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On a simple representation of a simple Lie algebra, there is a unique bilinear form called the Shapovalov form for which the actions of $E_i$ and $F_i$ are biadjoint, and some fixed highest weight vector has $\langle v_h,v_h\rangle=1$.

The representation has a distinguished collection of vectors $F_{i_1}\cdots F_{i_n}v_h$ for all sequences $\mathbf{i}$. One can calculate any inner product $\langle F_{i_1}\cdots F_{i_n}v_h, F_{j_1}\cdots F_{j_n}v_h\rangle$, by simply moving the $F_j$'s to become $E_j$'s on the other side, and commuting them past the $F_i$'s. This is not hard to do computationally, but the formulas one gets are not positive, which is annoying for my purposes.

Does anyone know of positive formulae for these inner products? What about their $q$-analogues for quantum groups?

EDIT: I should note, following Allen's comment: I'm pretty sure that I know a vector space that has the dimension which is this inner product. There's also a positivity proof using the canonical basis (all the elements I'm interested in are positive linear combinations of canonical basis elements).

I'm trying to show that a surjective map to this vector space is an isomorphism, and do so by finding a spanning set of the domain that has the right cardinality.

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Just to be clear, since [E_i,F_j] = \delta_{ij}H_i, the problem is just that F...Fv might be at a negative weight? Anyway, are you asking for a natural basis for the representation in which the bilinear form is given by a positive matrix? Then there is a (very not natural) answer: pick some very large number C, and let the basis consist of vectors of the form v + C^{-1} F...Fv, where v is your highest weight vector. If you'd rather have an integer matrix, let C be a very large integer, and take the basis to be things like Cv + F...Fv. –  Theo Johnson-Freyd Feb 27 '10 at 3:46
    
Is there a reason to expect them to be positive? –  Allen Knutson Feb 27 '10 at 4:23
    
Yes. I know a vector space that has that dimension. But it's not a vector space I have any hope of calculating a basis of and putting in bijection with something. –  Ben Webster Feb 27 '10 at 5:03
    
Just want to make sure I am not imagining things - are the inner products you describe are elements of the Gram matrix for the Shapovalov form? I think there are some elegant formulas for them in the affine case (there was some paper by David Hill a couple of years ago). Also, you might want to look at what they do when proving Kac formulas for Z(g)=S(h)^W where the Shapovalov form is used. –  Vladimir Dotsenko Feb 27 '10 at 9:54
    
These are the coefficients of a Gram matrix for the Shapovalov form. There's such a matrix for any collection of vectors, and I think the one I'm looking at is not one people would usually choose (it's not a basis, just a spanning set). The Hill paper is interesting (thanks for pointing it out), but I don't think it really overlaps much with the kind of information I'm looking for here (in particular, he was most interesting in determinants, whereas I want actual coefficients). –  Ben Webster Feb 28 '10 at 0:51

2 Answers 2

Ben, my paper on the Shapovalov form does give a generating series for the entries of a Gram matrix in Corollary 3.4, and those entries are evidently positive. It is not very hard to deduce a q-version of this generating series from the paper by Chari and Jing appearing in the references. This is not really what you want, though, since the calculations only apply to the basic representation.

In general, you are asking a hard question. The only place I can think to point you is Jantzen's thesis where he calculates the determinant of the Shapovalov form for irreducible representations of simple finite-dimensional Lie algebras. Of course, this isn't the easiest document to get and it is in German.

Otherwise, you might try to prove the Khovanov-Lauda conjecture on cyclotomic quotients of quiver Hecke algebras and see what that gets you. At any rate, there is a representation theoretic interpretation of the Shapovalov form on irreducible modules in type A in the "Graded Decomposition Numbers" paper of Brundan and Kleshchev.

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This may not be exactly what you want, but I'd suggest you look at Kashiwara's paper ``On crystal bass of the q-analogue of the universal enveloping algebra" (see MR1115118 ).

In section 2.5 Kashiwara discusses the quantum version of the Shapovalov form. More relevant to what I want to say is Proposition 3.4.4, which defines/proves existence of a modification of the Shapovalov form defined on $U_q^-$. Roughly what he is doing is the following: Consider the pairing of, for example, $F_1F_2F_1v_\lambda$ and $F_1^2 F_2v_\lambda$, using the Shapovalov form on $M(\lambda)$, and allow $\lambda$ to vary. What you get is $$P(\lambda)/(q-q^{-1})^3$$ where $P$ is some Laurent polynomial in the $K_i$ and $q$, and evaluation at $\lambda$ means setting $K_i$ equal to $q^{(\alpha_i, \lambda)}.$ $P$ has a well defined highest order term in the $K_i$, and the coefficient of this term is a Laurent polynomial in $q$. Let $(F_1F_2F_1, F_1^2 F_2)$ be that leading coefficient. This will be Kashiwara's inner product on $U_q^-$, up to a power of $q$. In general, when pairing two monomials in the $F_i$ applied to $v_\lambda$, the denominator in the above equation has a factor of $q_i-q_i^{-1}$ for each $F_i$ in the first monomial.

The way Kashiwara sets things up, it is clear that the inner product of two monomials in the $F_i$ is a Laurent polynomial in $q$ with positive coefficients. For monomials $m_1$ and $m_2$, the inner product will be zero unless $m_1$ and $m_2$ have the same weight (i.e. they are both the products of the same number of each $F_i$, but possibly in a different order). Furthermore, the sum of the coefficients is the number of ways of matching each $F_i$ in $m_1$ with an $F_i$ in $m_2$ for all $i$. This all follows from Equations (3.3.1) and (3.4.6) in Kashiwara's paper. I believe you can find the power of $q$ associated to a given matching by arranging the monomials correctly, drawing a line between each matching pair, and counting a contribution for each crossing in the resulting picture. So there should be a completely combinatorial formula.

As I said, this may well not be what you want. For instance, this construction does not depend on $\lambda$. But maybe it is related.

In case you are wondering about the connection with crystal bases (i.e. the title of Kashiwara's paper), Kashiwara shows that the inner product of any two elements in the crystal lattice $L(\infty)$ is regular at $q=0$, and a crystal basis is an orthonormal basis for the evaluation $(\cdot, \cdot)_0$ of the inner product at $q=0$. Of course for this to be true you need to get the powers of $q$ right, which I have not done here.

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Unfortunately, Ben wants an answer which depends very much on $\lambda$. The formula for the inner product on $U_q^-$ is actually quite easy: $(F_{i_1}\ldots F_{i_k},F_{j_1}\ldots F_{j_l})$ will be zero unless the $i$ sequence is a permutation of the $j$ sequence, and in that case, the answer will be $\frac{m}{(1-t^2)^k}$ where $m$ is the number of permutations sending the $j$ sequence to the $i$ sequence. For example, $(F_i^k,F_i^k)=\frac{k!}{(1-t^2)^k}$. This works because the Shapovalov form is a decategorification of the Hom bifunctor on Khovanov and Lauda's categorification of $U_q^-$. –  Ben Mar 3 '10 at 17:58
    
Uh, sorry, replace $m$ in that formula with the quantum-analog, which keeps track of the lengths of the permutations involved. Similarly, replace $k!$ with $[k]!$. –  Ben Mar 4 '10 at 1:06

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