Solution to Problem Set 1
October 1, 2015
Problem 1. By explicit computation with 2 × 2 matrices,
[x, y] = h,
(Check!) Thus
tively,

0
0
0
[h, x] = 2x,
[h, y] = −2x.
the matrices of ad(x), ad(h), ad(y) in this basis are, respec
−2 0
0 1 ,
0 0


2 0 0
0 0 0  ,
0 0 −2

0 0 0
−1 0 0 .
0 2 0

and
Problem 2.
Since x has n distinct eigenvalues a1 , . . . , an , it is diagonalizable. After
choosing a suitable basis in Cn , we may assume that x is diagonal, x =
diag(a1 , . . . , an ). We will now work in this basis.
The gl(n, C) has a natural basis {Eij : i, j = 1, . . . , n}, where Eij is the
elementary matrix with 1 in the (i, j)-position, and all other entries equal
to 0. Then x = a1 E11 + · · · + an Enn .
[x, Eij ] = aii [Eii , Eij ] + ajj [Ejj , Eij ] = (ai − aj )Eij .
Thus Eij ’s form a basis of eigenvectors for gl(n, C), and the eigenvalue associated to Eij is ai − aj .
Note: The word “precisely” in the statement of this problem means
that the eigenvalues are counted with multiplicity. It is not enough to show
that each ai − aj is an eigenvalue for adx, but you also need to know that
each occurs with the right multiplicity. Note that assertion of this problem
remains valid even if a1 , . . . , an are not distinct, as long as x is semisimple
(i.e., diagonalizable).
Problem 3. Since tr(AB) = tr(BA) for any two n × n-matrices A and
B, we have [gl(n, C), gl(n, C)] ⊂ sl(n, C).
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To prove the reverse inclusion, it suffices to show that the natural spanning set of sl(n, C) consisting of matrices Eij and Eii − Ejj where i 6= j, and
Eii − Ejj lies inside [gl(n, C), gl(n, C)]. Indeed, for i 6= j,
Eij = [Eik , Ekj ], ∀k 6= i, j, and
Eii − Ejj = [Eij , Eji ].
for any pair of distinct integers 1 ≤ i, j ≤ n.
Problem 4.
(1) Center of gl(n, C): For every matrix X, use X(i, j) to denote its
(i, j)-entry. If X is in the center of gl(n, C), first show it is diagonal.
If not, assume ∃i 6= j such that X(i, j) 6= 0. Then
[X, Ejj ](i, j) = (XEjj )(i, j) − (Ejj X)(i, j) = X(i, j) − 0 6= 0
contradiction.
Now assume X is diagonal, X = diag(a1 , . . . , an ). If ∃i 6= j such that
ai 6= aj , then
[X, Eij ](i, j) = (XEij )(i, j) − (Eij X)(i, j) = (ai − aj )Eij 6= 0
contradiction.
(2) Center of sl(n, C): If X commutes with all elements in sl(n, C),
then it commutes with all elements in gl(n, C) since gl(n, C) = sl(n, C)+CIn .
Hence X ∈ CIn and X ∈ sl(n, C) ∩ CIn = {0}.
Problem 5. We argue by induction on n.
P
j
i
base case: n = 0. Here 0!1 d0 (xy) = xy = i+j=0 di!x dj!y , and the formula
is true.
Assume the the following equivalent formula holds for n:
n X
n i n−i
d (xy) =
d xd y,
i
n
i=0
2
we show the formula for n + 1.
dn+1 (xy) = d(dn (xy))
n X
n i n−i
= d(
d xd y)
i
i=0
n n X
X
n i+1 n−i
n i n+1−i
=
d xd y +
d xd
y
i
i
i=0
i=0
n+1
n X n X
n i n+1−i
k
n+1−k
=
d xd
y+
d xd
y
k−1
i
i=0
k=1
n n X
X
n n+1 0
n
n i n+1−i
n 0 n+1
k
n+1−k
=
d
xd y +
d xd
y+
d xd
y+
d xd
y
n
k−1
i
0
i=1
k=1
n X
n
n
= dn+1 xd0 y +
+
di xdn+1−i y + d0 xdn+1 y
i−1
i
i=1
n X
n + 1 n+1 0
n + 1 i n+1−i
n + 1 0 n+1
=
d
xd y +
d xd
y+
d xd
y
n+1
i
0
i=1
n+1
X n + 1 =
di xdn+1−i y.
i
i=0
Problem 6. Let I be a non-zero ideal of sl(n, C). Our goal is to show
that I = sl(n, C). Note that the case, where n = 2 is settled in Section 2.1
see the Example at the bottom of p. 6. The folowing lemma is a consequence
of this example.
Lemma: Let a and b be distinct integers, between 1 and n and
L = SpanC (Eab , Eba , Eaa − Ebb ) ⊂ sl(n, C).
(a) If I ∩ L 6= (0), then L ⊂ I. In particular,
(b) If Eab ∈ I for some a 6= b then Eaa − Eab ∈ I.
(c) Conversely, if Eaa − Eab ∈ I, for some a 6= b, then Eab ∈ I and
Eba ∈ I.
Proof of the Lemma: L is easily seen to be a Lie subalgebra of sl(n, C)
isomorphic to sl(V ), where V is the 2-dimensional subspace of Cn spanned
by the standard basis vectors ei and ej . (Check!). Thus L is simple, L ∩ I
is a non-trivial ideal of L, and part (a) follows. Parts (b) and (c) are special
cases of (a).
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Now choose t1 , . . . , tn ∈ C such that t1 + · · · + tn = 0 but ti − tj are all
distinct for i 6= j. This can be done because the finitely many hyperplanes
xi − xj = xk − xj in Cn do not cover the hyperplane x1 + · · · + xn = 0.
Let h be the subalgebra of diagonal matrices in sl(n, C) and set
H = diag(t1 , . . . , tn ) ∈ h.
As we showed in Problem 2 above, the linear trasformation
ad(H) : gl(n, C) → gl(n, C)
is diagonalizable, with eigenvalues 0 of multiplicity n and ti − tj of multiplicity 1 for each i 6= j. The eigenspaces corresponding to these eigenvalues
are h and SpanC (Eij ), respectively.
Since I is invariant under H, it has a basis of eigenvectors of ad(H).
Denote this basis by h1 , . . . , hr , Ei1 j1 , . . . , Eis js , where h1 , . . . , hr ∈ h. Note
that a priori r may be 0 (i.e., I ∩ h = (0)) or s may be 0 (i.e., I ⊂ h).
Case 1: s 6= 0, i.e., Eij ∈ I for some i 6= j. Then Eik = [Eij , Ejk ] ∈ I for
any k 6= i, j. That is, Eik ∈ I for every k 6= i.
By Lemma (b), Eii − Ekk ∈ I for every k = 1, . . . , n. (Here i is fixed.)
Since these elements span h, we conclude that h ⊂ I. Thus Eaa − Ebb ∈ I for
every pair of distinct integers a, b between 1 and n. By Lemma (c), Eab ∈ I.
We have shown that h ⊂ I and Eab ∈ I for every a 6= b, Since these
elements span sl(n, C), we conclude that I = sl(n, C).
Case 2: s = 0, i.e., I ⊂ h. Then r ≥ 1. Let h1 = diag(a1 , . . . , an ),
where a1 + · · · + an = 0. Since h1 6= 0, ai 6= aj for some i 6= j. Now
[h1 , Eij ] = (ai − aj )Eij ∈ I. Thus Eij ∈ I, contradicting I ⊂ h.
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