Can an action of a compact Lie group be nontrivial if it is trivial on the boundary? - MathOverflow

Can an action of a compact Lie group be nontrivial if it is trivial on the boundary?

2011-05-15T21:06:09Z

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.

Answer by Allan Edmonds for Can an action of a compact Lie group be nontrivial if it is trivial on the boundary?

2011-05-15T23:58:29Z

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.