The Lie algebras of and are denoted and , respectively. Differentiating we conclude that consists of anti-Hermitian matrices: . Note that is anti-Hermitian if and only if is Hermitian.
Fact: if , then (where tr = trace). (Expanding by minors does the trick.) This makes one half of the following fact obvious: the Lie algebra for the Lie group of unimodular matrices consists of all the traceless matrices.
For the special case , things work out very nicely. Since
, one can write down the components for easily, and
equating them to , one concludes that consists of all matrices
of the form:
Next, an arbitary anti-Hermitian matrix looks like:
It turns out that this correspondence is a Lie algebra isomorphism. and are locally isomorphic, but not isomorphic-- as we will see next.
acts on via the adjoint representation. But we have a 1-1 correspondence between and , so we have a representation of in the group of real matrices. Let be an element of and be an element of . Note that . Since the map preserves determinants, it preserves norms when considered as acting on . So the adjoint representation maps into . Fact: it maps onto .
Incidentally, you can see directly that preserves anti-Hermiticity by writing it .
The kernel of the adjoint representation for is its center, which clearly contains -- and in fact, consists of just those two elements. So we have a 2-1 mapping . Our double cover! I'll look at the topology of this in a moment.
Physicists prefer to work with the Pauli spin matrices instead of the quaternions. The Pauli matrices are just the Hermitian counterparts to i, j, and k:
They form a basis (with 1) for the vector space of Hermitian
© 2001 Michael Weiss