# Splitting of Lie algebra extensions

Let $$\mathfrak{b}$$ and $$\mathfrak{g}$$ be finite dimensional Lie algebras and let $$(\tilde{\mathfrak{g}},j,\phi)$$ be a Lie algebra extension of $$\mathfrak{g}$$ by $$\mathfrak{b}$$, so we have a short exact sequence of Lie algebras$$0\to\mathfrak{b}\underbrace{\to}_{j}\tilde{\mathfrak{g}}\underbrace{\to}_{\phi}\mathfrak{g}\to0$$ as in https://www.staff.science.uu.nl/~ban00101/lecnotes/repq.pdf, page 15. Then, the author chooses a linear map $$\xi:\mathfrak{g}\to\tilde{\mathfrak{g}}$$ such that $$\phi\circ\xi=Id_{\mathfrak{g}}$$. If $$\xi$$ is a Lie algebra homomorphism, it means that the central extension splits but it is just assumed to be linear. Why does this map exist? I read on here that every short exact sequence of vector spaces splits, is there an intuitive reason why that is true?

• It suffices to define $\xi$ on a basis of $\frak{g}$. – Arnaud D. Oct 25 '19 at 12:46
• I have to define it using $\phi$ I suppose? – Lucas Smits Oct 25 '19 at 12:48
• If I understand correctly the question is unrelated to Lie algebras... – YCor Oct 25 '19 at 20:03
• I added an answer for vector spaces and $R$-modules. – Dietrich Burde Oct 26 '19 at 8:19

Extensions of vector spaces split, but not extensions of Lie algebras in general. So "this homomorphism $$\xi$$" need not exist. Consider an example, i.e., a non-split extension of the $$3$$-dimensional Heisenberg Lie algebra:
For vector spaces, or more generally for $$R$$-modules see here: