I am currently reading a paper, where the authors claim that for a finite regular cover $Y\to X$ of a sufficiently nice space $X$ (e.g. closed compact surface), the Deck transformation group $G$ acts on $\pi_1(X)$.
How would this action look like? Is it possible that this is a typo and they meant $\pi_1(Y)$ instead?
If not, I am wondering how the Deck transformation group would act on $\pi_1(X)$. Thanks for any reply!
The first case to think about could be to consider $Y \to X$ being a universal covering map. In this case, the fundamental group of $Y$ is trivial and the group of deck transformations is isomorphic to the fundamental group of $X$, so you can think of $G$ acting on $\pi_1(X)$ by using this isomorphism.
Now, you are asking about a finite, regular cover. Because of the regularity (sometimes called normality), we get that the group of deck transformations is a quotient of the fundamental group of $X$. To be more precise (I learned this with Hatcher, Algebraic Topology: see Section 1.3), given a covering $\phi:Y\to X$, it is regular precisely when $Deck\cong \pi_1(X)/\phi_*(\pi_1(Y))$. This suggests the notation "normal", because we would require that $\phi_*(\pi_1(Y))$ is a normal subgroup of $\pi_1(X)$ to make sense of the quotient. The finiteness relates to the index of $\phi_*(\pi_1(Y))$ in $\pi_1(X)$ being finite. This means $Deck$ is finite.
Unfortunately, we cannot say much more, so let´s try to devise an example to show what can go wrong. Let us consider the covering $p:\mathbb{S}^1\to\mathbb{S}^1$, $z\mapsto z^p$. We can think of the map making us travel the circle $p$ times faster, as if we had divided the covering circle in $p$ parts of equal length and mapped each one to the base circle. We get $Deck\cong \mathbb{Z}/p\mathbb{Z}\cong \mathbb{Z}_p$. We however cannot find a general natural way of acting on $\mathbb{Z}$ that is non trivial: if an element acts on a non-trivial way, because our acting group has order $p$, applying the action $p$ times needs to takes us back to where we started (compatibility condition of a group action). E.g. for $p=3$, if an element $a\in Deck$ acts via $a.g$ for $g\in \pi_1(X)$, we need $a^3.g=a.(a.(a.g))=g$ and you can check that is not possible given $\pi_1(X)\cong \mathbb{Z}$, unless $a.g=g$.