1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Tobias Kildetoft Oct 27 '17 at 13:29
  • $\begingroup$ @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? $\endgroup$ – YYF Oct 27 '17 at 13:47
  • $\begingroup$ Don't Cartan subalgebras form a large class of counterexamples? $\endgroup$ – user148212 Nov 6 '17 at 4:08
1
$\begingroup$

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 $\mathbb{R}^2$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.