# Finding a maximal tableau for a sum of Jucys-Murphy Elements

Let $X \in \mathbb{C}[S_n]$ be an element of the group algebra of $S_n$ expressible as the sum of some Jucys-Murphy elements. Then let $\lambda$ be any irreducible representation of $S_n$, with the usual basis given by the standard Young tableaux of shape $\lambda$. Each of these tableaux is of course an eigenvector for the action of $X$, with the eigenvalue easily computable from the contents of the relevant boxes.

My question is this: Given a particular $X$ and $\lambda$, is there an easy combinatorial way to construct the tableau $T$ that will have the maximal eigenvalue for the action of $X$? (Again, for the moment, I only care about $X$ that are sums of Jucys-Murphy elements - i.e. each has coefficient 0 or 1. Allowing arbitrary linear combinations of them would seem much harder....)

Ideally, it would be nice if there were a way to "build up" this tableau by adding the boxes $1, \ldots n$ in order, or by placing the numbers in the tableau from $n$ on down to 1, but I'm not convinced this is possible.

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One small observation: suppose that $\beta$ is the number in the rightmost removable box of a standard $\lambda$-tableau $t$ with greatest $X$-eigenvalue. Then either $L_\beta$ is a summand of $X$, or $X$ is a sum of $L_\alpha$ for $\alpha < \beta$. (Proof: take $\gamma > \beta$ with $\gamma$ minimal such that $L_\gamma$ appears in $X$. Then applying the cycle $(\gamma, \gamma-1, \ldots, \beta+1, \beta)$ to $t$ gives a tableau with greater $X$-eigenvalue.) – Mark Wildon Apr 5 '12 at 12:23
Thanks for the comment. In our thinking about the problem so far, we've mostly gotten lots of little ideas like this - algorithms that work most of the time but not always, special cases, etc.... – Matt Davis Apr 6 '12 at 20:54