# The Unapologetic Mathematician

## Mathematics for the interested outsider

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

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

June 13, 2011 -