John Stillwell - Naive Lie Theory.pdf - Index of

110 5 The tangent space
Conversely, if X has trace zero, then so has tX for any real t, soa
matrix X with trace zero gives a smooth path e tX in SL(n,C). This path
has tangent X at 1, so
sl(n,C)={X ∈ M n (C) :Tr(X)=0}.
We now show that the latter set of matrices is the complexification of
su(n), su(n)+isu(n). Since any X ∈ su(n) has trace zero, any member of
su(n)+isu(n) also has trace zero. Conversely, any X ∈ M n (C) with trace
zero can be written as
X = X 1 + iX 2 , where X 1 ,X 2 ∈ su(n).
We use the same trick as for u(n)+iu(n); namely, write
X = X − X T
2
+ i X + X T
.
2i
As before, X 1 = X−XT
2
and X 2 = X+XT
2i
are skew-Hermitian. But also, X 1
and X 2 have trace zero, because X has.
Thus, sl(N,C) ={X ∈ M n (C) :Tr(X) =0} = su(n)+isu(n), as
claimed.
Also, by an argument like that used above for gl(n,C), each X ∈sl(n,C)
corresponds to a unique ordered pair X 1 , X 2 of elements of su(n) such that
X = X 1 + iX 2 .
This equation therefore gives a 1-to-1 correspondence between the elements
X of sl(n,C) and the ordered pairs (X 1 ,X 2 ) such that X 1 ,X 2 ∈ su(n).
Exercises
5.6.1 Show that u(n) and su(n) are not vector spaces over C.
5.6.2 Check that X 1 = X−X T
2
and X 2 = X+X T
2i
are skew-Hermitian, and that X 1 and
X 2 have trace zero when X has.
5.6.3 Show that the groups GL(n,C) and SL(n,C) are unbounded (noncompact)
when the matrix with ( j,k)-entry (a jk + ib jk ) is identified with the point
(a 11 ,b 11 ,a 12 ,b 12 ,...,a 1n ,b 1n ,...,a nn ,b nn ) ∈ R 2n2
and distance between matrices is the usual distance between points in R 2n2 .