Say $f: A \rightarrow B$ is a $G$-equivariant map between finite dimensional $G$-representations, for some semi-simple Lie group $G$. If I know:
- $B$ decomposes into irreducibles as $B = B_1 \oplus ... \oplus B_k$
- $f(a) \in B_i$ for some $a \in A$, $f(a) \neq 0$
Must the decomposition of $A$ into irreducible $G$-reps contain a copy of $B_i$? Does the answer change depending on whether $f(a)$ is a highest weight vector for $B_i$?
It depends on the context, but the answer to your first question should be yes if you are thinking of complex semi-simple Lie groups. Indeed, this should be true in any setting where your representations decompose into irreducibles for purely formal reasons.
First a reduction: Schur's lemma holds in the context of Lie group representations, so every map to an irreducible $G$-rep is either zero or surjective. If $V = G\cdot a \subseteq A$ is the cyclic subrep generated by $a$, then of course $f$ defines a $G$-morphism $V \to B_i$ which is surjective (as long as $f(a) \neq 0$). From this we can see that $B_i$ is a quotient of $V$.
So your first question is equivalent to the following fundamental question: does every irrep which arises as a quotient of a $G$-representation $V$ actually arise as a subrepresentation? Equivalently, does every sub-representation have a complement? This is equivalent to asking for complete reducibility, as shown here.
For (edit: forgot to emphasize semisimple) Lie algebras, the answer is affirmative (this the fundamental result of Weyl) and the same basic result holds for compact Lie groups (one can construct complements by some kind of Reynolds averaging argument, or equivalently, one can normalize the representation to be a unitary representation and then construct complements via orthogonality). Since every semisimple complex Lie group has a compact real form, the result holds in that context too.