Surjective lie algebra homomorphism preserve center?

Question

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 don't think $Z(L_2)$ is in $\phi (Z(L_1))$ in general cases. Could someone help me to prove this?

Thank you in advance!

Answer 1

You're right, it's not true. A canonical source of examples would be the projection ${\mathfrak g}\to {\mathfrak g}/{\mathfrak z}({\mathfrak g})$ as long as you can make sure that ${\mathfrak z}({\mathfrak g}/{\mathfrak z}({\mathfrak g}))\neq 0$. A fortiori, this would be true if ${\mathfrak g}/{\mathfrak z}({\mathfrak g})$ is non-zero abelian, i.e. if for any $x,y\in{\mathfrak g}$ you have $[x,y]\in{\mathfrak z}({\mathfrak g})$, i.e. if any $3$-fold Lie bracket vanishes. Can you find such ${\mathfrak g}$?

Answer 2

We can see even without a computation that the quotient $\mathfrak{h}_3/Z(\mathfrak{h}_3)$ is abelian, because the Heisenberg Lie algebra $\mathfrak{h}_3$ is nilpotent and has $1$-dimensional center, so that the quotient is $2$-dimensional nilpotent. There is only one $2$-dimensional nilpotent Lie algebra - the abelian one.