Yeah my cats always sits next while working

speaking of work back to it haha

When people write $ f \in L^p+L^\infty$ what do they mean?

@Ted I'm not seeing the generator of $H^{0,1}$

i would imagine that they intend to express that f is a sum of an element of $L^p$ and $L^{\infty}$.

that f is of the form $f_1+f_2$ with $f_1$ in $L^p$ and $f_2$ essentially bounded

ie its ultraviolet singularities are $L^p$ summable

for fancy talk

Cheers :)

L^infty includes constants which generally are not in L^p. that could be the only purpose of including L^infty.

yup

there may be subtler purposes of including L^infty but i do not know of them. when i see L^p + L^infty i think that they just want those constant functions around.

I mean, there are essentially bounded guys that don't just differ from something in L^p by a constant too

this is true and i do not mean to disparage those functions. if there is a particular result i would be interested in its formulation. the problem with function spaces is that there are ten billion of them and you often have to read between the lines to see what someone is really using.

sometimes you cite a function space because it's convenient and not because you actually need it. it's just, that's what's in my bag of tricks, i'll grab that and use that. absolutely no other reason. but sometimes it is essential.

functional analysis gets junked up a bunch with this kind of stuff. people choosing and varying various parameters purely to fit into other literature. they don't explicitly say what they're actually using.

isnt that just most math

I find theres always a lack of words to accompany most written math

but then I go and commit the same crime

smart people speak with their fists instead of their words

maybe I'm mixing up the saying

smart people learn bjj

that's true. i walked into my phd qualifying exam and just beat everyone up. like it was a video game.

and when i was done it was over.

@Thorgott You need a $(0,1)$-form that is homogeneous of degree $0$ and we hope it'll be $\bar\partial$-closed. Try the easiest thing: $$\phi = \frac{z^1d\bar z^1 + z^2d\bar z^2}{\|z\|^2}.$$

(Oh, and then we hope next that it is not $\bar\partial$-exact.)

Anyone know "the jist" of this argument

not the details just the idea

$\sigma$ is the topology of the metric $d$

Could I have a small hint for this problem? Let $Y$ be a connected one-dimensional CW complex, $X$ a compact, path connected and locally path connected space, such that $\pi_1(X)$ is abelian. Prove that any map $f : X \rightarrow Y$ is either null homotopic, or factors through $S^1$

I already know that if $f$ isn't null-homotopic , $\pi_1(X)$ is abelian implies $f_{\ast} \pi_1(X) $ is isomorphic to the integers in $\pi_1(Y)$ (because its an abelian subgroup of a free-group) , trying to use this somehow to construct a map homotopic to $f$ that factors through $S^1$ to no avail

@TedShifrin Nice. Is there an easy way of seeing this?

Lift a non-nullhomotopic closed path, porridge.

Seeing what?

that it's non-exact

Oh, it is obviously exact upstairs, but cannot descend.

hmm, the obvious primitive doesn't descend, but who says another one can't

How can adding a global holomorphic function make it invariant?

@Sebastiano thank you for your kind comments, what a lovely positive person you are!

ah, I guess that will contradict Liouville after a bit of rearranging

Charge has to balance, so there have to be positive people to cancel us negative ones :)

Huh? @Thor

so would the idea be, there is a graph (one dimensional connected CW complex) $G$ that covers $Y$ with fundamental group $\mathbb{Z}$, and therefore it is homotopy equivalent to $S^1$ (after contracting away its maximal subtree, we are left with a wedge of circles, and there must only be one), so that $f$ is homotopic to a map that factors through $S^1$?

hmm nevermind, there's a gap in my thought

here im using the lifting criterion, since $f_{\ast} \pi_1(X,x_0) \subset p_{\ast} \pi_1(G,g_0)$, $f$ lifts to a map factoring through $G$, which is homotopic to $S^1$ , and so $f$ is homotopy equivalent to a map factoring through $S^1$

Oh, @porridge, I misread. It seems odd ... what if $Y$ is a figure 8?

right , I'm not claiming $S^1$ actually covers $Y$

but $\mathbb{Z}$ is a subgroup of $Y$'s fundamental group, so by galois correspondence there is a one dimensional CW complex with fundamental group $\mathbb{Z}$ covering $Y$, this CW complex won't be $S^1$, but it will be homotopy equivalent to $S^1$

@Thor Holo + $f(2z)=f(z) + c$ ... hmm ...

here's what I want to say: assume the descended form is exact, then a primitive pulls back to an invariant function $g$ on $\mathbb{C}^2\setminus\{0\}$, such that $f+g$ is $\overline{\partial}$-exact, i.e. holomorphic, where $f(z)=\frac{-1}{4\lVert z\rVert^2}$ is the obvious primitive. Then $f$ is bounded away from $0$ and $g$ is bounded by virtue of being pulled back from a compactum, so $f+g$ is bounded away from $0$, which can only happen for $f+g$ constant, but $f$ is not invariant.

also I meant to say which is *homotopy equivalent to $S^1$ and so $f$ is homotopic to a map factoring through $S^1$... got those mixed up :/

Your obvious primitive is wrong, Thor.

oh no, did I mess up

Plus, it should be $f+g$ invariant with $g$ holo?

equivalent scenario

Oh, my holo question cannot happen by Hartogs

A question on style, which is preferable? $\sqrt{\frac{\lambda}{\sqrt{\pi}}$ or $\frac{\sqrt{\lambda}}{\sqrt[4]{\pi}}$

I'm thinking "assume it's exact downstairs and pull back" and you're thinking "can I modify it so that it descends"

I agree its a weird question, whats more is that, if my proof works, it only shows $f$ is freely homotopic to a map that factors through $S^1$, not relative to any basepoint, for instance

Right.

I'm too far removed from topology at this point, porridge.

no worries, thanks for the hint anyway

@A-LevelStudent I was an optimist from an early age and my parents gave me a smile to share with everyone. I believe that our life is short and that here on this Earth of ours we should, each of us, leave beautiful memories of us.

@Andrew equally ugly.

Yeah I thought asmich

@A-LevelStudent I can always wish you all the best and that they can always go high and talented young people like you reach excellent results.

@Thor Did you see my Hartogs remark?

@TedShifrin Hi and my best regards from Sicily (Italy).

Is that suggesting I could reduce further huh?

All power to you, @Sebastiano

@TedShifrin I'm not Superman ahahahahahahahahaahahahahhaha

I was in Sicily when I was 7. Was supposed to go back 2 years ago, but didn't.

@Sebastiano My memories of you will certainly be lovely; you've lived up to your belief :)

All power to you, too, A-level.

@A-LevelStudent Ahahah....good words...I always wish you the best.

@TedShifrin I'm unfamiliar with that expression, what does it mean?

Sicily is beautiful to visit but living there is a big problem: there is a lack of work, educated people, respectful people, little desire to work; there are many young people with talent but who are not evaluated and considered. Only slackers go ahead in Sicilian society.

Wishing you all the best in your endeavors!

@Thorgott if you have some time , do you mind taking a look at my argument?

Oh, I see :) Thanks a lot @TedShifrin , you too !

Sebastiano Did you consider moving to the mainland?

Now you will find something in your profile before you go to bed. Up early tomorrow for school. Cordial greetings to all of you.

Night!

I've worked in so many places in Italy and always had a good time. But the love for the Sicilian land and for my parents was stronger. Now I work 25 km from my home. Good night.

unrelated, i think artins algebra is a really great book

Yes, I recommend it frequently!

ah, the right primitive is $\frac{1}{2}\log\lVert z\rVert$, I had forgotten how to properly differentiate

Are you sure about the $1/2$ (not that it matters)? I had $\log\|z\|^2$.

At any rate, it transforms by an additive factor under $z\rightsquigarrow 2z$. And no holomorphic function can cancel that out.

ah you're right, my factor was on the wrong side, it should be $2\log\lVert z\rVert$

it cancels out the $1/2$-factor in the complex differential operators

Easier always differentiating $\|z\|^2$, regardless.

This was a fun discussion on the Dolbeault cohomology, @Thor. Thanks :)

I'm still puzzling over the Hartogs remark.

Your point is the function would extend to $\mathbb{C}^2$ and then constancy of $g(2z)-g(z)$ yields a contradiction taking the limit as $z\rightarrow0$?

Hey guys!

Heya @Amin

and the rest o' y'all

How's everything going?

decent

Still livin' the dream

Oh wait, I just realized I misquoted the Frölicher inequality. It's $b_k\le\sum_{p+q=k}h^{p,q}$, not the other way round.

so I'm actually still far from done

Livin the dream

Livin the geodesic dream

good one

Livin the path with no acceleration dream

Sometimes I like to express the same thing in different language

@Thorgott Yes. The function has to be unbounded near $0$ by that functional equation.

Hi, Demonark.

ok, $H^{2,0}$ also trivially maps to $H^2$ and the map is injective by a similar argument as before. similarly, again, $H^{0,2}$ injects into $H^2$ with a disjoint image. so we do obtain $h^{2,0}+h^{0,2}\le b_2$.

it remains to find something useful to say about $H^{1,1}$

You are right that it should be $0$, but we don't have the proof yet.

hmm, is it possible to compute the Euler characteristic from Hodge numbers even without a Kähler hypothesis?

surely $\sum(-1)^{p+q}h^{p,q}$ has to be something meaningful

There's no relation between Hodge and topology without compact Kähler.

I've been asked to prove that a^p is congruent to a modulo p for some a in the integers and p in the primes. My first go-to was to try proof by contradiction: Suppose a^p is NOT congruent to a mod p. Then a^p - a \= pk (for some integer k). I'm not really sure where to go from here. I've collected what we know (or, at least, what I could think of): gcd(a^p - a, pk) = 1, and p does not divide a^p - a.

urgh, a shame

there has to be some argument

I'd post this as a question, but I don't think it would be appropriate. More importantly, I'm looking for a hint, I'm not interested in an answer.

@Thor Well, what does it give in this instance?

@KeithMadison This is in every textbook.

Contradiction is not a good idea. Induction is an elementary way to go. You'll have to prove a lemma along the way, but I'm not saying more.

0

we're supposed to have $h^{0,0}=h^{0,1}=1$ and $h^{1,0}=h^{2,0}=h^{0,2}=h^{1,1}=0$ and the rest determined by duality

I don't know the Hodge numbers for too many non-Kähler manifolds (even for many algebraic ones, it's non-obvious). The Iwasawa manifold is the next one to try.

The Frölicher spectral sequence still abuts to DeRham, so probably usual homological nonsense says you're right, @Thor.

I feel like my life would be a lot easier if I knew spectral sequences

I don't remember ever knowing anything about Euler characteristic being "conserved" as you go from one page to the next in a spectral sequence. paging @Balarka

Oh, but it should just be like the Euler characteristic of a complex is the same as the Euler characteristic of its homology. That's all that's going on. So it must be true.

I rest your case.

I forget the name of that lemma I just said. It shows up everywhere.

I know that lemma, but it only makes sense if the complex is finite-dimensional. I don't really see how to apply it to this double-complex to compare vertical and total cohomology.

Still finite-dimensional vector spaces once you get to $E_2$.

WTH is that thing called.

sigh

ok, I'll have to learn spectral sequences now

I will live to regret this

For someone of your categorical bent, it'll be easy.

Hmm, Hatcher doesn't name that algebraic lemma. But I remember some name goes with it.

hi... evening

i wonder, what is this series equals to? $\frac{1}{a}+\frac{1}{2}\cdot\frac{x}{a+2}+\frac{1\cdot 3}{2\cdot 4}\cdot \frac{x}{a+4}+\cdots$

sorry typo..

what is $\frac{1}{a}+\frac{1}{2}\cdot\frac{x}{a+2}+\frac{1\cdot 3}{2\cdot 4}\cdot \frac{x^2}{a+4}+\cdots$ equals to?

it looks similar to $1+\frac{1}{2}\cdot x+\frac{1\cdot 3}{2\cdot 4}\cdot x^2 +\cdots = \frac{1}{\sqrt{1-4x}}$ but i couldn't figure out how to simplify it...

yeah no, too many indices, I don't have time to think this through

but whatever, we got $h^{1,0}=0$ and $h^{0,1}\ge1$ successfully, which is enough to observe the failure of symmetry

@Ted however, do you have an example of a compact complex manifold for which Hodge decomposition fails?

You just did that.

Oh, maybe not.

Iwasawa is the standard example, I think.

yeah, Hodge decomposition still works for Hopf surface

more drastically, I found the claim that Hodge decomposition in degree 2 always holds for all compact complex surfaces (which would also imply the missing $h^{1,1}=0$)

aha, it does seem like Iwasawa is a counter-example, but those computations seem too involved for poor little me

These are not trivial issues.

Does Huybrechts discuss this?

not as far as I can tell

Hopf surfaces and Iwasawa manifolds are mentioned as examples, but he does not discuss their Dolbeault cohomologies

there is an example addressing that Iwasawa manifolds aren't "formal", whatever that means

Has to do with deformations.

yeah, I found a phd thesis computing the Dolbeault cohomologies of the Iawasa manifold and its deformations, but the only thing I got out of that was adding up the dimensions and seeing they don't match

good to at least know that an example exists, though, even if only to mention it in passing

I've never been through this stuff, personally.

I think the last step missing in the computation of the cohomology of Hopf surfaces may actually be non-trivial

I found an argument, but it quotes the Todd-Hirzebruch formula to obtain $h^{0,1}=1$

Hirzebruch's book is probably a good reference for you, regardless. You know it?

you mean topological methods in algebraic geometry?

Yup

looks like a great book, but also very scary

so many advanced topics