Let $X$ be a topological space ""good enough"" , let $\tilde{X}$ be its universal covering (suppose that $X$ has one), then for every connected covering $P$ on $X$ with fiber $F$, it is true that the group of deck transformation $Aut(\tilde{X})$ acts on $F$?

I want to solve this problem using the properties of $\tilde{X}$ to be connected and a normal covering and the fact that it covers all the connected coverings on $X$. So I don't want to mention explicitly the monodromy action of the fundamental group.

How can I do that?


1 Answer 1


Almost always, these are two different actions.

The monodromy action is by path lifting, lifting a given path at each of the points in the fiber and mapping each such point to the endpoint of the corresponding lift. This makes sense for any covering space $q:(Y,y_0) \to (X,x_0)$, where $Y$ is possibly disconnected.

To get a well-defined action by the deck group $G=G(\widetilde{X})$, which can be identified with $\pi_1(X,x_0)$ in the universal cover $p:(\widetilde{X},\widetilde{x}_0) \to (X,X_0)$, on the fiber of an intermediate connected normal cover $q:(Y,y_0) \to (X,x_0)$, certain elements need to identified: $G(Y) \cong G/N$, where $N= q_*\pi_1(Y,y_0)$. So, there is an action of $G$ on $F = q^{-1}(x_0)$ that factors through the action of $G/N$ on $F$.

Here's an example where these two actions are different:

Let $X = S^1 \vee S^1$ with oriented labeled edges $x$ and $y$, identified with generators of $\pi_1(X)$. And let $Y = \widetilde{X}$ be its universal cover, viewed as a graph (4-valent tree) with labeled oriented edges and its vertex set identified with $\pi_1(X)$.

Consider the pair of adjacent vertices $(1,y)$. The monodromy action by the element $x$ sends this pair to $(x,yx)$, which are not adjacent in $Y$. The deck group action by the element $x$ sends the pair to $(x,xy)$.

In general, the deck group action extends to a homeomorphism of $Y$. But the monodromy action need not extend continuously.

The above example is related to other actions: if $Y$ is a normal covering graph with fundamental group $N \trianglelefteq G$, where $G$ is a free group, then $Y$ is the right Cayley graph of the presentation $G/N$. The action of $F/N$ by deck transformations is the left action (by graph automorphisms). The monodromy action is the action on the vertex set by right multiplication (which does not extend to an action by graph automorphisms).

Exercise #27 in Chapter 1 of Hatcher's Algebraic Topology book is worth a look.

  • 1
    $\begingroup$ So if $π(X)$ is abelian then the two actions are same because in this case the left action is the same thing as right multiplication? $\endgroup$
    – user622044
    Dec 19, 2018 at 2:54
  • $\begingroup$ I am five years too late but I don't think this is necessarily true. Hatcher turns the right monodromy action on the fiber to a left action via $$[\alpha]\cdot f = f \cdot [\bar \alpha]^{-1}$$ for all $f\in p^{-1}(\{x_0\}), [\alpha]\in \pi_1(X,x_0)$. More explicitly, $[\gamma] \cdot f$ is equal to $\tilde{\bar\gamma}(1)$, where $\tilde{\bar\gamma}$ is the lift of $\bar\gamma$ starting at $f$ in which case the action of $x$ would send the pair $(1,y)$ to $(x^{-1},yx^{-1})$, no? (Assuming you are naming the vertices of the graph in a left-to-right manner) $\endgroup$
    – Y. Tamer
    Jan 19 at 23:42
  • $\begingroup$ If you're wondering when these two actions are the same (i.e. Hatcher exercise 1.3.27) here's an outline. First prove that both of these actions are sharply transitive, then prove that if two transitive actions agree on one point then they are the same. Lastly, prove the following: $\pi_1(X,x_0)$ is a $2-$group if and only if the actions agree on $\tilde {x_0}$. $\endgroup$
    – Y. Tamer
    Jan 20 at 1:05

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