Let $p:E \to X$ a topological covering of connected space $X$. Fix a basepoint $x_0$ of $X$ and denote by $\pi(X,x_0) $ the fundamental group of $X$.

The monodromy action of $\pi(X,x_0) $ on $p^{-1}(x_0)$ is defined by lifting a loop representant from $\pi(X,x_0) $ starting in a choosen $y_0 \in p^{-1}(x_0)$.

If $E,X$ are nice enough the lifting theorem garantees the uniqueness of this lift.

I often read that some extra conditions (...like locally path-connectedness and so on) for $X$ garantee also the existence of a simply connected universal cover $X_U$ such that every cover $E$ of $X$ with nice enough properties obtains a map $f_E: X_U \to E$.

By construction of $X_U$ there exist a map $\pi(X,x_0) \to Aut(X_U \vert X)$ taking into account the choice of base point $x_0$.

Futhermore $f_E$ induces a map $Aut(X_U \vert X) \to Aut(E \vert X)$ and therefore we obtain a map $\pi(X,x_0) \to Aut(E \vert X)$.

My question is if there exist a way to visualize or understand intuitively/geometrically what this map does.

Indeed the action of $\pi(X,x_0)$ of the fiber $p^{-1}(x_0)$ has a very intuitive & geometric interpretation as explaned above. Therefore I have keen interest to find out how the automorphism $a_{\gamma} \in Aut(E \vert X)$ induced by a loop $ \gamma \in \pi(X,x_0) $ looks like. Can it be visualized and what intuition lies behind that?

  • $\begingroup$ Here do you assume that $p:E\to X$ is a normal covering? I'm not entirely sure but I think the induced map as you mentioned do not exist if $E$ is not normal. Maybe you can describe how is that map defined? $\endgroup$ – lEm Apr 14 at 20:03
  • $\begingroup$ @lEm: I'm not sure. $X_U \to X$ is of course normal (=Galois). The only assumption that I demand for $E$ is that $E$ is an intermediate cover between $X_U$ and $X$. in other words $f_E$ exists as map of $X$-covers. If this condition can only be fulfilled when $E$ is normal that I assume it also. $\endgroup$ – Tim Grosskreutz Apr 14 at 20:19
  • $\begingroup$ What lEm meant was that if $E$ is not normal, then there is no induced map $Aut(X_U\mid X)\to Aut(E\mid X)$. For your question, assuming $E$ is indeed normal,then it's just $X_U/G$ for some discrete group $G$ acting very nicely on $X_U$, and this $G$ happens to be $\pi_1(X,x_0)/p_*\pi_1(E,*)$, so that gives some geometric way of viewing it $\endgroup$ – Max Apr 14 at 20:24
  • $\begingroup$ @Max: The universal cover $X_U$ consists of homotopy classes of paths starting at $x_0$ and the action of $\pi_0(X)$ is given by just composing these paths with the loops. So the action on the fundamental group on $E$ is just that of $X_U$ and then quotienting out $X_U \to X_U/G$. Formally it's clear. the problem is that generally $X_U$ is unfortunately a not really geometrically transparent space. When we quotient out some $G$ such that $E=X/G$ become a "nice visualisable" cover (for example connected double cover of $S^1$ or something else what we can "draw") and take an arbitrary point $\endgroup$ – Tim Grosskreutz Apr 14 at 21:16
  • $\begingroup$ @Max: such that $e \in E$ lying over a $x \in X$ what how does a $\gamma \in \pi_1(X)$ concretely act on it? Do you mean that in following way: Let $\alpha$ be a path from $x_0 \to x$ representing $e$ wrt $X_U \to X_U /G$. then we map $\gamma$ by $\alpha^*: \pi_0(X,x_0) \to \pi_0(X,x)$, lift $\alpha^*(\gamma)$ with initial point $e$ and take the final point of the lift? sounds good (if it is correct way) but the path $\alpha$ is stays still mysterious to me. $\endgroup$ – Tim Grosskreutz Apr 14 at 21:25

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