Recovering a covering space via monodromy representation

Question

Let $V$ be a connected real manifold with a fixed basepoint $q$. For a $d$-fold connected covering $\pi:\tilde{V}\to V$, its monodromy representation is defined as follows: Let $\pi^{-1}(q)= \{x_1,\dots,x_d\}$. For $[\gamma]\in \pi_1(V,q)$, let $\tilde{\gamma}_i$ be the lift of $\gamma$ with $\tilde{\gamma}_i(0)=x_i$. Then $\tilde{\gamma}_i(1)=x_{\sigma(i)}$ for some $\sigma(i)\in \{1,\dots,d\}$. Define $\rho:\pi_1(V,q)\to S_d$ by $[\gamma]\mapsto \sigma$. Using homotopy lifting property we can see that this map is well-defined. Also, if we define the group structure of $\pi_1(V,q)$ by $\alpha *\beta : =\beta\cdot \alpha$ (where $\beta\cdot \alpha$ is the usual concatenation of paths), then $\rho$ becomes a group homomorphism.

Conversely, suppose we are given a homomorphism $\rho:\pi_1(V,q)\to S_d$. Let $H$ be the subgroup $\{[\gamma]\in \pi_1(V,q):\rho([\gamma])(1)=1\}$ and let $\pi:(\tilde{V},\tilde{q})\to (V,q)$ be the covering corresponding to $H$ (so that $\pi_* \pi_1(\tilde{V},\tilde{q})=H$). Then why is the monodromy representation of $\pi$ equal to $\rho$ (up to conjugation)? This is asserted in p.89 of Miranda's book Algebraic Curves and Surfaces. I see that $H$ is an index $d$ subgroup, and thus $\pi$ is a $d$-fold covering, but I can't see why its monodromy is $\rho$.

Edit. Fix a basepoint $p\in \tilde{V}$ with $\pi(p)=q$. The group $\pi_1(V,q)$ acts on $\tilde{V}$ as follows: for $[\gamma]\in \pi_1(V,q)$ there is a unique deck transformation $\varphi_{\gamma}:\tilde{V}\to \tilde{V}$ such that $\varphi_{\gamma}(p)=\tilde{\gamma}(1)$ where $\tilde{\gamma}$ is the lift of $\gamma$ with $\tilde{\gamma}(0)=p$. For a point $x\in \tilde{V}$, define $[\gamma]\cdot x=\varphi_{\gamma}(x)$. This defines a well-defined action of $\pi_1(V,q)$ on $\tilde{V}$.

For a subgroup $H\subset \pi_1(V,q)$, the covering corresponding to $H$ is defined to be the orbit space $V_0/H$ where $V_0$ is the universal cover of $V$.

Answer 1

Strictly speaking, it doesn't make sense to say the monodromy action is equal to $\rho$ (even up to conjugation) as they are two actions on different sets. The correct notion of equivalence is isomorphism of group actions, defined as follows. If $G$ acts on both $X$ and $Y$, the two actions are isomorphic if there is a bijection $f: X \to Y$ such that $g\cdot f(x) = f(g\cdot x)$. You should convince yourself that this definition describes the notion of two actions being "the same".

If both sets are numbered, we sometimes conflate the actions with the isomorphic actions on $\{1,\ldots, d\}$, whereupon action isomorphism turns into conjugation. This can sometimes confuse matters, so let's be explicit: we are looking for a bijection between the fiber $\pi^{-1}(x)$ and $\{1,\ldots, d\}$ which commutes with the two actions of $\pi_1(V,q)$ (respectively, the monodromy action and the action $\rho$, given by $g\cdot x = \rho(g)(x)$).

The high-level explanation is that both the monodromy action and $\rho$ are isomorphic to the action of $\pi_1(V,q)$ on the cosets of $H$ (given by $b\cdot aH = (ba)H$).

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The monodromy action is isomorphic to the action on cosets.

I will compose paths right to left, so that $ah \in aH$ is first $h$, then $a$. Then the lift of $ah$ based at $\tilde{q}$ (which I denote $\widetilde{ah}_{\tilde{q}})$ is the concatenation of $\tilde{h}_{\tilde{q}}$ (a loop in $\tilde{V}$) and $\tilde{a}_{\tilde{q}}$. In particular, the lift of every $ah \in aH$ based at $\tilde{q}$ has the same endpoint, namely $\tilde{a}_{\tilde{q}}(1) \in \pi^{-1}(q)$. This gives a function $f:G/H \to \pi^{-1}(q)$ by $f(aH) = \tilde{a}_{\tilde{q}}(1)$. Any $y \in \pi^{-1}(q)$ can be connected to $\tilde{q}$ by a path $\tilde{\gamma}$, and $f(\pi(\tilde{\gamma}) H) = y$, so $f$ is moreover bijective.

$b \in \pi_1(V,q)$ acts on $x \in \pi^{-1}(q)$ by sending it to $\tilde{b}_x(1)$. When $x = f(aH) = \tilde{a}_{\tilde{q}}(1)$, we get $$b\cdot f(aH) =b\cdot \tilde{a}_{\tilde{q}}(1) = \tilde{b}_{\tilde{a}_{\tilde{q}}(1)}(1),$$ which is a complicated way of saying "first lift $a$ based at $q$, then lift $b$ based at the endpoint of $\tilde{a}_{\tilde{q}}$, and take the endpoint of that". But by uniqueness of lifts, this is the same as just taking the endpoint of $\widetilde{ba}_{\tilde{q}}$. Then $$b\cdot f(aH) = \tilde{b}_{\tilde{a}_{\tilde{q}}(1)}(1) = f((ba)H) = f(b\cdot aH).$$ Then the two actions are isomorphic, as claimed.

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The action on cosets is isomorphic to $\rho$.

The final piece of the puzzle is pure group theory—if $\rho: G \to S_n$ is a transitive* action (via $g\cdot x = \rho(g)(x)$), then it is isomorphic to the action on the cosets of the stabilizer $H = \text{Stab}(1)$. The isomorphism is given by $j(aH) = a \cdot 1$, and I'll leave it to you to verify it is indeed an isomorphism of actions (it's essentially a much easier version of the above).

Then $j\circ f^{-1}: \pi^{-1}(q) \to \{1,\ldots,d\}$ is an isomorphism between the monodromy action of $\tilde{V}$ and $\rho$.

*If $\rho$ is not transitive, then the statement is false, as the monodromy action of a connected covering space is always transitive.