In connection with string theory I encountered the following problem:

Given the set M_N of all semi-standard Young tableaux of size N (i.e. all fillings of Ferrers diagrams with natural numbers with weakly increasing rows and strictly increasing columns, the content of which sum to N).

The conjecture, which I cannot prove and for which I know no counterexample states:

M_N is the union of disjoint families, where each family consists of a father diagram and all his daughters, which are obtained from the father diagram by deletion of one box.

The conjecture holds for N<=8

http://www.itp.uni-hannover.de/~dragon/young.ps or


There numbers a,b are attached to the Ferrers diagrams, where a is the number of semi-standard Young Tableaux of size N and b the number of fathers, who remain after the daughters have filled up their families.

Light cone string theory proves that the conjecture holds for all M_N with less than 25 rows, i.e. for N < 325. But string theory provides no counterexample for N >= 325.

Any suggestions, e.g. how to compute the number a(N,lambda) of Young diagrams with size N and shape lambda, would be helpful.

Norbert Dragon

-- http://www.itp.uni-hannover.de/~dragon/

Superstition brings bad luck.

  • $\begingroup$ I do not understand the statement of the conjecture. How do you obtain the daughters precisely? $\endgroup$ May 6, 2013 at 16:10
  • $\begingroup$ The daughters are obtained from their father diagram by removing one box, e.g. the diagram 3,2,1 has daughters 2,2,1 and 3,1,1 and 3,2. The looked for families of father and daughter diagrams characterize the SO(D-2)representations in string theory which are needed for the construction of SO(D-1) representations. $\endgroup$ May 6, 2013 at 16:50

1 Answer 1


I use standard facts about symmetric functions that can be found, e.g, in Chapter 7 of Enumerative Combinatorics, vol. 2. Let $s_\lambda$ denote a Schur function and $p_1=s_1=x_1+x_2+\cdots$. Then $\frac{\partial s_\lambda}{\partial p_1}=\sum_\mu s_\mu$, where the $\mu$'s are obtained by removing a single box from $\lambda$. Moreover, the coefficient of $q^n$ in $s_\lambda(q,q^2,q^3,\dots)$ is equal to the number of SSYT with entries summing to $n$. Thus if we let $c_n$ be the coefficient of $q^n$ in the symmetric function $f$ defined by $$ f+\frac{\partial}{\partial p_1}f = \sum_\lambda s_\lambda\cdot s_\lambda(q,q^2,q^3,\dots), $$ and if $c_n=\sum a_{\mu,n}s_\mu$, then we need $a_{\mu,n}$ copies of the shape $\mu$ in a set of fathers generating all SSYT with entries summing to $n$. Hence we need to show that $a_{\mu,n}\geq 0$. Now $$ \sum_\lambda s_\lambda\cdot s_\lambda(q,q^2,q^3,\dots) = \exp \sum_{n\geq 1} \frac{q^n}{1-q^n}p_n. $$ This leads to a simple linear first-order differential equation with solution $$ f = (1-q) \sum_\lambda s_\lambda\cdot s_\lambda(q,q^2,q^3,\dots). $$ If $h_u$ denotes the hook length of the square $u$ of $\lambda$, then $$ s_\lambda(q,q^2,q^3,\dots) = \frac{q^{b(\lambda)}}{\prod_{u\in \lambda} (1-q^{h_u})}, $$ where $b_\lambda=\sum i\lambda_i$. Since there is always a hook length equal to one, the power series $(1-q)s_\lambda(q,q^2,q^3,\dots)$ will be a product of factors of the form $1/(1-q^h)$, $h\geq 1$, so will have nonnegative coefficients as desired. Thus we have not just an existence proof, but a precise generating function for the number of fathers of each shape.

  • $\begingroup$ For $\frac{\partial}{\partial p_1}s_{\lambda} = \sum_{\mu} s_{\mu}$ so these $\mu$ are symmetric skew schur funciton ? Is it true for differentiating with repsect to $p_n$? What about the $$\frac{\partial}{\partial p_1}$\frac{\partial}{\partial p_2}$ means it would be removing the $2,1$ subset from $\lambda$? $\endgroup$
    – GGT
    Mar 30, 2021 at 12:18
  • $\begingroup$ Applying $n\frac{\partial}{\partial p_n}$ corresponds to removing border strips $B$ of size $n$ weighted by $(-1)^{\mathrm{ht}(B)}$. This follows from EC2, Theorem 7.17.3, and the fact that multiplication by $p_n$ is adjoint to $n\frac{\partial}{\partial p_n}$. $\endgroup$ Mar 31, 2021 at 18:37
  • $\begingroup$ I get it now, I have the following question, so $s_{\lambda}(p_1, p_2..)$ written in power symmetric basis and if we specialise $p_i = x^i$ then we obtain $s_{(n)}\neq 0 $ and it's zero for all other $\lambda$. I guess it might follow from your theorem 7.21.2(?) so I am wondering if there is a closed formula for the linearly shifted schur function by the parameters $t_i$ that is $s_\lambda(p_1 +t_1 , p_2 +t_2 ,..)$ and then specialise $p_i= x^i$? I could derive it using the differential of schur function it's getting complicated(?) Some help. $\endgroup$
    – GGT
    Apr 4, 2021 at 11:49
  • $\begingroup$ Specializing $p_i=x^i$ is the same as the evaluation $p_i(x,0,0,\dots)$. There are then several ways to see that $s_\lambda(x,0,0,\dots)=0$ unless $\lambda=(n)$, e.g., from the combinatorial definition in terms of SSYT. As for $s_\lambda(p_1+t_1,\dots)$, let $t_i=p_i(z_1,z_2,\dots)$ (without loss of generality since the $t_i$'s remain algebraically independent). Let $y=(y_1,y_2,\dots)$. Then $\sum_\lambda s_\lambda(x+t_1,x^2+t_2,\dots)s_\lambda(y)$ $=(\sum x^nh_n(y)\left( \sum s_\lambda(z)s_\lambda(y)\right)$. Now use Pieri's rule and then extract the coefficient of $s_\lambda(y)$. $\endgroup$ Apr 5, 2021 at 16:21
  • $\begingroup$ βˆ‘πœ†π‘ πœ†(π‘₯+𝑑1,π‘₯2+𝑑2,…)π‘ πœ†(𝑦) =(βˆ‘π‘₯π‘›β„Žπ‘›(𝑦)(βˆ‘π‘ πœ†(𝑧)π‘ πœ†(𝑦)). So RHS of the equation I would expand the product of two Schur function in terms of Schur function, so writing it in terms of the combination of Littlewood Richardson coefficient? $\endgroup$
    – GGT
    Apr 11, 2021 at 0:31

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