By a faithful action of a topological group G on a topological manifold M, we mean a continuous injection $G\to \operatorname{Homeo}(M)$ (where $\operatorname{Homeo}(M)$ has the compact open topology). One equivalent way of expressing HilbertSmith Conj is:

Conjecture. There is no faithful action of $\mathbb Z_p$ on any connected $n$-manifold $M$.

My question is: why this is possible? for example $Z_p$ acts on 2-sphere as rotation along the $z-$axis. Why does this action does not count as faithful?


I think you are mistaking $\Bbb Z_p$ (which is actually the group of $p$-adic integers) for $\Bbb Z/p$ (the integers mod $p$). There is indeed a faithful $\Bbb Z/p$-action on $S^2$ given by rotating $2\pi k/p$ radians about the $z$-axis for $k \in \Bbb Z/p$.

The above example of a faithful $\Bbb Z/p$-action on $S^2$ does not contradict the Hilbert-Smith conjecture. A standard way of stating the Hilbert-Smith conjecture is as follows:

If a locally compact topological group $G$ acts faithfully on some connected $n$-manifold $M$, then $G$ is a Lie group.

We are fine with the case of $\Bbb Z/p$ since it is just a discrete Lie group.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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