Monodromy groups from Galois's viewpoint

Question

According to the Wikipedia article about monodromy, the monodromy group can be defined in terms of Galois theory in following way:

Let $F(x)$ denote the field of the rational functions in the variable $x$ over the field $F$, which is the field of fractions of the polynomial ring $F[x]$. An element $y = f(x)$ of $F(x)$ determines a finite field extension $[F(x) : F(y)]$.

This extension is generally not Galois but has Galois closure $L(f)$. The associated Galois group of the extension $[L(f) : F(y)]$ is called the monodromy group of $f$.

In the case of $F = \mathbb{C} $, Riemann surface theory enters and allows for the geometric interpretation given above. In the case that the extension $[\mathbb{C}(x) : \mathbb{C}(y)]$ is already Galois, the associated monodromy group is sometimes called a group of deck transformations.

This has connections with the Galois theory of covering spaces leading to the Riemann existence theorem.

Furthermore, it is explicitly remarked that for the field $F = \mathbb{C} $, this definition of monodromy coincides with the classical one in light of complex analysis and cover theory.

My question is: how, explicitly, do these two definitions of monodromy correspond to each other for $F = \mathbb{C} $? How do we obtain a cover map giving rise to the topological monodromy action from the field extensions above?

Answer 1

All of the following can be found in the third chapter of Szamuley's "Galois Groups and Fundamental Groups" but I will try to sum it up a bit:

I think the starting point of making this precise is the fact that the function field $\mathbb{C}(t)$ is isomorphic to the field of meromorphic functions on the Riemann sphere $ \textbf{P}^1(\mathbb{C})$. In fact, mapping a Riemann suface to its field of meromorphic functions, gives a contravariant functor $M(-)$ from the category of Riemann surfaces to the category of commutative rings. Now it is shown in Theorem 3.3.7 of Szamuely that if we restrict this functor to the category of connected compact Riemann surfaces equipped with map to $ \textbf{P}^1(\mathbb{C})$, then $M(-)$ induces a contravariant equivalence between this category and the category of finite field extensions of $M(\textbf{P}^1(\mathbb{C})) \cong \mathbb{C}(t)$.

Now it is shown before in Szamuely that if we have a compact Riemann surface and a map $\phi:X \rightarrow \textbf{P}^1(\mathbb{C})$, then there is a finite set of points $S \subseteq \textbf{P}^1(\mathbb{C})$ such that $\phi^\prime: \phi^{-1}( \textbf{P}^1(\mathbb{C}) \setminus S) \rightarrow \textbf{P}^1(\mathbb{C}) \setminus S$ is in fact a covering in the usual sense and this assertion also gives and equivalence of categories between compact Riemann surfaces mapping holomorphically onto $ \textbf{P}^1(\mathbb{C})$ and so called branched coverings of $ \textbf{P}^1(\mathbb{C})$ (i.e. continuous maps that are coverings outside of a finite set of points).

So we have identified finite field extensions of $\mathbb{C}(t)$ with branched coverings of $ \textbf{P}^1(\mathbb{C})$. It can also be seen that under this correspondences a finite Galois extension $\mathbb{C}(t) \hookrightarrow L$ corresponds to a Galois (branched) cover $Y \rightarrow \textbf{P}^1(\mathbb{C})$. So we get an isomorphism $Gal(L,\mathbb{C}(t)) \cong Aut(Y \rightarrow X)$ between the galois group and the automorphism group of a Galois covering. Now it is a standard fact in covering theory that a covering is already completely determined by the monodromy action. In particular we get an isomorphism between our Galois group and the automorphism group of the monodromy action on the fibre of our covering.

Answer 2

This may not be as explicit as what you're looking for, but:

Given a smooth projective curve $X$ over $\mathbb{C}$, its field of rational functions $k(X)$ is a finitely generated extension of transcendence degree 1 over $\mathbb{C}$.

On the other hand, section $I.6$ of Hartshorne's Algebraic Geometry describes a way to give the discrete valuations $\nu : K \to \mathbb{Z}$ on a finitely generated extension $K/\mathbb{C}$ of transcendence degree 1 the structure of a smooth projective curve $X$ over $\mathbb{C}$. By the way, this doesn't work in higher dimensions (which is why finding birational models with desired properties (minimal models, canonical models, etc.) is such an industry).

In fact, that section of Hartshorne shows taking the function field (the assignment $X \mapsto k(X)$ gives a contravariant equivalence of categories between smooth projective curves over $\mathbb{C}$ with dominant morphisms to finitely generated transcendence degree 1 extensions of $\mathbb{C}$. This equivalence of categories allows us to relate, for instance, automorphisms of an extension $k(X)/k(Y)$ with automorphisms of the corresponding dominant morphism $f: X \to Y$. In other words, the equivalence of categories gives us an isomorphism $$ \label{eq:1} \tag{1} \mathrm{Aut}(k(X)/k(Y)) \simeq \mathrm{Aut}(X/Y) $$

If $\{p_1, \dots, p_r\} \subset Y$ are the points where $f$ is ramified and $V : = Y \setminus \{p_1, \dots, p_r\}$, and if $U : = f^{-1}(V)$, then the restriction $f|_U: U \to V$ will be a (connected) covering map, and automorphisms of $X$ over $Y$ will restrict to deck transformations of $U$ over $V$: this gives a natural homomorphism $$ \label{eq:2} \tag{2} \mathrm{Aut}(X/Y) \xrightarrow{\mathrm{res}} \mathrm{Aut}(U/V) $$ To see that $\mathrm{res}$ is an injective, note that if $\varphi: X \to X$ is an automorphism of $X/Y$ and $\varphi|_U = \mathrm{id}|_U$ on $U$ then $\varphi = \mathrm{id}$ since $X$ is a variety and $U \subset X$ is a (dense) open. So $\mathrm{res}$ is injective.

On the other hand, given an automorphism $\psi$ of $U/V$, consider the composition $$ U \xrightarrow{\psi} U \xrightarrow{\iota} X $$

Where the second map is the inclusion. Since $X$ is projective and $X \setminus U$ has codimension 1, $\iota \circ \psi$ extends to a map $\bar{\psi}: X \to X$, and one can check $\bar{\psi}$ is an automorphism of $X/Y$. This shows $\mathrm{res}$ is surjective, hence an isomorphism. Putting \eqref{eq:1} and \eqref{eq:2} together gives an isomorphism $$ \mathrm{Aut}(k(X)/k(Y)) \simeq \mathrm{Aut}(U/V) $$