# On embeddings of Lie groups

Let $G$ be a Lie group with identity component $G_0$, such that $G_0$ embeds as a closed subgroup into some other connected Lie group $\widetilde{H}$. Then does there always exist a Lie group $H$ with identity component $H_0 = \widetilde{H}$, such that $G$ embeds as a closed subgroup into $H$ ?

As $G_0$ is always a normal subgroup of $G$, my idea was to examine the short exact sequence $0 \to G_0 \to G \to G/G_0 \to 0$, where $G/G_0$ is a countable discrete group, whose elements can be represented by the (countable many) components of $G$. If this sequence splits, we can identify $G = G_0 \rtimes_{\phi} F$ for $F$ some countable discrete group and $\phi: F \to Aut(G_0)$ some homomorphism. Now if there was a way to "extend" $\phi$ to a homomorphism $\tilde{\phi}: F \to Aut(\tilde{H})$ in such a way that for every $f \in F$, $\tilde{\phi}(f)|_{G_0} = \phi(f)$ (this is equivalent to requiring that every automorphism in the image of $\phi$ can be extended to an automorphism on $\widetilde{H}$), then $H := \widetilde{H} \rtimes_{\tilde{\phi}} F$ would be the required Lie group.

However, this is only a partial answer, since it does not always seem to be the case that this sequence splits (more generally, there seem to exist Lie groups that cannot be decomposed as the semi-direct product of a discrete group and its identity component). Moreover, even if we have such a splitting, i am uncertain of when exactly one can extend the homomorphism $\phi$ in the way described above.

## 1 Answer

HINT: It is not possible in general ( I changed the notations, they were counterintuitive to me)

Say we have $H$ connected imbedded into $G$ connected. Let $\mathfrak{S}$ a finite group of automorphisms of $H$. Consider the semidirect product $H \rtimes \mathfrak{S}$ that fits in the split exact sequence $$1 \to H \to H \rtimes \mathfrak{S} \to \mathfrak{S} \to 1$$ Recall that in $H \rtimes \mathfrak{S}$ the composition is $$(h_1, s_1) (h_2 s_2)= (h_1 \cdot s_1(h_2) , s_1 s_2)$$ The action of $\mathfrak{S}$ is in fact by conjugation. Since we have $$h_1 s_1 h_2 s_2 = h_1 (s_1 h_2 s_1^{-1}) s_1 s_2$$ inside $H\rtimes \mathfrak{S}$.

The question is whether we have a Lie group $\tilde G$ with identity component $G$ that contains $H\rtimes \mathfrak{S}$. If such a $\tilde G$ exists, $G$ will be a normal subgroup of $\tilde G$. The action of $\mathfrak{S}$ by conjugation on $H$ extends to an action of $\mathfrak{S}$ on $G$. But that is not always possible

As an example, consider the diagonal imbedding $H = (0, \infty)\times (0,\infty) \subset GL(2,\mathbb{R})_{+}$, and the action of $\mathbb{Z}/2$ on $H$, $(a,b)\mapsto (a^{-1}, b)$. This action does not extend to an action on $GL(2, \mathbb{R})_{+}$. Indeed, every automorphism of $GL(2, \mathbb{R})_{+}$ preserves or inverses the determinant.

• I don't see why being able to extend the automorphsims of the smaller connected group $H$ to the larger group $G$ is a necessary condition for finding a candidate group as in my question (although it is certainly a sufficient condition) – H1ghfiv3 Oct 13 '17 at 18:13
• @Berni Waterman: Inside $\tilde H$, conjugation by elements of $\mathfrak{S}$ is just the action of $\mathfrak{S}$. That conjugation better preserve $G$ inside $\tilde G$, since it preserves $H$ ( otherwise it would move it to other connected components of $\tilde G$ ). – orangeskid Oct 13 '17 at 18:18
• Okay, now the notation becomes confusing even to me :D I will try to digest what you mean tomorrow morning :) – H1ghfiv3 Oct 14 '17 at 0:09
• @Berni Waterman: Hi! I modified a bit the answer, added some details, and streamlined it. The key is looking at the semidirect product in two alternative ways. Hope it helps. – orangeskid Oct 14 '17 at 12:49
• @Berni Waterman: In fact it should work for any maximal compact subgroup of a complex reductive lie group ( whatever that means) – orangeskid Oct 15 '17 at 1:11