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 
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 
. As we might expect, this is the map 
— that is, 
. This is a diffeomorphism from 
back to itself, and in particular it has the identity 
as a fixed point: 
. Thus the derivative sends the tangent space at 
back to itself: 
. But we know that this tangent space is canonically isomorphic to the Lie algebra 
. That is, 
. So now we can define 
by 
. We call this the “adjoint representation” of .
To get even more specific, we can consider the adjoint representation of 
on its Lie algebra 
. I say that 
is just 
itself. That is, if we view as an open subset of 
then we can identify 
. The fact that and 
both commute means that 
, meaning that and are “the same” transformation, under this identification of these two vector spaces.
Put more simply: to calculate the adjoint action of 
on the element of 
corresponding to 
, it suffices to calculate the conjugate 
; then
)=\mathcal{I}_e(\tau_g(A))=\mathcal{I}_e(gAg^{-1})](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cleft%5B%5Cmathrm%7BAd%7D%28g%29%5Cright%5D%28%5Cmathcal%7BI%7D_e%28A%29%29%3D%5Cmathcal%7BI%7D_e%28%5Ctau_g%28A%29%29%3D%5Cmathcal%7BI%7D_e%28gAg%5E%7B-1%7D%29&bg=e6e6e6&fg=333333&s=0 "\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 John Armstrong |
Algebra, Differential Topology, Lie Groups, Topology
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