User allen knutson - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T08:33:47Z http://mathoverflow.net/feeds/user/391 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/131168/a-duality-on-partial-permutations A duality on partial permutations Allen Knutson 2013-05-19T20:48:24Z 2013-05-20T01:34:19Z <p>A <strong>partial permutation matrix</strong> $\pi$ is one with at most one 1 in any row and column (the rest 0s). Given one, we can cross out to the East and South (but not Southeast) of each 1. Some boxes get crossed out just horizontally, just vertically, neither, or both (such as the $1$s, which we think of as crossing themselves out both ways). Call these the <strong>SE hooks</strong> of $\pi$.</p> <p>Similarly, we can define the <strong>NW hooks</strong> of $\rho$. Call $(\pi,\rho)$ <strong>complementary</strong> if when the SE hooks of $\pi$ are overlaid on the NW hooks of $\rho$, each box is crossed out, and no $1$ crosses out any other. It is easy to see that for any $\pi$, there exists a unique complementary $\rho$, and $\pi \mapsto w_0 \rho w_0$ is an involution.</p> <blockquote> <p>Has this involution on partial permutations been studied before?</p> </blockquote> <p>For me, it has come up in studying conormal varieties to Schubert varieties.</p> <p>ADDED: Here's a familiar subcase. The <strong>diagram</strong> of $\pi$ is the set of boxes not crossed out in its SE hooks. If this is a partition in the NW corner, $\pi$ is called <strong>dominant</strong>, and every partition arises from a unique $\pi$. In this case, $w_0 \rho w_0$ is also dominant, and its diagram is (the rotation of) the complementary partition.</p> http://mathoverflow.net/questions/129857/relating-two-characterizations-of-mathfrak-sl-n-2-among-simple-lie-algeb Relating two characterizations of ${\mathfrak sl}_{n > 2}$ among simple Lie algebras Allen Knutson 2013-05-06T15:53:28Z 2013-05-18T14:38:29Z <p>Let $\mathfrak g$ be a finite-dimensional simple Lie algebra over $\mathbb C$.</p> <p>Theorem 1. The highest root is perpendicular to all but one simple root, except in the case ${\mathfrak g}={\mathfrak sl}_{n > 2}$, in which case it is perpendicular to all but two (the first and last).</p> <p>Theorem 2. The space $Hom({\mathfrak g}\otimes {\mathfrak g},{\mathfrak g})$ is $1$-dimensional (spanned by the Lie bracket), except for $\mathfrak g = {\mathfrak sl}_{n > 2}$, in which case it is $2$-dimensional. (The symmetric product is $(A,B)\mapsto$ the traceless part of $AB+BA$.)</p> <p>Bruce Fontaine showed me a proof of the second, using the first, via the geometric Satake correspondence. Much as I like that, I'm wondering if there is a more classical argument connecting these two facts.</p> <p>(Feel free to retag if inspired.) EDIT: I forgot that ${\mathfrak sl}_2$ is the exception that proves the rule.</p> http://mathoverflow.net/questions/130957/affine-weyl-group-and-affine-schubert-cells/130966#130966 Answer by Allen Knutson for affine weyl group and affine schubert cells Allen Knutson 2013-05-17T14:36:04Z 2013-05-17T14:56:05Z <p>If $G$ isn't simply connected, then $G(F)$ isn't connected, so $G(F)/I$ isn't connected. Your question is about resolving those $\overline{IwI}/I$ that live in components other than the one containing the basepoint $I/I$.</p> <p>On such a component $C$, there will still be a closed $I$-orbit $X_C$, automatically smooth. (In the familiar case, <code>$X_C = \{ I/I \}$</code>.) Generalize the Bott-Samelson-Demazure-Hansen resolution to $P_{\alpha_1} \times^I \cdots \times^I P_{\alpha_k} \times^I X_C \to C$. The LHS is smooth, and can be made to resolve any choice of $\overline{IwI}/I$, through suitable choice of $\vec \alpha$.</p> http://mathoverflow.net/questions/24748/hopf-fibration-inside-the-retraction-of-r4-minus-line-s2 Hopf fibration inside the retraction of R^4 minus line -> S^2? Allen Knutson 2010-05-15T13:08:49Z 2013-05-10T11:14:20Z <p>This was inspired by <a href="http://mathoverflow.net/questions/24671/why-do-the-homology-groups-capture-holes-in-a-space-better-than-the-homotopy-grou" rel="nofollow">this question</a>.</p> <p>Let $Y = {\mathbb R}^4 \setminus$a coordinate line, which retracts to ${\mathbb R}^3 \setminus$a point, which retracts to $S^2$.</p> <p>What is an explicit <em>immersion</em> $S^3 \to Y$, whose composition with the above retraction gives the Hopf fibration? </p> <p>My idea being, perhaps this would make clearer in what sense the $S^3$ is surrounding "a hole" in $S^2$.</p> http://mathoverflow.net/questions/129673/interpretation-of-multiplicity-of-a-point/129691#129691 Answer by Allen Knutson for Interpretation of multiplicity of a point Allen Knutson 2013-05-05T03:02:09Z 2013-05-05T03:02:09Z <p>I think more fundamental and interesting than the notion of "multiplicity" is the notion of "tangent cone". Given $x \in X = Spec \ R$ defined by an ideal $I$, it's an interesting fact that $I^\infty$ is small, so that $gr_I R := \oplus_n I^n/I^{n+1}$ is interesting to look at.</p> <p>(You can't do much with this if say, $R$ is the continuous functions on a compact Hausdorff space, as $I=I^2$.)</p> <p>I think of the degeneration of $R$ to $gr_I R$ (given by the Rees algebra) as follows: we stretch $X$ away from the point $x$, and take the stretch factor to $\infty$. When we're done, there's lots to ask. Is the limit reduced, irreducible, etc.? But in particular, since the ring is now graded, we can compute its growth, and the leading coefficient thereof is the "multiplicity".</p> http://mathoverflow.net/questions/128678/motivating-geometric-representation-theory/129646#129646 Answer by Allen Knutson for motivating geometric representation theory Allen Knutson 2013-05-04T14:01:22Z 2013-05-04T14:01:22Z <p>Consider triples $(\lambda,\mu,\nu)$ of dominant weights of $G$ such that the irrep $V_\nu$ occurs in $V_\lambda \otimes V_\mu$. Then this space of triples is closed under addition.</p> <p>Proof. An intertwiner can be identified with a $G$-invariant section of the $(\lambda^*,\mu,\nu)$ equivariant line bundle over $(G/B)^3$. Tensoring two such sections together, we get a third, <em>which is again nonzero</em> because $(G/B)^3$ is reduced and irreducible.</p> <p>(Moreover, this monoid is finitely generated, also not hard to prove with this approach.)</p> http://mathoverflow.net/questions/129044/defining-equations-of-a-flag-variety/129643#129643 Answer by Allen Knutson for Defining Equations of a Flag Variety Allen Knutson 2013-05-04T13:33:24Z 2013-05-04T13:33:24Z <p>Equations are for cutting schemes out of ambient schemes. Are you sure you want to cut your flag manifold out of projective space? There are easier places to find it.</p> <p>To begin with, you could embed $Flags(n) \to \prod_{k=1}^{n-1} Gr_k(n)$, taking the flag to its list of subspaces. Then the equations are "for each $i &lt; j$, the $i$-plane should be contained in the $j$-plane".</p> <p>You could Plücker embed each Grassmannian $Gr_k(n)$ into projective space, or you could regard it as $GL(k) \backslash \backslash M_{k\times n}$, i.e. look at row-spans of $k\times n$ matrices. Then the equations above say that when you stack your $i\times n$ matrix and $j\times n$, the resulting $(i+j)\times n$ matrix should only have rank $j$, so all $(j+1)\times (j+1)$ determinants should vanish. There's some equations.</p> <p>If you do Plücker embed, it means you only have the Plücker coordinates on those Grassmannians, and so you get Plücker relations between the Plücker coordinates of size $i$ and $j$. I find these much harder to remember than the determinants above.</p> <p>Once you've Plücker embedded the Grassmannians, then you can Veronese them each by different amounts, then Segre the whole thing together, and you get all the projecively normal embeddings of the flag manifold. It's interesting to note that all the equations encountered along the way are linear or quadratic (a theorem of Ramanathan for general $G$).</p> <p>Anyway one very good answer to your actual question is [Miller-Sturmfels], chapter 15 I think it is, as Victor Protsak suggested.</p> http://mathoverflow.net/questions/127779/higher-homotopy-groups-and-representation-theory/127830#127830 Answer by Allen Knutson for HIgher Homotopy Groups and Representation Theory Allen Knutson 2013-04-17T11:53:44Z 2013-04-17T11:53:44Z <p>If $G$ is compact, then we can think about its cohomology with Lie algebra cohomology, so be doing some kind of representation theory. And of course the first nontrivial homology is determined by the the first nontrivial homotopy. (Since $G$ is smooth, we can relate its homology and cohomology, too.)</p> <p>Specifically, if $G$ is simply connected (and compact), then $H^1 = H^2 = 0$, and $\dim H^3 =$ the number of simple factors. The $H^2$ result, one interprets as the statement that $G$ has no nontrivial central extensions. The $H^3$ (or $\pi_3$) result one usually interprets as saying that $LG$ (free loops) <em>does</em> have nontrivial central extensions.</p> http://mathoverflow.net/questions/127657/homotopy-classes-of-maps-to-lie-groups/127665#127665 Answer by Allen Knutson for Homotopy classes of maps to Lie groups Allen Knutson 2013-04-16T00:44:03Z 2013-04-16T00:44:03Z <p><a href="http://www.ams.org/bookstore?co1=AND&amp;co2=AND&amp;co3=AND&amp;d=BOOK&amp;f=G&amp;fn=105&amp;l=100&amp;op1=ADJ&amp;op2=ADJ&amp;op3=ADJ&amp;p=1&amp;pg1=&amp;pg2=&amp;pg3=ALLF&amp;r=4&amp;s1=&amp;s2=&amp;s3=mimura&amp;subject=genint" rel="nofollow">Topology of Lie Groups, I and II</a> certainly has the homotopy groups up a ways.</p> http://mathoverflow.net/questions/127293/stanley-reisner-ring-of-a-simplicial-complex-is-a-functor/127302#127302 Answer by Allen Knutson for Stanley-Reisner ring of a simplicial complex is a functor? Allen Knutson 2013-04-12T00:19:20Z 2013-04-14T13:42:35Z <p>I think I'd want to deal with <em>partially defined</em> functions $f: [n] \to [n']$, with the property that if $F$ is a face of $\Delta$, then $f(F)$ is a face of $\Delta'$. The linear extension of such an $f$ is a linear map from $K^n \to K^{n'}$, taking the $i$th basis vector to the $f(i)$th, or to zero if $f(i)$ is not defined.</p> <p>The Stanley-Reisner ideal is the functions vanishing on the union $X_\Delta \subseteq K^n$ of coordinate spaces $\bigcup_{F\in\Delta} K^F$, where $K^F$ denotes the linear span of {the $i$th basis vector : $i\in F$}. (Does that answer your second question?) Then the linear extension of $f$ takes $X_\Delta$ into $X_{\Delta'}$.</p> http://mathoverflow.net/questions/127176/halls-marriage-theorem-and-intervals Hall's Marriage Theorem and intervals Allen Knutson 2013-04-11T03:46:19Z 2013-04-11T11:15:27Z <p>In Hall's Marriage Theorem, we have a set $B$ of brides and $G$ of grooms, where each bride $b$ has an acceptable set $A_b \subseteq G$ of grooms. A <strong>matching</strong> $m:B\to G$ is an injection such that $m(b) \in A_b$ for each $b\in B$. (So each bride gets married, but some grooms may be out of luck.) Obviously, if there's some set $S \subseteq B$ of brides such that $|\cup_{b\in S} A_b| &lt; |S|$, then a matching is impossible; the theorem is that this is the <em>only</em> obstruction.</p> <p>I'm interested in the case that $G$ is an interval $[1,n]$ in $\mathbb N$, and each $A_b$ is a subinterval $[i,j]$. (Perhaps each bride is only willing to accept grooms within a certain range of heights.) I have been able to make the following refinement: if no matching is possible, then there is an interval $[x,y]$ such that <code>$|[x,y] \cap G| &lt; |\{b : A_b \subseteq [x,y]\}|$</code>.</p> <p>(This is an improvement in two ways -- it restricts the form of $S$, plus the left side is a priori larger than $|\cup_{b\in S} A_b|$.)</p> <blockquote> <p>Is this refinement known?</p> </blockquote> <p>This extension wasn't very difficult, but if it was known I'd rather give credit to its earlier discoverers.</p> http://mathoverflow.net/questions/126250/non-cohen-macaulay-varieties-and-groebner-degeneration/126256#126256 Answer by Allen Knutson for Non Cohen-Macaulay varieties and Groebner degeneration Allen Knutson 2013-04-02T11:05:30Z 2013-04-07T02:50:48Z <p>I will answer in the contrapositive. Let $X \subseteq \prod_i {\mathbb P}^{n_i}$ be irreducible of codimension $k$. If whenever $\sum k_i = k$, you can find subspaces $\prod_i {\mathbb P}^{n_i}$ that intersect $X$ in at most one point, then $X$ is called a "multiplicity-free subvariety".</p> <p>In this case Brion has proven that any degeneration of $X$ must still be Cohen-Macaulay: <a href="http://arxiv.org/abs/math/0211028" rel="nofollow">http://arxiv.org/abs/math/0211028</a> and in particular, $X$ itself must be!</p> <p>The case of the diagonal was studied by Cartwright and Sturmfels: <a href="http://arxiv.org/abs/0901.0212" rel="nofollow">http://arxiv.org/abs/0901.0212</a></p> http://mathoverflow.net/questions/126625/is-there-an-algorithm-to-decide-if-an-ideal-contains-monomials/126637#126637 Answer by Allen Knutson for Is there an algorithm to decide if an ideal contains monomials? Allen Knutson 2013-04-05T15:12:36Z 2013-04-06T13:18:05Z <p>Computing colon ideals is pretty quick. You could colon out the variables in order. If the ideal changes, record the variable that worked, and go back to the beginning of the list. Either you get to the unit ideal, in which case you found the lex-first monomial that's in the ideal, or you make it to the end of the list, in which case there's no monomial in your ideal.</p> <p>EDITS:</p> <ol> <li><p>There's no need to loop back to the beginning of the list. Do $x_1$ to completion, then $x_2$, and so on.</p></li> <li><p>Doing $x_i$ only requires a Gröbner basis for an elimination order for $x_i$, and the colon ideal will come with a Gröbner basis again, for free. So you don't even need a universal Gröbner basis, just $n$ elimination Gröbner bases.</p></li> <li><p>You could also do this by coloning out $x_1 x_2 \cdots x_n$, which is very close to François Brunault's <s>answer</s> comment. But my understanding is that colon ideals are computed using the elimination of a new variable, so I doubt this is actually faster than what I'm suggesting.</p></li> <li><p>You don't even need those full Gröbner bases, just ones that are Gröbner enough; see e.g. theorem 6.3 of <a href="http://arxiv.org/abs/dg-ga/9706003" rel="nofollow">http://arxiv.org/abs/dg-ga/9706003</a> where we use a criterion like this to compute a cohomology ring.</p></li> </ol> <p>ANOTHER:</p> <p>You could slice with random hyperplanes, and whenever you get isolated points, see if each of those points has some coordinate $=0$.</p> http://mathoverflow.net/questions/126516/simply-connected-algebraic-groups-and-reductive-subgroups-of-maximal-rank/126552#126552 Answer by Allen Knutson for Simply connected algebraic groups and reductive subgroups of maximal rank Allen Knutson 2013-04-04T19:28:16Z 2013-04-04T19:28:16Z <p>$G_2 \supset SO(4)$</p> http://mathoverflow.net/questions/126542/the-hilbert-function-of-an-intersection/126551#126551 Answer by Allen Knutson for The Hilbert function of an intersection Allen Knutson 2013-04-04T19:23:03Z 2013-04-04T19:23:03Z <p>It sure does help to assume complete intersection. It's nicer to describe the Hilbert series $H(t) := \sum_k h(k) t^k$ than the Hilbert function $h(k)$. For the polynomial ring itself, it's $H(t) = 1/(1-t)^{n+1}$. For the complete intersection, it's $(1-t^d)^r/(1-t)^{n+1}$ (with an obvious generalization if the degrees are different).</p> <p>If it's not a complete intersection, there's much less you can say -- you have to look at the syzygies of your generators, and the higher syzygies, and so on inclusion/exclusion.</p> <p>Going to $\sqrt{I}$, all that's easy to say is that the degree of $h(k)$ stays the same but the leading coefficient can drop.</p> http://mathoverflow.net/questions/26267/where-to-buy-premium-white-chalk-in-the-u-s-like-they-have-at-rims Where to buy premium white chalk in the U.S., like they have at RIMS? Allen Knutson 2010-05-28T14:08:44Z 2013-04-03T04:57:27Z <p>While not a research-level math question, I'm sure this is a question of interest to many research-level mathematicians, whose expertise I seek.</p> <p>At RIMS (in Kyoto) in 2005, they had the best white chalk I've seen anywhere. It's slightly larger than standard American chalk, harder, heavier, and most importantly covered with some enamel-like coating that one must rub through (on the end) to be able to write with. One's hands don't rub through the coating, and thus don't get chalky.</p> <p>Are there any U.S. manufacturers of such?</p> <p>EDIT: even though someone has given a link whereby to order this stuff from Japan, I would still be delighted to hear about American products that beat Binney &amp; Smith.</p> http://mathoverflow.net/questions/125606/other-variant-of-schur-polynomials-functions/125689#125689 Answer by Allen Knutson for Other Variant of Schur Polynomials/Functions Allen Knutson 2013-03-27T02:51:59Z 2013-03-27T02:51:59Z <p>The most general polynomials I'm familiar with, satisfying the vanishing and recursion properties you want, are the double Grothendieck polynomials <code>$\{G_\pi\}$</code>.</p> <p>There's one for every permutation $\pi$ of $\mathbb Z_+$ that moves only finitely many numbers. $G_\pi$ is a polynomial in two sets of variables, $x_1,x_2,\ldots$ and $y_1,y_2,\ldots$. The vanishing property is $G_\pi|_{y_i = x_{\rho(i)}} = 0$ for $\rho \not\geq \pi$. The recursion is defined using "Demazure operators" also known as "isobaric divided difference operators". I think the modern reference is Manivel's book on Schubert polynomials.</p> http://mathoverflow.net/questions/124265/non-uniqueness-of-smooth-compactification/124277#124277 Answer by Allen Knutson for Non-uniqueness of smooth compactification Allen Knutson 2013-03-11T23:54:17Z 2013-03-11T23:54:17Z <p>In characteristic 0, yes, yes, no. Check out Kollar's paper:</p> <p><a href="http://arxiv.org/abs/math/0508332" rel="nofollow">http://arxiv.org/abs/math/0508332</a></p> <p>In characteristic p, unknown.</p> http://mathoverflow.net/questions/123913/chern-numbers-via-euler-characteristics/123941#123941 Answer by Allen Knutson for Chern numbers via Euler characteristics? Allen Knutson 2013-03-08T05:14:20Z 2013-03-08T05:14:20Z <p>Let $B = Gr_{\dim E}(\infty)$ denote the classifying space for $\dim E$-bundles. Your $P(c_i(E))$ defines a class on $B$, which (if $P$ has integer coefficients, which you better've meant it to!) can pretty obviously be represented by an actual cycle, $\mathcal P \subseteq B$.</p> <p>Assume the classifying map $c : X \to B$ is transverse to $\mathcal P$. Then $c^{-1}(\mathcal P)$ is the space you're looking for. Too bad it's finite!</p> <p>Not quite. I want $c$ to be not just transverse to $\mathcal P$, but to meet it positively (unless you're okay with $\chi$ of a "negative point" being $-1$). Something like, $X$ should be algebraic, and the classifying map should be algebraic, some condition like $E$ being globally generated.</p> http://mathoverflow.net/questions/108098/grothendieck-ring-of-varieties-carrying-a-function Grothendieck ring of "varieties carrying a function" Allen Knutson 2012-09-25T20:45:45Z 2013-03-07T09:25:17Z <p>Fix a base ring $R$, and consider pairs $(X,f)$ where $X$ is a scheme of finite type over $R$ and $f:X\to R$ is an $R$-valued algebraic (not constructible!) function on $X$.</p> <p>I want to consider a Grothendieck $R$-algebra of such pairs, where if $X = Y \coprod Z$, then $[(X,f)] = [(Y,f|_Y)] + [(Z,f|_Z)]$, but also $[(X,f+g)] = [(X,f)] + [(X,g)]$ and $[(X,rf)] = r[(X,f)]$.</p> <blockquote> <p>Surely this is a standard extension of the usual notion of the Grothendieck ring of varieties (which only has $f=1$, and the first sort of relation)? If so, where can I read about it?</p> </blockquote> <p>Maybe I'm misreading the motivic integration survey literature (by K. Smith, and E. Looijenga), but it seems like they're insisting on constructible functions, not algebraic. Ordinarily when a construction like this isn't in the literature, I assume it's because it has too many relations and is $0$, but if $R = {\mathbb Z}$ it seems to me that this ring has many functionals, like $[(X,f)] \mapsto \sum_{x \in X_p} (f(x) \bmod p) \in {\mathbb Z}/p.$ (I don't see an analogue of $[X] \mapsto$ the Euler characteristic $\chi(X_{\mathbb C})$.)</p> <p>EDIT: One problem I see is that $({\mathbb A}^1, f(x)=x)$ is isomorphic under translation to $({\mathbb A}^1, f(x)=x+1)$. So $[({\mathbb A^1}, 1)] = [({\mathbb A}^1, (x+1)-x)] = [({\mathbb A}^1, x+1)] - [({\mathbb A}^1, x)] = 0$. Of course this fits with point-counting $\bmod p$.</p> http://mathoverflow.net/questions/123654/character-formula-for-demazure-modules/123662#123662 Answer by Allen Knutson for character formula for demazure modules Allen Knutson 2013-03-05T20:10:40Z 2013-03-05T20:10:40Z <p>Littelmann's, which gives a positive formula (counting Littelmann paths). His ICM address is here: <a href="http://www.mathunion.org/ICM/ICM1994.1/Main/icm1994.1.0298.0308.ocr.pdf" rel="nofollow">http://www.mathunion.org/ICM/ICM1994.1/Main/icm1994.1.0298.0308.ocr.pdf</a></p> <p>He proves its <em>validity</em> using Demazure operators -- I hope that doesn't disqualify it!</p> http://mathoverflow.net/questions/121958/representations-of-finite-subgroups-on-homology/121960#121960 Answer by Allen Knutson for Representations of Finite Subgroups on Homology Allen Knutson 2013-02-16T01:00:15Z 2013-02-16T01:00:15Z <p>Since $H\leq G$ and $G$ is connected, each $h\in H$ is connected to the identity, hence its action on $G$ is homologous to the identity.</p> <p>You're not going to get anything interesting here unless you either look at $N_G(H)/H$ acting on $H\backslash G$, or (equivalently) $N_G(H)$ acting on the $H$-equivariant cohomology of $G$, or something like that.</p> http://mathoverflow.net/questions/121299/edge-graph-of-the-polytope-of-a-bruhat-interval Edge graph of the polytope of a Bruhat interval Allen Knutson 2013-02-09T13:00:53Z 2013-02-11T00:45:37Z <p>Let $\Gamma$ be a Coxeter group on some generating set $S$, with reflection representation $V$. Then $\Gamma$ has two standard partial orders, the weak and strong Bruhat orders.</p> <p>Moreover, if $\lambda \in V$ is chosen generically (any free orbit will do), then the covering relations in weak order are given exactly by the edge graph of the convex hull of $\Gamma\cdot \lambda$.</p> <blockquote> <p>Let $[u,v]$ be an interval in strong Bruhat order. Has the edge graph of the polytope $hull([u,v]\cdot \lambda)$ been studied?</p> </blockquote> <p>For example, the polytope $hull([123,321] \cdot (1,2,3))$ is a hexagon, and the strong Bruhat cover $231 > 132$ defines an edge through the middle of this hexagon, so not a weak cover. Whereas the polytope $hull([132,321] \cdot (1,2,3))$ is a trapezoid, one edge of which connects $231$ and $132$.</p> <p>EDIT: perhaps I should admit the geometry here. If $\Gamma$ is a Weyl group of a Lie group $G$ -- and I am happy to make this assumption, albeit I want $G$ Kac-Moody -- and $V$ the corresponding weight lattice, and $\lambda$ a dominant weight, then $hull(W\cdot \lambda)$ is the moment polytope for $G/B$ bearing the Borel-Weil line bundle <code>${\mathcal L}_\lambda$</code>. Within $G/B$ we have the Richardson variety $\overline{BuB}/B \cap \overline{B_- vB}/B$, and $hull([u,v]\cdot \lambda)$ is the moment polytope of that.</p> http://mathoverflow.net/questions/120984/rep-theory-consequences-of-bott-weil-borel/121070#121070 Answer by Allen Knutson for Rep Theory Consequences of Bott--Weil--Borel Allen Knutson 2013-02-07T13:39:06Z 2013-02-07T13:39:06Z <p>What I'm writing here seems more like a contribution to a big-list than an "answer", but since you've already chosen one anyway...</p> <p>Say you're interested in which irreps $V_\nu$ occur in $V_\lambda \otimes V_\mu$. Let $C$ be the set of triples $(\lambda,\mu,\nu)$ for which this is true, a subset of $($the dominant weights$)^3$. Theorem: $C$ is closed under addition.</p> <p>The only proof I know is this: $(\lambda,\mu,\nu)$ is in $C$ iff the line bundle <code>${\mathcal L}_\lambda \boxtimes {\mathcal L}_\mu \boxtimes {\mathcal L}_\nu^*$</code> over $(G/B)^3$ has a nonzero $G$-invariant section. Given two nonzero sections for two different triples, we can tensor them together, <em>and the result will still be nonzero</em> because $(G/B)^3$ is reduced and irreducible. Hence the sum of the triples is again in $C$.</p> http://mathoverflow.net/questions/119665/geometric-description-of-the-algebra-of-central-functions-on-a-lie-group/119687#119687 Answer by Allen Knutson for "geometric" description of the algebra of central functions on a Lie group Allen Knutson 2013-01-23T19:12:42Z 2013-01-23T19:12:42Z <p>If $G$ is compact, then as explained in comments $G/ad$ is $T/W$. If further $G$ is simply-connected, hence a product of simple factors, then $T/W$ is a corresponding product of simplices.</p> <p>The point is that $T/W = ({\mathfrak t}/\Lambda)/W = {\mathfrak t}/(\Lambda \rtimes W)$, where $\Lambda$ is the kernel of the exponential map ${\mathfrak t}\to T$, and when $G$ is simply connected this latter group is an affine reflection group (the affine Weyl group, or at least, the product of those of the simple factors) whose fundamental chamber is the product of simplices.</p> http://mathoverflow.net/questions/119344/example-equivariant-mayer-vietoris-for-h-s1s2/119378#119378 Answer by Allen Knutson for Example equivariant Mayer-Vietoris for $H^{*}_{S^{1}}(S^{2})$ Allen Knutson 2013-01-20T00:21:39Z 2013-01-20T00:21:39Z <p>This is fine for $\mathbb Z$-coefficients; you don't need to go to $\mathbb R$.</p> <p>The part of the sequence you need is $$0\to H_{S^1}^0(S^2) \to H_{S^1}^0(U) \oplus H_{S^1}^0(V) \to H_{S^1}^0(U\cap V) \to H_{S^1}^1(S^2) \to H_{S^1}^1(U) \oplus H_{S^1}^1(V)$$ which you worked out to $$0\to H_{S^1}^0(S^2) \to {\mathbb Z}^2 \to {\mathbb Z} \to H_{S^1}^1(S^2) \to 0 \oplus 0$$ so it is enough to notice that the restriction map $H_{S^1}^0(V) \to H_{S^1}^0(U\cap V)$, aka $H^0({\mathbb CP}^{\infty}) \to H^0(pt)$, is onto. Then $H^0_{S^1}(S^2) = \mathbb Z$ and $H^1 = 0$.</p> http://mathoverflow.net/questions/119111/can-a-zerodivisor-reduce-both-the-depth-and-the-dimension/119122#119122 Answer by Allen Knutson for Can a zerodivisor reduce both the depth and the dimension? Allen Knutson 2013-01-16T23:59:01Z 2013-01-17T02:06:55Z <p>OOPS: As Sándor points out, I missed the assumption that the ring is local, in the following example of the wrong thing:</p> <p>"Inside 3-space, glue together a plane $y=0$ transversely with a parabola $z=0, x=y^2$ and a line $z=1, x=0$ meeting it in a separate point. This is reduced, and I'm pretty sure its depth is $1$. Because of the plane, its dimension is $2$.</p> <p>Now cut it with $x=0$, which cuts the plane to a line and the parabola to a double point leaving the line alone. Hence $x$ is a zero divisor in $k[x,y,z]/(\langle y\rangle \cap \langle z,x-y^2\rangle \cap \langle z-1,x\rangle)$, and cutting with it drops the dimension from $2$ to $1$.</p> <p>The resulting space is generically reduced, but not reduced, so I'm pretty sure its depth is $0$.</p> http://mathoverflow.net/questions/119121/representation-of-quotient-group/119125#119125 Answer by Allen Knutson for Representation of quotient group Allen Knutson 2013-01-17T00:24:07Z 2013-01-17T00:24:07Z <p>Yes, if $H$ is closed (i.e. not some irrational-flow subgroup inside the closed subgroup $Z(G)$).</p> http://mathoverflow.net/questions/116171/littlewood-richardson-coefficients-from-kostka-coefficients/116173#116173 Answer by Allen Knutson for Littlewood-Richardson Coefficients from Kostka coefficients Allen Knutson 2012-12-12T13:59:17Z 2012-12-12T13:59:17Z <p>I believe most of what you want is in <a href="http://front.math.ucdavis.edu/0308.5101" rel="nofollow">http://front.math.ucdavis.edu/0308.5101</a> , especially the polynomiality you're looking for. Note that that was first proven in [H. Derksen, J. Weyman] "On the Littlewood-Richardson polynomials," <a href="http://www.math.lsa.umich.edu/~hderksen/preprints/lrpoly.dvi" rel="nofollow">http://www.math.lsa.umich.edu/~hderksen/preprints/lrpoly.dvi</a> .</p> http://mathoverflow.net/questions/116034/moduli-space-of-polytopes/116036#116036 Answer by Allen Knutson for moduli space of polytopes Allen Knutson 2012-12-11T01:40:16Z 2012-12-11T01:40:16Z <p>One thing that is commonly done is to fix an initial polytope $P$, and consider all the polytopes whose fans are coarsenings of $P$'s fan. You can parametrize these by the space of convex piecewise-linear functions on $P$'s fan, to see that the moduli space itself forms a polyhedral cone.</p> <p>This is no good if you want to be able to turn the faces, just to breathe them in and out.</p> <p>I don't think the paper you cite will be of much use to you, unless you want a moduli space of polygons. The paper considers a space of polygons in 3-d with fixed edge lengths, but that comes with an involution "flip" whose fixed points are polygons in 2-d. But they're not convex, they self-intersect, etc. </p> http://mathoverflow.net/questions/131639/on-the-associated-graded-ring-to-a-section-ring Comment by Allen Knutson Allen Knutson 2013-05-24T03:40:55Z 2013-05-24T03:40:55Z I think all that's happening here is that you're filtering your $R$ by powers of an ideal $I$. No? In which case $R$ is Noetherian so $gr_I\ R$ is too. http://mathoverflow.net/questions/131449/motivation-for-frankls-conjecture Comment by Allen Knutson Allen Knutson 2013-05-22T14:15:12Z 2013-05-22T14:15:12Z If by &quot;motivation&quot; one means &quot;why should a mathematician care, how is it related to other mathematics&quot;, the evidence is that it's not, or we would have some way of attacking it. If one means &quot;why should one feel that it's true&quot;, I always sensed that taking unions `puts elements in' rather than taking them out, and having that $x$ is another sign of having elements in. http://mathoverflow.net/questions/131270/is-a-lie-group-equivariantly-formal-under-conjugation-by-a-maximal-torus Comment by Allen Knutson Allen Knutson 2013-05-22T11:48:24Z 2013-05-22T11:48:24Z No, I think it's the same -- in both cases the only interesting map is a certain $G/T$-bundle. I like your description a lot more, though! http://mathoverflow.net/questions/131364/how-much-of-character-theory-can-be-done-without-schurs-lemma-or-the-artin-wedde Comment by Allen Knutson Allen Knutson 2013-05-21T19:20:46Z 2013-05-21T19:20:46Z To be more precise, my suggestion was that your framework might be allowing in quantum groups at roots of unity, and their representations' &quot;quantum dimensions&quot;, alongside the things you actually want. http://mathoverflow.net/questions/131270/is-a-lie-group-equivariantly-formal-under-conjugation-by-a-maximal-torus Comment by Allen Knutson Allen Knutson 2013-05-21T01:51:45Z 2013-05-21T01:51:45Z Let's see, $H^*_{T_{Ad}}(G) = H^*_{T_{Ad}\times G}(G\times G) = H^*_{G_{Ad}}(G\times^T G) \from H^*_{G_{Ad}}(G)$ where the last is pullback along the multiplication map $G\times^T G \to G$. So I suspect you could use the representative from that last cohomology group. http://mathoverflow.net/questions/131168/a-duality-on-partial-permutations Comment by Allen Knutson Allen Knutson 2013-05-20T01:28:20Z 2013-05-20T01:28:20Z Absolutely right, I will correct the question. http://mathoverflow.net/questions/131156/how-can-i-randomly-draw-an-ensemble-of-unit-vectors-that-sum-to-zero Comment by Allen Knutson Allen Knutson 2013-05-19T18:14:21Z 2013-05-19T18:14:21Z Idea: pick $n$ randomly, subtract $1/n \sum v_i$ from each, then rescale each back to unit. I expect this would converge quickly, but don't have any reason to believe that it gets you exactly to the uniform distribution. http://mathoverflow.net/questions/131143/characterizing-a-certain-subset-of-isotropic-vectors Comment by Allen Knutson Allen Knutson 2013-05-19T13:49:56Z 2013-05-19T13:49:56Z What nondegeneracy condition do you want? At the moment $\mathcal S$ is closed under scaling, so contractible to the zero vector. http://mathoverflow.net/questions/131114/free-resolution-for-lie-algebras-reference Comment by Allen Knutson Allen Knutson 2013-05-19T03:10:10Z 2013-05-19T03:10:10Z Namely, <a href="http://mathoverflow.net/questions/130376/resolutions-of-lie-algebras" rel="nofollow" title="resolutions of lie algebras">mathoverflow.net/questions/130376/&hellip;</a> http://mathoverflow.net/questions/131054/how-to-detect-if-a-subgroup-lands-inside-an-orthogonal-group Comment by Allen Knutson Allen Knutson 2013-05-19T00:09:10Z 2013-05-19T00:09:10Z <a href="http://en.wikipedia.org/wiki/Frobenius%E2%80%93Schur_indicator" rel="nofollow">en.wikipedia.org/wiki/&hellip;</a> http://mathoverflow.net/questions/130654/a-catalog-of-faithful-representations-of-finite-groups/130657#130657 Comment by Allen Knutson Allen Knutson 2013-05-15T02:24:39Z 2013-05-15T02:24:39Z 10 seconds of playing with it shows me &quot;Choose the family of group&quot;, then &quot;Representations&quot;, and you should look at the first line of &quot;matrix representations&quot;. http://mathoverflow.net/questions/130537/reflection-lie-groups Comment by Allen Knutson Allen Knutson 2013-05-14T03:53:04Z 2013-05-14T03:53:04Z Indeed, if we compose $G \to U(V) \to U(V \otimes {\mathbb C}^2)$ then $G$ is not such a group. Hi Harm, welcome to MO! http://mathoverflow.net/questions/130445/ideal-generated-by-highest-weight-vectors Comment by Allen Knutson Allen Knutson 2013-05-13T03:45:28Z 2013-05-13T03:45:28Z May I assume $I_0$ homogeneous, and $G$ acting linearly? I'm just trying to understand the geometry so far. ${\mathbb P}V(I_0)$ is a $B$-invariant projective variety, and you're looking at the intersection of all its $G$-translates. That closed subscheme will contain some closed $G$-orbits, and it's enough to know that it's reduced along those... I doubt this is helping. http://mathoverflow.net/questions/130369/is-there-an-easy-way-to-write-down-the-singular-cohomology-of-a-hypersurface-in-a Comment by Allen Knutson Allen Knutson 2013-05-12T04:22:49Z 2013-05-12T04:22:49Z Since you're happy with Fano, this also specializes to &quot;write down the cohomology of a smooth anticanonical hypersurface&quot;, which is about the first sort of Calabi-Yau manifold physicists really grappled with. (After that they moved on to complete intersections, as you are, and beyond. But it was never easy.) http://mathoverflow.net/questions/130249/solve-the-singularities-of-parabolic-orbits-of-schubert-cells Comment by Allen Knutson Allen Knutson 2013-05-11T03:54:58Z 2013-05-11T03:54:58Z Why that's true! This sounds like a rather hard question.