stevengubkin/mathoverflow_text_arxiv_labels

5

For a linearly reductive (i.e. all representations are completely redudible) group [UNK], any quotient [UNK] is also linearly reductive. Given a representation [UNK] of [UNK], you get an induced representation of [UNK] (with the same underlying vector space). Then any decomposition of [UNK] as a representation of [UNK] is also a decomposition as a representation of [UNK]. Since [UNK]-invariant subspaces of [UNK] are the same thing as [UNK]-invariant subspace of [UNK], the notions of irreducible subrepresentations agree. So you get a complete decomposition of [UNK] as a representation of [UNK]. 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 [UNK] lifts to give you an element of the unipotent radical of [UNK], but I'm not sure.

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7

Can [UNK] have only digits 0 and 1, other than [UNK]? Pablo Solis asked this at a recent 20 questions seminar at Berkeley. Is there a positive integer [UNK], not of the form [UNK], such that the digits of [UNK] are all 0's and 1's? It seems very unlikely, but I don't have a proof. It's easy to see that such a number must end in 1 or 9, and then easy to see that it must end in 01, 49, 51 or 99, and you can continue recursively for as long as you like, determining possible "suffixes". Using this, I had a computer check for me that there are no such N up to about [UNK]. If you pretend that the digits of [UNK] are randomly distributed, and [UNK] has [UNK]-digits, there's a [UNK] chance of satisfying this condition. There are only [UNK] [UNK]-digit numbers, so you might expect a [UNK] chance of having a some [UNK]-digit number. This suggests we shouldn't expect to find anything. (If you try the same problem in other bases, where the probabilities are better, you do find a few: in base 5, 222112144, 22222111221444 and 100024441003001 work.)

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8

Is there a good version of Artin-Wedderburn for semisimple algebra objects? Artin-Wederburn says that if you have a semisimple algebra then it is a product of matrix algebras over division rings. Suppose that [UNK] 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 [UNK] is an algebra object in [UNK]. That is [UNK] is an object in [UNK] together with a multiplication map [UNK] and a unit etc. We call [UNK] semisimple if the category of [UNK]-module objects in [UNK] is semisimple. Is there some good analogue of Artin-Wedderburn?

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9

Let [UNK] be a finite field with [UNK] elements. Let [UNK] be the field of rational functions in two indeterminate variables over [UNK]. Consider the extension of [UNK] obtained by adjoining [UNK]-th roots of [UNK] and of [UNK]. More precisely, let [UNK] be an algebraic closure of [UNK]. In [UNK] we can solve the equation [UNK] in the variable [UNK]. Let [UNK] be a solution of this equation; so [UNK] is an element of [UNK] which satisfies [UNK]. Similarly find an element [UNK] which satisfies [UNK]. Consider [UNK]. [UNK] is a finite extension of [UNK], of order [UNK] as you can check. However there is no element of degree [UNK] in [UNK], and a primitive element would have to have degree [UNK]. 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 [UNK] are also simple, so we must use [UNK].

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21

Non-quasi separated morphisms What are some examples of morphisms of schemes which are not quasi separated?

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22

Homomorphism more than 3/4 the inverse Suppose [UNK] is a finite group and [UNK] is an automorphism of [UNK]. If [UNK] for more than [UNK] of the elements of [UNK], does it follow that [UNK] for all [UNK] in [UNK] 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 [UNK]. Claim: For [UNK] in [UNK], [UNK] is a subset of [UNK], the centralizer of [UNK]. Proof: For such [UNK], [UNK] and [UNK]. Now [UNK] So [UNK] and [UNK] commute. Since [UNK] is more than half of [UNK], so is [UNK]. So by Lagrange's Theorem, [UNK], and [UNK] is in the center of [UNK]. Thus [UNK] is a subset of the center, and it is more than half of [UNK]. So the center must be all of [UNK], that is [UNK] is commutative. Once [UNK] is commutative the problem is easy.

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28

Weil divisors on non Noetherian schemes 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?

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30

What is the universal property of normalization? 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.

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32

I think the point of this whole [UNK] business is the following. If [UNK] is the set of elements such that [UNK], then if we look at left multiplication on [UNK] by an element of [UNK], more than half the elements have to make back into [UNK]. Combining this with what we know about [UNK] it should follow that any [UNK] commutes with more than [UNK] the elements of [UNK], so if you say Lagrange's theorem enough times it should follow that [UNK] is abelian and [UNK] generates [UNK], which together imply the result.

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37

Normalization is right adjoint to the inclusion functor from the category of normal schemes into the category of reduced schemes. In other words, if [UNK] is the normalization of [UNK] and [UNK] is any morphism where [UNK] is a normal scheme, then [UNK] factors uniquely through [UNK].

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42

Is there a universal property for Witt vectors? Do the Witt vectors satisfy a universal property?

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44

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.

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45

Yes, I believe the quotient of a reductive group scheme is reductive. Perhaps it's possible to show directly as Anton was suggesting. I have an argument that shows that if G is geometrically reductive then G/H is geometrically reductive. The statement that G reductive implies G/H reductive follows from Haboush's theorem (ie. G is reductive if and only if G is geometrically reductive).

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49

For p-typical Witt vectors it is left adjoint to the functor 'reduction mod p' from the category of strict p-rings to the category of perfect F_p algebras. See 1.1 and 6 of Joe Rabinov's notes

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52

Complete theory with exactly n countable models? For [UNK] an integer greater than [UNK], Can one always get a complete theory over a finite language with exactly [UNK] models (up to isomorphism)? There’s a theorem that says that [UNK] is impossible. My understanding is this should be doable in a finite language, but I don’t know how. If you switch to a countable language, then you can do it as follows. To get [UNK] models, take the theory of unbounded dense linear orderings together with a sequence of increasing constants [UNK]. Then the [UNK]’s can either have no upper bound, an upper bound but no sup, or have a sup. This gives exactly [UNK] models. To get a number bigger than [UNK], we include a way to color all elements, and require that each color is unbounded and dense. (The [UNK]’s can be whatever color you like.) Then, we get one model for each color of the sup plus the two sup-less models.

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53

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).

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60

Can the valuative criteria for separatedness/properness be checked "formally"? 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?

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61

What is interesting/useful about Castelnuovo-Mumford regularity? What is interesting/useful about Castelnuovo-Mumford regularity?

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Witt vectors have several universal properties. For example the big Witt vectors are right adjoint to the forgetful functor from lambda rings to commutative rings. You find information like this in the book "Formal Groups and Applications" by M. Hazewinkel.

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69

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).

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74

How to understand character sheaves 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 How to understand these for a person with less than excellent representation theory background?

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76

What is a topos? According to Higher Topos Theory 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?

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77

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.

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79

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.

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82

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.

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What do epimorphisms of (commutative) rings look like? (Background: In any category, an epimorphism is a morphism [UNK] which is "surjective" in the following sense: for any two morphisms [UNK], if [UNK], then [UNK]. Roughly, "any two functions on [UNK] that agree on the image of [UNK] 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, [UNK] 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 [UNK] in a ring [UNK], the quotient map [UNK] is an epimorphism. It's also not hard to see that if [UNK] is a multiplicative subset, then the localization [UNK] is an epimorphism. Here's a proof to whet your appetite. If [UNK] are two homomorphisms that agree on [UNK], then for any element [UNK], we have [UNK] Also, if [UNK] is a finite collection of epimorphisms, where the [UNK] have disjoint support as [UNK]-modules, then [UNK] is an epimorphism. Is every epimorphism of rings some product of combinations of quotients and localizations? To put it another way, suppose [UNK] is an epimorphism of rings with no kernel which sends non-units to non-units and such that [UNK] has no idempotents. Must [UNK] be an isomorphism?

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A little searching turned up: Ring epimorphisms and C(X) by Michael Barr, W.D. Burgess and R. Raphael (article). They consider this question for rings of the form of continuous functions on a topological space. They quote the following characterisation of epimorphisms in the category of commutative rings: Proposition: A homomorphism f : A → B is an epimorphism if and only if for all b ∈ B there exist matrices C, D, E of sizes 1 × n, n × n, and n × 1 respectively, where (i) C and E have entries in B, (ii) D has entries in f(A), (iii) the entries of CD and of DE are elements of f(A) and (iv) b = CDE. (Such a triple is called a zig-zag for b.) This seems a little more complicated than localisation, though I haven't checked the details. They then go on to prove that 2.12: A subspace Y of a perfectly normal first countable space X induces an epimorphism if and only if it is locally closed. If I understand all the terminology correctly, then this implies that C([0,1],ℝ) → C((0,1),ℝ) is an epimorphism. There are plenty more references in that article, and it would be nice to have an actual zig-zag for this situation. But in the spirit of open-source mathematics, I thought I'd post this and see if someone (possibly me later on) can fill in the details. Added Later: The example I gave: C([0,1],ℝ) → C((0,1),ℝ) is a localisation. It is obtained by inverting all functions in C([0,1],ℝ) which are zero only at the end-points. Given a function f ∈ C((0,1),ℝ), there will be a function g ∈ C([0,1],ℝ) which is non-zero apart from at 0 and 1 and which goes to 0 at 0 and 1 faster enough that the product g f also goes to 0 at the end-points. Then g f is (the restriction of something in) C([0,1],ℝ) and g becomes invertible in C((0,1),ℝ). So f = g-1 (g f) is in the specified localisation of C([0,1],ℝ). Indeed, the Barr et. al. paper comments on the fact that in all the examples they consider (function rings), the zig-zag has length 1. I conjecture that if the zig-zags always have length 1 (for a particular function f: A → B), then B is formed by a localisation on A. A possibly stronger version of this conjecture would be that this is an if-and-only-if. In which case, finding a counter-example to Anton's conjecture would involve finding a case where there was a zig-zag of length 2. I suspect that a universal construction would be the best approach to finding one. In the spirit of wiki-ness and only doing a little at a time, I'll leave this here. Added Even Later: (Should I timestamp these? I know that the system does so, but is it useful to embed them in the edit?) Here's one direction for my conjecture above. If B = S-1A, then for b ∈ B, we have b = s-1a for some s ∈ S and a ∈ A. Then we put C = s-1, D = s, E = b = s-1 a. Then CD = 1, DE = a, D ∈ f(A), and CDE = b. So in a localisation, zig-zags have length 1.

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85

I'm kind of late to the party, but anyway: being reductive means having no nilradical. Now nilradical cannot become larger as you take quotient, ergo your question has a positive answer.

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86

What is the exact statement of "there are 27 lines on a cubic"? I think there was a theorem, like every cubic hypersurface in [UNK] has 27 lines on it. What is the exact statement and details?

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89

is there a good computer package for working with bicomplexes? I'm interested in working with bicomplexes of modules over polynomial rings, specifically tensoring them together, and the operation of taking cohomology in one direction, and then the other. Is there any computer algebra package which will do this efficiently?

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See the beautiful book of Fuchs and Tabachnikov, Mathematical Omnibus, for an answer.

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99

How can you tell if a space is homotopy equivalent to a manifold? 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?

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101

Atiyah-MacDonald, exercise 2.11 Let [UNK] be a commutative ring with [UNK] not equal to [UNK]. (The ring A is not necessarily a domain, and is not necessarily Noetherian.) Assume we have an injective map of free [UNK]-modules [UNK]. Must we have [UNK]? I believe the answer is yes. For instance, why is there no injective map from [UNK]? Say it's represented by a matrix [UNK]. Then clearly [UNK] is in the kernel. In the [UNK] case, we can look at the [UNK] matrix which represents it; call it [UNK]. Let [UNK] denote the determinant of the matrix obtained by deleting the [UNK]-th column. Let [UNK] be the vector [UNK]. Then [UNK] is in the kernel of our map, because the vector [UNK] has [UNK]-th component the determinant of the [UNK] matrix attained from [UNK] by repeating the [UNK]-th row twice. That almost finishes the proof, except it is possible that [UNK] is the zero vector. I would like to see either this argument finished, or, even better, a nicer proof. Thank you!

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Can one check formal smoothness using only one-variable Artin rings? Let [UNK] be a morphism of schemes over a field [UNK]. Can one check that [UNK] is formally smooth using only Artin rings of the form [UNK], where [UNK] is also a field? Considering cuspidal curves one can show that you do at least need arbitrarily large [UNK].

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106

Model category structures on categories of complexes in abelian categories 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). 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? 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.

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109

How do you see the genus of a curve, just looking at its function field? Yuhao asked in the 20-questions seminar: The genus of a curve is a birational invariant; the function field of a curve determines it up to birational equivelance. How do you see the genus directly from the field?

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Is there a category in which finite limits and directed colimits *don't* commute Andrew Critch asks at the 20-questions seminar: In Set and AbGrp (the categories of sets and abelian groups, respectively), finite limits commute with directed colimits. As an example, if you're working with sheaves of sets, you can take kernels (a type of limit) and stalks (a type of directed colimit) safely. Is there an example of a category where they don't commute? (Depending on how you choose to talk about topoi, this condition is sometimes an axiom, capturing the idea of "looking sufficiently like sets".)

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111

Can you describe the image of the exponential map B(H)->B(H). James Tener asks at the 20-questions seminar: The exponential map exp:B(H)->B(H) is just defined by its Taylor series. Can you describe its image?

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113

Can you explicitly write [UNK] as a disjoint union of two totally path disconnected sets? An anonymous question from the 20-questions seminar: Can you explicitly write [UNK] as a disjoint union of two totally path disconnected sets?

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115

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 [UNK] 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

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119

Assuming you work over [UNK], you can see it in the structure of the Galois group. The points of the curve can be seen as non-Archimedean valuations on the field, and you can see whether a map of curves is ramified by looking at valuations (each prime in the smaller field should have valuation 1 or 0 for any valuation of the bigger field). Basically, this means that around any point in the domain and its target, the completed local rings look like [UNK] and [UNK], and the map should always be [UNK], which [UNK] (so the map on completions is an isomorphism). There's a maximal unramified field extension, and the Galois group of this extension is the profinite completion of [UNK] in the topological sense, so the genus can be obtained by looking at the abelianization.

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120

Presumably the argument ilya was driving at is this: If one had a surjective map of group schemes [UNK], then consider the preimage of the nilradical of [UNK]. This is an algebraic group with a surjective map to a unipotent group. Since there are no group homomorphisms from reductive groups to unipotent ones, the nilradical of this preimage (which is contained in the nilradical of [UNK]) must surject onto the nilradical of [UNK]. So if the former is trivial, so is the latter.

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123

It's a standard result that the continuous locus is always G-delta. For each r>0, let U(r) be the set of points x such that some neighborhood of x maps into some ball of radius r. Then each U(r) is open, and the continuous locus is their intersection. Conversely, given a G-delta set, I'm pretty sure it's not hard to construct a function with that continuous locus, though I don't remember how off the top of my head.

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126

Dan, knowing your tastes, I think you will like Fulton-Harris very much. However, if I recall correctly, Fulton-Harris doesn't go into much depth about some important (and really cool) theorems in Lie groups, such as Peter-Weyl and Borel-Weil-Bott. But of course, you can learn these theorems elsewhere. I think the book "Compact Lie Groups" by Sepanski is nice, and it does cover P-W and B-W-B. I also found this note on B-W-B to be useful in the past: http://www-math.mit.edu/~lurie/papers/bwb.pdf

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129

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.

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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 ([UNK]). Your definition used the "bar-complex" resolution of the form [UNK] but k[t] has a much nicer resolution as a bimodule over itself, the Koszul resolution. This is of the form [UNK] with the map given by [UNK], so when you apply coinvariants, you get two copies of [UNK] 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 [UNK] 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.

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Is there a finite-index finite-depth II[UNK] subfactor which is more than [UNK]-super-transitive? Background: See Noah and Emily's posts on subfactors and planar algebras on the Secret Blogging Seminar. There are plenty of examples of [UNK]-super-transitive (3-ST) subfactors; Haagerup, [UNK], and others. There's exactly one known example of a [UNK]-ST subfactor, the Haagerup-Asaeda subfactor, and one [UNK]-ST subfactor, the extended Haagerup subfactor. Below index [UNK] there are the [UNK] and [UNK] families, which are arbitrarily super-transitive. Ignore those; I'm just interested above index [UNK]. Is there anything that's even more super-transitive?

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Zeta-function regularization of determinants and traces The short answer to my question may be a pointer to the right text. I will give all the background I know, and then ask my questions in list form. Let A be an operator (on an infinite-dimensional vector space). You might as well assume that its spectrum is all real and positive. In fact, I only care when the spectrum is discrete and grows polynomially, but I hear that this stuff works more generally. In general, A is not trace-class (the sum of the eigenvalues converges) or determinant-class (the product of the eigenvalues converges) — if the nth eigenvalue grows as np for some p>0, then it won't be. But there is a procedure to try to define a "trace" and "determinant" of A nevertheless. Let us hope that for large enough s, the operators A-s (=exp(-s log A), and log A makes sense if the spectrum of A is positive) are trace-class. If so, then we can define ζA(s) = tr(A-s); it is analytic for Re(s) large enough. Let's hope that it has a single-valued meromorphic continuation and that this function (which I will also call ζA(s)) is smooth near s=0 and s=-1. All these hopes hold when the eigenvalues of A grow polynomially, whence ζA(s) can be compared to the Riemann zeta function. Then we can immediately define the "regularized trace" TR A = ζA(0) and the "regularized determinant" DET A = exp(-ζA'(0)), where by ζA'(s) I mean the derivative of ζA(s) with respect to s. (If the eigenvalues λn are discrete, then ζA(s) = Σ λn-s, and so one would have TR A = Σ λn and DET A = Σ (log λn) λn-s |s=0, if they converged.) If A is trace- (determinant-) class, then TR A = tr A (DET A = det A). So, here are my questions: Is it true that exp TR A = DET exp A? Let A(t) be a smooth family of operators (t is a real variable). Is it true that d/dt [ log DET A(t) ] = TR( A-1 dA/dt )? (I can prove this when A-1dA/dt is trace-class.) Is DET multiplicative, so that DET(AB) = DET A DET B? (I can prove this using 1. and 2., or using the part of 2. that I can prove if B is determinant-class.) Is TR cyclic, i.e. TR(AB) = TR(BA)? Is TR linear, i.e. TR(A + B) = TR A + TR B? None of these are even obvious to me when A and B (or dA/dt) are simultaneously diagonalizable (except of course cyclicity), but of course in general they won't commute.

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Ribbon graph decomposition of the moduli space of curves 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?

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How do you show that [UNK] is contractible? Here I mean the version with all but finitely many components zero.

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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 [UNK] be the "shift everything down by 1" map. Then for any point [UNK], [UNK] is not a multiple of [UNK] and so the line between them does not go through the origin. We can therefore define a homotopy from the identity on [UNK] to [UNK] by taking the homotopy [UNK] and renormalising so that it is always on the sphere (incidentally, although you are working in [UNK], by talking about a sphere you implicitly have a norm). Then we simply contract the image of [UNK], which is a codimension 1 sphere, to a point not on it, say [UNK]. 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 [UNK] 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 [UNK]) you can still define the sphere as the quotient of [UNK] without the origin by the action of [UNK]. The argument above still works in this case. Added even later: Revisiting this in the light of the duplicate: Is [UNK] minus the zero function contractible?, the key property on [UNK] is that it be continuous, injective, have no eigenvectors, and be not surjective. These conditions imply the following: injective ⟹ the end-point of the homotopy is not the origin no eigenvalues ⟹ the homotopy does not pass through the origin en route not surjective ⟹ there is a point not in the image to which the image can be contracted 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.

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Is there an example of a variety over the complex numbers with no embedding into a smooth variety? Is there an example of a variety over the complex numbers with no embedding into a smooth variety?

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Theorem (EGA IV 13.1.3): Let [UNK] be a morphism of schemes, locally of finite type. Then [UNK] is upper semi-continuous. Corollary (Chevalley's upper semi-continuous theorem, EGA IV 13.1.5): Let [UNK] be proper, then: [UNK] is upper semi-continuous. Corollary (SGA3, ??): Let [UNK] be a group scheme, locally of finite type. Then [UNK] 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 [UNK] is the composition of the continuous function [UNK] and the upper semi-continuous function [UNK]. Concerning your application: The fiber dimensions of the stabilizer group scheme Stab/X is upper semi-continuous, but the "diagonal" [UNK] does not always have this property (unless it is proper, i.e., "[UNK] acts properly").

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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.

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Can a quotient ring R/J ever be flat over R? If R is a ring and J⊂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=R1⊕R2 and J=R1, 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. Spec(R) is connected) and J≠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→J→R→R/J→0. When you tensor with R/J, you get 0→ J/J2→R/J→R/J→0 where the map R/J→R/J is the identity map. If J≠J2, this sequence is not exact, contradicting flatness of R/J. But sometimes it happens that J=J2, like the case of the maximal ideal of the ring k[tq| q∈Q>0]. I can show that the quotient is not flat in that case (see this answer), 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.

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Another nice solution to a similar question is at http://katlas.math.toronto.edu/drorbn/index.php?title=0708-1300/the_unit_sphere_in_a_Hilbert_space_is_contractible: Let [UNK] and define [UNK]. Claim. [UNK] is contractible. Proof. For any [UNK] and any [UNK] define [UNK] for [UNK] and [UNK] for [UNK]. Observe that [UNK] is continuous and gives the desired retraction to the point [UNK].

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How to topologize X(R) when R is a topological ring? Given a topological ring [UNK], under what conditions and in what way, can one induce a topology on the [UNK]-points of a scheme [UNK]? For example, if [UNK] is [UNK] or [UNK], one has natural topology on the [UNK]-points. If [UNK] is a group scheme/A and [UNK] is [UNK]-algebra (still a topological ring), will the induced topology on [UNK]G[UNK]G[UNK]G\left(A_{K}[UNK] for any number field [UNK]. Is the induced topology on [UNK]G\left(A_{K}[UNK] be locally compact or satisfy other nice properties?

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Equivalent statements of the Riemann hypothesis in the Weil conjectures In the cohomological incarnation, the Riemann hypothesis part of the Weil conjectures for a smooth proper scheme of finite type over a finite field with [UNK] elements says that: the eigenvalues of Frobenius acting on the [UNK] are algebraic integers with complex absolute value [UNK]. For smooth proper curve [UNK] over [UNK] of genus g, the Riemann hypothesis is often stated in another way as [UNK] where [UNK] is the number of rational points on [UNK]. Why are these two statements equivalent? Is (are) there any corresponding inequality(ies) which are equivalent to the Riemann hypothesis in higher dimensions?

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The first step is to reformulate the zeta function in terms of l-adic cohomology. It is defined to be [UNK] where [UNK] is the number of points over [UNK] So [UNK] is #[UNK], and by the analogue of the Lefschetz fixed point theorem for l-adic cohomology, this implies [UNK]. We can substitute this in the expression for Z(u) and rearrange to get [UNK] where [UNK] is the dimension of the variety. The inner exponentials are actually [UNK], which relates zeros and poles of the zeta function to the eigenvalues of Frob* acting on [UNK]. Both [UNK] and [UNK] are 1-dimensional spaces, and we know the eigenvalues of Frob* acting on them: on [UNK] Frob* acts as the identity, so the eigenvalue is 1, and on [UNK] Frob* is multiplication by [UNK] because Frob is a finite map of degree [UNK]. For curves, [UNK] and the expression above for [UNK] becomes [UNK], where [UNK] is a polynomial of degree [UNK] whose zeros [UNK] (the eigenvalues of Frob* acting on H^1) have absolute value [UNK]. Then it's not hard to show (by logarithmically differentiating [UNK] and rearranging) that [UNK]. If we assume [UNK] (to prove this I guess you need some way to compare Zariski and etale cohomology), this shows that the bound on point counting is the same as the bound on the size of the eigenvalues of Frob on [UNK]. For higher dimensions, you could play the same sorts of games, but the expressions wouldn't be as nice.

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If [UNK] is arbitrary and [UNK] is an ideal of finite type such that [UNK] is a flat [UNK]-module, then [UNK] is open and closed. In fact, [UNK] is a finitely presented [UNK]-algebra and thus [UNK] is a flat monomorphism of finite presentation, hence an étale monomorphism, i.e., an open immersion (cf. EGA IV 17.9.1). If [UNK] is a noetherian ring then [UNK] is flat if and only if [UNK] is open and closed (every ideal is of finite type). If [UNK] 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 [UNK] is flat iff [UNK] is open and closed. Indeed, there is a result due to Lazard [Laz, Cor. 5.9] which states that the flatness of [UNK] implies that [UNK] is of finite type in this case. If [UNK] has an infinite number of minimal prime ideals, then it can happen that a flat closed immersion is not open. For example, let [UNK] be an absolutely flat ring with an infinite number of points (e.g. let [UNK] be the product of an infinite number of fields). Then [UNK] is zero-dimensional and every local ring is a field. However, there are non-open points (otherwise [UNK] 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: There exists a connected affine scheme [UNK], with an infinite number of irreducible components, and an ideal [UNK] such that [UNK] is flat but [UNK] 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) Edit: Corrected proof of 1). An open closed immersion is not necessarily an open immersion! (e.g. [UNK] 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).

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Is the long line paracompact? A manifold is usually defined as a second-countable hausdorff topological space which is locally homeomorphic to Rn. My understanding is that the reason "second-countable" is part of the definition is to make sure that the space is paracompact, 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 paracompact? Should I consider it a manifold?

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For connected locally Euclidean spaces, paracompactness is equivalent to second countable. Therefore if we truly insisted that our manifolds be second countable then the only thing that we would lose would be arbitrary coproducts. However, we don't insist that. Although it's often stated in the definition early on in a differential topology course or book, in my experience that's just because it's easier to explain than paracompactness (I tend to pick metrisable, myself). Of course, that early in a course we're probably not too concerned about arbitrary coproducts either. Moreover, because manifolds split into a coproduct of their connected components, we tend to deal with connected manifolds unless we really can't avoid it. And even when they aren't connected, they most often have countably many components. So in practise the distinction doesn't arise. I don't have it in front of me here, but I believe that an appendix to the first volume of Spivak's "Introduction to differential geometry" contains a proof of four equivalent conditions for locally Euclidean spaces (perhaps requiring Hausdorff). If I remember aright, the conditions are: paracompact, second countable, metrisable, and σ-compact. There was a paper on the arxiv on Monday, 0910.0885, which lists 107 conditions for a connected locally Euclidean Hausdorff space equivalent to that it be metrisable. Amongst them are paracompactness and second countable.

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Has anyone tabulated 2-knots? Would anyone like to try? 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 R^3, and count the triple points. Does anyone know if this has been attempted? Such a list could be quite useful.

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Compact Kaehler manifolds that are isomorphic as symplectic manifolds but not as complex manifolds (and vice-versa) 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... 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? 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?

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When are Hilbert schemes smooth? 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?

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Let S be a subset of the reals such that S∩[a,b] and Sc∩[a,b] cannot be written as a countable union of closed sets for any a<b. This can be done (this explicit example of a non-Borel set achieves this). Let ℚ be the rationals. Then, A=(Sxℚ)U(Scxℚc) and B=(Sxℚc)U(Scxℚ) should do it. The proof is as follows. Suppose that the curve t→(f(t),g(t)) lies in A, and consider a closed bounded interval I. As the curve lies in A, f(I)∩S = f(I∩g-1(ℚ))=∪x∈ℚf(I∩g-1(x)) is a union of countably many closed sets. By the choice of S, f(I) must be a single point. Hence, f is constant. Then, g is a continuous function mapping into either ℚ or ℚc, so is also constant. So A is totally path disconnected. The argument for B follows in the same way by exchanging S and Sc

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A very well-known condition is that the Hilbert scheme of a smooth surface is smooth. As David pointed out below, the Hilbert scheme of a smooth curve is smooth and equal to the symmetric product (since k[t] has only one finite dimension quotient of each dimension). I don't know of any other examples, but one of the versions of Murphy's Law in algebraic geometry is roughly "if you don't have a good reason for a Hilbert scheme to not be horrible, it will be as horrible as you can possibly imagine."

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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.

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Re 3: If you say projective, then yes. GAGA tells you that an analytic isomorphism is also an algebraic one. If you don't say projective, then no. See the appendix to Hartshorne for a family of nonisomorphic algebraic structures on C^2/Z^2.

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does a line bundle always have a degree 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 [UNK] 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 [UNK] 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 [UNK] that is degree 0 and has [UNK]) = 1?

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One generalization of degree is first Chern class: A Cartier divisor corresponds to a class in [UNK], and you take its image under the boundary map of the long exact sequence corresponding to the exponential exact sequence [UNK] 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 [UNK], 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 [UNK].

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Finding monochromatic rectangles in a countable coloring of [UNK] Given a countable coloring of the plane, is it always possible to find a monochromatic set of points [UNK] (the corners of a rectangle)?

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What is a TMF in topology? What is a topological modular form? How are they related to 'normal' (number-theoretic) modular forms?

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There is a Wikipedia entry on 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 [UNK]. There is a map of graded rings: [UNK] where [UNK] 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 [UNK] 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.

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What is the Theorem of the Cube? What is the "theorem of the cube" for abelian varieties? What is the statement and how should I think about it?

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This is equivalent to CH. Quoting "Problems and Theorems in Classical Set Theory" by Komjath and Totik, chapter 16, Continuum hypothesis: 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 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 R as a vector space over Q. Let A be some basis. Take any bijection A -> A + A, where + is disjoint sum. It induces a linear isomorphism f: R -> R * 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=(x1, y1), b=(x1+x2, y1), c=(x1, y1+y2), d=(x1+x2, y1+y2), certainly a+d=b+c. Using that isomorphism, f(a)+f(d)=f(b)+f(c) gives a monochromatic solution of quoted equation.

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If you have a line bundle trivial on 3 "surfaces" of a "cube" [UNK] where [UNK], [UNK], [UNK] are abelian varieties, then this line bundle in trivial on the whole "cube". See wikipedia.

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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).

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Which came first: the Fibonacci Numbers or the Golden Ratio? I know that the Fibonacci numbers converge to a ratio of .618, and that this ratio is found all throughout nature, etc. I suppose the best way to ask my question is: where was this .618 value first found? And what is the...significance?

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Golden ratio came first. Wikipedia has a rather thorough article on it. It's not nearly as pervasive in nature or architecture as people like to say it is. It will show up in anything with regular pentagons, though.

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Beilinson-Bernstein and Koszul duality 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?

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First you should know what a "derived scheme" (or "spectral scheme") is. Roughly, this is the same as an ordinary scheme, except instead of locally being the Spec of an ordinary commutative ring, it's locally the Spec of an E-infinity ring spectrum, or just E-infinity ring for short. An E-infinity ring is not the same as a ring spectrum; a ring spectrum is something that is "a ring up to homotopy", and an E-infinity ring is something that is "a ring up to coherent homotopy". As a topological space, Spec of an E-infinity ring A is defined to be the Spec of pi0(A), which is an ordinary commutative ring. The difference is that the sheaf of functions is no longer a sheaf of rings but a sheaf of E-infinity rings. This sheaf of E-infinity rings is (analogously to the structure sheaf for ordinary affine schemes) defined by Uf -> A[f-1], where Uf is a distinguished open subset, and the localization A[f-1] is now taken in the category (or rather infinity-category) of E-infinity rings (see section 2.2 of Lurie's survey for a characterization of this localization). There is a natural functor from derived schemes to ordinary schemes. It is (X, OX) -> (X, pi0(OX)), where pi0(OX) denotes the sheafification of the presheaf U -> pi0(OX(U)). Then Definition 4.1 of Lurie's survey article defines an elliptic curve over an E-infinity ring A: it is a commutative A-group E -> Spec A such that (E, pi0(OE)) -> Spec pi0(A) is an ordinary elliptic curve over pi0(A). I think one of the punchlines is that there is a derived Deligne-Mumford moduli stack of oriented derived elliptic curves, which becomes the ordinary Deligne-Mumford moduli stack of ordinary elliptic curves after hitting it with pi0. Taking global sections on this derived moduli stack is more or less tmf. I don't have a good short explanation of what an orientation is, but it's in the beginning of section 3 of the survey. Unfortunately I think a lot of the details of this stuff are still not available. (I think it is supposed to be in "DAG VII: Spectral Schemes"; there may be some hints and related material in "DAG V: Structured Spaces".)

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Are there interesting monoidal structures on representations of quantum affine algebras? Is there a good monoidal structure on a category of integrable representations of a quantum affine algebra? In the ordinary affine Kac-Moody case, there is the usual tensor product (symmetric, adds charges) and a fusion structure (braided, comes from G-bundles on curves, preserves central charge). In the quantum case, there is the usual tensor product (braided meromorphic-braided[UNK]), but all I see in the literature about fusion is vague comments that it can't exist. I guess my question should be "what is the major malfunction?" [UNK] Edit: The meromorphic property (in the sense of Soibelman's Meromorphic tensor categories) seems to be a first hint at problems, and I should have paid better attention to it.

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Eigenvalues of Laplacian What's the most natural way to establish the asymptotics of [UNK] on a compact Riemannian manifold [UNK] of dimension [UNK]? The asymptotics should be [UNK] (Perhaps one could consider first the case of a Kähler manifold? The Laplacian is particularly simple there.)

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Dualizing sheaf on singular curves I am trying to understand the stabilization map, which takes a prestable curve (a curve with some marked points, and at worst nodal singularities) and returns a stable curve (a curve with some marked points, at worst nodal singularities, and finite automorphisms). Essentially what the stabilization map does is it contracts all of the unstable components of the curve, that is, those with infinite automorphisms. There is furthermore supposed to be a map from the moduli of prestable curves to the moduli of stable curves, and therefore we need a nice description of the stabilization map which works well in flat families. One possible such description is supposed to be: Take the dualizing sheaf omegaC of the prestable curve C; then take L := omegaC(x1 + ... + xn), where the xi are the marked points; then some large power of L will be generated by global sections, and the stabilization of C is the image of C under the corresponding map to projective space. In particular, if C' is an unstable component of C, then L restricted to C' should be OC', as the image of C' should be a point. I have several dumb questions: Why is omegaC an invertible sheaf? Does it follow from, e.g., Hartshorne III.7.11? Why is some large power of L generated by global sections? How do we know that this power can be chosen to be constant in families? How do we see that L restricted to unstable C' is OC'? My more general question is: Given a singular curve, or at least a prestable curve, what explicit information can we deduce about the dualizing sheaf? For example, is there a way to figure out what the dualizing sheaf looks like when restricted to an irreducible component, or at least a smooth irreducible component?

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Kind of late to the party, but the (weak) contractibility follows from [UNK] for [UNK].

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What is Koszul duality? 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?

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Is every functor a composition of adjoint functors? My understanding of Ben's answer to this question 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?

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Yes. Better, it works for T1, too: T1 is the axiom that one-point sets are closed. Then since the set is finite, the complement of any point is also closed; the point is open. That's the discrete topology.

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The commenters on the wiki http://scratchpad.wikia.com/wiki/091006qa are very knowledgeable; I'm just expanding on Anonymous's answer. The exponential map is not surjective. If you look at Halmos's paper http://www.ams.org/mathscinet-getitem?mr=53391 (link requires academic access) you will see a wide variety of invertible maps that are not exponentials. Here is the simplest example. Take 0 < u < v and let D be the annulus u < |z| < v in the complex plane. Our Hilbert space will be the space of analytic functions f on D such that \integral |f(z)|^2 is finite. The operator is multiplication by z. To sketch Halmos' argument, H is complete because the property of being harmonic can be stated as a condition on integrals and thus passes through an L^2 limit. The logarithm of multiplication by z wants to be multiplication by \log z, but we can't define \log z on D without introducing a branch cut. Of course, D doesn't have to be this exact shape, any open region of the complex plane which has a function without a logarithm would work.

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Why are torsion points dense in an abelian variety? Hi everyone, let [UNK] be an abelian variety of dimension [UNK] over an algebraically closed field [UNK] of characteristic [UNK]. I'm trying to prove that the subgroup [UNK] which is the union of all torsion points [UNK] of order prime to [UNK] is Zariski dense in [UNK]. The statement would follow if the Zariski closure [UNK] (which by construction is a group variety) of [UNK] in [UNK] would again be an abelian variety of dimension [UNK], because assuming [UNK], the [UNK]-primary part of [UNK] would still be [UNK], while it SHOULD be of rank [UNK], contradiction. However, I fail to see why [UNK] should be irreducible. Does anyone see a way to salvage the argument, or a different, (simpler) argument?

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Let [UNK] be the connected component of the identity in [UNK]. Then [UNK] is a projective group variety, hence an abelian variety; let it have dimension [UNK]. Let [UNK], a finite group. Then the number of [UNK]-torsion points of [UNK] is at most [UNK]. For large [UNK], this is less than [UNK] if [UNK] is not [UNK].

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A permutation group is called regular if it is transitive and all stabilizers are trivial. Left multiplication yields a regular embedding of any group into its group of permutations (so the answer to your second question is "Yes").

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One application of the theorem of the cube is to study the map from an abelian variety A to its dual abelian variety; the map is defined in terms of line bundles and the key technical theorem one uses to prove anything (e.g. that the map to the dual is a homomorphism) is the theorem of the cube. See Mumford's Abelian Varieties book or Martin Olsson's notes from this summer's Hangzhou workshop.

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A reading list for topological quantum field theory? Can you suggest a reading list, or at least a few papers that you think would be useful, for a beginner in topological quantum field theory? I know what the curvature of a connection is, know basic algebraic topology, and have some basic background in quantum field theory. Perhaps others with different backgrounds will also be interested in a reading list on TQFTs, so feel free to ignore my background and suggest material at a variety of levels.

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Motivation for algebraic K-theory? 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.

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I've found the following articles useful in the past: Segal's notes: http://www.cgtp.duke.edu/ITP99/segal/ Atiyah's paper "Topological quantum field theories"

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There's a nice overview of many of the "higher categorical issues" in John Baez's paper Higher dimensional algebra and TQFT. It's also very friendly for beginners. For the details, from a perspective emphasising the 2- and 3-categories that make everything tick, you should read Bakalov-Kirillov Turaev's big blue book Dan Freed's paper are also very good. For a somewhat different perspective, try Kevin Walker's TQFT notes, on his webpage.

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If you considered varieties over Z instead of over C, you would have homomorphisms given by counting points over all the different finite fields.

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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.

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