Consider the simple Lie algebra $\mathfrak{sl}(2)$, with basis $H,E^+,E^-$, as usual. Let $\pi$ be a finite dimensional representation of this Lie algebra.
I want to show that the eigenspace of the highest weight $\Lambda$ of this representations is one dimensional.
I tried the following:
Let $w$ and $\tilde{w}$ be highest weight vectors. We have that $\pi(H)w=\Lambda w$ and $\pi(E^+)w=0$, and similarly for $\tilde w$. Then both $w$ and $\tilde w$ are in the kernel of $\pi(E^+)$. If this kernel is one dimensional, then $w$ and $\tilde w$ are multiples of each other and we are done. Now, I do know that the kernel of $E^+$ is one dimensional, but this does not mean anything about the kernel of its representation, right? For example, the kernel under the trivial $0$ representation is not one dimensional any more.
EDIT: The reason why I was trying to prove this is because it says in Fuchs book:
So that for irreducible modules the highest weight vector is unique up to multiplication by a scalar… (I forgot to refer irreducibility in the original post).
Best Answer
You can't prove it, since it is false. Take any finite-dimensional represention $\pi$ of $\mathfrak{sl}(2)$. And now consider the representaion $\pi^\star$ that it induces on $V\oplus V$. Then, whatever the dimension $d$ of the eigenspace of the highest weight $\Lambda$ of $\pi$ is, the dimension of the eigenspace of the highest weight $\Lambda$ of $\pi^\star$ is $2d$, and therefore it cannot possibly be equal to $1$.