0
$\begingroup$

Given topological spaces $X$ and $Y$ and a covering map $p: X \rightarrow Y$, we know that the group $\pi_1(Y,y_0)$, where $y_0\in Y$, acts on the fiber $F=p^{-1}(y_0)$.

Also, we know that the set of all action $G\times M \rightarrow M$ of a group $G$ on a set $M$ is in a bijective correspondence with the set of all group homomorphisms $G\rightarrow \text{Sym}(M)$. Here $\text{Sym}(M)$ is the symmetric group on the elements of $M$.

In particulat, the action of the fundamental group of $X$, as any action of a group on a set, by the above, induces the group homomorphism $\pi_1(Y,y_0)\rightarrow \text{Sym}(F)$. Similarly, by the above, any group homomorphism $\pi_1(Y,y_0)\rightarrow \text{Sym}(F)$ must induce the action of $\pi_1(Y,y_0)$ on the fiber $F$. But I was told that this is not true. The question is why? In such a case, what additional requirements one must impose on the homomorphism $\pi_1(Y,y_0)\rightarrow \text{Sym}(F)$ in order that it induce the action on the fiber? Is it true that the homomorphism must be transitive? Does it have something to do with the Riemann Existence theorem? (If you refer to this theorem, please use its statement presented here.)

|cite|improve this question
$\endgroup$
  • $\begingroup$ with all due respect. I am not sure about your question. if you have for instance a universal cover then the $\pi_1(Y)$ is the group of deck transformations and so permutes the fibers. But I can't understand at the moment what's your problem, excuse me. Can you help me? $\endgroup$ – user321268 Jun 23 '16 at 21:25
  • 1
    $\begingroup$ if $X$ for instance is connected, locally path-connected (and so is also $Y$) then this action is always free. If it is transitive (like the case where the covering is universal) then we call the cover regular. $\endgroup$ – user321268 Jun 23 '16 at 21:27
  • $\begingroup$ @mayer_vietoris It seems to me that it is always transitive. Regular transformations are such that some deck transformation (acting on $Y$) takes a point on $F$ to another point on $F$. But if you just choose two points on $F$ to be connected by a $\pi_1$-action, you only need $Y$ to be connected. $\endgroup$ – Peter Franek Jun 23 '16 at 21:44
  • $\begingroup$ @PeterFranek as far as I know the transitivity of an action is a special property among the covers. For instance, for sufficiently "nice" topological spaces regular covering spaces are in an one-to-one correspondence with the normal subgroups of its fundamental group. If you choose one space whose fundamental group is a simple group, then the above fails. $\endgroup$ – user321268 Jun 23 '16 at 21:48
  • $\begingroup$ Just choose a path in $Y$ connecting these two points $x,y\in F$ and project it to $X$: this gives you the element of $\pi_1$ that sends $x$ to $y$. (Not every action of $\pi_1$ on $F$ extends to a deck transformation) $\endgroup$ – Peter Franek Jun 23 '16 at 21:51
1
$\begingroup$

The first problem is that the homomorphism $\pi_1(Y, y_0) \rightarrow \mathrm{Sym}(F)$ could be too small. For instance, it could be the trivial homomorphism in a setting where the action of $\pi_1(Y,y_0)$ on $F$ is not always the identity permutation.

A minimal example of this is the (helical) double-cover of $S^1$. Here, $F$ is two points and both the identity and "swap" actions are induced by $\pi_1(Y, y_0)$. However, there are two homomorphisms $\pi_1(Y, y_0) \rightarrow \mathrm{Sym}(F)$. The faithful one (the one with trivial kernel) does induce the action. The trivial one does not.

Even more exciting, $\mathrm{Sym}(F)$ might be large enough to contain two copies of the action of $\pi_1(Y, y_0)$, only intersecting at the identity. Then the homomorphism $\pi_1(Y, y_0) \rightarrow \mathrm{Sym}(F)$ can have trivial kernel and still map $\pi_1(Y, y_0)$ to a bunch of elements of $\mathrm{Sym}(F)$ that have no relation to the permutations induced by the action. (This happens pretty quickly. Symmetric groups tend to have lots of conjugate, nearly disjoint (, small) subgroups.)

I haven't thought about conditions which would restrict the image of the homomorphism $\pi_1(Y, y_0) \rightarrow \mathrm{Sym}(F)$ to the subset of $\mathrm{Sym}(F)$ induced by the action of $\pi_1(Y, y_0)$, but I'm not convinced it can be much different than "the image of the homomorphism is exactly the set of permutations induced by the action".

|cite|improve this answer
$\endgroup$

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.