# Monodromy action

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?

• 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? – lEm Apr 14 at 20:03
• @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. – Tim Grosskreutz Apr 14 at 20:19
• 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 – Max Apr 14 at 20:24
• @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 – Tim Grosskreutz Apr 14 at 21:16
• @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. – Tim Grosskreutz Apr 14 at 21:25