Let $\mathfrak{g}$ and $\mathfrak{h}$ be two Lie algebras. A Lie algebra extension is a short exact sequence $$0\longrightarrow \mathfrak{h}\stackrel{\jmath}{\longrightarrow} \mathfrak{e}\stackrel{\pi}{\longrightarrow} \mathfrak{g}\longrightarrow 0.$$ This extension is said to be split if there is a Lie algebra subalgebra $\mathfrak{a}$ of $\mathfrak{e}$ such that $\mathfrak{e}=\mathfrak{a}\oplus \textrm{Ker}(\pi)$.

How can I show such an extension is split if and only if there is a Lie algebra homomorphism $\sigma:\mathfrak{g}\longrightarrow \mathfrak{e}$ such that $\pi\sigma=1_{\mathfrak{g}}$?

Of course $\mathfrak{a}$ should be taken as $\mathfrak{a}=\textrm{Im}(\sigma)$ but I can only show there is an isomorphism $$\textrm{Im}(\sigma)\oplus \textrm{Ker}(\pi)\longrightarrow \mathfrak{e}, (\sigma(x), \jmath(y))\longmapsto \sigma(x)+\jmath(y)$$ but I'd like to have the equality.



The symbol $\oplus$ actually has two different meanings: it can refer to internal direct sums or external direct sums. The external direct sum (of vector spaces, say) $A\oplus B$ is just the set $A\times B$ made into a vector space by letting the operations be defined separately on each coordinate. The internal direct sum $A\oplus B$, on the other hand, is only defined when $A$ and $B$ are both subspaces of a third vector space $C$ with the property that $A\cap B=0$. In this case, the internal direct sum $A\oplus B$ is just defined as $A+B$, i.e. the set of sums of elements of $A$ and elements of $B$. The connection between the two notions is that if $A,B\subseteq C$ have an internal direct sum, then there is an isomorphism from their external direct sum to the internal direct sum given by sending $(a,b)$ to $a+b$. (And conversely, the external direct sum is itself the internal direct sum of the subspaces $A\times 0$ and $0\times B$.)

In the case of your problem, in the statement $\mathfrak{e}=\mathfrak{a}\oplus \textrm{Ker}(\pi)$, the direct sum is meant to be an internal direct sum. That is, there is supposed to be a subalgebra $\mathfrak{a}$ of $\mathfrak{e}$ such that $\mathfrak{a}\cap\textrm{Ker}(\pi)=0$ and $\mathfrak{a}+ \textrm{Ker}(\pi)=\mathfrak{e}$. You should have no difficulty verifying that your choice $\mathfrak{a}=\textrm{Im}(\sigma)$ satisfies these conditions (since this is essentially equivalent to the isomorphism you have already given).

| cite | improve this answer | |

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.