Math 534, Fll 2015. Solutions to Problem Set 3
Problem 1. (5pts)
Let N be a nilpotent ideal of L of maximal dimension. If I is another
nilpotent ideal, then N is contained in N + I, which is also a nilpotent ideal.
By maximality of dim(N ), we have N + I = N and thus I ⊂ N . This proves
both maximality and uniqueness of N .
Problem 2 (1 on page 20). (5pts)
Let g = sl(V ) and R be its radical. By Lie’s theorem, there is a 0 6= v ∈ V
which is a common eigenvector for every element of R.
Now suppose g ∈ GL(V ) be an invertible linear map V → V . Then
conjugation by g, gives rise to a Lie algebra automorphism cg : sl(V ) →
sl(V ). (Check!) That is, cg (a) := gag −1 for every a ∈ sl(V ).
In particular, we have cg (R) = R. Hence, g(v) is also a common eigenvector for every element of R (check!). Since every non-zero w ∈ V can
be written as w = g(v), for a suitable g ∈ GL(V ), we conclude that every
0 6= w ∈ V is a common eigenvector for every element of R.
We claim that every a ∈ R is a scalar matrix. Let e1 , . . . , en be a basis
of V . Since every non-zero element of V is an eigenvector for a, a(ei ) = λi ei
for i = 1, . . . , n. Moreover, since ei + ej is also an eigenvector, we conclude
that λi = λj . This proves the claim.
Problem 3 (5 on page 21). (10pts)
Assume x, y ∈ g commute. Then xs commutes with y by proposition
4.2b. Since ys , yn are polynomials of y, xs commutes with ys , yn . Similarly
xs , xn , ys , yn all commute with each other.
Since xs , ys commute, they are diagonlized simultaneously, hence xs + ys
is semisimple. xn , yn are nilpotent and commuting, hence xkn = 0, ynl = 0
when k, l large enough. Then (xn + yn )k+l = 0 and xn + yn is nilpotent.
Now xs + ys semisimple and xn + yn nilpotent and xs + ys commutes
with xn + yn . By uniqueness of Jordan decomposition we have (x + y)s =
xs + ys , (x + y)n = xn +
yn . 1 0
0 1
Example: Let x =
and y =
. Here x is semisimple and
0 2
0 0
y is nilpotent, so xs = x, xn = 0, ys = 0, yn = y. However, they don’t
commute: xy =
y, but
yx = 2y.
1 1
Note that
= x + y, x is semisimple and y is nilpotent, but this
0 2
is not a Jordan decomposition. Since x + y has 2 different eigenvalues, x + y
1
is semisimple and
(x + y)s = x + y 6= xs + ys ,
(x + y)n = 0 6= xn + yn .
Problem 4. (10pts)
Consider the adjoint representation
ad : S → gl(L/S)
s 7→ {x
mod S 7→ [s, x]
mod S}
The image ad(S) is solvable subalgebra of gl(L/S). Since S is proper, L/S 6=
0. By Theorem 4.1, V = L/S has a common eigenvector of ad(S). In other
words, there exists a 0 6= v ∈ L/S such that ads(v) ∈ Cv. Since v 6= 0
in L/S, we can choose a representative v ∈
/ S. Then S 0 := S + Cv is a
subalgebra of L (check!), and dim(S 0 ) = dim(S) + 1.
S 0 may not be solvable. Indeed, if it always were, we would be able
to use this argument recursively, starting with S = (0), to show that L
itself is solvable. Equivalently, if S is a solvable subalgebra of L of maximal
dimension, then S 0 cannot be solvable.
Problem 5.
(a) (5pts) By Lie’s theorem, there exists a basis of Cn such that under this
basis L is contained in the algebra tn of upper triangular matrices. Thus
dim(L) ≤ dim(tn ) = n(n + 1)/2.
(b) (5pts) If two solvable subalgebras are of dimension n(n + 1)/2, By
Lie’s theorem, L1 is conjugate to a subalgebra of tn . Since dim(L1 ) =
n(n + 1)/2 = dim(tn ), L1 is conjugate to tn . By the same argument L2 is
conjugate to tn . Hence L1 and L2 are conjugate to each other.
(c) (5pts) Choose n ≥ 2, and let L1 = {X ∈ tn |Tr(X) = 0}, and
L2 = {X ∈ tn : X11 = 0}. Then dim(L1 ) = dim(L2 ) = n(n + 1)/2 − 1.
Every element of L1 is a singular matrix, where as some elements of L2 are
non-zingular. Hence, L1 and L2 are not conjugate.
Problem 6. (5pts)
Let Dij := Eij − Eji for any 1 ≤ i, j ≤ n. Then the matrices Dij :=
Eij − Eji , with 1 ≤ i < j ≤ n, form a basis for son . Since Dij = −Dji , the
Lie algebra structure on son is determined by the identities [Dij , Djl ] = Dil
if i, j, l are distinct, and [Dij , Dkl ] = 0 if i, j, l, k are distinct. In particular,
these identities tell us that [son , son ] = son , and hence, son is not solvable.
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Alternatively, we will show that son is not solvable using Cartain’s criterion as follows. Since D12 = −[D13 , D23 ] ∈ [son , son ], it suffices to show that
κ(D12 , D12 ) = Tr(ad(D12 )2 ) 6= 0. Note that ad(D12 ) takes D1j to −D2j ,
D2j to D1j , for any j ≥ 3, and all other Dij with i < j to 0. Thus
(
−Dij , if i = 1 or 2, and j > 2
2
ad(D12 ) (Dij ) =
0, if (i, j) = (1, 2) or 3 ≤ i < j ≤ n
In the basis Dij , 1 ≤ i < j ≤ n, ad(D12 )2 is represented by a diagonal matrix,
with the entries corresonding to D1j and D2j equal to −1, for 3 ≤ j ≤ n,
and all other entries equal to 0. Thus
κ(D12 , D12 ) = Tr(ad(D12 )2 ) = −2(n − 2) 6= 0,
as desired.
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