The Unapologetic Mathematician

Mathematics for the interested outsider

The Action on Tableaux and Tabloids

We’ve introduced Young tableaux and Young tabloids. We’ve also said that they carry symmetric group actions, but we never really said what they were.

So, let \lambda\vdash n be a partition, and consider the collection of Young tableaux of shape \lambda. I say that the symmetric group S_n acts on this set. Indeed, given a tableau t with entries t_{i,j}, and given a permutation \pi\in S_n, we define a new tableau \pi t by giving its entries:

\displaystyle(\pi t)_{i,j}=\pi(t_{i,j})

For example, we can calculate

\displaystyle (1\,2\,3)\begin{array}{cc}1&2\\3&\end{array}=\begin{array}{cc}2&3\\1&\end{array}

It should be clear that \pi t has an (i,j) entry if and only if t does, and so \pi t has the same shape as t does. Further, each number from 1 to n shows up exactly once as an entry in \pi t just as in t, and so \pi t really is a Young tableau of shape \lambda.

Since any tableau can be sent to any other by a judicious choice of permutation, this action is transitive. In fact, since there is exactly one permutation that works, the action is simply transitive.

Now, this action descends to an action on the set of Young tabloids of shape \lambda. That is, we can define \pi\{t\}=\{\pi t\}. The important thing here is to verify that the action is well-defined. But the action of the permutation \pi doesn’t care about the positions of the numbers within the tableau, so rearranging them has no effect. For example, we can check that

\displaystyle (1\,2\,3)\begin{array}{cc}2&1\\3&\end{array}=\begin{array}{cc}3&2\\1&\end{array}

Swapping the two numbers in the first row of the tableau before applying the permutation (1\,2\,3) gives the same result as first applying the permutation and then swapping the entries in the first row. Similarly, rearranging the entries in any row of any Young tableau commutes with the action of S_n. And so the action on Young tabloids is well-defined.

Given any two Young tabloids \{s\} and \{t\} of shape \lambda\vdash n, we have representative Young tableaux s and t. There is a unique \pi\in S_n such that \pi s=t, and so we have

\displaystyle\pi\{s\}=\{\pi s\}=\{t\}

Therefore the action on Young tabloids is transitive. It’s not simply transitive, though, since any permutation that only shuffles entries on the same row of a tableax t leaves the tabloid \{t\} the same.


December 13, 2010 - Posted by | Algebra, Representation Theory, Representations of Symmetric Groups


  1. […] that we have an action of on the Young tabloids of shape , we can consider the permutation representation that […]

    Pingback by Permutation Representations from Partitions « The Unapologetic Mathematician | December 14, 2010 | Reply

  2. […] with rows . Any permutation that just mixes up elements of leaves all but the first row alone when acting on . Since it leaves every element on the row where it started, we say that it stabilizes the rows […]

    Pingback by Row- and Column-Stabilizers « The Unapologetic Mathematician | December 22, 2010 | Reply

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