Lecture 22 – November 30, 2004

Prof. Victor Kač Scribes: Genya Zaytman and Yaim Cooper

Definition. An enveloping algebra of a Lie algebra 𝔤 is a pair (U, φ) where U is a unital associative algebra and φ is a Lie algebra homomorphism 𝔤 → U_{−} (where U_{−} denotes the Lie algebra structure on U given by [a, b] = ab − ba).

Example. Given a representation φ : 𝔤 → End V = U, we have an enveloping algebra (U, φ) of 𝔤.

Definition. The universal enveloping algebra of 𝔤 is an enveloping algebra (U(𝔤), φ) which is universal in the sense that for any other enveloping algebra (U, φ), there is a unique associative algebra homomorphism π : U(𝔤) → U, which makes the following diagram commute: (with respect to Lie algebra homomorphisms)

\xymatrix{𝔤 \ar[r]^{ψ} \ar[d]_{φ} & U(𝔤)_{−} \ar[dl]_{π} \\ U(𝔤)_{−}}

Theorem. For any Lie algebra 𝔤, a universal enveloping algebra exists and is unique.

Proof. 1. Uniqueness: Suppose there are two universal enveloping algebras, U(𝔤)_{−} and U′(𝔤)_{−}.

\xymatrix{𝔤 \ar[r]^{ψ} \ar[d]_{ψ′} & U(𝔤)_{−} \ar[dl]_{π′} \\ U(𝔤)′_{−} \ar[ur]_{π}}

But then π ∘ π′ : U(𝔤) → U(𝔤), and also π′ ∘ π : U′(𝔤) → U′(𝔤).

But

\xymatrix{𝔤 \ar[r]^{φ} \ar[d]_{ψ} & U(𝔤) \\ U(𝔤) \ar@{-->}[ur]^{π∘π′} \ar[ur]_{1}}

is a commuting diagram.

By uniqueness, we see that π ∘ π′ = 1, and by symmetry, π′ ∘ π = 1.

Existence: Let a_{i} be a basis for 𝔤. Let U(𝔤) be a unital associative algebra generated by a_{i}, with relations a_{i}a_{j} − a_{j}a_{i} = [a_{i}, a_{j}]. Let φ(a_{i}) = a_{i}.

Since we divided by the above relations, φ is a Lie algebra homomorphism. So, this is an enveloping algebra.

It is universal because:

\xymatrix{𝔤 \ar[r]^{ψ} \ar[d]_{φ} & U(𝔤)_{−} \ar[dl]_{π} \\ U(𝔤)_{−}}

This diagram commutes if we let π(a_{i_{1}}...a_{i_{s}}) = φ(a_{i_{1}})...φ(a_{i_{s}}).

Poincare-Birkhoff-Witt Theorem. The monomials a_{i_{1}}...a_{i_{s}} with i_{1} ≤ i_{2} ≤ ...i_{s} form a basis of U(𝔤). In particular, ψ : 𝔤 → U(𝔤)_{−} is an embedding.

Proof. I) These monomials span U(𝔤). Of course, the unordered monomials a_{j_{1}}..a_{j_{s}} span U(𝔤). We prove by induction on (s, number of inversions) that using relations [a_{i}, a_{j}] = a_{i}a_{j} − a_{j}a_{i}, we can bring this monomial to a linear combination of ordered monomials. If ...a_{j_{t}}a_{j_{t}+1}... with j_{t} > j_{t} + 1, we replace a_{j_{t}}a_{j_{t}+1} with [a_{j_{t}}, a_{j_{t}+1}] + a_{j_{t}+1}a_{j_{t}}, (where [a_{j_{t}}, a_{j_{t}+1}] is a linear combination of the generators a_{i}). We get a sum of monomials, where the number of factors is less than s in all but the last one, and in that one, the number of inversions drops by 1. Thus we can apply the inductive assumption.

II) Let 𝔅_{n} be the free vector space on generators u_{i_{1}}...u_{i_{n}}, i_{1} ≤ i_{2}... ≤ i_{n}. Let 𝔅 = 1 ⊕ 𝔅_{1} ⊕ 𝔅_{2} · · ·. We will show that there is a linear map σ : U(𝔤) → 𝔅 such that the image of the ordered monomials under σ is linearly independent, which will complete the theorem.

Let us define σ by

σ(1) = 1, σ(a_{i_{1}} ⋯ a_{i_{n}}) = u_{i_{1}} ⋯ u_{i_{n}} if i_{1} ≤ ⋯ ≤ i_{n}.  (1)

Finally, we want to have

σ(a_{j_{1}} ⋯ a_{j_{n}} − a_{j_{1}} ⋯ a_{j_{k+1}}a_{j_{k}} ⋯ a_{j_{n}}) = σ(a_{j_{1}} ⋯ [a_{j_{k}}, a_{j_{k+1}}] ⋯ a_{j_{n}}).  (2)

To show that such a linear map exists, we assume it has been defined for all monomials of degree less than or equal to n − 1 in U(𝔤), and show that it can be extended to a map on all monomials of degree less than or equal to n in U(𝔤). So, assuming σ has been defined for all monomials of degree less than or equal to n − 1 in U(𝔤), if a_{i_{1}} ⋯ a_{i_{n}} is an ordered monomial, let σ(a_{i_{1}} ⋯ a_{i_{n}}) = u_{i_{1}} ⋯ u_{i_{n}}.

If a_{i_{1}} ⋯ a_{i_{n}} is not ordered, suppose j_{k} > j_{k+1}. Then set

σ(a_{j_{1}} ⋯ a_{j_{n}}) = σ(a_{j_{1}} ⋯ a_{j_{k+1}}a_{j_{k}} ⋯ a_{j_{n}}) + σ(a_{j_{1}} ⋯ [a_{j_{k}}, a_{j_{k+1}}] ⋯ a_{j_{n}}).  (3)

We must check that this map is well defined, in that its independent of the choice of the pair (j_{k}, j_{k+1}). Suppose (j_{l}, j_{l+1}) is another pair with j_{l} > j_{l+1}. There are two cases: 1) l > k + 1 and 2) l = k + 1.

1) Set a_{j_{k}} = u, a_{j_{k+1}} = v, a_{j_{l}} = w, a_{j_{l+1}} = x. Then the induction hypothesis permits us to write for the right hand side of (3)

σ(⋯ vu ⋯ xw ⋯)+σ(⋯ vu … [wx] ⋯)+σ(⋯ [uv] … xw ⋯)+σ(⋯ [uv] … [wx] ⋯) (4)

If we start with (j_{l}, j_{l+1}), we obtain

σ(⋯ uv ⋯ xw ⋯) + σ(⋯ uv … [wx] ⋯) = σ(⋯ vu … xw ⋯)+ + σ(⋯ [uv] … xw ⋯) + σ(⋯ vu ⋯ [wx] ⋯) + σ(⋯ [uv] … [wx] ⋯) (5)

This is the same as the value obtained before.

2) Set a_{j_{k}} = u, a_{j_{k+1}} = v, a_{j_{l+1}} = w. Using the induction hypothesis, the right hand side of (3) becomes

σ(⋯ wvu ⋯) + σ(⋯ [vw]u ⋯) + σ(⋯ v[uw] ⋯) + σ(⋯ [uv]w ⋯)  (6)

Similarly, if we start with σ(⋯ wvu ⋯) + σ(⋯ u[vw]), we can wind up with

σ(⋯ wvu ⋯) + σ(⋯ w[uv] ⋯) + σ(⋯ [uw]v ⋯) + σ(⋯ u[vw] ⋯)  (7)

So we must show that σ applied to

(⋯ [vw]u ⋯) − (⋯ u[vw] ⋯) + (⋯ v[uw] ⋯)− − (⋯ [uw]v ⋯) + (⋯ [uv]w ⋯) − (⋯ w[uv] ⋯), (8)

a monomial of degree less than or equal to n − 1, gives 0.

But from the properties of σ on all monomials of degree less than or equal to n − 1 in U(𝔤), if (⋯ xy ⋯) is a monomial of degree less than or equal to n − 1,

σ(⋯ xy ⋯) − σ(⋯ yx ⋯) − σ(⋯ [xy] ⋯) = 0.  (9)

Hence σ applied to (8) gives

(⋯ [[vw]u] ⋯) + (⋯ [v[uw]] ⋯) + (⋯ [[uv]w] ⋯)  (10)

which is zero by the Jacobi identity and linearity of σ. Thus σ is well defined on monomials of degree less than or equal to n, as well, and we extend σ linearly to the space spanned by all monomials of degree less than or equal to n in U(𝔤). In this way, σ is defined on all of U(𝔤), and clearly, the image of the ordered monomials in U(𝔤) is linearly independent in 𝔅, as they are in bijection with the generators of 𝔅. Thus the proof is completed.

Remark. Any representation π of a Lie algebra 𝔤 in a vector space V extends to a represenation of U(𝔤) → End V (as associative algebras).

Definition. Given a representation π of 𝔤 over V, V can be considered a 𝔤-module by defining a binary product 𝔤 × V into V mapping g · v to π(𝔤)v. Thus the defining property of a 𝔤-module is: [a, b]v = abv − bav. By a homomorphism of 𝔤-modules we mean a linear map φ : V_{1} → V_{2} such that φ(gv) = gφ(v). An isomorphism is a homomorphism φ which is bijective.

Let 𝔤 be a finite dimensional Lie algebra with a fixed non-degenerate invariant symmetric bilinear form (., .). Chose a basis u_{i} of 𝔤 and let v_{i} be the dual basis, which means (u_{i}, v_{j}) = δ_{ij}.

Definition. The Casimir operator Ω = ∑_{i} u_{i}v_{i} ∈ U(𝔤).

Exercise 22.1. Ω is independent of the choice of the basis u_{i}.

Solution. Take another basis u_{i}′ with dual basis v_{i}′. Since u_{i} was a basis, we can write u_{i}′ = ∑_{j} a_{ij}u_{j}. In this basis, Ω′ = ∑_{i} u_{i}′v_{i}′.

Note that by definition, (u_{i}′, v_{k}′) = δ_{ij} = (∑_{j} a_{ij}u_{j}, v′_{k}) = ∑_{j}(a_{ij})(u_{j}, v′_{k}). Let the matrix A = ⟨a_{ij}⟩, and B = ⟨(u_{j}, v′_{k})⟩. Clearly, AB = I. Now consider I = BA = ∑_{k}(u_{j}, v′_{k})a_{ki}. This implies that ∑_{k}(u_{j}v′_{k})a_{ki} = δ_{ji} = (u_{j}, ∑_{k} a_{ki}v′_{k}). But since v_{i} is the unique vector such that (u_{j}, v_{i}) = δ_{ji}, it follows that v_{i} = ∑_{k} a_{ki}v′_{k}.

Finally, Ω′ = ∑_{i} u_{i}′v_{i}′ = ∑_{i}(∑_{j} a_{ij}u_{j})v_{i}′ = ∑_{j} ∑_{i} a_{ij}v_{i}′u_{j}. By the result of the previous paragraph, this equals ∑_{j} v_{j}u_{j}, which gives the desired result, Ω′ = ∑_{i} u_{i}v_{i} = Ω.

Lemma 1. (on dual bases) For a ∈ 𝔤 write [a, u_{i}] = ∑_{j} a_{ij}u_{j} and [b, u_{i}] = ∑_{j} b_{ij}u_{j}. Then a_{ij} = −b_{ji} (under the above assumptions).

Proof. We have: ([a, u_{i}], v_{k}) = ∑_{j} a_{ij}(u_{j}, v_{k}) = a_{ik} and similarly ([a, v_{i}], u_{k}) = b_{ik}. Hence a_{ik} = (a, [u_{i}, v_{k}]) and b_{ik} = (a, [v_{i}, u_{k}]). Therefore a_{ik} = −b_{ki}.

Definition. Let V be a 𝔤-module, where 𝔤 is a Lie algebra. A 1-cocycle is a linear map φ : 𝔤 ↦ V such that φ([a, b]) = aφ(b) − bφ(a). The space of 1-cocycles is denoted Z(𝔤, V).

Example of a 1-cocycle. The trivial 1-cocycle associated to v ∈ V is φ_{v}(a) = a · v.

Exercise 22.2. Check that φ_{v} is a 1-cocycle.

Solution. φ_{v}([a, b]) = [a, b]v = a(bv) − b(av) = aφ_{v}(b) − bφ_{v}(a)

The trivial 1-cocycles form a subspace B(𝔤, V) of Z(𝔤, V). Let H^{1}(𝔤, V) = Z(𝔤, V)/B(𝔤, V).

Main Theorem on Cohomology. If 𝔤 is a semi-simple Lie algebra over an algebraicly closed field of characteristic 0, and V is a finite dimensional 𝔤-module, then H^{1}(𝔤, V) = 0, i.e., every 1-cocycle is trivial.

Exercise 22.3. H^{1}(𝔤, V_{1} ⊕ V_{2}) = H^{1}(𝔤, V_{1}) ⊕ H^{1}(𝔤, V_{2}) where V_{1} ⊕ V_{2} denotes the direct sum of 𝔤-modules V_{1} and V_{2}.

Solution. First we show that Z(𝔤, V_{1} ⊕ V_{2}) = Z(𝔤, V_{1}) ⊕ Z(𝔤, V_{2}). It is clear that Z(𝔤, V_{1} ⊕ V_{2}) ⊃ Z(𝔤, V_{1})⊕Z(𝔤, V_{2}). Furthermore, every 1-cocycle φ ∈ Z(𝔤, V_{1}⊕V_{2}) can be decomposed as π_{1} ∘ φ ⊕ π_{2} ∘ φ ∈ Z(𝔤, V_{1}) ⊕ Z(𝔤, V_{2}).

It is also clear that B(𝔤, V_{1} ⊕ V_{2}) = B(𝔤, V_{1}) ⊕ B(𝔤, V_{2}) since φ_{v_{1}⊕v_{2}} = φ_{v_{1}} ⊕ φ_{v_{2}}.

Therefore H^{1}(𝔤, V_{1} ⊕ V_{2}) = Z(𝔤, V_{1} ⊕ V_{2})/B(𝔤, V_{1} ⊕ V_{2}) = Z(𝔤, V_{1}) ⊕ Z(𝔤, V_{2})/B(𝔤, V_{1}) ⊕ B(𝔤, V_{2}) = Z(𝔤, V_{1})/B(𝔤, V_{1}) ⊕ Z(𝔤, V_{2})/B(𝔤, V_{2}) = H^{1}(𝔤, V_{1}) ⊕ H^{1}(𝔤, V_{2}).

Lemma 2. In the situation of lemma 1. Let V be a 𝔤-module and φ : 𝔤 ↦ V a 1-cocycle. Then for any a ∈ 𝔤 we have: a ∑_{i} u_{i}φ(v_{i}) = Ωφ(a).

Proof. a ∑_{i} u_{i}φ(v_{i}) = ∑_{i}[a, u_{i}]φ(v_{i}) + ∑_{i} u_{i}aφ(v_{i}) = ∑_{i,j} a_{ij}u_{j}φ(v_{i}) + ∑_{i} u_{i}aφ(v_{i}) = ∑_{j} u_{j}φ(∑_{i} a_{ij}v_{i}) + ∑_{i} u_{i}aφ(v_{i}). Now ∑_{i} a_{ij}v_{i} = − ∑_{i} b_{ji}v_{i} = −[a, v_{i}], so a ∑_{i} u_{i}φ(v_{i}) = − ∑_{j} u_{j}φ([a, v_{i}]) + ∑_{j} u_{j}aφ(v_{j}) = ∑_{j} u_{j}(aφ(v_{j}) − φ([a, v_{i}])) = ∑_{j} u_{j}v_{j}φ(a) = Ωφ(a).

Corollary. 𝔤 commutes with Ω, i.e., in any 𝔤-module a(Ωv) = Ω(av) for all a ∈ 𝔤, v ∈ V.

Proof. Apply lemma 2 to the trivial cocycle φ_{v}(a) = a·v: a(Ωv) = a ∑_{i} u_{i}v_{i}(v) = Ω(av).

Proof of the Main Theorem on Cohomology. The proof is by induction on the dimension of the 𝔤-module V.

First note that we may assume that V is faithful, that is, aV = 0 implies a = 0 for a ∈ 𝔤. Indeed let 𝔤_{0} = {a ∈ 𝔤|aV = 0}. This is an ideal of 𝔤. Hence 𝔤_{0} and 𝔤/𝔤_{0} are again semi-simple Lie algebras. In particular [𝔤_{0}, 𝔤_{0}] = 𝔤_{0}. Let φ be a 1-cocycle of 𝔤 in V, i.e., φ([a, b]) = aφ(b) − bφ(a). If a, b ∈ 𝔤_{0}, we get φ([a, b]) = 0. So φ([𝔤_{0}, 𝔤_{0}]) = 0, therefore φ([a, b]) = 0. Hence φ : 𝔤/𝔤_{0} ↦ V, so we may replace 𝔤 by 𝔤/𝔤_{0}.

We want to apply lemma 2.

Take (a, b) = tr_{V} ab. It is non-degenerate since 𝔤 is semi-simple. Let {u_{i}} be a basis of 𝔤, {v_{i}} the dual basis, and Ω = ∑_{i} u_{i}v_{i} the Casmir operator. We decompose V = V_{0} ⊕ V_{1}, where V_{0} is the generalized eigenspace of Ω attached to 0 and V_{1} is the sum of all the other generalized eigenspaces. By the corollary V_{0} and V_{1} are 𝔤-invariant. So by Exercise 22.3 H^{1}(𝔤, V) = H^{1}(𝔤, V_{0}) ⊕ H^{1}(𝔤, V_{1}). If V_{0} and V_{1} are not both zero, by the induction hypothesis H^{1}(𝔤, V_{0}) = 0 and H^{1}(𝔤, V_{1}) = 0 and so H^{1}(𝔤, V) = 0. Hence we may assume V = V_{0} or V_{1}.

Case 1: V = V_{1}. So Ω is invertible. Let v = ∑_{i} u_{i}φ(v_{i}). Lemma 2 now states that a(v) = Ωφ(a). Hence φ(a) = Ω^{−1}a(v) = a(Ω^{−1}v). So φ = φ_{Ω^{−1}v} is a trivial cocycle.

Case 2: V = V_{0}. So Ω is a nilpotent operator. Hence tr_{V} (Ω) = 0, but tr_{V} (Ω) = tr_{V} ∑_{i} u_{i}v_{i} = ∑_{i}(u_{i}, v_{i}) = dim 𝔤. So 𝔤 = 0 and H^{1}(𝔤, V) = 0.