Surjective lie algebra homomorphism and diagonalisation of ad map

Question

I was wondering if anyone could check my solution to the following exercise (from Erdmann and Wildon's Introduction to Lie Algebras):

True or false: Let $\phi: L_1 \to L_2$ be a surjective Lie algebra homomorphism. Let $h \in L_1$. If ad $h$ is diagonalisable, then ad $\phi(h)$ is diagonalisable. What is different if $\phi$ is an isomorphism?

My solution

I claim the statement is true if $\phi$ is an isomorphism. Proof:

Let $\lambda_1, \cdots, \lambda_n$ be the n (not necessarily distinct) eigenvalues of ad $h$, and let $x_1, \cdots, x_n$ be associated linearly independant eigenvectors (which exist because ad $h$ is diagonalisable). Then $(x_1, \cdots, x_n)$ is a basis for $L_1$. So $(\phi(x_1), \cdots, \phi(x_n))$ is a basis for $L_2$. They are eigenvectors because we have ad $\phi(h)(\phi(x_i)) = [\phi(h), \phi(x_i)] = \phi([h, x_i]) = \phi(\lambda_ix_i) = \lambda_i\phi(x_i)$. So
ad $\phi(h)$ has a basis of eigenvectors, so is diagonalisable.

My guess is the statement would be false if $\phi$ is only surjective, because then there is no guarantee that the image of a set of eigenvectors of ad $h$ is linearly independant. However I can't come up with a counterexample; any ideas?

Answer 1

I'd say your proof shows in general that a Lie algebra homomorphism $\phi$ maps eigenvectors of $\mathrm{ad}(h)$ to eigenvectors of $\mathrm{ad}(\phi(h))$; besides, any surjective linear map $\phi$ between vector spaces maps any basis to a generating system. So if $x_1, ..., x_n$ is a basis of $L_1$ consisting of $\mathrm{ad}(h)$-eigenvectors, then $\phi(x_1), ..., \phi(x_n)$ are a generating system of $L_2$ consisting of $\mathrm{ad}(\phi(h))$-eigenvectors. True, they might not be linearly independent any more, but by basic linear algebra they contain some basis of $L_2$, and that basis still consists of $\mathrm{ad}(\phi(h))$-eigenvectors.