User ben webster - MathOverflow

Answer by Ben Webster for Thinking about the quadratic dual graphically

2013-06-16T12:19:24Z

It seems to me like the concept you're searching for is duality of matroids. Your simplices exactly correspond to the dependent sets of the matroid given by your monomials, assuming I understand your definition correctly. Incidentally, it's topologically better to look at the sets of variables that don't fit into a relation together (the independent sets), since these form an abstract simplicial complex. Knowing just these sets, you can figure out which sets are not in relations for $R^\perp$ very easily: those whose complements contain an independent set of maximal size for $R$. Anyways, lots is known about duality for matroids, though I'm not sure what it's likely to tell you about quadratic duality.

Answer by Ben Webster for A question about the proof of Beilinson-Bernstein localisation

2013-05-01T17:47:46Z

This may not be the most efficient way to get to the result, but here's how I would think about it:

The universal enveloping algebra can be identified with right $G$-invariant differential operators on the group $G$, via the map sending an element of the Lie algebra to the corresponding left translation vector field (and vice versa with left and right switched). In particular, the center of $U(\mathfrak{g})$ is given by bi-invariant differential operators on the group, in two different ways.

If I want to understand how elements of the center act on functions that satisfy $f(gb)=\lambda(b)f(g)$, then I should write them as $z=h(z) + n_1m_1+ \cdots+n_km_k$ where $m_i\in U(\mathfrak{n})\mathfrak{n}$ which is possible by the PBW theorem and the fact that central elements have weight 0 (here $h\colon Z(\mathfrak{g})\to U(\mathfrak{h})$ is the Harish-Chandra homomorphism). Thus, $z\cdot f= d\lambda (h(z)) f$; here I'm using $d\lambda$ to distinguish between characters of the group and Lie algebra.

Ok, I'm basically done, but I cheated a little here, since here I was using the map of the center to bi-invariant operators for the right action, and you really want the one that comes from the left action, which a priori might be different. Let me be lazy, and note that we've now shown that the map of the center factors through some character, and that this is induced by some ring map $Z(\mathfrak{g}) \to U(\mathfrak{h})$ (maybe not the HC homomorphism). Thus, it suffices to check it at a Zariski dense set of points; this follows from Borel-Weil, since we know how the center acts on the sections of $\mathcal{L}^\lambda$.

What's the "best" proof of quadratic reciprocity?

2009-10-20T13:00:34Z

For my purposes, you may want to interpret "best" as "clearest and easiest to understand for undergrads in a first number theory course," but don't feel too constrained.

Answer by Ben Webster for Springer Action on Centre of Parabolic Category O (after Brundan)

2013-04-13T19:16:46Z

David's comment is correct. This is proven in Section 3 of this paper of Stroppel: http://arxiv.org/abs/math.RT/0608234

Answer by Ben Webster for Endomorphisms in Category O and Schubert Classes

2013-04-13T18:59:27Z

1) I believe there is such a description, though I think its pretty debatable whether it is likely to tell you very much about Schubert calculus.

Category $\mathcal{O}$ has a nice collection of objects called "tilting modules"; these are distinguished by have a Verma and dual Verma filtration (actually all of these are self-dual); the indecomposables are indexed by the elements of the Weyl group (look at the lowest elements whose associated Verma or dual Verma appears in the filtration). The self-dual projective $P(w_0)=T(e)$ is an example of a tilting module.

Furthermore, these all have graded lifts in the graded version of category $\mathcal{O}$; in particular, there's a way of grading the Hom spaces between these objects so that the endomorphisms of $P(w_0)$ become $H^*(G/B)$ with the homological grading. If you choose the gradings correctly, the Hom spaces $Hom(T(e),T(w))$ and $Hom(T(w),T(e))$ have lowest degree $\ell(w)$ and dimension 1 in that degree. I believe the Schubert class for $w$ is (up to scalar) the composition of elements from these Hom spaces in lowest degree.

I won't give a detailed proof, but the point is that from Soergel's work you can identify $Hom(T(e),T(w))\cong Ext^\bullet(\mathbb{C}_{G/B},\mathbf{IC}_{S_w})$ and $Hom(T(w),T(e))\cong Ext^\bullet(\mathbf{IC}_{S_w},\mathbb{C}_{G/B})$ with composition being Yoneda product. This shows that any map $T(e)$ to $T(e)$ that factors through $T(w)$ is a sum of Schubert classes for $S_{w'}$ with $w'>w$ (it also shows the claim I made about Hom spaces). Thus, $S_w$ is (up to scalar) the only such element of degree $\ell(w)$.

2) Yes, if you're willing to think about $\mathrm{End}(P(w_0))$ via its canonical isomorphism with the center of category $\mathcal{O}$. It's the induced action on the center of a categorical braid group action. See Section 3 of this paper of Stroppel: http://arxiv.org/abs/math.RT/0608234

Answer by Ben Webster for Khovanov-Rozansky homology and spectral sequences

2013-03-19T03:55:30Z

Yes. I think the existence of these sequences is "general knowledge" amongst knot homologists, though I'll be honest, I'm not sure where they are written down at the moment. Had you asked me 10 minutes ago, I would have said they were in the paper of Rasmussen you cite, but they seem to not actually be written down (though they are secretly there).

The point is just that you always have spectral sequences from the homology for "more special" potentials to "less special" ones. Since 0 is special as things get, you always have a spectral sequence from HOMFLY homology to the homology attached to any potential. To get the spectral sequence from N to M for $N>M$, just consider the potential built from $p(x)=x^N+a_1x^{N-1}+\cdots + a_{N-M}x^M$ (using Rasmussen's notation); if we consider $a_i$ as free variables, we get sl_M homology with some boring junk, and if we set $a_M=0$, we get sl_N homology. Since the latter is a specialization of the former we get a spectral sequence.

Answer by Ben Webster for BGG-like resolutions and translations

2013-02-28T21:28:13Z

One way to think about this picture is that the blocks (EDIT: infinitesimal blocks) of the full category $\mathcal{O}$ in this picture aren't just equivalent; they are equivalent in a way that preserves the labeling of simple modules by elements of the Weyl group. So, any module that has some nice characterization using these labels and categorical properties must be preserved. So, the projective cover $P_{w\xi}$ of a simple $L_{w\xi}$ must be sent to $P_{w\nu}$.

Similarly, the Verma module $M_{w\xi}$ must be sent to $M_{w\nu}$ since it is largest quotient of $P_{w\xi}$ where $L_{w'\xi}$ for $w'<w$ in Bruhat order is not a composition factor.

You can describe a parabolic Verma module in a similar way; being in parabolic category $\mathcal{O}$ is just a question about what your composition factors are; you can only have $L_{w\xi}$ for $w$ shortest in a left coset of $W_{I}$ and longest in a right coset for the stabilizer of $\xi$. We can characterize $N_{w\xi}$ as the maximal quotient of $M_{w\xi}$ which only has these composition factors, and thus it must be sent to $N_{w\nu}$.

Answer by Ben Webster for A question on Lusztig's `graph with automorphism' construction?

2013-02-25T01:52:25Z

I think what you're looking for is the observation that if you do have a non-trivial gcd, then you can replace $\Gamma$ and $a$ with another pair that give the same answer and have gcd 1.

After all, let $\ell$ be the quotient $\operatorname{lcm}(d_1,\dots, d_n)/\operatorname{gcd}(d_1,\dots, d_n)$, and consider the automorphism $a^\ell$; this generates an automorphism of order $\operatorname{gcd}$ which acts freely. If you consider the quotient $\Gamma/a^\ell$, with the automorphism induced by $a$, that will give you the same Cartan matrix (since both $d_i$ and $(\alpha_i,\alpha_j)$ get divided by $ \operatorname{gcd}$), and has gcd 1.

Answer by Ben Webster for Covering of Verma modules by translation of a dominant Verma module

2013-02-16T02:51:57Z

This is a consequence of Theorem 3.3 in Gelfand and Gelfand's Tensor products of finite and infinite dimensional representations of semisimple Lie algebras; the projective cover of $M_{\lambda}$ is obtained by translating $M_\chi$ by an indecomposable projective functor.

Here's my "by hand" proof, which I'll leave here, since I wrote it out before bothering to look up the reference: The important point here is that $E\otimes M_\chi$ has a Verma filtration where $M_{\chi+\nu}$ appears with the multiplicity of the weight $\nu$ in $E$. The highest weight is at the bottom of the filtration, the lowest at the top.

Consider the case when $\lambda$ is dominant; this follows by the usual argument that translation functors give equivalences between blocks of category $\mathcal{O}$; Let $E$ be a f.d. representation with $\lambda-\chi$ extremal. Thus, $M_\lambda$ appears in the filtration on $E\otimes M_\chi$, and no other Vermas in the same block appear, so it's a summand and thus a quotient.

Thus, it suffices to replace $\chi$ with an arbitrary dominant weight; in particular, we can assume that $\chi-\lambda$ is dominant. Now choose $E$ to be the representation with lowest weight $\lambda-\chi$. Then $M_\lambda$ is the quotient of $E\otimes M_\chi$ by the submodule generated by all vectors of weight $>\lambda$. Thus we're done.

Is there a good notion of "induction" for representations of 2-categories?

2010-12-14T06:50:54Z

One of the most important observations in the representation theory of algebras is that if one has a subalgebra $A\subset B$, then these is a functor $B\otimes_A -\colon A\operatorname{-pmod}\to B\operatorname{-pmod}$ called induction.

This is a special case of a notion for representations of a category. (The representations of a category are the functors from that category into vector spaces.) In this case, one replaces the category with its idempotent completion (which doesn't change representations, since vector spaces are idempotent complete), which we'll assume for now is compactly generated. In this case, the induction of a representation $V$ is $$\operatorname{Ind} V(Y)=V(X)\otimes_{\operatorname{End}_A(X)}\operatorname{Hom}_B(X,Y).$$

Now, I can just as well talk about representations of 2-categories, which are 2-functors to the 2-category of categories (with whatever additions you like; additive and enriched over a field would be good choices).

Has anyone written up the basics of induction in this situation?

I don't strictly need to state things this way for what I want to do, but it would be better to give credit if someone else has done it and to have things "work out of the box."

What is the insight of Quillen's proof that all projective modules over a polynomial ring are free?

2010-03-28T04:30:28Z

One of the more misleadingly difficult theorems in mathematics is that all finitely generated projective modules over a polynomial ring are free. It involves some of the most basic notions in commutative algebra, and really sounds as though it should be easy (the graded case, for example, is easy), but it's not. The question at least goes as far back as Serre's FAC, but it wasn't proved until 1976, by Quillen EDIT: and also independently by Suslin.

I decided that this is the sort of fact that I should know a rough outline of how to prove, but the paper was not very helpful. Usually when someone kills off a famous conjecture in 5 pages, it's because they've developed some fantastic new piece of machinery people didn't have before. And, indeed, Quillen is famous for inventing some fancy and wonderful machinery, and the paper is only 5 pages long, but as far as I can tell, none of that fancy machinery actually appears in the proof.

So, what was it that Quillen saw, that Serre missed?

Answer by Ben Webster for can an nonzero IC sheaf have zero hypercohomology?

2012-11-19T19:04:55Z

Yes. Consider any local system (EDIT: of rank 1) over a characteristic 0 field on $\mathbb{C}^*$ with non-trivial monodromy. This satisfies all of (1), (2), (3) and (4). There are lots of ways to check that this has trivial cohomology; for example, if the monodromy has finite order, it's a summand of the pushforward from the constant sheaf, which has the same cohomology as the constant sheaf. If you want a projective example, an elliptic curve works.

Answer by Ben Webster for Triviality of Associated Bundles

2012-11-02T02:00:43Z

It's a quite common to think that a principal bundle is the same thing as a monoidal functor $Rep_G\to Vect(M)$ (this is part of "Tannakian philosophy" of describing objects related to G using the category of representations). I'm not finding any good references on line, but perhaps someone else can suggest one.

There's no non-trivial representation that will give a trivial functor, since the tautological bundle $EG \to BG$ gives an equivalence of tensor categories between $Rep_G$ and $Vect(BG)$.

Answer by Ben Webster for Question on "publication List" for applying to post-doctoral jobs

2012-10-26T16:20:46Z

While usually I don't like answering questions like this on MO, there is actually an important fact here specific to the mathematics community which would probably be missed on academia.stackexchange, or another non-mathematical site.

The answer to your 2) is:

There is no logic behind asking for separate publication lists for postdoctoral candidates. Job ads only ask for them because it is a default setting on MathJobs, and nobody bothers to uncheck the box. You can just copy whatever list of papers you put in your CV (and leave the list of papers in your CV) or you can add abstracts.

The question 1) is a bit more serious, and maybe better for academia.stackexchange. However, I would say that putting a paper on the arXiv is a serious signal that you think it is complete. To me, it means something.

Answer by Ben Webster for Good even grading and principal Levi type

2012-10-11T11:44:50Z

This was wrong. If you really want to read a wrong answer, look in the edit history.

Answer by Ben Webster for Kazhdan-Lusztig C-basis and categorification

2012-09-26T00:52:08Z

If you identify the Hecke algebra with the Grothendieck principal block of a graded lift of category $\mathcal{O}$ for the corresponding semi-simple Lie algebra (oddly, enough which of the two Langlands dual options you pick doesn't matter) such that $T_y$ is the class of a Verma module with highest weight $-y\rho-\rho$ (so $T_1$ corresponds to the anti-dominant Verma), then the $C'$-basis will match with the classes of tilting modules, and the $C$ basis with the simple modules (I tend to prefer matching the $C'$-basis with projectives, but then the $C$-basis is something horrible).

I'm not really sure where this is written down properly; philosophically, the point is that everything will be fixed in place once you decide what functor to send the bar involution to. If you send it the contragredient dual, then simples and tiltings are obvious sets of modules fixed by this duality, and they satisfy the triangularities you've written above.

Answer by Ben Webster for Brauer's permutation lemma -- extending to some other finite groups?

2010-04-13T13:49:34Z

There are many group actions on sets which are linearly equivalent but not equivalent as actions. In fact, every group other than the cyclic group has one. This follows from some easy linear algebra:

the number of irreducible reps over $\mathbb{Q}$ is the number of conjugacy classes of cyclic subgroups of $G$, (EDIT:there are at most this many since any two elements which generate conjugate cyclic groups have the same character in a rational representation; on the other hand, the characters of the inductions of the trivial from any set of cyclic groups, no two of which are conjugate, are linearly independent, so there are at least this many) and
the number of non-isomorphic transitive G-sets is the number of conjugacy classes of subgroups.

Thus, there must be an integer valued linear combination of transitive actions which has trivial character. Moving all the actions with negative coefficients to the other side of the equality, we get two different actions with the same character, and thus isomorphic representations.

I actually wrote a paper about this a few years back, which I think is a reasonable starting place for the subject, which actually has quite a long history, and a reasonably extensive literature.

How does an academic mathematician educate him/herself about job opportunities outside academia?

2010-05-05T01:16:45Z

One of the contradictions of being a math professor is that a big part of your job is to train people to do things which are quite different from what you do yourself professionally; this is especially true for undergrads, but to some measure also with grad students.

This is not particularly helpful when it comes to convincing students that they should major in mathematics; most mathematicians I know, myself included, are quite ignorant of what people will actually do if they get BAs in mathematics, and don't go to graduate school, or otherwise enter education, which is obviously something students will be very concerned about in these tough economic times.

Obviously, the correct response is to educate one's self on what the job opportunities are for people with BAs in math. I'm mostly interested in the US context, but would be happy to hear about other countries as well.

The best resource I know is the AMS Early Career Profiles page. This is a lot of links (some broken) to profiles of BA graduates in math that individual departments have put together. There some other reasonable links on this website. Is there anywhere else I should be looking?

Answer by Ben Webster for A remark in Jantzen's 'Lectures on Quantum Groups'

2012-09-06T15:10:37Z

Your argument in the second to last paragraph is wrong. The basis of the adjoint representation given by any basis vector in each root space, and the basis of $\mathfrak{h}$ given by the simple coroots $H_i=[E_i,F_i]$ has this property for any semi-simple Lie algebra.

The mistake you made was thinking that the basis had to be compatible with each $\mathfrak{sl}_2$ decomposition, and in particular that all but one basis vector in $\mathfrak{h}$ must be sent to zero by bracket with $E_i$ or $F_i$. This is not what Jantzen said (you're right that no basis with that property can exist except in products of $\mathfrak{sl}_2$'s); he only said that the bracket of one basis vector with $E_i$ or $F_i$ must be a multiple of another basis vector, and any basis of $\mathfrak{h}$ has that property. In fact, the only place where a basis vector from each root space and a completely arbitrary basis of $\mathfrak{h}$ will fail is that $[E_i,F_i]$ must be a multiple of a basis vector, which forces us to take the coroots.

Answer by Ben Webster for Question regarding a statement in `A proof of Jantzen conjectures'

2012-09-04T05:05:23Z

The problem is your statement: "The ordinary ( = non-mixed) $Ext^1$ group should be the extensions between the restrictions of $M_1$ and $M_2$ to the intersection $Y_1\cap Y_2$." This is absolutely not true in any generality I can think of. (EDIT: I should have said this doesn't work if you use *-restriction in both cases. The spectral sequence mentioned below that it does if you use *-restriction for one, and !-restriction for the other. The spectral sequence below applied to $X\supset Y_1\cup Y_2 \supset Y_1\cap Y_2$ shows this.)

The way one actually can calculate Ext groups using the geometry of the intersections is the spectral sequence given at the start of section 3.4 of Koszul duality patterns....

EDIT: Perhaps it's better to think of it this way: assume $Y_2\not\subset Y_1$ and let $j$ be the inclusion of $Y_2\setminus Y_1\cap Y_2$. Then any non-trivial extention $M_1 \to M \to M_2$ has a map $j_!j^!M_2\to M$ induced by the isomorphism $j^!M_2\cong j^!M$ which factors through the perverse truncation $H^p_0(j_!j^!M_2)$. As a map of perverse sheaves $H^p_0(j_!j^!M_2)\to M$ must be surjective since otherwise its image would split the exact sequence. Thus, $M_1$ must be a composition factor of $H^p_0(j_!j^!M_2)$ and so $Y_1\subset Y_2$.

MORE EDIT: If $M$ is a nontrivial extension of $M_1 \to M \to M_2$, it cannot have a subobject isomorphic to $M_2$. If it did, then then we would have an isomorphism $M\cong M_1\oplus M_2$ using the inclusion of $M_1$ we had before, and the inclusion of $M_2$ we just assumed existed.

Answer by Ben Webster for Module in category O not generated by a finite set of HWVs.

2012-08-28T22:03:23Z

This is false. Consider the contragradient dual on a Verma module $V_\lambda$. This is a module given by linear functions on $V_\lambda$ which kill all but finitely many weight spaces. This module has no highest weight vectors of weight other than $\lambda$; any non-zero vector $\xi$ of weight $< \lambda$ is non-zero on $F_i v$ for some $v$ (since $V_\lambda$ is generated in degree $\lambda$), and $E_i\xi\neq 0$ since it has non-zero value on $v$.

Thus, this module is generated by weight vectors if and only if it is generated by vectors of weight $\lambda$, but this is only possible if the Verma module is irreducible (otherwise, the dual of the simple quotient of the Verma is the proper submodule generated by the vectors of degree $\lambda$). Thus, this is a counter-example.

Answer by Ben Webster for Action of k* on a variety induces grading?

2012-08-15T11:10:49Z

Turning the action map of varieties into a map of rings, we get a ring map $\phi$ from $k[V]$ to $k[V][t,t^{-1}] $, the coordinate ring with an extra invertible variable (the coordinate on $k^*$) adjoined. Now, for any function $\phi(f)=\sum_{i\in \mathbb{Z}}f_it^i$ for some $f_i$'s, almost all of which are 0. Note that $f=\sum f_i$, which we obtain by restricting the function to $t=1$. Using associativity, applying $\phi$ again to the $f_i$'s is the same as applying pull-back by the multiplication map to t. Thus, as functions on $V\times k^*\times k^*$ (letting $t,u$ be the two coordinates)

$$\sum_{i\in \mathbb{Z}}\phi(f_i)u^i=\sum_{i\in \mathbb{Z}} f_i t^i u^i$$

since the pull-back of the coordinate by multiplication is just the product of the coordinates . Thus, $\phi(f_i)=f_it^i$.

We can define the grading by letting $f$ be homogeneous of degree $i$ if $\phi(f)=ft^i$. We have already seen that every element can be written uniquely as a sum of such elements (the $f_i$'s), and this is multiplicative since $\phi$ is a ring homomorphism.

Alternatively, we can note that we have proven that the span of the $f_i$'s is an finite-dimensional invariant subspace containing $f$, so we can apply your argument. In general, essentially the same argument shows that the action of any affine algebraic group on the coordinate ring of any affine variety by pull-back is a locally finite action: any function is contained in a finite-dimensional invariant subspace.

Answer by Ben Webster for About the intrinsic definition of the Weyl group of complex semisimple Lie algebras

2012-08-02T11:30:47Z

Let $g_r$ be the set of regular semi-simple elements of the Lie algebra, and $\tilde g_r$ be the set of these elements with a choice of Borel containing it. The Weyl group is the group of deck transformations of the cover $\tilde {g}_r\to {g}_r$.

Answer by Ben Webster for Points on Deligne-Lusztig varieties: Interpreting Borels in relative position as flags with conditions

2012-07-10T05:51:10Z

It might help if you specified what part of the translation your having trouble with (for example, it's unclear from what you wrote if you're comfortable with relative position).

For $SL_n$, a Borel corresponds to a flag in affine $n$-space over an extension of your field (essentially, the one you need to diagonalize the elements of a torus in the Borel). Applying the Frobenius means doing it to the coordinates on $n$ space, and looking at the resulting flag. Relative position w just means that the dimensions of the intersections of the pieces of the flag are the same as those between the standard flag, and that gotten by per muting the basis by $w$.

Is it written anywhere that open subvarieties of affine spaces have "completely impure" cohomology?

2012-05-28T21:39:19Z

Consider complex affine space $\mathbb{C}^n$ and let $U$ be a Zariski open subset of $\mathbb{C}^n$. By a celebrated result of Deligne, the cohomology $H^i(U)$ has a canonical Hodge structure. In particular, $H^i(U)$ has a weight filtration and a subalgebra of pure classes (since a cohomology class can't have lower weight than expected, only higher). I believe it's true that

The pure subalgebra of $H^i(U)$ is exactly the identity.

This is as far from being pure as possible.

What I hope to get from the collective intelligence of the internet is somewhere where this fact is written. I want to emphasize that what I am really hoping to get is a reference, since (as you can see below) I basically know how the proof should go.

In hopes of getting either confirmation or a mistake pointed out, let me write a proof:

By Alexander duality $\tilde H^i(U)\cong H_{n-i-1}^{BM}(X)$ where $X=\mathbb{C}^n\setminus U$. This is an isomorphism of Hodge structures after Tate twist by $n$. The weights of $H_{n-i-1}^{BM}(X)$ lie in $[-n+i+1,0]$, so those of $\tilde H^i(U)$ lie in $[i+1,n]$.

As a second-best request, does anyone know of a reference for the version of Alexander written above? It's dual to way things are usually written.

Answer by Ben Webster for Do fixed point sets in equivariant crepant resolutions have the same cohomology? How about for the specific case of Nakajima quiver varieties?

2012-05-17T03:29:34Z

For a Nakajima quiver variety satisfying Kirwan surjectivity on $H^2$ (which may well be all of them; I don't know a counterexample), all crepant resolutions are diffeomorphic, since they are quiver varieties themselves (you can check that you get every class in the Picard group from GIT reductions). If the $\mathbb{C}^*$ of interest to you comes from a linear action of the cotangent bundle you reduced to get the quiver variety, then you can make the diffeomorphism $S^1$-equivariant, and thus induce a diffeomorphism on fixed point sets (look at the proof of 3.4 in Proudfoot's thesis; that's for the abelian case, but the proof is the same.

Is the centralizer of a torus in a Kac-Moody algebra always a Borcherds algebra?

2012-04-18T15:43:48Z

If one has a finite dimension simple Lie algebra, one can easily calculate that taking the centralizer of a torus (or toral subalgebra), that is, summing the weight spaces that lie in some proper subspace of the dual Cartan, always gives a finite dimensional reductive Lie algebra; actually almost semi-simple, except that there is some central toral subalgebra (maybe bigger than the original torus).

For an affine Lie algebra, the picture is the same if you pick a subspace on which $(-,-)$ is positive definite, you get again something finite dimensional and reductive. If you pick a subspace containing $\delta$, however, things are a bit messier. Now you have an affine Lie algebra plus a central piece in every single one of the imaginary weight spaces; thus you have an infinite dimensional center. I would kind of like to think of this as a Borcherds algebra, where I add infinitely many rows and columns of zeros to the affine Cartan matrix.

For a hyperbolic Kac-Moody algebra, the picture is even worse; I still get a hyperbolic KM algebra attached to the root subsystem living in the subspace, but the "extra stuff" in imaginary weight spaces is much more complicated and not central anymore. This seems very complicated, but I hope to get some kind of handle on it:

Is it true that the centralizer of torus in a hyperbolic Kac-Moody algebra is a Borcherds algebra (necessarily infinite rank in most interesting cases)? Is there some nice description of its Cartan matrix?

Answer by Ben Webster for Tautological and normal bundles over flag manifolds and jet bundles

2012-04-18T13:03:27Z

The short answer is "yes." I'm not sure about the jet bundle stuff (not really my area), but everything else you've written is extremely well-known stuff in Lie theory. I've used them dozens of times in my own work. The quotient bundles you're calling "normal tautological bundle" i would just call "tautological bundle." In fact, it's well known enough that it's hard for me to give a reference, though you can try these notes of Brion or chapter 10 of Fulton's Young Tableaux (thought these are both for complex manifolds, not real ones; perhaps someone who knows the real theory better can help me with a reference). The line bundles on the flag variety indeed play an important role: they generate the Picard group of the variety and their Chern classes generate the cohomology, with all the relations in cohomology manifest from the fact that these filter a trivial bundle. They are also a special class of the line bundles that come up in discussions of Borel-Weil attached to weights of SL(n); these are ones attached to the weights which appear in the vector representation.

The notion of "flag bundles" you mention is also standard: you just quotient the frame bundle by upper-triangular matrices. This often comes up in discussion of the splitting principle.

Answer by Ben Webster for Quiver on tensor product

2012-04-18T00:10:03Z

There is a basis of the tensor product $A^{\otimes n}$ given by $\gamma_1\otimes \cdots \otimes \gamma_n$ where $\gamma_1,\dots, \gamma_n$ are a list of $n$ elements of $Q_1$ that are composable. That is, where concatenating $\gamma_1\cdots \gamma_n$ gives a path of length $n$. So, just as $A$ has a basis given by paths of length 1, $A^{\otimes n}$ has a basis given by paths of length $n$ in the original quiver.

I'm not sure what you mean by "a quiver on the tensor product" but you never use any quiver other than the original one.

Answer by Ben Webster for Different Lie group structures on a vector space with the same Lie algebra structure

2012-04-10T19:02:29Z

Sure. You'll always get isomorphic Lie groups, but they'll be different actual multiplication maps $\mu$. Just pick any Lie group diffeomorphic to $\mathbb{R}^n$ sending the identity to 0, and let $\phi:\mathbb{R}^n\to \mathbb{R}^n$ be any diffeomorphism whose Taylor series at 0 coincides with that of the identity, but which is not the identity, and let $\mu'=\phi^{-1}\circ \mu\circ \phi$. All examples will involve doing non-analytic funny business, since an analytic multiplication on $\mathbb{R}^n$ is determined from the Lie bracket by Campbell-Baker-Hausdorff.

EDIT: as Terry points out below, this isn't quite the right way to say things. However, no matter how you look at it, the issue is just one of picking strange coordinates on a Lie group where morally the multiplication really is recoverable from the Lie algebra structure.

Comment by Ben Webster

2013-05-09T12:05:50Z

Isn't this essentially the point of Noether's theorem?

Comment by Ben Webster

2013-05-08T02:48:56Z

Andre wants to consider the quotient of the universal enveloping algebra of Virasoro by the relation $k-c\cdot 1$ ($k$ is the central element of the Lie algebra, $c$ a scalar, and $1$ the identity in the UAE).

Comment by Ben Webster

2013-04-24T03:14:52Z

I'll just note, this question is not blatantly offensive; I just accidentally clicked the wrong reason to close. I can unilaterally reopen and recluse if people think the reason listed matters (it doesn't).

Comment by Ben Webster

2013-04-13T20:12:32Z

Mmm, not sure about that; I never really read that book. You should also be able to use the projective $P(w_0w)$ in place of the tilting $T(w)$; for some reason tilting modules felt better when I was writing the answer, but there's a functor that switches projectives and tiltings and induces the isomorphism $P(w_0)=T(e)$ so its essentially the same story.

Comment by Ben Webster

2013-04-13T19:14:27Z

I think the short answer is almost surely not. I don't think I have a truly killer argument that the other $T^*G/B$ is not a quiver variety, but it just feels all wrong. You would need to find a representation of a Lie group where the image of $U(\mathfrak{g})$ in endomorphisms of the representation was canonically isomorphic to $\mathbb{C}[W]$; I have no idea what that would be.

Comment by Ben Webster

2013-04-06T18:51:11Z

There's a construction called "multiproj" for rings with several $\mathbb N$ gradings. If you think of $S\otimes S$ as doubly graded and take multiproj, then you'll get the product. You can pass from a multiproj to a project by restricting to a generic ray (if you're lucky), which is what you did.

Comment by Ben Webster

2013-04-02T16:50:11Z

Last I knew, there were no rigorous mathematical constructions of categorified WRT invariants of 3 manifolds. Of course, lots of people would like to fix this, but at the moment, I don't think anyone knows, say, what replaces S-matrices.

Comment by Ben Webster

2013-03-30T14:31:49Z

Reza- This won't work outside $SU(2)$. In a compact Lie group of rank $>1$, you can find irrational geodesics in the torus.

Comment by Ben Webster

2013-03-01T23:37:48Z

I'm not really sure what question you're asking (perhaps the issue is alluded to in your handle). Geometric representation theorists have certainly noticed that D-modules are quantized coherent sheaves on the cotangent bundle and exploited this fact. So what about it?

Comment by Ben Webster

2013-02-24T20:55:58Z

Well if you just want a rough sense of which mathematics programs are well regarded USNWR is ok (grad-schools.usnews.rankingsandreviews.com/…). I wouldn't put too much stock in the difference between 30 and 40 on that list, but it will help you catch, for example, schools that are well regarded in general, but with a less prestigious math program.

Comment by Ben Webster

2013-02-17T15:32:13Z

Yemon- I guess I don't see what contradiction you think is there. Would you think it was strange to say "I'm perfectly happy with my house, but I sometimes wonder if I can find a nicer one"? On the other hand, I don't now what an answer to this question could really look like; the sort of places people at top 20 schools might think of as trading up (Harvard, MIT, IAS) do make tenured hires, though obviously not so many in the grand scheme of things.

Comment by Ben Webster

2013-02-16T16:28:01Z

Jim- Do you think there's something wrong with the direct proof given in my answer? The projective functors used there are translation functors. I also think the OP is just being sloppy about the distinction between translation and projective functors. I think a lot of people say "translation functors" nowadays when they really mean "projective functors."

Comment by Ben Webster

2013-02-16T16:20:44Z

Sorry about the typo (stupid auto-complete). Anyways, this is too complicated a story to explain in a comment; the elements $k$ behave like they are $q^{H}$ for $H$ in the Lie algebra. There are various answers to what the hell that really means, but you can just calculate without worrying, and everything will be fine.

Comment by Ben Webster

2013-02-16T14:55:50Z

These are the same Hopf algebra structure, written in terms of different generators. The $k$'s in the quantum group are exponentially of elements of the Lie algebra.

Comment by Ben Webster

2013-02-16T14:26:03Z

Yes, the description is in Aakumadula's comment above.