John Stillwell - Naive Lie Theory.pdf - Index of

122 6 Structure of Lie algebras
6.3 Classical non-simple Lie algebras
We know from Section 2.7 that SO(4) is not a simple group, so we expect
that so(4) is not a simple Lie algebra. We also know, from Section 5.6,
about the groups GL(n,C) and their subgroups SL(n,C). The subgroup
SL(n,C) is normal in GL(n,C) because it is the kernel of the homomorphism
det : GL(n,C) → C × .
It follows that GL(n,C) is not a simple group for any n, so we expect that
gl(n,C) is not a simple Lie algebra for any n. We now prove that these Lie
algebras are not simple by finding suitable ideals.
An ideal in gl(n,C)
We know from Section 5.6 that gl(n,C) =M n (C) (the space of all n × n
complex matrices), and sl(n,C) is the subspace of all matrices in M n (C)
with trace zero. This subspace is an ideal, because it is the kernel of a Lie
algebra homomorphism.
Consider the trace map
Tr : M n (C) → C.
The kernel of this map is certainly sl(n,C), but we have to check that this
map is a Lie algebra homomorphism. It is a vector space homomorphism
because
Tr(X +Y)=Tr(X)+Tr(Y ) and Tr(zX)=zTr(X) for any z ∈ C,
as is clear from the definition of trace.
Also, if we view C as the Lie algebra with trivial Lie bracket [u,v] =
uv − vu = 0, then Tr preserves the Lie bracket. This is due to the (slightly
less obvious) property that Tr(XY) =Tr(YX), which can be checked by
computing both sides (see Exercise 5.3.8). Assuming this property of Tr,
we have
Tr([X,Y ]) = Tr(XY −YX)
= Tr(XY) − Tr(YX)
= 0
=[Tr(X),Tr(Y )].
Thus Tr is a Lie bracket homomorphism and its kernel, sl(n,C), is necessarily
an ideal of M n (C)=gl(n,C).