Proof of Weyl's Theorem [Humphreys]

962 Views Asked by At

I don't really understand the first paragraph of the proof of Weyl's Theorem in Humphreys' Lie Algebra book (p. 28).

My problem is, that first of all I don't see, why (or in which sense) the exact sequence

$$0 \rightarrow W/W'\rightarrow V/W'\rightarrow \mathbb{F} \rightarrow 0$$

"splits". The author then claims that $\exists$ a one dimensional L-submodule of $V/W'$ complementary to $W/W'$. And this is exacly it. Why can he conclude that?

Thanks in advance from a new member ;)

1

There are 1 best solutions below

1
On BEST ANSWER

The exact sequence splits by induction hypothesis, because it is assumed already that such exact sequences of modules split in dimension less than $n$. In the beginning it is said that $\dim (V)=n$ and $\dim W=n-1$. A short exact sequence of modules splits, if the middle one is a semidirect product of the first with the third one (which has here dimension $1$, because $V\simeq W/F$, since we also have the short exact sequence $0\rightarrow W \rightarrow V\rightarrow F\rightarrow 0$).

That every such sequence "splits" is just another formulation of Weyl's theorem, namely that every $L$-submodule $W$ of $V$ possesses a module complement (so that $V$ is completely reducible).
Perhaps the proof of Weyl's theorem given at Climbing Mount Bourbaki makes this clearer.