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Let $\mathfrak{g}$ be a compact Lie algebra (a Lie algebra that admits a positive invariant bilinear form) and $\mathfrak{h}$ an abelian Lie algebra. Let $\rho\colon\mathfrak{h}\to\mathfrak{g}$ be a homomorphism of Lie algebras that takes some nonzero element $X$ to a nonzero central element $\rho(X)$ in $\mathfrak{g}$. Is it necessary that $\rho(\mathfrak{h})$ is central in $\mathfrak{g}$?

I've been thinking about this for days. I think it is false. However, I couldn't find a counterexample. Any help is appreciated.

YYF
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    Do you just require it for some element? Do you at least require that element to be non-zero? If just for one non-zero element (and let us assume it is sent to something non-zero as well to not make this trivial), then the question becomes of whether it is possible to have a compact Lie algebra with non-trivial center and an abelian subalgebra of larger dimension than the center of the Lie algebra, which seems like it should clearly be true, except I can never recall how strong being compact is, and whether it actually implies having trivial center. – Tobias Kildetoft Oct 27 '17 at 13:29
  • @TobiasKildetoft Thanks for your comments. I guess you already answered my question. So, if $\mathfrak{h}$ is a maximal abelian subalgebra in $\mathfrak{g}$, $\rho$ is the inclusion, and $X$ is just a nonzero central element which has to be in every maximal abelian subalgebra of $\mathfrak{g}$. Then, this would be a counterexample. Right? – YYF Oct 27 '17 at 13:47
  • Don't Cartan subalgebras form a large class of counterexamples? – user148212 Nov 06 '17 at 04:08

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Not necessarily take ${\cal H}$ to be any non commutative Lie algebra and $G={\cal H}\times \mathbb{R}^2$.

Consider the morphism $f:\mathbb{R}^2\rightarrow {\cal G}$ defined by $f(e_1)=(0,e_1)$ and $f(e_2)=u$ where $u$ is not in the centre of ${\cal H}$, here $(e_1,e_2)$ is a basis of $\mathbb{R}^2$.

Tsemo Aristide
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