Is a non-zero homomorphism f: V ---> W of L-representations V,W surjective if W is simple?

Question

This could be wrong but: If W is simple then V must be simple and if V is simple then its only subrepresentations are 0 and V which means the Ker(F)=0 and thus F must be injective. But how can I show it it is surjective?

Why must $V$ be simple? What about a projection $V:=W\oplus W \twoheadrightarrow W$? — Torsten Schoeneberg, Dec 07 '19 at 06:36

score 2 · Accepted Answer · answered Dec 07 '19 at 06:40

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$im(f) \subseteq W$ is a subrepresentation. So if $W$ is simple ...

answered Dec 07 '19 at 06:40

Torsten Schoeneberg

26,051

score 0 · Answer 2 · answered Dec 06 '19 at 14:27

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Let $W=sl(2,\mathbb{R})$, take a non zero element $x$ of $W$, $Vect(x)$ is a commutative algebra, and the canonical imbedding is not surjective.

answered Dec 06 '19 at 14:27

Tsemo Aristide

87,475

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But that embedding is not a homomorphism of representations. – Tobias Kildetoft Dec 06 '19 at 18:51

Joe Wolf · Answer 3 · 2019-12-07T16:01:26.027

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If $V, W$ are both simple L-modules Schur's lemma states that $f$ is either invertible or zero. However, since $f$ is by assumption non-zero, it must be invertible. Therefore, it is also surjective. However, this does not follow from your argumentation. See Torsten's reply for a (much) simpler argument.

edited Dec 07 '19 at 16:01

answered Dec 07 '19 at 01:04

Joe Wolf

117

So is it true then that if W is simple in this case then V must be also? – ConnectNumbers66 Dec 07 '19 at 13:41
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No, I don't see why that should be the case, see also Torsten's comment above. I've edited my answer to reflect this. – Joe Wolf Dec 07 '19 at 16:01

Is a non-zero homomorphism f: V ---> W of L-representations V,W surjective if W is simple?

3 Answers3