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Let $G$ be a Lie group (not necessarily connected) and let $H$ be a closed subgroup of $G$. I am after an algebraic (group theoretic) characterization of when the homogeneous space $G/H$ is connected.

I found the following necessary condition in Onishchik/Vinberg, Lie groups and algebraic groups: Let $G_0$ denote the connected component of the identity in $G$ then $G/G_0$ is a discrete group. If $G/H$ as above is connected then $G/G_0$ is isomorphic (as a group) to $H/(H\cap G_0)$.

Does anybody know if this is also sufficient? Onishchik/Vinberg do not give a proof. Any idea of how to go about that? I was quite surprised that I could not find anything relevant on this question in other books (Kobayashi/Nomizu, Helgason, Hilgert/Neeb, ...) and also a google search throws up nothing.

The above necessary condition seems to formalize the intuitive idea that $H$ needs to contain an element from every connected component of $G$ for $G/H$ to be connected.

Ultimately, I want to be able to decide this algorithmically (at least for matrix Lie groups), but this is not part of this question.

Edit: clarified wording, I did not know how to formally link the idea in paragraph four to the quotient group objects when I asked the question

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closed as unclear what you're asking by John Pardon, Anton Petrunin, Ryan Budney, Jeremy Rickard, Andrej Bauer Jan 31 '15 at 18:55

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    $\begingroup$ The group $H/(H\cap G_0)$ can be considered as a subgroup in $G/G_0$. The space is connected if $H/(H\cap G_0)=G/G_0$. Assume $G/G_0=\mathbb Z$ and $H/(H\cap G_0)=2{\cdot}\mathbb Z$; they are isomorphic but the space has two connected components. (Maybe you want to assume that $G$ is compact?) $\endgroup$ – Anton Petrunin Jan 30 '15 at 5:15
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    $\begingroup$ You gave the answer to your question in the 1st sentence of your 4th paragraph. What else would you like to know? $\endgroup$ – Ryan Budney Jan 30 '15 at 5:54
  • $\begingroup$ I did not know how to formally link the idea about components to the quotient group objects. Todd answers this below. Thanks also to Anton for pointing out the inclusion. $\endgroup$ – Jochen Trumpf Feb 2 '15 at 4:18
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As Ryan says, the idea is in the first sentence of the fourth paragraph, so it's just a matter of formalizing it into a proof.

There is a natural isomorphism $\pi_0(G) \cong G/G_0$ for any Lie group $G$. The inclusion $i: H \to G$ induces by functoriality a function $\pi_0(i): \pi_0(H) \to \pi_0(G)$; by the natural isomorphism it is given by a composite

$$H/H_0 \to H/H \cap G_0 \hookrightarrow G/G_0$$

where the first map is well-defined as $H_0 \subseteq H \cap G_0$ and is clearly an epi. Thus the composite gives the epi-mono factorization of $\pi_0(i)$, and the necessary condition cited in the question (as corrected by Anton Petrunin), which is that the inclusion is onto, is thus sufficient to guarantee that $\pi_0(i)$ is surjective, or that $H$ intersects each path component of $G$ in a nonempty set.

So given $g, g' \in G$, we may pick $h \in H$ so that $g$ and $g'h$ belong to the same path component. Pick a path $\alpha: I \to G$ connecting them; then the composite $I \stackrel{\alpha}{\to} G \stackrel{\pi}{\to} G/H$ is a path connecting $gH$ to $g'H$.

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