Let $g$ be some homogeneous Riemannian metric on $\mathbb R^3$ (i.e, a metric such that $(\mathbb R^3,g)$ is a homogeneous space). Consider its isometry group $G = Isom(\mathbb R^3,g)$.

A covering subgroup $H \subset G$ is a group acting freely and properly discontiously on $\mathbb R^3$, in other words, the natural projection $\pi: \mathbb R^3 \to \mathbb R^3/H$ is a covering map. Note that trivial group $H = \{e\}$ is always a covering subgroup of $G$.

Question 1: Does there exist a cocompact covering group $H \subseteq G$, i.e a covering group such that $\mathbb R^3/H$ is compact ?

Suppose $G$ contains a cocompact covering group. Then it is a classic result by Thurston that, up to conjugation by some diffeomorphism $\phi: \mathbb R^3 \to \mathbb R^3$, there are six Riemannian metrics $\{g_i\}_{i=1}^6$ on $\mathbb R^3$ (independent of the choice of $g$), so that $G$ is contained in precisely one of $Isom(\mathbb R^3,g_i)$.

Question 2: Is there some $G$ as above, such that (a conjugate of) $G$ is properly contained in one $Isom(\mathbb R^3,g_i)$, i.e $G \subsetneq Isom(\mathbb R^3,g_i)$.

If we consider $\mathbb R^2$ instead of $\mathbb R^3$, such a homogeneous metric cannot be found. For if $g$ is any homogeneous metric on $\mathbb R^2$, then as $\mathbb R^2$ is two dimesional, this means that $(\mathbb R^2,g)$ is also an isotropic space, i.e, $G$ acts transitively on the set of all tangent planes in $T\mathbb R^2$. It follows that $(\mathbb R^2,g)$ has constant curvature, implying that $G$ is either conjugate to $Isom(\mathbb R^2,g_{eukl})$ or $Isom(\mathbb R^2,g_{hyp})$, both of which are maximal.

EDIT: In an earlier version, I mistakenly assumed that every metric $g$ on $\mathbb R^3$ would induce a homogeneous structure, simply because every translation is always an isometry, regardless of the specific choice of $g$. This is, of course, blatantly false, as Professor Lee was so nice to remind me of, it's because the inner product induced by $g$ may very well vary from point to point (as it's the usual case when picking a random metric on a manifold).

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    $\begingroup$ If $g$ is an arbitrary metric on $\mathbb R^3$, the translations are almost certainly not isometries. $\endgroup$ – Jack Lee Oct 6 '17 at 14:48
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    $\begingroup$ The fact that the Jacobian is the identity in standard coordinates is irrelevant, unless the metric has the same matrix at different points. $\endgroup$ – Jack Lee Oct 6 '17 at 14:50
  • $\begingroup$ That is true, for some obscure reason, this simple observation went past me. I will edit accordingly $\endgroup$ – Berni Waterman Oct 6 '17 at 16:51
  • $\begingroup$ There is a 1-parameter family of solvable 3-dimensional Lie groups $L_t$; for generic value of the parameter $t$, $Isom(L_t)$ contains no discrete subgroups acting cocompactly. Here $L_t$ is equipped with a left-invariant Riemannian metric. $\endgroup$ – Moishe Kohan Oct 6 '17 at 17:51

Question 1: No. For instance, take the Riemannian metric on $R^3$ $$ dt^2 + e^{at}dx^2 + e^{bt}dy^2 $$ for generic values of $a, b>0$. This geometry is homogeneous (with a 3-dimensional solvable Lie group acting transitively) but has no discrete cocompact subgroups of the isometry group.

Question 2: Yes. For instance, start with the 3-dimensional real Heisenberg group $H_3$ and equip it with a suitable left-invariant Riemannian metric which is also invariant under $SO(2)$ (this is known as the Nil-geometry, see Thurston's book that you are reading). Then $H_3$ contains a discrete cocompact subgroup, while $H_3$ is strictly smaller than $Isom(H_3)$.

  • $\begingroup$ Thanks for answering my first question. I believe your second answer doesn't really answer my second question, as the $G$ in my question should always be the isometry group of some Riemannian metric (I think that you chose $G = H_3$, but $H_3$ is not the full isometry group of the left-invariant metric you chose). $\endgroup$ – Berni Waterman Oct 7 '17 at 20:17
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    $\begingroup$ @BerniWaterman: The group $H_3$ will be the isometry group of a randomly chosen left-invariant Riemannian metric on $H_3$; the Nil-metric on $H_3$ will have strictly larger isometry group. This is what your 2nd question is about. $\endgroup$ – Moishe Kohan Oct 7 '17 at 20:20
  • $\begingroup$ Just to clear things up: By $Isom(H_3)$, you mean the isometry group of $H_3$ with respect the Nil-metric (not with respect to the randomly chosen left-invariant one)? $\endgroup$ – Berni Waterman Oct 7 '17 at 20:24
  • $\begingroup$ @BerniWaterman: Right. $\endgroup$ – Moishe Kohan Oct 7 '17 at 20:54

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