The Unapologetic Mathematician

Mathematics for the interested outsider

Garnir Elements

Let A and B be two disjoint sets of positive integers. We’re mostly interested in the symmetric group S_{A\uplus B}, which shuffles around all the integers in both sets. But a particularly interesting subgroup is S_A\times S_B, which shuffles around the integers in A and B, but doesn’t mix the two together. Clearly, S_A\times S_B is a subgroup of S_{A\uplus B}.

Now, let \Pi\subseteq S_{A\uplus B} be a transversal collection of permutations for this subgroup. That is, we can decompose the group into cosets

\displaystyle S_{A\uplus B}=\biguplus\limits_{\pi\in\Pi}\pi(S_A\times S_B)

Then a “Garnir element” is

\displaystyle g_{A,B}=\Pi^-=\sum\limits_{\pi\in\Pi}\mathrm{sgn}(\pi)\pi\in\mathbb{C}[S_{A\uplus B}]

Now, the problem here is that we’ve written g_{A,B} as if it only depends on the sets A and B, when it clearly depends on the choice of the transversal \Pi. But we’ll leave this alone for the moment.

How can we come up with an explicit transversal in the first place? Well, consider the set of pairs of sets (A',B') so that \lvert A'\rvert=\lvert A\rvert, \lvert B'\rvert=\lvert B\rvert, and A'\uplus B'=A\uplus B. That is, each (A',B') is another way of breaking the same collection of integers up into two parts of the same sizes as A and B.

Any permutation \pi\in S_{A\uplus B} acts on the collection of such pairs of sets in the obvious way, sending (A',B') to (\pi(A'),\pi(B')), which is another such pair. In fact, it’s transitive, since we can always find some \pi with A'=\pi(A) and B'=\pi(B). If for each (A',B') we make just such a choice of \pi, then this collection of permutations gives us a transversal!

We can check this by first making sure we have the right number of elements. A pair (A',B') is determined by taking all \lvert A\uplus B\rvert integers and picking \lvert A\rvert to go into A'. That is, we have

\displaystyle\binom{\lvert A\uplus B\rvert}{\lvert A\rvert}=\frac{\lvert A\uplus B\rvert!}{\lvert A\rvert!\lvert B\rvert!}

pairs. But this is also the number of cosets of S_A\times S_B in S_{A\uplus B}!

Do we accidentally get two representatives \pi and \pi' for the same coset? If we did, then we’d have to have \pi^{-1}\pi'\in S_A\times S_B. But then \pi^{-1}\pi'(A,B)=(A,B), and thus \pi'(A,B)=\pi(A,B). But we only picked one permutation sending (A,B) to a given pair, so \pi'=\pi.

As an example, let A=\{5,6\} and B=\{2,4\}. Then we have six pairs of sets to consider

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

where in each case I’ve picked a \pi that sends (A,B) to (A',B'). This will give us the following Garnir element:

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

But, again, this is far from the only possible choice for this A and B.

January 14, 2011 Posted by | Algebra, Representation Theory, Representations of Symmetric Groups | 3 Comments

   

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