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The only connected 2 manifolds admitting a transitive action by a compact Lie group are the sphere projective plane and torus.

Let M be a connected three manifold which admits a transitive action by a compact Lie group. Then must it be the case that either M is the product of a circle with a surface that admits a transitive action by a compact Lie group $$ T^3, S^2 \times S^1, \mathbb{R}P^2 \times S^1 $$ or $M$ is $$ SU_2/\Gamma $$ for a finite subgroup $ \Gamma $ of $SU_2$?

EDIT: no it's not true the list is missing the mapping torus of the antipodal map of $ S^2 $.

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  • $\begingroup$ It seems that 3 dimensional homogeneous space are classified here $\endgroup$ Commented Dec 2, 2021 at 18:53

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Because you're assuming the group is compact, there's another method for classifying such spaces. This approach can be used in higher dimensions (though people often assume the resulting manifold is simply connected just to cut down on the work).

I'll also always be assuming things are connected: the components of a homogeneous space $G/H$ are always diffeomorphic, and the identity component of $G$ acts transitively on a connected component of $G/H$, so this is all ok.

Suppose $G$ is a connected compact Lie group, $H\subseteq G$ is a closed subgroup. Because $G$ is compact, we may equip the homogeneous space $G/H$ with a Riemannian metric for which the action by $G$ is isometric.

Now, consider the isotropy action of $H$ on $T_{eG} G/H$, where $e\in G$ denotes the identity element. Recall (see, e.g., this MSE question) that if an isometry of a (connected) Riemannian manifold fixes a point and acts as the identity at the tangent space at that point, then that isometry must be the identity. Elements of $h$ fix $e H\in G/H$, so, if such an $h$ acts trivially on $T_{eG}G/H$, we must have $h = e$.

Said another way, the isotropy action gives an injective map $H\rightarrow O(T_{eG} G/H)$ into the orthogonal group. Thus, we may view $H$ as a subgroup of $O(T_{eG} G/H)$.

I'll now specialize to the case where $G/H$ is $3$-dimensional, so $O(T_{eG} G/H)$ can be identified with $O(3)$. The connected subgroups of $SO(3)$ are well known: they are $\{e\}, SO(3)$, or a conjugate of the usual $SO(2)\subseteq SO(3)$. Note that $G/H$ is equivariantly diffeomorphic to $G/(gHg^{-1})$ for any $g\in G$, so, in terms of classification, we can assume the identity component $H^0$ of $H$ is given by one of $\{e\}, SO(2)$, or $SO(3)$.

We'll break into cases depending on $H^0$.

Case A

If $H^0 = \{e\}$, then $3 = \dim G - \dim H$ impies that $\dim G = 3$. From the classification of simply connected Lie groups, $G$ has a cover which is either isomorphic to $SU(2)$ or to $T^3$. In the first case $G = SU(2)$ or $SO(3)$ and $SO(3)/H = SU(2)/\pi^{-1}(H)$ with $\pi:SU(2)\rightarrow SO(3)$ the double cover, so you just get quotients of $SU(2)$. In the second case, you get quotients of $T^3$. However, $H\subseteq T^3$ is normal, so $T^3/H$ is an abelian Lie group of dimension $3$, so $T^3/H\cong T^3$ no matter what $H$ is (so long as $H^0 = \{e\}$.)

Case B

If $H^0 = SO(2)$, then $\dim G = 4$. From the classification of simply connected Lie groups, $G$ has a cover of the form $G = T^4$ or $G = SU(2)\times S^1$. If $G = T^4$, the argument from the previous paragraph establishes that $T^4/H\cong T^3$. So, we will assume that $G$ is covered by $G = SU(2)\times S^1$. In fact, we'll pull everything back: $G/H\cong (SU(2)\times S^1)/\pi^{-1}(H)$, so we'll actually work with $G= SU(2)times S^1$. (Note though, that by pulling back, $\pi^{-1}(H)$ need not act effectively on $(SU(2)\times S^1)/\pi^{-1}(H)$) any more.) Also, instead of writing $\pi^{-1}(H)$ everywhere, I'll abuse notation and just write $H$.

Lemma: Suppose $H$ acts diagonally on $G_1\times G_2$ and that the action of $H$ on $G_2$ is transitive with isotropy group $H'$. Then $(G_1\times G_2)/H$ is canonically diffeomorphic to $G_1/H'$.

Proof: Just check that the map $G_1/H'\rightarrow (G_1\times G_2)/H$ sending $g_1 H'$ to $(g_1,e)H$ is a diffeomorphism. $\square$

Using the lemma, it follows that if the projection of $H^0$ to the $S^1$ factor of $G$ is surjective, then we can write $G/H\cong SU(2)/H'$, where $H'$ is $0$-dimensional. These were classified in Case A). So, we may assume that the projection of $H^0$ to the $S^1$ factor of $G$ is trivial. That is, we may assume $H^0$ acts only on the $SU(2)$ factor of $G$, so $G/H^0\cong S^2\times S^1$.

So, we understand $H^0$ in this case, but what about $H$? Well, let's assume $H\neq H^0$, and pick $(h_1,h_2)\in H\setminus H_0$. Because the normalizer $N:=N_{SU(2)}(SO(2))$ has two components, and $h_1 \in N$, we know that $(h_1,h_2)^2$ acts as a rotation on just the $S^1$ factor of $G/H^0 \cong S^2\times S^1$. Hence, $(G/H^0)/\langle (h_1,h_2)^2\rangle \cong S^2\times S^1$, so we may as well assume that $(h_1,h_2)^2 \in H^0$, so that, in particular, $h_2 = \pm 1$. That is, we may as well assume that $H$ has precisely two components. If $h_1\in SO(2)$, then $G/H$ is diffeomorphic to $S^2\times S^1$ independent of $h_2$. If $h_1\notin SO(2)$ and $h_2 = 1$, we get $\mathbb{R}P^2\times S^1$, and if $h_1\notin SO(2)$ and $h_2 = -1$, we get the non-trivial $S^2$-bundle over $S^1$ (the quotient of $S^2\times S^1$ by the diagonal antipodal action).

This concludes case B.

Case C In this case, $H^0 = SO(3)$. Then $\dim G = 6$, so $G$ is covered by one of $T^6, SU(2)\times T^3$, or $SU(2)\times SU(2)$.

For $G=T^6$, because $H^0$ is non-abelian, there is no $H^0\subseteq G$, even up to cover.

For the other two, by replacing $H^0$ with a cover of $H^0$ (i.e., $SU(2)$), we can actually assume $G= SU(2)\times T^3$ or $G = SU(2)\times SU(2)$. However, in both cases, it is easy to see that the above Lemma applied, so we find $G/H\cong T^3/H'$ or $SU(2)/H'$ where $H'$ is $0$-dimensional, getting us back to Case A. This completes Case C.

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    $\begingroup$ The nice thing about this approach is that it doesn't rely on anything like geometrization. In particular, then same kind of analysis works in whatever dimension you want to work in (though, as the dimension increases, the number of possibilities increases dramatically). Because of this ramp up in cases, when people follow this approach, they typically assume $G/H$ is simply connected (so $H = H^0$) or focus on $G/H$ having certain restricted topological type (rational homotopy groups, cohomology ring, etc.), or that $G$ is restricted (semisimple, simple etc.) $\endgroup$ Commented Jan 23, 2022 at 23:37
  • $\begingroup$ I know that $ \text{Iso}(S^2 \times S^1) \cong \text{Iso}(S^2) \times \text{Iso}( S^1) \cong O_3 \times O_2 $ and $ \text{Iso}(\mathbb{R}P^2 \times S^1) \cong \text{Iso}(\mathbb{R}P^2) \times \text{Iso}( S^1) \cong SO_3 \times O_2 $. Am I right in thinking that the isometry group of the quotient of $ S^2 \times S^1 $ by the diagonal antipodal action is $ SO_3 \times SO_2 $? $\endgroup$ Commented Jan 26, 2022 at 17:54
  • $\begingroup$ @IanGershonTeixeira: My guess is that it would be $O_3\times O_2/\langle -1,-1\rangle$ with $-1$ denoting the antipodal map. In particular, I think it should have two components. $\endgroup$ Commented Jan 26, 2022 at 18:57
  • $\begingroup$ You are right of course! All these manifolds are Riemannian covered by $ S^2 \times S^1 $ with isometry group $ O_3 \times O_2 $ and so the isometry group of any quotient by a group $ \Gamma $ of isometries (the deck transformations for the covering) is just $ N(\Gamma)/\Gamma $ where $ N(\Gamma) $ is the normalizer in $ O_3 \times O_2 $ (the isometry group of the cover) But all the $ \Gamma $ here are central! So certainly normal and thus the quotient is as you say (and its another way to see $ \text{Iso}(\mathbb{R}P^2 \times S^1) \cong O_3 \times O_2 /(-1,1) \cong SO_3 \times O_2 $) $\endgroup$ Commented Jan 26, 2022 at 19:35
  • $\begingroup$ Normally it is illegal to use the normalizer formula here since $ S^2 \times S^1 $ is not the universal cover but I think it works here maybe something to do with subgroups being central? $\endgroup$ Commented Jan 26, 2022 at 22:25

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