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%\def\LeftHead{ \centerline{Doron Zeilberger}}
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\centerline
{\bf Proof of Theorem 10.3, Part 1 (VERSION  of JULY 31, 2014, 11:05am (my time) )}
\bigskip
{\it By DZ}
\bigskip

{\bf Theorem 10.3}
Let $\lambda\vdash n$ and $\tau\vdash n+1.$

1. Assume first that $\tau\ne (\mu,1)$, namely $\tau=(\tau_1,\ldots,\tau_k)$ where $\tau_k\ge 2$, then $$\sum_{\nu\in\lambda^+}\chi^\nu(\tau)=0.$$

2.  Let $\lambda,\mu\vdash n$ 
%and assume $\mu=(\mu^*,1)$. 
then
$$
\sum_{\nu\in\lambda^+}\chi^\nu (\mu,1)=(N_1(\mu)+1)\cdot \chi^\lambda (\mu),
$$
where $N_1(\mu)$ is the number of ones in $\mu$.

\bigskip
{\bf Remark}: A better way to state the theorem is:
$$
\sum_{\nu\in\lambda^+}\chi^\nu(\tau) \, = \, N_1(\tau) \cdot \chi^\lambda (\mu) \quad ,
$$
where $\mu=(\tau_1, \dots, \tau_{k-1},\tau_k-1)$ (and remove the last $0$ if $\tau_k=1$).
The proof is by induction on the number ones in  $\tau$, and  part 1 is the base case.

\bigskip

{\bf Important Definition} Let $F(x_1, \dots, x_n)$ be any Laurent polynomial. Then
the {\bf constant term} of $F$, denoted by $CT_{x_1, \dots, x_n} F$, or simply $CT F$ when there is
no ambiguity,  is the coefficient of $x_1^0 \cdots x_n^0$ in F.

{\bf Examples}:
$$
CT_{x_1,x_2} (3+x_1+x_2^{-1})=3 \quad ,
$$
$$
CT_{x_1} (3+x_1+x_2^{-1})=3+x_2^{-1} \quad .
$$

{\bf Proof of Part 1}:
One way to define the characters $\chi^{\lambda}(\mu)$ is via the formula mentioned
on the very last line of p. 114 of Maconad's ``Symmetric Functions and Hall Polynomials'' (2nd ed.).
Let's rephrase it in terms of constant terms.

{\bf Important Fact}: Let
$$
\Delta(x_1, \dots , x_n)= \prod_{1 \leq i < j \leq n} (x_i-x_j) \quad ,
$$
$$
p_r(x_1, \dots, x_n)= \sum_{i=1}^{n} x_i^r \quad,
$$
Let $n$ be a positive, integer and $\lambda$ and $\tau$ partitions of $n$.
Write $\tau=(\tau_1, \dots, \tau_k)$ and if $\lambda$ has $m<n$ parts,
define $\lambda_{m+1}=\dots=\lambda_n=0$. 
We have:
$$
\chi^{\lambda}(\tau) \, = \,
CT_{x_1, \dots , x_n} \,
\frac{\Delta(x_1, \dots, x_n) \prod_{i=1}^k p_{\tau_i}(x_1, \dots, x_n)}
{\prod_{i=1}^{n} x_i^{\lambda_i+n-i}}
  \quad .
$$
Now let's prove Theorem 10.3, part 1. If $|\tau|=n$, then $|\lambda|=n-1$, of course. 

Let's consider a typical member of $\lambda^{+}$. It is obtained by changing one of the parts,
$\lambda_i$ to $\lambda_i+1$. In order for this to be legal, we need $\lambda_{i-1}>\lambda_i$.
Doing this gives almost the same expression, but with the constant-termand multiplied
by $x_i^{-1}$ (the denominator gets multiplied by $x_i$). If it is not legal, then
 $\lambda_{i-1}=\lambda_i$, and the denominator (after multiplying it by $x_i$) has the degrees of $x_{i-1}$ and $x_i$ being the same.
Of course,  transposing $x_{i-1}$ and $x_{i}$ does not change the constant term,
but it makes the consant-termand the negative of the original (since $\Delta(x_1, \dots, x_n)$ is
{\bf anti-symmetric}), so the constant term equals to its negative and hence is 0 in this (illegal) case.


Hence, we have to prove  that, if $\tau_k \geq 2$, then
$$
CT_{x_1, \dots , x_n} \,
\frac{\Delta(x_1, \dots, x_n) \left ( \prod_{i=1}^k p_{\tau_i}(x_1, \dots, x_n) \right ) \cdot \left ( \, \sum_{i=1}^n x_i^{-1} \right ) }
{\prod_{i=1}^{n} x_i^{\lambda_i+n-i}} \, =0 \quad .
$$

But since  $\sum_{i=1}^n x_i^{-1}=e_{n-1}(x_1, \dots, x_n) \cdot \prod_{i=1}^{n} x_i^{-1}$, the left side equals
$$
CT_{x_1, \dots , x_n} \,
\frac
{
\Delta(x_1, \dots, x_n) \left ( \prod_{i=1}^k p_{\tau_i}(x_1, \dots, x_n) \right ) \cdot e_{n-1}(x_1, \dots, x_n)
}
{
\prod_{i=1}^{n} x_i^{\lambda_{i}+n-i+1}
}  \quad ,
$$
and we have to prove that it equals $0$. 
But since $\lambda$ is a partition of $n-1$, $\lambda_n=0$, and we claim
the much stronger result 

$$
CT_{x_n} \,
\frac{\Delta(x_1, \dots, x_n) \left ( \prod_{i=1}^k p_{\tau_i}(x_1, \dots, x_n) \right ) \cdot e_{n-1}(x_1, \dots, x_n) } 
{\prod_{i=1}^{n} x_i^{\lambda_i+n-i+1}} =0  \quad .
$$

The {\bf crucial fact} is that the parts of $\tau$ are all $ \geq 2$. Let $Junk_i (x_1, \dots, x_{n-1})$,
($i=0,1,2$) denote various polynomials that only depend on $x_1, \dots, x_{n-1}$.
We have
$$
 \prod_{i=1}^k p_{\tau_i}(x_1, \dots, x_n)= Junk_0+ O(x_n^2) \quad ,
$$
and we also have:
$$
\Delta(x_1, \dots, x_n)=Junk_1 \cdot \left ( \, \prod_{i=1}^{n-1} (x_i-x_n) \, \right ) =
Junk_1 \cdot \left ( \left ( \, \prod_{i=1}^{n-1} x_i \, \right ) \cdot x_n^0 \, - \, e_{n-2}(x_1, \dots, x_{n-1}) x_n + O(x_n^2) \right ) \quad .
$$
Also
$$
e_{n-1}(x_1, \dots, x_n)= \left ( \, \prod_{i=1}^{n-1} x_i \, \right ) \cdot  x_n^0 \, + \, e_{n-2}(x_1, \dots, x_{n-1}) \cdot x_n^1  \quad  .
$$
Now
$$
CT_{x_n} \, [ 
\frac{\Delta(x_1, \dots, x_n) \left ( \prod_{i=1}^k p_{\tau_i}(x_1, \dots, x_n) \right ) \cdot e_{n-1}(x_1, \dots, x_n) }
{\prod_{i=1}^{n} x_i^{\lambda_i+n-i+1}} ]
$$
$$
=Junk(x_1, \dots, x_{n-1}) \cdot CT_{x_n} [ (Junk_0+ O(x_n^2)) \cdot
$$
$$
\left (
 \left ( \left ( \, \prod_{i=1}^{n-1} x_i \, \right ) \cdot x_n^0 \, - \, e_{n-2}(x_1, \dots, x_{n-1}) x_n^1 + O(x_n^2) \right )
\cdot  \left ( \left ( \, \prod_{i=1}^{n-1} x_i \, \right ) \cdot  x_n^0 \, + \, e_{n-2}(x_1, \dots, x_{n-1}) x_n^1\right )  \cdot \frac{1}{x_n}
\right )]
$$
and, thanks to $(a-b)(a+b)=a^2-b^2$, this equals:
$$
Junk(x_1, \dots, x_{n-1}) \cdot CT_{x_n} [ (Junk_0+ O(x_n^2)) \cdot
$$
$$
\left (
 \left ( \left ( \, \prod_{i=1}^{n-1} x_i \, \right )^2 \cdot x_n^0 \, - \, (e_{n-2}(x_1, \dots, x_{n-1}))^2 x_n^2 + O(x_n^2) \right )
 \cdot \frac{1}{x_n}
\right ) ] \quad .
$$
and this is 0. $\halmos$.
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