# Semidirect product of Lie algebras and a corresponding Lie group

Let $$\mathfrak g$$ and $$\mathfrak h$$ be (finite-dimensional real) Lie algebras, and form their semidirect product $$\mathfrak g \oplus_\pi \mathfrak h$$ with respect to some homomorphism $$\pi : \mathfrak g \to \text{Der}(\mathfrak h)$$.

Suppose further that we have a linear representation of $$\mathfrak g \oplus_\pi \mathfrak h$$ (say as real matrices) so that $$e_i$$ and $$f_j$$ are bases of $$\mathfrak g$$ and $$\mathfrak h$$ respectively, and they satisfy the required commutator relations of $$\mathfrak g \oplus_\pi \mathfrak h$$. Then the canonical coordinates of the second kind $$(x,y) \mapsto \prod_i \text{Exp}(x_ie_i) \prod_j \text{Exp}(y_jf_j)$$ represent (not necessarily faithfully) a connected Lie group with Lie algebra $$\mathfrak g \oplus_\pi \mathfrak h$$.

Question: Is this Lie group automatically a semidirect product of Lie groups in general?

More generally, given a Lie group whose Lie algebra is a semidirect product, is the group necessarily a semidirect product as well?

• What is the definition of the ”Lie algebra semidirect product”? I have never seen it before. Thanks! – DanielC Nov 11 '19 at 0:21
• @DanielC One takes $\mathfrak g \oplus \mathfrak h$ as the underlying vector space and equips with with the Lie bracket such that $[X,Y] = \pi(X)Y$ when $X \in \mathfrak g$ and $Y \in \mathfrak h$. Of course one needs to show that such a Lie bracket exists and is unique for this definition to make sense, see e.g. Proposition 1.22 in Knapp's "Lie Groups, Beyond an Introduction". – MSDG Nov 11 '19 at 8:22

## 1 Answer

I think the answer is NO.

The reason is simply that the Lie algebra does not change if you consider a Lie Group or its universal cover or some its quotients by discrete subgroups of $$Z(G)$$.

Probably your statement is true if you suppose that G and H are simply connected and consider the only simply connected Lie group with Lie algebra $$\mathfrak{g} \otimes_\pi \mathfrak{h}$$ but actually I have no references to suggest.

• Thank you for the answer, I suspected that this might be the case. When the Lie groups are simply connected, then one can show that there is a unique action $\tau : G \to \text{Aut}(H)$ such that the simply connected semidirect product Lie group $G \ltimes_\tau H$ has Lie algebra $\mathfrak g \oplus_\pi \mathfrak h$, see e.g. Theorem 1.125 in Knapp's "Lie Groups, Beyond an Introduction". – MSDG Nov 11 '19 at 8:24