The Unapologetic Mathematician

Mathematics for the interested outsider

Coset Representations

Next up is a family of interesting representations that are also applicable to any group G. The main ingredient is a subgroup H — a subset of the elements of G so that the inverse of any element in H is also in H, and the product of any two elements of H is also in H.

Our next step is to use H to break G up into cosets. We consider g_1 and g_2 to be equivalent if g_2^{-1}g_1\in H. It’s easy to check that this is actually and equivalence relation (reflexive, symmetric, and transitive), and so it breaks G up into equivalence classes. We write the coset of all g'\in G that are equivalent to g as gH, and we write the collection of all cosets of H as G/H.

We should note that we don’t need to worry about H being a normal subgroup of G, since we only care about the set of cosets. We aren’t trying to make this set into a group — the quotient group — here.

Okay, now multiplication on the left by G shuffles around the cosets. That is, we have a group action of G on the quotient set G/H, and this gives us a permutation representation of G!

Let’s work out an example to see this a bit more explicitly. For our group, take the symmetric group S_3, and for our subgroup let H=\{e,(2\,3)\}. Indeed, H is closed under both inversion and multiplication. And we can break G up into cosets:

\displaystyle\begin{aligned}\{e,(1\,2),(1\,3),(2\,3),(1\,2\,3),(1\,3\,2)\}&=\{e,(2\,3)\}\uplus\{(1\,2),(1\,2\,3)\}\uplus\{(1\,3),(1\,3\,2)\}\\&=H\uplus(1\,2)H\uplus(1\,3)H\end{aligned}

where we have picked a “transversal” — one representative of each coset so that we can write them down more simply. It doesn’t matter whether we write (1\,2)H or (1\,2\,3)H, since both are really the same set. Now we can write down the multiplication table for the group action. It takes g_1\in G and g_2H\in G/H, and tells us which coset g_1g_2 falls in:

\displaystyle\begin{tabular}{c|rrr}&H&(1\,2)H&(1\,3)H\\\hline e&H&(1\,2)H&(1\,3)H\\(1\,2)&(1\,2)H&H&(1\,3)H\\(1\,3)&(1\,3)H&(1\,2)H&H\\(2\,3)&H&(1\,3)H&(1\,2)H\\(1\,2\,3)&(1\,2)H&(1\,3)H&H\\(1\,3\,2)&(1\,3)H&H&(1\,2)H\end{tabular}

This is our group action. Since there are three elements in the set G/H, the permutation representation we get will be three-dimensional. We can write down all the matrices just by working them out from this multiplication table:

\displaystyle\begin{aligned}\rho_H(e)&=\begin{pmatrix}1&0&0\\{0}&1&0\\{0}&0&1\end{pmatrix}\\\rho_H((1\,2))&=\begin{pmatrix}{0}&1&0\\1&0&0\\{0}&0&1\end{pmatrix}\\\rho_H((1\,3))&=\begin{pmatrix}{0}&0&1\\{0}&1&0\\1&0&0\end{pmatrix}\\\rho_H((2\,3))&=\begin{pmatrix}1&0&0\\{0}&0&1\\{0}&1&0\end{pmatrix}\\\rho_H((1\,2\,3))&=\begin{pmatrix}{0}&0&1\\1&0&0\\{0}&1&0\end{pmatrix}\\\rho_H((1\,3\,2))&=\begin{pmatrix}{0}&1&0\\{0}&0&1\\1&0&0\end{pmatrix}\\\end{aligned}

It turns out that these matrices are the same as we saw when writing down the defining representation of S_3. There’s a reason for this, which we will examine later.

As special cases, if H=\{e\}, then there is one coset for each element of G, and the coset representation is the same as the left regular representation. At the other extreme, if H=G, then there is only one coset and we get the trivial representation.

September 20, 2010 Posted by | Algebra, Group theory, Representation Theory | 5 Comments

   

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