Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $\phi: L_1 \rightarrow L_2"$ is a surjective Lie algebra homomorphism, is it true that $\phi (Z(L_1))=Z(L_2)$. I see that $\phi (Z(L_1))$ is in $Z(L_2)$, but if $\phi^{-1}(0)$ is not $0$, i.e $\phi$ not injective I think that the other inclusion is not true. Is that right? would a projection work as a counterexample?

share|cite|improve this question
up vote 3 down vote accepted

You are completely correct that $\varphi(Z(L_1))\subseteq Z(L_2)$ (as is easily seen by writing up what this means and using that the homomorphism is surjective).

You are also correct that in general, one need not get the entire center of $L_2$ this way. For a general example, one can take any nilpotent but not abelian Lie algebra (such as the Lie algebra consisting of $3\times 3$ strictly upper triangular matrices). As it is nilpotent, it has a non-trivial center. If one quotients by the center, one again gets a non-trivial nilpotent Lie algebra, which thus has a non-trivial center. But as the center of the original Lie algebra is sent to $0$ in this quotient, one cannot get the entire center of the quotient this way.

share|cite|improve this answer
It took me a while to see why if $L$ is nilpotent it has non-zero center (assuming is not abelian or $0$). I guess the argument is just: $[L^{k-1}L]=0$ so $L^{k-1}$ is in the center (where $k$ is the minimum st $L^k=0$) – inquisitor Feb 7 '13 at 18:58
@inquisitor Yes, exactly. – Tobias Kildetoft Feb 7 '13 at 19:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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