# The Unapologetic Mathematician

## Right Representations

In our discussions of representations so far we’ve been predominantly concerned with actions on the left. That is, we have a map $G\times V\to V$, linear in $V$, that satisfies the relation $g_1(g_2v)=(g_1g_2)v$. That is, the action of the product of two group elements is the composition of their actions.

But sometimes we’re interested in actions on the right. This is almost exactly the same, but with a map $V\times G\to V$, again linear in $V$, and this time the relation reads $(vg_1)g_2=v(g_1g_2)$. Again, the action of the product of two group elements is the composition of their actions, but now in the opposite order! Before we first acted by $g_2$ and then by $g_1$, but now we act first by $g_1$ and then by $g_2$. And so instead of a homomorphism $G\to\mathrm{End}(V)$, we have an anti-homomomorphism — a map from one group to another that reverses the order of multiplication.

We can extend the notation from last time. If the space $V$ carries a right representation of a group $G$, then we hang a tag on the right: $V_G$. If we have an action by another group $H$ on the right that commutes with the action of $G$, we write $V_{GH}$. And if $H$ instead acts on the left, we write ${}_HV_G$. Again, this can be read as a pair of commuting actions, or as a left action of $H$ on the right $G$-module $V_G$, or as a right action of $G$ on the left $G$-module ${}_HV$.

Pretty much everything we’ve discussed moves over to right representations without much trouble. On the occasions we’ll really need them I’ll clarify if there’s anything tricky, but I don’t want to waste a lot of time redoing everything. One exception that I will mention right away is the right regular representation, which (predictably enough) corresponds to the left regular representation. In fact, when I introduced that representation I even mentioned the right action in passing. At the time, I said that we can turn the natural antihomomorphism into a homomorphism by right-multiplying by the inverse of the group element. But if we’re willing to think of a right action on its own terms, we no longer need that trick.

So the group algebra $\mathbb{C}[G]$ — here considered just as a vector space — carries the left regular representation. The left action of a group element $h$ on a basis vector $\mathbf{g}$ is the basis vector $\mathbf{hg}$. It also carries the right regular representation. The right action of a group element $k$ on a basis vector $\mathbf{g}$ is the basis vector $\mathbf{gk}$. And it turns out that these two actions commute! Indeed, we can check

$\displaystyle(h\mathbf{g})k=\mathbf{hg}k=\mathbf{hgk}=h\mathbf{gk}=h(\mathbf{g}k)$

This might seem a little confusing at first, but remember that when $h$ shows up plain on the left it means the group element $h$ acting on the vector to its right. When it shows up in a boldface expression, that expression describes a basis vector in $\mathbb{C}[G]$. Overall, this tells us that we can start with the basis vector $\mathbf{g}$ and act first on the left by $h$ and then on the right by $k$, or we can act first on the right by $k$ and then on the left by $h$. Either way, we end up with the basis vector $\mathbf{hgk}$, which means that these two actions commute. Using our tags, we can thus write ${}_G\mathbb{C}[G]_G$.

November 2, 2010 -

## 1 Comment »

1. […] more explicit parallel between left and right representations: we have morphisms between right -modules just like we had between left -modules. I won’t […]

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