# The Unapologetic Mathematician

## Specht Modules

Now we have everything in place to define the representations we’re interested in. For any partition $\lambda$, the Specht module $S^\lambda$ is the submodule of the Young tabloid module $M^\lambda$ spanned by the polytabloids $e_t$ where $t$ runs over the Young tableaux $t$ of shape $\lambda$.

To see that the subspace spanned by the polytabloids is a submodule, we must see that it’s invariant under the action of $S_n$. We can use our relations to check this. Indeed, if $e_t$ is a polytabloid, then $\pi e_t=e_{\pi t}$ is another polytabloid, so the subspace spanned by the polytabloids is invariant under the action of $S_n$.

The most important fact about the Specht modules is that they’re cyclic. That is, we can generate one just by starting with a single vector and hitting it with all the elements in the group algebra $\mathbb{C}[S_n]$. Not all of the resulting vectors will be different, but among them we’ll get the whole Specht module. The term “cyclic” comes from group theory, where the cyclic groups are those from modular arithmetic, like $\mathbb{Z}_n=\mathbb{Z}/n\mathbb{Z}$. Many integers give the same residue class modulo $n$, but every residue class comes from some integer.

Anyway, in the case of Specht modules, we will show that the action of $S_n$ can take one vector and give a whole basis for $S^\lambda$. Then any vector in the Specht module can be written as a sum of basis vectors, and thus as the action of some algebra element from $\mathbb{C}[S_n]$ on our starting vector. But which starting vector will we choose? Well, any polytabloid will do. Indeed, if $e_s$ and $e_t$ are polytabloids, then there is some (not unique!) permutation $\pi$ so that $t=\pi s$. But then $e_t=e_{\pi s}=\pi e_s$, and so $e_t$ is in the $S_n$-orbit of $e_s$. Thus starting with $e_s$ we can get to every vector in $S^\lambda$ by the action of $\mathbb{C}[S_n]$.

December 27, 2010