# The Unapologetic Mathematician

## 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.

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February 3, 2011 -

## 2 Comments »

1. […] content , we can construct an intertwinor . Actually, we’ll actually go from to , but since we’ve seen that this is isomorphic to , it’s good enough. Anyway, first, we have to define the […]

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2. […] we’re implicitly using the fact that […]

Pingback by Intertwinors from Semistandard Tableaux Span, part 1 « The Unapologetic Mathematician | February 11, 2011 | Reply