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If $ \mathfrak g$ and $\ \mathfrak h$ are lie algebras and $\phi: \ \mathfrak g \rightarrow \ \mathfrak h$ is a lie algebra homomorphism. Show that the kernel of $\phi$ is an ideal of $\ \mathfrak g$.

Proof: The kernel of the lie algebra homomorphism is defined to be the set $\ker \phi= \{X \in \ \mathfrak g | \phi([X,H])=0 \}$. To show that $\ker \phi$ is an ideal we need to show that $[H,X] \in \ker \phi$ for every $X \in \ \mathfrak g$ and for every $ \ H \in \ker \phi$. Thus take $X \in \ker \phi $. Then $\phi([X,H])= [\phi(X),\phi(H)]=[0,\phi(H)]=0 \in \ker\phi.$ By surjectivity, we see that $[X,H] \in ker\phi$. Therefore $\ker\phi$ is an ideal.

Does this proof look correct? I'm not quite sure if I defined the $\ker\phi $ properly since I couldn't find an explicit form of it for lie algebra's.

Aweygan
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Alexander King
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  • $\ker\phi={X\in\mathfrak g :\phi(X)=0}$, and in your proof it looks like you should have $X\in\ker\phi$ and $H\in\mathfrak g$. – Aweygan Jun 09 '17 at 22:09

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This is not correct because $\ker\phi=\{X\in\mathfrak{g}\,|\,\phi(X)=0\}$.

It is an ideal because if $X\in\ker\phi$ and $Y\in\mathfrak g$, then $[X,Y]\in\ker\phi$, and this is true because $\phi\bigl([X,Y]\bigr)=\bigl[\phi(X),\phi(Y)\bigr]=\bigl[0,\phi(Y)\bigr]=0$.

José Carlos Santos
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