where a and b could be half integers as well?

but with $a, b$ in $\frac{1}{2} \Bbb Z$

right

gotcha

I'm doing the $y^2 + 5 = x^3$ has no integer solutions thang

yeah

passing to Z[(1+sqrt(-5)/2] instead of Z[sqrt(-5)]

@BalarkaSen I'm back

from my deadline fighting

@ÍgjøgnumMeg what cool NT did I miss

o/

just

pretending to be able to compute class numbers

@BalarkaSen play?

I think I reached my quota today lol

hahaha

@BalarkaSen just use your next day's quota

I have to prove $y^2 + 5 = x^3$ has no integer solutions for ANT1

i need to be awake enough to see the board bro

lmfao @BalarkaSen

How are you reducing to $\Bbb{Z}[\sqrt{21}]$ from $\Bbb{Z}[\sqrt{147}]$?

nah he's just saying all he was doing was computing the class group of Q(sqrt(21))

sqrt(147) is 7sqrt(3) my dude

Oh, ok

so what are we computing (read: looking up on LMFDB) now?

I guess $\Bbb Q(\sqrt{21})$

lol @ÍgjøgnumMeg forgot to check if it's square free?

ye

hahaha

lmfao

nice prank 10/10

rip rip

unintentional i promise

I'll just say that $\Bbb Z[\sqrt{21}]$ is the wrong ring

yeah

it is

Yeah

It's a non-maximal order

funny

@BalarkaSen What's the significance of this here?

banter

compute class group of $\Bbb Q(\sqrt{9})$

nothing i say is of significance

@ÍgjøgnumMeg just mod 7

I'm probably being dumb, but I don't think $\Bbb{Z}[\sqrt{21}]\supset \Bbb{Z}[\sqrt{147}]$?

@Ted I did the square root wrong

rofl

Oh ok haha

;)

but 21 is the more interesting question is all

It's 6am here

I thought you were using some fancy techniques

just look it up on LMFDB

you said that 5 times now

So we want to find the class number for $\Bbb{Q}(\sqrt{21})$

ok

you have a great site to cheat with ok we get it

try and get a paper published and it's just a list of class numbers for real quadratic fields copied out of LMFDB

@BalarkaSen just 3 actually

@Ted this should be quite easy since the Minkowski bound is just over 2

So the discrim is 21?

precision

true number theorist

discriminant is $21$ because $21 \equiv 1 \bmod 4$

So only $3\Bbb Z$ and $7\Bbb Z$ ramify

you don't need to care about 3 and 7 because 7>3>2

rofl

$\Bbb Q(\sqrt{21})$ has class number... spoilers :P

I don't know the minkoski bound :P

Minkowski bound is $\frac{n!}{n^n}\left(\frac{4}{\pi}\right)^{r_2}\sqrt{\lvert d_K\rvert}$

where $r_2$ is the number of pairs of complex embeddings of $K$

$n$ is degree of $K$

So there is 1 pair in our case?

$d_K$ is discriminant

so $\dfrac{2!}{2^2} \left(\dfrac4\pi\right)^0 \sqrt{21}$

@TedE no there's only real embeddings

So $(2/pi)\sqrt{21} \approx2.9$

Oh right

so sqrt(21)/2 ~ 2.29

@Ted it's real quadratic

Oops :P

Minkowski bound is $2.29$ ish so you need only look at primes lying over $2$

could check that $a^2 - 21b^2 = 2$ has no solution in half integers or smth

there are none

nise

What theorem are you using to make sense of 'you only need to look at primes lying over 2'

doesnt split

and by "reciprocity" you're just factoring $x^2-x-5$ over $\Bbb F_2$

How does the minkowski bound relate to the class number?

which is irreducible

so $e=1$, $f=2$, $g=1$

@Ted this is from Minkowski theory

@ÍgjøgnumMeg clearly superior to your "half-integer" method

applicable to monogenic number fields

and the fact that fractional ideals factorise uniquely into products of primes of your numbre field lel

@Leaky ye suprerieorer

@TedE wiki: every class in the ideal class group of K contains an integral ideal of norm not exceeding Minkowski's bound

@ÍgjøgnumMeg no this is such a cool trick

also screw you guys for distracting me

We did no such thing

rofl

i have to do topology

i have forgotten topology

You brought this on yourself for not checking if $147$ was square free

I might go to the petrol station and buy Leibniz cookies

@Ted :(

:D

wtf is a Leibniz cookie

and then made it worse by not being able to take square roots

They're small

@Balarka it's a cookie that invented calculus first

is it like

you eat the cookie

vey small cookies

and suddenly you have calculus

nah that's a Newton cookie

@BalarkaSen a topology is a collection of subsets of a given set that is closed under finite intersection and arbitrary union

@LeakyNun You're missing an axiom

no I'm not

@Leaky I posit that your definition is wrong

empty set bro

empty set is empty union

@ÍgjøgnumMeg I posit that you are wrong

closed under arbitrary intersection and finite union

smirk

no

(measure theory lol)

chunter

Zariski or smth

I'm just being a bellend

his open sets are closed

@ÍgjøgnumMeg isn't it an elliptic curve

I coined the word unabgeschloffen in my topology seminar a few weeks ago

so just find its Jacobian or something

@Leaky yeah it is

hahaha I think that would be too high powered for ANT 1

nobody cares

use etale cohomology if you can

the guy marking my pset would care

hahahaha

how do you make Pic^0(X) a variety naturally

tensor product

e l a b o r a t e

no that's the group law lmao

Show that the existence of a solution would imply the existence of an elliptic curve that was not modular

lmfao

Pic^0(X) isn't necessarily a variety

Depending on your definition of a variety

Picard scheme on nLab

for a Riemann surface X of genus g at least 1, Pic^0(X) is isomorphic to (C/Z^2)^g

so it's believable you can give it some abelian variety structure in a natural way

what if g=0

Well in that case its an abelian variety easily

@LeakyNun then it's zero. Boom!

@TedE Sure, but I think you can say something general for any arbitrary variety X

Also I don't actually know a direct isomorphism Pic^0(X) -> (C/Z^2)^g. I can derive it using Pic^0(X) = H^1(X, O_X) and running a sheaf cohomology long exact sequence

@BalarkaSen stacks project is not responding so...

An arbitrary smooth projective variety if you take variety to be integral separated finite type over an algrabaically closed field maybe

alas

No I am not a total loony

I mean a smooth projective variety over C

Then it seems believable

Brb gotta eat food

@BalarkaSen stacks project tag 0B9R

I can't even view it on firefox

maybe we'll have to wait until stacks is up again

ah this works: web.archive.org/web/20190622095032/https://…

tag 0B9R

also "the genus of $X$ is $g = \dim_k H^1(X, \mathcal O_X)$"

ye

I meant Pic(X) = H^1(X, O_X*) above

and the theorem that Pic is a scheme is 0B9Z

excited to learn some alg geo in semester 3 lol

which hopefully you can explain to me :P

Let me think for a bit before looking around. There should be some geometric way to understand this

what's the deal with Pic(E) again?

why is it isomorphic to E?

@ÍgjøgnumMeg @BalarkaSen

Points on E gives rise to divisors on E, and you can build a line bundle out of that

Addition of points correspond to tensor product of line bundles

And if you have like a line joining P, Q, R, then P + Q + R is a principal divisor (f)

f being the meromorphic function given by the line

principal divisors correspond to the trivial line bundle

so the group structures on E and Pic^0(E) are exactly the same

yeah how do you build a line bundle

here's how you do it for a Riemann surface for an instructive example. Let D be a divisor on X; so it's a collection of points with multiplicities on it

choose a cover of X such that every point is contained in a unique open set, so say D has points x1, ..., xn and they are contained in U1, .., Un

Choose a function fi on Ui which has (zero or pole, depending on sign) order ord(xi)

choose constant functions on all other open sets

Consider disjoint union of Ui x C and glue them along (Ui cap Uj) x C by the transition functions fi/fj : Ui cap Uj -> C^* = GL_1(C) - these form a cocycle by construction

this gives your line bundle, with a ready-made section defined by the fi's which vanishes on D with the exact multiplicities given

interesting

so if X=CP^1 and D=[0] then what do I do?

Good example. Take the two affine opens, and consider z on the lower hemisphere and 1 on the upper hemisphere

Verify that the line bundle you get is O(1)

Line bundles on CP^1 are classified by degree of the transition function C^* -> C^* given the standard cover by affine opens, in fact

that's why you have an O(n) for every n

Hi @TedShifrin

hmm

Ok, I am starting to see the picture

what's the picture

Consider the exponential sequence $1 \to \Bbb Z \to \mathcal{O}_X \to \mathcal{O}_X^* \to 1$, run the LES to get $0 = \mathcal{O}_X^*(X) = H^0(X, \mathcal{O}_X^*) \to H^1(X, \Bbb Z) \to H^1(X, \mathcal{O}_X) \to H^1(X, \mathcal{O}_X^*) \to H^2(X, \Bbb Z) = \Bbb Z$. The last map is actually the degree map $\text{Pic}(X) \to \Bbb Z$, so taking the kernel we have $0 \to H^1(X, \Bbb Z) \to H^1(X, \mathcal{O}_X) \to \text{Pic}^0(X) \to 1$.

By Serre duality, $H^1(X, \mathcal{O}_X) = H^0(X, \Omega_X^*)$, $\Omega_X$ being the sheaf of holomorphic 1-forms

So $\text{Pic}^0(X)$ is really $\Omega_X(X)^*/H_1(X, \Bbb Z)$, where I suppose $H^1(X, \Bbb Z)$ is embedded in $\Omega_X(X)^*$ by the nondegenerate pairing $\Omega_X \times H^1(X, \Bbb Z) \to \Bbb C$, integrating a 1-form over a 1-cycle

Fix a point $x \in X$. I will show there is a map $X \to \Omega_X^*$ which is well-defined upto $H^1(X, \Bbb Z)$. This is defined simply by taking any other point $y$, and integrating a form $\omega$ over a path joining $x$ and $y$.

This gives the map $X \to \Omega_X^*/H^1(X, \Bbb Z) = \text{Pic}^0(X)$

Somehow $\text{Pic}^0(X)$ has a nice variety structure, but I don't see why yet.

There should be some way to understand why a Riemann surface of genus g admits a map to a 2g-torus in a nice way

It's like abelianizing but at the level of the space

What does it do topologically? Take a surface of genus g, pinch to get a wedge of g tori, and embed it in the 2-skeleton of T^{2g}?

I realize I am using upper star for multiplicative sheaf and dual both... sigh

also i meant $H^0(X, \Omega_X)^*$

I'm still figuring out what on earth $1 \to \Bbb Z \to \mathcal O_X \to \mathcal O_X^\ast \to 1$ means

Oh that map $O_X \to O_X^*$ is taking a germ of a holomorphic function to exponential of it

$O_X^*$ is sheaf of nonvanishing holomorphic functions, under multiplication

Trivia: I am always queasy about writing the exponential short sequence sequence; really it should be $0 \to \Bbb Z \to \mathcal{O}_X \to \mathcal{O}_X^\times \to 1$ :P

I looked around a bit and apparently what I did was wrote down the Abel-Jacobi map

In dimension 1 the Picard group is the Jacobian variety, which I didn't know how to define before but it's exactly $\Omega_X^*/H_1(X; \Bbb Z)$

In general there's an Albanese variety, which is a universal abelian variety in the sense that any map from X to A factors through Alb(X) for any abelian variety A

It's exactly the abelianization huh

The Riemannian manifolds section explains this quite well!

am I missing something, I don't see why it's surjective

like what if $X = \Bbb C^\times$ and $f(z) = z$

log(z) is locally a holomorphic function whenever z is nonzero

you only care germ-level, right?

oh!

nvm I haven't seen sheafs for a long time

ok it's alright now

It's all good

what's $H^1(X,\Bbb Z)$?

So whenever you have a short exact sequence $1 \to F \to G \to H \to 1$ of sheaves on $X$, you have a long exact sequence of cohomology groups valued in those sheaves. In the case of the exponential sequence $\Bbb Z$ was the constant sheaf with values in $\Bbb Z$, the cohomology groups of the constant sheaf with values in $\Bbb Z$ are just plain vanilla cohomology groups with $\Bbb Z$ coefficient

So $H^1(X, \Bbb Z)$ is our friendly neighborhood singular cohomology group

(The theorem is that Cech cohomology with constant sheaf = singular cohomology with coefficient group of the sheaf)

oh right

what's the map $H^1(X, \mathcal{O}_X) \to \Omega_X^\ast(X)$?

Let $E$ be the sheaf of smooth $1$-forms on $X$. Since you can break up any $1$-form on $X$ in $(z, \overline{z})$-coordinates, you can write everything as $fdz + gd\overline{z}$. The differential map $d$ also breaks up as $\partial + \overline{\partial}$ in this process. So call $E^{(0, 1)}$ the sheaf of $(0, 1)$-forms, of the form $g d\overline{z}$.

For any smooth function $f$ on $X$, you can take $df = df/dz dz + df/d\overline{z} d\overline{z}$, so $\overline{\partial f} = df/d\overline{z} d\overline{z}$ is the $(0, 1)$-part

This gives a map $C^\infty_X \to E^{(0, 1)}$ at the sheaf level, which you can prove is surjective (a small PDE trick), and the kernel is $\mathcal{O}_X$ as $\overline{\partial} f = 0$ is exactly the CR equations

Run the sheaf LES on $0 \to \mathcal{O}_X \to C^\infty_X \to E^{(0, 1)} \to 0$, using that C^infty is a fine hence acyclic sheaf, to conclude $H^1(X, \mathcal{O}_X) = E^{(0, 1)}/\text{im}\,\overline{\partial}$

So $H^1(X, \mathcal{O}_X)$ can be thought as equivalence classes $[\eta]$ of germs of $(0, 1)$-forms. Using this you can define the pairing $H^1(X, \mathcal{O}_X) \times \Omega_X(X) \to \Bbb C$, $(\omega, [\eta]) \to \int_X \omega \wedge \eta$

The hard theorem of course is that this is a perfect pairing, which is what Serre proved (in a much greater generality)

That'll give your isomorphism $H^1(X, \mathcal{O}_X) \to \Omega_X(X)^*$

lol

Yeah there I meant the dual lol

cool

Honestly the proof of Serre duality in Forster is a living nightmare

I skip proofs so it's alright

You can't tell what's happening but at the end of the day you have your result anyway

I am just saying in case you read the good Hodge theory proof someday and teach it to me; future investment

how did Serre prove it

lol

Oh I think he proved in some massive generality over Cohen-Macaulay schemes

that's the statement Hartshorne has with Ext sheaves flying around everywhere

it's horrifying

math.uchicago.edu/~may/REU2013/REUPapers/Fung.pdf

this is pure horror

lol

Milne has a construction of the Jacobian variety in chapter 3 of his Abelian Variety notes

I should read that some time. It looks involved

He's thinking about the group of divisors $\text{Div}_n(X)$ of degree $n$ on a Riemann surface $X$, then observing that the map $X^n \to \text{Div}_n(X)$ given by $(x_1, \cdots, x_n) \mapsto \sum [x_i]$ is a bijection modulo $S_n$, so essentially $\text{Div}_n(X) \cong X^n/S_n$. This is the $n$-th symmetric power of $X$, $SP^n(X)$, which is well known to be a complex manifold

So $\text{Div}_n(X)/\text{Princ}(X)$ is a quotient of $\text{SP}^n(X)$, and he argues this is going to be smooth as well, but I don't follow.

Something something

I don't get it

Oh, I see, not what I said. For high enough $n$, compared to the genus of $X$, every divisor class in $\text{Pic}_n(X) = \text{Div}_n(X)/\text{Prin}(X)$ contains an effective divisor, so the map $X^n \to \text{Pic}_n(X)$ I described is surjective.

How is $(xy)^{2/3}$ (a function with discontinous partials) differentiable at (0,0)?

In particular $\text{SP}^n(X) \to \text{Pic}_n(X)$. Then there's some complicated argument to conclude $\text{Pin}_n(X)$ is also smooth; he assumes this has a section, in which case he constructs an appropriate fibered produt, so on (and then proceeds to further complications).

And then the key fact is $\text{Pic}_0(X)$ and $\text{Pic}_n(X)$ are still in bijection for any $n$, so you get natural variety structure on $\text{Pic}_0$ once you get in on $\text{Pic}_n$ for high enough $n$

So complicated!

almost one of the most tragic sentences I have ever read: "An example of a galactic algorithm is the fastest known way to multiply two numbers,[2] which is based on a 1729-dimensional Fourier transform.[3][4] This means it will not reach its stated efficiency until the numbers have at least $2^1729$ digits, which is vastly larger than the number of atoms in the known universe. So this algorithm is never used in practice. "

@Archer Having continuous partials is sufficient — but not necessary — for the function to be differentiable. The derivative at $(0,0)$ is $0$ and you can check the definition: $\lim_{(h,k)\to (0,0)} \frac{f(h,k)}{\sqrt{h^2+k^2}} = 0$.