# Summing Characters of the Symmetric Group over Derangements (Enumerative Combinatorics: Vol. II Ex. 7.63)

The following exercise is out of Stanley's Enumerative Combinatorics: Vol. II (Ex. 7.63):

For $\lambda \vdash n$ define $d_\lambda = \sum_{w \in D_n} \chi^\lambda(w)$ where $D_n$ is the set of all derangements in $S_n$. Show that

$\sum_{\lambda \vdash n} d_\lambda s_\lambda = \sum^n_{k=0}(-1)^{n-k}(n)_k h_{1^{n-k}k}$

To prove this, Stanley suggests using Murnaghan-Nakayama to arrive at $d_\lambda = n!s_\lambda |_{p_1=0,p_2=p_3= \cdots =0}$. Then by the Cauchy Identity and Proposition 7.7.4 in Stanley Vol. II: $\prod_{i,j}(1-x_iy_j)^{-1} = \sum_\lambda s_\lambda(x)s_\lambda(y) = \exp \sum_{n \geq 1}\frac{1}{n}p_n(x)p_n(y)$

Setting $p_1(y) = 0$ and $p_2(y) = p_3(y) = \cdots = 1$ we have:

$\sum_\lambda \frac{d_\lambda}{n!}s_\lambda(x) = \exp \sum_{n \geq 2}\frac{1}{n}p_n(x)$

Using the identity $\exp \sum_{n \geq 1}\frac{1}{n}p_n(x)p_n(y) = \sum_\lambda z_\lambda^{-1}p_\lambda(x)p_\lambda(y)$ and the fact that $h_n = \sum_{\lambda \vdash n}z_\lambda^{-1}p_\lambda$, it seems to me that we have $\exp \sum_{n \geq 2}\frac{1}{n}p_n(x) = e^{-h_1} \sum_{\lambda}z_\lambda^{-1}p_\lambda(x) = e^{-h_1}h_n$; however Stanley claims the RHS is $e^{-h_1}\sum_{n \geq 0}h_n$. I don't understand where this summation comes from. I'm sure it's something silly, but I've retraced my steps many times and can't find my mistake. Any help is greatly appreciated.

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If it's not too much trouble, please Tex the problem statement into your question to avoid a future dead link. –  Alex R. Apr 10 '13 at 5:32
There is a difference between $\sum_{\lambda}$ and $\sum_{\lambda\vdash n}$, and the difference is exactly the $\sum_{n}$, or did I get you wrong? –  darij grinberg Apr 10 '13 at 5:37
When you write notation such as $\sum_\lambda$ are you implying that you're summing over all possible tableaux shapes? If yes, then the $h_n$ formula is the sum over a fixed shape, whereas your sum just before your $e^{-h_1}h_n$ conclusion is probably over all shapes. –  Alex R. Apr 10 '13 at 5:38
You're both absolutely right. Thanks. –  Nathan Lindzey Apr 10 '13 at 6:13
just something else to discuss about, do you have any idea what is the implication of setting $$p_1(y) = 0, \quad p_2(y) = p_3(y) = \ldots = 1.$$ May I know why are you dealing with this problem? Any research-related problem that you are looking at now? –  terrylsc Apr 11 '13 at 6:41