User alexander braverman - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T19:51:04Z http://mathoverflow.net/feeds/user/3891 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120477/irreducibility-of-fibers-vs-irreducibility-of-fibered-product Irreducibility of fibers vs. irreducibility of fibered product Alexander Braverman 2013-02-01T00:00:59Z 2013-02-01T02:06:15Z <p>Let $f:X\to Y$ be a morphism of algebraic varieties over an algebraically closed field $k$ (I am ready to assume that $f$ is a smooth morphism, but that should not be necessary). I want to check that the generic fiber of $f$ is irreducible.</p> <p>$\mathbf{Question:}$ Assume that the fibered product $X\underset{Y}\times X$ is irreducible. Is it true that the generic fiber of $f$ is irreducible?</p> http://mathoverflow.net/questions/109260/good-even-grading-and-principal-levi-type Good even grading and principal Levi type Alexander Braverman 2012-10-09T23:46:23Z 2012-12-13T15:30:17Z <p>Let $\mathfrak g$ be a simple Lie algebra over $\mathbb C$ and let $e$ be a nilpotent element in it. In the theory of finite W-algebras one often encounters the following two conditions:</p> <p>1) $e$ is principal in some Levi subalgebra $\mathfrak l$ of $\mathfrak g$.</p> <p>2) There exists a good even grading on $\mathfrak g$ (recall that a grading on $\mathfrak g$ is called good for e if $e \in \mathfrak g_2$ and the linear map $ad~ e : \mathfrak g_j → \mathfrak g_{j+2}$ is injective for $j \leq −1$ and surjective for $j\geq −1$).</p> <p>My question is this: is there any relation between these conditions, or are they completely independent?</p> http://mathoverflow.net/questions/9154/how-to-think-about-parabolic-induction/115970#115970 Answer by Alexander Braverman for How to think about parabolic induction. Alexander Braverman 2012-12-10T12:25:03Z 2012-12-10T13:41:15Z <p>Usually (for $p$-adic groups, real groups or automorphic forms) you can write an explicit "intertwining operator" between the two inductions. This is a very basic construction in representation theory - as basis as parabolic induction itself in some sense. Usually it is given by some sort of integral which converges only in some range and one needs to work in order to prove its meromorphic continuation (it does have both zeros and poles; because of this parabolic induction is only generically independent of the choice of representation of $L$).</p> http://mathoverflow.net/questions/14612/cartan-involution-for-finite-w-algebras Cartan involution for finite W-algebras Alexander Braverman 2010-02-08T06:46:56Z 2012-07-08T20:23:02Z <p>Does anybody know if there is an analog of the Cartan (anti)involution for W-algebra associated to a nilpotent element e, which is principal in some Levi subalgebra of semi-simple Lie algebra g? Actually, I am more interested whether there exists an analog of the Shapovalov form on a Verma module for such a W-algebra.</p> http://mathoverflow.net/questions/95603/generalization-of-macdonald-polynomials Generalization of Macdonald polynomials? Alexander Braverman 2012-04-30T19:00:05Z 2012-04-30T19:00:05Z <p>Let $G$ be a semi-simple group with maximal torus $T$ and Weyl group $W$. It looks like from some geometric considerations I can define a family $P_{\lambda,\alpha}(q,t,z)$ of $W$-invariant polynomials of $z\in T$ which depend on two additional parameters $q$ and $t$. Here $\lambda$ is a dominant weight of $G$ and $\alpha$ is a positive element of the root lattice. Moreover, when $\alpha$ tends to $\infty$, $P_{\lambda,\alpha}$ tends to the corresponding Macdonald polynomial, but I don't know how to characterize $P_{\lambda,\alpha}$ for finite $\alpha$. </p> <p>Does anybody know if such polynomials exist in the literature?</p> http://mathoverflow.net/questions/58386/meromorphic-continuation-of-eisenstein-series Meromorphic continuation of Eisenstein series Alexander Braverman 2011-03-14T05:59:29Z 2012-04-15T15:41:27Z <p>I am interested what kind of (different) proofs of meromorphic continuation Eisenstein series (for general parabolic subgroups) exist in the literature? The only one I understand well, is Bernstein's proof using his "continuation principle" (it is probably unpublished) but apparently there are many others (especially I am interested in proofs which are different from the one in Langlands book)</p> http://mathoverflow.net/questions/88568/strong-kodaira-vanishing Strong Kodaira vanishing Alexander Braverman 2012-02-15T21:55:02Z 2012-02-16T18:03:40Z <p>Let $X$ be a smooth projective variety (say, over a field of characteristic zero). Let us say that strong Kodaira vanishing holds for $X$ if $$H^q(X,\Omega^p\otimes L)=0$$ for every $p\geq 0$, $q>0$ and an ample line bundle $L$ on $X$. </p> <p>My questions are now these:</p> <p>1) Does strong Kodaira vanishing hold for $X={\mathbb P}^N$?</p> <p>2) Does it hold for partial flag varieties of a semi-simple group $G$?</p> <p>3) What tools are there for proving that strong Kodaira vanishing holds for a given variety $X$? </p> http://mathoverflow.net/questions/88568/strong-kodaira-vanishing/88652#88652 Answer by Alexander Braverman for Strong Kodaira vanishing Alexander Braverman 2012-02-16T17:29:54Z 2012-02-16T17:29:54Z <p>It turns out that questions 1 and 2 are completely answered here <a href="http://arxiv.org/abs/alg-geom/9508009" rel="nofollow">http://arxiv.org/abs/alg-geom/9508009</a> (and some technique for 3 is there as well). In particular, the statement is true for ${\mathbb P}^N$ but not true for most flag varieties.</p> http://mathoverflow.net/questions/88567/decomposition-of-the-ring-of-functions-on-the-unipotent-radical-of-a-borel/88572#88572 Answer by Alexander Braverman for Decomposition of the ring of functions on the unipotent radical of a Borel Alexander Braverman 2012-02-15T22:42:00Z 2012-02-15T22:42:00Z <p>It is probably hard to give a complete description. A lot of partial information about this is contained in this paper of Kostant <a href="http://front.math.ucdavis.edu/1201.4494" rel="nofollow">http://front.math.ucdavis.edu/1201.4494</a> (in particular there is a description of the subring of invariant elements in $k[U]$; in fact Kostant works with the Lie algebra of $U$ instead of $U$ itself, but clearly they are isomorphic as $U$-varieties with respect to the adjoint action).</p> http://mathoverflow.net/questions/88142/bounding-the-size-of-stalks-of-ic-sheaves/88152#88152 Answer by Alexander Braverman for Bounding the size of stalks of IC sheaves Alexander Braverman 2012-02-10T22:11:20Z 2012-02-11T23:33:54Z <p>The answer to the first question is certainly "no": you can easily find example of X whose IC complex (for the trivial local system of rank 1) has (total) stalk of dimension $\geq 1$ (for instance this is true for most nilpotent orbit closures in a simple Lie algebra).</p> <p>Edit: Sorry I missed the assumption that X is smooth. But still I think the following is a counterexample. First, let $Y$ in ${\mathbb C}^n$ be a generic homogeneous hypersurface of degree d (it is smooth away from 0). Let $Z$be its projectivization. Then if I am not mistaken, the stalks of the IC sheaf of $Y$ at $0$ live in dimensions $-(n-1)$ and $-(n-2)$ and they are equal to $H^0(Z)$ and $H^1(Z)$ respectively (I am using perverse normalization). Now take $n=3$. Then $Z$ is a curve of degree $d$ in $\mathbb P^2$, so its genus is $g=\frac{(d-1)(d-2)}{2}$ and its $H^1$ has dimension $2g$. Now there is a finite map $\pi:Y\to \mathbb C^2$ of degree $d$ (take for example $Y$ to be given by the equation $x^d+y^d+z^d=0$ and consider the projection to $(x,y)$). Then $\pi_*$ of the constant sheaf is going to be equal to its Goresky-Macpherson extension from an open subset, where it will be equal to a local system $L$ of rank $d$. But the sum on the right hand side of your expression for $x=0$ is $1+(d-1)(d-2)$. </p> http://mathoverflow.net/questions/86853/what-is-the-relation-between-l-lafforgue-and-frenkel-gaitsgory-vilonen-results-o/87273#87273 Answer by Alexander Braverman for What is the relation between L. Lafforgue and Frenkel-Gaitsgory-Vilonen results on Langlands correspondence ? Alexander Braverman 2012-02-01T20:32:40Z 2012-02-01T20:32:40Z <p>Let me try to answer. [FGV] is only about unramified representations of the Galois group but they prove a stronger fact in this case (existence of certain "automorphic sheaf"). Lafforgue's result doesn't follow from there for several reasons:</p> <p>a) Formally [FGV] use Lafforgue, but this was actually taken care of by a later paper of Gaitsgory ("On the vanishing conjecture..."). So that is really not a problem now.</p> <p>b) Extending [FGV] to the ramified case is not trivial. I actually suspect that it can be done using the thesis of Jochen Heinloth but this has never been done (even the formulation is not completely clear in the ramified case)</p> <p>c) In the unramified case what follows immediately from [FGV] is that you can attach a cuspidal automorphic form to a Galois representation. It is not obvious to me that the converse statement follows (Lafforgue's argument actually goes in the opppsite direction: he proves that a cuspidal automorphic form corresponds to a Galois representation and then the converse statement follows immediately from the converse theorem of Piatetski-Shapiro et. al. and from the fact that you know everything about Galois L-functions in the functional field case).</p> http://mathoverflow.net/questions/87202/q-x-analog-of-n $(q,x)$-analog of $n!$ Alexander Braverman 2012-02-01T04:02:42Z 2012-02-01T07:48:23Z <p>While doing some work in geometric representation theory I have come across the following sequence of polynomials in two variables $(q,x)$ which I would like to denote by $n!_{q,x}$. For small $n$ these polynomials look as follows:</p> <p>$2!_{x,q}=x+q$</p> <p>$3!_{x,q}=x^3+x^2(q^2+q)+x(q^2+q)+q^3$</p> <p>$$4!_{x,q}=x^6+x^5(q^3+q^2+q)+x^4(q^4+q^3+2q^2+q)+x^3(q^5+q^4+2q^3+q^2+q)+$$ $$x^2(q^5+2q^4+q^3+q^2)+x(q^5+q^4+q^3)+q^6$$</p> <p>The polynomials are actually symmetric in $q$ and $x$ and when one puts $x=1$ one recovers the usual $q$-analog of $n!$ (in particular, when both $q$ and $x$ are 1, we get $n!$).</p> <p>My question is this: has anybody seen such polynomials before? What is the correct definition of those polynomials for general $n$? Any information will be greatly appreciated.</p> http://mathoverflow.net/questions/87191/reductive-groups-over-non-archimedean-local-fields/87196#87196 Answer by Alexander Braverman for Reductive groups over non archimedean local fields. Alexander Braverman 2012-02-01T03:09:09Z 2012-02-01T03:09:09Z <p>I think this is true for any affine variety $X$ over $F$: by Noether normalization lemma it can be represented as a finite cover of an affine space, for which the statement is clearly true (then take pre-image in $X$). </p> http://mathoverflow.net/questions/87171/tamagawa-number-for-functional-fields Tamagawa number for functional fields Alexander Braverman 2012-01-31T20:29:51Z 2012-02-01T02:41:41Z <p>Let $G$ be a split semi-simple simply connected group over a global field $F$ and let $\omega$ be a top-degree differential form on $G$ without zeroes (defined over $F$). It is well known that $\omega$ defines a measure on the adele group $G(\mathbb A)$. The Tamagawa number formula states (if I understand correctly) that</p> <p>1) If $F$ is a number field then the volume of $G(\mathbb A)/G(F)$ is 1</p> <p>2) If $F$ is a functional field isomorphic to $\mathbb F_q(X)$ where $X$ is a projective curve over $\mathbb F_q$ then the above volume is equal to $q^{(g-1)\dim G}$ where $g$ is the genus of $X$. </p> <p>My questions are the following:</p> <p>a) Do I understand the statements correctly?</p> <p>b) What is the reason why 1) and 2) look somewhat differently? Can one formulate the statement in a uniform way for all global fields?</p> <p>Edit: In fact my understanding was wrong. In 1) one needs to multiply by the volume of $(\mathbb A/F)^{\dim G}$ which is equal exactly to $q^{(g-1)\dim G}$ in the functional case. I was confused by the case $F=\mathbb Q$ where the above factor is 1.</p> http://mathoverflow.net/questions/87147/description-of-gl-3-u/87165#87165 Answer by Alexander Braverman for Description of $GL_3/U$ Alexander Braverman 2012-01-31T19:52:25Z 2012-01-31T19:52:25Z <p>Let $V$ be the basic (3-dimensional) representation of $GL(3)$. Then $SL(3)/U$ is the set of all pairs $x\in V, y\in V^*$ where $x$ and $y$ are non-zero and $(x,y)=0$. </p> <p>The quotient $GL(3)/U$ is non-canonically product of the above by $C^{\times}$. Canonically, you need to choose non-zero $x_i\in \Lambda^i(V)$ (for $i=1,2,3$) such that $x_i\wedge x_j=0$ for all $i$ and $j$ (note that if $x_3$ is fixed then $\Lambda^2(V)$ is canonically the same as $V^*$). This description generalizes immediately to any $GL(n)$ (and in fact to any $G$).</p> http://mathoverflow.net/questions/87044/normality-for-non-noetherian-schemes Normality for non-noetherian schemes Alexander Braverman 2012-01-30T17:22:04Z 2012-01-30T20:50:36Z <p>I am interested to know to what extent the notion of normality makes sense on a non-noetherian scheme.<br> Specifically, I can ask the following question: let $\pi:X\to Y$ be a formally smooth morphism of schemes. Assume that $Y$ is noetherian and normal. Let $U\subset Y$ be an open subset such that the complement has codimension $\geq 2$. Let $f$ be a regular function on $\pi^{-1}(U)$. </p> <p>$\mathbf{Question:}$ Is it true that $f$ extends to all of $X$?</p> http://mathoverflow.net/questions/86556/applications-of-artin-l-functions Applications of Artin L-functions Alexander Braverman 2012-01-24T18:16:14Z 2012-01-24T19:58:10Z <p>Does anybody know a good reference which gives examples of applications of Artin L-functions to "elementary" number theory? Many thanks!</p> http://mathoverflow.net/questions/86545/canonical-basis-via-gelfand-tsetlin-basis canonical basis via Gelfand-Tsetlin basis Alexander Braverman 2012-01-24T16:33:41Z 2012-01-24T16:33:41Z <p>Do there exist explicit formulas for the action of Lusztig's canonical basis of $U_q(\mathfrak n_+)$ in the Gelfand-Tsetlin basis of a Verma module for $sl(n)$?</p> http://mathoverflow.net/questions/82253/cohomology-vanishing-for-tensor-powers-of-tangent-bundle-on-the-flag-variety Cohomology vanishing for tensor powers of tangent bundle on the flag variety Alexander Braverman 2011-11-30T06:56:47Z 2011-11-30T19:46:25Z <p>Let $X$ denote the flag variety of a semi-simple group $G$ (in characteristic 0) and let $T_X$ denote its tangent bundle. I would like to ask the following question(s):</p> <p>1) Is it true that for any $n\geq 0$ we have $H^i(X,T_X^{\otimes n})=0$ for $i>0$?</p> <p>2) More generally, let $\lambda$ be a dominant weight of $G$ and let $\mathcal O(\lambda)$ be the corresponding line bundle on $X$. Is it true that $H^i(X, T_X^{\otimes n}\otimes \mathcal O(\lambda))=0$ for $i>0$?</p> <p>When tensor powers of $T_X$ are replaced by symmetric powers, this is known to be true (for example it is proved in a paper of Kumar, Lauritzen and Thomsen).</p> http://mathoverflow.net/questions/47315/extension-of-g-bundles extension of $G$-bundles Alexander Braverman 2010-11-25T09:33:45Z 2011-11-24T18:05:25Z <p>Let $S$ be a smooth surface (let's say over an algebraically closed field) and let $D$ be a smooth divisor in $S$. Let also $G$ be a connected algebraic group. Assume that we are given a principal $G$-bundle ${\mathcal F}$ on $S\backslash D$. Under what conditions can we extend it to all of $S$? Do I understand correctly, that this is always the case when the derived group $[G,G]$ is simply connected?</p> http://mathoverflow.net/questions/79624/equivariant-cohomology-of-nilpotent-orbits/79632#79632 Answer by Alexander Braverman for Equivariant cohomology of nilpotent orbits Alexander Braverman 2011-10-31T14:06:51Z 2011-10-31T14:11:52Z <p>First, since $\overline N$ is contractible its equivariant cohomology is the same as for $pt$. The Poincare pairing is uniquely determined by $\int_{\overline N} 1$ (since it is linear with respect to $H^*_G(pt)$). </p> <p>More precisely, any cohomology class of $\overline N$ has the form $\alpha\cdot 1$ where $\alpha$ is an equivariant cohomology class of $pt$ and $1$ denotes the unit cohomology class in $\overline N$ and we have $$\langle \alpha\cdot 1,\beta\cdot 1\rangle =\alpha\beta\int_{\overline N} 1.$$</p> <p>I don't know a good way to compute $\int_{\overline N} 1$ for arbitrary $N$ - other than replacing $\overline N$ by a resolution and using fixed point localization.</p> <p>By the way, if $\overline N$ is the minimal orbit, then ${\mathbb C}^2\times {\overline N}$ is the same as the Uhlenbeck space of $\mathbb C^2$ of second Chern class 1 - that should give you another way to compute that integral (is it obvious that you get the same answer?)</p> http://mathoverflow.net/questions/79397/reference-request-for-equivariant-cohomology-of-g/79413#79413 Answer by Alexander Braverman for Reference request for equivariant cohomology of G Alexander Braverman 2011-10-28T17:06:40Z 2011-10-28T17:06:40Z <p>This question has already been asked (and answered) here <a href="http://mathoverflow.net/questions/20671/what-is-the-equivariant-cohomology-of-a-group-acting-on-itself-by-conjugation" rel="nofollow">http://mathoverflow.net/questions/20671/what-is-the-equivariant-cohomology-of-a-group-acting-on-itself-by-conjugation</a></p> http://mathoverflow.net/questions/78999/continuous-cohomology-of-semi-simple-lie-group/79002#79002 Answer by Alexander Braverman for Continuous cohomology of semi-simple Lie group. Alexander Braverman 2011-10-24T17:41:36Z 2011-10-24T17:41:36Z <p>I think that it is true if you use "smooth" instead of continuous.</p> http://mathoverflow.net/questions/78717/cohomology-groups-of-homogeneous-spaces/78746#78746 Answer by Alexander Braverman for Cohomology groups of homogeneous spaces Alexander Braverman 2011-10-21T05:18:01Z 2011-10-21T05:36:02Z <p>I think that the best way to compute this cohomology is the following. The homogeneous spaces you are looking at are all of the form $X=G/M$ where $G$ is compact and $M$ is a connected subgroup of $G$ Let $T_M$ be a maximal torus of $M$, embedded into a maximal torus $T_G$ of $G$ and let $\mathfrak t_M$ $\mathfrak t_G$ be the corresponding (complexified) Lie algebras. Let us first look at the equivariant cohomology $H^*_G(X)$ (say, with $\mathbb C$-coefficients). It is obvious that it is the same as $H^*_M(pt)$ (here $pt$ denote "the point") which is known to be $Sym(\mathfrak t_M^*)^{W_M}$; here $Sym$ means "symmetric algebra", and $W_M$ means the Weyl group of $M$. By abstract nonsense it is clear that $H^*(X)$ </p> <p>is just $H^*_G(X)\underset{H^*_G(pt)}\otimes {\mathbb C}$, where $\otimes$ in principle means "derived tensor product". If $M$ and $G$ have the same rank, then you can show that $H^*_G(X)$ is always free over $H^*_G(pt)=Sym(\mathfrak t^*)^{W_G}$ (here I denote $\mathfrak t=\mathfrak t_M=\mathfrak t_G$) hence you finally get</p> <p>that $H^*(X)$ </p> <p>is equal to $Sym(\mathfrak t^*)^{W_M}\underset{Sym(\mathfrak t^*)^{W_G}}\otimes{\mathbb C}$.</p> <p>In the general case, you need to compute the above derived tensor product, which in every specific case is usually easy to do.</p> http://mathoverflow.net/questions/78132/is-there-an-analog-of-kodaira-vanishing-for-singular-varieties Is there an analog of Kodaira vanishing for singular varieties Alexander Braverman 2011-10-14T13:18:19Z 2011-10-14T15:50:21Z <p>I would like to know what kind of analogs of Kodaira vanishing theorem are valid for singular varieties. For example, is the following true: let $X$ be a projective Gorenstein variety and let $\omega_X$ be its canonical bundle. Is it true that $H^i(L\otimes \omega_X)=0$ for $i>0$ for an ample line bundle $L$?</p> http://mathoverflow.net/questions/77975/macdonald-polynomials-and-macdonald-positivity Macdonald polynomials and Macdonald positivity Alexander Braverman 2011-10-13T00:14:05Z 2011-10-13T01:49:07Z <p>I would like to have some order in my head about different version of Macdonald polynomials and positivity statements about them. I understand the following:</p> <p>1) There is a definition of Macdonald polynomials for any root system. These can be defined, for example as $W$-invariant polynomials on the torus $T$ of a semi-simple group $G$, which are orthogonal polynomials with respect to Macdonald scalar product and normalized in such a way that $$P_{\lambda}(q,t,x)=e^{\lambda}+\text{lower order terms}$$ where $\lambda$ is a dominant weight and $x\in T$.</p> <p>2) In type $A$ there is a notion of transformed Macdonald polynomials, which were extensively studied by Haiman. Haiman denotes them by $\tilde{H}_{\lambda}$ </p> <p>(here $\lambda$ is a partition, which can be thought of as a domonant weight of $GL(n)$); he proved the Macdonald positivity conjecture, which says that ${\tilde H}_{\lambda}(q,t,x)$ </p> <p>is a linear combination of Schur functions in $x$ whose coefficients are polynomials in $q$ and $t$ with non-negative integral coefficients. The definition of ${\widetilde H}_{\lambda}(q,t,x)$ appears for example on page 4 of <a href="http://math.berkeley.edu/~mhaiman/ftp/nfact/polygraph-jams.pdf" rel="nofollow">http://math.berkeley.edu/~mhaiman/ftp/nfact/polygraph-jams.pdf</a></p> <p>My questions are these:</p> <p>a) What is the relation between $P_{\lambda}$ and ${\tilde H}_{\lambda}$? It is not clear to me from the definition. </p> <p>b) Are there positivity statements for $P_{\lambda}$ itself? Or is there a version of the positivity conjecture for any root system?</p> http://mathoverflow.net/questions/77934/are-quivers-useful-outside-of-representation-theory/77946#77946 Answer by Alexander Braverman for Are quivers useful outside of Representation Theory? Alexander Braverman 2011-10-12T18:42:57Z 2011-10-12T23:50:09Z <p>As was mentioned above, many moduli spaces have a quiver description; one of the most famous example is given by Nakajima quiver varieties, which are defined for any quiver (and they serve as the main example of symplectic complex varieties which are resolutions of an affine variety), but when the quiver is the affine quiver of ADE type, they describe moduli spaces of torsion free sheaves on the quotients ${\mathbb C}^2/\Gamma$ where $\Gamma$ is a finite subgroup of $SL(2)$ (these are also known as ALE spaces). These moduli spaces are very important in many places in mathematics and physics (gauge theory) and quiver description is very useful when you want to tackle some explicit problems related to them.</p> http://mathoverflow.net/questions/77870/k-theory-and-k-theory-pushforward-in-topology-vs-in-algebraic-geometry/77879#77879 Answer by Alexander Braverman for K-theory and K-theory pushforward in topology vs. in algebraic geometry Alexander Braverman 2011-10-12T00:39:08Z 2011-10-12T00:39:08Z <p>Well, let me say something really straightforward. First, the map from $K^{alg}(X)$ to $K^{top}(X)$ is defined even if $X$ is not necessarily projective. Second, $f_!$ is defined only if $f$ is proper, and in that case I think it always agrees with the topological push-forward.</p> http://mathoverflow.net/questions/77187/is-the-ideal-of-a-closure-of-a-bruhat-cell-generated-by-generalized-minors/77208#77208 Answer by Alexander Braverman for Is the ideal of a closure of a Bruhat cell generated by generalized minors? Alexander Braverman 2011-10-05T06:25:37Z 2011-10-05T06:25:37Z <p>First of all, I think you need to write $w'\lambda&lt; w\lambda$ (look what happens when $w$ is 1).</p> <p>It seems to me that when $G$ is not $SL(n)$ the answer is no. For example assume that $w=1$. Then you know that the relations are generated by all matrix coefficients $\omega_{\eta,v}$ (your notations) where $\eta$ is a functional which vanishes on the lowest weight vector of $V$ and your generators correspond to $\eta$ being an extremal weight vector. But if the fundamental representations of $G$ are not minuscule I don't see how you get relations with $\eta$ not being an extremal weight vector in $V^*$ (for fundamental $V$) from those with extremal $\eta$ - this seems impossible for degree reasons (if you introduce the multigrading corresponding to $\lambda$).</p> http://mathoverflow.net/questions/77107/several-questions-on-semi-infinite-flag-manifold/77184#77184 Answer by Alexander Braverman for Several questions on semi infinite flag manifold Alexander Braverman 2011-10-04T23:18:39Z 2011-10-04T23:18:39Z <p>About defining the (ind)scheme structure: working with particular strata is basically never a good way to do this. What you need in order to define an algebro-geometric object is to define a functor from $Schemes$ to $Sets$ that it represents (it is enough to do it for affine schemes, i.e. it is enough to say what is an $R$-point of your space when $R$ is a ring). This is easy to do for semi-infinite flags. After you have done this, you can ask whether this functor is representable by a scheme or an ind-scheme (but I want to emphasize that this question doesn't make sense before you define the functor).</p> http://mathoverflow.net/questions/109260/good-even-grading-and-principal-levi-type/109293#109293 Comment by Alexander Braverman Alexander Braverman 2012-10-10T14:06:38Z 2012-10-10T14:06:38Z Thank you! I actually know most of the above, but what I wasn't able to figure out is this: how often does it happen that for $e$ which is regular in a Levi there is no good even grading? http://mathoverflow.net/questions/95603/generalization-of-macdonald-polynomials Comment by Alexander Braverman Alexander Braverman 2012-05-01T21:13:41Z 2012-05-01T21:13:41Z That's precisely what I don't know. In my situation I have a family of representations of $G\times {\mathbb C}^*\times{\mathbb C}^*$ depending on $\lambda$ and $\alpha$ such that their characters become Macdonald polynomials when $\alpha=\infty$. But otherwise I don't know anything about them. http://mathoverflow.net/questions/89582/what-can-one-say-about-a-smooth-variety-whose-lower-cohomology-is-trivial Comment by Alexander Braverman Alexander Braverman 2012-02-26T16:39:27Z 2012-02-26T16:39:27Z About &quot;classification&quot;: even if one assumes that the motive of $X$ is the same as the motive of an affine space, I don't think there is a reasonable classification. Also, there exist varieties whose motive is that of an affine space, but its Picard group is non-trivial. http://mathoverflow.net/questions/88991/exact-functor-and-representations-of-p-adic-groups Comment by Alexander Braverman Alexander Braverman 2012-02-20T04:26:33Z 2012-02-20T04:26:33Z If you only take the highest cohomology, then I don't see why it is a functor http://mathoverflow.net/questions/88567/decomposition-of-the-ring-of-functions-on-the-unipotent-radical-of-a-borel/88575#88575 Comment by Alexander Braverman Alexander Braverman 2012-02-16T19:41:16Z 2012-02-16T19:41:16Z Look at Kostant's paper that I mentioned above - he does exactly what you want for the $U$-invariants in $k[U]_C$. http://mathoverflow.net/questions/88568/strong-kodaira-vanishing/88654#88654 Comment by Alexander Braverman Alexander Braverman 2012-02-16T19:37:59Z 2012-02-16T19:37:59Z Thank you. There is actually a different argument in <a href="http://arxiv.org/abs/alg-geom/9508009" rel="nofollow">arxiv.org/abs/alg-geom/9508009</a> (using the lift of Frobenius mod $p^2$). http://mathoverflow.net/questions/88568/strong-kodaira-vanishing/88586#88586 Comment by Alexander Braverman Alexander Braverman 2012-02-16T17:14:43Z 2012-02-16T17:14:43Z Thank you. I wonder whether Frobenius splitting can be helpful here - it does prove everything for $q=0$ (and $q=\dim X$ is the usual Kodaira vanishing). http://mathoverflow.net/questions/88568/strong-kodaira-vanishing Comment by Alexander Braverman Alexander Braverman 2012-02-16T17:13:25Z 2012-02-16T17:13:25Z Sure, but I was not able to say that carefully for all $L$ (only for sufficiently positive ones) http://mathoverflow.net/questions/88142/bounding-the-size-of-stalks-of-ic-sheaves/88152#88152 Comment by Alexander Braverman Alexander Braverman 2012-02-11T04:20:33Z 2012-02-11T04:20:33Z A perverse sheaf is equal to its middle extension from an open subset iff that it and its dual have no stalks in dimensions $\geq 0$. This condition is preserved under direct image with respect to a finite map. http://mathoverflow.net/questions/88142/bounding-the-size-of-stalks-of-ic-sheaves/88152#88152 Comment by Alexander Braverman Alexander Braverman 2012-02-10T22:59:01Z 2012-02-10T22:59:01Z I edited my answer: I think now I can construct and honest counterexample. http://mathoverflow.net/questions/88142/bounding-the-size-of-stalks-of-ic-sheaves/88152#88152 Comment by Alexander Braverman Alexander Braverman 2012-02-10T22:25:33Z 2012-02-10T22:25:33Z Oh, sorry, I missed the assumption that $X$ is smooth http://mathoverflow.net/questions/86853/what-is-the-relation-between-l-lafforgue-and-frenkel-gaitsgory-vilonen-results-o/87273#87273 Comment by Alexander Braverman Alexander Braverman 2012-02-01T20:58:24Z 2012-02-01T20:58:24Z More precisely: if you only want to prove the &quot;classical&quot; (i.e. not geometric) Langlands conjecture for $GL(n)$ and if you are only interested to show that you can attach an automorphic form to a Galois representation, then it is enough to look at the tamely ramified case with unipotent monodromy. http://mathoverflow.net/questions/86853/what-is-the-relation-between-l-lafforgue-and-frenkel-gaitsgory-vilonen-results-o/87273#87273 Comment by Alexander Braverman Alexander Braverman 2012-02-01T20:55:49Z 2012-02-01T20:55:49Z Well, actually it is known that the tamely ramified case implies the general case. In fact, it is even enough to deal with the tamely ramified case with unipotent monodtromy - this will again imply everything (although for non-trivial reasons - you need to use the existence of global cycling lifting for $GL(n)$, which is known). http://mathoverflow.net/questions/87202/q-x-analog-of-n/87209#87209 Comment by Alexander Braverman Alexander Braverman 2012-02-01T18:04:56Z 2012-02-01T18:04:56Z Thanks a lot - that's exactly what I need! http://mathoverflow.net/questions/87202/q-x-analog-of-n Comment by Alexander Braverman Alexander Braverman 2012-02-01T07:20:23Z 2012-02-01T07:20:23Z Thanks! Can I ask you a stupid question: why is it obvious that it is symmetric with respect to $q$ and $x$?