Example of a manifold which is not a homogeneous space of any Lie group - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T07:30:18Z http://mathoverflow.net/feeds/question/89345 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89345/example-of-a-manifold-which-is-not-a-homogeneous-space-of-any-lie-group Example of a manifold which is not a homogeneous space of any Lie group MTS 2012-02-24T00:11:59Z 2012-02-25T23:01:51Z <p>Every manifold that I ever met in a differential geometry class was a homogeneous space: spheres, tori, Grassmannians, flag manifolds, Stiefel manifolds, etc. What is an example of a connected smooth manifold which is not a homogeneous space of any Lie group?</p> <p>The only candidates for examples I can come up with are two-dimensional compact surfaces of genus at least two. They don't seem to be obviously homogeneous, but I don't know how to prove that they are not. And if there are two-dimensional examples then there should be tons of higher-dimensional ones.</p> <p>The question can be trivially rephrased by asking for a manifold which does not carry a transitive action of a Lie group. Of course, the diffeomorphism group of a connected manifold acts transitively, but this is an infinite-dimensional group and so doesn't count as a Lie group for my purposes.</p> <p>Orientable examples would be nice, but nonorientable would be ok too.</p> http://mathoverflow.net/questions/89345/example-of-a-manifold-which-is-not-a-homogeneous-space-of-any-lie-group/89347#89347 Answer by Roberto Frigerio for Example of a manifold which is not a homogeneous space of any Lie group Roberto Frigerio 2012-02-24T00:29:28Z 2012-02-24T00:29:28Z <p>Here is a proof that any closed surface $S$ of genus at least 2 cannot support a homogeneous Riemannian metric. In fact, let $g$ be any such metric. Being homogeneous, $g$ has constant curvature, and e.g. by Gauss-Bonnet such a curvature must be negative. Therefore, $(S,g)$ is homotetic to a hyperbolic surface, and it is well-known that the isometry group of any closed hyperbolic surface is finite (in fact, if $h$ is the genus of $S$, then $(S,g)$ admits at most $84(h-1)$ isometries). </p> http://mathoverflow.net/questions/89345/example-of-a-manifold-which-is-not-a-homogeneous-space-of-any-lie-group/89352#89352 Answer by Tom Goodwillie for Example of a manifold which is not a homogeneous space of any Lie group Tom Goodwillie 2012-02-24T01:21:02Z 2012-02-24T16:59:44Z <p>$\pi_2$ of a Lie group is trivial, so $\pi_2(G/H)$ is isomorphic to a subgroup of $\pi_1(H)$, which is finitely generated (isomorphic to $\pi_1$ of a maximal compact subgroup of the identity component of $H$). But $\pi_2$ of a closed manifold is often not finitely generated. For example, the connected sum of two copies of $S^1\times S^2$ has as a retract a punctured $S^1\times S^2$, which is homotopy equivalent to $S^1\vee S^2$ and so has universal cover homotopy equivalent to an infinite wedge of copies of $S^2$.</p> <p>EDIT This ad hoc answer can be extended as follows: All I really used was that $\pi_2(G)$ and $\pi_1(G)$ are finitely generated. But $\pi_n(G)$ is finitely generated for all $n\ge 1$ (reduce to simply connected case and use homology), so $\pi_n(G/H)$ is finitely generated for $n\ge 2$. That leads to a lot more higher-dimensional non simply connected examples. </p> http://mathoverflow.net/questions/89345/example-of-a-manifold-which-is-not-a-homogeneous-space-of-any-lie-group/89358#89358 Answer by Paul Siegel for Example of a manifold which is not a homogeneous space of any Lie group Paul Siegel 2012-02-24T03:02:05Z 2012-02-24T03:02:05Z <p>Atiyah and Hirzebruch gave a rather dramatic answer to your question in their paper "Spin Manifolds and Group Actions": if $M$ is a compact smooth spin manifold of dimension $4k$ whose $\hat{A}$-genus is nonzero then no compact Lie group can act on $M$ <em>nontrivially</em>, let alone transitively! The proof uses Atiyah and Bott's Lefschetz fixed point theorem in a clever way.</p> <p>Unfortunately I don't have a simple example of such a manifold lying around, though I know that there are plenty of examples among 4-manifolds. It's possible that some 4 dimensional lens spaces would do the job.</p> http://mathoverflow.net/questions/89345/example-of-a-manifold-which-is-not-a-homogeneous-space-of-any-lie-group/89359#89359 Answer by Vitali Kapovitch for Example of a manifold which is not a homogeneous space of any Lie group Vitali Kapovitch 2012-02-24T03:03:49Z 2012-02-24T04:56:31Z <p>Apart from already mentioned non simply connected examples most simply connected manifolds are also not homogeneous. One easy criterion is that simply connected homogeneous spaces are rationally elliptic, i.e. they have finite dimensional total rational homotopy. That is because any connected Lie group is rationally homotopy equivalent to a product of odd dimensional spheres. so a homogeneous space is elliptic by a long exact homotopy sequence.</p> <p>Most simply connected manifolds are not rationally elliptic. For example the connected sum of more than two $CP^2$'s or $S^2\times S^2$'s. This is easily seen by looking at their minimal models. But even without computing minimal models it's known that an elliptic manifold $M^n$ has nonnegative Euler characteristic and has the total sum of its Betti numbers $\le 2^n$. So anything that violates either of these conditions such as the connected sum of several $S^3\times S^3$'s is definitely not rationally elliptic and hence can not be a homogeneous space or even a biquotient.</p> <p>As for higher genus surfaces it should not be hard to show that they can not be homogeneous spaces $G/H$ even if you don't assume that $G$ acts by isometries. If $G/H=S^2_g$ and the $G$ action is effective then for any proper normal $K\unlhd G$ which does not act transitively on $S^2_g$ we must have $K/(K\cap H)=S^1$. But then $G/K$ is also 1-dimensional and hence also a circle which is obviously impossible. This reduces the situation to the case of $G$ being simple which can also be easily ruled out for topological reasons.</p> http://mathoverflow.net/questions/89345/example-of-a-manifold-which-is-not-a-homogeneous-space-of-any-lie-group/89370#89370 Answer by Dylan Wilson for Example of a manifold which is not a homogeneous space of any Lie group Dylan Wilson 2012-02-24T06:54:00Z 2012-02-24T06:54:00Z <p>It is a result of Mostov that any compact homogeneous manifold must have nonnegative Euler characteristic:</p> <p><a href="http://www.ams.org/mathscinet-getitem?mr=2174096" rel="nofollow">http://www.ams.org/mathscinet-getitem?mr=2174096</a></p> <p>That should provide plenty of counterexamples. :)</p> http://mathoverflow.net/questions/89345/example-of-a-manifold-which-is-not-a-homogeneous-space-of-any-lie-group/89422#89422 Answer by Igor Belegradek for Example of a manifold which is not a homogeneous space of any Lie group Igor Belegradek 2012-02-24T16:50:19Z 2012-02-24T16:50:19Z <p>In the homotopy exact sequence of the fiber bundle $G\to G/H$ the group $\pi_i(G/H)$ sits between $\pi_i(G)$ and $\pi_{i-1}(H)$. For example, if $i=1$, then $\pi_1(G)$ is abelian, and $\pi_0(H)$ is finite (as $H$ is compact). Thus $\pi_1(G/H)$ has abelian subgroup of finite index. Surely there are lots of manifolds that do not have this property, e.g. any closed negatively curved manifold does not. Connected sum of several lens spaces is another example.</p> <p>By the way, it is my opinion that the class of compact homogeneous spaces is very rich, and their finer topological properties are still poorly understood. For example, classifying homogeneous spaces up to diffeomorphism in a given homotopy type is quite challenging, and it is not easy to cook up a homogeneous spaces with prescribed topological features.</p> http://mathoverflow.net/questions/89345/example-of-a-manifold-which-is-not-a-homogeneous-space-of-any-lie-group/89443#89443 Answer by Lee Mosher for Example of a manifold which is not a homogeneous space of any Lie group Lee Mosher 2012-02-24T20:05:43Z 2012-02-24T20:05:43Z <p>I would think that many examples from 3-manifold theory would work. Take any compact, oriented, irreducible 3-manifold $M$ whose torus decomposition is nontrivial and has at least one hyperbolic piece. Such examples at least have no locally homogeneous Riemannian metric, as a consequence of Thurston's analysis of the 8 geometries of 3-manifold theory. A specific example of this sort can be obtained from a hyperbolic knot complement in $S^3$ by deleting an open solid torus neighborhood of the knot and doubling across the resulting 2-torus boundary; the doubling torus produces a characteristic $Z^2$ subgroup of $\pi_1(M)$. These examples have universal cover homeomorphic to $R^3$, and so they have trivial $\pi_2$. By the homotopy exact sequence there would be a quotient group $\pi_1(G) / \pi_1(H)$ identified with a subgroup of $\pi_1(G/H)$ whose quotient set is $\pi_0(H)$. Perhaps, in order to get a proof, one can analyze this situation by considering the intersection of the $Z^2$ subgroup with the $\pi_1(G) / \pi_1(H)$ subgroup.</p>