The Unapologetic Mathematician

Mathematics for the interested outsider

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 - Posted by | Algebra, Representation Theory, Representations of Symmetric Groups


  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 […]

    Pingback by The Submodule Theorem « The Unapologetic Mathematician | January 4, 2011 | Reply

  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 […]

    Pingback by Consequences of the Submodule Theorem « The Unapologetic Mathematician | January 4, 2011 | Reply

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