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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!

2 Answers2

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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 ${\mathfrak g}$?

Hanno
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  • Dear Hanno, thank you for your answer. The most straightforward example I can think of is the 3-dimensional Heisenberg algebra. Am I correct? However, if I took Heisenberg algebra as an example, how can I show that Z(L/Z(L)) is not zero? thx! –  Oct 03 '16 at 08:11
  • @Tortuga.A: Exactly! For the $3$-dimensional Heisenberg algebra $L$ you can explicitly write down what $L(Z)$ is and see that $L/Z(L)$ is non-zero abelian. Give it a try, I'm sure you'll see it. – Hanno Oct 03 '16 at 08:17
  • Thank you for your reply! I write down the L(Z) is just , and L/Z(L) is {x+αz;for some x in L}..and [e+αz, f+αz] = αz. (where [e f]=z )Can this imply L/Z(L) commutes? thanks! –  Oct 03 '16 at 15:52
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    @Tortuga.A: If you know that $L$ is spanned by $e,f,z$ with $z$ central and $[e,f]=z$, then you have just computed that $L/Z(L)$ is abelian (because the self-bracket of $\overline{e}$ resp. $\overline{f}$ vanish). – Hanno Oct 03 '16 at 17:17
  • Sorry for the late reply. Thank you so much for your help! I totally understand now! –  Oct 10 '16 at 18:17
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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.

Dietrich Burde
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  • Thank you sir! But with the computation, I write down the L(Z) is just , and L/Z(L) is {x+αz;for some x in L}..and [e+αz, f+αz] = αz. (where [e f]=z )Can this imply L/Z(L) commutes? thanks! –  Oct 03 '16 at 15:53
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    Take a basis $(x,y,z)$ of $\mathfrak{h}_3$. Then $[x,y]=z$ and $Z(\mathfrak{h}_3)=\langle z\rangle$. So in the quotient $[\overline{x},\overline{y}]=\overline{z}=\overline{0}$, which is zero. So all brackets with elements in the quotient are zero. – Dietrich Burde Oct 03 '16 at 17:12
  • Sorry for the late reply. Thank you so much for your help! I totally understand now! –  Oct 10 '16 at 18:17