The Unapologetic Mathematician

Mathematics for the interested outsider

Intertwinors from Generalized Tableaux

Given any generalized Young tableau T with shape \lambda and content \mu, we can construct an intertwinor \theta_T:M^\lambda\to M^\mu. Actually, we’ll actually go from M^\lambda to \mathbb{C}[T_{\lambda\mu}, but since we’ve seen that this is isomorphic to M^\mu, it’s good enough. Anyway, first, we have to define the row-equivalence class \{T\} and column-equivalence class [T]. These are the same as for regular tableaux.

So, let t be our reference tableau and let \{t\} be the associated tabloid. We define

\displaystyle\theta_T\left(\{t\}\right)=\sum\limits_{S\in\{T\}}S

Continuing our example, with

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

we we define

\displaystyle\begin{aligned}\theta_T\left(\begin{array}{ccc}\cline{1-3}1&2&3\\\cline{1-3}4&5&\\\cline{1-2}\end{array}\right)&=\begin{array}{ccc}2&1&1\\3&2&\end{array}+\begin{array}{ccc}1&2&1\\3&2&\end{array}+\begin{array}{ccc}1&1&2\\3&2&\end{array}\\&+\begin{array}{ccc}2&1&1\\2&3&\end{array}+\begin{array}{ccc}1&2&1\\2&3&\end{array}+\begin{array}{ccc}1&1&2\\2&3&\end{array}\end{aligned}

Now, we extend in the only way possible. The module M^\lambda is cyclic, meaning that it can be generated by a single element and the action of \mathbb{C}[S_n]. In fact, any single tabloid will do as a generator, and in particular \{t\} generates M^\lambda.

So, any other module element in M^\lambda is of the form \pi\{t\} for some \pi\in\mathbb{C}[S_n]. And so if \theta_T is to be an intertwinor we must define

\displaystyle\theta_T\left(\pi\{t\}\right)=\pi\theta_T(\{t\})=\sum\limits_{S\in\{T\}}\pi S

Remember here that \pi acts on generalized tableaux by shuffling the entries by place, not by value. Thus in our example we find

\displaystyle\begin{aligned}\theta_T\left(\begin{array}{ccc}\cline{1-3}2&4&3\\\cline{1-3}1&5&\\\cline{1-2}\end{array}\right)&=\theta_T\left((1\,2\,4)\begin{array}{ccc}\cline{1-3}1&2&3\\\cline{1-3}4&5&\\\cline{1-2}\end{array}\right)\\&=(1\,2\,4)\theta_T\left(\begin{array}{ccc}\cline{1-3}1&2&3\\\cline{1-3}4&5&\\\cline{1-2}\end{array}\right)\\&=(1\,2\,4)\begin{array}{ccc}2&1&1\\3&2&\end{array}+(1\,2\,4)\begin{array}{ccc}1&2&1\\3&2&\end{array}+(1\,2\,4)\begin{array}{ccc}1&1&2\\3&2&\end{array}\\&+(1\,2\,4)\begin{array}{ccc}2&1&1\\2&3&\end{array}+(1\,2\,4)\begin{array}{ccc}1&2&1\\2&3&\end{array}+(1\,2\,4)\begin{array}{ccc}1&1&2\\2&3&\end{array}\\&=\begin{array}{ccc}3&2&1\\1&2&\end{array}+\begin{array}{ccc}3&1&1\\2&2&\end{array}+\begin{array}{ccc}3&1&2\\1&2&\end{array}\\&+\begin{array}{ccc}2&2&1\\1&3&\end{array}+\begin{array}{ccc}2&1&1\\2&3&\end{array}+\begin{array}{ccc}2&1&2\\1&3&\end{array}\end{aligned}

Now it shouldn’t be a surprise that since so much of our construction to this point has depended on an aribtrary choice of a reference tableau t, the linear combination of generalized tableaux on the right doesn’t quite seem like it comes from the tabloid on the left. But this is okay. Just relax and go with it.

February 5, 2011 Posted by | Algebra, Representation Theory, Representations of Symmetric Groups | 5 Comments

   

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