Let $s_{\lambda}$ denote the schur function and $\lambda$ is the partion of an integer. The schur function written in power sum symmetric basis apper as following. $\chi$ denote the character. \begin{equation} s_\lambda(p_1,p_2,p_3,\ldots) = \sum_{\nu} \frac{\chi^\lambda_\nu}{z_\nu} p_\nu = \sum_{\rho=(1^{r_1},2^{r_2},3^{r_3},\dots)}\chi^\lambda_\rho \prod_k \frac{p^{r_k}_k}{r_k! k^{r_k} } \end{equation}

If lambda is a partition of type $(l,1^{d-l})$ we call it a hook partition. We set $p_2,p_3,\ldots$ to zero then $$ s_{\lambda}(p_1,0,0,\ldots):=\frac{1}{dd!} \binom{d-1}{\ell-1}p_1^d$$ We get the above equation by replacing $\chi^{(l,1^{d-l})}_{\nu}$ interms of binomial coffecient and $\chi$ is nonzero only when $\nu$ is hook so such over $d$ hooks.

I want to study the following generating function $\sum_{d=1}^{\infty}\sum_{l=1}^{d}(-1)^{d-l}s_{l,1^{d-l}}(p_1,p_2\ldots)x^d$.

Now if we set $p_2,p_3,\ldots=0$ then the above generating function becomes \begin{align} \sum_{d=1}^{\infty}\sum_{l=1}^{d}s_{l,1^{d-l}}(p_1,0,0,\dots)x^d&=\sum_{d=1}^{\infty}\frac{1}{dd!}\sum_{\ell=1}^{d} (-1)^{(\ell+1)}\binom{d-1}{\ell-1}x^{d} \\ &=\sum_{d=1}^{\infty}\frac{1}{d^2}\sum_{l=1}^{d}(-1)^{l+1}\frac{1}{(\ell-1)!(d-\ell)!} \end{align} Let define $$P(x):=1+\sum_{d=1}^{\infty}\frac{1}{d!}x^dp_1^d$$, $$Q(x):=x+\sum_{d=1}^{\infty}(-1)^{d}\frac{1}{d!}x^{(d+1)}p_1^{(d+1)}$$. $$R(x):=\sum_{d=1}^{\infty}\frac{1}{d^2}x^d$$.

The generating function $\sum_{d=1}^{\infty}\sum_{l=1}^{d}s_{l,1^{d-l}}(p_1,0,0,\dots)x^d$ can be written as Cauchyproduct of $P(x)$ and $Q(x)$ that is $P(x)Q(x)$and Hadamard product of $R(x)$ and $P(x)R(x)$.

My questions are following

Let we have the sequence $p_1,p_2,0,0\ldots$ we have generating function $$ \sum_{d=1}^{\infty}\sum_{l=1}^{d}s_{l,1^{d-l}}(p_1,p_2,0,\ldots)x^d .$$ Does their exist $P(x)$,$Q(x)$ and $R(x)$ such that we could write it as above? If not why not ?

If we take the closer look $$P(x):=1+\sum_{d=1}^{\infty}s_{d}(p_1,0\ldots)x^d$$ and

$$ Q(x):=x+\sum_{d=1}^{\infty}s_{1^d}(p_1,0,\ldots)$$

If we change $P(x):=1+\sum_{d=1}^{\infty}s_{d}(p_1,p_2,0\ldots)x^d$ and $Q(x):=x+\sum_{d=1}^{\infty}s_{1^d}(p_1,p_2,0,\ldots)$ what is $R(x)$ ? such that it will give me the decomposition.


1 Answer 1


It follows from the Cauchy identity and Exercise 7.43 of Enumerative Combinatorics, vol. 2, that $$ 1+ (u+t)\sum_{1\leq l\leq d} s_{l,1^{d-l}}(x)u^{l-1} t^{d-l} = \prod_i\frac{1+tx_i}{1-ux_i} $$ $$ \qquad = \exp \sum_{n\geq 1}\frac 1n p_n(x)(u^n-(-t)^n). $$ If the decomposition you want actually exists then it should follow from the above formula, though I did not try to do this.

  • $\begingroup$ Left hand side of the equation you mean sum over $d$ from 1 to $\infty$ and R.H.S of the equation $n$ should be replaced by $d$? $\endgroup$
    – GGT
    Dec 21, 2017 at 1:46
  • $\begingroup$ Yes, $d$ from 1 to $\infty$. On the RHS $n$ is a dummy variable; it can be replaced by anything. $\endgroup$ Dec 21, 2017 at 3:04
  • $\begingroup$ Is there any similar expression for $\sum_{d=1}^{\infty}\sum_{l=1}^{d}(-1)^{l-1}s_{l,1^{d-l}}(p_1,p_2,0,\ldots)x^d $ ? In the above expression if I replace $u=t=x$ then I get a expression similar to mine but it's not alternating sum. $\endgroup$
    – GGT
    Dec 23, 2017 at 0:26
  • $\begingroup$ I had forgotten to put $(-1)^{l-1}$ before that's the expression I want to study. Replacing $u=-t$ makes the L.H.S of your equation vanish. So I can't use it? $\endgroup$
    – GGT
    Dec 23, 2017 at 0:28
  • $\begingroup$ I guess $-1+\sum_{d=1}^{\infty}\sum_{l=1}^{d}(-1)^{l-1}s_{l,1^{d-l}}(p_1,p_2,0,\ldots)x^d = -\qquad \exp \sum_{n\geq 1}\frac 1n p_n(x)(u^n-(-t)^n).$ $\endgroup$
    – GGT
    Dec 23, 2017 at 1:36

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