euclaise/Ferret_7B
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https://mathoverflow.net/questions/13 | 38 | Can someone suggest a good book for teaching myself about Lie groups? I study algebraic geometry and commutative algebra, and I like lots of examples. Thanks.
| https://mathoverflow.net/users/4 | Learning about Lie groups | There's also Fulton & Harris "Representation Theory" (a Springer GTM), which largely focusses on the representation theory of Lie algebras. Everything is developed via examples, so it works carefully through $sl\_2$, $sl\_3$ and $sl\_4$ before tackling $sl\_n$. By the time you get to the end, you've covered a lot, but might want to look elsewhere to see the "uniform statements". An excellent book.
| 26 | https://mathoverflow.net/users/3 | 15 | 1 |
https://mathoverflow.net/questions/12 | 21 | Is a quotient of a reductive group reductive?
**Edit** [Pete L. Clark]: As Minhyong Kim points out below, a more precise statement of the question is:
Is the quotient of a reductive linear group by a Zariski-closed normal subgroup reductive?
| https://mathoverflow.net/users/2 | Is a quotient of a reductive group reductive? | For a *linearly* reductive (i.e. all representations are completely redudible) group $G$, any quotient $G/H$ is also linearly reductive. Given a representation $V$ of $G/H$, you get an induced representation of $G$ (with the same underlying vector space). Then any decomposition of $V$ as a representation of $G$ is also a decomposition as a representation of $G/H$. Since $G$-invariant subspaces of $V$ are the same thing as $G/H$-invariant subspace of $V$, the notions of irreducible subrepresentations agree. So you get a complete decomposition of $V$ as a representation of $G/H$.
In characteristic zero, reductive is equivalent to linearly reductive, so you have your answer. In general, a group is reductive if its unipotent radical is trivial. I feel like you should be able to argue that an element of the unipotent radical of $G/H$ lifts to give you an element of the unipotent radical of $G$, but I'm not sure.
| 13 | https://mathoverflow.net/users/1 | 19 | 2 |
https://mathoverflow.net/questions/21 | 34 | What is an example of a finite field extension which is not generated by a single element?
Background: A finite field extension E of F is generated by a primitive element if and only if there are a finite number of intermediate extensions. See, for example, [Lang's Algebra, chapter V, Theorem 4.6].
| https://mathoverflow.net/users/1 | Finite extension of fields with no primitive element | Let $F$ be a finite field with $p$ elements. Let $K=F(x,y)$ be the field of rational functions in two indeterminate variables over $F$. Consider the extension of $K$ obtained by adjoining $p$-th roots of $x$ and of $y$. More precisely, let $k$ be an algebraic closure of $K$. In $k$ we can solve the equation $X^p=x$ in the variable $X$. Let $a$ be a solution of this equation; so $a$ is an element of $k$ which satisfies $a^p=x$. Similarly find an element $b$ which satisfies $b^p=y$.
Consider $L=K(a,b)$. $L$ is a finite extension of $K$, of order $p^2 $ as you can check. However there is no element of degree $p^2$ in $L$, and a primitive element would have to have degree $p^2$.
This example is, in a sense, the simplest possible. Separable finite extensions are simple (contain a primitive element), so we must use a non-perfect base field. Also, extensions of degree $p$ are also simple, so we must use $p^2$.
| 45 | https://mathoverflow.net/users/25 | 25 | 3 |
https://mathoverflow.net/questions/28 | 18 | I'm afraid my first question isn't a math puzzle per se, but rather question of math "presentation" . Basically I've been out of school for a year or two - so I'm a bit out of practice in writing up math papers. Recently I've found myself back at school for grad work and having to write equations again. Now I've already re-learned a good hatred for most equation editors that exist in word processors - but my "LaTex fu" is a bit weak to be writing up everything in notepad and then compiling it using LaTex. So I was wondering if anyone had found a good program to use a "friendly" interface while still being able to write LaTex style equations?
| https://mathoverflow.net/users/16 | LaTeX based document editors | <http://www.bakoma-tex.com/>
I suppose you are on Windows, if you are on \*nix you can try it with [wine](http://www.winehq.org/).
As I've read it is good, but it is not free.
| 9 | https://mathoverflow.net/users/14 | 29 | 4 |
https://mathoverflow.net/questions/26 | 62 | Can a (possibly infinite-dimensional) vector space ever be a finite union of proper subspaces?
If the ground field is finite, then any finite-dimensional vector space is finite as a set, so there are a finite number of 1-dimensional subspaces, and it is the union of those. So let's assume the ground field is infinite.
| https://mathoverflow.net/users/1 | Can a vector space over an infinite field be a finite union of proper subspaces? | You can prove by induction on n that:
An affine space over an infinite field $F$ is not the union of $n$ proper *affine* subspaces.
The inductive step goes like this: Pick one of the affine subspaces $V$. Pick an affine subspace of codimension one which contains it, $W$. Look at all the translates of $W$. Since $F$ is infinite, some translate $W'$ of $W$ is not on your list. Now restrict all other subspaces down to $W'$ and apply the inductive hypothesis.
This gives the tight bound that an $F$ affine space is not the union of $n$ proper subspaces if $|F|>n$. For vector spaces, one can get the tight bound $|F|\geq n$ by doing the first step and then applying the affine bound.
| 40 | https://mathoverflow.net/users/27 | 36 | 10 |
https://mathoverflow.net/questions/37 | 11 | What are some examples of morphisms of schemes which are not quasi separated?
| https://mathoverflow.net/users/28 | Non-quasi separated morphisms | Suppose $U\hookrightarrow X$ is a non-quasi-compact open immersion. Then you can glue two copies of $X$ together along $U$ (effectively doubling up the complement of $U$) to get a non-quasi-separated scheme $Y$.
By the assumption that $U\hookrightarrow X$ is not quasi-compact, there is some open affine $W$ of $X$ such that the intersection of $W$ and $U$ is not quasi-compact. So there are two copies of $W$ sitting inside of $Y$ (one for each copy of $X$). The intersection of these two is exactly $U\cap W$, which is not quasi-compact. So we found two open affines in $Y$ whose intersection is not quasi-compact, which shows that $Y$ is not quasi-separated.
Now we just have to find some non-quasi-compact open immersions. The complement of the origin in $\mathbb A^\infty$ is one (so $\mathbb A^\infty$ with a doubled origin is a non-quasi-separated scheme).
**Edit:** Here's another one that I don't completely understand, but gives a finite-dimensional example (zero-dimensional in fact). Consider $X=Spec(\overline{\mathbb Q}\otimes\_{\mathbb Q}\overline{\mathbb Q})$. Topologically, $X=Gal(\overline{\mathbb Q}/\mathbb Q)$, with the profinite topology (perhaps somebody could explain how to see this in a comment). In particular, any point is closed, but the complement $U$ is not quasi-compact, so we get another example of a non-quasi-compact open immersion, so $X$ with a doubled point is non-quasi-separated.
| 17 | https://mathoverflow.net/users/1 | 40 | 12 |
https://mathoverflow.net/questions/38 | 18 | Suppose $G$ is a finite group and $f$ is an automorphism of $G$. If $f(x)=x^{-1}$ for more than $\frac{3}{4}$ of the elements of $G$, does it follow that $f(x)=x^{-1}$ for all $x$ in $G\ ?$
I know the answer is "yes," but I don't know how to prove it.
---
Here is a nice solution posted by administrator, expanded a bit:
Let $S = \{ x \in G: f(x) = x^{-1}\}$.
>
> **Claim:** For $x$ in $S$, $S\cap x^{-1}S$ is a subset of $C(x)$, the
> centralizer of $x$.
>
>
>
*Proof:* For such $y$, $f(y) = y^{-1}$ and $f(xy) = (xy)^{-1}$. Now
$$x^{-1} y^{-1} = f(x)f(y) = f(xy) = (xy)^{-1} = y^{-1}x^{-1}.$$
So $x$ and $y$ commute.
Since $S\cap x^{-1}S$ is more than half of $G$, so is $C(x)$. So by Lagrange's Theorem, $C(x) = G$, and $x$ is in the center of $G$. Thus $S$ is a subset of the center, and it is more than half of $G$. So the center must be all of $G$, that is $G$ is commutative. Once $G$ is commutative the problem is easy.
| https://mathoverflow.net/users/27 | Homomorphism more than 3/4 the inverse | I think the point of this whole $3/4$ business is the following. If $G\_1$ is the set of elements such that $f(x) = x^{-1}$, then if we look at left multiplication on $G$ by an element of $G\_1$, more than half the elements have to make back into $G\_1$.
Combining this with what we know about $f$ it should follow that any $g \in G\_1$ commutes with more than $1/2$ the elements of $G$, so if you say Lagrange's theorem enough times it should follow that $G$ is abelian and $G\_1$ generates $G$, which together imply the result.
| 14 | https://mathoverflow.net/users/7 | 48 | 16 |
https://mathoverflow.net/questions/23 | 8 | Artin-Wederburn says that if you have a semisimple algebra then it is a product of matrix algebras over division rings.
Suppose that $C$ is a fusion category over the complex numbers (if you want to assume pivotal or similar things, that's fine, but don't assume symmetric or braided). Suppose that $X$ is an algebra object in $C$. That is $X$ is an object in $C$ together with a multiplication map $X \otimes X \rightarrow X$ and a unit etc. We call $X$ semisimple if the category of $X$-module objects in $C$ is semisimple. Is there some good analogue of Artin-Wedderburn?
| https://mathoverflow.net/users/22 | Is there a good version of Artin-Wedderburn for semisimple algebra objects? | I have a paper on that topic:
* *Finite, connected, semisimple, rigid tensor categories are linear*, Math. Res. Lett. **10** (2003), 411-421, doi:[10.4310/MRL.2003.v10.n4.a1](https://doi.org/10.4310/MRL.2003.v10.n4.a1), arXiv:[math/0209256](https://arxiv.org/abs/math/0209256).
| 4 | https://mathoverflow.net/users/1450 | 51 | 19 |
https://mathoverflow.net/questions/41 | 9 | Suppose I have a morphism $f:X\rightarrow Y$ such that the relative sheaf of differentials $\Omega\_{X/Y}$ is locally free. Does it follow that $f$ is smooth?
The answer is no, but for a silly reason. You could have some non-reducedness $\mathrm{Spec}\left(k\left[e\right]/e^{2}\right)$ over $\mathrm{Spec}\left(k\right)$ has a free sheaf of differentials, but isn't smooth). But what if you add the hypothesis that the rank of $\Omega\_{X/Y}$ is $\dim X-\dim Y$?
**Edit:** As Jonathan points out in his answer, I was careless with my counterexample. It only works if $\mathrm{char}\,k=2$.
| https://mathoverflow.net/users/1 | If $\Omega_{X/Y}$ is locally free of rank $\mathrm{dim}\left(X\right)-\mathrm{dim}\left(Y\right)$, is $X\rightarrow Y$ smooth? | I think Ishai's example is close, but one must be a little careful; the normalization of the node is a good example, but the normalization of the cusp is ramified, and the sheaf of relative differentials in that case is not even locally free.
The differential-wise condition you want is this: for the morphism morphism $f: X \to Y$ to be smooth, you need that the sequence
$$0 \to f^\* \Omega\_Y \to \Omega\_X \to \Omega\_{X/Y} \to 0$$
be exact and locally split (I can't find a reference that says this is sufficient, so it may not be). In the special case when $\dim X = \dim Y$,
$\Omega\_{X/Y}$ is 0 if and only if $f$ is unramified. But in this case $f^\* \Omega\_Y \to \Omega\_X$ can still fail to be injective.
| 5 | https://mathoverflow.net/users/37 | 52 | 20 |
https://mathoverflow.net/questions/46 | 32 | What is the universal property of normalization? I'm looking for an answer something like
>
> If X is a scheme and Y→X is its
> normalization, then the morphism
> Y→X has property P and any other
> morphism Z→X with property P
> factors uniquely through Y.
>
>
>
| https://mathoverflow.net/users/1 | What is the universal property of normalization? | I've realized that my answer is wrong. Here's the right answer: if $Z$ is a normal scheme and $f: Z \to X$ is a morphism such that each associated point of $Z$ maps to an associated point of $X$, then $f$ factors through $n$.
A counterexample that shows why what I said previously doesn't work: let $f$ be the inclusion of the node into the nodal curve. There is no unique lift of $f$ to the normalization of the nodal curve.
What's going on here: taking the total ring of fractions is not a functor for arbitrary morphisms of reduced rings. You need a morphism such that no NZD gets mapped to a ZD, which is equivalent (for Noetherian rings) to saying that the preimage of any associated prime is an associated prime.
| 20 | https://mathoverflow.net/users/37 | 56 | 23 |
https://mathoverflow.net/questions/27 | 3 | Burt Totaro has a result that for a certain class of algebraic stacks, having affine diagonal is equivalent to the stabilizers at closed points begin affine. Is there an example of this equivalence failing in general?
| https://mathoverflow.net/users/2 | Is there an example of an algebraic stack whose closed points have affine stabilizers but whose diagonal is not affine? | The answer is yes since a fiberwise condition (such as affine stabilizers) does not imply a global condition (such as affine diagonal) without extra hypotheses (such as having the resolution property). Think of quasi-finite+proper <=> finite.
There are (non-separated) *schemes* with non-affine diagonal, for example, two copies of the affine plane glued together outside the origin.
Also see the related [question](http://www.mathoverflow.net/questions/76/).
| 7 | https://mathoverflow.net/users/40 | 60 | 26 |
https://mathoverflow.net/questions/44 | 6 | Let X be an integral scheme that is separated (say over an affine scheme). Define a Weil divisor as a finite integral combination of height 1 points of X, where the height of a point of X is the dimension of its local ring. Let Z be a closed subscheme of X, Z not equal to X.
Is there an example of such an X and Z where Z contains infinitely many Weil divisors?
| https://mathoverflow.net/users/7 | Weil divisors on non Noetherian schemes | Yes, there are such examples (with *X* quasi-compact, otherwise it is trivial). There is a general result due to Hochster ("Prime Ideal structures in commutative rings", his thesis) which says that spectra of rings are exactly those topological spaces *X* such that:
1) *X* is Kolmogorov (*T*\_0).
2) *X* is quasi-compact.
3) The quasi-compact open subsets form an open basis.
4) *X* is quasi-separated, i.e., quasi-compact open subsets are closed under finite intersections.
5) Every non-empty irreducible closed subset has a generic point.
We will construct an irreducible topological space *X* satisfying 1-5 with underlying set 2^**N** U {*x*} such that 2^**N** is closed in *X* and totally disconnected and *X* has generic point *x*. If *X*=Spec(*A*), then the closed subvariety 2^**N** is a union of infinitely many Weil divisors. (*X* has dimension 1)
The topology on 2^**N** will be the product topology, i.e., that of the 2-adic integers (or if you prefer, the Cantor set). An open basis for this topology are cylinders, i.e., sets where a finite number of components are fixed. The cylinders are also closed. The space 2^**N** is Hausdorff, compact and totally disconnected, hence satisfies 1-5.
An open basis for the topology of *X* = 2^**N** U {*x*} is given by sets of the form *W* U {*x*} where *W* is a cylinder or the empty set. The quasi-compact open subsets of *X* are the finite unions of such sets and *X* satisfies 1-4. To see 2-4 note that if *V* is an open subset of *X* then
*V* is quasi-compact <=> The intersection of *V* and 2^**N** is clopen
Finally, the non-empty irreducible closed subsets of *X* are the singleton sets of 2^**N** and *X* itself and these all admit generic points so *X* satisfies 5.
**Remark**: *X* seems to be closely related to the spectrum of the Tate-algebra **Q**\_p<*x*>. The *canonical* topology on the closed points are exactly the *p*-adic integers. But the *Zariski* topology is completely different (**Q**\_p<*x*> is noetherian and regular of dimension 1).
| 13 | https://mathoverflow.net/users/40 | 70 | 32 |
https://mathoverflow.net/questions/72 | 25 | Can anyone provide an example of a real-valued function *f* with a convergent Taylor series that converges to a function that is not equal to *f* (not even locally)?
| https://mathoverflow.net/users/44 | What's an example of a function whose Taylor series converges to the wrong thing? | If you take the classic non-analytic smooth function: $e^{-1/t}$ for $t \gt 0$ and $0$ for $t \le 0$ then this has a Taylor series at $0$ which is, err, $0$. However, the function is non-zero for any positive number so it does not agree with its Taylor series in any neighbourhood of $0$.
| 31 | https://mathoverflow.net/users/45 | 73 | 33 |
https://mathoverflow.net/questions/76 | 8 | Give an example of a stack *X* with affine stabilizer groups and separated but not quasi-affine diagonal.
Remarks:
1) If *X* has finite stabilizer groups then the diagonal is quasi-finite and separated, hence quasi-affine (Zariski's MT).
2) If we drop the condition that the diagonal of *X* is separated, it is easy to find examples.
3) The stabilizer groups of *X* are affine if and only if they are quasi-affine.
| https://mathoverflow.net/users/40 | Stack with affine stabilizers but not quasi-affine diagonal | Here is an example:
In X13 of "Faisceaux amples sur les schemas en groupes", Raynaud provides an example of a group scheme *G -> S* in chacteristic 2 where
1. *S* is a local regular scheme of dimension 2.
2. *G -> S* is smooth, separated and quasi-compact.
3. The fibers of *G ->S* are affine and the generic fiber is connected.
such that *G -> S* is not quasi-projective.
Therefore, the classifying stack BG has affine stabilizers but does not have a quasi-affine diagonal.
On the other hand, in VII 2.2, Raynaud proves that if *G -> S* is a smooth, finitely presented group scheme such that
1. *S* is normal.
2. *G -> S* has connected fibers.
3. The maximal fibers are affine.
then *G -> S* is quasi-affine.
Question: Is the above statement true if (2) is weakened to require that the number of connected components over a fiber *s \in S* be prime the characteristic of the residue field *k(s)*?
Of course, one would really like to know if the statement is true if *G->S* is not necessarily flat so that one could apply it to the inertia stack.
On a related note, Raynaud also provides an example in VII3 of a smooth quasi-affine group scheme *G ->* **A^2** over a field k with connected fibers but which is not affine. The classifying stack BG gives an example of stack with affine and connected stabilizers but with non-affine inertia stack. In the example of a scheme with non-affine diagonal, the inertia is of course affine. It's also easy to provide examples of non-affine group schemes with affine but non-connected fibers (eg. the group scheme obtained by removing the non-identity element over the origin from **Z/2Z** *->* **A^2**).
| 9 | https://mathoverflow.net/users/42 | 92 | 41 |
https://mathoverflow.net/questions/79 | 9 | Suppose f:X→Y is a morphism of finite type between locally noetherian schemes. The valuative criterion for separatedness (resp. properness) says roughly that f is a separated (resp. proper) morphism if and only if the following condition holds:
>
> For any curve C in Y and for any lift
> of C-{p} to X, there is at most one
> (resp. exactly one) way to extend this
> to a lift of C to X.
>
>
>
More precisely,
>
> If C is the spectrum of a DVR with
> closed point p (a *very local* version
> of a curve: the intersection of all
> open neighborhoods of p on an "honest"
> curve), C→Y is a morphism, and
> C-{p}→X is a lift of that
> morphism along f, there is at most one
> (resp. exactly one) way to complete it to
> a lift C→X.
>
>
>
Does it suffice to check the valuative criteria on an even more local kind of object: the spectrum of a **complete** DVR? This would be quite nice because the only complete DVRs over a field k are rings of the form L[[t]] where L is an extension of k.
More generally, if you drop the hypotheses that f is of finite type and X and Y are locally noetherian, the usual valuative criteria must be verified for arbitrary valuation rings. Is it enough to check them for complete valuation rings?
| https://mathoverflow.net/users/1 | Can the valuative criteria for separatedness/properness be checked "formally"? | Yep, a quasi-compact morphism of schemes (resp. locally noetherian schemes) is universally closed if and only the existence part of the valuative criterion holds for complete valuation rings (resp. complete DVRs) with algebraically closed residue field.
This is in EGA (see II.7.3.8 and the remark II.7.3.9). Note that the separated hypothesis is not necessary there; for the valuative criterion of properness one needs to require that the morphism is quasi-separated.
This holds more generally for Artin stacks if one allows a field extension of the fraction field of the valuation ring (see LMB 7.3).
| 7 | https://mathoverflow.net/users/42 | 93 | 42 |
https://mathoverflow.net/questions/94 | 7 | [Darsh](http://math.berkeley.edu/~dranjan/) asked over at the [20 questions seminar](http://scratchpad.wikia.com/wiki/20qs):
Take a triangle in R^2 with coordinates at rational points. Can we find the smallest denominator point in the interior? (Consider denominator of an element of Q^2 to be the lcm of the denominators of the coordinates.) (Hint: you can do the 1-d version using continued fractions.)
| https://mathoverflow.net/users/3 | How can you find small denominators inside triangles? | Here's an example algorithm that finds the smallest denominator point in the interior:
1. Take the triangle's center and denote D to be its denominator.
2. Find all horizontal lines with y-coordinate's denominator not greater than D and that have a chance of intersecting your triangle.
3. Same for vertical lines.
4. Intersect these line families, select points inside your triangle and minimize their denominator.
This does look like an *unsatisfying* algorithm, but then your problem might benefit from being phrased in a different way, perhaps
* Can we find the smallest denominator point in the interior using some beautiful O(D) algorithm?
(where, presumably, D is the final answer)
| 2 | https://mathoverflow.net/users/65 | 100 | 45 |
https://mathoverflow.net/questions/95 | 15 | For Noetherian schemes this follows from Serre's criterion for affineness by a filtration argument.
| https://mathoverflow.net/users/2 | Is there an example of a scheme X whose reduction X_red is affine but X is not affine? | No, if *X* is any algebraic space such that *X*\_red is an affine scheme, then *X* is an affine scheme. This follows from Chevalley's theorem. For *X* noetherian scheme/alg. space this theorem is in EGA/Knutson. As you noted, this can also be showed using Serre's criterion for affineness or by an even simpler argument (see EGA I 5.1.9, first edition).
For *X* **non-noetherian**, the following general version of Chevalley's theorem is proved in my paper "Noetherian approximation of algebraic spaces and stacks" (arXiv:0904.0227):
**Theorem**: Let *W*->*X* be an integral and surjective morphism of algebraic spaces. If *W* is an affine scheme, then so is *X*.
Recall that any finite morphism is integral, in particular *X*\_red -> *X*. As a corollary, it follows that under the same assumptions, if *W* is a scheme then so is *X*.
| 18 | https://mathoverflow.net/users/40 | 102 | 46 |
https://mathoverflow.net/questions/99 | 12 | There's a well-known series of articles by Lusztig about **Character Sheaves**. They have important connections to many things in (geometric) representation theory, e.g. [0904.1247](http://arxiv.org/abs/0904.1247)
How to understand these for a person with less than excellent representation theory background?
| https://mathoverflow.net/users/65 | How to understand character sheaves | That's a pretty vague question. The vague answer is that all the operations (like induction from subgroups) that can be done for representations and characters can be done for sheaves, and doing these results in a category of sheaves on a group (like GL\_n over a finite field), which are close enough to the characters of representations to tell us something about them, but which also have more structure, since they are sheaves, not just functions.
Here's a somewhat more precise description: if you have a variety X which a group G acts on, then you can take the action of G on the cohomology of X. Better yet, you can get a sheaf on the group G, whose stalk over a group element g is the cohomology of the fixed points of g on X. The function sheaf correspondence sends this sheaf to the character of the representation on the cohomology of X (this follows from Lefschetz). Deligne and Lusztig defined certain varieties (the set of flags over F\_q in a given relative position to their conjugates by Frobenius) on which GL(n,F\_q) acts (actually, this works for any split simple algebraic group), and the corresponding sheaves (or rather the simple perverse constitutuents) are called character sheaves, and roughly capture the structure of the corresponding representations.
| 10 | https://mathoverflow.net/users/66 | 104 | 48 |
https://mathoverflow.net/questions/105 | 5 | This is pretty specific, but there are some experts around.
So, in Chari & Pressley, it's explained that in the standard \*-structure, every irreducible, finite-dimensional representation of a quantum group (at a generic parameter) is unitary. Is it written somewhere what the "right" unitary structure on a tensor product of these representations is?
I ask because if one categorifies such representations, one gets a unitary structure essentially for free, so it would extremely useful if someone had already written down one I could match up with.
| https://mathoverflow.net/users/66 | What is the "right" hermitian structure on tensor products of quantum group representations? | I know these are all about the root of unity case, but you might look at [this paper](http://www.ams.org/mathscinet-getitem?mr=1358983) by Kirillov, and [this one](http://www.ams.org/mathscinet-getitem?mr=1470857) by Wenzl.
| 4 | https://mathoverflow.net/users/3 | 106 | 49 |
https://mathoverflow.net/questions/101 | 26 | According to *Higher Topos Theory* [math/0608040](https://arxiv.org/abs/math/0608040) a **topos** is
>
> a category C which behaves like the
> category of sets, or (more generally)
> the category of sheaves of sets on a
> topological space.
>
>
>
Could one elaborate on that?
| https://mathoverflow.net/users/65 | What is a topos? | There are two concepts which both get called a *topos*, so it depends on who you ask. The more basic notion is that of an *elementary topos*, which can be characterized in several ways. The simple definition:
```
An elementary topos is a category C which has finite limits and power objects.
```
(A power object for A is an object P(A) such that morphisms B --> P(A) are in natural bijection with subobjects of A x B, so we could rephrase the condition "C has power objects" as "the functor Sub(A x -) is representable for every object A in C").
The issue with the simple definition is that it doesn't show you why these things are actually interesting. It turns out that a great deal follows from these axioms. For example, C also has finite colimits, exponential objects, has a representable limit-preserving functor P: C^op --> Doct where Doct the category of Heyting algebras such that if f: AxB --> A is the projection map for some objects A and B in C, then P(A) --> P(AxB) has both left and right adjoints considered as a morphism of Heyting algebras, etc etc. What the long-winded definition boils down to is "an elementary topos the the category of types in some world of intuitionistic logic." There's an incredible amount of material here; the best place to start is probably MacLane and Moerdijk's *Sheaves in Geometry and Logic.* The main reference work is Johnstone's as-yet-unfinished *Sketches of an Elephant,* but I certainly wouldn't start there.
The other major notion of *topos* is that of a *Grothendieck topos*, which is the category of sheaves of sets on some site (a site is a (decently nice) category with a structure called a *Grothendieck topology* which generalizes the notion of "open cover" in the category of open sets in a topological space). Grothendieck topoi are elementary topoi, but the converse is not true; Giraud's Theorem classifies precisely the conditions needed for an elementary topos to be a Grothendieck topos. Depending on your point of view, you might also look at *Sheaves in Geometry and Logic* for more info, or you might check out Grothendieck's SGA4 for the algebraic geometry take on things.
| 35 | https://mathoverflow.net/users/37 | 108 | 50 |
https://mathoverflow.net/questions/83 | 8 | Suppose a matrix equation $Ax = b$ has no solution ($b$ is not in the column space of $A$)
How can I find a vector $x^\prime$ so that $Ax^\prime$ is the closest possible vector to $b$?
| https://mathoverflow.net/users/44 | How to approximate a solution to a matrix equation? | I want to correct something in both Anton's ans las3rjock's answers. I hope that I am merely correcting bad notation.
The way to find x' (using the notation in the question) is, as correctly stated in each of the answers above, to derive the associated problem AT Ax' = AT b and solve that via Gaussian Elimination (also called Gauss-Jordan Elimination, the difference is technical and not important here). This is **guaranteed** to have a solution.
The fact that this is the correct solution to the problem relies on the properties of the "closest point" of a point to a subspace. We want the closest point to b on the subspace Im A. Call this b'. The properties of the "closest point" imply that the *difference*, b - b', is orthogonal to everything in Im A. It's simple to draw a picture to convince yourself that if c is in Im A and b - c is *not* orthogonal to everything in Im A then it is possible to "nudge" c a little, either away or towards the origin, to c' so that b - c' is shorter than b - c.
So b - b' is orthogonal to everything in Im A. Since being orthogonal to something is a linear condition, it is sufficient to check that b - b' is orthogonal to a spanning set for Im A, for which we can take the columns of A. As we are using the standard inner product, this means that for each column, say a of A, aT(b - b') = 0. Putting these together, we obtain the relation AT(b - b') = 0. As b' is in the image of A, there is some x' such that b' = A x', whence we see that x' satisfies ATb - ATA x' = 0, and get the desired formula on rearranging. This also guarantees the existence of a solution to this equation.
Using the fact that the closest point is the *unique* point b' in Im A such that b - b' is orthogonal to everything in Im A, we can run this argument backwards to see that if x' is a solution of AT A x' = AT b then Ax' is the closest point to b in Im A.
Where the above answers go wrong is to then talk about the matrix (AT A)-1 AT. The problem with this is that ATA **may not be invertible** (take A to be the 2 by 3 zero matrix). There is a matrix which *when ATA is invertible* is (ATA)-1AT and this is called the pseudo-inverse. Essentially, AT misses the kernel of ATA meaning that the *composition* is always well-defined but it might not be decomposable as the notation suggests.
This notation may be standard, of that I don't know, but if it is then it is bad notation because it suggests a property that may not hold. At the least, it should always carry a rider to make clear that the notation is merely suggestive and not to be taken literally.
| 9 | https://mathoverflow.net/users/45 | 126 | 58 |
https://mathoverflow.net/questions/80 | 16 | What is interesting/useful about Castelnuovo-Mumford regularity?
| https://mathoverflow.net/users/2 | What is interesting/useful about Castelnuovo-Mumford regularity? | Here's how I think about Castelnuovo-Mumford regularity. It's an invariant of an ideal (or module or sheaf) which provides a measure of how complicated that ideal (or module or sheaf) is. This invariant is related to free resolutions, and thus it measures complexity from that perspective.
Why is it interesting? One answer is that it can be used to provide an effective bound for two famous theorems. The first theorem I have in mind is that the Hilbert function of a graded ideal (or a finitely generated graded module) over the polynomial ring eventually agrees with the Hilbert polynomial of that ideal (or module). The second theorem I have in mind is Serre vanishing, which says that, given a coherent sheaf $\mathcal F$ on $\mathbb P^n$, there exists $d$ such that $H^i(\mathbb P^n, \mathcal F(e))=0$ for all $i>0$ and all $e>d$. These two theorems are related: if $M$ is a graded module of depth $> 0$, and $\mathcal F$ is the associated sheaf of $M$, then the Hilbert function of $M$ in degree $e$ equals $H^0(\mathbb P^n,\mathcal F(e))$.
An example where Castelnuovo-Mumford is particularly useful comes from the construction of the Hilbert scheme (I have heard that this is related to Mumford's original use, though I have no reference.) The basic point is that you can parametrize the set of ideals with a given Hilbert function by considering subloci of certain Grassmanians satisfying determinantal criteria, whereas it's less clear (at least to me) how to parametrize ideals with a given Hilbert polynomial.
Another great example where Castelnuovo-Mumford is useful is presented in Eisenbud "The Geometry of Syzygies" chapter 4, where he solves the interpolation problem for points in affine space.
| 21 | https://mathoverflow.net/users/4 | 127 | 59 |
https://mathoverflow.net/questions/115 | 22 | I think there was a theorem, like
* every cubic hypersurface in $\mathbb P^3$ has 27 lines on it.
What is the exact statement and details?
| https://mathoverflow.net/users/65 | What is the exact statement of "there are 27 lines on a cubic"? | The exact statement is that every smooth cubic surface in $\mathbb P^3$ (over an algebraically closed field) has exactly $27$ lines on it. Many books on algebraic geometry include a proof of this famous fact. The proof that I first learned comes from chapter V of Hartshorne, where cubic surfaces arise as the blowup of $\mathbb P^2$ at $6$ points, and where the formula $27=6+15+6$ is explained.
| 28 | https://mathoverflow.net/users/4 | 128 | 60 |
https://mathoverflow.net/questions/136 | 30 | Let $A$ be a commutative ring with $1$ not equal to $0$. (The ring A is not necessarily a domain, and is not necessarily Noetherian.) Assume we have an injective map of free $A$-modules $A^m \to A^n$. Must we have $m \le n$?
I believe the answer is yes. For instance, why is there no injective map from $A^2 \to A^1$? Say it's represented by a matrix $(a\_1, a\_2)$. Then clearly $(a\_2, -a\_1)$ is in the kernel. In the $A^{n+1} \to A^{n}$ case, we can look at the $n \times (n+1)$ matrix which represents it; call it $M$. Let $M\_i$ denote the determinant of the matrix obtained by deleting the $i$-th column. Let $v$ be the vector $(M\_1, -M\_2, ..., (-1)^nM\_{n+1})$. Then $v$ is in the kernel of our map, because the vector $Mv^T$ has $i$-th component the determinant of the $(n+1) \times (n+1)$ matrix attained from $M$ by repeating the $i$-th row twice.
That almost finishes the proof, except it is possible that $v$ is the zero vector. I would like to see either this argument finished, or, even better, a nicer proof.
Thank you!
| https://mathoverflow.net/users/71 | Atiyah-MacDonald, exercise 2.11 | Let M be the $n\times m$ matrix representing the injection $A^m \to A^n$. Define Di(M) to be the ideal generated by the determinants of all i-by-i minors of M. Let r be the largest possible integer such that Dr(M) has no annihilator (i.e. there is no nonzero element a∈A such that aDr(M)=0); I think r is usually called the *McCoy rank* of M.
Assume that $m>n$. We shall get a contradiction.
Choose a nonzero a∈A such that aDr+1(M)=0. By assumption, $a$ does not annihilate Dr(M), so there is some r-by-r minor that is not killed by $a$; we may assume it is the upper-left r-by-r minor. Thus, $r \leq n$, so that $r+1 \leq n+1\leq m$. Let M1, ..., Mr+1 be the cofactors of the upper-left (r+1)-by-(r+1) minor obtained by expanding along the bottom row. (This is well-defined even if the $r+1$-th row of $M$ does not exist, because these cofactors use only the first $r$ rows and the first $r+1$ columns of $A$, and $A$ has both since $r \leq n$ and $r+1 \leq m$.) Note, in particular, we know Mr+1 is the determinant of the upper-left r-by-r minor, so aMr+1≠0.
The claim is that the vector (aM1,...,aMr+1,0,0,...) (which we've already shown is non-zero) is in the kernel of M. To see that, note that when you dot this vector with the k-th row of M, you get $a$ times the determinant of the matrix obtained by replacing the bottom row of the upper-left (r+1)-by-(r+1) minor with the first r+1 entries in the k-th row. If k≤r, this determinant is zero because a row is repeated, and if k>r, this determinant is the determinant of some (r+1)-by-(r+1) minor, so it is annihilated by $a$. But since A is injective, the kernel of M is 0, and we have a contradiction.
| 14 | https://mathoverflow.net/users/1 | 142 | 63 |
https://mathoverflow.net/questions/129 | 37 | Is there some criterion for whether a space has the homotopy type of a closed manifold (smooth or topological)? Poincare duality is an obvious necessary condition, but it's almost certainly not sufficient. Are there any other special homotopical properties of manifolds?
| https://mathoverflow.net/users/75 | How can you tell if a space is homotopy equivalent to a manifold? | In surgery theory (which is basically a whole field of mathematics which tries to answer questions as the above), the next obstruction to the existence of a manifold in the homotopy type is that every finite complex with Poincaré duality is the base space of a certain distinguished fibration (Spivak normal fibration) whose fibre is homotopy equivalent to a sphere. (In order to get a unique such fibration, identify two fibrations if they are fiber homotopy equivalent or if one is obtained from the other by fiberwise suspension.)
For manifolds, this fibration is the spherization of the normal bundle, so the Spivak normal fibration comes from a vector bundle. This is invariant under homotopy equivalence.
Thus the next obstruction is: the Spivak normal fibration must come from a vector bundle.
If I remember right, then it was Novikov who first proved that for
simply-connected spaces of odd dimension at least 5, this is the only further obstruction.
In general, there is a further obstruction with values in a group $L\_n(\pi\_1,w)$ which depends on the fundamental group, first Stiefel-Whitney class and the dimension. See Lück's notes on surgery theory at <https://www.him.uni-bonn.de/lueck/data/ictp.pdf>
| 52 | https://mathoverflow.net/users/86 | 159 | 65 |
https://mathoverflow.net/questions/109 | 115 | (**Background:** In any category, an *epimorphism* is a morphism $f:X\to Y$ which is "surjective" in the following sense: for any two morphisms $g,h:Y\to Z$, if $g\circ f=h\circ f$, then $g=h$. Roughly, "any two functions on $Y$ that agree on the image of $X$ must agree." Even in categories where you have underlying sets, epimorphisms are not the same as surjections; for example, in the category of Hausdorff topological spaces, $f$ is an epimorphism if its image is dense.)
What do epimorphisms of (say commutative) rings look like? It's easy to verify that for any ideal $I$ in a ring $A$, the quotient map $A\to A/I$ is an epimorphism. It's also not hard to see that if $S\subset A$ is a multiplicative subset, then the localization $A\to S^{-1}A$ is an epimorphism. Here's a proof to whet your appetite.
>
> If $g,h:S^{-1}A\to B$ are two homomorphisms that agree on $A$, then for any element $s^{-1}a\in S^{-1}A$, we have
>
> $$g(s^{-1}a)=g(s)^{-1}g(a)=h(s)^{-1}h(a)=h(s^{-1}a)$$
>
>
>
Also, if $A\to B\_i$ is a finite collection of epimorphisms, where the $B\_i$ have disjoint support as $A$-modules, then $A\to\prod B\_i$ is an epimorphism.
Is every epimorphism of rings some product of combinations of quotients and localizations? To put it another way, suppose $f: A\to B$ is an epimorphism of rings with no kernel which sends non-units to non-units and such that $B$ has no idempotents. Must $f$ be an isomorphism?
| https://mathoverflow.net/users/1 | What do epimorphisms of (commutative) rings look like? | No, not every epimorphism of rings is a composition of localizations and surjections.
An epimorphism of commutative rings is the same thing as a *monomorphism* of affine schemes. Monomorphisms are not only embeddings, e.g., any localization is an epimorphism and the corresponding morphism of schemes is not a locally closed embedding.
**Example**: Let $C$ be the nodal affine cubic and let $X$ be its normalization. Pick any point $x$ above the node. Then $X\setminus\{x\}\to C$ is a monomorphism (see Proposition below). The corresponding homomorphism of rings is injective but not a localization.
**Proposition** (EGA IV 17.2.6): Let $f\colon X\to Y$ be a morphism *locally of finite type* between schemes. TFAE:
1. $f$ is a monomorphism.
2. Every fiber of $f$ is either an isomorphism or empty.
**Incorrect remark from 2009**: A *flat* epimorphism $A\to B$ is a localization if $A$ is normal and $\mathbb{Q}$-factorial. This is a result by D. Lazard and P. Samuel. [cf. Lazard, *Autour de la platitude*, IV, Prop 4.5]
**Correction of this remark** (May 2022):
1. A *flat* epimorphism $A\to B$ is a localization if $A$ is a normal noetherian domain with *torsion class group*. The result cited above proves this when $A$ is a Dedekind domain.
2. When $A$ is a Dedekind domain whose class group is not torsion, then there exists a flat epimorphism $A\to B$ of finite presentation (so an open immersion on Spec) which is not a localization [Lazard, *Autour de la platitude*, IV, Prop 4.6]. More generally, one can let $A$ be any normal domain whose Cartier class group is not torsion and let $\operatorname{Spec} B$ be the complement of a Cartier divisor $D$ whose class is not torsion.
3. The results 1-2 are best understood as follows. When $A$ is normal and locally $\mathbb{Q}$-factorial, then monomorphisms $A\to B$ correspond to subsets $U:=\operatorname{Spec} B\subseteq \operatorname{Spec} A$ such that $U$ is the complement of a, possibly infinite, union of irreducible divisors $\{D\_i\}\_{i\in I}$ [Raynaud, *Un critère d’effectivité de descente*, Cor. 2.7]. The complement of a Cartier divisor is always affine so it follows that $U$ is the intersection of the affine open subschemes $U\_J$ where $J\subseteq I$ is finite and $U\_J:=\operatorname{Spec} A\setminus \bigcup\_{i\in J} D\_i$. Equivalently, $B$ is the colimit (union if domain) of rings $B\_J$ such that $A\to B\_J$ is a monomorphism of finite type (an open immersion on Spec). When in addition $A$ has torsion class group, the $B\_J$ are localizations (in one element) and it follows that $B$ is a localization.
**Remark**: There was a [seminar](http://www.numdam.org/issues/SAC_1967-1968__2_) on epimorphisms of rings directed by P. Samuel in 1967-68. Raynaud's paper is part of this as well as articles by Lazard that later went into his thesis *Autour de la platitude*.
| 75 | https://mathoverflow.net/users/40 | 161 | 67 |
https://mathoverflow.net/questions/165 | 5 | Suppose $f:\mathbf{R}\to\mathbf{R}$ is a function. Let $S=\{x\in \mathbf{R}|f\text{ is continuous at }x\}$. Does $S$ have any nice properties?
Here are some observations about what $S$ could be:
* $S$ can be any closed set. For a closed set $S$, let $g$ be a continuous function whose vanishing locus is $S$ (for example, you could take $g(x)$ to be the distance of $x$ from $S$ if $S$ is non-empty). Then define
$$
f(x) =
\begin{cases}
g(x) &\text{if }x\in \mathbf{Q}\\
0 &\text{otherwise}.
\end{cases}
$$
Then the continuous locus of $f$ is exactly $S$.
* $S$ can be an open interval. For an open interval $S$, define
$$
f(x) =
\begin{cases}
0 &\text{if }x\in S \wedge x\in\mathbf{Q}\\
1 &\text{otherwise}.
\end{cases}
$$ Then the continuous locus of $f$ is exactly $S$.
* $S$ can be the complement of any countable set. Let $T =
\{t\_1,t\_2,t\_3,\ldots\}=\{t\_i\}\_{i\in\mathbf{N}}$ be a countable set, and let $\sum\_i a\_i$ be some absolutely convergent series all of whose terms are non-zero (like $a\_i=\frac{1}{2^i}$). Define
$$
f(x) = \sum\_{i\text{ s.t. }t\_i < x} a\_i.
$$
Then the continuous locus of $f$ is exactly the complement of $T$.
Here are some questions I'd like to know the answers to:
* Can $S$ be any open set?
* Can $S$ be non-measurable? (if $$
f(x) =
\begin{cases}
0 &\text{if }x\in S\\
1 &\text{otherwise}.
\end{cases}
$$
$f(x)=0$ if $x\in S$ and $f(x)=1$ otherwise, what will the continuous locus be?)
| https://mathoverflow.net/users/1 | Does the "continuous locus" of a function have any nice properties? | Yes, here's a quick proof that any given $G\_\delta$ (in $\mathbb{R}$) can be realized as the set of continuity points of some real-valued function.
Let $G$ be a given $G\_\delta$ set in $\mathbb{R}$, meaning $G = \cap\_{i=1}^\infty G\_i$, each $G\_i$ an open set. Define a function $f:\mathbb{R} \to \mathbb{R}$ as follows: $f(x)=0$ if $x$ is in $G$. If x is not in $G$, there is some $k$ such that $x$ is not in $G\_k$; let $k$ be minimal with that property. Define $f(x)=1/k$ if $x$ is rational and $f(x)=-1/k$ if $x$ is irrational.
If I'm not very much mistaken, $G$ is precisely the set of continuity points of this $f$. I'm happy to leave this as an exercise for now :-) Let me know if you're not sure how to do it, or - worse - if I'm just wrong about the construction.
| 6 | https://mathoverflow.net/users/25 | 174 | 76 |
https://mathoverflow.net/questions/166 | 20 | What are resources on invariant theory? Basically I've run into a need to teach myself some of the basics of invariant theory and was looking for a good place to start. I'd prefer online / freeish resources if anyone knows of any that can be trusted - otherwise if someone can recommend a good introductory book on the subject that would be most appreciated.
(Currently considering getting or perhaps convincing my library to order for me P. Olivers Classical Invariant Theory - but I thought I'd ask around first).
As a second question - two in one if you will - and perhaps this will help to let you know how little I know about invariant theory - can anyone quickly summarize the difference between Classical and Geometric Invariant Theory? Presently I am thinking what I need is the classical kind - but again more information is always helpful.
P.S. Geometric invariant theory may not have been the best tag for this question, but apparently new users can't create new tags, so I picked the closest one - anyone that has the power to change the tag to something more appropriate please do so.
| https://mathoverflow.net/users/16 | Resources on invariant theory | Usual invariant theory is dedicated to studying rings; a good example of a result from classical invariant theory is that the ring of invariant polynomials on any representation of a reductive group is finitely generated.
Geometric invariant theory is about constructing and studying the properties of certain kinds of quotients; a good example would be the moduli space of semi-stable vector bundles on an algebraic variety.
In my mind, the difference is this: Classical invariant theory is a collection of results about the interaction between group actions and commutative algebra. Geometric invariant theory is a technique for constructing interesting spaces.
| 7 | https://mathoverflow.net/users/66 | 175 | 77 |
https://mathoverflow.net/questions/178 | 14 | **Background**: the Hochschild homology of an associative algebra is the homology of the complex
$$ \ldots \longrightarrow A \otimes A \otimes A \longrightarrow A \otimes A \longrightarrow A$$
where the last two differentials are $$a \otimes b \otimes c \mapsto ab \otimes c-a \otimes bc + ca \otimes b$$ and $a \otimes b \mapsto ab-ba$, and you can guess the rest. More generally, it's "derived coinvariants": take a projective resolution of your algebra, then coinvariants of that.
For $k[t]$, the Hochschild homology is concentrated in degrees 0 and 1, and in both of those degrees it's $k[t]$. I know that I can go look up [Loday](http://www.ams.org/mathscinet-getitem?mr=1217970) (\S 3.2.2) and find a calculation, but I'd like a better explanation.
I know that the zero-th Hochschild homology $\operatorname{HH}\_0(k[t])$ must just be $k[t]$, because the zero-th Hochschild homology is just coinvariants, and $k[t]$ is commutative.
What I'd like is a "good" explanation for $\operatorname{HH}\_1(k[t])$.
**Edit**: Ben has a simple explanation below. Let me also rephrase the question, hoping for more. Here are a few things: If $A$ is semisimple, then $\operatorname{HH}\_{\ast}(A)$ is concentrated in degree 0. Is there something about $k[t]$ that ensures it's concentrated in degrees 0 and 1? Conversely, can I conclude anything about $A$ from the fact that $\operatorname{HH}\_{\ast}(A)$ is zero above $\ast=1$?
| https://mathoverflow.net/users/3 | Why is the Hochschild homology of k[t] just k[t] in degrees 0 and 1? | Another way to write the Hochschild homology is as follows:
take A as a *bimodule* over itself, take a free resolution as a *bimodule*, and then apply the functor of coinvariants ($M \mapsto M/\langle rm-mr|r\in A\rangle$).
Your definition used the "bar-complex" resolution of the form $\to A \otimes A \otimes A \to A \otimes A$
but k[t] has a much nicer resolution as a bimodule over itself, the Koszul resolution.
This is of the form $k[t] \otimes k[t] \to k[t] \otimes k[t]$ with the map given by $1 \otimes t - t \otimes 1$, so when you apply coinvariants, you get two copies of $k[t]$ with trivial differential.
Actually all Koszul algebras have a nice resolution of the diagonal bimodule, and thus its easier to compute their Hochschild homology, though in general, they don't always have trivial differential after applying coinvariants.
EDIT: For the later question, probably the best answers you'll get are from HKR, though just noting that the global dimension of $k[t] \otimes k[t]$ is 2 gets you halfway there.
EDIT AGAIN: Actually, any Koszul algebra has its Hochschild homology bounded above by its global dimension. This is clear from the existence of the diagonal Koszul resolution.
| 14 | https://mathoverflow.net/users/66 | 179 | 78 |
https://mathoverflow.net/questions/189 | 12 | How do you define the Hochschild (co)homology of a dg category or an A-infinity category? I've only seen it defined when the category is equivalent to a category of modules over a dg algebra; then the Hochschild (co)homology is just that of the algebra. But more generally?
| https://mathoverflow.net/users/83 | Definition of Hochschild (co)homology of a (dg or A-infinity) category | It's the self-Ext algebra of the identity functor from the category to itself. So, for an algebra, this reduces to the self-Ext algebra of the diagonal bimodule, since tensor product with that is the identity functor.
| 9 | https://mathoverflow.net/users/66 | 191 | 82 |
https://mathoverflow.net/questions/198 | 43 | Here I mean the version with all but finitely many components zero.
| https://mathoverflow.net/users/2 | How do you show that $S^{\infty}$ is contractible? | This is the swindle, isn't it?
There's an elegant way to phrase this with lots of sines and cosines, but working it all out is too much like hard work. Here's the quick and dirty way.
Let $T: S^\infty \to S^\infty$ be the "shift everything down by 1" map.
Then for any point $x \in S^\infty$, $T(x)$ is not a multiple of $x$ and so the line between them does not go through the origin. We can therefore define a homotopy from the identity on $S^\infty$ to $T$ by taking the homotopy $t x + (1 - t)T(x)$ and renormalising so that it is always on the sphere (incidentally, although you are working in $\ell^0$, by talking about a sphere you implicitly have a norm).
Then we simply contract the image of $T$, which is a codimension 1 sphere, to a point not on it, say $(1,0,0,0,0,...)$. Again, we can use 'orrible sines and cosines, but renormalising the direct path will do.
(Incidentally, there's nothing special about which space you are taking the sphere in. So long as your space is *stable* in the sense that $X \oplus \mathbb{R} \cong X$ then this works)
**Added a bit later:** Incidentally, if you want to work in a space that doesn't support a norm (such as an infinite product of copies of $\mathbb{R}$) you can still define the sphere as the quotient of $X$ without the origin by the action of $\mathbb{R}^+$. The argument above still works in this case.
**Added even later:** Revisiting this in the light of the duplicate: [Is $L^p(\mathbb{R})$ minus the zero function contractible?](https://mathoverflow.net/questions/38763/is-lp-mathbbr-minus-the-zero-function-contractible), the key property on $T$ is that it be **continuous**, **injective**, have **no eigenvectors**, and be **not surjective**. These conditions imply the following:
1. injective ⟹ the end-point of the homotopy is not the origin
2. no eigenvalues ⟹ the homotopy does not pass through the origin en route
3. not surjective ⟹ there is a point not in the image to which the image can be contracted
4. continuous ⟹ the homotopy is jointly continuous
Finally, there's no difference between the sphere and the space minus a point (indeed, without a norm the "space minus a point" is easier to deal with). Indeed, the homotopy described here actually works on the "space minus a point" and is just renormalised to work on the sphere.
| 55 | https://mathoverflow.net/users/45 | 199 | 87 |
https://mathoverflow.net/questions/195 | 15 | A morphism of schemes is formally smooth and locally of finite presentation iff it is smooth.
What happens if we drop the finitely presented hypothesis? Of course, locally of finite presentation is part of smoothness, so implicilty I am asking for the flatness to fail.
| https://mathoverflow.net/users/2 | Is there an example of a formally smooth morphism which is not smooth? | Here's an elementary example. For any field $k$, consider the ring $k[t^q|q\in\mathbb Q\_{>0}]$, which I'll abbreviate $k[t^q]$. I claim that the natural quotient $k[t^q]\to k$ given by sending $t^q$ to $0$ is formally smooth but *not flat*, and therefore not smooth.
First let's show it's formally smooth. Let $A$ be a ring with square-zero ideal $I\subseteq A$, and suppose we have maps $f:k[t^q]\to A$ and $g:k\to A/I$ making the following square commute (I drew it backwards because you're probably thinking of Spec of everything)
$$
\begin{array}{ccc}
A/I & \xleftarrow g & k \\
\uparrow & & \uparrow\\
A & \xleftarrow f & k[t^q]
\end{array}
$$
We'd like to show that there's a map $k\to A$ filling the diagram in. For any $q\in \mathbb Q\_{>0}$, note that $f(t^q)\in I$ by commutativity of the square, so $f(t^{2q})\in I^2=0$. But every $q$ is of the form $2q'$ for some $q'$, so we've shown that $f(t^q)=0$ for all $q\in \mathbb Q\_{>0}$. So $f$ factors through $k$, as desired.
Now let's show that $k$ is not flat over $k[t^q]$. Consider the exact sequence
$$0\to (t)\to k[t^q]\to k[t^q]/(t)\to 0.$$
When you tensor with $k$, you get
$$0\to k\to k\to k\to 0,$$
which is obviously not exact. So $k$ is not flat over $k[t^q]$.
| 28 | https://mathoverflow.net/users/1 | 200 | 88 |
https://mathoverflow.net/questions/193 | 29 | Suppose $f\colon X \to Y $ is a morphism of schemes. We can define a function on the topological space $Y$ by sending $y\in Y$ to the dimension of the fiber of $f$ over $y$.
>
> When is this function upper semi-continuous?
>
>
>
I have the following "concrete" application in mind. If an algebraic group $G$ acts on a scheme $X$, I'm pretty sure the stabilizer dimension is an upper semi-continuous function on $X$ (i.e. it can jump up on closed sub-schemes), but I don't know a proof. The stabilizers of points are the fibers of the map $\text{Stab}\to X$ in the following Cartesian square:
\begin{equation}
\require{AMScd}
\begin{CD}
\text{Stab} @>>> G \times X \\
@VVV @VV{\alpha}V \\
X @>{\Delta}>> X \times X.
\end{CD}
\end{equation}
where $\alpha\colon G\times X\to X\times X $ is given by $(g,x) \mapsto (g\cdot x,x)$, and $\Delta\colon X\to X\times X $ is the diagonal map $x\mapsto (x,x)$. It would be nice to have a condition satisfied by $\alpha\colon G\times X \to X\times X$ that would guarantee the upper semi-continuity of fiber dimension.
| https://mathoverflow.net/users/1 | When is fiber dimension upper semi-continuous? | **Theorem** (EGA IV 13.1.3): Let $f \colon X \to Y$ be a morphism of schemes, locally of finite type. Then
$$x \mapsto \dim\_x(X\_{f(x)})$$
is upper semi-continuous.
**Corollary** (Chevalley's upper semi-continuous theorem, EGA IV 13.1.5): Let $f \colon X \to Y$ be proper, then:
$$y \mapsto \dim(X\_y)$$
is upper semi-continuous.
**Corollary** (SGA3, ??): Let $X/Y$ be a group scheme, locally of finite type. Then
$$y \mapsto \dim(X\_y)$$
is upper semi-continuous.
*Proof:* The dimension of a group scheme over a field is the same as the dimension at the identity. Thus the function
$$y \mapsto \dim(X\_y)$$
is the composition of the continuous function $y \to e(y)$ and the upper semi-continuous function $x \mapsto \dim\_x(X\_{f(x)})$.
Concerning your application: The fiber dimensions of the stabilizer group scheme Stab/*X* is upper semi-continuous, but the "diagonal" $G \times X \to X \times X$ does not always have this property (unless it is proper, i.e., "$G$ acts properly").
| 35 | https://mathoverflow.net/users/40 | 202 | 89 |
https://mathoverflow.net/questions/201 | 18 | Is there an example of a variety over the complex numbers with no embedding into a smooth variety?
| https://mathoverflow.net/users/2 | Is there an example of a variety over the complex numbers with no embedding into a smooth variety? | Yes, there are even toric varieties. In fact, there are complete toric varieties with trivial picard group (see e.g., Eickelberg "Picard groups of compact toric varieties..." 1993). We have the following simple observations:
1) Any variety which can be embedded into a smooth variety has an ample family of line bundles.
2) A proper variety with trivial picard group cannot have an ample family of line bundles.
| 11 | https://mathoverflow.net/users/40 | 203 | 90 |
https://mathoverflow.net/questions/214 | 12 | Given a topological ring $R$, under what conditions and in what way, can one induce a topology on the $R$-points of a scheme $X$? For example, if $X$ is $P^n$ or $A^n$, one has natural topology on the $R$-points.
If $G$ is a group scheme/A and $R$ is $A$-algebra (still a topological ring), will the induced topology on $G\left(R\right)$ (as above) automatically make $G$ into a topological group.
For number theorists, if $G$ is an algebraic group/Q, we can consider the adelic points $G\left(A\_{K}\right)$ for any number field $K$. Is the induced topology on $G\left(A\_{K}\right)$ that of a restricted direct product? Under what conditions will $G\left(A\_{K}\right)$ be locally compact or satisfy other nice properties?
| https://mathoverflow.net/users/81 | How to topologize X(R) when R is a topological ring? | Brian Conrad has some notes on this on his website ("[Some notes on topologizing the adelic points of schemes, unifying the viewpoints of Grothendieck and Weil](http://math.stanford.edu/~conrad/papers/adelictop.pdf)"). The short version is that if X is affine, you can topologize X(R) in a natural, functorial way (specifically, the weakest topology such that the functions X(R)-->R induced by elements of the structure sheaf are continuous). If X isn't affine, you have to be more careful, because the units of R might not be open in X and might not be a topological group wrt the subspace topology. But those are the only problems, and if your ring doesn't have those problems, you can glue the naturally topologized affines and everything is functorial. Happily for number theorists, the adeles are fairly nice, and for a finite-type separated K-scheme X, X(A\_K) can be naturally topologized, and it is locally compact and Hausdorff.
| 17 | https://mathoverflow.net/users/88 | 219 | 97 |
https://mathoverflow.net/questions/117 | 8 | I'm interested in doing computations with certain non-commutative rings, most of which involve taking derived tensor products. Does anyone know of a computer algebra package which will find projective resolutions of complexes of modules over a finite-dimensional non-commutative ring, tensor with a bimodule, and do it all over again?
| https://mathoverflow.net/users/66 | Is there a good computer package for working with complexes over non-commutative rings? | I don't know whether Magma can handle all you ask for, but if I remember my coding for Magma correctly, at least the projective resolutions of modules over a non-commutative ring should be covered by that - for nice enough non-commutative rings. It's all been developed there as part of Jon F. Carlson's work on computing group cohomology rings.
If there is a system that does all you ask for, and does it efficiently, it is probably been written in connection to a group cohomology computation effort - which narrows the candidates down significantly: Magma and GAP do group cohomology rings, and SAGE now with the work of Simon King and David Green.
In contrast, I'm reasonably certain that Macaulay only does commutative things, and Singular doesn't have resolutions as a naturally occuring object at all.
Bergman might be able to deal with what you ask for, though.
To conclude: I'd recommend you to take a look at the homological algebra modules in Magma, GAP, SAGE and Bergman - I'd be highly surprised to see any other packages deal with the case you describe, and I'm not entirely convinced either of these do it well either.
| 6 | https://mathoverflow.net/users/102 | 220 | 98 |
https://mathoverflow.net/questions/181 | 10 | I realise this is a very vague question! I've heard people say that A∞ algebras are the right homotopy-theoretic generalization of usual associative algebras, because you can deform them. What exactly does this mean?
This roughly makes sense -- if you "deform" an associative algebra, it's generically going to stop being associative, but it will be "associative up to homotopy" in exactly the sense A∞ algebras are.
| https://mathoverflow.net/users/3 | What's the sense in which A_\infty algebras are "deformable"? | Well here's my shot: (skip to the punchline at the bottom if you want)
Take an associative algebra A and a k-local ring R (the formal power series over k, or the infinitesimal ring will do nicely).
The algebra A is naturally a homotopy algebra and so may be given by a degree -1 square-zero coderivative on the free coassociative coalgebra on A[1]. We write this coalgebra BA, the bar resolution. Note that in homotopy theory it often makes life easier if we forget any unit elements; BA is non-unital.
An A-infty R-deformation of A is now a square-zero coderivative on the coalgebra BA⊗R, such that the "obvious" diagram commutes (I could post this as an image when I'm permitted). The condition could alternatively by phrased as the following:
"such that it extends the original coderivative on BA".
So far this has all been definitions, my answer to your question comes next:
Consider now the cobar functor applied to the morphism BA⊗R→BA,
Ω(BA⊗R) ≅ (ΩBA)⊗R → ΩBA.
This is a proper algebra deformation, nothing infinity about it! Except... ΩBA is homotopy equivalent to A.
The short and snappy answer:
Infinity deformations are homotopy invariant, classical algebra deformations are not.
-------------------------------------------------------------------------------------
Edit: I should have added, if you would like me to expand on anything, I'm more than willing.
| 3 | https://mathoverflow.net/users/110 | 226 | 103 |
https://mathoverflow.net/questions/208 | 26 | If $R$ is a ring and $J\subset R$ is an ideal, can $R/J$ ever be a flat $R$-module? For algebraic geometers, the question is "can a closed immersion ever be flat?"
The answer is yes: take $J=0$. For a less trivial example, take $R=R\_1\oplus R\_2$ and $J=R\_1$, then $R/J$ is flat over $R$. Geometrically, this is the inclusion of a connected component, which is kind of cheating. If I add the hypotheses that $R$ has no idempotents (i.e. $\operatorname{Spec}(R)$ is connected) and $J\neq 0$, can $R/J$ ever be flat over $R$?
I think the answer is no, but I don't know how to prove it. Here's a failed attempt. Consider the exact sequence $0 \to J\to R\to R/J\to 0$. When you tensor with $R/J$, you get
>
> $$
> 0 \to J/J^2 \to R/J \to R/J \to 0
> $$
>
>
>
where the map $R/J\to R/J$ is the identity map. If $J\neq J^2$, this sequence is not exact, contradicting flatness of $R/J$.
But sometimes it happens that $J=J^2$, like the case of the maximal ideal of the ring $k[t^q| q\in\mathbf Q\_{>0}]$. I can show that the quotient is not flat in that case (see [this answer](https://mathoverflow.net/questions/195/is-there-an-example-of-a-formally-smooth-morphism-which-is-not-smooth/200#200)), but I had to do something clever.
I usually think about commutative rings, but if you have a non-commutative example, I'd love to see it.
| https://mathoverflow.net/users/1 | Can a quotient ring R/J ever be flat over R? | 1. If $A$ is arbitrary and $I$ is an ideal *of finite type* such that $A/I$ is a flat $A$-module, then $V(I)$ is open and closed. In fact, $A/I$ is a *finitely presented* $A$-algebra and thus $\operatorname{Spec}(A/I) \to \operatorname{Spec}(A)$ is a flat monomorphism of finite presentation, hence an étale monomorphism, i.e., an open immersion (cf. EGA IV 17.9.1).
2. If $A$ is a *noetherian* ring then $A/I$ is flat if and only if $V(I)$ is open and closed (every ideal is of finite type).
3. If $A$ is not noetherian but *has a finite number of minimal prime ideals* (i.e., the spectrum has a finite number of irreducible components), then it still holds that $A/I$ is flat iff $\operatorname{Spec}(A/I) \to \operatorname{Spec}(A)$ is open and closed. Indeed, there is a result due to Lazard [Laz, Cor. 5.9] which states that the flatness of $A/I$ implies that $I$ is of finite type in this case.
4. If $A$ has an *infinite number of minimal prime ideals*, then it can happen that a flat closed immersion is not open. For example, let $A$ be an absolutely flat ring with an infinite number of points (e.g. let $A$ be the product of an infinite number of fields). Then $A$ is zero-dimensional and every local ring is a field. However, there are non-open points (otherwise $\operatorname{Spec}(A)$ would be discrete and hence not quasi-compact). The inclusion of any such non-open point is a closed non-open immersion which is flat.
The example in 4) is totally disconnected, but there is also a connected example:
5. There exists a *connected* affine scheme $\operatorname{Spec}(A)$, with an infinite number of irreducible components, and an ideal $I$ such that $A/I$ is flat but $V(I)$ is not open. This follows from [Laz, 7.2 and 5.4].
[Laz] [Disconnexités des spectres d'anneaux et des préschémas (Bull SMF 95, 1967)](https://doi.org/10.24033/bsmf.1649)
**Edit**: Corrected proof of 1). An open closed immersion is not necessarily an open immersion! (e.g. $X\_{red} \to X$ is a closed immersion which is open but not an open immersion.)
**Edit**: Raynaud-Gruson only shows that flat+finite type => finite presentation when the spectrum has a finite number of *associated* points. Lazard proves that it is enough that the spectrum has a finite number of irreducible components. Added example 5).
| 27 | https://mathoverflow.net/users/40 | 227 | 104 |
https://mathoverflow.net/questions/210 | 5 | Hi,
The definition I have for a Path Algega of a quiver Q is that it is the algebra whose basis is formed by the oriented paths in Q, including the trivial ones. Apparently multiplication is given by concatenation of paths, and those that can't be concatenated are considered zero. That part I think I understand, but I am wondering if there is a non-formal interpretation of the addition operation on two elements of a path algebra someone could provide?
| https://mathoverflow.net/users/16 | Questions about Quivers | The short answer is no. You just have to think of them as formal sums, in the same way that you can only think of elements of a group algebra as formal sums.
What you can do is think of the path category of a quiver, which is the category whose objects are elements are vertices of the quiver, and whose morphisms are paths in the quiver. A representation of the path algebra as an algebra is essentially just a functor from this path category to vectors spaces.
| 9 | https://mathoverflow.net/users/66 | 231 | 105 |
https://mathoverflow.net/questions/229 | 11 | A manifold is usually defined as a [second-countable](http://en.wikipedia.org/wiki/Second-countable_space) [hausdorff](http://en.wikipedia.org/wiki/Hausdorff_space) topological space which is locally homeomorphic to **R**n. My understanding is that the reason "second-countable" is part of the definition is to make sure that the space is [paracompact](http://en.wikipedia.org/wiki/Paracompact_space), which you want so that you get locally finite partitions of unity. Once you have locally finite partitions of unity, basically anything you can multiply by a function can be constructed locally (any presheaf that is a module over the sheaf of functions is automatically a sheaf), a property you want manifolds to have.
But do we unnecessarily throw out some paracompact topological spaces which "should" be manifolds by requiring second-countability. A boring example is an uncountable disjoint union of manifolds, but there are other more interesting spaces that kind of look like they should be manifolds.
In particular, is the [long line](http://en.wikipedia.org/wiki/Long_line_%28topology%29) paracompact? Should I consider it a manifold?
| https://mathoverflow.net/users/1 | Is the long line paracompact? | Here's another proof, which shows that any connected paracompact locally Euclidean space X is second-countable. Cover X by Euclidean charts and take a locally finite refinement. Say an open set is good if it only intersects finitely many of the charts. Now take any point x and take a good neighborhood of it. The charts that intersect that good neighborhood can then themselves be covered by countably many good open sets. There are then only countably many charts intersecting those good open sets, and those charts can be covered by countably many good sets. Iterating this countably many times, you get an open set U associated to x which is covered by countably many charts such that if a chart intersects U, it is contained in U. It follows that the complement of U is also a union of charts, so by connectedness U is all of X. Thus X can be covered by countably many charts and is second-countable.
| 16 | https://mathoverflow.net/users/75 | 237 | 109 |
https://mathoverflow.net/questions/230 | 5 | The universal $R$-matrix of a quantized universal enveloping algebra is typically written as the product of two terms, one only involving elements of the Cartan, and one only involving elements of the upper and lower triangular pieces of $U\_q(g)$.
How do I think about commuting these individual pieces past things like coproducts? Are calculations like this written somewhere?
| https://mathoverflow.net/users/66 | How does one think about the "off-diagonal" part of the $R$-matrix? | Another way to think about this, which I learned from Mark Haiman's class, is that there are actually $4$ natural coproducts on the quantum group. $\Delta\left(E\right)$ could be $E\otimes K + 1 \otimes E$, or you could move the $K$ to the other factor, or you could invert $K$, or you could do both. The individual parts of the $R$-matrix move between these four different coproducts. In particular, the quasi-$R$-matrix (iirc) turns the $K$ into a $K^{-1}$.
| 4 | https://mathoverflow.net/users/22 | 242 | 111 |
https://mathoverflow.net/questions/243 | 10 | 1. What are some examples of compact Kaehler manifolds (or smooth complex projective varieties) that are not isomorphic as complex manifolds (or as varieties), but are isomorphic as symplectic manifolds (with the symplectic structure induced from the Kaehler structure)? Elliptic curves should be an example, but I can't think of any others. I'm sure there should be lots...
2. In the other direction, if I have two compact Kaehler manifolds (or smooth complex projective varieties) that are isomorphic as complex manifolds (or as varieties), then are they necessarily isomorphic as symplectic manifolds?
3. And one last question that just came to mind: If two smooth complex (projective, if need be) varieties are isomorphic as complex manifolds, then they are isomorphic as varieties?
| https://mathoverflow.net/users/83 | Compact Kaehler manifolds that are isomorphic as symplectic manifolds but not as complex manifolds (and vice-versa) | 1. Well, there are stupid examples like the fact that $\mathbb{P}^n$ has Kähler structures where any rational multiple of the hyperplane class is the Kähler class which are compatible with the standard complex structure (you just rescale the symplectic structure and metric). I think you should get similar examples with multi-parameter families on things like toric varieties with higher dimensional $H^2$.
2. I know some non-compact examples where you can deform the complex structure without changing the symplectic one. I don't know any compact examples, but they probably exist. The thing is, the only thing you can deform about a symplectic structure on a compact thing is its cohomology class (by the Moser trick), so anything with an big enough family of Kähler metrics will work.
3. This probably follows from GAGA, but you'd have to ask someone more expert than me to be sure. **Edit:** David's answer made me realize I forgot to say projective here. That's important.
| 4 | https://mathoverflow.net/users/66 | 248 | 114 |
https://mathoverflow.net/questions/241 | 17 | The *slope* of a vector bundle $E$ is defined as $\mu(E) = \deg(E)/\mathrm{rank}(E)$. Then a vector bundle $E$ is called *semistable* if $\mu(E') \leqslant \mu(E)$ for all proper sub-bundles $E'$. It is called *stable* if $\mu(E') < \mu(E)$.
I've heard that moduli spaces of stable and semistable vector bundles are somehow well-behaved, but I don't know why this is, nor do I know exactly what well-behaved should mean in this context. What goes wrong if we try to consider moduli of more general vector bundles? Moreover the definitions of slope and (semi)stable seem a bit artificial -- where do they come from?
I've also only seen the above definitions made in the context of vector bundles over curves. Why just curves? Does something stop working in higher dimensions or in greater generality?
| https://mathoverflow.net/users/83 | Why and how are moduli spaces of (semi)stable vector bundles well-behaved? | I believe "nice" here means "is a quasi-projective variety."
As for why, the reason is geometric invariant theory, which is roughly a way of looking at moduli problems, or actions of groups (which is roughly the same thing) and picking out a subset of the quotient (which itself is only nice as a stack) which is a quasi-projective variety. So there's a general definition of semi-stable points for any action of an affine algebraic group acting on a projective variety with choice of equivariant projective embedding (it depends on the choice of embedding) and the quotient of the semi-stable points is always a quasi-projective variety.
| 11 | https://mathoverflow.net/users/66 | 249 | 115 |
https://mathoverflow.net/questions/258 | 20 | For curves there is a very simple notion of degree of a line bundle or equivalently of a Weil or Cartier divisor. Even in any projective space $\mathbb P(V)$ divisors are cut out by hypersurfaces which are homogeneous polynomials of a certain degree.
Is there a more general notion of degree that applies to schemes with less structure?
Also, say you have a nice enough scheme $X$ so line bundles correspond to Cartier divisors under linear equivalence. In whatever the most general setting is so that the degree of a line bundle makes sense, is there an example of a line bundle $L \ne O\_X$ that is degree 0 and has $h^0(L$) = 1?
| https://mathoverflow.net/users/7 | does a line bundle always have a degree | One generalization of degree is first Chern class: A Cartier divisor corresponds to a class in $H^1(X;\mathcal{O}\_X^{\times})$, and you take its image under the boundary map of the long exact sequence corresponding to the exponential exact sequence $\mathbb{Z} \to \mathcal{O}\_X \to \mathcal{O}\_X^{\times}$ where the second map is taking exponential (if you want to work in the algebraic category, there is a fix for this, using the exact sequence $\mathbb{Z}/n\mathbb{Z} \to \mathcal{O}\_X^{\times} \to \mathcal{O}\_X^{\times}$, where the second map is nth power).
Geometrically, on a smooth thing, this means you take the sum of all the Weil divisors as a homology class, and then take the Poincare dual class in $H^2(X;\mathbb{Z})$.
| 25 | https://mathoverflow.net/users/66 | 260 | 121 |
https://mathoverflow.net/questions/259 | -2 | All the articles I've read regarding "Division by Zero" the main argument for it being an undefined operation, because all proofs lead to contradictions.
```
iff (0 / x) = (x / 0) = (0 / 0) = (0)
```
Irrespective of a proof, if the above rules were observed what field axioms of the real numbers would be violated, and how?
In regards to the multiplicative inverse of zero:
```
(0 * x) = (0) == (0 * y) = (0)
(0 * x) = (0 * y) // dividing both sides by 0 using the rules above results in
(0) = (0)
```
*Quick Reasoning:
Graphing Division by Zero shows two limits, as the graph tends towards these limits their combined projected values at these two limits negate each other.
ie: -infinity (as x tends towards 0 from below) and +infinity (as x tends towards 0 from above) either summate or negate each other to Zero.*
| https://mathoverflow.net/users/126 | Fundamental: Division by Zero | There's a thing called a *meadow* which is a (successful) attempt to make multiplicative inverses globally defined. What it does is instead of defining multiplicative inverses, it defines an operation $M \to M$, $x \mapsto x^{-1}$ with the property **not** that $xx^{-1} = 1$ but that $xx^{-1}x = x$. For any non-zero element then this agrees with the usual inverse but one can extend the inverse operation by defining $0^{-1} = 0$ and it works. I may be wrong, but I think that the result is that every field embeds in a meadow.
So providing you don't claim that $xx^{-1} = 1$ but rather $xx^{-1}x = x$ then you are absolutely fine with $0^{-1} = 0$.
| 7 | https://mathoverflow.net/users/45 | 261 | 122 |
https://mathoverflow.net/questions/263 | 33 | Given a filtered vector space (or module over a ring) $0=V\_{0}\subseteq V\_{1}\subseteq\cdots\subseteq V$, you can construct the *associated graded* vector space $\mathrm{gr}\left(V\right)=\oplus\_{i}V\_{i+1}/V\_{i}$. Does $\mathrm{gr}\left(V\right)$ satisfy a universal property? What is it?
Before anybody hastily says, "it's the universal graded vector space with a filtered map from $V$," let me point out that it's not so simple. A map of filtered vector spaces is a map of vector spaces which respects the filtration. It's clear what the map $V\_{i+1}\rightarrow V\_{i+1}/V\_{i}$ should be, but what would the map $\cup\_{i}V\_{i+1}\rightarrow\oplus\_{i}V\_{i+1}/V\_{i}$ be?
| https://mathoverflow.net/users/1 | What is the universal property of associated graded? | $\DeclareMathOperator{\Rees}{Rees}\DeclareMathOperator{\gr}{gr}$ The associated graded of a filtered $\mathbf R$-module $M$ is the universal $\mathbf R$-module with a map of the Rees module of $M$ over $\mathbf R[t]$ to $\gr M$.
Let me explain what the Rees module $\operatorname{Rees}(M)$ is: it's the submodule of $M[t,t^{-1}]$ which is generated as a $\mathbf{R}[t]$ module by $t^iM\_i$. Give this the obvious grading by degree of $t$. So $\Rees(M)/t\Rees(M)=gr M$, whereas $\Rees(M)/(t-1)\Rees(M)=M$ with the induced filtration. This is the thing that has a map to $\gr M$.
| 21 | https://mathoverflow.net/users/66 | 264 | 123 |
https://mathoverflow.net/questions/265 | 3 | Suppose $G$ is an affine algebraic group acting linearly on a vector space $V$. A point $v \in V$ is *stable* if the orbit $Gv$ is closed and $v$ is *regular* (the dimension of the stabilizer of $v$ is locally constant, or equivalently, [locally minimum](https://mathoverflow.net/questions/193/when-is-fiber-dimension-upper-semi-continuous/202#202)). I would really like to say this is equivalent to the orbit $Gv$ being closed and *not* being in the closure of another orbit.
Since the orbit of a regular point has locally maximum dimension, it can't be in the closure of another orbit. But is the converse true? If a point is not in the closure of an orbit larger than its own, is it regular?
The answer is no ... we have to throw in some hypotheses. If you consider the action of $\mathbf G\_a$ on $\mathbf A^2$ given by $t(x,y)=(x,tx+y)$, then all the orbits are closed, but points of the form $(0,y)$ are irregular. So let's throw in the hypothesis that $G$ *is linearly reductive*. I feel like we might also want to insist that $v\neq 0$, but I'm not sure about that.
Linearly reductive seems like a strange hypothesis, so feel free to modify it. I was thinking that you could somehow show that if v is not in the closure of a larger-dimensional orbit, then $\operatorname{span}(Gv)$ would be an invariant subspace with no complement, but I haven't been able to get this argument to work.
| https://mathoverflow.net/users/1 | Are irregular points of an action necessarily in the closure of a larger orbit? | Linearly reductive is a very good hypothesis; it tells you that the closed orbits of your group are the points of an affine variety (they are Spec of invariant polynomials). Maybe you can use some kind of semi-continuity of fiber dimension?
| 2 | https://mathoverflow.net/users/66 | 266 | 124 |
https://mathoverflow.net/questions/283 | 22 | What is a **topological modular form**? How are they related to 'normal' (number-theoretic) modular forms?
| https://mathoverflow.net/users/65 | What is a TMF in topology? | There is a Wikipedia entry on [topological modular forms](http://en.wikipedia.org/wiki/Topological_modular_forms), where you can see further references. The primitive version of Topological modular forms is as a generalized cohomology theory. Like K-theory and Cobordism theory it is a functor from spaces to graded rings, which satisfies the Eilenberg-Steenrod axioms, except for the dimension axiom.
In the case of TMF the ground ring is complicated, but essentially known. It is periodic, like K-theory, with period $24^2= 576$. There is a map of graded rings:
$TMF(pt) \rightarrow MF$
where $MF$ is the graded ring of integral modular forms (with appropriate grading based on the weight of the modular form). This map is NOT an isomorphism. It is neither surjective nor injective, but the kernel and cokernel are both torsion. In fact this torsion is only at the primes 2 and 3.
The more sophisticated version, and the one which makes makes the connection to modular forms more clear, is to view TMF as an $E\_\infty$ ring spectrum. The category of spectra is similar to the category of topological spaces, except the suspension functor is invertible. The things which represent cohomology theories live in spectra, and TMF is a ring object in spectra.
The connection to modular forms arises when you try to extend many constructions from algebraic geometry to this larger world of spectra. It is impossible to do the subject justice in a single post, but roughly you can look at the analog of elliptic curves over spectra. These have a moduli stack and the ``ring'' (i.e. ring spectrum) of functions on this stack is TMF.
| 13 | https://mathoverflow.net/users/184 | 288 | 131 |
https://mathoverflow.net/questions/138 | 5 | Let's say I have a field $\mathbb{K}$ and a flat family of $\mathbb{K}[t]$-modules $M$ over the formal disk $Spec \mathbb{K}[[h]]$.
Now, assume that $M/hM$ is torsion as a $\mathbb{K}[t]$-module (but NOT finitely generated). Can I conclude that $M[h^{-1}]$ is torsion as a $\mathbb{K}((h))[t]$-module?
| https://mathoverflow.net/users/66 | are deformations of torsion modules always torsion? | Let M = k[[h]][x] = \bigoplus\_{i=0}^{\infty}{ k[[h]]x^i }. We make this into a flat family of k[t] modules by setting t x^i = h x^{i+1}. Or in other words, p(h,t) \in k[[h]][t] acts by multiplication by p(h,xh) (wrt the natural ring structure on k[[h]][x]). Clearly M/hM = k[x] with the action by t equal to zero. Consider M[h^{-1}] = k((h))[x]. Since k((h))[x] is without zero divisors, and p(h,hx) is nonzero so long as p(h,t) was nonzero, we see that M[h^{-1}] is torsion free as a k((h))[t] module.
| 4 | https://mathoverflow.net/users/132 | 291 | 132 |
https://mathoverflow.net/questions/273 | 10 | Given a countable coloring of the plane, is it always possible to find a monochromatic set of points $\left\{ \left(x,y\right),\left(x+w,y\right),\left(x,y+h\right),\left(x+w,y+h\right)\right\} $ (the corners of a rectangle)?
| https://mathoverflow.net/users/151 | Finding monochromatic rectangles in a countable coloring of $R^{2}$ | This is equivalent to CH.
Quoting "Problems and Theorems in Classical Set Theory" by Komjath and Totik, chapter 16, Continuum hypothesis:$\DeclareMathOperator{\dist}{dist}$
>
> CH holds if and only if the plane can be decomposed into countably many parts none containing 4 different points $a$, $b$, $c$, $d$ such that $\dist(a,b)=\dist(c,d)$
>
>
>
This is a stronger requirement than your problem, so assuming CH the answer is no. Their solution, assuming CH is false, proves that there's a monochromatic rectangle.
---
Previous version, with added explanation about Hamel basis:
Using
>
> CH holds if and only if $\mathbf R$ can be colored by countably many colors such that the equation $x+y=u+v$ has no solution with different $x$, $y$, $u$, $v$ of the same color.
>
>
>
This gives a negative answer assuming CH. Explanation: consider $\mathbf R$ as a vector space over $\mathbf Q$. Let $A$ be some basis. Take any bijection $A \to A + A$, where $+$ is the disjoint sum. It induces a linear isomorphism $f:\mathbf R \to \mathbf R \times \mathbf R$. (You can think that there's a linear isomorphism between reals and complexes if that helps.) Then, if you were given a monochromatic rectangle $a=(x\_1, y\_1)$, $b=(x\_1+x\_2, y\_1)$, $c=(x\_1, y\_1+y\_2)$, $d=(x\_1+x\_2, y\_1+y\_2)$, certainly $a+d=b+c$. Using that isomorphism, $f(a)+f(d)=f(b)+f(c)$ gives a monochromatic solution of quoted equation.
| 12 | https://mathoverflow.net/users/158 | 292 | 133 |
https://mathoverflow.net/questions/290 | 12 | What is the "theorem of the cube" for abelian varieties? What is the statement and how should I think about it?
| https://mathoverflow.net/users/184 | What is the Theorem of the Cube? | If you have a line bundle trivial on 3 "surfaces" of a "cube" $A\times B\times C$ where $A$, $B$, $C$ are abelian varieties, then this line bundle in trivial on the whole "cube".
See [wikipedia](http://en.wikipedia.org/wiki/Theorem%20of%20the%20cube).
| 10 | https://mathoverflow.net/users/65 | 298 | 138 |
https://mathoverflow.net/questions/244 | 13 | I know that Hilbert schemes can be very singular. But are there any interesting and nontrivial Hilbert schemes that are smooth? Are there any necessary conditions or sufficient conditions for a Hilbert scheme to be smooth?
| https://mathoverflow.net/users/83 | When are Hilbert schemes smooth? | I don't know of many global conditions for a Hilbert scheme to be smooth/singular. Ben's answer probably gives the most interesting example of a smooth Hilbert scheme, namely the Hilbert scheme of n points on a smooth surface.
Here are two more examples of smooth Hilbert schemes.
1) The Hilbert scheme of hypersurfaces of degree d in PP^n. Such hypersurfaces are parametrized by homogeneous degree d polynomials in n+1 variables, and hence this Hilbert scheme is a projective space of dimension n+d choose d.
2) The Hilbert scheme of linear subpsace of dimension d of PP^n. This is just the Grassmanian Gr(d+1,n+1).
| 10 | https://mathoverflow.net/users/4 | 299 | 139 |
https://mathoverflow.net/questions/309 | 7 | For geometric representation theorists down here.
Consider the Beilinson-Bernstein theorem:
>
> Functor of global sections establishes
> the correspondence between twisted
> D-modules with fixed twist θ on the
> flag variety and g-representations
> with fixed central character. These
> are modules over the same algebra D[θ]
> = U /(Z −χ). This correspondence respects the structure of abelian
> category. It takes K-equivariant
> D-modules to (g, K)-admissible
> modules.
>
>
>
Why do people refer to its derived version as the Koszul duality?
How is this related to Soergel's conjecture?
| https://mathoverflow.net/users/65 | Beilinson-Bernstein and Koszul duality | Well, *I* don't call its derived version "Koszul duality." Koszul duality is a relationship between pairs of graded algebras (you could upgrade this to graded categories, but why bother?). Given a positively graded algebra A, consider the zero-degree part A0 as a module. Then B=Ext\*A(A0,A0) is a new bigraded algebra: it has a homological grade, and one induced by taking a graded free resolution of A0.
If these coincide, we say that A is Koszul and B is its Koszul dual. The functor of Ext\*A(A0,-) induces an equivalence between the derived categories of graded modules (with strange behavior on the grading!) over A and B (there's a version for ungraded modules of A, but those are sent to dg-modules of B and vice versa).
One of the strange facts of geometric representation theory is that categories which seem to a priori have no good reason to be Koszul dual actually are. For example, the categories of (g,N)-admissible modules with fixed generic central character for one Lie algebra is Koszul dual to the corresponding category for its Langlands dual. Soergel's conjecture is an version of this statement to (g,K) admissible modules for various symmetric subgroups K (which is much harder than the version for N).
What this has to do with Beilinson-Bernstein is that all these theorems are most naturally studied from a geometric perspective. For example, Koszulity tends to be related to purity of intersection cohomology.
| 10 | https://mathoverflow.net/users/66 | 311 | 145 |
https://mathoverflow.net/questions/315 | 4 | The decomposition of $L^{2}\left(S^{2}\right)$ under $SO\left(3,\mathbb{R}\right)$ is well-known.
Focus now on the hyperbolic plane $H$ presented as the quotient $SL\left(2,\mathbb{R}\right)/SO\left(2,\mathbb{R}\right)$. It is non-compact, therefore my understanding is that infinite-dimensional
representations of $SL\left(2,\mathbb{R}\right)$ will appear in the decomposition of $L^{2}\left(H\right)$.
(a) Is there an algebraic part of the spectrum and does it have a description
similar to the one in $L^{2}\left(S^{2}\right)$?
(b) How to classify the $SL\left(2,\mathbb{R}\right)$ representations and what is the whole spectrum?
(c) Consider $X\_{0}\left(1\right):=SL\left(2,\mathbb{Z}\right)\setminus H$. How does $L^{2}\left(X\_{0}\left(1\right)\right)$ decompose?
(d) The same for $X\_{0}\left(N\right):=\Gamma\_{0}\left(N\right)/H$. How does $L^{2}\left(X\_{0}\left(N\right)\right)$ decompose?
| https://mathoverflow.net/users/65 | Functions on hyperbolic space and modular curves | a) Weyl's unitary trick implies there are no nontrivial irreducible finite dimensional unitary representations of $SL\left(2,\mathbb{R}\right)$. This is basically the opposite of $SO\left(3\right)$.
b) [Wikipedia](http://en.wikipedia.org/wiki/Representation_theory_of_SL2(R)) has a classification of all unitary irreps. An irreducible representation given as a space of functions on H can be viewed as a massive particle state in relativistic QM on $R^{\left(1,2\right)}$.
c) I think you get real-analytic Eisenstein series and discrete series. Eisenstein series form a continuous spectrum, while discrete series give modular forms. You can find more in Gelbart's book "Automorphic forms on adele groups"
d) Same thing, except the Eisenstein series involve a summation over a smaller range of cosets of translation, and the modular forms are invariant under a smaller group. I am told that the Maass forms and holomorphic forms for congruence groups that I mentioned only give a countable collection of unitary representations, while the principal series has a continuous parameter.
| 3 | https://mathoverflow.net/users/121 | 318 | 148 |
https://mathoverflow.net/questions/335 | 24 | My understanding of Ben's answer to [this question](https://mathoverflow.net/questions/263/what-is-the-universal-property-of-associated-graded) is that even though associated graded is not an adjoint functor, it's not too bad because it is a composition of a right adjoint and a left adjoint.
But are such functors really "not that bad"? In particular, is it true that **any** functor be written as the composition of a right adjoint and a left adjoint?
| https://mathoverflow.net/users/1 | Is every functor a composition of adjoint functors? | The answer is no, because the nerve functor turns an adjoint pair of functors between categories into inverse homotopy equivalences between spaces (this is because of the existence of the unit and counit and the fact that nerve turns natural transformations into homotopies). In particular, this means that any functor whose nerve is not a homotopy equivalence cannot be a composite of adjoints. For a very simple example, you could take the functor from the 2-object discrete category to the terminal category.
| 43 | https://mathoverflow.net/users/75 | 336 | 155 |
https://mathoverflow.net/questions/134 | 9 | Consider a (smooth) bundle $E\to B$, and a (smooth) function $f: E\to\mathbf{R}$ on the total space. Then it makes sense to talk about the derivatives of $f$ along the fibers. Let $C$ be the subspace of $E$ consisting of all points for which all fiber-wise derivatives of $f$ vanish, so that upon intersecting with any fiber $C$ consists of the critical points of the restriction of $f$ to the fiber. If the fiber is $n$-dimensional, then $C$ is carved out by $n$ equations, and so generically has codimension $n$ in $E$.
Let's say that $c$ is a point in $C$ so that the second derivative of $f$ in the fiber is nondenegerate (i.e. has non-degenerate Hessian; i.e. $f$ restricts to a Morse function on the fiber through $c$). Does it follow that the projection $C\to B$ is a local diffeomorphism near $c$?
The answer is yes when everything is finite-dimensional (and I believe the statement is iff). I am interested in the case when $B$ is finite-dimensional but the fibers of $E$ are infinite-dimensional.
**Edit:** This is a response to Andrew's question below (since answering in the comments proves difficult).
I'm using the word "Morse" loosely, largely because I don't actually know Morse theory. I would suggest that the definition I give is better than what's traditionally used. What I actually mean is this:
Let $M$ be a smooth manifold and $f:M\to\mathbb{R}$ a smooth map. What type of object is the second derivative $f^{(2)}$? In general, you should not talk about it by itself, because it does not transform as a tensor, although the pair $(f^{(1)},f^{(2)})$ is a vector in the $2$-jet bundle over $M$. But if $c$ is a critical point of $f$, then $f^{(2)}(c)$ is naturally a symmetric bilinear form $\mathsf T\!\_cM \times \mathsf T\!\_cM \to\mathbf{R}$. Thus it is a map $\mathsf T\!\_cM\to \mathsf T^\ast\!\!\!\_c M$. All I ask is that this map have zero kernel.
But if this condition is too weak, whereas a reasonable stronger condition works, I'd love to hear it.
| https://mathoverflow.net/users/78 | Critical points on a fiber bundle | The part I'm still hesitant about is that the manifold you call $C$ is $n$-dimensional. Codimension arguments in the infinite dimensional setting are always a little sticky. So, I'm just going to treat it as an assumption.
Then the answer to your question is yes, and doesn't depend on what type of infinite dimensional vector spaces you use as your local model. Let's see why.
First you have a space $E$ which is a fiber bundle over $B$, via the projection $p$. Then at any point $x$ in $E$ we have the sequence of vector spaces,
$$
\mathsf T\!\_xF =\operatorname{ker}\mathrm{d}p\to \mathsf T\!\_xE \to\mathsf T\!\_{p(x)}B
$$
You also have a map $f: E \to R$, and hence a map $\mathrm{d}f: \mathsf T\!\_xE \to R =\mathsf T\!\_{f(x)}R$ for all points $x$ in $E$.
$C$ is defined to be the points in $E$ such that the restriction of $\mathrm{d}f$ to $\mathsf T\!\_{x}F$ vanishes. By assumption, $C$ is an $n$-dimensional manifold.
Now let's talk about second derivatives. There are two problems with second derivatives. The first arrises because we are working in the infinite dimensional setting. This means that we might not be able to think of the second derivative as a bilinear form due to possible convergence problems. However, as you pointed out, we can think of it as a map,
$$
\mathsf T\!\_x E\to \mathsf T^\ast\!\!\!\_x E.
$$
Now depending on the type of infinite dimensional manifold you are considering, this dual space may take on different meanings (banach dual, Frechet dual, etc). You at least get something in the algebraic dual (of possible non-continuous functionals).
The second problem with second derivatives is that they are usually not defined independent of a choice of coordinates. Again, as you pointed out, the pair ($\mathrm{d}f$, second derivative of $f$) transforms as a section of the second jet bundle, i.e. the second derivative doesn't change like a tensor but in an affine fashion depending on $\mathrm{d}f$.
This means that in general there is no intrinsic way to say the second derivative is "non-degenerate" at a random point of $E$. However, along $C$, a portion of the second derivative is still well defined, independently of coordinate choices. It is not the whole second derivative, but just the composite,
$$
H:\mathsf T\!\_cE\to \mathsf T^\ast\!\!\!\_c E \to \mathsf T^\ast\!\!\!\_c F
$$
(Here $H$ is for Hessian or some such thing). Under coordinate changes, $H$ transforms like a tensor because, along $C$, the restriction of $\mathrm{d}f$ to $\mathsf T\!\_c F$ vanishes.
Here is the key fact: The tangent space of $C$ is the kernel of $H$. This is an easy calculation, which can be done in local coordinates.
Finally, we are supposed look at the points of $C$ and we are supposed to consider the restriction of $H$ to $\mathsf T\!\_c F$. We want to consider a point $c$ in $C$, where the kernel of this restriction is zero.
But this condition is equivalent to saying that the intersection of $\mathsf T\!\_c F$ with the kernel of $H$ is zero, i.e.
* At the point $c$, $\mathsf T\!\_c F$ has zero intersection with $\mathsf T\!\_c C$.
Now since $\mathsf T\!\_c F$ is the kernel of $\mathrm{d}p$, this means the the restriction of $\mathrm{d}p$ to $\mathsf T\!\_c C$ also has zero kernel, and hence for dimension reasons is an isomorphism. Hence at the point $c$ in $C$, under your assumptions, the map $f: C \to B$ is a local diffeomorphism.
| 6 | https://mathoverflow.net/users/184 | 350 | 163 |
https://mathoverflow.net/questions/349 | 9 | Hi everyone,
let $A$ be an abelian variety of dimension $g$ over an algebraically closed field $k$ of characteristic $p\geqslant 0$. I'm trying to prove that the subgroup $A'$ which is the union of all torsion points $a\in A(k)$ of order prime to $p$ is Zariski dense in $A$.
The statement would follow if the Zariski closure $B$ (which by construction is a group variety) of $A'$ in $A$ would again be an abelian variety of dimension $d$, because assuming $d<g$, the $\ell$-primary part of $B$ would still be $(\mathbf Q\_\ell/\mathbf Z\_\ell)^{2 g}$, while it SHOULD be of rank $2d<2g$, contradiction.
However, I fail to see why $B$ should be irreducible. Does anyone see a way to salvage the argument, or a different, (simpler) argument?
| https://mathoverflow.net/users/259 | Why are torsion points dense in an abelian variety? | Let $C$ be the connected component of the identity in $B$. Then $C$ is a projective group variety, hence an abelian variety; let it have dimension $d$. Let $B/C=G$, a finite group. Then the number of $\ell$-torsion points of $B$ is at most $\left|G\right|\cdot\ell^{2d}$. For large $\ell$, this is less than $\ell^{2n}$ if $d$ is not $n$.
| 16 | https://mathoverflow.net/users/297 | 351 | 164 |
https://mathoverflow.net/questions/367 | 2 | I'm somewhat familiar with base change in scheme theory: sometimes a property of a morphism `X \to Y` survives a *base change* `f:Z\to Y`, meaning that `X \times_{Y} Z \to Z` also has this property.
>
> **Question:** Is the base change in number theory and derived algebraic geometry the same thing as above? What would be the examples?
>
>
>
| https://mathoverflow.net/users/65 | What is the base change in number theory? | In number theory, base change can also refer to an operation on automorphic representations. If L/K is an extension of number fields, and pi is an automorphic representation of a reductive group G over K, then pi should "lift" to a new automorphic representation of G over L. This is the sense of the phrase used in, e.g., Langlands' book "Base Change for GL(2)". The existence of a certain kind of base change for GL(2) was used to prove the modularity of some mod 3 Galois representations, which in turn played a role in proving Fermat's last theorem.
| 4 | https://mathoverflow.net/users/271 | 370 | 176 |
https://mathoverflow.net/questions/372 | 18 | There are two constructions that look quite similar to me: the derived category of an abelian category, and the homotopy category of a model category. Is there any explicit relationship between these two constructions? (This question is related to, and indeed the inspiration for, one of my previous [questions](https://math.stackexchange.com/questions/141/model-category-structures-on-categories-of-complexes-in-abelian-categories).)
| https://mathoverflow.net/users/83 | Derived categories and homotopy categories | Yes. The former is a special case of the latter. There is a model category structure on the category of (say bounded) chain complexes of objects in your given abelian category. The weak equivalences are the quasi-isomorphisms, and the homotopy category is the derived category.
In the case of R-modules, for a ring R, this is explained in detail in
[this paper by Dwyer-Spalinski](http://hopf.math.purdue.edu/Dwyer-Spalinski/theories.pdf).
| 14 | https://mathoverflow.net/users/184 | 373 | 178 |
https://mathoverflow.net/questions/354 | 3 | Any quantized universal enveloping algebra (in fact, any toplogically quasi-triangular Hopf algebra) has an (in its completion) an element u called the Drinfeld element which gives an isomorphism from a representation to its double dual.
However, most people prefer to use a different pivotal structure on the category of representations of the quantized universal enveloping algebra, where u is replaced by g=v-1u. Several obvious references don't seem to have a formula for this element, even though my dim recollection is that it is very simple. Is there anywhere where this is written down properly?
| https://mathoverflow.net/users/66 | What is a formula for the "group-like Drinfeld element"? | Up to sign conventions it's just K\_\rho, where \rho is the Weyl vector (half the sum of the positive roots). It's easy to check that this does satisfy the basic properties of g, namely that it is grouplike and that conjugation by it acts by S^2. Thus it gives a pivotal structure. (You still need to check that it actually is the g that comes from the usual ribbon element, which is a bit more delicate.)
| 4 | https://mathoverflow.net/users/22 | 378 | 182 |
https://mathoverflow.net/questions/329 | 70 | Okay, let's make sure I'm on the same page with those who know homological algebra.
What is **Koszul duality** in general?
What does it mean that *categories* are Koszul dual (I guess representations of Koszul dual algebras are the examples?) What are examples of "categories which seem to a priori have no good reason to be Koszul dual actually are" [Koszul dual] other than (g, R)-admissible modules?
| https://mathoverflow.net/users/65 | What is Koszul duality? | Let me try to give a more down to earth answer:
First, it's important to understand there are a lot of algebras whose derived categories are equivalent in surprising ways. Morita equivalences (equivalences between the abelian categories of modules) are kind of boring, especially for finite dimensional algebras; essentially the only thing you can do is change the dimensions of objects.
The way you see this is that if A-mod and B-mod are equivalent, then the image of A as a module over itself is a projective generator of B-mod, and for a finite-dimensional algebra, essentially the only thing you can do is take several copies of the indecomposible projectives of B.
On the other hand, if you take the derived category of dg-modules over A (the dg part of this is not a huge deal; it's just that they're very close to, but a bit better behaved than actual derived/triangulated categories which I consider something of a historical mistake, which should be replaced with dg/A-infinity versions), this is equivalent to the category of dg-modules over the endomorphism algebra (this is in the dg sense, so it's a dg-algebra whose cohomology is the Ext algebra) of *any* generating object. There are a *lot* more generating objects than projective generators, so there are a lot of derived equivalences.
In particular, you can take your favorite finite dimensional algebra A, and the most obvious not-very-projective generating object: the sum of all the simples. Call this L.
As I mentioned, there's an equivalence $A-dg-mod = \mathrm{Ext}(L,L)-dg-mod$, just given by taking $\mathrm{Ext}(L,-)$.
Now, in general, $\mathrm{Ext}(L,L)$ is an absolutely horrible object (ask Mikael Vejdemo-Johansson about doing this for group algebras over finite fields some time), but sometimes it turns out to be nice. For example, if you start with A being the exterior algebra, you'll get a polynomial ring on the dual vector space. Another (closely related) example is that the cohomology of a reductive group (over C) is Koszul dual to the cohomology of its classifying space (here you see a hint of this delooping mentioned in Scott's answer).
One thing that could help you make sure that $\mathrm{Ext}(L,L)$ is nice is if your algebra is graded. Then $\mathrm{Ext}(L,L)$ inherits an "internal" grading in addition to its homological one. If these coincide, then $B=\mathrm{Ext}(L,L)$ is forced to be formal (if it had any interesting A-infinity operations, they would break the grading), so you're dealing with a derived equivalence between actual algebras, though you have to be a bit careful about the dg-issues. You've found that the derived category of usual modules over A is equivalent to dg-modules over B (with its unique grading) and vice versa. You can fix this by taking graded modules on both sides.
As for more examples...well, some collaborators and I [found some cool examples](https://arxiv.org/abs/0806.3256)
coming from the combinatorics of hyperplane arrangements.
| 55 | https://mathoverflow.net/users/66 | 394 | 194 |
https://mathoverflow.net/questions/395 | 81 | I'd like to ask if people can point me towards good books or notes to learn some basic differential geometry. I work in representation theory mostly and have found that sometimes my background is insufficient.
| https://mathoverflow.net/users/135 | Reading list for basic differential geometry? | To Kevin's excellent list I would add Guillemin and Pollack's very readable, very friendly introduction that still gets to the essential matters. Read "Malcolm's" [review](http://www.amazon.com/review/R2N732504Q648/ref=cm_cr_rdp_perm) of it in Amazon, I agree with it completely.
Milnor's "Topology from the Differentiable Viewpoint" takes off in a slightly different direction BUT it's short, it's fantastic and it's Milnor (it was also the first book I ever purchased on Amazon!)
| 31 | https://mathoverflow.net/users/25 | 397 | 196 |
https://mathoverflow.net/questions/389 | 18 | 3d TQFTs constructed from modular tensor categories don't in general give an honest 3d TQFT, instead they have an "anomaly." My vague understanding from Kevin Walker's talks and from skimming Freed-Hopkins-Lurie-Teleman is that what's really going on is that its a 4d TQFT that's almost boring on the 4d part and that's what the "anomaly" means. Does anyone know how to make this more precise?
| https://mathoverflow.net/users/22 | What's the right way to think about "anomalies" in 3d TQFTs? | You can find some of the details in a draft version of my TQFT notes on my web page. Here's a summary.
Given an n-category with strong duality (by which I mean, more or less, pivotal if $n=2$ and a higher dimensional version of pivotal for $n > 2$; this is stronger than what Lurie means by an n-category "with duals"), there's a standard procedure to construct an n+1-dimensional TQFT. This procedure works for free for the 0-through n-dimensional parts of the (extended) n+1-dimensional TQFT. In these dimensions, we need no additional assumptions on the n-cat, and there is no need to choose a decomposition of the manifolds, so there are no combinatorial topology moves to check. The construction is manifestly invariant.
To construct the top, n+1-dimensional part of the TQFT, we need to make some finiteness assumptions on the n-category. (This corresponds to the top-dimensional part of what Lurie et al mean by "fully dualizable".) If the n-category satisfies these assumption, then we get, for each handle decomposition of the n+1-manifold, a state sum type expression for the path integral of the n+1-manifold. It is not hard to show that this state sum is invariant under handle slides and handle cancelation, so we get a well-defined invariant of n+1-manifolds that interacts with the rest of the TQFT (via gluing formulas) in the correct way.
(Small technicality: the path integral construction depends on a choice of element in the Hilbert space of the n-sphere, corresponding to the path integral of the n+1-ball. Multiplying this choice by $\lambda$ changes the path integral by $\lambda^\chi$, where $\chi$ is the Euler characteristic of the n+1-manifold.)
A modular tensor category is a 3-category with strong duality and the right sort of finiteness properties, so we can apply the above construction to get a 3+1-dimensional TQFT.
In dimension 3 the vector space we construct is an old-fashioned skein module: finite linear combinations of ribbon graphs in M^3 modulo local relations. (Actually, the dual of this vector space.) If M is closed this in 1-dimensional. More generally, if M has boundary then it has the same dimension as the Witten-Reshetikhin-Turaev vector space associated to the boundary of M.
In dimension 4, the type of state sum we get depends on the type of handle decomposition. For a general handle decomposition we get the Crane-Yetter state sum. For 2-handles attached to a 4-ball along a framed link L we get the Reshetikhin-Turaev surgery formula for L. For a 4-dimensional neighborhood of a 2-complex we get the Turaev "shadow" state sum. For a closed 4-manifold we find that the path integral is equal $a^\chi b^\sigma$, where $\chi$ is the Euler characteristic and $\sigma$ is the signature of the 4-manifold. By choosing $\lambda$ above appropriately we can make a=1. b is related to the central charge of the MTC (or to the value of the RT surgery formula on framing +-1 unknot). For a 4-manifold with boundary we find that the state sum computes the Witten-Reshetikhin-Turaev invariant of the boundary of the 4-manifold.
In dimension 2, the 1-category associated to a closed surface is a full matrix category; i.e. it is Morita trivial. For a surface with boundary k circles the category is Morita equivalent to k copies of the MTC thought of as a 1-category.
In all of the above cases, we find that the TQFT invariant of $Z(X)$, where dim(X) = 2 or 3 or 4, depends strongly on the boundary of $X$ but only weakly (i.e. only up to bordism) in the interior of $X$. So we can define a new 2+1 dimensional TQFT $Z'$ via the formula
$$Z'(Y) := Z(\mathrm{boundary}^{-1}(Y))$$
This TQFT has an anomaly, since we need to enhance Y with enough extra structure to pick out an inverse-boundary, up to bordism.
| 11 | https://mathoverflow.net/users/284 | 401 | 199 |
https://mathoverflow.net/questions/423 | 9 | Somebody answered this question instead of the question [here](https://mathoverflow.net/questions/410/what-is-an-example-of-a-smooth-variety-over-a-finite-field-fp-which-does-not-emb), so I am asking this with the hope that they will cut and paste their solution.
| https://mathoverflow.net/users/2 | What is an example of a smooth variety over a finite field F_p which does not lift to Z_p? | Examples are also in my paper "Murphy's Law in Algebraic Geometry", which you can get from my [preprints page](http://math.stanford.edu/~vakil/preprints.html)
Here is a short (not quite complete) description of a construction, with two explanations of why it works. I hope I am remembering this correctly!
In characteristic $>2$, consider the blow up of $\mathbf{P}^2$ at the $\mathbf{F}\_p$-valued points of the plane. Take a Galois cover of this surface, with Galois group $(\mathbf{Z}/2)^3$, branched only over the proper transform of the lines, and the transform of another high degree curve with no $\mathbf{F}\_p$-points.
Then you can check that this surface violates the numerical constraints of the Bogomolov-Miyaoka-Yau inequality, which holds in characteristic zero; hence it doesn't lift. (This is in a paper by Rob Easton.) Alternatively, show that deformations of this surface must always preserve that Galois cover structure, which in turn must preserve the data of the branch locus back in $\mathbf{P}^2$, meaning that any deformation must preserve the data of those $p^2+p+1$ lines meeting $p+1$ to a point, which forces you to live over $\mathbf{Z}/p$.
The two papers mentioned above give more exotic behavior too (of different sorts in the two papers), e.g. you an find a surface that lifts to $\mathbf{Z}/p^{10}$ but still not to $\mathbf{Z}\_p$.
| 17 | https://mathoverflow.net/users/299 | 429 | 213 |
https://mathoverflow.net/questions/141 | 12 | Section 2.3 of Hovey's Model Categories book defines a model category structure on Ch(R-Mod), the category of chain complexes of R-modules, where R is a ring. Lemma 2.3.6 then essentially states (I think) that taking projective resolutions of a module corresponds to taking cofibrant replacements of the module, at least in nice cases (e.g. when the projective resolution is bounded below). There is of course also a "dual" model category structure which gives the "dual" result for injective resolutions and fibrant replacements (Theorem 2.3.13).
1. I think the results in Hovey are proven for not-necessarily-commutative rings. Do things become nicer if we restrict our attention to commutative rings only?
2. Do these results generalize? For example, is there an analogous model category structure and an analogous result for Ch(OX-Mod), the category of chain complexes of OX-modules, where X is a scheme? More generally, how about for Ch(A), where A is an abelian category?
If the answers to these questions are known, then I assume they would be "standard", but I don't know a reference.
---
**I've re-asked my question in a different form [here](https://math.stackexchange.com/questions/372/derived-categories-and-homotopy-categories).**
| https://mathoverflow.net/users/83 | Model category structures on categories of complexes in abelian categories | I don't think the existence of the dual "injective" model structure merits an "of course," since its generators are much less obvious to construct. However, it turns out that injective model structures actually exist in more generality than projective ones, for instance they exist for most categories of sheaves. I believe this was originally proven by Joyal, but it was put in an abstract context by Hovey and Gillespie.
The basic idea is that model structures on Ch(A) correspond to well-behaved "cotorsion pairs" on A itself. The projective model structure comes from the (projective objects, all objects) cotorsion pair (which is well-behaved for R-modules, but not for sheaves), and the injective one comes from (all objects, injective objects). There is also e.g. a *flat* model structure coming from (flat objects, cotorsion objects) which is monoidal and thus useful for deriving tensor products. A good introduction, which I believe has references to most of the literature, is Hovey's paper [Cotorsion pairs and model categories](http://www.math.uic.edu/~bshipley/hovey.pdf).
| 10 | https://mathoverflow.net/users/49 | 456 | 226 |
https://mathoverflow.net/questions/455 | 12 | Say G is a profinite group with elements of arbitrarily large order. Do elements of infinite order exist (A) if we assume G is abelian? (B) in general?
A start for (A): we can ask the same question for the closure of the torsion subgroup of G (a subgroup since G is abelian), so WLOG we can assume the torsion subgroup is dense in G.
| https://mathoverflow.net/users/84526 | Elements of infinite order in a profinite group | (B) is probably difficult since it is listed as an open problem in the book *Profinite Groups* by Ribes and Zalesski (2009). [Question 4.8.5b (p. 401): "Is a torsion profinite group necessarily of finite exponent?"]
| 11 | https://mathoverflow.net/users/93 | 459 | 228 |
https://mathoverflow.net/questions/461 | 74 | There is a function on $\mathbb{Z}/2\mathbb{Z}$-cohomology called *Steenrod squaring*: $Sq^i:H^k(X,\mathbb{Z}/2\mathbb{Z}) \to H^{k+i}(X,\mathbb{Z}/2\mathbb{Z})$. (Coefficient group suppressed from here on out.) Its notable axiom (besides things like naturality), and the reason for its name, is that if $a\in H^k(X)$, then $Sq^k(a)=a \cup a \in H^{2k}(X)$ (this is the cup product). A particularly interesting application which I've come across is that, for a vector bundle $E$, the $i^{th}$ Stiefel-Whitney class is given by $w\_i(E)=\phi^{-1} \circ Sq^i \circ \phi(1)$, where $\phi$ is the Thom isomorphism.
I haven't found much more than an axiomatic characterization for these squaring maps, and I'm having trouble getting a real grip on what they're doing. I've been told that $Sq^1$ corresponds to the "Bockstein homomorphism" of the exact sequence $0 \to \mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$. Explicitly, if we denote by $C$ the chain group of the space $X$, we apply the exact covariant functor $Hom(C,-)$ to this short exact sequence, take cohomology, then the connecting homomorphisms $H^i(X)\to H^i(X)$ are exactly $Sq^1$. This is nice, but still somewhat mysterious to me. Does anyone have any good ideas or references for how to think about these maps?
| https://mathoverflow.net/users/303 | understanding Steenrod squares | Here's one way to understand them. The external cup square $a \otimes a \in H^{2n}(X \times X)$ of $a \in H^n(X)$ induces a map $f:X \times X \to K(Z\_2, 2n)$. It can be show that this map factors through a map $g:(X \times X) \times\_{Z\_2} EZ\_2 \to K(2n)$, where $Z\_2$ acts on the product by permuting the factors and $EZ\_2$ can be taken to just be $S^\infty$. If you unravel what this means, it says that our original map $f$ was homotopic to the map obtained by first switching the coordinates and then applying $f$. It also says that this homotopy, when applied twice to get a homotopy from $f$ to itself, is homotopic to the identity homotopy, and we similarly have a whole series higher "coherence" homotopies. Now $X \times BZ\_2$ maps to $(X \times X) \times\_{Z\_2} EZ\_2$ as the diagonal, so we get a map $X \times BZ2 \to K(2n)$. But $BZ\_2$'s cohomology is just $Z\_2[t]$, so this gives a cohomology class $Sq(a) \in H^\*(X)[t]$ of degree $2n$. If we write $Sq(a)=\sum s(i) t^i$, it can be shown that $s(i)=Sq^{n-i}a$.
What does this mean? Well, if our map $f$ actually *was* invariant under switching the factors (which you might think it ought to be, given that it appears to be defined symmetrically in the two factors), we could take $g$ to be just the projection onto $X \times X$ followed by $f$. This would mean that $Sq(a)$ comes from just projecting away the $BZ\_2$ and then using $a^2$, i.e. $Sq^n(a)=a^2$ and $Sq^i(a)=0$ for all other $i$. Thus the nonvanishing of the lower Steenrod squares somehow measures how the cup product, while *homotopy*-commutative (in terms of the induced maps to Eilenberg-MacLane spaces), cannot be straightened to be actually commutative. Indeed, in the universal example $X=K(Z\_2,n)$, the map $f$ is exactly the universal map representing the cup product of two cohomology classes of degree $n$.
Some somewhat terse notes on this can be found [here](http://my.harvard.edu/icb/icb.do?keyword=k54584&pageid=icb.page237558); see particularly part III. (Sorry, the link is now dead.)
| 56 | https://mathoverflow.net/users/75 | 462 | 230 |
https://mathoverflow.net/questions/437 | 10 | It's well-known that every natural number can be written as a sum of 4 squares of integers.
Has there been any recent progress about the similar problem for the cubes, 4-th powers and so on? I believe this was proven to be representable using some N — that depends on the power — and what's the story about it?
| https://mathoverflow.net/users/65 | Sums of cubes and more | Arguably, the problem "what is the least g=g(k) such that *every* integer can be written as the sum of g k-th powers" is less interesting than the version that ignores the random stuff that's happening with a finite number of special cases.
Namely, the "real" question should be "what is the least G=G(k) such that for some N, every integer greater than N can be represented as the sum of G k-th powers".
For example, every number is the sum of 19 4th powers, but every number greater than 13,792 is actually the sum of just 16 4th powers. The "16" was known for quite some time; the verification that 13,792 is the last offender is quite recent (I found the value on Wikipedia, btw).
Evaluating G(k) is harder than evaluating g(k), and most of the actual values are still not known. I don't think there was tremendous recent progress on this front, although there certainly is progress on things like bounds, number of representations, etc.
You should look at Wooley's survey [here](http://www.maths.bris.ac.uk/~matdw/2002%20wps.pdf) (I haven't read it yet).
| 9 | https://mathoverflow.net/users/25 | 463 | 231 |
https://mathoverflow.net/questions/468 | 21 | The Dynkin diagrams An, Dn, E6,
E7, E8 can be characterized among finite simple connected
graphs by the property that their eigenvalues (that is, the eigenvalues of their adjacency matrices) all have absolute value
strictly less than 2. What does this condition have to do with the
ADE classification, quiver representations, and so forth?
| https://mathoverflow.net/users/290 | Why are Dynkin diagrams characterized by their eigenvalues? | Yes. For example, with quiver representations, we have a formula
$\chi(M,N)=\dim Hom(M,N)-\dim Ext^1(M,N) = \sum d\_i(M)d\_i(N) - \sum\_{i \to j} d\_i(M)d\_j(N).$
where $d\_i(M)$ is the dimension of M at node i.
The proof is to check that it's true for simples, and then note that the category of representations of the path algebra of a quiver has global dimension 1.
So, what you've noted above is that this is positive definite if and only if the graph is Dynkin. Well, what's good about being positive definite? For one thing, if an object has trivial Ext^1 with itself, then it is rigid, it has no deformations. On the other hand, it also must have $\chi(M,M)>0$, since Hom always has positive dimension, and $Ext\_1(M,M)=0$.
Thus, if our quiver is not Dynkin, it has dimension vectors where no module can be rigid. On the other hand, if you work a bit harder, you can show Gabriel's theorem:
>
> if the graph is Dynkin, every dimension vector has a unique rigid module and this is indecomposible if and only if $\chi(M,M)=1$, that is if $M$ is a positive root of the root system.
>
>
>
| 13 | https://mathoverflow.net/users/66 | 471 | 237 |
https://mathoverflow.net/questions/476 | 24 | Last week Yan Zhang asked me the following: is there a way to realize vector spaces as categories so that adjoint functors between pairs of vector spaces become adjoint linear operators in the usual sense?
It seems as if one needs to declare an inner product by fiat for this to work out. An obvious approach is to take the objects to be vectors and hom(v, w) to be the inner product (so the category should be enriched over C). But I don't see how composition works out here, and Yan says he tried this and it didn't work out as cleanly as he wanted. In this setup I guess we want the category to be additive and the biproduct to be vector addition, but I have no idea whether this actually happens. I think John Baez's ideas about categorified linear algebra, especially categorified Hilbert spaces, are relevant here but I don't understand them well enough to see how they work out.
Anyone who actually knows some category theory care to clear things up?
| https://mathoverflow.net/users/290 | Can adjoint linear transformations be naturally realized as adjoint functors? | There's a canonical way of going the other way, starting with two linear categories with nice finiteness properties, with adjoint functors between them and getting a pair of vector spaces with adjoint linear transformations. The vector spaces are generated by formal symbols for each object in the category, and the inner product between any objects is the dimension of the Hom space (so Hom spaces had better be finite dimensional). Note that this doesn't have to be symmetric.
Functors give linear transformations, and adjoint functors are adjoint in the usual sense.
You can soup up this construction when you have some more structures on your category. For example, if you have a direct sum, then you can impose the relation $[A+B]=[A]+[B]$, and everything will work fine.
If your category is abelian, you can take Grothendieck group, where $[A]+[C]=[B]$ for every short exact sequence $0\to A \to B \to C\to 0$, but then you have to be much more careful about the fact that lots of functors (including Hom with objects in the category!) aren't exact: they don't send short exact sequences to short exact sequences. You need to use derived functors to fix this.
There's no canonical way of going the direction you asked, though in practice we have a very good record of being able to and I don't know of any really good examples of there being two equal natural seeming but different such constructions.
| 12 | https://mathoverflow.net/users/66 | 477 | 240 |
https://mathoverflow.net/questions/473 | 5 | Maximal Biclique: A complete bipartite subgraph, that isn't a subgraph of another complete bipartite subgraph.
Given a bipartite graph $G=(V\_{1}\cup V\_{2}, E)$ where $|V\_{1}|=|V\_{2}|$ with probability $p$ of there being an edge from any $a\in V\_{1}$ to any $b\in V\_{2}$, what is the expected number of maximal bicliques?
What I have worked out is the upper and lower bounds:
Lower Bound: 1 or 2. If $|E|$ is divisible by $n$, then $\frac{|E|}{n}$ nodes can be connected to completely to $\frac{|E|}{n}$ other nodes, making one maximal biclique. Otherwise, connect $\frac{|E|}{n}$ nodes completely to $\frac{|E|}{n}$ nodes and connect one node to $|E|(mod\ n)$ nodes.
Upper Bound: There are $2^n$ unique subsets, the empty set and the entire set not included, leaves $2^n-2$ subsets. Therefore, there can be at most $2^n-2$ maximal bicliques. This upper bound is achievable when there are $n^2-n$ edges (I can prove this if anyone wants me to).
Both of these results are also easily extended to bipartite graphs where $|V\_{1}|\neq |V\_{2}|$.
The upper and lower bounds are both fairly trivial for the most part and it's the expected number of maximal bicliques that I've had the most trouble with. I've done a little work analyzing simple cases and brute forcing the expected number for small values of $n$ (I suppose I could write a program to brute force larger values of $n$), but it hasn't amounted to anything worth saying.
I'd appreciate any suggestions for methods of attacking the problem or references that I might find useful.
| https://mathoverflow.net/users/256 | What is the expected number of maximal bicliques in a random bipartite graph? | The expected clique number (i.e. size of the maximum clique) in a random graph (all edge probabilities = 1/2) is around 2log\_2(n) where n is the number of vertices. You can find the proof in Alon and Spencer's "The Probabilistic Method", chapter 4. My guess is that a similar method would apply in the bipartite case, with a similar result. It also shouldn't be hard to extend the result to general p.
I you want to understand how to solve this kind of problems, I can't think of any better way than to carefully study the first few chapters of "The Probabilistic Method".
| 2 | https://mathoverflow.net/users/25 | 480 | 243 |
https://mathoverflow.net/questions/332 | 6 | This is also the space of real, symmetric bilinear forms in $\Bbb R^n$.
| https://mathoverflow.net/users/238 | Does the space of $n \times n$, positive-definite, self-adjoint, real matrices have a better name? | Two possible answers:
* Standard jargon is **SPD** (for "symmetric positive-definite").
* This isn't exactly a "name," but the n x n symmetric positive-definite matrices are exactly those matrices A such that the bilinear function (x, y) -> yTAx defines an inner product on **R**n. Conversely, every bilinear function is of that form for some A, so with some abuse of terminology, you could equate the set of those matrices with the set of inner products on **R**n.
There are many other ways to characterize SPD matrices, but that's the only one I can think of at the moment that can be summarized as a single noun phrase.
| 9 | https://mathoverflow.net/users/302 | 485 | 248 |
https://mathoverflow.net/questions/374 | 16 | Deligne's conjecture states that the Hochschild cochain complex of an A-infinity algebra is an algebra over the operad of chains on the topological little discs operad.
Of course the conjecture has been proven, but proofs aside -- what are some moral reasons why we should believe in Deligne's conjecture?
| https://mathoverflow.net/users/83 | Deligne's conjecture (the little discs operad one) | Here is a super simple way to think about it: The Hochshcild cochains on $A$ is derived $Hom$ over $A \otimes A^{op}$ of $A$ into$ A$. This has two multiplications, given by composition and multiplication in the target respectively.
These commute, or intertwine, so the result has an action of the tensor product of two $A$-infinity operads, which is (at least if you chose the right $A$-infinity operads) an $E\_2$ operad.
| 17 | https://mathoverflow.net/users/910 | 490 | 253 |
https://mathoverflow.net/questions/414 | 8 | Fix a prime $p$. The Teichmuller representative associated to a $p$-adic integer $a$ is the unique root of $x^p - x$ in $Z\_p$ congruent to $a$ mod $p$. One can identify this representative with the limit, as $n$ tends to infinity, of $a^{p^n}$.
Now let $a\_1, a\_2, \ldots, a\_k$ be the roots of an irreducible monic polynomial in $Z\_p[x]$. One can show that the limit, as $n$ tends to infinity, of $a\_1^{p^n} + a\_2^{p^n} + ... + a\_k^{p^n}$ also exists as a $p$-adic integer. Is there a characterization of this $p$-adic integer analogous to the above characterization?
| https://mathoverflow.net/users/290 | Generalized Teichmuller representatives | I think I can give a characterization of your limit as a sum of Teichmüller representatives.
Let $q = p^f$ be some power of $p$. Let $Z\_q = W(F\_q)$ be the valuation ring of the unramified extension of $Q\_p$ of degree $f$. Then for any $a$ in $Z\_q$, there is a unique root of $x^q - x$ in $Z\_q$ congruent to $a$ mod $p$. One can identify this with the limit, as n tends to infinity, of $a^{q^n}$.
I've never seen this before, but I guess you can do the same thing even if your extension is ramified. Let $R$ be some finite extension of $Z\_p$. Let $F\_q$ denote its residue field. Then for any $a$ in $R$, there is a unique root of $x^q - x$ in $R$ congruent to $a$ mod $p^{1/e}$, where $e$ is the ramification index. Again, it can be identified with the limit of $a^{q^n}$.
Assuming the limit you mentioned exists, it is the same as the limit of $a\_1^{q^n} + \cdots + a\_k^{q^n}$. And then this limit is the sum of the Teichmüller representatives that I just described.
| 1 | https://mathoverflow.net/users/71 | 497 | 258 |
https://mathoverflow.net/questions/494 | 2 | So I am (barely) familiar with the construction of the theta function of an integral lattice $L$. The theta function, as I understand it, is defined as the function which takes a variable $z$ and spits out the sum of $q$ raised to the $N(v)$ power, where the sum is over all vectors $v$ in the lattice, $q$ is equal to $exp(2 \pi i z)$, and $N(v)$ is the norm squared. We can regard this as a formal sum for the sake of this question. Let's call it $T(z)$.
I know there all kinds of identities about this function. For example if the lattice is even and unimodular, then $T(z)$ is a modular form. I am curious about the related function $T(-z)$. Is there an easy way to relate it to $T(z)$?
What I'm really really interested in is the quotient $T(z)/T(-z)$. I know there are people who know this stuff way better then I do, can anyone help? Is it easier in the unimodular case?
| https://mathoverflow.net/users/184 | Theta Functions and Cousins | I'm having trouble understanding this, because under your definition, T(-z) requires T to be defined on the lower half-plane. The series doesn't converge there.
Also, N(v) should be half of the squared norm. Equivalently, N(v) should be the value of the quadratic form that defines the inner product on the lattice.
I can think of one interpretation, where the theta function is actually a [Jacobi theta function](http://en.wikipedia.org/wiki/Jacobi_theta_functions#Jacobi_theta_function), which is a [Jacobi form](http://en.wikipedia.org/wiki/Jacobi_form), i.e., a section of a line bundle on the universal elliptic curve. This can be viewed as a function T(t,z) on HxC, with some invariance properties under translation by lattice elements in C and SL(2,Z) transformations in H. Then you can negate the z variable.
Eichler and Zagier have a book on Jacobi forms, called *The theory of Jacobi forms*.
| 2 | https://mathoverflow.net/users/121 | 498 | 259 |
https://mathoverflow.net/questions/500 | 21 | Say I have two finite groups $G$ and $H$ which aren't isomorphic but have the same character table (for example, the quaternion group and the symmetries of the square). Does this mean that the corresponding categories of finite dimensional complex representations are isomorphic (ignoring the forgetful functor to vector spaces), or just that the corresponding representation rings are?
| https://mathoverflow.net/users/321 | Finite groups with the same character table | In the particular case of the non-abelian groups of order 8, their categories of modules are not equivalent as monoidal categories. That they're not equivalent as pivotal categories can be proved by looking at the Frobenius-Schur indicator (I learned this from a paper of Susan Montgomery). That they're not equivalent even as monoidal categories can be proved by counting the fiber functors to vector spaces and seeing that one has more in one case (I can't remember which paper I saw this in, but almost surely Pavel Etingof was one of the coauthors).
| 14 | https://mathoverflow.net/users/22 | 506 | 266 |
https://mathoverflow.net/questions/446 | 44 | So ... what is the Fourier transform? What does it do? Why is it useful (both in math and in engineering, physics, etc)?
(Answers at any level of sophistication are welcome.)
| https://mathoverflow.net/users/83 | Fourier transform for dummies | One of the main uses of Fourier transforms is to diagonalize convolutions. In fact, many of the most useful properties of the Fourier transform can be summarized in the sentence "*the Fourier transform is a unitary change of basis for functions (or distributions) that diagonalizes all convolution operators.*" I've been ambiguous about the domain of the functions and the inner product. The domain is an abelian group, and the inner product is the L2 inner product with respect to Haar measure. (There are more general definitions of the Fourier transform, but I won't attempt to deal with those.)
I think a good way to motivate the definition of convolution (and thus eventually of the Fourier transform) starts with probability theory. Let's say we have an abelian group (G, +, -, 0) and two independent random variables X and Y that take values in G, and we are interested in the value of X + Y. For simplicity, let's assume G = {x1, ..., xn} is finite. For example, X and Y could be (possibly biased) six-sided dice, which we can roll to get two independent elements of **Z/6Z**. The sum of the die rolls mod 6 gives another element of the group.
For x ∈ G, let f(x) be the probability P(X = x), and let g(x) = P(Y = x). What we care about is h(x) := P(X + Y = x). We can compute this as a sum of joint probabilities:
h(x) = P(X + Y = x) = Σy+z=xP(X = y & Y = z)
However, since X and Y are independent, P(X = y & Y = z) = P(X = y)P(Y = z) = f(y)g(z), so the sum is actually
h(x) = Σy+z=xf(y)g(z) = Σy∈Gf(y)g(x-y).
This is called the *convolution* of f and g and denoted by f\*g. In words, *the convolution of two probability distributions is the probability distribution of the sum of two independent random variables having those respective distributions.* From that, one can deduce easily that convolution satisfies nice properties: commutativity, associativity, and the existence of an identity. Moreover, convolution has the same relationship to addition and scalar multiplication as pointwise multiplication does (namely, bilinearity). In the finite setting, there's also an obvious L2 inner product on distributions, with respect to which, for each f, the transformation g -> f \* g is normal. Since such transformations also commute, recalling a big theorem from finite-dimensional linear algebra, we know there's an orthonormal basis with respect to which all of them are diagonal. It's not difficult to deduce then that in such a basis, convolution must be represented by coordinatewise multiplication. That basis is the Fourier basis, and the process of obtaining the coordinates in the Fourier basis from coordinates in the standard basis (the values f(x) for x ∈ G) is the Fourier transform. Since both bases are orthonormal, that transformation is unitary.
If G is infinite, then much of the above has to be modified, but a lot of it still works. (Most importantly, for now, the intuition works.) For example, if G = **R**n, then the sum Σy∈Gf(y)g(x-y) must be replaced by the integral ∫y∈Gf(y)g(x-y)dy to define convolution, or even more generally, by Haar integration over G. The Fourier "basis" still has the important property of representing convolution by "coordinatewise" (or pointwise) multiplication and therefore of diagonalizing all convolution operators.
The fact that the Fourier transform diagonalizes convolutions has more implications than may appear at first. Sometimes, as above, the operation of convolution is itself of interest, but sometimes one of the arguments (say f) is fixed, and we want to study the transformation T(g) := f\*g as a linear transformation of g. A lot of common operators fall into this category. For example:
* Translation: T(g)(x) = g(x-a) for some fixed a. This is convolution with a "unit mass" at a.
* Differentiation: T(g)(x) = g'(x). This is convolution with the derivative of a negative unit mass at 0.
* Indefinite integration (say on **R**): T(g)(x) = ∫x-infinityg(t)dt. This is convolution with the Heaviside step function.
In the Fourier basis, *all* of those are therefore represented by pointwise multiplication by an appropriate function (namely the Fourier transform of the respective convolution kernel). That makes Fourier analysis very useful, for example, in studying differential operators.
| 46 | https://mathoverflow.net/users/302 | 517 | 272 |
https://mathoverflow.net/questions/518 | 8 | Fix $n$ and a field $k$ of characteristic zero. Let $G$ be the pro-algebraic group of automorphims of $k[[x\_1,...x\_n]]$. Let $G\_0$ be the subgroup of automorphisms preserving the closed point (note that for general $T$, $G\_0(T)$ can be a proper subgroup of $G(T)$). Let $X$ be a regular variety over $k$ and let $P$ be the principal $G$ bundle of formal coordinate systems, naturally a $G$ torsor over $X$. I hear that there is a connection between $P$ and $D\_X$-modules. what is this connection?
| https://mathoverflow.net/users/132 | what is the connection between D-modules and coordinate bundles? | Assume X is n-dimensional and regular. Then there is a functor from G-modules V to DX-modules, given by an associated bundle construction. Take the trivial (ind-)bundle on P with fiber V, and quotient by the action of G on P and V. If you replace G with G0 and P with the canonical G0-torsor, the same construction yields an OX-module. The extra structure of a G-action lets you identify infinitesimally nearby fibers.
| 4 | https://mathoverflow.net/users/121 | 523 | 275 |
https://mathoverflow.net/questions/515 | 20 | We know that modules over skewfields are free. Is the converse true? In other words, is it true that a nontrivial ring over which every module is free is a skewfield?
If the ring A is commutative, then writing that for any proper ideal I of A, the A-module A/I is free yields the result. What about the general case?
| https://mathoverflow.net/users/336 | Rings over which every module is free | The non-commutative case isn't very different. Assume for reasonableness that your ring is Noetherian (I refuse to think about non-Noetherian rings on principle). Your argument gives that all two-sided ideals just be the whole algebra or trivial. So, your algebra is simple.
Another consequence is that every exact sequence splits. Thus, if you have a pair of left ideals $I \subset J$, you can choose left ideals $K\_1,K\_2$ such that $J=I \oplus K\_1$ and $R=J \oplus K\_2$. Thus, if you had a infinite descending chain of ideals, you would have a complementary ascending chain of ideals, which is impossible, since your ring is Noetherian. Thus, your ring is Artinian.
By Artin-Wedderburn, your algebra is a matrix algebra over a skew field. But every matrix ring except 1 x 1 has a non-free projective module. So A is a skew-field.
| 13 | https://mathoverflow.net/users/66 | 524 | 276 |
https://mathoverflow.net/questions/239 | 10 | I'd love to have a list of 'small' $2$-knots, for some sense of small. It's not clear what one should filter by, but there are two obvious candidates
* Write a movie presentation, and count the frames.
* Project the $2$-knot to $\Bbb R^3$, and count the triple points.
Does anyone know if this has been attempted? Such a list could be quite useful.
| https://mathoverflow.net/users/3 | Has anyone tabulated 2-knots? Would anyone like to try? | There is a table in Yoshikowa's paper based upon the representation as graphs with certain markings of vertices. The markings indicate which directions are the $A$ and $B$ smoothings. The table is quite small.
Kamada's book on Surface braids has a nice table of $2$-knots that don't have triple points. There are tons of these. So your candidate for small is sort of strange. Braid index might be better.
Quandle cocycles are the only known ways to get lower bounds for triple points, and there are not nearly enough $3$-cocycles that have been calculated.
Last spring Dennis Roseman told me a neat way to generate "random" $2$-knots. I don't think he has written anything down about it though.
So if you look at movies, you have to be careful about what you define as an event. In the CRS point of view each of Reid. I,II,III, birth, death, saddle, switchback, and psi (pitchfork) moves is an event, but exchanges of distant critical points are also events. Any such event leaves its trace on the chart.
There are other notions of simplicity. For example, there is a notion of thickness. Project into $3$-space and subsequently onto the plane (that gives the chart). Take a spear perpendicular to the plane and see how many times it generically passes through the surface. Find the maximum of these, and then minimize over all projections.
Probably the real reason for making a census would be to discover new invariants. So it is a good problem in that sense.
| 9 | https://mathoverflow.net/users/36108 | 535 | 283 |
https://mathoverflow.net/questions/416 | 15 | Hi everyone,
let X be a variety over a field k, S an integral scheme such that the function field K of S is contained in k. An S-scheme **X** is called model of X/k if **X** x\_S k = X, i.e. if the generic fiber of **X** over S is isomorphic to X.
* Are there general conditions on X, S, k, such that **X** exists?
* If X is smooth and projective, what are the conditions, such that there is a smooth model **X**?
* Any good references that go into models and reduction in general, and not only in the case of curves?
| https://mathoverflow.net/users/259 | Existence of (smooth) models | I would also be very interested if someone could point to good references for this question, but I am not yet allowed to vote for the question.
Nekovar's survey article on the Beilinson conjectures from the early 90s mentions some results for varieties over Q. He says in section 5.3 that given a smooth projective variety over Q, there always exists a proper flat model over Z, but that a regular such model is rarely known to exist. However, in the published version of the same survey, there is an added note at the very end of the article saying that "Spivakovsky recently announced a general result on resolution of singularities, which implies that a regular proper flat model of X mentioned in 5.3 always exist". However, I have never seen this result of Spivakovsky mentioned anywhere else, so I doubt that it is true. Does anyone else know more about this?
The survey is available here:
<http://people.math.jussieu.fr/~nekovar/pu/mot.pdf>
For the published version, google "Serre Jannsen Motives", click at the Google Books link, and then search for "Spivakovsky" within the book.
In general for the case when k is a number field and S is the ring of integers in k, it seems unreasonable to ask for general theorems about the existence of smooth models, although one could maybe hope for something about regular models. For example, I think there are no nontrivial smooth and proper curves over Spec(Z) at all (although there are smooth and proper such schemes of higher dimension, and there are also smooth and proper such schemes in any dimension, including 1, over rings of integers in other number fields).
| 13 | https://mathoverflow.net/users/349 | 537 | 285 |
https://mathoverflow.net/questions/364 | 95 | I'm looking for a big-picture treatment of algebraic K-theory and why it's important. I've seen various abstract definitions (Quillen's plus and Q constructions, some spectral constructions like Waldhausen's) and a lot of work devoted to calculation in special cases, e.g., extracting information about K-theory from Hochschild and cyclic homology. As far as I can tell, K-theory is extremely difficult to compute, it yields deep information about a category, and in some cases, this produces highly nontrivial results in arithmetic or manifold topology. I've been unable to piece these results into a coherent picture of why one would think K-theory is the right tool to use, or why someone would want to know that, e.g., K22(Z) has an element of order 691. Explanations and pointers to readable literature would be greatly appreciated.
| https://mathoverflow.net/users/121 | Motivation for algebraic K-theory? | Algebraic K-theory originated in classical materials that connected class groups, unit groups and determinants, Brauer groups, and related things for rings of integers, fields, etc, and includes a lot of local-to-global principles. But that's the original motivation and not the way the work in the field is currently going - from your question it seems like you're asking about a motivation for "higher" algebraic K-theory.
From the perspective of homotopy theory, algebraic K-theory has a certain universality. A category with a symmetric monoidal structure has a classifying space, or nerve, that precisely inherits a "coherent" multiplication (an E\_oo-space structure, to be exact), and such an object has a naturally associated group completion. This is the K-theory object of the category, and K-theory is in some sense the universal functor that takes a category with a symmetric monoidal structure and turns it into an additive structure. The K-theory of the category of finite sets captures stable homotopy groups of spheres. The K-theory of the category of vector spaces (with appropriately topologized spaces of endomorphisms) captures complex or real topological K-theory. The K-theory of certain categories associated to manifolds yields very sensitive information about differentiable structures.
One perspective on rings is that you should study them via their module categories, and algebraic K-theory is a universal thing that does this. The Q-construction and Waldhausen's S.-construction are souped up to include extra structure like universally turning a family of maps into equivalences, or universally splitting certain notions of exact sequence. But these are extra.
It's also applicable to dg-rings or structured ring spectra, and is one of the few ways we have to extract arithmetic data out of some of those.
And yes, it's very hard to compute, in some sense because it is universal. But it generalizes a lot of the phenomena that were useful in extracting arithmetic information from rings in the lower algebraic K-groups and so I think it's generally accepted as the "right" generalization.
This is all vague stuff but I hope I can at least make you feel that some of us study it not just because "it's there".
| 63 | https://mathoverflow.net/users/360 | 541 | 288 |
https://mathoverflow.net/questions/546 | 69 | In a [recent blog post](https://terrytao.wordpress.com/2009/10/02/at-the-austms-conference/) Terry Tao mentions in passing that:
"Class groups...are arithmetic analogues of the (abelianised) fundamental groups in topology, with Galois groups serving as the analogue of the full fundamental group."
Can anyone explain to me exactly in what sense are Galois and fundamental groups analogous?
| https://mathoverflow.net/users/361 | Galois groups vs. fundamental groups | You should think of coverings of manifolds as analogous to field extensions. Once you accept this, then the fundamental group and absolute Galois group play the same role; coverings correspond to subgroups of the former and field extensions to subgroups of the latter (though for the absolute Galois group you have to consider its topology).
This can be made precise in algebraic geometry: if you have a covering map of projective algebraic varieties, then the function field of the target embeds into the function field of the domain by pullback, and this is a finite degree unramified field extension.
You can think of lifting paths downstairs as being a bit like algebraic number theory: each closed path downstairs has an inverse image that's a union of paths. If the covering is Galois, then each component will cover the original with the same degree, but otherwise maybe not. You can think of the conjugacy class of the path as the "Frobenius" whose orbit type on the set of preimages of a point determines the "splitting into primes."
There's even a version of the theory of L-functions given by considering the spectrum of the Laplacian for a metric on the varieties.
| 63 | https://mathoverflow.net/users/66 | 548 | 292 |
https://mathoverflow.net/questions/551 | 35 | A statement referring to an infinite set can sometimes be logically rephrased using only finite sets/objects. For example, "The set of primes is infinite" <-> "There is no largest prime". Pleasantly, the proof of this statement does not seem to need infinity either (assume a largest prime, contradiction).
What reason is there, other than convenience or curiosity, to adjoin infinite sets to our universe by axiomatically declaring that one exists?
Specifically:
>
> What is an example of a theorem in ZF or ZFC which 1) does not refer to infinite sets, but 2) cannot be proven if the axiom of infinity is excluded?
>
>
>
(See [Zermelo–Fraenkel set theory](http://en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_theory#The_axioms) for the axiom of infinity in context.)
| https://mathoverflow.net/users/84526 | Does finite mathematics need the axiom of infinity? | ZF - infinity + not infinity is bi-interpretable with Peano Arithmetic. Bi-interpretable means that a model of either one can view a subset of itself as a model of the other (all in a definable way). So ZF - infinity can't prove anything that PA wouldn't prove.
There are some fairly natural statements which are independent of PA but provable in ZF. In fact, they're provable in theories much weaker than ZF. The first convincing example was the [Paris-Harrington Theorem](http://en.wikipedia.org/wiki/Paris%2DHarrington_theorem), which proved that a certain Ramsey-like property is independent of PA. Another good example is [Goodstein Sequences](http://en.wikipedia.org/wiki/Goodstein%27s_theorem) which Anton mentioned.
| 35 | https://mathoverflow.net/users/27 | 555 | 296 |
https://mathoverflow.net/questions/410 | 15 | Such an example of course could not be projective and would not itself lift to Z\_p. The context is that one can compute p-adic cohomology of a variety X over a finite field F\_p via the cohomology of an embedding of X into a smooth Z\_p scheme.
This is similar in spirit to my questions [here](https://mathoverflow.net/questions/201/) and [here](https://mathoverflow.net/questions/423/what-is-an-example-of-a-smooth-variety-over-a-finite-field-fp-which-does-not-lif) (but a different question than the second link).
| https://mathoverflow.net/users/2 | What is an example of a smooth variety over a finite field F_p which does not embed into a smooth scheme over Z_p? | EDIT 7/15/14 I was just looking back at this old answer, and I don't think I ever answered the stated question. I can't delete an accepted answer, but I'll point at that, as far as I can tell, the Vakil reference I give also only address the question of deforming $X$ over $\mathbb{Z}\_p$, not of embedding it in some larger flat family over $\mathbb{Z}\_p$.
EDIT Oops! David Brown points out below that I misread the question. I was answering the question of finding a smooth scheme which does not deform in a smooth family over Z\_p.
Well, to make up for that, I'll point to some references which definitely contain answers. Look at section 2.3 of Ravi Vakil's paper *[Murphy's Law in algebraic geometry: Badly-behaved deformation spaces](https://arxiv.org/abs/math/0411469)* for some history, and several good references. Moreover, Ravi describes how to build an explicit cover of P^2 in characteristic p which does not deform to characteristic 0. Basically, the idea is to take a collection of lines in P^2 which doesn't deform to characteristic 0 and take a branched cover over those lines. For example, you could take that p^2+p+1 lines that have coefficients in F\_p.
| 5 | https://mathoverflow.net/users/297 | 561 | 301 |
https://mathoverflow.net/questions/563 | 3 | That's very poor wording, so let me be more precise. Suppose $L$ is an unambiguous regular language on an alphabet $\{a\_1, \dots, a\_n\}$, and suppose to each letter of the alphabet we associate two non-negative integers $(x\_i,y\_i)$ which are not both zero. Associate to a word $w$ the sum of the pairs of integers associated to each of its letters; call this $M(w) = (x, y)$.
Let $L'$ be the language consisting of all words such that $M(w) = (x, x)$ for some $x$. Is $L'$ an unambiguous context-free language?
| https://mathoverflow.net/users/290 | Is the "diagonal" of a regular language always context-free? | Yes.
There's no reason to have two nonnegative integers, you can just use one integer xi-yi. Then you care about whether the sum is zero. The language K of things which sum to zero is recognized by a push down automata -- the stack is always just a bunch of +1 tokens or -1 tokens corresponding to the current sum. Since K is recognized by a push down automata, it is context free.
The language you are interested in is L intersect K. The intersection of a regular language and a context free language is always context free.
| 4 | https://mathoverflow.net/users/27 | 566 | 304 |
https://mathoverflow.net/questions/559 | 8 | This question is inspired by [What is an example of a function on M\_g?](https://mathoverflow.net/questions/453/what-is-an-example-of-a-function-on-mg) . Consider Mg, the moduli space of genus g curves, NOT compactified. When g is 3 or greater, this is not affine. Does anyone know a good description of the ring of global functions on Mg, or of the spectrum of this ring?
| https://mathoverflow.net/users/297 | What is the affinization of M_g? | By exercise 2.10 of *Moduli of Curves* by Harris and Morrison there exists a complete curve through any two points of M\_g (this follows from the fact that the `boundary of the Hodge theory' compactification of M\_g has codimension at least 2 and that M\_g is quasi-projective). It follows the only global functions on M\_g are constant.
| 10 | https://mathoverflow.net/users/2 | 569 | 307 |
https://mathoverflow.net/questions/55 | 14 | I've read that an elliptic curve is supersingular if and only if its endomorphism ring is an order in a quaternion algebra. Does anyone have a simple explanation of this (or a good reference)?
| https://mathoverflow.net/users/32 | Supersingular elliptic curves | Let $k$ be an algebraically closed field, and let $E/k$ be an elliptic curve. In general, *how do we know the structure of $\mathrm{End}(E)$*?
We know the following two facts in all cases: (1) considered as an additive group, $\mathrm{End}(E)$ is free abelian on 1, 2 or 4 generators; and (2) $\mathrm{End}(E) \otimes\_Z Q$ is a division algebra. The first fact comes from considering homology (more on this momentarily), and the second comes from the theory of the dual isogeny.
* In the case of rank=1, we must have $\mathrm{End}(E)$ an order in $Q$, i.e. $\mathrm{End}(E)=Z$.
* In the case of rank=2, we must have $\mathrm{End}(E)$ an order in a quadratic field $F/Q$. This field must be imaginary, because its norm map is identified with $\lambda \mapsto \lambda \lambda^\vee$, and the latter is positive definite.
* In the case of rank=4, we must have End(E) an order in a quaternion algebra R/Q.
The last case gets *ruled out* when $\operatorname{char}(k)=0$. We don't see it in the familiar characteristic zero setting, so we think of it as strange, but there really is nothing unnatural about it. The "simple explanation" that you seek is, most bluntly, that it *does not get ruled out*, since there is no replacement for $H\_1(E,Z)$ in positive characteristic.
By the way, here is the reason it gets ruled out in characteristic zero: Without loss of generality by the Lefschetz principle, we may declare that k=C. Assume that $\mathrm{End}(E)$ is an order in $R$. The hard step in proving (1) above is showing that $\mathrm{End}(E)$ acts faithfully on the first homology group $H\_1(E,Z)$. Granted this, $\mathrm{End}(E)$ embeds as a free rank four $Z$-submodule of $\mathrm{End}(H\_1(E,Z)) = M\_2(Z)$. Tensoring with $Q$ we get that $R = M\_1(Q)$, and hence $M\_1(Q)$ would be a division algebra, which is false.
The reason I mention this argument is that, even though when $\operatorname{char}(k)=p>0$ the argument fails as stated (since one can't make $k=C$ and access $H\_1(E,Z)$), one can still modify it to get information about $R$. As a substitute for $H\_1(E,Z)$, one instead takes a prime $\ell$ not equal to $p$, and considers the $\ell$-adic Tate module $T\_\ell(E) = \varprojlim\_n E[\ell^n]$, with transition maps given by multiplication by \ell. (This gadget is a free $Z\_\ell$-module of rank 2 whether $\operatorname{char}(k)=0$ or not, and when $k=C$ it is canonically identified with $H\_1(E,Z\_\ell) = H\_1(E,Z) \otimes\_Z Z\_\ell$, which motivates its use as a substitute.) Considering again the faithfulness of the action of $\mathrm{End}(E)$, we have that $\operatorname{End}(E) \otimes\_Z Z\_\ell$ embeds into $\mathrm{End}(T\_\ell(E)) = M\_2(Z\_\ell)$, and therefore $R \otimes\_Q Q\_\ell = M\_2(Q\_\ell)$. By definition, this means that the quaternion algebra $R$ is "split at $\ell$". Now we invoke a by-product of global class field theory, which is the determination of all quaternion algebras over $Q$. They are parameterized by nonempty finite sets of even cardinality, consisting of prime numbers and possibly the symbol $\infty$. There is a unique quaternion algebra that is split at exactly those primes not occurring in the parameterizing set. Since for all $\ell$ not equal to $p$ we know that $R$ is split at $p$, the only possibility for the set associated to $R$ is $\{p,\infty\}$. Thus we know, on the nose, *which* quaternion algebra $R = \mathrm{End}(E) \otimes\_Z Q$ is.
| 17 | https://mathoverflow.net/users/367 | 571 | 308 |
https://mathoverflow.net/questions/124 | 39 | Harold Williams, Pablo Solis, and I were chatting and the following question came up.
In Lie group land (where you're doing differential geometry), given a finite-dimensional Lie algebra *g*, you can find a faithful representation *g* → End(V) by Ado's theorem. Then you can take the group generated by the exponentiation of the image to get a Lie group G⊆GL(V) whose Lie algebra is *g*. I think this is correct, but please do tell me if there's a mistake.
This argument relies on the exponential map, which we don't have an the algebraic setting. Is there some other argument to show that any finite-dimensional Lie algebra *g* is the Lie algebra of some *algebraic* group (a closed subgroup of GL(V) cut out by polynomials)?
| https://mathoverflow.net/users/1 | Is every finite-dimensional Lie algebra the Lie algebra of an algebraic group? | A Lie subalgebra of $\mathfrak{gl}(n,k)$ which is the Lie algebra of an algebraic subgroup of $GL(n,k)$ is called an algebraic subalgebra. Apparently there are Lie subalgebras which are not algebraic, even in characteristic zero. If $\mathfrak{g}$ is the Lie algebra of an affine algebraic group then it must be ad-algebraic, ie. its image in $\operatorname{End}(\mathfrak{g})$ under the adjoint representation must be an algebraic subalgebra. An example of a non-ad-algebraic Lie algebra is given on pg. 385 of [Lie Algebras and Algebraic Groups](http://books.google.com/books?id=ntKhAutD8I0C&pg=PA385&lpg=PA385&dq=Lie+Algebras+and+Algebraic+Groups,+by+Tauvel+and+Yu.&source=bl&ots=4V46-zVx1n&sig=Wiq7bS5rJocA-k09uJVYWXDdqTw&hl=en&ei=ZHbXSvmiG4HysgOx2cCLBg&sa=X&oi=book_result&ct=result&resnum=1&ved=0CA4Q6AEwAA#v=onepage&q=&f=false), by Tauvel and Yu.
| 26 | https://mathoverflow.net/users/371 | 576 | 313 |
https://mathoverflow.net/questions/519 | 8 | A linear series on a curve C is a line bundle L together with a subspace V of the global sections of L. Eisenbud and Harris develeoped a theory of limit linear series which explans how (L, V) degenerate when C does. I was under the impression that one of the nice things about limit linear series is it allows you to prove statements by induction on the genus of a curve. What are some examples of such statements that you can prove like this?
| https://mathoverflow.net/users/28 | Limit Linear Series | Not sure off the top of my head whether any of the Eisenbud-Harris papers literally used induction. One example is my paper "Linked Grassmannians and crude limit linear series" which gives a simple inductive proof of the Brill-Noether theorem using limit linear series. One can give similar arguments for existence of certain maps with prescribed ramification in characteristic p (see Theorem 7.1 of version 1 on the arxiv of "Linear series and existence of branched covers).
But more broadly, one might say that the "inductive" structure of limit linear series is encapsulated by the fact that one can describe limit linear series component by component (in contrast to the situation for higher rank vector bundles, where one has to worry about gluing maps). Many arguments can either be phrased in terms of induction or by degenerating immediately to a maximally degenerate case (e.g., a comb curve, or a chain of elliptic curves), and are essentially the same either way. But the point either way is that the machinery reduces to studying linear series on the individual components of the degeneration, which have smaller genus.
| 8 | https://mathoverflow.net/users/378 | 577 | 314 |
https://mathoverflow.net/questions/433 | 5 | First of the questions about derived algebraic geometry from a noobie.
The way I understand it, every discrete ring $R$ corresponds to some ring spectrum whose $\pi\_0$ is $R$. Now consider $p$-adic numbers. They are a limit of discrete rings — what should correspond to them? How to generalize?
(+ what could be good tags for derived algebraic geometry? I was considering: e-infinity, infty-structures, math-0703204, derived-alggeom, derived-spaces, infty-topology, a-infinity-algebras (2 currently tagged))
| https://mathoverflow.net/users/65 | $E_\infty$ spectrum corresponding to $\Bbb Z_p$ | There is indeed an Eilenberg-Maclane spectrum $H\Bbb Z\_p$. It is equivalent to the $p$-completion of $H\Bbb Z$ and the (homotopy) limit of the $H\Bbb Z/p^k$. However, it doesn't remember the topology.
If you want to remember the topology on the p-adics you need to do something more complicated, such as view the inverse system $\{H\Bbb Z/p^k\}$ as a pro-object in spectra rather than actually taking the limit. This is something like what you would do if you wanted to talk about formal schemes.
| 10 | https://mathoverflow.net/users/360 | 591 | 325 |
https://mathoverflow.net/questions/204 | 5 | Can anybody give me an example of a "naturally-occurring" algebraic category $C$ in which:
1. $C$ has two non-isomorphic objects $A$ and $B$ which are bi-embeddable via monic maps; but
2. $C$ does NOT have any infinite collection $A\_{1}$, $A\_{2}$, ... of objects which are pairwise bi-embeddable (via monic maps) and pairwise nonisomorphic?
Alternatively: can anybody give a reason why, under some reasonable hypothesis about the category $C$, property 1 should imply that 2 fails ("there is an infinite collection of pairwise bi-embeddable, pairwise nonisomorphic objects")?
| https://mathoverflow.net/users/93 | Bi-embeddability vs. isomorphism | I don't know if this counts as a "naturally-occuring" algebraic category but as I scanned through my internal list of categories of modules over a ring, I spotted that the following has (1) and (2).
Take $G$ to be the "$p$-adic dihedral group" that is a semidirect product of the p-adic integers and a cyclic group of order $2$ where the latter acts by $-1$ on the former. Then form the completed group algebra (Iwasawa algebra) with coefficients in the field with $p$ elements, $R=F\_{p}\left[\left[G\right]\right]$.
Now there are precisely two isomorphism classes of indecomposable projective modules [$P\_{1}$] and [$P\_{2}$] and $P\_{1}$ and $P\_{2}$ are bi-embeddable via monic maps (see section 9.6 of <http://www.dpmms.cam.ac.uk/~sjw47/char.pdf> for the proof of both these claims).
It follows that if we take the category of all finitely generated projective modules over $R$ then it consists of objects that are direct sums of m copies of $P\_{1}$ and $n$ copies of $P\_{2}$, say. Two such objects will be bi-embeddable if and only if they have the same value of $m+n$ thus whilst there are arbitrarily large finite collections of pairwise bi-embeddable but pairwise nonisomorphic objects in this category there are no infinite ones.
I am sure there will be many more examples along these lines, I just happen to spend a lot of time thinking about Iwasawa algebras.
| 4 | https://mathoverflow.net/users/345 | 602 | 333 |
https://mathoverflow.net/questions/587 | 9 | Can anyone recommend some good reading for Fourier analysis (and the Fourier transform) over finite abelian groups? I've found it given brief descriptions in both books on representation theory and on regular Fourier analysis (the best so far in the chapter in Tao and Vu's Additive combinatorics), but can't find a dedicated, detailed exposition of this area proper.
| https://mathoverflow.net/users/385 | Reading for finite Fourier analysis | There's Audrey Terras' "[Fourier Analysis on Finite Groups and Applications](http://rads.stackoverflow.com/amzn/click/0521457181)", which covers both the commutative and non-commutative settings (about half the book focuses on finite abelian groups). It's a fairly good introductory text, although it doesn't run very deep.
| 6 | https://mathoverflow.net/users/25 | 609 | 336 |
https://mathoverflow.net/questions/590 | 7 | The Lyndon-Hochschild(-Serre) spectral sequence applies to group extensions in a manner analogous to the Serre-Leray spectral sequence applied to a fibration.
Does anyone know of a good description (or reference) of the transgression maps in the Lyndon-Hochschild spectral sequence? MacLane describes them in terms of an additive relation, but I don't find this helpful in computing them.
More generally, I don't know how to calculate the differentials in this spectral sequence. In the Serre spectral sequence, I can see how an exact couple arises and the differentials are straightforward to see if not easy to calculate. But the LHSS arises from a double complex and I'm not sure how to get an exact couple from this.
| https://mathoverflow.net/users/343 | Differentials in the Lyndon-Hochschild spectral sequence | If you only need the first couple of transgressions, there is a nice description of them in the paper
[MR0641328 (83a:18021)
Huebschmann, Johannes
Automorphisms of group extensions and differentials in the Lyndon\mhy Hochschild\mhy Serre spectral sequence.
J. Algebra 72 (1981), no. 2, 296--334.](https://core.ac.uk/download/pdf/82585136.pdf)
For concrete calculations, this description is sometimes easier than the abstract nonsense description from the double complex.
| 8 | https://mathoverflow.net/users/317 | 611 | 337 |
https://mathoverflow.net/questions/190 | 11 | What is a ribbon graph? What is the ribbon graph decomposition of the moduli space of curves? What are some good references for this material?
| https://mathoverflow.net/users/83 | Ribbon graph decomposition of the moduli space of curves | There are a number of essentially different ways to get coordinates on moduli space using ribbon graphs. The survey
MR0963064 (90a:32026)
Harer, John L.
The cohomology of the moduli space of curves. Theory of moduli (Montecatini Terme, 1985), 138--221,
Lecture Notes in Math., 1337, Springer, Berlin, 1988.
by Harer gives an inspiring account of one of them. Another useful survey is the article "Lambda Lengths" by Penner, which can be downloaded at the bottom of the following webpage :
<http://www.ctqm.au.dk/events/2006/August/>
Penner is doing something slightly different (his spaces have certain decorations), but it is still quite useful.
| 8 | https://mathoverflow.net/users/317 | 613 | 339 |
https://mathoverflow.net/questions/625 | 5 | An identical relation on a group G is a word w in Fr, the free group on r elements (for some r), such that evaluating w on any r-tuple of elements of G yields the identity (this just means substituting elements of g for the variables in w, and evaluating the product).
Does the complete set of identical relations characterize a finite group? That is, are there two finite groups with precisely the same set of identical relations?
| https://mathoverflow.net/users/25 | Two finite groups with the same identical relations? | Z/2 and Z/2 x Z/2 have the same identical relations: those words w such that every variable in w appears an even number of times. (This is obviously sufficient for w to be an identical relation. If w has a variable occurring an odd number of times, assign it a nonzero value and the other variables zero to obtain a nonzero value for w.)
| 13 | https://mathoverflow.net/users/126667 | 629 | 351 |
https://mathoverflow.net/questions/582 | 25 | I've occasionally heard it stated (most notably on Terry Tao's blog) that "the Cauchy-Schwarz inequality can be viewed as a quantitative strengthening of the pigeonhole principle." I've certainly seen the inequality put to good use, but I haven't seen anything to make me believe that statement on the same level that I believe that the probabilistic method can be used as a (vast) strengthening of pigeonhole.
So, *how* exactly can Cauchy-Schwarz be seen as a quantitative version of the pigeonhole principle? And for extra pigeonholey goodness, are there similarly powered-up versions of the principle's other generalizations? (Linear algebra arguments [particularly dimension arguments], the probabilistic method, etc.)
| https://mathoverflow.net/users/382 | Cauchy-Schwarz and pigeonhole | My own interpretation (which I guess is pretty similar to the one above):
Suppose you have $r$ pigeons and $n$ holes, and want to minimize the number of pairs of pigeons in the same hole. This can easily be seen as equivalent to minimizing the sum of the squares of the number of pigeons in each hole.
Classical Cauchy Schwarz: $x\_1^2+...+x\_n^2 \ge\displaystyle\frac1n(x\_1+...+x\_n)^2$.
Discrete Cauchy Schwarz: If you must place an integer number of pigeons in each hole, the number of pairs of same-hole pigeons is minimized when you distribute the pigeons as close to evenly as possible subject to this constraint.
Pigeonhole: In the case $r=m+1$, the most even split is $(2,1,1,...,1)$, which has a pair of pigeons in the same hole.
| 24 | https://mathoverflow.net/users/405 | 633 | 354 |
https://mathoverflow.net/questions/610 | 5 | Suppose we call a subset S of $F^{n}$ ($F$ is the field with two elements) good if for any $x$ and $y$ (possibly $x=y$) we have $[x,y]=1$ where $[ , ]$ denotes the obvious bilinear form on F. What's the best bound as a function of $n$ on the size of $S$? In particular, do we have $|S|$ less than or equal to $n$ (or just a linear function in $n$)?
| https://mathoverflow.net/users/399 | Linear Algebra Over $F_{2}$ | Andy's example is actually within a constant factor of optimal. To see this, imagine adding on an additional coordinate equal to 1 to each element of S. These new elements now satisfy [x',y']=0.
Now let W be the subspace of F^{n+1} spanned by the augmentation of S. The above relation is equivalent to saying that W is contained in its own orthogonal complement. Since the dimension of a space and that of its complement add to n+1, we know the dimension of W is at most (n+1)/2, so W contains at most 2^{(n+1)/2} vectors.
This result (and proof) is one I've heard referred to as the Eventown theorem and attributed to Berlekamp.
As an interesting side note, if we replaced the condition [x,y]=1 for all x,y by one saying that [x,y]=1 if and only if x does not equal y, then we'd have a bound of n on the size of S (this is Berlekamp's oddtown theorem).
| 6 | https://mathoverflow.net/users/405 | 651 | 365 |
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