Bourbaki

Lemma 1

Tag 09V6, permanent. Cite it as [Bourbaki, Tag 09V6].

Lie Groups and Lie Algebras, chapter I, § 2. Enveloping algebra of a Lie algebra

For every integer p≥0p \geq 0, there exists a unique homomorphism fpf_p of the K-module g⊗KPpg \otimes_K P_p into the K-module P satisfying the following conditions:

(Ap)fp(xλ⊗zM)=zλzMfor λ≤M,zM∈Pp;(Bp)fp(xλ⊗zM)−zλzM∈Pqfor zM∈Pq,q≤p;(Cp)fp(xλ⊗fp(xμ⊗zN))=fp(xμ⊗fp(xλ⊗zN))+fp([xλ,xμ]⊗zN) \begin{align*} (A_p) \quad & f_p(x_\lambda \otimes z_M) = z_\lambda z_M \quad \text{for } \lambda \leq M, z_M \in P_p; \\ (B_p) \quad & f_p(x_\lambda \otimes z_M) - z_\lambda z_M \in P_q \quad \text{for } z_M \in P_q, q \leq p; \\ (C_p) \quad & f_p(x_\lambda \otimes f_p(x_\mu \otimes z_N)) = f_p(x_\mu \otimes f_p(x_\lambda \otimes z_N)) + f_p([x_\lambda, x_\mu] \otimes z_N) \end{align*}
for zN∈Pp−1z_N \in P_{p-1}. (The terms appearing in (Cp) are meaningful by (B_p).)*

Moreover, the restriction of fpf_p to g⊗Pp−1g \otimes P_{p-1} coincides with fp−1f_{p-1}.

The last assertion follows from the others since the restriction of fpf_p to g⊗KPp−1g \otimes_K P_{p-1} satisfies conditions (A_{p-1}), (B_{p-1}) and (C_{p-1}). We shall prove the existence and uniqueness of fpf_p by induction on pp. For p=0p = 0, condition (A_0) gives f0(xλ⊗1)=zλf_0(x_\lambda \otimes 1) = z_\lambda and conditions (B_0) and (C_0) are then obviously satisfied. Suppose now that the existence and uniqueness of fp−1f_{p-1} are proved. We show that fp−1f_{p-1} admits a unique extension fpf_p to g⊗KPpg \otimes_K P_p satisfying conditions (A_p), (B_p) and (C_p).

We must define fp(xλ⊗zM)f_p(x_\lambda \otimes z_M) for an increasing sequence M of pp elements.

If λ≤M\lambda \leq M, the value is given by condition (A_p). Otherwise, M can be written uniquely in the form (μ,N\mu, N), where μ<λ,μ≤N\mu < \lambda, \mu \leq N. Then

zM=zμzN=fp−1(xμ⊗zN) z_M = z_\mu z_N = f_{p-1}(x_\mu \otimes z_N)
by (A_{p-1}), so that the left hand side of (C_p) is fp(xλ⊗zM)f_p(x_\lambda \otimes z_M). Now the right hand side of (C_p) is already defined: for (B_{p-1}) allows us to write:
fp(xλ⊗zN)=fp−1(xλ⊗zN)=zλzN+w f_p(x_\lambda \otimes z_N) = f_{p-1}(x_\lambda \otimes z_N) = z_\lambda z_N + w
where w∈Pp−1w \in P_{p-1}; hence the right hand side of (C_p) becomes:
zλzμzN+fp−1(xμ⊗w)+fp−1([xλ,xμ]⊗zN). z_\lambda z_\mu z_N + f_{p-1}(x_\mu \otimes w) + f_{p-1}([x_\lambda, x_\mu] \otimes z_N).

Thus fpf_p is defined uniquely and obviously satisfies conditions (A_p) and (B_p). Condition (C_p) is satisfied if μ<λ,μ≤N\mu < \lambda, \mu \leq N. As [xμ,xλ]=−[xλ,xμ][x_\mu, x_\lambda] = -[x_\lambda, x_\mu], condition (C_p) is also satisfied for λ<μ,λ≤N\lambda < \mu, \lambda \leq N. As (C_p) is trivially satisfied for λ=μ\lambda = \mu, (C_p) is therefore satisfied if λ≤N\lambda \leq N or μ≤N\mu \leq N. If none of these inequalities holds, N=(ν,Q)N = (\nu, Q), where ν≤Q,ν<λ,ν<μ\nu \leq Q, \nu < \lambda, \nu < \mu. Writing henceforth to abbreviate fp(x⊗z)=xzf_p(x \otimes z) = xz for x∈gx \in g and z∈Ppz \in P_p, we have by the induction hypothesis:

xμzN=xμ(xνzQ)=xν(xμzQ)+[xμ,xν]zQ. x_\mu z_N = x_\mu(x_\nu z_Q) = x_\nu(x_\mu z_Q) + [x_\mu, x_\nu]z_Q.

Now zμzQz_\mu z_Q is of the form zμzQ+wz_\mu z_Q + w, where w∈Pp−2w \in P_{p-2}. (C_p) can be applied to xλ(xν(zμzQ))x_\lambda(x_\nu(z_\mu z_Q)), since ν≤Q\nu \leq Q and ν<μ\nu < \mu, and to xλ(xνw)x_\lambda(x_\nu w) by the induction hypothesis, and hence to xλ(xν(xμzQ))x_\lambda(x_\nu(x_\mu z_Q)). Hence:

xλ(xμzN)=xν(xλ(xμzQ))+[xλ,xν](xμzQ)+[xμ,xν](xλzQ)+[xλ,[xμ,xν]]zQ. x_\lambda(x_\mu z_N) = x_\nu(x_\lambda(x_\mu z_Q)) + [x_\lambda, x_\nu](x_\mu z_Q) + [x_\mu, x_\nu](x_\lambda z_Q) + [x_\lambda, [x_\mu, x_\nu]]z_Q.

Exchanging λ\lambda and μ\mu and subtracting:

xλ(xμzN)−xμ(xλzN)=xv(xλ(xμzQ)−xμ(xλzQ))+[xλ,[xμ,xv]]zQ−[xμ,[xλ,xv]]zQ=xv([xλ,xμ]zQ)+[xλ,[xμ,xv]]zQ+[xμ,[xv,xλ]]zQ=[xλ,xμ](xvzQ)+([xv,[xλ,xμ]]+[xλ,[xμ,xv]]+[xμ,[xv,xλ]])zQ x_{\lambda}(x_{\mu}z_{N}) - x_{\mu}(x_{\lambda}z_{N}) = x_{v}(x_{\lambda}(x_{\mu}z_{Q}) - x_{\mu}(x_{\lambda}z_{Q})) + [x_{\lambda}, [x_{\mu}, x_{v}]]z_{Q} - [x_{\mu}, [x_{\lambda}, x_{v}]]z_{Q} = x_{v}([x_{\lambda}, x_{\mu}]z_{Q}) + [x_{\lambda}, [x_{\mu}, x_{v}]]z_{Q} + [x_{\mu}, [x_{v}, x_{\lambda}]]z_{Q} = [x_{\lambda}, x_{\mu}](x_{v}z_{Q}) + ([x_{v}, [x_{\lambda}, x_{\mu}]] + [x_{\lambda}, [x_{\mu}, x_{v}]] + [x_{\mu}, [x_{v}, x_{\lambda}]])z_{Q}

and hence by the Jacobi identity

xλ(xμzN)−xμ(xλzN)=[xλ,xμ]zN x_{\lambda}(x_{\mu}z_{N}) - x_{\mu}(x_{\lambda}z_{N}) = [x_{\lambda}, x_{\mu}]z_{N}

which completes the proof of Lemma 1.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Cites

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Transcribed from the printed volume. Not translated.