Meanwhile, @Danu has disappeared ... Maybe he finished with his question.

No, I haven't.

Rats.

I'm just sitting here confused.

@TedShifrin Can I pay in upvotes/likes?

not now though, it's pretty late and it wouldn't be very wise to arrive with too little sleep to tomorrow's abstract algebra exam

Indeed, g'night, @Alessandro.

I only take cash.

good night to everyone

Me too, @MikeM.

Well, actually, I'll take payment through PayPal :P

Even though that organization seems to be run by a Trump supporter.

Indeed?

Peter Thiel is a nutcase.

Seems there's no shortage thereof.

@Danu: Why are you confuzled now?

Elon Musk is involved in PayPal too, wikipedia says.

@TedShifrin I just don't understand why it's true

Or what it has to do at all with that isomorphism

Which it? The isomorphism?

I thought Musk got kicked out of PayPal

Because it's going to give you an immersion (injective derivative)?

whaaa?

What's the derivative of the map into the Albanese?

Lets first go back to that integral

I'm supposed to understand this

language

Linearly independent $\omega$s will give you linearly independent tangent directions in the Albanese. Isn't that useful?

Shall I switch to French, @MikeM?

@MikeMiller You feel like the word "hell" is offensive? Welp...

s'il vous plait

Bien, d'accord.

Nous pouvons continuer en Francais, bien sur

You're confuzled enough in English, Danu.

Maybe it transcends language. We need Dutch?

My English is better than my Dutch.

Huh.

I haven't spoken Dutch much during the past 5.5 years.

My BSc. was also English-taught

Anyhow, what about this?

So that integral stuff is just $\int \sum f_i dx_i=\int \sum d F_i$ and then taking the derivative again, correct?

You can't write $dF_i$, kiddo.

Yeah, that's what I thought, but I don't see what else to do lol

I made a comment up there about how you actually prove it.

I don't get it---you said you pick a path that goes in the $x_i$ direction from $x$?

Along the $i$th coordinate axis ... the way you compute the partial with respect to $x_i$. :P

Quand est-ce que nous commencons parler en francais?

Are you talking about a path to integrate along, or for computing the limit that appears in the derivative?

Il est interdit de faire des fautes de grammaire, @dsillman :P

I thought you meant the former.

The latter, @Danu.

Which I don't understand.

Okay.

You were trying to understand why the Fund. Thm. of Calculus?

?

OK, now I'm confused. Are we trying to prove that the derivative is $\omega$ or are we moving on?

Generally a question mark doesn't qualify as a question.

Sleep, Balarka.

It qualifies as "I can't understand what that previous message meant" :P

@TedShifrin Former.

So I need to see that if I take the partial of $\int_{x_0}^x \sum f_i\,dx_i $ with respect to $x_i$ I get precisely the function $f_i$.

@BalarkaSen You learned this the hard way.

Sigh. Yes.

Therefore I take a path that moves on a line segment on the $i$th coordinate axis from $x$.

@TedShifrin Yeah, I'm doing that now.

Yeah, I now know what you meant---writing it out.

Or you could look at my multivariable math book, @Danu :P

Which I don't have access to :P

Our library doesn't even carry it

No, you need to know how to do this stuff.

Ted, I've always wondered what your profile picture is a picture of

I know it, in normal context. I am just having a bad time.

The new avatar, you mean, @dsillman?

I'm probably not good enough at it

but meh, it's fine usually

Yeah. It looks like an interesting solid

Yup. It's a great exercise :)

I give you a hint: It's a surface of revolution. What's revolving?

But what is it? What is it called?

He's posing a puzzle to you.

Hmmmm hang on

It hasn't a name of which I'm aware. But it has some very neat classical geometry happening in it.

I can see that

It like goes upward

Then a new segment goes diagonally horizontal

And then another goes back down

Well, there are cones on either end. What's in the middle?

@TedShifrin Why is the birational automorphism group of something of general type finite?

It's three line segments

What kind of surface is it, @dsillman?

I don't know how I would quantitatively describe it. Probably with like matrices or something

So is this okay? $$\lim_{h\to 0}\frac{1}{h}\Bigg(\int_{x_0}^{x+hx_i}\omega-\int_{x_0}^x\omega\Bigg)=(\dots) \int \limits_{x_0}^{x+hx_i} f_i dx_i-\int_{x_0}^x f_i d x_i$$

where $(...)$ is the limit stuff

Um, I don't remember, @MikeM.

@BalarkaSen Yes, you misunderstood it to mean "the opposite of general type"

Now if you parenthesize it correctly and combine integrals you should recognize the usual FTC in one variable, @Danu.

So no holomorphic $n$-forms, @MikeM.

General type is analogous to $g\ge 2$ for curves, @Balarka.

Oh, ok.

If we had a 1-parameter group of automorphisms, could we produce a non-trivial form?

@TedShifrin Yeah, that's what I had. So this should just be the definition of the derivative of the antiderivative of $f_i$ (w.r.t $x_i$)

So then back to $f_i$

I haven't thought about this in centuries, @MikeM, sorry.

@TedShifrin It could be discrete but infinite.

Right, @Danu.

Yeah, okay. Sorry for this (though I almost feel more sorry for myself for embarrassing myself like that in front of you all)

Moving on.

Now understand my comment about getting a $g$-dimensional subspace of the tangent space of the Albanese, @Danu.

Does $\mathbb R^*$ have any meaning? or is it likely a typo?

Nonzero, @Axoren.

or dual space ... depending on context.

I embarrass myself pretty often in front of everyone I don't want to embarrass myself.

It doesn't matter @Balarka---better on the internet than in front of my supervisor ;D

@MikeM: I think that sort of thing is in Kobayashi's book, and certainly in alg geo somewhere.

@TedShifrin Thanks. It's probably still a typo, but it might be less of one with that in mind.

What's the context, @Axoren?

True, @Danu

(# people Ted is helping =3)

Balarka quit on geometry, @Danu. He's supposed to be asleep.

I think he averages at 3

I was asked to show that $S^\star$ is a convex subset of $\mathbb R^*$, where $S^\star$ is the polar of a subset $S$ of $\mathbb R^n$

@TedShifrin Speaker just invoked it. Terry asked what we were all wondering: "Can a surface of general type have a positive entropy automorphism?"

@Danu He does multitask a lot.

Well, invoked it 10m ago.

Oh, sounds like a typo, @Axoren.

So @TedShifrin you mean the comment that linearly independent $\omega$ give linearly independent directions in $T\text{Alb}$?

Yes, @Danu.

It MUST be a typo, because $S^\star$ is definitely a subset of $\mathbb R^n$

Or is it $(\Bbb R^n)^*$, the dual space?

Which only one such subset is also a subset of the nonzero reals.

The polar may be defined in terms of linear functionals, @Axoren.

Right, hyperplanes ...

@TedShifrin In this case, it's defined over dot-product with members of $S$.

Oh, ok, using the isomorphism. It should be $\Bbb R^n$.

sorry, had to fix my internet. just got back

"What kind of surface is it"

hm

Are vectors of a vector space also members of the dual space?

I'm not sure how I'd say

I never really thought about it, but they also define linear functions of vectors to scalars, by the dot product.

Yeah, @Axoren, it can be done in terms of separating hyperplanes.

Have you been lately to a zoo of common surfaces, @dsillman?

So I guess you just made me compute the derivative (I still find it more natural to call it the differential but oh well) of the Albanese map, @TedShifrin. Is that right?

@Axoren No, but you can use an inner product to give an isomorphism.

Yes, @Danu.

Okay, I'm starting to understand stuff.

How terrible :D

@TedShifrin Is that just a matter of them having a different identity even though being functionally the same? I understand that $f(\vec x) = y'x$ is different from $y'$, but I just want to make sure it's for the reason I'm thinking.

Do you mean transpose?

Yeah, sorry. My professor doesn't like the $\ ^T$ notation

I used to like it so much; I've been unlearned.

The identification of $\Bbb R^n$ with $\Bbb R^n{}^*$ is given by transposing, yes.

Because $y'x = x\cdot y$.

If we identify $1\times 1$ matrices with the scalars.

Alright. I've been trying to get the hang of the dual space. It wasn't a concept I managed to get to in my linear algebra learning and it was never explicitly used until now.

I'm having a hard time translating this into the right words. If I have linearly independent $\omega_i$'s, then...

But our middle vector space is these, right?

So $D\text{alb}$ sends $\partial_k$ to $\omega\mapsto \partial_k \int \omega$ right

Sounds right.

@TedShifrin I'm going to depart for now, but I wanted to ask you about a question I'd ask you (answer whenever): Would you be able to explain the Frechet derivative and why it was useful (and even worked in the first place)? The Gateaux derivative makes more sense as to why it actually functions as a derivative.

I'm still not sure how to piece this together to what I want

Good-bye, all.

I never remember all the definitions, @Axoren. You'll have to remind me sometime.

Will do

@Danu: Remind me what we're trying to do.

I want an injective/surjective/bijective map $H^0(X,\Omega_X)\to H^0(\text{Alb},\Omega_\text{Alb})$

No, not quite.

The left is a vector space, the right is a torus.

Oh so sorry

just typo

But then what I said is right.

But global holomorphic $1$-forms on a torus come from the associated vector space.

How can you see that $T(z) = \frac{z-i}{z+i}$ will map the circle $|z-1|=1$ to another circle?

All linear fractional maps map circles to circles (or lines). Have you not proved this, @Lozansky?

Have you learned about cross-ratios?

Yes

But how do they know it's not a line?

So LFTs preserve cross-ratio, and circles/lines are characterized by real cross-ratio.

Is it because there are no singularities?

Does any point on the original circle get sent to infinity?

No okay that explains it

So @Danu, I'm claiming that the LHS is the tangent space to the Alb, so dimensions match up. Can we give a map?

@TedShifrin Okay, so what I need is $g$ linearly independent (at every point) elements of $H^0(X,\Omega_X)^*$, with (doubly) periodic things

So you need the general fact that the only holomorphic $1$-forms on a torus descend from constant-coefficient $1$-forms on the vector space.

I know that on $\Bbb C$, doubly periodic things are constant...

There you go.

And I guess then you just apply this to each "direction"

So lift a holomorphic $1$-form and argue.

You have $n$-fold doubly-periodic functions as coefficients of your 1-form upstairs.

So It has to be constant as we vary each coordinate separately right

Yup.

So then it's constant

Okay, so now we just need an injective map $H^0(X,\Omega_X)$ to its dual

But we're on compact Kahler right so doesn't the metric just work?

Well, we're supposed to see complex-linear isomorphisms, I assume,

There may actually be a missing dual in the original statement. This is always confusing.

So wait

Why aren't we using the Albanese map now?

Because I translated to $X$ instead of to Alb?

we should probably use that map like what you said before

Oh, there probably is a missing dual in what we were doing.

What page of the book is this on?

135, prop. 3.3.8

oh, he just says "induces a bijection"

So he's just pulling back forms on the Albanese by the Albanese map and counting dimensions.

But he writes $\cong$

Well he said bijection, not isomorphism. There may be a duality in there. Don't waste any more time on this.

This is one of the issues with complex geometry, that sometimes there are unexpected dualizations :P

Anyhow ... move on!

Still, though, we haven't established injectivity or anything

Dimension counting only suffices once you have an inclusion or something

No, my tangent vector comment establishes immersion or something.

But what's an immersion? $D\text{alb}$??

No, the map $X \to$ Alb.

Thats $\text{alb}$

OK.

But it's not so obvious to me that $\partial_k\mapsto(\omega\mapsto \partial_k\int \omega)$ is injective

What if there's just global forms that don't feature all $dz_k$'s?

That doesn't happen, of course.

Or else you'd have the wrong-dimensional torus.

But what I'm saying isn't quite right, anyhow.

So is it clear to you why there is a bijection, after all?

Are the partial derivatives of an integrable function necessarily integrable?

Yes, $H^0(X,\Omega^1)$ is (up to duality?) the tangent space of the Albanese.

No, @Simeon. That's why you need Sobolev spaces.

What if we're dealing with a holomorphic function?

@Simeon: On what sort of domain are you doing this?

And what do you mean by integrable?

I'm not sure :) I need to clear my thoughts before asking.

OK, @Danu, I'm quitting for now. Have fun!

Hmm.. Fun... :)

Thanks a lot for all the help!

hey @0celo7

hello

would you like to discuss a analysis question ?

If a the partial derivative of a continuous function is continuous with respect to the variable it was differentiated, is it continuous as a multivariable function?

as always, depends what it is

ok here it is

@Simeon probably not.

Let $X = (X,||.||_X)$ and $Y = (Y,||.||_Y)$ be banach spaces. Let T be a closed linear operator from X to Y. Consider the normed space $Z = (X, ||.||_u)$ where $||x||_u = ||x||_X + ||Tx||_Y$.

I would like to prove that Z is a banach space.

Closed in the usual topological sense or something special here?

so it is a linear map from X to Y which sends closed sets to closed sets.

Yes, that's what I mean by "usual."

yeah

Is closed necessary?

I am not sure. It seems we don't need closed

I must say I've not seen closed map as a hypothesis yet

Outside of quotient spaces

Maybe some other things that I'm forgetting.

hm

What do you have so far?

I am considering maybe some cauchy sequence

in Z

Sure, how else are you gonna prove it's complete :P

I was thinking of using something fancy

but maybe direct way would be better

Oh?

Show me fancy

but I couldn't find something fancy haha

haha

So let $\{z_n\}$ be a cauchy sequence in Z.

hm just a sec

So you have $$||z_n-z_m||_u<\epsilon$$

yeah

but then we can expand this

Or $||z_n-z_m||_X+||Tx_n-Tx_m||_Y<\epsilon$

yeah

Is $T$ bounded?

but that T is linear right ?

Does closed imply bounded?

@Adeek I already used that.

hm

Is a closed linear map bounded?

hm let me think

hm wait we need some fancy theorems

What book are you looking at?

I guess we have the open mapping theorem which states

my prof notes

If we have an open linear map

then it is automatically surjective

I would if that would help

hm

It is?

Any linear map between two normed space which is open is automatically surjective

yess

@0celo7 so what you said is true

@Adeek Proof?

banach closed theorem

Banach closed theorem states that T is closed iff it is bounded.

It does?

yeah

I'm reading wiki

I don't think wiki says that

@0celo7

Well

I need to learn some functional analysis

but I have no time

well your are learning some with me atm :P

I still have nuclear physics homework and then a quantum mechanics midterm tomorrow

oh

good luck

I don't think having bound on T will help though @0celo7 ?