Bourbaki

Lemma 3

Tag 0145, permanent. Cite it as [Bourbaki, Tag 0145].

Lie Groups and Lie Algebras, chapter VIII, § 7. Finite dimensional modules over a split semi-simple Lie algebra

Let G be an abelian group written additively, and φ:Fa→G\varphi :\mathscr{F}_{\mathfrak{a}}\rightarrow G a map; by abuse of notation, we denote by φ(F)\varphi (F) the image under φ\varphi of the class of any finite dimensional a\mathfrak{a}-module F. Assume that, for any exact sequence

0⟶F′⟶F⟶F′′⟶0 0\longrightarrow F'\longrightarrow F\longrightarrow F''\longrightarrow 0

of finite dimensional a\mathfrak{a}-modules, we have φ(F)=φ(F′)+φ(F′′)\varphi (F) =\varphi (F') +\varphi (F''). Then, there exists a unique homomorphism θ:R(a)→G\theta :\mathscr{R}(\mathfrak{a})\rightarrow G such that θ([F])=φ(F)\theta ([F]) =\varphi (F) for every finite dimensional a\mathfrak{a}-module F.

There exists a unique homomorphism θ\theta from R(a)\mathscr{R}(\mathfrak{a}) to G such that θ([E])=φ(E)\theta ([E]) =\varphi (E) for every finite dimensional simple a\mathfrak{a}-module E. Let F be a finite dimensional a\mathfrak{a}-module, and (Fn,Fn−1,...,F0)(F_n,F_{n-1}, . . . ,F_0) a Jordan-Hölder series of F; if n>0n >0, we have, by induction on nn,

θ([F])=∑i=1nθ([Fi/Fi−1])=∑i=1nφ(Fi/Fi−1)=φ(F) \theta ([F]) =\sum_{i=1}^n\theta ([F_i/F_{i-1}]) =\sum_{i=1}^n\varphi (F_i/F_{i-1}) =\varphi (F)

If n=0n= 0 then [F] = 0 so θ([F])=0\theta ([F]) = 0; on the other hand, by considering the exact sequence 0⟶0⟶0⟶0⟶00\longrightarrow 0\longrightarrow 0\longrightarrow 0\longrightarrow 0 we see that φ(0)=0\varphi (0) = 0.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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