# The Unapologetic Mathematician

## Garnir Elements from Tableaux

There are, predictably enough, certain Garnir elements we’re particularly interested in. These come from Young tableaux, and will be useful to us as we move forward.

Given a tableau $t$, let $A$ be a subset of the entries in the $j$th column of $t$, and let $B$ be a subset of the entries in the $j+1$st column. We can come up with Garnir elements associated to this choice of $A$ and $B$, but — as we pointed out last time — we need some way of picking which particular transversal elements to use. For each summand in $g_{A,B}$, we separate $A\uplus B$ into a pair of sets $(A',B')$, but we have flexibility in how we order the elements of $A'$ and $B'$. Our answer in this case is to always pick the permutation that puts the elements of $A\uplus B$ into increasing order as we move down the columns of $t$.

For example, consider the tableau

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

This tableau has a “row descent” in the second row: a pair of adjacent entries in the row where the larger entry is on the left instead of the right. Let $A$ be the entry on the left along with all the entries below it in its column — $\{5,6\}$ — and let $B$ be the entry on the right along with all the entries above it in its column — $\{2,4\}$. We look at all six ways of rearranging the collection $\{2,4,5,6\}$ into two subsets of two elements each (we listed then last time, actually) and choose permutations that keep entries in increasing order as we move down the columns.

$\displaystyle\begin{array}{cccc}A'&B'&\pi&\pi t\\\hline\\\{5,6\}&\{2,4\}&e&\begin{array}{ccc}1&2&3\\5&4&\\6&&\end{array}\\\{4,6\}&\{2,5\}&(4\,5)&\begin{array}{ccc}1&2&3\\4&5&\\6&&\end{array}\\\{2,6\}&\{5,4\}&(2\,4\,5)&\begin{array}{ccc}1&4&3\\2&5&\\6&&\end{array}\\\{5,4\}&\{2,6\}&(4\,6\,5)&\begin{array}{ccc}1&2&3\\4&6&\\5&&\end{array}\\\{5,2\}&\{6,4\}&(2\,4\,6\,5)&\begin{array}{ccc}1&4&3\\2&6&\\5&&\end{array}\\\{2,4\}&\{5,6\}&(2\,5)(4\,6)&\begin{array}{ccc}1&5&3\\2&6&\\4&&\end{array}\end{array}$

Notice that we’ve picked different permutations this time, and so we get a different Garnir element:

$\displaystyle g_{A,B}=e-(4\,5)+(2\,4\,5)+(4\,6\,5)-(2\,4\,6\,5)+(2\,5)(4\,6)$

Also, note that only the first of these tableaux has the descent in the second row, although some now have descents in the first row. Slightly less obvious is the fact that $g_{A,B}e_t=0$, and so we can write

$\displaystyle e_t=(4\,5)e_t-(2\,4\,5)e_t-(4\,6\,5)e_t+(2\,4\,6\,5)_t-(2\,5)(4\,6)e_t$

Thus we can rewrite this polytabloid that has a row descent in terms of a bunch of other polytabloids that don’t have it and are “more standard”, in a sense we’ll define later.

January 17, 2011