Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm dealing with actions of the circle over differentiable manifolds. In the book I'm reading, they use the fact that an action of $S^1$ over a disk has to be equivalent (there has to exist an equivariant diffeomorphism) to a rotation. Can someone give me a hint to prove this? Or maybe a reference where I can consult this result? It seems to be a "well known fact" but I haven't been able to find a place where they prove it.

share|improve this question
2  
Pick any point. Its orbit under $S^1$ is a loop (winding around an integer number of times in a fixed direction) or a point. The different orbits can't cross each other.... –  Hurkyl Jul 4 '12 at 23:45
    
I think this might settle it for the case of the disk of dimension 2 but I'm failing to see this argument applied to the higher dimensional cases. I feel the orbits have too much space, so ones could go one way and others other way. I feel something more has to be said to conclude it is a rotation. –  Chu Jul 5 '12 at 5:09
1  
Ah, I hadn't realized you were interested in other dimensions. My intuition is that if the action sends a point around a loop, then the point drags a tubular neighborhood along with it. And the rotation within that tube drags a larger tubular neighborhood with it, and so forth. But I'm much less confident in turning that intuition into a proof; before trying to make it work I would first try to stumble about with homotopy groups. –  Hurkyl Jul 5 '12 at 11:40
    
What is a disk of dimension higher than 2? –  timur Aug 18 '12 at 2:13
2  
Dumb question: Is it obvious that every $S^1$ action on a disc as a fixed point? The fact that the disc is a manifold with boundary is causing me difficulty. –  Jason DeVito Aug 18 '12 at 2:51

1 Answer 1

up vote 5 down vote accepted
+50

Your question is a little imprecise but I'll take it to be the assertion that a smooth action of $S^1$ on $D^n$, i.e. a smooth homomorphism $S^1 \to Diff(D^n)$ is conjugate via a diffeomorphism of $D^n$ to a homomorphism $S^1 \to SO_n$.

First off, the generator of the motion is a vector field on $D^n$ which is tangent to $\partial D^n = S^{n-1}$. Since its tangent on the boundary, you can perturb the vector field near the boundary to be outward-pointing. Poincare-Hopf kicks in and tells you this vector field needs to have a zero on the interior, so your original vector field has a zero on the interior.

So in the orbit decomposition of $D^n$ you have a non-empty fixed point set. So all we need to do is show the fixed point set is isotopic to a linear subspace -- once you have that you know that the disc $D^n$ is just an $S^1$-equivariant tubular neighbourhood of that fixed point set, and so its characterized by its behaviour near the fixed point set, which is linear.

Hmm, this part seems likely to not be true. Apparently there are actions of circles on discs with precisely two fixed points in the interior. Ref I don't have access to the paper from home so I haven't looked at anything other than the first page, but this appears to be a non-linear action. Perhaps in the reference you're referring to they're content with a local result rather than a global result? You might want to check the wording carefully. Locally it's certainly true by the above argument.

share|improve this answer
    
Thank you for the correction. Yes, that is precisely what I wanted. Since your answer, I found the reference I needed. Apparently this is a theorem due to Brouwer and independently Kérékjartó. A modern exposition is here: arxiv.org/pdf/math/0303256 Thank you very much for your time, and I hope it wasn't too much trouble getting around my imprecise statement. –  Chu Aug 29 '12 at 4:32
    
Your references are for only the $2$-dimensional case. –  Ryan Budney Aug 29 '12 at 18:27

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.