John Stillwell - Naive Lie Theory.pdf - Index of

6.4 Simplicity of sl(n,C) and su(n) 125
Simplicity of sl(n,C). For each n, sl(n,C) is a simple Lie algebra.
Proof. If X =(x ij ) is any n × n matrix, then Xe ij has all columns zero
except the jth, which is occupied by the ith column of X, and−e ij X has
all rows zero except the ith, which is occupied by −(row j) of X.
Therefore, since [X,e ij ]=Xe ij − e ij X,wehave
⎛ ⎞
x 1i
.
x i−1,i
column j of [X,e ij ]=
x ii − x jj
,
x i+1,i
⎜ ⎟
⎝ . ⎠
and
row i of [X,e ij ]= ( −x j1 ... −x j, j−1 x ii − x jj −x j, j+1 ... −x jn
)
,
and all other entries of [X,e ij ] are zero. In the (i, j)-position, where the
shifted row and column cross, we get the element x ii − x jj .
We now use such bracketing to show that an ideal I with a nonzero
member X includes all the basis elements of sl(n,C), soI = sl(n,C).
Case (i): X has nonzero entry x ji for some i ≠ j.
Multiply [X,e ij ] by e ij on the right. This destroys all columns except
the ith, whose only nonzero element is −x ji in the (i,i)-position, moving it
to the (i, j)-position (because column i is moved to column j position).
Now multiply [X,e ij ] by −e ij on the left. This destroys all rows except
the jth, whose only nonzero element is x ji at the ( j, j)-position, moving it
to the (i, j)-position and changing its sign (because row j is moved to row
i position, with a sign change).
It follows that [X,e ij ]e ij −e ij [X,e ij ]=[[X,e ij ],e ij ] contains the nonzero
element −2x ji at the (i, j)-position, and zeros elsewhere.
Thus the ideal I containing X also contains e ij . By further bracketing
we can show that all thebasiselementsofsl(n,C) are in I. For
a start, if e ij ∈ I then e ji ∈ I, because the calculation above shows that
[[e ij ,e ji ],e ji ]=−2e ji . The other basis elements can be obtained by using
the result
{
eik if i ≠ k,
[e ij ,e jk ]=
e ii − e jj if i = k,
x ni