4
$\begingroup$

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.

$\endgroup$
7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 '17 at 1:46
  • $\begingroup$ Yes, $d$ from 1 to $\infty$. On the RHS $n$ is a dummy variable; it can be replaced by anything. $\endgroup$ – Richard Stanley Dec 21 '17 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 '17 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 '17 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 '17 at 1:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.