The Unapologetic Mathematician

Mathematics for the interested outsider

Standard Tableaux

So we’ve described the Specht modules, and we’ve shown that they give us a complete set of irreducible representations for the symmetric groups. But we haven’t described them very explicitly, and we certainl can’t say much about them. There’s still work to be done.

We say that a Young tableau t is “standard” if its rows and columns are all increasing sequences. In this case, we also say that the Young tabloid \{t\} and the polytabloid e_t=\kappa_t\{t\} are standard.

Recall that we had a canonical Young tableau for each shape \lambda that listed the numbers from 1 to n in each row from top to bottom, as in

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

It should be clear that this canonical tableau is standard, so there is always at least one standard tableau for each shape. There may be more, of course. For example:

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

Clearly, any two distinct standard tableaux s^\lambda and t^\lambda give rise to distinct tabloids \{s\} and \{t\}. Indeed, if \{s\}=\{t\}, then s and t would have to be row-equivalent. But only one Young tableau in any row-equivalence class has increasing rows, and only that one even has a chance to be standard. Thus if s and t are row-equivalent standard tableaux, they must be equal.

What’s not immediately clear is that the standard polytabloids e_s and e_t are distinct. Further, it turns out that the collection of standard polytabloids e_t of shape \lambda is actually independent, and furnishes a basis for the Specht module S^\lambda. This is our next major goal.

January 5, 2011 Posted by | Algebra, Representation Theory, Representations of Symmetric Groups | 10 Comments

   

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