The Unapologetic Mathematician

Mathematics for the interested outsider

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})

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June 13, 2011 - Posted by | Algebra, Differential Topology, Lie Groups, Topology

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