Download LIE ALGEBRAS M4P46/M5P46 - PROBLEM SHEET 1 Recall: n(n

LIE ALGEBRAS M4P46/M5P46 - PROBLEM SHEET 1
Recall: n(n) := {(aij ) ∈ gl(n) : aij = 0 when i ≥ j}; t(n) = {(aij ) ∈ gl(n) : aij =
0 for i > j}; the derived and lower central series are defined inductively by respectively g(0) := g, g(i) := [g(i−1) , g(i−1) ]; g0 := g, gi := [g, gi−1 ].
(1) show that the Lie algebras sl(2) and o(3) are isomorphic over C.
Solution. Recall that sl(2) is a 3-dimensional Lie algebra spanned by vectors
H, X+ , X− such that [H, X+ ] = 2X+ , [H, X− ] = −2X− , [X+ , X− ] = H. We
need to find three linearly independent vectors in o(3) which fulfill the same commutation relations.
We start by finding a vector A ∈ 0(3) such that adA : g → g has eigenvalues 0,
2 and −2. The matrix of adRx in the basis Rx , Ry , Rz is


0 0 0


 0 0 −1  ,
0 1 0
and its characteristic polynomial is X 3 + X = X(X 2 + 1). It follows that the
eigenvalues of adRx are 0, i, −i, and a direct computation shows that Y− :=
Ry − iRz and Y+ := Ry + iRz are the eigenvectors for the eigenvalues i and −i.
It follows that Y+ and Y− are eigenvectors for eigenvalues 2 and −2 of 2i·adRx =
ad2iRx . We have
[Y+ , Y− ] = [Ry − iRz , Ry + iRz ] =
= [Ry , Ry ] − i[Rz , Ry ] + i[Ry , Rz ] + [Rz , Rz ] = 2i[Ry , Rz ] = 2iRx ,
so the vectors A := 2iRx , Y+ and Y− fulfill the relations [A, Y+ ] = adA (Y+ ) =
2Y+ , [A, Y− ] = adA (Y− ) = −2Y− , [Y+ , Y− ] = A as desired. They are linearly
independent because Ry and Rz are contained in the linear span of Y+ and Y− . (2) (a) check the Jacobi identity for (R3 , ×)
(b) chech that the structure of Lie algebra on the quotient algebra g/a is
well-defined
Solution. (a) We need to check that the map J : R3 × R3 × R3 → R3 given by
J(X, Y, Z) := X × (Y × Z) + Y × (Z × X) + Z × (X × Y ) is identically zero. Since
this map is tri-linear it is enough to check it when X, Y, Z ∈ {e1 , e2 , e3 }, where
e1 , e2 , e3 is the standard basis of R3 .
This can be checked directly (one needs to check 8 equalities); but we can use
symmetries of the equation to limit the work. Note that J is anti-symmetric (i.e.
σ is a permutation on three elements then J(X, Y, Z) = sgn(σ) · J(σ(X, Y, Z)), so
it is enough to check J(ea , eb , ec ) = 0 for a ≤ b ≤ c. Since for every two 2-element
subsets I, J ⊂ {1, 2, 3} there exists a permutation which sends I to J we need
to check just three cases: J(e1 , e1 , e1 ), J(e1 , e1 , e2 ), and J(e1 , e2 , e3 ). The first is
trivial; the second is
J(e1 , e1 , e2 ) = e1 ×(e1 ×e2 )+e1 ×(e2 ×e1 )+e2 ×(e1 ×e1 ) = e1 ×e3 +e1 ×(−e3 )+0 = 0,
1
2
LIE ALGEBRAS M4P46/M5P46 - PROBLEM SHEET 1
and the third:
J(e1 , e2 , e3 ) = e1 ×(e2 ×e3 )+e2 ×(e3 ×e1 )+e3 ×(e1 ×e2 ) = e1 ×e1 +e2 ×e2 +e3 ×e3 = 0.
(b) Recall the the Lie bracket is defined by [x + a, y + f raka] := [x, y] + a. If
x+a = x0 +a and y+a = y 0 +a then x−x0 , y−y 0 ∈ a and so [x0 −x, y], [x0 , y 0 −y] ∈ a.
Therefore [x, y] + a = [x, y] + [x0 − x, y] + [x0 , y 0 − y] + a = [x0 , y] + [x0 , y 0 − y] + a =
[x0 , y 0 ] + a (in the last two equalities we used the bilinearity of the bracket).
(3) Consider the Lie algebra of the affine transformations of the form A(x) =
ax + b, a, b ∈ C, with the standard Lie bracket [A, B] := A ◦ B − B ◦ A. Is
it nilpotent? Is it solvable?
Proof. Solution Let the algebra in question be denoted by g, and for c ∈ C let
Fc (x) = c.
Let A(x) = ax+b, C(x) = cx+d. Then [A, C](x) = a(cx+d)+b−c(ax+b)+d =
ad − bc. It follows that g0 = {Fc : c ∈ C}. Since the latter is abelian, we have
g(2) = 0 and so g is solvable.
On the other hand, [A, Fc ](x) = c + d − c = d, so g2 = g0 and so g is not
nilpotent.
(4) (a) Show that subalgebras of nilpotent (resp. solvable) Lie algebras are
nilpotent (resp. solvable)
(b) Show that every element of n(n) is nilpotent and give an example of
a nilpotent subalgebra of gl(n) which doesn’t contain nilpotent transformations.
Solution. (a) It is enough to show that g(k) ⊂ gk which we prove by induction.
Certainly it is true for k = 1; assuem it is true for k − 1, and let x ∈ g(k) , i.e.
P
x = ni=1 [ai , bi ], where ai , bi ∈ g (k−1) , so in particular, by the inductive assumption,
we have bi ∈ g k−1 , and so [ai , bi ] ∈ g k , and therefore x ∈ g k .
(b) The algebra gl(1) = C is abelian, and hence nilpotent, but consists only of
diagonal matrices, and the latter are not nilpotent.
Let e1 , . . . en be the standard basis of Cn , and let Vk be the subspace of Cn
spanned by e1 , . . . , ek . Let A ∈ n(n); then A(Vk ) ⊂ Vk−1 and since Vn = Cn ,
V0 = {0}, we get An (C n ) ⊂ {0}, so in particular A is nilpotent.
(5) Show that quotients of nilpotent (resp. solvable) Lie algebras are nilpotent
(resp. solvable)
Solution. Let g be a Lie algebra and let a ⊂ g be an ideal, and let π : g → g/a be
the quotient map. Since the arguments are very similar we show the solution only
for the case when g is solvable.
Let us show by induction that (g/a)(k) ⊂ {x + a : x ∈ g(k) }. It is clearly true for
P
k = 0. Assume it is true for k − 1, and let x ∈ (g/a)(k) , i.e. x = ni=1 [ai + a, bi + a],
and by inductive assumption we can assume ai , bi ∈ g(k−1) . But by the definition of
the bracket
structure on theP
quotient we have π([a,
b]) = [π(a), P
π(b)] for all a, b ∈ g,
Pn
Pn
n
so x = i=1 [π(ai ), π(bi )] = i=1 π([ai , bi ]) = π( i=1 [ai , bi ]) = ni=1 [ai , bi ] + a, and
P
so the claim follows since ni=1 [ai , bi ] ∈ g(k) .
LIE ALGEBRAS M4P46/M5P46 - PROBLEM SHEET 1
3
of the bracket structure on g/a we have that for every ideal b such that a ⊂ b
we have π(b)0 = π(b0 ). It follows by induction that (g/a)(k) = (π(g))(k) , and the
claim follows.
(6) Compute the derived and the lower central series of sl(n)
Solution. Let ei be the standard basis of Cn , let Eij be the matrix such that
Eij (ej ) = ei and Eij (ek ) = 0 for k 6= j, and let Hij = Eii − Ejj . The matrices
Eij , Hij , where i 6= j, span sl(n). We claim that g0 = g (and so both the lower and
the derived central series are constant and equal to g.)
Indeed, direct computation shows that [Eij , Eji ] = Hij and [Hij , Eij ] = 2Eij , so
for all i, j (with i 6= j) we have Hij , Eji ∈ sl(2), and the claim follows.
(7) Compute the derived and the lower central series of t(n)
Solution. We use the notation from the previous problem.
For k = 0, . . . , n define Mk to be the set of matrices spanned by those Eij such
that j − i ≥ k. Clearly t(n) = M0
For k, l ∈ N let δkl be equal to 1 iff k = l. Direct computation shows that
(0.1)
[Eij , Ekl ] = δjk Eil − δli Ekj ,
and in particualar, for l 6= k,
[Ekk , Ekl ] = Ekl ,
and
[Ekk , Ell ] = 0
and that [M0 , M1 ] = M1 , i.e. M0 , M1 , M1 , . . .
It follows that M00 = M1 ,
central series of (t n).
for any k, l.
is the lower
As to the derived series, observe that if j − i ≥ a and l − k ≥ b than (j = k
implies l − i ≥ a + b) and (l = i implies j − k ≥ a + b). This together with
(0.1) implies that [Mk , Mk ] = M2k . In particular the derived series of t(n) is
M0 , M1 , M2 , M4 , M8 , . . ..
(8) Compute the derived and the lower central series of n(n)
Solution. Using the same method as in the last paragraph of the solution of the
previous problem we get that the lower central series is M1 , M2 , M3 , . . ., and the
derived series is M1 , M2 , M4 , . . ..