3
$\begingroup$

I read on page 4 here that the Kostka coefficients $K_{\lambda,\mu}$ are specializations of the Littlewood-Richardson coefficients $c^\tau_{\sigma,\lambda}$ by specializing $\sigma,\tau$ depending on $\mu$ in a simple manner (certain sums of parts of $\mu$).

Is there a similar specialization/translation for Kostka coefficients obtained from skew shapes, $K_{\lambda,\mu}^\nu$ where $\lambda/\nu$ is a skew shape?

Motivation: I would like to see if polynomiality of the map $n \mapsto K_{n \lambda, nw}^{n \nu}$ implies polynomiality for a similar map with LW-coefficients.

$\endgroup$

2 Answers 2

7
$\begingroup$

I prefer to write $K_{\lambda/\nu,\mu}$ for $K^\nu_{\lambda,\mu}$. Using standard symmetric function notation, we have $$ K_{\lambda/\nu,\mu}=\langle s_{\lambda/\nu},h_\mu\rangle = \langle s_\lambda,s_\nu h_\mu\rangle. $$ Let $\rho/\sigma$ be a skew shape which is a disjoint union of shapes $\nu, (\mu_1), (\mu_2), \dots$. Here $(\mu_i)$ is a single row of length $\mu_i$. By "disjoint union," I mean that none of the shapes has a square in the same row or in the same column as a square of another of the shapes. Thus $s_{\rho/\sigma} = s_\nu h_\mu$, so $$ K_{\lambda/\nu,\mu}=\langle s_\lambda,s_{\rho/\sigma}\rangle = \langle s_\lambda s_\sigma, s_\rho\rangle, $$ an ordinary Littlewood-Richardson coefficient.

$\endgroup$
5
  • $\begingroup$ Thanky you, that answers my first question! $\endgroup$ Dec 13, 2012 at 6:51
  • $\begingroup$ I don't really follow the first identity; Macdonald (5.14) says that $K_{\lambda−\mu,\nu}=\langle s_{\lambda/\mu},h_\nu\rangle$ but this is surely different from $K_{\lambda/\mu,\nu}$ which counts the number of skew tableaux of shape $\lambda/\mu$ and weight $\nu$? $\endgroup$ Jun 5, 2013 at 18:59
  • $\begingroup$ $K_{\lambda-\mu,\nu}$ is simply another notation for $K_{\lambda/\mu,\nu}$. In general $\langle f,h_\mu\rangle$ is the coefficient of $m_\mu$ when $f$ is expanded in the basis of monomial symmetric functions. $\endgroup$ Jun 5, 2013 at 19:40
  • $\begingroup$ @Richard Stanley Ah, that explains it, but then I must be clearer with stating what I seek; The number of tableaux of shape $\lambda-\mu$ with weight $\nu$, (which I denote $K_{\lambda-\mu,\nu}$) is not the same as the number of skew tableaux of shape $\lambda/\mu$ and weight $\nu,$ which I denote $K_{\lambda/\mu,\nu}.$ (The latter has three free "parameter" partitions, while the first has essentially two.) What I would like is to be able to translate between $K_{\lambda/\mu,\nu}$ and $c_{\tau/\simga,\rho}.$ $\endgroup$ Jun 6, 2013 at 10:42
  • $\begingroup$ Never mind, I think I managed to get the statement that I need, thank you for your effort! $\endgroup$ Jun 6, 2013 at 13:22
2
$\begingroup$

I believe most of what you want is in https://arxiv.org/abs/math/0308101 , especially the polynomiality you're looking for. Note that that was first proven in [H. Derksen, J. Weyman] "On the Littlewood-Richardson polynomials," http://www.math.lsa.umich.edu/~hderksen/preprints/lrpoly.dvi .

$\endgroup$
2
  • $\begingroup$ Yes, I am aware that the function is polynomial in n. The question is if it is easy to see if polynomiality for the Kostka map easily implies polynomiality for LW-coefficients. (The reverse implication should be quite easy, I think). The reason I ask for this, is that I think I have a very short proof of the polynomiality of the map $n \to K_{n\lambda,nw}^{n\mu}$ and it would be interesting to see if this easily implies polynomiality for LW-coefficients. $\endgroup$ Dec 12, 2012 at 20:14
  • $\begingroup$ You mean, again, LR-coefficients, right? The point is that the Steinberg/Klimyk tensor product rule expresses the LR coefficients as an alternating sum (over the Weyl group) of weight multiplicities (here, Kostka numbers). $\endgroup$ Dec 13, 2012 at 1:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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