Riemann surfaces from meromorphic functions

MatheinBoulomenos

Jun 27, 2018 23:00

@MikeMiller okay, so the starting point for me was this exercise we did in NT that goes as follows: Let $k$ be any field, then any discrete valuation on $k(t)$ is either given by the $f$-adic valuation where $f$ is an irreducible polynomial (works like the p-adic dude) or by the "valuation at infinity" which sends $f/g$ to $\operatorname{deg}(g) - \operatorname{deg}(f)$ (this corresponds to measuring the vanishing/pole order at infinity for $k=\Bbb C$ and $K$ meromorphic functions on the sphere)

Ted Shifrin

Understood 1000%, Mike.

MatheinBoulomenos

oh I forgot a condition

it's trivial when restricted to $k$

that's the function field analog of Ostrowski's theorem

Mike Miller

Allow me to head to my laptop to read, I will be back in 2 or 3 minutes - you can keep writing

MatheinBoulomenos

but for $k=\Bbb C$ any irreducible polynomial is of course just $z-\lambda$, so we get a bijection between points on a sphere and discrete valuations on its function field

So let's do this construction first for the Riemann sphere and then talk about general compact Riemann surfaces later

Mike Miller

I'm with you

MatheinBoulomenos

Jun 27, 2018 23:04

Suppose we're given the function field $\Bbb C(t)$ (just as an abstract field, we don't know the generator $t$, but we know how $\Bbb C$ sits inside that

then we can define the Riemann sphere as a set as equivalence classes of discrete valuations (or you can just take normalized ones) that are trivial on $\Bbb C$

Mike Miller

Not sure how to parse. You have a field $K/\Bbb C$ which you know is 1-generated but you have not chosen the generator?

MatheinBoulomenos

yes

just there exists one

but we don't need it for the construction

Mike Miller

@MatheinBoulomenos I guess what I really mean here is 'why?'

MatheinBoulomenos

@MikeMiller because discrete valuations "see" the points in the algebraic structure of the field

by the Ostrowski result, every discrete valuation measures zeroes or poles at a unique point

Mike Miller

No, I'm fine with that, but why is this the Riemann sphere?

You've identified the function field to start somehow?

MatheinBoulomenos

Jun 27, 2018 23:07

I'm starting with the function field

Mike Miller

Oh, I'm sorry

I see now what you were saying above

MatheinBoulomenos

and I want to build a Riemann surface out of that

Mike Miller

$\text{val}(k) = \Bbb P^1(k)$

MatheinBoulomenos

well, valuations on $k(t)$ that are trivial on $k$, but yeah

Mike Miller

I was just being lazy in notation, I got you

MatheinBoulomenos

Jun 27, 2018 23:08

okay

So the idea is if we have this set of "points", we can make sense of elements in our abstract field as actual functions on that

Mike Miller

Okay, so this is how we define $\Bbb P^1$

Is the point that function fields for other Riemann surfaces are going to be pretty much arbitrarily complex?

MatheinBoulomenos

their valuations are going to be as simple as for the Riemann sphere :)

but I want to reduce that to the sphere

Mike Miller

Hmm

MatheinBoulomenos

so before I get into extension of valuations theory, I want to do the construction for the sphere

Mike Miller

Wait, is every Riemann surface rational?

Surely not

MatheinBoulomenos

Jun 27, 2018 23:11

It's a finite ramified covering of the sphere, which means that the function field is a finite extension of $\Bbb{C}(x)$

that's enough

Mike Miller

Ah I see, ok, that's what I wanted

It's virtually rational ;)

MatheinBoulomenos

valuations behave in controllable ways under finite (separable) extensions

Mike Miller

Aha!

MatheinBoulomenos

But I wanted to talk about how to evaluate our elements of the abstract field at points first

Mike Miller

Ok

MatheinBoulomenos

Jun 27, 2018 23:13

So we have an element of $f \in K$ and we want to get an actual function from it on our "set of points", so we want to define from $f$ a function that sends valuations to points in $\Bbb C \cup \infty$

no that's wrong

now it should make sense

So with $f$ fixed, and a valuation $v$, if $v(f) < 0$, $f$ has a pole, so $f$ should send $v$ to $\infty$

Mike Miller

Ok, sure

MatheinBoulomenos

if $v(f) \geq 0$, then $f$ is an element of the valution ring $\{ g \in K \mid v(g) \geq 0\}$ which is a local ring with maximal ideal $\{ g \in K \mid v(g) > 0\}$

the quotient of that is isomorphic to $\Bbb C$ (and in a canonical way, as $\Bbb C$ is a subring of everything)

Then the value in that case is just the image under the quotient map $\{ g \in K \mid v(g) \geq 0\}/\{ g \in K \mid v(g) > 0\} = \Bbb C$

Mike Miller

Wow!

Wow!!

MatheinBoulomenos

So now when we can treat the elements of $K$ as actual functions, we can just take the initital topology wrt to the standard topology on $\Bbb C \cup \infty$ to get a topology for our set of valuations

I haven't been able to prove that this gives us the topology we want but I'm quite sure this should work out

Mike Miller

If you can prove that at points where the map isn't branched I believe it should be not hard in general

MatheinBoulomenos

Jun 27, 2018 23:23

and then for general Riemann surfaces, the idea is that the function field is a finite extension of $\Bbb C(z)$ and you use that you can understand how discrete valuations behave under finite extensions to see that the discrete valuations (if we start with a Riemann surface) correspond to zero-or-pole order again

loch

I think at the very least you can probably recover the zariaki topology?

MatheinBoulomenos

@loch yeah, if you take the initital topology wrt Zariski topology on $\Bbb P^1$, then that's what you should get

Mike Miller

@loch I think that a circular proof works: If you start with $K = \text{Func}(\Sigma)$ one can check that the topology is the manifold topology

Leaky Nun

@loch what do I get if I start with an arbitrary ring?

taking the functions and modulo the obvious relation

loch

Yeah - i also dont think you used anything special about complex numbers so far - so i am unsure how you can get the euclidean topology without something like analytification

@LeakyNun well now you cant really make sense of adding two functions defined on different opens - since their intersection might be disjoint :p

Mike Miller

Jun 27, 2018 23:27

He said initial topology w/r/t standard topology on $\Bbb{CP}^1$

MatheinBoulomenos

@loch but we can take the analytic topology on $\Bbb C \cup \infty$, the set in which our functions take values when we take the inital topology!

loch

Oh

Mike Miller

It is easy to see how to analytify the latter fella

And we transfer along that

I think it's quite clever

Leaky Nun

@loch oh we need it to be reduced?

loch

Yeah i think that makes sense

MatheinBoulomenos

Jun 27, 2018 23:29

yeah and if you have a discrete valuation on a larger field than $\Bbb C(z)$ (but still a finite extension), you restrict it down to $\Bbb C(z)$ that gives you a discrete valuation on that, which we understand and then there's some general theory that tells you how discrete valuations behave under finite extensions (I learned that in ANT, which is probably the place most persons encounter this valuation stuff)
This valuation stuff also describes ramification: the ramification of a point depends on the number of ways you can extend the valuation, relative to the degree of extension

loch

@LeakyNun you need your scheme to be irreducible

Leaky Nun

what does that mean downstairs?

loch

Nonreduced is fine.. you just dont get a field

There is a unique minimal prime

Leaky Nun

hmm

MatheinBoulomenos

@MikeMiller yeah that was just some algebraic musings that I thought up for RS without really knowing the theory, I think the usual thing you do is going from function fields to algebraic curves and then analytification? This way is direct, without needing algebraicity

Mike Miller

Jun 27, 2018 23:32

@MatheinBoulomenos I wonder if it's possible to construct the Teichmuller space (or at least the moduli space of Riemann surfaces) algebraically, starting with the function field

$Mathematics$ Mathematics

Participants