Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I have a proof that given a partition $\lambda=(\lambda_1,\dots,\lambda_l)$ then the number of semi-standard Young tableaux of shape $\lambda$ with entries in $1,2,\dots, n$ is given by

$$\frac{1}{1!2!\cdots (n-1)!} \prod_{1\leq i\lt j\leq n} (\lambda_i-i)-(\lambda_j-j).$$ (We define $\lambda_j:=0$ if $j \gt l.$)

The product is also recognized as a Vandermonde determinant.

There are plenty of product formulas (over boxes in the tableau) and determinant formulas (but not in Vandermonde form, as far as I can tell) for the number of such SSYTs, but I have not seen a this particular one in the literature or in any article I've come across.

Is this formula known? Is this formula of any interest?

share|improve this question
4  
Probably known, sorry to disappoint. The number of semistandard Young tableaux of shape $\lambda$ is the value of the $\lambda$-Schur polynomial (which I consider as a polynomial in $n$ variables) at $\left(1,1,...,1\right)$ (because the $\lambda$-Schur polynomial is a generating function for semistandard Young tableaux of shape $\lambda$). But the value of the $\lambda$-Schur polynomial at $\left(1,1,...,1\right)$ is the dimension of the $\lambda$-Schur module of $\mathrm{GL}_n\mathbb C$ (since the $\lambda$-Schur polynomial is the character of this $\lambda$-Schur module). ... –  darij grinberg Sep 7 '12 at 14:00
5  
... By Weyl's character formula (Theorem 4.63 in Pavel Etingof's representation theory notes www-math.mit.edu/~etingof/replect.pdf ), the latter dimension is $\prod\limits_{1\leq i < j \leq n} \dfrac{\lambda_i - \lambda_j + j - i}{j-i} = \dfrac{1}{1! 2! ... \left(n-1\right)!} \prod\limits_{1\leq i < j \leq n} \left(\left(\lambda_i - i\right) - \left(\lambda_j - j\right)\right)$. Combining these observations, we get your formula. I suspect that some text has this more explicitly. –  darij grinberg Sep 7 '12 at 14:02
    
@darij grinberg: Yes, I am aware that it is the evaluation of the Schur polynomial at (1,...1). Do you have a reference? I'd love to put this identity at Wikipedia for example, since, in my humble opinion, this is a very straightforward way to compute these numbers. –  Per Alexandersson Sep 7 '12 at 14:03
1  
One reference is the case $q=1$ of Enumerative Combinatorics, vol. 2, (7.105) on page 375. This goes back to Littlewood and Richardson. See the discussion on page 403 of EC2. –  Richard Stanley Sep 8 '12 at 0:28
1  
I believe that this formula is due to MacMahon. –  Ira Gessel May 23 at 22:18

1 Answer 1

up vote 2 down vote accepted

So, to answer this question, the formula is known. It is also (now) on the wikipedia page about Schur polynomials, so the next person will find it easily.

share|improve this answer

Your Answer

 
discard

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.