(+) Direct sum
~> My attempt at an arrow (mapping to symbol)
1/21/99
Lie Algebras
1) Definitions
2) Examples
3) Classification
4) Representations
5) Applications
Note: In this class we will mainly study Lie algebras over the field C.
Def: A Lie algebra L is a vector space over C with a bilinear
product [ . , . ] from L X L to L satisfying
1) [x, y] = - [y, x] (antisymmetry)
2) [x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0 (Jacobi identity)
Ex: Let L = {all n by n matrices over C} = gl(n, C). Define [A, B] = AB - BA where AB is the usual matrix product. The product here is frequently known as the commutator. We must check our properties. Antisymmetry is obvious because -(AB - BA) = BA - AB. For the Jacobi identity, [A, [B, C]] + [B, [A, C]] + [C, [A, B]] = [A, (BC - CB)] + [B, (CA - AC)] + [C, (AB - BA)] = [A, BC] - [A, CB] + [B, CA] - [B, AC] + [C, AB] - [C, BA] = ABC - BCA - ACB + CBA - CAB - BAC + ACB + CAB - ABC - CBA + BAC = 0.
Ex: (A) Let L = {all n by n matrices with trace zero over C} = sl(n, C). We just need to check that L is closed under the commutator, because the properties hold from being a subalgebra of gl(n, C). However trace [AB] = trace(AB - BA) = trace(AB) - trace(BA) = trace(AB) - trace(AB) = 0 for any A, B. Thus [AB] is in sl(n, C). We sometimes refer to Lie algebras of the from sl(n, C) as type A.
Ex: (B and D) Let L = {all n by n skew matrices over C} = O(n, C). These are matrices where XT = -X. Again we check that L is closed under the commutator. Suppose A and B are in O(n, C), so AT = -A and BT = -B. Then [AB]T = (AB - BA)T = (AB)T - (BA)T = BTAT - ATBT = BA - AB = - (AB - BA) = - [AB] and hence [AB] is in O(n, C). We call Lie algebras of type O(n, C) for n odd, type B. For n even we call them type D.
Ex: (C) Let J be the following 2n by 2n (nondegenerate skew) matrix
|
|
|
|
|
|
Ex: Consider L = sl(2, C) which has dimension 3. Let x be the matrix with 1 in the upper right corner, y be the matrix with 1 in the lower left corner and h be the matrix with 1, -1 down the diagonal. Then x, y, h is a basis for L. We can write a multiplication table for the bracket operation as follows:
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Def: Suppose L is a Lie algebra. A representation of L on a vector space V is a linear map p: L ~> End(V) satisfying p([xy]) = p(y) - p(y)p(x) = [p(x), p(y)].
Def: Suppose W is a subspace of V such that p(x)w is in W for all w in W, x in L. Then by restriction we get a map p: L ~> End(W). This gives a representation of L on W called a subrepresentation of p. We call W a p(L) invariant subspace.
Def: A representation of p on V is called irreducible if there are no non-trivial (not equal to 0 or V) invariant subspaces.
Question: What are all the irreducible finite dimensional representations
of sl(2, C)? Well if p: sl(2, C)
~> End(V) is a representation, then we have linear transformations p(x),
p(y),and p(h).
We will examine the eigenvectors of p(h) to
answer this question next time.
1/25/99
Ex: What are all the irreducible finite dimensional representations
of sl(2, C)? We choose a basis as before x,y,z with [xy] =
h, [hx] = 2x, [hy] = -2y. Rember that an irreducible representation
V means that there does not exist a W subspace of V (W not equal to 0 or
V) so that p(x)w is in W for all w in W, x in
L (there are no nontrivial L-invariant subspaces). Suppose (p
,V) is an irreducible representation. Consider p(h)
in End(V), a linear transformation over C. As C is
algebraically closed p(h) has at least one eigenvalue
and corresponding eigenvector v. Let W = span{eigenvectors of p(h)}.
We will show W to be an L invariant subspace thus forcing it to be V.
As p is linear we only need check that p(x)
and p(y) take p(h)
eigenvectors to other eigenvectors. Suppose p(h)v
= lv (l in C).
Then p(h)(p(x)v)
= p(h)p(x)v = (p([hx])+p(x)p(h))v
= p(2x)v+p(x)lv
= 2p(x)v+lp(x)v =
(2 + l)(p(x)v) thus p(x)v
is an eigenvector of p(h) with eigenvalue l
+ 2. We have shown that if v is in W then p(x)v
is in W. We can do the same for p(y)v
and conclude that if v is in W then p(y)v is
in W. Thus we have that p(L)W is in W.
Next notice that p(h) is diagonalizable thus
we can write V = (+)l in S
Vl where our Vls
are nonzero l-eigenspaces for p(h)
and S is a finite subset of C. Now there is some l
in S such that l + 2 is not in S. We can
pick a nonzero eigenvector vl
in Vl. Then we know
p(h)vl
= lvl
and p(x)vl
= 0. We define vl-2
= p(y)vl,
vl-4 = p(y)vl,...
as long as vl-2i is nonzero.
We get in this way vl,
vl-2, vl-4,
..., vl-2k nonzero vectors
so that p(y)vl-2k
= 0. The terminology is to call the p(h)
eigenvectors weight vectors. If p(x)v
= 0 then v is a highest weight vector. If p(y)v
= 0 then v is a lowest weight vector.
Let W = span{vl, vl-2,
vl-4, ..., vl-2k}.
We want to show that W is L-invariant. Now by definition W is p(y)
invariant, and being eigenvectors of p(h), W
is p(h) invariant as well. We only need
to show W is p(x) invariant. We must compute
p(x)vl-2i.
Well we can look at the simple cases first. p(x)vl
= 0. p(x)vl-2
= p(x)p(y)vl
= (p([xy]) + p(y)p(x))vl
= (p(h) + p(y)p(x))vl
= p(h)vl
+ p(y)0 = lvl.
To find the pattern we try the next case. p(x)vl-4
= p(x)p(y)vl-2
= (p([xy]) + p(y)p(x))vl-2
= (p(h) + p(y)p(x))vl-2
= p(h)vl-2
+ p(y)lvl
= (l - 2)vl-2
+ p(y)lvl
= (l - 2)vl-2
+ lvl-2
= (l + (l - 2))vl-2.
Trying the next case will reveal p(x)vl-6
= (l + (l - 2) +
(l - 4))vl-4
allowing us to see the pattern p(x)vl-2(i+1)
= (l + (l - 2) +
... + (l - 2i))vl-2i
which can be proved by induction. We can simplify (l
+ (l - 2) + ... + (l -
2i)) = (i + 1)l - (i + 1)i = (i + 1)(l
- i) and conclude p(x)vl-2(i+1)
= (i + 1)(l - i)vl-2i.
This shows W to be p(x) invariant thus p(L)
invariant.
As W = span{vl, vl-2,
vl-4, ..., vl-2k}
is p(L) invariant we get V = W = Cvl
(+) Cvl-2 (+) ...
(+) Cvl-2k.
We also have the formulas p(h)vl-2i
= (l-2i)vl-2i,
p(y)vl-2i
= vl-2(i+1), and p(x)vl-2i
= i(l-i+1)vl-2(i+1).
This gives us the matrices for p(h), p(y),and
p(x) under the basis {vl,
vl-2, vl-4,
..., vl-2k}. We can
gain even more information, however.
Now l has to satisfy a condition. We can
use the definition of a representation to write p(h)vl-2k
= p([xy])vl-2k
= (p(x)p(y) - p(x)p(y))vl-2k.
As p(y)vl-2k
= 0 and by the relations above we get (l - 2k)vl-2k
= p(h)vl-2k
= (p(x)p(y) - p(x)p(y))vl-2k
= -p(y)p(x)vl-2k
= -p(y)k(l - k +
1)vl-2(k-1) = -k(l
- k + 1)vl-2.
Thus (l - 2k) = -k(l -
k + 1) or l(k + 1) = k(k + 1). As the
smallest value of k is zero, k + 1 is nonzero and we get l
= k. We have shown l is a nonnegative
integer. Also V now equals span{vk, vk-2,
vk-4, ..., vk-2k}
so S, our set of weights, equals {k, k-2, ..., -k}.
What we have shown is that if we have some irreducible representation,
then it has this structure. We have not yet proved such a structure
exists. This part is easy. For each k > 0 we may have an irreducible
representation of dimension k + 1 with weights Sk such that p(h),
p(y),and p(x) have
the following matrices.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1/28/99
Def: Suppose L and L' are Lie algebras. A linear map f from L to L' is called a homomorphism if f([xy])=[f(x)f(y)].
Def: Suppose K is a subspace of a Lie algebra L. Then K
is called a subalgebra if for any x,y in K then [xy] is also in K.
K
is called an ideal if for any x in L, y in K then [xy] is also in K.
Note: Every ideal in L is a subalgebra of L.
Lemma: If f : L --> L' is a homomorphism then Im(f)={f(x) : x in L} is a subalgebra of L and Ker(f)={x in L : f(x)=0} is an ideal in L.
Proof: Say x,y are in Im(f). Then for some x',y' in L f(x') = x and f(y') = y. Then f([x'y'])=[f(x')f(y')]=[xy] in Im(f). Thus Im(f) is a subalgebra of L'. Next say y is in Ker(f) and x is in L. Then f([xy])=[f(x)f(y)]=[f(x)0]=0 so [xy] is in Ker(f) and Ker(f) is an ideal of L. n
Lemma: If K is an ideal of L then the coset space L/K={x+K: x in L} is a Lie algebra with [x+K, y+K] = [xy] + K.
Proof: W must show [x+K, y+K] = [xy]+K is well defined and the properties will follow from the bracket of L. Suppose x+K = x'+K, then x-x' is in K. Then [xy]-[x'y]=[x-x',y] in K as K is an ideal and so [x'y]+K = [x'y]+[xy]-[x'y]+K = [xy]+K. n
Def: The Lie algebra L/K is called the quotient Lie algebra.
Theorem: (Fundamental Theorem of Lie Algebras)
1) Suppose f: L --> L is a homomorphism
of Lie algebras, then L/Ker(f) is isomorphic
to Im(f).
2) If I is an ideal of L so I is contained in Ker(f) then there is
a unique y such that f(x)=y(p(x)).
3) Suppose I and J are ideals of L with I an ideal of J. Then
J/I is an ideal of L/I and (L/I)/(J/I) is isomorphic to L/J.
4) Suppose I and J are ideals of L, then I+J={x+y : x in I, y
in J} is an ideal of L, I intersect J is an ideal of L and (I+J)/J is isomorphic
to I/(I intersect J).
Proof: Exercise.
Corollary: There is a 1-1 correspondence between the ideals of L/I and the ideals of L containing I.
Proof: Let J be an ideal in L containing I. Then let f map J to the ideal J/I of L/I (which we know exists and is an ideal from the third theorem). Suppose f(J) = 0, meaning the zero ideal in L/I. Then J is in I, but as J is an ideal containing I, J can only be I. Thus Ker(f) ={J} and f is 1-1. Next let K be an ideal of L/I. Consider p-1(K) = {x in L : x + I is in K}. This is an ideal in L containing I which gets sent back to K by f. This shows f is surjective and we are done. n
Def: If L is a Lie algebra we define L(1) = [L, L] = {[xy] : x,y in L}, L(2) = [L(1), L(1)], ... , L(k) = [L(k-1), L(k-1)]. This series is called the derived series of L. We call L(k) the kth derived algebra of L.
Corollary: If L is a Lie algebra, every term in the derived series of L is an ideal of L.
Lemma: For any ideals I, J of L, [IJ] is also an ideal of L.
Proof: Take z in L, x in I, y in J. Then [xy] is in [IJ]. Here [z[xy]] = [[zx]y] + [x[zy]]. As I is an ideal, [zx] is in I, so [[zx]y] is in [IJ]. As J is an ideal, [zy] is in J, so [x[zy]] is in [IJ]. Hence [z[xy]] is in [IJ] and we are done. n
Corollary: If L is a Lie algebra, every term in the derived series of L is an ideal of L.
Def: If L is a Lie algebra we define L1 = [L, L] = {[xy] : x,y in L}, L2 = [L, L1], ... , Lk = [L, Lk-1]. This series is called the descending central series of L.
Note: For all k, L(k) is an ideal of Lk which is an ideal of L.
Def: A Lie algebra is called solvable if L(k) = 0 for some k. A Lie algebra is called nilpotent if Lk = 0 for some k.
Note: As L(k) is an ideal of Lk, if Lk = 0 for some k then L(k) = 0 for the same k. Thus if L is nilpotent L is solvable.
Note: Consider any L generated by one element x. Thus L = Cx. We can call such Lie algebras cyclic. As [xx] = 0, we know that cyclic Lie algebras are abelian, meaning [LL] = 0. Abelian Lie algebras are nilpotent as L1 = 0. Finally nilpotent Lie algebras are solvable. Thus we get the following chain which also occurs in Group Theory: Cyclic ~> Abelian ~> Nilpotent ~> Solvable.
Ex: Let L = {n by n upper triangular matrices (excluding the digonal)}. Call the main diagonal (entries of form aii) the first diagonal, the diagonal above that the second, and so on. If x and y and in L, x has zeros on the first p-diagonals, and y has zeros on the first q diagonals then xy has zeros on the first p + q diagonals. We get this from the formula EijEkl = djkEil. Now Eij is on diagonal (j - i) + 1 (> p), Ekl is on diagonal (l - i) + 1 (> q), and Eil is on diagonal (l-i) + 1 with l - i + 1 > p + q. Similarly this is true for yx, so it is true for xy - yx. Thus Lk had k + 1zero diagonals and L is nilpotent. Moreover if K is a subalgebra of L then Kk < Lk so any subalgebra of L is nilpotent.
Ex: Let K = {{n by n upper triangular matrices (including the digonal)}. If A, B are in K then AB and BA in K may have nonzero entries on the diagonal. However, the diagonal of AB equals the diagonal of BA so AB - BA will have zeros on the main diagonal and thus is in L from our last example. Now [K[KK]] = [KK] so K is not nilpotent. However as [[KK],[KK]] does have more zero diagonals by our last example. In fact K(k) has zeros on the first 2k-1 diagonals. Thus K is solvable. We can also conclude that any subalgebra of K is also solvable.
Note: Over C these example are all the solvable and nilpotent
Lie algebras. This is a consequence of Engel's theorem, Lie's theorem,
and Ado's theorem.
Click here to return.