0

1

The starting point is that it is known that the Hankel determinants for the Catalan sequence give the number of nested sequences of Dyck paths. I would like to promote this to symmetric functions. This is motivated by some representation theory.

The naive idea is to start with the sequence of symmetric functions $s_{n,n}$ and take the Hankel determinants using the inner product (that is product in the group ring of $S(2n)$) instead of the usual outer product. However this doesn't make sense.

Take the $2 \times 2$ case. Then the naive determinant is $$ \left|\begin{array}{cc} s_{n-1,n-1} & s_{n,n} \\ s_{n,n} & s_{n+1,n+1}\end{array}\right|$$ The inner product of the two diagonal terms is defined but the inner product of the two off-diagonal terms is not.

The idea that I want to test is that this is $\sum_\lambda s_\lambda$ where the sum is over the conjugates of the partitions $4^a2^{n-2a}$.

Any suggestions on how to fix this? If this does get fixed then I would like to know how to calculate the result. The difficulty is that I have not seen an implementation of the inner product in the computer algebra systems I use, Magma and Sage (which I think both use the same source for symmetric functions).

flag
That would be Hankel. – Charles Matthews Jun 16 2010 at 8:54
There is some standard technique for the Hankel determinants; see Christian Krattenthaler's "Advanced Determinant Calculus" and "Advanced Determinant Calculus: A Compliment" available from the author's page and arXiv. – Wadim Zudilin Jun 16 2010 at 9:15

1 Answer

1

It is unlikely to obtain for such a determinant the sum of all Schur functions indexed by partitions of $2n$ with four parts all even or all odd. Indeed, this sum is already equal to the inner product $s_{n,n}\ast s_{n,n}$ (see arXiv:0809.3469).

About your second question: you can compute the inner product of symmetric functions in SAGE using the "kronecker_product" command. For instance, compute the inner product of Schur functions $s_{6,2}\ast s_{5,3}$ as follows:

s=SymmetricFunctionAlgebra(QQ,basis='schur')

s([6,2]).kronecker_product(s([5,3]))

You may also use Maple with John Stembridge's package SF and the command "itensor".

link|flag
That's interesting. The answer I "want" is not all four parts even, just the differences. However I didn't know the result you state. – Bruce Westbury Jun 16 2010 at 10:04
I copied the description of the expansion from the preprint I gave in reference, but it's the same. "All parts even or all parts odd" also means "all differences even". – Emmanuel Briand Jun 16 2010 at 10:40
You are right. I have changed the post. There is still a difficulty in that $[2,2,1,1]$ appears in the proposed answer but does not have all parts even or all parts odd. – Bruce Westbury Jun 16 2010 at 11:23
The reference you gave appears to attribute the result to: MR2548921 Garsia, A. ; Wallach, N. ; Xin, G. ; Zabrocki, M. Hilbert series of invariants, constant terms and Kostka-Foulkes polynomials. Discrete Math. 309 (2009), no. 16, 5206--5230. However I didn't see it. Have I missed it? – Bruce Westbury Jun 16 2010 at 11:26
There is no problem with your link. The reference I have given is [9] in this paper - except there it is called a "preprint". They give this as the reference for $s_{n,n}*s_{n,n}$ but I can't see it. Either I need better glasses or the published paper is different to the preprint. – Bruce Westbury Jun 16 2010 at 12:42
show 1 more comment

Your Answer

Get an OpenID
or

Not the answer you're looking for? Browse other questions tagged or ask your own question.