The Unapologetic Mathematician

Mathematics for the interested outsider

“Straightening” a Polytabloid

Let’s look at one example of “straightening” out a polytabloid to show it’s in the span of the standard polytabloids, using the Garnir elements.

We’ll start with one we’ve already partially worked out:

\displaystyle t=\begin{array}{ccc}1&2&3\\5&4&\\6&&\end{array}

Now, it’s slightly abusive to the notation, but we’ll just write a tableau t and know that we actually mean the polytabloid e_t in our linear combinations. Using this, we’ve seen that we can write

\displaystyle\begin{array}{ccc}1&2&3\\5&4&\\6&&\end{array}=\begin{array}{ccc}1&2&3\\4&5&\\6&&\end{array}-\begin{array}{ccc}1&4&3\\2&5&\\6&&\end{array}-\begin{array}{ccc}1&2&3\\4&6&\\5&&\end{array}+\begin{array}{ccc}1&4&3\\2&6&\\5&&\end{array}-\begin{array}{ccc}1&5&3\\2&6&\\4&&\end{array}

Now, by the way we’ve selected our Garnir elements, we know that none of these can have any column descents. And we also know that they can’t have a row descent in the same place e_t did. Indeed, the only three that have a row descent all have it between the second and third entries of the first row. So now let’s look at

\displaystyle u=\begin{array}{ccc}1&4&3\\2&5&\\6&&\end{array}

We can write down another table, just like before:

\displaystyle\begin{array}{cccc}A'&B'&\pi&\pi u\\\hline\\\{4,5\}&\{3\}&e&\begin{array}{ccc}1&4&3\\2&5&\\6&&\end{array}\\\{3,5\}&\{4\}&(3\,4)&\begin{array}{ccc}1&3&4\\2&5&\\6&&\end{array}\\\{3,4\}&\{5\}&(3\,5\,4)&\begin{array}{ccc}1&3&5\\2&4&\\6&&\end{array}\end{array}

which lets us write

\displaystyle\begin{array}{ccc}1&4&3\\2&5&\\6&&\end{array}=\begin{array}{ccc}1&3&4\\2&5&\\6&&\end{array}-\begin{array}{ccc}1&3&5\\2&4&\\6&&\end{array}

Similarly we can write

\displaystyle\begin{array}{ccc}1&4&3\\2&6&\\5&&\end{array}=\begin{array}{ccc}1&3&4\\2&6&\\5&&\end{array}-\begin{array}{ccc}1&3&6\\2&4&\\5&&\end{array}

and

\displaystyle\begin{array}{ccc}1&5&3\\2&6&\\4&&\end{array}=\begin{array}{ccc}1&3&5\\2&6&\\4&&\end{array}-\begin{array}{ccc}1&3&6\\2&5&\\4&&\end{array}

Putting these all together, we conclude that

\displaystyle\begin{aligned}\begin{array}{ccc}1&2&3\\5&4&\\6&&\end{array}=&\begin{array}{ccc}1&2&3\\4&5&\\6&&\end{array}-\begin{array}{ccc}1&3&4\\2&5&\\6&&\end{array}+\begin{array}{ccc}1&3&5\\2&4&\\6&&\end{array}-\begin{array}{ccc}1&2&3\\4&6&\\5&&\end{array}\\+&\begin{array}{ccc}1&3&4\\2&6&\\5&&\end{array}-\begin{array}{ccc}1&3&6\\2&4&\\5&&\end{array}-\begin{array}{ccc}1&3&5\\2&6&\\4&&\end{array}+\begin{array}{ccc}1&3&6\\2&5&\\4&&\end{array}\end{aligned}

All of these tabloids are standard, and so we see that our original — nonstandard — e_t is in the span of the standard polytabloids.

January 25, 2011 Posted by | Algebra, Representation Theory, Representations of Symmetric Groups | Leave a comment

   

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