Mathematics - 2018-06-27 (page 4 of 4)

loch

22:05

@LeakyNun the fraction field = field of all regular functions defined on some open set of your scheme- is that what youre thinking of?

Leaky Nun

22:26

@loch yes

Mike Miller

@MatheinBoulomenos I am impressed that Miranda seems to dodge all analysis and actually prove everything

I will have to look at some point to see how Serre duality works

Maybe he manages to take the sheafy proof and write it down very cleanly or something because they're curves

MatheinBoulomenos

@MikeMiller I really want to read Miranda as my first serious encounter with Riemann surfaces, hopefully soon

we did mention some basic stuff in modular forms and single complex variable 1&2, but not much

Hey @Ted

Ted Shifrin

Heya Mathein

Mike Miller

Where I am right now the library has all of the GSMs, GTMs, EMSs, etc in order

So I am looking through them to see what's good reading

MatheinBoulomenos

@MikeMiller pretty cool

Mike Miller

22:39

GSM 37: Farkas-Kra - Theta Constants, Riemann Surfaces, and Number Theory seems pretty cool

MatheinBoulomenos

the title seems cool

Mike Miller

The hyperbolic version of Elliptic function thelry

With applications at the end to partition functions

MatheinBoulomenos

In the preface, Miranda mentions an "adelic proof" of Riemann Roch which sounds really cool

I know adeles from NT

apparently he broke down the adelic proof

Mike Miller

I like adeles

Long ago at least

MatheinBoulomenos

it makes sense that you have adeles for compact riemann surfaces because of the function field analogy etc. and it also works for function fields over finite fields, but you don't really use that your coefficient field is finite

Mike Miller

22:46

I see!

MatheinBoulomenos

I think I came up with the construction how you can reconstruct a Riemann surface from its function field, but I can't really prove it gives you a complex manifold

loch

What is the adele ring of a compact riemann surface? I only know the one from nt

MatheinBoulomenos

and it's inspired by an exercise from an ANT class actually

@loch I imagine restricted product over completions of all valuations which are trivial on $\Bbb C$

@MikeMiller do you want to hear how I think going from the function field to the RS works?

I mean meromorphic functions by function field, I'm not going over algebraic curves as an intermediate step or something like that

loch

Hmm not really familiar valuations on function fields of a riemann surface - if im not being dumb though i think there are uncountably many of them ? (As opposed to countably many in the case of number fields / function fields over finite fields) - i suppose the definition still makes sense

MatheinBoulomenos

@loch the valuations correspond to points on the RS! you can prove that every valuation that is trivial on constant functions is given by measuring poles or zeroes at a point

loch

22:53

Oh yes of course

:p

MatheinBoulomenos

and it actually follows from the theory of extensions of valuations you would do in NT and the function field analog of Ostrowski which works for rational functions over any base field

Mike Miller

@MatheinBoulomenos What time is it for you? I would listen maybe later or tomorrow

MatheinBoulomenos

@MikeMiller 1:00 am

Ted Shifrin

for Mathein that's usually mid-afternoon :P

MatheinBoulomenos

but I'm not really sleepy

Mike Miller

22:55

Well, lecture starts in about an hour. I was going to have breakfast but nothing looks appetizing

So I suppose I have time

Ted Shifrin

breakfast at 4 PM?

or are you at a conference somewhere?

Mike Miller

I am at a summer school in Japan

Ted Shifrin

oh wow !! good for you!

Mike Miller

It is 8am

MatheinBoulomenos

@MikeMiller really cool!

Ted Shifrin

22:57

better eat something ... seriously

Mike Miller

We are out in the country on a nature preserve

The building is entirely made of wood

But the people are very good and it is a rejuvenating experience

I am a bit (what word to use...?) Disenchanted lately and it is nice to just talk about math with people and share pictures and ideas

Daminark

Hey everyone!

MatheinBoulomenos

Hi @Daminark

Ted Shifrin

Hi Demonark

MatheinBoulomenos

@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

23:00

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

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

23:05

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

I'm starting with the function field

Mike Miller

23:07

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

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

23:08

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

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

23:11

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

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

23:16

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

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

23:23

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

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

MatheinBoulomenos

23:27

@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

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

23:29

@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

@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

I guess more likely this constructs a moduli space of branched covers, which is like Riemann surface equipped with nonzero meromorphic function

Leaky Nun

23:33

@loch so if $\mathfrak p$ is the minimal prime, then I just get $A_{\mathfrak p}$?

Ted Shifrin

You can do that with the $j$-invariant for elliptic curves, @MikeM, but I dunno how you get the analytic parameters explicitly into the function field in general.

But I'm probably being silly.

Leaky Nun

hi @Ted

Mike Miller

@TedShifrin No sillier than I.

MatheinBoulomenos

@Ted have you read my construction? do you think it works?

Also hi @Ted

Ted Shifrin

Nah, Mathein, I'm chatting elsewhere and not following your lecture series :)

MatheinBoulomenos

23:35

@MikeMiller hmm, I don't know about that

@TedShifrin my idea was to (re)construct a Riemann surface from its function field by looking at discrete valuations as points and then showing how you can view an element in the field as a function on the set of the discrete valuations

and then take the initial topology wrt the analytic topology on $\Bbb P^1$

loch

@LeakyNun yep

MatheinBoulomenos

oh no wait, it's final topology actually

Leaky Nun

@loch can you give me a ring with two disjoint nonempty opens?

Mike Miller

Whatever, I assumed you chose the right one hehe

MatheinBoulomenos

no, initial is right, lol

loch

23:37

Product of rings

Leaky Nun

does $\Bbb Z^2$ work?

loch

Uh sure

Leaky Nun

hmm

loch

Two points is probably simpler- but why not :p

Leaky Nun

if $D(f) \cap D(g)$ is empty then $fg=1$???

Ted Shifrin

23:39

That sounds wrong.

Leaky Nun

there's nowhere that both f and g vanish on

Xander Henderson

Ugh... the internet sucks... I just want to know if I can eat the leaves off of my tree... WHY DOESN'T THE INTERNET KNOW IF IT IS SAFE?

loch

Aside- you can prove your spec is disconnected iff your ring can be written as a product

Leaky Nun

oh

$fg=0$

so I should pick $D(1,0)$ and $D(0,1)$

MatheinBoulomenos

@TedShifrin why do you think that it sounds wrong?

Leaky Nun

23:40

@MatheinBoulomenos because the correct one is fg=0

Ted Shifrin

No, @Leaky, the intersection is not where the product vanishes.

Leaky Nun

oh?

Ted Shifrin

The union is where the product vanishes.

Leaky Nun

D is not vanish

Ted Shifrin

Canonical construction of reducible.

Oh, ...

Leaky Nun

23:41

D for Does not vanish :P

MatheinBoulomenos

@Ted oh, you didn't mean my construction with "that sounds wrong"

Ted Shifrin

serves me right for butting in.

no, @Mathein.

I'm too rusty on this kind of stuff to be of any value.

(pun intended)

Leaky Nun

there's nowhere that fg does not vanish <=> fg vanishes everywhere <=> fg=0

MatheinBoulomenos

@MikeMiller this construction doesn't really choose a particular meromorphic function on the surface I think. I need to use the existence of a meromorphic function, i.e. the existence of an embedding of $\Bbb{C}(z)$ into the function field (such that it's a finite extension) to show that given a Riemann surface, valuations actually correspond to points, but once I have that I can forget about the choice of function again and just work with valuations on an abstract field

Mike Miller

@Mathei The complex structure is only dependent on the function field?

MatheinBoulomenos

23:43

add some "non-constant"s in this statements for it to make sense

@MikeMiller hmm, no I fix the embedding of $\Bbb C$, that's true

the category of fields is finitely generated extensions of $\Bbb C$ of transcendence degree $1$

Mike Miller

But somehow maybe the choices should be parsmeterizable

Or the difference not too heavy

@Ted Trust me, I get it

MatheinBoulomenos

ah, I don't really know about this moduli space/ Teichmüller stuff

Mike Miller

Me neither, but I suppose that's a bit blase to say if I am asking about it

loch

@LeakyNun * fg is nilpotent

Mike Miller

@Ted My brain has rotted enough already I am not looking forward to 40 years from now

Well, there are many reasons I am not looking forward to 2060

Leaky Nun

23:50

@loch we have projection maps $\pi_i : R_1 \times R_2 \to R_i$

Ted Shifrin

@MikeM ... I don't even remember your name! :D

Leaky Nun

whose upstairs is $\operatorname{Spec}(R_i) \to \operatorname{Spec}(R_1) \sqcup \operatorname{Spec}(R_2)$

Mike Miller

It's hard, with all these pseudonyms

Leaky Nun

@loch are you saying that if I have a global section in $R_1 \times R_2$, then I actually have two independent functions defined on $R_1$ and $R_2$?

@MatheinBoulomenos can you make morphism of sheaves concrete? What I mean is that if I want to make sheaf concrete, I would consider the sheaf of analytic functions on C

i.e. on each open set we have a ring of functions defined there

also I feel like $\mathcal O_{\Bbb C}(B(x,r))$ are all non-canonically isomorphic

Mike Miller

Holomorphic maps between Riemann surfaces induce sheaf homomorphisms

Though one must be careful what sheaves

Leaky Nun

23:56

so e.g. $z \mapsto z^2 : \Bbb C \to \Bbb C$?

ok not Riemann, but whatever

loch

@LeakyNun basically yes - although i guess it's probably nicer to phrase things more precisely

what do you mean by making a morphism of sheaves concrete?

anyway there is actually a third part of the iff, which will probably help you (if you are trying to show the iff)

Leaky Nun

in the same sense of making a sheaf concrete

which iff?

loch

spec A is disconnected <=> A = A_1 x A_2 <=> A has a non-0,1 idempotent

lol what does it mean to make a sheaf concrete?

MatheinBoulomenos

I don't understand the question, in what sense concrete?

Leaky Nun

just give an example

loch

23:59

oh i lied

Leaky Nun

@loch I've heard the morally correct proof of 1 => 3

maybe you also have

loch

i meant A has non-zero $e_1,e_2$ such that $e_1e_2 = 0, e_1^2 = e_1, e_2^2=e_2, e_1+e_2=1$

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