# The Unapologetic Mathematician

## Standard Polytabloids are Independent

Now we’re all set to show that the polytabloids that come from standard tableaux are linearly independent. This is half of showing that they form a basis of our Specht modules. We’ll actually use a lemma that applies to any vector space $V$ with an ordered basis $e_\alpha$. Here $\alpha$ indexes some set $B$ of basis vectors which has some partial order $\preceq$.

So, let $v_1,\dots,v_m$ be vectors in $V$, and suppose that for each $v_i$ we can pick some basis vector $e_{\alpha_i}$ which shows up with a nonzero coefficient in $v_i$ subject to the following two conditions. First, for each $i$ the basis element $e_{\alpha_i}$ should be the maximum of all the basis vectors having nonzero coefficients in $v_i$. Second, the $e_{\alpha_i}$ are all distinct.

We should note that the first of these conditions actually places some restrictions on what vectors the $v_i$ can be in the first place. For each one, the collection of basis vectors with nonzero coefficients must have a maximum. That is, there must be some basis vector in the collection which is actually bigger (according to the partial order $\preceq$) than all the others in the collection. It’s not sufficient for $e_{\alpha_i}$ to be maximal, which only means that there is no larger index in the collection. The difference is similar to that between local maxima and a global maximum for a real-valued function.

This distinction should be kept in mind, since now we’re going to shuffle the order of the $v_i$ so that $e_{\alpha_1}$ is maximal among the basis elements $e_{\alpha_i}$. That is, none of the other $e_{\alpha_i}$ should be bigger than $e_{\alpha_1}$, although some may be incomparable with it. Now I say that $e_{\alpha_i}$ cannot have a nonzero coefficient in any other of the $v_i$. Indeed, if it had a nonzero coefficient in, say, $v_2$, then by assumption we would have $e_{\alpha_1}\prec e_{\alpha_2}$, which contradicts the maximality of $e_{\alpha_1}$. Thus in any linear combination $\displaystyle c_1v_1+\dots+c_mv_m=0$

we must have $c_1=0$, since there is no other way to cancel off all the occurrences of $e_{\alpha_1}$. Removing $v_1$ from the collection, we can repeat the reasoning with the remaining vectors until we get down to a single one, which is trivially independent.

So in the case we care about the space is the Young tabloid module $M^\lambda$, with the basis of Young tabloids having the dominance ordering. In particular, we consider for our $v_i$ the collection of polytabloids $e_t$ where $t$ is a standard tableau. In this case, we know that $\{t\}$ is the maximum of all the tabloids showing up as summands in $e_t$. And these standard tabloids are all distinct, since they arise from distinct standard tableaux. Thus our lemma shows that not only are the standard polytabloids $e_t$ distinct, they are actually linearly independent vectors in $M^\lambda$.

January 13, 2011

## The Maximality of Standard Tableaux

Standard tableaux have a certain maximality property with respect to the dominance order on tabloids. Specifically, if $t$ is standard and $\{s\}$ appears as a summand in the polytabloid $e_t$, then $\{t\}\trianglerighteq\{s\}$.

Any such $\{s\}$ comes from $s=\pi t$, where $\pi\in C_t$. We will make our induction on the number of “column inversions” in $s$. That is, the number of pairs of entries $k that are in the same column of $s$, but which are “out of order”, in the sense that $k$ is in a lower row than $l$.

Given any such pair, the dominance lemma tells us that $\{s\}\triangleleft(k\,l)\{s\}$. That is, by “untwisting” the column inversion, we can move up the dominance order while preserving the columns. It should also be clear that $(k\,l)\{s\}$ has fewer column inversions than $\{s\}$ does. But if we undo all the column inversions, the tableau we’re left with must be standard. That is, it must be $\{t\}$ itself.

January 13, 2011