I want to study the field $\mathcal{M}(\Sigma_g)$ as Galois extension over $\mathcal{M}(S^2)$ when we have the information about the Galois group of this extension. Can we say the degree of extension of this extension or anything related to $\mathcal{M}(\Sigma_g)$?
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The order of the galois group is equal to the degree of the extension. – Ahr Feb 15 '21 at 19:10
1 Answers
The field of meromorphic functions on a Riemann surface depends not only on the genus of the surface $S$ but also its complex structure. Every closed Riemann surface is a smooth projective curve over $\newcommand{\CC}{\mathbb{C}}\CC$. (This is a consequence of the Riemann existence theorem; more generally, Chow's theorem generalizes this to higher-dimensional closed complex manifolds.) In other words, every meromorphic function on a closed Riemann surface is rational.
By the theory of algebraic curves (see, for example, Hartshorne's Algebraic Geometry, section I.6), the category of smooth projective curves over $\CC$ with surjective regular maps is equivalent to the category of function fields of transcendence degree $1$ over $\CC$ with $\CC$-homomorphisms. This tells us two relevant things:
- Two smooth projective curves have isomorphic function fields if and only if the curves themselves are isomorphic, i.e., there's a regular (locally polynomial) map between them with regular inverse.
- Embeddings of $\CC(t)$, the function field of the complex projective line $\mathbb{P}^1$, into the function field of a smooth projective curve $X$ correspond to surjective regular maps from $X$ to $\mathbb{P}^1$, and the degree of the function field as a field extension of $\CC(t)$ is the degree of the regular map.
In general, given a smooth projective curve $X$, there are many surjective regular maps $X \to \mathbb{P}^1$. The smallest degree of such a map is called the gonality of $X$, and is an important geometric invariant of the curve.
The relation between gonality and genus is somewhat complicated. For example:
- A smooth projective curve $X$ over $\CC$ has gonality 1 if and only $X$ has genus 0.
- If $X$ has genus 1 or 2, then $X$ has gonality 2. (A curve of gonality 2 is called hyperelliptic, or elliptic in the special case of genus 1. Elliptic curves are a huge area of study in their own right.)
- If $X$ has genus 3, then $X$ is either hyperelliptic (and is the completion of the affine curve given by an equation $y^2 = f(x)$ with $\deg(f) = 8$), or is given by a quartic equation in $\mathbb{P}^2$. In the latter case, $X$ has gonality 3 (such a curve is called trigonal).
The Galois groups can vary as well: Among the genus 3 curves of gonality 3, there are the Picard curves, given in an affine patch by an equation of the form $y^3 = f(x)$ with $\deg(f) = 4$, whose function fields are cyclic cubic extensions of $\CC(t)$. But a "generic" genus 3 curve's function field is an $S_3$ cubic extension of $\CC(t)$.
The picture gets increasingly complicated as the genus increases. In general, for genus $\geq 3$, the gonality $\operatorname{gon}(C)$ and the genus $g(C)$ are related by the inequality $$\operatorname{gon}(C) \leq \left\lfloor \frac{g(C) + 3}{2} \right\rfloor,$$ which is an equality for "generic" curve of any given genus (more precisely: the space of curves of a $g$ that have smaller gonality form a lower-dimensional subvariety of the moduli space of all curves of genus $g$). And of course many different Galois groups can occur among curves of a given gonality, just as many different Galois groups occur among number fields of a given degree.
This is an active area of research with plenty of open problems; there are many interesting questions you can ask about the gonality of algebraic curves. One other keyword that's useful for finding more about this is Brill–Noether theory.
EDIT: I should also mention that, unlike in the number field case, the degree over $\CC(t)$ isn't an intrinsic property of the function field but rather depends on the embedding. (Gonality is only concerned with the least possible degree.) So, for example, the function field $K$ of the elliptic curve with Weierstrass equation $y^2 = x^3 + 1$ (elliptic curve 36.a4 in the LMFDB) is isomorphic to the quadratic extension $\CC(t)(\sqrt{t^3 + 1})$ by projection onto the $x$-coordinate, but also to the cubic extension $\CC(t)(\sqrt[3]{t^2 - 1})$ by projection onto the $y$-coordinate, and many other realizations are possible. (As an even simpler example, every nonzero rational function gives an embedding of $\CC(t)$ into itself.)
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Is there any other technic that involves less algebraic geometry? I am trying to find the answer using topology or complex analysis. – Lokenath Kundu Feb 18 '21 at 17:08
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You can phrase it all in terms of Riemann surfaces if you prefer, but since there's this equivalence of categories between closed Riemann surfaces and smooth projective algebraic curves over $\mathbb{C}$, the answer will be the same. In particular, the field of meromorphic functions on a Riemann surface determines the Riemann surface up to biholomorphism (isomorphism as complex manifolds), and the minimum degree of $\mathcal{M}(S)$ over $\mathcal{M}(\mathbb{CP}^1)$ is the gonality of $S$. – Daniel Hast Feb 18 '21 at 17:17
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If you kindly help me to understand the field structure in terms of Riemann surface or suggest me a good reference that phrases the problem in terms of Riemann surfaces will be very helpful. – Lokenath Kundu Feb 18 '21 at 17:28
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A few good references are Forster's "Lectures on Riemann surfaces", Miranda's "Algebraic curves and Riemann surfaces", and Griffiths' "Introduction to Algebraic Curves". See also this MSE post: https://math.stackexchange.com/q/407004/41415 – Daniel Hast Feb 18 '21 at 17:38
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First of all, I want to say thank you. Now I am not finding any specific way of constructing irreducible polynomial of higher degree like 168 over $\mathbb{C}(t)$. Any kind of idea or suggestion would be very helpful. Thank you in advance. – Lokenath Kundu Mar 19 '21 at 05:47
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Maybe that should be a separate question (since I don't really see the connection to this one)? But you'd also need to be more precise about what you mean—for example, since $\mathbb{C}$ is algebraically closed, only linear polynomials in $\mathbb{C}[t]$ are irreducible. – Daniel Hast Mar 19 '21 at 14:25