The h Bar

alarge

00:02

@0537 Not sure if rigour is the word I'd use. As far as I can tell, in the paper they "just" take an optimization problem, recast it in spline form, and throw it at a solver. Not to say that it's not cool (I think it is), but probably would leave someone more mathematically inclined with a bit of an empty feeling inside.

0537

00:16

@alarge I suppose you're right.

how could it become more rigorous?

alarge

@0537 Well, they don't have any theorems or proofs.

0537

@alarge right.

alarge

What the mathematical guys interested in numerical analysis would be looking at is convergence and accuracy of the algorithms and all that jazz.

Now that stuff is usually not interesting for the target audience of the paper, engineers, at least not at the level of rigour required in mathematics.

0537

I guess.

=(

alarge

e.g. hindawi.com/journals/jam/2012/808216 (just happened to be the first paper to come up on a search, not saying the paper is good or bad or even representative)

@0537 But if you are interested in things that LOOK cool, and require a lot of engineering effort, read papers presented at SIGGRAPH. They're often pretty damn neat.

vzn

04:34

spking of siggraph, got to go in person many yrs ago. a related area: pixar has opened up a lot of their internal docs recently. see eg khanacademy.org/partner-content/pixar or graphics.pixar.com/library

vzn

04:47

@0537 re Two-Level Free-Form and Axial Deformation for Exploratory Aerodynamic Shape Optimization, by Gagnon/ Zingg, thx for sharing. think alarge undervalues this paper on a (maybe) too-fast skim. it seems to involve aerodynamic simulation and optimization of 3d object shapes (wings) wrt aerodynamic properties such as lift/ drag. possibly involving optimizing the objects wrt (nearly) fluid dynamic simulations. impressive! and it would be an understatement to call it merely "engineering"...

it seems to invoke many deep physics aspects in the analysis.

as for "empty feeling inside", lol. no mathematician is going to be able to "derive" optimal shapes of wings. and there would be few thms applicable. but what could have more practical significance?

0celo7

06:45

@0537 >optimization problem without any rigorous PDE stuff

lol

Random Americans on British CNN...

kevin Tah N.

10:36

can someone tell me what an orbifold is? Not an overly mathy disquisition. Just something understandable, may be some examples.

Secret

11:37

@kevinTahN. I think most quantum chemical background physicists (which are also the abstract algebra guys)) in this chat have went to holidays, I don't think they will be other's input very soon (and there might be less orbifod people here available to answer that question). However I can still try...

@Ghost do you know spectroscopy so you can help me on a conceptual problem on molecular electronic transitions?

Ghost

am afraid I know ittle of that

Secret

ok nvm then

kevin Tah N.

12:01

lol

Bass

@kevinTahN. as far as I understand it, it's a quotient space of a manifold under a group action that can be a manifold itself (but there are examples where it isn't a manifold)

example: let $M$ be the euclidean space $\mathbb R^2$, and consider the group action of $\mathbb R_+$ that translates points in $\mathbb R^2$ along their x axis, so a number $a\in\mathbb R$ acts on $M$ by $\mathbb R\ni (x,y)\mapsto (x+a,y)$

in this case, the orbifold $M/G$ is the set of lines that are parallel to the x-axis

because this is the set of orbits that the points in $M$ take under the group operation

in this case, $M/G\simeq\mathbb R$, so the orbifold is itself a manifold

Secret

@Bass Do you know spectroscopy, to b able to help me?

Bass

12:18

another example, the group action of $G=\mathbb R_*$ that acts on $\mathbb R$ by scaling the points, i.e. the group action of $a\in\mathbb R$ is $\mathbb R^n\ni \{x_i\}\mapsto\{a x_i\}$. In this case, the orbits are the straight lines through $0$, so the orbifold $M/G$ is the real projective space $\mathbb RP^{n-1}$

does that make sense? I'm not sure on the topic, so please correct me if that does not fit your definitions of orbifolds

you can just imagine an orbifold to be the "orbit manifold"

@Secret unfortunately no.. just write your questions in the chat or in a PSE question, I'm sure somebody will answer it.. you don't have to @ the people..

kevin Tah N.

@Bass thanks, this makes sense now

Secret

I have, and probably because so many people are on holidays, there isn't many answers and I suspect some have missed the question as they read the chat, hence hy the @

I planned to wrote it on PSE tonight

Bass

@kevinTahN. however, I think one has to be careful. In the second example, the orbit of the point 0 is trivial, so if you don't remove 0, the orbifold is the real projective space PLUS the origin, which is not a manifold. But the orbifold of $\mathbb R^n \setminus \{0\}$ is a manifold.

kevin Tah N.

@I was just going to ask when it is not a manifold

Huy

@kevinTahN. a lot can go wrong when you quotient. sometimes the resulting space won't even be Hausdorff

Bass

12:28

I think it's a manifold when all points transform "about the same" under the group operation. As soon as you have some special "singular" points that transform in a different way, then the orbifold is most likely not a manifold. But I don't know the exact criteria.

Huy

I think if the action is not by isometries, you might end up without a Riemannian metric

kevin Tah N.

wow!

0celo7

@kevinTahN. One of the canonical non-manifold orbifolds is $\mathbb{R}^2/\sim$ where $\sim$ is the identification of points by a rotation of $\pi/3$.

Or something like that.

jesus TeX

sorry for the 15 pings :P

kevin Tah N.

@0celo7 the 15 pings were epic :P

0celo7

@Huy Yes, this is important in GR. Hausdorffness is an important criterion for physical spacetime.

Huy

12:30

having something not Hausdorff is almost nowhere desirable I think :P

Bass

@Huy well I think my second example from above is an example where the orbifold isn't even Hausdorff. The orbits of the origin and another point cannot be separated by two disjoint open sets can they? Because every open set containing the orbit of the origin must be the whole orbifold..

kevin Tah N.

is para compact important? lol

0celo7

It is very important.

You need it to define integration (over "large" subsets of the manifold) and the metric.

kevin Tah N.

I thought so

0celo7

I think Geroch showed that the existence of a LC connection is equivalent to paracompactness...or something in that vein.

@kevinTahN. If you assume a partition of unity and a metric exist, you can forget about paracompactness.

Huy

12:34

@Bass: why does it have to be the whole orbifold?

0celo7

Actually just the partition of unity...it's used to define both the metric and the integral.

Huy

@0celo7: you assume partition of unity? shouldn't it be the other way around?

0celo7

@kevinTahN. Although...I swear some spacetime topology proof in Hawking-Ellis uses paracompactness.

@Huy Just take it for granted.

Huy

I've never seen it done in this order

0celo7

The proof that paracompactness $\implies$ PoU is boring.

So just assume that proof.

Huy

12:36

ah, you assume boring proofs?

:P

0celo7

Yes.

Huy

that's an interesting approach to doing maths

0celo7

@Huy But!

If you assume the Levi-Civita connection exists, you can prove the manifold is paracompact.

Or something like that.

Huy

well that is also way too much of an assumption

that's like saying "if you assume the floor is wet from a lot of rain you can prove it has rained"

0celo7

More like

Bass

12:38

@Huy because every environment (open set) of the origin intersects every line through the origin. So, in the orbifold topology, an environment of the origin must contain every line, thus the whole projective space. If that doesn't make sense, how do you define the topology of the orbifold?

0celo7

If you assume that you can do GR you can derive the mathematical structure of spacetime

kevin Tah N.

@0celo7 lol Hawking-Ellis literature scare me to shits :D

Huy

@Bass in English it's usually called neighbourhood :P

0celo7

environment?

Huy

@Bass that makes sense though

@0celo7: I think the "relations" are: 2nd countable, $\sigma$-compact and countable atlas are equivalent; each of these implies paracompact, and paracompact implied partition of unity

Bass

12:39

ok :D (I thought environment sounds smelly)

0celo7

@Huy fucking $\sigma$ compact

don't remember what that is and don't care

Huy

when it's a union of compact sets

it's like the easiest definition

0celo7

lalalalallala

is $\mathbb{R}$ a union of compact sets

Huy

lol

of course

$[0,1]$ is compact

0celo7

is it

I don't know BW

Huy

12:40

in the usual topology it is

0celo7

proof?

I don't know Bolzano-Weierstrass

Or is that the other one

Heine Borel

yeah I don't know that one

Huy

Heine Borel

which one is that? closed + (totally) bounded <=> compact?

if yes, which definition of compactness would you like to use?

0celo7

closed interval is compact

but there might be different formulations

I'm using the one from Wald's GR book

Huy

I haven't read it

0celo7

I could look at Lee but I'm lazy

Huy

12:43

tell me whatever definition you want

0celo7

every open cover has a finite subcover

is that not the standard one?

Huy

yes but there are several

0celo7

Paracompactness of a $C^3$ manifold $\Leftrightarrow$ existence of a $C^1$ connection

@Huy want the ref?

Huy

no

0celo7

@Huy hmm, what are some others?

Huy

12:47

sequential

BW: sequential <=> closed & bounded

0celo7

Geroch,R.P . (1968), 'Spinor structure of space-times in General Relativity. I', J. Math. Phys. 9,1739-44.

@Huy Hausdorff, Lorentz metric $\Leftrightarrow$ paracompact

Huy

ok

0celo7

@kevinTahN. So, really, paracompactness is a physical condition necessary for GR.

kevin Tah N.

so GR takes place on a haussdorff paracompact topological thing with a metric

0celo7

yes

kevin Tah N.

12:53

epic!

0celo7

but not any metric!

a Lorentz metric

kevin Tah N.

ah yes

0celo7

this is harder to obtain than a Riemannian metric

and there are some 4dim manifolds that do not have one

kevin Tah N.

can you give a simple example?

0celo7

$S^4$

kevin Tah N.

12:54

yes

of course

0celo7

to get a Lorentz metric one needs a nowhere vanishing vector field $v$

then one takes some Riemannian metric $g$

Valery Saharov

hi. is this possible to reopen physics.stackexchange.com/questions/224413/… ?

0celo7

then $g':=g-2\frac{g(v,.)\otimes g(v,.)}{g(v,v)}$ is a Lorentz metric, @kevinTahN.

@kevinTahN. in two dimensions the torus does not have a Lorentz metric IIRC

Huy

why

kevin Tah N.

hmm

0celo7

13:01

@Huy $\chi(T^2)=0$

Huy

and you need if for

?

0celo7

use Poincare-Hopf

there is no such $v$ on the torus

Huy

I don't know what a Lorentz metric is

I'm just wondering

0celo7

literally defined it above

Huy

that's just some weird expression

like I can understand what that means

0celo7

13:03

a metric with one negative eigenvalue

technically not a metric

a section of the symmetric rank two tensor bundle

with one negative and three positive eigenvalues

Huy

what does one need Lorentz metric for

I only know Riemannian

0celo7

@Huy relativity

Huy

is that the + + + - thing

0celo7

yes

kevin Tah N.

yes

Huy

13:06

I've only seen SR in a crappy physics course

where it was just a matrix

without anything deep behind

kevin Tah N.

@Huy me too, lol

0celo7

well I just told you the deepness behind it

Huy

very deep 0celo7

kevin Tah N.

yup

0celo7

so what is the standard notation for the symmetric tensor bundle

we have $\Lambda^k(M)$ for antisymmetric

Huy

13:07

idk

did you know that the eigenvalues of the Laplacians don't determine the structure of a manifold

I found that surprising

0celo7

I'm trying to read Jost which talks a lot about eigenvalues of the Laplacian

don't know enough PDE for that section tho

@Huy you really don't like the Spectre theme song

I think it's pretty good

Huy

I've stumbled across a Milnor paper of 1 page the other day where he gave an example of two manifolds that were not isometric but the Laplacians have same eigenvalues

@0celo7 I never said I didn't like it

@0celo7 chat.stackexchange.com/transcript/message/26198337#26198337

0celo7

Spectre

Huy

oh

yeah I hate it

0celo7

wtf

Huy

13:12

but I'd expect you to like it

you have no taste in music

0celo7

@Huy is this some fact that most people think they know or something

Huy

@0celo7: no but it is reasonable to try relate behaviour of Laplacian with geometric structure

0celo7

Jost says there are whole books devoted to studying the Laplacian

Huy

yes

0celo7

crazy stuff

Huy

13:14

PDE and FA is all about the Laplacian

yuggib

a free quantum particle is all about the Laplacian

0celo7

is it

Huy

study of the exterior derivative leads to de Rham which relates topological and smooth structure and the Laplacian goes even deeper

0celo7

@yuggib Proof?

kevin Tah N.

so I found a somewhat equivalent definition on page 3 of this people.mpim-bonn.mpg.de/hwbllmnn/archiv/dg2srm02.pdf

Huy

13:16

cuz then you relate even geometry with topology and the smooth structure

Atiyah Singer for example

yuggib

@0celo7 trivial

0celo7

I want to read a book on that, @Huy

Need some FA first!

Huy

@0celo7 probably useful

0celo7

@yuggib pls explain

yuggib

non-relativistic quantum mechanics is all about the Laplacian and its (maybe singular) perturbations

0celo7

13:18

huh

yuggib

@0celo7 the hamiltonian operator of the free particle is the Laplacian

0celo7

is it

yuggib

I assumed you knew that...

0celo7

would a German middle schooler know that

yuggib

probably yes

0celo7

13:19

really

yuggib

of course not

Huy

:D

0celo7

tfw went to German school thru 8th grade

fml

they never taught us any QM

Huy

I'd like to teach some QM

but that would require me to understand it first

:(

yuggib

anyways: Laplacian and quantum mechanics are strictly related

Huy

13:21

I can only solve some super easy Schrödinger and do some fancy annihilation tricks

0celo7

but isn't the Laplacian on $\mathbb{R}^3$ boring

yuggib

why?

Huy

why

yuggib

try to describe $-\Delta + V(x)$

0celo7

because engineers know about it

Huy

13:21

you still have many PDEs on R^3 with Laplacian

what do they know about it

0celo7

@yuggib I am no quantum physicist

yuggib

where $V(x)\in L^2_{loc}(\mathbb{R}^3,\mathbb{R}_+)$

Huy

many engineers don't know much more maths than looking up Fourier transforms

0celo7

*Laplace

Huy

that too

yuggib

13:22

hint: that ^ is a pretty reasonable potential

0celo7

@yuggib I don't know any QM

Huy

I know some teleportation stuff

0celo7

nothing beyond Shankar, really

yuggib

@0celo7 so? you said that the laplacian is boring

Huy

always Alice and Bob

0celo7

13:23

@yuggib it is

yuggib

and I am not talking about physics at all

0celo7

doesn't mean I can analyze $\Delta-V$

yuggib

so why is it boring?

0celo7

cuz

@Huy how did you end up doing that $L^2(\Gamma(TM))$ thing

$(X,Y):=\int_M X^\flat\wedge\star Y^\flat$ works too

Huy

@0celo7: still have to bring everything in order in my tex file

I don't wanna hodge star

0celo7

13:27

@Huy lame

Huy

no u

Secret

I need acuriousmind. He is the best quantum guy (who also knows a bit of abstract algebra and chemsitry) in town in this chat that is readily reachable. The hoildays prevent me from reaching him

0celo7

a bit

Huy

in town in this chat ?

0celo7

ACM doesn't know anything, really

yuggib

13:29

@Secret what is a quantum guy?

kevin Tah N.

lol

yuggib

one that behaves like a wave?

0celo7

@Secret you know @yuggib does QM for a living

kevin Tah N.

neh obeys superposition

Huy

one that collapses when you touch him

kevin Tah N.

13:30

good one

0celo7

@Huy heh because he's a noob

yuggib

@0celo7 Let's say that I do (mathematics disguised as) QM for a living

0celo7

Is the derivative of the geodesic flow $\in \Gamma(TTM)$?

Huy

T_T

Secret

but from my experience, he often answers quantum questions nicely throughout the site

@0celo7 I know, the problem is my latest 3 questions are quantum chemistry and spectroscopy. I have asked yuggib on that question before

I also tried to ask my peers who are also quantum chemists, but it seems quantum chemists rarely get into that hardcore QM stuff that is needed to address those questions

0celo7

13:32

@Huy a vector field on $TM$ is in $\Gamma(TTM)$, right?

yuggib

@Secret define "hardcore QM stuff"

Huy

@0celo7: a vector field is $ \in \Gamma(TM)$

ah on TM

0celo7

$\mathrm{i}\dot\psi=H\psi$

Huy

yeah probably

0celo7

pretty damn hardcore

does $TTM$ have any interesting properties

I know that categorical geometry book talks about it

Huy

13:34

how would I know

I never go so deep

TM is enough for me really

Secret

yuggib had helped me on understand the group theory part of the things, but a lot of us here are not known to have the spectroscopy background needed for that question (so far I have asked 5 people on that question already, Kevin said he might got an answer, but he want to see what others said first)

@yuggib
Something like using operators and scrodinger equations to descibe all kinds of relaxation process in an electronic system with many electornic levels, and the system in question is a multielectorn system

yuggib

@0celo7 charcterization of a strongly continuous unitary group via its generator on a Hilbert space

Huy

I know some EPR and Bell's basis shit

is that hardcore QM

yuggib

pretty starightforward

0celo7

@yuggib hmm

if I do graduate level PDE will I start talking like that

Huy

13:36

no if you do PDEs you'll just talk about regularity

and you'll end up saying smoothing operator instead of Green's function

yuggib

@Huy hint: strongly continuous group is all about regularity

Huy

and no physicsist ever will understand you

0celo7

@Huy wow so what happens if there's a swiss balarka sen who does geometry and he asks you about $TTM$ and you say "god dammit do your homework I never went that deep" and he goes home cries and takes a bottle of pills

kevin Tah N.

@yuggib straight forward ey lol

0celo7

???

Huy

13:37

@0celo7: I hit him in the face

0celo7

@Huy what about engineers

Huy

I hit those too

0celo7

what

Huy

no they won't even understand Green's function

most of them

0celo7

wow why all the engineer hate here

Huy

13:39

no hate

just truth

0celo7

you want to hit them!

Huy

that's what my parents did when I asked them questions about differential geometry they couldn't answer

0celo7

are they engineers

Huy

my dad is

0celo7

I can ask my dad the engineer about calculus and he can't answer

Huy

13:40

my dad can't either

he's retired though

0celo7

^

Huy

and he won some awards at IBM for being my dad

so I guess he knows some shit

0celo7

@yuggib how do I Sobolev space

Huy

close $C_c^\infty$ wrt $H^1$ norm

trivial

yuggib

@Secret I call that physical modelling, not hardcore QM...

0celo7

13:41

what does tht mean

Huy

learn topology

is what it really means

yuggib

@0celo7 what do you want to know about Sobolev?

0celo7

all of it

Huy

read the book I sent you from Einsiedler

FA is all about Sobolev spaces

yuggib

you need only Sobolev based on $L^2$ or on any space?

on $L^2$ is somewhat easier

0celo7

13:43

hmm, is $\langle X,Y\rangle=(\mathrm{d}X,\mathrm{d}Y)+(\mathrm{d}^\dagger X,\mathrm{d}^\dagger Y)+(X,Y)$ also an $L^2$ norm?

what about if we throw in some Laplacians too

yuggib

first of all, it is an inner product

0celo7

@Huy don't have it any more

also not reading it until I know some measure theory

yuggib

and complete inner product spaces are actually isomorphic to some $L^2$

0celo7

@yuggib uh Straumann was talking about Sobolev on spin bundles :/

let's do something easier

yuggib

that surely is some generalization using some f***ing Haar measure

Huy

13:45

why the hate towards Haar

yuggib

damn geometers

Huy

Haar measures are coo

l

0celo7

Theorem. $C^\infty_0(\Omega)$ is dense in $L^p(\Omega)$.

Huy

trivial

0celo7

$\Omega\subset \mathbb{R}^n$

@Huy :(

Huy

13:46

I play the guitar now

0celo7

help me with the proof pls

Think of me as one of your students

Huy

then your face would be all different

0celo7

?

Huy

see starred messages

yuggib

@0celo7 define the $L^p$ space and we can talk :-P

Secret

13:47

Problem is, while they have transition moment integrals to explain how electronic rotational and vibrational transition occurs, they don't write down operators that describes the various relaxation pathways.

I tried to solve this myself but I am clueless on what type of operators will be sensible for the job so I can investigate

Since quantum physicists deal with these kinds of stuff all the time, I felt like a more physicists oriented will know the answer to my question

Huy

just define it as the closure wrt to Lp norm

that's the easiest way to prove the theorem

:D

0celo7

@yuggib uh $||f||_{L^p(\Omega)}$ is finite

yuggib

@0celo7 how is that norm defined?

is $f$ a fucntion?

0celo7

$(\int_\Omega|f(x)|^p\mathrm{d}x)^{1/p}$

@yuggib probably

ACuriousMind

It should be a representant of an equivalence class of functions :P

0celo7

13:50

@ACuriousMind is that really so important

set of measure zero and all that crap

what does that even mean

Huy

YES

Secret

@ACuriousMind do you know spectroscoppy to help me on understand a concept?

ACuriousMind

@0celo7 Yes, because on functions the $L^p$ "norm" isn't a norm.

Secret

in that question I tagged you earlier?

0celo7

@ACuriousMind why

I really need to take analysis :(

ACuriousMind

13:52

@Secret Did admittedly not read it (since I have about 2 dozens pings from the last two days I have to catch up on), but I don't know anything specific about spectroscopy, just a few bits I might remember from when I had to do it once

Huy

@0celo7: or just think for a second which property could fail

0celo7

@Huy wtf is a norm, anyway

Huy

................................................................................‌...............

ACuriousMind

@0celo7 It should not be hard to check the properties of the norm for failure when you know that taking the equiv. classes gets rid of that problem.

0celo7

idk what the equivalence classes even are

Secret

13:53

@ACuriousMind ok nvm then

(Posting the question on the main site now (I'll tell you guys if I survived in a few days time)

@kevinTahN. You can post your answer there

0celo7

what is a set of measure zero?

Huy

@0celo7: a set whose measure is zero

0celo7

trol

yuggib

@0celo7 Lebesgue measure

Huy

I go play some Taylor Swift on my guitar now

0celo7

13:54

@yuggib what's that

Secret

If you integrate a measure zero set, it contributes nothing to the integral (If my memory serves)

they act like zeros

ACuriousMind

@0celo7 ...if you don't know what, you shouldn't be reading about FA, really

0celo7

@ACuriousMind no clue, honestly

@ACuriousMind I'm not!

yuggib

@0celo7 the "usual" measure on $\mathbb{R}^n$

0celo7

it's a riem geom book ;(

yuggib

13:55

@0celo7 then you should simply skip that theorem until you know some measure theory

0celo7

@ACuriousMind well @Huy said I should learn FA

@ACuriousMind I don't see which property fails

Huy

the first already does

0celo7

no...

that's a simple property of integrals

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