User harold williams - MathOverflow

How do I know the derived category is NOT abelian?

2010-02-18T02:27:19Z

I have heard the claim that the derived category of an abelian category is in general additive but not abelian. If this is true there should be some toy example of a (co)kernel that should be there but isn't, or something to that effect (for that matter, I could ask the same question just about the homotopy category).

Unless I'm mistaken, the derived category of a semisimple category is just a ℤ-graded version of the original category, which should still be abelian. So even though I have no reason to doubt that this is a really special case, it would still be nice to have an illustrative counterexample for, say, abelian groups.

Conformal Welding Reference

2011-02-17T05:26:47Z

I'm looking for a reference for the following fact: given two Riemann surfaces and an identification of their boundaries, once I topologically glue the surfaces together there exists a unique conformal structure on my new surface that is compatible with the conformal structures I started with.

Non-Lie Subgroups

2009-10-28T22:41:33Z

A result of Borel and Lichnerowicz states that the holonomy group of a connection on a principal G-bundle is a Lie subgroup of G (Cartan had earlier asserted this, but apparently without proof).
This restriction, that it be a Lie subgroup, allows for a lot of poorly-behaved subgroups, for example a line with irrational slope on a torus. This subgroup comes from a perfectly fine immersion of the Lie group ℝ, but it's not closed in the induced topology of the torus.

As an example of something that's not a Lie subgroup, let G= ℝ, consider an uncountable set of ℚ-independent points, none of which are rational, and consider the subgroup they generate. If this were a Lie subgroup it would be the image of an uncountable discrete space (there can't be anything 1-dimensional, since we left out the rationals), which wouldn't be second countable, hence not a manifold and not a Lie group.

This seems like a pretty contrived example, and I suspect there is more content to 'being a Lie subgroup' than having countably many components. However, I can't seem to pin down something that would illustrate this. Can anyone give me an example of a connected subgroup of a Lie group that is not a Lie subgroup?

Poisson Ind-Varieties

2011-11-08T00:50:12Z

I am looking for any places in the literature where an author has had occasion to consider Poisson structures on infinite-dimensional algebro-geometric objects, e.g. ind-varieties or proalgebraic groups (or more specifically Kac-Moody groups). There are various analytic instances I know of where one can find a discussion of Poisson structures on infinite-dimensional manifolds modeled on topological vector spaces, but I know of no such algebraic discussions and would be interested to hear if someone knows of circumstances where such things have come up.

koszul duality and algebras over operads

2010-09-22T02:41:49Z

Given a pair of Koszul dual algebras, say $S^*(V)$ and $\bigwedge^*(V^*)$ for some vector space $V$, one obtains a triangulated equivalence between their bounded derived categories of finitely-generated graded modules.

Given a pair of Koszul dual operads, say the Lie and commutative operads, what is the precise analogue of a derived equivalence between their categories of algebras?

Cohomology of the Moduli of G-bundles on a Curve

2010-11-30T04:20:24Z

For a simple complex group G and Riemann surface X, are the (integral, if possible) cohomology groups of the moduli of holomorphic G-bundles on X written down somewhere, either explicitly or implicitly? If not, are there some specific cases where these groups have been calculated?

Galois Groups vs. Fundamental Groups

2009-10-15T02:22:19Z

In a recent blog post 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?

Chain Complexes and Linear Infinity-Categories

2010-09-27T02:53:46Z

A statement I heard recently is that "chain complexes are the same thing as strict linear $\infty$-categories". Can someone explain how to see this?

Duals of Abelian Categories

2010-09-23T08:13:56Z

The dual of an abelian category is again abelian, since the axioms are all preserved by the reversing of arrows. For example, the category of finite-dimensional vector spaces over a field is easily seen to be dual to itself, since we can just take linear duals of vector spaces. However, this is the only example where I know a 'concrete' description of both an abelian category and its dual, where by concrete I mean describing the category as a variant of, say, modules over a ring or sheaves on some space.

What are other examples of dual pairs of abelian categories which can both be described 'nicely'? For example, is there a concrete description of the dual of the category of abelian groups?

K-Theory and the Stack of Vector Bundles

2010-04-21T04:40:12Z

I have some understanding that vector bundles provide a basic, familiar example of what I should call a stack. Namely, consider the functor $Vect$ that assigns to a space $X$ the set of isomorphism classes of vector bundles on $X$. This functor isn't local, in the sense that the isoclass of a vector bundle isn't determined by its restriction to an open cover, but rather by gluing data on overlapping sets in a cover. Since for any space $Y$ a map $X \to Y$ is determined by what it does when restricted to a cover of $X$, this tells us there is no space $Y$ that represents the functor $Vect$ in this fashion. However, I can also consider $Vect$ as a stack, which assigns to $X$ the groupoid of vector bundles on $X$. This gadget is fancy enough to understand how vector bundles glue together, and so recovers the locality missing from our earlier functor.

In K-theory, we attach to a space $X$ a ring $K(X)$ whose underlying group is the the free abelian group on the set of isoclasses of vector bundles on $X$, mod short exact sequences. It turns out that one can describe $K(X)$ as the set of homotopy classes of maps from $X$ to $\mathbb{Z} \times BU(\infty)$.

At this point my meager understanding of K-theory seems to be contradict what I said in the first paragraph. The fact that $K(X)$ has a classifying space seems at odds with the observation that vector bundles aren't determined by their restrictions to open covers, whereas maps to another space are. Is something wrong with what I've said so far? If not, perhaps there isn't a contradiction because either 1) $K(X)$ isn't quite the set of isoclasses of vector bundles, but rather a group completion thereof, or 2) we're looking at homotopy classes of maps to $\mathbb{Z} \times BU(\infty)$, so what I said in the first paragraph doesn't apply?

Images and Monomorphisms of Schemes

2010-03-30T19:38:04Z

If $X$ is an object in an arbitrary category, there is a natural definition of a subobject of $X$ as a monomorphism into $X$ (or really an equivalence class of monomorphisms). If $X$ is a scheme, however, the term 'subscheme' is conventionally reserved only for locally closed immersions (as in EGA I.4.1.2). There are certainly many monomorphisms of schemes that in this sense aren't subschemes, for example the inclusion of Spec of a local ring such as $Spec K[x]_{(x)} \to Spec K[x]$ .

When we restrict 'subscheme' to mean 'locally closed immersion', defining images of schemes becomes problematic. A sensible definition, in any category, of the image of a morphism is the minimal subobject through which it factors. Using the above definition of subscheme, there are perfectly well-behaved examples of morphisms of schemes that don't have images in this sense. For example, consider the morphism $\mathbb A^2_K \to \mathbb A^2_K $ induced by the ring homomorphism $(x,y) \mapsto (x,xy)$ ; the set-theoretic image is the union of the origin and the complement of the $y$-axis, and there is no minimal locally closed set containing this.

There is, however, always a minimal closed immersion through which a given morphism factors, and so if one defines the scheme-theoretic image in this sense, it always exists. My question is that, if we let our notion of 'subscheme' include all monomorphisms, would the resulting notion of 'scheme-theoretic image' always exist? In other words, is there always a minimal monomorphism of schemes through which a given morphism factors? Say, in the above example? If I hand you a constructible subset of a scheme, can you only find a monomorphism onto that set if it's locally closed?

As a 'softer' question, can someone explain why we don't want to call general monomorphisms subschemes? In particular, suppose I have a morphism that is a submersion onto a locally-but-not-globally closed subscheme. It seems much more sensible to call that locally closed subscheme the image, rather than its global closure.

HOMFLY and homology; also superalgebras

2009-10-22T06:48:37Z

My understanding is that an analogy along the following lines is (roughly) true:

"The Alexander polynomial is to knot Floer homology is to gl(1|1)

as the Jones polynomial is to Khovanov homology is to sl(2)

as a-lot-of-other-specializations-of-HOMFLY are to Khovanov-Rozansky homology are to sl(n)."

1) To what extent is it possible to add another line that starts with the (unspecialized) HOMFLY polynomial? I think there is a triply-graded complex that I can put here (and that maybe this is what I should be calling Khovanov-Rozansky homology? or at least is also due to them?), but is there an analogous object to put in place of the Lie (super-)algebras appearing above?

2) Why is gl(1|1) here? That seems weird.

Answer by Harold Williams for Noncommutative smooth manifolds

2010-03-03T07:50:50Z

I believe the closest answer is in Connes' On the Spectral Characterization of Manifolds. The main theorem is that if a (commutative) spectral triple (A,H,D) satisfies a list of certain nice properties, then A is the algebra of smooth functions on a compact oriented smooth manifold. I'm not sure this really separates the smooth structure and metric data, but hopefully the reference is still useful.

Answer by Harold Williams for What does "quantization is not a functor" really mean?

2009-12-26T08:37:30Z

Here is one precise statement of how quantization is not a functor:

5) There is no functor from the classical category $\mathcal C$ of Poisson manifolds and Poisson maps to the quantum category $\mathcal Q$ of Hilbert spaces and unitary operators that is consistent with the cotangent bundle/$\frac12$-density relation (explained below).

The result is due to Van Hove, in "Sur le probleme des relations entre les transformations unitaires de la mecanique quantique et les transformations canoniques de la mecaniques classique." This is an old paper and I can't find a link for it, but the reference I found it in is Weinstein's "Lectures on Symplectic Manifolds."

By "cotangent bundle/$\frac12$-density relation" I mean the following: if $\mathcal M$ is the category of smooth manifolds and diffeomorphisms, we have a cotangent functor $\mathcal M \to \mathcal C$. This assigns to each manifold its cotangent bundle with the canonical symplectic structure, and to each diffeomorphism the induced symplectomorphism of cotangent bundles.

We also have a natural functor $\mathcal M \to \mathcal Q$. For any smooth manifold $X$ consider the bundle of complex $\frac12$-densities on $X$. (What is the bundle of complex $s$-densities? Well, the fiber over a point $x \in X$ is the set of functions $\delta_x: \bigwedge^{top} T_xX \to \mathbb{C}$ such that $\delta(cv) = |c|^{s}\delta(v)$.) If $\delta^1$ and $\delta^2$ are smooth compactly-supported $\frac12$-densities, their pointwise product $\delta^1 \bar{\delta^2}$ is a compactly supported 1-density which we can integrate to get a complex number. This turns the space of all such sections into a pre-Hilbert space, the completion of which is what our functor assigns to the manifold $X$. As we would hope for, the canonical nature of the construction lets us assign unitary operators between Hilbert spaces to diffeomorphisms between smooth manifolds, hence is functorial.

(Note: If we choose a volume form on $X$, the above procedure produces something isomorphic with the space of $L^2$ functions on $X$ with respect to this form, but to get something functorial we want a canonical construction.)

From this pair of functors $\mathcal M \to \mathcal C$ and $\mathcal M \to \mathcal Q$ we get a product functor $\mathcal M \to \mathcal{C} \times \mathcal{Q}$. The image of this functor is a subcategory of $\mathcal C \times \mathcal Q$ which we will call the "cotangent bundle/$\frac12$-density relation." (The word relation is meant in the same sense that an ordinary relation between two sets is a subset of their product).

Now we can clarify just what is meant by our original statement: there is no functor $\mathcal C \to \mathcal Q$ whose graph contains the cotangent bundle/$\frac12$-density relation. The reasons why this is a desirable condition come from physics and are beyond me, but roughly speaking I think the point is that there exists a good idea of what a quantization functor is supposed to do to cotangent bundles.

Answer by Harold Williams for Do orbits and stable loci of group actions have natural scheme structures?

2009-10-28T21:00:05Z

For (1), I think the moral of your remark is that the orbit is the scheme-theoretic image of G×∗→X. If you insist on viewing the 'actual' image as the orbit, well then your definition of what an 'orbit' is will only feel as natural as your definition of what the 'actual' image is.

For example, suppose we want to say something about 'actual' orbits such as 'this orbit is contained in the closure that orbit.' If I take orbit to mean scheme-theoretic image, then I can restate this as 'this orbit is contained in that orbit', which sounds more natural anyway.

Computations in Knot Homology Theories

2009-10-27T21:33:49Z

1) Relative to one another, how computable are the various knot homology theories? For example, how many crossings can we allow a knot and still hope to compute its Khovanov homology, versus its knot Floer or Khovanov-Rozansky homologies? (The latter two seem to be generally unlisted at the Atlas; KF at least has a page about how it can be computed, but KR seems totally absent. If I'm just not seeing the right link, feel free to let me know)

2) Do the algorithms by which these invariants are computed share common features, or are they really very specific to the particular homology being computed?

For example, people computing KF homology draw square pictures that look very different from the pictures drawn by people doing Khovanov homology. On a less superficial level, KF algorithms (as far as I can tell) appear to be 'global' in an essential way, while Khovanov calculations can be done locally, breaking a knot into smaller pieces and then working with the pieces. So I'm led to believe the computations involved are different in a really fundamental way, but would be interested to see some combinatorial connection (as opposed to a topological relationship). I have no idea how KR homology is computed, so have no idea how closely related the computations involved are to, say, ordinary Khovanov homology.

Answer by Harold Williams for Reading list for basic differential geometry?

2009-10-20T01:24:09Z

"Lectures on Differential Geometry" by Chern, Chen, and Lam is an excellent book, and one which truly addresses differential geometry rather than differential topology alone.

Bad Categorical Quotients

2009-10-16T21:04:32Z

Let G be an algebraic group acting on a scheme X. Then f: X --> Y is a categorical quotient if it is constant on G-orbits and any other G-invariant morphism factors through it in a unique fashion. We say f is a 'good' categorical quotient if:

1) f is a surjective open submersion (i.e. the topology on Y is induced from X).

2) for any open U ⊂ Y, the induced map from functions on U to G-invariant functions on f^-1(U) is an isomorphism.

Does anyone know an example of a 'bad' categorical quotient (by which I mean...well...a not good one).

Comment by Harold Williams

2010-09-26T01:25:32Z

thanks for the great answer!

Comment by Harold Williams

2010-05-23T00:44:08Z

Perhaps the last remark may be incorrect? Jim's list above claims, for example, that the Coxeter number for B_n is 2n but the dual Coxeter number for C_n is n+1.

Comment by Harold Williams

2010-05-15T00:02:12Z

Wow, thanks for the long and thoughtful answer!

Comment by Harold Williams

2010-04-21T15:56:59Z

Theo: my intent was for the phrase 'mod short exact sequences' to be a good enough approximation to a precise definition for purposes of the question :)

Comment by Harold Williams

2010-04-21T06:30:59Z

Kevin: Ahhh I somehow didnt see your post before I commented above, but this link is the perfect answer to my question - thanks!

Comment by Harold Williams

2010-04-21T06:26:13Z

Thanks for the comments! I see now that the function Andy denoted by 'f' in general isn't injective once we mod out by homotopy, so the fact that the functor I called Vect doesn't have this property isn't an obstruction to representability.

Comment by Harold Williams

2009-10-22T20:33:30Z

@ paragraph 1: of course, the notion of 'in a category C, objects have underlying sets' means I have a certain functor F: C --> Set, and the notion of surjectivity of a morhpism f in C is then whether or not F(f) is a categorical epimorphism in Set.

Comment by Harold Williams

2009-10-22T20:18:49Z

so $#8484; is html for Z (i.e. the integers), but apparently comments don't like html, and apparently I cant edit comments. boo.

Comment by Harold Williams

2009-10-22T20:15:47Z

sl(1) is smaller than sl(2) :) That being said you're correct in the following sense: there is really a correspondence here between ℤ and a bunch of algebras, where n>0 ~ sl(n), n<0 ~ sl(-n), and n=0 ~ gl(1|1). So even though sl(1) is kind of a silly think to think about, sl(0) is a really silly thing to think about, and gl(1|1) steps in and prevents us from having to be so silly. Now a) this is really cool, and b) I can rephrase (part of) my question as "why is gl(1|1) the correct substitute for sl(0)?"

Comment by Harold Williams

2009-10-16T22:52:57Z

Correct; it's sub- not im-