# Lie Algebra Homomorphisms for Lie Subgroups

Let $G$ be a Liegroup and $H$ a Lie subgroup of $G$. Then we find a Liegroup homomorphism $i \colon H \to G$ and the induced map $i_* \colon \mathfrak{h} \to \mathfrak{g}$ between the corresponding Lie algebras $\mathfrak{g}$ and $\mathfrak{h}$ is a Lie-algebra homomorphism.

Is it always possible to find a map $$\pi \colon \mathfrak{g} \to \mathfrak{h}$$ such that $\pi \circ i_* = \operatorname{id}_{\mathfrak{h}}$ and such that $\pi$ is a Lie algebra homomorphism?

If not, are there some concrete assumptions (on the Lie groups or Lie algebras) under which such a map $\pi$ exists?

Such a Lie algebra homomorphism is called a section. It need not exist in general. It means that the short exact sequence $$0\rightarrow \mathfrak{a} \rightarrow \mathfrak{g} \xrightarrow{\pi} \mathfrak{h} \rightarrow 1$$ splits. This is equivalent to saying that $\mathfrak{g}$ is a semidirect product of $\mathfrak{h}$ and $\mathfrak{a}$, i.e., $\mathfrak{g}\cong \mathfrak{h}\ltimes \mathfrak{a}$. Assuming that $\mathfrak{a}$ is abelian, the assumption to assure that the sequence splits is, that the cohomology group $H^2(\mathfrak{h},\mathfrak{a})$ is trivial.
• Thank you for your quick response. Do we need, that $\mathfrak{a}$ is also a Lie subalgebra of $\mathfrak{g}$? And do you have some references to read a bit about this splittings? – Olorin Feb 19 '16 at 16:25
• @DietrichBurde: Assuming there is a non-degenerate-bilinearform $\langle \cdot, \cdot \rangle$ on $\mathfrak{g}$, which is non-degenerate when restricted to $\mathfrak{g}_1 \times \mathfrak{g}_1$. If $\pi$ is our orthogonal projection, then $\pi$ is a section, am I right? – Olorin Feb 19 '16 at 16:55