Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $G$ be a compact Lie group acting on a connected topological manifold $M$ with boundary. Suppose the action on one boundary component is trivial. Does it follow that the action on the whole of $M$ is trivial as well?

If $M$ and the action map $G\times M\to M$ are smooth, it is not too difficult to show that the answer is positive. Indeed, let $X$ be the set of all fixed points $x\in M$ of the action such that the action of $G$ on $T_xM$ is trivial. This set is closed, so it suffices to show it is open and non-empty (since $M$ is connected). To do so take a Riemannian metric on $M$ and average it to get a $G$-invariant metric. Using this one can show that $X$ contains the boundary component on which $G$ acts identically, so $X$ is non-empty. Moreover, if $x\in X$, then any $g\in G$ acts identically on a neighborhood of $x$ since $g(exp(v))=exp(dg_x v)$ for all $v$ in a sufficiently small neighborhood of $0\in T_xM$.

However, this argument uses smoothness and it is not clear if it can be adapted to the topological case.

share|improve this question
    
In applications: do you really ever have Lie groups acting non-smoothly on non-smooth manifolds? +1 for the question, BTW. –  Theo Johnson-Freyd May 15 '11 at 23:00
    
Theo -- thanks. Re applications: I don't have any off hand, but I think there may be examples of the following sort: take a finite group acting on freely and continuously on a smooth manifold; then it may turn out that the quotient will have properties (e.e. non-smoothable?) which quotients by smooth actions can't have. –  algori May 15 '11 at 23:21

1 Answer 1

Yes, it follows that the action of $G$ on all of $M$ is trivial. In brief this follows from what is known as "local Smith theory." Replace M by the union of $M$ and an open boundary collar on which $G$ acts as the product of the action on the boundary with a trivial action in the collar parameter. Then one would have an action on a connected manifold without boundary that is the identity on an open set. If $G$ is a $p$-group for some prime $p$ then local Smith theory says that each component of the fixed point set has the local mod $p$ (Cech) cohomology $H^{*}(F,F-\{x\};\mathbb{Z}_{p})$ groups of a manifold. It follows that the component of the fixed set containing the fixed boundary collar is all of $M$, plus the collar. For a general compact Lie group $G$ the kernel $K$ of the action is a closed subgroup that contains all elements of prime power order. This is enough to imply that $K=G$, i.e., that $G$ acts trivially.

share|improve this answer
    
Dear Allan -- thanks a lot for the answer. However, I was wondering if you could clarify a couple of points: 1. why can we assume that the group acts trivially on the collar parameter? 2. is there a reference for the local Smith theory? –  algori May 16 '11 at 13:32

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.