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Schur functors are functors from the category of vector spaces to itself. If we take an operator $M: V->V$ and apply a Schur functor to it and then calculate trace $Tr(M^{\Lambda})$ we will get Schur polynomial in the eigenvalues of $M$.

Question Can one generalize (deform) Schur functors, such that $Tr(M^{\Lambda})$ will give polynomials which generalize (deform) Schur polynomials e.g. Hall-Littlewood polynomial, or Jack polynomial and most generally Macdonald polynomials ?

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Have you seen arxiv.org/abs/q-alg/9503012 ? –  Gjergji Zaimi Jun 4 '12 at 9:05
    
@Gjergii Thank you I know, may be I forget something now, but it seems to me it is not the answer. Why do I need intertwiner? is it natural ? It does seems to me so. Morever you will need to take very specific representation to obtain Calogero model (which corresponds to Jack polynoms, respectively in q-case to Macdonalds)... –  Alexander Chervov Jun 4 '12 at 14:14

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It seems to me that this is answered, perhaps in a boring way, by Haiman's work on the $n!$-conjecture (now a theorem due to Haiman). For any partition $\lambda$, Haiman constructs a finite dimensional graded module $C_{\lambda}$ for $\mathbb{C}[x,y][S_n]$. (The group elements commute with $x$ and $y$, and $x$ and $y$ commute with each other.) The doubly graded Frobenius character of $C_{\lambda}$ is the $\lambda$-Macdonald polynomial.

Now just use Schur-Weyl duality: Define the functor $F_{\lambda}$ from vector spaces to vector spaces by $$V \mapsto V^{\otimes |\lambda|} \otimes_{\mathbb{C}[S_n]} C_{\lambda}.$$ The result is a doubly graded $\mathbb{C}[x,y]$ module which is the sum of Schur functors corresponding to the Macdonald polynomial.

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@David thank you very much ! It is unexpected and interesting form me ! (What is "Frobenius character") Actually I hoped something like a deformation e.g. vector spaces will be the same but morphisms somehow deformed. What do you think is it possible have something like this ? Morally this should correspond to quantum groups - take V as basic representation of U_q(gl). Send "V" to irrep of U_q(gl) which corresponds to Young diagram Lambda. But it is not clear for me how to make functor from this... –  Alexander Chervov Jun 4 '12 at 14:23
    
The "Frobenius character" is the standard map which sends an $S_n$-representation to a symmetric polynomial. The funny thing is that it is NOT a character -- in particular, the Frobenius character of a tensor product is not the product of the Frobenius characters. The relationship goes through Schur-Weyl duality: If $M$ is an $S_n$-rep with Frob. character $f$, then $V^{\otimes n} \otimes_{k[S_n]} M$, as a $GL(V)$-rep, has character $f$. –  David Speyer Jun 4 '12 at 15:10
    
@David is it the same thing which is called "characteristic map" ? arxiv.org/abs/1112.0620 Characteristic maps for the Brauer algebra A. I. Molev, N. Rozhkovskaya –  Alexander Chervov Jun 4 '12 at 16:18

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