# The Unapologetic Mathematician

## Corollaries of the Sign Lemma

The results we showed last time have a few immediate consequences we will have use of.

First, let $t^\lambda$ and $s^\mu$ are two Young tableaux of shapes $\lambda$ and $\mu$, respectively, where $\lambda\vdash n$ and $\mu\vdash n$. If $\kappa_t\{s\}\neq0$ — where $\kappa_t$ is the group algebra element we’ve defined — then $\lambda$ dominates $\mu$.

To see this, let $b$ and $c$ be two entries in the same row of $s$. They cannot be in the same column of $t$, since if they were then the swap $(b\,c)$ would be in the column-stabilizer $C_t$. Then we could conclude that $\kappa_t\{s\}=C_t^-\{s\}=0$, which we assumed not to be the case. But if no two entries from the same row of $s^\mu$ are in the same column of $t^\lambda$, the dominance lemma tells us that $\lambda\trianglerighteq\mu$.

Now if it turns out that $\lambda=\mu$ it’s not surprising that $\lambda\trianglerighteq\mu$. Luckily in that situation we can say something interesting:

$\displaystyle\kappa_t\{s\}=\pm e_t=\kappa_t\{t\}$

Indeed, we must have $\{s\}=\pi\{t\}$ for some $\pi\in C_t$, basically by the same reasoning that led to the dominance lemma in the first place. Indeed, the thing that would obstruct finding such a $\pi$ is having two entries in some column of $t$ needing to go on the same row of $s$, which we know doesn’t happen. And so we calculate

\displaystyle\begin{aligned}\kappa_t\{s\}&=\kappa_t\pi\{t\}\\&=\mathrm{sgn}(\pi)\kappa_t\{t\}\\&=\pm e_t\end{aligned}

Now if $u\in M^\mu$ is any vector in the Specht module, and if $t^\mu$ is a tableau of shape $\mu$, then $\kappa_t u$ is some multiple of $e_t$. Indeed, we can write

$\displaystyle u=\sum\limits_ic_i\{s_i\}$

were the $s_i$ are $\mu$-tableaux. For each one of these, we have $\kappa_t\{s_i\}=\pm e_t$. Thus we find

$\displaystyle\kappa_tu=\sum\limits_i\pm c_ie_t$

which is a multiple of $e_t$, as asserted.

December 31, 2010 -

1. […] see this, let be any vector, and let be a -tableau. Our last result showed us that for some . First, we’ll assume that there is some pair of and so that . In […]

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2. […] Now, the can’t all be zero, so we must have at least one -tableau so that . But then our corollary of the sign lemma tells us that , as we […]

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