# Weyl's theorem and Representations

Let $$L$$ be a semisimple Lie algebra and let $$(V,\varphi)$$ be a finite-dimensional $$L$$-module representation. Our main goal is to prove that $$\varphi$$ is completely reducible. Consider an $$L$$-submodule of $$V$$ of codimension one, let $$0 \longrightarrow W \longrightarrow V \longrightarrow F \longrightarrow 0$$ be an exact sequence (where $$F$$ is an $$L$$-module). From the book of James Humphreys called "Introduction to Lie Algebras and Representation Theory", I have understood the following steps:

1. We take another proper submodule of $$W$$ denoted by $$W'$$ such that the exact sequence $$0 \longrightarrow W/W' \longrightarrow V/W' \longrightarrow F \longrightarrow 0$$ splits, so there exists a one dimensional $$L$$-submodule of $$V/W'$$ (say $$\tilde{W}/W'$$) complementary to $$W/W'$$.
2. We proceed by induction on the dimension of $$W$$, so we get an exact sequence $$0 \longrightarrow W' \longrightarrow \tilde{W} \longrightarrow F \longrightarrow 0$$ which splits. It follows easily that $$V=W \oplus X$$, where $$X$$ is a submodule complementary to $$W'$$ in $$\tilde{W}$$.
3. We suppose that $$W$$ is irreducible, so we may use Schur's lemma on $$c \vert_{W}$$ to say that $$\operatorname{Ker} c$$ is an $$L$$-submodule of $$V$$, where $$c$$ is the endomorphism of $$V$$ defined in 6.2.

The other parts of the proof are very hard, and I didn't understand them. Can someone help me to figure out those parts? If there is another comprehensible method, can someone share it with us?

• I don’t have a copy of Humpreys on hand right now, so I can’t check (though could you say where in the book this is?), but it seems odd without context to start with talk of codimension 1 submodules, as there is a priori no reason why any codimension 1 subspace should be a sub module. Sep 15, 2019 at 0:01
• It's pages 28 and 29 (in my edition from grad school, at least). Humphreys starts with the special case where $V$ has an $L$-submodule $W$ of codimension 1 which allows use of a previous lemma. Sep 15, 2019 at 0:43
• @Aaron: The point of the proof is to reduce the general case to the codimension 1 case. Anyway it's not my proof. Sep 15, 2019 at 20:27
• @Brian: It's usually safer to refer to sections in a textbook rather than to pages. For instance, in my book the Weyl theorem on complete reducibility is treated in 6.3. Sep 15, 2019 at 20:32
• See answers to this question. Sep 16, 2019 at 10:40

Maybe it would clarify matters if I gave a little more background, in community wiki format.

The basic idea of this algebraic proof goes back to a short paper by Richard Brauer (1936) in German in Mathematische Zeitschrift, an interesting time in his life when he had been expelled from his professorship in Berlin and took up a position in Toronto. (This particular paper has no marginal additions, as do some other papers; I bought a used copy of his collected papers in three volumes which used to belong to him.)

Brauer's proof is of course less natural than the original proof by Weyl, but it applies to all semisimple Lie algebras over an algebraically closed field of characteristic 0 and may be the simplest algebraic proof. An attempt was made to simplify the proof, in a more recent textbook by K. Erdmann and her recent student M. Wildon Introduction to Lie algebras (MSN), published by Springer in 2006 as an undergraduate text. This result is not essential for their further work but occurs in Theorem 9.16 with an unconvincing proof. This was acknowledged by them in a list of errata, and affects Exercises 9.15 (cf. 9.16) as well perhaps as Lemma 10.7. (A good exercise is to track down the specific fault in the proof.) The book itself is perhaps an attractive alternative to mine, having a more leisurely pace and more examples but covering much less material in a similar number of pages.

The Brauer method is used in Bourbaki's Chapter 1 (of their Groupes de Lie et algèbres de Lie) as well as Jacobson's 1962 book Lie algebras (now in a Dover reprint). But as stated above, Weyl's 1925 proof is more natural in the Lie group context, imitating the finite group method.

• Just to add to this: The proof of Weyl's Theorem in Appendix B of my book with Karin Erdmann is, I believe, correct. (In fact we used Jim Humphreys' book quite closely here.) The error mentioned above is a mistake in the proof of the existence of the abstract Jordan decomposition (Theorem 9.16). We tried to avoid using Weyl's Theorem, but Jim pointed out that our proof is flawed. This is a shame, because having the abstract Jordan decomposition to hand would be very useful for showing complete reducibility (this was the idea in Exercise 9.15) Sep 16, 2019 at 15:01
• Sep 16, 2019 at 20:27
• @LSpice: Yes, but I was probably following Bourbaki's version. Sep 17, 2019 at 11:20

Since you also ask for another method, perhaps you may try Hans Samelson's approach in his textbook "Notes on Lie Algebras" (I have an older edition but it has been republished by Springer). It is in Chapter III Section 4. Roughly, the idea is to prove first a lemma of Whitehead that for the Lie algebra $$L$$ acting on a finite dimensional vector space $$V$$ and a linear function $$f:L\rightarrow V$$ satisfying that $$f[x,y]=xf(y)-yf(x)$$ (a cocycle condition) there exists a vector $$v\in V$$ such that $$f(x)=xv$$ for all $$x\in L$$ (a coboundary condition). Then, the proof of Weyl's complete reducibility proceeds to split a canonical epimorphism in a more direct fashion. You may enjoy reading this approach.

• Probably I've conflated this argument when referring to Jacobson, who does much the same thing. My impression is that this method is quite close to Brauer's, but I'll have to reread these sources. Sep 17, 2019 at 11:24
• According to the historical note in Bourbaki's Lie Groups and Lie Algebras (Vol. 1) in page 424 of the English translation, Brauer's article predates Whitehead's paper of 1937, "Certain equations in the algebra of a semi-simple infinitesimal group", Quart. J. Math. (2) 8 (1937), 220--237. In the historical note, Bourbaki also says that the proof in his book is Brauer's. Sep 17, 2019 at 21:53