The Unapologetic Mathematician

Mathematics for the interested outsider

The Submodule Theorem

Sorry for the delay.

Let U\subseteq M^\mu be a submodule of one of the Young tabloid modules. Then I say that either U contains the Specht module S^\mu\subseteq M^\mu, or it is contained in the orthogonal complement {S^\mu}^\perp\subseteq M^\mu. In particular, each Specht module S^\mu is irreducible, since any nontrivial submodule U\subseteq S^\mu cannot be contained in the orthogonal complement, and so it must also contain — and thus be equal to — S^\mu.

To see this, let u\in U\subseteq M^\mu be any vector, and let t^\mu be a \mu-tableau. Our last result showed us that \kappa_tu=ce_t for some c\in\mathbb{C}. First, we’ll assume that there is some pair of u and t so that c\neq0. In this case, e_t=\frac{1}{c}\kappa_tu\in U. Since e_t generates all of S^\mu — the Specht modules are cyclic — we must have S^\mu\subseteq U.

On the other hand, what if there is no such pair of u and t? That is, \kappa_tu=0 for all vectors u\in U and \mu-tableaux t. We can calculate

\displaystyle\begin{aligned}\langle u,e_t\rangle&=\langle u,\kappa_t\{t\}\rangle\\&=\langle\kappa_tu,\{t\}\rangle\\&=\langle0,\{t\}\rangle\\&=0\end{aligned}

So each vector u\in U is orthogonal to each polytabloid e_t. Since the polytabloids span S^\mu, we must have U\subseteq{S^\mu}^\perp.

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January 4, 2011 - Posted by | Algebra, Representation Theory, Representations of Symmetric Groups

1 Comment »

  1. […] of the Submodule Theorem We have a number of immediate consequences of the submodule theorem. First, and most important, the Specht modules form a complete list of irreducible modules for the […]

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


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