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I would like to know examples of smooth compact connected manifolds, on which there exists an effective smooth circle action preserving a positive smooth volume, besides the simple example: $[0,1]^d \times \mathbb{S}^1$. Is there a classification of these manifolds?

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Mustafa --what exactly do you mean by an "effective action of a circle"? –  algori Sep 4 '12 at 16:08
    
Your example has a boundary. Do you want to allow boundary? –  Ben McKay Sep 4 '12 at 18:54
    
An effective action of a group $G$ on $M$ is an action such that there is a set of full measure $F \subset M$ such that for any $x \in F$, there is $g \in G$ such that $gx \not\eq x$. Yes, I allow boundary. –  Mustafa Sep 5 '12 at 12:33
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5 Answers 5

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As already pointed out, whenever you have a compact group $G$ acting on a manifold $M$, this action can be made isometric by constructing a metric on $M$ via a standard averaging process. In particular, if you have a (effective) circle action on $M$, you can find a metric $g$ on $M$ such that this action is isometric. In other words, the isometry group of $(M,g)$ contains a circle. This should provide you with plenty of examples: intuitively, a manifold that carries a circle action is one that has (at least) a "circle" worth of symmetries.

For example, take $M$ to be any rotationally symmetric surface in $\mathbb R^3$, e.g., a round sphere $S^2$ or a torus $S^1\times S^1$. Then $M$, by definition, has an isometric (effective) circle action. Moreover, these are the only (orientable) manifolds of dimension $2$ with an effective circle action. This follows from the fact that if we have a torus action on $M$, then the Euler characteristic of $M$ is $\chi(M)=\chi(M^T)$, where $M^T$ is the fixed point set of the torus. Since $M^T$ is a totally geodesic submanifold of $M$, it can be a circle or a finite collection of points. Thus, $\chi(M)\geq 0$, which means $M$ is homeomorphic to $S^2$, $\mathbb RP^2$, $S^1\times S^1$ or the Klein bottle $K^2$, according to orientability and $\chi(M)=2,1,0$, respectively. This is the classification you ask for in dimension equal to $2$.

Classifying compact connected manifolds of any dimension with a circle action and nothing else seems rather too general. More interesting problems are, for example, classifying manifolds with an isometric circle action that also have, e.g., some curvature condition. For example, if $\dim M=4$ and $M$ has a metric of positive curvature that has an isometric circle action, then Hsiang-Kleiner proved that $M$ must be homeomorphic to $S^4$ or $\mathbb C P^2$.

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I was primarily looking for elementary and "concrete" examples, that you provided in dimension 2. In dimension 3, Seifert manifolds can be examples, as mentioned by Liviu. The problem of classification may be too general indeed. –  Mustafa Sep 5 '12 at 13:22
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This is more of a comment than an answer. Any simplectic manifold with an effective circle action preserving the form is an example. So any smooth projective toric variety is an example of a manifold with an effective circle action preserving a volume form.

Edit: I forgot to say what the volume is: it's the top power of the symplectic form.

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@Mustafa: I am sure Professor Lerman is a busy man, and is not obliged to answer your questions, so the correct response is "thank you!" –  Igor Rivin Sep 5 '12 at 18:41
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The $3$-manifolds with effective $S^1$-actions are called Seifert manifolds and are completely classified. Essentially, all of them can be obtained from a circle bundle over a surface by removing tubular neighborhoods of several fibers and gluing them back in a clever way. For details you can check P. Orlik's book Seifert manifolds, Lecture Notes in Math. Springer Verlag, 1972.

The classification in higher dimensions is more complicated and I don't know if it is complete. For certain $S^1$ actions (called hamiltonian) on $4$-manifolds there is a wealth of results; see e.g. Y. Karshon's paper Periodic Hamiltonian flows on four dimensional manifolds, Mem. Amer. Math. Soc. 672 (1999) (see short version here.)

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In the answer of Will, one can exchange manifold $N$ to an orbifold with singularities modeled on $\mathbb R^{n-1}/\mathbb Z_k$ and $\mathbb R^{n}/\mathbb S^1$. Then his answer works for effective action. You can get a lot of examples this way, but I do not think you can call it a "classification".

On the other hand assume you have a manifold $M$ and want to know if it admits an effective $\mathbb S^1$-action. In this case Will's answer is useless and as far as I know no good answer is known.

If there is an action on $M$ then

  • There are some restrictions on $\pi_1(M)$; for example compact hyperbolic manifolds do not admit smooth $\mathbb S^1$-actions;
  • If $M$ is simply connected and spin then $\hat A(M)$ has to vanish.

I do not know any other restriction.

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Assuming you mean: An action of the group $\mathbb R/\mathbb Z$ without stabilizer:

If you look at the quotient of the manifold $M$ by the action, you get a manifold $N$ with a circle bundle $M$. Circle bundles are classified by their Chern class, which is an element of $H^2(N,\mathbb Z)$ coming from the exact sequence $H^1(\mathbb Z) \to H^1(\mathbb R) \to H^1(\mathbb R/\mathbb Z) \to H^2(\mathbb Z)$.

To find a volume preserved by the circle action, take any volume and average it with its pullbacks along the circle action.

So the classification is any smooth compact connected manifold $N$ plus an element of $H^2(N,\mathbb Z)$.

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@Will: It seems the OP is not assuming the action is free, so the orbit space need not be a manifold. Take, e.g., the rotation action of $S^1$ on $\mathbb R^2$, whose quotient is $[0,+\infty)$. Regarding the assumption of the action being effective, every action can be made effective by taking a quotient by the innefective kernel (the intersection of all isotropy subgroups). In other words, this assumption only takes out trivial examples, where parts of the group just fix the entire thing. –  Renato G Bettiol Sep 4 '12 at 18:49
    
If that's the case, then this answer is indeed only valid for the special case when the action is free. Non-free actions are obviously much more complicated. –  Will Sawin Sep 4 '12 at 19:17
    
@Will: I guess Anton's answer solves the issue I pointed out; your argument is valid also for orbifolds and not only manifolds. Sorry for the confusion. –  Renato G Bettiol Sep 4 '12 at 19:52
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