User spg - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T04:22:22Z http://mathoverflow.net/feeds/user/15933 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/77975/macdonald-polynomials-and-macdonald-positivity/77981#77981 Answer by SPG for Macdonald polynomials and Macdonald positivity SPG 2011-10-13T01:49:07Z 2011-10-13T01:49:07Z <p>As far as I know there is no positivity statement for $P_\lambda$ as a linear combination of characters (which are the analog of Schur functions in general type). There is a very general statement that $P_\lambda$ expands positively as a sum of monomial symmetric functions; this is proved in a root-system uniform way by Ram and Yip in</p> <p><a href="http://arxiv.org/abs/0803.1146" rel="nofollow">http://arxiv.org/abs/0803.1146</a></p> <p>The proof they give is for the "constant parameters" situation, but the technique works in general. There is some inefficiency in their formula (in type A it can be "compressed" to a formula of Haglund, Haiman and Loehr in <a href="http://arxiv.org/abs/math/0409538" rel="nofollow">http://arxiv.org/abs/math/0409538</a>) which is the subject of current work. </p> <p>Presumably character-positivity of $P_\lambda$ actually fails already in type A, and in general type there is no notion of plethystic substitution to save us!</p> http://mathoverflow.net/questions/73246/why-is-mathbbr-infty-defined-the-way-it-is/73414#73414 Answer by SPG for Why is $\mathbb{R}^{\infty}$ defined the way it is? SPG 2011-08-22T15:33:50Z 2011-08-22T15:33:50Z <p>Perhaps the following is so obvious that no one saw fit to mention it in a comment:</p> <p>Let $X$ be a paracompact topological space. With the definition the way it is, homotopy classes of maps from $X$ to the $n$-Grassmannian are in bijection with isoclasses of rank $n$ vector bundles on $X$. (If you haven't got to them yet, this is thms 5.6 and 5.7). This is what really gets used in application to Stiefel-Whitney classes. </p> <p>I don't know what is classified by the other definition of the $n$-Grassmannian you suggest ($n$ dimensional subspaces of the product of infinitely many copies of the reals). But certainly for your purposes (working through Milnor+Stasheff) this is the point. </p> <p>The fact that the usual Grassmannian is a direct limit is used in several technical lemmas in chapter 5, but I have never tried pushing the construction you suggest through. </p> http://mathoverflow.net/questions/72268/less-discriminating-discriminants/72382#72382 Answer by SPG for Less discriminating discriminants SPG 2011-08-08T18:31:46Z 2011-08-08T18:31:46Z <p>For partitions of the special form $\lambda=(k,1,1,\dots,1)$ (i.e., hook shapes) there is a very explicit description of the ideal of definition of the corresponding set of polynomials (i.e., those polynomials with a root of order $\geq k$). Namely, it is the span of the set of (symmetric) Jack polynomials $f_\mu$, with parameter specialized to $-1/k$, and where $\mu$ ranges over all partitions satisfying</p> <p>$$\mu_i-\mu_{i+k-1} \geq 2$$ for all $i$. This is the main result of the paper <a href="http://arxiv.org/pdf/math/0112127" rel="nofollow">http://arxiv.org/pdf/math/0112127</a> by Feigin, Jimbo, Miwa, and Mukhin.</p> <p>For the application to degree, one might note that this computes the Hilbert function of this ideal quite explicitly in combinatorial terms, though for your problem you are probably interested in the grading where each elementary symmetric function has degree one, and I don't know, off the top of my head, how to extract that information. Maybe I'll post a separate question so we can see if anywhere else here knows.</p> http://mathoverflow.net/questions/72198/algebraic-geometric-measure-theory/72202#72202 Answer by SPG for Algebraic geometric measure theory SPG 2011-08-05T18:51:27Z 2011-08-07T03:35:39Z <p>There is an explicit upper bound based on a 2-d version of the Crofton formula. Namely, the area of $B \cap V$ is the integral of the number of points of intersection $W \cap (B \cap V)$ over the space of all affine 2-planes $W \subseteq \mathbb{R}^{2n}$. Since the real algebraic variety $V$ has degree $\leq d^2$ the number of points of intersection is at most $d^2$. So an upper bound is $d^2$ times the measure of the space of affine $2$-planes meeting $B$. It seems to me that, unless I have misunderstood, the bound on the coefficients is unnecssary.</p> http://mathoverflow.net/questions/70575/why-is-there-such-a-close-resemblance-between-the-unitary-representation-theory-o/70588#70588 Answer by SPG for Why is there such a close resemblance between the unitary representation theory of the Virasoro algebra and that of the Temperley-Lieb algebra? SPG 2011-07-18T01:37:45Z 2011-07-18T19:51:46Z <p>The Cherednik algebra has a similar classification into discrete and unitary series: see arXiv:1106.5094 and arXiv:0901.4595. Strictly speaking, these papers classify the unitary irreducibles in category O. I don't know whether there is a larger category in which contravariant forms will exist, but anyway for the symmetric group category O will be closely tied to affine Lie algebras (thus to Virasoro) by the Arakawa-Suzuki functor, and to Hecke (thus TL algebras) by the Knizhnik-Zamolodchikov functor (which actually identifies O with the category of q-Schur modules for most values of the parameter). Maybe the Cherednik algebra can serve as a bridge between them: Etingof conjectures (true by case by case check for the symmetric group) that KZ of a unitary module is unitary, and it is true (again case by case) that via Arakawa-Suzuki the unitary modules (i.e. integrable modules) for affine $gl_n$ correspond to unitary modules for the Cherednik algebra. </p> <p>At least for the symmetric group, the question of when there is a faithful unitary module in O is not very interesting: there is always one (either $L_c(triv)$ or $L_c(sign)$ will work). But if one is to make the connection to TL and the Virasoro algebra work probably one needs more detail.</p> <p>Every Cherednik algebra module is in particular a module over a ring C[V] of polynomial functions on a vector space V, and its support is a subvariety of V. The faithful unitaries should be the unitaries with full support (I have not checked this, though one direction is obvious).</p> <p>In the (much simpler) case of the Cherednik algebra of the symmetric group $S_n$, the algebra depends on one parameter c, which we may assume positive. The irreducibles in O are indexed by irreducible $S_n$-modules, and therefore by partitions of n. Writing $a(\lambda)$ for the largest hook length of the partition $\lambda$ and $b(\lambda)$ for a certain smaller hook length (see the paper of Etingof/Stoica for the precise def'ns), the corresponding irreducible $L_c(\lambda)$ is unitary iff $\lambda=(1^n)$ (corresponding to the sign representation), or $c \leq a(\lambda)$ or $c=1/m$ for a positive integer $m$ with $m \leq b(\lambda)$. The continuous part of the unitary set is precisely the closure of the set where the corresponding standard module is irreducible and unitary (this much is not surprising: the condition for the contravariant form to be positive definite on the standard module is open, and it's obviously pos. def. at $0$).</p> <p>The module $L_c(\lambda)$ has full support iff: $c$ is not rational or $c=k/m$ and the partition is $m$-regular: the differences $\lambda_i-\lambda_{i+1}$ are strictly less than $m$. Thus $L_c(\lambda)$ is unitary of full support iff (1) $\lambda=(1^n)$, (2) $\lambda=(n)$ and $0 \leq c &lt; 1/n$, (3) $\lambda \neq (n),(1^n)$ is a rectangle and $c \in [0,1/a(\lambda)]$ or $c=1/m$ for a positive integer <code>$m$</code> with <code>$m&lt;b(\lambda)$</code>, (4) <code>$\lambda$</code> is not a rectangle and <code>$c \in [0,1/a(\lambda)]$</code> or <code>$c=1/m$</code> for a positive integer <code>$m$</code> with <code>$m \leq b(\lambda)$</code>.</p> <p>Taking the $n \rightarrow \infty$ limit of all this should be possible; I am running out of time again. </p> http://mathoverflow.net/questions/68378/clifford-algebra-non-zero/68466#68466 Answer by SPG for Clifford algebra non-zero SPG 2011-06-22T01:09:52Z 2011-06-22T01:36:05Z <p>Here is a braindead way to generate PBW theorems like this one:</p> <p>One is given a presentation of an algebra, which allows one to put words in the generators $x_1,x_2,\dots,x_n$ into a "normal form" (in the case of the Clifford algebra, if $x_1,x_2,\dots,x_n$ is a basis of $V$, then a normal form might be words $x_{i_1} x_{i_2} \cdots x_{i_p}$ with $1 \leq i_1 &lt; i_2 &lt; \cdots &lt; i_p \leq n$). One suspects that the set of words in normal form is actually a basis for the algebra. So one constructs the regular representation of the algebra in question. If the theorem is to be true, there is no choice about this: it is the vector space spanned by words in normal form, and the left (and right) multiplication operators are determined by the relations. Usually the easiest way to write down the formulas is recursively. One is then left to check that these operators satisfy the defining relations. This is "just linear algebra", but the computations in any particular example (or class of examples) may get messy. When it works, it usually works in arbitrary characteristic (and even integrally).</p> <p>In the case of the Clifford algebra and similar algebras (e.g. enveloping algebras of Lie algebras, symplectic reflection algebras and their generalizations) this all works without too much difficulty. It also works for the Hecke algebra attached to a Coxeter system, though to make the calculations manageable in a case-free fashion it's good to use the Bourbaki trick of employing both the right and left representations simultaneously. There is even a general theorem here, which usually goes by the name "Bergman diamond lemma" (but in the cases I'd care most about, checking that its conditions are satisfied is just about the same level of difficulty as doing the work directly).</p> http://mathoverflow.net/questions/77948/an-isomorphism-of-2-schur-modules Comment by SPG SPG 2011-10-13T13:32:37Z 2011-10-13T13:32:37Z I plead: use the notation &quot;\oplus 2&quot; in the exponent! http://mathoverflow.net/questions/76942/what-motivates-modern-algebraic-geometry-for-a-combinatorial-constructive-algebra/76960#76960 Comment by SPG SPG 2011-10-05T13:30:51Z 2011-10-05T13:30:51Z KConrad, I don't think we have a fundamental disagreement here. You are right, the Garsia-Haiman modules $D_\mu$ give a satisfying rep.thy. explanation for positivity, and it's amazing that 3 of the 4 known proofs (Haiman via geometry of Hilbert schemes, Grojnowski-Haiman, and now Gordon, via Hodge theory) use so much modern machinery. Producing an explicit bigraded basis might also be possible via AG or RT; at any rate this is an obvious challenge for people working in the area. Haiman himself regards this as important: his (former) student Sami Assaf has been working on a CO approach. http://mathoverflow.net/questions/76942/what-motivates-modern-algebraic-geometry-for-a-combinatorial-constructive-algebra/76960#76960 Comment by SPG SPG 2011-10-04T22:13:11Z 2011-10-04T22:13:11Z BTW, I think the &quot;real&quot; thm proved by Haiman in this connection is the identification of $S_n$-Hilb of $\mathbb{C}^{2n}$ with the Hilbert scheme of points in the plane. This is of course of interest for its own sake---it's just that I don't see that the combinatorial corollary is all that one might hope for. http://mathoverflow.net/questions/76942/what-motivates-modern-algebraic-geometry-for-a-combinatorial-constructive-algebra/76960#76960 Comment by SPG SPG 2011-10-04T22:08:51Z 2011-10-04T22:08:51Z Alex, thanks for clarifying my assertion somewhat. KConrad, I don't see a binary choice here, and it's not a matter of taste: Haiman's theorem is a very good theorem. An even better theorem would give an effective procedure for calculating the expansion of Macdonald polynomials in terms of Schur functions, with coefficients that have (1) a concrete geometric and/or representation theoretic interpretation that (2) visibly deforms the usual tableaux combinatorics (which should be recovered by setting q=1=t). Unfortunately, this is not the type of result that is typically provided by AG... http://mathoverflow.net/questions/76942/what-motivates-modern-algebraic-geometry-for-a-combinatorial-constructive-algebra/76960#76960 Comment by SPG SPG 2011-10-02T14:17:16Z 2011-10-02T14:17:16Z It is of course true that this is a nice concrete application of modern algebraic geometry. But people interested in explicit formulas have the right to expect more than these techniques give: we don't want an abstract reason for positivity as much as a manifestly positive formula. http://mathoverflow.net/questions/74664/some-weird-system-of-inequalities-in-nonnegative-integers Comment by SPG SPG 2011-09-06T16:24:28Z 2011-09-06T16:24:28Z Do you mean to have $i \leq j \leq k \leq l$ in the first line? http://mathoverflow.net/questions/73260/characterizing-zeros-of-schur-functions-over-mathbbrn-or-mathbbcn Comment by SPG SPG 2011-08-30T15:48:29Z 2011-08-30T15:48:29Z Ahmed, I don't understand the definition of $Z_{d,n}$. Is it the set of common zeros of all Schur functions of degree $d$ in $n$ variables? Nor do I understand the assumption that follows: what is &quot;the&quot; Schur function whose zeros are assumed to generate a free abelian group of rank equal to the number of variables? It seems to me that this can basically never hold for a single polynomial (assuming $n&gt;1$ and you mean zeros in $\CC^n$, anyway). http://mathoverflow.net/questions/73246/why-is-mathbbr-infty-defined-the-way-it-is/73414#73414 Comment by SPG SPG 2011-08-29T14:43:02Z 2011-08-29T14:43:02Z In fact, that's the only sentence in the OP's question with a question mark after it, so I'm even more puzzled by your assertion that you don't see anything that implies we are comparing the direct sum with the direct product. http://mathoverflow.net/questions/73246/why-is-mathbbr-infty-defined-the-way-it-is/73414#73414 Comment by SPG SPG 2011-08-29T14:40:13Z 2011-08-29T14:40:13Z From the question: &quot;My question is why do we insist that only finitely many of the xi are non-zero for each (x1,x2,x3,â€¦)âˆˆRâˆž?&quot; If we do not insist this, we get the direct product, do we not? This seems unambiguous to me. http://mathoverflow.net/questions/73564/when-can-we-lift-an-action-of-lie-algebra/73570#73570 Comment by SPG SPG 2011-08-24T15:04:37Z 2011-08-24T15:04:37Z Presumably (this applies to Giuseppe's answer also, I believe) the OP is interested in infinite dimensional $V$. For instance, what is the representation of $SL_2$ that corresponds to the Verma module $M(0)$ for $sl_2$? I don't think there is one. http://mathoverflow.net/questions/73246/why-is-mathbbr-infty-defined-the-way-it-is/73414#73414 Comment by SPG SPG 2011-08-24T12:55:14Z 2011-08-24T12:55:14Z Andrew, I thought the question was pretty clear (and don't understand all the fuss): why define the Grassmannian of n-planes using the direct sum instead of direct product? The cleanest possible answer is: because using direct product doesn't classify vector bundles on paracompact spaces. Unfortunately, it's not clear to me that this is true, but it might be worth thinking about for a little while for someone who is interested. http://mathoverflow.net/questions/73246/why-is-mathbbr-infty-defined-the-way-it-is/73414#73414 Comment by SPG SPG 2011-08-23T13:01:18Z 2011-08-23T13:01:18Z Andrew, I'm not sure our understandings of the question are the same. I thought the question was, &quot;Why use the direct sum of countably many real lines instead of the direct product?&quot;. Your comment seems to indicate that you think the questioner was asking why Milnor and Stasheff use the particular model they do rather than any other model. I have to admit, I think the answer to your version of the question is easy: because it's technically convenient. After all, if your goal is thms 5.6 and 5.7, getting any model, by hook or by crook, is fine, and the one they use is the obvious one. http://mathoverflow.net/questions/73246/why-is-mathbbr-infty-defined-the-way-it-is/73414#73414 Comment by SPG SPG 2011-08-23T12:49:57Z 2011-08-23T12:49:57Z Andrew, it's not clear to me that the other definition will have the same homotopy type. Is it clear to you? http://mathoverflow.net/questions/72707/what-locally-euclidean-topological-spaces-are-embeded-in-mathbbrn Comment by SPG SPG 2011-08-11T20:45:53Z 2011-08-11T20:45:53Z What's wrong with just writing &quot;Locally Euclidean topological space&quot;? Not much longer than &quot;topological manifold&quot;, and not ambiguous. http://mathoverflow.net/questions/21791/general-equilibrium-for-mathematicians/37917#37917 Comment by SPG SPG 2011-08-11T18:38:10Z 2011-08-11T18:38:10Z ...''just a fixed point theorem.'' I don't find this criticism very trenchant. One of the fundamental problems of mathematics is to solve equations (of various sorts). A result that proves that a system of equations has at least one solution is a good result. How good depends on how non-obvious the existence of the solution is. Maybe von Neumann's criticism was somewhat more detailed? How non-obvious was Nash's result at the time?