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17:00
alright, I'll think about that exercise, sounds very useful
So we need to find a $\bar\partial$-closed (0,1)-form. Presumably we'll have $|z|^2$ in the denominator.
I'll let you play and report back to me :)
One hint for the exercise: The fact that it's a surface is key. You will want to use integrals ...
17:14
Is anyone experienced in the Laplace-equation?
Depends what you want to do.
I have a more specific question about the equation.
Why is our integral randing from 0 to INFINITY? I understand the zero part, but why infinity?
What integral?
Why not choose a set number or constant value that you want to focus within?
Hold on, let me find it for you
You need to give more context here.
17:16
Can you post pictures in here?
You talking about the d'Alembert formula for the one-dimensional wave equation?
Yes, you can post pictures.
How?
I am fairly new to this chat function
There is an upload button next to the box you type in.
Maybe it takes a certain amount of rep. I dunno.
Yea, I dont see it
And you have to be on a computer, not phone.
17:17
Can I just send you the link?
I am on computer
The picture of the equation is on this site
Oh geez. You even asked the question wrong. We're not talking about the Laplace equation. We're talking about the Laplace transform.
The answer to your question is that it comes from residues and complex analysis.
Ohh, snap, sorry :-)
I am fairly experienced in complex analysis
It's a cousin of the Fourier transform.
The Fourier transform is an integral from $-\infty$ to $\infty$.
However, would it not make more sense to have the upper limit be a constant value that you choose?
That way you can check what happens at the upper limit you are focusing on?
@TedShifrin it does
17:20
Does your question make sense for the Fourier transform? You can't arbitrarily say you're only going to care about frequencies up to 10.
if it's a constant value that you choose, it's a family of transforms. some of them may not have the right properties.
I would say that none of them will.
@TedShifrin You talking to me? xD
Yes, I was talking to you.
We both are.
I am talking about Laplace Transform, not Fourtier Transform
17:22
you can certainly write down any integrals that you like and call them transforms. but i would look at a list of properties of the laplace transforms and their proofs and think about how a person choosing their own finite upper bound would affect those proofs.
Laplace is a special case of Fourier, really.
i think of the laplace and fourier transforms in terms of their properties. i don't really ask why the formula is what it is. i understand the properties to be important. if i don't get something with those properties, i can't use it.
Because, let's say you want to focus on a frequency between 0 and x, where "x" is your upper limit, that way, you can see what happens when it surpasses your upper limit.
@TedShifrin I did not know that
i don't think you can get the full picture of the frequency domain without integrating to infinity.
@TedShifrin to this question
@leslietownes Of course, I totally understand it. Just curious if anyone would choose a upper-limit
17:24
i don't have a huge amount of intuition about this. i had a prof who said the fourier transform is a tunnel where you climb into it and then on the other side there's results, and nobody knows why. which probably isn't a good bit of professoring but does feel a lot like the fourier transform to me.
You will get horrendous, horrendous formulas if you do that. Just try doing the Laplace transform of the derivative if you don't have infinity there.
hi all. I had a question on differential equations
Well, a Laplace transform is TRANSFORMING a set of functions to another set of functions, so your prof did get the climbing right haha
satan himself having a question on differential equations. that seems right.
@leslietownes Don't answer him, he will use the answer to destroy humanity
17:26
$(x^2D^2 - 2)y= x^2+1/x$
the general solution can be written as the sum of the homogenous soln and one particular solution.
but i want to destroy humanity. that's why i'm here. i'm picking up clues for how to do it.
solving the homogenous equation is easy (this is the euler-cauchy form), we can let y=x^m.
mostly i plan on asking humanity questions about profit percentages. i also have a system for pronouncing binary numbers.
we find that $y_{h}= ax^2 + b/x$
now, for finding one particular solution, I tried using variation of parameters
I got very close to the answer, but theres a sign error thats driving me crazy...
that seems like something worth trying.
17:29
the wronskian W is -3
if i let $y_{p} = ux^2 + v/x$, then
i love wronski beat. i have all of their records and singles. my favorite is 'smalltown boy.' very good exemplar of 80s synth pop
$u= 1/3 \int (1/x + 1/x^4)dx$
since $u= \int (-1/w* (1/x) * R dx)$
@leslietownes Shaaddddup.
it's occurring to me that some of the stuff i say on here is somewhat stupid.
4
since $R+ 1+ 1/x^3$. (the standard form is $D^2- 2/x^2)y= R$ )
so, we can solve for $u$= $ lnx/3 -1/(9x^3))$
similarly, $v= \int ( 1/w* x^2*R )dx$
we can solve for $v= -x^3/9 -lnx/3 $
so $y_{p}= x^2u + v/x= -x^2/9 -1/(9x) + 1/3*lnx (x^2 - 1/x)$
adding this to $y_{h}$, and letting $a-1/9 = A$ , and $b-1/9= B$, we get:
$$ y= Ax^2 + B/x + 1/3*lnx (x^2-1/x) $$
the given answer is:
$$y= Ax^2 + B/x + 1/3*lnx (x^2+1/x)$$
where did I go wrong :(
17:42
have you popped this into wolfram alpha? it is entirely possible that the given answer is wrong. i was just editing a document where i mis-named a patent, i was off on the number by about 1000. because i was thinking about another patent.
oh god
i forgot about WA
brb
i don't see an obvious reason why those answers could be expected to be the same, so at the moment i suspect that at most one of you are right.
but it might be you.
hello. some of that is undesirable.
we can all calm down and resume a discussion of, perhaps mathematics, or at least birds. something pleasant.
some nice relaxing double tap maybe?
17:47
18 messages moved to Trashcan
mumble mumble something about cleanup
Thanks, @Vogel.
sorry for the invitations to the trash
it's fine. you know me well enough that you wouldn't be surprised to learn how often i've been invited to the trash.
how is it that a mod came here immediately after someone started trolling randomly?
17:48
i flagged it maybe?
I flagged them too
@TedShifrin oh
@LucasHenrique oh
lots of flags and a mod that is glued to their laptop :D
chat flags are network wide
so multiple flagging alerts all moderators?
I almost never ask for help externally.
17:49
@satan29 no, I just happened to see it :)
Ted, I think I never asked you this:
ah cool
What is your profile picture about?
i can't ask for help externally. sometimes i ask my wife to ask her therapist something. we're both trying to figure me out.
What do you think, Lucas?
17:50
Looks like a rotation of a cube
I mean, the solid of revolution
Exactly right.
Hmm, cool!
Dumb question: is the usual Euclidean topology on $\mathbb{R}^n$ the product topology on $\mathbb{R}$?
The question is: How do you know what the surfaces are?
Yes, same topology.
@TedShifrin I suppose you prove that showing that open balls are open in the product topology and rectangles are open in the Euclidean topology, right?
Thus they are the same, etc.
@TedShifrin sorry, what do you mean?
Right.
17:53
that's right
You see cones on either end. What's in the middle?
i can't figure that out. is the middle part a ruled surface? or are the lines just a software artefact
I don't know. I gave a shot since I've seen a gif with that rotation
here's another picture
It looks like the ruled surface string construction I think was in Geometry and the Imagination. Idk though
17:55
i once had a hyperboloid trash can. i tried to use it to instruct students on ruled surfaces until i realized it probably sounds crazy to college students to suggest that a trash can is meaningful. so i stopped.
@Lucas: If you think about the cube, you see that that middle surface has to be in fact doubly ruled (two lines through each point), not just ruled. There's only a few doubly ruled surfaces — planes and hyperboloids of one sheet and saddle surfaces (the latter two are projectively equivalent).
Yes, @Quin, that's right.
satan, if my wife's therapist hasn't broken through i don't know what chance you have, but thanks. it did make me laugh.
i'm pleased that i recognized it as a potentially ruled surface.
this whole thing is turning me into a geometer. i guess i'm OK with that.
18:00
"This whole thing"?
mostly the chat. maybe signs from G-d. maybe i should have been a geometer the whole time.
Well, let's not get too carried away!
let's manipulate symbols and mostly act like we don't know what we're doing. geometry is the cheat sheet but it isn't real. symbols are real.
18:14
being real is kinda redundant
you can't define reality without being circular, so it's questionable if any kind of math is real or not, after all everything is symbols
a friend told me one day: "mathematics is just applied computer science". and i can't even disagree
I don't even know what "real" means. I lived in a complex world.
hippies.
I totally disagree.
mathematics is definitely not applied computer science.
ted and i keep agreeing, it's upsetting me.
18:16
Because very little of mathematics is related to computer science at all.
not just applied computer science, i should say. some of it might be. but it's not just that.
logic is just spicy computer science, math is based on logic, thus it's applied computer science
by computer science i mean formal languages
one issue is how computers actually approach problems. i can easily define sets in mathematics that a computer would not be able to delineate or explore. although i can reason about them.
Sorry, @Leslie. I had to star that. And @Thor and I have agreed surprisingly much, too. I must be becoming weak.
That's like saying that all of mathematics is arithmetic of integers, @Lucas. It's total bull.
@leslietownes oh, i'm not being silly. being literal, every turing machine can only accept finite sets.
@TedShifrin it's total bull, but it's not wrong
:p
18:19
i don't deny that a computer from the 1970s could perform more computations than me faster than i could. but i have better opinions. i have vibes. do you know about vibes?
I won't even continue this discussion. It's stupid.
computers don't know about vibes.
@leslietownes you can't define vibes
@TedShifrin don't be so serious, come on. obviously you won't be talking about C*-algebras and stuff on computer science, it was a joke
there are some parallels in computer science, but that is about it.
18:22
i'm happy to see anybody reference c star algebras. i can't be annoyed by that. let's spread the good news about c star algebras.
I will only talk about $C^*$-algebras when I'm dead and buried.
(Now I'm channeling Leslie.)
sometimes i forget which one of us is talking.
Update @Thor: I have the obvious generator of $H^{0,1}$. I presume I could prove that there can't be more than one. There is presumably a single generator of $H^{1,1}$ as well.
@TedShifrin What do you call computer science?
I don't.
18:33
Ok
@Ted So I think the first part goes like this: let $\alpha$ be a holomorphic $(1,0)$-form, i.e. $\overline{\partial}\alpha=0$, then $\partial\alpha$ is a $(2,0)$-form and its conjugate $\overline{\partial}\overline{\alpha}$ is a $(0,2)$-form, so that $\partial\alpha\wedge\overline{\partial}\overline{\alpha}=\partial(\alpha\wedge\overline{\partial}\overline{\alpha})=d(\alpha\wedge\overline{\partial}\overline{\alpha})$ is a $(2,2)$-form whose integral vanishes by Stokes. On the other hand, locally we can write $\partial\alpha=fdz^1\wedge dz^2$ for some $f$ and $\overline{\partial}\overline{\a
Whoa. That's like a whole tôme.
@Thorgott why you always doin this
the exercise turned out to be slightly involved
Did you show that the $\phi_j,\bar\phi_j$ are linearly independent?
18:40
yeah, that's implicit in the above
also $H^{1,1}=0$
My exercise sheet had a hint for the second inequality. Something like the SES of sheaves $0\to\Bbb C\to\mathscr O\to d\mathscr O\to 0$.
I haven't yet figured out why, but the inequality $\sum_{p+q=k}h^{p,q}\le b_k$ always holds
and the Hopf surface has $b_2=0$ of course
Hmm, you certainly can't get $H^{1,1}$ in general, and I think it's even false for Hopf.
Oh, maybe my "generator" is exact.
@TedShifrin Ah, I'm not fancy enough for that.
Well, get fancy!
18:42
this general inequality I just posted together with the surface-specific one from the exercise should yield $h^{0,1}=1$ for the Hopf surface
and then we obtain the complete Hodge diamond by Serre duality
Yeah, I believe that. And I thought it should have $H^{1,1}\ne 0$.
I prefer the elementary linear algebra to sheaves :P
sheaf cohomology still scars me
Oh, no, I guess $H^{1,1}=0$ is right.
How can I love sheaf cohomology and you, the categorical algebraist, hate it?
Of course, I hate derived functors.
I don't hate it, I just haven't acquainted myself enough with it yet
Anyhow, the Dolbeault cohomology of Calabi-Eckmann manifolds (of which Hopf surfaces are a special case) is all about spectral sequences. So you can get to that in a bit.
18:47
spectral sequences are another thing I haven't learned yet
I need to up my homological algebra
if you run spectral sequences on the Dolbeault double complex, you apparently get what's called the Frölicher spectral sequence and the inequality $\sum_{p+q=k}h^{p,q}\le b_k$ is supposed to be a consequence thereof
Right.
I even had to put a spectral sequence in my thesis.
Well, an appendix thereof.
Like magic. Say "spectral sequence" and — poof — a @Balarka appears!
Hello.
Thor and I have been having "fun" with Hopf surfaces.
What's a Hopf surface
$\Bbb C^2-\{0\}/z\sim 2z$
18:53
Oh yeah that's a good one
Easiest example of non-Kähler surface.
So Thor is computing Dolbeault cohomology.
$S^1 \times S^3$, not enough $b_2$ for Kahler
I know 0 Dolbeault cohomology
You can scroll up and get some good exercises :P
18:54
I'll pass
LOL ... party-pooper.
yeah, not Kähler due to $b_2$ is easy
I'm computing Dolbeault cohomology, because you can explicitly observe that the Hodge diamond in this case has neither vertical nor horizontal symmetry
But now we see that $h^{1,0}\ne h^{0,1}$, so super naughty.
Hodge diamond lol
That's what it's called :)
18:55
the Hodge diamond is cool, so much symmetry
Or not.
at least we still have Serre duality even without Kähler hypothesis
Yes, that's totally sheaf theory, no harmonic theory.
Serre duality always works on any complex manifold twisted by a holomorphic bundle, no?
Hmm, the proof I did in my course years ago used harmonic forms.
18:59
if $\alpha$ is $\overline{\partial}$-harmonic, then $\ast\overline{\alpha}$ is $\overline{\partial}$-harmonic and $\int_X\alpha\wedge\ast\overline{\alpha}=\lVert\alpha\rVert^2$
pure harmonicity
I have heard the harmonic forms version but I do not know details. I understand it in terms of coherent duality
but to answer the question, yes
Which is just sheaves
But relating $d$-Laplacian to $\bar\partial$-Laplacian uses Kähler.
I'm gonna have to rethink this from 123 years ago.
19:01
ok mr. Verdier duality
I remember there was something called Kahler identities or something
Magic
well, it just depends on what you apply the theory to
Yes, part of that whole story.
the general fact is that elliptic differential operators between sections of bundles induce decompositions in image plus kernel of adjoint
I don't know the general holomorphic bundle stuff, but I assume you will use a modified $\overline{\partial}$-Laplacian in that case
because there's always a $\overline{\partial}$, but not a $\partial$, on holomorphic vector bundles, right
Yes, that's the general Hodge decomposition. Maybe Serre uses just the $\bar\partial$ Laplacian and I don't have to relate it to the usual. That seems right.
19:03
I have no clue what Thorgott is talking about anymore
yeah, because the usual $\overline{\partial}$-Laplacian on forms computes Dolbeault cohomology
Right, there's no $\partial$ on holomorphic vector bundles unless you define a connection. Just like there's no $d$ on smooth vector bundles.
Kähler just gives $\Delta_d=2\Delta_{\overline{\partial}}$, so that you can relate the Hodge theory of both Laplacians and obtain the Hodge decomposition
right.
@BalarkaSen I don't know what I'm talking about either
all I'm doing is linear algebra with fancier symbols
19:05
And I rarely know what Balarka is talking about anymore.
the hard part of all this is the elliptic PDE theory
Kodaira vanishing is sooo cool, though.
and I conveniently ignore all the elliptic PDE theory
Give talk explaining all of this later @Thorgott
All your knowledge is belong to us
sure, I can tell you what little I know about cohomology of Kähler manifolds
you'll have to consult Mike for any of the analysis tho
19:08
And Balarka should talk about the Fröhlicher spectral sequence.
Has anyone seen Mike here in ages? He hasn't even been in touch with me about his linear algebra class in weeks.
He's busy with classes I imagine
And life, I imagine.
We talk elsewhere briefly
@TedShifrin True
Well, say hi for me. (I used to tell him to say to you when you had disappeared. I don't know if he ever did.)
Yes, he did. I will!
19:10
In case you folks knew Pedro years ago, he's about to defend his Ph.D. in two weeks! Everyone is growing up!
5
@TedShifrin confusingly, it actually is Frölicher and not Fröhlicher
I feel very old :D
Oops. I knew that. Thanks, @Thor.
@TedShifrin I can learn
You know enough about $(p,q)$-forms; there isn't much to know.
19:13
Just the $\partial$ and $\bar\partial$ operators and $d=\partial+\bar\partial$.
also talk about Bott-Chern and Aeppli cohomology
I think cohomology of Kähler manifolds is actually just pure algebra once you have the $\partial\overline{\partial}$-Lemma
there's apparently a general theory of double complexes satisfying a $\partial\overline{\partial}$-Lemma and the vertical and total cohomology are related in the same way then, I think
Bott-Chern is just a generalization of Gauss-Bonnet and higher-dimensional residues.
Now you are doing algebra again, Thor.
@Balarka: You saw my remark about Pedro, I presume. Perhaps no one else here remembers him.
@TedShifrin Oh, I missed this (multitasking). That's very cool!
Fine, I can learn and then give a talk. Set up a MSE mailing list
@Thorgott has to start
The tricky part is how. I don't have Zoom for more than two people for more than 45 minutes or something.
This is not good for typing in chat.
Yeah, good point.
19:21
Faculty/grad students in academic institutions have access to class Zooms, but I don't know if you guys do.
I don't, but someone must.
my law firm has a pretty expansive zoom license. everyone can piggyback on it. what could possibly go wrong.
Well, I don't think that a math lecture fits your law firm. And you surely aren't interested in these topics :P
cohomology is a scam. like the "vaccines" for the coronavirus, 5g cellular networks, and, oh, let's say, almond milk.
Even in $C^*$-algebras cohomology must show up.
OR maybe even hypercohomology.
19:26
Yeah Connes stuff
yes there's actually a good amount of it. and k-theory.
Cyclic cohomology
Yup, and K-theory, which is super-abstract nonsense.
i learned too much of it, but because i'm not a SHEEP i forgot all of it.
i learned k-theory for $C^*$ algebras and then someone told me you could also do it for topological spaces. i was like, "no way!"
19:28
And, what happened next, like?
ted has caught on to my californianism.
Yes, Leslie. You've become a valley girl.
then i learned enough of it to compute, like, one k-group during my qualifying examination.
it was like legally blonde except math school instead of law school and me instead of reese witherspoon. i emerged triumphant.
@TedShifrin I don't think I understand what you're getting at
19:30
Maybe you're thinking of something else Bott-Chern.
Bott-Chern cohomology, $d$-closed $(p,q)$-forms modulo $\partial\overline{\partial}$-exact $(p,q)$-forms
Oh, OK, I was thinking of their residue theorem.
Probably all intermingled.
ah, I don't know that one
Hi @TedShifrin
is that pronounced "d, d-bar"? i'm honestly asking. never heard it said out loud.
19:32
Hi Everyone
@TedShifrin: To give an update about some math, I am beginning to think the symplectic camel theorem is a classical uncertainty principle.
But I am not sure how exactly. I plan to explore.
apropos k-theory, I was looking at a proof that even-dimensional spheres of dimension >6 don't admit almost complex structures, but I didn't understand it cause it was full of K-theory. the 4k-dimensional case is a lot nicer.
@leslie del-del-bar
that makes more sense.
No, I say "d" and "d-bar," never "del". Oh well.
4k dimension follows from that signature identity or?
19:34
you don't distinguish between $d$ and $\partial$ in speech?
@TedShifrin I decided I want to also work in mathematical physics I ended up buying coursera plus
it's even easier than Hirzebruch
i mean if you'd just ask me to read it i would have gone with d. but if there's another d going around in the exact same context, del might be a good idea.
and enrolled first in thermodynamics been acing it so far
I will take thermodynamics, Mechanics, E&M, QM and take advanced classes after that.
Everyone says the "d-bar" operator. If necessary to distinguish $\partial$ and $d$ in speech, one might refer to the first as "del."
19:35
I say d bar
point is that almost complex structure implies $TS^{4k}$ is a complex bundle, but then you obtain that $(-1)^kp_k(TS^{4k})=2e(TS^{4k})$ from the classic formula comparing Pontryagin and Chern classes of a complex bundle and the fact that all cohomologies but the highest one vanish. However, the former is $0$ by stable triviality of $TS^{4k}$, but the latter is non-zero by Poincaré-Hopf, since $\chi(S^{4k})=2$.
Aha got it
@Thor: "Hermitian Vector Bundles and the Equidistribution of the zeroes of their holomorphic sections," Acta Math 114 (1965) ... and "Some formulas related to complex transgression 1970. There's also Bott Residue Formula.
i'm going to call it wiggly backwards b. and wiggly backwards b with a line over it.
19:38
@leslietownes did you research in C* operator ?
yes. i can't vouch for my reputation in the area, but at least i pretended to do that.
cool do you know if we can study K-theory of C* operators using algebraic cycles has it ever been done before ?
@Karim: That's a lot of physics to learn.
just out of curiosity
You should look at all of Griffiths's work on algebraic cycles, Karim.
19:41
wegge-olesen had a very accessible book about the k-theory of C*-algebras. i don't have anything in my mind that would supplement it. atiyah also had a good book, although it might have been topological spaces.
There's a huge volume of it.
@TedShifrin Yeah I want to also work in mathematical physics. I put 2 hours per day.
On top of 5 pages of Griffiths Harris every day?
@TedShifrin cool I will look at his work
@TedShifrin Yeah my wife doesn't mind me working all the time as long as she gets to foster animals haha
i've been fostering some fruit flies around the pears i forgot to eat on the kitchen counter.
19:45
Fostering animals is super important. Good for her.
it counts as volunteer work, i'm deducting it for tax purposes.
Not you, @leslie.
you don't know that he's not fostering fruit flies. it's probably cats or dogs or rabbits or something, but it might be fruit flies. in which case it would be as super important as what i'm doing by neglecting my pears.
It is cats and dogs
well, that's my argument busted. i give up.
19:49
Is anybody good at plotting $f(n)=\exp\big(\Lambda(\log n)\big).$ where $\Lambda$ is the Von Mangoldt function
it's nice to foster cats and dogs. we have our cat because somebody fostered her and then gave her to a shelter, and now she's here bothering us all the time. good times were had by all.
@geocalc33 what do you mean "good at plotting", put it into a CAS
Wolfram alpha doesn't let me plot the Von Mangoldt function
the Mangoldt Lambda needs an interger as an input, Lambda(log(n) ) doesnt work
that is cool @leslietownes
19:58
i think i may have posted photos of my cat. we have no idea who fostered her, when she was born, or where she came from, but she has been a great comfort to us.
@s.harp yeah you're right - I'm trying to define an analogous Von Mangoldt function assuming that the primes $p$ were mapped to $e^p$
@geocalc33 anyway you can do a ListLogLinearPlot to compress the x-scale logarithmically

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