The Adjoint Representation

Since Lie groups are groups, they have representations — homomorphisms to the general linear group of some vector space or another. But since $GL(V)$ is a Lie group, we can use this additional structure as well. And so we say that a representation of a Lie group should not only be a group homomorphism, but a smooth map of manifolds as well.

As a first example, we define a representation that every Lie group has: the adjoint representation. To define it, we start by defining conjugation by $g\in G$ . As we might expect, this is the map $\tau_g=L_g\circ R_{g^{-1}}:G\to G$ — that is, $\tau_g(h)=ghg^{-1}$ . This is a diffeomorphism from $G$ back to itself, and in particular it has the identity $e\in G$ as a fixed point: $\tau_g(e)=e$ . Thus the derivative sends the tangent space at $e$ back to itself: $\tau_{g*e}:\mathcal{T}_eG\to\mathcal{T}_eG$ . But we know that this tangent space is canonically isomorphic to the Lie algebra $\mathfrak{g}$ . That is, $\tau_g\in GL(\mathfrak{g})$ . So now we can define $\mathrm{Ad}:G\to GL(\mathfrak{g})$ by $\mathrm{Ad}(g)=\tau_g$ . We call this the “adjoint representation” of $G$ .

To get even more specific, we can consider the adjoint representation of $GL_n(\mathbb{R})$ on its Lie algebra $\mathfrak{gl}_n(\mathbb{R})\cong M_n(\mathbb{R})$ . I say that $\mathrm{Ad}_g$ is just $\tau_g$ itself. That is, if we view $GL_n(\mathbb{R})$ as an open subset of $M_n(\mathbb{R})$ then we can identify $\mathcal{I}_e:M_n(\mathbb{R})\cong\mathfrak{gl}_n(\mathbb{R})$ . The fact that $\tau_g$ and $\mathrm{Ad}(g)$ both commute means that $\mathcal{I}_e\circ\tau_g=\mathrm{Ad}(g)\circ\mathcal{I}_e$ , meaning that $\tau_g$ and $\mathrm{Ad}(g)$ are “the same” transformation, under this identification of these two vector spaces.

Put more simply: to calculate the adjoint action of $g\in GL_n(\mathbb{R})$ on the element of $\mathfrak{gl}_n(\mathbb{R})$ corresponding to $A\in M_n(\mathbb{R})$ , it suffices to calculate the conjugate $gAg^{-1}$ ; then

$\displaystyle\left[\mathrm{Ad}(g)\right](\mathcal{I}_e(A))=\mathcal{I}_e(\tau_g(A))=\mathcal{I}_e(gAg^{-1})$

About this weblog

This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.

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The Unapologetic Mathematician

Mathematics for the interested outsider