Constructing 6-fold cover of $S^1 \vee S^1$ with deck transformation group $\cong S_6$

Question

So i'm thinking that this will be a cover space of maximum possible symmetry. Will a "necklace" of 6 circles work? Any tips appreciated

Answer 1

The comments are correct that this construction is impossible, however the reason is flawed.

First of all, the action of the deck transformation group $G$ of a connected covering map $f : X \to Y$ a free action, meaning that the action of a nontrivial element of $G$ has no fixed points --- so no, a deck transformation cannot take $y_2$ to $y_3$ whilst fixing $y_1$.

Second, given $x \in X$ and $y=f(x) \in Y$, the action of $G$ on the subset $f^{-1}(y) \subset X$ need not be transitive. In fact, that action is transitive if and only if the covering map $f : X \to Y$ is regular, if and only if the image of the induced homomorphism $\pi_1(X,x) \to \pi_1(Y,f(x))$ is a normal subgroup of $\pi_1(Y,f(x))$. However, this only makes the construction more impossible (if that means anything), because it implies that the cardinality of $f^{-1}(x)$ is greater than or equal to the order of the group $G$.

So, in a covering map of degree $6$, the set $f^{-1}(x)$ has cardinality $6$, and there is no free action of a group of order $6! = 120$ on a set of cardinality $6$.

For proofs of the various things discussed in this answer, I suggest sitting down with a good solid course on the relationship between fundamental groups and covering spaces, as in Hatcher.

Answer 2

There is a $6!$-fold connected normal cover of $S^1\vee S^1$ with deck transformation group $S_6$. Recall that $S_6$ has a presentation with two generators, hence there is a normal subgroup $N\subset F_2$ of the free group on two generators. Let's identify $F_2=\pi_1(S^1\vee S^1)$. Hence, there is a homomorphism $f:\pi_1(S^1\vee S^1)\to S_6$ through the quotient map $F_2\to F_2/N$.

By standard facts about covering spaces (as seen in Hatcher) there is then a connected normal cover of $S^1\vee S^1$ associated to the normal subgroup $N$ for which $S_6$ is the group of deck transformations.

You might know this covering space as the Cayley graph of $S_6$ with the chosen generating set. In fact, the two-generator Cayley graphs of $S_6$ encompass the set of all connected normal covers of $S^1\vee S^1$ with deck transformation group isomorphic to $S_6$.

(William in the comments points out that there is a $6$-fold disconnected cover of $S^1\vee S^1$ with $S_6$ as its group of deck transformations: the disjoint union $\coprod_{i=1}^6S^1\vee S^1$.)