# 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 -

1. [...] « Previous | [...]

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2. [...] Sign Lemma As we move towards proving the useful properties of Specht modules, we will find the following collection of results helpful. Through them all, let be a subgroup, [...]

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3. [...] if is any vector in the Specht module, and if is a tableau of shape , then is some multiple of . Indeed, we can [...]

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4. [...] be a submodule of one of the Young tabloid modules. Then I say that either contains the Specht module , or it is contained in the orthogonal complement . In particular, each Specht module is [...]

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

5. [...] 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 symmetric group . We know that they’re [...]

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

6. [...] Specht Modules [...]

7. [...] standard tableaux are linearly independent. This is half of showing that they form a basis of our Specht modules. We’ll actually use a lemma that applies to any vector space with an ordered basis . Here [...]

Pingback by Standard Polytabloids are Independent « The Unapologetic Mathematician | January 13, 2011 | Reply

8. [...] defined the Specht module as the subspace of the Young tabloid module spanned by polytabloids of shape . But these [...]

Pingback by Standard Polytabloids Span Specht Modules « The Unapologetic Mathematician | January 21, 2011 | Reply

9. [...] that we have a canonical basis for our Specht modules composed of standard polytabloids it gives us a matrix representation of for each . We really only [...]

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10. [...] Generalized Tableaux We want to take our intertwinors and restrict them to the Specht modules. If the generalized tableau has shape and content , we get an intertwinor . This will eventually [...]

Pingback by Semistandard Generalized Tableaux « The Unapologetic Mathematician | February 8, 2011 | Reply