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Unitary Matrices Up: Lie Algebras

The Adjoint Representation

Elements of $SO(3)$ act on ${\bf R}^3$, or equivalently on orthonormal bases of ${\bf R}^3$ (frames).2 But $SO(3)$ can also be regarded as a set of transformations on the vector space $so(3)$, as we will see in a moment. Intuitively, the triple ($\bf\hat{x}$, $\bf\hat{y}$, $\bf\hat{z}$) takes the place of the standard frame for ${\bf R}^3$.

General fact: for any Lie group $G$, there is a homomorphism (also known as a representation) of $G$ into the group of non-singular linear transformations on the vector space $T(G)$, with kernel $Z(G)$, the center of $G$.

Here's how it goes. For any group $G$, we have the group of inner automorphism ${\rm Inn}(G)$ and a homomorphism $G \rightarrow {\rm Inn}(G)$ defined by $g\mapsto \iota_g$, where $\iota_g(h)=ghg^{-1}$. The kernel is $Z(G)$. The automorphism $\iota_g$ is furthermore determined completely by its effects on any set of generators for $G$.

Now take $G$ to be a Lie group. Let's consider the effect of $\iota_g$ on an ``infinitesimal'' generator ${\bf 1}+\epsilon h$, where $h\in T(G)$.

g({\bf 1}+\epsilon h)g^{-1} = {\bf 1}+ \epsilon ghg^{-1}

Or in terms of derivatives, if $x(t)$ is our prototypical smooth curve through 1, then the derivative of $gx(t)g^{-1}$ at $t=0$ is $gx'(0)g^{-1}$. So the vector space $T(G)$ is closed under the map $h\mapsto ghg^{-1}$. (Remember that both $G$ and $T(G)$ are sets of matrices.) $Z(G)$ is clearly contained in the kernel, and in fact: $Z(G)$ is the kernel.

For $SO(3)$, this is rather intuitive. Suppose $h\in SO(3)$ is a rotation about the axis determined by vector $v\in{\bf R}^3$. Then $ghg^{-1}$ is a rotation about the axis $gv$: $(ghg^{-1})(gv)=ghv=gv$. If we think of $h$ as an infinitesimal rotation, then we see that the action of $SO(3)$ on $so(3)$ given by $\iota_g$ looks just like the action of $SO(3)$ on ${\bf R}^3$.

Only in three dimensions do things work out so neatly. $SO(2)$ is abelian, and the adjoint representation for abelian Lie groups is boring-- the trivial homomorphism. And the vector space $so(4)$ has dimension 6, so the adjoint representation gives an imbedding of $SO(4)$ in the group of non-singular $6\times 6$ matrices.

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© 2001 Michael Weiss