How to check if a cover is normal given the action on a fiber.

Question

Assume you have a space $X$ and a homomorphism $r:\pi_1X\to S_n$. Then $r$ defines a n-sheeted covering space $Y$ of $X$. My question is, can you immediatly determine if $Y$ is a normal covering using only $r$ (without computing $\pi_1Y\subset\pi_1X $ and checking if it is a normal subgroup.

Example: $X=S^1\vee S^1$, and let $a,b$ denote the two generators of $\pi_1 X$. Look at the r defined via $a\mapsto (12),\ b\mapsto(123).$ This gives rise to a three sheeted cover $Y$ over $X$. Is this cover normal?

Again, I can check it because in this case it is very easy to draw $Y$ and it is possible to calculate $\pi_1Y$ by using Van Kampen. My question is - can you see if $Y$ is normal or not only by looking at $r$.

Answer 1

Let $p:Y\rightarrow X$ be the covering. Then the orbits of the $\pi_1(X)$ action (via $r$) correspond to the connected components of $Y$. So let's assume this action is transitive, so that $Y$ is connected [I don't know what normality would mean when $Y$ was disconnected].

Then $p_\ast\pi_1(Y)\le\pi_1(X)$, and $rp_\ast\pi_1(Y)$ is a point stabilizer, since the point stabilizer is precisely which loops lift to loops in $Y$.

So for $p_\ast\pi_1(Y)$ to be normal, the point stabilizer of $r\pi_1(X)$ must be normal. But then it stabilizes all points (since point stabilizers are conjugate), and so $rp_\ast\pi_1(Y)$ must be the identity subgroup.

Thus, $p:Y\rightarrow X$ is a normal covering precisely when the action of $\pi_1(X)$ (via $r$) is regular. I guess that's one reason why a normal covering is also called regular.

Answer 2

Let $G=\pi_1(X)$ and $H=p_*(\pi_1(Y))$. The group $G$ acts on the set $G/H$ of cosets by left multiplication. Deck transformations of $Y$ correspond to $\operatorname{Aut}_G(G/H)$, which are functions $G/H\to G/H$ that commute with the $G$ action. The group $\operatorname{Aut}_G(G/H)$ is isomorphic to $N(H)/H$, where $N(H)$ is the normalizer of $H$ in $G$. (This automorphism group corresponds to the group of deck transformations of $Y\to X$.)

In your specific problem, we have $[G:H]=3$. The homomorphism $f:\langle a,b\rangle\to S_3$ given by $a\mapsto (1\ 2)$ and $b\mapsto (1\ 2\ 3)$ corresponds to a group action $G\to \operatorname{Bij}(G/H)$ through some identification of the set $G/H$ with the three-element set $\{1,2,3\}$, whatever $H$ might be. In any case, $H$ is normal iff $\lvert\operatorname{Aut}_G(G/H)\rvert=3$. That is, iff $\lvert\operatorname{Aut}_G(\{1,2,3\})\rvert=3$, with $G$ acting on $\{1,2,3\}$ via $f$.

Suppose $h:\{1,2,3\}\to\{1,2,3\}$ is such an automorphism. Then $ah(1)=h(a1)=h(2)$, and since this means $h(1),h(2)\in \{1,2\}$, we deduce $h(3)=3$. Already this means there are at most two automorphisms, so $H$ is not normal.

It is not clear if this is any better than constructing the covering space and seeing the lack of symmetry.