MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 4 down vote accepted

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.

share|cite|improve this answer
Thanky you, that answers my first question! – Per Alexandersson Dec 13 '12 at 6:51
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 $\nu$? – Per Alexandersson Jun 5 '13 at 18:59
$K_{\lambda-\mu,\nu}$ 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. – Richard Stanley Jun 5 '13 at 19:40
@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 $c_{\tau/\simga,\rho}.$ – Per Alexandersson Jun 6 '13 at 10:42
Never mind, I think I managed to get the statement that I need, thank you for your effort! – Per Alexandersson Jun 6 '13 at 13:22

I believe most of what you want is in , especially the polynomiality you're looking for. Note that that was first proven in [H. Derksen, J. Weyman] "On the Littlewood-Richardson polynomials," .

share|cite|improve this answer
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. – Per Alexandersson Dec 12 '12 at 20:14
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). – Allen Knutson Dec 13 '12 at 1:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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