Splitting of Lie algebra extensions: why does this linear map exist

exact-sequencelie-algebraslinear algebra

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$$
0\to\mathfrak{b}\xrightarrow{\;\;j\;\;}\tilde{\mathfrak{g}}\xrightarrow{\;\;\phi\;\;}\mathfrak{g}\to0
$$

as in page 15 of these notes by Erik van den Ban. 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?

Best Answer

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:

Non-split extension of Lie algebras?

For vector spaces, or more generally for $R$-modules see here:

Why any short exact sequence of vector spaces may be seen as a direct sum?

A question about split short exact sequence of modules

Related Question