The h Bar

@DanielSank Can you represent the lowpass filter as a system of SDEs? Like you would for e.g. RLC as a system of ODEs effectively. Then the integration would work out as usual (should be a trivial generalization of the ODE).

AccidentalFourierTransform

12

Q: Product coproduct sign

I want to have a product sign analogously to $A \amalg B$. I tried $A \Pi B$, but as $\Pi$ isn't a binary operator, this does not look that good. Can you help me?

math-mode

0celo7

not that I could write down the definition of a coproduct without checking Hatcher...

ACuriousMind

@0celo7 It's that of the product, with all arrows reversed :)

0celo7

@ACuriousMind that's about as helpful as telling @Slereah cohomology is the dual of homology ;)

@ACuriousMind My prof thinks it's strange I prefer cohomology to homology. Is it?

ACuriousMind

@0celo7 To a topologist it is

0celo7

8:30 PM

Why?

ACuriousMind

homology is very natural in terms of cycles

0celo7

Granted, I don't really understand what cohomology is telling me

@ACuriousMind sure

ACuriousMind

But cohomology is...linear functions on cycles. That's not geometric

0celo7

But the arrows go the wrong way in homology

DanielSank

Guys, I gotta say, this is one of the nicer chats I've been in on SE.

Some of the other ones... yeeesh.

0celo7

8:31 PM

@DanielSank aww, we <3 you too

Which one is not nice?

AccidentalFourierTransform

$Mathematics$ The JEE LaunchPad

We love Physics, Chemistry and Mathematics. And we are prepari...

DanielSank

@0celo7 Electrical engineering seems awfully hostile.

Two people in one day responded to me with comments that served little purpose other than to stroke their egos.

After I pointed out that someone was wasting time stroking their ego, we got this:

> @MickLH I don't think there is much left when people want to ban words like "victim" and replace it by some terms like "experiencing person"

wat?

dmckee

8:46 PM

Seen on the site:

> As a theorist, I'm not aware of any experimental method to detect the de Broglie wavelength of particles.

::facepalm::

@dmckee Our site?

Yeah. Our site.

Uh oh.

@ACuriousMind Unless, of course, you work on a different version of cohomology than singular.

ACuriousMind

@BalarkaSen sure, but singular is love, singular is life :P

Balarka Sen

8:49 PM

yeah man

@0celo7 Homology is essentially a singular version of cobordism theory, so it's easy to visualize

Cohomology, on the other hand ...

DanielSank

Wow, guys go take a look at Electrical Engineering.

Especially the mods might find it interesting. We're a really tame chat by comparison.

AccidentalFourierTransform

link please :-)

DanielSank

chat.stackexchange.com/transcript/15

We were talking about git and some guy starting playing some "you don't have kids so you don't know what life is like" card.

Now they're talking smack about liberals etc.

ACuriousMind

Tip: Link to the transcript, not the room, otherwise one enters the room when clicking on the link

DanielSank

oops

How to do?

ACuriousMind

8:56 PM

@DanielSank Replace "rooms" by "transcript" in the link

It's also accessible from the "room" menu of the room

Wow!

@ACuriousMind noted

amusing

@ACuriousMind if you wanna edit the link please go ahead. I can't anymore.

Balarka Sen

I probably won't go in that room tho

DanielSank

8:57 PM

Man, I wish having some rep would give you a slightly longer edit period in chat.

@BalarkaSen Well, in my capacity as a physicist I am presently doing some electrical engineering.

^ Hint hint to all those around here who think physics and engineering are neatly separable.

0celo7

@ACuriousMind I think you mean de Rham

DanielSank

Spoiler alert: they're not.

AccidentalFourierTransform

in Electrical Engineering, 53 secs ago, by PlasmaHH

@AccidentalFourierTransform or as these days you would probably hear: tää tääää, tää tääää, tää täää

help

me dont understand

DanielSank

Let them be.

0celo7

@dmckee ok, I'm gonna bite. How do you detect it?

DanielSank

8:59 PM

They seem sick.

Balarka Sen

all drunk

ACuriousMind

@DanielSank Eh, don't be mean just because they are.

DanielSank

@ACuriousMind Mean?

Balarka Sen

@0celo7 de Rham is too complicated. forms contain more info than homology alone

*co

DanielSank

I really meant it. People who take opportunities to lash out at others for no reason are not well.

@BalarkaSen I like forms tho.

ACuriousMind

9:02 PM

@DanielSank Calling someone "sick" is mean in my perception. If you actually mean that as an assessment of mental health, please don't diagnose people over the internet.

DanielSank

@ACuriousMind Oh for... whatever.

Balarka Sen

@DanielSank Yeah, they're great. But I have this strange feeling that I never completely understand what they mean.

Personal failing

D3075

@ACuriousMind why do we always say fields in qft and other field theories have to be smooth functions?

why not just differentiable? is there a reason?

AccidentalFourierTransform

we dont say that because its not true

:-P

ACuriousMind

@D3075 'cause we wanna take derivatives of them and don't really know how many :P

D3075

9:06 PM

oh ok

@AccidentalFourierTransform example?

i probably won't understand your example btw xD

AccidentalFourierTransform

in qft or in other field theories?

D3075

qft

AccidentalFourierTransform

in the operator formalism the fields are distributions, not (smooth) functions

in the path integral formalism you integrate over non-smooth functions too

0celo7

@DanielSank :)

AccidentalFourierTransform

(the contribution of smooth functions is typically negligible IIRC)

D3075

9:09 PM

okay thx

i'll keep that in mind

0celo7

@BalarkaSen Hmm? De Rham is isomorphic to singular

Or are you talking about harmonics?

ACuriousMind

@AccidentalFourierTransform One can, for the actual Wiener path integral measure in 1d, actually show that the differentiable functions are of measure zero :O

Balarka Sen

de Rham admits a natural differential graded algebra structure, which singular doesn't.

ACuriousMind

Oh, I heard a nice talk involving dg-algebras today

0celo7

What even is that

ACuriousMind

9:12 PM

I should find a good resource for the mathematical BV formalism sometime...

0celo7

Functions of bounded variation?

ACuriousMind

Batalin-Vilkowski

0celo7

Algebra?

ACuriousMind

algebraic physics, if you wish

The BV formalism is a way to formulate field theory

0celo7

High school physics?

2

AccidentalFourierTransform

9:14 PM

@ACuriousMind I really want to learn about that :-(

0celo7

@BalarkaSen so what does de Rham give you that singular doesn't?

ACuriousMind

@AccidentalFourierTransform Several talks in our physical mathematics seminar dealt with it recently, I have a decent high-level understanding but nothing detailed

AccidentalFourierTransform

:: is envious ::

ACuriousMind

I think physicists tend to call it the "antifield formalism" for some reason

0celo7

Ah

That's in Weinberg

Made approximately zero sense

AccidentalFourierTransform

9:17 PM

^^^

ACuriousMind

It's also in Henneaux/Teitelboim

but there it's fullblown BV-BRST

0celo7

Haven't heard that in a while

AccidentalFourierTransform

@ACuriousMind and, again, made almost approximately zero sense

something about ghosts of ghosts

and ghosts of ghosts of ghosts

ACuriousMind

The ghosts are pure BRST, nothing to do with BV

0celo7

Ok what is with these prodigies in my engineering classes

This guy has specific heat values memorized

Who the hell does that?

AccidentalFourierTransform

9:20 PM

@ACuriousMind ok then it made literally zero sense :-P

4180

0celo7

Ignore that.

Slereah

@AccidentalFourierTransform you integrate only on non-smooth functions, even

Balarka Sen

@0celo7 I think I was wrong about my previous statement. Singular cochain complex does admit a commutative differential graded algebra structure, since cup product can be defined on cochains.

Oh well!

0celo7

@BalarkaSen I was thinking about cup, but I don't know what a differential graded algebra is.

ACuriousMind

@AccidentalFourierTransform Ghosts of ghosts are what you get when the gauge transformations themselves can have gauge transformations

AccidentalFourierTransform

9:25 PM

@Slereah wat

Slereah

Smooth functions have measure $0$

Balarka Sen

It's basically a cochain complex with a cup product and a differential which is compatible with the cup product. Don't worry about it

Anyway, it is true that forms contain more information than cohomology.

AccidentalFourierTransform

@ACuriousMind e.g. in sugra, right?

Balarka Sen

A nice example I recently got to know: If a closed manifold admits a nowhere zero form then it fibers over the circle.

AccidentalFourierTransform

$h_{\mu\nu}\to h_{\mu\nu}+\partial_{(\mu}A_{\nu)}$, and $A_\mu\to A_\mu+\partial_\mu B$

Slereah

9:28 PM

@ACuriousMind Double ghosts???? D:

0celo7

@BalarkaSen form of what degree?

Balarka Sen

1-form, sorry

ACuriousMind

@AccidentalFourierTransform Simplest example is whenever you have a higher gauge field, like the Kalb-Ramond field in ST or the C-field in M-theory

0celo7

Speaking of forms, I really need to read de Rham's book

AccidentalFourierTransform

"Simplest"

Slereah

9:31 PM

Is there any gauge simpler than $U(1)$

DanielSank

Yeah, wire gauge.

ACuriousMind

They are $\mathrm{U}(1)$ theories, but the gauge field is not a 1-form, but a higher p-form

0celo7

@BalarkaSen how does the proof go?

Slereah

@DanielSank you silly man

0celo7

It see what You're saying now

DanielSank

9:32 PM

@Slereah I was in the Silliman college at Yale...

0celo7

Sorry I'm walking and my fingers are cold. It's hard to type

ACuriousMind

The gauge trafo $C\mapsto C + \mathrm{d}\Lambda$ has higher trafos $\Lambda \mapsto \Lambda + \mathrm{d}\chi$.

AccidentalFourierTransform

but that doesnt work in $U(1)$, right?

ACuriousMind

Sure it does

$C$ is a three-form

AccidentalFourierTransform

hmm

ok so $\Lambda^2$ and $\chi^1$

but $\Lambda$ is defined up to a closed form, so $\Lambda\to\Lambda+\mathrm d\chi$ is trivial

0celo7

9:35 PM

@BalarkaSen using a metric, doesn't that mean any closed manifold with zero Euler characteristic fibers over the circle?

Balarka Sen

@0celo7 Sorry, I messed again. I meant a closed 1-form.

ACuriousMind

@AccidentalFourierTransform Yes and no - it's trivial and that's why it's a gauge transformation

AccidentalFourierTransform

yeah reading it back it made no sense lol

ACuriousMind

Like, $A\mapsto A+\mathrm{d}\Lambda$ is also trivial, since it does not change $F$

0celo7

@BalarkaSen ok, how does the proof go?

DanielSank

9:36 PM

Anyone here who has to deal with mechanical drawings, this is really useful: gdandtbasics.com/feature-control-frame

Balarka Sen

@0celo7 Yeah, you get a closed 1-form by dualizing a nonzero vector field on the manifold

0celo7

I doubt it'll be closed...

Balarka Sen

@0celo7 I don't know yet but I suspect what's happening is this. Look at the kernel of $\omega$ on $TM$: that a codimension 1 distribution. Because $\omega$ is closed, that's integrable. So you have a codimension 1 foliation.

0celo7

Yeah, it won't. In 3 dimensions you need it to be irrotational

Unless I'm missing something important I doubt that's always the case

Balarka Sen

since $\omega$ is nowhere zero, you get a nowhere zero transverse vector field on the foliation.

Flowing along that tells you nearby leaves are diffeomorphic. Aka, you have a codim 1 foliation with diffeomorphic leaves

Now that's not sufficient, because you can get things like the irrational foliation on the torus. So I think a generic perturbation of this foliation should be a fiber bundle over the circle

@0celo7 Ah, yes, of course.

0celo7

9:49 PM

@BalarkaSen Oh, so an actual fiber bundle?

Balarka Sen

Yes.

0celo7

@BalarkaSen Hmm. Is there a more rigorous argument here or is it supposed to be obvious?

Oh, compact manifold

Yeah, sure

That argument is used in morse theory

Balarka Sen

Ya, it's sort of like that gradient flow argument.

0celo7

Gradient flow is also something I want to understand

@BalarkaSen generic perturbation?

@DanielSank what's this about contour integration?

Balarka Sen

You can perturb a foliation; foliate a ball cross manifold so that each slice is a union of leaves

0celo7

10:01 PM

sorry, foliate a ball cross manifold?

Balarka Sen

I don't understand the question

You just foliate $D^k \times M$

0celo7

@BalarkaSen yeah, but how is that a "perturbation" of the original one?

Balarka Sen

By restricting it to be the original foliation on $0 \times M$

It's just a homotopy idea

0celo7

@BalarkaSen what's the reference?

I still dunno what you're trying to tell me

Balarka Sen

Candel-Conlon, the book I am reading right now

I don't know what's so hard about what I am saying but whatever

DanielSank

10:09 PM

@0celo7 Talk about it in the EE room.

0celo7

@DanielSank But why?

DanielSank

@0celo7 Because it's funny.

0celo7

@BalarkaSen Def. 2.4.5?

Balarka Sen

probably? I don't have it in front of me

0celo7

@BalarkaSen What is $C^{1,0+}$?

Balarka Sen

10:16 PM

who cares. C^1 manifold with C^0 foliation I guess

or something.

DanielSank

I can't get over how cool this is: $$\int_0^T f(t) dW_t = \mathcal{N} \left(0, \int_0^T f(t)^2 \, dt \right) \, .$$

0celo7

That doesn't seem like the right attitude for learning the subject...

Balarka Sen

i don't care about technicalities

DanielSank

^ Mah man!

Balarka Sen

everything is smooth, every foliation is smooth. i'm happy

0celo7

10:20 PM

@DanielSank what is that?

DanielSank

@0celo7 It's a stochastic integral. Allow me to explain.

Suppose I have a Gaussian distribution $\mathcal{N}(0, \sigma)$.

That reads: Normal (Gaussian) distribution with mean 0 and std deviation $\sigma$.

Now suppose I have a variable $x$ which is formed by sampling the Gaussian twice and summing the results.

What, dear @0celo7, is the distribution of $x$?

AccidentalFourierTransform

every integrable function is also square-integrable too, right?

0celo7

Wait, I'm searching through my complex analysis book for a good integral

AccidentalFourierTransform

wait no, $1/\sqrt x$

0celo7

@AccidentalFourierTransform No.

Yes.

AccidentalFourierTransform

10:21 PM

and if it is non-singular?

0celo7

Do you mean bounded?

AccidentalFourierTransform

yes

Balarka Sen

@0celo7 Ok, it means the foliation is integral to a C^0 subbundle.

0celo7

Compact support?

AccidentalFourierTransform

nope

0celo7

10:23 PM

@AccidentalFourierTransform I'm thinking of a function that has integral equal to some series

But the square of the series diverges

...does that even happen?

DanielSank

@0celo7 talk about stochastic integrals later?

0celo7

@DanielSank ahhh

I can't find an integral

DanielSank

ahhh what?

0celo7

Is a scattering amplitude cheating?

DanielSank

@0celo7 Generally speaking, no, scattering amplitude is not cheating.

I don't even know what that means.

0celo7

10:28 PM

@AccidentalFourierTransform For a bounded set you have $L^2\subset L^1$

AccidentalFourierTransform

cool thanks <3

0celo7

@AccidentalFourierTransform well you said no compact support

You have $L^2\not\subset L^1$ in general but I can't think of a bounded function example

AccidentalFourierTransform

bounded set means compact support? I thought you meant that for bounded functions ...

0celo7

Also, you want a function in $L^1-L^2$ so that doesn't even help

@AccidentalFourierTransform no

my brain is failing right now

AccidentalFourierTransform

whatever, I just need some integrals to converge. I think Im going to go with "for $f$ a Schwartz function..." and forget about the details

0celo7

10:31 PM

@AccidentalFourierTransform Oh, yeah, I'm stupid.

Still no, that's not even integrable.

AccidentalFourierTransform

lol

I just bit a hang-nail and Im losing a lot of blood

0celo7

@AccidentalFourierTransform If $|f|\le M$, then $\int f^2=\int |f|^2\le M\int |f|$.

AccidentalFourierTransform

I cant think right now

@0celo7 ah thats it

wait

yes

ok, thanks $\varepsilon >$

0celo7

If $f$ is complex you omit the first $=$

@DanielSank Ok, educate me

@DanielSank Yes

@DanielSank Ugh. Convolution?

I'd have to get out my probability notes

DanielSank

@0celo7 Well, the results winds up being that you sum the $\sigma^2$'s.

0celo7

10:37 PM

@DanielSank But is convolution the correct approach?

DanielSank

@0celo7 yeah

0celo7

Then the convolution integral would spit out a Gaussian I imagine

Complete the square and all that

DanielSank

@0celo7 Nah just use Fourier transforms. Convolutions are products in Fourier land.

0celo7

Right, and fourier transform of Gaussian is Gaussian

DanielSank

Anyway, so $x$ has distribution $\mathcal{N}(0, 2 \sigma^2)$.

Ok so now suppose we are summing infinitely many Gaussians.

Suppose we invent notation $\int_0^T dW_t = \mathcal{N}(0, T)$.

(I just changed the notation so the second argument in $\mathcal{N}$ is the $\sigma^2$ instead of $\sigma$. Sorry.

Slereah

10:40 PM

are we talking about wieners

@Slereah Yes.

good old wieners

@DanielSank ok

@0celo7 Ok so now ask yourself what you get form $$\int_0^T f(t) dW_t$$

Just kind of think about it and see what you expect.

0celo7

I have no clue what that symbol is supposed to mean...

DanielSank

10:42 PM

@0celo7 which?

0celo7

$\int_0^T f(t)dW_t$

Slereah

Isn't it just the gaussian measure

DanielSank

I means add a continuum of Gaussian distributed random variables, indexed by $t$, each one weighted by $f(t)$.

@Slereah I don't know what that means.

Slereah

$$dW_t \approx \prod e^{-x(t)^2} dt$$

Or something

0celo7

@BernardoMeurer haha holy shit Fantano gave it a "not good"

Slereah

10:44 PM

Or is it $dx$

0celo7

@DanielSank Umm, assuming $f(t)$ modified each variance by $f^2$ (?), then would it be $\int_0^T f(t)^2dt$?

in the $\sigma^2$ slot

AccidentalFourierTransform

(is there any standard name for the set of bounded and integrable functions?)

Slereah

isn't that just the set of test functions?

Oh wait, I guess they don't have to be smooth

0celo7

@AccidentalFourierTransform not that I know of

I lied

DanielSank

@0celo7 Which is what I wrote.

0celo7

10:48 PM

@AccidentalFourierTransform $L^1\cap L^\infty$

@DanielSank ok, but what about it

what is it telling me?

Slereah

Is $L^\infty$ the set of bounded functions?

AccidentalFourierTransform

hmm I think im gonna go with "bounded and integrable function". $f\in L^1\cap L^\infty$ feels slightly pedantic :-P

DanielSank

It tells you how to make sense of stochastic integral notation, which allows you to solve some problems easily.

AccidentalFourierTransform

$||f||_\infty=\max f<\infty$?

DanielSank

@0celo7 see here.

0celo7

10:50 PM

@AccidentalFourierTransform less pedantic?

AccidentalFourierTransform

ok xoxo

0celo7

@AccidentalFourierTransform $||f||_\infty=\mathrm{ess\,sup}\,(|f|)=\inf\{M\in\bar{\Bbb R}:\mu([|f|\ge M])=0\}.$

Slereah

What is the ass sup

AccidentalFourierTransform

thats my def. of $\max$

0celo7

essential supremum

@AccidentalFourierTransform I doubt it, but ok

AccidentalFourierTransform

10:52 PM

YOU DONT KNOW ME

0celo7

I think I've got you figured out

@Slereah It's the set of functions that are a.e. equal to a bounded function.

Slereah

Sounds like you almost everywhere answered his question then

0celo7

@AccidentalFourierTransform so what are you trying to do?

analysis?

AccidentalFourierTransform

lol I wish

Im playing around with basic field theory

and I want to make sure that all Fourier transforms are well-defined and all the generators of symmetries are well-defined

nothing fancy

0celo7

So you know the Stone theorem?

AccidentalFourierTransform

10:57 PM

as I said, in the end I might go with "everything is a Schwartz function" and forget about the details

classical field theory

dmckee

@0celo7 Diffractive scattering. Any kind of diffractive scattering is sensitive to wave-length. Do it with massive particles and you are measuring the de Broglie wavelength of the particle.

And that is why people need to do an advanced lab course just as badly as they need demanding whiteboard electives.

AccidentalFourierTransform

youre just bragging

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