bio | website | mathoverflow.net |
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location | ||
age | ||
visits | member for | 3 years, 4 months |
seen | Sep 9 '13 at 12:53 | |
stats | profile views | 283 |
Feb 9 |
awarded | Yearling |
Oct 14 |
awarded | Nice Answer |
Jun 22 |
awarded | Yearling |
Oct 13 |
comment |
An isomorphism of 2-Schur modules
I plead: use the notation "\oplus 2" in the exponent! |
Oct 13 |
answered | Macdonald polynomials and Macdonald positivity |
Oct 5 |
comment |
What motivates modern algebraic geometry for a combinatorial/constructive algebraist?
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. |
Oct 4 |
comment |
What motivates modern algebraic geometry for a combinatorial/constructive algebraist?
BTW, I think the "real" 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. |
Oct 4 |
comment |
What motivates modern algebraic geometry for a combinatorial/constructive algebraist?
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... |
Oct 2 |
comment |
What motivates modern algebraic geometry for a combinatorial/constructive algebraist?
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. |
Sep 6 |
comment |
Some weird “system” of inequalities in nonnegative integers.
Do you mean to have $i \leq j \leq k \leq l$ in the first line? |
Aug 30 |
comment |
Characterizing zeros of schur functions over $\mathbb{R^n}$ or $\mathbb{C^n}$
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 "the" 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 $n>1$ and you mean zeros in $\CC^n$, anyway). |
Aug 30 |
awarded | Critic |
Aug 29 |
comment |
Why is $\mathbb{R}^{\infty}$ defined the way it is?
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. |
Aug 29 |
comment |
Why is $\mathbb{R}^{\infty}$ defined the way it is?
From the question: "My question is why do we insist that only finitely many of the xi are non-zero for each (x1,x2,x3,â€¦)âˆˆRâˆž?" If we do not insist this, we get the direct product, do we not? This seems unambiguous to me. |
Aug 24 |
comment |
when can we lift an action of Lie algebra?
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. |
Aug 24 |
comment |
Why is $\mathbb{R}^{\infty}$ defined the way it is?
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. |
Aug 23 |
comment |
Why is $\mathbb{R}^{\infty}$ defined the way it is?
Andrew, I'm not sure our understandings of the question are the same. I thought the question was, "Why use the direct sum of countably many real lines instead of the direct product?". 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. |
Aug 23 |
comment |
Why is $\mathbb{R}^{\infty}$ defined the way it is?
Andrew, it's not clear to me that the other definition will have the same homotopy type. Is it clear to you? |
Aug 22 |
answered | Why is $\mathbb{R}^{\infty}$ defined the way it is? |
Aug 11 |
comment |
General Equilibrium for Mathematicians
...''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? |