Contents
1. Day 2 1
1.1. Homomorphisms of Lie algebras 1
1.2. Representations 2
1.3. Examples 4
1.4. Reducible representations 4
2. Day 3 5
2.1. Representations ofsl(2) 5
2.2. Enveloping algebra 7
2.3. Killing form and Casimir operator 8
1. Day 2
• (a) Homomorphisms of Lie algebras
• (b) Representations
• (c) Reducible representations
• (d) Examples (adjoint representation)
1.1. Homomorphisms of Lie algebras. Let gand hbe two Lie algebras overK. A homomorphismφ:g → his aK-linear map such that
[φ(X), φ(Y)] =φ([X, Y]).
One has the expected isomorphism theorems.
Theorem 1.1. (a) Let φ : g → h be a homomorphism. Then ker(φ) is a Lie ideal in g.
(b) Let k ⊂ g be a Lie ideal in g. Then the bracket on g defines a Lie algebra structure on the quotient space g/k, and g/k is the universal object for Lie algebra homomorphisms fromg that are trivial on k.
(c) Suppose k ⊂h⊂g is a sequence of ideals of g. Then h/k is an ideal of g/k and there is an isomorphism
(g/k)/(h/k) −→g/h.∼
(d) If k and hare ideals of g then so are k+hand k∩h, and there is an isomorphism of Lie algebras
k/k∩h−→(k∼ +h)/h.
1
One already knows the isomorphisms for vector spaces, and one has to show that the maps respect the Lie algebra structure. But this is easy.
Ifφ:G → H is a homomorphism of Lie groups, thendφ:TG,e →TH,e is a homomorphism of Lie algebras (and we have seen that the converse holds in the semisimple case.
1.2. Representations.
Definition 1.2. LetV be a vector space over K, and endowEnd(V) with the structure of Lie algebra defined by the bracket. Letg be a Lie algebra overK. A representation ofg on V is a homomorphism
φ:g → End(V) of Lie algebras.
The definition immediately gives rise to the usual categorical construc- tions: subrepresentation, quotient representation, etc. For example
Definition 1.3. Let φ : g →End(V), φ0 : g → End(V0) be two repre- sentations of V. A homomorphism f : (φ, V)→ (φ0, V0) is a linear map f : V → V0 that commutes with the action of g (it makes the obvious tri- angle commute).
In the above definition, we can defineφ⊕φ0 :g→ End(V⊕V0) by setting (φ⊕φ0)(X)(v, v0) = (φ(X)(v), φ0(X)v0), X ∈g, v∈V, v0 ∈V0. If f : (φ, V) →(φ0, V0) is a homomorphism then so is λf for any λ ∈ K.
ThusHom(V, V0) (we drop theφ’s) is aK-vector space. Representations of gthus form aK-linear additive category; and it is even an abelian category.
Moreover, there is an action onV ⊗V0:
(φ⊗φ0)(X)(v⊗v0) =φ(X)(v)⊗v0+v⊗φ0(X)v0, X ∈g, v∈V, v0∈V0. Proof. We have to show that
(φ⊗φ0)([X, Y]) = [(φ⊗φ0)(X),(φ⊗φ0)(Y)].
In other words
φ([X, Y])(v)⊗v0+v⊗φ0([X, Y])v0
=(φ⊗φ0)(X)(φ(Y)v⊗v0+v⊗φ(Y)v0)−(φ⊗φ0)(Y)(φ(X)(v)⊗v0+v⊗φ0(X)v0).
(1.4)
This is a simple computation; however, the left-hand side has four terms upon expansion, whereas the right-hand side has eight terms, so some terms
must cancel in order to obtain an equality.
So Rep(g) is a monoidal category.
There is also the contragredient: ifV∨ =Hom(V, K) thenφ∨(X)(f)(v) =
−f(φ(X)(v)). More generally, onHom(V, V0) we define
s(X)(f)(v) =φ0(X)(f(v))−f(φ(X)v), f ∈Hom(V, V0), v∈V
The lettersmeans nothing and is only introduced for convenience in the next few lines. The proof that this is a representation deserves writing out, because there is a delicate point in the middle of the proof. Here is what has to be checked: for all X, Y ∈g, all f ∈Hom(V, V0), and all v ∈V, we have the relation
s([X, Y])(f)(v) =s(X)(s(Y)(f))(v)−s(Y)(s(X)(f))(v).
The left hand side is given by φ0([X, Y])(f(v))−f(φ([X, Y])(v)). Since φ and φ0 are both representations, each of these terms can be expanded. We have the unproblematic term
φ0([X, Y])(f(v)) =φ0(X)◦φ0(Y)(f(v))−φ0(Y)◦φ0(X)(f(v)) and then
−f(φ(X)◦φ(Y)v) +f(φ(Y)◦φ(X)v) (sincef is a linear map).
The right hand side has two parts, of which the first is
s(X)(φ0(Y)(f(v))−f(φ(Y)v))
=φ0(X)◦φ0(Y)(f(v))−φ0(Y)(f(φ(X)v))−φ0(X)f(φ(Y)v) +f(φ(Y)◦φ(X)v).
(1.5)
Note that theφ(X) has been inserted between theφ(Y) and thev, and not in front ofφ(Y)!!! This is becauseXis acting on the functionv7→f(φ(Y)v).
One should take the time to understand why this is the correct formula.
The second part of the right hand side is obtained from the first by ex- changing the role ofX andY, but appears with a−sign:
−s(Y)(s(X)(f))(v)
=−φ0(Y)◦φ0(X)(f(v)) +φ0(X)(f(φ(Y)v)) +φ0(Y)f(φ(X)v)−f(φ(X)◦φ(Y)v).
(1.6)
Adding the two parts of the right hand side, one sees that the mixed terms cancel, and what remains is
φ0(X)◦φ0(Y)(f(v))+f(φ(Y)◦φ(X)v)−φ0(Y)◦φ0(X)(f(v))−f(φ(X)◦φ(Y)v).
This coincides with the left-hand side and completes the proof that s([X, Y]) = [s(X), s(Y)].
Proposition 1.7. Let V, V0 be two finite-dimensional representations of g.
Then the natural isomorphism of vector spaces
V∨⊗V0 −→Hom(V, V∼ 0);λ⊗v0 7→[v7→λ(v)v0] is an isomorphism of representations.
Proof. Exercise.
In particular,
Endg(V) =Hom(V, V)g={∈HomK(V, V)|φ(X)f(v) =f((φ(X)v))∀X ∈g, v ∈V}.
1.3. Examples.
Example 1.8. Let V = K, and φ : g → K = End(K) be the zero map.
This is the trivial representation.
More generally, if (φ, V) is any representation, we set Vg ={v∈V |Xv= 0∀X∈g}.
Example 1.9. The map ad : g → End(g) is a representation, called the adjoint representation.
Proof. We have to show that ad[X,Y] = [adX, adY], in other words for all Z ∈g,
[[X, Y], Z] = [X,[Y, Z]]−[Y,[X, Z]].
Equivalently,
0 = [[X, Y], Z]+[[Y, Z], X]+[Y,[X, Z]] = [[X, Y], Z]+[[Y, Z], X]−[[X, Z], Y]
= [[X, Y], Z] + [[Y, Z], X] + [[Z, X], Y]
and you recognize once again the Jacobi identity.
1.4. Reducible representations. A representation (φ, V) isirreducibleif there is no subrepresentation W ⊂V other thanW = 0 or W =V. We say V is reducible if it is not irreducible, and V is completely reducible if V is isomorphic to ⊕i∈IVi with eachVi an irreducible representation.
A representation (φ, V) is indecomposable if it is not isomorphic to the direct sum of two or more representations. A representation can be reducible but indecomposable. For example, ifg=K, the one-dimensional abelian Lie algebra, letXbe a generator,V =K2, andφ(X) =e=
0 1
0 0
∈sl(2, K).
Then
kerφ(X) =K·
1
0
is the only non-trivial subrepresentation. This is the typical example.
In general, if K is algebraically closed, and g = K as above, then any representation (φ, V) has a one-dimensional invariant subspace. Indeed, φ(X) has a non-zero eigenvector v such that φ(X)(v) = λv. Then (φ, V) is completely reducible if and only if φ(X) is a diagonalizable matrix. The classification of representations ofgis then given by the Jordan decomposi- tion.
Definition 1.10. The Lie algebragis semisimple if every representation of g is completely reducible.
Lemma 1.11(Schur’s Lemma). Let K be algebraically closed and let(φ, V) be a finite-dimensional irreducible representation of g. Then Endg(V) = Hom(V, V)g=K·IdV.
Proof. Letf ∈Endg(V). SinceK is algebraically closed, f has a non-zero eigenvector v say f(v) =λv. Let Vλ ⊂ V be the λ-eigenspace of V. Since V is irreducible it suffices to show thatVλ is invariant underg. But sincef
commutes withg this is clear.
Lemma 1.12. Let (φ, V) be a finite-dimensional representation of g. The following conditions are equivalent.
(a) If W ⊂ V is a subrepresentation then there is a subrepresentation W0 ⊂V such that V =W ⊕W0.
(b) V is completely reducible.
(c) V is the sum of its irreducible subrepresentations.
Proof. This is an elementary induction on dimension but it is worth seeing it once.
(a)⇒(b): Induction on dimV. IfV is irreducible, in particular if dimV ≤ 1, there is nothing to show. Suppose it is known up to dimV −1. Let 0 6= W ⊂ V be an irreducible representation, which we may assume to be of codimension > 0. Write V = W ⊕V0 with V0 ⊂ V of codimension
= dimW > 0. It suffices to show that V0 satisfies condition (a). Let V0 ⊃W0 and let V00 =W +W0 =W ⊕W0 ⊂V. Then V =V00⊕W00 for someW00⊂V, with codimension = dimV0/W0. Consider pr:V → V0 with kernelW, and letU =pr(W00). SinceW00∩W = 0, dimU = dimW00. Since W00∩W0 = 0, U ∩W0 = 0, and therefore V0 =W0⊕U. Thus V0 satisfies (a), and so is a sum of irreducibles by induction; thus so isV.
(b)⇒ (c): Indeed, V is a direct sum of some of its irreducible subrepre- sentations, so this is a tautology.
(c) ⇒ (a): Let W ⊂ V be a subrepresentation and find W0. We may assume W of positive codimension a. By (c) there is an irreducible sub- representation U ⊂ V that is not contained in W. Thus U ∩W = 0 and W0=W⊕U ⊂V is of codimension< a. By induction it has a supplement:
V =W0⊕U0 =W ⊕(U ⊕U0) and we are done.
