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

For a partition $\lambda \vdash n$, define $\dim \lambda$ to be the number of standard Young tableaux of shape $\lambda$, and $\dim \lambda/(k)$ as the number of standard Young tableaux with $1,2,\ldots,k$ in the first row. It is not hard to see that $\dim \lambda/(k)$ is equal to the number of skew standard tableaux of shape $\lambda/(k)$ if $k \leq \lambda_1$ and $0$ if $k > \lambda_1$.

We write $\lambda \unrhd \beta$ to mean that $\beta$ precedes $\lambda$ in the standard dominance ordering of partitions.

I have an inelegant proof of the following statement: For every $\lambda \vdash n$ and $\beta \vdash n$ where $\lambda \unrhd \beta$, and any positive integer $k$, we have

$$ \frac{\dim \lambda / (k)}{\dim \lambda} \geq \frac{\dim \beta / (k)}{\dim \beta}. $$

This seems to me like a statement that should be in the literature, yet I could not find it. Does anyone have a reference for (or a clean proof of) this statement?

share|cite|improve this question
"This seems to me like a statement that should be in the literature". I can't say I have seen too many inequalities about Young tableaux in literature... – darij grinberg Jan 12 '13 at 2:19
I agree with darij grinberg's comment. Do you know if there is a refinement to a chain of results: $\frac{\dim \lambda / (k)}{\dim \beta / (k)} \ge \frac{\dim \lambda / (k-1)}{\dim \beta / (k-1)}$ for $k \ge 1$? – Mark Wildon Jan 12 '13 at 19:36
Thanks, darij and Mark! I haven't seen too many inequalities either, but this is most definitely not my area of expertise. Mark, I did not consider the refinement you propose, so I have no idea if it holds or not (though my intuition says that it does). – KEW Jan 12 '13 at 21:37
Somehow this reminds me of a related Schur polynomial majorization style inequality that once Gjergji Zaimi had posted on MO....cannot find that link right now.. – Suvrit Oct 17 '14 at 16:44
I think Suvrit means this question: – Ryan O'Donnell Aug 2 at 17:46

Your Answer


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

Browse other questions tagged or ask your own question.