The Unapologetic Mathematician

Mathematics for the interested outsider

Modules of Generalized Young Tableaux

We can obviously create vector spaces out of generalized Young tableaux. Given the collection T_{\lambda\mu} of tableaux of shape \lambda and content \mu, we get the vector space \mathbb{C}[T_{\lambda\mu}]. We want to turn this into an S_n-module.

First, given any tabloid \{s\} of shape \mu, we can product a (generalized) tableau T\in T_{\lambda\mu} by defining T(i) to be the number of the row in s that contains the entry i. As an example, consider the tabloid

\displaystyle\{s\}=\begin{array}{cc}\cline{1-2}2&3\\\cline{1-2}1&5\\\cline{1-2}4&\\\cline{1-1}\end{array}

This gives us the function T(2)=T(3)=1, T(1)=T(5)=2, and T(4)=3. If \lambda=(3,2) and we use the usual reference tableau t, this gives us the generalized tabloid

\displaystyle T=\begin{array}{ccc}T(1)&T(2)&T(3)\\T(4)&T(5)&\end{array}=\begin{array}{ccc}2&1&1\\3&2&\end{array}

The shape of T is obviously \lambda, and it’s easy to see that the content is exactly \mu. Indeed, there are \mu_i entries in T with the value i, just as there are \mu_i entries in the first row of \{s\}.

It should also be clear that this correspondence is a bijection. That is, given any generalized tableau T of shape \lambda and content \mu we can get a tabloid of shape \mu by turning T into a function and then putting k on row i of \{s\} if T(k)=i.

That means that the basis of generalized tableaux T_{\lambda\mu} of the vector space \mathbb{C}[T_{\lambda\mu}] is in bijection with the basis of \mu-tabloids of the vector space M^\mu. And this space carries an action of S_n — the linear extension of the action on tabloids. We want to pull this action across the bijection we just set up to get an action on T_{\lambda\mu}.

On the one hand, this is as easy as saying it: if T corresponds to \{s\}, we define \pi T to be the generalized tableau corresponding to \pi\{s\} and we’re done. To be a bit more explicit, we define \pi T by considering it as a function and setting

\displaystyle\left[\pi T\right](i)=T(\pi^{-1}i)

So, for example, if

\displaystyle T=\begin{array}{ccc}T(1)&T(2)&T(3)\\T(4)&T(5)&\end{array}

then we can calculate

\displaystyle (1\,2\,4)T=\begin{array}{ccc}T(4)&T(1)&T(3)\\T(2)&T(5)&\end{array}

Even more explicitly, if

\displaystyle T=\begin{array}{ccc}2&1&1\\3&2&\end{array}

then we calculate

\displaystyle (1\,2\,4)T=\begin{array}{ccc}3&2&1\\1&2&\end{array}

We should be clear about a major distinction here: the permutation \pi\in S_n acts on the entries in \{s\} — replacing i by \pi i — but it acts on the places in T — moving T(i) to the position of T(\pi i).

If we write the correspondence as \theta(\{s\})=T, then for \theta to be an intertwinor we need \theta(\pi\{s\})=\pi T. This forces

\displaystyle\begin{aligned}\left[\pi T\right](i)&=\text{row number of }i\text{ in }\pi\{s\}\\&=\text{row number of }\pi^{-1}i\text{ in }\{s\}\\&=T(\pi^{-1}i)\end{aligned}

and so this explicit action is forced on us.

The really interesting thing is that when we use this action on the generalized tableaux in T_{\lambda\mu}, we always get a module \mathbb{C}[T_{\lambda\mu}]\cong M^\mu, no matter what shape \lambda we start with.

February 3, 2011 Posted by | Algebra, Representation Theory, Representations of Symmetric Groups | 2 Comments

   

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