User chuck hague - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T15:50:55Z http://mathoverflow.net/feeds/user/1528 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/131639/on-the-associated-graded-ring-to-a-section-ring On the associated graded ring to a section ring Chuck Hague 2013-05-23T19:42:31Z 2013-05-23T20:21:18Z <p>Consider a nonsingular projective variety $X$ over an algebraically closed field $k$ and let $Y \subseteq X$ be a nonsingular closed subvariety. Let $\mathcal I \subseteq \mathcal O_X$ be the ideal sheaf of $Y$ and let $\mathcal L$ be a very ample invertible sheaf on $X$. If it makes the setup any easier we can assume that $X$ is projectively normal under the associated embedding to some projective space. Set $$R(\mathcal L) := \bigoplus_{n \geq 0} \Gamma( X, \mathcal L^n ),$$ the section ring associated to $\mathcal L$. Then for each $m \geq 0$ we have the homogeneous ideal $$R_m(\mathcal L) := \bigoplus_{n \geq 0} \Gamma( X, \mathcal I^m \otimes_{\mathcal O_X} \mathcal L^n )$$ of the ring $R(\mathcal L)$. These ideals define a decreasing multiplicative filtration on $R(\mathcal L)$ so we can consider the associated (bi-)graded ring $gr R(\mathcal L)$.</p> <p>Here is my question: Under what conditions can we conclude that this graded ring is Noetherian? Is this always true? Unfortunately I do not have a deep knowledge of commutative algebra so I have gotten stuck on this point. If it makes the answer easier we can even assume that $Y$ is a point.</p> http://mathoverflow.net/questions/129972/quasi-minuscule-representations/129978#129978 Answer by Chuck Hague for quasi-minuscule representations Chuck Hague 2013-05-07T15:21:52Z 2013-05-07T15:21:52Z <p>There is a list <a href="http://books.google.com/books?id=Np7y-LVcwSwC&amp;pg=PA221" rel="nofollow">here</a>.</p> http://mathoverflow.net/questions/129044/defining-equations-of-a-flag-variety/129116#129116 Answer by Chuck Hague for Defining Equations of a Flag Variety Chuck Hague 2013-04-29T16:28:49Z 2013-04-29T16:47:58Z <p>One point of view on equations defining flag varieties (and more generally Schubert varieties) is the following, which puts the construction of defining equations in a representation-theoretic context. The following idea works over an arbitrary algebraically closed field, so let's fix an algebraically closed field $k$ of any characteristic. Let $G$ be a semisimple algebraic group over $k$ and denote by $X = G/B$ the flag variety of $G$. For regular dominant $\lambda$ let $V(\lambda)$ denote the Weyl module for $G$ of highest weight $\lambda$. Then we have the standard embedding $X \hookrightarrow \mathbb P(V(\lambda))$ and we would like to understand the defining relations for this embedding.</p> <p>Now, it is known that $X$ is projectively normal under this embedding. So the homogeneous coordinate ring of $X$ is given by <code>$$R = \bigoplus_{n \geq 0} \Gamma \big( X, \mathcal O(n)|_X \big) = \bigoplus_{n \geq 0} H^0( -n\lambda ) = \bigoplus_{n \geq 0} V(n\lambda)^*,$$</code> where $H^0(-\lambda)$ is the induced module for $G$ of lowest weight $-\lambda$. On the other hand, we have $$T := \bigoplus_{n \geq 0} \Gamma \big( \mathbb P(V(\lambda)), \mathcal O(n) \big) = \bigoplus_{n \geq 0} S^n H^0(-\lambda) ,$$ the homogeneous coordinate ring of $\mathbb P(V(\lambda))$. So to understand the defining equations of $X$ we need to find the kernel of the restriction morphism $T \twoheadrightarrow R$. This is in general a subtle question! It amounts to understanding for each $n \geq 0$ the kernel of the natural $G$-equivariant morphism $S^n H^0(-\lambda) \twoheadrightarrow H^0(-n\lambda)$ given by multiplication in the ring $R$, which is a nontrivial (but interesting!) representation-theoretic problem. </p> <p>There are a number of approaches to this problem. I am far from an expert in this area, so I can't give very precise details, but I know of at least two ways to handle this. One is representation-theoretically: since the map $S^n H^0(-\lambda) \twoheadrightarrow H^0(-n\lambda)$ is a $G$-module map one can try to analyze the kernel representation-theoretically. However, if you want some explicit, hands-on formulas, I believe this is where standard monomial theory comes in. In standard monomial theory, a nice basis of these modules is constructed, and I believe you can then explicitly describe this kernel in a combinatorial fashion using that construction. There are by now many papers on standard monomial theory, including some general overview articles, so a Google search should bring up more reading in that direction if you are interested.</p> http://mathoverflow.net/questions/128522/stratifications-and-cohomology-computations/128622#128622 Answer by Chuck Hague for Stratifications and Cohomology Computations Chuck Hague 2013-04-24T14:34:06Z 2013-04-24T14:34:06Z <p>I don't know if this is the direction you're interested in, but the book "An Introduction to Intersection Homology Theory" by Kirwan and Woolf is a nice readable book that has a lot about stratifications and their connection to topological invariants. The last section of the book deals with the particular case of the flag variety for a semisimple Lie group (and the famous Beilinson-Bernstein correspondence), so that might be helpful. (You may also want to look at the book "D-modules, perverse sheaves, and representation theory" by Hotta, Takeuchi and Tanisaki that focuses on the connection to Lie groups, their representations, and homogeneous spaces.)</p> http://mathoverflow.net/questions/126722/why-does-the-naive-choice-of-homogeneous-coordinate-ring-of-a-product-of-projecti Why does the naive choice of homogeneous coordinate ring of a product of projective schemes not work? Chuck Hague 2013-04-06T17:52:05Z 2013-04-06T19:01:18Z <p>Let $S$ be a graded ring with $A := S_0$. Set $X := \textrm{Proj} (S)$. Then the projective coordinate ring of $X \times_A X$ is the graded ring $\bigoplus_{n \geq 0} S_n \otimes_A S_n$, cf Hartshorne, Exercise II.5.11. (This matches with geometric intuition because this ring corresponds [at least when $S$ is nice enough, eg if $X$ is projective over $A$ and $S$ is the homogeneous coordinate ring of an embedding into $\mathbf P^m_A$] to the section ring of a very ample sheaf on $X \times_A X$ coming from the exterior product of very ample sheaves on $X$.)</p> <p>However, there is another natural graded ring that we can construct here, namely $S \otimes_A S$ with the obvious grading coming from total degree. This ring is clearly not the same as the graded ring in the first paragraph above. Naively, though, by analogy to the affine case one might still expect that $\textrm{Proj}(S \otimes_A S) \cong X \times_A X$. Is this true? Probably this is not true, although I don't know how to show it; so a followup question is: what scheme does this produce?</p> http://mathoverflow.net/questions/125943/degree-of-a-commutator-in-a-hyperalgebra-or-enveloping-algebra Degree of a commutator in a hyperalgebra or enveloping algebra Chuck Hague 2013-03-29T21:07:31Z 2013-03-29T21:34:40Z <p>Consider a semisimple algebraic group $G$ over an algebraically closed field of arbitrary characteristic and let $\bar U(G)$ denote its hyperalgebra (ie, the restricted Hopf dual of the coordinate ring of $G$). In characteristic 0, $\bar U(G)$ is just the enveloping algebra of Lie($G$) but this is not the case in positive characteristic. There is a standard degree filtration $\bar U_{\leq n}(G)$ on $\bar U(G)$. For an element $X \in \bar U(G)$ let us define the degree $\textrm{deg}(X)$ of $X$ to be the minimum $n$ such that $X \in \bar U_{\leq n}(G)$.</p> <p>It is a straightforward verification that for $X$ of degree n and $Y$ of degree m we have $XY - YX \in \bar U_{\leq n + m -1}(G)$ (cf for example Jantzen's Representations of Algebraic Groups). What I am wondering is when we achieve the maximum degree; eg, when is it the case that $\textrm{deg}(XY - YX) = n + m -1$?</p> <p><S>I would not be surprised if this always holds in characteristic 0, ie when we are just considering the enveloping algebra. However, it is clear that this equality does not always hold in positive characteristic since there are zero divisors in $\bar U(G)$ in that case. Nevertheless, I would be happy if there was a weaker statement that could be made, perhaps something like: $\textrm{deg}(XY - YX) = n + m -1$ when $X,Y$ are basis element monomials (with respect to some ordering of Lie($G$)) such that $XY \neq 0$. </S></p> <p>EDIT: As Bruce Westbury points out below, a lot of what I wrote above is silly and I've struck it out. Perhaps my question still has merit though.</p> http://mathoverflow.net/questions/109225/on-q-demazure-operators On q-Demazure operators Chuck Hague 2012-10-09T13:46:18Z 2013-02-11T14:50:13Z <p><strong>Setup</strong></p> <p>Let $G$ be a semisimple algebraic group over an algebraically closed field of arbitrary characteristic with Borel subgroup $B$. Let $\Lambda$ denote the weight lattice of $G$; we write elements of the group ring $\mathbb Z[\Lambda]$ of $\Lambda$ as linear combinations of elements of the form $e^\lambda$, $\lambda \in \Lambda$. In particular, characters of finite-dimensional $B$-modules are elements of $\mathbb Z[\Lambda]$.</p> <p>For dominant $\lambda \in \Lambda$ let $V(\lambda)$ denote the Weyl module for $G$ with highest weight $\lambda$ and for any element $w$ in the Weyl group $W$ of $G$ let $V_w(\lambda)$ denote the Demazure submodule of $V(\lambda)$ associated to $w$; this is the $B$-submodule of $V(\lambda)$ generated by an extremal vector of weight $w\lambda$. (Remark in particular that $V_{w_0}(\lambda) = V(\lambda)$).</p> <p>For any simple root $\alpha_i$ of $G$ with associated simple reflection $s_i \in W$ define the Demazure operator $D_{s_i} : \mathbb Z[\Lambda] \to \mathbb Z[\Lambda]$ by $$D_{s_i}(e^\lambda) = \frac{ e^\lambda - e^{s_i \lambda - \alpha_i} }{ 1-e^{-\alpha_i} } .$$ It is easy to see that this is well-defined. For any word $\mathfrak w = (s_{i_1}, \ldots, s_{i_k})$ of simple reflections in $W$ we have a Demazure operator $D_{\mathfrak w}$ defined by the obvious composition. </p> <p>We now have the following theorem: Choose $w \in W$ and let $\mathfrak w$ be any (not necessarily reduced!) word of simple reflections representing $w$. Then the character of $V_w(\lambda)$ is $D_{\mathfrak w}(e^\lambda)$. [A reference for this is, say, section 3.3 of Brion-Kumar's Frobenius splitting book].</p> <p><strong>Question</strong></p> <p>Has anyone studied $q$-analogues of these Demazure operators? In light of recent work on $q$-character formulas and Kazhdan-Lusztig polynomials, this seems like a natural combinatorial thing to consider. For example, I would (perhaps naively) expect that an appropriate $q$-analogue of the Demazure operators computes, say, the $q$-analog of weight multiplicity considered by Kazhdan-Lusztig, R. Brylinski, Joseph, and others. (Also, I don't know much about the path model or crystal bases, but it seems as though there may be a connection to those as well).</p> http://mathoverflow.net/questions/120984/rep-theory-consequences-of-bott-weil-borel/121086#121086 Answer by Chuck Hague for Rep Theory Consequences of Bott--Weil--Borel Chuck Hague 2013-02-07T15:44:48Z 2013-02-07T15:51:49Z <p>Check out the book Frobenius Splitting Methods in Representation Theory by Brion and Kumar. There are lots of representation-theoretic results in positive characteristic in that book that rely crucially on the geometric realization of induced representations. For many of the representation-theoretic facts proved by Frobenius splitting methods it is true that one can also write down a non-geometric proof, but the Frobenius splitting proofs are almost always simpler and case-free (see for example chapter 4 of Brion-Kumar regarding the good filtration property for modules over a reductive group). </p> <p>And even if you're only interested in complex representations rather than positive-characteristic representations, this technique is still useful because one can base-change facts from positive characteristic to characteristic 0. There is at least one purely representation-theoretic statement over the complex numbers (regarding the so-called generalized Brylinski-Kostant filtration) that I know only a geometric proof of, using Frobenius splitting methods in positive characteristic and then base-changing.</p> <p>One thing that somewhat complicates this whole picture is that often when a representation-theoretic result is first proved by geometric means it is natural to then look for an algebraic proof, so you'll often find proofs that come in both flavors. (See for example R. Brylinski's original paper on the BK-filtration, which crucially uses geometry to prove a purely representation-theoretic fact; her paper was followed by a paper of Joseph and Heckenberger which gives an alternate algebraic proof of her results. Another example is the algebraic Frobenius splitting technique devised by Kumar and Littelmann.)</p> http://mathoverflow.net/questions/118647/on-the-blow-up-along-the-diagonal-in-a-product On the blow-up along the diagonal in a product Chuck Hague 2013-01-11T16:58:35Z 2013-01-13T21:04:29Z <p>Let $X$ be a smooth variety and consider the diagonal $\Delta \subseteq X \times X$. It seems to be well-known that the exceptional divisor in the blow-up of $X \times X$ along $\Delta$ is isomorphic to the projectivized tangent bundle $\mathbb P(\mathcal T_X)$ of $X$ but I can't find a reference or a proof; where might I find one?</p> http://mathoverflow.net/questions/116913/on-local-parameters-at-the-origin-in-an-algebraic-group On local parameters at the origin in an algebraic group Chuck Hague 2012-12-20T20:47:55Z 2012-12-20T21:52:55Z <p>Let $k$ be an algebraically closed field and $G$ an algebraic group over $k$ which is also a $k$-variety (so $G$ is integral, etc). Let $I$ be the ideal defining the identity $e \in G$ and let ${ t_1, \ldots, t_n } \subseteq I$ be a set of local parameters at $e$. Since $e$ is a smooth point, the $t_i$ are algebraically independent and we naturally obtain a $k$-algebra embedding $k[t_1, \ldots, t_n] \hookrightarrow k[G]$. Let $A$ denote the subalgebra of $k[G]$ thus defined.</p> <p>I do not expect that $A$ will be a sub-Hopf algebra of $k[G]$ (in particular, I see no reason why the comultiplication should preserve $A$), but I don't know any concrete examples. So I have two questions: (1) Is there a specific example of an algebraic group $G$ as above such that $A$ is not a sub-Hopf algebra of $k[G]$? (2) Are there nice conditions on $G$ under which $A$ will be a sub-Hopf algebra of $k[G]$?</p> http://mathoverflow.net/questions/112214/on-the-divided-power-ring-over-the-integers On the divided power ring over the integers Chuck Hague 2012-11-12T20:04:51Z 2012-11-13T14:17:32Z <p>Consider the divided-power ring $A := \mathbb Z \langle x_1, \ldots, x_n \rangle$ consisting of $\mathbb Z$-linear combinations of divided-power monomials of the form $x_1^{(a_1)} \cdots x_n^{(a_n)}$; this can be defined as the subring of the polynomial ring $\mathbb Q[ x_1, \ldots, x_n ]$ which is generated as a $\mathbb Z$-algebra by the elements $x_i^{(m)} := \displaystyle \frac 1 {m!} x_i^m$. </p> <p>Next consider the lattice $V \subseteq A$ generated by $x_1, \ldots, x_n$; that is, $V$ is the subspace of $A$ consisting of degree-1 polynomials. Then divided powers of elements of $V$ are elements of $\mathbb Q[ x_1, \ldots, x_n ]$, and one can check that they are in fact in $A$. Now my question is the following: Do the divided powers of elements of $V$ generate $A$ as an abelian group? (This is motivated by the fact that for a field $k$, the divided-power ring $k\langle x_1, \ldots, x_n \rangle := A \otimes_{\mathbb Z} k$ is spanned as a $k$-vector space by the divided powers of degree-1 polynomials). EDIT: Following Scott Carnahan's answer below, I need to take $k$ algebraically closed here.</p> http://mathoverflow.net/questions/104932/kostants-theorem-on-invariant-polynomials-in-positive-characteristic Kostant's theorem on invariant polynomials in positive characteristic Chuck Hague 2012-08-17T18:41:23Z 2012-09-15T05:44:06Z <p>Let $k$ be an algebraically closed field and let $G$ be a reductive linear algebraic group over $k$ with Lie algebra $\mathfrak g$. If the characteristic of $k$ is $0$ then, by a classical result of Kostant, $k[\mathfrak g]$ is free over the subalgebra $k[\mathfrak g]^G$. Is there an analog of this result when the characteristic of $k$ is positive?</p> http://mathoverflow.net/questions/103813/the-killing-form-on-quantized-enveloping-algebras-and-reduction-to-the-classical The Killing form on quantized enveloping algebras and reduction to the classical case Chuck Hague 2012-08-02T19:22:43Z 2012-08-02T19:22:43Z <p>Let $U_q$ be the quantized enveloping algebra associated to a semisimple Lie algebra $\mathfrak g$. It is a result due to Tanisaki (see <a href="http://www.worldscientific.com/doi/abs/10.1142/S0217751X92004117" rel="nofollow">here</a>; also see Chapter 6 of Jantzen's book Lectures on Quantum Groups) that there is an ad-invariant nondegenerate bilinear form on $U_q$. By the standard base-change rigamarole, from an appropriate $\mathbb Z[q,q^{-1}]$-subalgebra of $U_q$ we can obtain the enveloping algebra $\bar U_{\mathbb C}$ of $\mathfrak g$ over $\mathbb C$ and, for a field $k$ of positive characteristic, the hyperalgebra $\bar U_k$ of a linear algebraic group over $k$ associated to $\mathfrak g$.</p> <p>My question is: can we base-change the bilinear form on $U_q$ to obtain ad-invariant bilinear forms on $\bar U_{\mathbb C}$ and $\bar U_k$? Or is there an easier direct construction that can be made without going through the quantum construction first?</p> http://mathoverflow.net/questions/103533/on-an-interesting-subalgebra-of-the-functions-on-the-cotangent-bundle-of-the-flag On an interesting subalgebra of the functions on the cotangent bundle of the flag variety Chuck Hague 2012-07-30T19:17:19Z 2012-07-30T20:52:12Z <p><strong>Setup</strong></p> <p>Let $G$ be a semisimple algebraic group over a field $k$ of characteristic $p$ where $p = 0$ or $p > 0$ is a good prime for $G$. Fix a Borel subgroup $B \subseteq G$ corresponding to the positive roots. Set $\mathfrak g := \textrm{Lie}(G)$. Let $U \subseteq B$ be the unipotent radical and let $U^- \subseteq G$ be the opposite unipotent radical. Set $\mathfrak n := \textrm{Lie}(U)$ and $\mathfrak n^- = \textrm{Lie}(U^-)$.</p> <p>Let me first recall the following general fact about sections of bundles on the flag variety $G/B$. Let $V$ be any $B$-module. Then we have a left $G$-action on $k[G] \otimes V$ given by the left action in $k[G]$ and we have a right $B$-action given by $(f \otimes v).b = f.b \otimes b^{-1}.v$. The global sections of the $G$-equivariant bundle $G \times^B V$ on $G/B$ now identify with the $G$-submodule $(k[G]\otimes V)^B \subseteq k[G] \otimes V$ of invariants under the right $B$-action. It is also known that the projection map $(k[G] \otimes V)^B \to k[U^-] \otimes V$ given by restriction of functions in the first coordinate is an inclusion which corresponds geometrically to the open inclusion $(G \times^B V) |_{\mathfrak B} \hookrightarrow G \times^B V$, where $\mathfrak B$ is the big cell $U^-B \subseteq G/B$.</p> <p>In particular let us take $V = \mathfrak n$ so that we have the cotangent bundle $\mathcal T^* := G \times^B \mathfrak n$ of $G/B$. Denote by $\mathcal N \subseteq \mathfrak g$ the nullcone of nilpotent elements. <code>$\mathcal T^*$</code> is a resolution of singularities of $\mathcal N$ and there is an isomorphism $k[\mathcal T^*] \cong k[\mathcal N]$. Also, there is a natural $B$-equivariant surjection $\mathcal N \twoheadrightarrow \mathfrak g / \mathfrak b \cong \mathfrak n^-$ induced by the projection $\mathfrak g \twoheadrightarrow \mathfrak g / \mathfrak b$. Thus we obtain a $B$-equivariant inclusion <code>$$k[\mathfrak n^-] \hookrightarrow k[\mathcal T^*] .$$</code> Putting this all together, we obtain an interesting algebra inclusion $$i: k[\mathfrak n^-] \hookrightarrow k[U^-] \otimes k[\mathfrak n] .$$</p> <p><strong>Remark</strong></p> <p>It's possible that the above description is overcomplicated, since the inclusion $i$ is just the comorphism of the variety morphism $U^- \times \mathfrak n \to \mathfrak n^-$ given by the composition of the action morphism $U^- \times \mathfrak n \to \mathfrak g, u \times X \mapsto u.X$ and the projection $\mathfrak g \twoheadrightarrow \mathfrak n^-$. However, when described in this way it's not a priori clear to me that $i$ is a dominant morphism, so perhaps the above description is necessary.</p> <p><strong>Question</strong></p> <p>It seems highly likely that someone has considered this algebra inclusion before. If so, I would like a reference. In particular, it would be nice to have some sort of concrete description of the image of the algebra morphism $i$, perhaps in terms of generating eigenfunctions for $k[U^-]$ and $k[\mathfrak n]$.</p> http://mathoverflow.net/questions/102647/kostant-partition-function-asymptotics-and-specifics/102686#102686 Answer by Chuck Hague for Kostant partition function: asymptotics and specifics Chuck Hague 2012-07-19T17:17:46Z 2012-07-20T17:36:34Z <p>To expand on Jim's answer, it's a result of Kostant that for any weight $\mu$ and any dominant weight $\lambda$ the dimension of the $\mu$-weight space of the induced module $H^0(\lambda)$ with highest weight $\lambda$ is given by $$\sum_{w \in W}(-1)^{l(w)} \mathfrak P( w(\lambda + \rho) - \mu - \rho ) .$$ (Remark that equivalently this gives the dimension of the $\mu$-weight space of the Weyl module with highest weight $\lambda$, and in characteristic 0 the induced and Weyl modules of highest weight $\lambda$ are isomorphic to each other and simple so this gives weight multiplicities for simple modules in characteristic 0 too). As Jim mentions, this formula should imply the result you want for the 0-weight space of $H^0(r\rho)$. [EDIT: Or perhaps not -- see Jim's updated answer] [SECOND EDIT: It's not true in general that the dimension of the $0$-weight space of $St_r$ is $\mathfrak P( (p^r-1)\rho )$.].</p> <p>Furthermore, there are a lot of interesting facts about the Kostant partition function and its $q$-analog; in particular, the Kostant partition function is equal to the evaluation at 1 of certain Kazhdan-Lusztig polynomials, and it's also connected to facts about nilpotent orbits. If you want to know more about this story, the paper "Limits of weight spaces, Lusztig's $q$-analogs, and fiberings of adjoint orbits" by R.K. Brylinski is a good paper to look at.</p> http://mathoverflow.net/questions/95274/homogenous-bundles/95379#95379 Answer by Chuck Hague for homogenous bundles Chuck Hague 2012-04-27T17:06:18Z 2012-04-27T17:06:18Z <p>Section I.5 ("Quotients and associated sheaves") of Jantzen's Representations of Algebraic Groups is (at least in my mind) a standard resource for this question. (Here is a <a href="http://books.google.com/books?id=UEqTJd1SMqsC&amp;lpg=PP1&amp;dq=representations%2520of%2520algebraic%2520groups&amp;pg=PA65#v=onepage&amp;q=representations%2520of%2520algebraic%2520groups&amp;f=false" rel="nofollow">Google books link</a>). He considers your question in full generality there. In particular, he proves (cf I.5.6.(8)) that if $G$ is an algebraic group over a field $k$ and $H$ is a closed subgroup scheme of $G$ then $G/H$ is a scheme. (Here the definition of $G/H$ agrees with what you think it should mean over a field, but in general the definition of $G/H$ is given categorically, cf the definition of the <em>quotient faisceau</em> $X/G$ for any $G$-space $X$ in I.5.5).</p> http://mathoverflow.net/questions/93026/frobenius-splitting-over-non-algebraically-closed-fields Frobenius splitting over non-algebraically closed fields Chuck Hague 2012-04-03T18:31:00Z 2012-04-05T13:31:13Z <p>Let $X$ be a scheme over an algebraically closed field $k$ of positive characteristic $p$. Recall that the absolute Frobenius morphism $F : X \to X$ is the map which is the identity on points and the $p^{th}$ power morphism on functions. Recall also that we say that $X$ is Frobenius split if there is an $\mathcal O_X$-linear morphism splitting the $p^{th}$ power morphism $\mathcal O_X \to F_* \mathcal O_X$.</p> <p>Now, whenever one sees the definition of Frobenius splitting, it is always stated for an algebraically closed field $k$. However, the definitions above make perfectly good sense for any scheme over $\mathbb F_p$, and in fact many Frobenius-split schemes, eg flag varieties, are "split over $\mathbb F_p$" in the sense that the Frobenius splitting is the appropriate base-change of a morphism $F_* \mathcal O_X \to \mathcal O_X$ for a scheme $X$ over $\mathbb F_p$. (Although I defined the Frobenius morphism above for schemes over $k$ it also makes sense for any scheme over $\mathbb F_p$). My question is: Why is the definition of Frobenius splitting always stated for schemes over an algebraically closed field, when one can state it more generally for schemes over any field containing $\mathbb F_p$? Is this just convention or is there a deeper reason?</p> http://mathoverflow.net/questions/92030/generators-and-relations-for-the-enveloping-algebra-of-a-unipotent-radical Generators and relations for the enveloping algebra of a unipotent radical Chuck Hague 2012-03-23T18:49:51Z 2012-03-23T18:49:51Z <p>Let $G$ be a semisimple algebraic group over an algebraically closed field $k$ of characteristic 0 and let $B \subseteq G$ be a Borel subgroup with unipotent radical $U$. Let $P \supseteq B$ be a parabolic subgroup with unipotent radical $U_P \subseteq U$. Set $\mathfrak n := \textrm{Lie}(U)$ and $\mathfrak n_P := \textrm{Lie}(U_P)$. Then we have the enveloping algebras $U(\mathfrak n)$ and $U(\mathfrak n_P)$ associated to $\mathfrak n$ and $\mathfrak n_P$.</p> <p>Let $\alpha_1, \ldots, \alpha_\ell$ denote the simple roots of $G$. It is well-known that $U(\mathfrak n)$ is generated as a $k$-algebra by generators $E_{\alpha_i}^{(n)}$ for $1 \leq i \leq \ell$ and $n \geq 0$. (Here we are using the divided-power notation $E_{\alpha_i}^{(n)} = \frac 1 {n!} E_{\alpha_i}^n$). Furthermore, the relations between these generators are given by the Serre relations $$\sum_{a+b = -\alpha_j(\alpha_i^\vee)+1} (-1)^a E_{\alpha_i}^{(a)} E_{\alpha_j} E_{\alpha_i}^{(b)} = 0 .$$ Succinctly, using the adjoint action $*$ of $U(\mathfrak n)$ on itself, we can write the Serre relations as $$E_{\alpha_i}^{(c)}*E_{\alpha_j} = 0 \textrm{ for } c > -\alpha_j(\alpha_i^\vee) .$$</p> <p>Now to my question: Is there a nice presentation of $U(\mathfrak n_P)$ by generators and relations given in a similar fashion?</p> http://mathoverflow.net/questions/91786/minimal-relative-schubert-modules/91841#91841 Answer by Chuck Hague for Minimal relative Schubert modules Chuck Hague 2012-03-21T17:52:57Z 2012-03-22T16:02:19Z <p>This is not really an answer to your question -- more of a long comment, I suppose -- but there is a somewhat intuitive way to understand what these modules look like, representation-theoretically speaking. It's easier to describe what the duals to these modules look like, so let me do that. First, for any weight $\mu$, let $\mu^+$ denote the unique dominant element in the Weyl group orbit of $-\mu$ and let $V(\mu^+)$ denote the Weyl module for $G$ of highest weight $\mu^+$ (i.e., $V(\mu^+) = H^0(-w_0\mu^+)^*$). Then the Joseph module <code>$P(\mu)^*$</code> is just the $B$-submodule of $V(\mu^+)$ generated by any nonzero weight vector of weight $-\mu$. (Dually, this now describes the surjection $H^0(-w_0\mu^+) \twoheadrightarrow P(\mu)$).</p> <p>Now set $$\lbrace \mu_1, \ldots, \mu_r \rbrace := \lbrace s \mu : s \in W \textrm{ is a simple reflection and } s \mu &lt; \mu \rbrace .$$ Let $I(\mu)$ be the $B$-submodule of $V(\mu^+)$ generated by $P(\mu_i)^*$, $1 \leq i \leq r$. Equivalently, $I(\mu)$ is the $B$-submodule of $V(\mu^+)$ generated by nonzero weight vectors of weights $-\mu_1, \ldots, -\mu_r$. Then $Q(\mu)^*$ fits into an exact sequence <code>$$0 \to I(\mu) \to P(\mu)^* \to Q(\mu)^* \to 0 .$$</code> Remark that we can also describe $I(\mu)$ as the submodule of <code>$V(\mu^+)$</code> generated by all Joseph modules properly contained in <code>$P(\mu)^*$</code>.</p> <p>As for the evaluation map $\varepsilon$, it can be described as follows. Let <code>$V \subseteq V(\mu^+)$</code> denote the highest weight subspace of weight <code>$\mu^+$</code>. Then <code>$\varepsilon : P(\mu) \twoheadrightarrow k_{-\mu^+}$</code> is dual to the inclusion <code>$V \hookrightarrow P(\mu)^*$</code> of the highest weight subspace. (Remark also that <code>$V = P(-\mu^+)^*$</code>).</p> http://mathoverflow.net/questions/91383/on-a-resolution-of-sections-of-line-bundles-on-the-cotangent-bundle-of-a-flag-var On a resolution of sections of line bundles on the cotangent bundle of a flag variety Chuck Hague 2012-03-16T16:18:01Z 2012-03-16T16:18:01Z <p><strong>Background</strong></p> <p>Let $G$ be a semisimple algebraic group over an algebraically closed field $k$ of characteristic 0. Let $B \subseteq G$ be a Borel subgroup and let $U \subseteq B$ be its unipotent radical. Let $\mathfrak g$, $\mathfrak b$, and $\mathfrak n$ be the corresponding Lie algebras. Also let $\mathcal T^*$ denote the cotangent bundle of the flag variety $G/B$; then <code>$\mathcal T^*$</code> is isomorphic to the homogeneous bundle $G \times^B \mathfrak n$.</p> <p>For any representation $M$ of $B$ let $\mathcal L(M)$ denote the homogeneous line bundle on $G/B$ with fiber $M$ and let $\mathcal L'(M)$ denote the pullback of this bundle to $\mathcal T^*$. In particular, for any weight $\lambda$ of $G$, we have the bundles $\mathcal L(\lambda)$ and $\mathcal L'(\lambda)$. Let $H^0(\lambda)$ denote the global sections of $\mathcal L(\lambda)$.</p> <p>By the projection formula, we have $$H^0( \mathcal T^*, \mathcal L'(M) ) \cong H^0 \big( G/B, \mathcal L( k[\mathfrak n] ) \otimes \mathcal L(M) \big) .$$ In particular, since $k[\mathfrak n]$ is a graded $B$-module, <code>$H^0( \mathcal T^*, \mathcal L'(M) )$</code> naturally has the structure of a graded $G$-module.</p> <p><strong>Question</strong></p> <p>Let $\lambda$ be a weight of $G$. The natural $B$-equivariant surjection $k[\mathfrak g] \twoheadrightarrow k[\mathfrak n]$ corresponding to the inclusion $\mathfrak n \hookrightarrow \mathfrak g$ induces a sheaf surjection $\mathcal L(k[\mathfrak g]) \twoheadrightarrow \mathcal L(k[\mathfrak n])$ and hence, twisting by $\lambda$ and taking global sections, we obtain a $G$-equivariant gradation-preserving morphism $$k[\mathfrak g] \otimes H^0(\lambda) \cong H^0( G/B, \mathcal L(k[\mathfrak g]) \otimes \mathcal L(\lambda) ) \to H^0( G/B, \mathcal L(k[\mathfrak n]) \otimes \mathcal L(\lambda) ) \cong H^0( \mathcal T^*, \mathcal L'(\lambda) ) .$$ In his paper "Line Bundles On The Cotangent Bundle Of The Flag Variety," Broer proves that this morphism is surjective when $\lambda$ is dominant. He also remarks that there is an alternate proof of this fact using the following standard fact from algebraic geometry: If $i : X \to \mathbb P^n_A$ is a morphism of $A$-schemes, then $i^*( \mathcal O(1) )$ is an invertible sheaf on $X$ which is globally generated by the pullback of the global sections of $\mathcal O(1)$.</p> <p>My question is: what is this alternate proof? It's mysterious to me where this remark comes from. I assume that one should use the inclusion $$G \times^B \mathfrak n \hookrightarrow G \times^B \mathfrak g \cong G/B \times \mathfrak g,$$ but I can't seem to figure the rest out. (My motivation is that I'd like to give a proof of this fact when $k$ is a field of arbitrary characteristic).</p> http://mathoverflow.net/questions/89795/symmetrization-for-hyperalgebras-in-positive-characteristic Symmetrization for hyperalgebras in positive characteristic Chuck Hague 2012-02-28T20:55:52Z 2012-02-29T04:02:00Z <p>Let $G$ be an algebraic group over an algebraically closed field $k$ of arbitrary characteristic and let $U$ be the hyperalgebra of $G$. Recall that $U$ is defined as the subspace of the full linear dual of $k[G]$ consisting of elements that vanish on some power of the ideal $I \subseteq k[G]$ defining the identity in $G$. Then one has the standard degree filtration $U_{\leq n}$ on $U$ given by setting $$U_{\leq n} := ( k[G]/I^{n+1} )^* .$$</p> <p>Set $\mathfrak g :=$ Lie$(G)$. In characteristic 0, $U$ is just the enveloping algebra of $\mathfrak g$, and the PBW theorem tells us that the ring gr $U$ is isomorphic to $S(\mathfrak g)$. Moreover, there is a $G$-equivariant symmetrization isomorphism $S(\mathfrak g) \to U$ of vector spaces over $k$, where we take the conjugation action of $G$ on $U$ (note that this is NOT an algebra isomorphism, however). Remark that the symmetrization isomorphism is given explicitly by $$x_1 \cdots x_n \mapsto \frac 1 {n!} \sum_{\sigma \in S_n} x_{\sigma(1)} \cdots x_{\sigma(n)} .$$</p> <p>Now assume that char $k = p > 0$. In this case, gr $U$ is a commutative algebra -- let's call it $S'(\mathfrak g)$ -- that is <em>not</em> a polynomial ring. (One way of seeing this is that the ring $S'(\mathfrak g)$ is infinitely generated as a $k$-algebra). Note that $S'(\mathfrak g)$ carries a natural $G$-module structure coming from the conjugation $G$-action on $U$. I'm wondering if one still can construct a $G$-equivariant "symmetrization" isomorphism $S'(\mathfrak g) \to U$ of $k$-vector spaces in this case as well. (Clearly the formula for the symmetrization morphism in characteristic 0 doesn't work, since $n!$ isn't invertible in $k$ for $n \geq p$).</p> http://mathoverflow.net/questions/89584/about-g-modules-versus-lieg-modules-for-algebraic-groups/89589#89589 Answer by Chuck Hague for About $G$-modules versus $Lie(G)$-modules for algebraic groups Chuck Hague 2012-02-26T16:45:03Z 2012-02-26T16:57:00Z <p>I'm sure that there are a number of ways of answering this in the affine case; here's one. Let's assume $G$ affine and let's say that $k$ is our algebraically closed field. The following argument may not be the slickest one, but it has the benefit of working in arbitrary characteristic with the enveloping algebra replaced by the hyperalgebra. (In positive characteristic, instead of considering a Lie($G$)-stable vector we'd want to consider a hyperalgebra-stable vector. It is <em>not</em> true in positive characteristic that a Lie($G$)-stable vector is $G$-stable).</p> <p>Now, a $G$-module structure on $M$ is given by a comodule morphism $$c : M \to k[G] \otimes M .$$ Let $U(G)$ denote the enveloping algebra of Lie($G$); equivalently, this is the so-called hyperalgebra of $G$ (hyperalgebra = enveloping algebra when char$(k) = 0$, but not in positive characteristic). We will view $U(G)$ as a subspace of the full linear dual of $k[G]$; namely, $U(G)$ is the subspace of $k[G]^*$ consisting of elements that vanish on some power of the ideal defining the identity. (You can look in, say, Jantzen's book Representations of Algebraic Groups for more details). Note that $v \in M$ is $G$-stable if and only if $c(v) = 1 \otimes v$.</p> <p>Now, the action of $U(G)$ on $M$ also comes from the comorphism $c$. Namely, for $v \in M$, if $c(v) = \sum f_i \otimes v_i$ then for $X \in U(G)$ we have $$X.v = \sum X(f_i) \cdot v_i ,$$ where $X(f_i)$ is the dual action of $X$ on $f_i \in k[G]$. Let $U(G)^+$ denote the augmentation ideal; this is the two-sided ideal of $U(G)$ generated by the Lie algebra inside of $U(G)$. Then $v \in M$ is killed by Lie($G$) if and only if $X.v = 0$ for all $X \in U(G)^+$. (This part only works in characteristic 0; there are slight modifications to be made in positive characteristic). </p> <p>So now let's assume that $v \in M$ is killed by Lie$(G)$. Let's write $$c(v) = \sum f_i \otimes v_i .$$ Without loss of generality we may assume that the $v_i$ are $k$-linearly independent. Since $X.v = 0$ for all $X \in U(G)^+$ this implies $$\sum X(f_i) \cdot v_i = 0$$ for all $X \in U(G)^+$. Since the $v_i$ are linearly independent, this means that $X(f_i) = 0$ for all $i$ and for all $X \in U(G)^+$. Now, any element $f \in k[G]$ that is killed by all elements of $U(G)^+$ must be constant. Hence we have $$c(v) = \sum 1 \otimes v_i .$$ But one of the properties of the comodule morphism is that $(\epsilon \otimes Id_M) \circ c = 1 \otimes Id_M$, where $\epsilon$ is the augmentation of $k[G]$. Hence $$\sum 1 \otimes v_i = 1 \otimes v$$ and $v$ is $G$-stable.</p> http://mathoverflow.net/questions/88567/decomposition-of-the-ring-of-functions-on-the-unipotent-radical-of-a-borel Decomposition of the ring of functions on the unipotent radical of a Borel Chuck Hague 2012-02-15T21:53:27Z 2012-02-16T13:26:57Z <p><strong>Background</strong></p> <p>Let $G$ be a semisimple linear algebraic group over an algebraically closed field $k$ of arbitrary characteristic. Let $B \subseteq G$ be a Borel subgroup of $G$ and let $U \subseteq B$ be its unipotent radical. Consider $k[U]$ as a $U$-module under left multiplication in $U$; let's call this module $k[U]_L$. By, say, identifying $U$ with the big cell in the flag variety $G/B^-$, it is not hard to see that $k[U]_L$ is isomorphic as a $U$-module to a direct limit of standard modules $H^0(\lambda)$ for $G$. (In characteristic 0 one can also see this by identifying $k[U]$ with the dual zero Verma module for the enveloping algebra of $G$, cf <a href="http://www.ams.org/journals/jams/2000-13-04/S0894-0347-00-00347-7/home.html" rel="nofollow">this paper</a>). In fact there is even a natural ring structure on this direct limit such that this isomorphism is a ring isomorphism.</p> <p>EDIT: A reference for this fact is Lemma 2.5 and the discussion following it in "<a href="http://www.kryakin.com/files/Invent_mat_%282_8%29/123/123_06.pdf" rel="nofollow">The Nil Hecke Ring and Singularity of Schubert Varieties</a>," by Shrawan Kumar. Although the construction is done there over $\mathbb C$, it works over any algebraically closed field, and although Kumar only proves that there is a $T$-equivariant morphism, it is easy to check that the morphism he constructs is $U$-equivariant.</p> <p><strong>Question</strong></p> <p>Now consider $k[U]$ as a $U$-module under the <em>conjugation</em> action of $U$ on itself; let's call this module $k[U]_C$. Is there a nice description of $k[U]_C$ as a $U$-module in terms of $G$-modules in a way analogous to the description of $k[U]_L$?</p> http://mathoverflow.net/questions/83379/does-there-exist-a-canonical-degree-filtration-on-quantum-groups/84175#84175 Answer by Chuck Hague for Does there exist a canonical "degree" filtration on quantum groups? Chuck Hague 2011-12-23T18:55:21Z 2011-12-23T18:55:21Z <p>Nobody has answered this yet, so maybe I'll expand on my comment above, with the caveat that I'm no expert in this area. I believe the answer to your question is yes; the reference for all of this is Lusztig's paper <a href="http://www.math.toronto.edu/lzhang/Seminars/References/Lusztig-QGrpRootsUnity.pdf" rel="nofollow">Quantum Groups at Roots of 1</a>.</p> <p>Let $\Delta^+$ denote the positive roots of $G$ and let $\ell$ be the rank of $G$. Given a quantum group $U_q(\mathfrak g)$ defined over $\mathbb Q(v)$, Lusztig defines a set of $\mathbb Q(v)$-algebra generators of $U_q(\mathfrak g)$ given by elements $E_\beta^{(n)}$, $F_\beta^{(n)}$, $K_i$, $K_i^{-1}$, and <code>$\begin{bmatrix} K_i \\ n \end{bmatrix}$</code> for $\beta \in \Delta^+$, $n \geq 0$, and $1 \leq i \leq \ell$. (The proof that this indeed gives a set of algebra generators is nontrivial and takes most of the paper). If we put $K_i$ and $K_i^{-1}$ in degree 1 and the other elements in degree $n$ then the explicit relations given in Lusztig's paper show that this indeed defines a degree filtration on $U_q(\mathfrak g)$.</p> <p>Furthermore, when one goes through the standard quantum rigamarole for base change (take the $\mathbb Z[v,v^{-1}]$-subalgebra generated by the elements above, base change to your favorite field considered as a $\mathbb Z[v,v^{-1}]$-algebra, and mod out by $K_i - 1$ and $K_i^{-1} - 1$) these elements go to the standard elements in the hyperalgebra, so in particular when one plays this game in characteristic zero, we do get the standard filtration on the enveloping algebra $U(\mathfrak g)$.</p> http://mathoverflow.net/questions/84117/springer-isomorphisms-and-parabolics Springer isomorphisms and parabolics Chuck Hague 2011-12-22T19:47:55Z 2011-12-23T12:23:48Z <p>Let $G$ be a semisimple, simply-connected algebraic group over an algebraically closed field $k$ of positive characteristic. Fix a Borel subgroup $B \subseteq G$ with unipotent radical $U$. Also let $P$ be a parabolic subgroup of $G$ containing $B$ and let $L$ be its Levi factor. Denote by $U_P \subseteq U$ the unipotent radical of $P$ and set $U_L := U \cap L$.</p> <p>Let $\mathfrak U \subseteq G$ denote the unipotent variety and let $\mathfrak R \subseteq \textrm{Lie}(G)$ denote the nilpotent cone. If $p$ is a good prime for $G$ then there is a $G$-equivariant Springer isomorphism $\phi : \mathfrak U \to \mathfrak R$ that restricts to an isomorphism $U \to \textrm{Lie}(U)$. My question is: Does $\phi$ restrict to isomorphisms $U_P \to \textrm{Lie}(U_P)$ and $U_L \to \textrm{Lie}(U_L)$? If this does not always happen, are there conditions on the parabolic $P$ under which it will be true?</p> http://mathoverflow.net/questions/82253/cohomology-vanishing-for-tensor-powers-of-tangent-bundle-on-the-flag-variety/82306#82306 Answer by Chuck Hague for Cohomology vanishing for tensor powers of tangent bundle on the flag variety Chuck Hague 2011-11-30T19:36:17Z 2011-11-30T19:46:25Z <p>I don't have a complete answer, but let me just note that your question 2) would be true if there is a Frobenius splitting of the cotangent bundle $T^*_{X^n}$ of $X^n$ which compatibly splits the diagonal copy $\Delta$ of $T^*_X$, the cotangent bundle of $X$. Since <code>$T^*_{X^n}$</code> is the cotangent bundle of the flag variety of the semisimple algebraic group $G^n$, there is a candidate for this splitting, namely the splitting of Kumar, Lauritzen, and Thomsen that you mention. I don't know, though, if their splitting compatibly splits $\Delta$; that is an interesting question.</p> http://mathoverflow.net/questions/81611/picard-groups-of-fiber-products/81624#81624 Answer by Chuck Hague for Picard groups of (fiber) products Chuck Hague 2011-11-22T16:38:08Z 2011-11-22T16:38:08Z <p>A weakening of 2) is an exercise (III.12.6) in Hartshorne: Let $X$ be an integral projective scheme over an algebraically closed field $k$ and assume that $H^1(X, \mathcal O_X) = 0$. Let $T$ be a connected scheme of finite type over $k$. Then $\textrm{Pic}(X) \times \textrm{Pic}(T) \cong \textrm{Pic}(X \times T)$ under the obvious morphism. However, it is not true in general that $\textrm{Pic}(X) \times \textrm{Pic}(T) \cong \textrm{Pic}(X \times T)$ for two arbitrary $k$-schemes $X$ and $T$, cf exercise IV.4.10 in Hartshorne.</p> <p>In general, you might be interested in exercises III.12.4 and III.12.5 of Hartshorne as well; they give more results on Picard groups in this context.</p> http://mathoverflow.net/questions/80861/coherent-cohomology-of-g-u-g-reductive-group-b-tu-borel-subgroup/80912#80912 Answer by Chuck Hague for Coherent cohomology of G/U, G = reductive group, B = TU Borel subgroup Chuck Hague 2011-11-14T17:15:09Z 2011-11-14T17:49:03Z <p>Chris Brav's answer gives a nice description of the cohomology in the $D$-module context. Just to expand a bit on that, I'd like to give a direct description, and also say a word about positive characteristic, since the paper of Levasseur-Stafford only discusses the characteristic-0 case.</p> <p>As Jason Starr mentions above, we can decompose this cohomology in terms of line bundles on $G/B$. Fix an isomorphism $B/U \cong T$; in particular, this gives $T$ and $k[T]$ structures of $B$-modules. For any $B$-module $M$ let $\mathcal L(M)$ denote the $G$-equivariant bundle on $G/B$ with fiber $M$. Then we have a $G$-equivariant isomorphism $$H^*(G/U, \mathcal O_{G/U}) \cong H^* \big(G/B, \mathcal L ( k[T] ) \big) .$$ So, we basically want to understand the structure of $k[T]$ as a $B$-module.</p> <p>Let $X(T)$ denote the character group of $T$. Then $k[T] \cong k( X(T) )$, the group algebra of $X(T)$ over $k$. As Jason pointed out, we now get a direct sum of line bundles on $G/B$ corresponding to the elements of $X(T)$. However, note that we may not get <em>all</em> of the line bundles on $G/B$, since $G$ might not be simply-connected; $X(T)$ may be a proper subset of the full weight lattice of $G$. Here isogeny will play a role. (In Levasseur-Stafford, for example, they assume $G$ to be simply connected). In any event, the characteristic 0 story will now follow from Borel-Weil. The postive-characteristic answer, on the other hand, is still an open question, since the full cohomology of line bundles on $G/B$ isn't completely known there (although a lot is known, cf Jantzen's book "Representations of Algebraic Groups").</p> <p>Regardless of characteristic, though, we have a nice description of the global sections. Let $X^+(T)$ denote the set of dominant weights in $X(T)$; then we get $$H^0 \big(G/B, \mathcal L ( k[T] ) \big) \cong \bigoplus_{\mu \in X^+(T)} H^0( G/B, \mathcal L(\mu) ) ,$$ a direct sum of standard modules for $G$. In characteristic 0 these modules are all simple, but they are not all simple in positive characteristic.</p> http://mathoverflow.net/questions/80635/what-information-is-contained-in-the-kazhdan-lusztig-polynomials/80696#80696 Answer by Chuck Hague for What information is contained in the Kazhdan-Lusztig polynomials? Chuck Hague 2011-11-11T17:00:09Z 2011-11-14T16:50:26Z <p>Certain Kazhdan-Lusztig polynomials compute the so-called Brylinski-Kostant filtration on weight spaces of irreducible representations. They also compute multiplicities of irreducible modules occurring in global sections of line bundles on cotangent bundles of flag varieties. These two ideas are related.</p> <p>In more detail: Let $G$ be a semisimple algebraic group over $\mathbb C$. Fix a Borel subgroup $B \subseteq G$ with unipotent radical $U$ and choose a principal nilpotent element $X \in \textrm{Lie}(U)$ (a principal nilpotent element is one whose $G$-orbit is dense in the subvariety of all nilpotent elements). Let $\lambda, \mu$ be dominant weights and consider the $\mu$-weight space $V_\mu(\lambda)$ of the irreducible representation $V(\lambda)$ of $G$. Then there is a filtration -- the Brylinski-Kostant filtration -- on $V_\mu(\lambda)$ coming from the action of $X$ on $V(\lambda)$: define $\mathcal F^n( V_\mu(\lambda) )$ to be the subspace of $V_\mu(\lambda)$ consisting of vectors killed by n+1 applications of $X$. Since $X$ is nilpotent, this does indeed give a filtration on $V_\mu(\lambda)$.</p> <p>Now, an important theorem due to R. Brylinski (1) is that certain Kazhdan-Lusztig polynomials give the dimensions of the subspaces in this filtration. This is a very interesting theorem: to prove it, Brylinski actually proves an intermediate geometric theorem relating Kazhdan-Lusztig polynomials to twisted functions on G/T (which are connected to degrees of sections of line bundles on the cotangent bundle of G/B), so that along the way she gives yet another interpretation of Kazhdan-Lusztig polynomials. In particular, they compute multiplicities of irreducible modules occurring in global sections of line bundles on the cotangent bundle of G/B. They also give information on the degrees of sections of these bundles (I will just refer to Brylinski's paper for the appropriate definition of "degree").</p> <p>Here I will humbly submit my own work: in (2) I extended some of Brylinski's results to the case where the nilpotent element is not necessarily principal. In this case, certain Kazhdan-Lusztig polynomials also appear in an analogous filtration. Further, the full cotangent bundle of G/B is replaced by an appropriate subbundle E of the cotangent bundle, and one has the following result: certain Kazhdan-Lusztig polynomials compute multiplicities of irreducibles occurring in global sections of line bundles on E. (Unfortunately the full general statement that one would like to make is still a conjecture, due to difficult technical issues involving cohomology vanishing of these bundles).</p> <p><strong>Remark:</strong> Both Brylinskis have made fundamental contributions to the theory of Kazhdan-Lusztig polynomials. Jean-Luc Brylinski and M. Kashiwara, and independently Beilinson and Bernstein, proved the Kazhdan-Lusztig conjectures (this is the interpretation mentioned by Jan in his question); Ranee Brylinski, Jean-Luc's wife, gave the interpretation I've described above.</p> <p><strong>References:</strong></p> <p>(1) Brylinski, R. K. Limits of weight spaces, Lusztig's $q$-analogs, and fiberings of adjoint orbits. J. Amer. Math. Soc., 1989, Vol. 3, pp. 517--533.</p> <p>(2) Available on the ArXiV <a href="http://arxiv.org/abs/0803.3424" rel="nofollow">here</a>.</p> http://mathoverflow.net/questions/80558/reference-needed-for-representation-theory-of-direct-products-of-algebraic-groups/80610#80610 Answer by Chuck Hague for Reference needed for representation theory of direct products of algebraic groups over a field (of arbitrary characteristic) Chuck Hague 2011-11-10T17:29:16Z 2011-11-10T17:29:16Z <p>As Vladimir mentions, the statement of your theorem is unclear. However, since you have written the $G$- and $H$-actions as tensor product actions where each acts nontrivially on one tensor factor and trivially on the other, it appears that you are making the following claim (please correct me if I'm wrong):</p> <blockquote> <p>Any representation of $G \times H$ is isomorphic to a representation of the form $V \otimes W$, where $V$ is a representation of $G$ and $W$ is a representation of $H$, and where we let the $G$-action on $W$ and the $H$-action on $V$ be trivial.</p> </blockquote> <p>If this is the statement you intend to make, it is false. Indeed, if we take two representations $M$, $M'$ of $G$ and two representations $N$, $N'$ of $H$, then in general the $G \times H$-module $( M \otimes N ) \oplus ( M' \otimes N' )$ will not be isomorphic to any module of the form $V \otimes W$ as described above.</p> http://mathoverflow.net/questions/131639/on-the-associated-graded-ring-to-a-section-ring Comment by Chuck Hague Chuck Hague 2013-05-24T14:28:39Z 2013-05-24T14:28:39Z That makes sense, although it's not clear to me why the multiplication map $R_1(\mathcal L)^{\otimes m} \to R_m(\mathcal L)$ should be surjective (but I might be missing something easy here). http://mathoverflow.net/questions/129218/algebraic-stratifications-of-g-varieties Comment by Chuck Hague Chuck Hague 2013-04-30T16:54:20Z 2013-04-30T16:54:20Z This is true in the case of partial flag varieties, cf the comments at the beginning of section 3.12 in this paper: <a href="http://arxiv.org/pdf/0809.4785.pdf" rel="nofollow">arxiv.org/pdf/0809.4785.pdf</a> http://mathoverflow.net/questions/126722/why-does-the-naive-choice-of-homogeneous-coordinate-ring-of-a-product-of-projecti/126728#126728 Comment by Chuck Hague Chuck Hague 2013-04-06T19:57:34Z 2013-04-06T19:57:34Z Very nice answer -- thanks! http://mathoverflow.net/questions/126722/why-does-the-naive-choice-of-homogeneous-coordinate-ring-of-a-product-of-projecti Comment by Chuck Hague Chuck Hague 2013-04-06T18:36:34Z 2013-04-06T18:36:34Z Ah yes, that makes sense. So I guess this boils down to the obvious fact that a product of affine cones over a projective scheme will never be an affine cone over the product (eg by dimension considerations), which immediately shows that the naive hypothesis is false. http://mathoverflow.net/questions/125943/degree-of-a-commutator-in-a-hyperalgebra-or-enveloping-algebra Comment by Chuck Hague Chuck Hague 2013-03-29T21:31:14Z 2013-03-29T21:31:14Z Oh yikes, yes. That's a good point. I obviously haven't thought this through well enough. http://mathoverflow.net/questions/119850/can-one-embedd-the-projectivezed-tangent-space-of-cp2-in-a-projective-space Comment by Chuck Hague Chuck Hague 2013-01-25T17:57:34Z 2013-01-25T17:57:34Z Sorry -- I should have been more clear. I'm using the fact that if E is an algebraic vector bundle on a variety then indeed the projectivization of E is also a variety (this is contained in the $\textbf{Proj}$ construction, cf Section II.7 of Hartshorne). Since the tangent space to $\mathbb P^n$ is an algebraic bundle the result follows. (As for the projectiveness, one doesn't require a variety to be projective for it to embed in projective space -- eg, every affine variety will embed in projective space too. If we are talking about a closed embedding, of course the answer is different.) http://mathoverflow.net/questions/119850/can-one-embedd-the-projectivezed-tangent-space-of-cp2-in-a-projective-space Comment by Chuck Hague Chuck Hague 2013-01-25T17:13:38Z 2013-01-25T17:13:38Z Since every complex algebraic variety can be embedded in a projective space, unless I'm missing something here the answer to your question is trivially yes. http://mathoverflow.net/questions/118661/good-book-on-representation-theory-of-gln/118669#118669 Comment by Chuck Hague Chuck Hague 2013-01-14T17:38:33Z 2013-01-14T17:38:33Z I would second the recommendation for Fulton and Harris. It covers the basics of the representation theory of $GL(n)$ in a friendly and accessible way. http://mathoverflow.net/questions/118647/on-the-blow-up-along-the-diagonal-in-a-product/118840#118840 Comment by Chuck Hague Chuck Hague 2013-01-14T17:29:14Z 2013-01-14T17:29:14Z Great, thanks! That's very helpful. http://mathoverflow.net/questions/112214/on-the-divided-power-ring-over-the-integers/112246#112246 Comment by Chuck Hague Chuck Hague 2012-11-13T14:16:33Z 2012-11-13T14:16:33Z Very nice answer! I think that my above statement is true for an algebraically closed field, but somehow I'd tricked myself into thinking it worked over any field. http://mathoverflow.net/questions/109225/on-q-demazure-operators Comment by Chuck Hague Chuck Hague 2012-10-09T18:22:54Z 2012-10-09T18:22:54Z Ah, that's useful -- a Mathscinet search gives a few papers following up on Kumar's question that might be applicable. http://mathoverflow.net/questions/109225/on-q-demazure-operators Comment by Chuck Hague Chuck Hague 2012-10-09T16:07:46Z 2012-10-09T16:07:46Z Thanks for your comments -- I was vague in my statements because I didn't know how far the idea of Demazure operators went, but it seems as though they occur in many places. I didn't know about the quantum Schubert calculus; I'll look at that. My interest is in the non-quantum (classical) case (ie, &quot;q-ifying&quot; the classical formulas), but I'm also interested in the quantum case if it can shed light on the classical case. http://mathoverflow.net/questions/107504/an-equivalence-between-mathcald-xm-textmod-and-mathcald-xm1 Comment by Chuck Hague Chuck Hague 2012-09-24T17:14:28Z 2012-09-24T17:14:28Z I don't know a lot about this story, but you may want to look at two papers by Haastert: &quot;On direct and inverse images of D-modules in prime characteristic&quot; and &quot;UÌˆber Differentialoperatoren und D-Moduln in positiver Charakteristik.&quot; In the latter paper he considers modules over various such rings $\mathcal D^m_X$ of differential operators and proves that for a smooth variety there is an equivalence of categories between $\mathcal D^m_X-\textrm{mod}$ and $\mathcal O_{X^{(m)}}-\textrm{mod}$. http://mathoverflow.net/questions/106230/schubert-varieties-which-admit-small-resolutions-of-singularities Comment by Chuck Hague Chuck Hague 2012-09-03T15:34:02Z 2012-09-03T15:34:02Z The book &quot;Singular Loci of Schubert Varieties&quot; by Billey and Lakshmibai has some results about small resolutions (cf in particular section 9.1); you might want to take a look there. http://mathoverflow.net/questions/106014/does-g-h-quotient-of-a-real-semisimple-lie-group-by-a-cartan-subgroup-have-a Comment by Chuck Hague Chuck Hague 2012-08-31T13:19:38Z 2012-08-31T13:19:38Z @solbap: One needs to be a bit careful here - the cotangent bundle of $G/B$ is not a homogeneous $G$-space and hence will never be of the form $G/H$.