John Stillwell - Naive Lie Theory.pdf - Index of

124 6 Structure of Lie algebras
We let
⎛
⎞
x 11 x 12 ... x 1n
x 21 x 22 ... x 2n
X = ⎜ .
. ⎟
⎝ .
. ⎠
x n1 x n2 ... x nn
be any element of gl(n,C), and consider its Lie bracket with e ij , the matrix with
1 as its (i, j)-entry and zeros elsewhere.
6.3.1 Describe Xe ij and e ij X. Hence show that the trace of [X,e ij ] is x ji − x ji = 0.
6.3.2 Deduce from Exercise 6.3.1 that Tr([X,Y]) = 0foranyX,Y ∈ gl(n,C).
6.3.3 Deduce from Exercise 6.3.2 that sl(n,C) is an ideal of gl(n,C).
Another example of a non-simple Lie algebra is u(n), the algebra of n × n
skew-hermitian matrices.
6.3.4 Find a 1-dimensional ideal I in u(n), andshowthatI is the tangent space
of Z(U(n)).
6.3.5 Also show that the Z(U(n)) is the image, under the exponential map, of the
ideal I in Exercise 6.3.4.
6.4 Simplicity of sl(n,C) andsu(n)
We saw in Section 5.6 that sl(n,C) consists of all n × n complex matrices
with trace zero. This set of matrices is a vector space over C, and it has
a natural basis consisting of the matrices e ij for i ≠ j and e ii − e nn for
i = 1,2,...,n−1, where e ij is the matrix with 1 as its (i, j)-entry and zeros
elsewhere. These matrices span sl(n,C). In fact, for any X ∈ sl(n,C),
n−1
X =(x ij )=∑ x ij e ij + ∑ x ii (e ii − e nn )
i≠ j i=1
because x nn = −x 11 −x 22 −···−x n−1,n−1 for the trace of X to be zero. Also,
X is the zero matrix only if all the coefficients are zero, so the matrices e ij
for i ≠ j and e ii − e nn for i = 1,2,...,n − 1 are linearly independent.
These basis elements are convenient for Lie algebra calculations because
the Lie bracket of any X with an e ij has few nonzero entries. This
enables us to take any nonzero member of an ideal I and manipulate it
to find a nonzero multiple of each basis element in I, thus showing that
sl(n,C) contains no nontrivial ideals.