rest of the string link. Possibly, these unknots can be erased, and any future RM that involved them ignored, while still obtaining the same final braid diagram.<|endoftext|> TITLE: Is the forcing relation defined for mathematical formulas? QUESTION [7 upvotes]: Meta-matematical formulas of the language of set-theory (which are not sets, but just sequences of signs) should not be confused with mathematical ones (i.e. formulas coded as sets, e.g. finite sequences of natural numbers or even just natural numbers, if we wish). For instance, for each meta-mathematical formula $\phi$, we can define its relativization $\phi^M$ to any class $M$, but the relation $M\vDash \phi[v]$ in the model-theoretic sense can be only defined when $M$ is a set, but not a proper class. The impossibility of defining $M\vDash\phi[v]$ for proper classes is due to two kinds of reasons: on one hand, if so, we could prove Con ZFC within ZFC, and technically, the problem is that a "typical" recurrence definition of $M\vDash \phi[v]$ would require having defined $M\vDash \psi[w]$ for each subformula $\psi$ of $\phi$ and for each valuation $w$ from the set of free variables of $\psi$ to $M$, and such valuations form a proper class if $M$ is a proper class, and this does not fit the recursion theorem. My first question is that I suspect that this is also valid for the definition of the forcing relation $p\Vdash \phi$, i.e.: Question 1: Can the forcing relation $p\Vdash \phi$ be defined for mathematical formulas $\phi$ or just for meta-mathematical ones? I suspect that only for meta-mathematical ones because in order to define $p\Vdash \forall x\phi(x)$ we must assume that $p\Vdash \phi(\tau)$ is defined for each $\mathbb P$-name $\tau$, and the class of $\mathbb P$-names is a proper class. This question is related to theorem III 2.11 of Shelah's book Proper and Improper Forcing (second edition). It states that if $\lambda$ is an uncountable cardinal, $N\prec H(\lambda)$ is an elementary submodel of the set of all sets whose transitive closuse has cardinality less than $\lambda$, and $\mathbb P\in N$ is a pre-ordered set, then, for each generic filter $G$ over $V$ we have $N[G]\prec H^{V[G]}(\lambda)$. The proof is clear, but in order to show that the definition of elementary submodel holds for a formula $\phi$ (in fact, the Tarski-Vaught criterion is considered instead of the definition), the forcing relation $p\Vdash \psi$ is used for a formula $\psi$ involving $\phi$. Hence, if the answer to Question 1 is that it only makes sense for meta-mathematical formulas, we conclude that $N[G]$ is an elementary submodel just in the weak meta-mathematical sense that we have a family of theorems, each one stating that an arbitrary meta-mathematical formula is absolute for $N[G]-H^{V[G]}(\lambda)$. So, my second question is: Question 2: Under the hypotheses of Shelah's theorem, can we conclude that $N[G]\prec H^{V[G]}(\lambda)$ in the usual model-theoretic sense (for mathematical formulas) or just in the weak meta-mathematical sense? I feel that the answer should be positive even if the answer to question 1 is negative. REPLY [5 votes]: To answer Question 2, I think your intuition is right. Assume $N \prec H_\lambda$, where $\lambda$ is a regular cardinal bigger than $2^\mathbb{P}$, where $\mathbb{P} \in N$ is a partial order. (Maybe Shelah has a more subtle argument where we assume less about $\lambda$, not sure.) Then $H_\lambda$ is a model of $ZFC$ minus powerset, and we can consider in $V[G]$ the model $H_\lambda[G]$, defined the same way as usual. We have: For any generic $G \subseteq \mathbb{P}$, $H_\lambda^{V[G]} = H_\lambda[G]$. $G$ is generic over $V$ iff $G$ is generic over $H_\lambda$. The basic forcing lemmas, including the "truth lemma," go through for forcing with $\mathbb{P}$ over $H_\lambda$. The point is that by working in $V[G]$, we can treat $H_\lambda[G]$ as an ordinary set model and carry out the argument that $N[G] \prec H_\lambda[G]$ within the language of set theory in $V[G]$, rather than as a meta-mathematical scheme.<|endoftext|> TITLE: Maximal regions with given diameter QUESTION [5 upvotes]: Let us call a bounded region $D$ in the plane maximal if the conditions $D\subset D'$ and $\mathrm{diam} D'=\mathrm{diam} D$ imply $D'=D$. Is it possible to describe all maximal regions? The only examples I know are discs and Reuleaux triangles. If a complete description is difficult, can one prove some properties of maximal regions? For example, I suppose that they must be convex. Do they have piecewise-analytic boundaries? This is inspired by my answer to another MO question REPLY [4 votes]: I am pretty sure that the condition you state is equivalent to $D$ being of constant width, not just in the plane, but in every dimension. See, for instance http://www.ciem.unican.es/encuentros/banach/2012/moreno1.pdf Other references: Dalla, Leoni; Tamvakis, N. K. Sets of constant width and diametrically complete sets in normed spaces. Bull. Soc. Math. Grèce (N.S.) 26 (1985), 27–39, MR0854917. Moreno, José Pedro; Schneider, Rolf Diametrically complete sets in Minkowski spaces. Israel J. Math. 191 (2012), no. 2, 701–720, MR3011492.<|endoftext|> TITLE: Real-rootedness, interlacing, root-bounds of a sequence of polynomials QUESTION [11 upvotes]: Problem: the number $a(n,k)$ is defined by the following recurrence \begin{equation} a(n,k)=(k+1)(k+2)\, a(n-1, k)+\frac{(k+1)(k+2)(k+3)}{k} \,a(n-1, k-1), \end{equation} with $a(1,1)=1$ and $a(n,k)=0$ if $k<1$ or $k>n$. For fixed $n$, the generating polynomial of $a(n,k)$ is defined as $A_n(x)=\sum_{k=1}^n a(n,k)x^k$. The recurence above is equivalent to the following differential equation $$ \frac{\mathrm{d}}{\mathrm{d}x} A_{n+1}(x) =24 A_n(x)+(36x+6)\frac{\mathrm{d}}{\mathrm{d}x} A_{n}(x)+(12x^2+6x)\frac{\mathrm{d}^2}{\mathrm{d}x^2} A_{n}(x)+(x^3+x^2)\frac{\mathrm{d}^3}{\mathrm{d}x^3} A_n(x). $$ Question 1: all roots of $A_n(x)$ are real? Question 2: $A_n(x)$ interlaces $A_{n+1}(x)$? i.e. $$ b_1 \leq a_1 \leq b_2 \leq a_2 \leq \cdots \leq b_n \leq a_n \leq b_{n+1}, $$ where $\{a_i\}$ and $\{b_j\}$ are roots of $A_n(x)$ and $A_{n+1}(x)$, respectively. Question 3: all roots of $A_n(x)$ are located in $(-1,0]$? These three statements are verified to be true for $n\leq 50$. Some examples of $A_n(x)$ are given \begin{eqnarray*} A_1(x) &=& x, \\ A_2(x) &=& 6x + 30 x^2,\\ A_3(x) &=& 36x + 540 x^2+ 1200 x^3,\\ A_4(x) &=& 216x + 7560 x^2+ 45600 x^3 +63000 x^4. \end{eqnarray*} Background of this problem: this number arises from certain graph enumeration problem. One can easily prove the log-concavity of $\{a(n,k)\}_k$ by induction. Many literatures on real-rootedness or interlacing deal with recurrences with polynomial coefficients or first and second order differential equations, not for this example. Any things about these numbers and polynomials would also be appreciated. Progress: as pointed out by Per Alexandersson (see below reply), the linear differential operator $p\mapsto 24 p +(36x+6)p' + (12x^2+6x)p'' + (x^3+x^2)p'''$ preserves real-rootedness. This show that if $A_n(x)$ has only real zeros, so does the right hand side of the differential equation. Notice that if a polynomial $f$ has only real zeros, the primitive integral $\int f dx$ of $f$ could have complex zeros in general. I am wondering, under what conditions of $f$, primitive integral $\int f dx$ has only real zeros. (Assume that the constant term of $\int f dx$ is zero.) REPLY [3 votes]: Well, if you can show that the linear differential operator $p\mapsto 24 p +(36x+6)p' + (12x^2+6x)p'' + (x^3+x^2)p'''$ preserves real-rootedness, then you are a bit closer. Now, this can be attacked by proving that the symbol of this operator is stable, see http://www.math.kth.se/~dirocco/ML2011/CIAMWORKSHOP/25/Branden.pdf It will be some work; you end up with a 2-variable polynomial, and you want to show that this is non-zero whenever both variables have positive real part. Now, it is quite easy to see that in "the limit", i.e, for large degrees of your polynomial $p$, this operator preserves real-rootedness, since then only the $(x^3+x^2)p'''$ term matters. And $(x^3+x^2)$ has roots 0 and -1, so it is not surprising that you get [-1,0] as the special interval.<|endoftext|> TITLE: SDP formulation of noisy low rank matrix completion QUESTION [6 upvotes]: Exact low rank matrix completion using nuclear norm minimization can be formulated as a semidefinite program (SDP). Following the notation in the paper, a convex problem for noisy matrix completion can be: $$\text{minimize} \,\, \|X\|_* \quad \text{subject to} \quad \|\mathcal{P}_\Omega(X)-\mathcal{P}_\Omega(M)\|_2\le\epsilon \tag{*}$$ where $\mathcal{P}_\Omega(X)$ is the projection of $X$ onto the set of observed entries $\Omega$. This problem can be formulated as the SDP below SDP in $X, W_1, W_2$ $$\begin{array}{rl} \text{minimize} & \text{trace}(W_1)+\text{trace}(W_2)\\ \text{subject to} & \left[ \begin{array}{cc} \epsilon I & \mathcal{P}_\Omega(X-M) \\ \mathcal{P}_\Omega(X-M)^T & \epsilon I \end{array} \right] \succeq 0\\ & \left[ \begin{array}{cc} W_1 & X \\ X^T & W_2 \end{array} \right] \succeq 0\end{array}$$ The question is when the $2$-norm constraint of problem (*) is replaced by the Frobenius norm constraint, i.e., $$\text{minimize} \,\, \|X\|_* \quad \text{subject to} \quad \|\mathcal{P}_\Omega(X)-\mathcal{P}_\Omega(M)\|_F \le \epsilon$$ it can still be cast as a SDP as stated in here and many other papers. Could anyone tell me how to formulate this problem as a SDP? Edit: follow up question. REPLY [8 votes]: The Frobenius norm does not cause a problem. Remember, the Frobenius norm of a matrix $X$ is actually nothing more than the 2-norm of the vector formed by stacking the columns of $X$ on top of each other. And for vectors, the vector 2-norm and the matrix 2-norm coincide. So if $\mathcal{Q}$ implements this "stacking" isomorphism, we have $$\|\mathcal{P}_\Omega(X-M)\|_F=\|\mathcal{Q}(\mathcal{P}_\Omega(P-M))\|_2$$ and $$\|\mathcal{P}_\Omega(X-M)\|_F\leq\epsilon \quad\Longleftrightarrow\quad \begin{bmatrix} \epsilon I & \mathcal{Q}(\mathcal{P}_\Omega(X-M)) \\ \mathcal{Q}(\mathcal{P}_\Omega(X-M))^T & \epsilon \end{bmatrix} \succeq 0$$ In fact, handling the Frobenius norm is actually somewhat easier than the spectral norm in practice, because most solvers handle vector 2-norms more directly and efficiently. See second-order cone programming or "semidefinite quadratic linear programming" (SQLP) for more background.<|endoftext|> TITLE: Is there a nice "minimum" of two symmetric operators? QUESTION [7 upvotes]: Let $A$ and $B$ be two bounded symmetric positive operators in Hilbert space, such that $A-B$ is trace class. If needed, $A$ and $B$ may be assumed reasonably "small", let's say, Hilbert-Schmidt. Does there exist another symmetric operator $C$, $0 \le C \le A$, $0 \le C \le B$, such that both $A-C$ and $B-C$ are trace class? REPLY [9 votes]: Let $P=\left[\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}\right]$ and $Q(\phi)=\left[\begin{matrix} \cos^2(\phi) & \cos(\phi)\sin(\phi) \\ \cos(\phi)\sin(\phi) & \sin^2(\phi) \end{matrix}\right]$. Then $P$ and $Q$ are orthogonal projections and if $\phi\neq 0$, then the only operator $T$ which satisfies $0 \leq T \leq P$ and $0 \leq T \leq Q(\phi)$ is zero. Now, set $A := \oplus_{n \in \mathbb N} n^{-1} P$ and $B:= \oplus_{n \in \mathbb N} n^{-1} Q(\pi/n)$. Then, $A,B$ are Hilbert-Schmidt (and not trace-class), $A-B$ is trace-class and the $2$-dimensional phenomenon from above shows that every operator $C$ with $0 \leq C \leq A$ and $0 \leq C \leq B$ must be zero. In particular, $A-C$ and $B-C$ are not trace-class. Thus, the answer to your question is no.<|endoftext|> TITLE: Number of partitions whose blocks form arithmetic progressions QUESTION [6 upvotes]: As is known, the set $\{1,\ldots,n\}$ has $2^n$ many subsets and $B_n$ (the $n$th Bell number) many partitions, where clearly $B_n<2^{2^n}$ and it is actually known that $B_n TITLE: Ramanujan's problem 754 still open? QUESTION [10 upvotes]: In addition to the MO question The Ramanujan Problems. , I would like to ask the following. Problem 754 from the list of the Ramanujan's problems ( http://www.imsc.res.in/~rao/ramanujan/collectedpapers/question/q754.htm ) asks to show that $$e^xx^{-x}\pi^{-1/2}\Gamma(1+x)=(8x^3+4x^2+x+E)^{1/6},$$ where E lies between $\frac{1}{100}$ and $\frac{1}{30}$ for all positive values of $x$. According to http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.39.9228 (The Problems Submitted by Ramanujan to the Journal of the Indian Mathematical Society, by Bruce C. Berndt, Youn--seo Choi and Soon-Yi Kang) this problem was still open in 1997. What is the status of this problem at present? REPLY [17 votes]: Perhaps I should elevate my comment to an answer. The problem is solved in Ekatherina A. Karatsuba, On the asymptotic representation of the Euler gamma function by Ramanujan, J. Comput. Appl. Math. 135 (2001), no. 2, 225–240, MR1850542 (2002i:33004). A version of this paper is available here.<|endoftext|> TITLE: Lines in image; are they significant to prime numbers if so how? QUESTION [5 upvotes]: Amateur math question. I was playing around generating some 2D images, and wondered what it would look like if placed $P_{i}$ dots on a circle with diameter of $i$ for increasing values of $i$, where $P_{i}$ is the $(i + 1)$-th prime - with $P_{0}$ being 1. One interesting thing about the image is that there appear to be contours emanating from the center of the image. In particular there appear to be 3 distinguishable meta axes running though the center of the image. One top to bottom, the other two at about 30 degrees to the horizontal. I can explain the one running top to bottom easily, and also the general left to right symmetry. But I can't explain the other two. Why are they there? Note the first image has had some pre-processing and a threshold to show the contours more clearly. Note code used to generate points is like: primes = []; radius = 75; get_primes( primes, radius + 1 ); for( i = 0; i <= radius; i++ ) prime = primes[i]; for( j = 1; j <= prime; j++ ) y = i * cos( 2 * M_PI * ( j /prime ) ); x = i * sin( 2 * M_PI * ( j/prime ) ); printf( "%5f\t%5f\n", x, y ); end; end; Appendix 1: Original image, and one with 0 <= i < = 1000: Appendix 2: I used PHP script to generate the dots, then gnuplot to generate image, and then GIMP for further processing. See http://pastebin.com/0Hfb6Kdq for the simple PHP and gnuplot code. REPLY [8 votes]: The vertical line going up comes from when $j=prime$, so $x=0$, $y=i$. The vertical line going down comes from the fact that $prime/2$ is always fairly far ($1/2$) away from an integer, so $x=0$, $y=-i$, is white for $i$ sufficiently small. There are also repeating patterns that occur nearby and approximately parallel to the axis. These correspond to formulas like $j=prime/2+1/2$, $j=primes/2+1$, etc. The other axes are explained in a similar way. They correspond to $ prime/6$, $prime/3$, $2prime/3$, $5prime/6$, which are always $1/3$ away from an integer. $1/3<1/2$ so this is a less powerful pattern that lasts a shorter amount of time. Also, you see an oscillation based on which integer is closer, which depends on whether $p\equiv 1$ or $p \equiv 2$ modulo $3$. If you shrunk the radius of the dots, then more primes would appear, and I believe you would start to see a more distinct pattern on the horizontal axis corresponding to $prime/4$. As the dots get smaller more and more patterns would appear. It might be interesting to make the radius of the dots vary proportionally to $i/prime$ so they always cover the same proportion of the circle.<|endoftext|> TITLE: Nonmonotonicity of expected distance of a random walk QUESTION [5 upvotes]: What's the simplest example of a reversible random walk $X_n$ on an infinite vertex-transitive graph such that the expected distance from the origin is not increasing, i.e. there exists $n$ such that: $\mathbb{E} \,d(X_0, X_{n+1}) < \mathbb{E}\,d(X_0, X_n)$? EDIT: Douglas Zare provided an example below, but the random walk is not simple (we are giving bigger weight to one of the outgoing edges) and the nonmonotonicity is essentially due to a parity issue, which is not that interesting. So let me reformulate the question: does there exist a simple random walk on an infinite vertex-transitive graph (preferably: a Cayley graph) where nonmonotonicity would not arise due to some periodicity issue? The phenomenon I have in mind is that some graphs have "dead ends", i.e. vertices from which every step takes the random walker closer to the origin (equivalently: geodesics to such vertices cannot be extended). But I don't know whether this obstruction can be serious enough so that on average the distance will be nonmonotonous. REPLY [3 votes]: Here are two examples with bipartite graphs. Let the vertices be the integers. Take steps of $\pm(2n+1)$ with probability $2^{-n-2}$. This is bipartite and all vertices are connected to each vertex in the other part. At odd time steps, the distance from the origin is always $1$. At even time steps, the distance from the origin is usually $2$, with a small probability of $0$. For example, $\mathbb{E}d(x_0,x_2) = 1/6(0)+5/6(2) = 5/3 \gt \mathbb{E}d(x_0,x_3) = 1$ so the third step doesn't increase the average distance. Let the vertices be $\lbrace n \rbrace_{n\in \mathbb Z} \cup \lbrace n' \rbrace_{n\in \mathbb Z}$. Let $n$ move to $n'$ with probability $p$ close to $1$, and to $(n-1)'$ or $(n+1)'$ with low probability. Symmetrically, let $n'$ move to $n$ with probability $p$ and to $(n-1)$ or $(n+1)$ with low probability. Taking two steps usually returns to the origin. $\mathbb{E}d(x_0,x_1) = 1$. $\mathbb{E}d(x_0,x_2) \le 2(1-p^2)$ which is less than $1$ for $p \gt 1/\sqrt{2}$ so the second step doesn't increase the average distance.<|endoftext|> TITLE: Why are they called Specht Modules? QUESTION [12 upvotes]: I know that the simple modules of $\mathbb{C}S_n$ are called Specht Modules, and they are named after the German Mathematician Wilhelm Specht because he studied them, but I think these modules were studied before him, for example by Frobenius. Could anyone please tell me what is Specht's contribution to this area that made people name these modules after him? **I was reluctant about asking this question here or in MathsStackExchange, but finally, I posted it here, if you think it is too naive to be here, please let me know I will happily move it to there. REPLY [18 votes]: The question is interesting though perhaps not strictly "research-level". Terminology in mathematics develops a bit haphazardly, and sometimes things get misleading names. In this case the work of Specht around 1935 did place the representations of symmetric groups in the then-modern setting of module theory. But the notion of "Specht module" seems to have emerged around 1970 in the rapidly developing work on representations of the groups in prime characteristic. Work of M.H. Peel and especially work of Gordon James popularized the notion. In particular, James exploited the fact that the characteristic 0 Specht modules have a fairly natural reduction mod $p$ for any prime $p$. This is somewhat analogous to the algebraic group situation, where "Weyl modules" come by such reductions and then have a unique distinguished composition factor. By now the literature on symmetric group representations in prime characteristic is quite extensive, with the term "Specht module" being ubiquitous. In the classical characteristic 0 theory, no such language is usually needed. Anyway, one might make a case for the terminology "James module" here, but it's too late for that. Specht himself had no special influence on the modular theory.<|endoftext|> TITLE: Maximum distance between two consecutive points of N random points on a unit length line QUESTION [7 upvotes]: I have encountered a seemingly simple question on distances of random points. Place N points randomly and uniformly on the line segment [0..1]. How to derive the expectation (or the distribution) of the "maximum" distance between two consecutive points? Just a reference would be very helpful. Thanks, REPLY [3 votes]: For $n$ random points on a circle, Theorem 1 of this article provides an expression for the $m$th moment of the maximum distance, namely $$ \mathbb{E}\left[\left(d_{\max}\right)^m\right]=\frac{(n-1)!}{(n+m-1)!}\sum_{\stackrel{r_1+2r_2+\ldots+m r_m}{r_i\in\mathbb{N}_0}}{\frac{m!}{r_1!1^{r_1}r_2!2^{r_2}\cdots r_m! m^{r_m}}H_{n,1}^{r_1}H_{n,2}^{r_2}\cdots H_{n,m}^{r_m}}, $$ where $H_{n,k}=\sum_{j=1}^n{j^{-k}}$ is the $n$th harmonic number of order $k$.<|endoftext|> TITLE: K3 surface with $D_{14}$ singular fiber QUESTION [5 upvotes]: Let $X$ be an elliptic K3 surface with $D_{14}$ singular fiber. Do you know an explicit equation for such $X$? Also, how many disjoint sections such fibration admits? Any reference would be greatly appreciated. REPLY [6 votes]: [In comments guest2014 amended the question to ask not for a $D_{14}$ fiber but for $I^*_{14}$, a.k.a. $\tilde D_{18}$] The elliptic surface $$ X : y^2 = x^3 + (t^3+2t) x^2 - 2(t^2+1)x + t $$ over ${\bf C}(t)$ has a $I^*_{14}$ fiber at $t=\infty$. (Note that the right-hand side is a cubic in $x$ whose discriminant $4t^4 + 13 t^2 + 32$ has degree only $4$ in $t$, where a generic K3 surface would have discriminant of degree $24$.) $X$ is the unique such surface with a section, up to isomorphism. The group of sections is trivial: the zero-section and the components of the $\tilde D_{18}$ fiber already generate a lattice ${\rm II}_{1,1} \oplus D_{18} \langle -1 \rangle$ of rank $20$ in ${\rm NS}(X)$, so the Mordell-Weil rank is zero (else the Picard number would exceed $20$), and if there were a $2$-torsion section then the Neron-Severi lattice would be ${\rm II}_{1,1} \oplus L \langle -1 \rangle$ for some even unimodular lattice $L$ of rank $18$, which is impossible because $18 \not\equiv 0 \bmod 8$. [added a bit later: For the situation in positive characteristic see Schütt's paper, which cites Shioda's note Shioda, T.: The elliptic K3 surfaces with a maximal singular fibre, C. R. Acad. Sci. Paris, Ser. I, 337 (2003), 461-466. for the description of elliptic surfaces with a maximal singular fiber in characteristic zero.]<|endoftext|> TITLE: Which automated theorem provers can address the combinatorics of periods in strings? QUESTION [5 upvotes]: Five years ago, I made a conjecture on the number of correlation classes that are exhibited by pairs of words in an alphabet of a given size. I later speculated that the conjecture could be tackled via automated theorem proving. I would now like to make some progress on this conjecture, by finding out whether there are any automated theorem provers capable of at least formulating a simpler statement along the same lines. The simpler statement that I have chosen is Theorem 1 of Halava, V., Harju, T., & Ilie, L. (2000). Periods and binary words. Journal of Combinatorial Theory, Series A, 89(2), 298-303. This theorem is a restatement of part of Theorem 5.1 of Guibas, L. J., & Odlyzko, A. M. (1981). Periods in strings. Journal of Combinatorial Theory, Series A, 30(1), 19-42: Theorem 1: For any alphabet $A$ and any word $w \in A^*$, there exists a word $w' \in \{0,1\}^*$ such that $\mathcal{P}(w') = \mathcal{P}(w).$ Here $\mathcal{P}(w)$ is the set of periods of the word $w$: An integer $p,$ with $1 \leqslant p \leqslant |w|,$ is called a period of $w$ if $w_i = w_{i+p},$ for all $1 \leqslant i \leqslant |w|-p.$ My questions are: Which automated theorem provers are capable of formulating Theorem 1 with its proper semantics (i.e arbitrary finite strings over an arbitrary finite alphabet)? Which automated theorem provers are cable of proving Theorem 1, along the lines of Guibas and Odlyzko, or Halava, Harju and Ilie, or in an otherwise intelligible manner? How much work would be involved in creating such a formulation and proof? Would this be an undergraduate student assignment, a PhD project, or a decades-long project for a research team? Or would it require the development of new types of theorem provers? REPLY [3 votes]: I think that Any of the standard theorem provers could formulate your statement without much trouble. There would be various choices about what kind of data structures to use (implicitly or explicitly), and the right choice might make a big difference to the ease of writing proofs. I had a quick look at the paper of Halava et al, which is five pages long and is written explicitly as an algorithm to find $w'$. I think that anyone who was reasonably fluent with a proof assistant could encode and verify that proof without much trouble. One would expect the formalised proof to be several times longer than the paper. For this kind of work, I don't think there is much reason to prefer any one of the standard systems over the others. I do not think that any existing proof assistant has any chance of finding such a proof for itself, unless you can cast the problem into a form that can be attacked by a SAT solver. I don't really know anything about SAT solvers myself but this post by Gowers seems relevant and interesting. If you need to enumerate a bunch of cases to consider, I would be more inclined to use a system like Maple or Mathematica (or a general programming language) to do the initial enumeration. Later on, you could feed the results into a proof assistant to validate the overall logic if necessary.<|endoftext|> TITLE: Vaught's conjecture for partial orders QUESTION [8 upvotes]: In ``Steel, John R. On Vaught's conjecture. Cabal Seminar 76–77, pp. 193–208'' the following is proved: Theorem. Let $\phi\in L_{\omega_1,\omega}.$ If every model of $\phi$ is a tree, then $\phi$ has either $\leq \aleph_0$ models or perfectly many countable models. In page 206 of the paper the following is stated: It would be natural to attempt to extend the above theorem from tree to arbitrary partial orders. However Arnold Miller has shown Vaught's conjecture for partial orders is equivalent to the full conjecture. Question 1. How Miller's result can be proved? Added remarks and new questions: According to Enayat's answer, even the Vaught's conjecture for lattices, implies the full Vaught's conjecture. So it is natural to ask how much we can weaken our assumptions, for example Question 2. Does Vaught's conjecture for Boolean algebras imply the full Vaugh's conjecture? In general: Question 3. For which structures it is known that the full Vaught's conjecture is implied by the Vaught's conjecture for those structures? REPLY [3 votes]: As for question 2: according to http://spot.colorado.edu/~szendrei/BLAST2010/kach.pdf, Vaught’s conjecture holds for Boolean algebras (and many other classes of theories).<|endoftext|> TITLE: About the pro-algebraic group structure of $G(\mathbb{C}[[t]])$ QUESTION [5 upvotes]: I hope this is not too elementary! Let $G$ be a algebraic reductive group over $\mathbb{C}$. The group $G(\mathbb{C}[[t]])$ can be given the structure of a pro algebraic group as follows. Let $l\in \mathbb{N}$ and $J^l:=\mathbb{C}[[t]] /t^l\mathbb{C}[[t]])$. Then $G(J^l)$ is known to be an algebraic group and $G(\mathbb{C}[[t]])$ can be identified with $\varprojlim_l G(J^l)$. So my question is: Why is $G(J^l)$ an algebraic group. I mean I can prove this with dumb calculations. However, I wonder if there is an nice argument to see this? REPLY [13 votes]: As hinted by S. Carnahan, this is a particular case of a general construction, known as Weil restriction, Greenberg functor, or arc space, depending on the context. Let $X$ be a scheme of finite type over a field $k$. Then there is a scheme $\mathcal L(X)$, which is a projective limit of schemes $\mathcal L_m(X)$ of finite type over $k$, such that $\mathcal L(X)(F)=X(F[[t]])$ and $\mathcal L_m(X)(F)=X(F[[t]]/(t^m))$, for every $k$-algebra $F$. Most of the construction is, in fact, relatively easy. Begin with $X=\mathbf A^1$. Then, it suffices to take $\mathcal L_m(X)=\mathbf A^m$, the identification of $\mathcal L_m(X)(F)$ with $X(F[[t]]/(t^m))$ begin given by $(x_0,\dots,x_{m-1})\mapsto x_0+x_1t+\dots+x_{m-1}t^{m-1}$. Then $\mathcal L(X)=\mathbf A^\infty=\mathop{\mathrm {Spec}}(T_0,T_1,T_2,\dots)$. This generalizes readily to $X=\mathbf A^n$ (take the $n$th power of the preceding schemes). Now, if $X$ is a closed subscheme of $\mathbf A^n$, with ideal $I=(P,\dots)$, one can expand $P(x_0+x_1t+\dots+x_{m-1}t^{m-1})=P_0(x)+P_1(x)t+\dots+P_{m-1}(x)t^{m-1} \pmod {t^m}$ and $\mathcal L_m(X)$ is viewed as a closed subscheme of $\mathcal L_m(\mathbf A^n)$ by adding the equations $P_0=\dots=P_{m-1}=0$ for every polynomial $P\in I$ (or in a generating subset of $I$). In your particular case, where $X$ is an (affine) algebraic group, this is all you need. If $U$ is an affine open subscheme of $X$, then $\mathcal L_m(U)$ identifies as an open affine subscheme of $\mathcal L_m(X)$. This will allow to define $\mathcal L_m(X)$ in general by gluing $\mathcal L_m(U)$, for affine open subschemes $U$ of $X$. One then takes the limit $\mathcal L(X)=\varprojlim_m \mathcal L_m(X)$, which exists as a scheme, because the transition morphisms $\mathcal L_{m+1}(X)\to \mathcal L_m(X)$ are affine. Finally, the formula $\mathcal L(X)(F)=X(F[[t]])$ is easy if $F$ is a field, or if $X$ is affine (this is all you need), and is relatively easy if $X$ is quasiprojective. The general case is due to B. Bhatt (private communication). His proof used techniques of derived algebraic geometry (a theorem of Lurie/Brandenburg-Chiravasitu) and existence of enough perfect complexes (Thomason-Trobaugh). EDIT: I realize that I did not fully answer your question, which was why one gets a pro-algebraic group when one begins with an algebraic group. The point is that this construction is functorial and commutes with products. Consequently, if $X$ is a group scheme, then one gets morphisms $\mathcal L_m(X)\times_k \mathcal L_m(X)\to \mathcal L_m(X)$ and $\mathcal L(X)\times_k\mathcal L(X)\to\mathcal L(X)$ which give $\mathcal L_m(X)$ a structure of algebraic group, and $\mathcal L(X)$ a structure of group scheme, projective limit of algebraic groups.<|endoftext|> TITLE: Seeming contradiction about P vs NP between graphclasses.org and at least two papers about clique in even-hole-free graphs QUESTION [7 upvotes]: I believe correctness about clique in even-hole-free graphs of graphclasses.org and the paper Vertex elimination orderings for hereditary graph classes, Pierre Aboulker, Pierre Charbit, Nicolas Trotignon, Kristina Vuskovic would imply $P=NP$. Accordinth to graphclasses: Clique: NP-complete by IS on complement, the comeplement is related to IS on 2-subdivision citing S. Poljak, A note on the stable sets and coloring of graphs Comment. Math. Univ. Carolin. 15 (1974) 307-309 ZMath 0284.05105 According to the paper, p. 4: (Improve) Maximum weighted clique in even-hole-free graphs in time $O(nm)$. Related paper is LexBFS, structure and algorithms, Pierre Aboulker, Pierre Charbit, Maria Chudnovsky, Nicolas Trotignon, Kristina Vuskovic which claims the same. The sources appear credible to me, though an error is certainly possible. Added Third paper gives $O(n^2m)$ algorithm for MWC. On graph classes, to get an explanation about complexity click on [+]Details. What went wrong? REPLY [14 votes]: On http://www.graphclasses.org, an even hole means at least 6 vertices. If you open the details of the forbidden subgraphs list of the page for even-hole-free graphs and click on the link for even-hole, you'll see the definition. In the articles you refer to, an even hole means at least 4 vertices. The even-hole-free of the articles is what on graphclasses.org is called even-cycle-free. Hence, P=NP is still open.<|endoftext|> TITLE: A map from a curve $C$ to $\mathbb{P}^n$ which represents a generator of $H^2(C,\mathbb{Z})$. QUESTION [5 upvotes]: Let $C$ be a smooth complex curve. By the standard algebraic topology (recall that $K(\mathbb{Z},2)=\mathbb{P}^\infty$), there exists a map $\phi:C\rightarrow \mathbb{P}^n$ for sufficiently large $n$ that corresponds to a generator of $H^2(C,\mathbb{Z})$. How can one construct the map $\phi$ and can it be algebraic? REPLY [7 votes]: It cannot be algebraic, unless $C\cong \mathbb{P}^1$. What you ask is that $\phi^*:H^2(\mathbb{P}^n,\mathbb{Z})\rightarrow H^2(C,\mathbb{Z})$ takes the class $h$ of a hyperplane section of $\mathbb{P}^n$ to the class of a point $p$ of $C$. But if $\phi$ is algebraic, we have $\phi^*h=rd[p]$, where $r$ is the degree of the finite map $C\rightarrow \phi(C)$ and $d=\deg(\phi(C))$. If $r=d=1$, $\phi$ is an isomorphism onto a line in $\mathbb{P}^n$.<|endoftext|> TITLE: (Short) Exact sequences with no commutative diagram between them QUESTION [32 upvotes]: This question was asked by a student (in a slightly different form), and I was unable to answer it properly. I think it's quite interesting. The problem is to produce an example of the following situation: find a short exact sequence $$ 0 \to X_1 \to X_2 \to X_3 \to 0$$ (in some category of your choice), and a second exact sequence $$ 0 \to Y_1 \to Y_2 \to Y_3 \to 0$$ such that there are isomorphisms $X_n \cong Y_n$ for all $n$, BUT in such a way that there is no commutative diagram whatsoever between the two sequences, with the vertical maps being isomorphisms. It is impossible to find such an example in the category of vector spaces, or of finitely-generated abelian groups. I don't know about the general case, though. I would be grateful for any example, but would be disappointed if the chosen category were constructed specifically to answer the problem. EDIT / COMMENT: in the first version of this question I was asking for sequences which could potentially be infinite. Some great examples came in the comments straightaway. I'm very thankful for them, but I recall only now that the student's original question was about short exact sequences, so I've edited accordingly. (I'm sorry for the confusion, the student asked me this question several months ago, and I was posting only now, for some reason... and got it wrong. I'm very happy to know about the examples involving infinite sequences, though.) REPLY [4 votes]: Here is an answer involving finitely generated groups in the quotient, and free groups in the kernel and the total group. Suppose that $G$ is an infinite, finitely generated group. Let $F_n = \langle s_1,...,s_n \rangle$ be the rank $n$ free group, and let $f : F_n \to G$ be a surjective, noninjective homomorphism; this is a fancy way of saying that $f(s_1),...,f(s_n)$ is a generating set of $G$ but not a free basis. The kernel of $f$, being an infinite index, nontrivial, normal subgroup of $F_n$, must be of infinite rank. So, you get a short exact sequence $$1 \to F_\infty \to F_n \xrightarrow{f} G \to 1 $$ Now we just have to dream up some $G$ and two homomorphisms $f,g : F_n \to G$ for which the two short exact sequences are inequivalent. This problem is a translation of a well-studied (but nonetheless still mysterious) problem in group theory: the two short exact sequences are equivalent if and only if the generating sets $f(s_1),...,f(s_n)$ and $g(s_1),...,g(s_n)$ are ``Nielsen equivalent''. This paper (by Lars Louder with most cases attributed to Zieschang) shows that if $G$ is the fundamental group of a closed surface of genus $g \ge 2$ then any generating set of cardinality $2g$ is Nielsen equivalent to the "standard" one given by the presentation $$\pi_1(G) = \langle a_1,b_1,...,a_g,b_g \, | \,\, [a_1,b_1] [a_2,b_2] ... [a_g,b_g] = 1\rangle $$ So we won't get an answer from surface groups. On the other hand the paper of Lustig and Moriah (in Topology (1991)) gives examples where $G$ is a Fuchsian group with torsion (fundamental group of a closed 2-D orbifold) and two generating sets of the same cardinality, both of which look "standard", but which are not Nielsen equivalent. This paper (by Kapovich and Weidmann) has an introduction with a survey of examples.<|endoftext|> TITLE: How much do the classes of geodesic metrics and Green metrics on a Gromov hyperbolic group differ? QUESTION [9 upvotes]: Background Let $G$ be a Gromov hyperbolic group. If $G$ acts properly discontinuously and cocompactly on a proper geodesic metric space $X$, then any bijective identification of $G$ with its orbit induces a metric on $X$ (changing the orbit leads to a bounded perturbation of the metric). I will call such metrics on $G$ geodesic. This class includes for instance the word metrics associated to finite generating sets. On the other hand, let $\mu$ be a symmetric probability measure on $G$. Denote by $F(x,y)$ the probability that the corresponding random walk starting at $x$ will ever hit $y$. Following Blachere, Haissinsky, & Mathieu, we assume two technical conditions, satisfied by all finitely supported measures: $\mu$ has exponential moment, i.e. $\sum_{g\in G} e^{\lambda \lvert{g}\rvert} \mu(g) < \infty$ for some $\lambda > 0$ (the length of $g$ is considered with respect to any word metric on $G$) for any $r>0$ there exists a constant $C(r)$ such that $F(x,y) \leq C(r) F(x,v)F(v,y)$ whenever $v$ is within distance $r$ from a geodesic segment joining $x$ and $y$ in a fixed Cayley graph of $G$ Under these assumptions the Green metric $d_G(x,y) = -\log F(x,y)$ is hyperbolic, left-invariant and quasi-isometric to the word metrics through the identity map. (This is the metric on $G$ with respect to which the hitting probability of the random walk is a quasi-conformal measure on the boundary). Finally, we say that two metrics $d_1, d_2$ on $G$ are roughly similar if there exists $\alpha > 0$ such that the function $\lvert d_1-\alpha d_2 \rvert$ is bounded. This is an appropriate equivalence relation if we are interested in the measurable structure on the boundary, as roughly similar metrics give rise to Hölder equivalent visual metrics and the same class of Patterson-Sullivan measures. Questions Above, I described two large classes of metrics on a Gromov hyperbolic group. My question is: how different are they in terms of rough similarity? (Blachere, Haissinsky, & Mathieu say that the Green metrics are usually not geodesic, but they don't provide more details on this matter). More precisely: Are there any nice examples of metrics in one of these classes, for which it is easy to see that they are not in the other class? Is it known how large is the intersection of these two classes of metrics (considered up to rough similarity)? REPLY [4 votes]: In general, for a finitely supported measure $\mu$ whose support generates $\Gamma$, the Green metric and the word metric are not equivalent (in the sense you give, that is they are not roughly similar). Actually, you only need superexponential moments and not finite support. More precisely, if the two metrics are roughly similar, then $\Gamma$ is virtually free. This is Theorem 1.5 in the paper of Gouëzel, Matheus and Maucourant entropy and drit in word hyperbolic groups (Inventiones, 2018). A rough argument is as follows. Equivalence means that $|d_1-ad_2|\leq C$ for some $a$ and some $C$. Suppose that $C=0$. Then, you have $d_1=ad_2$ and so the horofunction boundaries for both metric must coincide. However, the horofunction boundary for the word metric must be totally disconnected, whereas the horofunction for the Green metric (the so-called Martin boundary) is not, since it coincides with the Gromov boundary (except for the particular case where $\Gamma$ is virtually free). You can make this discussion precise, even when $C\neq 0$, using some asymptotic version of the above arguments. One way to do so is to look at the Martin cocyle and to compare it with the Buseman cocyle for the word distance. For free groups, it can happen that the two metrics are roughly similar. As you say, they are roughly similar if and only if the corresponding Patterson-Sullivan measures are equivalent, if and only if the Hausdorff dimension of the hitting measure on the Gromov boundary is maximal (this is contained in the paper by Blachere Haïssinsky and Mathieu you refer to). So basically, you have to compare the Hausdorff dimension of the Gromov boundary and the Hausdorff dimension of the hitting measure for the random walk. For probability measures supported on the standard set of generators on a free group, computations can be made and the answer is as follows: the hitting measure has maximal dimension if and only if the random walk is the simple random walk (each generator has the same weigth). This is described with many details in Ledrappier's survey : Some asymptotic properties of random walks on free groups. However, for general probability measures with an exponential moment, this is still widely open to know if the two metrics are roughly similar, even in free groups. Also, I don't think that we even know the answer for (non-nearest neighbor) finitely supported random walks on free groups.<|endoftext|> TITLE: Exponential mapping versus flow QUESTION [5 upvotes]: In Hamilton's article on the Nash-Moser Theorem, he gives the map that maps a vector field $X$ to its flow $e^{tX}$ in $\mathrm{Diff}(M)$ as an example where the implicit function theorem in Frechet spaces fails: The derivative of this map at the zero vector field is the identity, yet the map is not surjective in any neighborhood of the zero section in the space of vector fields. However, when thinking about this example, I realized that I was not sure how to construct the manifold structure on $\mathrm{Diff}(M)$. The first idea would be to choose a Riemannian metric on $M$ and define a map from vector fields to $\mathrm{Diff}(M)$, $X \mapsto \phi_X$ by setting $$ \phi_X(p) := \exp_p(X(p))$$ using the Riemannian exponential map. The claim would then be that this map is a diffeomorphism on some neighborhood of the zero section. However, it seems to me that again, one would use an Implicit function type of statement to make this conlusion. So why does the latter work and the first doesn't? Or rather, the map $X \mapsto e^{tX}|_{t=1}$ is certainly injective for $X$ small enough (right?) so one could just use this map to define the manifold structure in the first place (or why does this fail?). REPLY [6 votes]: You could definitely define a smooth structure on $\text{Diff}(M)$ in the vicinity of the identity by declaring your maps $X \mapsto \phi_X$ or $X \mapsto e^{tX}|_{t=1}$ to be diffeomorphisms onto. However, the topology induced by this procedure is not the natural topology which one would expect, i.e. the subspace topology induced from the $C^\infty$-compact open topology on $C^\infty(M, M)$. The failure of local surjectivity of $\text{Vec}(M) \to \text{Diff}(M)$ says exactly that reasonable small vector fields does not necessarily have flows which are close to the identity. While this is sometimes inconvenient, in my opinion it is better to live with it instead of refining the topology to something 'unnatural'.<|endoftext|> TITLE: Weakly Compact Cardinal and Iterability QUESTION [7 upvotes]: In $\textit{Set Theory}$ by Jech 1978 edition, in the proof of Lemma 32.5 which you can hopefully see at the Google book link. In the course of the proof using the tree property, he produces from any weakly compact cardinal $\kappa$ a non principal $L_\alpha$-ultrafilter $U$ on $\kappa$, which is $L_\alpha$-$\kappa$ complete, and moreover the intersection of $\kappa$ many elements of $U$ (taken in $V$) is nonempty. The latter implies that countable intersection (taken in $V$) is nonempty. This fact implies iterability for example by 19.13 of Kanamori. So $L_\alpha$ with this ultrafilter can be iterated of length $\text{Ord}$ However, $0^\sharp$ follows from the existence of a mouse (iterable premouse). But $0^\sharp$ has stronger consistency strength than a weakly compact. Is this $(L_\alpha, U)$ not a pre-mouse? I have found varying definition of pre-mouse. One definition has the addition condition that $\kappa$ should be the largest cardinal of $L_\alpha$. This seems like it may not hold since at the beginning of the proof, the proof chooses an $\alpha$ such that $L_\alpha \models ZF^-$. I would like to have the $\text{Ord}$ length iteration for what I am trying to do; however, I am troubled by whether or not this would give an iterable premouse and hence imply sharps. Thanks for any clarification you can provide. REPLY [8 votes]: I would like to elaborate a little on Joel's answer. To carry out the iterated ultrapower construction with an ultrafilter that is external to the model, as it is in the case of $L_\alpha$ and $U$, requires making extra assumptions. In the case of an ultrafilter that is an element of your model, the successor stages of the construction simply use the image of the ultrafilter from the previous stage. But we cannot apply $j$ to $U$ because $U$ is not an element of $L_\alpha$. To define the successor stage ultrafilter we need that $U$ is "partially internal" to $M$. This translates into the technical requirement that $U$ is weakly amenable: for every $X\in L_\alpha$ such that $|X|^{L_\alpha}=\kappa$, we have $X\cap U\in L_\alpha$. Weak amenability is equivalent to $L_\alpha$ and its ultrapower having the same subsets of $\kappa$, as Joel mentioned. It is easy to see that assuming that every $A\subseteq\kappa$ is contained in a transitive model $M$ of size $\kappa$ for which there is a weakly amenable $M$-ultrafilter on $\kappa$ (measuring sets in $M$) with a well-founded ultrapower is a much stronger large cardinal assumption than a weakly compact cardinal. I call such cardinals 1-iterable. Assuming that such weakly amenable $M$-ultrafilter is additionally countably complete characterizes Ramsey cardinals, which imply $0^\sharp$. Meanwhile, it is possible to have $M$-ultrafilters with varying degrees of iterability. An $\alpha$-iterable cardinal is characterized by the existence of weakly amenable $M$-ultrafilter with $\alpha$-many well-founded iterated ultrapowers and $\alpha$-iterable cardinals form a hierarchy of strength below Ramsey cardinals. We introduced and studied these cardinals together with Philip Welch.<|endoftext|> TITLE: Classical algebraic varieties VS $k$-schemes VS schemes QUESTION [16 upvotes]: We know that there is an equivalence of categories between the two following categories: $1)$ Classical varieties over $k$, where $k$ is an algebraically closed field. (Informally I mean locally ringed spaces formed by patching affine irriducible algebraic sets over $k$. This is the definition of algebraic variety present for example in Perrin's book.) $2)$ Schemes over $k$ which are integral and of finite type over $k$ Now I need an example, if it does exist, of two non isomorphic (as classical varieties) algebraic sets that are non isomorphic as $k$-schemes (this is obvious by the previous functor) but isomorphic in the category of schemes. To be more specific I need two non isomorphic algebraic sets such that when I use the above functor $1)\longrightarrow 2)$ and then I forget the structure morphism on Spec $k$, I finally obtain two isomorphic schemes. Thanks in advance REPLY [33 votes]: Let's say $k=\mathbb{C}$ (although something like this should work over any algebraically closed field). Let $V_1=\mathbb{P}^1-\{0,1,\infty,\pi\}$ and $V_2 = \mathbb{P}^1-\{0,1,\infty,e\}$. One can see that $V_1$ and $V_2$ are not isomorphic as varieties/schemes over $\mathbb{C}$: such an isomorphism would extend to a map $\mathbb{P}^1\to\mathbb{P}^1$ taking $\{0,1,\infty,\pi\}$ to $\{0,1,\infty,e\}$, which cannot exist because the cross ratio of any permutation of $\{0,1,\infty,\pi\}$ does not equal the cross ratio of $\{0,1,\infty,e\}$. However, there exists an abstract field automorphism $\sigma$ of $\mathbb{C}$ taking $\pi$ to $e$, so there will be an isomorphism of schemes $V_1\to V_2$ which is $\sigma$-semilinear.<|endoftext|> TITLE: Varieties where every algebra is free QUESTION [21 upvotes]: I'd like to know more about varieties (in the sense of universal algebra) where every algebra is free. Another way to state the condition is that the comparison functor from the Kleisli category to the Elienberg-Moore category is an equivalence. For example, every object is free in The category of sets The category of pointed sets The category of vector spaces (over a specified field), or more generally, the category of modules over a division ring (Added 3/14/14) The category of affine spaces (vector spaces without a zero) and affine maps (linear maps + translations) over a division ring $R$. The algebraic structure is given by, for each $r \in R$, a ternary operation $f_r(x,y,z)$ meaning essentially $r(x-y)+z$, with appropriate relations to specify this. In the vector space case this example is mentioned in the paper John Baldwin links to below. Is there a name for this property? Over at the n-Category Café, Zhen Lin suggested the term "absolutely free", but I gather this has a different meaning in universal algebra. Has this property been studied in the literature? Are there other good examples? It seems like a very restrictive condition: is it restrictive enough to obtain some kind of structure theory for varieties with this property? In the commutative algebra case: If all the modules over a ring $k$ are free, then is $k$ necessarily a division ring? EDIT (2/19/14) The Masked Avenger mentions below that this property can be parsed in terms of categoricity in the sense of model theory. This reminds me that on the n-Category Café, Zhen Lin mentioned there should be an approach in terms of elimination of imaginaries. If anybody could flesh out the model-theoretic aspects I'd really appreciate it. Perhaps the topic has been well-covered model-theoretically? I think the linear case has been clarified by multiple people. Benjamin Steinberg has some interesting results related to the classification aspect; any further observations would be great. I'm still looking for a name for this property[3/14/14: "pantofree" sounds joke-y to my ear, but maybe it is apt after all...], and still looking for further interesting examples. Maybe I'll also mention: one variation that might be interesting is to require only that finitely generated algebras be free. REPLY [18 votes]: If there are no constant symbols in the language, $\mathcal L$, then the nontrivial varieties where every algebra is free are exactly the varieties term equivalent to sets or to affine spaces over a division ring. Here is why. If there are no constants in the language, $F_{\mathcal V}(\emptyset)$ is empty, so $F_{\mathcal V}(1)$ must be the 1-element algebra. This forces the variety to be idempotent. $\mathcal V$ is $\kappa$-categorical for $\kappa\geq |\mathcal L|$. This forces the first-order theory of infinite members of $\mathcal V$ to be complete, so $\mathcal V$ is a minimal variety. The minimal idempotent varieties were partially classified in my paper Almost all minimal idempotent varieties are congruence modular, Algebra Universalis 44 (2000), 39-45. Each one must be equivalent to the variety of sets, the variety of semilattices, a variety of affine modules over a simple ring, or must be congruence distributive. So what is left is to rule out the nonabelian ones and to show that a variety of affine modules over a simple ring where ever member is free is actually affine over a division ring. The division ring conclusion can be reached in various ways, e.g., by referring to the answer by Mariano Suárez-Alvarez. (You can shorten that argument a bit by noting that $\mathcal V$ has a simple member, and for a simple module to be free the ring must be a division ring.) What is still left is to rule out the varieties equivalent to semilattices and the congruence distributive varieties. This can be done directly or we can cite papers of Baldwin + coauthors. A variety equivalent to semilattices is locally finite, so by his paper with Lachlan (cited in Baldwin's answer), which applies to locally finite varieties whose infinite members are free, $\mathcal V$ is totally categorical. These varieties are known. A congruence distributive variety is congruence modular, so by his paper with McKenzie called "Counting models in universal Horn classes", Algebra Universalis 15 (1982), 359-384, the $\kappa$-categoricity and the congruence modularity of $\mathcal V$ jointly imply that $\mathcal V$ is abelian. (Their paper assumes a countable language, but this part of their argument doesn't require it.) I don't know how to do the case where there are constants in the language. What can be shown is that there is at most one constant up to equivalence, that $F_{\mathcal V}(1)$ is abelian and simple, and that $\mathcal V = SPP_U(F_{\mathcal V}(1))$ is a minimal variety that is minimal as a quasivariety. The only examples I know are the varieties of pointed sets and the varieties of vector spaces over a division ring. EDIT (8/16/15) I know more now. Steve Givant solved the main question posed here (Which varieties have the property that all members are free?) in his 1975 PhD thesis. The answer is: only those varieties term equivalent to sets, pointed sets, vector spaces over a division ring or affine spaces over a division ring. His methods do not seem to apply to the variation suggested by Tim: For which varieties are the finitely generated members free? However Emil Kiss, Agnes Szendrei and I just worked out the answer to that question (it is the same answer): sets, pointed sets, vectors spaces and affine spaces. I just submitted a short note on this to the arxiv.<|endoftext|> TITLE: Finite field Szemeredi-Trotter theorem with unequal number of points and lines QUESTION [11 upvotes]: My question concerns the Szemerédi-Trotter theorem in $\mathbb{F}_q^2$. If we have $m$ points and $n$ lines in $\mathbb{F}_q^2$, then by Cauchy-Schwarz the number of point-line incidences is as most $$\min(mn^{1/2}+n,\ m^{1/2}n+m). \qquad\qquad (1)$$ This is the brown line in the figure below. We also know that if $m=n$, and $q^{\delta}0$. This is the red dot in the figure below. Equation $(2)$ allows us to beat the bound $(1)$ provided $m$ is very close to $n$---just increase either $m$ or $n$ slightly so that $m=n$, and then apply $(2)$. This is the red line in the figure below. My question is: are there any results that allow us to do better than the envelope of the brown and red lines---i.e. can we obtain any sort of bound that looks like the blue line below? My hope is that we can somehow interpolate between the red dot (in the figure below) and the two endpoint cases (the brown dots at $(1/2, 1/2)$ and at $(2,2)$. Does anyone know of a result of this form, or does anyone have ideas on how to obtain such a result? [edit: updated the picture to have correct numerology, and corrected (1)] REPLY [4 votes]: In case anyone stumbles across this thread, there is now a nice paper by Sophie Stevens (https://arxiv.org/abs/1609.06284) which improves upon the Cauchy-Schwarz bound for all ranges of $m$ and $n$ with $m^{1/2} \leq n \leq m^2$.<|endoftext|> TITLE: Are totally degenerate chains null-homologous? QUESTION [12 upvotes]: Let $X$ be a CW complex. Suppose $\gamma\in C_n(X)$ is a cycle which is a sum of maps $\sigma:\Delta^n\to X$ which factor as $\Delta^n\to\mathbb R^{n-1}\to X$. Does it follow that $\gamma$ is null-homologous? More generally, we could ask the following which also seems likely to be true but hard to prove: Let $L\subseteq C_\bullet(X)$ be the subcomplex generated by $\sigma$ and $d\sigma$ for all maps $\sigma:\Delta^n\to X$ which factor as $\mathbb R^{n-1}\to X$. Is $L$ acyclic? REPLY [4 votes]: First, we may as well allow the manifolds $M^{n-1}$ to have a boundary, in which case we can replace $M^{n-1}$ by a closed neighbourhood of the image of $\Delta^n$ if necessary, and thus assume that $M^{n-1}$ is compact. In this context, the union of the images of all the sets $M^{n-1}$ will be contained in a finite subcomplex of $X$, so we may assume that $X$ itself is finite, of dimension $d$ say. We can argue by induction on $d$. The case $d TITLE: An integral related to the Euler gamma function QUESTION [11 upvotes]: The question is from the paper http://arxiv.org/abs/1312.7115 (A curious formula related to the Euler Gamma function, by Bakir Farhi): is it possible to express the integral $$\eta=2\int\limits_0^1 \ln{(\Gamma(x))}\cdot \sin{(2\pi x)}\,dx= 0.7687478924\ldots$$ in terms of the known mathematical constants as $\pi,\,e,\,\gamma,\;\ln{\pi},\, \ln{2},\,\Gamma{(1/4)}\ldots$? It is shown in the paper that $$\frac{\ln{1}}{1}-\frac{\ln{3}}{3}+\frac{\ln{5}}{5}-\ldots=\pi\ln{(\Gamma{(1/4)})}-\frac{\pi^2}{4}\eta-\frac{\pi}{2}\ln{\pi}-\frac{\pi}{4}\ln{2}.$$ So the question actually asks whether one can give a closed-form expression for the series in the l.h.s. in terms of the known mathematical constants. REPLY [2 votes]: In fact, both the experimental result by Joro and the (very nice) asymptotic analysis by A. Shamov, above, are correct. I'd like to advice those more interested readers that I've included this topic and some generalizations in my most recent work on integrals and series involving the log-Gamma function (to appear in arXiv website). F. M. S. Lima<|endoftext|> TITLE: Is the space of rapidly decreasing (non-smooth) functions nuclear? QUESTION [9 upvotes]: We denote by $\mathcal{S}(\mathbb{R})$ the space of smooth and rapidly decreasing functions. We define on $\mathcal{S}(\mathbb{R})$ the family of semi-norms $$\lVert \varphi \lVert_{n,m} = \lVert (1+|\cdot|^m) \varphi^{(n)} \lVert_\infty.$$ This family of semi-norms defines a topology $\tau$ on $\mathcal{S}(\mathbb{R})$. The space $(\mathcal{S}(\mathbb{R}),\tau)$ is then well-known to be nuclear. If $\mathcal{M}(\mathbb{R})$ is the space of measurable functions in $\mathbb{R}$, I define the space $\mathcal{R}(\mathbb{R})$ of rapidly decreasing functions (without smoothness condition) by $$\mathcal{R}(\mathbb{R}) = \{ \varphi \in \mathcal{M}(\mathbb{R}), \ \lVert \varphi \lVert_{0,m} < \infty \}.$$ Moreover, the family of (semi-)norms $(\lVert \cdot \lVert_{0,m})_{m\in \mathbb{N}}$ defines a topology $\tau'$ on $\mathcal{R}(\mathbb{R})$. Question: Is $(\mathcal{R}(\mathbb{R}),\tau')$ a nuclear space? NB. The proofs I found for the nuclearity of $\mathcal{S}(\mathbb{R})$ are essentially using the facts that the space $s$ of sequences with quick decay is nuclear and $\mathcal{S}(\mathbb{R})$ is homeomorphic to $s$. Several authors are claiming that the nuclear structure is strongly related with the condition of infinite smoothness. The space $\mathcal{D}(\mathbb{R})$ of compactly supported and smooth function (with the inductive topology coming from the spaces $\mathcal{D}([-n,n])$), and the space $\mathcal{C}^\infty(\mathbb{R})$ (projective limit of the spaces $\mathcal{D}([-n,n])$) are other famous examples of nuclear structure that go in that sense. This suggests that the space $\mathcal{R}(\mathbb{R})$ may be not nuclear or at least that the usual technics are not available. REPLY [12 votes]: Your space contains an isomorphic copy of $L^\infty([0,1])$ (consider the family of elements with support in the interval) and so is not nuclear. By the way an explanation of the relationship between smoothness and nuclearity is that many spaces of test functions are generated in a natural way by differential operators as described in a short but important article by A. Pietsch (Math. Ann. vol. 164). This article contains a simple criterion for such spaces to be nuclear, based on the spectral properties of the operators as unbounded, self-adjoint ones on Hilbert space and displays some of the standard spaces as explicit examples.<|endoftext|> TITLE: Supercuspidal with Iwahori fixed vector QUESTION [5 upvotes]: Let $F$ be a local field. Is there a reference for the following fact: No supercuspidal representation of $GL_2(F)$ has an Iwahori-fixed vector? I have a proof, by I'd prefer a reference, because it is not enlightening. Rough sketch of proof: We can easily see that Iwahori-fixed vector implies depth-zero and that depth-zero supercuspidal are induced from $GL_2(o)$ times the center, hence correspond modulo central characters to supercuspidal of $GL_2(o/p)$. The proof of the second conclusion is somewhat messy in my exposition. REPLY [4 votes]: I quote from one of my papers (On Bernstein's presentation of Iwahori-Hecke algebras and representations of split reductive groups over non-Archimedean local fields, Bulletin of the Kerala Mathematics Association, Special issue on Harmonic Analysis and Quantum Groups, December 2005, also available from http://arxiv.org/abs/math.GR/0506094) "Casselman, using some techniques of Jacquet, showed that under the correspondence described by Borel, the irreducible admissible representations of G(F) coming from irreducible H-modules are precisely those that occur as subquotients (or equivalently, subrepresentations) of the unramified principal series of G(F ) [Cas80, Proposition 2.6]. It is implicit in Proposition 2.5 of this article that the Jacquet module of such a representation with respect to a minimal parabolic subgroup corresponds to restriction to a certain commutative subalgebra of H (later identified by Bernstein)." The reference is to: [Cas80] W. Casselman. The unramified principal series of p-adic groups. I. The spherical function. Compositio Math., 40(3):387–406, 1980.<|endoftext|> TITLE: Objections to and arguments for the simplicity of all Riemann zeros QUESTION [8 upvotes]: It seems to be that the simplicity of all the zeros is quite widely accepted as a working hypotheses, and it is known that a positive proportion are as such. Titchmarsh explains in the last chapter on RH that the weak Mertens hypothesis, $M(x)=O(x^{1/2})$, implies that all zeros are simple. Though I think this is conjectured to be false by many authors. Is there a weaker bound whose optimality is known to imply that not all zeros are simple? Given the classical estimate $N(\sigma, T)=O(T^{4\sigma(1-\sigma)+\epsilon})$ combined with the fact that the number of solutions to $\zeta(s)=z$ in any rectangle of height $T$ contained in $1/2\leq \sigma\leq 1$ is proportional to $T$, for any $z\neq 0$, simplicity of the zeros suggests that the derivative must vanish vastly more frequently in each such rectangle else one could presumably tie the number of such solutions to the number of zeros up to height $T$, for all sufficiently small $|z|$, by Rouche's theorem. One might therefore conjecture that $ \inf|\rho_{\zeta}-\rho_{\zeta'}|=0$ (is this known perhaps?). An old theorem of MacDonald (Whittaker and Watson, p131) says that the number of zeros of an analytic function on the interior of a simple closed curve on which the modulus of $f$ is constant exceeds that of it's derivative by one. It can also be seen that the derivative necessarily vanishes on the largest such curve by the open and inverse mapping theorems. In the case of $\zeta(s)$ I'd expect that the largest of such curves around each zero would become small for large $T$, so I wonder if something in this direction is known? Anyway, as Titchmarsh points out, $\Omega$-theorems for $M(x)$ seem to be more difficult to obtain than those for the prime counting functions and, since a significant distinction between the problems is that of the order of the zeros, I would like to enquire specifically about anything that is known in either direction, on any other hypotheses. REPLY [12 votes]: Quite a lot is known in the literature about zeros of $\zeta(s)$ and $\zeta^{\prime}(s)$, and in particular one can show that zeros of $\zeta^{\prime}(s)$ do get close to zeros of $\zeta(s)$. However, I am skeptical that these results have any implications for $M(x)$. The number of zeros of $\zeta^{\prime}(s)$ with ordinates in $(0,T)$ is $$ = \frac{T}{2\pi} \log \frac{T}{4\pi e} + O(\log T). $$ This is due to Bruce Berndt, and is an application of the argument principle. Note that it is slightly different from the corresponding formula for $\zeta(s)$. The zeros of $\zeta^{\prime}$ are all expected to lie in the half plane $\sigma \ge 1/2$ -- this is equivalent to RH (proved first by Speiser). Levinson and Montgomery (Acta Math. 1974) investigated this further, and their work was the precursor to Levinson's famous result on the critical zeros of $\zeta(s)$. Levinson and Montgomery showed that most of the zeros of $\zeta^{\prime}(s)$ (with ordinate in $(0,T)$) lie in the region $|\sigma -1/2| \le w(T) \log \log T/\log T$ for any function $w(T)$ tending to infinity with $T$. In fact, most of the zeros of $\zeta^{\prime}$ probably lie closer to the critical line, on the scale of $1/\log T$ from $1/2$. For work in this direction, see papers of Conrey and Ghosh, Soundararajan, Yitang Zhang, Farmer and Ki, and most recently Radziwill (http://arxiv.org/abs/1301.3232, which will give the other references). From above one knows that there are plenty of zeros of $\zeta^{\prime}(s)$ lying very close to the half line. From zero density results, we also know that there are plenty of zeros of $\zeta(s)$ lying close to the half line. For example, we can guarantee that there are at most $T/\log T$ zeros of $\zeta(s)$ with $|\sigma- 1/2| \ge 100 \log \log T/\log T$, say. The only thing is to rule out a conspiracy separating ordinates of such zeros of $\zeta$ and $\zeta^{\prime}$. A straightforward way to do this is to use Littlewood's theorem that the gaps between consecutive ordinates of zeros of $\zeta(s)$ tends to zero (at the rate of $1/\log \log \log |\gamma|$). First find, using the results above, $\gg T$ zeros of $\zeta^{\prime}$ (with ordinates in $[T,2T]$ say) such that these zeros all lie near the critical line, and their ordinates are spaced a distance $1$ apart. Next by Littlewood, for each such zero of $\zeta^{\prime}$, find a zero of zeta whose ordinate is close to it. Lastly, note that by the zero density theorem for zeta, at most $T/\log T$ of these corresponding zeros of zeta can be far from the $1/2$ line. That completes the proof. The lazy argument above shows that $\min |\rho_{\zeta} -\rho_{\zeta^{\prime}}| \ll 1/\log \log \log T$ for zeros around ordinate $T$. By working harder one should be able to get $\ll 1/\log T$, and one probably cannot prove anything better. It is believed that zeros of $\zeta$ can get very close to each other, and so plausibly the truth here is that $|\rho_{\zeta} -\rho_{\zeta^{\prime}}|$ can get as small as $T^{-\frac 13+o(1)}$. This is very speculative, and even with this I don't see any implication for $M(x)$.<|endoftext|> TITLE: Ergodic theory and dynamical systems books references QUESTION [10 upvotes]: I am arranging a weekly meeting of 2 hours with postgraduate students in ergodic theory (for a period of 3 weeks). I am asking here for an advice of a book (or maybe a set of papers) to look at during our reading meetings. We would like to discuss some topic of current research (say not older than 5 years), but we do not want to be too specific, because is an activity extra to our own research. At the same time, we would like to build a useful basis for further research. For me the the classical references are: An Introduction to Ergodic Theory (Graduate Texts in Mathematics) by Peter Walters. Ergodic Theory (Cambridge Studies in Advanced Mathematics) by Karl E. Petersen. Introduction to the Modern Theory of Dynamical Systems (Encyclopedia of Mathematics and its Applications) by Anatole Katok and Boris Hasselblatt. Equilibrium States and the Ergodic Theory of Anosov Diffeomorphisms (Lecture Notes in Mathematics) by Robert Edward Bowen, Jean-René Chazottes and David Ruelle. Ergodic Theory: With a View Towards Number Theory (Graduate Texts in Mathematics) by Thomas Ward, Manfred Einsiedler. Many thanks. REPLY [2 votes]: Personally, I like Mañe's book Teoría Ergódica. I do think it's a classical book full of exercises. Also, it's orientated to start studying smooth ergodic theory. On the other hand the book has loads of mistakes (which makes it interesting to read, you realise that you are understanding everything when you spot the mistakes) and, as far as I know, both versions (English and Portuguese) are sold out. Also, Dynamics Beyond Uniform Hyperbolicity by Bonatti, Diaz and Viana can provide a good set of open problems to work in, but it is a bit hard to read.<|endoftext|> TITLE: Equivariant algebraic K-theory of affine space QUESTION [7 upvotes]: Unlike algebraic K-theory, equivariant K-theory of affine space (over a field $k$) can be quite nontrivial, depending on the action of the group in question. For example, if one takes the standard action of $\mathbb G_m$ on $\mathbb A^n$, then the $\mathbb G_m$-equivariant K-theory of $\mathbb A^n$ can be computed using a localization sequence corresponding to the inclusion of the origin inside $\mathbb A^n$. (In the localization sequence in equivariant K-theory, one uses that the quotient of $\mathbb A^n \backslash \{0\}$ is $\mathbb P^{n-1}$ and uses the projective bundle theorem, for example, to proceed.) How does one compute $K_i^G(\mathbb A^n)$ in other cases? Is there a general procedure for doing so? For example, take $G = \mathbb G_m \times \cdots \times \mathbb G_m$ (product of $n$ copies) and consider the diagonal action on $\mathbb A^n$. How does one compute $K_i^G(\mathbb A^n)$ in this case? Is it easier to write down the description of $K_0^G(\mathbb A^n)$? REPLY [7 votes]: In fact, just like in the case of non-equivariant algebraic $K$-theory, there is a homotopy-invariance for equivariant $K$-theory (of nonsingular varieties). Addressing your question specifically, for any action of an algebraic group $G$ on affine space, the answer is: $$ K^G_i({\Bbb A}^n) = R(G) \otimes K_i(k), $$ where $k$ is the ground field and $R(G) = K^G_0(k)$ is the representation ring. If $G$ is a (split) torus, the representation ring is just the group ring of the character group; choosing a basis $x_1,\ldots,x_n$ of characters identifies $R(G)$ with the Laurent polynomial ring ${\Bbb Z}[x_1^\pm,\ldots,x_n^\pm]$. (The general statement is that when $f\colon X \to Y$ is a $G$-equivariant affine bundle, there is an isomorphism $$f^*\colon G^G_*(Y) \to G^G_*(X),$$ where $G^G_*$ denotes equivariant $K$-theory of coherent sheaves. For nonsingular varieties $X$ one has $G^G_*(X) = K^G_*(X)$. See Theorem 4.1 of Thomason's foundational paper "Algebraic K-theory of group scheme actions".)<|endoftext|> TITLE: Two questions about sphere bundles QUESTION [10 upvotes]: I would like to better understand the relationship between different notions of orientable sphere bundle. Let me say that a locally trivial fiber bundle $\pi\colon E\to M$ with fiber $S^n$ and structure group $G$ is a topological sphere bundle if $G=\mathrm{Homeo}^+(S^n)$, smooth sphere bundle if $G=\mathrm{Diffeo}^+(S^n)$, linear sphere bundle if $G=\mathrm{Gl}^+(n,\mathbb{R})$. It is known that there exist smooth sphere bundles that are not equivalent (as smooth sphere bundles) to linear sphere bundles (see e.g. the following posts here on MO): Examples of sphere bundles Is it true that all sphere bundles are boundaries of disk bundles? For example, in an answer to the second question linked above R. Budney shows that there are smooth sphere bundles over $S^2$ which are not linear. Here comes my first question: Is it possible to construct a smooth sphere bundle over $S^1$ which is not smoothly equivalent to a linear one? One could take an exotic diffeomorphism $f\colon S^n\to S^n$, and consider the mapping cone of $f$, which is a smooth sphere bundle. Is this bundle linear? The second question is about topological bundles: Do there exist a topological sphere bundle which is not topologically isomorphic to a linear one (or even to a smooth one)? REPLY [10 votes]: Recall that linear, smooth, and topological $S^k$-bundles over a finite complex $X$ are classified by the sets of homotopy classes $[X, BO_{k+1}]$, $[X, B\mathrm{Diff}(S^k)]$ and $[X, B\mathrm{Homeo}(S^k)]$. There are obvious maps between the sets, and we are interested in their cokernels. For example, if $X=S^1$, then the smooth nonlinear bundles are elements of the cokernel of $\pi_1(BO_{k+1})\to \pi_1(B\mathrm{Diff}(S^k))$. By the exact homotopy sequence the cokernel is isomorphic to $\pi_0(\mathrm{Diff}(S^k)/O_{k+1})$. The space $\mathrm{Diff}(S^k)/O_{k+1}$ is simply $\mathrm{Diff}(D^k, \partial)$, the diffeomorphisms of the disk that are the identity on the boundary. The latter space is classically studied, and by results of Smale and Cerf the group $\pi_0(\mathrm{Diff}(D^k, \partial))$ is isomorphic to the group of homotopy $(k+1)$-spheres, and hence it is often nonzero. Thus there are lots of smooth non-linear bundles over a circle. EDIT: the crossed out text below is nonsense, and I retract it. What one actually needs is that $\mathrm{Diff}(S^k)\to \mathrm{Homeo}(S^k)$ is not $\pi_i$-surjective for some $i$, and at the moment I do not know any $k, i$ for which this is true. Non-smoothable sphere bundles over spheres also exist because the space $\mathrm{Homeo}(S^k)/\mathrm{Diff}(S^k)$ is not contractible, for otherwise the inclusion $\mathrm{Diff}(S^k)\to \mathrm{Homeo}(S^k)$ would be a $\pi_i$-isomorphism for all $i$, which it is not for some $k$ (this is due to Alexander's trick, see page 3 of the reference 2 below). Some references to online sources which you may find useful, and which contains references to the above claims): A. Hatcher's survey: http://www.math.cornell.edu/~hatcher/Papers/Diff%28M%292012.pdf P.L. Antonelli, D. Burghelea, P.J. Kahn, http://www.sciencedirect.com/science/article/pii/0040938372900213 D. Crowley, T. Schick: http://front.math.ucdavis.edu/1204.6474<|endoftext|> TITLE: Integer Solutions of $x+y^n = y + x^m$ for $n < m$ QUESTION [13 upvotes]: I found 8 of them and believe there is no more: $$2+3^2=3+2^3$$ $$2+6^2=6+2^5$$ $$6+15^2=15+6^3$$ $$3+16^2=16+3^5$$ $$3+13^3=13+3^7$$ $$2+91^2=91+2^{13}$$ $$5+280^2=280+5^7$$ $$30+4930^2=4930+30^5$$ I call the solution a principal pipe of order 2. Any idea to attack this equation? If $n=x=2$, it becomes $2+y^2=y+2^m$. It is Ramanujan and Nagell Equation, and there are only 3 solutions. If $n=2$ and $x \geq 3$, then we can use the similar way to prove that there is no other solution for $x < 10000$ with help of computer. If $n=3$, and we assume $x$ and $y$ are prime, and $6\mid (m-1)$, then the solution is $3+13^3=13+3^7 = 2200$. I believe there is no solution for $n>3$. I am particularly interested in solutions that both $x$ and $y$ are odd primes. REPLY [11 votes]: This problem was considered in a paper of Mignotte and Petho {Publ. Math. Debrecen 1999) and subsequently in one of mine [Canad. Math. J 2001], where there is a conjecture that the equation $a^x-b^y=c$ has, for fixed positive integers $a, b$ and $c$, with $a, b \geq 2$, at most one solution in positive integer exponents $x$ and $y$, with precisely $11$ exceptions. This is provable if $c$ is suitably large or suitably small, but, as far as I know, is still open in general. This problem arose from four classical papers of Pillai.<|endoftext|> TITLE: Monstrous Moonshine for Thompson group $Th$? QUESTION [27 upvotes]: I. As a background, in Traces of Singular Moduli (p.2), Zagier defines the modular form of weight 3/2, $$g(\tau) = \frac{\eta^2(\tau)}{\eta(2\tau)}\frac{E_4(4\tau)}{\eta^6(4\tau)}=\vartheta_4(\tau)\, \eta^2(4\tau)\,\sqrt[3]{j(4\tau)}$$ which has the nice q-expansion (A027652, negated terms), $$g(\tau) = 1/q - 2 + 248q^3 - 492q^4 +(15^3+744)q^7 + \dots + (5280^3+744)q^{67} + \dots + (640320^3+744)q^{163}+\dots$$ However, one can use other Eisenstein series $E_k(\tau)$ as the one below. II. In a paper by Bruinier (p.6), Borcherds defines a modular form of weight 1/2. First let, $$K(\tau) = \tfrac{1}{16}\big(\vartheta_3^4(\tau) - \tfrac{1}{8}\vartheta_2^4(\tau)\big)\,\vartheta_2^4(\tau)\vartheta_3(\tau)\vartheta_4^4(\tau)$$ then, $$b(\tau) = 60\vartheta_3(\tau)+\frac{K(\tau)E_6(4\tau)}{\eta^{24}(4\tau)} = 60\vartheta_3(\tau)+\frac{K(\tau)\sqrt{j(4\tau)-1728}}{\eta^{12}(4\tau)} $$ This has the q-expansion (A013953), $$b(\tau) = 1/q^3 + 4 - 240q + (\color{blue}{27000}-240)q^4 - \color{blue}{85995}q^5 + \color{blue}{1707264}q^8 - (\color{blue}{4096000}+240)q^9 + \color{blue}{44330496}q^{12}-91951146q^{13}+\dots$$ I noticed that the blue numbers appear in the degrees of irreducible representations of the Thompson group $Th$, given by the finite sequence of 48 integers (A003916), $$1, 248, 4123, \color{blue}{27000}, 30628, 30875, 61256, \color{blue}{85995}, 147250, 767637, 779247, 957125, \color{blue}{1707264}, 2450240, 2572752, 3376737, \color{blue}{4096000}, 4123000, 4881384, 4936750,\dots\color{blue}{44330496},\dots 91171899, 111321000, 190373976.$$ with repeated terms deleted for brevity. Questions: If it is not coincidence, what is the reason? In Bruinier (p.6) he says one can use $E_k(\tau)$ for $k=4,6,8,10,14$. What are the other functions for higher $k$? Edit: As pointed out by S. Carnahan below, there is already a known moonshine for $Th$. A partial(?) list of others from Griess' "Happy Family" can be found in Monstrous Moonshine, such as for the Higman-Sims $HS$ and so on. (Are there more?) REPLY [5 votes]: Your observation, which was expanded into a concrete conjecture last year by Harvey and Rayhaun, is now a theorem. See M. Griffin, M. Mertens, "A proof of the Thompson Moonshine conjecture". This is not to say that the story is complete. While we now know that a suitable Thompson module exists, there is still no explicit construction or conceptual (e.g., physical) explanation for its existence.<|endoftext|> TITLE: What is the shape of large cardinal tree in implication strength order? QUESTION [8 upvotes]: There are two natural orders on large cardinal axioms. (a) Consistency strength order $\sigma \leq_C \theta \Longleftrightarrow ZFC\vdash Con(ZFC+\theta)\longrightarrow Con(ZFC+\sigma)$ (b) Implication strength order $\sigma \leq_I \theta \Longleftrightarrow ZFC\vdash \theta\longrightarrow \sigma$ It is well-known that these are not same and many large cardinals (e.g. Woodin cardinals) have different positions in large cardinal tree when we endow it with different orderings. Question 1. Are there any other important orderings on the tree of large cardinal axioms? Question 2. What is the shape of large cardinal tree in implication strength order? Is there any diagram somewhere in the texts which summarizes the results on direct implication of large cardinal axioms? I hope somebody give me an explicit diagram which shows the differences between large cardinal tree in implication and consistency strength orders or at least based on the answers one be able to design a diagram for large cardinal tree in implication order and post it as an answer which summarizes the answers of the others. REPLY [3 votes]: There is also: c) interpretability strength order: $\sigma \leq_F \theta \Longleftrightarrow \exists f\ \forall \psi\ ZFC\vdash Con(ZFC+\theta+f(\psi))\longrightarrow Con(ZFC+\sigma+\psi)$ for some suitable interpretation $f$. This article on Independence and Large Cardinals gives more exposition. A linear order is also likely here. If we found $\theta$ and $\phi$ to be incompatible, we would probably say "$\theta$ is a large cardinal axiom, about the size of the set-theoretic universe, and $\phi$ is more about the structure of the universe, so it's not really a large cardinal axiom."<|endoftext|> TITLE: Detecting Monodromy in Integrable Systems QUESTION [9 upvotes]: Suppose I have a completely integrable system on a symplectic manifold $(M^{2n},\omega)$ with momentum map $H:M \rightarrow \mathbb{R}^n$ that has compact, connected fibers. Further, suppose I know the set of regular values is not simply connected. Question Short of computing local action-angle coordinates or the period lattice, how can we compute the monodromy of this system? Or simply detect whether it is non-trivial? The only way I know to do this is to compute the transition functions of the period lattice bundle explicitly (a la Cushman and Bates), which seems to be roughly equivalent to several other approaches (such as computing local action angle coordinates). Are there other known tricks for doing this, or famous examples I should be aware of? For example, if I can realize my system as a Lax pair, are there `algebraic' ways to detect monodromy? Edit: cf. http://arxiv.org/pdf/1401.3630.pdf it appears that (at least in 2 degrees of freedom) if you have an isolated focus singular value in the base, then (in the Hamiltonian case) the monodromy around that point can be computed from the number of singular points in the fiber. So one strategy is to study the topology of the singular fibers. Apparently this approach breaks down if the system is non-Hamiltonian. Does this approach generalize to Hamiltonian systems with arbitrary degrees of freedom? For example, in degree 3 we would be looking at monodromy around a critical line, so we might need to study the topology of a bundle of singular tori over this line. REPLY [2 votes]: After reading a bit into the problem, it seems that 'Cushman's principle' is the answer I am looking for. In the case of the spherical pendulum, one can observe with Morse theory that the energy level sets change topology as one passes through the isolated critical value. This indicates that a pull-back of the torus bundle to a circle around the critical value should be non-trivial, which is precisely the situation of non-trivial topological monodromy. This is a nice observation because doing Morse theory in this situation is much more simple than computing action-angle coordinates. On the other hand, this does not compute the monodromy.<|endoftext|> TITLE: Self-adjoint operator QUESTION [6 upvotes]: Assume that $B$ is a self-adjoint operator and $\alpha\in(0,1)$. I need a reference for the following equality $$B^{-\alpha}=\frac{\sin\alpha \pi}{\pi}\int_0^\infty \lambda^{-\alpha}(\lambda+B)^{-1}\mathrm d\lambda.$$ REPLY [3 votes]: See pp. 231-232 in Birman M.S., Solomyak M.Z. Spectral Theory of Self-Adjoint Operators in Hilbert Space (Reidel, 1987).<|endoftext|> TITLE: Assessing effectiveness of (epsilon, delta) definitions QUESTION [10 upvotes]: There is much discussion both in the education community and the mathematics community concerning the challenge of (epsilon, delta) type definitions in calculus and the student reception of them. The mathematical community often holds an upbeat opinion on the success of student reception of this (see examples below), whereas the education community often stresses difficulties and their "baffling" and "inhibitive" effect (see below). A typical educational perspective on this was recently expressed by Paul Dawkins in the following terms: 2.3. Student difficulties with real analysis definitions. The concepts of limit and continuity have posed well-documented difficulties for students both at the calculus and analysis level of instructions (e.g. Cornu, 1991; Cottrill et al., 1996; Ferrini-Mundy & Graham, 1994; Tall & Vinner, 1981; Williams, 1991). Researchers identified difficulties stemming from a number of issues: the language of limits (Cornu, 1991; Williams, 1991), multiple quantification in the formal definition (Dubinsky, Elderman, & Gong, 1988; Dubinsky & Yiparaki, 2000; Swinyard & Lockwood, 2007), implicit dependencies among quantities in the definition (Roh & Lee, 2011a, 2011b), and persistent notions pertaining to the existence of infinitesimal quantities (Ely, 2010). Limits and continuity are often couched as formalizations of approaching and connectedness respectively. However, the standard, formal definitions display much more subtlety and complexity. That complexity often baffles students who cannot perceive the necessity for so many moving parts. Thus learning the concepts and formal definitions in real analysis are fraught both with need to acquire proficiency with conceptual tools such as quantification and to help students perceive conceptual necessity for these tools. This means students often cannot coordinate their concept image with the concept definition, inhibiting their acculturation to advanced mathematical practice, which emphasizes concept definitions. See the entire article (note that the online article provides links to the papers cited above). To summarize, in the field of education, researches decidedly have not come to the conclusion that epsilon, delta definitions are either "simple", "clear", or "common sense". Meanwhile, mathematicians often express contrary sentiments. Two examples are given below. ...one cannot teach the concept of limit without using the epsilon-delta definition. Teaching such ideas intuitively does not make it easier for the student it makes it harder to understand. Bertrand Russell has called the rigorous definition of limit and convergence the greatest achievement of the human intellect in 2000 years! The Greeks were puzzled by paradoxes involving motion; now they all become clear, because we have complete understanding of limits and convergence. Without the proper definition, things are difficult. With the definition, they are simple and clear. see Kleinfeld, Margaret; Calculus: Reformed or Deformed? Amer. Math. Monthly 103 (1996), no. 3, 230-232. I always tell my calculus students that mathematics is not esoteric: It is common sense. (Even the notorious epsilon, delta definition of limit is common sense, and moreover is central to the important practical problems of approximation and estimation.) see Bishop, Errett; Book Review: Elementary calculus. Bull. Amer. Math. Soc. 83 (1977), no. 2, 205--208. Having presented some published references, I would like to complement the published references with anecdotal evidence drawn from this very page, namely the sentiment that epsilon-delta definition is immediately appealing to students with a certain clarity of mind, etc. Such sentiments I believe are common and reveal a belief that these convoluted definitions are "natural" and a failure to follow them possibly constitutes an absence of a "clarity of mind". When one compares the upbeat assessment found in the mathematics community and the somber assessments common in the education community, sometimes one wonders whether they are talking about the same thing. How does one bridge the gap between the two assessments? Are they perhaps dealing with distinct student populations? Are there perhaps education studies providing more upbeat assessments than Dawkins' article would suggest? Note 1. A number of editors have commented by now on the subject of alternatives. Thus, Neil Strickland notes that "People may think that rigorous analysis is valuable despite this [teaching difficulty], and that certain other proposed teaching methods are no better, but those are different questions." In response, I would like to mention that the comparison of distinct approaches to teaching analysis is a fascinating subject, on which I have first hand experience and much to say. However, this subject is not that of this question. Note 2. Evidence from classroom experience comparing the two approaches has been presented in this recent publication based on the opinion poll of the students involved. REPLY [6 votes]: Maybe this is in part due to what we might call Gromov's zero-one law (from this paper on pg. 64, and also quoted in the conclusions here)... ...this common and unfortunate fact of the lack of an adequate presentation of basic ideas and motivations of almost any mathematical theory is, probably, due to the binary nature of mathematical perception: either you have no inkling of an idea or, once you have understood it, this very idea appears so embarrassingly obvious that you feel reluctant to say it aloud; moreover, once your mind switches from the state of darkness to the light, all memory of the dark state is erased and it becomes impossible to conceive the existence of another mind for which the idea appears nonobvious... Just in case... I'm not saying the math education people don't understand limits :).<|endoftext|> TITLE: When are arithmetic and geometric monodromy equal? QUESTION [6 upvotes]: Let $f: Y\to X$ be a finite separable morphism of curves over the finite field $\mathbb{F}_q$. Is there a simple condition under which the arithmetic and geometric monodromy of the covering are equal (i.e. $\mathbb{F}_q$ is algebraically closed in the Galois closure of the extension $\mathbb{F}_q(Y)/\mathbb{F}_q(X)$)? It is relatively easy to construct examples for which that condition is not satisfied: for example, a Kummer covering of $\mathbb{P}^1$ of degree $n$ coprime to $q-1$. However, I have the vague feeling that for many "natural" coverings I come across, arithmetic and geometric monodromy are indeed equal (although I'm not sure how I'd define "natural" here). Is it possible to give a simple criterion? I'm especially interested in the case when $X$ is an elliptic curve (and if that helps, the characteristic of $\mathbb{F}_q$ is much larger than the degree/genus of everything involved). REPLY [6 votes]: I have worked on reasonably similar problems to this, so hopefully what I have to say will be of some use. In general there is no easy way to do this, as far as I know. There are techniques for proving this kind of statement, and you could phrase an individual technique as a criterion, but it would be unhelpful, because most covers won't satisfy the criterion. This is the sort of statement that is usually true, but can be false for arbitrarily subtle reasons. That said, here is how a typical proof of this might look: First, lower bound the geometric monodromy group. The most powerful technique to do this is by evaluating the local monodromy at each ramification point. This is usually much easier to understand then the full monodromy group, and gives you conjugacy classes in the group. You can also use more global facts about the cover - if you can prove it is geometrically irreducible, for instance, then the monodromy must be transitive. Second, upper bound the arithmetic monodromy group. If there are any automorphisms of your cover defined over $\mathbb F_q$, the arithmetic monodromy must commute with them. If your cover is defined by adding some extra structure to a moduli space, then the arithmetic monodromy group must act on that structure. If there are any intermediate curves you know exist between $X$ and $Y$, you can use those to both lower bound and upper bound the groups. This might follow from the way they are constructed. Can you compute the discriminant? Third, use group theory to close the gap between the two groups. The key fact here is that the geometric monodromy group is a normal subgroup of the arithmetic monodromy group, with cyclic quotient. You hope to prove that no pair of groups satisfying all the conditions you set up beforehand can exist. Alternately, prove that one can exist, and try to come up with clever new arguments to rule them out! This advice might not be too helpful, because it basically says "solve this math problem like you would any other math problem, using math," but I hope that something I said here clicks with something about the "natural" coverings you are defining.<|endoftext|> TITLE: (Un)distorted subgroups in the mapping class group: reference required. QUESTION [6 upvotes]: Let $S$ be an orientable surface with negative Euler characteristic. Can somebody provide a reference for the following well-known results: the cyclic subgroup generated by a pseudo-Anosov element in the mapping class $MCG(S)$ is undistorted in the mapping class group. the cyclic subgroup generated by a Dehn twist of the mapping class group is undistorted. $\pi_1(S)$ is exponentially distorted in the mapping class group. Thank you in advance. REPLY [8 votes]: The following paper proves that all infinite cyclic subgroups of the mapping class group are undistorted (subsuming questions 1 and 2): Farb, Benson; Lubotzky, Alexander; Minsky, Yair, Rank-1 phenomena for mapping class groups. Duke Math. J. 106 (2001), no. 3, 581–597. For your third question, you don't specify which surface group you are talking about. I'm going to assume that you are asking about the "point-pushing subgroup", in which case the result you are looking for is in the following paper: Broaddus, Nathan; Farb, Benson; Putman, Andrew, Irreducible Sp-representations and subgroup distortion in the mapping class group. Comment. Math. Helv. 86 (2011), no. 3, 537–556.<|endoftext|> TITLE: Normalizer of SL_2(Z) in GL_2(R) QUESTION [6 upvotes]: What is the normalizer of ${\mathrm{SL}}_2({\Bbb Z})$ in ${\mathrm{GL}}_2({\Bbb R})$? Namely ${\mathrm{N}}_{{\mathrm{GL}}_2({\Bbb R})}({\mathrm{SL}}_2({\Bbb Z}))$? REPLY [4 votes]: Although it goes far beyond the question as posed (for which other answers provide a variety of elementary solutions), there is a broader context for this number-theoretic question. The setting involves a global field $K$, ring $R$ of $S$-integers for a non-empty finite set $S$ of places of $K$ (containing the archimedean places), arbitrary extension field $F$ of $K$ (such as $R = \mathbf{Z}$ and $F = \mathbf{R}$), and Chevalley group $G$ over $R$ (e.g., ${\rm{SL}}_n$, ${\rm{GL}}_n$, ${\rm{Sp}}_{2g}$, ${\rm{PGL}}_n$, etc.). To formulate a general proposition about normalizers in this setting, we need some more notation. Let $G' = \mathscr{D}(G)$ be the derived group, $Z_G \subset G$ the center, and $G^{\rm{ad}} = G/Z_G = G'/Z_{G'}$ (quotients in the scheme sense that is well-suited to working over rings such as $R$). For example, if $G = {\rm{GL}}_n$ then $G^{\rm{ad}} = {\rm{PGL}}_n$ and $G' = {\rm{SL}}_n$, but note that ${\rm{PGL}}_n(R)$ is strictly larger than ${\rm{GL}}_n(R)/R^{\times}$ when the class group of $R$ has nontrivial $n$-torsion. We will use the strong approximation theorem over global fields and the Iwasawa decomposition over non-archimedean local fields (all applied to split connected semisimple groups) to prove: Proposition: An element $g \in G(F)$ satisfies $gG'(R)g^{-1} \subset G'(R)$ if and only if the image of $g$ in $G^{\rm{ad}}(F)$ lies in $G^{\rm{ad}}(R)$. The sufficiency is obvious due to the natural action of $G^{\rm{ad}}$ on $G$ over $R$, so we need to prove the necessity. Remark: The condition that $g$ lands in $G^{\rm{ad}}(R)$ is equivalent to saying that $g$ lies in $G(R)Z_G(F)$ when $R$ is a UFD (or when $G$ is semisimple), but it is a strictly weaker condition on $g$ when $G = {\rm{GL}}_n$ and $R$ has nontrivial $n$-torsion in its class group. In this respect, the case $R = \mathbf{Z}$ might give the wrong impression as to what happens over more general number fields. Remark: In the special case $G = {\rm{GL}}_n$, everything that follows can be made entirely explicit in terms of matrices and requires no heavy machinery (e.g., strong approximation for ${\rm{SL}}_n$ can be deduced from strong approximation for adele rings, and the Iwasawa decomposition is just a reformulation of the structure theorem for modules over a PID). To prove the Proposition, we first show a weaker assertion: Lemma: The image of $g$ in $G^{\rm{ad}}(F)$ lies in $G^{\rm{ad}}(K)$. Proof: The key point is to exploit descent theory by considering the automorphism functor of $G_K$ over $K$-algebras (such as $F$ and $F \otimes_K F$). Fix a split maximal $R$-torus $T$ in $G$ (provided by the very definition of Chevalley groups, and which can be taken to be the diagonal subgroup when $G = {\rm{GL}}_n$), so we get closed root subgroups $U_a$ in $G'$ over $R$ for every root $a \in \Phi(G,T)$ (described in terms of elementary matrices when $G = {\rm{GL}}_n$). The subset $U_a(R) \subset U_a(K)$ is Zariski-dense in $(U_a)_K$ for all $a$ (as $U_a$ is an affine line over $R$), and the subgroups $U_a(K)$ generate a subgroup of $G'(K)$ that is Zariski-dense in the connected semisimple $G'_K$. Thus, for any $K$-algebra $A$ we can detect equality for a pair of $A$-automorphisms of $G'_A$ by checking on the subset $G'(R) \subset G'(A)$. Let $j_1, j_2:F \rightrightarrows F \otimes_K F$ denote the two natural scalar extensions $x \mapsto x \otimes 1, 1 \otimes x$. Thus, $g$-conjugation $c_g$ on $G'_F$ satisfies $(j_1)^{\ast}(c_g) = (j_2)^{\ast}(c_g)$ as automorphisms of the $F \otimes_K F$-group $G'_{F \otimes_K F}$ since these automorphisms coincide on $G'(R)$ (as $j_1$ and $j_2$ coincide on $K$, let alone on $R$). It follows by descent theory that the $F$-automorphism $c_g$ of $G_F$ descends to a $K$-automorphism of $G'_K$. But we have naturally $G^{\rm{ad}}_K \simeq {G'}^{\rm{ad}}_K$ as $K$-groups via the action of $G_K$ on $G'_K$, so an element of $G^{\rm{ad}}(F)$ whose action on $G'_F$ is defined over $K$ must arise from $G^{\rm{ad}}(K)$. QED Lemma It suffices to show that any $\overline{g} \in G^{\rm{ad}}(K) = {G'}^{\rm{ad}}(K)$ whose conjugation action on $G'_K$ carries $G'(R)$ into itself lies in the subgroup ${G'}^{\rm{ad}}(R) \subset {G'}^{\rm{ad}}(K)$. In particular, we can now rename $G'$ as $G$ to reduce to the case when $G$ is semisimple (for the purpose of proving our refined claim that has nothing to do with elements of $G'(R)$, but rather with elements of ${G'}^{\rm{ad}}(R)$). Since $G^{\rm{ad}}$ is identified with a closed subgroup scheme of the automorphism scheme ${\rm{Aut}}_{G/R}$ over $R$, a $K$-point of $G^{\rm{ad}}$ extends to an $R$-point if and only if the associated $K$-automorphism of $G_K$ extends to an $R$-automorphism of $G$. Now it suffices to show for any semisimple Chevalley group $G$ over $R$ that any $K$-automorphism $f$ of $G_K$ which carries $G(R)$ into itself extends (necessarily uniquely) to an $R$-automorphism of $G$. Letting $q:\widetilde{G} \rightarrow G$ be the simply connected central cover (a central isogeny over $R$, with $\widetilde{G}$ a simply connected Chevalley group), by canonicity there is a unique $R$-automorphism $\widetilde{f}$ of $\widetilde{G}$ over $f$. Thus, $\widetilde{f}(\widetilde{G}(R)) \subset \widetilde{G}(K) \cap q^{-1}(G(R)) = \widetilde{G}(R)$, where the final equality for the Dedekind $R$ expresses the valuation criterion of properness for the elementary case of the finite map $q$. Hence, we may replace $(G, f)$ with $(\widetilde{G}, \widetilde{f})$ to reduce to the case that the semisimple Chevalley group $G$ over $R$ is simply connected (e.g., ${\rm{SL}}_n$ or ${\rm{Sp}}_{2g}$). So far we have not used anything about $R$ beyond that it is a Dedekind domain (and the Dedekind property was only used in the final step above with the valuative criterion). It is time to bring in the arithmetic input in order to "localize" out problem (despite the hypothesis in terms of $G(R)$ appearing to be rather "global"). Recall the definition of $R$ as a ring of $S$-integers. By the strong approximation theorem for simply connected groups, applied by omitting any place in $S$ (at all of which $G$ is locally split), $G(R)$ has dense image in $G(R_v)$ for all $v \not\in S$; this is classical for $G = {\rm{SL}}_n$ and $R = \mathbf{Z}$. Hence, the closure of $G(R)$ in $G(K_v)$ is $G(R_v)$ for all $v \not\in S$, so the $K_v$-automorphism $f_{K_v}$ of $G_{K_v}$ carries $G(R_v)$ into itself for all such $v$. Our task is to show that the $K$-automorphism $f$ of $G_K$ extends to an $R$-automorphism of $G$ with $R = O_{K,S}$, and for this purpose it suffices to check that $f_{K_v}$ extends to an $R_v$-automorphism of $G_{R_v}$ for all $v \not\in S$. Now our problem is intrinsic to the local situation: by renaming $(K_v, R_v)$ as $(K, R)$ we're given a non-archimedean local field $K$ with valuation ring $R$ and an automorphism $f$ of $G_K$ for a simply connected Chevalley group $G$ over $R$ such that $f(G(R)) \subset G(R)$ and we claim that $f$ extends to an $R$-automorphism of $G$. Fix a split maximal $R$-torus $T$ of $G$ and a Borel $R$-subgroup $B$ of $G$ containing $T$. Let $\Phi^+ = \Phi(B,T)$ and let $\Delta$ be the base of $\Phi^+$, so $T^{\rm{ad}} := T/Z_G$ is a split maximal $R$-torus of $G^{\rm{ad}}$ whose character group admits $\Delta$ as a $\mathbf{Z}$-basis. In particular, a point in $T^{\rm{ad}}(K)$ lies in $T^{\rm{ad}}(R)$ if and only if its image in $K^{\times}$ under each $a \in \Delta$ lies in $R^{\times}$. For each $a \in \Delta$, let $X_a$ be a basis of the $R$-line ${\rm{Lie}}(U_a)$. Since ${\rm{Aut}}_{G/R}/G^{\rm{ad}}$ is identified with the finite constant $R$-group of automorphisms preserving the pinning $(B,T, \{X_a\}_{a \in \Delta})$ of $G$ over $R$, we can adjust $f$ by the $K$-fiber of a "pinned" $R$-automorphism of $G$ so that $f$ arises from $G^{\rm{ad}}(K)$. The Iwasawa decomposition gives $G^{\rm{ad}}(K) = G^{\rm{ad}}(R) T^{\rm{ad}}(K) G^{\rm{ad}}(R)$, so we may assume that $f$ arises from the action of some $t \in T^{\rm{ad}}(K)$. Thus, $f$ preserves every root group $(U_a)_K$ over $K$, inducing the scaling $t^a$ on this affine line over $K$. But this carries $(U_a)(K) \cap G(R) = U_a(R)$ into itself for all $a \in \Phi(G,T)$, which is to say that $t^a$-multiplication on $K$ preserves an $R$-line for all $a$, so $t^a \in R$ for all $a$. Applying this to $-a$ as well, we see that $t^a \in R^{\times}$ for all $a$. By considering $a \in \Delta$, we conclude that $t \in T^{\rm{ad}}(R)$, so $f \in G^{\rm{ad}}(R)$ as desired. QED Proposition<|endoftext|> TITLE: Dual of a smooth hypersurface QUESTION [5 upvotes]: Is it true that the dual of a smooth hypersurface $X$ of $\mathbb{P}^n$ of degree $d\ge 2$ is a hypersurface? If yes, could you give me a simple proof ? Or a reference? Note that in this case, the dual is birational to $X$. REPLY [5 votes]: A reference for this statement is Dolgachev's beautiful book Classical Algebraic Geometry, Chapter 1.<|endoftext|> TITLE: Can we use unparameterized chains to calculate singular homology? QUESTION [12 upvotes]: Most models of singular chains on a topological space $X$ use maps from some particular collection of "nice" objects, such as the standard simplices $\Delta^n$, the standard cubes $[0,1]^n$, etc. Is there a model of singular chains on a topological space $X$ which uses unparameterized maps from a similar such collection of nice objects? By unparameterized, I mean that a generator $\sigma:(\text{nice object})\to X$ is the same as the generator $\sigma\circ a$ for any $a\in\operatorname{Aut}(\text{nice object})$. My motivation for this question is the following. If $\pi:E\to B$ is a principal $S^1$-bundle, I would like to have the Gysin map $\pi^!:H_\ast(B)\to H_{\ast+1}(E)$ defined on the chain level. Intuitively, I should send a simplex $\sigma:\Delta^n\to B$ to the associated map $\Delta^n\times S^1\to E$ (note that $\sigma^\ast E$ is trivial since $\Delta^n$ is contractible). Of course, there is no canonical trivialization, though, so this map $\Delta^n\times S^1\to E$ is only defined up to an automorphism of the domain. Hence the desire to have chains generated by unparameterized maps. REPLY [2 votes]: I write this as an answer because it is a bit longer than a comment. 1.If $X$ is a subset of some $\newcommand{\bR}{\mathbb{R}}$ Euclidean space $\bR^n$ which additionally is a locally Lipschitz retract of one of its neighborhood, then you can use integral flat currents to represent cycles in this fashion. This is not quite what you want because the objects representing such such chains are rectifiable sets which can be quite wild. Additionally, the maps bewteen your topological spaces need to be locally Lipschitz. 2.If the category of spaces you are interested in is smaller, subanalytic sets or, more generally, sets definable in some $o$-minimal category, then you can work with a smaller collection of chains. (All such sets live in some finite dimensional Euclidean space.) As generators of your chain complex you can use the currents of integration defined by the oriented real analytic manifolds $S\subset X$. Such submanifolds can be triangulated and everything is generated by the currents of integration along simplices $\Delta\subset X$ such that their interiors are smooth manifolds. The Gysin map is simple in this case: it sets the current defined by $\Delta$ to the current defined by $\pi^{-1}(\Delta)$. This may seem like a restrictive category, but many spaces that admit local Kuranishi-type descriptions belong to this category, especially if they are compact. Hardt has the complete description of this approach, though it may be difficult to digest at first. 3.Sometime the high-brow road may pay dividends. By that I mean thinking of homology as cohomology with coefficients in the dualizing complex. Many operations is can be emulated in this language. Have a look at Iversen's "Cohomology of sheaves". It might give you some ideas.<|endoftext|> TITLE: counterexample to the Chern number inequality on Fano manifold QUESTION [8 upvotes]: We know that if an $n$-dimensional Fano manifold admits a Kahler-Einstein metric, it satisfies the following Chern number inequality $$nc_1^n\leq 2(n+1)c_2c_1^{n-2}.$$ My question is whether there exists a Fano manifold which does not satisfy this inequality (of course in this case this Fano manifold can not have any K-E metric). I calculated some examples which have been famously known that they don't support K-E metrics and find out that all of them still satisfy this inequality. Thanks in advance. REPLY [7 votes]: I think a counterexample is given by $\mathbb{P}_{P}(\mathcal{O}_{P}\oplus \mathcal{O}_{P}(n-1))$ for $n\geq 4$, with $P:=\mathbb{P}^{n-1}$. The computation is a bit long but here are the main steps (I hope I didn't make mistakes). Let $h\in H^2(X,\mathbb{Z})$ be the class of the tautological bundle, and $f$ the pull back of the class of $\mathcal{O}_{P}(1)$. Standard computations give $$c_1(X)=2h+f\qquad c_2(X)= 2n\,hf-\binom{n}{2}f^2\, \quad\mbox{ hence }$$ $$\Delta :=2(n+1)c_2-nc_1^2=4n\,hf - n^3f^2\ .$$ I want to prove $\Delta .c_1^{n-2}<0$. This is a sum of terms $\Delta .h^pf^{n-2-p}$ with positive coefficients. Note that $h^2=(n-1)hf$ (Chern class relation), so $h^pf^{n-2-p}=(n-1)^{p-1}hf^{n-3}$ for $p\geq 1$. Then $$\Delta .hf^{n-3}=4n(n-1)-n^3\qquad \Delta .f^{n-2}=4n\ .$$ For $n\geq 4$ all the terms are highly negative except $\Delta .f^{n-2}$, one sees immediately that the sum is negative.<|endoftext|> TITLE: The universal algebra of a $\sigma$-algebra QUESTION [6 upvotes]: I am searching for the 'dual' algebraic structure of a $\sigma$-algebra. The notion of duality is like in the case of the Boolean algebra and set algebra. If $X$ is a set, the complement and intersection on the power set of $X$ is called a set algebra and the series of equations that define a Boolean algebra is the dual of this structure. I found this link that seems related to my question: Is there such a thing as the sigma-completion of a Boolean algebra? but still it does not solve my problem. REPLY [4 votes]: In the paper [1], Sikorski constructs a duality that generalizes Stone duality to certain $\sigma$-complete Boolean algebras and more generally certain $\kappa$-complete Boolean algebras. I shall outline the duality mentioned in Sikorski's paper here. Suppose that $\lambda$ is a cardinal. Then a Boolean algebra $B$ is said to be $\lambda$-complete if the least upper bound $\bigvee R$ exists whenever $|R|<\lambda$. A filter $Z$ on a $\lambda$-complete Boolean algebra $B$ is said to be a $\lambda$-complete filter if whenever $|R|<\lambda$ and $R\subseteq Z$, then $\bigwedge R\in Z$ as well. We shall call a $\lambda$-complete Boolean algebra strongly $\lambda$-representable if every $\lambda$-complete filter can be extended to a $\lambda$-complete ultrafilter. A $P_{\lambda}$-space is a completely regular space such that the intersection of less than $\lambda$ many open sets is open, and a topological space $X$ is said to be $\lambda$-compact if every open cover of $X$ has a subcover of cardinality less than $\lambda$. Sikorski gave a correspondence between all $\lambda$-compact $P_{\lambda}$-spaces and all $\lambda$-representable $\lambda$-complete Boolean algebras. The proof of this result is exactly the same as the proof of the duality between Boolean algebras and compact totally disconnected spaces. Sikorski, Roman Remarks on some topological spaces of high power. Fund. Math. 37, (1950). 125–136.<|endoftext|> TITLE: Reference requested: Random walk on groups QUESTION [9 upvotes]: I am looking for a good reference to learn about random walks on groups (either finite groups or Lie groups). Ideally, I would like a reference for general theory of random walks on groups that is self-contained in terms of the probability theory it relies on, while remaining condensed and precise. As an analogy, something like Atiyah-Macdonald's book in commutative algebra (i.e. short, but very precise and somehow complete) would be great. Thank you very much. REPLY [6 votes]: My book http://link.springer.com/book/10.1007%2F978-1-4614-0776-8 on group representation theory has a chapter on this.<|endoftext|> TITLE: Easy to state applications of dimension theory in algebraic geometry QUESTION [7 upvotes]: Dimension theory is quite a sophisticated topic (at least for me), it is fully settled in Shafarevich's book on the first 100 pages. Shafarevich gives two nice applications of the theory. 1) A proof of Tzen's theorem 2) A proof of existence of a line on a cubic surface in $\mathbb P^3$. Both applications are geometric statements. I would like to learn about some other applications that are easy to state. I am asking this partially because of an introductory course in algebraic geometry that I am teaching. I would like to give some motivation for developing the dimension theory. REPLY [5 votes]: This isn't a very direct answer, but one of the things I use all the time in algebraic geometry as opposed to non-algebraic is that if $X$ is dense in $Y$, then $\dim(Y\setminus X) < \dim X$. The topologist's sine curve is a counterexample to this in non-algebraic geometry.<|endoftext|> TITLE: What is/are the best bound/s on the sum of squares of degrees in a graph? QUESTION [6 upvotes]: Let $G$ be a graph with degrees $d_{1},\ldots,d_{n}$. I am interested in upper bounds on $$ \sum_{i=1}^{n}{d_{i}^{2}}. $$ An example is de Caen's bound: $$ \sum_{i=1}^{n}{d_{i}^{2}} \leq e(\frac{2e}{n-1}+n-2), $$ where $e$ is the number of edges. His bound is very elegant but far from tight. I have read many papers which give other bounds but now I am a bit confused and would love to know if someone can supply a definitive answer. P.S. The quantity in question is also known as the "first Zagreb index", mostly in chemistry-related circles. REPLY [4 votes]: There is a simple spectral upper bound: namely, denoting by $A$ the adjacency matrix of $G$, and by $\mathbf 1$ the $n$-dimensional all-$1$ vector, we have $$ \sum_{I=1}^n d_i^2=\|A{\mathbf 1}\|^2 \le \|A\|^2 \|{\mathbf 1}\|^2 = n\lambda_{\rm max}^2, $$ where $\lambda_{\rm max}$ is the largest eigenvalue of $G$. This bound is sharp, say, for regular graphs.<|endoftext|> TITLE: Salmon's proof that tangents to a cubic from a point on it have the same cross-ratio QUESTION [6 upvotes]: In Higher plane curves, nr 167, Salmon proves that the cross-ration of the four tangents to a non-singular plane cubic, drawn from a point on the curve, is independent of the point. A proof can be found in Van der Waerden's Einführung in die algebraische Geometrie. But can anyone explain in modern terms Salmon's proof? It runs as follows: If from two consecutive points $O$, $O'$ of the curve we draw the two sets of tangents $OA$, $OB$, $OC$, $OD$; $O'A$, $O'B$, $O'C$, $O'D$, any tangent $OA$ intersects the consecutive tangent $O'A$ in its point of contact. Now the four points of contact $A$, $B$, $C$, $D$ lie on the polar conic of $0$, which also touches the cubic at the point (Art. 64) ; hence the six points $OO'ABCD$ lie on the same conic, and therefore the anharmonic ratio of the pencil $\{O.ABCD\}$ is the same as that of the pencil $\{O'.ABCD\}$. Since then this ratio remains the same when we pass from one point of the curve to the consecutive one, we learn that the anharmonic ratio is constant of the pencil formed by the four tangents which can be drawn from any point of the curve. REPLY [3 votes]: Consider a local parametrization $O(t), A(t), ...$, with $O = O(0), A = A(0), ...$. Salmon is essentially checking that $\frac{d}{dt}(O(t)A(t),O(t)B(t);O(t)C(t),O(t)D(t)) = 0$ at $t = 0$, where $(OA,OB;OC,OD)$ denotes the anharmonic ratio of the pencil $\{O.ABCD\}$. Since $A(t)$ stays on the line $OA$ to first order, and similarly for $B(t), C(t), D(t)$, we have $\frac{d}{dt}(O(t)A(t),O(t)B(t);O(t)C(t),O(t)D(t))|_{t=0} = \frac{d}{dt}(O(t)A,O(t)B;O(t)C,O(t)D)|_{t=0}$. This will be zero if and only if $O(t)$ stays on the conic $OABCD$ to first order, which Salmon checks using the fact that for $O$ on the cubic curve the polar conic of $O$ passes through $O, A, B, C, D$ and is tangent to the cubic curve at $O$.<|endoftext|> TITLE: Why does this setting imply that a category is Grothendieck? QUESTION [11 upvotes]: I came across the following Lemma in Mitsuyasu Hashimoto's Equivariant Twisted Inverses; it is Lemma 11.2 on page 107 of this pdf. Let $\mathcal{A}$ be an abelian category which satisfies the (AB3) condition (i.e., is cocomplete) and $\mathcal{B}$ a Grothendieck category. Assume we have an adjoint pair $$ F \colon \mathcal{A} \rightleftarrows \mathcal{B} :G$$ such that $F$ is faithful exact and $G$ preserves filtered colimits. Then $\mathcal{A}$ is also Grothendieck. Sadly, the proof is not given there but at least should work similar to the one before, where "Grothendieck" is replaced by "locally noetherian". Since one easily deduces the exactness of filtered colimits in $\mathcal{A}$, it remains to show that $\mathcal{A}$ admits a generator. But somehow I am confused between the notion of a generator/generating set in the sense of a separator (detecting non-trivial morphisms) and being able to write an object as a filtered colimit over a diagram involving only the generator. What is the connection between these notions? Possible application: This statement would immediately give that the category of comodules over a flat Hopf algebroid $(A,\Gamma)$ is Grothendieck -- take $F$ to be the forgetful functor from $\Gamma$-comodules to $A$-modules and let $G$ be the cofree comodule. REPLY [5 votes]: Here's a rather convoluted proof of this lemma. It suffices to show that $\mathcal A$ is accessible, that is, $\mathcal A\simeq Ind_\kappa(\mathcal A_0)$ for some small category $\mathcal A_0$ (the sum of all objects in $\mathcal A_0$ is then a generator in $\mathcal A$). Since $F$ is faithful and exact, the adjunction is comonadic (Barr-Beck), and the comonad preserves filtered colimits by the assumption on $G$. In particular, the comonad is accessible, i.e., it preserves $\kappa$-filtered colimits for some regular cardinal $\kappa$. It is well-known that coalgebras for an accessible comonad on a locally presentable category form a locally presentable category: see Prop A.1 in http://www.math.harvard.edu/~eriehl/coalgebraic.pdf (this is basically a special case of the Makkai-Paré theorem stating that lax limits of accessible categories are accessible, Theorem 2.77 in Adámek-Rosicky).<|endoftext|> TITLE: Graphs with dangling edges QUESTION [5 upvotes]: In conventional Graph Theory the role of Nodes and Edges is skewed: nodes are perfectly ok being aloof, but poor edges are always drawn between existing nodes (that is, the two maps from EDGES to NODES are total). Now, I am curious to know if there is somewhere in the Mare Magnum of mathematics an extension of Graph Theory, let us call it provisionally the Theory Partial Graphs (it may have already a well-established name, in which case I apologize), which contemplates dangling edges (ie edges which either stand alone, or are attached only to one node). To be more specific, I would like to see some refs on TYPED partial graphs: dangling edges hare equipped with types, and there is an algebra of sorts which tells which dangling nodes can "merge" with other kindred typed dangling nodes to make up a genuine edge of the graph (think of nodes which protrude some edges like an octopus, with the possibility of two arms joining and forming a perfectly traditional edge). Anything along the lines above? PS Although I do not have any specific preconceptions around the way this theory may appear (for instance, as some kind of algebra, or as a chapter of topology), ideally it would be codes within a categorical framework REPLY [2 votes]: A graph with dangling edges can be simply a bipartitioned graph where the set of vertices splits into a union $\ V:=A\cup B\ $ of disjoint sets $\ A\ B, $ and vertices $\ b\in B\ $ can have at the most two neighbors (from $\ A).\ $ The partition into $\ A\ B\ $ has to be specified in the general case if we want a unique representation of the bi-partitioned graph as a graph with dangling edges so that vertices from $\ B\ $ would be interpreted as the dangling edges.<|endoftext|> TITLE: Complete Boolean algebra not isomorphic to a $\sigma$-algebra QUESTION [18 upvotes]: Does there exist a complete Boolean algebra that is not isomorphic to any $\sigma$-algebra? If so, what is an easy or canonical example or construction? REPLY [10 votes]: Here is yet another way to see this. In a $\sigma$-algebra of sets, countable intersections an unions are computed setwise, which means that we always have $$\bigwedge_{n<\omega} \bigvee_{m < \omega} A_{m,n} = \bigcap_{m<\omega} \bigcup_{n < \omega} A_{m,n} = \bigcup_{f:\omega\to\omega} \bigcap_{m<\omega} A_{m,f(m)},$$ where the right hand side is to be considered in a purely set theoretic basis. In any complete boolean algebra, we have $$\bigwedge_{n<\omega} \bigvee_{m < \omega} A_{m,n} \geq \bigvee_{f:\omega\to\omega} \bigwedge_{m<\omega} A_{m,f(m)}.$$ In the case of a $\sigma$-algebra of sets, we have $$\bigvee_{m<\omega} A_{m,f(m)} = \bigcap_{m<\omega} A_{m,f(m)}$$ for every $f:\omega\to\omega$, hence the set theoretic identity above shows that in a complete $\sigma$-algebra we must have $$\bigwedge_{n<\omega} \bigvee_{m < \omega} A_{m,n} = \bigvee_{f:\omega\to\omega} \bigwedge_{m<\omega} A_{m,f(m)}.$$ There are plenty of complete Boolean algebras that do not satisfy this distributive identity. In fact, from the perspective of forcing, this distributive property for a complete atomless Boolean algebra is equivalent to not adding new reals, so any complete atomless Boolean algebra that adds a new real is not isomorphic to any $\sigma$-algebra of sets. This includes Cohen forcing as Joel pointed out, Random forcing as Simon pointed out, and a whole lot more...<|endoftext|> TITLE: Elliptic curves over $\mathbb{Q}$ with no rational torsion and $\mu$-invariant equal to 1 at $p=3$ QUESTION [5 upvotes]: How to find out examples over elliptic curves over $\mathbb{Q}$ with no rational torsion and $\mu$-invariant equal to 1 at $p=3$ $?$ REPLY [9 votes]: EDIT: There was a problem with my original answer. Details below the bottom line. If you have any elliptic curve $E/{\mathbb Q}$ with a point of order 3, then we have an exact sequence of Galois modules: $$ 0 \to {\mathbb Z}/3{\mathbb Z} \to E[3] \to \mu_3 \to 0. $$ Here the ${\mathbb Z}/3{\mathbb Z}$ term arises from the point of order 3 defined over ${\mathbb Q}$. The $\mu_3$ term is then forced upon us since the determinant of this Galois representation is the mod 3 cyclotomic character. Now here is the trick. There is an isogenous curve $E'/{\mathbb Q}$ such that $$ 0 \to \mu_3 \to E'[3] \to {\mathbb Z}/3{\mathbb Z} \to 0 $$ (i.e. the two outer terms have switched!), and moreover this sequence is not split. I'll say in a moment how to find this curve. But let me first point out that this new curve should have $\mu$-invariant equal to 1 -- this follows from Greenberg's conjecture on $\mu$-invariants. The $\mu$-invariant should be the smallest $n$ such that $E[p^n]$ has a cyclic sub which is odd and ramified. In this case, the sub of $\mu_3$ (which is odd and ramified) gives that $\mu(E')$ is at least 1. Since the sequence is not split, there is no cyclic sub of size 9 (and so conjecturally the curve has $\mu$-invariant 1). (EDIT: Nope. This isn't true. See below. I should be assuming that 3-isogeny class for $E$ has only two curves in it.) Now, how to find this curve? Just take $E$ and mod out by that ${\mathbb Z}/3{\mathbb Z}$ sub of $E[3]$. The resulting curve then has $\mu_3$ as a sub and ${\mathbb Z}/3{\mathbb Z}$ as quotient. If this extension is non-split, we are done. If not, mod out by ${\mathbb Z}/3{\mathbb Z}$ again. And keep repeating until the extension is non-split. (If this continued indefinitely, then the Tate module of $E$ at 3 would be reducible which it is not.) EDIT: OK. The problem is that just because the sequence $$ 0 \to \mu_3 \to E'[3] \to {\mathbb Z}/3{\mathbb Z} \to 0 $$ is not split does not mean that $E'[9]$ doesn't contain a cyclic sub of size 9. In fact, if the process described above of modding out by ${\mathbf Z}/3{\mathbf Z}$ takes two steps, then it will contain such a sub and have $\mu$-invariant strictly greater than 1. In fact, this problem comes up in the example of conductor 19 in Jeff H's answer. In this case, there are 3 isogenous curves of conductor 19 -- 19a1, 19a2, and 19a3 as in Cremona's tables. I'll call them $E_1$, $E_2$ and $E_3$ respectively. For $E_1$ we have $E_1[3] \cong \mu_3 \times {\mathbb Z}/3{\mathbb Z}$. Modding out $E_1$ by the $\mu_3$ gives $E_3$, and for $E_3$ we have a non-split extension. $$ 0 \to {\mathbb Z}/3{\mathbb Z} \to E_3[3] \to \mu_3 \to 0. $$ Modding out $E_1$ by the ${\mathbb Z}/3{\mathbb Z}$ gives $E_2$, and for $E_2$ we have a non-split extension. $$ 0 \to \mu_3 \to E_2[3] \to {\mathbb Z}/3{\mathbb Z} \to 0. $$ So if we had started my solution above with $E = E_3$ (which is a curve with a point of order 3), we would have to first mod out by ${\mathbb Z}/3{\mathbb Z}$ and we arrive at $E_1$ which again has a point of order 3. So we again mod out by ${\mathbb Z}/3{\mathbb Z}$ and we arrive at $E_2$ which has no point of order 3. However, $E_2$ has a cyclic isogeny of degree 9 to $E_3$ whose kernel is odd and ramified. Thus $\mu(E_2)$ is at least 2 (and exactly 2 by Greenberg's conjecture). And every other curve in this isogeny class has a point of order 3. OK. How to fix this? Well the problem is that third curve in the isogeny class. So instead, we need a curve with a point of order 3 whose 3-isogeny class has size 2. This way we are guaranteed from the start that there is no cyclic sub of size 9. After a quick peek at Cremona's tables, I see that 44a2 works. It has $\mu$-invariant equal to 1 and no 3-torsion.<|endoftext|> TITLE: Modular form on $\Gamma_0(N)$ QUESTION [7 upvotes]: I recently asked this question on Math.StackExchange with no answer so far. So I thought maybe I can find an answer here. Let $M(k,\Gamma_0(N))$ be a space of modular forms of weight $k$ on $\Gamma_0(N)$. Each $f \in M(k,\Gamma_0(N))$ has a Fourier expansion of the form $$ f(\tau)=\sum_{n\in \mathbb{N}}a(n)\,q^n \quad \text{where}\quad q=\mathrm{e}^{2\pi \mathrm{i}\tau} . $$ Now, let $g(\tau)$ be a function obtained from $f(\tau)$ by omitting all $a(n)$ such that $\gcd(n,N)\neq 1$, i.e. $$ g(\tau)=\sum_{\substack{n\in \mathbb{N}\\(n,N)=1}}a(n)\,q^n . $$ Question: Is $g(\tau)$ a modular form? What is its level? Many thanks. REPLY [5 votes]: $$ g(\tau)=f(\tau)\otimes \left(\tfrac{N^2}{\cdot}\right)=\sum_{n\in \mathbb{N}}\left(\tfrac{N^2}{n}\right)a(n)\,q^n \quad \text{where}\left(\tfrac{N^2}{\cdot}\right) \text{ is the Kronecker symbol}.$$ This means that $g(\tau)$ is just the twist of $f(\tau)$ by a principle character. Indeed we have $$g(\tau)\in M(k,\Gamma_0(N^3)).$$<|endoftext|> TITLE: Extension of $C^*$ isomorphism to $W^*$ isomorphism QUESTION [5 upvotes]: Let $\mathfrak{A}$ be $C^*$algebra, and $\pi$ its faithful representation on Hilbert space $\mathcal{H}$. Bicommutant $\mathfrak{B}=\pi(\mathfrak{A})''$ is the von Neumann algebra generated by $\pi(\mathfrak{A})$. Let $\phi$ be ${}^*$-automorphism of $\mathfrak{A}$. Under what conditions the ${}^*$-automorphism $\gamma: \pi(\mathfrak{A})\rightarrow \pi(\mathfrak{A})$, $\gamma = \pi\circ\phi\circ\pi^{-1}$ of the algebra $\pi(\mathfrak{A})$ might be extended to ${}^*$-automorphism of $\mathfrak{B}$. EDIT: Let us assume in addition that $\forall_{A\in\mathfrak{A}}\gamma(\pi(A)) = \lim_{n\rightarrow\infty}\gamma_n(\pi(A))$, where $\gamma_n(B)=U_n B U_n^{-1}$, $U_n$ - sequence of unitary operators in $\mathcal{H}$, but $\gamma$ is not implementable by unitary operator. REPLY [4 votes]: Here is a partial answer to your question, in the paper: Derivations and automorphisms of operator algebras by Richard V. Kadison and John R.Ringrose, Commun.math.Phys. 4,32-63(1967) It is proved that each derivation on your represented algebra extends to its weak closure provided the representation is faithful. Furthur, they use this result to conclude that if an automorphism is within norm distance 2 to the identity automorphism, then it is in the connected component of the group of automorphism and therefore also extends to an ( actually inner) automorphism of the weak closure. You should be able to find the necessary details there.<|endoftext|> TITLE: Existence of injective operators with dense range QUESTION [7 upvotes]: Given two separable (infinite dimensional) Banach spaces $X$ and $Y$, it is not difficult to show that there exists an injective (bounded linear) operator $T:X\to Y$ with range dense in $Y$. See S. Goldberg and A.H. Kruse. "The Existence of Compact Linear Maps Between Banach Spaces$. Proc. A.M.S. 13 (1962), 808-811. I am interested in the existence of such an operator when $Y$ is a hereditarily indecomposable (H.I.) Banach space and $X=Y\times Y$. Since H.I. spaces can be embedded as closed subspaces of $\ell_\infty$ (see the Introduction of the Argyros-Tolias Memoir A.M.S. 806, 2004), we can consider the following related questions: Let $X$ be a Banach space isomorphic to a subspace of $\ell_\infty$. ($Q_1$) Is it posible to find an injective operator $T:X\to X\times X$ with dense range? ($Q_2$) Is it posible replacing the initial $X$ by a closed subspace $X_0$ of $X$? REPLY [6 votes]: In Argyros, Spiros A.; Arvanitakis, Alexander D.; Tolias, Andreas G. Saturated extensions, the attractors method and hereditarily James tree spaces. Methods in Banach space theory, 1–90, London Math. Soc. Lecture Note Ser., 337, Cambridge Univ. Press, Cambridge, 2006 the authors construct a separable space $Z$ so that $X:= Z^*$ is non separable, hereditarily indecomposable (HI), and every strictly singular operator on $X$ is weakly compact and hence has separable range. Of course, $X$ embeds isometrically into $\ell_\infty$ because $Z$, being separable, is a quotient of $\ell_1$. There is no operator from this $X$ into $X \oplus X$ that has dense range. Indeed, if $T$ is such an operator, then letting $P_i$ for $i= 1,2$ be the projections onto the factors of $X \oplus X$, we see that $P_iT$ is of the form $\lambda_i I + W_i$ with $W_i$ strictly singular by the usual Gowers-Maurey HI theory. Both operators $P_iT$ have dense range, so neither $\lambda_i$ is zero because both $W_i$ have separable range. To see that $T$ cannot have dense range, observe that $T^*$ is not injective. Indeed, take a norm one $x^*$ in $X^*$ which vanishes on $W_1X+W_2X$. Then $T^* (\lambda_1^{-1}x^*, - \lambda_2^{-1}x^*)= 0$. There is also no injective dense range operator from $X\oplus X$ into $X$. For suppose that $S(x_1,x_2) = S_1 x_1 + S_2 x_2$ is such an operator. Write $S_i= \lambda_i I + W_i$ with $W_i$ strictly singular. Since the $W_i$ have separable range and $S$ has dense range, one of the $\lambda_i$, say $\lambda_1$, is not zero, and hence $S_1$ is Fredholm of index zero. If $S_2$ has finite rank, then clearly $S$ is not injective, but otherwise the infinite dimensional $S_2X$ intersects nontrivially the finite codimensional $S_1X$, which again implies that $S$ is not injective.<|endoftext|> TITLE: Skolem Hulls in $H_{\omega_2}$ QUESTION [8 upvotes]: I put this on stack exchange over a week ago with no answer, so let's try here. Consider a model of the form $\mathfrak{A} = (H_{\omega_2}, \in, \lhd, f_0, f_1, ...)$, some expansion of $H_{\omega_2}$ in a countable language, with $\lhd$ giving a well-order. Does there exist an infinite set $z$ of uncountable ordinals, such that for every $y \subseteq z$, $Sk^{\mathfrak{A}}(y) \cap z = y$? Perhaps the following observation will be useful. Note that the following are equivalent for a given infinite $z \subseteq \omega_2$. For all $y \subseteq z$, $Sk^{\mathfrak{A}}(y) \cap z = y$. For all finite $y \subseteq z$, $Sk^{\mathfrak{A}}(y) \cap z = y$. REPLY [11 votes]: No. For sufficiently nasty $f_i$'s, you can't even get a 3-element $z$ with the independence property that you specified. Choose, for each ordinal $\alpha<\omega_2$, a one-to-one map into $\omega_1$, and assemble all these maps into a single binary function $f$, so that, for each fixed $\alpha<\omega_2$, the function $\beta\mapsto f(\alpha,\beta)$ maps ordinals $\beta<\alpha$ one-to-one to countable ordinals. (I don't care what $f(\alpha,\beta)$ is when $\alpha\leq\beta$.) Similarly, let $g$ be a binary function such that, for each fixed $\alpha<\omega_1$, the function $\beta\mapsto g(\alpha,\beta)$ maps the ordinals $\beta<\alpha$ one-to-one to natural numbers. Now suppose $z$ is a 3-element set of ordinals, say $\xi<\eta<\theta$. Let $y\subseteq z$ consist of $\theta$ and whichever of $\xi$ and $\eta$ has the larger image under $f(\theta,-)$. Call that larger image $\rho$, and let $\sigma<\rho$ be the image of the other one of $\xi$ and $\eta$. Then the Skolem hull of $y$, in $(H_{\omega_2},\in,\lhd,f,g)$ contains all of the following: $\rho$ (because it's closed under $f$, which maps the pair of elements of $y$ to $\rho$), $g(\rho,\sigma)$ (because this is a natural number and therefore definable in $H_{\omega_2}$), $\sigma$ (because it's definable from $g(\rho,\sigma)$ and $\rho$, using $g$), and the element of $z-y$ (because it maps to $\sigma$ under $f(\theta,-)$ and is therefore definable from $\sigma$ and $\theta$ using $f$). So the Skolem hull of $y$, intersected with $z$, is not only $y$ but all of $z$.<|endoftext|> TITLE: Is a left topological group which is a manifold a topological group? QUESTION [15 upvotes]: Let $G$ be a left topological group, i.e. a topological space with group operation such that left multiplication $L_g : x \mapsto gx$ is continuous (but right multiplication and inversion are not required to be). Assume also $G$ to be a topological manifold. Does this imply that $G$ is a topological group? EDIT 1: as pointed out by Yves Cornulier, the aswer is no if $G$ is not required to be connected. Hence, I would like to know if the statement still holds when $G$ is assumed to be connected. EDIT 2: as discussed in the comments, we are also assuming $G$ to be paracompact, with countable atlas say. REPLY [20 votes]: Here is a counter-example with $G$ homeomorphic to $\mathbb R^2$. Let $f:\mathbb R\to\mathbb R$ be a discontinous additive homomorphism (constructed using a Hamel basis of $\mathbb R$ over $\mathbb Q$). Define a group operation $*$ on $\mathbb R^2$ by $$ (x,y)*(x',y') = (x+x'e^{f(y)},y+y') . $$ This groups is a semidirect product of $\mathbb R$ and $\mathbb R$ with respect to the action where $y\in\mathbb R$ acts on $\mathbb R$ by multiplication to $e^{f(y)}$. The formula is continuous in $(x',y')$ but not in $(x,y)$.<|endoftext|> TITLE: Probability that random cubic polynomials meet in a square QUESTION [12 upvotes]: Let $p_1(x)$ and $p_2(x)$ be cubic polynomials with random coefficients in $[-1,1]$. I wanted to compute the probability that $p_1$ and $p_2$ share at least one point within the square $[-1,1]^2$. Of course this is the same as the probability that $p_1(x)=p_2(x)$ for $x \in [-1,1]$ and $p_1(x) \in [-1,1]$. I am not seeing this as a straightforward computation based on known root distribution results for random polynomials, but I admit to ignorance in this area. Crude experimentation suggests the probability might not be far from $\frac{1}{2}$. Below, $5$ out of $9$ instances meet within the $[-1,1]^2$ square.       My interest derives from graphics and cubic splines. I wanted to understand how likely it would be that the (relatively expensive) root computation would be necessary to compute certain graphic representations, e.g., visibility from above. REPLY [3 votes]: If $p_i(x)=a_ix^3+b_{i,2}x^2 + b_{i,1}x + b_{i,2}$ for $i\in\{1,2\}$ then a sufficient condition for $\exists (x,y)\in [-1,1]^2$ with $y=p_1(x)=p_2(x)$ is that (writing $a=a_1-a_2$, $b_j=b_{1,j}-b_{2,j}$) \begin{eqnarray} \tag{1}\sum_{j=0}^2 |b_{j}|\le |a|\quad\text{and}\\ \tag{2}|a_i| + \sum_{j=0}^2 |b_{i,j}| \le 1\quad\exists i\in\{1,2\} \end{eqnarray} and this gives a nonzero lower bound of approximately $0.02$ on the probability (an 8-fold multiple integral which it is tricky to get Mathematica to do). Indeed, it is easy to see that with probability 1 there exist $(x,y)\in\mathbb R^2$ with $y=p_1(x)=p_2(x)$; consider one such pair. If $|x|\ge 1$ then we have by (1) $$ |ax|\le |b_2| + \frac{|b_1|}{|x|} + \frac{|b_2|}{|x|^2} \le \sum_j |b_j|\le |a| $$ so $|x|\le 1$ after all. To ensure $|y|\le 1$ we use (2).<|endoftext|> TITLE: Can the monoidal structure on manifolds be strictified? QUESTION [6 upvotes]: I'm asking this question purely out of curiosity. Let $\{M_\alpha\}_{\alpha\in A}$ be a collection of closed smooth manifolds, with exactly one in every diffeomorphism class of closed smooth manifold. Hence for each $\alpha,\beta\in A$, there exists a unique $\alpha\ast\beta\in A$ for which $M_{\alpha\ast\beta}$ is diffeomorphic to $M_\alpha\times M_\beta$. Hence we get an binary operation $\ast:A\times A\to A$ which is associative and commutative. Can we choose diffeomorphisms $\phi_{\alpha\beta}:M_\alpha\times M_\beta\to M_{\alpha\ast\beta}$ which are associative? what about associative and commutative? To be precise, associativity means the following diagram commutes: $$\begin{matrix}M_\alpha\times M_\beta\times M_\gamma&\rightarrow&M_{\alpha\ast\beta}\times M_\gamma\cr\downarrow&&\downarrow\cr M_\alpha\times M_{\beta\ast\gamma}&\rightarrow&M_{\alpha\ast\beta\ast\gamma}\end{matrix}$$ and commutativity means the following diagram commutes: $$\begin{matrix}M_\alpha\times M_\beta&\rightarrow&M_{\alpha\ast\beta}\cr\downarrow&&\downarrow\cr M_\beta\times M_\alpha&\rightarrow&M_{\beta\ast\alpha}\end{matrix}$$ REPLY [4 votes]: You cannot have commutativity even for zero-dimensional manifolds, as you will see by considering the case where $\alpha=\beta$ and $M_\alpha=M_\beta=\{0,1\}$. For associativity, my guess is that the monoid of isomorphism classes is probably a free abelian monoid (so we have "unique factorization"). If that is true then you can certainly arrange associativity. It would be a bit fiddly to write down a precise proof and I do not have time just now. However, you will get the idea if you first consider a category in which all objects are isomorphic to $A^i$ for some fixed object $A$ and some $i\in\mathbb{N}$, and then consider a category in which every object is of the form $A^i\times B^j$.<|endoftext|> TITLE: Boundaries of relatively hyperbolic groups QUESTION [5 upvotes]: When the interior of an n-manifold $M$ has a pinched negative curvature metric of finite volume, then its fundamental group $\Gamma=\pi_1M$ is relatively hyperbolic relative to the parabolic groups $\pi_1\partial M$, so one can consider $\partial\Gamma$ which is defined as the ideal boundary of the coned off Cayley-Graph (i.e. the Cayley-Graph of $\pi_1M$ coned off over the right cosets of $\pi_1\partial M$). Question: Is $\partial\Gamma$ homeomorphic to the (n-1)-sphere and, if yes, where has this been proved? In the closed case this is immediate from the quasiisometry between $\Gamma$ and the universal covering $\widetilde{M}$. In the cusped case of course $\Gamma$ is quasiisometric to the complement of the horoballs in $\widetilde{M}$, but the cones over the parabolic subgroups seem not to be quasiisometric to the horoballs, so it is not clear to me how to adapt the argument. REPLY [7 votes]: As pointed out in the comments, the boundary of the coned-off graph won't be a sphere in this case. In fact, it won't even be compact. Bowditch's paper on relatively hyperbolic groups (recently published in IJAC) gives some details about the relationship between the two boundaries in the last few sections. In particular see Propositions 8.5 and 9.1. To summarize Bowditch's results: If $\hat{\Gamma}$ is the coned-off Cayley graph for a relatively hyperbolic pair, and $B$ is the standard boundary for the relatively hyperbolic pair, then $\partial\hat{\Gamma}$ sits inside $B$ as a dense subset. The complement is a countable (also dense) set which can be identified with the set of maximal parabolic subgroups. In the case you are interested in $B$ is an $(n-1)$-sphere. Thus $\partial\hat{\Gamma}$ is an $(n-1)$-sphere with a countable dense subset removed. (EDIT: I was going to add this as a comment, but I don't have enough reputation. Let $X$ be the space obtained by coning off the boundary of $M$, and let $Y$ be the universal cover of $X$. The space $Y$ isn't quasi-isometric to $\hat{\Gamma}$. Rather the normal closure $N$ of the parabolic subgroups acts on $\hat{\Gamma}$, with a quotient quasi-isometric to $Y$.)<|endoftext|> TITLE: A ZFC construction to get a proper extension which is a $\omega_1$-model QUESTION [5 upvotes]: In $V$, let me call a set theory structure A is a $\omega_1$ model if the $\omega_1$ of $A$ is the same as the $\omega_1$ in $V$ (up to isomorphism). The question I would like to ask is the following: Given any transitive structure $(B,\in)$ containing $\omega_1$ in $V$, is it always possible to find an $\omega_1$ model $A$ such that $B\prec A$ and $B\cap H(\omega_2)$ is in the well-founded part of $A$. Some remark: Lowenheim-Skolem or Compactness theorem can be employed to construct an extension. But in general this extension is not an $\omega_1$ model. To get one $\omega_1$ model, one can either use some large cardinal or force a generic ultrafilter to construct an ultrapower which is well founded below $\omega_1$. But I am not sure how to get this using ZFC alone. On the other hand, it is known that the existence of $\omega$-model is correct between transitive ZFC models and thus can be constructed under ZFC. I know very little about model theory of set theory. So it would be very helpful if anyone can provide some reference about this topic. REPLY [4 votes]: In this generality, this is impossible, if you want $B\prec A$ to be a proper extension. For a counterexample, let $B=V_{\omega_2}$, the rank initial segment of the universe of height $\omega_2$. I claim that there can be no proper extension $\langle B,\in\rangle\prec\langle A,\hat\in\rangle$, where the well-founded part of $A$ includes $H_{\omega_2}$. Suppose that we have such an $A$, and let $\beta$ be any of the new ordinals in $A$, that are not in $B$. (There must be such a new ordinal, since $B$ thinks every set has an ordinal rank, and for the ranks up to $\omega_2$, the original structure $B$ already had all the sets of that rank.) So $\beta$ is larger than every ordinal in $B$. But since $B$ thinks every ordinal has size $\omega_1$, it follows that $A$ thinks it has a bijection of $\beta$ with $\omega_1$. But it is correct about $\omega_1$, and so we can form an actual bijection of the objects that $A$ thinks are below $\beta$ with the actual $\omega_1$. This is impossible, since the former set includes all of $\omega_2$. Contradiction. In the comments, you inquire whether one can find a counterexample $B$ of size $\omega_1$, and the answer is yes. Let $B=L_{\omega_1+1}$, which is a transitive set that contains $\omega_1$ as an element (I mean to use true $\omega_1$ here, which might be larger than $\omega_1^L$), and indeed $\omega_1$ is the largest ordinal of $B$. Suppose that $\langle B,\in\rangle\prec \langle A,\hat\in\rangle$ is a nontrivial extension, where $A$ is an $\omega_1$-model and $B\cap H_{\omega_2}=B$ is in the well-founded part of $A$. It follows that $A$ is well-founded up to and including $\omega_1$, and by elementarity, since $\omega_1$ is the largest ordinal of $B$, it must also be the largest ordinal of $A$. So the ordinals of $A$ are precisely $\omega_1+1$. Since also $A\models V=L$, it follows by elementarity that $A=L_{\omega_1+1}$, which is to say, that $A=B$, and so this isn't a nontrivial extension after all. Contradiction. Perhaps this example shows that you haven't quite asked the question you intended to ask.<|endoftext|> TITLE: If Gaussian measures on a Hilbert space converge weakly to 0, how do their covariance operators converge? QUESTION [6 upvotes]: Suppose we have a sequence of Gaussian measures $N(0, S(n))$ supported on a Hilbert space $H$ and we know that the sequence converges weakly to the delta measure at $0$, what are the necessary and sufficient conditions on convergence of the sequence of operators(trace class) $S(n)$? They should converge to $0$. Should the convergence be uniform, strong, weak? REPLY [5 votes]: Denote the Gaussian random vectors by $X(n)$. Clearly, $\mathrm{tr} \, S(n) = \mathsf{E} \, \Vert X(n) \Vert^2$, so $\mathrm{tr} \, S(n) \to 0$ is certainly sufficient for weak convergence to $0$. And in fact it's also necessary. Indeed, $\Vert X(n) \Vert^2$ is a quadratic functional of a Gaussian, and for those convergence in probability is known to be equivalent to convergence in $L^2$ (this is also true for any polynomials of bounded degree, cf. Corollary 6.11 of Janson "Gaussian Hilbert spaces").<|endoftext|> TITLE: The boundary of a domain whose interior is diffeomorphic to the ball QUESTION [10 upvotes]: We know that there are compact manifolds with diffeomorphic interiors but their boundaries are not homeomorphic (see the question Manifolds with homeomorphic interiors). My question is about a very special case: Assume $D$ is a bounded open set in $R^n$ with smooth boundary. If $D$ is diffeomorphic to the unit ball, is $\bar D$, the closure of $D$, diffeomorphic to the closed unit ball? REPLY [10 votes]: The case $n=4$ is open as far as I know. The case $n=3$ follows since $\mathbb R^3$ is irreducible, so it contains no fake 3-disk, i.e. $\bar D$ must be the standard disk. The case $n=5$ is equivalent to the smooth $4$-dimensional Poincare conjecture (which is still open). Here is why: Any homotopy $4$-sphere embeds smoothly into $\mathbb R^5$ (Sketch: homology $4$-sphere bounds a contractible smooth manifold $C$ [Kervaire, "Smooth homology spheres and their fundamental groups", Theorem 3]. In our case $\partial C$ is simply-connected, so attaching a collar on the boundary one gets a contractible $5$-manifold that is simply-connected at infinity, and hence it is diffeomorphic to $\mathbb R^5$ by a result of Stallings). Any embedding of the standard $4$-sphere into $\mathbb R^5$ bounds a standard disk, see [Smale, "Differentiable and Combinatorial Structures on Manifolds", Corollary 1.3]. What Smale actually states is that any embedded $S^{n-1}$ in $\mathbb R^n$ bounds a standard disk unless $n=4$ or $7$. This was before he proved the h-cobodorsm theorem hence he excludes $7$. Finally, as mentioned in comments if $n>5$, then $\bar D$ is diffeomorphic to the standard disk by the h-cobordism theorem (sketch: since by assumption $D$ is simply-connected at infinity, $\partial D$ is a homotopy sphere and $\bar D$ is a contractible smooth manifold, so removing a small ball in its interior results in h-cobordism between then standard sphere and the embedded one. Proving that this is an h-cobordism involves standard excision considerations in homology).<|endoftext|> TITLE: Fibrations with isomorphic fibers, but not Zariski locally trivial QUESTION [6 upvotes]: (I posted this same question on MSE. Sorry if it is too elementary.) I am looking for examples of fibrations $f:X\to Y$ where the fibers are all isomorphic, but $f$ is not Zariski locally trivial. In particular, I am interested in understanding how much such examples are "rare". (I believe they are not that rare) First of all, by fibration I mean a proper flat surjective morphism of (complex) varieties. But I am not sure this is the correct definition of fibration in Algebraic Geometry; in that case, any correction is much appreciated. By $f:X\to Y$ being Zariski locally trivial, I mean there is a variety $F$ such that every point in the base $Y$ has a Zariski open neighborhood $U$ such that $f^{-1}(U)\to U$ is isomorphic to the projection $F\times U\to U$. Here $F$ is called the fiber of $f$ (in particular, Zariski locally trivial fibrations do have isomorphic fibers). One example I came up with is that of an étale cover of curves: the fibers are discrete of the same size, hence isomorphic, but it is not Zariski locally trivial in general. Remark. Sometimes a fibration is required to have connected fibers; if this was the correct definition of a fibration, my example would not be an example. Probably there are many important examples that I am missing here. I would very much appreciate if you could help me to fill in this picture! Thank you. REPLY [7 votes]: They are not rare. A general construction goes as follows. Start with a finite (étale) group (scheme) $G$ which acts on varieties $F$ and $\tilde Y$, where the second action has no fixed points. Let $Y =\tilde Y/G$. Then $(F\times \tilde Y)/G\to Y$ has all its fibres isomorphic to $F$. It is locally trivial in the étale topology, but not usually in the Zariski topology. This includes the above examples, and many others.<|endoftext|> TITLE: Moduli of flag varieties QUESTION [5 upvotes]: I work over an algebraically closed field $k$ of characteristic zero. Recall that a flag variety is a projective variety which is a homogeneous space for some semisimple algebraic group. Every flag variety is of the form $G/P$, where $G$ is a semisimple algebraic group and $P$ is a parabolic subgroup. It seems to me that this data is "discrete", so I expect flag varieties to have discrete moduli. Moreover flag varieties are Fano varieties, and it is known that there are only finitely many deformation types of Fano varieties of fixed dimension. This leads to my question. For each $n \in \mathbb{N}$, are there are only finitely flag varieties of dimension $n$, up to isomorphism? A proof/disproof or a reference for this would be much appreciated. By "up to isomorphism" I mean up to isomorphism as an algebraic variety. REPLY [9 votes]: Yes, there are only finitely many. One only needs to observe that for a given group, the variety $G/P$ has dimension at least the rank of the group $G$ (the dimension of a maximal torus). You can see this by inspecting cases by hand, but there's also a conceptual reason: the maximal torus of the adjoint group $G$ acts on $G/P$ faithfully with isolated fixed points, so it acts faithfully on the tangent space of some fixed point. Since a faithful module over a torus must have dimension $\geq$ that of the torus, this establishes the desired result. Thus, for a given dimension $n$, there can only be flag manifolds for groups with rank $\leq n$, there are finitely many of these, and only finitely many flag manifolds for each one. Q.E.D.<|endoftext|> TITLE: Reference for invariance of essential spectrum under relatively compact perturbations QUESTION [12 upvotes]: I'm looking for a proof of the following statement: Let $X$ be a Banach space and $T$ be a closed map on $X$. For any relatively compact map $A$ the essential spectrum of $T$ and $T+A$ are the same. This is proven in Kato's Perturbation Theory in IV - Stability Theorems- Section 5. The proof goes via the theory of semi-Fredholm operators, however I would like to know of any other references/proof techniques/books I can look at to see this proof. (I don't like Kato.) The corresponding result in Hilbert spaces (Weyls theorem) I've seen several proofs for, so I am looking for the result for Banach spaces. REPLY [3 votes]: Claim: If $T$ is a closed Fredholm operator acting on a complex Banach space $X$ and $A$ is a relatively $T$-compact operator, then the fact that $T+A$ is Fredholm can be derived from the stability under compact perturbations of the bounded Fredholm operators: Indeed, denoting $X_T$ the domain $D(T)$ with the graph norm $\|x\|_T :=\|x\| +\|Tx\|$, and $j_T$ the inclusion on $X_T$ into $X$, then $Tj_T$ is a bounded Fredholm operator and $Aj_T$ is a compact operator (note that $D(T)\subset D(A)$). Thus $(T+A)j_T$ is bounded Fredholm, hence $T+A$ Fredholm. The essential spectrum of $T$ is $\sigma_e(T):=\{\lambda : T-\lambda I \textrm{ Fredhom }\}$. The fact that $\sigma_e(T+A)=\sigma_e(T)$ is obtained by applying the Claim to $T-\lambda I$. In S. Goldberg, Unbounded linear operators, McGraw-Hill 1966, you can find additional details.<|endoftext|> TITLE: Bound on Minimal Length of Vectors in Lattice and its Dual Lattice QUESTION [6 upvotes]: Let $\Lambda$ be a lattice in $\mathbb{R}^n$ and $\Lambda^\ast$ its dual lattice. Let $d=\min_{v\in\Lambda} (v,v)$ and $d^\ast =\min_{v\in\Lambda^\ast} (v,v)$ be the minimal squared lengths of vectors in the two lattices. I'm interested in lattices in which both $d$ and $d^\ast$ are large. For $n=24$, I know of the self-dual Leech lattice which has $d=d^\ast=4$. I'd like to find lattices with $d,d^\ast\ge 8$. Is this at all possible? Self-duality is not so important for me, so if I can lower $n$ at the expense of self-duality that's OK. REPLY [3 votes]: The product $d\times d^*$ cannot be "so" large, as a consequence of the so-called Transference theorems Particularly, Thm. 2.1. of the paper shows that $d\times d^* \leq n^2$ (hence in your example, $n$ should be at least $8$). So the best you can do in terms of the product is $\Omega(n^2)$. The aforemetioned Conway-Thompson Theorem (see [1], p. 46) shows that indeed it is possible to achieve $\Omega(n^2)$. In the case of $d, d^* \geq 8$, we have the lower bound $n \geq 8$ and the achievable upper bound $72$ (due to the unimodular table). Nevertheless, there is still a huge gap between both bounds. I am not aware of any result that shows that smaller $n$ are possible. [1]. J. Milnor and D. Husemoller, ``Symmetric Bilinear Forms,'' Springer-Verlag, New York, 1973<|endoftext|> TITLE: Topological fundamental group of spec(R) QUESTION [5 upvotes]: Let $R$ be a commutative ring with identity. Assume that $X = Spec(R)$ with the Zariski topology. When is this space path connected? And also we want to know the topological fundamental group of the space $X$. How can we think about these questions? REPLY [5 votes]: The point is that you may be asking the 'wrong' question. The use of paths is inappropriate in this algebraic geometric context, and hence also ideas such as 'path connected' and 'fundmental group(oid)'although there are analogues. The classical fundamental group classifies covering spaces of $X$, and that is a useful property to generalise. Look at Grothendieck's SGA1 (if you can read French) for the original material on this, but there are lots of more recent sources 'out there' (including some well written surveys, some done by various Masters students, that are a good read and get you to the point fairly quickly without a lot of generality! One such that I have used is 'M. A. D. Robalo, 2009, Galois Theory towards Dessins d’Enfants, Master’s thesis, Instituto Superior Technico, Lisboa'. Another very good source is by Dubuc and de la Vega, (and which can be found on the ArXiv as math.CT/0009145.) There are analogues of the fundamental groupoid and of covering spaces and then you can ask if the fundamental groupoid is connected (corresponding to path connectedness in the classical topological case) That fundemantal groupoid is usually thought of as a profinite groupoid, and that may encode the topological information that you are hoping for in the second part of the question. (I should add that it is best to consider general schemes with the étale topology, but to start with that may not be necessary.) I hope this helps.<|endoftext|> TITLE: Is there an oscillating analog of the Gaussian distribution? QUESTION [7 upvotes]: It frequently happens that, in some famillies of polynomials with positive coefficients, the coefficients of large polynomials look like a bell curve and tend to the distribution function of the Gaussian law. One obvious example is given by the familly $(1+x)^n$. I know some examples of famillies of polynomials where the coefficients are not positive, but where plotting the coefficients give a nice oscillating curve, as the one displayed below. My question is the following: Is there any known such oscillating function, with some kind of universal property ? In other words, what is the function one can see in this picture ? As far as I can tell, this does not seem to be given by a product of $\exp(-x^2)$ by trigonometric functions, because the local maxima do not fit very well on a parabola. EDIT: Here is a graph of the log of the absolute value of the coefficients, compared with a parabola. REPLY [5 votes]: Another ... the Hermite functions. Hermite polynomials $H_n(x)$ times the density $\exp(-x^2)$. Here is $H_{20}(x)\exp(-x^2)$<|endoftext|> TITLE: Restriction of "$\pi_{1}$" to topological groups QUESTION [12 upvotes]: Let $G$ and $H$ be two topological groups. Assume that $\phi:\pi_{1}(G) \to \pi_{1}(H)$ is a group homomorphism. Is there a continuous function $f:G\to H$ such that $f_{*}=\phi$? REPLY [22 votes]: No. Take $G=SO(5)$ and $H=SO(3)$, both of which have fundamental group $\mathbb{Z}/2$. I claim that there is no continuous map $f: G\to H$ which induces the identity homomorphism. If there were, then $f$ would induce a nontrivial homomorphism $f_\ast: H_1(G;\mathbb{Z}/2)\to H_1(H;\mathbb{Z}/2)$, and a graded ring homomorphism $f^\ast: H^\ast(H;\mathbb{Z}/2)\to H^\ast(G;\mathbb{Z}/2)$ which is nontrivial on $H^1$. To see that no such ring homomorphism exists, recall (see Hatcher's Algebraic Topology, section 3.D) that $$H^\ast(H;\mathbb{Z}/2)\cong \mathbb{Z}/2[\alpha_1]/(\alpha_1^4)$$ and $$H^\ast(G;\mathbb{Z}/2)\cong \mathbb{Z}/2[\beta_1]/(\beta_1^8)\otimes\mathbb{Z}/2[\beta_3]/(\beta_3^2),$$ where each $\alpha_i,\beta_i$ is in degree $i$.<|endoftext|> TITLE: Kaplansky's idempotent conjecture for Thompson's group F QUESTION [8 upvotes]: Let $K$ be a field and $G$ be a torsion-free group. Kaplansky's idempotent conjecture states that the group ring $K[G]$ does not contain any non-trivial idempotent, i.e. if $x^2=x$ then $x=0$ or $x=1$. Is Kaplansky's idempotent conjecture known for Thompson's group $F$? REPLY [6 votes]: They satisfy Baum-Connes conjecture, so by surjevtivity of the assembly map, the Thompson's group $F$ satisfies even the stronger conjecture of Kaplansky and Kadison: the reduced group $C^*$-algebra has no idempotents or projections.<|endoftext|> TITLE: Green's function for *GJMS* operator QUESTION [5 upvotes]: Consider a Riemannian manifold $(M^n, g)$ of dimension $n$ with a metric $g$. We assume $M$ to be closed (compact without boundary). Let's not assume any hypothesis on the Yamabe invariant of the manifold. A GJMS operator on $M^n$ is a differential operator of order $n$ with principal symbol $(-\Delta_g)^{(n/2)}$. If the Yamabe invariant on $M^n$ is positive then the kernel of this operator is just the constants (for this statement cfr Gursky - "Conformal invariants and nonlinear elliptic equations", as for the Green's function in this case you can check Ndiaye - "Constant $Q$-curvature metrics in arbitrary dimension"). In general the constants on $M^n$ are just a subset of the kernel. Does anyone know if there exist a Green function for such operator in this case? I cannot find any references for this case. Any help? REPLY [2 votes]: I believe that the answer is Yes by way of the method of the parametrix. This equation is elliptic (the principal symbol is injective, being a power of the principal symbol of the Laplacian) so the parametrix is generated by a pseudo-differential operator.<|endoftext|> TITLE: Closed form for 3j-symbol ratios QUESTION [5 upvotes]: I am working on the spherical harmonic decomposition of cosmic microwave background maps, therefore I often deal with functions that are proportional to Wigner 3J symbols/Clebsch–Gordan coefficients. I would be very grateful if you could share with me a closed form of the ratio between $$ \begin{pmatrix} l_1 &l_2 &l_3\\ 0&2&-2 \end{pmatrix} $$ and $$ \begin{pmatrix} l_1 &l_2 &l_3\\ 0&0&0 \end{pmatrix} \;, $$ for $l_1+l_2+l_3$ even, if it exists. Ideally, I would like an expansion of the first 3j-symbol in terms of the second one, i.e. first_3j = second_3j * ( 1 + ... ) Thank you for your consideration. Best wishes, Guido REPLY [4 votes]: The following recursion relation $$ C(m_2+1,m_3-1)\begin{pmatrix}l_1 &l_2 &l_3\\m_1&m_2+1&m_3-1\end{pmatrix}+ D(m_2,m_3)\begin{pmatrix}l_1 &l_2 &l_3\\m_1&m_2&m_3\end{pmatrix}+$$ $$ C(m_2,m_3)\begin{pmatrix}l_1 &l_2 &l_3\\m_1&m_2-1&m_3+1\end{pmatrix}=0, \tag{1}$$ where $$C(m_2,m_3)=\sqrt{(l_2-m_2+1)(l_2+m_2)(l_3+m_3+1)(l_3-m_3)},\tag{2}$$ and $$D(m_2,m_3)=l_2(l_2+1)+l_3(l_3+1)-l_1(l_1+1)+2m_2m_3,\tag{3}$$ can be found (formula 9) in http://scitation.aip.org/content/aip/journal/jmp/16/10/10.1063/1.522426?ver=pdfcov (Exact recursive evaluation of 3j‐ and 6j‐coefficients for quantum‐mechanical coupling of angular momenta, by Klaus Schulten and Roy G. Gordon). For $m_1=m_2=m_3=0$, we get from (1): $$ C(1)\begin{pmatrix}l_1 &l_2 &l_3\\0&1&-1\end{pmatrix}+ D(0)\begin{pmatrix}l_1 &l_2 &l_3\\0&0&0\end{pmatrix}+ C(0)\begin{pmatrix}l_1 &l_2 &l_3\\0&-1&1\end{pmatrix}=0. \tag{4}$$ Using the symmetry relation $$\begin{pmatrix}l_1 &l_2 &l_3\\0&-1&1\end{pmatrix}=(-1)^{l_1+l_2+l_3} \begin{pmatrix}l_1 &l_2 &l_3\\0&1&-1\end{pmatrix},$$ (4) can be rewritten as follows $$[C(1)+(-1)^{l_1+l_2+l_3}C(0)]\begin{pmatrix}l_1 &l_2 &l_3\\0&1&-1\end{pmatrix}+D(0)\begin{pmatrix}l_1 &l_2 &l_3\\0&0&0\end{pmatrix}=0.\tag{5}$$ Here, according to (2) and (3), $$C(0)=\sqrt{l_2l_3(l_2+1)(l_3+1)},\;\; C(1)=C(0),$$ and $$D(0)=l_2(l_2+1)+l_3(l_3+1)-l_1(l_1+1).$$ Now let us assume $m_1=0,m_2=1,m_3=-1$ in (1). We get $$ \tilde C(2)\begin{pmatrix}l_1 &l_2 &l_3\\0&2&-2\end{pmatrix}+ \tilde D(1)\begin{pmatrix}l_1 &l_2 &l_3\\0&1&-1\end{pmatrix}+ \tilde C(1)\begin{pmatrix}l_1 &l_2 &l_3\\0&0&0\end{pmatrix}=0. $$ so that $$\begin{pmatrix}l_1 &l_2 &l_3\\0&1&-1\end{pmatrix}=-\frac{\tilde C(2)} {\tilde D(1)}\begin{pmatrix}l_1 &l_2 &l_3\\0&2&-2\end{pmatrix} -\frac{\tilde C(1)} {\tilde D(1)}\begin{pmatrix}l_1 &l_2 &l_3\\0&0&0\end{pmatrix}.$$ Substituting this into (5), we get after a little algebra $$\frac{\begin{pmatrix}l_1 &l_2 &l_3\\0&2&-2\end{pmatrix}} {\begin{pmatrix}l_1 &l_2 &l_3\\0&0&0\end{pmatrix}}=\frac{1}{\tilde C(2)}\left[\frac{D(0)\tilde D(1)}{C(1)+(-1)^{l_1+l_2+l3}C(0)}-\tilde C(1)\right ].$$ It remains to take into account that $l_1+l_2+l_3$ is even and calculate from (2) and (3) (for $m_3=-1$) $$\tilde C(1)=C(1),\;\;\tilde C(2)=\sqrt{(l_2-1)(l_2+2)(l_3-1)(l_3+2)},\;\;\; \tilde D(1)=D(0)-2.$$ Finally, $$\frac{\begin{pmatrix}l_1 &l_2 &l_3\\0&2&-2\end{pmatrix}} {\begin{pmatrix}l_1 &l_2 &l_3\\0&0&0\end{pmatrix}}=\frac{1}{\tilde C(2)}\left[\frac{D(0)[D(0)-2]}{2C(0)}-C(0)\right ].$$<|endoftext|> TITLE: Reference request: embedding the hyperbolic triangulation in $\mathbb{R}^3$ QUESTION [6 upvotes]: Let $T_d$ be the infinite valence $d$ triangulation of the hyperbolic plane, where each triangle is equilateral and $d \ge 7$. Question: Is there an isometric embedding from $T_d \to \mathbb{R}^3$? Here an isometric embedding means that each triangle is flat, with Gaussian curvature only at the vertices. I've been told that Thurston and others have tackled this problem and failed, so apparently it is fairly hard. Actual question: Are there references discussing existing attempts? The problem is likely very different for even vs. odd $d$, since for even $d$ we can almost fold along geodesics down to a single triangle. That is, given any $\epsilon > 0$, there is an almost isometric embedding of $T_{2d}$ into an $\epsilon$ neighborhood of a single triangle, where the isometry is off by a factor of $1 \pm \epsilon$. For odd $d$, at least, my intuition is that it is completely unreasonable that such an embedding exists: the number of triangles in a ball grows exponentially with the radius due to the exponential growth of hyperbolic space, which results in an exponential number of arbitrarily closely layered sheets. Here is the radius 5 portion of $T_7$ in the Poincare disk model, together with an embedding into $\mathbb{R}^3$: REPLY [2 votes]: Ian Agol's example in the comments generalizes -- you can probably tile any triply periodic surface with equilateral triangles (after some fiddling you might even make them flat triangles) and the tiling will admit a covering map from a tiling of the hyperbolic plane because of the negative Gaussian curvature. I am not sure if you can get all $T_d$ this way (the tilings you get don't necessarily have constant valence). I learned about this connection between triply periodic surfaces and hyperbolic tilings from this paper by Ramsden, Robins and Hyde. They also have a nice website with more information. They consider more general tilings, not just of equilateral triangles. As I said, to make these into the flat equilateral triangular tilings you are looking for you may need to do some fiddling around. Here's a picture I just found of $T_8$ on Schwarz's D-surface by Doug Dunham.<|endoftext|> TITLE: Historical quotation search: Equations/formulae in (Latin?) prose, before modern symbolic notation QUESTION [11 upvotes]: I have been trying, without success, to find a vaguely-remembered quotation: the quadratic equation (or perhaps the quadratic formula), given in (Latin?) prose, along lines like “Consider that quantity, which, when multiplied by itself and under multiplication by the first constant, …” The purpose is just to contrast with the clarity and compactness of modern algebraic notation — so it doesn’t need to be the quadratic equation or formula, and it doesn’t need to be in Latin; any similar historical quotation to illustrate the point would do. (Two relevant meta threads: Are history of maths questions acceptable?, and Area 51 proposal: history of science and maths. Also, this question as a sample question on that area 51 proposal.) REPLY [5 votes]: A better known figure than the 16th century Stifel is the 17th century Pierre de Fermat who still used notation that was not fully symbolic. Thus, in his work anticipating the calculus, he used the terms aequalitat and adaequalitat rather than the familiar equality symbol. An example may be found in section 8.8 of this text.<|endoftext|> TITLE: More than $n$ approximately orthonormal vectors in $R^n$ QUESTION [13 upvotes]: This question was asked at math.stackexchange, where it got several upvotes but no answers. It is impossible to find $n+1$ mutually orthonormal vectors in $R^n$. However, it is well established that the central angle between legs of a regular simplex with $n$-dimensional volume goes as $\theta = \mathrm{arccos}(-1/n)$. This approaches $90$ degrees as $n \rightarrow \infty$, so since there are $n+1$ vertices of a simplex with $n$-dimensional volume, we can conclude Given $\epsilon > 0$, there exists a $n$ such that we can find $n+1~$ approximately mutually orthogonal vectors in $\mathbb{R}^n$, up to tolerance $\epsilon$. (Unit vectors $u$ and $v$ are said to be approximately orthogonal to tolerance $\epsilon$ if their inner product satisfies $\langle u,v \rangle < \epsilon$) My question is a natural generalization of this - if we can squeeze $n+1$ approximately mutually orthonormal vectors into $\mathbb{R}^n$ for $n$ sufficiently large, how many more vectors can we squeeze in? $n+2$? $n+m$ for any $m$? $2n$? $n^2$? $e^n$? Actually the $n+m$ case is easy to construct from the $n+1$ case. Given $\epsilon$, one finds the $k$ such that you can have $k+1$ $\epsilon$-approximate mutually orthogonal unit vectors in $\mathbb{R}^k$. Call these vectors $v_1, v_2, ..., v_k$. Then you could squeeze $mk+m$ vectors in $\mathbb{R}^{mk}$, by using the vectors $$\begin{bmatrix} v_1 \\ 0 \\ \vdots \\ 0 \end{bmatrix}, \begin{bmatrix} v_2 \\ 0 \\ \vdots \\ 0 \end{bmatrix}, \begin{bmatrix} v_{k+1} \\ 0 \\ \vdots \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ v_1 \\ \vdots \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ v_2 \\ \vdots \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ v_{k+1} \\ \vdots \\ 0 \end{bmatrix}, \dots \begin{bmatrix} 0 \\ 0 \\ \vdots \\ v_1 \\ \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ \vdots \\ v_2 \\ \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ \vdots \\ v_{k+1} \\ \end{bmatrix}. $$ So, setting $n = mk$, we have found an $n$ such that we can fit $n + m$ $\epsilon$-orthogonal unit vectors in $\mathbb{R}^n$. I haven't been able to construct anything stronger than $n+m$, but I also haven't been able to show that this is the upper bound. REPLY [4 votes]: Terry Tao has a nice blog post on a 'cheap version' of the Kabatjanskii-Levenstein bound mentioned in Lucia's answer, using the so-called 'tensor product trick'.<|endoftext|> TITLE: Independence results about independence results QUESTION [12 upvotes]: Define $Ind(ZFC,\sigma)$ to be the assertion "The sentence $\sigma$ is independent from ZFC". I am looking for theorems in the form $Ind(ZFC, Ind (ZFC,\sigma))$. Are there such theorems in set theory? Of course the first order inexpressibility is a problem in these theorems and many of them are theorem schemas. REPLY [10 votes]: Almost all instances of independence are instances of your further phenomenon, for the following general reason. Namely, if there is a model of ZFC which believes that "$\sigma$ is independent of ZFC" is true, then there is another model of ZFC which believes that it is false, since any assertion of independence fails in a model of ZFC plus $\neg$Con(ZFC), as there are no independent assertions for an inconsistent theory. Another way to say this is that if it is consistent with ZFC that your sentence is independent, then this assertion is itself never provable in ZFC, because to prove an instance of independence is to prove consistency, which is impossible by the incompleteness theorem. One might like to say that all instances of independence are instances of your further phenomenon, but unfortunately, we cannot quite prove that without further hypotheses. Indeed, it is consistent with ZFC that there are independent statements, but no instances of Ind(ZFC,Ind(ZFC,$\sigma$)). To see this, consider any model $M$ of ZFC+Con(ZFC)+$\neg$Con(ZFC+Con(ZFC)). Such a model exists by the incompleteness theorem, and in this model we'll have Con(ZFC) and hence all kinds of independent statements, but inside $M$ we will have no models of ZFC+Con(ZFC), and thus $M$ thinks that there are no models of Ind(ZFC,$\sigma$) for any $\sigma$, and hence $M$ thinks Ind(ZFC,Ind(ZFC,$\sigma$)) always fails.<|endoftext|> TITLE: Representations of $\mathrm{SL}(2)$ in characteristic 2 QUESTION [10 upvotes]: $\DeclareMathOperator\SL{SL}$In characteristic zero one can use the Clebsch-Gordan rule to decompose tensor products of $\SL(2)$-modules. In characteristic $p$, things are more complicated. I am interested in the special case $S^dV\otimes V$ (where $V$ is the 2-dimensional standard representation) for fields $k$ of characteristic $p>0$. In fact, I mainly want to know about $d=3$. If one computes the Clebsch-Gordan isomorphism explicitly, one can see that the denominator is $(d+1)$. So there will be a problem for $p|(d+1)$. What is known in this case? I'd be happy just to know the case $d=3$, especially an explicit composition series and whether one still has some nice direct sum decompositions into representations of smaller dimension (I realize that these will no longer be simple modules as in characteristic $0$). I'd also like to know references about how the invariant theory of $\SL(2)$ works in positive characteristic. REPLY [4 votes]: There is unfortunately no "formula" for tensor products in prime characteristic. Instead you can derive a list of composition factors $L(\lambda)$ (with multiplicity) by recursion. When $p=2$ there are only two simple modules with restricted highest weights (abbreviated by non-negative integers), namely the trivial module $L(0)$ of dimension 1 and the natural module $L(1)$ of dimension 2. After this you need to rely on Steinberg's twisted tensor product theorem relative to a $p$-adic expansion of the highest weight. For instance, $L(2) \cong L(1)^{(1)}$, the first Frobenius twist of the natural module (still having dimension 2). In your specific example, the recursion is easy to carry out: peel off the composition factor of highest weight and see what weights remain. Here yuu are looking at the tensor product of the natural module $L(1)^{(1)}$ with the "induced" module $H^0(3)$ of dimension 4 as in Jnntzen's book (polynomials in two variables of homogeneous degree 3) which is actually simple when $p=2$, isomorphic to $L(1) \otimes L(1)^{(1)}$. (These factors are the respective Steinberg modules for the first and second Frobenius kernels.) From the recursion one arrives at a list of composition factors (having total dimension 8): $L(1)^{(2)}, \: L(1)^{(1)} \text{ twice }, L(0)\: \text{ twice}.$ Here at least there is a recursive method, but getting the precise module structure can be quite tricky. In this special case, you might take advantage of the fact that you are tensoring with a projective module for a certain Frobenius kernel. But in general it's complicated even in rank 1. ADDED: For some recent work on decomposition of tensor products into indecomposables, see for example a paper by Doty and Henke, Decomposition of tensor products of modular irreducibles for SL$_2$, Q. J. Math. 56 (2005), no. 2, 189-207 (arXiv:math/0205186). This involves tilting modules and their Frobenius twists. FURTHER COMMENTS: I intended to mention something about the extra question raised by Lloyd on invariant theory in prime characteristic. There is a classical result describing the ring of invariants for the general linear groups over a finite field, or for $\mathrm{SL}_n(\mathbb{F}_q)$, which goes back to L.E. Dickson in 1911. This has been reworked a number of times, for instance in an article by R. Steinberg, On Dickson’s theorem on invariants, J. Fac. Sci. Univ. Tokyo Sect. IA Math. 34 (1987), no. 3, 699–707. (It's especially fitting to recall Steinberg's diverse contributions, since he died very recently on his 92nd birthday.) In the special case dicussed here, Jantzen (unpublished) showed how to recover Dickson's theorem from the well-understood representation theory of the groups. This led me to investigate the general case in a similar spirit, though it remains to be seen whether we will know enough about representation theory to recover the full theorem this way. Anyway, my short paper contains an assortment of references to the literature (including work of Donkin on tilting modules and a paper by Wilkerson motivated by algebraic topology): Another look at Dickson's invariants for finite linear groups, Comm. Algebra 22 (1994), no. 12, 4773-4779.<|endoftext|> TITLE: What kind of random matrices have rapidly decaying singular values? QUESTION [23 upvotes]: I've been told that in machine learning it's common to compute the singular value decomposition of matrices in order to throw out all information in the matrix except that corresponding to, say, the $k$ largest singular values for some reasonable $k$; moreover, in practice it happens that the singular values decay rapidly and so all but the $k$ largest singular values contain very little information anyway. What kind of distribution on random matrices has this kind of distribution of singular values? For example, classical results (Theorems 5.31 and 5.32) imply that for an $N \times n$ matrix ($N \ge n$) whose entries are, say, i.i.d. Gaussian with zero mean and unit variance, the largest singular value is around $\sqrt{N} + \sqrt{n}$ with high probability and the smallest singular value is around $\sqrt{N} - \sqrt{n}$ with high probability; in particular, for $N \gg n$ all of the singular values are pretty similar. This is a very different distribution from the above, so it seems reasonable to infer that the matrices people encounter in machine learning don't behave like they have i.i.d. entries. Probably the independence assumption is the least reasonable one - for example, in an incidence matrix describing which users of Netflix have watched which movies, one expects lots of correlations between similar movies as well as between similar users; that is, one expects the incidence matrix to be well approximated by a low-rank matrix (which is precisely what we're doing above). I guess one also expects power-law (log-Gaussian?) distributions on, say, both the statistics of how many movies users have watched and how many users have watched movies. Have random models with this kind of behavior been studied, and do their singular values behave as expected? REPLY [11 votes]: I hope I understood the OP correctly, in case not please let me know. And I will discuss the case of eigenvalue instead of singular value without much loss of generality. In case you are only interested in the distribution of eigenvalues of an i.i.d. sample covariance matrix(which is also a random matrix but underlying distribution is always of dimension one), the correct reference should be [Anderson] instead of [Johnstone]. [Johnstone] dealt with a much more general case where the underlying distribution is of multi-dimension and especially with double-Wishart situation and its associated concentration bound. On the other hand [Anderson] did not consider $n(p),m(p)$ but only $m,n$ as separated parameters that tend to $\infty$, which is not suitable in "machine learning era". $\blacksquare$1.Theorem 5.31 in [Vershynin] and Theorem 5.32 as corollary. Unlike what the OP claimed, Bai's result also requires finite fourth moment [link1] which is not quite easy to verify in some applications, say the censored data. Assuming for the moment that the fourth moment of the underlying distribution is finite and consider the simple Gaussian regression for example, we usually study $N\approx n$ where $N$ is the sample size and $n$ is the number of variables in the model. You are right on the point if $N\gg n$ then all eigenvalues concentrate on a certain neighborhood of $\sqrt{N}$. However, a challenge in the field is that when $n$ is large and $N\approx n$. Intuitively, when the dimension of the regression model is high and the data is sparse. In this situation, some "machine learning" methods were proposed to select variables [Hastie&Tibshirani]. Among these methods, one is called sparse PCA which penalized sparse input and "forces the eigenvalues to deviate from $N$ with positive probability" if you want to view in Bai's law (Theorem 5.31). So it is not that "so it seems reasonable to infer that the (random) matrices people encounter in machine learning don't behave like they have i.i.d. entries." instead researchers will treat i.i.d. in companion with consideration of the dimension of the space that the working model lied in. Dependent data will look almost like i.i.d data if the dimension is high. [Pastur et.al] The consideration of dimension is not unusual even in classical statistic, some admissible models will become inadmissible if the dimension increase, for example Stein's effect [link2]. $\blacksquare$2. What kind of distribution on random matrices has this kind of distribution of singular values(eigenvalues)? 2.1 Behavior of eigenvalues of dependent Gaussian random matrices The independent assumption is not necessary as you already noted. However, even if we dropped the independence we can still obtain a concentration bound for the largest eigenvalue. Consider $N(\boldsymbol{0},\Sigma)$ instead of $N(\boldsymbol{0},I)$. Again we consider regression setting, the following result tells you that without independence between variables we can still impose a bound on the covariance matrix which is required to proceed PCA or other variable selection techniques using eigenvalue decomposition. Notice that $Wishart(I,\bullet)$ is not a serious constraint as commented by [Johnstone] p.3. Thm 1 in [Johnstone](modified) Assume $p$ is even and that $p, m = m(p)$ and $n= n(p)\rightarrow\infty$ together in such a way that $\lim_{p\rightarrow\infty}\frac{\min(n,p)}{(m+n)}>0,\lim_{p\rightarrow\infty}\frac{p}{m}<1$. For each $s_{0}\in\mathbb{R}$, there exists $C > 0$ such that for $s\geq s_0$, $$|Pr\{W_p\leq\mu_p+\sigma_p\cdot s\}-F_1(s)|0,\lim_{p\rightarrow\infty}\frac{p}{m}<1$$. If it decays rapidly, we have some existing methods to penalized it as mentioned in part 1. $\bullet$ The eigenvalues of product uniform distributions (on Grassmanians of dim $p$) will not decay rapidly under certain constraints. Therefore it is safe to say that in most applications, as long as $m(p),n(p)$(usually sample sizes) grows with $p$(usually model dimension) in a reasonable manner, the eigenvalues will neither concentrate severely NOR change their relative magnitude too dramatically. On the other hand, if a random matrix is actually generated from a model with varying (increasing) sample size $N$, then it will most likely have a set of rapidly decaying eigenvalues under mild constraints as discussed above. In case where the growth conditions are not satisfied, the word "sparse" come into play. $\blacksquare$3. Return to OP's last question. Therefore the focus of (statistics) research is not on which family of random models have rapidly decaying eigenvalues such that we can avoid using variable selection on these "morbid models"; the focus of (some) statistical methods are on how to wrap up these possible morbidity and make variable selection based on such eigenvalues still work. Later I have a nice discussion with some expert in the field, who pointed out that the eigenvalue problem can be framed as the distribution of roots of a polynomial with random coefficients(coefficients as random variables) and simultaneously a kernel if the covariance matrix is p.d. The distribution of eigenvalues of a random matrix is described in terms of the distribution of roots of a determinant equation, which turns out to be a random polynomial[Deift]. And by now I think your question that Have random models with this kind of behavior been studied, and do their singular values behave as expected? is addressed. Reference [Anderson]Anderson, Theodore Wilbur. "Asymptotic theory for principal component analysis." The Annals of Mathematical Statistics 34.1 (1963): 122-148. [link1]https://math.stackexchange.com/questions/1467581/what-does-it-mean-for-a-probability-distribution-to-have-a-finite-fourth-moment [Vershynin]http://www-personal.umich.edu/~Vershyninv/papers/non-asymptotic-rmt-plain.pdf [Hastie&Tibshirani]Trevor Hastie and Robert Tibshirani "Statistical Learning with Sparsity: The Lasso and Generalizations." [link2]https://stats.stackexchange.com/questions/263879/is-an-admissible-minimax-rule-unique [Johnstone]Johnstone, Iain M. "Multivariate analysis and Jacobi ensembles: Largest eigenvalue, Tracy–Widom limits and rates of convergence." Annals of statistics 36.6 (2008): 2638.(Slides: http://www.stat.harvard.edu/NESS10/Johnstone-slides.pdf) [Tracy&Widom]Tracy, Craig A., and Harold Widom. "On orthogonal and symplectic matrix ensembles." Communications in Mathematical Physics 177.3 (1996): 727-754. [Absil et.al]Absil, P-A., Alan Edelman, and Plamen Koev. "On the largest principal angle between random subspaces." Linear Algebra and its applications 414.1 (2006): 288-294. [link3]angle between subspaces [Jiang]Jiang, Tiefeng. "Approximation of Haar distributed matrices and limiting distributions of eigenvalues of Jacobi ensembles." Probability theory and related fields 144.1-2 (2009): 221-246. [Diaconis&Shahshahani]Diaconis, Persi, and Mehrdad Shahshahani. "On the eigenvalues of random matrices." Journal of Applied Probability (1994): 49-62.http://statweb.stanford.edu/~cgates/PERSI/papers/random_matrices.pdf [Pastur et.al]Pastur, Leonid Andreevich, Mariya Shcherbina, and Mariya Shcherbina. Eigenvalue distribution of large random matrices. Vol. 171. Providence, RI: American Mathematical Society, 2011. [Deift]Deift, Percy. Orthogonal polynomials and random matrices: a Riemann-Hilbert approach. Vol. 3. American Mathematical Soc., 2000. [Bandeira&van Handel]Bandeira, Afonso S., and Ramon van Handel. "Sharp nonasymptotic bounds on the norm of random matrices with independent entries." The Annals of Probability 44.4 (2016): 2479-2506.<|endoftext|> TITLE: Is it ever a good idea to use Keisler-Shelah to show elementary equivalence? QUESTION [14 upvotes]: The most useful way I know to show that two structures are elementarily equivalent is Ehrenfeucht-Fraisse games. These are quite nice and intuitive, and even when I can't use them to solve my problem E-F games usually give me a much better understanding of the structures I'm working with. In principle, though, one could also use the Keisler-Shelah ultrapower theorem - "Two structures are elementarily equivalent iff they have isomorphic ultrapowers by some ultrafilter" (see Ultrafilters arising from Keisler-Shelah ultrapower characterisation of elementary equivalence; one direction is trivial, Keisler proved the other direction under GCH, and Shelah brought the proof down to ZFC). However, I know of no setting in which this is actually a better way to approach the specific problem of showing elementary equivalence. So, my question is: If we want to show $\mathcal{A}\equiv\mathcal{B}$, is it ever efficient to go through the Shelah-Keisler ultrapower theorem? I'm asking about Keisler-Shelah specifically, as opposed to some other result about elementary equivalence, because any K-S based approach to establishing elementary equivalence would presumably involve some set-theoretic combinatorics, and I'm generally interested in set theory cropping up in "concrete"(ish) questions. My suspicion is that the answer is "no," and that the work involved in showing that two structures have isomorphic ultrapowers would always subsume the work involved in establishing elementary equivalence, but I have no real evidence for this. REPLY [3 votes]: You may also look at Lelek’s conjecture, where Shelah's theorem is used in an essential way (the questions at the end of the paper ask if one can give an easier (more direct) proof of the results without using Shelah’s theorem). Of course this answer is not along the lines you want, but possibly the other direction.<|endoftext|> TITLE: What are benefits of foundations with more than one sort of objects? QUESTION [5 upvotes]: Amongst different foundations of mathematics, $ZF$ and $NF$ are talking about "sets" but $MK$ and $GB$ are talking about two sorts of objects "sets" and "classes". What are benefits of studying the axiomatic systems with more than one sort of objects even if they are conservative relative to $ZF$? Is it just because theorems about proper classes have simpler forms in such systems in comparison with corresponding theorem schemas in $ZF$ or it is because we lose some useful information and power of the theorems on proper classes when we reduce them to $ZF$? If the last is true, what are the examples of such harmful reductions? Does the same phenomenon happen for sets as same as proper classes too? In the other words, do $GB$ or $MK$ in some sense tell us "more" about sets and proper classes than $ZF$? REPLY [6 votes]: I would like to address the part of the question about whether allowing classes beyond those that are definable over the model adds to our understanding of the properties of sets/classes. For instance, the famous Kunen inconsistency showing that there cannot be an elementary embedding $j:V\to M$ is nearly trivial to prove if we assume that $j$ is definable. Kunen's sophisticated argument is required to show that there cannot be such a class embedding in any model of ${\rm GBC}$. The existence of classes beyond those that are definable can also be useful in model theoretic ultrapower constructions. If $M$ is a (set) model of ${\rm GBC}$, we can construct ultrapowers of $M$ by using an ultrafilter on its proper classes. In particular, this method allows us to build end-extensions of models of set theory. By controlling what types of classes exist in $M$, we can obtain end extensions with different desired properties.<|endoftext|> TITLE: Asymptotic expansion of modified Bessel function $K_\alpha$ QUESTION [7 upvotes]: An integral representation for the Bessel function $K_\alpha$ for real $x>0$ is given by $$K_\alpha(x)=\frac{1}{2}\int_{-\infty}^{\infty}e^{\alpha h(t)}dt$$ where $h(\alpha)=t-(\frac{x}{\alpha})\cosh t$ Do you know any reference where its shown that using some generalization of Laplace method $$K_\alpha(x)\sim(\frac{\pi}{2\alpha})^{\frac{1}{2}}(\frac{2\alpha}{ex})^\alpha$$ as $\alpha\rightarrow\infty$ ? I have difficulties doing the calculation myself, first we cann see that the integrand has a maximum at $$t=\sinh^{-1}(\alpha/x)\sim\log(2\alpha/x)$$ which rises the idea using the substitution $t=\log(2\alpha/x)+c$ but I it does not help me. REPLY [6 votes]: abbreviate $t_0={\rm arsinh}(\alpha/x),\;\;x_0=x\sqrt{1+\alpha^2/x^2}>0$ expand the exponent around the saddle point, to second order: $$\alpha[t-(x/\alpha)\cosh t]=-x_0+\alpha t_0-\tfrac{1}{2}x_0(t-t_0)^2+{\rm order}(t-t_0)^3$$ carry out the Gaussian integration: $$\int_{-\infty}^{\infty}\tfrac{1}{2}\exp\left(-x_0+\alpha t_0-\tfrac{1}{2}x_0(t-t_0)^2\right)dt=\sqrt{\frac{\pi}{2x_0}}\exp(-x_0+\alpha t_0)$$ take the limit $\alpha\rightarrow\infty$ and you're done: $x_0\rightarrow\alpha$, $t_0\rightarrow\ln(2\alpha/x)$, so $$K_\alpha(x)\rightarrow\sqrt{\frac{\pi}{2\alpha}}e^{-\alpha}(2\alpha/x)^\alpha$$<|endoftext|> TITLE: Characteristic Classes in Geometric Representation Theory QUESTION [11 upvotes]: Geometric respectively topological methods are widely applied in representation theory. As far as I know mainly cohomological methods are used. I wonder if there are concrete applications of the theory of characteristic classes in geometric representation theory. I am mainly interested in the following situation. Let $G$ be a complex reductive group and $B$ a Borel subgroup. There are various ways to give the intersection cohomology groups of the Schubert varieties $X_w$ of $G/B$ some representation theoretic meaning. On the other hand Goresky and MacPherson introduced a theory of characteristic classes (the $L$-class) in the context of intersection cohomology of certain singular spaces. Have these $L$-classes been computed for Schubert varieties. Do they contain or measure any interesting representation theoretic information? REPLY [7 votes]: Let me attempt a partial answer, trying to estimate the information contained in these characteristic classes. Since the explanations below are a bit lengthy, I begin with the short version: I think that these characteristic classes could have representation-theoretic applications, but they likely do not provide more information than what is also available by other means. Setting - characteristic classes for singular varieties Let us first examine the setting, and figure out how we could understand the $L$-class. As explained in (S. Yokura: On Cappell-Shaneson's homology $L$-classes of singular algebraic varieties. Trans AMS. 347 (1995), 1005--1012), the $L$-class of Cappell-Shaneson is the unique transformation $L_\ast:\Omega(-)\to H_\bullet(-,\mathbb{Q})$ of covariant functors which agrees with Hirzebruch's $L$-class when $X$ is smooth. Here, $\Omega(X)$ is an algebraic group of cobordism classes of self-dual complexes on $X$. The Goresky-MacPherson $L$-class of the question is obtained by applying the Cappell-Shaneson $L$-class transformation to the class of the intersection complex of $X$. Now at this point, we can ask several slightly more precise questions: for $X=G/B$ or a Schubert variety in there, what is $\Omega(X)$, what is the map $L_\ast(X):\Omega(X)\to H_\bullet(X,\mathbb{Q})$ and what/how can we describe the class of the intersection complex in $\Omega(X)$? I do not exactly know which of these questions you wanted to have answered, and I actually would not know the answer to any of them. Instead of answering the specific question, let me look at a slightly different, more general questions. In fact, the Cappell-Shaneson $L$-class transformation is only a special case of a motivic Hirzebruch transformation, cf. J.-P. Brasselet, J. Schürmann, S. Yokura: Hirzebruch classes and motivic Chern classes for singular spaces. J. Topol. Anal. 2 (2010) 1--55. The motivic Hirzebruch class transformation is a map $$ T_{y\ast}:K_0(\operatorname{Var}/X)\to H_\bullet(X)\otimes\mathbb{Q}[y]. $$ This can be specialized to all sorts of characteristic classes, Chern-Schwartz-MacPherson classes, $L$-classes, Todd classes, etc. Now the new, more general question is: for $X=G/B$ or a Schubert variety in there, can we define a class in $K_0(\operatorname{Var}/X)$ - a universal motivic characteristic class - which specializes to the intersection complex? How can we describe/compute it? And what does it mean? Grothendieck group of varieties over $G/B$ Before that, it may be helpful to say a word about $K_0(\operatorname{Var}/X)$. This is the Grothendieck ring of quasiprojective varieties over $X$, for the application to the $L$-class we may better look at compact complex analytic spaces. The relations that need to be divided out come from the blow-up square, cf. F. Bittner. The universal Euler characteristic for varieties of characteristic zero. Compos. Math. 140 (2004), 1011--1032. Universal characteristic classes Now, for $X=G/B$, what class in $K_0(\operatorname{Var}/X)$ could best specialize to the intersection complexes of Schubert varieties? Claim 1: all my bets are on the classes of Bott-Samelson resolutions of Schubert varieties. Claim 2: in fact, my guess is that the abelian subgroup of $K_0(\operatorname{Var}/X)$ generated by Bott-Samelson varieties is isomorphic to the split $K_0$ of the category of pure weight 0 stratified mixed Tate motives on $X$ (resp. the category of Soergel modules). So, let's sum up. With a reasonable leap of faith, the information of the class of the Bott-Samelson resolution in $K_0(\operatorname{Var}/X)$ would seem to be closely related to the information that is contained in the Kazhdan-Lusztig basis in the Hecke algebra. (Ok, I'll try to make this more precise later, but let it stand as it is for now. Making this precise would probably mean doing some of these things equivariantly - talking about equivariant characteristic classes, Soergel bimodules categorifying the Hecke algebra and the relation between equivariant cohomology and the nil-Hecke algebra... too much for now) Now that we have a vague idea about the universal characteristic class, all that (any specialization of) the motivic Hirzebruch transformation does from this point on only forgets some of this information. This is why I said in the very beginning that it is very likely that there are representation-theoretic applications for these characteristic classes. However, the most complete characteristic class - in $K_0(\operatorname{Var}/X)$ - seems to be pretty close to what has been used all along in Kazhdan-Lusztig theory. Computations in the literature Slightly different homology characteristic classes (Chern-Schwartz-MacPherson classes) for Schubert varieties in Grassmannians have been worked out in the following papers: P. Aluffi and L.C. Mihalcea. Chern classes of Schubert cells and varieties. J. Algebraic Geom. 18 (2009), 63--100. Benjamin Jones: Singular Chern classes of Schubert varieties via small resolutions. IMRN 2010, 1371--1416. The first paper does the computation via Bott-Samelson resolution (as explained above), and you can see from the computations with Young tableaux that this is pretty close to applications in geometric representation theory. I hope this is useful for you.<|endoftext|> TITLE: Consecutive square values of cubic polynomials QUESTION [21 upvotes]: Let $P(x)$ be a cubic polynomial with integer coefficients. Does there exist a constant $c$ such that at least one of the following values $P(0),P(1),...,P(c)$ is not a square? It is known that the number of integral points of $y^2 = P(x)$ is bounded by a constant dependent on the coefficients of $P(x)$ and I am wondering whether in this particular case, it is possible to find a constant independent of the coefficients. Any links to existing literature or maybe an explanation of why such a problem would be impossible to solve with current tools would be of a great help too. REPLY [9 votes]: [EDITED to fix typos, show smoothness of $V_c$, and extend the exhaustive-search result] Solutions of $P(i) = y_i^2$ ($0 \leq i \leq c$) with $P(X) = \sum_{j=0}^3 a_j X^j$ are parametrized up to scaling by a threefold $V_c$ in projective space ${\bf P}^c$ that's the complete intersection of $c-3$ quadrics $$ y_i^2 - 4 y_{i+1}^2 + 6 y_{i+2}^2 - 4 y_{i+3}^2 + y_{i+4}^2 = 0 \phantom{0\infty} (0 \leq i \leq c-4), $$ minus the points on the hyperplane $y_0^2 - 3 y_1^2 + 3 y_2^2 - y_3^2 = 0$ where the leading coefficient $a_3$ vanishes. We expect plenty of points for $c < 7$, a sparse but still infinite set of points for $c = 7$, and only finitely many points for $c > 7$ except possibly on a proper subvariety. This last part is a special case of the Bombieri-Lang conjecture, and if we assume this conjecture for $V_8$ then we can probably use the forgetful maps $V_c \rightarrow V_8$ for $c>8$ to prove (more directly than using Caporaso-Harris-Mazur as Michael Zieve proposed) that some $V_c$ has no rational points except on the hyperplane $a_3=0$. [EDIT René's comment raises the question of whether this complete intersection is smooth. The answer is yes in characteristic zero. Any linear combination of the differentials of the quadrics $y_i^2 - 4 y_{i+1}^2 + 6 y_{i+2}^2 - 4 y_{i+3}^2 + y_{i-4}^2$ has the form $(a_0 y_0, a_1 y_1, \ldots, a_c y_c)$ with $\sum_{m=0}^c Q(m) a_m = 0$ for any polynomial $Q(m)$ of degree at most $3$. Therefore in any nonzero combination at least $5$ of the $a_m$ do not vanish. Therefore at a singularity at least $5$ of the $y_i$ must be zero. But this is impossible because $y_i^2$ are valuees of a cubic polynomial at distinct points $i=0,1,\ldots,c$, and at most $3$ of those can be zero unless the polynomial vanishes identically.] An exhaustive search for rational points on $V_7$ with $a_3 \neq 0$ and $0 \leq y_2,y_3,y_4,y_5 < 1024$ finds only the following $22$ [EDIT extended from $1024 = 2^{10}$ to $1536 = 3 \cdot 2^9$, and found eight more solutions, for a total of $30$], up to the symmetry $y_i \leftrightarrow y_{7-i}$: 13 7 1 1 5 7 7 1 53 21 7 29 45 53 49 3 1586 847 24 73 610 861 868 221 139 23 31 115 173 209 217 181 1061 577 35 73 469 721 883 935 31 52 47 34 35 64 107 158 821 433 49 127 355 479 473 79 139 83 71 97 125 139 127 41 359 19 79 299 439 509 481 229 163 124 107 110 121 128 121 82 169 157 119 67 55 131 233 349 368 247 134 35 76 163 242 311 826 481 164 1 286 451 544 539 323 223 167 167 197 223 223 167 595 379 187 17 109 205 251 217 973 109 239 817 1195 1387 1319 679 497 323 247 283 353 397 377 203 676 499 266 121 416 799 1226 1691 34 369 332 185 138 419 784 1203 2258 1259 356 235 838 1183 1216 521 1393 927 449 229 705 1273 1873 2499 2836 1597 470 223 1016 1471 1562 925 1373 889 475 211 317 497 581 475 1179 728 581 750 977 1124 1113 794 2027 1315 673 199 499 917 1255 1487 1027 1064 749 50 343 1252 2239 3334 2573 1629 791 65 507 923 1121 969 961 896 817 802 925 1196 1579 2042 1297 1082 827 608 613 922 1393 1948 1144 1135 854 413 388 1099 1970 2951 none of these lifts to a rational point on $V_8$ (neither $P(-1)$ nor $P(8)$ is a square); possibly $c=8$ is already small enough th make a solution impossible.<|endoftext|> TITLE: Gabber's proof of Br' = Br for quasiprojective schemes QUESTION [14 upvotes]: In a note by deJong showing the cohomological and ordinary Brauer groups coincide for separated quasicompact schemes with ample line bundle, it is mentioned that Gabber had an unpublished proof of the main result therein, but by a different method. From elsewhere I read that this proof dates back to the 90s. Is this proof available anywhere? If not, could someone give an outline? REPLY [9 votes]: $\newcommand{\Z}{\mathbb{Z}}\newcommand{\G}{\mathbb{G}}$ I emailed Johan de Jong, and he sent me the following, which I reproduce with his permission (lightly edited to only keep mathematical content). Gabber's proof is not written up anywhere, from what de Jong told me. The argument is different because I lectured about the proof from my write-up in front of Gabber and he then subsequently told me it is different. After the lecture he also explain his proof. Roughly what he does is two things: (1) Prove it for regular quasi-projective schemes. This involves carefully choosing a maximal order over the scheme which is Azumaya by modifying along an ample divisor (on some affine open you can already do the thing). (2) Prove the following amazing theorem: Suppose that $X$ is a scheme which has an ample invertible sheaf. Suppose that $\alpha$ is a cohomology class in $H^i(X, \Z/n\Z)$ for some $n > 0$ and since $i \ge 0$. Then one can find a quasi-projective scheme $Y$ smooth over $\Z$ and a morphism $X \to Y$ and a $\beta \in H^i(Y, \Z/n\Z)$ which pulls back to $\alpha$. Moreover, the same thing can be done with torsion classes in $H^2(X, \G_m)$. To do this you may immediately assume that $X$ is of finite type over $\Z$ by a limit argument. Then you can embed $X$ into some smooth $P$, e.g., projective space. Now on an open you can sort of extend the class (after replacing $P$ by a blow up in closed subscheme and maybe an etale neighbourhood). Then, and this was all explained to me at a party later in the day [...] you kind of keep blowing up until $\alpha$ extends.<|endoftext|> TITLE: Method to compute fundamental solutions which are distributions QUESTION [7 upvotes]: The Malgrange-Ehrenpreis theorem tells us that there is a fundamental solution for any linear differential operator of constants coefficients. The original proof was not constructive (it was based on the Hahn-Banach theorem). Lars Hörmander first gave a method (Hörmander's staircase) to explicitely compute fundamental solutions. Nevertheless, the solutions his method provides are usually "too bad", for they are not even tempered distributions. Therefore, a natural question arises: Is there a general method to explicitely construct fundamental solutions which are actually tempered distributions? REPLY [5 votes]: The fact that there always exists a temperate fundamental solution was proved by Hormander here in 1958. I don't know if this answers some part of your question. Concerning regularity, well, this is an interesting question (most of the classical theory of PDEs is concerned with it). Let me also mention a nice paper by Rosay who reproves Malgrange-Ehrenpreis in the $L^2$ setting with an elementary proof, but still non constructive<|endoftext|> TITLE: Why do parameter spaces (often) behave like what they parameterize? QUESTION [7 upvotes]: From the my limited experience it seems that often a parameter space (miraculously) has some of the properties of the elements it parameterizes. For example: The parameter space of all plane conics is an algebraic variety. The moduli space of genus $g$ curves $\mathcal{M}_{g}$ is itself an algebraic variety. The Mandelbrot set is connected and can be thought of as the parameter space: $$\{c\in \mathbb{C}\; : \; J_{f_c}\; \text{is connected}\}$$ where $J_{f_c}$ is the Julia set of the polynomial $f_c=x^2+c$. The Teichmuller space parameterizes complex structures on a surface $X$ and itself is a complex manifold. I am wondering if there is a reason for this occurring. State a bit more formally my question is: Question: Why is it the case that if parameter space $M$ parameterizes spaces of with property $P$ then $M$ (often) also has property $P$? Note I say often as there are examples where this does not happen. For example, there exists quadratic polynomials whose Julia set is not locally connect and there exists quadratic polynomials whose Julia set is locally connected, but of course the Mandelbrot set is either locally connected or not. REPLY [9 votes]: I disagree with the premise: I think the correct statement is "moduli spaces in algebraic/complex geometry are unusually nice, in that they are also spaces with algebraic/complex structure." By way of contrast: The "moduli space" of topological manifolds of dimension $d$ is a discrete set and has no interesting topological structure. The moduli space of smooth structures on a topological manifold is usually discrete (eg the 24 smooth structures on $S^7$). I have the impression that the reverse is true for $\mathbb{R}^4$ -- that the space of exotic $\mathbb{R}^4$'s is infinite dimensional -- but this is very far from my expertise and I couldn't find a paper stating it. The moduli space of Riemannian metrics on a topological manifold is infinite dimensional. I think it is quite amazing that, in complex and algebraic geometry, moduli spaces are so often finite dimensional but not discrete. REPLY [6 votes]: Moduli spaces should represent moduli functors. Moduli functors should describe what it means to have an "$X$-parameterized family" of objects in terms of a suitable kind of map $E \to X$, the idea being that the fibers of this map are the family. To say that such a functor is represented by a moduli space $M$ is to say that giving such a map is equivalent to giving a map $X \to M$; in particular, it's natural to ask for $M$ to be an object in the same category that $X$ and $E$ (and the fibers of the map $E \to X$) are. And even if a moduli functor is not, strictly speaking, representable by an object in the category you care about, morally speaking it describes a "generalized object" in the category anyway. Formally, the category of presheaves on a category $C$ is the free cocompletion of $C$, so any presheaf on $X$ (including a moduli functor) can be thought of as a formal colimit of objects in $C$. (There's an additional complications in the case of $M_g$, which is a stack: this reflects the fact that the moduli functor in this case should be thought of as landing in groupoids, not sets.)<|endoftext|> TITLE: Image of a hypersurface under a map $\mathbb CP^n\to \mathbb CP^n$ QUESTION [5 upvotes]: Let $H$ be a degree $d$ hypersurface in $\mathbb CP^n$ defined by an explicit equation $F=0$. Let $\varphi: \mathbb P^n \to \mathbb P^n$ be an explicit degree $m$ morphism. In this case $\varphi(H)$ is a degree $d^{n-1}m$ hypersurface. Is there an algorithm to calculate the coefficients of the degree $d^{n-1}m$ polynomial that defines this hypersuface? REPLY [3 votes]: As suggested by Alex in his comment, this is a classical problem involving elimination theory. The relevant result is the following, see [Greuel - Pfister, A Singular introduction to commutative algebra, Proposition A.7.12 p. 505]. Proposition. Let $f=(f_0: \ldots :f_m)\colon \mathbb{P}^n \longrightarrow \mathbb{P}^m$ be a morphism, with $f_i \in \mathbb{K}[x_0, \ldots, x_n]$ homogeneous polynomials of the same degree without common zeroes. Moreover, let $I \subset \mathbb{K}[x_0, \ldots, x_n]$ be a homogeneous ideal, defining the projective variety $X=V(I) \subset \mathbb{P}^n$, and set $$J= \langle I, \, f_0-y_0, \ldots, f_m-y_m \rangle \cap \mathbb{K}[y_0, \ldots, y_m].$$ Then $$f(X)=V(J)\subset \mathbb{P}^m.$$ This can be explicitly determined by using Groebner bases. The computation is usually unpractical by hands, but it can be done with all the most common Computer Algebra Systems. For instance, many explicit computations with $\textrm{Singular}$ can be found in the book by Greuel and Pfister quoted above.<|endoftext|> TITLE: Integral formula for Euler class QUESTION [20 upvotes]: The tangent bundles of closed hyperbolic surfaces have flat $PSL(2,\mathbb{R})$ connections showing that there can be no integral formula for the Euler class such connections. This contrasts the situation for the Orthogonal group where the Pfaffian applied to curvature integrates to the Euler class. Can anyone suggest a conceptual reason for the absence of a local formula in the case where the connection fails to preserve a metric? Is it related to the fact that the Euler class is unstable? REPLY [8 votes]: I think the point is the theorem, due to Cartan, that the universal Chern-Weil homomorphism $$CW_G:Sym^{k} (\mathfrak{g})^G \to H^{2k}(BG; \mathbb{R})$$ from invariant forms on the Lie algebra to the cohomology of the classifying space is an isomorphism once $G$ is compact. If $G$ is complex reductive (such as $GL_n (\mathbb{C})$), then it follows by the unitary trick that $CW_G$ is an isomorphism as well. But if $G$ is neither compact nor reductive, there is no reason to expect an expression of a general characteristic class to have an expression in terms of the curvature. In fact, Milnor's example shows that $CW_G$ is not surjective for $G=PSL_2 (\mathbb{R})$. I vaguely remember that if $G$ is the $3$-dimensional Heisenberg group (with center $S^1$), then $CW_G$ is the zero map in positive degrees, even though both source and target are nonzero.<|endoftext|> TITLE: What are the difference between modeling with stochastic differential equations (SDE) and ordinary differential equations (ODE) with a random force? QUESTION [11 upvotes]: There are lots of differences between SDE and ODE. From the theoretical point of view an also from the numerical algorithms used for simulations. But I am interested in knowing if there is a point when you can use a ODE with a random force to model a SDE. For example, in which conditions the SDE: $$ d X_t = -\theta X_t dt + \sigma d W_t $$ could be represented by an ODE like: $$ \frac{dX(t)}{dt} = -\theta X(t) + \sigma \xi(t) $$ Where $W_t$ is a Weiner process and $\xi(t)$ is an equivalent random force. Thanks REPLY [2 votes]: From a numerical standpoint, although you can interpret SDEs as non-autonomous ODEs, the forcing term is not differentiable (it's "almost" Holder continuous $\alpha=1/2$ a.s.) and thus the error estimates one uses to derive numerical methods for ODEs usually do not apply (for example, Runge-Kutta order $k$ methods assume $k$ derivatives in their derivations). In fact, in just about any case where $g$ is not a constant, the order of convergence for deterministic methods on SODEs ends up being $1/2$, which is clearly sub-optimal. Thus instead the numerical methods for SODEs have to use the properties of the "Ito derivative" and stochastic Taylor series to derive new methods for which the higher orders of the error analyses apply. However, what you pointed out is where a lot of the intuition comes from. If you read work from researchers like Hairer or Kloden, they frequently build the heuristics for understanding the system by approaching them as non-autonomous dynamical systems where the forcing term has not so nice qualities (i.e. Holder $1/2$ or in the case of space-time white noise, $1/4$). Thus in some sense you can understand things heuristically like "the deterministic methods are only order $1/2$ because the forcing term is Holder continuous $1/2$", and so understanding the SDE as a type of ODE has value in that sense.<|endoftext|> TITLE: What can representations of affine Weyl groups do? QUESTION [7 upvotes]: In Carter's Finite Groups of Lie type and Lusztig's Characters of Reductive Groups over a Finite Field, the representations of Weyl groups are helpful in finding the representations of algebraic groups. But what can the representations of affine Weyl groups do? Are they important only in their own right, or only when we are interested in them? REPLY [3 votes]: As noted the question is rather broad, with somewhat different answers possible depending on which field you are working over. Moreover, definitive answers are yet to be found in some directions. In any case, the algebraic groups of special interest here are the semisimple (or more generally reductive) ones. In the classical Cartan-Weyl theory of finite dimensional representations in characteristic 0, it's clear that the Weyl group $W$ plays a key role. The representations of $W$ get involved less directly, at first via the finite groups of Lie type and their ordinary characters (as in Carter's book and much of Lusztig's related work). Affine Weyl groups arise separately in several theories. The oldest appearance is in the structure theory of compact Lie groups, but here the representations or Hecke algebras don't come up. Starting with the work of Iwahori and Matsumoto in the 1960s, the $p$-adic groups (initially just in characteristic 0) have been studied profitably in terms of Hecke algebras and their representations, based on the use of an auxiliary affine Weyl group. This is emphasized in Prasad's answer here. A much less understood area involves the finite dimensional (rational) representations of a semisimple algebraic group $G$ in prime characteristic. Here there has been an evolution of ideas toward more involvement of affine Weyl groups relative to powers of the characteristic $p>0$: my preliminary work on "linkage" around 1970 was followed by Verma's formalization in terms of affine Weyl groups. His paper in the proceedings of the 1971 Budapest summer school on Lie groups is called The role of affine Weyl groups in the representation theory of algebraic Chevalley groups and their Lie algebras. This was followed in the 1970s by Jantzen's important work, but the introduction of Hecke algebra methods came only with the publication in 1979 of the landmark paper by Kazhdan and Lusztig. Soon Lusztig made connections with Jantzen's work in characteristic $p$ and proposed his own conjecture for certain irreducible characters of $G$ based on the formalism of Kazhdan-Lusztig polynomials arising from base change in the (dual) affine Hecke algebra. Much further refinement has gone on, but as Geordie Williamson has observed indirectly, there is still a problem in handling primes which are not "sufficiently large" relative to $G$. (Primes smaller than the Coxeter number have been left in limbo for a long time.) In any case, there is not yet a clear explanation in characteristic $p$ of the precise way in which all the affine Weyl groups relative to powers of $p$ will impact the representation theory of $G$. Even though the affine Weyl groups and their representations are somewhat detached from $G$, they are clearly influential in the known multiplicities essential to the finite dimensional representations of $G$.<|endoftext|> TITLE: Examples of elliptic curves over $\mathbb{Q}$ QUESTION [16 upvotes]: I need examples of two non-isogenous elliptic curves $E_{1}, E_{2}$ over $\mathbb{Q}$ having the following 2 properties - 1) $E_{1}, E_{2}$ have no rational torsion points. 2) $E_1[9] \cong E_2[9]$ as Gal$(\overline{\mathbb{Q}} / \mathbb{Q})$ modules. I will request you to kindly suggest how to start finding out examples having the above two properties or suggest some examples. If necessary, I am comfortable with basic Sage computations. REPLY [26 votes]: In general, for any integer $N$ and any fixed elliptic curve $E$, the elliptic curves $E'$ for which $E[N]\cong E'[N]$ as Galois modules (and such that the isomorphism respects the Weil pairing) are parametrised by a twist $Y_E(N)$ of the modular curve $Y(N)$. Tom Fisher has worked out equations of these twists for $N=7,9$ and 11. He explicitly writes down examples of such pairs of elliptic curves (and the very first example in section 7.2 has trivial torsion), and shows that there are infinitely many pairs satisfying your condition 2), together with the additional condition that the isomorphism preserves (respectively "anti-preserves") the Weil pairing. Since he gives his families in parametric form, you may be able to restrict the parameter to ensure that the torsion subgroup is trivial. If you want to go and search for such pairs yourself, you can start with any elliptic curve with trivial torsion, and search for points on $Y_E(N)$, using Tom Fisher's equation and brute force. Edit: Tom also has a fairly large table of pairs of curves satisfying your condition 2). Edit II, regarding $N=3$: The (compactifications of) the twists $Y_E(3)$ have genus 0, and moreover, they each have a rational point, given by $E$ itself. So for any fixed $E$, you can find infinitely many $E'$ with isomorphic 3-torsion, all you need to know is the equation of $Y_E(3)$. For that, see the papers of Rubin and Silverberg, referenced on the first page of Tom Fisher's paper.<|endoftext|> TITLE: Lieb: Stability of matter, problem with variational method QUESTION [5 upvotes]: I am trying to recalculate 'Stability of Matter' Paper from Lieb. I have a problem at the last step of the proof at page 3, where Lieb calculates the minimizer. I will explain my ideas and my problem so far. I have the following functional to be minimized on $\rho\in L^1(\mathbb R^d)$. Here $\lambda$ is a Lagrange multiplier and $\rho\geq 0$. $h(\rho) = \frac{1}{C_d}\left(\int_{\mathbb R^d} dx \rho(x)^\frac{d}{d-2}\right)^\frac{d-2}{d} - Z \int_{\mathbb R^d} dx \frac{\rho (x)}{|x|} +\lambda\left( \int_{\mathbb R^d} dx \rho(x) -1\right)$ My idea is the following: $\frac{\text{d}}{\text{d}t}h(\rho_m +t \eta)|_{t=0} =0$, where $\rho_m$ is the minimizer and $\eta\in C_0^\infty(\mathbb R^d)$ arbitrary. Then I obtain the following: $0 = \int dx \,\eta(x) \left(\frac{1}{C_d}\|\rho_m\|_\frac{d}{d-2}^\frac{-2}{d-2}\rho_m(x)^\frac{2}{d-2} - \frac{Z}{|x|} + \lambda\right)$ Then by the fundamental lemma of variations it follows that: $\frac{1}{C_d}\|\rho_m\|_\frac{d}{d-2}^\frac{-2}{d-2} \rho_m(x)^\frac{2}{d-2} - \frac{Z}{|x|} + \lambda = 0$ Now I use that $\rho_m\geq 0$ and therefore $\frac{Z}{|x|} - \lambda\geq 0$. Since one also sees that $\lambda$ has to be bigger or equal 0 as otherwise $\rho_m$ wouldn't be in $L^1$. I conclude that $\rho_m$ has the following form: $\rho_m (x) =\begin{cases} C_d^\frac{d-2}{2} \|\rho_m\|_\frac{d}{d-2} \left(\frac{Z}{|x|} - \lambda \right)^\frac{d-2}{2} & \text{ if} |x|\leq \frac{Z}{\lambda} \\ 0 & \text{ if} |x|> \frac{Z}{\lambda} \end{cases}$ Now comes the point of my concerns: This $\rho_m$ has compact support and therefore cannot really satisfy $0 = \int dx \,\eta(x) \left(\frac{1}{C_d}\|\rho_m\|_\frac{d}{d-2}^\frac{-2}{d-2}\rho_m(x)^\frac{2}{d-2} - \frac{Z}{|x|} + \lambda\right)$. Is there a reason why this $\rho_m$ is however the correct minimizer? Best wishes :) REPLY [5 votes]: You are minimizing under the constraint that $\rho \ge 0$. Hence your variation $\rho_m + t \eta$ might not be admissible. (Say if $\eta \ne 0$ and $\vert t\vert$ is large enough.) The trick is to consider either any $\eta$ with $t \ge 0$, wich will give you the inequality $$ \frac{1}{C_d} \Vert \rho_m \Vert_{d/(d - 2)}^{-2/(d - 2)} \rho_m (x)^{2/(d - 2)} - \frac{Z}{\vert x \vert} + \lambda \ge 0. $$ or consider $\eta \in L^{\infty} (\mathbb{R}^d) \cap L^1 (\mathbb{R}^d)$ such that $$ \inf \{\rho_m (x) : \eta (x) \ne 0\} > 0. $$ Then, there exists $\tau > 0$ such that $u + t \eta \ge 0$ for each $t \in (-\tau, \tau)$. One concludes therefrom that if $\rho_m (x) > 0$, then $$ \frac{1}{C_d} \Vert \rho_m \Vert_{d/(d - 2)}^{-2/(d - 2)} \rho_m (x)^{2/(d - 2)} - \frac{Z}{\vert x \vert} + \lambda = 0. $$ This brings you to the desired solution, and the function satisfies the necessary conditions of extremum even if its support is compact. Since the regularity of $\rho_m$ is not known a priori, it is important to take nonsmooth variations $\eta$.<|endoftext|> TITLE: 1-Wasserstein distance between two multivariate normal QUESTION [23 upvotes]: The $p$-Wasserstein between two measures $\nu_1$ and $\nu_2$ on $X$ is given by $$d_p(\nu_{1},\nu_{2})=\underset{\pi\in\Gamma(\nu_{1},\nu_{2})}{\inf}\int_{\mathbf{\mathcal{X}}^{2}}d(x,y)^p\pi(dx,dy)$$ where $\Gamma(\nu_{1},\nu_{2})$ is the set of all couplings between $\nu_1$ and $\nu_2$. For $X=\mathbb{R}^d$ and $d$ being the euclidean distance the optimal transport between $\nu_{1}=N(m,V)$ and $\nu_{2}=N(n,U)$ is well known for $p=2$ see e.g. On Wasserstein geometry of the space of Gaussian measures by Asuka Takatsu. However what is known for $p=1$ for the Euclidean distance or other "reasonable" metrics? I am interested in explicit formulas or sharp bounds. REPLY [3 votes]: For $p=1$ one can bound the 1-Wasserstein metric by $$|m-n| + \sqrt{\sum_{i=1}^{d} \left[ \left( \sqrt{\lambda_i} - \sqrt{\gamma_i}\right)^2 + 2\sqrt{\lambda_i\gamma_i}(1-v_i\cdot u_i) \right]}$$ when $\lambda_i$ and $\gamma_i$ are the $i^{th}$ eigen values of $U$ and $V$ respectively, $v_1,\ldots,v_d$ and $u_1,\ldots,u_d$ are the corresponding orthonormal basis of eigen-vectors. See Chafai & Malrieu Lemma 2.4. Although this bound seems close-in nature to the $p=2$ bound, I'm not sure if it is sharp.<|endoftext|> TITLE: Cohomology of relative motives QUESTION [9 upvotes]: Notation Let $S$ be a scheme, proper over a field $k$. Let $\mathrm{SmPr}_{S}$ denote the category of smooth projective $S$-schemes. Let $\mathcal{M}_{S}$ denote the category of relative Chow motives over the base $S$. Let $h_{S} \colon \mathrm{SmPr}_{S}^{\mathrm{op}} \to \mathcal{M}_{S}$ be the functor assigning to $X/S$ its Chow motive, and to a morphism $X \to Y$ the transpose of the (relative) graph in $Y \times_{S} X$. Motivation This might pretty long. Scroll down if you want to read the question (-; If $f \colon S \to S'$ is smooth projective (and a morphism of $k$-schemes), there is a pushforward functor $f_{*} \colon \mathcal{M}_{S} \to \mathcal{M}_{S'}$. It maps $(X, p, m)$ to $(X, j_{*}(p), m)$, where $j$ is the canonical map $X \times_{S} X \to X \times_{S'} X$. One can check that this is well defined (see e.g., [MNP, Cor 8.1.7]). It is easy to check that the following diagram commutes. $$ \begin{array}{ccc} \mathrm{SmPr}_{S}^{\mathrm{op}} & \stackrel{f \circ \_}{\longrightarrow} & \mathrm{SmPr}_{S'}^{\mathrm{op}} \\ \quad\downarrow h_{S} & & \downarrow h_{S'} \\ \mathcal{M}_{S} & \stackrel{f_{*}}{\longrightarrow} & \mathcal{M}_{S'} \end{array} $$ Suppose $\ell$ is a prime different from $\mathrm{char}(k)$. We can define a cohomology functor $$ \begin{array}{rrll} \mathrm{H}_{S} \colon & \mathcal{M}_{S} & \longrightarrow & \{ \text{$\mathbb{Q}_{\ell}$-sheaves on $S$} \} \\ & ((g \colon X \to S), p, m) & \longmapsto & \mathrm{Im}(p_{*}|\mathrm{R}g_{*} \mathbb{Q}_{\ell}[2m])(m) \end{array} $$ (Obviously, it also makes sense to do this with $\mathbb{Q}$-coefficients if $S$ is a complex scheme.) To make this more explicit, let $\bar{s} \to S$ be a geometric point. A $\mathbb{Q}_{\ell}$-sheaf on $S$ is the same as a $\pi(S, \bar{s})$-representation (with $\mathbb{Q}_{\ell}$-coefficients). Also $(\mathrm{R}g_{*}\mathbb{Q}_{\ell})_{\bar{s}}$ is $\mathrm{H}^{\bullet}(X_{\bar{s}}, \mathbb{Q}_{\ell})$. Pulling back $p$ along $X_{\bar{s}} \times X_{\bar{s}} \to X \times_{S} X$ gives an autocorrespondence $p_{\bar{s}}$ on $X_{\bar{s}}$. This shows that we can think about $\mathrm{Im}(p_{*}|\mathrm{R}g_{*} \mathbb{Q}_{\ell}[2m])(m)$ as $\mathrm{Im}(p_{\bar{s},*}|\mathrm{H}^{\bullet+2m}(X_{\bar{s}}, \mathbb{Q}_{\ell}))(m)$. Question A natural question to ask is whether the following diagram commutes. $$ \begin{array}{ccc} \mathcal{M}_{S} & \stackrel{f_{*}}{\longrightarrow} & \mathcal{M}_{S'} \\ \quad\downarrow h_{S} & & \downarrow h_{S'} \\ \{ \text{$\mathbb{Q}_{\ell}$-sheaves on $S$} \} & \stackrel{\mathrm{R}f_{*}}{\longrightarrow} & \{ \text{$\mathbb{Q}_{\ell}$-sheaves on $S'$} \} \end{array} $$ It commutes on objects, because $\mathrm{R}f_{*}\mathrm{R}g_{*} = \mathrm{R}(f \circ g)_{*}$. However, I have no clue how to show that it commutes on morphisms. I would be very happy with an answer to: $\frac{1}{2}$Q: Does the above diagram commute on morphisms that come from $\mathrm{SmPr}_{S}$? And I would be even more happy if one can show that the diagram actually commutes: Q: Does the above diagram commute on morphisms in general? Spelling out $\frac{1}{2}$Q If $X$ is an $S$-scheme, denote the structure morphism $X \to S$ with $g_{X/S}$. If $\phi/S \colon X/S \to Y/S$ is a morphism of $S$-schemes, we also have a morphism $\phi/S' \colon X/S' \to Y/S'$ of $S'$-schemes. We have induced morphisms $$ (\phi/S)^{*} \colon \mathrm{R}(g_{Y/S})_{*}\mathbb{Q}_{\ell} \to \mathrm{R}(g_{X/S})_{*}\mathbb{Q}_{\ell} $$ and $$ (\phi/S')^{*} \colon \mathrm{R}(g_{Y/S'})_{*}\mathbb{Q}_{\ell} \to \mathrm{R}(g_{X/S'})_{*}\mathbb{Q}_{\ell}. $$ Applying the $\mathrm{R}f_{*}$-functor to the first morphism, we get a morphism $\mathrm{R}f_{*}(\phi/S)^{*} \colon \mathrm{R}f_{*}\mathrm{R}(g_{Y/S})_{*}\mathbb{Q}_{\ell} \to \mathrm{R}f_{*}\mathrm{R}(g_{X/S})_{*}\mathbb{Q}_{\ell}.$ But, of course we can rewrite the source and target, to get $$ \mathrm{R}f_{*}(\phi/S)^{*} \colon \mathrm{R}(g_{Y/S'})_{*}\mathbb{Q}_{\ell} \to \mathrm{R}(g_{X/S'})_{*}\mathbb{Q}_{\ell}. $$ $\frac{1}{2}$Q': Do we have $\mathrm{R}f_{*}(\phi/S)^{*} = (\phi/S')^{*}$? References [MNP] — Murre, Nagel, Peters. Pure Motives. (2013) REPLY [4 votes]: Yes it commutes. To prove this you should express the map $(\phi/S)^\ast$ in "six functors" language. All functors below are derived. What's needed is that $\newcommand{\Q}{\mathbf Q_\ell} \renewcommand{\S}{\mathrm{pt}}\Q$ is pulled back from $\S = \mathrm{Spec}(k)$. Then $$ g_{X/S,\ast} \Q= g_{X/S,\ast}g_{X/\S}^\ast \Q = g_{Y/S,\ast}\phi_\ast\phi^\ast g_{Y/\S}^\ast \Q $$ which receives a map from $g_{Y/S,\ast}g_{Y/\S}^\ast \Q$ because of the unit of the adjunction $1 \to \phi_\ast\phi^\ast$. This map is exactly $(\phi/S)^\ast$. If we express $(\phi/S')^\ast$ in the same way and apply $f_\ast$ then we get the equality you want, precisely because $S \to S'$ is a map of $k$-schemes.<|endoftext|> TITLE: Degeneration of the Hodge spectral sequence for scheme over truncated Witt ring? QUESTION [11 upvotes]: Let $f:X\to S$ be a smooth proper morphism of schemes. If $S$ is of characteristic zero (i.e., $S$ is a $\mathbb{Q}$-scheme), then Deligne has shown: The Hodge-De Rham spectral sequence $E^{a,b}_1=R^af∗Ω^b(X/S)⇒H^{a+b}_{DR}(X/S)$ degenerates in $E_1$. This is also true when $S=Spec(k)$ with $k$ a perfect field of characteristic $p$, and $X$ has a smooth $W_2(k)$-lifing. Here $W_2(k)$ is the length 2 truncated Witt ring with values in $k$. My question is when $S= Spec(W_2(k))$, and $X$ has a smooth $W_3(k)$-lifting, does the statement of $E_1$ degeneration still hold (maybe one needs more conditions)? Has anyone worked on this kind of question before? REPLY [9 votes]: First of all, the statement of Deligne-Illusie as you've given it is slightly wrong; there is only degeneration in degree $\leq \dim(X)$ (and thus full degeneration if $\dim(X)\leq p$). Yukiyoshi Nakkajima proves an analogous result of the sort you'd like (extending work of Ogus) under the condition that the smooth proper scheme $X/W_n(k), n\geq 2$ admits a Frobenius lift. Here is the statement: Theorem. Let $X$ be a smooth proper scheme over $W_n(k)$ ($k$ perfect of characteristic $p$, $n\geq 2$). If $X$ admits a lift of the Frobenius of $X\otimes_{W_n(k)} k$ and has a smooth lifting $\tilde{X}$ over $W_{n+1}(k)$, then the Hodge-de Rahm spectral sequence $$E^{i,j}_1=H^j(X, \Omega^i_{X/W_n(k)})\implies H^{i+j}_{dR}(X/W_n(k))$$ degenerates at $E_1$. Note that unlike the Deligne-Illusie result, there is no condition on the dimension of $X$ or the degree of the Hodge cohomology; rather the whole spectral sequence degenerates. On the other hand, this result requires the (strong) condition of a Frobenius lift, which is vacuous for Deligne-Illusie (where $n=1$). Ogus proves a similar result (with a degree condition) in 8.2.6 of "F-crystals, Griffiths Transversality, and the Hodge Decomposition."<|endoftext|> TITLE: ζ(-n) and "powers" of Grandi's series QUESTION [10 upvotes]: For n a non-negative integer, $ζ(-n)$ can be interpreted as assigning a value to the (divergent) series $1^n+2^n+3^n+4^n+\cdots$ A value can also be assigned to the related series ${n+0 \choose n}+{n+1 \choose n}+{n+2 \choose n}+{n+3 \choose n}+\cdots$ by comparing its value with that of "powers" of Grandi's series $1-1+1-1+\cdots$ Concretely (but not rigorously): Considering $1+x+x^2+x^3+x^4+\cdots$ as a formal power series or by analytic continuation, it can be identified with $\frac{1}{1-x}$. This gives a well-defined value except at $x=1$, but then we can use the fact that, where the terms are defined, $\frac{1}{1-x} - \frac{2x}{1-x^2} = \frac{1}{1+x}$ or, in power series, $(1+x+x^2+x^3+\cdots)-2x(1+x^2+x^4+x^6+\cdots)=1-x+x^2-x^3+\cdots$ and setting x = 1, $(1+1+1+1+\cdots) - 2(1+1+1+1+\cdots) = 1-1+1-1+\cdots$ where the right-hand side is already assigned the value of $1/2$, making $1+1+1+1+\cdots = -1/2$. This agrees with the value given by $ζ(0)$. Similarly, $\frac{1}{(1-x)^2} - \frac{4x}{(1-x^2)^2} = \frac{1}{(1+x)^2}$ or, in power series, $(1+2x+3x^2+4x^3+\cdots)-4x(1+2x^2+3x^4+4x^6+\cdots)=1-2x+3x^2-4x^3+\cdots$ and setting x = 1, $(1+2+3+4+\cdots) - 4(1+2+3+4+\cdots) = 1-2+3-4+\cdots$ where the right-hand side is already assigned the value of $(1/2)^2 = 1/4$, making $1+2+3+4+\cdots = -1/12$. Again, this agrees with the value given by $ζ(-1)$. This is all extremely hand-wavy. In particular, the argument above only works because we chose to look at ("powers" of) Grandi's series $1-1+1-1+\cdots$ and not some other geometric series $1+y+y^2+y^3+\cdots$ for $y≠-1$. My question then is: why does this work for Grandi's series? Surely it isn't all just a coincidence? REPLY [7 votes]: [3/30 -- Not sure anyone is paying attention to this any more, but I attempted to fill in the gaps mentioned in the comments] I was sort of of hoping an expert would come by, but I'll give it a shot. The short answer is that Grandi's series and its "powers" aren't very divergent. In particular, Tao's smoothed asymptotics have no terms with a positive power of $N$ and give the same answer as evaluating the power series at 1. $$ \sum (-1)^n \sim \frac{1}{2} + O\left(\frac{1}{N}\right)\ . $$ (I'm using the sim symbol to indicate the smoothed asymptotics.) Similarly, we have $$ \sum n (-1)^n \sim \frac{1}{4} + O\left(\frac{1}{N}\right)\ . $$ More generally, we have $$ \sum n^m (-1)^n \sim -\frac{2^{m+1}-1}{m+1}B_{m+1} + O\left(\frac{1}{N}\right)\ . $$ These sums are Cesaro summable $(C,k)$ for high enough $k$ and the sum is the same as the Abel sum (i.e., evaluating the power series at 1 -- the proof is in Hardy's book, for example) which can be evaluated in terms of Eulerian numbers. $(C,k)$ summability is also equivalent to Riesz summability (also in Hardy's book) which means that there are no divergent terms with the smoothing $\eta(x)=(1-x)_{+}^k$. However, I don't think this is sufficient, so a proof of this formula for sufficiently nice $\eta(x)$ is given below. Moving to the the sums of integer powers, with the smoothed asymptotics, they have the form $$ C + aN^s + O\left(\frac{1}{N}\right)\ . $$ Then, as shown in the blog post, C is the value you get via zeta function regularization. What you're doing in your question is manipulating the sums to give something where the divergent terms cancel as in the two "Grandi-like" sums above. Therefore, what remains gives an identity you can solve for C, i.e., the value of the zeta function. Using the asymptotics in the linked blog above, it's easy to see this happen explicitly: $$ \sum 1 \sim -\frac{1}{2} + aN + O\left(\frac{1}{N}\right)\ , $$ $$ \sum n \sim -\frac{1}{12} + bN^2 + O\left(\frac{1}{N}\right)\ , $$ $$ \sum n^2 \sim cN^3 + O\left(\frac{1}{N}\right)\ . $$ So, starting with $$ (1 + \dots) - 2(1 + \dots) = 1 - 1 + \dots\ , $$ you get $$ C + aN - 2\left(C + a\frac{N}{2}\right) + O\left(\frac{1}{N}\right) = \frac{1}{2} + O\left(\frac{1}{N}\right) $$ or $$ -C = 1/2 $$ confirming that $\zeta(0)=-1/2$. The sum for $\zeta(-1)$ works the same where the $4(N/2)^2$ divergence in the second term cancels the $N^2$ divergence in the first. And you can get $\zeta(-2)=0$ using the identity: $$ \frac{1+x}{(1-x)^3} - \frac{8x(1+x^2)}{(1-x^2)^3} = \frac{1-x}{(1+x)^3}\ . $$ As above, to do the general case, you can use the Abel summation: $$ \sum_n (-1)^n n^m \sim -\frac{1}{2^{m+1}} \sum (-1)^i \left<{m \atop i}\right> = -\frac{2^{m+1}-1}{m+1}B_{m+1}\ . $$ and divide to get the usual answer for the zeta function evaluated at negative integers. Finally, the promised proof that the smoothed alternating sums of powers give the same answer as Abel summation up to terms that vanish as $N\to\infty$. First, generalizing a statement in Tao's blog post, it's easy to show that: $$ \sum (-1)^n f(n) = \frac{1}{2^{k-1}} \left((-1)^k \sum (\Delta^k f)(2n) + \sum_{l=0}^{k-1} (-1)^l f(l) \sum_{i=1}^{\infty} \left({k \atop l + 2i}\right)\right)\ . $$ For $k$ large enough and with $\eta(x)$ highly differentiable, supported on $[0,1]$ and $\eta(0)=1$, we can apply this to $f(n) = \eta(n/N)(-1)^n n^m$. Given the assumptions on $k$ and $\eta(x)$, the first term will always go as some negative power of $N$, so all that remains to be shown is that the second sum gives the same answer as the Abelian sum up to terms of order $1/N$. This can be done through a brute force calculation. Here's a quick sketch (assuming I didn't miss anything): (1) Since $\eta(0)=1$, $\eta(l/N) = 1 + O(1/N)$, so it can be set to 1. (2) Let the limits of $l$ go up to $k$ and $i$ down to zero. (3) Replace $2i$ by $i$ by multiplying the sum by $(-1)^i + 1$. (4) Change variables so $l+i$ is one variable and $l$ is the other. (5) One of the sums will go to zero for large enough $k$ because it is a high order finite difference of a lower degree polynomial. (6) For the other sum, use the explicit formula for the sum $$ \sum_{n=0}^k (-1)^n n^m $$ in terms of Eulerian numbers. (7) More things vanish because they end up being high order finite differences. (8) End up with the same alternating sum of Eulerian numbers as in the Abel summation above.<|endoftext|> TITLE: A curious identity related to finite fields QUESTION [39 upvotes]: To three elements $a_1$, $a_2$, $a_3$ in the finite field $\mathbb F_q$ of $q$ elements we associate the number $N(a_1,a_2,a_3)$ of elements $a_0\in \mathbb F_q$ such that the polynomial $x^4+a_3x^3+a_2x^2+a_1x+a_0$ splits into four distinct linear factors over $\mathbb F_q$. The number $$\sum_{(a_1,a_2,a_3)\in\mathbb F_q^3}{N(a_1,a_2,a_3)\choose 2}$$ counts then the number of minimal pairs in the Craig lattice $C_{q-1,3}$ (for $q$ a prime number). Experimentally (for all primes up to 2000) this number seems to be given by $$\frac{1}{1152}q(q-1)(q^3-21q^2+171q-c_q)$$ (the factors $q$ and $(q-1)$ are easy to explain through the action of the affine group) where $$c_q=\left\lbrace\begin{array}{ll} 455\qquad&q\equiv 5\pmod{24}\\ 511&q\equiv 7\pmod{24}\\ 583&q\equiv 13\pmod{24}\\ 383&q\equiv 23\pmod{24}\end{array}\right.$$ and no nice formula seems to exist for the remaining cases (which after exclusion of powers of $2$ and $3$ form a subgroup of the multiplicative group $(\mathbb Z/24\mathbb Z)^\ast$). Is there an explanation for these identities? Remark: One can of course define similarly $N(a_1,a_2)$ or $N(a_1,a_2,a_3,a_4)$. Nice formulae for $\sum{N(\ast)\choose 2}$ exist for all primes in the first case (and are easy to prove using quadratic reciprocity). I could see nothing in the second case (computations become however quite heavy and I could not get very far). REPLY [25 votes]: For prime $q \geq 5$ write the count as $$ \frac1{1152} q (q-1) (q^3 - 21q^2 + 171 q - c_q) $$ where $$ c_q = 483 + 36 \left(\frac{-1}{q}\right) + 64 \left(\frac{-3}{q}\right) + \delta_q. $$ Then for $(\frac{-2}{q}) = -1$ Ronald Bacher's calculations indicate $\delta_q=0$. If $(\frac{-2}{q}) = +1$ then $q$ can be written as $m^2 + 2n^2$, uniquely up to changing $(m,n)$ to $(\pm m, \pm n)$, and we have $$ \delta_q = 24(m^2 - 2n^2) + 192 + 72 \left(\frac{-1}{q}\right). $$ The explanation is as follows. Start as did Will Sawin by considering the variety of $(s_1,s_2,s_3,s_4,t_1,t_2,t_3,t_4)$ such that for $i=1,2,3$ the $i$-th elementary symmetric function of the $s$'s equals the $i$-th elem.sym.fn. of the $t$'s. We may apply any $aX+b$ transformation to all $8$ variables, which explains the $q(q-1)$ factor. (The factor $1152 = 2 \cdot 4!^2$ is from coordinate permutations that respect the partition of the $8$ variables into two sets of $4$.) In odd characteristic, there's a unique representative with $\sum_{i=1}^4 s_i = \sum_{i=1}^4 t_i = 0$; this takes care of the translations, and then we mod out by scalars by going to projective space. We end up with the complete intersection of a quadric and a sextic in ${\bf P}^5$. This threefold, call it ${\cal M}$, turns out to be rational. (This has probably been known for some time, because ${\cal M}$ classifies perfect multigrades of order $4$, and such things have been studied since the mid-19th century, see the Prouhet-Tarry-Escott problem; I outline a proof below.) However, by requiring that all coordinates be distinct we're removing some divisor ${\cal D}$ on this threefold, so the final count decreases by the outcome of an inclusion-exclusion formula whose terms are point counts over some subvarieties of ${\cal M}$. Most of these sub varieties are rational curves, or points that may be defined over ${\bf Q}(i)$ or ${\bf Q}(\sqrt{-3})$, the latter explaining the appearance of Legendre symbols $(\frac{-1}{p})$, $(\frac{-3}{p})$ in the counting formula. But the two-dimensional components of ${\cal D}$ are isomorphic K3 surfaces, arising as a complete intersection of a quadric and a cubic in ${\bf P}^4$; and those components make a more complicated contribution. Fortunately these K3 surfaces are "singular" (i.e. their Picard number attains the maximum of $20$ for a K3 surface in characteristic zero) $-$ I computed that they're birational with the universal elliptic curve over $X_1(8)$ $-$ and it is known that the point-count of this singular K3 surface can be given by a formula that involves $m^2-2n^2$ when $(\frac{-2}{q}) = +1$. To show that $\cal M$ is rational, it is convenient to apply a linear change of variables from the "$A_3$" coordinates $s_i,t_i$ to "$D_3$" coordinates, say $a,b,c$ and $d,e,f$, with $$ s_i = a+b+c, \phantom+ a-b-c, \phantom+ -a+b-c, \phantom+ -a-b+c $$ and likewise $t_i = d+e+f, \phantom. d-e-f, \phantom. -d+e-f, \phantom. -d-e+f$. Then $\sum_{i=1}^4 s_i = \sum_{i=1}^4 t_i = 0$ holds automatically, and the quadric and cubic become simply $$ a^2+b^2+c^2 = d^2+e^2+f^2, \phantom\infty abc = def. $$ Let $d=pa$ and $e=qb$. Then $f=(pq)^{-1}c$, and the quadric becomes a conic in the $(a:b:c)$ plane with coefficients depending on $p,q$: $$ (p^2-1)a^2 + (q^2-1)b^2 + ((pq)^{-2}-1) c^2 = 0. $$ So $\cal M$ is birational to a conic bundle over the $(p,q)$ plane, and this conic bundle has a section $(a:b:c:d:e:f) = (1:p:pq:p:pq:1)$ which lets us birationally identify $\cal M$ with the product of the $(p,q)$ plane with ${\bf P}^1$. This is a rational threefold, QED.<|endoftext|> TITLE: Locally finite compact groups QUESTION [5 upvotes]: I assume all tolpological groups here to be Hausdorff. A group is called locally finite if every finitely generated subgroup is finite. What can be said about a locally finite compact group? Must it be finite? Is there any structure theorem for this kind of groups? For the larger class of torsion compact groups structure theorems exist. In particular, They tell us that these groups are profinite. So, locally finite compact groups are profinite as well. But, can one prove more? REPLY [4 votes]: To summarize some of what is stated in the comments above, there is a related question (posed as Question 4.8.5b on p. 156 of Ribes-Zalesskii) of whether torsion profinite groups have bounded exponent. As noted in the comments, your question is equivalent to this question due to Zelmanov's solution of the restricted Burnside problem, which implies that finitely generated profinite torsion groups are finite (see again p. 156 of Ribes-Zalesskii for discussion). In turn, the case of whether profinite torsion groups have bounded exponent may be reduced to the case of pro-$p$ torsion groups by a theorem of Wilson (4.8.5d in Ribes-Zalesskii, which suffices for the finitely generated case).<|endoftext|> TITLE: Derivation on real analytic manifolds QUESTION [19 upvotes]: Let $M$ be a real analytic manifold. By $C^{\omega}(M)$ we mean the algebra of all analytic functions from $M$ to $\mathbb{R}$. Assume that $D$ is a derivation on $C^{\omega}(M)$ . Is there a global real analytic vector field $X$ on $M$ such that $D(f)=X.f$ for all $f\in C^{\omega}(M)$? The motivation: The smooth version of this statment is true but the proof is based on usage of non analytic functions REPLY [10 votes]: $\def\RR{\mathbb{R}}\def\CC{\mathbb{C}}\def\NN{\mathbb{N}}$I'm going to try once more. As my previous flawed answers should have made obvious, the real analytic category is not my home, so read with caution. Conveniently, the two big results I need are stated as Corollary 5.43 and 5.44 here: Cor. 5.43 $M$ has a proper real analytic embedding into some $\RR^n$. I'll write $(z_1, \ldots, z_n)$ for the coordinates on $\RR^n$. Cor. 5.44 (specialized and rephrased) Let $p$ be a point of $M$ and let $M$ be embedded in $\RR^n$ as above. Let $U$ be a neighborhood of $p$ in $\RR^n$. Let $f: M \cup U \to \RR$ be a function which is real analytic on $U$ and on $M$. Let $d$ be a nonnegative integer. Then there is a real analytic function $g: \RR^n \to \RR$ such that $g|_M=f$ and $(g-f)|_U$ vanishes to order $d$ at $p$. We also need the following: Easy Lemma: Let $f: \RR^n \to \RR$ be a real analytic function for which $f(z_1, \ldots, z_{n-1},0)$ is identically zero. Then $f(z_1, z_2, \ldots, z_{n-1}, z_n)/z_n$ extends to a real analytic function $\RR^n \to \RR$. Proof This is a local statement, and can be easily checked locally using convergent power series. $\square$ Key Lemma: Let $g: \RR^n \to \RR$ be a real analytic function vanishing to order $d$ at $0$. Then we can write $g(z)$ as a finite sum $\sum z^a p_a(z)$ where $z^a$ are monomials of degree $d$ and $p_a(z)$ are real analytic functions. Proof By induction on $d$ and $n$. The base case $d=0$ is trivial: Write $g(z) = 1 \cdot g(z)$; the base case $n=0$ is vacuous. We assume $d$ and $n>0$. Define $g(z_1, \ldots, z_n)$ by $$f(z_1, \ldots, z_n) = f(z_1, \ldots, z_{n-1}, 0) + z_n g(z_1, \ldots, z_{n-1}).$$ By the Easy Lemma, $g$ is analytic and, by a basic computation, it vanishes to order $d-1$. By induction on $d$, we can write $g = \sum z^c r_c(z)$ where the monomials $z^c$ have degree $d-1$. By induction on the number of variables, we can write $f(z_1, \ldots, z_{n-1},0)$ as $\sum z^b p_b(z_1, \ldots, z_{n-1})$ where the monomial $z^b$ have degree $d$. So $$f(z_1, \ldots, z_n) = \sum z^b p_b(z_1, \ldots, z_{n-1}) + \sum (z^c z_n) q_c(z_1,\ldots, z_n),$$ an expression of the desired form. $\square$ We now prove the result. Lemma Let $f$ be a real analytic function on $M$ that vanishes to second order at $p$. Then $(D f)(p)=0$. Proof Embed $M$ in $\RR^n$ by Cor 5.43. Extend $f$ to a function on $\RR^n$ also vanishing to order $2$ at $p$ and, without loss of generality, translate $p$ to $0$. By the Key Lemma, we can write $f(z) = \sum z_i z_j c_{ij}(z)$. Then $$(D f)(0) = \sum (D z_i)(0) \cdot 0 \cdot c_{ij}(0) + \sum 0 \cdot (D z_j)(0) \cdot c_{ij}(0) + \sum 0 \cdot 0 \cdot (D c_{ij})(0) =0. \quad \square$$ Lemma $(D f)(p)$ depends only on $(df)(p)$. Proof Suppose that $(d f_1)(p) = (d f_2)(p)$. Define $q(z)$ by the equation $f_2(z) = f_1(z) + (f_2(p)-f_1(p)) + q(z)$, then $q$ vanishes to order $2$ at $p$ so $(D q)(p) =0$. Also, $D$ of a constant function is $0$. So $(D f_1)(p) = (D f_2)(p)$. $\square$ The previous lemma shows that there is a well defined map $v(p) : T^{\ast}_p M \to \RR$ so that $(D f)(p) = v(p)(d f)$. (Well, we also need to see that $C^{\omega}(M) \to T^{\ast}_p M$ is surjective, but that is obvious because the functions $z_i$ span the cotangent space.) It is easy to see that $v(p)$ is linear. So $v(p)$ is an element of $(T_{\ast})_p M$. We have now created a section $v: M \to T_{\ast} M$ so that $(D f)(p) = \langle v(p), df \rangle$. It remains to see that $v$ is real analytic. This question is local. Let $(z_1, \ldots, z_d)$ give local coordinates on $M$ near some point $p$. Then we can write $$v(p) = \sum_{i=1}^d v_i(p) \frac{\partial}{\partial z_i}.$$ We need to show that the functions $v_i$ are real analytic. Since $v_i = D z_i$, this is clear. QED.<|endoftext|> TITLE: Reverse mathematics of meromorphic functions on Riemann surfaces QUESTION [10 upvotes]: Various sources touch briefly on the reverse mathematics of measure theory and complex analysis. But I have found none on the uniformization theorem for Riemann surfaces or the existence of non-constant meromorphic functions. The proofs of those results use very penetrating analysis but I do not know if they need high logical strength. For references on the usual proofs see The existence of meromorphic functions on Riemann surfaces. Is the reverse math of these theorems known? REPLY [10 votes]: I do not know about the full uniformization theorem, but reverse mathematics of Riemann mapping theorem (by applying non-standard second-order arithmetics) is done in MR3129726 Horihata, Yoshihiro; Yokoyama, Keita(J-AIST-SIF) Nonstandard second-order arithmetic and Riemann's mapping theorem. (English summary) Ann. Pure Appl. Logic 165 (2014), no. 2, 520–551. References of this paper may also be useful. Regarding uniformization theorem from the viewpoint of computability theory, see MR2983724 Rettinger, Robert(D-HGNMC) Compactness and the effectivity of uniformization. (English summary) How the world computes, 626–625, Lecture Notes in Comput. Sci., 7318, Springer, Heidelberg, 2012.<|endoftext|> TITLE: stability of linear systems of quadrics QUESTION [5 upvotes]: A pencil of quadrics in $\mathbb{P}^n$ is a line in $\mathbb{P}^N$, where $N=\frac{n(n+3)}{2}$. So the space of pencil of quadrics is the Grassmannian $Gr(2,N+1)$. The group $SL_{n+1}(\mathbb{C})$ acts on $Gr(2,N+1)$ by $T\circ (a_1A+a_2B)=a_1T^tAT+a_2T^tBT$ for any $T\in SL_{n+1}(\mathbb{C})$ and pencil $a_1A+a_2B$.(Here, $A,B$: symmetric $(n+1)\times (n+1)$ matrices corresponding to quadrics.) Thinking $Gr(2,N+1)$ as embedded into projective space via $Pl\ddot{u}cker$ mapping we have the notion of stability of GIT for pencils of quadrics. And $D(a_1,a_2):=det(a_1A+a_2B)$ is a polynomial of degree $n+1$ in the two variables. $D(a_1,a_2)$ has roots in $\mathbb{P}^1$. There is natural $SL_{2}(\mathbb{C})$ action. From this paper, "stability of pencils of quadrics in $\mathbb{P}^n$"="stability of $n+1$ points in $\mathbb{P}^1$." I want to know I can tell same(or similar) thing about "stability of linear systems(net, web or any linear system) of quadrics". Can You give any results or references? REPLY [3 votes]: This is a partial answer. There are many papers on nets of quadrics (see e.g. work of Debarre and Beauville). Here the discriminant locus is a plane curve. If the base-locus of the net is non-singular, then this curve is stable, but need not be non-singular in general (e.g. singular discriminant curves arise from a net generated by sufficiently general diagonal quadrics). I don't know whether the base-locus being stable implies that the discriminant locus is stable, or conversely. People are certainly interested in the case where the base-locus is singular however, again I would recommend the papers by the above authors as a good starting place. For higher-dimensional linear systems of quadrics, things get increasingly more complicated and much less is known. I would be surprised if the results you wanted were known in such cases. Certainly the case of nets is the best place to look first. REPLY [3 votes]: The paper "Stability of genus five canonical curves " by Fedorchuk and Smyth (arXiv:1302.4804) contains a detailed analysis of the (semi-)stability of nets of quadrics in $\mathbb{P}^4$, and completely describes the GIT quotient. In particular, they show that the (semi-)stability of a net cannot be read off the (semi-)stability of the disriminant locus (see for instance Remark 3.21). [Fedorchuk and Smyth's motivation is to construct a birational model of $\overline{M}_5$, and to give a modular interpretation of it. This work is part of the general idea to try to run explicitely the MMP for moduli spaces of curves.]<|endoftext|> TITLE: Obtaining a lightface pointclass from a boldface one QUESTION [9 upvotes]: Define a pointclass to be: boldface inductive-like if it is $\mathbb{R}$-parameterized, has the scale property, and is closed under $\wedge$, $\vee$, $\forall^\mathbb{R}$, $\exists^\mathbb{R}$, and preimages by continuous functions, and lightface inductive-like if it is $\omega$-parameterized, has the scale property, and is closed under $\wedge$, $\vee$, $\forall^\mathbb{R}$, $\exists^\mathbb{R}$, and preimages by continuous functions. Given a lightface inductive-like pointclass $\Gamma$ we can define the boldface inductive-like pointclass ${\bf \Gamma} = \bigcup_{x \in \mathbb{R}} \Gamma(x)$ as usual (this does not have much to do with the particulars of "inductive-like.") My question is, can we do the reverse? That is, given a boldface inductive-like pointclass ${\bf \Gamma}$ can we find a lightface inductive-like pointclass $\Gamma$ such that ${\bf \Gamma} = \bigcup_{x \in \mathbb{R}} \Gamma(x)$? (We wouldn't expect uniqueness; if $\Gamma$ works then so does $\Gamma(z)$ for any real $z$.) I have heard that Howard Becker proved something like this, but I can't find it and I don't know what kind of pointclasses it was for (probably more general than "inductive-like," which is merely the case I happen to be interested in at the moment.) For a motivating example, consider the case where $\mathsf{AD}$ holds and $\bf \Gamma$ is the pointclass of $\kappa$-Suslin sets. If $\bf \Gamma$ is closed under $\forall^\mathbb{R}$ then it is boldface inductive-like, but as far as I know there is no canonical choice of lightface pointclass that corresponds to it. REPLY [4 votes]: See Lemma 3.4 and preceding paragraph of my paper "A characterization of jump operators", JSL vol. 53, 1988. Howard Becker<|endoftext|> TITLE: Clarification about Tits' article in the Corvallis QUESTION [5 upvotes]: I am studying Tits' article in the Corvallis wherein he defines the apartment in the general case (not necessarily split). I wish to know what he means about the filtration of the groups $U_a(K)$ (Section 1.4, page 32); in particular the following sentences: Let $\alpha(a,u)$ denote the affine function on $A$ whose vector part is $a$ and whose vanishing hyperplane is the fixed point set of $r(u)$ and let $\Phi'$ be the set of all affine functions whose vector part belongs to $\Phi$. For $\alpha \in \Phi'$, we set $X_\alpha = \{u \in U_a(K) : u=1 \text{ or } \alpha(a,u)\geq \alpha\}$. Here, $a$ is a root and $u$ is a non-identity element of $U_a(K)$. Also, $r(u)$ is the value (additive valuation) of the unique element in the intersection $U_{-a}uU_{-a} \cap N_G(S(K))$. I am attaching the Google Books link (p.32) below. http://books.google.com/books?id=ob-oRBICenMC&printsec=frontcover&source=gbs_ge_summary_r&cad=0#v=onepage&q=Filtration%20of%20the%20groups&f=false I understand he wants to index subgroups in a systematic way but the exposition seems a bit convoluted. REPLY [5 votes]: I think Tits could use better notation and define terminology more accurately. This Corvallis article, being the standard reference must have, and will continue to confuse many. I finally understand what's going on so let me elaborate. Notations as in Tits' article. One has an exact sequence $\require{AMScd} \begin{CD} 0 @>>> \Lambda = Z(K)/Z_c @>>> \tilde{W}=N(K)/Z_c @>>> {}^\nu\tilde{W} = N(K)/Z(K) @>>> 0 \\ \end{CD} $ which gives rise to an action of the normalizer $N(K)$ on the affine space $A$ (which is $V$ as a set) is given from the following commuting diagram (which can be found in Landvogt's book ``A compactification of the Bruhat-Tits building"). $\require{AMScd} \begin{CD} 0 @>>> \Lambda @>>> \tilde{W} @>>> {}^\nu\tilde{W} @>>> 0 \\ & @VV{\nu}V @VVV @VV\text{reflection}V & \\ 0 @>>> V @>>> \text{Aff}(A) @>>>GL(V) @>>> 1 \end{CD} $ Any element $n \in N(K)$ gives an affine action on the apartment $A$ as follows. Using the top exact sequence, one writes $n = a.z$ with $a \in {}^\nu\tilde{W}$ and $z \in Z(K)$. What Tits calls the "vector part" $r_a$ (notation abused as $a$ sometimes) is the reflection on $V$ induced by the the Weyl group element $a$. The translation is then $\nu(z)$. The map $\nu$, initially defined for $S(K)$ extends uniquely to $Z(K)$ because their character groups are commensurable. (I recommend the example of $G=SU(3)$ to see that $S(K)$ and $Z(K)$ need not be equal.) The affine action of $N(K)$ on $A$ can be computed knowing the reflection (vector part) and translation $\nu(z)$. (The example of $GL_{n,D}$ computed on the last line on p.39 is instructive.) Choosing a non-identity element $u\in U_a(K)$ gives an element $m(u) \in N(K)$ acting on $A$ as above. How do we define the affine function (functional, rather) $$ \alpha(a,u) : A \to \mathbb R?$$ Well, $\alpha(a,u) = \langle a, \circ \rangle + c$ for some real number $c$. If only we could find $c$! The map $r(u) = \nu(m(u)) : A \to A$ has a fixed point set which is a hyperplane, say $H_u$. Solving the equation $$ \langle a, x \rangle + c = 0 $$ for some (every) $x \in H_u$ gives the value of $c$. I highly recommend the example 1.14 of $GL_{n,D}$.<|endoftext|> TITLE: Wander distance of self-avoiding walk that backs out of culs-de-sac QUESTION [11 upvotes]: Suppose a self-avoiding walk on $\mathbb{Z}^2$, with random steps each one unit long, backs out of culs-de-sac, but retaining the lattice points on which it stepped marked as unavailable for future steps. For example, in (a) below, a cul-de-sac is reached. Unwalking two steps allows the path to escape and walk in another direction in (b); but note those two points are marked. Eight steps further another cul-de-sac is reached (c). And so on.   The walk can continue forever, because once it unwinds to a point on the bounding box, it is free to step outside that box. My question is: What is the growth rate of the distance of the path endpoint from the origin, with respect to the total length of (number of steps in) the path? Perhaps this model has been studied? Here's a longer example:           (Added.) I can't resist one more, extending the above to 11,371 steps:           (Added 29Agu14.) Permit me to point to the new paper j.c. found and includes in a comment to his (knowledgeable) answer. That paper calls these walks SKSAWs: Smart Kinetic Self-Avoiding Walks. REPLY [6 votes]: This process is quite natural and has in fact been studied under the name of the infinitely growing self-avoiding walk in a paper by Kremer and Lyklema (see also this paper by Birshtein, Buldyrev and Elyashevitch who called it the "infinitely prolonging walk"). Later Weinrib and Trugman wrote a paper introducing smart kinetic walks (which are IGSAWs which can close up on themselves) and connected the whole family to walks around critical percolation clusters - thus your process should be very much related to the critical percolation exploration process. See also this paper by Coniglio, Jan, Majid and Stanley. I think similar arguments are described in sections 0.4 and onwards of Lawler's book "Conformally Invariant Processes in the Plane". Coniglio et al quote results from percolation which imply that if $N$ is the number of steps taken which don't go into the cul-de-sacs, then the root mean squared end to end distance scales like $N^{4/7}$. Here's a picture of the percolation explorer on the hexagonal lattice, from this page: The critical percolation explorer is conjectured to have SLE${}_6$ as its scaling limit, and so that in some sense should be the scaling limit of your curves as well. This was famously proved by Smirnov for the hexagonal lattice and is still open for the square lattice.<|endoftext|> TITLE: When is it $C(X)$? QUESTION [5 upvotes]: Suppose that $\tilde{X}$ is a compact space. If $C(\tilde{X})$ is isometrically isomorphic to the second dual of a Banach space, does there exist a locally compact space $X$ such that $C(\tilde{X})=C_0(X)^{**}$? P.S. Note that the converse is always true, namely if $X$ is a locally compact space, there is a compact space $\tilde{X}$ such that $C_0(X)^{**}=C(\tilde{X})$. The Banach space $C_0(X)^{**}$ turns into a Banach algebra with the first (second) Arens product. This is a non-trivial product on $C_0(X)^{**}$. REPLY [5 votes]: This is a detailed version of a previous comment that gives a partial answer. If $K$ is a compact space and $C(K)$ is isometrically isomorphic to $X^{**}$, then $X$ is a $\mathcal{L}_{\infty}$-space. Hence $X^{**}$ is injective. It was proved by R. Haydon [Israel J. Math. 31 (1978), 142-152] that an injective bidual space is isomorphic to $\ell_\infty(\Gamma)$ for some set $\Gamma$. Hence $C(K)$ is isomorphic to $c_0(\Gamma)^{**}$. I do not know if $C(K)$ is isometrically isomorphic to $c_0(\Gamma)^{**}$.<|endoftext|> TITLE: Relations between Stiefel-Whitney classes QUESTION [8 upvotes]: I need to know all relations between Stiefel-Whitney classes for closed manifolds of dimensions 3 and 4. Unfortunately, I found the literature on the subject quite confusing. The answer for all dimensions appears to be contained in E. H. Brown and F. P. Peterson, Bull. AMS 69 (1963), p. 228, but I found it rather cryptic. REPLY [18 votes]: What is cryptic? All relations follow from $u_i=0$ for $2i>\dim X$, where $$ u=\operatorname{Sq}^{-1}w=1+w_1+(w_2+w_1^2)+w_1w_2+(w_4+w_1w_3+w_2^2+w_1^4)+\ldots $$ is the total Wu class. (The reason is the fact that $(\operatorname{Sq}^px)[X]=(u_p\smile x)[X]$ for any class $x\in H^{n-p}(X)$, $n=\dim X$.) Explicitly, in small dimensions we have: $w_1^2+w_2=w_1w_2=0$ for $\dim X=3$, $w_1w_2=w_1^4+w_2^2+w_1w_3+w_4=0$ for $\dim X=4$. Of course, these relations generate an ideal in the algebra of Stiefel--Whitney classes invariant under the Steenrod operations. Hence, in dimension $3$ we also have $w_3=w_1^3=0$, i.e., all classes of dimension $3$ are trivial (any $3$-manifold is null cobordant, Rokhlin's theorem). In dimension $4$, we get, in addition, $w_1w_3=w_1^2w_2=0$.<|endoftext|> TITLE: Are there non-diagonal proofs for Cantor's continuum and Godel's incompletness theorems? QUESTION [12 upvotes]: There is a formal definition for the notion of a formal proof. Question 1. Is there any formal definition for the notion of a diagonal formal proof? Consider the following theorems both proved by diagonal proofs. Theorem 1. (Cantor) $2^{\aleph_{0}}>\aleph_{0}$ Theorem 2. (Godel) $ZFC\nvdash Con(ZFC)$ Question 2. Are there any known non-diagonal proofs for these theorems? There are proofs for non-existence of formal proofs for some mathematical statements. Now Assuming an affirmative answer for the question 1, Question 3. Are there proofs for non-existence of non-diagonal formal proofs for some mathematical statements? Are Cantor and Godel's theorems in this category? REPLY [4 votes]: About 2 (2nd Gödel) see http://andrescaicedo.files.wordpress.com/2010/11/2ndincompleteness1.pdf, quoted in What are some proofs of Godel's Theorem which are *essentially different* from the original proof? (the zoo of Gödel's proofs).<|endoftext|> TITLE: Cosimplicial commutative rings in stable homotopical algebra QUESTION [8 upvotes]: When doing "derived commutative algebra" over a discrete commutative ring $R$ that doesn't contain $\mathbb{Q}$, it's fairly well known that you generally have two flavors of "totally commutative object," given by simplicial commutative $R$-algebras on the one hand, and $E_\infty$-algebras in connective complexes over $R$ (equivalently connective $E_\infty$-algebra spectra over $HR$) on the other. The former category maps to the latter, and there are various ways to describe its image, as detailed in the answers to the question "What is a simplicial commutative ring from the point of view of homotopy theory? " I'm interested in the dual question involving cosimplicial commutative $R$-algebras and coconnective $E_\infty$-algebra spectra over $HR$. There seems again to be a natural functor given by co-Dold-Kan, but I have no feeling for the objects it picks out. Are there similar intrinsic characterizations of the image of this functor as in the connective case? I'd be happy to know anything about this, even/especially when $R$ is a finite field. REPLY [4 votes]: Take $R = \mathbb{F}_2$. This isn't anywhere near a characterization, but here is a necessary condition that the $E_\infty$-ring associated to a cosimplicial $\mathbb{F}_2$-algebra must satisfy (that is not always satisfied for coconnective $E_\infty$-algebras under $\mathbb{F}_2$). Namely, if $R$ arises as the totalization $R = \mathrm{Tot} (R^\bullet)$ of a cosimplicial diagram $R^\bullet$ of discrete commutative $\mathbb{F}_2$-algebras, then $R$ is the homotopy inverse limit of the tower $\{ \mathrm{Tot}_n(R^\bullet)\}$. Observe that each $\mathrm{Tot}_n(R^\bullet)$ has homotopy groups concentrated in degrees $[-n, 0]$. Moreover, the map $R \to \mathrm{Tot}_n(R^\bullet)$ induces an isomorphism on homotopy groups in the range $(-n, 0]$. A consequence is that any $x \in \pi_{-n-1}(R)$ is mapped to zero in $\pi_*(\mathrm{Tot}_n(R^\bullet))$. Let ${Q^2}$ be the second Dyer-Lashof operation, which raises the degree by two. If $Q^2$ carries $x$ to a nonzero element in $\pi_{-n + 1} (R)$, we obtain a contradiction by naturality of the Dyer-Lashof operations and the map $R \to \mathrm{Tot}_n(R)$. In the most common class of examples of coconnective $E_\infty$ $\mathbb{F}_2$-algebras (namely, cochain algebras on spaces), $Q^2$ acts by zero, but this is not necessary. Consider the free $E_\infty$-ring on a generator in degree $-3$; this is $\bigoplus_{n \geq 0} (\mathbb{F}_2^{\otimes n})_{h \Sigma_n}$. Now truncate this by throwing away everything with $n \geq 3$ and making the multiplication "cube zero." Finally, replace the top term, i.e., $(\mathbb{F}_2^{\otimes 2})_{h \Sigma_2}$, with its $\tau_{\leq 0}$ (which it maps to). The result is an $E_\infty$-ring under $\mathbb{F}_2$, but $Q^2$ of the class in degree $-3$ is not zero. Therefore, this $E_\infty$-ring does not arise as a cosimplicial $\mathbb{F}_2$-algebra.<|endoftext|> TITLE: general position lemma for tangent lines of an algebraic curve QUESTION [5 upvotes]: Let $X$ be a smooth irreducible algebraic curve in $\mathbb{P} V$. The general position lemma states that the points given by general hyperplane section of $X$ are "in general position". I'm considering a similar situation for tangent lines. For each point of the general hyperplane section, the tangent line at that point corresponds to a point in $\mathbb{P} ( \wedge^2 V)$. Is it true that these tensors in $\wedge^2V$ span the whole ambient space $\wedge^2 V$? (Of course, one should assume that the degree of $X$ is bigger than $\dim (\wedge^2 V)$). I suspect if this question is related to (the surjectivity of) the Gauss map. REPLY [2 votes]: No, that is not true. Let $V$ be the $4$-dimensional vector space with ordered basis $(\mathbf{e}_0,\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3)$, and let $[X_0,X_1,X_2,X_3]$ be the corresponding homogeneous coordinates on $\mathbb{P}V$. Let $[T_0,T_1]$ be homogeneous coordinates on $\mathbb{P}^1$. For every integer $r\geq 3$, consider the morphism, $$ u_r:\mathbb{P}^1 \to \mathbb{P}V, \ \ [T_0,T_1] \mapsto [T_0^r\mathbf{e}_0 + T_0^{r-1}T_1\mathbf{e}_1 + T_0T_1^{r-1}\mathbf{e}_2 + T_1^r\mathbf{e}_3].$$ The corresponding Gauss map is $$\widetilde{u}_r: \mathbb{P}^1 \to \mathbb{P}(\bigwedge^2V),$$ $$[T_0,T_1] \mapsto [rT_0^{2r-2}\mathbf{e}_0\wedge \mathbf{e}_1 + r(r-1)T_0^rT_1^{r-2}\mathbf{e}_0\wedge\mathbf{e}_2 + r^2T_0^{r-1}T_1^{r-1}\mathbf{e}_0\wedge \mathbf{e}_3$$ $$ + r(r-2)T_0^{r-1}T_1^{r-1}\mathbf{e}_1\wedge \mathbf{e}_2 + r(r-1)T_0^{r-2}T_1^r\mathbf{e}_1\wedge\mathbf{e}_3 + rT_1^{2r-2}\mathbf{e}_2\wedge \mathbf{e}_3]. $$ In particular, the coefficients of $\mathbf{e}_0\wedge\mathbf{e}_3$ and $\mathbf{e}_1\wedge\mathbf{e}_2$ are proportionate, i.e., the following "linear form" vanishes on the image curve, $$(r-2)X_0\wedge X_3 - rX_1\wedge X_2.$$ Notice that $r$ can be arbitrarily positive.<|endoftext|> TITLE: Expressive power of first-order category theory QUESTION [5 upvotes]: Given the signature $\lbrace \mathsf{dom}, \mathsf{cod}, \mathsf{id},\circ \rbrace$ and the axioms of category theory – which are expressible in the signature's first-order (FO) language – I wonder which (relevant) properties of categories as a whole which properties of and relations between individuals (= objects and morphisms) inside a category which other concepts and constructions are expressible in category theory's first-order language – and which are genuinely not.1 (For graph theory there is a highly elaborate investigation of expressibility of properties in different languages - FO, SO, MSO - basically driven by Bruno Courcelle.) The following properties are easily seen to be FO-expressible: being a initial/terminal object being a product/coproduct being a monic/epic (morphism) being a groupoid (category) What - among other things - is not clear to me is how being a subobject of $A$ might be FO-expressible? Officially a subobject of a given object $A$ is a specific equivalence class of morphisms with codomain $A$ (as a subclass of individuals this is not an FO-definable individual) which corresponds naturally to an isomorphism class of objects (ditto). But maybe being an element of this isomorphism class is FO-expressible, eventually? This becomes interesting when we ask if being connected is a FO-expressible property of objects (= graphs) in the category of finite graphs: Connected graphs are the noninitial objects in the category of finite graphs that can not be expressed as a coproduct of two nonempty subobjects. (Note: As a property of graphs as a whole connectedness is provably not expressible by a closed formula in the FO-language of graphs with signature $\lbrace R\ \rbrace$) On the other hand: Is the property of being a category of finite graphs expressible by a closed formula in the FO-language of categories? 1 Properties of categories "as a whole" are expressed by closed formulas. $n$-ary relations between individuals are expressed by formulas with $n$ free individual variables. REPLY [4 votes]: No, being the category of finite graphs can not be expressed by a formula, or even a set of formulas, in the FO language of category theory. Any set of formulas satisfied by that category also has models of other cardinalities, by the Lowenheim-Skolem theorem.<|endoftext|> TITLE: Hölder's inequality for matrices QUESTION [23 upvotes]: I was wondering if the Hölder's inequality was true for matrix induced norms, i.e. if $$\|AB\|_1 \leq \|A\|_p\|B\|_q, \quad\forall p,q \in [1,\infty] \text{ s.t. } \tfrac{1}{p}+\tfrac{1}{q} = 1.$$ But it seems that this does not hold in general, in fact $$A = \begin{bmatrix}1 & 2\\ 0 & 0 \end{bmatrix}, \; B = \begin{bmatrix}1 & 0\\ 2 & 0 \end{bmatrix}, \; p = 1, \; q = \infty $$ is a simple counterexample, and it is not hard to find similar ones for other choices of $p$ and $q$. Question Does a Hölder-like inequality hold for matrix induced norms? REPLY [25 votes]: There are (at least two) "generalizations" of Hölder inequality to the non-commutative case. One is the so called tracial matrix Hölder inequality: $$ |\langle A, B \rangle_{HS} |= |\mathrm{Tr} (A^\dagger B) | \le \| A\|_p \,\, \| B\|_q $$ where $\| A\|_p$ is the Schatten $p$-norm and $1/p+1/q=1$. You can find a proof in Bernhard Baumgartner, An Inequality for the trace of matrix products, using absolute values. Another generalization is very similar to what you wrote and reads $$ \parallel|AB|\parallel \, \le\, \parallel |A|^p\parallel^{1/p} \,\, \parallel|B|^q \parallel^{1/q} $$ where $|M|:=(M^\dagger M)^\frac12$ and it holds whenever $ \parallel \cdot \parallel$ is a unitarily invariant norm. You can find a proof in the book of Bhatia Matrix Analysis.<|endoftext|> TITLE: Oscillatory integrals of algebraic functions QUESTION [6 upvotes]: Consider an algebraic function $\phi$ on $R^{d}$. By this I mean that there exists a polynomial $P$ with coefficients in $R[x_1,...,x_d]$ (coefficients are polynomials!) such that $P(\phi) = 0$ Let $r$ be a real number >0, I would like to know whether $1/r^{d}$ times $\int_{[-r,r]^d} e^{i \phi}$ tends to zero when $r$ tends to infinity. Of course it would be enough to show that this integral is bounded by $r^{d-\epsilon}$... I am not at all familiar with the subject, but from the literature I looked at, I understand that this is the case when $\phi$ itself is a polynomial. I was wondering whether this holds true for any algebraic function. REPLY [3 votes]: The answer is no. Consider $\phi(x)=1+1/(x^2 +1)$. Then as $r\rightarrow\infty$ your integral approaches $e^i$. If you want instead your function $\phi$ to be integral over the polynomial ring, meaning it satisfies a monic polynomial equation, then the answer is yes as long as $\phi$ is not a constant function. To see this, note that you can by partial integration replace the integrand with $e^{i\phi(x)}\frac{\phi''(x)}{\phi'(x)^2}$. Now if $\phi(x)$ grows like $x^a$ then $\frac{\phi''(x)}{\phi'(x)^2}$ grows like $x^{-a}$. The integrality assumption forces $a>0$, which completes the proof. Note that I assume you want $\phi$ real valued, else all hell breaks loose.<|endoftext|> TITLE: Showing that $2c_1(f_*\mathscr O_X)=-f_*R_f$ on curves, maybe by local fields QUESTION [9 upvotes]: I originally asked this question on Mathematic StackExchange, but it did not seem to be attracting any attention, so now I am trying mathoverflow. I hope it is not too simple or unappropriate a question for this site, and if it is, then please may those in power feel free to delete it. I would like a way to see that $2c_1(f_*\mathscr O_X)=-f_*R_f,$ where $R_f$ is the ramification divisor of a finite morphism of smooth curves $f\colon X\to Y$ over an algebraically closed field of characteristic $0.$ I am aware that this follows directly from Grothendieck-Riemann-Roch and that it is precisely what is proven in exercise IV. 2.6 of Hartshorne's Algebraic Geometry, where it boils down to the local duality for a finite etale morphism. I would like to have a way of seeing it different from those two. For instance this paper states it is an easy calculation imitating Chapter 3.6, Proposition 13 of Serre's Local Fields. I do not see how that applies, so I would be very grateful for an explanation of how methods of local fields can be applied to local calculations in algebraic geometry in general, and specifically how to derive the above equality. That is not to say that I would be any lesss happy with a proof of the above equality by some entirely different methods. Thanks in advance for any help! REPLY [12 votes]: I looked at that section of Local Fields, and I believe this is about the trace pairing. For a finite, flat morphism of Dedekind schemes, $f:X\to Y$, the pushforward sheaf $f_*\mathcal{O}_X$ is a locally free $\mathcal{O}_Y$-module, let's say of (constant) rank $n$. It is also an algebra, i.e., multiplication is a $\mathcal{O}_Y$-module homomorphism, $$ \theta_f : f_*\mathcal{O}_X\otimes_{\mathcal{O}_Y} f_*\mathcal{O}_X \to f_*\mathcal{O}_X.$$ By adjunction of tensor and hom, this is equivalent to an $\mathcal{O}_Y$-module homomorphism, $$ \widetilde{\theta}_f:f_*\mathcal{O}_X \to \textit{Hom}_{\mathcal{O}_Y}(f_*\mathcal{O}_X,f_*\mathcal{O}_X).$$ But the target module, being End of a locally free sheaf, has a trace map in the usual sense of trace of matrices, $$\text{Tr}_{f_*\mathcal{O}_Y}:\textit{Hom}_{\mathcal{O}_Y}(f_*\mathcal{O}_X,f_*\mathcal{O}_X) \to \mathcal{O}_Y.$$ The composition with $\widetilde{\theta}_f$ is a trace map for $f$, $$ \text{Tr}_f:f_*\mathcal{O}_X \to \mathcal{O}_Y.$$ The composition of this trace map with $\theta_f$ is a $\mathcal{O}_Y$-bilinear pairing on the locally free $\mathcal{O}_Y$-module $f_*\mathcal{O}_X$, $$ T_f: f_*\mathcal{O}_X\otimes_{\mathcal{O}_Y} f_*\mathcal{O}_X \to \mathcal{O}_Y.$$ This pairing is often called the "trace pairing". Again, by adjunction of tensor and hom, it is equivalent to an $\mathcal{O}_Y$-module homomorphism, $$\widetilde{T}_f:f_*\mathcal{O}_X \to \textit{Hom}_{\mathcal{O}_Y}(f_*\mathcal{O}_X,\mathcal{O}_Y).$$ In particular, taking the associated $n^{\text{th}}$ exterior power of both sides, i.e., taking the determinant, gives a morphism of invertible $\mathcal{O}_Y$-modules, $$\text{det}(\widetilde{T}_f): \text{det}(f_*\mathcal{O}_X) \to \textit{Hom}_{\mathcal{O}_Y}(\text{det}(f_*\mathcal{O}_X),\mathcal{O}_Y).$$ If $f$ is generically étale, then $\text{det}(\widetilde{T}_f)$ is injective. In this case, by definition, the determinant of $f$ is defined to be the unique invertible ideal sheaf $\mathfrak{d}_f$ of $\mathcal{O}_Y$ such that $$ \text{Image}(\text{det}(\widetilde{T}_f)) = \textit{Hom}_{\mathcal{O}_Y}(\text{det}(f_*\mathcal{O}_X),\mathcal{O}_Y)\otimes_{\mathcal{O}_Y}\mathfrak{d}_f,$$ as subsheaves of $\textit{Hom}_{\mathcal{O}_Y}(\text{det}(f_*\mathcal{O}_X),\mathcal{O}_Y)$, i.e., $\text{det}(\widetilde{T}_f)$ is an isomorphism, $$ \text{det}(\widetilde{T}_f): \text{det}(f_*\mathcal{O}_X) \xrightarrow{\cong} \textit{Hom}_{\mathcal{O}_Y}(\text{det}(f_*\mathcal{O}_X),\mathcal{O}_Y)\otimes_{\mathcal{O}_Y}\mathfrak{d}_f.$$ On the other hand, the different of $f$, $\mathfrak{D}_f$, is defined to be the maximal minimal invertible ideal sheaf on $X$, $$\iota:\mathfrak{D}_f \hookrightarrow \mathcal{O}_X,$$ such that for the associated transpose, $$\iota^\dagger: \mathcal{O}_X \to \textit{Hom}_{\mathcal{O}_X}(\mathfrak{D}_f,\mathcal{O}_X), $$ with its induced pushforward map under $f_*$, $$f_*\iota^\dagger: f_*\mathcal{O}_X \to f_*\textit{Hom}_{\mathcal{O}_X}(\mathfrak{D}_f,\mathcal{O}_X), $$ the $\mathcal{O}_Y$-module homomorphism $\text{Tr}_f$ factors through $f_*\iota^\dagger$, i.e., there exists a (unique) $\mathcal{O}_Y$-module homomorphism, $$\tau_f:f_*\textit{Hom}_{\mathcal{O}_X}(\mathfrak{D}_f,\mathcal{O}_X) \to \mathcal{O}_Y,$$ such that $\text{Tr}_f$ equals $\tau_f\circ f_*\iota^\dagger$. The ramification divisor of $f$, $R_f$, is the unique effective Cartier divisor of $X$ whose associated invertible sheaf equals $\mathfrak{D}_f$. One of the basic computations, Chapter III.3, Proposition 6 on p. 50 of Local Fields, states that the pushforward Cartier divisor $f_*R_f$ equals the effective Cartier divisor on $Y$ associated to the ideal sheaf $\mathfrak{d}_f$, i.e., $$\mathfrak{d}_f = \text{Nm}_f(\mathfrak{D}_f),$$ where $\text{Nm}_f$ is the norm associated to $f$. In particular, the $\mathcal{O}_Y$-module homomorphism $\text{det}(\widetilde{T}_f)$ above now becomes an isomorphism, $$ \text{det}(\widetilde{T}_f):\text{det}(f_*\mathcal{O}_X) \xrightarrow{\cong} \textit{Hom}_{\mathcal{O}_Y}(\text{det}(f_*\mathcal{O}_X),\mathcal{O}_Y)(-f_*R_f).$$ Using adjunction of tensor and hom one last time, this is an isomorphism of invertible $\mathcal{O}_Y$-modules, $$\text{det}(f_*\mathcal{O}_X) \otimes_{\mathcal{O}_Y} \text{det}(f_*\mathcal{O}_X) \xrightarrow{\cong} \mathcal{O}_Y(-f_*R_f).$$ This is the identity that you asked about.<|endoftext|> TITLE: Open subgroups of the etale fundamental group of $P^1_\mathbb Q\setminus\{0,\infty\}$ QUESTION [5 upvotes]: Let $G$ be the etale fundamental group of $P^1_\mathbb Q\setminus\{0,\infty\}$. Then $G$ is isomorphic to a semidirect product of $\widehat {\mathbb Z}(1)$ by $ Gal_\mathbb Q$. Is it true that every open subgroups of $G$ splits as a semidirect product of $\widehat {\mathbb Z}(1)$ by $ Gal_K$ for some finte extension $K$ of $\mathbb Q$? In general, is there a nice description of the extensions $U$: $$1\to \widehat {\mathbb Z}(1)\to U\to Gal_K\to 1?$$ REPLY [2 votes]: The first question seem OK. Here is a sketch. I might have overlooked something. If $H\subset G$ is an open subgroup, then the image of $H$ in $Gal_\mathbb{Q}$ is open so necessarily of the form $Gal_K$. So we get an extension $$1\to H\cap \hat{\mathbb{Z}}\to H\to Gal_K\to 1$$ The kernel $ H\cap \hat{\mathbb{Z}}\subset \hat{\mathbb{Z}}$ is closed and of finite index, so it is $N\hat{\mathbb{Z}}$ for some integer $N>0$. Let $H'\subset G$ be the preimage $Gal_K$. The kernel of $H'\to Gal_K$ is $\hat{\mathbb{Z}}$. We have that $H\subset H'$ and $H\cap \ker(H\to Gal_K) = N\hat{\mathbb{Z}}$. We can identify $H'=\pi_1^{et}(\mathbb{G}_{m,K})$. Let $H''\subset H'$ be the subgroup corresponding to the étale cover $\mathbb{G}_{m,K}\to \mathbb{G}_{m,K}$ given by multiplication by $N$. As a subgroup $H\subset H'$ coincides with $H''$. Thus $H\cong \pi_1^{et}(\mathbb{G}_{m,K})$ which a semi direct product of $Gal_K$ with $\hat{\mathbb{Z}}(1)$.<|endoftext|> TITLE: A weak-mixing, zero entropy measure on the 2-shift which gives equal weight to both symbols QUESTION [7 upvotes]: I am currently sketching a paper in the general area of symbolic dynamics in which I would like to be able to use the following fact: Proposition (proposed): there exists a shift-invariant measure $\mu$ on $\{0,1\}^{\mathbb{Z}}$ such that $\mu$ is weak-mixing and has zero entropy with respect to the shift, and such that $$\mu\left(\left\{(x_i)\in \{0,1\}^{\mathbb{Z}} \colon x_0=0\right\}\right)=\mu\left(\left\{(x_i)\in \{0,1\}^{\mathbb{Z}} \colon x_0=1\right\}\right)=\frac{1}{2}.$$ I would like to know if anyone can suggest a reference to an article or textbook which proves that such a measure exists, or failing that if anyone can think of a very quick and crisp proof. My grounds for believing that the above proposition should be true are that in the set of all shift-invariant probability measures on $\{0,1\}^{\mathbb{Z}}$ equipped with the weak-* topology, there is a dense $G_\delta$ subset all of whose elements are weak-mixing and have zero entropy (see my earlier question for details). It would be somewhat bizarre if this residual set somehow completely missed the one-codimensional affine subspace of measures satisfying the above equation. (It is not hard to show that there are weak-mixing, zero-entropy measures for which the two quantities in the above equation are arbitrarily close to one half, or to construct weak-mixing measures which satisfy the above equation and have arbitrarily small entropy, but I would like to be able to go the whole distance. I am indifferent to the matter of whether or not $\mu$ is also strong-mixing, but in order to have zero entropy it is well-known that it cannot be Bernoulli or Kolmogorov.) Edited to add: all of the answers given below have been extremely helpful. After some thought I have decided to accept Tom's answer since in my opinion it most exactly answers the question as specified, but this is not to overlook the fact that Anthony and RW's answers are also very educational and are somewhat broader in their implications. REPLY [5 votes]: To prevent any further beating around the bush let me explain Anthony's answer - as it is in fact much easier than what he actually wrote. Take any weakly mixing zero entropy transformation $T$ on a probability space $(X,m)$, and take a subset $A\subset X$ with $m(A)=1/2$. Then the symbolic coding of $T$ determined by the partition $(A,X\setminus A)$ satisfies OP's Proposition.<|endoftext|> TITLE: Does a small-area sphere in a 3-manifold bound a small ball? QUESTION [16 upvotes]: Let $M$ be a closed riemannian 3-manifold. I think that the following fact should be true and should have a relatively simple proof, but I cannot figure it out. For every $\varepsilon>0$ there is a $\delta>0$ such that every smooth 2-sphere in $M$ of area smaller than $\delta$ bounds a ball of volume smaller than $\varepsilon$. Roughly, small-area spheres must bound small-volume balls. Note that: If $M\neq S^3$ then $M$ contains spheres that bound regions of arbitrarily small volume that are not balls (just take a spine of $M$ and small regular neighborhoods of it). It suffices to prove that the 2-sphere is contained in a small-volume ball and invoke Alexander theorem. The same fact stated for 3-spheres in $S^4$ would imply the (open) Schoenflies problem (every 3-sphere bounds a 4-ball), since every 3-sphere in $S^4$ can be shrinked to have arbitrarily small area. It is not true in general that a torus of small area is contained in a ball (pick neighborhoods of a homotopically non-trivial knot). REPLY [6 votes]: This is a direct corollary of Federer -Flemming deformation Lemma saying that vary small area sphere can be homotoped to 1-skeleton of the fixed triangulation of the manifold. The dimension assumption is irrelevant. Proof of this Lemma can be found somewhere in Federer's book (I will chase down the precise reference when I can).<|endoftext|> TITLE: Why is the right permutohedron order (aka weak order) on $S_n$ a lattice? QUESTION [25 upvotes]: This is one of those things I never expected to be hard until I tried to prove it. Why is the right permutohedron order (a.k.a. weak Bruhat order, a.k.a. weak order -- not to be confused with the strong Bruhat order) on the symmetric group $S_n$ a lattice? Details: Let $n$ be a nonnegative integer. Consider the symmetric group $S_n$, with multiplication defined by $\left(\sigma\pi\right)\left(i\right)=\sigma\left(\pi\left(i\right)\right)$ for all $\sigma$ and $\pi$ in $S_n$ and all $i \in \left\lbrace 1,2,\cdots ,n \right\rbrace$. The right permutohedron order is a partial order on the set $S_n$ and can be defined in the following equivalent ways: Two permutations $u$ and $v$ in $S_n$ satisfy $u \leq v$ in the right permutohedron order if and only if the length of the permutation $v^{-1} u$ equals the length of $v$ minus the length of $u$. Here, the length (also known as "Coxeter length") of a permutation is its number of inversions. Two permutations $u$ and $v$ in $S_n$ satisfy $u \leq v$ in the right permutohedron order if and only if every pair $\left(i, j\right)$ of elements of $\{ 1, 2, \cdots, n \}$ such that $i < j$ and $u^{-1}\left(i\right) > u^{-1}\left(j\right)$ also satisfies $v^{-1}\left(i\right) > v^{-1}\left(j\right)$. (In more vivid terms, this condition states that whenever two integers $i$ and $j$ satisfy $i < j$ but $i$ stands to the right of $j$ in the one-line notation of the permutation $u$, the integer $i$ must also stand to the right of $j$ in the one-line notation of the permutation $v$.) A permutation $v \in S_n$ covers a permutation $u \in S_n$ in the right permutohedron order if and only if we have $v = u \cdot \left(i, i + 1\right)$ for some $i \in \{ 1, 2, \cdots, n - 1 \}$ satisfying $u\left(i\right) < u\left(i + 1\right)$. Here, $\left(i, i + 1\right)$ denotes the transposition switching $i$ with $i + 1$. (I have mostly quoted these definitions from a part of Sage documentation I've written a while ago. A "left permutohedron order" also exists, but differs from the right one merely by swapping a permutation with its inverse.) It is easy to prove the equivalence of the above three definitions using nothing but elementary reasoning about inversions and bubblesort. This made me believe that everything about the permutohedron order is simple. Now I have read in some sources (which all give either no or badly accessible references) that the poset $S_n$ with the right permutohedron order is a lattice. This is related to the Tamari lattice. (Specifically, there is an injection from the Tamari lattice to the permutohedron-ordered $S_n$ sending each binary search tree to a certain 132-avoiding permutation obtained from a postfix reading of the tree, and there is a surjection in the other direction sending each permutation to its binary search tree. If I am not mistaken, these two maps form a Galois connection.) But I am not able to prove the lattice property! I see some obstructions to the existence of overly simple proofs: The strong Bruhat order is not a lattice. One might think that the meet of two permutations $u$ and $v$ will be a permutation $p$ whose coinversion set (= the set of all pairs $\left(i, j\right)$ of elements of $\{ 1, 2, \cdots, n \}$ such that $i < j$ and $p^{-1}\left(i\right) > p^{-1}\left(j\right)$) will be the intersection of the coinversion sets of $u$ and $v$. This is not the case. A permutation having such a coinversion set might not exist, and the meet has a smaller coinversion set. In particular, it is not always possible to obtain the meet of $u$ and $v$ by bubblesorting each of $u$ and $v$ without ever killing inversions which are common to $u$ and $v$. The permutohedron-order lattice is not distributive. REPLY [3 votes]: A proof is in George Markowsky, "Permutation Lattices Revisited," Mathematical Social Sciences 27 (1994), 59-72. http://www.umcs.maine.edu/~markov/permutationlattices.pdf<|endoftext|> TITLE: Is Van der Waerden's function elementary QUESTION [6 upvotes]: Van der Waerden's function was proved to have elementary upper bound on growth rate. Is the Van der Waerden's function itself elementary in the sense of Kalmar? REPLY [8 votes]: Yes, this should follow from the elementary bound. The point is that having a Kalmar elementary time bound is "closed under" searches through exponentially large collections. Suppose $N=W(r,k)$ is least such that if the integers $\{1, 2, \dots, N\}$ are colored, each with one of $r$ different colors, then there are at least $k$ integers in arithmetic progression all of the same color. Say we have an elementary upper bound $b$ on $N$. Then starting with $n=0$ we just try all $n\le b$, try all possible colorings ($r^n$ many) try all possible $k$-tuples (less than $n^k$ many) and check whether they are same-colored and in arithmetic progression. Once we have an $n$ such that for all colorings we have found such a progression, then $n=N$.<|endoftext|> TITLE: Can Haar measure fail to be bi-invariant without conjugation shrinking a set? QUESTION [7 upvotes]: (This is a slightly reformatted and clarified version of my question from math.SE, since I believe the answer there is wrong and its poster has not responded to my comment in over two weeks.) Let $\: \langle \hspace{-0.02 in}G,\hspace{-0.04 in}\cdot,\hspace{-0.04 in}\mathcal{T}\hspace{.02 in}\rangle \:$ be a locally compact Hausdorff topological group, and $\mu$ be a left Haar measure on $\: \langle \hspace{-0.02 in}G,\hspace{-0.04 in}\cdot,\hspace{-0.04 in}\mathcal{T}\hspace{.02 in}\rangle$. Suppose that $\mu$ is not right-invariant. Does it follow that there is a Borel subset $B$ of $G$ and an element $g$ of $G$ such that $\:\operatorname{closure}\hspace{.015 in}(B)\:$ is compact and for $C$ given by $$C= \{ \; h\in G \:\: : \:\: g^{\hspace{-0.02 in}-1} h\hspace{.02 in}g \: \in \: B \; \}$$ which satisfies $\: C\subset B \;\;$ and $\;\; \operatorname{interior}\hspace{.02 in}(B\hspace{-0.04 in}-\hspace{-0.04 in}C\hspace{.02 in}) \neq \emptyset$ ? (In other words, is there a topological group such that the Haar measures are not bi-invariant but there is no "single-conjugacy" explanation for that fact?) REPLY [7 votes]: I restate your question: is it true that for every non-unimodular locally compact group $G$, there exists a Borel subset $B$ with compact closure and $g\in G$ such that $gBg^{-1}\subset B$ and $B\smallsetminus gBg^{-1}$ has non-empty interior? The answer is no: here is a simple and typical counterexample. Fix $c>1$ and consider the semidirect product $\mathrm{SOL}_c=\mathbf{R}^2\rtimes\mathbf{R}$, where the action of $t\in\mathbf{R}$ is given by $t\cdot (x,y)=(e^tx,e^{-ct}y)$. This group is not unimodular (while $\mathrm{SOL}_1$ is unimodular). Denote by $p$ the projection $\mathrm{SOL}_c\to\mathbf{R}$ modulo the normal $\mathbf{R}^2$. Let us check that your condition fails; assume we have $B$, $g$ satisfying the conditions. Then there exists $t_0$ such that $B'=p^{-1}(\{t_0\})\cap B$ contains at least 2 elements. Since $p^{-1}(\{t_0\})$ is invariant by conjugation, we have $gB'g^{-1}\subset B'$. If $p(g)\neq 0$, then the action of $g$ on $p^{-1}(\{t_0\})$ is an affine transformation whose linear part has two eigenvalues distinct from 1, and thus has a unique fixed point; it is actually conjugate to the linear map $\mathrm{diag}(e^t,e^{-ct})$ for some $t$ and thus the positive orbit of any non-fixed point is unbounded, contradicting $gB'g^{-1}\subset B'$ ($B'$ being bounded and having at least 2 points). Hence $p(g)=0$. Now we see that the conjugation by $g$ acts as a nontrivial translation on each $p^{-1}(\{t\})$ for $t\neq 0$. We deduce that $B$ is contained in $p^{-1}(\{0\})$, contradicting that it has non-empty interior.<|endoftext|> TITLE: quantitative version of the rigidity of the 2-sphere QUESTION [11 upvotes]: I am looking for a quantitaive version of the following theorem: A compact surface with $K\equiv 1$ is isometric to the round sphere. Of course I get the Berger, Brendle-Schoen Theorem which insures that if we are $1/4$ pinching, the surface is diffeomorphic to the sphere. I also know the result of de-Lellis and Muller which asserts that if the $L^2$-norm of the trace-less second fundamental form is small enough, the surface is close (as an immersed object) to the round sphere. Unfortunately it does not seem to imply the following: if $max K/min K$ is small enough, we are close to a round sphere. Does any one know something in that spirit? Thanks REPLY [7 votes]: All hypersurfaces are $n$-dimensional. Suppose $M$ and $N$ are two smooth, strictly convex hypersurfaces, such that $B_r(o)\subset M,N\subset B_R(o)$, where $B_x(y)$ denotes a ball centred at $x$ of radius $y.$ If $$|\frac{1}{K_M}-\frac{1}{K_L}|\leq \varepsilon,$$ then $d_H(M,N')\leq \gamma \varepsilon^{\frac{1}{n+1}}$. Here, $d_H$ denotes the Hausdorff distance and $N'$ is a suitable translation of $N,$ and $\gamma$ depends only on $n,r,R$ (see Schneider, ``CONVEX BODIES: THE BRUNN–MINKOWSKI THEORY"" Theorem 8.5.4). Now suppose the Gauss curvature of the hypersurface $M$ satisfies $a\leq \frac{1}{K}\leq b$. First, We translate $M$ such that its centroid is at the origin $o.$ Second, by Lemma 3 of Cheng-Yau "On the Regularity of the Solution of the n-Dimensional Minkowski Problem", there are $r,R$ depending only on $a,b$ such that $$B_r(o)\subset M \subset B_R(o).$$Third, we may increase $R$ and decrease $r$ such that $$B_r(o)\subset B_{\sqrt[n]{\frac{a+b}{2}}}(o)\subset B_R(o).$$ Fourth, note that $$|\frac{1}{K}-\frac{a+b}{2}|\leq \frac{b-a}{2}.$$ Therefore, $$d_H(M',B_{\sqrt[n]{\frac{a+b}{2}}}(o))\leq \gamma \left(\frac{b-a}{2}\right)^\frac{1}{n+1},$$ where $M'$ is a translate of $M$ and $\gamma$ only depends on $n,a,b.$<|endoftext|> TITLE: An algebraic number is not a root of unity? QUESTION [18 upvotes]: This problem is related to my study of the Burau representation of the braid group $B_3$: I was trying to show that certain "congruence subgroups" are of infinite index. There is an approach that boils it down to the following question: Let $\xi$ be a primitive root of unity of degree $2n>12$ (just in case). Then the roots $\lambda$ of $\chi(\lambda):=\lambda^2+(\xi^2-\xi+1)\lambda+\xi^2$ are not roots of unity. I was particularly interested in the cases $n=7$ or $9$. In both cases, $\deg\Phi_{2n}=6$ (here $\Phi$ is the cyclotomic polynomial) and, hence, $\mathbb{Q}[\lambda]$ has degree at most $12$. There are finitely many (although quite a few) cyclotomic polynomials $\Phi_k$ with $\deg\Phi_k\le12$. For each such polynomial, one computes the $\lambda$-resultant $R(\xi)$ of $\Phi_k$ and $\chi$ and checks that $R(\xi)\ne0\bmod\Phi_{2n}(\xi)$. Clearly, this approach should work for any given $n$, but I have no idea how this can be done "in general". So, here is the question: Is there a smarter way to prove that an algebraic number is not a root of unity? REPLY [11 votes]: Just noticed this question. I think the following is an even more elementary/self-contained proof. First, let $\eta = \xi u$ (so $u$ is a root of unity iff $\eta$ is), and divide by $\xi^2$ to get $$ u^2 + (\xi-1+\xi^{-1}) u + 1 = 0, $$ whose two roots are inverses of each other, so either both of them are roots of unity or neither is. Now since all primitive $(2n)$-th roots of unity are conjugate, $u$ is a root of unity for one of them iff it is a root of unity for all of them. But then $$ \xi - 1 + \xi^{-1} = -(u+u^{-1}) \geq -2, $$ which holds iff $\xi = e^{2\pi i r}$ with $|r| \leq 1/3$. But the primitive $(2n)$-th roots of unity are precisely $e^{2 \pi i r}$ for $r=m/2n$ with $m \in [-n,n]$ coprime to $n$. So it comes down to checking that there is such $r > 2n/3$. But $r$ can be as large as $n-1$ or $n-2$ according as $n$ is even or odd, and that exceeds $2n/3$ once $n>7$, QED. Conversely, if $n$ is one of the few small numbers such that $2n$ is not coprime with any $r \in (n/3, 2n/3)$ then all conjugates of $u$ are algebraic integers of absolute value 1, and thus roots of unity by a theorem of Kronecker; you presumably observed this for $n=5$.<|endoftext|> TITLE: A question about cardinal functions QUESTION [6 upvotes]: In the context of topological groups, one has $w(X)=d(X)\chi(X)$, where $X$ is a topological group and $w$, $d$ and $\chi$ are the weight, the density and the character, respectively. Since $d(X)\leq nw(X)\leq w(X)$, where $nw(X)$ is the least cardinality of a network on $X$, we also have that $w(X)=nw(X)\chi(X)$. Is the hypothesis of $X$ to be a topological group needed to guarantee that $w(X)=nw(X)\chi(X)$? If not, which separation axioms (or other hypotheses) are needed? Thank you. REPLY [6 votes]: None of the standard separation axioms is enough to guarantee $nw(X) \cdot \chi(X)=w(X)$. In fact, in the exercise section of Engelking's chapter on compactness, you can find this construction of a perfectly normal space, known as the Bow-tie Space. Define a topology on the plane by leaving neighborhoods of all points off the x-axis untouched and declaring that a neighborhood of a point $x$ on the x-axis consists of $x$ plus the intersection of the ball of radius $\epsilon$ centered at $x$ with the region bounded by the two lines passing through $x$ and having slopes $\epsilon$ and $-\epsilon$ respectively, for $\epsilon>0$. The Bow-tie Space is first countable and has a countable network because it's the union of two spaces with a countable base: indeed, the topology induced by this space on both the x-axis and its complement in the plane is the Euclidean one. But its weight is continuum. To see that, let $\mathcal{B}$ be any base. For every $B \in \mathcal{B}$, let $S(B)$ be the set of all points of the x axis such that the vertical line passing through them intersects $B$ in exactly one point. $S(B)$ is a discrete subset of the x-axis (with the Euclidean topology), and hence it's countable. But the $S(B)$'s cover the x axis and hence there have to be continuum many distinct $B$'s. Ramiro's observations that network weight coincides with weight for compact and metric spaces can be generalized to locally compact and developable spaces, respectively (a "developable space" is a topological space having a sequence $\{\mathcal{U}_n: n < \omega \}$ of open covers such that $\{st(x, \mathcal{U}_n): n < \omega \}$ is a local base at $x$, for every point $x \in X$, where $st(x, \mathcal{U})=\bigcup \{U \in \mathcal{U}: x \in U \}$. The importance of this class lies mainly in a famous topological problem which turned out to be independent of the usual axioms of set theory: does there exist a normal developable non-metrizable space?). I think it would be interesting to know whether your equality is true for homogeneous spaces.<|endoftext|> TITLE: Are $D^b_{coh}(X)$ and $D^b(Coh(X))$ derived equivalent? QUESTION [10 upvotes]: Let $X$ be a variety. Let $D^b(Coh(X))$ be the derived category of bounded complexes of coherent sheaves on $X$, and $D^b_{coh}(X)$ be the derived category of bounded complexes of sheaves of $\mathcal{O}_X$-modules with coherent sheaves as cohomologies. Similarly, we have $D^b_{qc}(X)$ and $D^b(Qco(X))$ by replacing $coherent$ sheaves to $quasi$-$coherent$ sheaves. It is proved in "Residues and Duality" by Hartshorne (Chapt. II Corollary 7.19) that $D^b_{qc}(X)$ and $D^b(Qco(X))$ are derived equivalent. I was wondering if the same thing is still true for $D^b_{Coh}(X)$ and $D^b(coh(X))$? Hartshorne's proof seems could not be generalized to this case because I feel that any quasi-coherent module might not be embedded to a quasi-coherent injective module. REPLY [11 votes]: For X noetherian this is still true. (Proposition 3.5 in Daniel Huybrechts' book)<|endoftext|> TITLE: Why non-compact Calabi-Yau surfaces are not self-mirror? QUESTION [7 upvotes]: By the work of Gross and Bernard-Matessi, in dimension 3 $T$-duality should be understood as an exchange of positive and negative local model of Lagrangian torus fibrations, at least in its topological sense. Since in dimension 2, all local models of a Lagrangian torus fibration are generic, so we should expect that $X$ and $X^\vee$ should be topologically the same, just like the case of elliptic $K3$ surfaces. The philosophy of Gross and Bernard-Matessi should be able to extend to certain noncompact Calabi-Yau manifolds. For example $X=K_{\mathbb{P}^2}$, in this case Gross has constructed a special Lagrangian fibration $f$ away from an anticanonical divisor $D$, and the discriminant is a trivalent graph. It should be regarded as a positive Lagrangian fibration in the sense of Bernard-Matessi and the critical locus of the dual Lagrangian fibration $f^\vee$ on the mirror of $K_{\mathbb{P}^2}$ should be topologically the mirror curve in the sense of Hori-Vafa. If we want to include the divisor $D$ in the fibration, then $f$ will no longer be special Lagrangian since the volume form $\Omega$ will be singular along $D$. So if we include $D$ in the fibration, which can also be regarded as $D$ is fibered by Lagrangian tori with lower dimension $f|_D:D\rightarrow\partial B$, then the base will be a manifold with boundary, which differs from the case considered by Gross and Bernard-Matessi a little bit. Now let's take away the divisor $D$ and consider the simplest case of $\mathbb{C}^2\setminus D$, by the work of Auroux, it's mirror should be $\mathbb{C}^2$ blow up a point, also taking away a anticanonical divisor $D^\vee$. On the mirror there is an explicit special Lagrangian torus fibration constructed by using symplectic reduction, and the affine structure on $B^\vee$ has only one interior singularity. Away from the boundary, it should be the same with the singular affine structure on the base $B$ of the special Lagrangian fibration of $\mathbb{C}^2\setminus D$. But in this case, $X\setminus D$ and $X^\vee\setminus D^\vee$ are not diffeomorphic with each other. Similar phenomenon happens for other ALE spaces given by resolution of $\mathbb{C}^2/\mathbb{Z}_m$, although $X$ and $X^\vee$ are homologically similar in the sense that $H_2(X,\mathbb{Z})\cong H_2(X^\vee,\mathbb{Z})$. Also notice that these spaces are hyperkahler for dimensional reasons, and hyperkahler rotation is expected to produce diffeomorphic mirrors. Here the divisors $D$ and $D^\vee$ are taken away to avoid the consideration of the superpotentials. If we also include $D$ and $D^\vee$, then $B$ will be diffeomorphic to $\mathbb{R}\times\mathbb{R}_{\geq0}$, and $B^\vee$ is diffeomorphic to $\mathbb{R}_{\geq0}\times\mathbb{R}_{\geq0}$, even the affine structures differ from each other. REPLY [3 votes]: I agree with the above answer. In a bit more detail, let E be a smooth conic in $\mathbb{C}^2$, of the form $xy=1$. Then the mirror given by Auroux's construction states that the mirror to $\mathbb{C}^2\setminus E$ is the complement of some divisor $D$ in $\widehat{\mathbb{C}^2}$. Well, if we say that the point is $(0,1)$, the blow up can be written explicitly as the zero locus of $ut_1=(z-1)t_2$ in $\mathbb{C}^2 \times P^1$. The relevant divisor $D$ is then the union of $t_2=0$ and $z=0$. Thus we have something like $$ uv=z-1 $$ where z is in $\mathbb{C}^*$. This is clearly the same as what we started with. In fact, these two spaces are even algebraically equivalent in this example, which would not be true if you consider the analogous construction for $A_m$ surfaces for $m \geq 1$. You can see this example studied in a lot of detail in the thesis work of Pascaleff. All examples, following Auroux, Gross-Keel-Hacking, ... suggest that for surfaces, if the surface is "log Calabi-Yau," the slogan that hyperkahler rotation is mirror symmetry does seem to work, at least to some extent (for log CY $A_m$ surfaces as above, the mirror is given by a hyperkahler rotation, in general the mirror will at least be diffeomorphic to the original variety). Otherwise, it needs to be corrected along the lines that you suggest (superpotentials, birational transformations, deformations ...). There seems to be an analogue of this for higher dimensional hyperkahler manifolds, but there aren't as many examples which have been explored in the literature.<|endoftext|> TITLE: Groups which are only defined up to conjugation QUESTION [14 upvotes]: I'm trying to understand what the right way is to think about "groups which are only well-defined up to conjugation." Since this is somewhat vague let me clarify it by pointing out the main examples I have in mind: "The" absolute Galois group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ which depends on a choice of algebraic closure and so is only well-defined up to conjugation. Here the usual approach is to study finite dimnesional representations of this group, since the isomorphism class of the representation is a conjugation invariant. The (outer) automorphism group of "the" hyperfinite $\mathrm{II}_1$ factor. Again this depends on a choice of hyperfinite $\mathrm{II}_1$ factor, and so is only well-defined up to conjugation. This group is simple, so the kinds of techniques you use to study the absolute Galois group won't work. As pointed out in comments, and in email by Qiaochu, one way to think about these examples is that instead of a group you have a connected groupoid. If you pick a basepoint a connected groupoid gives a group, but without picking a basepoint you have something somewhat different from a group. But these two examples have something additional in common which is that one doesn't have easy direct access to the points. Or even if you do find a particular way to write down a base point, there are many natural choices of basepoint and there's no constructive way to write down the path between basepoints. Is there a good way to think or talk about this which will make more concrete the intuiton that you're banning yourself from ever fixing basepoints or fixing a path between two points? (The original version of this question was NARQ, so I've edited and rewritten it to try to improve it.) REPLY [3 votes]: I think I can say something about why this problem rears its head particularly in the study of $\operatorname{Gal}( \overline{\mathbb Q}|\mathbb Q)$. As Qiaochu and others have pointed out, a group-up-to-conjugation is more or less a connected groupoid, and this interpretation meets up with the topological intuition. As HJRW points out, the fact that the Galois group appears to us as a groupoid does not distinguish it much from other symmetry groups. Those are naturally groupoids, and we must fix a basepoint to speak of elements of the group. Note also that there is no difficulty in fixing a basepoint of the Galois goup! Simply take the algebraic closure of $\mathbb Q$ inside $\mathbb C$. Grothendieck's dessins d'enfants are essentially a way of studying the Galois group via this basepoint. Instead, I think a key problem is that there is more than one good way of choosing the basepoint. It might be better to work with a more general notion of a basepoint. A basepoint is a functor from the one-object one-morphism category to the groupoid. More generally, we can consider functors from an arbitrary category to the groupoid. For each prime $p$, we have a functor from the groupoid of algebraic closures of $\mathbb Q_p$ to the groupoid of algebraic closures of $\mathbb Q$ - just take all the algebraic elements. Since the Galois group of $\mathbb Q_p$ is not so hard to understand (a least compared to the Galois group of $\mathbb Q$), having one of these is a lot like having a basepoint. Why is this problematic? Because different phenomena are most naturally studied using different basepoints. Remember that, inverse Galois problem aside, we mostly don't study the Galois group as an abstract group. Instead, we study how it relates to other things we care about: Number fields are subgroups of the Galois group, but more importantly, we can relate that subgroup to the arithmetic of that number field. Almost everywhere a Galois group appears in number theory, it's being related to lots of other interesting things. These interesting things usually have aspects that only make sense at a given place. So to study the Galois group, we really need to study it at each place idependently. While these different basepoints exist perfectly constructively, the paths between them do not, and these paths are crucial to get an understanding of the Galois group. So I think there is no mathematical difference with the group of symmetries of the cube, but a metamathematical difference. I know nothing about von Neumann algebras, but I would be interested to hear if there is a similar picture for the hyperfinite $\operatorname{II}_1$ factor. Finally, note that one reason Galois representations are so important is that we have a good tool for constructing Galois representations, that being etale cohomology of algebraic varieties. This is why we consider Galois representations over $\mathbb Q_l$, rather than another field. We also have powerful tools like L-functions and automorphic forms for describing properties of Galois representations. These all make Galois representations a fruitful field, whereas dessin d'enfants has not been so fruitful because there are not so many powerful tools - but there is nothing intrinsic about the origin of the Galois group that forces it to be that way. Maybe we just haven't discovered the right tools yet!<|endoftext|> TITLE: How can I prove that a particular family of graphs is integral? QUESTION [13 upvotes]: I'm working with an infinite family of graphs that seems to always have all integral eigenvalues, and I'd like to find some way to prove that (if it's true). Call the graphs $G_{n,k}$ and define them as follows: For $n, k>1$, the vertices are sequences of length $n$ with exactly one $0$ and the remaining terms taken from $1, 2, \dots, k$. $G_{n,k}$ thus has $nk^{n-1}$ vertices. There are two ways vertices $x$ and $y$ can be adjacent: If they differ in exactly one term For $i \ne j$, if $x_i=y_j=0$, then $x_k=y_k$ for all $k \ne i,j$ An an example, here's a picture of the graph $G_{3,2}$: A counting argument will show that $G_{n,k}$ is $(n-1)(2k-1)$-regular. As another example, $G_{2,k} \cong K_{2k}$. I've looked at over a dozen examples with various small choices for $n$ and $k$ and the eigenvalues (found using Matlab) have always followed these patterns: There will be $\lceil \frac{3n-2}{2} \rceil$ distinct eigenvalues The maximum eigenvalue will be $(n-1)(2k-1)$, then there will be $\lceil \frac{n-1}{2} \rceil$ jumps of size $2k$ between the next biggest distinct eigenvalues, then jumps of size $k$ down to the smallest eigenvalue of $1-n$. For example, the eigenvalues of $G_{3,2}$ are $6, 2^{(3)}, 0^{(2)}, -2^{(6)}$. I've read a few papers on families of integral graphs and it seems like you need a good understanding of the structure of the graphs or their adjacency matrices to show that they are integral. So in this case, is the definition of the graphs alone enough to understand why they might be integral? Are there techniques I can use to try to prove integrality? Thank you. REPLY [15 votes]: $\def\CC{\mathbb{C}}$The specturm is integral. The following trick is very useful in computing spectra of highly symmetric graphs. Let $G$ be a finite graph, let $\Gamma$ be a group of symmetries of $G$, and let $V$ be the set of vertices of $G$. Let $A$ be the adjacency matrix, so $A : \CC^V \to \CC^V$. Then $A$ commutes with the action of $\Gamma$ on $\CC^V$. Therefore, if $\CC^V = \bigoplus W_i$ is the decomposition of $\CC^V$ into isotypic summands, then the spectrum of $A$ is the disjoint union of the spectra of $A$ restricted to the $W_i$. In our case, we'll take $\Gamma = (\mathbb{Z}/k)^n$. For $(g_1, \ldots, g_n) \in \Gamma$ and $(x_1, \ldots, x_n)$ in $G_{n,k}$, we have $\left( (g_1, \ldots, g_n) \ast (x_1, \ldots, x_n) \right)_j = 0$ if $x_j=0$ and $\left( (g_1, \ldots, g_n) \ast (x_1, \ldots, x_n) \right)_j = g_j+x_j \bmod k$ if $x_j \neq 0$. Let $\chi: (g_1, \ldots, g_n) \mapsto \exp(\frac{2 \pi i}{k} \sum g_i c_i)$ be a character of $\Gamma$, where the $c_i$ in $\mathbb{Z}/k$. We will compute the action of $A$ on the corresponding eigenspace of $\CC^V$. Reorder the coordinates so that $c_1=c_2=\cdots=c_j=0$ and the other $c_i$ are not $0$. The corresponding eigenspace has dimension $j$: For $1 \leq r \leq j$, there is a one dimensional space of functions in $\CC^V$ which transform by this character and have the $0$ entry in position $r$. I get that $A$ acts on this $j$ dimensional eigenspace by the matrix whose diagonal entries are $(j-1) (k-1) + (n-j) (-1)= jk -k-n+1$ and whose off diagonal entries are $k$. This matrix has one eigenvalue of $2jk - 2k+n+1$ and all the others are $jk-2k+n+1$. In particular, they are integers.<|endoftext|> TITLE: Non-Polish Lebesgue probability space? QUESTION [5 upvotes]: Any Lebesgue probability space is mod. 0 isomorphic to some Polish probability space (with $\sigma$-algebra the completion of Borel algebra, and some Borel probability). I would like to see an example of a Lebesgue probability space which is not isomorphic to such a Polish space. In other word, I want a Lebesgue probability space $(X,\mathcal A,\mu)$ for which there is no topology $\mathcal T$ such that $\mathcal A$ is the completion of the $\sigma$-algebra generated by $T$ and $X$, endowed with $T$, is a Polish space. (For me a Lebesgue probability space is a complete probability space $(X,\mathcal A,\mu)$ such that there is some Hausdorff, second countable topology $\mathcal T$ which turns $\mu$ into an inner regular Borel probability measure on the completion of $\sigma(\mathcal T)$). Edit. I just noticed that I forgot the most important hypothesis that $\mu$ must be inner regular in the definition of Lebesgue probability space. Sorry about that. So in particular $\mu$ is a compact measure. Also this definition is in fact equivalent to Rohlin's original definition. REPLY [4 votes]: When requesting inner regularity, you hit the ZFC - undecidable Cantor's continuum problem. First note that a Lebesgue space has cardinality at most continuum since a countable collection of subsets separates points. A uncountable polish space has exactly cardinality of the continuum since it has a perfect subset (Cantor - Bendixon). If the continuum hypothesis is false, in the compact metric $[0,1]$ with Dirac's delta measure concentrated in 0 take a subspace $X$ of cardinality $\aleph_1$ containing 0. You have a Radon probability measure on the metric second countable $X$ which cannot support polish topologies since it is uncountable but not of continuum cardinality. If the continuum hypothesis is true, on a Lebesgue space $X$ take a "concassage" (Bourbaki, chap. IX); as measure space, $X$ is direct sum of compact (and metrisable, being second countable) subsets of positive measure and a remainder of (essential) measure 0. The remainder is at most countable or of continuum cardinality, hence one can put on it a polish (even compact metric) topology; the sum is countable (the sum of measures converges to 1), so one obtains a polish (even locally compact second countable) space. [You might want to ask the set theory experts about what happens in the ZF models of Soloway resp. Selah, where dependent choice holds (so one still has the basic theorems about measures, like countable additivity of Lebesgue measure on the real line) but every subset of a Euclidean space is Lebesgue measurable resp. has the Baire property, so a uncountable subset of the continuum has continuum cardinality but cannot be well ordered]<|endoftext|> TITLE: Triangle with largest perimeter in a convex region QUESTION [11 upvotes]: What is the largest value of $r$ such that the following statement is always true? "Let $C$ be a convex region with area $1$. There must exist a triangle contained in $C$ whose perimeter is at least $r$." I don't need the actual largest value of $r$, but a lower bound would be nice. Using the fact that any convex region with unit area must contain a line segment of length $2/\sqrt{\pi}$, it is clear, for example, that $r\geq4/\sqrt{\pi}$ . REPLY [11 votes]: $\bullet$ Every convex region of area $1$ contains a triangle of area at least as large as the area of the equilateral triangle inscribed in the circle of area $1$. Moreover, if the region does not contain a triangle of larger area, then it is an ellipse. [E. Sas, Über eine Extremaleigenschaft der Ellipsen, Compositio Math. 6 (1939), 468–470]. (Actually, Sas proves a more general statement, for $n$-gons with $n\ge3$.) $\bullet$ Among all triangles of a given area the equilateral one, and no other, is of minimum perimeter. $\bullet$ If the maximum area triangle inscribed in an ellipse is equilateral, then the ellipse is a circle. Putting it all together: Theorem. If $C$ is a convex region with area $1$, then there exists a triangle contained in $C$ whose perimeter is at least as large as the maximum perimeter of a triangle inscribed in the circle of area $1$, and if $C$ does not contain a triangle of perimeter larger than that, then $C$ must be a circle.<|endoftext|> TITLE: Randomly walking a leashed dog QUESTION [16 upvotes]: Let a human $h(t)$ random walk on $\mathbb{Z}^2$ by taking a unit-length step at every time step $t$. A dog $d(t)$ on a leash of length $\lambda$ follows $h(t)$, also taking a unit-length step at every time step, but always retaining a Manhattan distance $|x|+|y| \le \lambda$, where $(x,y)=h(t)-d(t)$. So $d(t)$ remains inside the diamond-shaped $L_1$-ball centered on $h(t)$. Just as in real life, it is natural to expect the dog to be closer to the periphery of the $L_1$-ball than to its master, constantly tugging on a taut leash. Indeed when I simulate this process, this situation obtains. Here is an example of a 100,000-step walk, with a leash of length $\lambda=50$, showing a histogram of the $L_1$-distance of $h(t)-d(t)$:       The growth is roughly linear in distance, but is perhaps falling off below linear. My question is: Is there some logic to suggest this distribution should be linear, or instead, sublinear, with respect to distance? Here is the human path (blue) and the dog's wandering (red) which the above histogram summarizes: REPLY [17 votes]: The position of the dog relative to the human is a Markov chain, which is symmetric by inspection (there are three cases to consider: interior squares, "edge squares," and "corner squares"). This Markov chain splits up into two irreducible ergodic components corresponding to even and odd squares. Finally, any symmetric, irreducible, ergodic Markov chain has uniform stationary distribution. Since the number of squares at distance $d$ is $4d$ (except for $d = 0$), it follows that the distribution of the distances will be linear (again except for $d = 0$). Here's how to show symmetry. If you're in an interior square, you move in each cardinal direction with probability $\frac{1}{16}$ and each diagonal direction with probability $\frac{2}{16}$. If you're in an edge square, you move diagonally off the edge with probability $\frac{2}{16}$, cardinally off the edge in each of two directions with probability $\frac{1}{16}$, and along the edge in each direction with probability $\frac{3}{16}$. If you're in a corner square, you move diagonally to an edge in each of two directions with probability $\frac{3}{16}$, and into the interior in one cardinal direction with probability $\frac{1}{16}$. We then observe that the probabilities of moving to each square from each other square are equal to the probabilities of moving the other way. (e.g., if we're on an edge next to a corner, we move into the corner with probability $\frac{3}{16}$ and then move out to that same edge square with the same probability.) REPLY [6 votes]: Consider the random variable $r(t)=d(t)-h(t)$. Now $r(t)$ is a random walk inside the "sphere" of "radius" $\lambda$. The possible steps of $r(t)$ are no longer just the four neighboring cells, but this isn't a crucial difference. Essentially $r(t)$ is just doing a random walk inside the square (constrained not to step outside of it). It should be elementary to show that after running for many iterations, $r(t)$ becomes evenly distributed in the square (well, except for some silly parity issues: $r(t)$ only visits squares whose coordinate sum is even, and at time $t$ the coordinate sum is $\equiv 2t\mod 4$). Now the linear growth of the distance distribution is easy to explain, since the number of points of distance $x$ from zero is a linear function of $x$.<|endoftext|> TITLE: Does there exist a relative compactification with flat boundary? QUESTION [6 upvotes]: Let $S$ be the spectrum of a DVR, $X/S$ a flat finite type affine scheme. Fix a closed immersion $X\subseteq \mathbb{A}^n_S$ and consider the closure $\bar X$ of $X$ in $\mathbb{P}^n_S$. Then $\bar{X}$ is flat over $S$, as $X$ is dense in it. However, the boundary $Y = \bar X - X$ does not have to be flat over $S$ (see the examples below). Does every point of the closed fiber of $X$ have an open neighborhood $X'$ which can be compactified to a projective $\bar X'/S$ with $Y' := \bar X' - X'$ flat over $S$? Example. Take $X$ to be the generic point of $S$, $\bar X = S$ and $Y$ is the closed point of $S$. But as the closed fiber $X$ is empty, the condition is trivially satisfied. Example. Let $X \subseteq{A}^2_S$ be given by the equation $x = \pi y^2$ where $\pi$ is a uniformizer, then $\bar X \subseteq \mathbb{P}^n$ is given by $xz = \pi y^2$ and $Y$ is given by $z = 0 = \pi y^2$, a union of the $z=0$ line in the closed fiber and the section $(1:0:0)$, so not flat. Of course, $X$ is isomorphic to $\mathbb{A}^1_S$, so we can find a compactification for which $Y$ is flat, namely $\bar X = \mathbb{P}^1_S$. REPLY [6 votes]: I believe the answer is no. All of the following will be relative to an algebraically closed field $k$. You may assume $k$ has characteristic $0$, but I think that is unnecessary (use the work of Stefan Schröer). Let $g',g''>2$ be positive, distinct integers. Let $g$ be $g'+g''$. Let $\overline{M}_g$ denote the Deligne-Mumford moduli stack of stable curves of (arithmetic) genus $g$. Let $\overline{M}_g^o\subset \overline{M}_g$ be the maximal open substack that is a scheme. In particular $\overline{M}_g^o$ intersects the boundary divisor $\Delta_{g',g''}$ since $g'\neq g''$; denote the intersection by $\Delta_{g',g''}^o$. Denote the universal curve over $\overline{M}_g^o$ by $$\pi:\mathcal{C}\to \overline{M}_g^o.$$ Over $\Delta_{g',g''}^o$, the restriction of the universal curve is a union of two proper closed subschemes, $\mathcal{C}'$ and $\mathcal{C}''$, intersecting along divisors, $Z'\subset \mathcal{C}'$, respectively $Z''\subset \mathcal{C}''$, such that $$(\pi':\mathcal{C}'\to \Delta_{g',g''}^o,Z'), \ \text{resp.} \ (\pi'':\mathcal{C}''\to \Delta_{g',g''},Z''),$$ is a family of stable, $1$-pointed curves of (arithmetic) genus $g'$, resp. $g''$. Let $\delta_{g',g''}$ denote the generic point of $\Delta_{g',g''}^o$, let $R$ be $\mathcal{O}_{\overline{M}_g^o,\delta_{g',g''}}$, the local ring of $\overline{M}_g^o$ at this codimension $1$ point. Of course $R$ is a DVR. Let $S$ be $\text{Spec}(R)$ with its natural morphism to $\overline{M}_g^o$. Denote the closed point of $S$ by $0$, and denote the generic point by $\eta$. Let $$ \pi_S:\mathcal{C}_S\to S $$ be the base change of $\pi$ over $S$. Denote the closed fiber by $C_0$, with its two irreducible components $C'_0$ and $C''_0$ as above. Let $X$ be $\mathcal{C}_S \setminus C''_0$. By way of contradiction, assume that there exists a Zariski open subset $X'\subset X$ that intersects $C'_0$ and that has a projective compactification $\overline{X}'_S$ over $S$ whose boundary $Y'$ is flat over $S$. In particular, $Y'$ is finite over $S$. Denote by $$u:\widetilde{X}\to \overline{X}'$$ the normalization of $\widetilde{X}$. In particular, the generic fiber $\widetilde{X}_\eta$ is isomorphic to the generic fiber $\widetilde{C}_\eta$, i.e., $\widetilde{X}_\eta \cong \mathcal{C}_\eta$. Moreover, because the closed fiber $\widetilde{X}_0$ is some compactification of an open subset of the genus $g'>0$ curve $C'_0$, in particular $\widetilde{X}_0$ contains no genus $0$ curves. Moreover, $\widetilde{C}_S$ is regular. Thus, by Abhyankar's lemma, etc., the isomorphism of generic fibers extends to an $S$-morphism, $$ v :\mathcal{C}_S \to \widetilde{X}, $$ that is automatically projective. Moreover, this morphism contracts the irreducible component $C''_0$ of the closed fiber of $\mathcal{C}_S$. In particular, for every invertible sheaf $\mathcal{L}$ on $\widetilde{X}$, $v^*\mathcal{L}$ restricts to the trivial invertible sheaf on $C''_0$. In particular, since $\widetilde{X}$ is projective, there exists an invertible sheaf $\mathcal{L}$ that is ample, so that $v^*\mathcal{L}$ is big. This is impossible. By Franchetta's conjecture / Harer's theorem, the Picard group of $\mathcal{C}_\eta$ is generated by the relative dualizing sheaf, and the Picard group of $\mathcal{C}_S$ is generated by the relative dualizing sheaf together with the invertible sheaf $\mathcal{O}_{\mathcal{C}_S}(\underline{C'}_0)$. The restrictions of these invertible sheaves to $C''_0$ are $\omega_{C''_0}(\underline{Z}'')$ and $\mathcal{O}_{C''_0}(\underline{Z}'')$. But these are linearly independent in $\text{Pic}(C''_0)$, i.e., the universal curve over the generic point of $\Delta_{g',g''}$. Since $v^*\mathcal{L}$ has trivial restriction to $C''_0$, $v^*\mathcal{L}$ is trivial. In particular, the restriction of $v^*\mathcal{L}$ to the generic fiber $\widetilde{C}_\eta$ is trivial. However, by construction, it is also big. This is a contradiction, proving that $\overline{X}'$ does not exist. Observe the following points. First, $S$ is not Henselian. Second, the residue field of $S$ is not algebraically closed. Third, this argument only works if $\overline{X}'$ is a projective scheme; the argument fails if $\overline{X}'$ is only a proper algebraic space. So, if you really need something like this to hold, there are still directions to explore.<|endoftext|> TITLE: Finding roots of a recursively given polynomial sequence / eigenvalues of an almost Toeplitz matix QUESTION [5 upvotes]: I would like to find the roots of the polynomial sequence given by a recurrence relation as follows: $V_0(x) = 1-a^2$ $V_1(x) = 1-a^2 - x$ $V_{k \geq 2}(x) = (1+a^2 - x)V_{k-1}(x) - a^2V_{k-2}(x)$ I know that $a \in (0,1)$. Another formulation of the problem could be to find the eigenvalues of the tridiagonal symmetric matrix $C_n$ $C_n = \left[ \begin{matrix} 1-a^2 & -a\sqrt{1-a^2} & & \\ -a\sqrt{1-a^2} & 1+a^2 & -a & \\ & -a & 1+a^2 & \ddots \\ & & \ddots & \ddots \end{matrix} \right]$ Since the matrix is symmetric, the roots are going to be real. I also know that the matrices $C_n$ are positive definite. This is almost like a Chebyshev recursion, but a little bit perturbed. A bonus question: I have some vague memories about the proof of the roots of Chebyshev polynomials that involved an argument that the roots of the consecutive polynomials in the sequence separate each other, and from this, with some additional tools the roots were derived. If someone can point to a location where I can find that proof, it would be really helpful. Edit after the solutions: Based on the trigonometric equations derived by Pietro Mejer, I suspect that there is no closed formula that describes the roots. REPLY [5 votes]: For $k\in\mathbb{N}$ and $0 TITLE: Forms of algebraic varieties QUESTION [8 upvotes]: Let $X$ be an algebraic variety (say, projective, irreducible and smooth), defined over a field $K$, and let $L$ be a Galois extension. I am interested in algebraic varieties $Y$, defined over $K$, such that there exists an isomorphism $\psi\colon X_L\to Y_L$ defined over $L$ (but not over $K$ in general). The map $\psi$ induces a 1-co cycle from $Gal(L/K)$ to the group $Aut(X_L) $ of $L$-automorphisms, and it is easy to check that the image in $H^1 (Gal, Aut(X_L))$ only depends on the class of $Y$, up to $K$-isomorphisms. My question is the following: is every element of $H^1$ obtained? In other words, we have a map from the set of $L$-forms of $X$ to $H^1 (..)$, the map is injective but is it surjective? If no, could it be true in some natural cases, if yes I would be happy to see a reference. REPLY [5 votes]: In case you are interested in seeing a situation where the map is not surjective, consider the set of rational maps of degree $f:\mathbb{P}^1\to\mathbb{P}^1$, modulo the conjugation action of $\phi\in\text{PGL}_2$, i.e., $f^\phi=\phi\circ f\circ\phi^{-1}$. This is the natural action if one is interested in dynamics (iteration of the map $f$). The automorphism group of $f$ is $\text{Aut}(f)=\{\phi\in\text{PGL}_2:f^\phi=f\}$. Two maps $f_1$ and $f_2$ defined over $K$ are isomorphic over $\bar K$ if $f_1=f_2^\phi$ for some $\phi\in\text{PGL}_2(\bar K)$, and the set of twists of $f$ is the set of maps that are $\bar K$-isomorphic to $f$, modulo $K$-isomorphism. Just as in the case you're looking at, the set of twists injects into the cohomology group $H^1(G_{\bar K/K},\text{Aut}(f))$. But the image is not surjective. The image turns out to be exactly the elements of $H^1(G_{\bar K/K},\text{Aut}(f))$ that become trivial in $H^1(G_{\bar K/K},\text{PGL}_2(\bar K))$. You can read about this in the following places: The Arithmetic of Dynamical Systems, Springer, Section 4.9, specifically Theorem 4.79. The Field of Definition for Dynamical Systems, Compositio Math. 98 (1995), 269-304.<|endoftext|> TITLE: General additive function of probability QUESTION [5 upvotes]: Let $H$ be a function of finite sequences of probabilities (non-negative numbers summing up to 1) into real numbers, such that: $H$ is continuous, $H$ is symmetric w.r.t. the order of its arguments, $H$ is additive, i.e. $H(P \otimes Q) = H(P) + H(Q)$, where $P = \{p_i\}_i$, $Q = \{q_i\}_i$, $P \otimes Q = \{p_i q_j\}_{i,j}$. (A physicist would call the last property extensive.) One class of examples is Rényi entropy $H_q$, along with linear combinations of $H_q$ (for different $q$s), what was already pointed out in the original paper: Alfréd Rényi, On Measures of Entropy and Information (1961) [pdf here] Which other, if any, functions $H$ fulfill the above postulates? REPLY [2 votes]: Here is a partial answer. As has been hinted at in the comments, one should expect that the space of functions under question strongly depends on the continuity assumption made. For example, the Rényi max-entropy $H_0$ and more generally all $H_\alpha$ with $\alpha\leq 0$ are not continuous as one of the probabilities tends to $0$. There are also subtler continuity properties that one could impose to exclude certain functions. For example, $H_\alpha$ is continuous with respect to the statistical distance if and only if $\alpha>1$. Or: the change in $H_\alpha(p)$ upon adding a fixed $\varepsilon$-tail to $p$ can be bounded in a $p$-independent manner if and only if $\alpha\leq 1$. What the OP might have had in mind is the following simple continuity requirement: a function of probabilities $F$ is declared continuous if and only if it its restriction to every probability simplex $$ \Delta_n = \{(p_1,\ldots,p_n)\:|\: p_i\geq 0,\: \sum_i p_i=1\} $$ is continuous. Moreover, it is also natural to assume that $F$ is invariant under adding outcomes of zero probability, $$ F(p_1,\ldots,p_n)=F(p_1,\ldots,p_n,0). $$ With this additional assumption, $F$ becomes a function on the inductive limit of the simplices $\Delta_1\subset\Delta_2\subset\Delta_3\ldots$. Let me capture these requirements in a definition: Definition: $F$ is an information measure if it is additive, continuous on every $\Delta_n$, and invariant under permutations and adding zero probabilities. All the Rényi entropies $H_\alpha$ with $\alpha>0$ are information measures, and so is every linear combination. More generally, every integral \begin{equation} F(p) = \int_0^\infty H_\alpha(p) \mu(d\alpha) \end{equation} for a regular measure $\mu$ on $(0,\infty]$ is an information measure. But there are others as well: instead of integrating the Rényi entropies over $\alpha$, we can also differentiate and obtain new information measures. To see what this gives, it makes things easier to get rid of the conventional $(1-\alpha)^{-1}$ factor in the definition of the Rényi entropies, and instead consider the family of information measures $$ Z_\alpha(p):= \log \sum_i p_i^\alpha = (1-\alpha) H_\alpha(p). $$ This is a bit easier to differentiate, and we obtain another family of information measures, $$ \frac{d Z_\alpha(p)}{d\alpha} = \frac{\sum_i p_i^\alpha \log p_i}{\sum_i p_i^\alpha}. $$ This recovers the Shannon entropy at $\alpha=1$. It can alternatively be written as $$ \frac{d Z_\alpha(p)}{d\alpha} = Z_\alpha(p) + \sum_i \frac{p_i^\alpha}{\sum_j p_j^\alpha} \log\left(\frac{p_i^\alpha}{\sum_j p_j^\alpha}\right), $$ where the second term is exactly the Shannon entropy of the 'quenched' distribution $p^\alpha$ defined in terms of taking all probabilities to the $\alpha$-th power and renormalizing. $$ \frac{d Z_\alpha(p)}{d\alpha} = Z_\alpha(p) - S(p^\alpha). $$ (More generally, plugging $p^\alpha$ into any information measure $p\mapsto F(p)$ results again in an information measure, but this doesn't give anything new for the non-Shannon Rényi entropies.) By differentiating again and putting $\alpha=1$ for simplicity, one obtains another interesting information measure: $$ \frac{d^2 Z_\alpha(p)}{d\alpha^2}\bigg|_{\alpha=1}=\sum_i p_i \left(\log p_i\right)^2 - \left(\sum_i p_i \log p_i\right)^2. $$ This is related to the OP's question on higher moments of the surprisal: the second and higher derivatives of $Z_\alpha(p)$ result in second and higher powers of $-\log p_i$. TL;DR. So then, what about the original question? While not every information measure that I am aware of is an integral of Rényi entropies with respect to a regular measure, it is still conceivable that every information measure is the integral of Rényi entropies with respect to a certain kind of distribution $T$, $$ F(p) = \int_0^\infty H_\alpha(p) T(\alpha) d\alpha. $$ Taking $T$ to be a suitable linear combination of derivatives of the Dirac $\delta$ recovers the above examples. Someone who knows more about distributions may be able to say what kind of distribution $T$ needs to be for the integral to exist, and then we can formulate a precise classification conjecture in terms of distributions.<|endoftext|> TITLE: Does $h^1(D)=0$ imply numerical connectedness on K3 surfaces? QUESTION [5 upvotes]: Let $X$ be a complex K3 surface and $D$ an effective divisor on $X$. We shall say: $D$ is connected if its support is connected. $D$ is numerically connected if for any non-trivial effective decomposition $D\sim D_1+D_2$ we have $D_1\cdot D_2\geq1$. Notice that these two notions are not equivalent (if $D$ is not reduced), e.g. $D=2E$ where $E$ is an elliptic curve, is numerically disconnected. Assume $h^1(D)=0$. The ideal sheaf sequence of $D$ yields $h^0(\mathcal{O}_D)=h^0(\mathcal{O}_X)=1$ in cohomology. Hence $D$ is connected. My question is: Does $h^1(D)=0$ imply numerical connectedness? REPLY [5 votes]: Interesting question. I think the answer is yes, let me try to prove it. As you noticed, the ideal sheaf sequence shows that $h^1(D)=0$ is equivalent to the fact that $H^0({\mathcal O}_D)$ is 1-dimensional generated by the constant function $1$. Considering, for every effective decomposition $D=A+B$, the exact sequence $$ 0 \rightarrow {\mathcal O}_B(-A) \rightarrow {\mathcal O}_D \rightarrow {\mathcal O}_A \rightarrow 0 $$ we deduce then that $H^0({\mathcal O}_D \rightarrow {\mathcal O}_A)$ is injective and therefore, since by Riemann-Roch $B$ has arithmetic genus $1+\frac{B^2}2$, $$0=h^0({\mathcal O}_B(-A))\geq \chi({\mathcal O}_B(-A))=-AB-\frac{B^2}{2}$$ so $B^2 \geq -2AB$ and similarly $A^2\geq -2AB$. Assume by contradiction $AB\leq 0$: then $A^2B^2 -(AB)^2 \geq 3(AB)^2 \geq 0$. This implies, by the index theorem, that $AB=0$ and $A$ and $B$ are proportional. Let us then choose a primitive $C \in Pic(X)$ such that $A=aC$, $B=bC$; obviously $C^2=0$ and by Riemann-Roch $\chi(A)=\chi(B)=\chi(C)=\chi(D)=2$. Up to replacing $C$ by $-C$, $C$ is effective, $a$ and $b$ are positive and $h^2(A)=h^2(B)=h^2(C)=h^2(D)=0$. Then $h^0(C) \geq 2$, so $D=(a+b)C \geq 2C \Rightarrow h^0(D) \geq 3 \Rightarrow h^1(D) \neq 0$, a contradiction.<|endoftext|> TITLE: Cocycle condition for equivariant sheaves QUESTION [6 upvotes]: Let $G$ be an affine group that acts on a variety $X$. Equivariant sheaves on $X$ could be defined in the following way. Consider the simplicial space $X_\bullet$ : $X_n := G^n \times X$, $s_0:X_0 \to X_1$ defined by $s_0(x)=(1,x)$ and $d_i:X_n \to X_{n-1}$ defined by $d_0(g_1,\dots,g_n,x) = (g_2,\dots,g_n,g_1^{-1}x)$, $d_i(g_1,\dots,g_n,x) = (g_1,\dots,g_ig_{i+1},\dots,g_n,x)$ if $0 TITLE: Noncommutative geometry and category theory QUESTION [5 upvotes]: The point in which one starts to talk about noncommutative geometry is the Gelfand Najmark theorem. It establishes an equivalence of the catgeories of commutative (non)unital $C^*$-algebras and the opposite category of the (locally) compact Hausdorff spaces. However the theory goes much much further: from the noncommutative analog of the measure theory (von Neumann algebras) to spectral triples which are noncommutative generalisations of (spinor) Riemanian manifolds. The paradigm of the noncommutative geometry is to look at the "space" through some (posiibly noncommutative) algebra which serves as the space of "functions on the space". However I have never heard about the precise definition of noncommutative space in such a way that these spaces forms a category. From the other hand, I've read that $\mathbb{C}$ and $M_n(\mathbb{C})$ and more general the algebra of compact operators should decribe the same "space" while many invariants (in the spirit of homology theories) for these algebras are the same. It suggest that the naive definition of isomorphism between "noncommutative spaces" defined as the isomorphism of the corresponding $C^*$-algebras seems to be to strong. So what is the most resonable way to define objects and morphisms of noncommutative (say topological) spaces? REPLY [6 votes]: If you want Morita equivalences to be the natural notion of equivalence, on the purely algebraic level you can work with the bimodule $2$-category. This is the $2$-category whose objects are rings, morphisms $R \to S$ are $(R, S)$-bimodules $_RM_S$ (composition is given by tensor product), $2$-morphisms are given by $(R, S)$-bimodule homomorphisms. Any $2$-category has a notion of equivalence, and in this $2$-category equivalence is precisely Morita equivalence. (You can also think of this $2$-category as the $2$-category whose objects are the categories of right $R$-modules over rings $R$, whose morphisms are cocontinuous functors between module categories, and whose $2$-morphisms are natural transformations between these. This is $2$-equivalent to the above by the Eilenberg-Watts theorem.) There should be a version of this $2$-category in the C*-world although I don't know what the precise definitions are. In the above definition we can replace rings with $\text{Ab}$-enriched categories and bimodules with $\text{Ab}$-enriched profunctors. We can also replace the bimodule categories above with derived bimodule categories; this is the natural setting for defining things like Hochschild (co)homology, which is actually derived Morita invariant, not just Morita invariant. Combining these two, we can talk about dg-categories and dg-bimodules between them; some authors use dg-categories as the objects of noncommutative (derived? algebraic?) geometry. See, for example, the nLab.<|endoftext|> TITLE: Is the "continuous on compact subsets" characterization of measurable functions actually useful? QUESTION [10 upvotes]: According to Lusin's theorem (and the slightly weaker converse of that result), measurable functions on locally compact topological spaces that are equipped with a regular measure may be characterized as follows: Definition. Let $X$ be a locally compact space and $\mu$ be a regular measure on $X$. We call $f:X\to\mathbb C$ measurable if for every $\varepsilon>0$ and for every compact set $K\subset X$, there exists a compact set $L\subset X$ such that $\mu(K\setminus L)<\varepsilon$, and $f$ restricted to $L$ is continuous. In fact, Bourbaki defines a measurable functions as such in their volume on integration. To me, this definition of a measurable function seems much less intuitive than the one I am used to, where we require that the preimage of every measurable set be measurable. My question is therefore the following: Are there any situations in which this definition of measurability is clearly advantageous or easier to work with? REPLY [4 votes]: Tautological answer: the cases where you know the measure of compact sets (as in section 3 of chapter IX of Bourbaki), but not the class of all measurable sets (which you then can define using the above definition applied to the indicator function of the set). So the next question is: should one take the topology (or even only the compact subsets), and not the class of all measurable set, as starting point? Answer: perhaps. To understand the answer, one can look at the review of Bourbaki's integration by G.A.Edgar, pag. 82-84 of Mathematical intelligencer, 1981 (I have no month in my copy of that review). It is worth reading; among other things one reads: There are valid motives for doing measure theory using Radon measures instead of abstract measures. An important one is the problem of disintegration of measures (or of "regular conditional probabilities"). [...] Another problem with abstract measures concerns the connection between point-maps and set-maps. [...] A third shortcoming of abstract measures is called the "image-measure catastrophe" by L.Schwartz [in his book "Radon measures on arbitrary topological spaces and cylindrical measures"]. [...] On the other hand, there are also many reasons to use abstract measures instead of Radon measures. For these reasons, he cites the preceding reviews of Bourbaki's integration by Halmos and by Munroe in the bull. AMS: 59 (1953) and 65 (1958); you can find them easily on line. He cites another reason, the "barycenter catastrophe". Even A.Weil himself was not completely satisfied with Bourbaki's integration (but for reasons different from the requester); the see Weil's comments in the third volume of his collected papers, where he refers to a compact system of sets as starting point and Schwartz's book for some more details. I only remark that Weil's starting point would provide a connection with one concept categorical topologist studied around 1970: compactly generated (weak) Hausdorff spaces as convenient setting for topology (convenient meaning that the exponential law is available)<|endoftext|> TITLE: What is the name of this type of groups? QUESTION [6 upvotes]: Suppose $A$ is a finite set and $\Sigma=A\cup A^{-1}$. Let $L\subseteq \Sigma^{\ast}$ be a regular language on the alphabet $\Sigma$. Is there a common name for the group $G$ presented as: $$G=\langle A: L\rangle ?$$ Is there a reference text or article on this type of groups? REPLY [7 votes]: By the comment of Derek Holt we can argue for the context-free case as follows: For any $s\in \Sigma^{\ast}$, let $|s|$ denotes the length of $s$ in the monoid $\Sigma^{\ast}$. By the pumping lemma, there exists a $p\geq 1$ such that for any $s\in L$ with $|s|\geq p$ we have $s=uvxyz$, $u, y\neq 1$ and $|vxy|\leq p$ and for all $i\geq 0$, $uv^ixy^iz\in L$. If we let $i=0$, then in $G$, we have $uxz=1$, and hence for any $i\geq 0$, $$ z^{-1}x^{-1}v^ixy^iz=1$$ which is equivalent to $x^{-1}vxy=1$. Hence every relation $s=1$ with $|s|\geq p$ can be replaced by a relation of the form $x^{-1}vxy=1$. Note that $|x^{-1}vxy|\leq p+1$ and hence $$G=\langle A: s\in L, |s|\leq p+1\rangle$$ so $G$ is f.p.<|endoftext|> TITLE: Distances between boundaries in a hyperbolic pants QUESTION [7 upvotes]: Consider a hyperbolic pair of pants with totally-geodesic boundaries of lengths $l_i$ for $i \in \{1,2,3\}$. For any two distinct boundary components, is the length of the shortest geodesic connecting them already determined? If so, is there a simple formula for these lengths as a function of the $l_i$? REPLY [5 votes]: Yes. Cutting along the shortest geodesics $\alpha, \beta, \gamma$ between the pants boundaries produces two congruent right angled hexagons. They are congruent because a right angled hexagon is determined by a triple of sides (in this case, $\alpha, \beta, \gamma).$ This means that each of the hexagons has sides $\alpha, l_1/2, \beta, l_2/2, \gamma, l_3/2.$ Now, the odd sides of a right angled hexagon can be determined from the even sides. See, for example, Fenchel's Elementary Geometry in Hyperbolic Space, available on Google Books, but in any case, the relevant formula (for the hexagon $a, \beta, c, \alpha, b, \gamma$) is: $$\cosh c = \sinh a \sinh b \cosh \gamma - \cosh a \cosh b.$$<|endoftext|> TITLE: $p$-adic periods QUESTION [18 upvotes]: For a variety $X$ defined over $\mathbb{Q}$, there's a (functorial) comparison isomorphism $$ H^i_{dR}(X)\otimes\mathbb{C}\to H^i_B(X,\mathbb{Q})\otimes\mathbb{C}. $$ If we pick $\mathbb{Q}$-bases for $H^i_{dR}(X)$ and $H^i_B(X,\mathbb{Q})$, the matrix entries of the comparison are complex numbers called periods of $X$, or more specifically periods of the motive $h^i(X)$. Is there a comparison of cohomology theories that leads to a definition of 'period' for $p$-adic numbers, so that some but not all elements of $\mathbb{Q}_p$ are periods? I have a specific example in mind: Riemann zeta values $\zeta(k)$, $k\geq 2$ an integer, are periods. An analogue in $\mathbb{Q}_p$ are the values $\zeta_p(k):=L_p(k,\omega_p^{1-k})$ of the $p$-adic $L$-function of Kubota-Leopoldt, where $\omega_p$ is the Teichmuller character. I'm hoping there's a sense in which $\zeta_p(k)$ is a period. For $X$ a sufficiently nice variety over $\mathbb{Q}_p$, if I understand correctly there's a comparison isomorphism $$ H^i_{dR}(X)\otimes B \to H^i_{et}(X,\mathbb{Q}_p)\otimes B, $$ where $B$ is Fontaine's ring of $p$-adic periods. Elements of $B$ which show up as matrix coefficients with respect to $\mathbb{Q}_p$ bases of $H^i_{dR}(X)$ and $H^i_{et}(X,\mathbb{Q}_p)$ should also be called periods. Since both cohomologies in this comparison started out defined over $\mathbb{Q}_p$, every element of $\mathbb{Q}_p$ is a period in this sense. REPLY [2 votes]: There is a notion of $p$-adic period coming from the Frobenius action on $p$-adic cohomology. Suppose $X$ is a smooth variety over $\mathbb{Q}$, and let $p$ be a prime. If there is a smooth model $\tilde{X}$ of $X$ over $\mathbb{Z}_{(p)}$, then $H_{dR}^*(X)\otimes\mathbb{Q}_p$ depends only on the special fiber $\tilde{X}_p:=\tilde{X}\times \mathrm{Spec}\,\mathbb{F}_p$. There are several different cohomology theories to explain this: Monsky-Washnitzer, crystalline, rigid. In particular, the Frobenius endomorphism of $\tilde{X}_p$ induces a linear map $$ F_p:H_{dR}^*(X)\otimes\mathbb{Q}_p\to H_{dR}^*(X)\otimes\mathbb{Q}_p. $$ If we choose a basis for $H^*_{dR}(X)$, we may represent $F_p$ by a square matrix with entries in $\mathbb{Q}_p$, and $\mathbb{Q}$-linear combinations of matrix entries might be viewed as $p$-adic periods of $X$. The $p$-adic zeta values $\zeta_p(k)$ (and more generally $p$-adic multiple zeta values) are $p$-adic periods in this sense. There is a discussion of $p$-adic periods at the end of Section 5.4 of Francis Brown's Notes on motivic periods. Another reference (given by Nils Matthes in the comments) is Go Yamashita's Bounds for the dimensions of $p$-adic multiple $L$-value spaces, in particular Remark 3.9.<|endoftext|> TITLE: Is every graph the center of some other graph? QUESTION [14 upvotes]: The center of a graph $G$ is the set of vertices that minimize the largest distance to vertices in $G$, e.g., in the graph below, that radius is $4$:           Define the center $C$ as the subgraph of $G$ induced by those vertices. I seek to learn constraints on $C$. Is it the case that every graph $C$ is the center of some graph $G$? Or are there constraints on the possible structures of $C$? (Addendum 5Mar14.) Joe Malkevitch asked (personal communication): Is every plane graph the center of some other plane graph? REPLY [22 votes]: I'm just copying my comment above, as it seems to answer the question. For any graph $C$ there exists a graph for which this is the center. Introduce two new vertices $A$ and $B$. Connect all the vertices in $C$ to $A$; and connect all the vertices in $C$ to $B$. From $A$ draw a long line with many vertices on it (the line doesn't have any other edges to $C$), and similarly from $B$ draw a long line with many vertices on it (same number of vertices as the line from $A$, and this number is larger than the maximum length between two vertices in $C$). Now the vertices in $C$ form the center of this new graph.<|endoftext|> TITLE: Is the Modularity Theorem (currently) effective? QUESTION [28 upvotes]: The Modularity Theorem says every elliptic curve over $\mathbb{Q}$ can be gotten from the classic modular curve $X_0(N)$ by a rational map. Here $N$ is the conductor, easily calculable from a polynomial for the curve. Are the coefficients of the map calculable? Unless this has changed lately, the proofs rely on some non-effective results. For example, Brian Conrad's comment at Can you get Siegel's theorem "for free" from modularity and Mazur's Eisenstein Ideal paper? points out that proofs of the modularity theorem (at least at that time) use the Shafarevich conjecture, which was proved by Faltings as a consequence of the once Mordell conjecture -- now Faltings's theorem. At least up to Levin http://arxiv.org/abs/1109.6070 the Shafarevich conjecture is not (yet) effective. But I do not know how that conjecture is used for Modularity. See also Damian Rössler's comment at Effective proofs of Siegel's theorem using arithmetic geometry. To clarify: effective proof is not the same thing as constructive proof. As Noam Elkies's and Qiaochu Yuan's answers say this theorem is effective because you can calculate the coefficients once you know the curve is modular, so the relevant searches will return results. A constructive proof would also require effectiveness at each step to show all the relevant searches indeed return results. That would be a further question. REPLY [30 votes]: Yes. Once you know $E$ is modular, a dominant map $\varphi: X_0(N) \rightarrow E$ can be computed effectively. That's because one can effectively compute (a bound on) $\deg\varphi$. Of course you need to compute a model for $X_0(N)$ for the question to make sense, but we know how to do that and to write $q$-expansions for the coordinates. By integrating the $q$-expansion of the modular form associated to $E$, and using the Weierstrass model of $E$, you can find $q$-expansions of the functions $x$ and $y$ on $E$ to arbitrary precision. This is already implemented in gp, see elltaniyama. Once you have $x$ and $y$ to enough precision you can solve linear equations in undetermined coefficients to recognize $x,y$ as rational functions in your coordinates on $X_0(N)$.<|endoftext|> TITLE: Besicovitch Almost Periodic Functions a subspace of what? QUESTION [5 upvotes]: The common example of a nonseparable Hilbert space comes from the collection of Besicovitch almost periodic function spaces. Starting with $L^p_{\text{loc}}(\mathbb{R})$ we look at those elements finite under the seminorm $$ \|f\|_{M^p} = \limsup_{R\to\infty} \left(\frac1{2R}\int_{-R}^R \left|f(x)\right|^pdx\right)^{1/p} $$ Modding out (and completing?) by the zero elements nets us a Banach space for $1 \le p <\infty$, and a Hilbert space with the natural inner product for $p=2$. We'll call this space $M^p$. We can see that $M^2$ is an example of a nonseparable Hilbert space because the collection $e^{i\xi x}$ is orthonormal for all $\xi \in \mathbb{R}$. We can look at the subspace $B^p\subseteq M^p$ of elements spanned by these functions, called the Besicovitch almost periodic functions. We can see that $B^2\neq M^2$ since there are functions like $$ f(x) = \left\{\begin{align}1 \ \ \ \ \ x\ge0 \\ -1 \ \ \ \ \ x < 0\end{align}\right. $$ which is orthogonal to all $e^{i\xi x}$, and $\|f\|_2 = 1$. Question: I can't seem to find any discussion of $M^p$ independent from $B^p$. Is there a standard name for $M^p$? Is there a convenient description of an orthonormal basis for $M^2$? REPLY [2 votes]: OK. Perhaps a reason $M^p$ is not often studied is: it is not even a vector space. (using the original defintion with lim not limsup.) Define functions $f$ and $g$ as follows: $f(x)=0$ if $x<1$, $f(x)=1$ if $x \ge 1$ and $\{x\}< 1/2$; here, $\{x\} = x-\lfloor x\rfloor$ is the fractional part $f(x)=-1$ if $x \ge 1$ and $\{x\} \ge 1/2$. $g(x)=0$ if $x<1$, $g(x)=f(x)$ if $x \ge 1$ and $\lfloor \log_2(x)\rfloor$ is even, $g(x)=-f(x)$ if $x \ge 1$ and $\lfloor \log_2(x)\rfloor$ is odd. Some graphs: $f(x)$ $g(x)$ $f(x)+g(x)$ But note: $$ \lim_{R\to\infty} \left(\frac{1}{2R}\int_{-R}^R |f(x)|^2\right)^{1/2} = \lim_{R\to\infty} \left(\frac{1}{2R}\int_{-R}^R |g(x)|^2\right)^{1/2} = \frac{1}{\sqrt{2}} $$ both exist, while $$ \lim_{R\to\infty} \left(\frac{1}{2R}\int_{-R}^R |f(x)+g(x)|^2\right)^{1/2} $$ does not exist. In fact (do some computations): $$ \limsup_{R\to\infty} \left(\frac{1}{2R}\int_{-R}^R |f(x)+g(x)|^2\right)^{1/2} =\frac{2}{\sqrt{3}}, $$ $$ \liminf_{R\to\infty} \left(\frac{1}{2R}\int_{-R}^R |f(x)+g(x)|^2\right)^{1/2} =\frac{\sqrt{2}}{\sqrt{3}} $$ added With the "limsup" definition, as suggested by Jean Van Schaftingen, we contradict the parallelogram law, since $$ \|f\|_{M^2} = \|g\|_{M^2} = \frac{1}{\sqrt{2}},\qquad \|f+g\|_{M^2} = \|f-g\|_{M^2} = \frac{2}{\sqrt{3}} $$<|endoftext|> TITLE: "Gross-Zagier" formulae outside of number theory QUESTION [58 upvotes]: The Gross-Zagier formula and various variations of it form the starting point in most of the existing results towards the Birch and Swinnerton-Dyer conjecture. It relates the value at $1$ of the derivative of the $L$-function of an elliptic curve $E$ to the canonical height of a special rational point on $E$. Let me describe the Gross-Zagier formula in the simplest case. Setup. Let $E/\mathbf Q$ be an elliptic curve. According to the modularity theorem, there exists a finite $\mathbf Q$-morphism $p :X_0(N) \to E$ mapping the cusp $\infty$ to the origin of $E$, where $N$ is the conductor of $E$. Let $K\subseteq \mathbf C$ be an imaginary quadratic field, other than $\mathbf Q[i]$ or $\mathbf Q[\sqrt{-3}]$, in which the prime factors of $N$ split. Then we can choose an ideal $\mathcal J$ in $\mathcal O_K$, such that $\mathcal O_K/J \simeq \mathbf Z/N\mathbf Z$. Viewing $\mathcal J$ as a lattice in $\mathbf C$, we can form the elliptic curves $C_\mathcal J = \mathbf C/\mathcal J$ and $C_K = \mathbf C/\mathcal O_K$. While they are a priori elliptic curves over $\mathbf C$, we know from the theory of complex multiplication that they are actually defined over the Hilbert class field $H$ of $K$. Anyways, the inclusion $\mathcal J \subseteq \mathcal O_K$ induces by passage to the quotient an isogeny $C_\mathcal J \to C_K$ of degree $N$, which is precisely the kind of gadget that $X_0(N)$ parametrizes as a moduli space. Thus we get a point in $X_0(N)(H)$. We can take the image of this point via the modular parametrization $p$ to get a point $y_J \in E(H)$. Taking the sum of its Galois conjugates down to $K$ we get a point $y_K \in E(K)$, which, it turns out, doesn't depend on the choice of $\mathcal J$, up to sign and up to torsion. This means that its Néron-Tate height $\hat{h}(y_K)$ is a well-defined non-negative real number, which, by the non-degeneracy of the height pairing, is zero if and only if $y_K$ is a torsion point on $E$. With this setup, Gross and Zagier proved: Theorem: $$L'(E/K, 1) = \hat{h}(y_K) \frac{\iint_{E(\mathbf C)} \omega \wedge i\omega}{\sqrt D},$$ where $\omega$ is the invariant differential on $E$ (suitably normalized), and $D$ is the discriminant of $K$. Since the factor $\frac{\iint_{E(\mathbf C)} \omega \wedge i\omega}{\sqrt D}$ is never zero, we have: Corollary: Suppose that $L(E/K, s)$ has a simple zero at $s=1$. Then $y_K$ is a point of infinite order in $E(K)$, and in particular, $\text{Rank}_\mathbf{Z}E(K)>0$. Thus, we have an example of a situation where the vanishing of $L(E/K, s)$ at $s=1$ implies the existence of an (explicit!) point of infinite order in $E(K)$ -- a behavior which is, of course, predicted by the BSD conjecture (minus the "explicit" part, which comes as a surprise). If we believe BSD, it is natural to make the following conjecture: Conjecture (Gross-Zagier): Suppose that $L'(E/K, 1)$ is nonzero, or equivalently, that $\hat{h}(y_K)$ is nonzero, or still equivalently, that $y_K$ is non-torsion. Then $E(K)$ has rank exactly one, and $\left$ has finite index in it. This conjecture was proven by Kolyvagin. In any case, since $\hat{h}(y_K) = \left$ where $\left<,\right>$ is the Néron-Tate height pairing on $E$, we can rewrite the Gross-Zagier formula as $$L'(E/K, 1) = \left \times C$$ where $C$ is the nonzero constant of the formula. Let me now tell you about something completely different, namely the Minakshisundaram-Pleijel zeta function. (I don't know enough about this, so please forgive any inaccuracies.) Given a compact Riemannian manifold $M$, one has the Laplace-Beltrami operator $\Delta$ on $M$ which generalizes the familiar Laplacian on $\mathbf R^n$. The spectrum of this operator is an important invariant of $M$, and it is encoded in the M-P zeta function $$\zeta(\Delta, s) = \sum_{n=1}^\infty |\lambda_n|^{-s}.$$ It admits a meromorphic continuation to the whole plane, and is holomorphic at $s=0$. Its Taylor coefficients at $0$ around contain geometric data about $M$. Let us specialize to the case where $M$ is a surface; then $$\zeta'(\Delta, 0) = \frac{1}{12}\int_M K dA$$ where $K$ is the Gaussian curvature. According to the Gauss-Bonnet theorem, this is essentially the Euler characteristic, so: $$\zeta'(\Delta, 0) = \chi(M) \times C, \qquad C\neq 0.$$ But there is a more suggestive way of writing the Euler characteristic. The diagonal $D \subseteq M \times M$ determines a cohomology class on $M \times M$; moreover it lies in the piece of the cohomology of $M\times M$ which is self-dual with respect to Poincaré duality, so the expression $\left$ makes sense (it is the "self-intersection" number of the diagonal). As is well-known, $\left$ is precisely the Euler characteristic of $M$. So, we can write $$\zeta'(\Delta , 0) = \left \times C.$$ Corollary: If $\zeta'(\Delta, 0) \neq 0$, then $D$ is a nontrivial cohomology class on $M\times M$. Thus, we have a completely different example of a situation where the non-vanishing of the derivative of a zeta function implies the existence of an explicit non-trivial cohomology class which "accounts" for the non-vanishing (in the first case, viewing a rational point on $E$ as a degree $0$ Galois cohomology class). As René points out, $D$ is always nontrivial, for trivial reasons. What is interesting to me is not so much the nontriviality of $D$, but rather that one can deduce this nontriviality from $\zeta'(\Delta, 0)\neq 0$. Another important difference that I should point out betweeen the two situations is that $\zeta'(\Delta, 0)\neq 0$ does not imply $\zeta(\Delta, 0)= 0$, because there is no analogue of BSD. In fact, $\zeta(\Delta, 0)$ is essentially $\mathrm{vol}(M)$. Nevertheless, I am still curious. Questions: Is the similarity between the two formulas, and between their Corollaries, coincidental? (This question probably doesn't have a precise answer, but I feel that it's worth asking.) And, assuming that the answer to this question is not completely disappointing... Have things like this been pointed out before? Are there more examples of formulas outside of number theory which bear resemblance to the Gross-Zagier formula? REPLY [2 votes]: Roughly speaking, you're really looking at quite similar objects if you take the following perspective: a. The Gross-Zagier formula computes the first derivative of $L(E,s)$. By modularity, one may associate to the elliptic curve a modular form $f$ such that $L(E,s)=L(f,s)$. The resulting formula is due to Waldspurger, was reinterpreted by Jacquet as a `relative trace formula'. In this setting one has the Petersson norm of $f$ instead of the height pairing, and similar the other constants have parallel interpretations. b. The M-P formula that you mention is Selberg's Trace Formula, with a special choice of test function. When you take the derivative with respect to the $s$ parameter, it is again a trace formula (ignoring possible analytic difficulties). For a compact Riemann surface, indeed the main term is the volume factor, but if you pass to a noncompact setting other terms appear. Jacquet's Relative Trace Formula is so-called in comparison to Selberg's. Indeed, the $L^2$ spectrum of your manifold is built out of modular forms, or more generally, automorphic forms. So to answer your questions: No, they are not coincidental! A fruitful approach has been relating special values of (derivatives) of such L-functions to arithmetic geometry, particular ranks of Chow groups or, if you will, motivic cohomology. Attached to this are conjectures due to Deligne, Beilinson, Bloch-Kato, etc. In the more general sense above, yes, this have been pointed out and very profitably explored. Still, there is much left to be proved!<|endoftext|> TITLE: Asymptotic behavior of $\{\text{#}(a,x,b,y)\in \mathbb{N^4}| \text{ }ax+by=n\}$ for large $n$ QUESTION [10 upvotes]: For a fixed natural number $n$ is it possible to obtain an asymptotic for the number of solutions $(a,x,b,y)\in\mathbb{N^4}$ to $ax+by=n$ or equivalently an asymptotic for $\sum_{k=1}^{n-1}d(k)d(n-k)$? Using some heuristics I got that possibly: $$\sum_{k=1}^{n-1}d(k)d(n-k)\sim \frac{3}{\pi^2}\sigma(n)\ln(n)^2$$ I wrote: $$\sum\limits_{k=1}^{n-1}d(k)d(n-k)=\sum\limits_{{(a,x,b,y)\in \mathbb{N^4}}\atop{ax+by=n} }1=\sum\limits_{d\mid n}\sum\limits_{{(a,x,b,y)\in \mathbb{N^4}}\atop{(a,b)=d,ax+by=n}}1 =\sum\limits_{d\mid n}\sum_{{(a,x,b,y)\in \mathbb{N^4}}\atop(a,b)=1,{ax+by=\frac{n}{d}}}1$$ $$=\sum\limits_{d\mid n}\sum\limits_{{(a,b)\in \mathbb{N^2}}\atop{(a,b)=1}}\sum\limits_{{(x,y)\in \mathbb{N^2}}\atop{ax+by=\frac{n}{d}}}1$$ Now I considered a fixed natural number $m$ and looked at the solutions $(x,y)$ to $ax+by=m$ Then noted that $x\leq \frac{m-b}{a}$ and that the solutions $x$ would be of the form: $x\equiv c+a,c+2b,c+3b,c+4b,...c+kb$, so I wrote $c+kb\leq\frac{m-b}{a}\implies k\leq \frac{m-b}{ab}-\frac{c}{b}$ So the largest such $k$ that should satisfy this is $\lfloor{\frac{m-b}{ab}-\frac{c}{a}}\rfloor\approx \lfloor{\frac{m}{ab}}\rfloor$ Then went back and wrote: $$\sum\limits_{k=1}^{n-1}d(k)d(n-k)\approx \sum\limits_{d\mid n}\sum\limits_{{(a,b)\in \mathbb{N^2}}\atop(a,b)=1}\lfloor{\frac{n/d}{ab}}\rfloor\approx \sum\limits_{d\mid n}\frac{n}{d}\sum\limits_{{ab\leq n,(a,b)=1}\atop(a,b)\in \mathbb{N^2}}\frac{1}{ab}$$ $$=\sigma(n)\sum\limits_{ab\leq n\atop(a,b)=1}\frac{1}{ab}=\sigma(n)\sum\limits_{k\leq \sqrt{n}}\frac{\mu(k)}{k^2}\sum\limits_{ab\leq \frac{n}{k^2}}\frac{1}{ab}=\sigma(n)\sum\limits_{k\leq \sqrt{n}}\frac{\mu(k)}{k^2}\sum\limits_{j\leq \frac{n}{k^2}}\frac{d(j)}{j}$$ $$\approx\sigma(n)\sum\limits_{k\leq \sqrt{n}}\frac{\mu(k)}{k^2}\frac{\ln(n/k^2)^2}{2}\sim\frac{\sigma(n)\ln(n)^2}{2}\sum_{k\ge 1}\frac{\mu(k)}{k^2}=\frac{3}{\pi^2}\sigma(n)\ln(n)^2$$ But this doesn't seem right either because I know that, $$\sum_{ax+by\leq m\atop(a,x,b,y)\in \mathbb{N^4}}1=\frac{m^2\ln(m)^2}{2}+(2\gamma-\frac{3}{2})m^2\ln(m)+(2\gamma^2-\frac{5}{2}\gamma-\frac{\pi^2}{12}+\frac{3}{2})m^2+O(m^{1+\theta}\ln(m))$$ Where $\theta$ is the same $\theta$ in Dirichlet's divisor problem. Thus I would expect that if $\sum\limits_{k=1}^{n-1}d(k)d(n-k)\sim C'\sigma(n)\ln(n)^2$ for some constant $C'$ that we would then have that $C'=\frac{6}{\pi^2}$, which it isn't according to the last heuristic. So I'm not really sure how to proceed, and would really appreciate any help in finding an asymptotic expansion for $\sum\limits_{k=1}^{n-1}d(k)d(n-k)$. REPLY [13 votes]: This is a binary additive divisor problem (also known as the shifted convolution problem) and related questions are to understand asymptotics for $\sum_{n\le x} d(n) d(n+k)$, or $\sum_{n\le x} \lambda_f(n) \lambda_f(n+k)$ where $\lambda_f$ denotes the Fourier coefficients of a cusp form. For your particular problem, Ingham first showed the asymptotic formula $$ \sum_{k=1}^{n-1} d(k) d(n-k) \sim \frac{6}{\pi^2} \sigma(n) (\log n)^2. $$ See for example Motohashi's paper http://archive.numdam.org/ARCHIVE/ASENS/ASENS_1994_4_27_5/ASENS_1994_4_27_5_529_0/ASENS_1994_4_27_5_529_0.pdf which discusses this asymptotic formula, with strong error terms.<|endoftext|> TITLE: Hamming weight of Fibonacci numbers QUESTION [9 upvotes]: The Hamming weight $w(n)$ is the number of 1s in $n$ when written in binary. Is there some effective bound on Fibonacci numbers $F_n$ with $w(F_n)\le x$ for a given $x$? Clearly only $F_0=0$ has weight 0, and it's not hard to show (e.g. by Carmichael's theorem) that only $F_1=F_2=1, F_3=2, F_6=8$ have Hamming weight 1. (Pedantry: I'm ignoring negative indices.) It seems that only $F_4=3, F_5=5, F_9=34, F_{12}=144$ have weight 2, but I cannot even prove this. It seems that the fact that $\varphi$ is a Pisot number, and in particular $F_n=\varphi^n/\sqrt5+o(1)$, should/may be related. REPLY [13 votes]: The case $x=2$ is still tractable. If $F_n = 2^e + 2^f$ with $e e$, try each $f$ with $e < f_0 < f$, and then once $f \geq f_0$ we use the condition $F_n = 2^e + 2^f \equiv 2^e \bmod 2^f$ to get a congruence condition on $n$, and then reach a contradiction by considering $F_n$ modulo some odd prime (usually $3$, but with one much larger exception). $e=0$: We take $f_0 = 4$. Trying $f=1$ and $f=2$ yields the Fibonacci numbers $F_4=3$ and $F_5=5$, and $f=3$ yields the non-Fibonacci number $9$. Once $f \geq 4$ we have $F_n \equiv 1 \bmod 16$. But $F_n \bmod 16$ is periodic with period $24$, and it turns out that the remainder is $1$ only for $n \equiv 1, 2, 23 \bmod 24$. But $F_n \bmod 3$ has period $8$, which is a factor of $24$; and $F_1 = F_2 = F_{-1} = 1$. We deduce $F_n \equiv 1 \bmod 3$. Hence $2^f \equiv 0 \bmod 3$, which is impossible. $e=1$: The Fibonacci numbers $F_n$ congruent to $2 \bmod 4$ are those with $n \equiv 3 \bmod 6$, and these always turn out to be $2 \bmod 32$. Thus $f \geq 5$, and $f=5$ yields the Fibonacci number $34 = F_9$. We claim that this is the only possibility, using $f_0 = 6$. Once $f \geq 6$ we have $F_n \equiv 2 \bmod 64$, and then $n \equiv \pm 3 \bmod 24$. But (again thanks to $8$-periodicity mod $3$) this implies $F_n \equiv 2 \bmod 3$, so once more we reach a contradiction from the congruence $2^f \equiv 0 \bmod 3$. $e=2$: impossible because $F_n$ is never $2 \bmod 4$. $e=3$: We take $f_0=5$. Since $2^3 + 2^4 = 24$ is not a Fibonacci number, we may assume $f \geq 5$, and then $F_n \equiv 8 \bmod 32$. This is equivalent to $n \equiv 6 \bmod 24$, which again yields a contradiction mod $3$ since $2^f = F_n - 2^e$ would have to be a multiple of $3$. $e=4$: This is the hardest case: because $f=7$ yields $144 = F_{12}$, it is not enough to use congruences that can be deduced from $F_n \equiv 2 \bmod 2^7$, and we must take $f_0 > 7$. It turns out that $f_0 = 9$ works. Then $f=5,6,8$ yield the non-Fibonacci $48, 80, 272$. Once $f \geq 9$ we must have $F_n \equiv 16 \bmod 2^9$. Now $F_n \bmod 2^9$ has period $768$, but the condition $F_n \equiv 16 \bmod 2^9$ determines $n \bmod 384$ (half of $768$), and we compute $n \equiv -84 \bmod 384$. Now $n \bmod 384$ determines $F_n$ modulo the prime $4481$ (the period is $128$), and we find $F_n \equiv 2284 \bmod 4481$, whence $2^f = F_n - 2^e \equiv 2284 - 16 = 2268 \bmod 4481$. But this is impossible because $2$ is a fourth power (even an $8$th power) mod $4481$, and $2268$ is not. $\langle$ /EDIT $\rangle$ But I doubt that one can prove that such a technique can work for all $x$...<|endoftext|> TITLE: Function Approximation in c.c.c Forcings without AC in Ground Model QUESTION [6 upvotes]: Consider the following basic theorem. Theorem. If $M$ is a c.t.m of ZFC and $\mathbb{P}$ a c.c.c forcing notion in $M$ and $G$ a $\mathbb{P}$ - generic filter on $M$ then for all $A,B$ in $M$ and for all function $f:A\longrightarrow B$ in $M[G]$ there is a function $F:A\longrightarrow P(B)$ in $M$ such that $\forall a\in A~~~f(a)\in F(a)~\wedge~(|F(a)|\leq \aleph_{0})^{M}$. $F$ is an approximation for $f$ in $M$. But in classic proof we use $AC$ in order to prove countability of the set $F(a)$ in $M$ for each given $a\in A$. Question. If we remove $AC$ from the ground model $M$ can we guarantee the countability of the set $F(a)$ for each $a\in A$? Can we put an upper bound for the size of the sets $F(a)$ at all? REPLY [6 votes]: This is a delicate issue, because antichains break easily when you're not using the axiom of choice. More to the point, it is consistent that there is a partial order which satisfies the countable chain condition, simply because every antichain is finite and there are no infinite maximal antichains. In other partial orders there might not be maximal antichains at all. While we're at it, the maximality principle in forcing is equivalent to the axiom of choice, so that one fails to (I mean the fact that $p\Vdash\exists\tau\varphi(\tau)\iff \exists\dot\tau\ p\Vdash\varphi(\dot\tau)$ which is often used in forcing). Moreover, in peculiar forcings as I mentioned above it is generally expected for the maximality principle to fail. Making this a double-trouble problem. But then again, not all is lost. As Mohammad points out, the cardinality of the forcing poset is an upper bound. And that upper bound works out for every function, and every two sets in the ground model. While this is not necessarily an $\aleph$ number, it is still an upper bound in some sense. If $1_P\Vdash\dot f\colon\check A\to\check B$, then every condition must agree that $\dot f$ is a name for a function. Therefore if $p\Vdash\dot f(\check a)=\check b$, then this $b$ is unique. So if we define $F(a)=\{b\in B\mid\exists p\ p\Vdash\dot f(\check a)=\check b\}$, then there is a surjection from $P$ onto $F(a)$, if $p$ doesn't decide the value of $\dot f(\check a)$ then map it to some fixed element; otherwise map it to the value it decided upon. Of course, in the absence of choice, the existence of a surjection need not imply the existence of an injection, so if you want an upper bound in the sense of injections ($\leq$) rather than surjections ($\leq^*$) you need to consider $2^P$ instead. The above can be circumvented if $P$ is already a complete Boolean algebra. In that case we define an injection from $F(a)$ into $P$ defined by $b\mapsto\sum\{p\mid p\Vdash\dot f(\check a)=\check b\}$. Since $P$ is a complete Boolean algebra this condition exists, and it is unique. As Andreas Blass points out, in the case of a complete Boolean algebra which satisfy c.c.c. the different values of $\dot f(\check a)$ form an antichain, which is therefore countable.<|endoftext|> TITLE: Do coherent toposes descend along open surjection? QUESTION [7 upvotes]: Let $f:\mathcal{L} \rightarrow \mathcal{S}$ be a geometric morphism between two toposes. Let $g:\mathcal{T}\rightarrow \mathcal{S}$ be an open surjection (of toposes) and assume that the map $\mathcal{T} \times_{\mathcal{S}} \mathcal{L} \rightarrow \mathcal{T}$ turn $\mathcal{T} \times_{\mathcal{S}} \mathcal{L}$ into a coherent $\mathcal{T}$-topos. Does $\mathcal{L}$ is a coherent $\mathcal{S}$-topos ? I assume that all the toposes involved are Grothendieck toposes, or at least that the geometric morphisms are bounded. Also does it work under different assumptions on the map $g:\mathcal{T}\rightarrow \mathcal{S}$ ? (like if it is a Proper surjection, or an Hyperconnected map) Note: By cohenrent toposes, I mean as defined in section D3.3 of the elephant. Theorem C5.1.7 of the elephant cover a large number of similar "descent" properties for other type of toposes but does not mention Coherent toposes. There is in the work of Moerdijk and Vermeulen (Here and Here) a few things about relative coherent morphisms which goes in this direction, but they do not talk about this kind of descent properties... REPLY [2 votes]: I recently found an answer for coherent locales, which was enough for my purpose although the argument can probably be expanded to an arbitrary coherent geometric morphism (see at the end, if someone can be bring some precision about this it would be very interesting for me). Let $f : \mathcal{T} \rightarrow \mathcal{S}$ be an open surjection. Let $\mathcal{L} \rightarrow \mathcal{S} $ be a locale in $\mathcal{S}$ such that, internally in $\mathcal{T}$, $f^* \mathcal{L}$ is coherent. We will prove that $\mathcal{L}$ is already coherent in $\mathcal{S}$. Internally in $\mathcal{T}$, $f^* \mathcal{L}$ is then the stone spectrum of a distributive lattice $I$, and $I$ can be chosen canonically: it is the set of open compact subspaces of $f^* \mathcal{L}$. In particular, when $\gamma: \mathcal{M} \rightarrow \mathcal{N}$ is an isomorphism between two coherent locales it induces an isomorphism between the corresponding distributive lattice, and hence the descent data on $f^* \mathcal{L}$ gives rise to a descent data on the corresponding distributive lattice $I$. (We are also using the fact that the pullback of a spectrum of a distributive lattice is the spectrum of the pullback of the distributive lattice) Open surjections are of effective descent for sets (objects) hence $I$ (and its descent data) are of the form $f^* I'$ for some $I'$ a distributive lattice (in the base topos). The construction $I \mapsto \text{spec } I$ is compatible with pullback along geometric morphisms hence the isomorphism between $f^* \mathcal{L}$ and $\text{spec } I = f^* \text{spec } I'$ (which is compatible with the descent data simply because we used the isomorphism to construct them) descend into an isomorphism between $\mathcal{L}$ and $\text{spec } I'$ over $\mathcal{S}$, proving that $\mathcal{L}$ is coherent. So one has the answer for coherent locales, and the open surjection can be replaced by any geometric morphism which is of effective descent for locales (claim: a geometric morphism which is of effective descent for locales is also of effective descent for objects). This technique can maybe be extended to coherent toposes, but one needs an argument of descent for categories (and for "week descent data") because the distributive lattice of compact open subspace will be replaced by the pre-topos of coherent objects. I have never seen such a result of descent for categories stated but it will probably be not too difficult to obtain at least for hyperconnected geometric morphism.<|endoftext|> TITLE: Change of coordinates for Teichmüller space of the 4-holed sphere QUESTION [7 upvotes]: The diagram below indicates 2 ways to use Fenchel-Nielsen coordinates to parameterize the Teichmüller space of conformal structures on the 4-holed sphere with totally-geodesic boundary, corresponding to the two pants decompositions, as indicated by this diagram: Here $a,b,c,d,x,y$ are lengths of the indicated geodesics in a hyperbolic instance of the conformal structure, and $\alpha,\beta$ are the twisting parameters for the gluing. What is the coordinate transformation implied by these decompositions? That is, what is the formula for $y$ and $\beta$ as a function of $a,b,c,d,x,\alpha$? REPLY [8 votes]: See the paper "Effects of a change of pants decompositions on their Fenchel-Nielsen coordinates" by Takayuki Okai, published in Kobe J. Math. You can also find a version where some of the boundary components are cusps in the paper "The behaviour of Fenchel-Nielsen distance under a change of pants decomposition" by Alessandrini, Liu, Papadopoulos, and Su, published in Comm. Anal. Geom. and also available on the ArXiv. Added by @JamieVicary: here are the equations as given in the Alessandrini, Liu and Papadopoulos paper in the notation of this question, with $a'=\text{cosh}(a/2)$, $b'=\text{cosh}(b/2)$, $c'=\text{cosh}(c/2)$, $d'=\text{cosh}(d/2)$, $x'=\text{cosh}(x/2)$, $y'=\text{cosh}(y/2)$, $\alpha' = \text{cosh}(\alpha)$ and $\beta' = \text{cosh}(\beta)$: $$\begin{align} x'&=y' ^{-2} ( a'c'+d'b' + y'(a'd' + c'b') + \beta'(y'^2 + 2 a'b'y' + a'^2 + b'^2 - 1] ^{1/2}(y'^2 + 2 c'd'y' + c'^2 + d'^2)^{1/2}) \\ \alpha'&=(a'^2 + c'^2 + 2 a' c' x' + x'^2-1) ^{-1/2} ( d'^2 + b'^2 + 2 d' b' x' + x'^2 - 1) ^{-1/2} ((x'^2 - 1)y' - a'b' - c'd' - x'(a'd' + c'b')) \end{align}$$<|endoftext|> TITLE: Minimize norm of a polynomial around a circle (count the solutions) QUESTION [11 upvotes]: I already posted this question at MSE here, but as it received no significative feedback for a while I cross-post it here. I also noticed a related question here on MO (which does not answer my present question). Let $P=\sum_{k=0}^n a_kX^k$ ba a polynomial of degree $n \gt 0$, and let $r\gt 0$. Suppose that $P$ is not the monomial $a_nX^n$, in other words there is at least an $i TITLE: Starting Hilbert's Program on the other end QUESTION [8 upvotes]: The idea of Hilbert's program was to start with a simple finitary logic and proof the consistency of more complex systems in this system. Of course, this turned out to be problematic. Even when absolute consistency is replaced by relative consistency. But I just wondered, if someone attempted to tried it the other way. So, what is the simplest logical system X, such that X can prove the consistency of ZFC relative to X? In other words, are there things in ZFC that can be dropped from foundational point of view? Is there any scientific article about this? Lucas REPLY [9 votes]: You can drop choice: ZF proves that ZFC and ZF are equiconsistent. You can also drop excluded middle from your logic, in various ways, e.g.: IZF proves that IZF and ZF are equiconsistent. See here.<|endoftext|> TITLE: Morley Phenomena for Special Families of Reals QUESTION [6 upvotes]: Vaught's Conjecture is a dual form of Continuum Hypothesis in model theory. It asserts that for each complete consistent theory $T$ in a countable language if $I(T,\aleph_{0})>\aleph_{0}$ then $I(T,\aleph_0)=2^{\aleph_{0}}$. The best known result here is proved by Morley. For each complete theory $T$ in a countable language if $I(T,\aleph_{0})>\aleph_{1}$ then $I(T,\aleph_0)=2^{\aleph_{0}}$. We call this kind of overspill from $\aleph_1$ to $2^{\aleph_0}$, a Morley phenomena. Question. For which non-trivial properties $P$ is the following true? If $\mathcal{F}$ is a family of reals (subsets of $\omega$) with $|\mathcal{F}|>\aleph_{1}$ and each set $X\in \mathcal{F}$ satisfies the property $P$ then $|\mathcal{F}|=2^{\aleph_0}$. In the other words, for which kind of reals do we have a Morley phenomena for families of that particular type of reals? Is there any correspondence between these type of reals and countable models of a complete theory in a countable language? REPLY [6 votes]: The classical example is ${\mathbf\Pi}^1_1$ sets of reals. We say that $X\subset \mathbb R$ is ${\mathbf\Pi}^1_1$ if the complement of the projection of a Borel set in ${\mathbb R}^n$. Any such set of cardinality greater than $\aleph_1$ has cardinality continuum.<|endoftext|> TITLE: Root space decomposition QUESTION [5 upvotes]: What is the root system for the special unitary lie algebra $\mathfrak{su}(p, q)$. Remind that these are matrices of the form $\left( \begin{array}{cc} X & Y \\ \overline{Y}^t & Z \\ \end{array} \right) $, where $\overline{X}^t=-X$, $\overline{Z}^t=-Z$, and $tr(X)+tr(Z)=0$. Also is there any reference regarding the action of the Lie group $SU(p, q)$ on complex projective space? Thanks. REPLY [7 votes]: $\mathrm{SU}(p,q)$ is known as "type AIII", see e.g. Goodman-Wallach, Helgason, or Knapp which may have the most details. For its action on projective space and other flag manifolds a classic reference is Wolf. (If $pq\ne0$ then by Witt's theorem $\mathrm{SU}(p,q)$ has three orbits $P_+$, $P_-$, $P_0$ in projective space, consisting of the lines on which the defining hermitian form is positive, resp. negative, resp. zero. The former two are open while the latter one is closed.)<|endoftext|> TITLE: How to show twisted complexes over a DG category is again a DG category? QUESTION [7 upvotes]: In Bondal and Kapranov's paper enhanced triangulated categories, a twisted complex over a DG category $A$ is a set $\{(E_i)_{i\in \mathbb Z}, q_{ij}: E_i\to E_j\}$, where $E_i$ are objects in $A$, equal to zero for almost all $i$, and the $q_{ij}$ are morphisms in $A$ of degree $i-j+1$ satisfying $dq_{ij}+\sum_k q_{kj}q_{ik}=0$. Let $C=\{E_i,q_{ij}\}$ and $C'=\{E_i', q_{ij}'\}$ be two twisted complexes. Put $ Hom^k(C,C')=\bigoplus_{l+j-i=k} Hom_A^l(E_i,E_j') $ and, for any $f\in Hom_A^l(E_i,E_j')$ $df=d_Af+\sum_m (q_{jm}f+(-1)^{l(i-m+1)}fq_{mi}),$ where $d_A$ is the differential of the DG category $A$. It seems to me that the differential defined on the twisted complexes does not satisfy $d^2=0$ and the graded Leibniz rule. How do we check it is indeed a well-defined DG category? REPLY [3 votes]: Hope the answer will be still of some importance. Twisted complexes are really not well-defined by Bondal and Kapranov. Fernando Muro, in the new category you need to pay attention to the new grading of $f$, which is $k$, but the old Leibniz rule uses the degree $l = k+i-j \neq k$, so Xingting's argument was correct. To define twisted complexes well, you need to consider a dg category $A$ and formally add there finite direct sums and shifts. So you obtain the category $A^+$. Then consider an object of the form $(X_1 \oplus \dots \oplus X_n , \alpha)$, such that $X_i \in A^+$, $\alpha \in \mathrm{End}^1 (X_1 \oplus \dots \oplus X_n)$, $\mathrm{d}\alpha + \alpha^2 = 0$ and $\alpha$ is upper-triangular (if $\alpha = (\alpha_{ij})_{1\leq i,j \leq n}$, $\alpha_{ij} \in \mathrm{Hom}^1 (X_i, X_j)$, then $\alpha_{ij}=0$ whenever $i\geq j$). Now everything is fine and you should have obtained in some sense the category $A^+$ with cones. Edit: Oops. I've forgotten to define the differential. Suppose there are two objects $(X, \alpha)$ and $(Y, \beta)$ and suppose there is a morphism $f \in \mathrm{Hom} ((X, \alpha),(Y, \beta)) = \mathrm{Hom}_{A^+} (X,Y)$. Then the differential is defined as follows: $df = d_{A^+} f + \beta f - (-1)^{\mathrm{deg}f} f \alpha$. Grading in the Hom-complexes in the category of twisted complexes is the same as that in $A^+$ and the equality $\mathrm d ^2 = 0$ is easily verified. Leibniz rule is now just obvious.<|endoftext|> TITLE: Using math software to show that the following groups are infinite? QUESTION [6 upvotes]: I would like to show that the following finitely presented group in 3 generators $P, Q, R$ is infinite in certain cases: $$P^p, Q^q, R^r, (PQ)^2, (QR)^2, (PQR)^2, (QR^{r/2+1})^a (RQR^{r/2})^b$$ For example when $p=3, q=4, r=4$ and $a=5$, $b=1$ or when $p=5, q=3, r=6$ and $a=2, b=1$. I gave it to gap via F:=FreeGroup("P","Q","R"); G:=F/[rels]; and tried IsInfinite but that never finished. Then I tried F:=FreeMonoid("P","Q","R"); G:=F/[[rel,Identity(F)],...]; and ReducedConfluentRewritingSystem with a couple of different reduction orderings but that never finished either. Are there some other methods I can use? REPLY [7 votes]: I think it might be more sensible to ask questions like this on the GAP forum, but I can help you with the first of your examples. I haven't tried the second. A standard method of trying to prove finitely presented groups infinite is to look for subgroups of low index and compute their abelian invariants, hoping that you will find a subgroup with an infinite abelian quotient. To look for subgroups of low index, there is the LowIndexSubgroups command, which finds all subgroups up to a given index, and also GQuotients, which looks for homomorphisms onto specific groups. In your first example, the first approach found me subgroups up to index $12$, and the one of index $12$ turned out to be the second derived group. I then tried systematically looking for homomorhisms of $G''$ onto simple groups. I found homomorphisms onto ${\rm PSL}(2,13)$ and ${\rm PSL}(2,25)$. The kernel of the first of these has a large (and probably infinite) $2$-quotient, but no infinite abelian quotient. I was luckier with the second. Here are the actual GAP commands that prove infiniteness. gap> F := FreeGroup(3);; gap> P:=F.1;; Q:=F.2;; R:=F.3;; gap> p:=3;; q:=4;; r:=4;; a:=5;; b:=1;; gap> rels:=[P^p,Q^q,R^r,(P*Q)^2,(Q*R)^2,(P*Q*R)^2, > (Q*R^(r/2 + 1))^a * (R*Q*R^(r/2))^b ];; gap> G := F/rels; gap> D := CommutatorSubgroup(G,G);; gap> DD := CommutatorSubgroup(D,D);; gap> Index(G,DD); 12 gap> homs := GQuotients(DD,PSL(2,25));; gap> K := Kernel(homs[1]);; gap> AbelianInvariants(K); [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ]<|endoftext|> TITLE: The best upper bound for the number of involutions in a finite non-abelian simple group QUESTION [6 upvotes]: Let $G$ be a finite non-abelian simple group and $t$ is equal to the number of involutions of $G$. We know that $t<|G|/3$ or $3t+1 \leq |G|$. Is this the best upper bound for the number of involutions of $G$? and if not what is it (if there is any)? REPLY [18 votes]: If $n$ denotes the number of involutions of the non-Abelian simple group $G,$ then we have $n + 1 = \sum_{\chi} \nu(\chi)\chi(1),$ where $\chi$ runs over the irreducible characters of $G$ and $\nu(\chi) \in \{0,1,-1\}$ denotes the Frobenius-Schur indicator of $\chi.$ Hence, by Cauchy-Schwarz, we have $n < \sqrt{k(G)} \sqrt{|G|},$ where $k(G)$ denotes the number of conjugacy classes of $G.$ By a Theorem of Fulman and Guralnick, which uses the classification of finite simple groups, we have $k(G) < |G|^{0.41},$ so that $n < |G|^{0.705}.$ This is certainly less than $\frac{|G|}{4}$ when $|G| > 256.$ The only non-Abelian simple groups of order less than $256$ are $A_{5}$ and ${\rm PSL}(2,7),$ which have respectively $\frac{|G|}{4}$ and $\frac{|G|}{8}$ involutions. Hence every finite non-Abelian simple group $G$ has at most $\frac{|G|}{4}$ involutions and $4$ can't be replaced by any larger constant. However, the upper bound $|G|^{0.705}$ for the number of involutions in $G$ is asymptotically much stronger ( and I suspect this is far from best possible). In fact, it is possible to prove (without using the classification of finite simple groups- I believe that R. Brauer knew this fact) that for any $\varepsilon > 0,$ there are only finitely many finite simple groups $G$ which have more than $\varepsilon |G|$ involutions. A more careful analysis of the argument at the beginning shows that if $d$ is the smallest degree of a non-trivial complex irreducible character of $G,$ then $G$ has less than $\frac{|G|}{d}$ involutions. But by Jordan's theorem on linear groups, for any integer $d >1,$ only finitely many non-Abelian simple groups have a non-trivial complex irreducible character of degree $d$ or less. In particular only finitely many non-Abelian simple groups have a non-trivial irreducible character of degree at most $\varepsilon^{-1},$ so only finitely many non-Abelian simple groups $G$ have more than $\varepsilon |G|$ involutions.<|endoftext|> TITLE: Mean of i.i.d Random Variables With No Expected Value QUESTION [12 upvotes]: Let $X$ be an integer-valued random variable and let $X_n$ be the sum of $n$ independent realizations of $X$. I would like to understand the behavior of $X_n/n$ for large $n$ in some cases where $X$ has no expected value and therefore the central limit theorem (along with Chebyshev's Theorem, etc.) does not apply. For a concrete example, let $X$ have the following distribution: $$Prob(X=0)=1/3\qquad Prob(X=4^k)=1/4^{k+1}\qquad Prob(X=-4^{k})=1/4^{k+1}$$ with all other probabilities zero. Note that $X$ has no expected value so we can't use the central limit theorem. Still, I'd like to have a good way to estimate $Prob(X_n/n > M)$ for a given large $n$ and $M$. Note also that in this example $Prob(X_n/n=M)$ can depend on number theoretic properties of $M$ (and in particular will not be monotonic in $M$), but I'm hoping that when $n$ is big and we consider $Prob(X_n/n > M)$, this sort of thing will wash out. I've done some relevant calculations, but I wonder whether I'm missing either some knowledge or some insight that would make this easier to understand. REPLY [13 votes]: We can attack this problem by using Fourier transforms (i.e. characteristic functions). I'll consider the example in the problem where $X$ is a random variable taking the value $0$ with probability $1/3$, and $4^{k}$ and $-4^{k}$ with probability $1/4^{k+1}$ (for $k=0$, $1$, $\ldots$). I'll show that the probability that $|X_n|/n >M$ is about a constant times $1/M$ (more precise result below). Fix a smooth function $\Phi$ compactly supported in $[-1,1]$ and approximating the characteristic function of that interval. Concretely, suppose $\epsilon$ is small and $\Phi(x)=1$ on $[-1+\epsilon, 1-\epsilon]$ and is between $0$ and $1$ on the rest of $[-1,1]$. Since $\Phi$ is smooth, its Fourier transform ${\hat \Phi}(\xi) = \int_{-\infty}^{\infty}\Phi(x) e^{-2\pi i x\xi} dx$ has rapid decay for $|\xi|$ large. Now let $n$ and $M$ be large and consider $$ {\Bbb E}\Big(\Phi\Big(\frac{X_n}{nM}\Big)\Big). $$ Note that $$ \text{Prob} (|X_n| >nM) \le 1 -{\Bbb E}(\Phi(X_n/(nM))) \le \text{Prob}(|X_n| > (1-\epsilon)nM), $$ and so our problem is to understand the expectation above. By Fourier inversion, $$ {\Bbb E}(\Phi(X_n/(nM))) = \int_{-\infty}^{\infty} {\hat \Phi}(\xi) {\Bbb E}\Big( e^{2\pi i \xi X_n/(nM)}\Big) d\xi = \int_{-\infty}^{\infty} {\hat \Phi}(\xi) \Big( {\Bbb E}\Big( e^{2\pi i \xi X/(nM)}\Big)\Big)^{n} d\xi. $$ Now we compute that $$ {\Bbb E}\Big( e^{2\pi i \xi X/(nM)}\Big) = \frac{1}{3} + 2 \sum_{k=0}^{\infty} \frac{1}{4^{k+1}} \cos \Big( \frac{2\pi \xi 4^k}{nM}\Big) = 1 - 2\sum_{k=0}^{\infty} \frac{1}{4^{k+1}}\Big (1-\cos \Big( \frac{2\pi \xi 4^k}{nM}\Big) \Big). $$ Now using that $(1-\cos(x)) = O(\min(x^2, 1))$ we see that $$ \sum_{k=0}^{\infty} \frac{1}{4^{k+1}}\Big (1-\cos \Big( \frac{2\pi \xi 4^k}{nM}\Big) \Big) = O\Big( \frac{|\xi|}{nM}\Big). $$ Therefore, using $(1-x)^n = 1-nx +O(n^2 x^2)$ for $0\le x\le 1$, $$ {\Bbb E}(\Phi(X_n/(nM))) = \int_{-\infty}^{\infty} {\hat \Phi}(\xi) \Big( 1- 2n \sum_{k=0}^{\infty} \frac{1}{4^{k+1}}\Big (1-\cos \Big( \frac{2\pi \xi 4^k}{nM}\Big) \Big) + O\Big(\frac{\xi^2}{M^2}\Big) \Big) d\xi. $$ Since ${\hat \Phi}$ has rapid decay, the error term above is $O(1/M^2)$ (with the implied constant depending only on the fixed function $\Phi$). Using Fourier inversion, we conclude that $$ {\Bbb E}({\Phi }(X_n/(nM))) = \Phi(0) - n \sum_{k=0}^{\infty} \frac{1}{4^{k+1}} \Big(2 \Phi(0) - \Phi\Big(\frac{4^k }{nM} \Big) -\Phi\Big(-\frac{4^k}{nM}\Big)\Big) + O\Big(\frac{1}{M^2}\Big). $$ Since $\Phi(0)=1$, we get $$ 1- {\Bbb E}({\Phi }(X_n/(nM))) = n \sum_{k=0}^{\infty} \frac{1}{4^{k+1}} \Big(2 - \Phi\Big(\frac{4^k }{nM} \Big) -\Phi\Big(-\frac{4^k}{nM}\Big)\Big) + O\Big(\frac{1}{M^2}\Big). $$ By our choice for $\Phi$, the main term above is $$ \ge 2n \sum_{k, 4^{k} \ge nM} \frac{1}{4^{k+1}}, $$ and is $$ \le 2n \sum_{k, 4^{k} \ge (1-\epsilon) nM} \frac{1}{4^{k+1}}. $$ Thus we have obtained a good understanding of the probability that $|X_n|/n$ is large. Note also that the precise answer will have discontinuities when $nM$ gets near a power of $4$, but in any case the probability is about a constant times $1/M$.<|endoftext|> TITLE: Finite groups for which the element orders form an arithmetic progression QUESTION [5 upvotes]: Which are the finite groups $G$ such that the element orders of $G$ form an arithmetic progression? Several remarks: $S_3$, $A_4$ and any $p$-group of exponent $p$ satisfy this property. If $G$ satisfies this property and $p_1 TITLE: What do algebraic theories with strictly terminal trivial models look like? QUESTION [5 upvotes]: By algebraic theory I mean one in the sense of Lawvere, i.e. a collection of finitary operations, including projections, together with a multi-composition satisfying the obvious axioms. (I believe universal algebraists call these abstract clones?) A strict terminal object in a category is a terminal object $1$ such that every morphism with domain $1$ is an isomorphism. A well-known example of a category with strictly terminal objects is the category of (unital) rings: the only ring homomorphisms with domain $\{ 0 \}$ are isomorphisms. It is a fact that any model of an algebraic theory whose underlying set is a singleton must be a terminal object in the corresponding category of models. My question is, When are terminal objects in the category of models for an algebraic theory strict? It is easy to generalise the proof that the algebraic theory of rings has the strict terminal object property: indeed, for any algebraic theory with two constants $0$ and $1$ and a binary operation for which $0$ is a (left) absorbing element and $1$ is a (left) unit, the trivial model is strictly terminal. Another easy observation is that any algebraic theory that "interprets" an algebraic theory with the strict terminal object property must itself have the strict terminal object property. (Here, $T$ interprets $T'$ if there is a map from the operations of $T'$ to the operations of $T$ that preserves the multi-composition and projections.) So a natural follow-up question is, Is there a "universal" algebraic theory with the strict terminal object property? REPLY [4 votes]: I believe these would be the algebraic theories where there is a finite set of identities of closed terms that together imply all identities. (If there are no closed terms, just add a fresh constant symbol with no new axioms.) This is the case for rings, where the single identity $0 = 1$ implies that all ring identities hold. It is easy to see that if there is such a finite set of identities of closed terms, then the terminal algebra $1$ must be strict. For the converse, let $\mathcal F$ be the collection of all finite sets of identities of closed terms. For each $\Sigma \in \mathcal F$, suppose that $A_\Sigma$ is a nontrivial algebra that satisfies $\Sigma$. Let $\mathfrak{U}$ be an ultrafilter on $F$ that contains all the sets $\{\Sigma \in \mathcal F : a=b \in \Sigma\}$, where $a = b$ denotes a fixed identity between closed terms $a$ and $b$. The ultraproduct $A^\ast = \prod_{\Sigma \in \mathcal F} A_\Sigma/\mathfrak{U}$ is nontrivial since each $A_\Sigma$ is nontrivial. However, it does admit a homomorphism from from the terminal algebra $1$ since all closed terms evaluate to the same thing in $A^\ast$ and hence the map sending the unique element of $1$ to that element of $A^\ast$ is a homomorphism. Adding a new constant symbol works when the theory has no closed terms becase the terminal algebra is strict if and only if it is the only algebra with a trivial subalgebra. The new constant can be thought as a generator for a subalgebra and the criterion above boils down to the existence of finitely many identities of that generator that together imply that the whole algebra is trivial.<|endoftext|> TITLE: rationality question while dealing with an isogeny QUESTION [5 upvotes]: I don't think that the following is known, but before going to other things, I would like to know what can be said about it. Thanks in advance for any relevant comment ! So here is the situation. Let $k'\subset k$ be a finite field extension. Take affine algebraic group schemes defined over $k'$ together with a central isogeny $\pi : \tilde{G}\rightarrow G$ (i.e. $\pi $ is surjective on underlying topological space, and is a finite flat morphism such that $ker$ $\pi\subset Z(G)$). Given a subgroup $\Gamma \leq \tilde{G}(k)$, assume that $\Gamma _{down}=\pi_{k}(\Gamma )$ lies in fact already in $G(k')$. Can we conclude that $\pi_{k'}^{-1}(\Gamma_{down})$ is big enough compared to $\Gamma $ (more precisely, that its intersection with $\Gamma $ is of finite index in $\Gamma $ ?) Ok, now more precision about assumption (but I would be interested to have counter-examples if not in that situation, especially for the first assumption) : 1) $k$ is a local field, and $k'$ a closed subfield 2) $\tilde{G}$ (resp. $G$) is absolutey simple simply connected (resp. adjoint) Also, let me stress that I am mainly interetsed in the positive characteristic case, i.e. $k$ a finite extension of $\mathbb{F}_{p}$((T)), and that I do not assume that the extension is Galois (but I would be interested to know what Galois cohomology can bring to the matter, I'm not at all familiar with that theory). EDIT (prompted by the answer of user 76758) I'm very happy to have forgotten to mention the crucial assumption on $\Gamma $ since I got this very illuminating answers without it. But now, here it is : $\Gamma $ is in fact assumed to be an open compact subgroup of $\tilde{G}(k)$ (in the topology given by the local field $k$). Also, the purpose of all that was to prove the following claim : if $\Gamma $ is an open compact subgroup of $\tilde{G}(k)$ avoiding the center (as above, $\tilde{G}$ absolutely simple simply connected), then any closed normal subgroup of $\Gamma $ is of finite index. I finally found a Paper of Carl Riehm, "The congruence subgroup problem over local fields", which completely settles the question. But still, I would be interesting to know what happens in the above situation when $\Gamma $ is open compact (and if the finite index fact is true in this case, do we need the compactness assumption ?). REPLY [7 votes]: The answer is negative (even under all of the given hypotheses); this expresses a standard difficulty in the arithmetic aspects of connected semisimple groups over local function fields in contrast with $p$-adic fields. As S.Carnahan notes, it is equivalent to ask that the subgroup $\Gamma_{\rm{down}}$ of $G(k')$ contains the image of its preimage on $k'$-points with finite index. This in turn is equivalent to the image of $\Gamma_{\rm{down}}$ under the connecting map $\delta:G(k') \rightarrow {\rm{H}}^1(k', \ker \pi)$ having finite image. The answer is generally negative because ${\rm{H}}^1(k',\mu_p) = {k'}^{\times}/({k'}^{\times})^p$ is infinite whenever $k'$ is a local function field of characteristic $p$ even though $[{k'}^{1/p}:k'] = p < \infty$ in such cases. To be specific, consider the central isogeny $\pi_0: {\rm{SL}}_p \rightarrow {\rm{PGL}}_p$ between absolutely simple and connected semisimple groups over $\mathbf{F}_p$ (with ${\rm{SL}}_p$ simply connected and ${\rm{PGL}}_p$ of adjoint type), so the induced map $D \rightarrow \overline{D}$ between split diagonal tori is a direct product of the $p$-power endomorphism of ${\rm{GL}}_1$ and the identity map on ${\rm{GL}}_1^{p-2}$. So if we let $k'$ be any local function field of characteristic $p > 0$, $k = {k'}^{1/p}$ (a finite extension!), and $\pi$ the base change of $\pi_0$ over $k'$ then for the group $\Gamma = k^{\times}$ of $k$-points in a 1-parameter subgroup ${\rm{GL}}_1 \subset D$ multiplied by $p$ under $\pi_0$ we have $\Gamma_{\rm{down}} = {k'}^{\times}$ and the image of the preimage of $\Gamma_{\rm{down}}$ on $k'$-points is $({k'}^{\times})^p$, so the quotient is the group ${k'}^{\times}/({k'}^{\times})^p$ is infinite. (In this case the connecting map $\delta$ carrie $\Gamma_{\rm{down}}$ onto the group ${\rm{H}}^1(k', \ker \pi) = {\rm{H}}^1(k', \mu_p)$ that is infinite.) Equivalently, the preimage of $\Gamma_{\rm{down}}$ on $k'$-points is the subgroup ${k'}^{\times} = (k^{\times})^p \subset k^{\times} = \Gamma$ that has infinite index in $\Gamma$. If some motivation were given for the question then it might be clearer if there is an appropriate salvage to the question. But it is hard to tell what (if anything) to "fix" since the OP doesn't indicate where the question is coming from. (There are ways around the above difficulty in practice, depending on what the aim is.)<|endoftext|> TITLE: Homotopy of localisations of colimits QUESTION [7 upvotes]: Let $X_k$ be a family of spectra equipped with maps $f_k: X_k \to X_{k+1}$. If $Y$ is a compact object (such as a sphere), then I can compute homotopy classes of maps from $Y$ into the homotopy colimit $X_\infty := hocolim \; X_k$ of the sequence by $$[Y, X_\infty] = \lim_{k \to \infty} [Y, X_k]$$ Now if I assume that the $X_k$ are local for Morava K-theory $K(n)$, then it need not be the case that the homotopy colimit is, too. In order to force this to be true, I can simply re-localise $X_\infty$. My question is: is there a similar formula for $[Y, L_{K(n)} X_\infty]$? Does it help that the $X_k$ were local to begin with? I'm happy to assume that the $X_k$ have finite $K(n)_*$, if that makes life any easier. REPLY [6 votes]: There are two natural finiteness conditions that you might impose on $Y$. The stronger one says that the Morava $E$-theory of $Y$ is finite in each degree, or equivalently that $Y$ is a retract of $L_{K(n)}Y_0$ for some finite spectrum $Y_0$ with $E(n-1)_*Y_0=0$. If this holds, then $[Y,L_{K(n)}X_\infty]$ is the colimit of the terms $[Y,X_k]$, the key point being that $DY\wedge X_\infty$ is already $K(n)$-local (even though $X_\infty$ itself is not). If $Y$ does not have the above smallness property then it will rarely be true that $[Y,L_{K(n)}X_\infty]$ is the colimit of $[Y,X_k]$. For the canonical counterexample, let $\{M_k\}$ be a tower of generalised Moore spectra of the usual kind, so $$ BP_*(M_k)=BP_*/(v_0^{a_0},\dotsc,v_{n-1}^{a_{n-1}}) $$ with the exponents $a_i$ tending to infinity as $k$ increases. Put $Y=L_{K(n)}S^0$ and $X_k=F(M_k,L_{K(n)}S^0)$. In this case it is known that $X_\infty=L_{K(n)}S^0$ so $[Y,X_\infty]$ contains $\mathbb{Z}_p$, but each group $[Y,X_k]$ is finite so the colimit is torsion.<|endoftext|> TITLE: Silly me & Van der Waerden conjecture QUESTION [17 upvotes]: So I walked into this very innocent-looking combinatorics problem, and quite soon I ended up with the problem to prove that any doubly stochastic $n \times n$ matrix has a non-zero permanent. Now clearly, this follows from the Van der Waerden conjecture (which is now a theorem), which give a lower (positive) bound for the permanent.. However, in my case, it feels like overkill to reference this theorem, so I wonder if there is some elementary argument that shows that the permanent of a doubly stochastic matrix is positive. (Although, the lower bound mentioned above converges to 0, as the matrix size grows, so it must be non-trivial...). Or, is proving that the permanent is non-zero "as hard as" proving the lower bound? REPLY [16 votes]: Let me cite here a famous result that is equivalent to Hall's theorem, and from which the positivity of the permanent of a DS matrix follows. Theorem (Frobenius-König). The permanent of an $n\times n$ nonnegative matrix $A$ is zero if and only if $A$ has an $r\times s$ zero submatrix with $r+s=n+1$. From this theorem a brief argument shows that for a DS matrix $A$, we must have $\text{per}\ A > 0$.<|endoftext|> TITLE: Understanding the definition of the quotient stack $[X/G]$ QUESTION [14 upvotes]: I'm trying to understand the definition of the quotient stack $[X/G]$ as defined in Frank Neumann's Algebraic Stacks and Moduli of Vector Bundles. Explicitly, let $G$ be an affine smooth group $S$-scheme with right action $\rho:X\times G\to X$ on a noetherian $S$-scheme $X$. The quotient stack $[X/G]$ is the pseudofunctor $$ [X/G]:(\mathsf{Sch}/S)^{\operatorname{op}}\to\mathsf{Grpds} $$ defined as follows. For an $S$-scheme $U$ let $[X/G](U)$ be the category whose objects are diagrams $$ U\xleftarrow{\pi}E\xrightarrow{\alpha}X $$ where $\pi$ is a principal $G$-bundle and $\alpha$ is a $G$-equivariant morphism. The morphisms in $[X/G](U)$ are the isomorphisms of principal $G$-bundles commuting with the $G$-equivariant morphisms. For a morphism of $S$-schemes $f:U^\prime\to U$ let $$ [X/G](f):[X/G](U)\to[X/G](U^\prime) $$ be the functor induced by pullbacks of principal $G$-bundles. It is not clear to me how to determine when $[X/G]$ is representable in general. My questions are: Question 1. Is there a sufficient condition we can impose on $\rho$ to ensure that $[X/G]$ is representable? The Wikipedia article on quotient stacks says something about when the categorical quotient $X/G$ exists the canonical map $[X/G]\to\operatorname{Hom}(-,X/G)$ need not necessarily be an isomorhpism. This is somewhat opaque to me as I'm not even sure how $[X/G]\to\operatorname{Hom}(-,X/G)$ is defined. Question 2. What is an example where $[X/G]$ is not representable? If these questions are too broad, I'd be very grateful if someone could point me in the direction of a good reference. REPLY [6 votes]: It follows from the definitions that $[X/G]$ is representable by a scheme $S$ if and only if $X$ is a $G$-torsor over $S$, i.e. the natural map $G\times X\rightarrow X\times _SX$ given by $(g,x)\mapsto (x,gx)$ is an isomorphism. Example 0.4 (Chapter 0, §3, page 11 of 3rd edition) in Mumford's GIT is a variety $X$ with an action of $G=SL(2)$ with trivial stabilizers and geometric quotient $\mathbb{A}^1$, but such that $X$ is not a $G$-torsor over $\mathbb{A}^1$, so $[X/G]$ is not representable.<|endoftext|> TITLE: Wanted: a nontrivial weakly inadmissible Heegaard diagram QUESTION [12 upvotes]: This is a question asked by a student in my lecture. After drawing pictures for awhile, I thought it was a good one. I seek a nontrivial example of a pointed Heegaard diagram $(\Sigma,\mathbf{\alpha},\mathbf{\beta},z)$ of a closed, oriented 3-manifold that fails to be weakly admissible in the sense of Heegaard Floer homology. What I call the trivial example is the diagram of $S^1\times S^2$ consisting of two parallel circles in the torus, with $z$ appropriately chosen so that there is a positive domain given by an annulus, bounded by these circles, whose index is 0. However, there are no generators of $\widehat{CF}$ in this diagram. For all I know, any Heegaard diagram that has generators is weakly admissible! Lipshitz's index formula (Lemma 4.11 of http://arxiv.org/abs/math/0502404) seems to imply that a domain in $\widehat\pi_2(\mathbf{x},\mathbf{x})$ that violates weak admissibility either needs lots of acute corners (assuming acute corners are even available to periodic domains, it seems their negative contribution to the index should be more than canceled by $2n_{\mathbf{x}}$), or high genus (with similar cancellation coming from any entries of $\mathbf{x}$ in its interior). REPLY [3 votes]: If you have any 3-manifold $Y$ with positive $b_1$, and any (generic) Heegaard diagram $\mathcal{H} = (\Sigma, \alpha, \beta)$ for it, you can place the basepoint $z$ so that $(\Sigma, \alpha, \beta,z)$ is not admissible. In fact, consider any domain $P_0$ in $\mathcal{H}$ whose boundary is a nontrivial sum of $\alpha$- and $\beta$-curves (this would be called periodic, if the usual definition of periodicity didn't require that the multiplicity of the basepoint be 0), and let $m$ be the minimal multiplicity of $P_0$, attained at a region $D$. Consider $P = P_0 - m\cdot \Sigma$, and place the basepoint $z$ in $D$. Clearly, $P$ is a nonzero periodic domain in $(\Sigma, \alpha, \beta,z)$ which has only non-negative multiplicities. Here's what happens for the "standard" admissible-looking diagram for $S^1\times S^2$ you refer to. (I'm linking to an SVG picture, that I can't convert at the moment -- if anyone wants to edit, feel free to do it). The "standard" periodic domain is the difference of the two bigonal regions. Adding the whole torus and placing the basepoint in the 0-multiplicity bigon yields the linked picture. Colour indicates multiplicity: white means 0, lighter gray 1, darker gray 2. EDIT: as pointed out by the OP in a comment below, the domain above has index 2, while we'd like to have an index-0 periodic domain. I think I have an example for this phenomenon (again, a diagram for $S^2\times S^1$): Consider the domain $P_0$ shown in the picture below (yellow means multiplicity -1, white multiplicity 0, and gray multiplicity 1), and generator $x$ corresponding to the two thick black points, and the green basepoint $z$. The index of $P_0$ as a domain in $\pi_2(x,x)$ is -2 (the Euler number is the Euler characteristic of the doubly-pointed torus, which is -2), and the multiplicity at $x$ is zero. If you add the entire surface, you get a periodic domain of index 0 with only non-negative multiplicities. Notice that there is a lot of redundancy, both in the number of generators and the genus of the diagram, but I wouldn't even try to formulate a question where this is addressed.<|endoftext|> TITLE: What is a good poster for a math conference? QUESTION [25 upvotes]: I'm going to participate to a conference and they ask me to do a poster on my research. I've never made a poster for a conference/seen a poster session in a conference. So what is important? What do you want to see in a poster (references, basic definitions and ideas)? What are the things that must not appear in a poster (completed proofs?)? What is a poster session? PS : My subject of research is about PDE (theoretical point of view : existence of a solution, regularity, unicity) and I'm going to a conference about Numerical Analysis (mostly) and PDE. REPLY [7 votes]: Know your audience so that you can communicate to them most effectively. Text should be large enough to be seen from 5 feet away. The pieces should be organized in a way that leads the viewer through the display. Make illustrations simple and bold. The display should be self-explanatory so that you are free to talk. Keep displays simple and text brief; a viewer should "get it" in 30 seconds. You can provide in-depth information in a handout. A neutral colored poster on matte board is more pleasing to the eye than one on a bright colored background. Organize your material and edit your content to eliminate distracting visual noise. When in doubt, edit out; make sure every item is necessary. Take a note pad and pen for notes, extra thumbtacks, pins, tape or glue. Here are some templates of posters. you can use of online latex to make a nice poster with Latex . https://www.overleaf.com/gallery/tagged/poster And http://www.latextemplates.com/cat/conference-posters Here is good advise to make a scientific poster http://www.personal.psu.edu/drs18/postershow/<|endoftext|> TITLE: Vector bundles on vector bundles QUESTION [19 upvotes]: Are all vector bundles on a given vector bundle the pull back of a vector bundle on the base? In more detail: let $X$ be a space and $p:E\rightarrow X$ a vector bundle over $X$. Let $\iota: X \rightarrow E$ denote the inclusion given by the zero section. Then for $V$ a vector bundle over $E$, we have the restriction $\iota^* V$ a vector bundle over $X$, while for $W$ vector bundle over $X$ we can consider the pull back $p^*V$ to $E$. On one hand, it is straightforward that $\iota^* p^* W = W$. When is it true that $p^* \iota^* V$ is isomorphic to $V$? (so that any vector bundle on $E$ is the pull back of a vector bundle on the base) Clearly the question depends on which setting we are in. For $X$ a topological space and $E$ a topological vector bundle this is clearly true, since $\iota\circ p$ is homotopic to the identity in $E$. This can be shown using the classifying space for vector bundles and the fact that homotopic maps give the same vector bundle. For $X$ a manifold and $E$ a smooth vector bundle this is true as well, using the fact that they are isomorphic at the topological level. What happens in the holomorphic and algebraic setting? Say $X$ a complex manifold, $E$ a holomorphic vector bundle and considering the vector bundles $V$ and $p^*\iota^* V$ up to equivalence of holomorphic vector bundles. Or $X$ a smooth algebraic variety (over $\mathbb{C}$ if needed), $E$ and algebraic vector bundle and equivalence of algebraic vector bundles. I thought it had to be true, but then I realized that for $X$ equal to a point (in the algebraic setting) this is Quillen-Suslin theorem, so that one need to be more cautious... Another remark: for $V$ of rank one this is true in both complex and algebraic setting, since there is an isomorphism between the Chow groups of $X$ and $E$. REPLY [6 votes]: Swan's paper "On Seminormality" shows that whenever the ring $R$ fails to be seminormal, there is an algebraic counterexample with $X=Spec(R)$, $E$ trivial and rank one over $X$, and $V$ rank one over $E$. Seminormality means that whenever $b^2=c^3\in R$, it follows that $b=a^3,c=a^2$ for some $a\in R$. In particular, the coordinate ring of the cusp $k[X^2,X^3]$ (where $k$ is, say, any field) is not seminormal.<|endoftext|> TITLE: Which hyperbolic tilings are Cayley graphs? QUESTION [5 upvotes]: I realise the question is easy but after asking to a few people (and never getting a clear answer), I thought it could be instructive to ask it here: Given a regular tiling of the hyperbolic plane is there a criterion to say it is not a Cayley graph? By regular tiling, I mean that the graph is vertex-transitive and there are only finitely many different type of faces up to isometry. Easier question: what about the tilings with only one type of face (a triangle) with $n$ of them meeting at each vertex, $n>6$. REPLY [5 votes]: Actually the requirement that your graph is vertex transitive is very strong. Any such tiling is "close" to the Cayley graph of a hyperbolic surface group, i.e. the fundamental group of a closed surface, which is not a sphere, torus, projective plane, or Klein bottle. Unfortunately the best description I can give you is algorithmic. First of all let's assume that the tiling doesn't have any ideal polygons. I am presuming that you are interested in Cayley graphs of finitely generated groups, so the graph should be locally finite. Also every vertex should be adjacent to an even number of edges. Let $\Gamma$ be the automorphism group of some tiling $\mathcal T \subset \mathbb{H}^2$ without ideal polygons. Any $g \in \Gamma$ that fixes a point must be of finite order, therefore $\Gamma$ acts properly discontinuously and cocompactly by isometries on $\mathbb{H}^2$, and $\Gamma$ is the fundamental group of a hyperbolic orbifold $\mathcal{O}$ without boundary. If there is a group $H\leq \Gamma$ that acts freely and vertex transitively of $\mathcal T$, then $\mathcal T$ is a Cayley graph of $H$, which is the fundamental group of a closed surface. This $H$ would be the subgroup corresponding to some "good" (technical term signifying a manifold) cover of $\mathcal{O}$ of minimal degree. In particular $[\Gamma: H]<\infty$. Although any automorphism group $\Gamma$ has "good" subgroups $H$ that act freely on the tiling, it may be that none of these subgroups are vertex transitive. It follows that if a tiling $\mathcal T$ that is the Cayley graph of a group, then there must exist some closed hyperbolic surface $\Sigma$ that can be decomposed as a union of polygons with a single vertex (i.e. all the corners of the polygons are at the same point.) Call this tiling of $\Sigma$ $\mathcal P$. The tiling $\mathcal T$ is obtained as the universal cover of $\mathcal P$. Since the number of edges that meet at a vertex of $\mathcal T$ give an upper bound for the rank of $\pi_1(\Sigma)$ this bounds the number of surfaces to check, so your property can be recognized algorithmically. To answer your questions about triangles, an Euler characteristic of surfaces calculation gives us $n = 3F$ where $F\geq 4$ is any even number.<|endoftext|> TITLE: Positivity question on K3 surfaces QUESTION [5 upvotes]: Let $X$ be a smooth projective complex K3 surface and $L, D$ two effective divisors, $L^2\geq0$ and $D^2\geq0$. (Q1). do we have $L\cdot D\geq0$ ? If either one has positive self-intersection, the answer is yes by index theorem. Assume $L^2=D^2=0$. If either one is basepoint free then it is a multiple of an elliptic curve and so the answer is yes. It remains to consider the case when $L$ and $D$ are both not basepoint free. I did not manage to show this, so I tried to construct a simple counterexample where $L$ and $D$ both share the same rational curve as fixed part, but I could not find any. Is there any? REPLY [3 votes]: Let me prove that $(L,D)\geq 0$. Consider the set of all (1,1)-classes which satisfy $\eta^2\geq 0$. It is a union of two components $P_+$ and $P_-$, intersecting in 0. If $\eta, \eta'$ are in the same component ($P_+$ or $P_-$), they intersect non-negatively, by Hodge index theorem. Let $P_+$ be a component containing a Kahler class. It remains to prove that any effective class that satisfies $\eta^2=0$ lies in $P_+$. However, for each $a\in P_-$ and $b\in P_+$, one has $(a,b) \leq 0$, because vectors in $P_+$ intersect positively by Hodge index theorem. Therefore, an intersection of any class in $P_-$ and the Kaehler class is non-positive. Therefore, $P_-$ does not contain effective classes.<|endoftext|> TITLE: coends of stable infinity categories QUESTION [9 upvotes]: Let $\mathcal{I}$ be a small ordinary category that I would like to think of as a diagram category. (If it helps: In my application $\mathcal{I}$ has only one object, i.e. comes from a monoid). Denote the $\infty$-category of small stable $\infty$-categories and exact functors by $Cat^{\rm ex}_{\infty}$. Suppose I have given a functor $$ F \colon \mathcal{I} \times \mathcal{I}^{\rm op} \to Cat^{\rm ex}_{\infty} $$ where I am secretly regarding the category on the left as an $\infty$-category. Are coends defined for $\infty$-functors as above? Does the coend of $F$ as above always exist? If the answer to the above questions is "yes": Can we describe the weak Kan complex underlying the coend in some way? REPLY [7 votes]: There is some discussion of coends in $\infty$-categories in my PhD thesis (Sheffield, 2010). I think that, given that the target category $\mathrm{Cat}_\infty^\mathrm{ex}$ is cocomplete, this gives you one way to do it. I have another way, which is arguably rather less fussy, but it is currently not written up in a presentable fashion.<|endoftext|> TITLE: Properties of singularities that are preserved by categorical quotients QUESTION [7 upvotes]: Let $G$ be a reductive group acting on an affine singular variety $X$, and let $X/G$ be the categorical quotient. I know that if $X$ has rational singularities, then so does $X/G$ (http://link.springer.com/article/10.1007%2FBF01405091). I am curious about other properties of singularities that pass to the quotient. For example, if $X$ is Gorenstein, must $X/G$ be Gorenstein? What if $X$ has canonical singularities? More generally, I'd be happy for a reference where these questions are discussed. REPLY [10 votes]: Neither Gorenstein nor canonical are preserved. Already Gorenstein is destroyed for the action of $\mathbb{Z}/3\mathbb{Z}$ on $\mathbb{A}^2$ acting by $(x,y)\mapsto (\zeta\cdot x,\zeta\cdot y)$, where $\zeta$ is a primitive cube root of $1$. Also, the whole point of the Reid -- Shepherd-Barron -- Tai criterion is to determine when a quotient singularity is canonical. Typically it is not canonical.<|endoftext|> TITLE: Renewal systems: Intrinsic ergodicity and a question related to the Adler's conjecture QUESTION [6 upvotes]: Consider the alphabet $\mathcal{A} = \{0,1\}$ and consider a finite set of words $W = \{\omega_1, \ldots , \omega_n\}$ over $\mathcal{A}$. Then the renewal system $\Sigma_{W}$ generated by $W$ is formed by bi-infinite concatenations of words of $W$. My first question is, does every renewal system is intrinsically ergodic? Secondly, is there a one sided version of the definition of renewal subshift? On the other hand, Adler asked the following question: Is every transitive subshift of finite type topologically conjugated to a renewal system? To the best of my knowledge the conjecture stills open. Is the conjecture open when the alphabet is $\{0,1\}$? My third question is: Are there any examples of renewal systems $\Sigma_W \subset \mathcal{A}^{\mathbb{Z}}$ that are subshifts of finite type and viceversa? REPLY [6 votes]: There are some examples related to your third question in "Renewal Systems, Sharp-Eyed Snakes, and Shifts of Finite Type" by Johnson and Madden, Amer. Math. Monthly 109 (2002), 258-272. A long time ago Goldberger, Smorodinsky, and I showed that for every possible entropy of a shift of finite type (or, what amounts to the same thing, for the logarithm of every Perron number), there is a renewal system with that entropy. However, as far as I know, Adler's question is still open.<|endoftext|> TITLE: The trace of a wedge product of matrices QUESTION [7 upvotes]: I'm trying understand a computation on page 371 of Besse's book on Einstein Manifolds. I already know the curvature operator $R:\bigwedge^2\to\bigwedge^2$ may be written in block diagonal form relative to the direct sum decomposition: $$R=\begin{pmatrix}A&B\\C&D\end{pmatrix}$$ where $A=A^*,C=B^*,D=D^*$. And, $$\begin{pmatrix}A&0\\0&D\end{pmatrix}-s/12=W,\mbox{the Weyl tensor}$$ The two components of the Weyl tensor $W^+=A-s/12,W^-=D-s/12$ are called the self-dual and the anti-self-dual parts respectively. Now, pay attetion in this part: $$p_1(M)=-\frac{1}{8\pi^2}\int_MTr(R\wedge R)$$ where $R$ is considered as a matrix of 2-forms. Since $B$ and $B^*$ are acting on orthogonal spaces(this part is OK), $$\begin{matrix}Tr(R\wedge R)&=&Tr(A\wedge A)+Tr(D\wedge D)\\&=&-2(|W^+|^2-|W^-|^2)\omega_g\end{matrix}$$ because $\alpha\wedge\alpha=|\alpha|^2\omega_g$ if $\alpha$ is sel-dual. My questions are: 1) What $Tr(R\wedge R)$ stand for? 2) How compute $Tr(R\wedge R)$, and why the term -2 appears in the expression? I greatly appreciate any response, observation or correction! REPLY [5 votes]: The confusion is caused by using $R$ to denote two different things. In Section 13.6, Besse introduces $R$ as a $6$-by-$6$ matrix of (scalar) curvature coefficients, which is the matrix of the linear transformation $\Lambda^2\to\Lambda^2$ induced by the curvature operator with respect to a basis adapted to the standard self-dual-and-anti-self-dual splitting of $\Lambda^2$. When you square this $R$ (which is the same as $R\wedge R$ since the entries are scalars), you will get a symmetric $6$-by-$6$ matrix whose trace is quadratic in the coefficients of the curvature matrix (and, in fact, it will be a positive definite quadratic form on the space of curvature tensors, not what you want at all). Note that you could see that the entries of $A$ and $D$ couldn't be $2$-forms because, if they were, it wouldn't make any sense to subtract $s/12$, which is clearly a scalar. (Actually, it doesn't make any sense anyway because $A$ and $D$ are $3$-by-$3$ matrices, but, never mind. This was only meant to be a heuristic indication of the real formulae anyway.) Meanwhile, in that passage of Section 13.8, Besse is instead using $R$ to denote a skew-symmetric $4$-by-$4$ matrix of $2$-forms, or, more invariantly, a $2$-form with values in the skew-symmetric endomorphisms of the tangent bundle of $M$. (The entries of $A$, $B$, $C$, and $D$ appears as coefficients of the $2$-form entries in this $R$.) At this point, $R\wedge R$ is a symmetric $4$-by-$4$ matrix of $4$-forms, and the trace is a $4$-form on $M$. As for the appearance of the factors of $2$, that is wrapped up with Besse's choice of the norms on $W_\pm$ as spaces of linear transformations $\Lambda^2_\pm\to\Lambda^2_\pm$. These norms are unique up to a choice of a constant, but one has to make that choice consistently in order for the formulae to work out correctly. Added remark (at the request of the OP): The point is that, with respect to any local orthonormal coframing, the first Pontrjagin form has to be of the form $p_1 = Q(R)\omega$ where $Q(R)$ is quadratic in the entries of the $6$-by-$6$ matrix $R$ and $\omega$ is the local volume form associated to the coframing. Because of the way that $\mathrm{SO}(4)$ acts on $R$ under change of oriented coframe, $p_1$ must be of the form $$ p_1 = (c_1\ s^2 + c_2\ |B|^2 + c_3\ |W_+|^2 + c_4\ |W_-|^2)\ \omega $$ for some universal constants $c_i$ (since $R$ breaks up into 4 inequivalent representation of $\mathrm{SO}(4)$). Now, $p_1$ doesn't depend on a choice of orientation, nor does $s$ nor $B$, but moving to a different orientation will switch $W_+$ and $W_-$ and will also replace $\omega$ by $-\omega$. It follows that the only way the right hand side will be uninfluenced by the choice of orientation is if it is of the form $c\bigl(|W_+|^2-|W_-|^2\bigr)\ \omega$ for some universal constant $c$. The constant $c$ must be positive because $p_1(\mathbb{CP}^2)=3$, and $\mathbb{CP}^2$ has $W_- = 0$. (You can now determine that $4\pi^2c = 1$ by simply doing the integral on $\mathbb{CP}^2$, if you want. Of course, this depends on which norm you chose for $W_\pm$; I'm assuming you choose it to agree with Besse's choice. :) ) I think that one is expected to write out the translation between the two meanings and figure out which choice of norm on $W_\pm$ is meant when one is learning the subject. Remark: The notation in Besse's Sections 13.6 and 13.8 appears to be drawn, with little modification, from the famous 1978 paper "Self-Duality in Four-dimensional Riemannian geometry" by Atiyah, Hitchin, and Singer. However, the use of $\mathrm{tr}(R\wedge R)$ in the formula in Besse 13.8 seems to have been spliced in from the standard formula for the first Pontrjagin form; it is not in AHS where, instead, a different expression appears. I suspect that Besse also copied the choice of norm for the Weyl tensor from AHS, since they get the very same coefficient.<|endoftext|> TITLE: Does the centroid depend continuously on the curve? QUESTION [10 upvotes]: Let $\gamma$ be a piecewise smooth curve in $\mathbb{R}^n$. Recall that the centroid of $\gamma$ is the point $(\overline{x}, \overline{y})$ where $\overline{x}$ is the average value of $x$ on $\gamma$ and $\overline{y}$ is the average value of $y$ on $\gamma$: $$\overline{x} = \frac{1}{\text{Length}(\gamma)} \int_\gamma x\, d\gamma, \hspace{1cm} \overline{y} = \frac{1}{\text{Length}(\gamma)} \int_\gamma y\, d\gamma$$ My question is: if $\gamma_n$ is a sequence of piecewise smooth curves which converge uniformly to a piecewise smooth curve $\gamma$, is it true that $(\overline{x_n}, \overline{y_n}) \to (\overline{x}, \overline{y})$? If it is more convenient to replace "piecewise smooth" with "rectifiable" or something else, I don't mind. A hint that this might not be completely trivial is the observation that $\text{Length}(\gamma_n)$ need not converge to $\text{Length}(\gamma)$: the standard example is a sequence of finer and finer staircase curves converging uniformly to a diagonal line. However, the sequence of centroids does converge to the right limit in this example. REPLY [2 votes]: It is possible to have $\ \overline{x_n}\rightarrow \overline x\ $ while $\ \overline{y_n}\not\rightarrow\overline y\ $ (so that the convergence holds just for one coordinate but not for both). Just turn @WlodekKuperberg's example by half of the right angle (by $\ 45^\circ\ $ or, professionally speaking, by $\ \frac\pi 4$).<|endoftext|> TITLE: A model of Krivine QUESTION [5 upvotes]: In a paper by J.-L. Krivine, Modèles de ZF+AC dans lesquels tout ensemble de réels définissable en termes d'ordinaux est mesurable-Lebesgue [C. R. Acad. Sci. Paris Sér. A-B 269 (1969), A549–A552, MR0253894], he presents a model of ZFC in which every set of reals definable from ordinals is Lebesgue measurable. My question is: He seems not to assume the existence of an inaccessible cardinal in the ground model, is this correct? In Theorem 3 of that paper, he concludes that every set of reals definable from ordinals is Lebesgue measurable. Does this continue to hold if we replace definable from ordinals by definable from a countable sequence of ordinals? Does Theorem 3 of that paper still hold if we replace "Lebesgue measurable" by "having the property of Baire"? REPLY [5 votes]: Yes, that is correct. Krivine's observation is that by collapsing the continuum (or any larger cardinal) to $\aleph_0$ using finite conditions then any set of reals definable from ground model parameters (e.g. ordinals) is Lebesgue measurable in the extension. No, not at all countable sequences of ordinals are in the ground model. Yes, the argument works just as well replacing the use of random forcing by Cohen forcing at the end and the conclusion then is regarding to the Baire property instead of Lebesgue measurability.<|endoftext|> TITLE: Classifying two-faces of four-polytopes QUESTION [8 upvotes]: Motivation: This question is related to my study of hyperbolic Coxeter polytopes. In general, if one put some restrictions on the type of their dihedral angles (say, all dihedral angles are equal to $\pi/2$), an the type of the polytope (all its vertices are proper, or all are ideal, or of mixed type), then one obtains some restrictions on the combinatorics. At the moment, I try to see if there are compact right-angled four-dimensional polytopes with volume bounded above by some (reasonably small) constant. I have come to the following (probably, quite technical) question (or, better say, the following type of questions). Question: Is there a convex simple polytope (in $\mathbb{R}^4$) which has $n=2,3,...,12$ hexagonal two-faces, while all other two-faces are pentagons. Or, more generally, let $n_k \geq 0$ be the number of $k$-gonal faces ($k\geq 6$). Does there exist a convex simple polytope (in $\mathbb{R}^4$) which has given amounts $n_6$, $n_7$, ..., of, respectively, $6$-, $7$-, ...-gonal two-faces, while all other two-faces are pentagons? I'm curious to know if there are general methods to study these (or similar) problems. Also, computational methods can be of interest (e.g. one can impose an upper bound on the number of three-facets of such a polytope and try to get a computer seek possible polytopes). Thank you in advance for any help/references/etc! REPLY [3 votes]: I dont know the anser to the specific question. It seems that for the study of hyperbolic Coxeter polytopes even if using some properties of general simple 4-polytope one needs to use the very restricted nature of these Coxeter polytopes. Overall the question of prescribing the sizes of 2-faces of simple (and general) 4-polytopes (and d-polytope) is very interesting. For $d>4$ it is known that there is always a 2-face with three or four edges and this is easy for the simple case. For 4-polytopes it is conjectured (by Igor Pak) that if all 2-faces has at least 5 sides then the number of $k$-faces is at least that of the 120-cell. (This is quite delicate as it does not extend to duals of triangulations of homology spheres.) The following answer https://mathoverflow.net/a/24340/1532 to Characterizing faces of 3-dimensional polyhedra. (Related to Victor Eberhard's Theorem [1890]:) fives some related results and questions. To attack this question we can try to use The Dehn Sommerville relations to study the sequence $n_3,n_4,\dots, $ and to estimate the average number of sides in a polygon. This number is below 6 and it can be expressed in terms of two parameters of the dual polytope $g_1(P^*)$ (essentially the number of vertices) and $g_2(P^*)$. Understanding the sequences $(n_3(P),n_4(P),\dots )$ for 4-polytopes (and even very special classes like stacked polytopes) and in particular the cases where $n_3=n_4=0$ is indeed very interesting.<|endoftext|> TITLE: Property of $L$ Relating to Reflection QUESTION [5 upvotes]: The idea of the question is whether it is ever possible that $L$ is so nice in the sense that $\{L_\alpha\}$ does not incorrectly "guess" a bigger inaccessible than $L$ really has, as long as $L_\alpha \models ZFC + V = L$. More precisely, is it ever possible that $L \models "\kappa$ is the largest inaccessible" and for all $\alpha > \kappa$ with $L_\alpha \models ZFC + V = L$ (or perhaps large finite fragments), $L_\alpha \models "\kappa$ is largest inaccessible". Are there assumption that one can put on $V$ so the it's $L$ has the above property? By reflection, there are unboundedly many $\alpha$ such that $L_\alpha \models "\kappa$ is largest inaccessible". However, it appear that potentially there are many $\alpha$'s which could "temporarily" think some ordinal greater than $\kappa$ is an inaccessible cardinal. Thanks for any information or clarification. REPLY [7 votes]: The answer to the "More precisely ..." question is that yes, it may be the case that $L$ has a single inaccessible cardinal $\kappa$ and that for no larger $\beta >\kappa$ is there any $\alpha$ with $L_\alpha\models$``$\beta$ is inaccessible.'' For example if $\kappa$ is the first inaccessible in $L$ (if such exists) and there is a least $\delta$ with $L_\delta\models ZFC$ (if such exists) then $L_\delta$ is a model for this property: namely for every $\alpha \in (\kappa,\delta)$ we have $L_\alpha\models \kappa $ is the largest inaccessible. The restrictive assumption to put on $V$ to ensure that $L$ has this property is just to rule it out by fiat: that `There is no transitive model of $ZFC + \exists \kappa(\kappa $ is inaccessible)''.<|endoftext|> TITLE: Is $\lfloor \log(n!)\rfloor \alpha$ equidistributed on the unit circle? QUESTION [12 upvotes]: In this question $\lfloor a\rfloor$ means the greatest integer not exceeding $a$. Using van der Corput's inequalities one is able to show that $\log(n!)\alpha$ is equidistributed on the unit circle ($\mathbb{R}\setminus\mathbb{Z}$) for all $\alpha \in \mathbb{R}^{+}$. I am however having trouble trying to show that $\lfloor\log(n!)\rfloor\alpha$ is equidistributed on the unit circle for $\alpha \in \mathbb{R}\setminus\mathbb{Q}$. The reason I believe this result to hold is due to the research here, showing that $\lfloor\log(n!)\rfloor$ is a good sequence for mean $L^{2}$ convergence. Any ideas of how one could try and prove this would be much appreciated. I suppose the more general question is: "Are there any results or methods that help in proving equidistribution (or good sequences for mean $L^{2}$ convergence) for functions that are rounded down?" REPLY [15 votes]: Yes. This follows from Theorem 3.2 of my paper with Michael Boshernitzan, Gregori Kolesnik and Máté Wierdl, 'Ergodic Averaging Sequences'.<|endoftext|> TITLE: Are left adjoints a left adjoint? QUESTION [6 upvotes]: Let $\mathcal C$ be a strict, locally small 2-category. Consider a subcategory $\mathcal L$ of $\mathcal C$ such that $\mathcal L$ has the same objects as $\mathcal C$, and the arrows of $\mathcal L$ are the left adjoints in the 2-category $\mathcal C$. Let $U:\mathcal L\to\mathcal C$ be the inclusion functor. When is $U$ a (left or right) adjoint? REPLY [3 votes]: Your question is impossible to answer in full generality. There are, however, some interesting special cases. I shall focus on your motivating example (I think, I have written about this on many occasions, but cannot recall the exact sources at the moment). If $\mathbb{C}$ is a (finitely complete, locally small) regular category, then you may construct a category $\mathit{Rel}(\mathbb{C})$ of canonical internal relations in $\mathbb{C}$. Moreover, $\mathit{Rel}(\mathbb{C})$ has a 2-categorical structure induced by the natural ordering of monomorphisms, and $\mathbb{C}$ is equivalent to the (2-)subcategory of $\mathit{Rel}(\mathbb{C})$ consisting of morphisms that have right adjoint. On the other hand, the inclusion: $$J \colon \mathbb{C} \rightarrow \mathit{Rel}(\mathbb{C})$$ has a right adjoint: $$P \colon \mathit{Rel}(\mathbb{C}) \rightarrow \mathbb{C}$$ iff $\mathbb{C}$ is an elementary topos. Indeed, one may easily verify, that $P$ has to be the internal power functor: $$\hom_{\mathit{Rel}(\mathbb{C})}(A, B) \approx \hom_\mathbb{C}(A, P(B))$$ More generally, if $p$ is a regular fibration, then under some mild conditions, $p$ has a generic object iff the canonical embedding: $$J \colon \mathit{Map}(\mathit{Rel}(p)) \rightarrow \mathit{Rel}(p)$$ has a right adjoint, where $\mathit{Rel}(p)$ is the category of $p$-internal relations, and $\mathit{Map}(\mathit{Rel}(p))$ is the subcategory of $\mathit{Rel}(p)$ consitiong of morphisms that have right adjoints. The link with the previous example appears, when we consider for $p$ the canonical subobject fibration $\mathit{sub}_\mathbb{C}$ of a regular category $\mathbb{C}$ (the canonical subobject fibration of a regular category is regular, and the generic object of a subobject fibration corresponds to the subobject classifier of the category). Another interesting case, is when instead of a subobject fibration, we consider a fibration of regular subobjects --- which leads to the definition of a quasitopos. Another view of the above generalization is to consider an allegory $\mathbb{A}$. The subcategory $\mathit{Map}(\mathbb{A})$ of $\mathbb{A}$ consisting of maps that have right adjoints is called the category of maps in $\mathbb{A}$. An allegory $\mathbb{A}$ is a "power allegory" iff the canonical inclusion: $$J \colon \mathit{Map}(\mathbb{A}) \rightarrow \mathbb{A}$$ has right adjoint $P$. As you may see, your question for mere 2-posetal categories (i.e. $\hom(X, Y)$ is a poset for every $X, Y$) of relations is highly non-trivial. In some sense, the concept of proarrow equipment (pointed by Tim in his comment) is a 2-categorification of the concept of an allegory. Category $\mathbf{Cat}$ can be reconstructed, up to Cauchy completion (one should not expect more by categorical methods, because $\mathbf{Prof}$ does not distinguish between a category and its Cauchy completion --- i.e. they are equivalent in $\mathbf{Prof}$), from $\mathbf{Prof}$ by taking profunctors that have right adjoints. The inclusion: $$J \colon \mathbf{Cat}_\mathit{CC} \rightarrow \mathbf{Prof}$$ does not have right adjoint due to the size issues. The same is true for the inclusion: $$J \colon \mathbf{Cat}_\mathit{CC}(\mathbb{C}) \rightarrow \mathbf{Prof}(\mathbb{C})$$ of Cauchy complete $\mathbb{C}$-internal categories to $\mathbb{C}$-internal profunctors, for any non-trivial category $\mathbb{C}$.<|endoftext|> TITLE: Log forms and Tate classes QUESTION [7 upvotes]: Let $X$ be a smooth finite type variety over $\mathbb{C}$. Suppose that $\theta$ is a closed algebraic $1$-form whose cohomology class is weight $2$. Can we always express $\theta$ as $$\theta = \sum_i c_i \ d \log f_i$$ where the $c_i$ are constants and the $f_i$ are nowhere vanishing algebraic functions? As long as I'm at it, I'll ask for something more. Let $L(X)$ be the sub-$\mathbb{C}$-algebra of $H^{\ast}(X)$ consisting of forms of degree $p$ and weight $2p$. Is $L(X)$ always generated in degree $1$? I just remembered that I know a counterexample to this. $\mathbb{P}^2 \setminus (\mbox{nodal cubic})$ has betti numbers $(1,0,1)$, and the $H^2$ is in weight $4$, but there is no $H^1$ and, for that matter, no units in the coordinate ring. Motivation: In an earlier question, we showed that representations of cohomology classes as polynomials in $d \log$ forms are unique. This question is about which forms are of that type. It looks a little bit like a variant of the Hodge conjecture, but hopefully much easier! REPLY [5 votes]: What is true is that the exponential map gives surjection $$H^0(\mathcal{O}_X^*)\to Hom_{MHS}(\mathbb{Z},H^1(X,\mathbb{Z}(1))$$ Classes on the right can be interpreted as weight two classes represented by $1$-forms logarthmic at infinity with integral periods; the map can be identified with $d\log$. I'm not sure if this exactly what you want, but it is at least close to it. The argument is not hard, but I'm too lazy to try to reproduce it here. So at the risk of self advertising, you can look at second page of my paper "Beilinson-Hodge cycles on semiabelian varieties" with Manish Kumar. Here's a link http://front.math.ucdavis.edu/0808.2990<|endoftext|> TITLE: Sobolev spaces and geometry QUESTION [12 upvotes]: This is a very naive question, is there a way to geometrically understand Sobolev spaces without going through analysis and PDE's? To my knowledge, Sobolev spaces where created precisely to study PDE's so it might be a bit nonsensical to not want to go through that path, but perhaps one can understand these spaces via some construction on manifolds or something analogous? In case the answer to the above is "No." then I would ask what you would consider to be the nicest use of Sobolev spaces to geometry (including solving a particular PDE and stuff like that, of course)? Also, it would be nice to know a bit of the history behind the modern usage of Sobolev spaces... Thanks! REPLY [5 votes]: Unlike their Holder cousins, most Sobolev spaces are reflexive Banach spaces. Reflexivity is a highly desirable feature for variational problems because it gives a little bit of compactness enough so you can prove the existence of minimizers (or more general critical points) of various energy functionals. Such critical points satisfy the Euler-Lagrange equations and thus you obtain existence of (weak) solutions of many important equations in geometry or physics. Given that their norms have an integral description, the Sobolev spaces are tailor made for energy functionals described by integrals. The famous and for a while controversial Dirichlet principle states that any function $u$ defined on a compact smooth domain $\newcommand{\bR}{\mathbb{R}}$ $\Omega\subset \bR^N$ which is zero on the boundary of $\Omega$ and minimizes the energy functional $$E(u)=\int_\Omega\left(\frac{1}{2}|\nabla u(x)|^2 -f(x) u(x)\right) dx $$ must be a solution of the Poisson problem $\Delta u=f$ in $\Omega$ satisfying the boundary conditions $u=0$ on $\partial \Omega$. Weierstrass pointed out a major flaw in the classical understanding of this principle by constructing, in a special case, a minimizer of this energy functional which is not twice differentiable so the Laplacian does not make sense. That stopped things in their tracks for a while until Hilbert, in his famous 1900 Paris Conference talk included this in the list of his famous 27 problems. In particular, he hinted to a way out by stating that any variational problem has a solution provided that, if need be, we suitably define the concept of solution. You can read more about this and see many applications of Sobolev spaces to geometry in these lectures.<|endoftext|> TITLE: How to find a tetrahedron that covers four points? QUESTION [7 upvotes]: I’m looking for an explicit formula for the vertices of a regular tetrahedron that covers four given points. In particular: Given four distinct real numbers $a_1$, $a_2$, $a_3$, $a_4$, is there a simple formula for four complex numbers $z_1$, $z_2$, $z_3$, $z_4$ such that the four points $(a_i,z_i)$, $i=1,\ldots,4$, in ${\Bbb R}\times{\Bbb C}$ form the vertices of a regular tetrahedron? That is, $|a_i-a_j|^2+|z_i-z_j|^2$ is independent of choice of distinct $i$, $j$. Same question but with $a_i$ complex. As an example of the type of thing I’m looking for, for three real numbers numbers $a$, $b$, $c$, $(a,(b-c)/\sqrt 3), (b,(c-a)/\sqrt 3), (c,(a-b)/\sqrt 3)$ are the vertices of an equilateral triangle. I've tried things like $(a,(b+c\omega+d\overline\omega)/2)$ (where $\omega$ is a third root of unity) which looks to be on the right track but... I just hope someone knows the answer. Thanks in advance. REPLY [2 votes]: A regular tetrahedron is inscribed in a cube. Finding a regular tetrahedron whose vertices project to $a_1,a_2,a_3,a_4$ is equivalent to finding a cube so that one vertex projects to $a_0 = \frac{1}{2}(a_1+a_2+a_3-a_4)$ and three adjacent vertices project to $a_1, a_2, a_3$. Translate the values so that $a_0 = 0$ (equivalently, translate so that $a_4 = a_1+a_2+a_3$). We want $3$ orthogonal vectors of equal length whose first coordinates are $a_1, a_2, a_3$. Rescale so that $a_1^2+a_2^2+a_3^2=1$. Equivalently (transposing), we want an orthogonal matrix whose first column is $\vec v =\begin{smallmatrix}a_1 \\a_2 \\a_3\end{smallmatrix}$. For example, we can reflect $\vec e_1=\begin{smallmatrix} 1 \\ 0 \\ 0\end{smallmatrix}$ to $\vec v$ (assuming $\vec v \ne \vec e_1$). Let $\vec w = \|\vec v - \vec e_1 \|^{-1} (\vec v - \vec e_1) = \begin{smallmatrix} w_1 \\ w_2 \\ w_3 \end{smallmatrix}.$ The reflection $\vec u \mapsto \vec u - 2(\vec u \cdot \vec w) \vec w$ corresponds to the orthogonal matrix $$\begin{pmatrix} a_1 & -2w_1w_2 & -2 w_1w_3 \\ a_2 & 1-2w_2^2 & -2 w_2w_3 \\ a_3 & -2w_2w_3& 1-2w_3^2\end{pmatrix}.$$ So, the points $(a1,-2w_1w_2,-2w_1w_3)$,$(a_2,1-2w_2^2,-2w_2w_3)$,$(a_3,-2w_2w_3,1-2w_3^2)$, and $(a_1+a_2+a_3, 1-2w_2(w_1+w_2+w_3),1-2w_3(w_1+w_2+w_3))$ are the vertices of a regular tetrahedron.<|endoftext|> TITLE: Why is "naive" definition of non-commutative spectrum bad? QUESTION [15 upvotes]: It is well-known that the category of affine schemes is equivalent to the opposit category of commutative unital rings. So naively, one would think that the same should hold in non-commutative setting. So my question is: Why can't we define the category of "affine non-commutative schemes" by simply putting it to be $\mathsf{Alg}_k^{op}$, the opposit of the category of unital $k$-algebras? I was told that it is not a reasonable way of defining non-commutative schemes, but I don't really understand why. Can someone, please, explain to me, why this approach is bad? I hope this question is not too silly. Thank you very much for your help! REPLY [19 votes]: I have also wondered about this question and recently came across some papers that seem to answer it. First of all, the paper Manuel L. Reyes, Obstructing extensions of the functor Spec to noncommutative rings, Israel J. Math. 192 (2012), no. 2, 667–698, arXiv. shows that there are problems with extending the Spec functor from commutative rings to noncommutative rings. In particular, any functor Spec : Rings -> Spaces whose restriction to CRings coincides with the usual spectrum functor must map the matrix algebras $\mathrm{M}_n(\mathbf{C})$ ($n \ge 3$) to the empty space. One can still hope to have a spectrum in the category of locales or toposes. However, the following paper shows the same result for functors valued in either of these categories. Benno van den Berg and Chris Heunen, Extending obstructions to noncommutative functorial spectra, 2012, pdf. Finally, one can try to represent noncommutative rings by sheaves on Rings with respect to some subcanonical Grothendieck topology that extends the Zariski topology. This also fails, as described in the paper Manuel L. Reyes, Sheaves that fail to represent matrix rings, 2014, arXiv. However this approach can be made to work, after modifying the notion of descent. For this, Rosenberg has developed the notion of a Q-category. He realized that, though there is no suitable Grothendieck topology on Ring, it does admit the structure of a Q-category. See, for example, Maxim Kontsevich, Alexander Rosenberg, Noncommutative spaces, 2004, pdf. or various other preprints of Alexander Rosenberg. Interestingly, the work of Rosenberg predates the above "no-go" results by about a decade.<|endoftext|> TITLE: Does being special on a club imply being special? QUESTION [8 upvotes]: Let $T$ be an Aronszajn-tree, $C\subset \omega_1$ a club set and $f:\bigcup\limits_{\alpha\in C}T_\alpha\longrightarrow \mathbb Q$ a strictly increasing function (where $T_\alpha$ is the $\alpha$-level of $T$). Is $T$ special (i.e. there exists such an $f$ defined on the whole $T$)? I suspect that this is true and that it is a known fact, but I have not found any reference. REPLY [6 votes]: This is a classic result that I learned by reading The Souslin Problem by Devlin and Johnsbråten (Lecture Notes in Mathematics 405). First, recall that another way to think of special trees is that they are the $\omega_1$-trees that admit a cover by countably many antichains. Suppose $C = \{\gamma_\alpha:\alpha\lt\omega_1\}$ is closed unbounded with $\gamma_0 = 0$ and that $\bigcup_{\alpha\lt\omega_1} T(\gamma_\alpha) = \bigcup_{n\lt\omega} A_n$ where each $A_n$ is an antichain. For $t \in T(\gamma_\alpha)$ let $\{t_m : m \lt \omega\}$ be the set of extensions of $t$ with height less than $\gamma_{\alpha+1}$. Observe that each $B_{m,n} = \{t_m : t \in A_n\}$ is an antichain in $T$ and $T = \bigcup_{m,n\lt\omega} B_{m,n}$.<|endoftext|> TITLE: Finite morphisms to projective space QUESTION [8 upvotes]: Let $X$ be a projective variety of dimension n. Then there exists a finite surjective morphism $X \to \mathbf P^n$. Let $d$ be the minimal degree of such a finite surjective morphism. Let $d^\prime > d$. Question. Does there exist a finite surjective morphism $X\to \mathbf P^n$ of degree $d^\prime$? The answer is yes for multiples of $d$, but what about the general case? Also, by Riemann-Roch the answer is yes for $n=1$. REPLY [13 votes]: The answer is no in general. For an easy $2$-dimensional case, take for example $X=\mathbb{P}^1\times \mathbb{P}^1$, in which case $d=2$. Then, only the multiples of $d$ are possible for $d'$. Indeed, take a morphism $X\to \mathbb{P}^2$, and denote by $D$ the divisor of its linear system, which satisfies $D^2=d'$. The Picard group of $X$ is generated by $f_1,f_2$, the two fibres of the two projections, and $D=af_1+bf_2$, so $d'=D^2=2ab$.<|endoftext|> TITLE: Simultaneous Orthogonal basis for $L^2(\mathbb{R}^n)$ and $H^1(\mathbb{R}^n)$ QUESTION [8 upvotes]: Given a smooth bounded set $U\subset \mathbb{R}^n$, there is a simultaneous orthogonal basis for $L^2(U)$ and $H^1_0(U)$ by the existence of eigenvectors to the Laplacian in a bounded domain, which particularly requires boundedness for compactness of the solution operator of the corresponding elliptic problem. Is it possible to construct or does there exist a simultaneous orthogonal basis for $L^2(\mathbb{R}^n)$ and $H^1(\mathbb{R}^n)$ as well? I thought it might be possible to use a basis for $L^2(U)$ where U is a cube and then by translations construct an orthogonal basis for $L^2(\mathbb{R}^n)$. I do not know if that will also be orthogonal basis for $H^1(\mathbb{R}^n)$ or if there will be some edge effects creating trouble. As an attempt, I split $\mathbb{R}^n$ into the integer lattice $U_k:=U+k$ for $k\in\mathbb{Z}^n$, and where $U$ is the unit cube. For $L^2(U_k)$, there is an orthonormal basis $\{e_l^k; k\in\mathbb{Z}^n, l\in\mathbb{Z}\}$ which are also eigenvectors of the Laplacian $-\Delta$, and therefore, it also forms an orthogonal basis for $H_0^1(U_k)$. Now, we may extend each $e_n^k$ outside $U_k$ by $0$ so that it belongs to $H^1(\mathbb{R}^n)$. These $\{e_l^k; k\in\mathbb{Z}^n, l\in\mathbb{Z}\}$ form an orthonormal basis for $L^2(\mathbb{R}^n)$ but not an orthogonal basis for $H^1(\mathbb{R}^n)$. If I had instead a simultaneous orthogonal basis for $L^2(U)$ and $H^1(U)$, it would not be possible to extend outside $U$ by zero and other extensions would not preserve orthogonality. I wanted to know because I was reading the existence of solutions to wave equations as given in Evans's book on Partial Differential Equations using the Galerkin Method and at one point it requires this simultaneous basis for $L^2(U)$ and $H^1_0(U)$, which is available only for bounded smooth domains $U$ and I wonder if that proof could be extended for existence in $[0,T]\times \mathbb{R}^n$. I have asked this on math.stackexchange earlier. REPLY [2 votes]: I your aim is to apply the Galerkin method, you do not need simultaneous orthonormal basis. An inspection of Evans’ proof shows that you need a sequence of linear maps $(P_n)_{n \in \mathbb{N}}$ such that for each $m \in \mathbb{N}$, the range of $P_n$ is finite-dimensional, for each $u \in L^2 (U)$, $$ \lim_{m \to \infty} \Vert P_m (u) - u \Vert_{L^2 (U)} = 0 $$ there exists $C > 0$ such that for each $m \in \mathbb{N}$ and $u \in L^2 (U)$, $$ \Vert P_m (u) \Vert_{L^2 (U)} \le C \Vert u \Vert_{L^2 (U)}, $$ for each $u \in H^1_0 (U)$, $$ \lim_{m \to \infty} \Vert P_m (u) - u \Vert_{H^1_0 (U)} = 0 $$ there exists $C > 0$ such that for each $m \in \mathbb{N}$ and $u \in H^1_0 (U)$, $$ \Vert P_m (u) \Vert_{H^1_0 (U)} \le C \Vert u \Vert_{H^1_0 (U)}. $$ (These assumptions are in fact redundant in view of the uniform boundedness principle and the density of $H^1_0 (U)$ in $L^2 (U)$.) To construct such a family, consider a function $\eta \in C^1 (\mathbb{R}^n;[0, 1])$ such that $\eta = 0$ in $\mathbb{R}^n \setminus B_1 (0)$ and $\eta = 1$ in $B_{1/2} (0)$, set $\eta_\ell (x) = \eta (x/\ell)$ and define the linear map $$ Q_\ell (u) = \eta_\ell u. $$ Observe that - for each $\ell \in \mathbb{N}$, $Q_\ell (L^2 (\mathbb{R}^n)) \subset L^2 (B_{\ell})$ and $Q_\ell (H^1_0 (\mathbb{R}^n)) \subset H^1_0 (B_{\ell})$ - for each $u \in L^2 (U)$, $$ \lim_{\ell \to \infty} \Vert Q_\ell (u) - u \Vert_{L^2 (U)} = 0 $$ - for each $\ell \in \mathbb{N}$ and $u \in L^2 (U)$, $$ \Vert Q_\ell (u) \Vert_{L^2 (U)} \le \Vert u \Vert_{L^2 (U)}, $$ - for each $u \in H^1_0 (U)$, $$ \lim_{\ell \to \infty} \Vert Q_\ell (u) - u \Vert_{H^1_0 (U)} = 0 $$ - for each $\ell \in \mathbb{N}$ and $u \in H^1_0 (U)$, $$ \Vert Q_\ell (u) \Vert_{H^1_0 (U)} \le (1 + \Vert \nabla \eta \Vert_{L^\infty}) \Vert u \Vert_{H^1_0 (U)}. $$ In a second step, consider a basis $(w_{k, \ell})_{k \in \mathbb{N}}$ of eigenvectors of $-\Delta$ in $B_{\ell}$ with Dirichlet boundary conditions, define $$ Q_{k, \ell} (u)= \sum_{j = 1}^k (w_{k, j}, Q_\ell (u)) w_{k, j}, $$ and observe that - for each $k, \ell \in \mathbb{N}$, $Q_\ell$ has finite range, - for each $u \in L^2 (U)$, $$ \lim_{\ell \to \infty} \lim_{k \to \infty} \Vert Q_{k, \ell} (u) - u \Vert_{L^2 (U)} = 0 $$ - for each $k, \ell \in \mathbb{N}$ and $u \in L^2 (U)$, $$ \Vert Q_{k,\ell} (u) \Vert_{L^2 (U)} \le \Vert u \Vert_{L^2 (U)}, $$ - for each $u \in H^1_0 (U)$, $$ \lim_{\ell \to \infty} \lim_{k \to \infty} \Vert Q_\ell (u) - u \Vert_{H^1_0 (U)} = 0 $$ - for each $k, \ell \in \mathbb{N}$ and $u \in H^1_0 (U)$, $$ \Vert Q_{k, \ell} (u) \Vert_{H^1_0 (U)} \le (1 + \Vert \nabla \eta \Vert_{L^\infty}) \Vert u \Vert_{H^1_0 (U)}. $$ Since $H^1_0 (\mathbb{R}^n)$ and $L^2 (\mathbb{R}^n)$ are separable, conclude with a diagonal argument.<|endoftext|> TITLE: Is there any relation between the simplicial $S^1$ and the Hochschild homology of a noncommutative algebras QUESTION [11 upvotes]: Let $k$ be the base field and $A$ be a unital associative $k$-algebra. Let's review the Hochschild homology theory: we have the Hochschild chain comple $C_{\cdot}(A)$ where $$ C_n(A):=A^{\otimes n+1} $$ and the differntial $b: C_n(A)\rightarrow C_{n-1}(A)$ is defined to be $$ b(a_0\otimes\ldots\otimes a_n):=\sum_{i=0}^{n-1}(-1)^i a_0\otimes \ldots\otimes a_ia_{i+1}\otimes\ldots\otimes a_n. $$ The Hochschild homology $HH^{\cdot}(A)$ is defined to be the homology of the complex $(C_{\cdot}(A),b)$. Moreover, $C_{\cdot}(A)$ is a simplicial module where $d_i: C_n(A)\rightarrow C_{n-1}(A)$ is defined to be $$ d_i(a_0\otimes\ldots\otimes a_n):=a_0\otimes \ldots\otimes a_ia_{i+1}\otimes\ldots\otimes a_n $$ and $b=\sum_{i=0}^{n-1}(-1)^id_i$ according to standard simplicial homology theory. On the other hand, we have a simplicial model of $S^1$, denoted by $S^1_{\cdot}$ (see Loday's Free loop space and homology). Where $$ S^1_0=\{*\}\\ S^1_1=\{s_0(*),\tau\} $$ where $d_0\tau=*$ and $d_1\tau=*$. For $n>1$ $$ S^1_n=\{s_0^n(*),s_{n-1}\ldots\widehat{s_{i-1}}\ldots s_0\tau,i=1,\ldots,n\}. $$ We see that each $S^1_n$ consists of $n+1$ elements. Anyway, it is not difficult to check that its geometric realization $|S^1_{\cdot}|$ is homeomorphic to $S^1$. $\textbf{If the algebra A is commutative}$, the Hochschild chain complex $C_{\cdot}(A)$ can be considered as the tensor product of $S^1_{\cdot}$ and $A$, and the $d_i$'s in $C_{\cdot}(A)$ are induced by the $d_i$'s in $S^1_{\cdot}$. For example, both $d_0$ and $d_1$ have to map the two elements in $S^1_1$ to the same element $*$ in $S^1_0$ and both of them correspond to $$ a_0\otimes a_1\mapsto a_0a_1. $$ Here we don't need to distinguish $a_0a_1$ and $a_1a_0$ since $A$ is commutative. $\textbf{My question}$ is: when $A$ is NOT commutative, could we also get the simplicial module $C_{\cdot}(A)$ from $S^1_{\cdot}$? REPLY [8 votes]: The article "Hochschild and cyclic homology via functor homology," by Pirashvili and Richter, gives precisely such a description. In this paper they describe a category of "noncommutative sets," isomorphic to one described earlier by Fiedorowicz and Loday. The objects are finite sets, and maps are maps of finite sets together with a total order on each preimage. Associative algebras are not tensored over finite sets, but they are tensored over noncommutative sets, and if the algebra is commutative this agrees with the tensor over the underlying set. They then show that the simplicial set $S^1_\bullet$ lifts to the structure of a simplicial noncommutative set, and the Hochschild complex of an associative algebra $A$ comes from the levelwise tensor of $A$ with $S^1_\bullet$.<|endoftext|> TITLE: How many Complexity Classes do you know? QUESTION [10 upvotes]: We can read about the main complexity classes in textbooks and online in Wikipedia: http://en.wikipedia.org/wiki/Computational_complexity_theory However, in papers, there are a lot of important new classes which are rarely found, such as $\textsf{PPAD}$. How many complexity classes do you know? Could you please give a diagram to show the relations between them? REPLY [10 votes]: The top part of the Computability Zoo (r.e., recursive, and beyond) is covered in more detail in the Computability Menagerie.<|endoftext|> TITLE: Cokernel of the stable J-homomorphism at odd primes QUESTION [8 upvotes]: Where can one learn about odd-primary components of the cokernel of the stable J-homomorphism? According to wonderful Wikipedia article on Homotopy groups of spheres, the "hard" part of the stable stem is the cokernel of $J$. I am not an expert and have trouble finding what is known. After an extensive search all I found was some low-dimensional computations (in $k$th stem for $k\le 17$) which do not work for my current purpose. Theorem 1.1.14 in Ravenel's "green book" book which gives some infinite families. Embarassingly, I do not even understand the statement of the theorem and cannot locate its proof (which is probably implicit in Section 4 of Chapter 4). Theorem 1.1.14 says in particular "For $p\ge 3$ the $p$-component of $\mathrm{coker}\, J$ has the following generators in dimensions $\le 3pq − 6$ (where $q = 2p − 2$), each with order $p$", and then it goes on to list two generators in $(pq-2)$ stem and $(pq+q-3)$ stem, and six of their products. Does this mean that the $p$-component of $\mathrm{coker}\, J$ in the $k$th stem (with $k\le 3pq-6$) is zero unless $\pi_k^S$ contains one of the either elements mentioned above? Or does this merely mean that these eight elements are nonzero? Is there a more comprehensive account of what is known (preferably with proofs or references)? Again, I am after infinite families of nonzero elements. REPLY [2 votes]: I wish to record some other families of elements in $\mathrm{coker}(J)$ arising from computations of Toda and of Oka of stable homotopy groups at odd primes. These computations aren't merely for small stems (as I incorrectly believed from reading Toda's book and looking at tables in Hatcher or Ravenel texts). Since the image of the stable J-homomorphism is known explicitly at any prime $p$, we can determine $\mathrm{coker}(J)$ in Toda-Oka range. As is explained in Appendix B of Milnor-Stasheff, the image of the J-homomorphism in the $k$th stem is a cyclic group $\mathrm{Im}(J_k)$ whose $p$-component $\mathrm{Im}(J_k: p)$ is as follows for an odd prime $p$ (the case $p=2$ is slighly different but is just as easy to describe): If $\frac{k+1}{2(p-1)}\notin\mathbb N$, then $\mathrm{Im}(J_k: p)$ is zero. If $\frac{k+1}{2(p-1)}\in\mathbb N$, then $\mathrm{Im}(J_k: p)$ is isomorphic to $\mathbb Z_{p^{r+1}}$ where $p^r$ is the largest power of $p$ that divides $\frac{k+1}{2(p-1)}$. Toda computed the $p$-component of the $k$th stem for $k<2p^2(p-1)-3$. I won't analyse $\mathrm{coker}(J_k)$ for his range except for one obvious example: If $k=2p(p-1)^2-1$, Toda shows that the $k$th stem has $p$ component $\mathbb Z_p\times\mathbb Z_{p^2}$, which is non-cyclic, and hence $\mathrm{coker}(J_k)$ is nontrivial; in fact $\mathrm{coker}(J_k: p)$ is $\mathbb Z_p$. Oka in a series of papers, see here and references therein, extended Toda's range and constructed for each $p>3$, some elements (in his notations $\phi$, $\mu$, $\beta$) such that $\mathrm{coker}(J_k: p)$ is $\mathbb Z_{p^2}$. In particular, for the $\beta$-elements the degree $k$ is even, and for $\phi$-elements $\frac{k+3}{2(p-1)}\in\mathbb N$, so in these cases the image of $J$ is zero at $p$. I do not know any examples where $\mathrm{coker}(J_k: p)$ has elements of order $>p^2$, and wonder if this is due to natural limitations of Toda-Oka range, or is there some other explanation of this phenomenon?<|endoftext|> TITLE: Failure of Cantor-Bernstein for the Levy Collapse QUESTION [6 upvotes]: Related to this question, is it possible to give an example of the failure of Cantor-Bernstein for complete embeddings of forcing notions involving the Levy Collapse $Col(\omega,<\kappa)$? Suppose $\kappa$ is a large cardinal. Does there exist some forcing $\mathbb{P}$ such that $\mathbb{P} \triangleleft Col(\omega,<\kappa)$, and $Col(\omega,<\kappa) \triangleleft \mathbb{P}$, but $\mathbb{P}$ is not equivalent to $Col(\omega,<\kappa)$? The analogous question for $Col(\mu,<\kappa)$, $\mu$ regular, is also of interest. Clarification: By $\mathbb{P} \triangleleft \mathbb{Q}$, I mean there exists a complete embedding (aka regular embedding) $e : \mathbb{P} \to \mathcal{B}(\mathbb{Q})$, or equivalently a projection map $\pi : \mathcal{B}(\mathbb{Q}) \to \mathcal{B}(\mathbb{P})$. REPLY [2 votes]: If you mean as in the other question that $\mathbb{P}\triangleleft\mathbb{Q}$ just in case every generic filter for $\mathbb{Q}$ adds a generic filter for $\mathbb{P}$, then the answer is yes, because Jonas Reitz's idea on the other question is also applicable here. Namely, let $\mathbb{P}$ be the lottery sum $\text{Col}(\omega,\lt\kappa)\oplus\mathbb{Q}$, where $\mathbb{Q}$ is some much larger poset above, such as $\text{Col}(\omega,\lt\theta)$, which definitely adds a generic for $\text{Col}(\omega,\lt\kappa)$. The point now is that forcing with $\mathbb{P}$ will definitely add a generic for $\text{Col}(\omega,\lt\kappa)$, regardless of which choice is made in the lottery, and conversely every generic for $\text{Col}(\omega,\lt\kappa)$ is also generic for $\mathbb{P}$, by opting for the first term in the lottery. But they are not forcing equivalent. The same idea works with $\text{Col}(\mu,\lt\kappa)$, and indeed, with essentially any forcing notion at all. But I find it more likely that you intend $\mathbb{P}\triangleleft\mathbb{Q}$ just in case there is a complete embedding of $\mathbb{P}$ into the (Boolean completion) of $\mathbb{Q}$, which in many ways is the natural relation to consider with forcing. In this case, your question is related to the question whether the Lévy collapse $\text{Col}(\omega,\lt\kappa)$ up to an inaccessible cardinal $\kappa$ is characterized up to forcing equivalence as the unique $\kappa$-c.c. forcing notion of size $\kappa$ that collapses all smaller cardinals to $\omega$. Namely, if this characterization is correct, then whenever $\mathbb{P}\triangleleft\text{Col}(\omega,\lt\kappa)\triangleleft\mathbb{P}$, we would see that $\mathbb{P}$ would be $\kappa$-c.c. and have a size $\kappa$ dense set by the first relation, and it would collapse all smaller cardinals by the second relation, and so by the assumed characterization it would be forcing equivalent to $\text{Col}(\omega,\lt\kappa)$. Thus, if this characterization of the Lévy collapse is correct, then there are no counterexamples of the kind you seek. The analogous characterization is true in the case of $\text{Col}(\omega,\theta)$, which is in fact the unique poset of size $|\theta|$ necessarily collapsing $\theta$ to $\omega$, and a proof of this can be found for example as (folklore) lemma 18 of my paper Structural connections between a forcing class and its modal logic. I have always expected that this characterization extends to the Levy collapse in the way I have suggested, but I do not have a proof of this.<|endoftext|> TITLE: Determine or estimate the number of maximal triangle-free graphs on $n$ vertices QUESTION [18 upvotes]: Among the collections of the open problems of Paul Erdős on the website of Professor Fan Chung, there is one called "number of triangle-free graphs". http://www.math.ucsd.edu/~erdosproblems/erdos/newproblems/NumberOfTriangleFreeGraphs.html Open Problem: Determine or estimate the number of maximal triangle-free graphs on $n$ vertices. Is any one working on this problem? Any known related results? Now I am considering about using a "connection game" method to solve this problem: Given a set of $n$ players, each one chooses to connect to other nodes, if any two neighbors of it connect a new edge(which means there would form a triangle), it has to choose delete either edges with those two neighbors. Then the question is how many different topology connections does it have? Any comments on this method? REPLY [5 votes]: This question was recently solved by Balogh & Petrickova. Douglas Zare's bound is tight (apart from the $o(1)$ term). See http://arxiv.org/abs/1409.8123<|endoftext|> TITLE: Can every discrete martingale be embedded in a continuous martingale? QUESTION [10 upvotes]: Let $(X_k)_{k=0,1,..., n}$ be a discrete martingale defined on some probability space $(\Omega,\mathcal{F},\mathbb{P})$. I would like to know whether there exists a (continuous) martingale $(\tilde{X}_t)_{0\le t\le n}$ defined on some probability space such that $(X_0, X_1,\dots, X_n)$ and $(\tilde{X}_0, \tilde{X}_1,\dots, \tilde{X}_n)$ have the same law? Thanks a lot for the reply! REPLY [5 votes]: Martingales on the Brownian filtration are continuous, so we construct one with the same law on discrete times as $X$. We construct this process in a similar way to how one might simulate $X$, given its conditional distributions and a sequence of normals. For simplicity, assume $X_0$ is constant. Let $(W_t)_{t\ge 0}$ be a Brownian motion on $(C[0,\infty), \mathcal{W}, \mathbb{Q})$, with natural filtration $(\mathcal{W}_t)_{t\ge 0}$. For $i=1,\cdots, n$, let $$N_i = W_i - W_{i-1}.$$ Claim: For $k=0,\cdots,n$, there is a $\sigma(N_1,\cdots,N_k)$-measurable random variable $Y_k$ such that $(X_0,\cdots, X_n)$ has the same law as $(Y_0,\cdots, Y_n)$. $Y$ has the same law as $X$, so it is a martingale (on its natural filtration). Also it is (we may check) a discrete-time martingale on the Brownian filtration, $$Y_k = \mathbb{E}[Y_n|\mathcal{W}_k],$$ so we may extend the definition of $Y_k$ to all $k$ in $[0,n]$. $Y$, being a martingale on the Brownian filtration, is continuous. Proof of claim: Without loss of generality, assume $X_k$ is the $k$th coordinate map on a probability space $(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n), \mathbb{P})$. Let $\phi$ be the CDF of a standard normal random variable. For $i=0,\cdots,N-1$, let $\mathcal{G}_i =\sigma(X_1,\cdots, X_{i})$, and let $Q_i:\Omega\times \mathcal{F}\to[0,1]$ be the regular conditional probability of $\mathbb{P}$ with respect to $\mathcal{G}_{i-1}$. Let $$F_{X_i|X_0,\cdots, X_i}(x| x_0, \cdots, x_{i-1}) = Q((x_0,\cdots, x_i,0,\cdots,0), \{X_i\le x_i\}).$$ For $G \in\mathcal{G}$, $\omega\mapsto Q(\omega,G)$ is $\mathcal{G}$-measurable, so $Q$ only depends on the first $i$ components of $\omega$. Denote $F_{X_i|X_0,\cdots, X_{i-1}}$ by $F_i$. Let $$G_i(y| x_0, \cdots, x_{i-1}) = \inf\{x:F_i(x| x_0,\cdots, x_{i-1})\ge y\}$$ Let $Y_0$ equal the constant $X_0$ and $$Y_i = G_i(\phi(N_i)| Y_0,\cdots, Y_{i-1})$$ $X_0$ has the same law as $Y_0$. Suppose $(X_0, \cdots,X_{i-1})$ has the same law as $(Y_0, \cdots,Y_{i-1})$. Fix $x_0, \cdots, x_i$. $$ \begin{align}\{Y_i\le x_i\} &= \{G_i(\phi(N_i), Y_0,\cdots, Y_{i-1}) \le x_i\}\\ &= \{F_i(x_i| Y_0,\cdots, Y_{i-1})\ge \phi(N_i)\}\\ &= \{Q((Y_0,\cdots,Y_{i-1}, x_i, 0,\cdots, 0), \{X_i\le x_i\})\ge \phi(N_i)\} \end{align} $$ Similarly, $$ \begin{align}\mathbb{Q}[Y_0\le x_0, \cdots, Y_i\le x_i] &= \mathbb{Q}[Y_0\le x_0,\cdots,Y_{i-1}\le x_{i-1}, Q((Y_0,\cdots,Y_{i-1}), \{X_i\le x_i\})\ge \phi(N_i)]\\ &= \mathbb{E}[Q((Y_0,\cdots, Y_{i-1}),\{X_i\le x_i\})\mathbf{1}_{Y_0\le x_0}\cdots\mathbf{1}_{Y_{i-1}\le x_{i-1}}]\\ &= \mathbb{E}[Q((X_0,\cdots, X_{i-1}),\{X_i\le x_i\})\mathbf{1}_{X_0\le x_0}\cdots\mathbf{1}_{X_{i-1}\le x_{i-1}}]\\ &= \mathbb{P}[X_0\le x_0, \cdots, X_i\le x_i] \end{align} $$<|endoftext|> TITLE: Who made the famous error in calculation that 'wasted' the final years of his life? QUESTION [26 upvotes]: Sorry, I am merely a Middle School maths teacher at an Australian secondary school. I remember reading years ago about a famous mathematician (18th or 19th Century?) who calculated table upon table of numbers (pi? prime numbers?) but made an error somewhere along the way and, based on that error, all subsequent numbers in the table were incorrect. Tragically, he continued to make that error for the last few years of his life without realizing it. I'd love to know his name but also I wondered if anyone could calculate the probability of his making the mistake in the first place. Obviously, he wasn't going to make the error in the first 10 seconds of doing the sums so why did he make the error when he did? Is there some type of critical mass beyond which creating an error becomes all but inevitable? Thank-you in advance for any replies I receive. Michael McLean REPLY [47 votes]: Was it William Shanks? He calculated $\pi$ to 707 digits, but he made a mistake in the 528th digit.<|endoftext|> TITLE: $\text{Rep}(D(G))$ as representation category of a vertex operator algebra QUESTION [17 upvotes]: The category of representations $\text{Rep}(D(G))$ of the quantum double of a finite group is well-known to be a modular tensor category. Can these modular tensor categories also be obtained as representation categories of vertex operator algebras? REPLY [6 votes]: A lot has happened in the last four years, and we now have lots of positive results. The current state of knowledge is given in Evans-Gannon, "Reconstruction and Local Extensions for Twisted Group Doubles, and Permutation Orbifolds". In particular, if $G$ is a finite solvable group, then $D(G)$ (and more generally, any twist $D^\omega(G)$) is the representation category of some vertex operator algebra (in particular, the fixed points of a $G$-action on some holomorphic vertex operator algebra). For non-solvable groups, the result you want would follow from the conjectured regularity of fixed points (i.e., a suitable generalization of C-Miyamoto). Oddly enough, it turns out that permutation orbifolds can have non-trivial twists. This is discussed in the Evans-Gannon paper, and earlier in Johnson-Freyd's "The Moonshine Anomaly".<|endoftext|> TITLE: Does this 'alternating' Euler product converge for all $\Re(s) > 0$? QUESTION [5 upvotes]: Does the following 'alternating' Euler product, with $p_n$ the $n$-th prime number, converge for $\Re(s)>0$ ? $$\displaystyle \prod_{n=1}^\infty \left( \dfrac{1}{1-\frac{1}{p_{n}^{s}}} \right)^{(-1)^n}$$ Based on numerical evidence, I dare to conjecture that this is indeed the case (note that I could not find any zeros), but keen to find approaches towards a proof. Note that I also tried other triggers than $n$ to 'flip the factors', like for instance prime congruence to either $p_n \pmod 6 =1$ or $5$ and $p_n \pmod 4 =1$ or $3$. Even tried the Möbius function $\mu(n)$ as the exponent of $(-1)$, but did not observe any convergence in the domain $\Re(s)\le 1$. So, using $n$ to flip any other prime factor, seems a delicate choice to make convergence work or fail in this domain. P.S.: This question loosely builds on: Equality of an alternating infinite product and an infinite sum REPLY [11 votes]: As Jacob mentioned, the question comes down to the conditional convergence of $\sum p^{-s}(-1)^n$ (for essentially the reason that $\prod(1-x)$ converges and is nonzero if $\sum x$ converges and $\sum x^2$ absolutely converges. Now, if you look in Titchmarsh's Theory of Functions, there is a theorem that the region of conditional convergence of a Dirichlet series is a half plane (along with part of its boundary). This is proven by showing that convergence at some $s$ gives convergence on every wedge with vertex at $s$ and otherwise strictly to the right. Thus since the series converges for every real $x>0$ (alternating series), it also converges in the right half plane.<|endoftext|> TITLE: geometric interpretation and differences of Gorenstein rings, Complete intersections and regular rings QUESTION [12 upvotes]: Let $R$ be a local Noetherian ring. What is the geometric interpretation of: 1- Gorenstein rings 2- Complete intersections 3- Regular rings? and how can I realize differences by geometric interpretation? REPLY [16 votes]: Regular rings have clear and famous geometric interpretation: They correspond to non-singular (affine) varieties. A (local) complete intersection rings can be thought of as a local ring at a point, such that at this point the variety can be defined by minimal number of relations, i.e. a regular sequence. In more geometric flavor a local complete intersection subscheme is the one whose normal bundle is still a vector bundle (a property similar to the non-singular case) but it can be singular. Gorenstein rings have more complicated geometric meaning: They are local rings of affine varieties which may be singular at the point, but the maximal $R$-regular sequence in the maximal ideal i.e. the associated complete intersection subscheme is an irreducible space. In general Gorenstein and complete intersection (and Cohen-Macaulay) local rings are substitutes of the regular local rings. It means that they might be singular, but their singularities is as nice as it can be. So we have the following strict inclusions: Regular rings $\subset$ CI rings $\subset$ Gorenstein rings The simplest example of a complete intersection ring which is not regular is the ring $k[x]/(x^{2})$. The maximal ideal is the ideal $(\overline{x})$ and of course $x$ is a regular element in $k[x]$. For a ring which is Gorenstein but not CI, consider the $0$-dimensional local ring $k[x,y]$$/(x^{2},y^{2},z^{2}$$,xz,yz, z^{2}-xy)$. The unique maximal ideal here is the ideal $(x,y,z)$ but you can see that the ideal $(x^{2},y^{2}, xz,yz, z^{2}-xy)_{(\overline{x}, \overline{y}, \overline{z})}$ is not generated by a regular sequence.<|endoftext|> TITLE: Positivity of coefficients of a power series QUESTION [9 upvotes]: How does one check for the positivity of coefficients of a rational function,say, for example $\frac{p_1(x,t)}{(1-xt)(1-x^2t)(1-x^3t)}$ where $p_1(x,t) = 1 + tx + 2t^2x^2 - 3x^3t^2 -x^5t^3 - x^{10}t^4$? In general, let $p_1(x,t)/p_2(x,t)$ be a rational function over field $k$,suppose the power series expansion of $1/p_2(x,t)$ has non-negative coefficients. If $p_1(x,t)$ is a polynomial with some negative coefficients, when can we say that the expansion of $p_1(x,t)/p_2(x,t)$ has non-negative coefficients? REPLY [6 votes]: This seems to be a hard question. See this paper by Laffey et al (2012) for extensive references.<|endoftext|> TITLE: Why is the derived tensor product only defined for bounded above derived categories? QUESTION [5 upvotes]: In "Residues and Duality" by Hartshorne, the derived tensor $\otimes$ only defined for the bounded above categories (see Chapter I, section 4), that is one has $$\otimes: D^{-}(X) \times D^{-}(X) \to D(X), $$ where $X$ is a variety, and $D^{-}(X), D(X)$ are derived category of bounded above complex of $\mathcal{O}_X$-modules and derived category of complex of $\mathcal{O}_X$-modules respectively . The main reason it is defined for the bounded above complex is that Hartshorne used the fact that every sheaf of module has a flat resolution. However, why cannot one use the injective resolution for the derived tensor, and defined it for bounded below derived cateogry $D^{+}(X)$? In fact, the later is the case I need (to be precise, I need $F\otimes : D^{+}(X) \to D(X)$, where $F$ is a sheaf of module), and I checked every conditions for the existence of derived functor, it seems that everything is satisfied. REPLY [16 votes]: I would like to add my 2 euro ¢. In a nutshell: your information is outdated. "Residues and Duality" by Hartshorne was the first available text on derived categories of coherent sheaves and Grothedieck duality but it was published in 1966. In 1988, Spaltenstein showed in ("Resolutions of unbounded complexes", Compositio Math. 65 (1988), no. 2, 121–154) that one can use unbounded homotopically flat resolutions of sheaves to get a derived tensor product of unbounded complexes. Similarly, you may use homotopically injective resolutions of sheaves to derive functors on the right. A very readable and useful update of duality for coherent sheaves using this developments is Joe Lipman's text "Notes on Derived Functors and Grothendieck Duality" in Springer Lecture Notes, no. 1960 (2009), 1–259. There is a lot of extra literature on variants for Grotehdieck abelian categories, differential graded algebras and so on.<|endoftext|> TITLE: Reference for statement that almost every $n$-element partial order has trivial automorphism group QUESTION [5 upvotes]: I'm looking for a reference for the statement that almost every partial order on $n$ elements has trivial automorphism group. I've been told that this is a folklore result. Does anyone know of a good reference? REPLY [3 votes]: Prömel (1987) proves a more general statement of rigidity for many classes of structures. In particular he has: Corollary 2.3. Let $P^u(n)$ denote the number of unlabeled partial orders on an $n$-element set. Then there exists a constant $s$ such that for all $n$ $$ P^u(n) \le \frac{P(n)}{n!} \left(1 + \frac{s}{2^{n/4}} \right), $$ and as a consequence, Corollary 2.3.a. Almost all partial orders are rigid, i.e., have no non-trivial automorphism. Prömel, Hans Jürgen, Counting unlabeled structures, J. Comb. Theory, Ser. A 44, 83-93 (1987). ZBL0618.05029.<|endoftext|> TITLE: Is there a polynomial time approximation scheme for the feedback arc set problem for the class of tournaments? QUESTION [5 upvotes]: A tournament is an orientation of a complete graph. A feedback arc set is a set of arcs in a digraph whose removal leave the digraph acyclic. The feedback arc set problem consists in finding a feedback arc set of minimum size. A polynomial time approximation scheme is an algorithm which takes an instance of an optimization problem and a parameter $ \epsilon > 0 $ and, in polynomial time, produces a solution that is within a factor $ 1+\epsilon $ of being optimal. The feedback arc set problem has been proved NP-hard. See [ACM, A, CTY, C]. It was shown in [RS] that the feedback arc set problem is fixed parameter tractable for tournaments. Bibliography *[AA] N. Ailon, N. Alon, link, Inform. and Comput. 205 (8) (2007) 1117–1129. [ACM] N. Alion, M. Charikar, A. Newman, Aggregating inconsistent information: Ranking and clustering, in: Proceedings of the 37th Symposium on the Theory of Computing, STOC, ACM Press, 2005, pp. 684–693. [A] N. Alon, Ranking tournaments, SIAM J. Discrete Math. 20 (2006) 137–142. [CTY] P. Charbit, P. Thomassé, A. Yeo, The minimum Feedback arc set problem is NP-hard for tournaments, Combin. Probab. Comput. 16 (1) (2007) 1–4. [C] V. Conitzer, Computing Slater rankings using similarities among candidates, in: Proceedings, The Twenty-First National Conference on Artificial Intelligence and the Eighteenth Innovative Applications of Artificial Intelligence Conference, July 16–20, AAAI Press, Boston, Massachusetts, USA, 2006. [RS] V. Raman, S. Saurabh, Parameterized complexity of directed feedback arc set problems in tournaments, in: Algorithms and Data Structures, in: Lecture Notes in Computer Science, vol. 2748, Springer, Berlin, 2003, pp. 484–492. REPLY [3 votes]: Not only a PTAS is known for this problem. It is also possible to compute a PTAS, even without seeing the entire adjacency matrix ! In 2011, Ailon has showed that by a smart choice of queries you can compute a $(1+\epsilon)$-approximation while reading only $O(\epsilon^{-6}\cdot n\cdot log^5n)$ entries (while having the entire matrix means making $O(n^2)$ queries) of the weight matrix $W$ (which becomes the adjacency matrix for unweighted instances),.<|endoftext|> TITLE: Extended Deformation Theory (dg-Lie algebra principle in positive characteristic?) QUESTION [11 upvotes]: Recently, I looked at articles that make use of Deligne's idea that "in characteristic 0 every deformation problem is governed by a differential graded Lie algebra" as explained first in Goldman-Millson (1988). So I wonder what precisely goes wrong in (mixed) characteristic p>2? REPLY [8 votes]: More precisely, the "Deligne principle" of deformation theory (but actually one could add a lot of other names) is that every deformation problem corresponds to a deformation functor, which in turn is defined by a certain dg Lie algebra controlling the deformations. More precisely, from any dg Lie algebra $g$ and any commutative algebra $A$ you obtain a new dg Lie algebra $g\otimes_{\mathbb{K}}A$, and this gives rise to a functor $$ MC(g\otimes-):\textit{Com-Alg}\rightarrow \textit{Sets} $$ which is your deformation functor. Now, you would like to define equivalences classes of deformations over $A$ by a certain equivalence relation in $MC(g\otimes A)$. For this, you use the fact that the degree $0$ part of your dg Lie algebra $g\otimes A$ is a Lie algebra that you can exponentiate into an algebraic group $G(A)$ called the gauge group (under nilpotence assumption). The gauge group acts on the Maurer-Cartan elements, and the equivalences classes are given by the quotient under this action. I think you should at least encounter some troubles if you try to exponentiate your Lie algebra in positive characteristic. As far as I know, you have to use exponential power series. Moreover, in positive characteristic $p$ you have to use restricted Lie algebras or Lie $p$-algebra (in order to take into account the operation coming from the Frobenius map), and consider nilpotent. I do not know any reference about deformation theory "à la Deligne" in positive characteristic using restricted Lie algebras.<|endoftext|> TITLE: Invariant subsets of $z \mapsto z^2$ QUESTION [11 upvotes]: Where can I find an explicit construction of closed invariant subsets of the map $z \mapsto z^2$ on the unit circle? Furstenberg mentions that there are continuum of such disjoint minimal sets but does not provide specific details. REPLY [5 votes]: Let's consider the semicircle $I_\theta=[\theta,\theta+1/2]$ for each $\theta\in\mathbb{T}$. Then there is a unique minimal ordered subset $K_\theta\subset I_\theta$, in the sense that the restriction of $\tau:x\mapsto 2x$ on $K_\theta$ is order-preserving. So the rotation number $\rho(\theta)=\rho(K_\theta,\tau)$ is well defined. It is proved by Bullett and Sentenac that $\text{Im}(\rho)=\mathbb{T}$. In particular, for every rotation number $\rho\in\mathbb{T}$, there exists a minimal subset $K_\theta$ with rotation number $\rho$ (so they form a continuum family of minimal subset). Example 1. For all $\frac{1}{6}<\theta<\frac{1}{3}$, we have $K_\theta=\{\frac{1}{3}, \frac{2}{3}\}\subset I_\theta$ and $\rho(\theta)=\frac{1}{2}$. Example 2. For all $\frac{9}{62}<\theta<\frac{5}{31}$, we have $K_\theta=\{\frac{5}{31}, \frac{10}{31}, \frac{20}{31}, \frac{9}{31}, \frac{18}{31}\}\subset I_\theta$ and $\rho(\theta)=\frac{2}{5}$. The unique invariant measure supported on $K_\theta$ is called a Sturmian measure. So I think this construction is related to Professor Anthony Quas's answer.<|endoftext|> TITLE: Sum of digits of repeating end of reciprocal of prime over period is $\frac{9}{2}$ QUESTION [5 upvotes]: Take a prime other than 2,3 or 5 and look at the part of it that repeats in base 10. Is it true that the sum of the digits in the end divided by the period(number of repeated digits id always $\frac{9}{2}$ Can we extend this result to other bases? Why does this happen? Regards. This is a crosspost from math.se REPLY [11 votes]: Note first that the counterexamples to user4140's observation involve decimal expansions with odd period length. In the case of even period length, what user4140 has observed is a consequence of Midy's theorem, which says the following: let p be a prime not dividing 10, and suppose that the repeating part of the decimal expansion of 1/p has even length. Corresponding digits in the first and second half of the repeating part then add to 9. (For example, we have 1/7 = 0.142857 and 1+8 = 4+5 = 2+7 = 9.) The proof of Midy's theorem is not hard and essentially comes down to the fact that modulo p there are exactly two numbers whose square is 1, namely 1 and -1. The same argument applies to the decimal expansion of 1/n whenever there are exactly two numbers whose square is 1 modulo n. (The n with this property are thus n = 4 and n = a power of an odd prime.) The argument also applies to base-b expansions, where "b-1" plays the role of "9". REPLY [3 votes]: The reason why this seems to be true for lots of cases is that it is true when the period of the fraction is even (hence different from 3) and $p$ is coprime with $10$ (hence different from $2$ or $5$). This is a known result which follows from Fermat's little theorem and actually the fraction can have any numerator $r$, since it doesn't affect the result. Here is the proof: For a prime $p$ coprime with $10$ Fermat's theorem guarantees the existence of an integer $c$ such that $10^{p-1}-1=cp$. Hence, supposing the period is even, say, $2t$, one has: $$\frac{r}{p}=\frac{rc}{10^{2t}-1}$$ Assume wlog that $r < p$. The right hand side is the sum of a geometric series showing that, expressed in base $10$, the period is exactly $rc$. Let $M$ and $N$ be the two halves of the block of digits. It is enough to prove that $M+N=10^t-1$, since then the sum of the digits of the period (which is precisely $M+N$) will be exactly $9t$ (as $10^t-1$ consists of $t$ nines in base $10$). Now, we had $(10^{t}-1)(10^{t}+1)=cp$, and it is easy to see that $p$ does not divide $10^{t}-1$ (otherwise, the number of digits in the period cannot be $2t$, since it is always the smallest integer $k$ such that $p$ divides $10^k-1$). Hence, $p$ divides $10^k+1$. The equation displayed above now implies: $$\frac{r}{p}=\frac{M.10^t+N}{(10^{t}-1)(10^{t}+1)}$$ and so: $$\frac{r(10^{t}+1)}{p}=M+\frac{M+N}{10^t-1}$$ while $M, N \leq 10^t-1$. Since the left hand side is an integer, so is the fraction on the right hand side, and the previous inequalities proves that $M+N$ is at most $2(10^t-1)$, so the integer in question is $1$ or $2$. But it cannot be $2$, since otherwise the period would consists only of nines, which is absurd. That finishes the proof. The same idea of the proof can be adapted to show a similar statement for a general base $b$, in which case one just has to choose a prime $p$ coprime with $b$ to be able to use Fermat's theorem.<|endoftext|> TITLE: What are all the stationary and pointwise independent random processes? QUESTION [6 upvotes]: In the 60's, I. Gel'fand introduced the concept of generalized stochastic processes (Ch. III, Vol. 4 of his work on Generalized functions). For a generalized stochastic process $\Phi$, he defines the concepts of stationarity ($\Phi(\varphi)$ and $\Phi(\varphi(\cdot - t_0))$ have the same law) and of independence at every point (the random variable $\Phi(\varphi_1)$ and $\Phi(\varphi_2)$ are independent if $\varphi_1$ and $\varphi_2$ have disjoint supports). Gel'fand especially introduces the complete class of Lévy white noises as generalized stochastic processes with characteristic functional of the form $$L(\varphi) =\exp\left( \int f(\varphi(t)) \mathrm{d}t \right),$$ with $f$ a function that has a L\'evy-Khintchine representation. Obviously, white noises are not the only stationary and independent at every point processes (ex: the weak derivative of a white noise). I am interested by a characterization of stationary and independent at every point processes. Especially, Gel'fand conjectured the following result. Conjecture: If a generalized stochastic process $\Phi$ is stationary and independent at every point, then the characteristic function $L$ of $\Phi$ has the form $$L(\varphi) =\exp\left( \int f(\varphi(t),\varphi^{(1)}(t),\cdots, \varphi^{(n)}(t)) \mathrm{d}t \right),$$ with $f$ a continuous function from $\mathbb{R}^{n+1}$ to $\mathbb{C}$ with $f(0)=0$. Is that result true? In order to express a kind of reciprocal result, can we characterize the functions $f$ such that the previous functional is a characteristic functional (main problem: its positive-definiteness)? Some people are extensively studied the positive-definiteness of functionals but wasn't able to find references that are clearly answering my question. Thanks for your attention. REPLY [2 votes]: I would suggest the reference [1], where J.N. Pandey addresses this problem. [1] J.N. Pandey, ``On the positive definiteness of a functional'', Canad. Math. Bull., Vol 22 (2), 1979<|endoftext|> TITLE: Normalization of a curve and push forward of vector bundles QUESTION [5 upvotes]: Let $C$ be a projective curve (over an algebraically closed field, not necessarily of characteristic zero) which is smooth except for exact one node. Let $\pi:\tilde{C} \to C$ be its normalization. Let $\mathcal{V}$ be a locally free sheaf over $\tilde{C}$. Is it true that the push forward, $\pi_*(\mathcal{V})$ is a locally free sheaf? If not, are there any general conditions (for example on the rank of $\mathcal{V}$) so that this holds true? REPLY [3 votes]: Quoting answer_bot: "No and yes for the rank zero vector bundle." In other words, the answer is "no" except for one case: the rank zero vector bundle. That means the answer is already "no" for every rank 1 vector bundle. To user46578, I recommend that you contemplate what happens for the trivial rank 1 vector bundle on $\widetilde{C}$. If you still have trouble, perhaps you should ask on Math StackExchange.<|endoftext|> TITLE: intersection of finitely many maximal ideals QUESTION [5 upvotes]: For what commutative rings with infinitely many maximal ideals we can say that the intersection of any combination of finitely many maximal ideals is not zero? Obviously it holds for Dedekind domains because a product of fin many is not zero and the product is contained in the intersection. I would like some weaker condition... Thanks REPLY [11 votes]: If finitely many distinct maximal ideals $\mathfrak{m}_1, ..., \mathfrak{m}_r$ of a commutative ring $R$ have intersection $0$, then by the Chinese remainder theorem the ring is isomorphic to the finite product of fields $\prod_{i=1}^r R/\mathfrak{m}_i$; in particular, it has only finitely many maximal ideals. So the answer to your question is: "For all."<|endoftext|> TITLE: Perfect set property implies $\omega_1$ is a limit cardinal in $L$ QUESTION [5 upvotes]: Specker proved in 1957 that if in $V$ every set of real numbers has the perfect set property, than in $L$, $\omega_1^V$ is actually a limit cardinal. The original proof is in German, and I've been looking for an English account of the proof. I couldn't find it in all the usual places (usually just a reference). I do remember it not being that complicated when it was given in a course I attended a couple of years ago, so it seems strange that I couldn't find it in the books. Can anyone help me find such English version of the proof (or give it here if it is short enough)? REPLY [8 votes]: The usual proof (as in Kanamori's book, section 11) is as follows: Work in $\mathsf{ZF}$. Note first, with Bernstein, that if $\omega_1\le\mathfrak c$, then there is a set of reals without the perfect set property: Either $\omega_1=\mathfrak c$, so $\mathbb R$ can be well-ordered, and we can build Bernstein sets using the usual transfinite recursion, or else $\omega_1<\mathfrak c$, and any set of reals of size $\omega_1$ lacks the perfect set property (that there are exactly continuum many perfect sets, and that each perfect set contains a copy of the Cantor set and therefore has size $\mathfrak c$ are provable in $\mathsf{ZF}$). Now, under the assumption that all sets of reals have the perfect set property, we argue that $\omega_1$ is a limit cardinal in $L[r]$ for all reals $r$: Suppose otherwise, so for some real $r$ and some $\kappa$, $\omega_1=\kappa^+$ in $L[r]$. Let $s$ be a real coding $r$ and a well-ordering of $\omega$ in type $\kappa$. In $L[s]$, we have that $\omega_1$ is computed correctly. But now we see that $\omega_1\le\mathfrak c$, as witnessed by $\mathbb R^{L[s]}$. As Asaf points out in the comments, if $\omega_1$ is regular in $V$, this gives us that it is inaccessible in $L[r]$ for all reals $r$, but it is equiconsistent with $\mathsf{ZF}$ that $\omega_1$ is singular and yet the perfect set property holds, see John Truss. Models of set theory containing many perfect sets. Ann. Math. Logic, 7, (1974), 197–219. MR0369068 (51 #5304).<|endoftext|> TITLE: Making Hironaka's theorem explicit for hypersurfaces QUESTION [8 upvotes]: Given a smooth hypersurface $H$ in $\mathbb{C}^n$, a theorem of Hironaka promises that one can find a strict normal crossings compactification $\bar{H}$ inside of a projective variety $X$. For me, this result is the definition of a "black box fact". What are some techniques for computing such a compactification in practice? More practically, I am interested in obtaining a strict normal crossings compactification of the Koras-Russell cubic which is the hypersurface in $\mathbb{C}^4$ cut out by the equation: $$ x + x^2y + z^2 + t^3 = 0 $$ I should confess that I haven't really tried in this example. I could just compactify the hypersurface in say $\mathbb{C}P^n$ (or some weighted projective space) and then start blowing things up and if I am lucky enough eventually arrive at something smooth. But I'm wondering how professional algebraic geometers go about doing these things in a more systematic way. REPLY [5 votes]: By now there are more tractable proofs of resolution of singularities than Hironaka's, so it no longer has to be a black box. A relatively elementary approach is described in Kollár's Lectures on Resolution of Singularities. There are many references there to algorithmic resolutions. In particular, in the case of hypersurfaces there is even a Maple program to do it. Check out Bodnár-Schicho's paper Automated Resolution of Singularities for Hypersurfaces and the references in there. (I'm not trying to be comprehensive in this answer to provide all the references, because there are lots of them. But probably most references that are at least 6-8 years old are referenced in the above two sources.)<|endoftext|> TITLE: Classical and Quantum Chern-Simons Theory QUESTION [22 upvotes]: Please excuse a sloppy question from an old user who hasn't been here in a long time. I think the expertise here is such that it can be answered anyway. Let $\Sigma$ be a two-manifold and $M$ a moduli space of flat connections on it with some gauge group $G$. $M$ will carry a determinant line bundle $L$. In a number of situations, if we pick a holomorphic structure on $\Sigma$, then we will get one on $M$ and $L$. Now, let's assume that $\Sigma$ is the boundary of a 3-manifold $B$. I would like to understand the process whereby Chern-Simons theory on $B$ gives rise to a section $v$ of $L$ over $M$. Now, I'm told this was first described in the paper of Witten on the Jones polynomial. There, by some process I don't really understand, there is a path integral formalism that fits together into a 3D TQFT so that $v$ is simply the image of the vacuum vector under the map induced by $B$. On the other hand, if you look at treatments like Lecture 4 in https://www.ma.utexas.edu/users/dafr/OldTQFTLectures.pdf I get the impression that one can get such sections in a very elementary manner by using just the classical Chern-Simons functional. (Last displayed formula on page 41 of the notes.) I seem to find the same thing in more recent treatments, such as papers of Andersen (which I've hardly looked into at all). So I thought I would ask if this understanding is indeed correct and, if so, what the relation is between the classical and quantum constructions of sections. REPLY [3 votes]: This question has been answered at PhysicsOverflow.<|endoftext|> TITLE: Hochschild homology of upper triangular matrix algebra? QUESTION [8 upvotes]: Let $K$ be a field and $A$ the associative unital $K$-algebra of all $n\times n$ upper triangular matrices with entries in $K$. What is $\dim_K$ of its hochschild homology $HH_k(A;A)$? Is there any software to compute Hochschild homology? REPLY [12 votes]: Sasha's argument is pretty technological. You can really do this almost by hand, though. Let $A$ be the algebra in question, let $r$ be its Jacobson radical (that is, the subspace of strictly upper triangular matrices), and let $E$ be the subalgebra of the diagonal matrices in $A$ (which is a complement to $r$) Notice that both $A$ and $r$ are $E$-bimodules. The algebra $A$ has a projective resolution as a bimodule of the form $A\otimes_E r^{\otimes_E *}\otimes_E A$ which looks exactly like the Hochschild resolution but the inner copies of $A$ have been replaced by $r$, and all tensor products involved are over $E$ and not over the base field; the differentials in the complex have exactly the same formula as the usual Hochschild differential. This can be checked easily —it is a nice exercise— but you can find the details in a nice paper by Claude Cibils on square-zero algebras, if I recall correctly. (This is like the reduced Hochschild resolution, but instead of killing the copy of $k$ inside $A$, we kill the whole of $E$; almost anything useful that one wants to do equires that we be aware of this complex, so it is important to keep it at hand) Now, $HH_*(A)$ is $Tor^{A^e}_*(A,A)$, so it can be computed as the homology of the complex obtained from $A\otimes_E r^{\otimes_E *}\otimes_E A$ by applying the functor $A\otimes_{A^e}(\mathord-)$. You should explicitly describe this complex: its only non-zero term is the $0$th one, so its homology is very, very easy to compute! (The same thing can be done for every triangular algebra, that is, every algebra whose ordinary quiver is acyclic) There are various programs people have written to compute Hochschild homology and cohomology; for example, I understand that Ed Green and his students have written code to do non-comm. Groebner bases on quotients of paths algebras and, probably, to compute (co)homology, and there are others (I have written code to handle very special cases, for example) I am not aware of any other approach apart from «try to be smart about n.c. Groebner bases, work hard for a minimal resolution, and then just do linear algebra.» The first step is pretty well understood for quotients of path algebras, say; the second one can be don algorithmically, I think; the third one is of course very well understood. In practice, interesting examples tend to result in huuuuge computations, and the result is somewhat unenlightening.<|endoftext|> TITLE: A simple proof that parallelizable oriented closed manifolds are oriented boundaries? QUESTION [28 upvotes]: So let $M$ be a smooth closed orientable real manifold such that $M$ is parallelizable, i.e., the tangent space $TM$ of $M$ is trivial. From the triviality of $TM$ we get that the Stiefel-Whitney and Pontryagin classes are trivial and therefore the Stiefel-Whitney and Pontryagin numbers are all equal to $0$. We have the following amazing theorem of C.T.C. Wall: (Wall's theorem) A closed orientable real manifold $M$ is the boundary of a compact oriented manifold (with boundary) iff its Stiefel-Whitney and Pontryagin numbers are trivial. Q: Using the stronger additional assumption that $M$ is parallelizable, is it possible to give a simple proof that $M$ is the boundary of an oriented manifold? REPLY [17 votes]: I sketch the proof of Buoncristiano and Hacon: Let $M$ be a parallelizable manifold of dimension $m$. Let $N$ be $M \times M \setminus U$, where $U$ is a tubular neighbourhood of the diagonal (invariant under the natural involution on $M\times M$.) The involution on $N$ can be induced from the antipodal involution on the sphere $S^q$ for a sufficiently big $q$ (i.e., one may chose a $\mathbb Z/2$-equivariant embedding $N\hookrightarrow S^q$). The boundary of $N$ is $M \times S^{m-1}$, and the involution on the boundary can be induced from that on $S^{m-1}$. So factorizing out by the involution the manifold $N$ we get a manifold $N'$, its map $f$ to $\mathbb{RP}^q$, and the boundary of $N'$ is mapped into $\mathbb{RP}^{m-1} \subset \mathbb{RP}^q$. Take an $\mathbb{RP}^{q-m+1}$ in $\mathbb{RP}^q$ that intersects $\mathbb{RP}^{m-1}$ in a single point. If both $f$ and its restriction to the boundary are transverse to this $\mathbb{RP}^{m-1}$ (this can be supposed) then $f^{-1}(\mathbb{RP}^{q-m+1})$ is a manifold with boundary $M$. Q.E.D.<|endoftext|> TITLE: How useful/pervasive are differential forms in surface theory? QUESTION [17 upvotes]: Every year I teach an introductory class on the differential geometry of surfaces, including numerical aspects (e.g., how to solve PDEs on surfaces). Historically this class has included an introduction to exterior calculus, describing (for example) differential operators on curved surfaces in terms of the exterior derivative and the Hodge star. I teach the class this way because that's how I myself learned differential geometry, but am starting to wonder how much value this approach really has for the students. Exterior calculus takes considerable effort to absorb, and students are often left wondering why we don't use simpler, more conventional language to discuss the same ideas. Take the gradient, for example. One could say that the gradient of a scalar field $\phi$ on a curved surface is the unique vector field $\nabla\phi$ satisfying $$ \langle \nabla\phi, u \rangle = D_u \phi, $$ where $\langle \cdot, \cdot \rangle$ is the Riemannian metric and $D_{\cdot}$ is the directional derivative. From there, it takes additional work to introduce the exterior derivative $d$ and the sharp operator $\sharp$ so that one can write the ever so slightly more succinct relationship $$ \nabla \phi = (d\phi)^\sharp. $$ But what is really gained by having this slightly more succinct expression? My question is an earnest one. I actually want to know, especially from people who do surface theory, why they bother to use exterior calculus and not a more elementary language. Or to put it another way: Are there clear examples where it's beneficial to use exterior calculus, as opposed to simpler language (e.g., writing everything explicitly in terms of the inner product)? How pervasive is exterior calculus in surface theory? E.g., roughly what percent of new results about surfaces would one estimate are expressed in the language of differential forms? Likewise, how much are new results being written in a more classical language? What did the historical development look like? I.e., why has exterior calculus become more (or less) popular over time? In other words, since this is not my main area, I am trying to get some broad cultural knowledge about how pervasive this tool is, and how likely it is to benefit my students. Thanks! REPLY [18 votes]: Well, there's really not a whole lot more to say beyond what Deane already wrote. He certainly hit the main points, but maybe I can expand a bit on what he wrote and comment on my own experience over the years both learning and teaching differential geometry. Maybe I should say a little bit about my own education: Originally, I learned differential geometry from O'Neill's Elementary Differential Geometry, which introduces differential forms and makes very good use of them. I went on to read a series of papers in differential geometry that used differential forms extensively, particularly many beautiful papers of S.-s. Chern and his students, and his postdoctoral advisor, the great Élie Cartan. So differential forms became very natural to me, but when I started teaching differential geometry, I found that they were always a little bit of a barrier to students, who had to take some time getting used to them. I experimented with teaching the curves and surfaces course without them, using, for example, do Carmo's wonderful book Differential Geometry of Curves and Surfaces, which avoids differential forms in favor of a classical vector calculus in local coordinates approach and manages to cover a lot of great material from the classical literature. Ultimately, though, I became convinced that this was not an efficient way to go, and have reverted to teaching the curves and surfaces course using differential forms, but spending a little extra time at the beginning to redevelop vector calculus using forms, which has its own benefits. Your example using differential forms to get a minor improvement in notation does not give any indication of the efficiency of actually using differential forms (mainly because you haven't really used the exterior derivative). Here's a better example: The construction of the Gauss curvature invariant of a metric $g$ on a surface. Locally write $g = {\omega_1}^2 +{\omega_2}^2$ where $\omega_1$ and $\omega_2$ are a ($g$-orthonormal) basis of $1$-forms. There is then a unique $1$-form $\phi$ that satisfies $\mathrm{d}\omega_1 = -\phi\wedge\omega_2$ and $\mathrm{d}\omega_2 = \phi\wedge\omega_1$, and there is a unique function $K$ such that $\mathrm{d}\phi = K\ \omega_1\wedge\omega_2$. Then $K$ clearly depends on $2$ derivatives of the coframing $\omega_i$, but it turns out to depend only $g$ and not on the choice of coframing. Here is why: If $g = {\bar\omega_1}^2 +{\bar\omega_2}^2$, then $\bar\omega_1 = \cos\theta\ \omega_1 + \sin\theta\ \omega_2$ and $\bar\omega_2 = \pm(-\sin\theta\ \omega_1 + \cos\theta\ \omega_2)$ for some function $\theta$ on the domain of the coframing. Then a simple computation yields $\bar\phi = \pm(\phi+\mathrm{d}\theta )$, so $$\bar K\ \bar\omega_1\wedge\bar\omega_2 = \mathrm{d}\bar\phi = \pm \mathrm{d}\phi = \pm K\ \omega_1\wedge\omega_2 = K\ \bar\omega_1\wedge\bar\omega_2\,, $$ so $\bar K = K$ depends only on the metric $g$. (It's hard to imagine any construction of $K$ and proof that it is invariantly defined that starts with writing $$ g = E(x,y)\ \mathrm{d}x^2 + 2F(x,y)\ \mathrm{d}x\mathrm{d}y + G(x,y)\ \mathrm{d}y^2 $$ and goes through the process of defining the Christoffel symbols and then $K(x,y)$ and then proving that the result doesn't depend on the choice of coordinates that isn't considerably longer than this.) Moreover, you get easy proofs of fundamental results: For example: If $K$ vanishes identically, then $\mathrm{d}\phi = 0$, so, locally $\phi = -\mathrm{d}\theta$ for some function $\theta$. Using this to define $\bar\omega_i$ as above, we get $\mathrm{d}\bar\omega_i = 0$, so $\bar\omega_i = \mathrm{d}x_i$ for some local functions $x_1$ and $x_2$, so $g = {\mathrm{d}x_1}^2 + {\mathrm{d}x_2}^2$, i.e., $g$ is locally flat. Similarly, there are easy proofs that $K\equiv 1$ implies that $g$ is locally isometric to the unit $2$-sphere, and lots of other such results. Of course, as Deane pointed out, it was Cartan who advocated and popularized the use of differential forms in surface theory and throughout differential geometry. Beginning in the 1890s and continuing through the 1940s, he wrote many highly influential papers and books using differential forms that effectively demonstrated their efficacy. The number of results that he was able to derive using them and the efficiency with which he did it is still astonishing today, and it made an enormous impression on the differential geometers of the time. Throughout the 20th century, his followers continued to develop and apply the techniques in higher dimensions, but also in classical surface theory. Many results from the classical literature were rewritten in forms language and extended. A typical (and beautiful) example is the paper by Chern and Terng, An analogue of Bäcklund's theorem in affine geometry (Rocky Mountain Journal of Mathematics (1980), 105–124, in which they give a much shorter proof of Bäcklund's theorem (about Euclidean surfaces with $K=-1$) using differential forms and discover a corresponding version in affine geometry. (One can give lots of examples of this kind, of course.) I have no way of reliably estimating the percentage of current research papers about curves and surfaces that use differential forms, but it is fairly high, simply because the language is very efficient; I can say that 100% of my own papers, even the curves and surfaces papers, use differential forms. I can't even imagine translating some of them out into classical vector calculus notation without them becoming much longer and essential unintelligible. The great efficiency of using differential forms accounts for their popularity; it's a pervasive language throughout differential geometry now. The resistance to introducing differential forms is that there seems to be an extra level of abstraction beyond ordinary vector calculus, but part of that is caused by the common feeling that you have to introduce tensors and all kinds of abstract concepts in order to define them. However, that's not really so if they are introduced the right way. Once students understand that $1$-forms measure velocities, $2$-forms measure areas, $3$-forms measure volumes, etc., they learn the rules pretty quickly and come to appreciate that the exterior algebra keeps track of what are otherwise messy formulas involving determinants and cross-products, etc. Anyone who has struggled to memorize the formulas for div, grad, and curl in cylindrical and spherical coordinates can appreciate the simplicity of the exterior derivative.<|endoftext|> TITLE: In what generality does Eilenberg-Watts hold? QUESTION [13 upvotes]: In homological algebra, the Eilenberg-Watts theorem states that if $F\colon\text{Mod}_R\to\text{Mod}_S$ is a right-exact coproduct preserving functor of modules, then $F\cong-\otimes_R F(R).$ The proof which you find in Rotman relies on the existence of a generator for $\text{Mod}_R$, the existence free resolutions, and the five lemma, all three things which are particular to the abelian category of modules over a ring. I'm now wondering whether this theorem will hold for a general category $C$, enriched over $\mathcal{V}.$ The module category of $C$ is the functor category $\hom(C,\mathcal{V})$, which is also the category of presheaves over $C$, which enjoys the nice property of being the free cocompletion of $C$. So a cocontinuous functor out of $\text{Mod}_C$ is completely determined by its values on $C$. Hence the $\mathcal{V}$-category of cocontinuous functors $\text{Mod}_C\to\text{Mod}_D$ is isomorphic to the functors $C\to\text{Mod}_D,$ and by the hom-tensor adjunction, this is functors $C^\text{op}\otimes D\to\mathcal{V}$, also known as bimodules (or spans, correspondences, profunctors, distributors, it has a lot of names). Perhaps even more easily seen, by the coYoneda lemma, every presheaf is a colimit of representable presheaves, which just explicitly says that so cocontinuous functor $F\cong -\otimes_C F(Y)$, where $Y$ is the Yoneda embedding. So my question is, is this a correct proof of the Eilenberg-Watts theorem? Are the assumptions of a generator, the existence of free resolutions, and the validity of the five lemma really not necessary for this result? REPLY [10 votes]: My advisor Mark Hovey recently updated a paper from 2009 in which he proved the Eilenberg-Watts Theorem in a very general case, namely for model categories. The arxiv version is here. Note that there's a big difference between version 1 and version 2, and in particular version 2 is much shorter. I recommend version 1 for those of a more algebraic dint. At the Oregon WCATSS13 conference, we also discussed this result for $\infty$-categories. If you want more information, I recommend emailing the organizers, as they have access to the abstracts and notes from most of the talks. In particular, we explicitly related Hovey's results to the results in Schwede and Shipley's paper Stable Model Categories are Categories of Modules. So this seems very related to your second paragraph, especially in situations arising from homotopy theory. Hope that helps!<|endoftext|> TITLE: Triple bubble conjecture: Natural candidate? QUESTION [6 upvotes]: Is there a standard natural candidate surface for the shape that encloses three given volumes in $\mathbb{R}^3$ and has minimal surface area? I know the planar triple bubble conjecture was proved by Wacharin Wichiramala in 2004 in a paper that I cannot access now ("Proof of the planar triple bubble conjecture." Journal fur die Reine und Angewandte Mathematik 567 (2004): 1-49). I am guessing this paper may describe the candidate in $\mathbb{R}^3$. Perhaps "the" is already incorrect, in that there may be several candidate surfaces? If so, the question could be narrowed to three equal volumes. Ultimately I am seeking an image of the candidate(s), but any help here would be appreciated—Thanks! Thanks for everyone's help! (I posted this summary at the same time j.c. was providing his comprehensive answer.) Here is an image made by John Sullivan: "a standard cluster of three bubbles, where the surfaces are spherical. Any three desired volumes can be achieved by a Möbius transformation of this cluster":         And here is a photograph found at this link:         (Added 9Mar14) j.c. found the original photo of the above and noticed it actually has a tiny 4th bubble! REPLY [11 votes]: Problem 2 in the list of open problems that Douglas Zare linked to answers the question (namely that there is a standard candidate, and it is even called the standard triple bubble). I quote it here with a few interspersed comments of my own. Problem 2 (Sullivan) We construct the standard clusters of k bubbles in $\mathbb{R}^n$ ($k\leq n+1$) as follows. Start with a regular k-simplex in $S^{k-1}$, and lift this to $S^n$ along longitudes. Then consider different stereographic projections of this complex into Euclidean space; they have $S^{n-k}$ symmetry. [The following is an image of the standard triple bubble in $\mathbb{R}^3$ as in Lucia's answer. It is by John Sullivan and it is on this page of his website.] Conj. For any k prescribed volumes, there is a unique standard k-bubble with those volumes [Here is a copy of the paper by Montesinos proving existence (a result of Brakke's) and uniqueness of the standard clusters (Conj. 1). I am hosting a copy in my dropbox since it seems International Press has removed their online copy.] Conj. This standard k-bubble is uniquely area minimizing. [As far as I know this is still open aside from the triple bubble case in the plane (Wichiramala (I'll host the paper for now)) and the double bubble case for $\mathbb{R}^n$ for all $n$ (Reichhardt and another proof by Lawlor).] Conj. For n=3, these are the only strictly stable k-bubbles ($k\leq 4$). Here we allow bubbles with disconnected regions; strictly stable means that the second variation is positive definite. [This paper of Morgan and Wichiramala proves that double bubbles (and two single bubbles) in 2D are the only stable bubbles with fixed area. See also this paper of Cicalese, Leonardi and Maggi.] Conj. There is a strictly stable cluster of six bubbles with a non-spherical interface, but no such cluster has fewer bubbles. ... [I include an image from the same page of Sullivan's as above.].<|endoftext|> TITLE: Consistency of Analysis (second order arithmetic) QUESTION [12 upvotes]: Is there a proof of the consistency of Analysis (second order arithmetic), which is similar to Gentzen's proof of the consistency of arithmetic? Update: Which (different) methods can be used to prove the consistency of Analysis? and where can I find such proofs? REPLY [14 votes]: As Noah says, the direct successor of Gentzen's method, cut-elimination, has been generalized up to $\Pi^1_2$-comprehension. This was shown separately by Rathjen and Arai; the full results have never been published, but fragments have appeared in various papers. Rathjen published "An ordinal analysis of parameter free $\Pi^1_2$-comprehension", which covers a fairly strong subtheory of $\Pi^1_2$-comprehension. The strongest results of Arai's I'm finding actually published only go up to $\Pi_3$ reflection, "Proof theory for theories of ordinals II: $\Pi_3$-reflection". My paper, "Ordinal analysis by transformations", is rather vague about the comparison between the system it's analyzing and usual hierarchy, but I now think the system it analyzes is around the strength of parameter free $\Pi^1_2$-comprehension. However the consistency of analysis has been shown by other means. First, as Carl mentions, Spector's proved the consistency of analysis in "Provably recursive functionals of analysis: a consistency proof of analysis by an extension of principles formulated in current intuitionistic mathematics". Spector's proof is considered non-constructive (it uses bar recursion). Girard gave a constructive consistency proof of analysis: he proves strong normalization proof for System F, which is an equivalent system. This proof can be found written up quite nicely in his book, Proofs and Types. REPLY [9 votes]: Although the proof-theoretic ordinal of second-order arithmetic is very hard to determine, there is another standard method for the proving consistency of arithmetic: Gödel's Dialectica interpretation. This was originally used by Gödel to give a different relative consistency proof of Peano arithmetic by reducing its consistency to the consistency of a quantifier-free theory of functionals of finite type known as system $T$. This work was later extended by Spector and Howard to give a relative consistency proof for second-order arithmetic. The weaker system used is the same system $T$ augmented with bar recursion. The details are spelled out in section 6 of Gödel's Functional ("Dialectica") Interpretation by Jeremy Avigad and Solomon Feferman from the Handbook of Proof Theory. Although this is not a Gentzen-style analysis, it does have a certain analogy. Gentzen showed that the consistency of Peano Arithmetic reduces to that of a weak theory augmented with transfinite induction. The Dialactica-style relative consistency proof for second-order arithmetic reduces its consistency to that of a (different) weaker theory $T$ augmented with bar recursion, which can be seen as a scheme for constructing objects by transfinite recursions. The induction scheme dual to bar recursion, bar induction, is a kind of transfinite induction scheme. The proof also gives a characterization of the provably total computable functions of second-order arithmetic, much like the consistency proof for Peano arithmetic does.<|endoftext|> TITLE: Formal completion of the normal bundle QUESTION [10 upvotes]: Let me for simplicity start with affine case. If $X=\operatorname{Spec}(A)$ is an affine variety $Z \subset X$ is a closed affine subvariety $Z=\operatorname{Spec}(A/I)$. What conditions are sufficient to say that completion of the (total space of) normal bundle $N_{Z/X}$ along zero section is isomorphic to the completion of $X$ along $Z$? REPLY [6 votes]: The formal completion of $X$ along $Z$ is always isomorphic to that of the normal bundle $N_{Z \subset X}$ along its zero section if $X= \mathrm{Spec}(A)$ and $Z = \mathrm{Spec}(A/I)$ are smooth and affine over some base $k$. If the affineness or smoothness assumption is dropped, it is easy to give counterexamples. It's easier to work in the world of algebra, so I will show that there exists an isomorphism of $k$-algebras $$\widehat{\mathrm{Sym}}_{A/I}(I/I^2) \to \widehat{A}$$ compatible with the projection down to $A/I$. Here $\widehat{A}$ is the $I$-adic completion of $A$, and $\widehat{\mathrm{Sym}}_{A/I}(I/I^2)$ is the completion of the symmetric algebra $\mathrm{Sym}_{A/I}(I/I^2)$ along the augmentation ideal. By the infinitesimal criterion for smoothness, one can find a compatible system of $k$-algebra maps $\epsilon_n:A/I \to A/I^n$ lifting the projection $A/I^n \to A/I$. In the limit, we obtain a $k$-algebra map $\epsilon:A/I \to \widehat{A}$ lifting the projection $\widehat{A} \to A/I$. Using $\epsilon$, we view $\widehat{A}$ and each quotient $A/I^n$ as $A/I$-algebras. Now the $A/I$-module $I/I^2$ is a projective module as $X$ and $Z$ are smooth. There is an obvious $A/I$-module map $I/I^2 \to A/I^2 \simeq \widehat{A}/I\widehat{A}^2$. By projectivity, one can find an $A/I$-module map $I/I^2 \to \widehat{A}$ lifting the previous one. By the universal property of the symmetric algebra, this extends to an $A/I$-algebra map $\mathrm{Sym}_{A/I}(I/I^2) \to \widehat{A}$. As both sides are augmented over $A/I$, this naturally extends to an $A/I$-algebra map $\widehat{\mathrm{Sym}}_{A/I}(I/I^2) \to \widehat{A}$; by filtering both sides suitably, one can check that this is an isomorphism. REPLY [2 votes]: This is exactly the same answer as that of anonymous (which was 12 minutes earlier). So please read that one. Suppose that $R$ is a ring and that $X$ is a scheme over $R$. I think the question is whether there exists an $R$-algebra ismorphism $$ A^\wedge \cong A/I[[I/I^2]] $$ where the left hand side is the $I$-adic completion of $A$ and the right hand side is completion of the symmetric algebra on $I/I^2$. Let's assume for simplicity that $I$ is finitely generated, so that we know that $A^\wedge$ is complete. To get such an isomorphism, the first thing we want is section $\sigma : A/I \to A^\wedge$ to the canonical map $A^\wedge \to A/I$. Besides $\sigma$ we want a map $\tau : I/I^2 \to A^\wedge$ which is $A/I$-linear (here we are using $\sigma$), maps into the kernel $IA^\wedge$ of the canonical map $A^\wedge \to A/I$ and then induces an isomorphism $I/I^2 \to IA^\wedge/I^2A^\wedge$. Given $\sigma$ and $\tau$ we can at least define an $R$-algebra map $$ A/I[[I/I^2]] \longrightarrow A^\wedge $$ In general the existence of $\sigma$ and $\tau$ is not good enough to imply that this map is an isomorphism. But if you assume for example that $I$ is a quasi-regular ideal then it is an isomorphism (follows more or less immediately from the definition of quasi-regular ideals). If both $A$ and $A/I$ are smooth over $R$, then $\sigma$ and $\tau$ exist and the resulting map is an isomorphism. The existence of $\sigma$ for example follows because $A/I$ is formally smooth over $R$, hence we can successively lift the map $\text{id} : A/I \to A/I$ to $A/I \to A/I^n$ and taking the limit gives $\sigma$. Then existence of $\tau$ comes from the fact that in this case the module $I/I^2$ is a finite projective $A/I$-module, hence there are no obstructions to lifting the linear map $I/I^2 \to A/I^2$ to $I/I^2 \to A/I^n$ and we can take the limit. Finally, the ideal defining a closed immersion of schemes smooth over $R$ is a regular ideal, see for example the more general Lemma Tag 067T. If you are unfamiliar with some of the terms used above please try using the search function of the Stacks project to find their definitions. Moreover, some of the statements used above about completions of rings aren't entirely trivial and can be found there as well.<|endoftext|> TITLE: Main conjecture for elliptic curves QUESTION [9 upvotes]: Suppose $E$ is an elliptic curve defined over $\mathbb{Q}$ with good ordinary reduction at a prime $p$. Then one can define nonnegative integers $ \lambda_{E}^{alg} $, $ \mu_{E}^{alg} $, $ \lambda_{E}^{an} $ and $ \mu_{E}^{an} $. The "algebraic" Iwasawa invariants $ \lambda_{E}^{alg} $ and $ \mu_{E}^{alg} $ are defined in terms of the structure of the $p$-primary subgroup $ Sel_{E}(\mathbb{Q}_{\infty})_{p} $ of the Selmer group for $E$ over the cyclotomic $ \mathbb{Z}_{p} $-extension $ \mathbb{Q}_{\infty} $ of $\mathbb{Q}$. The definition of the "analytic" invariants $ \lambda_{E}^{an} $ and $ \mu_{E}^{an} $ is in terms of the $p$-adic $L$-function for $E$ constructed by Mazur and Swinnerton-Dyer. Now the Main Conjecture (Mazur) implies that $ \mu_{E}^{alg}=\mu_{E}^{an} $ and $ \lambda_{E}^{alg}=\lambda_{E}^{an} $. I want to know 1) What are the results proved till now towards proving the Main Conjecture $?$ 2) For a particular elliptic curve over $\mathbb{Q}$ having good ordinary reduction at a prime $ p $, are there any methods to check that it satisfies the Main Conjecture $?$ EDIT: Prof. D Loeffler has mentioned that the main conjecture is now a theorem if the image of the mod $p$ Galois representation of $E$ is the whole of $GL_2(\mathbf{F}_p)$. But what happens if the residual representation of $E$ is not irreducible and $E$ has a $p$-isogeny $?$ REPLY [9 votes]: The main conjecture is a theorem if the image of the mod $p$ Galois representation of E is the whole of $GL_2(\mathbf{F}_p)$. The full statement of the conjecture, which implies what you wrote about lambda and mu invariants but is quite a bit stronger, is the claim that the p-adic L-function generates the characteristic ideal of the Selmer group. This was proved in two steps. The statement that the p-adic L-function lies in the char ideal of Selmer group (the Selmer group is not too large) was proved by Kato in 2004, using his Euler system; see his paper in Asterisque 295. This implies that $\mu^{alg} \le \mu^{an}$ and $\lambda^{alg} \le \lambda^{an}$. The reverse inclusion, that the p-adic L-function divides the characteristic ideal of Selmer (the Selmer group is not too small) was proved by Skinner and Urban in this 2014 paper.<|endoftext|> TITLE: Postnikov's algebraic reconstruction of cohomology from homotopy invariants QUESTION [17 upvotes]: In his short paper (1951) and longer monograph (1955), Postnikov introduced what I believe are now called Postnikov systems or towers. It is my understanding that Postnikov systems have since then been widely adopted, as a way of totally encoding the homotopy type of a topological space. Expressed algebraically, a Postnikov system of a topological space $X$ consists of its homotopy groups $\pi_i = \pi_i(X)$, where each of the higher groups $\pi_i$ ($i\ge 2$) has the structure of a $\pi_1$-module, together with a group cohomology class $[k_1] \in H^{3}(\pi_1,\pi_2)$ and higher "cohomology classes" (that are somewhat more difficult to describe algebraically) $[k_i]$ with coefficients in $\pi_{i+1}$ for each $i\ge 2$. The main application of these systems, in Postnikov's original work, was to reconstruct the cohomology of the space $X$ from its homotopy invariants in a purely algebraic way. The reconstruction allowed also for cohomology with coefficients in a local system, with the local system algebraically presented as some $\pi_1$-module. In that work, this result was presented as a generalization of the earlier work (1945) of Eilenberg and MacLane (and of course others) on the algebraic reconstruction of the cohomology of spaces that are aspherical except in one degree. Now my question. Is there a modern reference for the reconstruction of the cohomology (desirably with local coefficients) of a topological space from its Postnikov sequence? While Postnikov sequences themselves are treated in many places, I've not been able to find the cohomology reconstruction theorem anywhere except in Postnikov's original longer monograph (in Russian). REPLY [7 votes]: One question which has puzzled me is, just to take the $2$-type, how does one specify an element of $H^3(\pi_1,\pi_2)$? In 1972, Philip Higgins and I discussed with Saunders Mac Lane the possibility of a $2$-dimensional van Kampen theorem to calculate the homotopy $2$-type of a union, and he explained why he had decided this was impossible. Calculating $\pi_1$ is OK, by the van Kampen theorem. Then you have to calculate $\pi_2$, as a $\pi_1$-module; then the $k$-invariant! Absurd! Now the $2$-type is also described by a crossed module, by a theorem of Mac Lane and Whitehead. Philip and I proved in 1975, in a paper published in 1978, that under reasonable circumstances, the crossed module of a union is given by a pushout of crossed modules. That seems some kind of answer! Of course, there is still a problem of calculating $\pi_2$ from this pushout, and even more of calculating the $k$-invariant. But at least the topology has been translated into algebra; the theorem generalises strongly a theorem of Whitehead on free crossed modules, and has enabled specific, even computer, calculations of $2$-types. For more detail, see this 2011 EMS Tract vol 15. In effect, crossed modules are a useful algebraic model of homotopy $2$-types, and it is certainly useful to consider what one wants from an algebraic model. Thus Loday worried that his cat$^n$-group model (called by him $n$-cat groups in his 1982 paper) of homotopy $(n+1)$-types was "purely formal", so was very happy with our van Kampen theorem for $n$-cubes of spaces which enabled some new computations, for example of homotopy $3$-types of a suspension of a $K(G,1)$. More applications and related algebra were developed by Ellis and Steiner, in a 1987 JPAA paper. It also seems true that strict $n$-fold groupoids model weak homotopy $n$-types. How does one use this? and how is it related to Postnikov systems? To add a bit to a possible answer to the question on cohomology, Graham Ellis has a paper on the cohomology of crossed modules. Let the debate continue! May 7, 2019 Looking at my answer, I first updated two links and then realised there is more to be said on modelling homotopy types, part of which is in this paper, published in 2018. The general problem of relating cohomology to models of homotopy types seems very difficult. May 23, 2019 I should also have referred you to Loday's paper ‘Spaces with finitely many nontrivial homotopy groups’. J. Pure Appl. Algebra 24 (2) (1982) 179–202. and Ellis, G. J. and Steiner, R. ‘Higher-dimensional crossed modules and the homotopy groups of (n + 1)-ads’. J. Pure Appl. Algebra 46 (2-3) (1987) 117–136.<|endoftext|> TITLE: Nontrivial upper bounds on proof-theoretic ordinals of strong theories: do we have any? QUESTION [16 upvotes]: Motivated by Consistency of Analysis (second order arithmetic) and Proof-Theoretic Ordinal of ZFC or Consistent ZFC Extensions?, I have the following question: Are there any examples of strong theories for which we have nontrivial upper bounds on the proof-theoretic ordinal, but do not know the exact value yet? By "strong theory," I mean a theory $T$ such that for no theory $T'$ extending $T$ do we know the exact proof-theoretic ordinal of $T'$. Let me explain why I am making this restriction: the easiest way to have a nontrivial upper bound on the proof-theoretic ordinal of a theory is to have it be a subtheory of an already-analyzed theory. Such theories, however, may still have unknown proof-theoretic ordinals by virtue of being strange: e.g., I'm sure there are some fragments of $ATR_0$ which are vaguely interesting, but for which finding the proof-theoretic ordinal would require some serious new work. This is not what I'm looking for. Basically, what I understand of the process of finding proof-theoretic ordinals is that we "work from below," and try to build up a system of notations which exhaust the $T$-provably-well-founded recursive ordinals. I'm curious if there are any techniques for "working from above," other than actually computing the proof-theoretic ordinal of some stronger theory. REPLY [13 votes]: I think the short answer is no, and I'm not sure there really could be: I don't think we "know" any ordinals above the ordinal of $\Pi^1_2-CA$ but below $\omega_1^{CK}$. Theoretically someone could write down a large notation for, say, $\Pi^1_3-CA$ and then discover it was too big, but in practice the hard part seems to be writing down notations that are sufficiently large. And we don't seem to have a source for those other than getting the ordinals of yet stronger theories.<|endoftext|> TITLE: coherent modules QUESTION [6 upvotes]: Let $R$ be a nontrivial ring. A right $R$-module $M$ is called coherent if ${\rm Ker} (f)$ is finitely generated for any $R$-module homomorphism $f: L\to M$ with $L$ finitely generated. It is well-known that the full subcategory ${\rm Coh}(R)$ of ${\rm Mod}(R)$ consisting of coherent modules is an abelian category. I wonder that is it possible that ${\rm Coh}(R)$ is the trivial abelian category? In another word, does there exist a ring $R$ over which the zero module is the only coherent right module? REPLY [6 votes]: Yes, for example the ring $R = k[x_1, x_2, x_3, \ldots]/(x_ix_j)$ where $k$ is a field. Namely, every finite submodule of a coherent module is coherent and so every nonzero coherent module contains a nonzero coherent module of the form $R/I$. Then $I$ has to be finitely generated hence $x_i \not \in I$ for some $i$. Then $R/I$ contains a copy of $k$ (namely, the submodule generated by $x_i \bmod I$) which is not a coherent module over $R$. Hence $R/I$ is not coherent.<|endoftext|> TITLE: Fourier expansion of Takagi-function (everywhere non differentiable function). QUESTION [7 upvotes]: Let us consider Takagi-function defined by $T(x) \colon\!= \sum_{n=0}^{\infty}s(2^nx)/2^n$, where $s(x) \colon\!\!= \underset{n \in {\Bbb Z}}{\mathrm{min}} \,|x-n|$. $T(x)$ has its period $1$, so I am trying to develop its Fourier expansion $S(T)(x) \colon\!=\sum_{n= -\infty}^{\infty}\left(a_n\,sin(2 \pi n x) + b_n\,cos(2 \pi n x)\right)$. Q: Where do we have $T(x) = S(T)(x)$? For the case of Weierstrass function $W(x) \colon\!= \sum_{n=0}^{\infty}a^n cos(b^n \pi x)$ with $ab > 1 + 3\pi/2$, $ 0 < a < 1$ and $b$ an odd integer, it is already expanded in Fourier series with period $2$ by shape. By Carlson, we know $T(x) = S(T)(x)$ except for sets of Lebesgue measure $0$. I would like to know whether $T(x) = S(T)(x)$ holds everywhere for Takagi-function as well. REPLY [5 votes]: Yes. The triangle wave $s(x):=\min_{k\in\mathbb{Z}}\big|x-k\big|$ has an absolutely convergent Fourier series $$s(x)=\frac{1}{4}-\frac{2}{\pi^2}\sum_{k=0}^\infty\frac{1}{(2k+1)^2}\cos\big(2\pi (2k+1)x\big)\, ,\qquad x\in\mathbb{R}.$$ Therefore $$T(x):=\sum_{n=0}^\infty2^{-n}s(2^nx)= \frac{1}{4}\sum_{n=0}^\infty2^{-n}-\frac{2}{\pi^2}\sum_{n=0}^\infty\sum_{k=0}^\infty \frac{1}{2^n(2k+1)^2}\cos\big(2\pi 2^n(2k+1)x\big)\, .$$ By absolute convergence we can reorder the double sum. Since every positive integer $m\ge1$ writes uniquely as $m=2^n(2k+1)$ for nonnegative integers $n$ and $k$, we then obtain an absolutely convergent Fourier series for $T(x)$: $$T(x)=\sum_{m=0}^\infty a_m\cos(2\pi m x)$$ with $a_0=1/2$ and for $m\ge 1$ $$a_m:=-\frac{2^{\nu(m)+1}}{\pi^2m^2}\, , $$ where $2^{\nu(m)}$ is the maximum power of $2$ that divides $m$.<|endoftext|> TITLE: Riesz's representation theorem for non-locally compact spaces QUESTION [12 upvotes]: Every version of Riesz's representation theorem (the one expressing linear functionals as integrals) that I have found so far assumes that the underlying topological space is locally-compact. (For instance, Rudin in "Real and Complex Analysis" uses this assumption in the proof of Urysohn's lemma, upon which he bases the proof of Riesz's theorem.) Nevertheless, in an answer on MO, @jbc (inactive on MO since 2013 and with no real name or contact information available) claims that the theorem can be stated in much more generality, for Hausdorff completely regular spaces (which suggests that Urysohn's lemma is indeed the essential ingredient here). Unfortunately, no detail or bibliography is given in that answer. Currently, I have only found a practically useless theorem in Bourbaki's "Elements of Mathematics - Integration II", chapter IX, paragraph 5, page IX.59, and a very difficult to use version in "Bounded Continuous Functions On A Completely Regular Space" by Dennis F. Sentilles ("Transactions of the AMS", vol.168, June 1972, page 311, theorems 9.1.c and 9.2). My question is: does anybody know of a "definitive", recent statement concerning the dual of the algebra of bounded continuous functions on non-locally-compact spaces? What is lost when one gives up local-compactness? (Please notice that I am not interested in the algebra of functions with compact support or vanishing at infinity.) REPLY [2 votes]: Heinz König has proved a very general version of the the Riesz Representation Theorem on an arbitrary Hausdorff topological space. See Chapter 5 of the book: H. König, Measure and Integration: An Advanced Course in Basic Procedures and Applications, Springer, 1997, corr. reprint 2009, pp. XXI+260. The monograph H. König, Measure and Integration: Publications 1997–2011, Birkhäuser, Springer Basel, 2012, pp. XI+512 is a collection of some related papers of Konig that appeared after the book, including some on Konig's representation theorems. The following article provides a nice overview of the machinery König develops. H. König, Measure and integral: new foundations after one hundred years. (English summary) Functional analysis and evolution equations, 405–422, Birkhäuser, Basel, 2008.<|endoftext|> TITLE: Necessary condition on Calabi-Yau manfiold to be a hypersurface in a Fano manifold QUESTION [8 upvotes]: Let $X$ be a smooth projective Calabi-Yau threefold. Are there any known obstructions to it being a member of a base-point-free linear system in a nef-Fano fourfold? What, in anything, is known regarding a similar question in one dimension less (K3 in Fano threefolds)? REPLY [3 votes]: For the K3/Fano threefolds case you might have a look at Beauville's paper Fano threefolds and K3 surfaces. Proceedings of the Fano Conference 175-184, Univ. di Torino (2004).<|endoftext|> TITLE: Angular equidistribution of lattice points on circles QUESTION [11 upvotes]: The following question is well known: Consider representations of a given integer as sums of two squares, i.e. solutions to $a^2 + b^2 = n$ in $a,b\in\mathbb Z$ with $n$ fixed. As $n \to \infty$, are the normalized points $\left(\frac a {\sqrt n}, \frac b {\sqrt n}\right)$ uniformly distributed on the unit circle? An equivalent formulation is given by: Denote by $\Delta(n)$ the discrepancy $$\Delta(n) = \sup_{\Gamma\text{ arc on }\mathcal S^1} \left|\frac {\text{number of points for }n\text{ in }\Gamma}{\text{number of points for }n} - \frac {\text{Length}(\Gamma)} {2\pi}\right|$$ Then as $n \to \infty$, is it true that $\Delta(n) \to 0$? The answer is "it depends on what you mean by $n \to \infty$". Obviously, for many $n$ there are no points at all or a small number of points. Even if we require the number of points to grow to infinity, a counterexample was constructed by J. Cilleruelo in The distribution of the lattice points on circles, J. Number Theory 43 (1993), no. 2, p. 198-202. An answer in the positive direction is given by Erdős and Hall in On the angular distribution of Gaussian integers with fixed norm, Discrete Math 200 (1999), p. 87-94. The formulation is something like this (which I reproduce here in a simplified fashion): For "almost all" integers $n \le x$ that are representable as a sum of two squares, we have $\Delta(n) \le \log ^{-\kappa} x$, where $\kappa > 0$ is an absolute constant. A similar result (with essentially the same analysis) was given earlier by Kátai and Környei in On the distribution of lattice points on circles, Ann. Univ. Sci. Budapest., Sect. Math. 19 (1977), p. 87-91. I have two questions: The analysis done by Erdős and Hall and by Kátai and Környei is based on an averaging argument. This means they say nothing about which $n$ give low discrepancy, only that most $n$ do. Is there a known result that gives criteria on $n$ to ensure low discrepancy? (In contrast, Cilleruelo's counterexample construction gives a criterion to ensure high discrepancy.) It seems odd to me that this result is so recent (1999 or even 1977): It has been done in higher dimensions, which seems a lot harder, in 1959 (Pommerenke), and in this MathOverflow question a reference to a similar problem was given from 1920. Is this "folklore question" - if so, what is "new" about the more recent results? Could you give me a reference for the simplest way to solve the problem if what I want is to show that for "many sequences of integers" $n \to \infty$ (preferably, with description) we have $\Delta(n) \to 0$, and I don't care about the speed of convergence? On the other hand, if this indeed is a new result - how come it wasn't known in the 1950s? REPLY [8 votes]: If $p\equiv 1 \pmod 4$ then we may write $p=a^2+b^2$ in a unique way with $a$, $b$ both positive and $b< a$. Corresponding to such a representation, write $a+bi = \sqrt{p} e^{i\theta(p)}$ with $\theta(p) \in (0,\pi/4)$. If now $n=p_1\cdots p_k$ (assume for simplicity that $n$ is odd and square free) where the $p_j$'s are $1\pmod 4$, then by multiplying $\sqrt{p_j}e^{\pm i\theta({p_j})}$ we find many representations of $n$ as a sum of two squares, and the angles corresponding to these representations are $\sum_{j} \epsilon_j \theta(p_j)$ where $\epsilon_j=\pm 1$. We can also multiply these representations by $\pm 1$ and $\pm i$ at the end, but that clearly doesn't affect our discussion. The problem then is about the equidistribution of the angles $\sum_{j=1}^{k} \epsilon_j \theta(p_j)$ when we vary $\epsilon_j$ over all the $2^k$ possible choices of sign. Now suppose there is some subset of the $\theta(p_j)$ (say $\{ \theta(p_1), \ldots, \theta(p_\ell)\}$) which is equidistributed in $(0,\pi/4)$ (that is $\ell$ is large, and the discrepancy of these points for arcs in $(0,\pi/4)$ is small). Then using Weyl's criterion (or the Erdos-Turan discrepancy bound) one can see that the points $\sum_{j=1}^{\ell} \epsilon_j \theta(p_j)$ will get equidistributed $\pmod{2\pi}$ (that is the discrepancy will be small). Since the points in $\sum_{j=1}^{k} \epsilon_j \theta(p_j)$ are $2^{k-\ell}$ translates of this equidistributed set, it follows that they too are equidistributed. This gives a criterion for when you should expect equidistribution of angles, and is essentially the criterion used in Cilleruelo's paper referenced in my comment. Since most numbers have a lot of prime factors, and $\theta(p)$ is equidistributed in $(0,\pi/4)$ as $p$ runs over all primes, this criterion can be used to show angular equidistribution for almost all numbers that are sums of two squares. As a concrete example, if $N= \prod_{p\le z, p\equiv 1\pmod 4} p$ then it follows that the angles corresponding to representations of $N$ as a sum of two squares are equidistributed (as $z\to \infty$). On the other hand, if the points $\theta(p_j)$ (for $1\le j\le k$) all accumulate near rational numbers with the same denominator, then one will not get equidistribution. Thus for example if all the $\theta(p_j)$ are small, then one has non-uniform distribution, and this is what Cilleruelo does to produce examples with high discrepancy.<|endoftext|> TITLE: Are finite-dimensional representations of groups of type $\text{FP}_{\infty}$? QUESTION [5 upvotes]: Let $G$ be a group (possibly infinite) and $k$ be a field. A module $M$ over $k[G]$ is said to be of type $\text{FP}_{\infty}(k)$ if it has a projective resolution each of whose terms is finitely generated. We say that $G$ itself is of type $\text{FP}_{\infty}(k)$ if the trivial $k[G]$-module $k$ is of type $\text{FP}_{\infty}(k)$. Assume that $G$ is a group of type $\text{FP}_{\infty}(k)$ and that $M$ is a $k[G]$-module that is finite-dimensional over $k$ (in other words, $M$ is a finite-dimensional representation of $G$). Must $M$ be of type $\text{FP}_{\infty}(k)$? If not, are there stronger finiteness properties that we can put on $G$ to assure that this holds (for instance, having a compact $K(G,1)$)? REPLY [7 votes]: Cleaned up answer (6/11/18) after comments of Andy Putman. The answer is yes. Thm 2 of https://www.tandfonline.com/doi/abs/10.1080/00927870600796110 shows that if G is $FP_\infty$ over $k$, then $kG$ has a free resolution as a bimodule by finitely generated free bimodules in each dimension. If you tensor this resolution with $M$ over $kG$ you get a free resolution of $M$ with the finiteness properties you want. Tensoring with $M$ gives a resolution because its homology is $Tor^{kG}(M,kG)$. It is easy to check that $(kG\otimes_k kG)\otimes_{kG} M\cong kG^{\dim M}$ as a left $kG$-module so that the free resolution is finitely generated in each degree. The basis as a $kG$-module is the tensors $1\otimes 1\otimes b$ with $b$ running over a basis of $M$.<|endoftext|> TITLE: Are n-truncated quasicategories a model for n-categories? QUESTION [12 upvotes]: In Higher Topos Theory, Lurie gives a description of $n$-categories (section 2.3.4) in terms of quasicategories. In other words, he gives a definition (2.3.4.1) of an $\infty$-categorical notion of $n$-category, as well necessary and sufficient conditions for a given quasicatogory to be equivalent to an honest $n$-category (as he defines it) (2.3.4.18). For me, the important characterization is in terms of mapping spaces. That is, a quasicategory is equivalent to an $n$-category if its mapping spaces are all $n-1$-truncated. So I have two questions: In section 5.5.6 of Higher Topos Theory, Lurie describes truncated objects of quasicategories (generalizing the notion of truncated spaces, for instance as in the construction of the Postnikov tower). Does Lurie's characterization of $n$-categories as described above coincide with the $n$-truncated objects of the $(\infty,1)$-category of $(\infty,1)$-categories? I believe I have a rough proof of this, but I'm not certain, and it seems like the sort of the thing that would already be written up somewhere. If the above is true, can we get at $E_n$-objects by looking at the images of $E_\infty$-objects under these truncation maps? Of course, the above must come with certain caveats. For instance, by $E_1$-object I would mean a monoid with an $A_\infty$-multiplication that was also commutative up to no higher homotopy or something like that. I'm not really sure if that question makes sense, but I guess ultimately what I'm asking is, if we're only interested in commutativity up to degree $n$-homotopies, can we get everything we need by looking inside of $n$-truncated $\infty$-categories? REPLY [11 votes]: Let $C$ be an $\infty$-category, and $n\geq -1$. The following are equivalent: $C$ is $n$-truncated. The $\infty$-groupoids $\def\Map{\operatorname{Map}}\Map(\Delta^0,C)$ and $\Map(\Delta^1,C)$ are $n$-truncated. (Remember that $\Map(B,C)$ is the maximal Kan complex inside $\operatorname{Fun}(B,C)$.) ($n\geq0$ only) For all pairs of objects $x,y$ in $C$, the $\infty$-groupoid $\Map_C(x,y)$ is $n$-truncated and $\def\Aut{\operatorname{Aut}}\Aut_C(x)$ is $(n-1)$-truncated. (Here $\Aut_C(x)\subseteq \Map_C(x,x)$ is the subobject of self-equivalences.) (1) $\Rightarrow$ (2) is immediate. (2) $\Rightarrow$ (1) is because $\mathrm{Cat}_\infty$ is generated under colimits by $\Delta^0$ and $\Delta^1$. (2) $\Leftrightarrow$ (3) is via the fiber sequences $$ \Map_C(x,y) \to \Map(\Delta^1,C) \to \Map(\Delta^0,C)\times \Map(\Delta^0,C). $$ associated to each pair of objects $(x,y)$ and $$ \Aut_C(x) \to * \to \Map(\Delta^0,C) $$ associated to each object $x$ (This last one is why you need $n\geq 0$.)<|endoftext|> TITLE: Counting categories with at most $n$ morphisms QUESTION [21 upvotes]: There are a number of results which count the number of possible algebraic structures on a set of $n$ elements. Notable previous MO questions are for example here and here. The analogous question for finite categories does not seem to enjoy the same attention as other algebraic structures. The only source I could find was the following OEIS entry. Simply put the first question is as follows: How many categories with at most $n$ morphisms are there up to isomorphism? Up to equivalence? Comparing with the analogous OEIS entry for monoids, we see that for small $n$, "almost all" categories are monoids. In general, under some appropriate weighting scheme (which?), are almost all categories monoids, in the sense that the quotient $$\frac{\text{Number of monoids with at most }n \text{ elements}}{\text{Number of categories with at most }n \text{ morphisms}}\rightarrow 1$$ as $n\rightarrow \infty$ ? Tom Leinster asks a similar question about groups on the $n$-category Café. REPLY [6 votes]: Note: I originally posted an answer claiming the opposite, and then deleted it because it was wrong. I have since reworked it and made this post community wiki. This doesn't directly answer your question, but you may be interested to know that the conjecture is true in the infinite case. In fact, for every infinite $\kappa$ there are $2^\kappa$ categories with $\leq\kappa$ morphisms, up to equivalence or isomorphism, $2^\kappa$ of which are monoids. Of course, in the infinite case, it's more general to ask how many categories there are with $<\kappa$ morphisms rather than $\leq \kappa$ morphisms, but I think this question is more complicated. On the one hand, $2^\kappa$ is an obvious upper bound on the number of categories with $\leq \kappa$ morphisms: since composition is a partial binary operation on the set of morphisms, there are at most $\sum_{\lambda \leq \kappa} 2^{\lambda \times \lambda \times \lambda} \leq \kappa 2^\kappa = 2^\kappa$ categories with $\leq \kappa$ morphisms. On the other hand, it's easy to find $\kappa$-many pairwise-inequivalent connected categories with $\leq\kappa$ morphisms: for instance, it suffices to consider the oridnals $<\kappa$. By taking disjoint unions of these, we get $2^\kappa$ pairwise-inequivalent categories with $\leq \kappa$ morphisms. We can tweak this construction to get $2^\kappa$ pairwise-inequivalent monoids. For each $S\subseteq \kappa$, we can adjoin a bottom element to $\coprod_{\lambda \in S} \lambda$ to get a semilattice $(\coprod_{\lambda \in S} \lambda)_{\bot}$, and then consider the corresponding monoid under $\vee$. We obtain $2^\kappa$ monoids this way by taking different $S$, and they are still pairwise nonisomorphic as monoids, because a monoid isomoprhism entails a semilattice isomorphism, which entails an isomorphism of the original disjoint-union categories, which are clearly pairwise inequivalent.<|endoftext|> TITLE: Is there a holomorphic Morse-Bott lemma? QUESTION [11 upvotes]: It asks for a generalization of the question in the post Normal form for a holomorphic Morse function Suppose $f$ is a holomorphic function on a complex manifold $M$ which has Bott type critical points, i.e., $df$ vanishes along a complex submanifold $S\subset M$ with nondegenerate Hessian in the normal direction to $S$. For each $p \in S$, can we find a holomorphic local coordinates $z_1, \ldots, z_s, z_{s+1}, \ldots, z_m$ such that locally $$f(z_1, \ldots, z_m) = \sum_{j = s+1}^m z_j^2 + C,$$ where $s = {\rm dim} S$? REPLY [2 votes]: An invitation to Morse theory is very good, but it can be a bit heavy. I managed to find a very simple version of the lemma that is directly in the form you requested (I recently made use of this myself in my thesis). It is a link to the PhD thesis of Matthew Petro, University of Wisconsin, from 2008. A preview is on available on google books, and the Holomorphic Morse-Bott Lemma is Lemma 3.8 in the text; the link should take you straight to it. Out of curiosity, what are you using it for?<|endoftext|> TITLE: Is 'the' homotopy colimit of a sequence of exact triangles an exact triangle? QUESTION [9 upvotes]: Let $$ \begin{array}{rccccl} A_0&\to& B_0&\to& C_0&\to\\ \downarrow & &\downarrow&&\downarrow\\ A_1&\to& B_1&\to& C_1&\to\\ \downarrow & &\downarrow&&\downarrow\\ \vdots & &\vdots&&\vdots\\ \end{array} $$ be a commutative diagram in a triangulated category such that all the rows are exact triangles. One defines the homotopy colimit of a sequence $A_0\to A_1\to\ldots$ as a third object in an exact triangle $$ \coprod A_n\xrightarrow{1-shift}\coprod A_n\to \operatorname{hocolim}_nA_* $$ which is unique up to (non-unique) isomorphism. For a morphism of diagrams $A_*\to B_*$ (as above), one gets a (non-unique) morphism $\operatorname{hocolim}_nA_*\to \operatorname{hocolim}_nB_*$. Given the diagram above, is there a sequence of morphisms $$ \operatorname{hocolim}_nA_*\to \operatorname{hocolim}_nB_*\to \operatorname{hocolim}_nC_*\to $$ which is an exact triangle? I am especially interested in the case that $B_*$ is a constant diagram. I was trying to use the 9-lemma for triangulated categories, but this explicitely constructs the morphisms $\coprod C_n\to \coprod C_n\to \tilde C$ and I cannot prove, why the first map has to be $1-shift$ and therefore why $\tilde C$ has to be the homotopy colimit of $C_0\to C_1\to\ldots$. This question is strongly related, I think. REPLY [5 votes]: The answer is yes if you assume that your triangulated category is the basis of a triangulated derivator, see Appendix 2 in MR3031826 Keller, Bernhard(F-PARIS7-IMJ); Nicolás, Pedro(E-MURC2ED) Weight structures and simple dg modules for positive dg algebras. (English summary) Int. Math. Res. Not. IMRN 2013, no. 5, 1028–1078. Among all possible enhancements of triangulated categories, derivators are the closest to the original concept. I mean, it's known how to construct a triangulated derivator out of any other kind of enhancement, but not the other way around. The reason that derivators suffice is that the poset $\mathbb N$ is a free category.<|endoftext|> TITLE: Source for derogatory quote about graph theory QUESTION [9 upvotes]: (Edited in accordance with suggestions in comments.) I remember once I read a quote that sounded like "graph theory is the scum of topology" (please approximate). I can not find it on the web, and I can not remember who said this. Or did I dream it after a heavy meal? REPLY [12 votes]: Graph theory is the slum of topology... You may have read it in the first paragraph of the very first opinion of Prof. Zeilberger: Topology: The slum of combinatorics OR "Don't show off too much, your specialty will soon be trivialized" He attributes it therein to a certain Whitehead too (but it is not clear from his text to which Whitehead he is referring).<|endoftext|> TITLE: "Paradoxes" in $\mathbb{R}^n$ QUESTION [24 upvotes]: One may think of this question as a duplicate of this one. I see it more like an extension. The "inscribed sphere paradox" discussed in the aforementioned question states that if you inscribe a sphere in an $n$-dimensional cube of side $1$, then the volume of the sphere goes to $0$ as $n \to \infty$, while the volume of the cube remains the same. A more involving paradox is Hamming's Four Circle paradox, also described as an answer to that post. A more straightforward paradox (also discussed earlier by R. C. Hamming) is the fact that the angle $\theta$ between the diagonal of a cube $(1,1,\ldots,1)$ and any direction $(0,\ldots,1,0,\ldots,0)$ satisfies $$\cos \theta = \frac{1}{\sqrt{n}} \to 0 \mbox{ as } n \to \infty.$$ This means that, as $n$ increases, the diagonal is almost perpendicular to all $(0,\ldots,1,0,\ldots,0)$ (almost lying in (all!) corresponding hyperplanes). My question is: are there any other elementary examples of these so-called "paradoxes" (for instance, for other objects than sphere/cubes)? I am thinking more of elementary examples in which the intuition from simple plane geometry ($\mathbb{R}^2$) fails miserably in $\mathbb{R}^n$, particularly when $n \to \infty$. REPLY [13 votes]: For the $n$-simplex $$\Delta_n = \{x\in\mathbb{R}^{n+1}\ :\ x_i\geq 0, \sum x_i = 1\},$$ its "midpoint" $m = [1,\dots, 1]/(n+1)$, and its corners $e_k$ it holds that $$\|m - e_k\|\to \infty$$ while the distance of the midpoint to the $n-1$ dimensional faces goes to zero (both for $n\to\infty$). I found it quite counterintuitive that for a convex body like the simplex it can happen that its corners move apart from the midpoint while its "sides" move closer to it. But actually, this is somehow that standard picture of a high dimensional convex body as I learned from these slides by Roman Vershynin (who attributes this finding to V. Milman):<|endoftext|> TITLE: What set theoretical questions could never be answered by Turing machines of arbitrary cardinality? QUESTION [9 upvotes]: Let us assume that there are Turing machines of arbitrary cardinality, by that I mean they can have input tapes of any arbitrarily high cardinality and compute for a number of steps also of arbitrarily high cardinality. Those machines are, in principle, much more powerful than Hamkins infinite time Turing machines. Basically what I mean is, assume we are not limited by any kind of finiteness so we can construct sets and verify their properties without any cardinal limitation. Assuming we are gods that can watch the results of those computations, it is still not clear to me if questions such as the existence of $0^\#$ (or $V=L$) would be answered by such a machine or if those questions are for ever undecidable (Gödel?) and so you will be always able to define two consistent theories, one in which V=L and one in which $V\not =L$). In the latter case, what makes our hypothetical machine unable to answer the question? (I mean, what makes a property not being able to be tested as either true or false if you are allowed an arbitrarily high infinite number of cells and time steps?) Edit by P.W.: Without rewriting the question, I believe the following is something like the intention of the original question: allowing a Turing Machine to act transfinitely (as the Hamkins-Kidder Infinite Time Turing machine does) but supplying it with an On-length tape (rather than an omega order-type length as they do) tape, and making some sensible assumptions about its behaviour at limit stages, can such a machine 'decide' or otherwise settle questions, such as $V=L$ or whether $0^\#$ exists? Or, alternatively what can such machines 'compute' or write on their tape(s)? This does have a clear answer. UPDATE by OP: Thanks to all for your answers and comments. With the risk of going into the ridicule, let me assume that our hypothetical machine (do not call it Turing as it could be misleading), however it works, should be able to build all finite sets of an arbitrary cardinal $\kappa$ and decide if it is a Ramsey cardinal (R)? (Am I asking too much of such a machine?). If so, the statement V=L might be "(super-)undecidable" (that is, if a Ramsey cardinal does not exist the machine never halts), but it must be either true or false. Thus, assuming we are gods (and so we have oracles for our machines' halting problems), we could say that V=L is either true or false, not just an arbitrary axiom (I know that in such a case we can always have models where V=L is true (when $\kappa$=lowest R), but that is not what I mean). PS: I know it most likely doesn't matter the original motivation in this forum, but the question is related to an argument about Max Tegmark's mathematical universe hypothesis (which to me should be the same as Joel Hamkins multiverse "if ..."), and if he is too restrictive in assuming only Gödel computable mathematical structures. REPLY [4 votes]: As already explained by Joel and Philip, the machines you describe basically compute $L$. Still, there are natural principles connected with these machines that lead out of $L$: By a theorem of Sacks, if a real $x$ is (Turing)-computable relative to all oracles in a set of positive Lebesgue measure, then $x$ is recursive. Intuitively, this means that for a non-computable real $x$, randomly chosing an oracle is not likely to help in computing $x$. The analogous principle for Ordinal Turing Machines turns out to be independent from $ZFC$ and is in particular false in $L$: In fact, the halting problem for Ordinal Turing Machines is solvable relative to all but countably many real oracles in $L$.<|endoftext|> TITLE: Genus of Tutte-Coxeter Graph QUESTION [5 upvotes]: What is the genus of the Tutte-Coxeter graph -- the incidence graph of the GQ of order 2? Seems like it should be well known, since nearly every other parameter for that graph is known, but I can find no reference to cite. REPLY [2 votes]: Really just a long comment with some pictures. An embedding onto a genus 4 surface can also be seen directly although in the end we probably need sage to establish the bound is sharp. Here is a description of the graph followed by a picture (produced by sage), {0: [1, 17, 29],1: [0, 2, 22],2: [1, 3, 9],3: [2, 4, 26], 4: [3, 5, 13], 5: [4, 6, 18], 6: [5, 7, 23], 7: [6, 8, 28], 8: [7, 9, 15], 9: [2, 8, 10], 10: [9, 11, 19], 11: [10, 12, 24], 12: [11, 29, 13], 13: [4, 14, 12], 14: [13, 15, 21], 15: [8, 14, 16], 16: [15, 17, 25], 17: [0, 16, 18], 18: [5, 19, 17], 19: [10, 20, 18], 20: [19, 21, 27], 21: [14, 20, 22], 22: [1, 23, 21], 23: [6, 24, 22], 24: [11, 23, 25], 25: [16, 24, 26], 26: [3, 25, 27], 27: [20, 26, 28], 28: [7, 29, 27], 29: [0, 28, 12]} The edges [j,j+1 mod 30] form a $C_{30}$ subgraph this $C_{30}$ together with the edges $[0,17],[4,13],[20,27]$ and can be embedded in a disk $D_0$ such that the $C_{30}$ is embedded in the boundary of $D_0$. Attach a second disk $D_1$ to the boundary of $D_0$ in this disk embed the edges $[1,22],[5,18],[8,15]$. Likewise, attach disks $D_2$, $D_3$, $D_4$ with edge sets $[2,9],[12,29],[16,25]$; $[3,26],[6,23],[10,19]$; $[7,28],[11,24],[14,21]$. (The groups of 3 represent 3 non-intersecting edges in the graph that are not part of the specified $C_{30}$.) Next embed this quotient space of the five disks into $\mathbb{R}^3$ and take a regular neighborhood N of this space. (Note we can choose an embedding such that the space has the cross section below.) Actually, the graph embeds in the boundary of the neighborhood of this cross section, which is a genus 4 surface. Furthermore, since the girth of the graph is 8, the Euler characteristic of any surface that this graph embeds into is less than $-2.75$, so by hand we can see the genus is either 3 or 4.<|endoftext|> TITLE: Normal polytopes - counterexample? QUESTION [6 upvotes]: An integral polytope $P$ is normal if all lattice points inside the integer dilation $kP$ can be expressed as $p_1+p_2+\dots+p_k$, where $p_i \in P$ are lattice points. I am looking for an example $P$ for which the above is true for $k=2$, but fails for higher $k$. A related question: is there a number $M$, that only depends on the dimension of $P$, such that if the above holds for $k\leq M$, then $P$ is integral? REPLY [12 votes]: Examples of this sort (and worse) are given, for dimension 5 and up, in http://scholar.google.com/scholar?cluster=14055290405510744870&hl=en&oi=scholarr David Handelman [me], Effectiveness of an affine invariant for indecomposable integral polytopes, J Pure and Applied Algebra 66 (1990) 165–184, section 3, pp 16ff. Specifically, this gives integral polytopes $K$ such that $e(K \cap Z^d) = eK \cap Z^d$ [the first denotes the sum of $e$ points in $K \cap Z^d$) for $e \leq d/2$, but for no $e > d/2$, and when $e \geq d/2$, $eK$ is projectively faithful (that is, its set of lattice points generates the standard copy of $Z^d$ as an abelian group) inside Euclidean space of dimension $d$. When the dimension is 6 or more, we can also assume $K$ itself is projectively faithful. Towards the related question, any $M \geq d-1$ will do, as in my answer to Lattice points in dilated polytopes and sumsets.<|endoftext|> TITLE: Microlocalizing Hochschild homology QUESTION [13 upvotes]: A recent paper of Ben-Zvi and Nadler gives a general formalism for understanding "dimensions" in sheaf theories. Without getting too far into details, amongst other things, this formalism allows us to study the associated Hochschild classes of D-modules on stacks, which naturally live in the deRham cohomology of the associated loop stack. So, for example, for equivariant D-modules on a variety $X$ for a group $G$, the Hochschild classes of equivariant D-modules live in the equivariant cohomology of the inertia stack $I(X)=\{(x,g)\in X\times G | g\cdot x=x\}$. In the case of a smooth variety, it's known how to enrich this picture by microlocalizing it (see, for example, the paper of Kashiwara and Schapira) and in fact to generalize beyond D-modules to modules of sheaves on other deformation quantizations. This is extremely valuable, since one can define Hochschild classes supported on subvarieties in the cotangent bundle which are non-zero but pushforward to zero in the full cotangent bundle. Thus, this gives a much richer picture. In this case however, the loop space makes no interesting appearance, since a smooth variety is its own loop space. Is it well understood how to unify these two stories? How does one think about the Hochschild homology of (say) equivariant D-modules on a smooth variety microlocally? REPLY [6 votes]: One answer to Ben's question is to calculate the Hochschild homology of the category of $D$-modules $D_\Lambda(X)$ with fixed microsupport $\Lambda\subset T^*X$. The functor $D_\Lambda(X)\to D(X)$ has a continuous right adjoint, so defines a map $HH_*(D_\Lambda(X))\to HH_*(D(X))$. Thus given a $D$-module with $\Lambda$ microsupport we have its characteristic class with supports and its image in de Rham cohomology which is its usual characteristic class (but may be zero as you point out). Formally one can apply this idea in the other settings Ben considers, but one needs to calculate the $HH_*$ of the categories of sheaves with supports geometrically, perhaps along the lines Tom indicates. I haven't thought through the calculation of the relevant Hochschild homology - one needs to think through the properties of the adjoint functor above (enforcing $\Lambda$-support) which are worked out (in the coherent setting) in Arinkin-Gaitsgory, but I don't think it's easy. But in applications one often sees microsupport enforced by equivariance - i.e., you might be considering $D(X/H)$ for some group $H$ with finitely many orbits (e.g., the flag variety with $B$ or a a symmetic subgroup, or general spherical varieties), so that $\Lambda$ is the union of the conormals to the orbits. In this case we are asking to calculate $HH_*(D(X/H))$ and the induced map (from pullback under $X\to X/H$) to $HH_*(D(X))$. Note that the category $D(X/H)$ recovers its nonequivariant cousin, the category $D_\Lambda(X)$, from its structure as a category over $BH$ so I think we can ``deequvariantize" this calculation to answer the question for these special $\Lambda$'s. This is very geometric: by the results Ben quotes, the characters of equivariant $D$-modules are given by Borel-Moore homology of the inertia stack of $X/H$, i.e., of the stabilizers of the $H$-action. These are I believe related by linear Koszul duality to the characteristic classes with support that Ben is talking about (see the papers of Mirkovic-Riche).<|endoftext|> TITLE: Sources of Theorem drafts by the original author QUESTION [7 upvotes]: When I look at first time to a theorem and I try to understand it or when I try to memorise a useful theorem I always have difficulties (I am not the only one. For example: I read a question: I always have trouble memorizing theorems. Does anybody have any good tips? ) However, I have learn from the theorems that I am able to prove, that what I am doing is not following the theorem itself but a rather very clear non too abstract idea (usually a concrete example) I have on mind preconceived somehow. I wonder to know if there is any sources of pictures that big mathematicians has drawn when proving an important theorem. I would like to see what was the concrete idea (usually rather simple) that they have on mind. It is not difficult some time to write a nice representation of the theorem after is conceived, as usually everybody does, but I am interesting in other way around. REPLY [4 votes]: Short answer: I believe Paul Cohen's piece on his own development of forcing provides an answer to your question. I will elaborate further below the break, but you can find his article written up as: Cohen, P. (2002). The discovery of forcing. Rocky Mountain Journal of Mathematics, 32(4). Elaboration: A brief idea of the "big picture" seen by Cohen (here I assume, perhaps incorrectly, that you mean "big idea" rather than a literal drawing) is as follows: $1.$ Believe that philosophical arguments can be applied to questions in Number Theory; $2.$ Believe there are many different models of mathematics (cf. the Löwenheim–Skolem theorem); and $3.$ Believe in the existence of some sort of "decision procedure" that simplifies complex statements until they are decidable. These "big picture" ideas are what evolved and interacted with one another in leading Cohen to his proof of the independence of the Continuum Hypothesis. It might be worth remarking that another important belief, more related to affect but also mathematically relevant, held by Cohen was that: $4.$ The Continuum Hypothesis was the sort of problem Cohen would like to tackle, and he was determined not to lose the chance to resolve such an outstanding question. Here are some quotations by Cohen to support items $1$ to $4$, respectively: $1.$ In set theory when dealing with fundamental questions, one often has a kind of philosophical basis or conviction, rooted in intuition, which will suggest the technical development of theorems. $ $ Basically [the Continuum Hypothesis] was not really an enormously involved combinatorial problem; it was a philosophical idea. $2.$ We will never speak about proofs but only about models. and, expatiating further, The existence of many possible models of mathematics is difficult to accept upon first encounter, so that a possible reaction may very well be that somehow axiomatic set theory does not correspond to an intuitive picture of the mathematical universe, and that these results are not really part of normal mathematics . . . . I can assure you that, even in my own work, one of the most difficult parts of proving independence results was to overcome the psychological fear of thinking about the existence of various models of set theory as being natural objects in mathematics about which one could use natural mathematical intuition. On the relation of $1$ and $2$, Cohen remarks: Of course, in the final form, it is very difficult to separate what is theoretic and what is syntactical. As I struggled to make these ideas precise, I vacillated between two approaches: the model theoretic, which I regarded as roughly more mathematical, and the syntactical-­forcing, which I thought as more philosophical. $3.$ If you actually wrote down the rules of deduction—why couldn’t you in principle get a decision procedure? I had in mind a kind of procedure which would gradually reduce statements to simpler and simpler statements. Similarly, Because of my interest in number theory, however, I did become spontaneously interested in the idea of finding a decision procedure for certain identities... I saw that the first problem would be to develop some kind of formal system and then make an inductive analysis of the complexity of statements. In a remarkable twist this crude idea was to resurface in the method of ‘forcing’ that I invented in my proof of the independence of the continuum hypothesis. $4.$ This concerns, in particular, Friedberg's solution to Post's Problem using the priority method, and parallels between this and Cohen's work on ~CH using forcing. See the earlier MO question here, and, e.g., Cohen's remark: It was exactly the kind of thing I would like to have done. I mentally resolved that I would not let an opportunity like that pass me again. You can find the points made above in a slightly more (or less?) organized write-up here. As a final, general remark: Having a big picture in mind when proving something occurs, of course, outside of mathematics as well. The best exposition of this (of which I am aware) is Howard Gruber's (1974) book Darwin on Man, which explores how Charles Darwin developed his theory of evolution. Though Origin of Species is structured as a presentation of overwhelming evidence that simply led Darwin to his final theory, the reality is that he was constantly theorizing along the way. Gruber explores this by going through previously unread notebooks of Darwin's, and includes a discussion of a literal picture that turns up repeatedly - in different forms - of a branching tree. Although this books concerns work outside of mathematics, I think reading it can be both informative about "creative thinking" in general, and can also provide a way for those who are mathematically-inclined to try and write up the biography of a mathematical theorem and the big ideas/big picture behind it.<|endoftext|> TITLE: Ground Axiom and behaviors of continuum function QUESTION [12 upvotes]: The Ground Axiom ($GA$) is the assertion that the universe of sets ($V$) is not a forcing extension of any inner model $W$ by nontrivial forcing $P\in W$. Is $GA$ consistent with any possible behavior of continuum function $\kappa\mapsto 2^{\kappa}$? It seems in models of $GA$ like $L$ and some other canonical models the growth speed of continuum function is too low (e.g. $L\models GCH$). So the natural question is: What is the consistency situation for faster growth speeds of $\kappa\mapsto 2^{\kappa}$? REPLY [12 votes]: In the paper The ground axiom is consistent with $V\ne{\rm HOD}$ (J. D. Hamkins, J. Reitz, W.H. Woodin, PAMS 136(8):2008), we prove that the ground axiom is consistent with $V\neq\text{HOD}$, and remark there that: The proof of Theorem 1 is flexible and generalizes in a variety of ways. For example, not much was used about the specific iteration $\mathbb{P}$. For establishing GA, we needed to know only that the stage $\gamma$ forcing was ${\lt}\gamma$-closed, and we could easily have accommodated $\text{Add}(\gamma,\gamma^{++})$ or occasionally $\text{Coll}(\gamma,\gamma^{+})$, for example, without any difficulty in the argument. Also, we needn't have forced specifically at every regular cardinal stage $\gamma$, but could have forced at regular cardinals in some other unbounded pattern. Thus, the argument establishes that after forcing over $L$ with any of the usual reverse Easton iterations of closed forcing, one obtains the Ground Axiom in the extension. Thus, with this kind of forcing, one can achieve a wide spectrum of patterns in the continuum function. Although the usual Easton forcing itself is a product, rather than an iteration, and will therefore definitely not result in the ground axiom, nevertheless one may transform the usual Easton product into an iteration, by taking big chunks of the forcing at a time. That is, given an Easton function $E$, one looks at sufficient closure points of $E$, and performs the iteration, which performs the usual Easton product between these closure points. The result will be a model of GA with the desired Easton function as the continuum function. I don't know if the details of this argument, however, have ever been written down, and if you were inclined in that direction, I would encourage you to do it (feel free to contact me).<|endoftext|> TITLE: Number of representations of an integer as an (arbitrary) sum of products QUESTION [11 upvotes]: If $n$ is a positive integer, let $r(n)$ denote the number of representations of $n$ as a sum of products of pairs of positive integers. (Here, the order of the terms in the sum does not matter, but products with the same answer are regarded as different, even if they contain the same two numbers in a different order.) For example, r(1) = 1; r(2) = 3: 2 = 2*1 = 1*2 = 1*1 + 1*1; r(3) = 5: 3 = 3*1 = 1*3 = 2*1 + 1*1 = 1*2 + 1*1 = 1*1 + 1*1 + 1*1. I would very much like to know whether the asymptotic of $r(n)$ is known, or whether it can be derived easily from known results/methods. (Of course, the problem can be rephrased in terms of $\Lambda$-partitions. Recall that if $\Lambda$ is a non-decreasing sequence of positive integers, $$\Lambda_1 \leq \Lambda_2 \leq \ldots,$$ with $\Lambda_k \to \infty$ as $k \to \infty$, a $\Lambda$-partition of $n$ is a representation of $n$ as a sum $$n = \sum_{k=1}^{\infty} m_k \Lambda_k,$$ with $m_k \in \mathbb{N} \cup \{0\}$ for all $k$. If $\Lambda$ is the sequence with $d(s)$ copies of $s$ for every positive integer $s$, where $d(s)$ denotes the number of divisors of $s$, then $r(n)$ is the number of $\Lambda$-partitions of $n$.) The generating function of $r(n)$ is given by $$\sum_{n=0}^{\infty} r(n) z^n = \prod_{(k,l) \in \mathbb{N}^2} (1-z^{kl})^{-1},$$ and so the theorem of N. A. Brigham in [A General Asymptotic Formula for Partition Functions, Proc. Amer. Math. Soc., 1950] gives the asymptotic of $\log r(n)$: $$\log r(n) \sim \sqrt{2\zeta(2) n \log n} = \pi \sqrt{\tfrac{1}{3} n \log n}.$$ Unfortunately, the more precise Tauberian theorem of Ingham [A Tauberian Theorem for Partitions, Ann. Math, 1941], which gives the asymptotic of $a(n)$ for a wide class of partition functions $a(n)$, does not seem to be applicable, since his function $R(u)$ (in his Theorem 2) grows like $u \log u$ in the above case, rather than like $u^{\beta}$+(small error) for some fixed $\beta >0$. The associated Dirichlet series $$D(s) = \sum_{(i,j)\in \mathbb{N}^2} s^{-ij} = \zeta(s)^2$$ is the square of the Riemann zeta function. This has a pole of order 2 at $s=1$, so the theorem of Meinardus on $\Lambda$-partitions is not applicable, either. I am not familiar enough with the area to know whether the Hardy-Littlewood circle method, or some other method, can be adapted to work in the case of $r(n)$. Any help would be greatly appreciated! REPLY [8 votes]: Put $F(z)= \prod_{n=1}^{\infty} (1-z^n)^{-d(n)}$, where $d(n)$ is the number of divisors of $n$. The problem asks for the asymptotics for $$ R(N) = \frac{1}{2\pi i} \int_{|z|=r} F(z) z^{-N} \frac{dz}{z}, $$ where the integral is taken over any circle with radius $r<1$. A standard way of obtaining asymptotics in such contexts is to use the saddle point method. Note that $$ \log F(z) = \sum_{n=1}^{\infty} d(n) \log (1-z^n)^{-1} = \sum_{n=1}^{\infty} z^{n} \sum_{ab=n} \frac{d(a)}{b}. $$ One chooses the radius $r$ so as to minimize $F(r)r^{-N}$: this is attained when $$ N= r\frac{F^{\prime}}{F}(r) =\sum_{n=1}^{\infty} r^{n} \sum_{ab=n} ad(a). $$ For such a value of $r$, the quantity $F(re^{i\theta})(re^{i\theta})^{-N}$ for small values of $\theta$ will look like $F(r)r^{-N}e^{-\frac{\theta^2}{2} F_2(r)}$ where $$ F_2(r) = \sum_{n=1}^{\infty} nr^{n} \sum_{ab=n} ad(a). $$ Then our integral is asymptotically the same as $$ \frac{1}{2\pi} \int_{-\pi}^{\pi} F(r)r^{-N} e^{-\frac{\theta^2}{2} F_2(r)} d \theta \sim \frac{F(r) r^{-N}}{\sqrt{2\pi F_2(r)}}. $$ This is the desired asymptotic for $R(N)$, once we gain some understanding of how $r$ depends on $N$, and how the corresponding $F(r)$ and $F_2(r)$ behave. Put $r=e^{-1/X}$. Then using Mellin transforms, our equation for $r$ becomes (for any $c>2$) $$ N = \sum_{n=1}^{\infty} e^{-n/X} \sum_{ab=n} ad(a) = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} X^s \Gamma(s) \zeta(s) \zeta(s-1)^2 ds. $$ The above can be evaluated by moving the line of integration to the left and computing the residues of the poles at $s=2$ (double pole), $s=1$ and $s=0$. Note that the integrand is regular at all other points (because the other poles of $\Gamma$ will cancel with corresponding trivial zeros of $\zeta(s)$), so the resulting asymptotic will be very accurate with an error of $O(X^{-A})$ for any natural number $A$. Thus a computation shows that $$ N= \zeta(2) X^2 \Big(\log X + \gamma + \frac{\zeta^{\prime}}{\zeta}(2)\Big) + \frac{X}{4} + \zeta(0) \zeta(-1)^2 + O(X^{-A}). \tag{1} $$ A similar calculation shows that $$ \log F(r) = \zeta(2) X \Big(\log X+ \gamma +\frac{\zeta^{\prime}}{\zeta}(2) \Big) + \frac 14 \log X + 2\zeta^{\prime}(0) \zeta(0). $$ (Note that $\zeta(0) \zeta(-1)^2 =-1/288$ and $2\zeta^{\prime}(0)\zeta(0) = \log \sqrt{2\pi}$.) Finally we have $$ F_2(r) =\sum_{n=1}^{\infty} nr^{n} \sum_{ab=n} ad(a) \sim 2XN. $$ Putting these together we get $$ R(N) \sim \frac{1}{\sqrt{2} X^{\frac 14} N^{\frac 12}} \exp\Big(\frac{2N}{X} -\frac 14\Big). $$ Here $X$ is given in terms of $N$ by (1), and it is easy to see that $X \sim \sqrt{12N/(\pi^2 \log N)}$; one can of course compute $X$ much more precisely, and it is a close relative of the Lambert $W$-function. Warning: I think the constants above are correct, but wouldn't swear to them. In any case the method of proof should be clear. For details along these lines for the partition function, see for example the course notes linked in my comment above.<|endoftext|> TITLE: Explicit isomorphism for quaternion algebras over $\mathbb{Q}$? QUESTION [9 upvotes]: It is known that the isomorphism class of a quaternion algebra $A=\binom{a,b}{K}$ over a number field $K$ is determined by the finite set of places $v$ of $K$ where $A\otimes_K K_v$ is a division algebra, equivalently at which the projective curve $ax^2+by^2-z^2$ fails to have a $K_v$-rational point. Given two quaternion algebras $\binom{a,b}{\mathbb{Q}}$ and $\binom{c,d}{\mathbb{Q}}$ over $\mathbb{Q}$ that are ramified at the same places, is there a known algorithm to construct an explicit isomorphism? (I.e. to find $I,J\in \binom{a,b}{\mathbb{Q}}$ such that $I^2=c, J^2=d, IJ=-JI$?) If so, I would love some references. REPLY [8 votes]: Timo Hanke has written on this problem in the generality of cyclic algebras (http://arxiv.org/abs/math/0702681). He shows that it is equivalent to the solution of a norm equation, which has an algorithmic solution over global fields. In your case, this would in general involve the solution of a norm equation over a biquadratic field; I don't know how well this would fare in practice. In the special case of quaternion algebras that you are inquiring about, there is a link to quadratic forms, and as Keith Conrad mentions, it is equivalent to find a zero of a quadratic form in six variables. This goes back to Albert, who looked at this form in detail in order to prove that there was an "honest" (non-quaternion) biquaternion algebra. A good reference for this is section 16 of the "Book of Involutions" or section XII.2 of Lam's "Introduction to Quadratic Forms over Fields". It may seem like just a reformulation of the problem, but it turns out there are algorithmic methods to find points on quadrics over number fields that are quite efficient in practice: the buzzword here is "indefinite LLL", and Watkins (http://magma.maths.usyd.edu.au/~watkins/papers/illl.pdf) explains what Magma does to accomplish this task. In general, here is an idea I kicked around once which at least reduces the problem to solve norm equations over quadratic extensions (instead of biquadratic extensions). This might be only of theoretic/algorithmic interest, but at least it potentially generalizes. We wish to test if $A \cong B$ over a global field $F$ (say of characteristic not $2$ for now), and if so, to find an explicit isomorphism. If $A=(a,b)$ and $B=(c,d)$ and $a=c$, then there is an isomorphism if and only if $b/d$ is a norm from $\mathbb{Q}(\sqrt{a})$, and this can be accomplished algorithmically by a norm equation over this field; so it is enough to reduce to this case. To find a common subfield $K=\mathbb{Q}(\sqrt{a})$ in $A,B$, one can simply pick one (choose $K$ such that $K_v$ is not split at all places $v$ ramified in $A$ and $B$, e.g., take $a=-\mathrm{lcm}(ab,cd)$ if $\gcd(a,b)=\gcd(c,d)=1$, so this step does not even require factoring). The problem of embedding a quadratic field $K$ in a quaternion algebra is equivalent to (checking if $K$ splits the algebra and so) to a norm equation (a standard result, see e.g. http://www.math.dartmouth.edu/~jvoight/articles/quatalgs-060513.pdf). Once the field is embedded, we can diagonalize the quadratic form to reduce to the case where $a=c$. This approach probably generalizes to cyclic algebras (of any characteristic), and if it is interesting to you, it is something I would be happy to work out with you.<|endoftext|> TITLE: Trapping a convex body by a finite set of points QUESTION [23 upvotes]: In $\mathbb{R}^n$, let $K$ be a convex body and $T$ a finite set of points disjoint from the interior of $K$. Say that $T$ traps $K$ if there is no continuous motion of $K$ carrying $K$ arbitrarily far away from its original position during which no point of $T$ penetrates the interior of $K$. Is there a finite number $k=k(n)$ such that each $n$-dimensional convex body $K$ can be trapped by a set of at most $k$ points? If so, what is the smallest such $k$? Observe that the $3$-cube can be trapped by six points, but not by five (for the $n$-cube it's $2n$ points, but not $2n-1$). The $n$-ball can be trapped by $n+1$ points, and it seems that no $n$-dimensional convex body can be trapped by $n$ points. The main question is: Can every convex body in $\mathbb{R}^3$ be trapped by six points? More generally: Can every convex body in $\mathbb{R}^n$ be trapped by $2n$ points? Footnote 1. Here is a variation of the problem, perhaps easier to handle: restrict the motions of $K$ to parallel translations. Footnote 2. For $n=2$, a closely related problem, namely of immobilizing the body with a finite set of points, has been studied and solved: four points always suffice. Reference will be provided upon request (I will have to look it up). REPLY [4 votes]: I will assume we are not allowing re-orientation of the body (rotation). If one wishes to determine if some arbitrary rigid body $K$ can pass some obstacle $T$ without touching it, one can reduce the problem to determining if some point $k \in K$ can pass by the Minkowski sum of $K$ and $T$ (which I will denote $K+T$) without touching it. In this case, if we wish to determine if $K$ can escape to infinity, assuming $K$ is bounded and finite, we need only determine if $K+T$ disconnects $\mathbb{R}^n$ into at least two connected components. Since $T$ is a set of points, $K + T$ is simply taking $|T|$ copies of $K$, where each copy is centered at a point in $T$ (you can pick any point in $K$ to be the "center" of $K$, WLOG). Our question then becomes: what is the shape that maximizes the number of copies we would need to union together (without rotations) to disconnect $\mathbb{R}^n$ into two non-empty sets, and how many copies would that be? I am fairly certain that proving that the $n$-cube is this shape, and that it requires $2n$ copies is a hop and a skip away from this, but I'm at a loss for a sufficiently formal argument. Something about how the faces are orthogonal and can only restrict one direction in one dimension at a time or something, which is somehow the worst-case. There's probably a similar argument for n+1 being a lower bound and tetrahedra. Unfortunately, this is where my knowledge stops. de Berg et al's Computational Geometry: Algorithms and Applications covers the Minkowski Sum stuff very well in Chapter 13: Robot Motion Planning.<|endoftext|> TITLE: Internal logic of the topos of simplicial sets QUESTION [30 upvotes]: I am looking for a closed statement (i.e. not depending on any parameter objects) which is true in the internal logic of the topos of simplicial sets, but is not an intuitionistic tautology. Ideally, I would like it to be a simple universal statement of propositional logic (e.g. "for all propositions P, Q, and R, blah blah", like LEM or de Morgan's law). (To clarify: this has nothing to do with higher topoi or homotopy type theory; it's purely a question about ordinary 1-categorical 1-topos theory.) REPLY [13 votes]: I will augment François Dorais’s answer with an exact identification of the propositional logic. Let $D_n$ be the free distributive lattice with a top, considered as a finite Heyting algebra. Algebras are a nuisance to work with, so let us compute the dual Kripke frame $F_n$, which validates the same formulas, but is much smaller and simpler. Since we are dealing with a finite algebra, the elements of $F_n$ are just the $\lor$-irreducible (which entails $\bot$-irreducible, i.e., nonzero) elements of $D_n$, ordered upside down. If $\{x_i:i\in[n]\}$ is the set of free generators, each element of $D_n$ can be written in a disjunctive normal form $$a=\bigvee_{i\in I}\bigwedge_{j\in J_i}x_j,$$ where $J_i\subseteq[n]$, $I\ne\varnothing$, and $\{J_i:i\in I\}$ is an antichain. Clearly, if $a$ is $\lor$-irreducible, we must have $|I|=1$, thus $a$ is of the form $$a_J=\bigwedge_{j\in J}x_j$$ for some $J\subseteq[n]$. Moreover, we must have $J\ne[n]$ so that $a_J\ne\bot$. Conversely, it is not difficult to check that all $a_J$ with $J\ne[n]$ are indeed $\lor$-irreducible. Since $$a_J\le a_{J'}\iff J\supseteq J',$$ $F_n$ is just the powerset Boolean algebra without top: $$F_n\simeq\langle\mathcal P([n])\smallsetminus\{[n]\},{\subseteq}\rangle.$$ These Kripke frames are known as Medvedev frames, and the logic defined by $\{F_n:n\in\mathbb N\}$ is Medvedev’s logic (aka logic of finite problems), denoted LM or ML, based on Medvedev [1]. See e.g. Chagrov&Zakharyaschev [2,§2.9]. Thus, the logic of $\{D_n:n\in\mathbb N\}$ is also Medvedev’s logic. Despite its seemingly simple definition, this logic is shrouded in mystery: it is particularly scandalous that half a century after its discovery, it is still an open problem if the logic is recursively axiomatizable. Let me just mention that apart from the Kreisel–Putnam axiom $$(\neg p\to q\lor r)\to(\neg p\to q)\lor(\neg p\to r),$$ Medvedev’s logic is also known to include Scott’s axiom $$((\neg\neg p\to p)\to p\lor\neg p)\to\neg\neg p\lor\neg p,$$ and an axiom identified by Andrews (reported in Gabbay [3]), which in fact implies both the Kreisel–Putnam and Scott axioms: $$((\neg p\to q)\to r\lor s)\to(p\to r)\lor(q\to s).$$ On the other hand, Medvedev’s logic is included in the logic of weak excluded middle $\mathrm{KC}=\mathrm{IPC}+(\neg p\lor\neg\neg p)$. Medvedev’s logic has many interesting properties. In particular, it is the largest logic with the disjunction property that extends Kreisel–Putnam logic (Levin [4], Maksimova [5], cf. [2,Thm. 15.18]), and it is the only known superintuitionistic logic that simultaneously has the disjunction property and is structurally complete (Prucnal [6,7]). References: [1] Yuriĭ T. Medvedev: Finite problems. Doklady Akademii Nauk SSSR 142 (1962), no. 5, pp. 1015–1018, http://mi.mathnet.ru/eng/dan26117 (in Russian). English translation: Soviet Mathematics, Doklady 3 (1962), pp. 227–230. [2] Alexander Chagrov and Michael Zakharyaschev: Modal logic. Oxford Logic Guides vol. 35, Oxford University Press, 1997. [3] Dov Gabbay: Semantical investigations in Heyting’s intuitionistic logic. Synthese Library vol. 148, Springer, 1981, doi: 10.1007/978-94-017-2977-2. [4] Leonid A. Levin: Some syntactic theorems on the calculus of finite problems of Yu. T. Medvedev. Doklady Akademii Nauk SSSR 185 (1969), no. 1, pp. 32–33, http://mi.mathnet.ru/eng/dan34473 (in Russian). English translation: Soviet Mathematics, Doklady 10 (1969), 288–290. [5] Larisa L. Maksimova: On maximal intermediate logics with the disjunction property. Studia Logica 45 (1986), no. 1, pp. 69–75, doi: BF01881550. [6] Tadeusz Prucnal: Structural completeness of Medvedev’s propositional calculus. Reports on Mathematical Logic 6 (1976), pp. 103–105. [7] Tadeusz Prucnal: On two problems of Harvey Friedman. Studia Logica 38 (1979), no. 3, pp. 247–262, doi: 10.1007/BF00405383.<|endoftext|> TITLE: Are the asymptotics of A003238 known? QUESTION [14 upvotes]: Sequence A003238 of the OEIS counts ``rooted trees with $n$ vertices in which vertices at the same level have the same degree.'' The sequence, $a$, begins 1, 1, 2, 3, 5, 6, 10, 11, 16, ... and it is of course easy to generate as many terms as you like via recurrences (basically $a(n)$ is the sum of $a(d)$ as $d$ runs over divisors of $n-1$). In the formula section of the OEIS page we find: Conjecture : $\log(a(n))$ is asymptotic to $c \log(n)^2$ where $0.4 < c < 0.5$ (Benoit Cloitre, Apr 13 2004) What I'm interested in is both the state of this conjecture, and more generally, methods for analysing the asymptotics of sequences defined by ``similar'' recurrences - either globally (as above) or in the average sense i.e. asymptotics of things like $(1/n)\sum_{i=1}^n a(n)$. REPLY [22 votes]: I'll show that $$ \log a(n) \sim \frac{(\log n)^2}{\log 4} \approx 0.7213\ldots (\log n)^2. $$ So the range for the constant given in the conjecture is false, but an asymptotic of that general shape holds. One can obtain more precise asymptotics by working harder with the argument below. Roughly speaking what the argument says is that $a(n)$ behaves (in rough order) like the sequence $b(n)$ defined by $b(n+1)=b(n)$ if $n$ is odd, and $b(n+1)=b(n)+b(n/2)$ if $n$ is even. Put $A(x) = \sum_{j\le x} a(j)$. Then $$ A(n+1) = \sum_{j\le n+1} \sum_{d|(j-1)} a(d) =\sum_{j\le n+1} \sum_{d|(j-1)} a((j-1)/d) = \sum_{d} \sum_{j\le n, d|j} a(j/d) = \sum_d A(n/d). $$ Rearranging we get $$ a(n+1) = A(n+1)-A(n) = \sum_{2\le d\le n} A(n/d). $$ By the monotonicity of $a(n)$, it follows that $$ a(n+1) \le \sum_{2\le d\le n} \frac{n}{d} a(\lfloor n/2\rfloor) \le ( n\log n) a(\lfloor n/2\rfloor). $$ Iterating this inequality leads to the estimate $$ \log a(n) \le (1+o(1)) \frac{(\log n)^2}{2\log 2}. $$ Similarly, keeping just the $d=2$ term in our identity we get $$ a(n+1) \ge A(n/2) \ge \frac{n}{2\log n} a(\lfloor n/2 - n/(2\log n) \rfloor), $$ and iterating this gives the desired lower bound $$ \log a(n) \ge (1+o(1)) \frac{(\log n)^2}{2\log 2}. $$<|endoftext|> TITLE: Probability of Brownian motion to have a zero in an interval QUESTION [7 upvotes]: I have what should be a very simple questions for Brownian motion experts... Let $[a,b]$ be a given time interval. Let $f(x)$ be the probability that a linear Brownian motion with initial value $x$ at time $t=0$ has a zero in the interval $[a,b]$. I want to argue that $f(x)$ is maximal for $x=0$. This seems intuitively clear but I cannot figure out a simple proof of this (other than writing the exact expression for $f$ which is a double integral and analyzing its variations via long calculations). I would be interested, in order of preference, by such a simple proof or by a reference where this lemma is proven. Thanks! REPLY [3 votes]: There is an easy way to do it via calculation. Start by supposing that $x\geq 0$ Then we're asking for the probability that $x+W_t$ reaches $0$ in the interval $[a,b]$ where $W_t$ is a standard Brownian motion. This the same as $$\begin{eqnarray} P(\min_{a\leq t\leq b}x+W_t<0~and~\max_{a\leq t\leq b}x+W_t>0)=P(\min_{a\leq t\leq b}W_t<-x~and~\max_{a\leq t\leq b}W_t>-x) \end{eqnarray}$$ Then $$\begin{eqnarray} P(\min_{a\leq t\leq b}W_t<-x)&=& E[1\{\min_{a\leq t\leq b}W_t<-x \}| W_a]\\ &=& E[1\{\min_{a\leq t\leq b}W_{t-a}<-x-W_a \}| W_a]\\ &=& E[1\{\min_{0\leq t\leq b-a}W_{t}<-x-W_a \}| W_a]\\ &=& E[1\{-|W_{b-a}|<-x-W_a \}| W_a]\\ \end{eqnarray}$$ Where we used the fact that $W_t-W_a$ has the same law as $W_{t-a}$ and $\min_{0\leq t\leq b-a}W_{t}$ has the same law as $-|W_{b-a}|$ then we have $$P(\min_{a\leq t\leq b}W_t<-x)=2\int_R \frac{\exp(-y^2/2a)}{\sqrt{2\pi a}}N(-y-x)dy $$ Where $N$ is the Gaussian cumulative. Now it is easy to see that the right hand side is decreasing as function of $x$. We treat then $x\leq 0$ in the same way by symmetry of the Brownian motion.<|endoftext|> TITLE: Several simple questions on the geometry of higher stacks QUESTION [13 upvotes]: I'm trying to understand definition/work out some examples. So, there are some simple questions about higher stacks. For the simplicity assume that we are working with higher DM (Deligne-Mumford) stacks over $\mathbb{C}$. Is there some analogue of Keel-Mori theorem about existance of coarse moduli space for higher stacks? It is well known that ussual DM (1-)stacks with a point as a coarse moduli are quotients of a point by a finite group $G$ acting trivially, and coherent sheaves on it are just representations of $G$. So, what are the higher DM stacks, whose coarse moduli is just a point? What are the categories of coherent sheaves on such stacks? What are the higher quotient stacks? What are the coherent sheaves on them? For example, what are the quotients of $\mathbb{A}^1$? It is well known that ussual DM (1-)stacks etale-locally are quotient stacks. Is there some analogue for higher stacks? REPLY [7 votes]: I don't have any knowledge about most of your questions, but I wanted to point out that asking for coherent sheaves is not a very good way to test higher stacks. By analogy, if I'm interested in rings of functions I won't get much of a sense for schemes or stacks - the ring of functions on a scheme or stack $X$ is the same as that of the affine scheme $Aff(X)=Spec \Gamma(O_X)$, the affinization of $X$. The same can be said for categories of (quasi)coherent sheaves -- they will factor through the "1-affinization" of $X$, which is the stack recovered from $X$ by Tannakian reconstruction -- see this remarkable paper and DAG VIII for more on these notions. For example the group stack BG for G an abelian group doesn't have interesting actions on vector spaces (homomorphisms to the scheme $GL_n(R)$ , so there aren't any nontrivial coherent sheaves on $BBG$.<|endoftext|> TITLE: Why the name "variety" and the notation "V" for zeroes of polynomials? QUESTION [17 upvotes]: The following questions came to my mind while preparing the notes for the first class of (my first) course on algebraic geometry. Question 1: Is there any motivation for choosing the term "variety" for zeroes of polynomials? For example, I can sort of guess/understand the logic behind the term "manifold": 2-fold, 3-fold, ... $\rightarrow$ many-fold. For the term "variety" I don't see any such clear explanation. Question 2: Why write $V(f_1, \ldots, f_k)$ for zero sets of polynomials $f_1, \ldots, f_k$? Is it because of the term "variety"? Some (newer) texts use $Z(f_1, \ldots, f_k)$ - I thought $Z$ was meant to convey "zeroes". Does $V$ stand for "zeroes" in some other languages (a cursory look at the German and French Wikipedia pages for algebraic variety did not help)? REPLY [21 votes]: Also in Italian "varietà" is the term for both. Starting from this, I looked at the Italian Wikipedia webpage for varieties which, at the end, has this remark about the origin of the term: In italiano si traduce con varietà il termine tedesco Mannigfaltigkeit, che compare per la prima volta nella tesi di dottorato del 1851 di Bernhard Riemann, Grundlagen für eine allgemeine Theorie der Functionen einer veränderlichen complexen Grösse. Riemann si pone il problema di introdurre delle "grandezze molteplicemente estese", aventi cioè "più dimensioni", e le definisce usando quel termine. Analizzando il termine come parola composta, Mannig-faltig-keit, si riconosce in essa un parallelo con il termine latino multi-plic-itas, sicché lo si potrebbe tradurre letteralmente come 'molteplicità'. That can be approximatively translated as In Italian we translate with "varietà" the German word "Mannigfaltigkeit", which appears for the first time in the 1851 doctoral thesis of Bernhard Riemann, Grundlagen für eine allgemeine Theorie der Functionen einer veränderlichen complexen Grösse. Riemann introduced some "dimensions with multiple extensions", having "more dimensions", and he defines them using that term. Analiyzing the term as a compound word, "Mannig-faltig-keit", we can identify a parallelism with the Latin word "multi-plic-itas", so that it can be literally translated as "multiplicity". So, at least in Italian, the term comes from "variety" and "multiplicity" in the sense of "diversity".<|endoftext|> TITLE: Average number of distinguished leaves in a binary tree QUESTION [5 upvotes]: By a binary tree, I mean in this question a full rooted binary tree in which left and right child are labeled. A leaf of such a tree is a vertex of degree at most 1 (most references would probably consider that a leaf is a vertex of degree exactly 1) and an internal vertex is a vertex of degree at least 2. With these conventions, there are $C_n=\frac{1}{n+1}$${2n}\choose {n}$ binary trees with $n\in\mathbb N$ internal vertices (and $n+1$ leaves). We say that a leaf $x$ of a binary tree $T$ is distinguished if $\sigma(x)$ is equal to $x$ for all graph automorphism $\sigma\in\operatorname{Aut}(T)$. For instance, the unique binary trees with 0 internal vertex has 1 distinguished leaf, the unique binary tree with 1 internal vertex has no distinguished leaves, the 2 binary trees with 2 internal vertices have each 1 distinguished leaf, the 5 binary trees with 3 internal vertices have on average $2/5$ distinguished leaves and the 14 binary trees with 4 internal vertices have on average $15/7$ distinguished leaves. Among these 14 binary trees, 8 have 3 distinguished leaves and 6 have a unique distinguished leaf. Is there a known formula or an asymptotic expansion of the number of distinguished leaves as $n$ goes to $+\infty$? Is there a way to estimate the proportion of binary trees with $n+1$ leaves having at most (or equivalently, at least) $d$ distinguished leaves? The motivation behind this question, and its perhaps surprising tags, comes from syntax. A possible formalization of the syntax of natural languages (commonly associated with the names of N.Chomsky and R.Kayne) is to assume that sentences are built by recursive applications of a binary operation operating on a set containing (but not limited to) lexical items. The process yields a binary tree as output with some leaves attached to lexical items which is then believed to be transformed into a linear string of morphemes. It is believed that in order for such a tree to be linearized, it has to satisfy a condition which is close (though actually not equivalent) to the condition of having lexical items only at distinguished leaves. One of the aims of the question is to get an idea of the relative size of the hypothesized underlying binary tree compared to the size of the sentence measured in the usual way. REPLY [4 votes]: Start with the corrected version of Roland's formula. (Actually, I now think both he and I had correct formulas, it is just that my $n$ is the number of leaves and his is the number of internal vertices. But I worked this out with mine, so I'm not rewriting it.) $$D_n(t) = \sum_{k=1}^{n-1} D_k(t) D_{n-k}(t) + [ n \equiv 0 \bmod 2 ] (D_{n/2}(1) - D_{n/2}(t^2)) \quad D_1=t.$$ The selection of $D_0$ is arbitrary as it never enters the recursion. Put $$F(x,t) = \sum_{n=1}^{\infty} D_n(t) x^n.$$ Then $$F(x,t) = t x + F(x,t)^2 + F(x^2,1) - F(x^2, t^2).$$ Also, we know that $F(x,1)$ is the Catalan generating function $C(x) := (1-\sqrt{1-4x})/2$. Writing primes for derivative with respect to $t$, we have $$F'(x,1) = x + 2 F'(x,1) F(x,1) - 2 F'(x^2,1).$$ Putting $G(x) = F'(x,1)$, we have $$G(x) = x + 2 G(x) C(x) - 2 G(x^2)$$ or $$G(x) = \frac{x - 2 G(x^2) }{1-2 C(x)}= \frac{x-2G(x^2)}{\sqrt{1-4x}}.$$ Plugging this formula into itself, $$G(x) = \frac{x}{\sqrt{1-4x}}- \frac{2 x^2}{\sqrt{(1-4x)(1-4 x^2)}} + \frac{4 x^4}{\sqrt{(1-4x)(1-4x^2)(1-4 x^4)}} - \cdots.$$ As $x \to 1/4^{-}$, the product $G(x) \sqrt{1-4x}$ approaches $$\frac{1}{4} - \frac{2}{4^2 \sqrt{1-4/4^2}} + \frac{4}{4^4 \sqrt{(1-4/4^2)(1-4/4^4)}} - \frac{8}{4^8 \sqrt{(1-4/4^2)(1-4/4^4)(1-4/4^8)}} + \cdots$$ Call this sum $\beta$. I doubt that it has a closed form, but it converges very rapidly to $\approx 0.1238$. In fact, the sum is alternating, so we can bound it between consecutive partial sums: going to $3$ and $4$ terms gives $0.123847$ and $0.123776$ if I didn't make any error keying it in. So $$G(x) \approx \frac{\beta}{\sqrt{1-4x}} \ \mbox{as} \ x \to 1/4^{-}.$$ Meanwhile, $$x \frac{dC}{dx} = \frac{x}{\sqrt{1-4x}} \approx \frac{1/4}{\sqrt{1-4x}} \ \mbox{as} \ x \to 1/4^{-}.$$ You want to compute the ratio of the coefficient of $x^n$ in $G(x)$ and in $x \frac{d C}{d x}$. This suggests (and methods from books such as Wilf's Generatingfunctionology should be able to make this rigorous) that the limit is $4 \beta$.<|endoftext|> TITLE: Escape Time of Fractional Brownian Motion QUESTION [8 upvotes]: Let $B(t)$ be Brownian motion with $B(0)=x>0$ and let $A>x$. It is well known that the expected time for $B(t)$ to escape the interval $[0,A]$ is equal to $x(A-x)$. Is the expected time known if the process is a fractional Brownian motion with Hurst exponent $H\in (0,1)$ and $H\neq\frac{1}{2}$? REPLY [5 votes]: Since martingale methods fail, not much is know about the distribution of hitting times for the fractional Brownian with parameter $H \neq 1/2$. The hitting time of a level is studied by Decreusefond and Nualart in http://arxiv.org/abs/math/0606086 They get an upper bound for the Laplace transform of the hitting time. I doubt there is an "explicit" formula for the expected escape time of an interval, but you may be able to obtain an upper bound using the techniques of the above paper.<|endoftext|> TITLE: Adjoining adjoints in a 2-category QUESTION [10 upvotes]: For a given 2-category $C$, does there exist a faithful and locally faithful 2-functor $C \to C^*$, such that the image of every 1-morphism of $C$ has a right adjoint in $C^*$? Below are some of my thoughts. Although I'll be happy to other ideas too. There are two ways which I believe give a (not very explicit) construction of a right adjoint adjoining 2-functor. As far as I can see both give the universal such 2-functor, so there should be an isomorphism between them. 1) To construct $C^*$, first take an underlying 2-graph of $C$. Consider it as a 2-computad, and add to it 1-cells which will be the right adjoints, and 2-cells which will be units and counits for the adjunctions. Then take the free 2-category on the resulting 2-computad, and factor it out by an appropriate congruence, which comes from the structure of the original 2-category and the triangle equalities. There is an obvious $C \to C^*$. 2) Let $C_0$ be the underlying 1-category of $C$. Take the universal right adjoint adjoining 2-functor $C_0 \to C'$. The explicit construction of $C'$ is in R. J. M. Dawson, R. Paré, D. A. Pronk, Adjoining adjoints, Advances inbMathematics 178 (2003), pp. 99-140. Then let $C^*$ be the pushout of $C \leftarrow C_0 \to C'$ in 2-Cat. We obtain $C \to C^*$. The question is whether $C \to C^*$ is faithful and locally faithful. It is shown in the three author paper that $C_0 \to C'$ is faithful and locally faithful. So in (2) we are taking a pushout of faithful and locally faithful 2-functors. The same question can be considered in which one wants to adjoin adjoints only to a given class of 1-morphisms of $C$. REPLY [2 votes]: I think, that under some mild conditions, there is a more natural and explicit construction (I am writing this off the top of my head, you have to carefully check my statements). The idea is that "adjoining a right adjoint to a morphism, makes the morphism a relation". First, let us consider an easier (1-dimensional) case, when $\mathbb{C}$ is 2-discrete. If $\mathbb{C}$ is regular, then there is a faithful functor: $$\mathbb{C} \rightarrow \mathit{Rel}(\mathbb{C})$$ that embeds $\mathbb{C}$ into its 2-posetal category of internal relations $\mathit{Rel}(\mathbb{C})$ and has the property that the image of every morphism has a right adjoint (in fact the converse is also true --- if a morphism in $\mathit{Rel}(\mathbb{C})$ has a right adjoint, then it comes from $\mathbb{C}$). An internal relation in $\mathbb{C}$ is a span of morphisms $A \leftarrow R \rightarrow B$, where the legs are jointly monic. If we take a pushout of such a span, then we obtain a cospan representation of a relation. A 2-dimensional analogue of an internal relation is a codiscrete cofibred cospan. So, let us assume, that $\mathbb{C}$ is a cofibrational 2-category. Then one may consider the 2-category $\mathit{Mod}(\mathbb{C})$ of codiscrete cofibred cospans in $\mathbb{C}$ with the (co)fibrational composition (i.e. the dual of discrete fibred spans in $\mathbb{C}^{op}$). There is an embedding: $$\mathbb{C} \rightarrow \mathit{Mod}(\mathbb{C})$$ sending a morphism $f \colon A \rightarrow B$ to the cocomma cospan (i.e. collage) over $A \overset{id}\leftarrow A \overset{f}\rightarrow B$ which has a right adjoint in $\mathit{Mod}(\mathbb{C})$. (If $\mathbb{C}$ is fibrational, one may also consider the dual construction --- the embedding of $\mathbb{C}$ into the 2-category $\mathit{DFib}(\mathbb{C})$ of discrete fibred spans in $\mathbb{C}$.)<|endoftext|> TITLE: Fractal-like structures arising from the action of a group on $\mathbb{Z}^2$ QUESTION [29 upvotes]: Let $G := \langle a, b, c \rangle < {\rm Sym}(\mathbb{Z}^2)$ be the group generated by the permutation $$ a: \ (m,n) \ \mapsto \ (m-n,m) $$ of order $6$ and the involutions $$ b: \ (m,n) \ \mapsto \ \begin{cases} (m,2n+1) & \text{if} \ n \equiv 0(\text{mod} \ 2), \\ (m,(n-1)/2) & \text{if} \ n \equiv 1(\text{mod} \ 4), \\ (m,n) & \text{if} \ n \equiv 3(\text{mod} \ 4) \\ \end{cases} $$ and $$ c: \ (m,n) \ \mapsto \ \begin{cases} (m,2n+3) & \text{if} \ n \equiv 0(\text{mod} \ 2), \\ (m,(n-3)/2) & \text{if} \ n \equiv 3(\text{mod} \ 4), \\ (m,n) & \text{if} \ n \equiv 1(\text{mod} \ 4). \\ \end{cases} $$ Drawing spheres of radius $r$ about $(0,0)$ for "large enough" $r$ reveals fractal-like structures. Added on July 28, 2014: A video showing more pictures is now available on YouTube here. The video starts with a sequence showing entire spheres of small radii, i.e. from $r = 8$ to $r = 24$, and continues with pictures showing smaller parts of spheres of larger radii up to $r = 45$. Monochrome pictures show only one sphere, respectively, a part thereof; colored pictures show multiple spheres in different colors. Added on March 16, 2014: Pictures of the spheres of radius $30$ and $36$ can be downloaded here: Radius 30 (3487 x 3079 pixels, 111KB), Radius 36 (10375 x 9103 pixels, 693KB). Sample snippets of the large -- about $200$ megapixels at $r = 38$ to about $3$ gigapixels at $r = 45$ -- pictures are (black pixel = belongs to sphere, white pixel = doesn't belong to sphere): The images above show parts of the spheres of radius $38$, $40$ and $45$. Question: How can the observed patterns be explained? Remark 1: The cardinalities of the spheres of radii $r = 0, \dots, 45$ about $(0,0)$ are 1, 2, 4, 8, 14, 26, 39, 68, 114, 188, 289, 404, 560, 827, 1341, 2052, 3158, 4540, 6091, 8630, 12241, 17739, 27727, 41846, 61234, 86647, 117806, 163795, 233939, 340659, 523862, 768739, 1110855, 1569204, 2148377, 2994661, 4287462, 6195498, 9389566, 13568954, 19542862, 27619364, 38048372, 53304607, 76433012, 109839303. The entire sphere of radius $20$ looks as follows (the overall shape of the larger spheres is roughly similar): Added: At a scale of about $1:100$, the entire sphere of radius $45$ looks as follows: Remark 2: In the notation of this and this question, $b$ and $c$ induce on the second coordinate the class transpositions $\tau_{0(2),1(4)}$ and $\tau_{0(2),3(4)}$, respectively. Further, in the notation of (1) and (2) we have $G < {\rm RCWA}(\mathbb{Z}^2)$. REPLY [10 votes]: This phenomenon can be understood as follows: Let $\Gamma < Sym(\mathbb{Z}^{2})$ be a transitive hyperbolic group (and non virtually cyclic). Through this faithful transitive action of $\Gamma$ on $\mathbb{Z}^{2}$ (and orbit of (0,0)), the Cayley graph of $\Gamma$ (Gromov hyperbolic geometry) is projected (in a manner not too degenerate) on $\mathbb{Z}^2$ (euclidean geometry). Then the emergence of explosive and fractal-like structures on the spheres is not surprising. So any (non virtually cyclic) hyperbolic group would generate beautiful fractal-like pictures. Example: $\Gamma = \langle a,b \rangle < Sym(\mathbb{Z}^{2})$ such that $a: (m,n) \to (m+1,m+n)$ $b: (m,n) \to (m+n,n+1)$ To be confirmed: $\Gamma$ acts transitively on $\mathbb{Z}^{2}$. Remark: $\Gamma$ is neither free nor hyperbolic nor torsion-free (see the comments below). The cardinalities of the spheres of radii $r=0,\dots,14$ about $(0,0)$ are : $1,4,11,20,47,100,238,512,1124,2540,5569,12101,26208,56720,122600$ The entire sphere of radius $14$: The entire ball of radius $14$ with a rainbow gradient according to the spheres: Remark: I don't know if your group $G$ is hyperbolic, but we could have the same understanding for all the finitely generated groups of exponential growth.<|endoftext|> TITLE: Are there geometrically formal manifolds, which are not rationally elliptic? QUESTION [25 upvotes]: Formality of a space is meant in the sense of Sullivan, i.e. a space $X$ is called formal, if its commutative differential graded algebra of piecewise linear differential forms $(A_{PL}(X),d)$ is weakly equivalent (There is zig-zag of quasi-isomoprhisms.) to the rational cohomology algebra $(H^*(X,\mathbb{Q}),0)$ of $X$. Since I'm only interested in manifolds, this is equivalent to the case of weak equivalence between the de Rham algebra of differential forms $(A_{DR}(X),d)$ and the real cohomology algebra of $X$. A compact oriented manifold $X$ is called geometrically formal, if it admits a formal metric, i.e. a Riemannian metric $g$, such that the wedge product of harmonic forms is harmonic again. The typical class of examples are compact symmetric spaces. By using Hodge decomposition, it's easy to see, that geometric formality implies formality. The converse is not true, geometric formality is much stronger than formality. A formal metric has topological consequences, which go beyond formality. One can for example show that the Betti numbers of a geometric formal manifold $X$ are bounded above by the one of a torus in the same dimension, e.g. $$b_i(X)\le b_i(T^{dim(X)})={dim(X) \choose i}.$$ This bound on the Betti numbers is also known for rationally elliptic spaces $X$, that are spaces whose total dimensions of their rational cohomology $H^*(X,\mathbb{Q})$ and their rational homotopy $\pi_*(X)\otimes\mathbb{Q}$ are finite. For compact manifolds, this is clearly equivalent to having finite dimensional rational homotopy. There are examples of rationally elliptic compact simply-connected manifolds, which are not geometrically formal. (Some of them generalized symmetric spaces, see On formality of generalised symmetric spaces by D. Kotschick and S. Terzic) Are there examples of compact oriented manifolds, which are geometrically formal, but not rationally elliptic? Are there even examples of simply connected ones? Edit: As a sidenote, beside symmetric spaces, one can show, that manifolds with nonnegative curvature operatore are geometric formal. But since those have nonnegative sectional curvature, they won't serve as counterexamples easily (at least in the simply connected case), because of the Bott conjecture, which states that all simply connected nonnegative sectional curved manifolds are rationally elliptic. REPLY [9 votes]: Consider complex conjugation $\mathbb{C}\mathbb{P}^2 \to \mathbb{C}\mathbb{P}^2$. Perturb this map by taking an affine chart $\mathbb{C}^2$ and setting $f(z) = \rho(|z|)\cdot \bar{z}$, where $\rho$ is a smooth map to $\mathbb{R}_{\geq 0}$ that is 0 for $|z| < 1$, and 1 for $|z| > 2$, say. The map $f\colon \mathbb{C}\mathbb{P}^2 \to \mathbb{C}\mathbb{P}^2$ is orientation-preserving and its effect on $H^2(\mathbb{C}\mathbb{P}^2)$ is that of negative the identity. Now, take two copies of $\mathbb{C}\mathbb{P}^2$, cut out small neighborhoods of the balls in the corresponding affine charts where $\rho$ makes the map $f$ constant, and glue the resulting manifolds together to obtain $\mathbb{C}\mathbb{P}^2 \# \mathbb{C}\mathbb{P}^2$ along with an orientation-preserving diffeomorphism $\mathbb{C}\mathbb{P}^2 \# \mathbb{C}\mathbb{P}^2 \to \mathbb{C}\mathbb{P}^2 \# \mathbb{C}\mathbb{P}^2$ whose effect on second cohomology is given by $(\begin{smallmatrix} -1 & 0 \\ 0 & -1 \end{smallmatrix})$. Repeat this procedure on $\mathbb{C}\mathbb{P}^2 \# \mathbb{C}\mathbb{P}^2$ and $\mathbb{C}\mathbb{P}^2$ to obtain an orientation-preserving diffeomorphism $g \colon (\mathbb{C}\mathbb{P}^2)^{\#3} \to (\mathbb{C}\mathbb{P}^2)^{\#3}$ whose induced map on second cohomology is $$\begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix}.$$ Now consider the mapping torus $M$ of $g$. As can be seen from the long exact sequence in homotopy for the fibration $(\mathbb{C}\mathbb{P}^2)^{\#3} \to M \to S^1$, the rational homotopy $\pi_*(M) \otimes \mathbb{Q}$ has infinite rank. From the long exact sequence in relative cohomology for the inclusion $(\mathbb{C}\mathbb{P}^2)^{\#3} \hookrightarrow M$ we see that $H^2(M, \mathbb{Q}) = 0$. Therefore $M$ has Betti numbers $b_1 = 1, b_2 = 0, b_3 = 0, b_4 = 1$, and so by Theorem 8 in Kotschick's ``On products of harmonic forms'' (https://arxiv.org/abs/math/0004009) it admits a formal metric. Note that the manifold $M$ has the rational cohomology ring of $S^1 \times S^4$ and so, since geometric formality implies formality, the minimal model of $M$ is given by $$\Lambda(x_1, x_4, y_7; \ \ dx_1 = dx_4 = 0, \ dy_7 = x_4^2),$$ where subscripts denote degrees. It would then seem from this model that $M$ is in fact rationally elliptic. However, $M$ is not a nilpotent space (consider the action of $\pi_1$ on $\pi_2$), and so the ranks of the rational homotopy groups cannot be read off immediately from the amount of generators in the model. Also note that "rationally elliptic" is a concept that, at least in Félix, Halperin, Thomas, is defined only for simply connected spaces. So the above example doesn't truly answer your question.<|endoftext|> TITLE: Second moment for the number of divisors function QUESTION [5 upvotes]: Let $d(n)$ be the number of divisors of an integer $n$. Does there exists a bound for $\sum_{k\leq n}d^2(k)$? I saw in a paper of Barry and Louboutin that the asymptotics is $\frac1{\pi^2}nlog^3n$ but there was no proof nor a reference or any indication that this is the best possible bound. Any idea where can I find this? Thanks. REPLY [9 votes]: Inspired by Gerry Myerson's comment, I present below an elementary argument that does not use any complex analysis. Our starting point is the identity $$ d^2(k)=\sum_{k=l^2m}\mu(l)d_4(m), $$ where $d_4(m)$ enumerates the number of decompositions $m=m_1m_2m_3m_4$. By the multiplicative nature of this equation, it suffices to verify it when $m=p^r$ is a prime power. In this case the identity boils down to $$ (r+1)^2=\binom{r+3}{3}-\binom{r+1}{3}, $$ which is straightforward to check. With the above identity at hand, we rewrite $$ \sum_{k\leq n}d^2(k)=\sum_{l^2m\leq n}\mu(l)d_4(m)=\sum_{l\leq\sqrt{n}}\mu(l)\sum_{m\leq\frac{n}{l^2}}d_4(m). $$ We approximate the inner sum by Dirichlet's hyperbola method (cf. Exercise 2 in Section 1.5 of Iwaniec-Kowalski: Analytic Number Theory), and we obtain $$ \sum_{k\leq n}d^2(k)=\sum_{l\leq\sqrt{n}}\mu(l)\frac{n}{6l^2}\left\{\log^3\left(\frac{n}{l^2}\right)+O\left(\log^2\left(\frac{n}{l^2}\right)\right)\right\}. $$ It follows that $$ \sum_{k\leq n}d^2(k)=\frac{n\log^3(n)}{6}\left(\sum_{l\leq\sqrt{n}}\frac{\mu(l)}{l^2}\right)+O\left(n\log^2(n)\right) =\frac{n\log^3(n)}{6\zeta(2)}+O\left(n\log^2(n)\right).$$ Finally, using $\zeta(2)=\pi^2/6$ as a black box, we obtain the formula asked by the OP.<|endoftext|> TITLE: More asymptotics for trees QUESTION [7 upvotes]: This is a follow up to my recent question on the asymptotics of A003238. Lucia gave a fine answer to that question, but as I hinted the 'real' problem I have in mind is slightly different, and I've not been able to massage that answer into a solution for this case. On the other hand it feels like it should be easier or known how to handle the resulting problem, so here we are again. Consider rooted trees where all vertices at the same level have the same number of children and this is $\geq 3$ except for leaves. If we let $s(n)$ denote the number of such trees with $n$ nodes then we get a sequence for $n \geq 1$: 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, ... Note the first 2 occurs at $n = 13$ corresponding to either a root with 12 children, or a root with three children, each of which has three children. Obviously, the $s$ sequence doesn't have an asymptotic form (since e.g. $s(p+1) = 1$ for all odd primes $p$). However, letting $$ S(x) = \sum_{n \leq x} s(n) $$ and rearranging the order of summation as in Lucia's solution gives: $$ S(n+1) = 1 + \sum_{d \geq 3} S(n/d) $$ (plus the obvious extra conditions, $S(0) = 0$, $S(x) = S(\lfloor x \rfloor)$.) So, what are the asymptotics of $S$? REPLY [3 votes]: I'd like to promote Lucia's comment to an answer if I could but apparently I can't. I'll just fill in a few of the details. The basic idea is to pretend that $S(x) = C x^{\alpha}$. Plug in to the recursion and ignore terms that are $o(x^{\alpha})$ and you get $\zeta(\alpha) - 1 - 2^{-\alpha} = 1$. Now suppose that $\beta < \alpha$. Choose a "suitably large" $N$ and a constant $C$ such that $S(n) \geq C n^{\beta}$ for all $n \leq N$. Then it's easy to show that that inequality remains true for all $n$ (in fact it gets better - but this doesn't really matter, since we could just use a slightly larger $\beta$ instead). So in particular $n^\beta = o(S(n))$ for all $\beta < \alpha$. Similarly $S(n) = o(n^{\gamma})$ for all $\gamma > \alpha$. Now it does seem that in fact $S(n) = C n^{\alpha} (1 + o(1))$ for $C \approx 0.19165$, and I don't quite see how to get that level of detail, but frankly I don't really need it.<|endoftext|> TITLE: Hochschild homology of quiver algebras QUESTION [5 upvotes]: Let $K$ be a field and $\Gamma$ a quiver (=multidigraph) and $K[\Gamma]$ its quiver algebra (free $K$-module on the set of all paths of length $\geq0$ where multiplication is concatenation if endpoints match and $0$ otherwise). I'm looking for families of $\Gamma$ for which Hochschild homology $HH_\ast(K[\Gamma])$ hasn't yet been computed. If $\Gamma$ is the directed $n$-cycle (vertices $1,\ldots,n$ and edges $(1,2),(2,3),\ldots,(n-1,n),(n,1)$), or if $\Gamma$ is the $n$-loop (vertex $1$ and edges $(1,1),\ldots,(1,1)$ with $n$ copies), what is $\dim_KHH_k(K[\Gamma])$? How about if $\Gamma$ is the one-point-wedge of finitely many directed cycles $C_{n_1}\vee\ldots\vee C_{n_k}$? REPLY [8 votes]: The Hochschild homology of all path algebras $A=k\Gamma$ is known. I will assume the quiver is finite. They are hereditary algebras, so $HH_p(A)$ is zero as soon as $p>0$ provided $A$ is finite dimensional (which in this case is the same thing as $\Gamma$ being acyclic): this follows from the theorem of Keller that Dag Madsen mentioned the other day to you. In general, there is a very short and simple projective resolution of $A$ as a bimodule, from which one can actually compute with ease. Let $E\subseteq A$ be the subalgebra spanned by the vertices and $V\subseteq A$ be the subspace spanned by the arrows. Then both $A$ and $V$ are $E$-bimodules in a natural way, so we can consider the complex $$0\to A\otimes_EV\otimes_E A\xrightarrow{d}A\otimes_EA$$ with $d$ the unique $A$-bimodule map such that $d(1\otimes v\otimes 1)=v\otimes 1-1\otimes v$. You should be able to check that this is in fact a projective resolution of $A$ as an $A$-bimodule, and use it to compute $HH_*(A)$. This is to answer Dag's question in the comment. Let's use the above resolution $P$ to compute $HH_*(A)=\operatorname{Tor}_*^{A^e}(A,A)$, so we have to compute $H(A\otimes_{A^e}P)$. The complex $A\otimes_{A^e}P$ is, up to hopefully obvious identifications, $$0\to A\otimes_E V\otimes_E\to A\otimes_E\tag{$\star$}$$ where the notation $A\otimes_E$ means $A\otimes_{E^e}E$ and the differential is the $k$-linear map such that $d(a\otimes v)=[a,v]$. Now $A$ has the set of paths in $\Gamma$ as basis, $A\otimes_E$ has the set of closed paths in $\Gamma$ as basis, and $A\otimes_EV\otimes_E$ has the set of elementary tensors of the form $u\otimes\alpha$ with $u$ and $\alpha$ a path and an arrow in $\Gamma$, respectively, such that $u\alpha$ is a closed path; the image of the map $d$ is then generated by the elements $ua-au$ for all such elementary tensors. If $u$ is a closed path in $\Gamma$ of positive length, then for all $n\in\mathbb N$ we have $u^n\in A\otimes_E$. There is a $k$-linear map $\epsilon:A\otimes_E\to k$ which maps every closed path to $1$; this map clearly vanishes on the image of $d$ and then, as $\epsilon(u^n)=1$, we see that $u^n$ is not a boundary. Moreover, the set $\{u^n:n\geq1\}$ is linearly independent modulo boundaries because the image of $d$ is a homogeneous subspace of $A\otimes_E$ for the grading by length. It follows that $HH_0(A)$ is infinite dimensional as soon as there is a cycle of positive length in $\Gamma$. Let $\ell\geq1$ and let us look at the degree $\ell$ component of the complex ($\star$), again for the length graduation, we have $$0\to A_{\ell-1}\otimes_E V\otimes_E\xrightarrow{d_\ell} A_\ell\otimes_E$$ where $A_i$ denotes the subspace spanned by paths of length $i$. The dimension of $A_\ell\otimes_E$ is the number of closed paths of length $\ell$, and the dimension of $A_\ell\otimes_EV\otimes_E$ is the number of pairs $(v,\alpha)$ with $v$ a path, $\alpha$ and arrow and $v\alpha$ a closed path. These two numbers are the same: there is an bijective map from closed paths of length $\ell$ to pairs $(u,\alpha)$ of this form --- simply split off the arrow. It follows that the kernel of $d_\ell$ is zero iff its cokernel is zero. Since we know that $HH_0(A)$ is infinite dimensional, it follows that so is $HH_1(A)$.<|endoftext|> TITLE: Intersecting Sets of Pythagorean Triples with Common Hypotenuses QUESTION [6 upvotes]: For any $r\in\mathbb{N}$, let $A_r$ denote the set of all natural numbers that are potentially a side of a Pythagorean triple with hypotenuse $r$. Given any $N\in\mathbb{N}$, does there exist $r,s$ such that $A_r\cap A_s$ contains at least $N$ elements? I've been looking at the parameterization of Pythagorean triples as $k(m^2-n^2,2mn,m^2+n^2)$, and considering numbers $r$ that can be written as the sum of two squares in a large number of ways. But the components of this sum seem rather unpredictable, and I'm unsure how to proceed in this fashion. REPLY [2 votes]: Take integers that are sums of two squares; this set is multiplicatively closed because of the identity $(a^2+b^2)(c^2+d^2) = (ad-bc)^2 + (ac+bd)^2$ Now interchanging $a$ and $b$, the new rhs gives another expression for the same LHS as sum of two squares. Now this explodes when a number is a product of many terms which are individually sum of two squares. And primes 1 mod 4 are by Fermat sum of two squares, and a number that is a product of many such primes will provide examples you need.<|endoftext|> TITLE: Hall-Littlewood functions and functions on the nilpotent cone QUESTION [9 upvotes]: The following observation between the spaces of global sections of line bundles on the nilpotent cone and the Hall-Littlewood polynomials is made in a recent physics preprint 1403.0585. Is this a known mathematical fact? Pick a simple Lie algebra $\mathfrak{g}$ over $\mathbb{C}$. Let $\mathcal{N}$ be the nilpotent cone of $\mathfrak{g}$. Given an element $w$ in the positive Weyl chamber of the weight lattice of $\mathfrak{g}$, there is a natural ad $\mathfrak{g}$-equivariant line bundle $L_w$ on $\mathcal{N}$. Let $V_w=H^0(\mathcal{N},L_w)$. $V_w$ has a natural action of $G\times \mathbb{C}^\times$ (where $G$ is a Lie group for $\mathfrak{g}$). Then the character of $V_w$ as a $G\times \mathbb{C}^\times$ representation is (essentially) given by the Hall-Littlewood polynomial $P_w(x,q)$ of type $\mathfrak{g}$ where $(x,q)\in G\times \mathbb{C}^\times$. I must confess that I don't know how to construct $L_w$ for general $\mathfrak{g}$. For type $A$, it goes as follows. Let $\mathfrak{g}=\mathfrak{sl}(N)$. The nilpotent cone $\mathcal{N}$ is a holomorphic symplectic quotient of a linear space (in particular, a quiver variety in the sense of Nakajima). In particular, it has the form $\mathcal{N}=X / GL(1)\times GL(2) \times \cdots GL(N-1)$ Then the line bundle $L_w$ is $L_w = (X\times \mathbb{C} )/GL(1)\times GL(2) \times \cdots GL(N-1)$ where $GL(1)\times GL(2) \times \cdots GL(N-1)$ acts on $\mathbb{C}$ by a one-dimensional representation $\chi_w: GL(1)\times GL(2) \times \cdots GL(N-1)\to \mathbb{C}^\times$ determined by $w$ (a weight vector in the positive Weyl chamber of the weight lattice of $\mathfrak{sl}(N)$. I would also appreciate if you could tell me how $L_w$ can be constructed for other $\mathfrak{g}$, as part of the question. REPLY [5 votes]: I think Steven's answer addresses whether this was already known (yes), but let me try to clear up what's going with these line bundles. In Nakajima's picture, you're describing the nilcone in an inherently Springery way: you consider the quiver $\mathbb{C}^1\to \mathbb{C}^2 \to \mathbb{C}^3\to \cdots \to \mathbb{C}^N$, double it and consider the action of $GL(1)\times \cdots \times GL(N-1)$ acting by pre- and post-composition. If $X$ is the map going right, and $Y$ the map going left (let's just take the sum of all the arrows; we can distinguish them by what space they act on), the moment map condition says that $[X,Y]=0$. In particular, the composition $XY|_{\mathbb{C}^N}$ is nilpotent, since $(XY)^{N+1}=X^{N+1}Y^{N+1}=0$. Note that this map is unchanged by the acton of $GL(1) \times \cdots \times GL(N-1)$. So, this is what you mean by getting the nilcone in the Nakajima way. Your line bundles on the other hand, correspond to the determinants of the tautological bundles associated to the smaller spaces $\mathbb{C}^j$. So, the fundamental problem with talking about line bundles on the nilcone is that for those to make sense, you would have to start with the nilpotent and find the smaller spaces (whatever that means). Which is impossible. This is a sign that you are doing the wrong thing. The problem is you never say what kind of quotient you're taking of the space X; if you want the nilcone, it's a very non-free quotient, which doesn't bode well for the line bundles. Instead, you have to throw out "bad points," specifically, you must require that $X$ is injective at each step. Thus, I can think of the quiver as actually a flag, and the action of $GL(1)\times \cdots \times GL(N-1)$ as relating all the different ways of parametrizing the spaces in the flag. Thus, modding out by them leaves us with a flag, and we still have $XY$, which is now a nilpotent transformation preserving the flag (and as it happens, all the other $Y$'s are determined by this one; they're just its restriction to the steps of the flag). Thus, we've found the Springer resolution of the nilcone. In arbitrary types, the line bundles on $T^*G/B$ have a nice description; they're in bijection with integral weights of the torus. In the case of $SL(N)$, the tautological bundles in the quiver become tautological bundles in the flag variety, so you can describe one in terms of the other. It certainly looks to me like Steven's story becomes yours in this case.<|endoftext|> TITLE: Study of convex polytopes via commutative algebra QUESTION [5 upvotes]: Let $P \subset \mathbb{R}^d$ be any convex polytope with integral vertices, and let $M$ be the additive submonoid of $\mathbb{R}^{d+1}$ which is generated by $\{ (v,1) : v \in P \cap \mathbb{Z}^d \}$. Then $M$ is an affine monoid (i.e. finitely generated, torsionfree and cancellative) and all combinatorial data of $P$ are completely determined by the isomorphism class of $M$. By choosing an arbitrary commutative ring $R$, and considering the monoid ring $R[M]$ instead of $M$ we make the whole machinery of commutative algebra available for the study of $P$. There are many interesting connections between $P$ and $R[M]$. For example, the face lattice of $P$ is anti-isomorphic to the lattice of monomial prime ideals of $R[M]$. So theoretically, we can determine faces of $P$ by studying certain prime ideals of $R[M]$. But does this also work in practice? Are there examples of polytopes $P$ (given by integral vertices) where the algebraic viewpoint makes it easier to study the combinatorics of $P$ (e.g. the face lattice, combinatorial automorphism group, etc.)? Are there famous polytopes which were defined by affine monoids and whose properties were proved by considering monoid rings? In short: I would really like to see examples where polyhedral geometry benefits from commutative algebra. Thank you in advance! REPLY [4 votes]: The Sturmfels correspondence is my favorite example. It relates the toric ideal of a point configuration (which could be the vertices of your polytope) to a regular triangulation of the point configuration. It is beautifully described, e.g., in Rekha Thomas' book Lectures in Geometric Combinatorics.<|endoftext|> TITLE: When does Merten's product theorem accurately estimate the number of coprimes in an interval? QUESTION [6 upvotes]: Assume an arbitrary $x$ and let $z$ be smaller than $y$, where $y$ is the length of the interval $[x,x+y]$. What I would like to know is: Let $W(z)=\prod_{p\leq z}\left(1-\frac{1}{p}\right)$. For what values of $z=z(y)$ is $$ y W(z) \sim \textrm{e}^{-\gamma} \frac{y}{\log z} $$ an accurate estimate of the number of integers coprime to all primes $p\leq z$ in the interval $[x,x+y]$? A trivial answer—since the set of coprimes to $p\leq z$ is periodic with period $P(z)=\prod_{p\leq z} p$—is to choose $z$ so that $y/P(z) p \rightarrow \infty$. From numerical experiments, however, it seems that the far more moderate constraint $z/y\rightarrow 0$ is sufficient. So when can we indeed assume that Merten's product theorem correctly estimates the number of coprimes in an interval? EDIT 1: To be specific, the observation I make numerically is that if we look at the interval $[x,x+y]$ in regions where $x$ is much larger than $y$, but still much smaller than $P(z)$, Merten's product theorem seems to accurately describe the density of coprimes to $p1$ is a constant, and we let $y\rightarrow \infty$, the density of coprimes to $p