John Stillwell - Naive Lie Theory.pdf - Index of

5.6 Complexification 109
Now we show why gl(n,C) is the complexification of u(n):
gl(n,C)=M n (C)=u(n)+iu(n).
It is clear that any member of u(n)+iu(n) is in M n (C). So it remains to
show that any X ∈ M n (C) can be written in the form
X = X 1 + iX 2 where X 1 ,X 2 ∈ u(n), (*)
that is, where X 1 and X 2 are skew-Hermitian. There is a surprisingly simple
waytodothis:
X = X − X T
+ i X + X T
.
2 2i
We leave it as an exercise to check that X 1 = X−XT
2
and X 2 = X+X T
2i
satisfy
T T
X 1 + X 1 = 0 = X2 + X 2 , which completes the proof.
As a matter of fact, for each X ∈ gl(N,C) the equation (*) has a unique
solution with X 1 ,X 2 ∈ u(n). One solves (*) by first taking the conjugate
transpose of both sides, then forming
X + X T = X 1 + X 1
T + i(X2 − X 2
T )
= i(X 2 − X 2
T ) because X1 + X 1
T = 0
= 2iX 2 because X 2 + X 2
T = 0.
X − X T T T
= X 1 − X 1 + i(X2 + X 2 )
T
T
= X 1 − X 1 because X 2 + X 2 = 0
= 2X 1 because X 1 + X 1
T = 0.
Thus X 1 = X−XT
2
and X 2 = X+X T
2i
that satisfy (*).
are in fact the only values X 1 ,X 2 ∈ u(n)
SL(n,C) and its Lie algebra sl(n,C)
The group SL(n,C) is the subgroup of GL(n,C) consisting of the n × n
complex matrices A with det(A)=1. The tangent vectors of SL(n,C) are
among the tangent vectors X of GL(n,C), but they satisfy the additional
condition Tr(X)=0. This is because e X ∈ GL(n,C) and
det(e X )=e Tr(X) = 1 ⇔ Tr(X)=0.