2
$\begingroup$

The fact that Kostka numbers equals to Littlewood-Richardson coefficients for some partitions is already known $\colon$ \begin{align} K_{\lambda \mu} = c_{\sigma \lambda}^\tau \end{align} where $\tau_i = \mu_{i} + \mu_{i+1} +\cdots, \sigma_i = \mu_{i+1} + \mu_{i+2} + \cdots$.

For example, it is written in Here, page 3.

But I have not found a proof of this claim. So if you know paper or book in which this claim's proof is written, then please tell me it. If possible I would like to know a proof which construct a bijection between semi-standard tableaux and LR-tableaux.

Improve this question
$\endgroup$
7
  • $\begingroup$ That's a good question. There is an excellent answer by R. Stanley to a more general question here: mathoverflow.net/questions/116171/… $\endgroup$
    – Per Alexandersson
    Commented Nov 5, 2018 at 13:17
  • 4
    $\begingroup$ Possible duplicate of Skew Kostka coefficients from Littlewood-Richardson Coefficients $\endgroup$
    – Timothy Chow
    Commented Nov 5, 2018 at 15:53
  • 1
    $\begingroup$ What do you mean by "$\tau_i = (\mu_{i}, \mu_{i+1},\dots), \sigma_i = (\mu_{i+1}, \mu_{i+2},\dots)$"? Either you're defining partitions or you're defining entries of a partition. $\endgroup$
    – darij grinberg
    Commented Nov 5, 2018 at 17:16
  • $\begingroup$ @ darij grinberg I defined $\tau_i$ and $\sigma_i$ as entries of a prtition, namely Set $\mu = (\mu_1, \mu_2, \dots, \mu_k)$, then we define $\tau = (\tau_1, \dots, \tau_k)$, $\sigma = (\sigma_1, \dots, \sigma_{k-1})$. where $\tau_i = (\mu_i + \mu_{i+1} +\cdots)$, $\sigma_i = (\mu_{i+1} + \mu_{i+2} +\cdots)$. Thank you for your pointing it out, I have modified the question. $\endgroup$
    – s_nrsw
    Commented Nov 6, 2018 at 4:10
  • $\begingroup$ @Per Alexandersson Thank you. The answer is probably what I want to know (I still don't have completely understanding). By the way, is there a bijection proof of this formula ? $\endgroup$
    – s_nrsw
    Commented Nov 6, 2018 at 6:00

0

Reset to default

You must log in to answer this question.