Mathematics - 2016-05-03 (page 2 of 3)

user19405892

6:00 PM

can a set be closed and not bounded?

Eric Stucky

What do you think?

user19405892

no since if it weren't bounded then it wouldn't be closed

right?

Mike Miller

Whether or not that's true, it's not an argument (you just wrote down the converse, without giving a reason for it).

s.harp

Hi, does anybody know what $\mathcal O(-3)$ is? It appears in the context of a bundle over the complex projective 2d space $\mathcal O(-3) \to \mathbb P^2$

Semiclassical

@eric what're you up to lately?

Eric Stucky

6:11 PM

jeje

The end of the semester is hitting me like a truck

Semiclassical

ahh, that sucks

Eric Stucky

because I basically took a very long break for most of april

s.harp

The total space of that bundle is apparently called "local $\mathbb P^2$" (and is a known thing)

Eric Stucky

I mean, yeah it sucks but it's not really surprising XD

I'm giving a talk :P

s.harp

unfortunatley searchign google for O(-3) and local P2 gives absolutely nothing related to mathematics :P

Eric Stucky

6:13 PM

that's most of what's on my mind right now

Semiclassical

ah, nice. when/where/what?

Akiva Weinberger

@user19405892 Yes.

[0,\infty) is a closed subset of the reals.

Eric Stucky

thursday / somewhere in Murphy / matrix-tree theorems :)

Semiclassical

hmm!

Mike Miller

@s.harp First, there's a line bundle called the tautological line bundle over $\Bbb P^2$. You know this guy?

Akiva Weinberger

6:14 PM

If a set is closed and bounded, it is called compact. (That is, if it's a subset of a Euclidean space R^n.) @user19405892

s.harp

No, I don't

Mike Miller

Start simpler: Do you know what line bundles are?

s.harp

yes

Mike Miller

Ok, consider the subset $E = \{(\ell,x) \in \Bbb P^2 \times \Bbb C^3 \mid x \in \ell\}$.

There's a map $E \to \Bbb P^2$, whose fiber is a line $\Bbb C$. One checks that this defines a holomorphic line bundle.

This is called the tautological line bundle because cmon look at it that's tautological

s.harp

Ok, yes, I vaguely remember seeing that one before

Semiclassical

6:17 PM

that sounds more self-referential than tautological :p

Mike Miller

It's just "if $\Bbb P^2$ is a space of lines, then consider the bundle of lines over it whose fibers are, uh, the line"

s.harp

yes

Mike Miller

This is called $E = \mathcal O(-1)$. The tensor products $\otimes_k E$ are called $\mathcal O(-k)$. Its inverse in the group of holomorphic line bundles is called $\mathcal O(1)$.

Semiclassical

is this K-theory-ish stuff?

Mike Miller

Just line bundles

s.harp

6:19 PM

Ok, thanks

Semiclassical

ok then

Joe Stavitsky

Is there some property of exponents that says x^(3/2)=sqrt(x)^3 and not sqrt(x^3)?

s.harp

The tensor product of two line bundles is again a line bundle right?

Semiclassical

@JoeStavitsky if $x>0$, all three expressions are identical.

Mike Miller

Yeah, @s.harp, the best way I know how to describe the tensor product is to tensor the transition functions to $GL_n$; for $GL_1$, tensor product is just multiplication

Joe Stavitsky

6:20 PM

@Semiclassical, was that for me?

Mike Miller

These are all of the holomorphic line bundles over $\Bbb P^2$ up to isomorphism. They're named such, because given a generic meromorphic section of $\mathcal O(-k)$, its zero/pole set gives a divisor, which defines an element of $H_2(\Bbb P^2;\Bbb Z)$

Given the canonical isomorphism $H_2(\Bbb P^2;\Bbb Z) \cong \Bbb Z$ (so that the generator is $\Bbb{P}^1 \subset \Bbb P^2$), the appropriate element is $-k$

s.harp

What do you mean with "divisor"?

Mike Miller

for $\mathcal O(-1)$, see this as follows: define $(x:y:z) \mapsto (x/z,y/z,1)$. there are no zeroes, and the pole set is the line $\Bbb{CP}^1$

linear combination of codim 1 subvarieties. you add up the zero sets and subtract the pole sets

Joe Stavitsky

@Semiclassical, if so, how do calc 2 trig subs work unambiguously? If you raise u^2+a^2 to the third power, you can't do pythagorean identity

s.harp

In your specific example the poles are $(x:y:0)$ which can be identified with $\mathbb{CP}^1$ via $(x:y:0) \mapsto (x:y)$?

Mike Miller

6:24 PM

right, when I say $\Bbb P^1 \subset \Bbb P^2$ I specifically mean that subvariety, yeah

Balarka Sen

what's a generic section?

Semiclassical

well, one issue is that trig functions aren't always $>0$. so that could produce some problems. but I don't really see what you actually mean without an example

user19405892

@AkivaWeinberger My question is is does a set being closed imply it is bounded?

Balarka Sen

or maybe I wouldn't want to know.

Mike Miller

what do you think a generic section is?

s.harp

6:25 PM

ok thanks @MikeMiller, I know very little algebraic geometry (or whatever this is :-) ), this was helpful

Mike Miller

sure, gladly. I have no idea why $\mathcal O(-3)$ is important.

Akiva Weinberger

@user19405892 No.

Mike Miller

oh, yeah I do @s.harp

Balarka Sen

@MikeMiller no idea, to be honest.

Mike Miller

i'm skeptical but oh well

given a complex manifold/algebraic variety, there's its holomorphic cotangent bundle $T^*M$; this is a holomorphic vector bundle of rank $n$. you can define $K_X = \Lambda^n T^*M$. this is a holomorphic line bundle, called the "canonical bundle"

user19405892

6:28 PM

@AkivaWeinberger Give me an example where not.

Joe Stavitsky

@Semiclassical, like so - ctrlv.in/747824 - clearly if first you raise to the numerator you will go nowhere

Mike Miller

@user19405892 He literally just did.

@s.harp ah the canonical bundle is actually $\Lambda^n TM$. I'm not so expert at this stuff

Anyway for $\Bbb P^n$ that's $\mathcal O(-n-1)$.

s.harp

ah

Mike Miller

no I was right the first time

it's the cotangent bundle. sorry!

the canonical bundle's sections are (possibly degenerate) volume forms; if there is a holomorphic section, you have a holomorphic volume form

Semiclassical

@JoeStavitsky i don't see an issue. for example, subbing $x=\tan\theta$ in $\frac{1}{(1+x^2)^{5/2}}$ gives $\frac{1}{(\sec^2\theta)^{5/2}} = \cos^5\theta$. nothing particularly strange there.

user19405892

6:30 PM

@AkivaWeinberger Are you saying $\mathbb{R}^n$ is a counterexample?

Mike Miller

for $\Bbb P^n$, we see that we cannot possibly have a holomorphic volume form; we'll have poles, because $-n-1 < 0$.

Joe Stavitsky

o right duh the sqrt is nothing special

s.harp

it looks like no matter how hard I try I cannot escape geometry and topology

Mike Miller

@Semiclassical $|\cos^5 \theta|$ :)

Semiclassical

pffff

Mike Miller

6:32 PM

mm, but I guess we're only doing this for angles such that $\cos > 0$, because we implicitly used an arctan up above

so you win

@s.harp you're some flavor of physics-liker, yes?

s.harp

I'm supposed to be a physics student^

Semiclassical

i should go grab coffee, my brain is just kind of dead right now

Mike Miller

That is not true.

You should review the definitions.

user19405892

sorry i forgot it is closed

Mike Miller

@s.harp Is this physics or "physics"?

@Semiclassical You would probably be into this local P^2 stuff since I just googled it and there are Picard-Fuchs equations everywhere.

Semiclassical

6:37 PM

ah, nice

what did you google specifically?

Mike Miller

local projective plane

Semiclassical

(if it involves the word 'haupftmodules' i don't have a freaking clue)

user19405892

The definition of closed is that every limit point is contained in the set. In $\mathbb{R}$ for example $\infty$ is a limit point but it is not contained in $\mathbb{R}$ so how is it closed?

Mike Miller

@user19405892 To talk about closed sets, you need to be living inside some bigger space. What space are we talking about when we say "$\Bbb R$ is closed"?

Semiclassical

if you mean the 'stability conditions' paper, it's above my head quite quickly

Mike Miller

6:39 PM

me too

s.harp

@MikeMiller its probably "physics", although maybe I will go do some applied stuff if this doesn't work out

Mike Miller

tsk tsk

user19405892

$\mathbb{R}^2$?

Mike Miller

@s.harp on the other hand "physics" makes what I do possible, so I can't tsk too hard :)

s.harp

are you tsking at the applied things or at the "physics"? :-)

Mike Miller

6:41 PM

the latter, though not in any seriousness

Semiclassical

@MikeMiller I kind've wish I did understand what's going on in that 'stability conditions' paper, if only because I did run into Bridgeland's papers on stability conditions in triangulated categories when delving into quadratic/strebel differentials stuff

Mike Miller

do you think about triangulated categories??

Semiclassical

oh god no

well

i've run into the triangulations themselves, i suppose

Mike Miller

my god, all is lost for you

Semiclassical

hah

Mike Miller

6:42 PM

oh, no no no, triangulated categories and triangulated surfaces are completely unrelated objects

Semiclassical

hmm. am i mixing up things

Mike Miller

my best friend does (something like) triangulated categories

he's a loon

s.harp

I wonder actually, in my mathematical physics seminars we have lots of strange arguments imported from physics and used to prove some mathematical theorems, like showing Gauß bonnet with path integral methods. Do mathematicians also use these physics methods or are they restricted to those that call themselves mathematical physicists?

Semiclassical

ah, yeah, i'm conflating things

this is the paper i had run into

and i really only looked at the quadratic differentials stuff

user19405892

@MikeMiller Is $\infty$ a limit point of $\mathbb{R}$?

Balarka Sen

6:45 PM

is $\infty$ even an element of $\Bbb R$?

user19405892

no

limit points don't need to be elements of the set

Mike Miller

Unfortunately, @user19405892, there are a few levels of confusion. There is a threshold after which one can try to bring someone out of confusion. But in this discussion (and in previous) you have tended to seem so confused about the very nature of what you're even asking (and then do not clarify for some minutes after someone prods) that I don't think I can do anything to help. So I'm not going to try to engage.

Balarka Sen

limit point of $\Bbb R$ inside what then?

Mike Miller

@s.harp I think there are a group of mathematicians that understand path integrals, but I don't, and that group tends to be filled with category theorists.

I think somehow the right notion of chern-simons theory on 3-manifolds and path integrals and whatnot is Reshitikhin-Turaev theory, which involves modular categories? One day someone will explain this to me.

That is not today.

user19405892

I am just trying to prove that $\mathbb{R}^2$ is closed

Semiclassical

6:48 PM

path integrals are weird

s.harp

I wish they didn't exist

user19405892

but for example in $\mathbb{R}$ we have $\infty$ is a limit point (?) of $\mathbb{R}$ but it is not contained in it so I am wondering how it is closed

Mike Miller

@s.harp do gauge theory like us mathematicians and you'll never see them!

Balarka Sen

You're confused, @user19405892.

You need to read the definitions.

Semiclassical

my main problem with path integrals is that when i saw it in coursework, it was in the context of QFT as taught by particle physicists

user19405892

6:50 PM

I have read the definitions and the defintion of a closed set is a set which contains all of its limit points

Semiclassical

and that's utterly painful

Mike Miller

isn't that the only place they appear?

s.harp

they can be done properly in stochastical systems, ie brownian motion and wiener process or something

Semiclassical

funny you should ask that. the book i have out next to me (for rather random reasons) is Altland and Simons book on "Condensed Matter Field Theory"

s.harp

when I heard it in QFT every second term in the derivation was a divergent quantity, the others were terms I did not know what their actual definition was

Semiclassical

6:52 PM

the third and fourth chapters of which are on the Feynman path integral and the functional field integral i.e. many-body path integral

Mike Miller

The physicists tend to say a bunch of nonsense and then be right

so I believe them and move on

Semiclassical

plus, as s.harp says, there's applications to nonequillibrium systems i.e. Keldysh formalism. my advisor has done a lot of work in that area, though i've stayed away from it

Balarka Sen

Someone should tell me what a path integral is at some point of time. But that time is not this time.

Mike Miller

There is no such thing as a path integral.

No need to worry about it!

Semiclassical

problem is, there's a difference between explaining the physical concept of a path integral, and actually making sense of it mathematically

the former isn't hard, the latter is

Mike Miller

6:55 PM

@Semiclassical I think the physical notion makes perfect "sense". It's just that the sense is nonsense :)

Semiclassical

hah

as an organizing principle of field theory, path integrals are great

as something that is actually well-defined, not so much :/

Balarka Sen

Better to learn what path integral means than watch Guardians of Galaxy.

Constructive way of wasting time.

Semiclassical

bottom line for me is: quantum mechanics is easy, quantum field theory is hard

s.harp

I think the book "Functional Integration And Quantum Physics" by Barry Simon is seen as the best rigorous treatment. Unfortunately it was very dense and I could not read it.

Mike Miller

The path integral means nothing, @BalarkaSen :)

Balarka Sen

6:58 PM

how do I know you're not lying?

Mike Miller

The physicists will back me up.

s.harp

The path integral is like drugs, if you do it and like it everything is great but if you don't like it and everybody around is doing it you don't feel too happy

DeNiSkA

'$∃ $' <----- what does this symbol mean?

Mike Miller

"Spatula"

s.harp

@DeNiSkA it means "there exists"

Balarka Sen

7:00 PM

lol

DeNiSkA

oh ho ho!

s.harp

ok, library is kicking me out, good bye guys :P

Balarka Sen

@s.harp i am happy with not drugging myself though.

Semiclassical

i'll object to the phrase "the path integral means nothing", if only because of how genuinely useful it has been

Balarka Sen

so Mike was lying

Semiclassical

7:01 PM

if i can use it to get physically meaningful results and predictions, it's not meaningless

the problem is that it doesn't have a sound mathematical underpinning.

Mike Miller

@BalarkaSen If I asked you to define the average value of a function on $\Bbb R$, what would you tell me?

Balarka Sen

that's nonsense unless $f$ is integrable on $\Bbb R$?

Mike Miller

Assume it's integrable, whatever.

Semiclassical

this MO question from a few years ago is relevant:

30

Q: Mathematics of path integral: state of the art

I was told that one of the most efficient tools (e.g. in terms of computations relevant to physics, but also in terms of guessing heuristically mathematical facts) that physicists use is the so called "Feynmann path integral", which, as far as I understand, means "integrating" a functional (actio...

Mike Miller

don't click that yet

Balarka Sen

7:08 PM

Ah, it's nonsense if it's not compactly supported I guess. I need $\int_{\Bbb R} f$ divided by the volume of the domain it's nonzero on.

Or at least I don't know what it means to say if $f$ is not compactly supported.

Mike Miller

@BalarkaSen Well, even then, I'm not sure I would say that's the average value "on $\Bbb R$", because that domain is larger than the zero set.

user19405892

what are some examples of figures with multiple centers of symmetry?

Mike Miller

The problem here, more or less, is that $\Bbb R$ does not have a translation-invariant probability measure.

Akiva Weinberger

@user19405892 $\Bbb R$ is closed in $\Bbb R$. (It's also closed in $\Bbb R^2$.)

The thing is, a set is not open or closed on its own. It needs a topological space containing it.

Mike Miller

@BalarkaSen Now, consider an infinite-dimensional hilbert space $\mathcal H$. Perhaps it represents the space of all quantum states.

Mambo

7:10 PM

IT HAS

Mike Miller

Let $f: \mathcal H \to \Bbb C$, say, be some smooth function, maybe energy.

The path integral of $f$ is the average value over all states.

Akiva Weinberger

$(0,1]$ is a closed subset of $(0,2)$ (with the subspace topology) but not a closed subset of $\Bbb R$, for example. @user19405892

Mike Miller

Therefore, the path integral is nonsense. :)

user19405892

@AkivaWeinberger Since $\infty$ is a limit point of $\mathbb{R}$, how is it closed if it isn't contained in $\mathbb{R}$?

Balarka Sen

lol!

Akiva Weinberger

7:12 PM

@user19405892 $\Bbb R$ is closed in $\Bbb R$, because it has no limit points in $\Bbb R$.

It is not closed in $\Bbb R\cup\{\infty\}$, though.

It depends on what the topological space is.

Semiclassical

the following comment to one of the answers to that MO question is relevant:

Mambo

@AkivaWeinberger Are you speaking of one point compactification here

Balarka Sen

@MikeMiller What makes physicists think it makes sense physically?

Akiva Weinberger

@Mambo I was thinking of the order topology, but either works for the example

Semiclassical

"It is possible to rigorously define the path integral. What you can't rigorously define is the 'Lebesgue' measure that physicists write as $\mathcal{D}\phi$. When a physicist writes $\frac{1}{Z}e^{iS(\phi)}\mathcal{D}\phi$, you should understand this as a notational shorthand; they're telling you roughly what form the finite-dimensional approximations you use to define the path integral ought to take. "

Mike Miller

7:15 PM

I don't know anything about what the physicists think.

Balarka Sen

fair enough.

Semiclassical

re: finite-dimensional approximations, i'd include in that the physicist penchant for assuming periodic boundary conditions in space and time

Mike Miller

finite dimensional approximations are Good and i like them

Semiclassical

the average value of a function on the real line may not make sense, for example, but the average value on the unit circle does.

Juan Sebastian Lozano

So, I saw this presentation I didn't entirely understand, but you can redefine $D\phi$ without the notion of measure, and instead using cohomology, and then the path integral becomes well defined.

I'll try to find the paper/notes it was based on

Mambo

7:17 PM

@AkivaWeinberger How are you going to define open sets in order topology for $\Bbb R \cup \{\infty\}$?

Mike Miller

i don't think that really helps me understand finite-dimensional approximation, @Semiclassical

Balarka Sen

@MikeMiller Off-topic: do you care much about Bowie's songs?

Mike Miller

I think the better analogy is $[-n,n]$

Semiclassical

that's fair.

n?

Mike Miller

some integer

you're approximating all of $\Bbb R$ by intervals of finite length

yes, very much so @Balarka

user19405892

7:18 PM

@AkivaWeinberger What are some examples of bounded figures that have multiple centers of symmetry?

Balarka Sen

I listened to Space Oddity and I was impressed.

Semiclassical

eh, that seems to me like a matter of boundary conditions

Akiva Weinberger

@Mambo Just define $a<\infty$ for all real $a$ and construct the order topology

@user19405892 I don't know what that means, sorry

Mike Miller

@Semiclassical I don't understand your point

the albums i keep on my phone are ziggy, diamond dogs, scary monsters, and his last album

Mambo

How is any neighborhood of $\infty$ looks like?

user19405892

7:19 PM

It is known that some bounded figures have multiple centers of symmetry so I am wondering which ones exist

Akiva Weinberger

@Mambo The basis is all sets of the form $\{x:a<x<b\}$ for $a,b\in\Bbb R\cup\{\infty\}$. The neighborhoods of infinity are the sets containing $(a,\infty)\cup\{\infty\}$ for real $a$.

anon

@user19405892 do you plan to explain what "center of symmetry" means in some precise terms?

Semiclassical

well, I take your point to be that while there's no notion of a probability measure on $\mathbb{R}$, you can consider the sequence of average values generated by the intervals $[-n,n]$

and while that may or may not converge as $n\to\infty$, it is at least a well-defined sequence

Balarka Sen

haha, nice. i liked space oddity very much because it had the same kind of tune and topic of songs of an old band i like to listen to has.

i look forward to listening more of his stuff.

Juan Sebastian Lozano

ncatlab.org/nlab/show/BV-BRST+formalism#HomologicalIntegration

Mike Miller

7:23 PM

yeah @Semiclassical

Balarka Sen

so one can define average value of f on R as a... sequence?

how's that used?

i mean it's an interesting idea but i am not sure what i'd do with it.

Mike Miller

as the limit of that sequence...

might get you answers you find counter intuitive tho

Balarka Sen

yes, that's what i am worried about

Semiclassical

and i guess my point is that, if I have a function which decays to zero sufficiently fast at infinity, I can find an approximation of it over a sufficiently large interval $[-L,L]$ with periodic boundary conditions

Mike Miller

no idea what that means

Semiclassical

7:26 PM

i.e. $f(x)\approx f_L(x)$ with $f(-L)=f(L)$

Balarka Sen

Oh I see. $[-n, n]$, not $[n, n+1]$. horrible misread.

Semiclassical

well, suppose I have a Gaussian $e^{-x^2}$

user19405892

@anon Hi anon, yes a center of symmetry $O$ of a closed, bounded figure is a point that such that for any collinear points $X,O,Y$ with $X,Y$ on the figure, $OX = OY$

Balarka Sen

so yeah of course that matches with the average value of compactly supported objects.

Mike Miller

then my suggestion says that the average value is zero

@BalarkaSen no... it very much does not...

my average value of a compactly supported function is zero

which, really, of course it is

CRUDBUMP: Ohio Shape feat. KC Green

►

Semiclassical

7:28 PM

one way I could get a sensible 'periodization' of $e^{-x^2}$ would be to replace it with $\sum_n e^{-(x-nL)^2}$

Balarka Sen

average value in the sense of $1/\text{vol}(\Omega) \int_{\Bbb R} f$ is not zero. do you mean the limit thing?

Semiclassical

and if $L$ isn't too small (large enough that the Gaussians don't overlap much), then I've got a function which is periodic on $[-L/2,L/2]$ and which approximates the original function on this interval to exponential precision

anon

@user19405892 if your figure is a single point, that means every single point O in the plane is vacuously a center of symmetry no?

Mike Miller

@BalarkaSen yes, the limit thing is zero.

Semiclassical

the advantage is that now I can use Fourier series to develop a sequence of finite-mode approximations

Balarka Sen

7:29 PM

um um um.

Mike Miller

you keep saying average value as if it's clear that the right place to take the 'average value' is strictly the domain over which it's not zero; that's not clear to me

Balarka Sen

oh crud you're quotienting by vol([-n, n]) = ${2n}$ each time.

I see.

Semiclassical

so now I'm able to study the Gaussian by Fourier analysis

it's in that sense which I say that periodic boundary conditions give rise to finite-dimensional approximations

Balarka Sen

@MikeMiller I suppose you have a point. You'd have to add up the contributions from the whole zero set, which I am not doing.

Mike Miller

also an important song, for when everyone here is done with Ohio Shape

Crudbump "NSA"

►

user19405892

7:31 PM

@anon I meant $O$ needs to lie inside

Balarka Sen

In that sense, yes, average value should be 0.

I am just a narrow-minded person I suppose.

Mike Miller

the first counterintuitive thing is that we learn that the average value of $1/(x^2+1)$ is zero

user19405892

So not every point in the plane is a center of symmetry of a point

Balarka Sen

whereas it should be $1$, indeed.

urk.

Semiclassical

...jeeze, how many conversations are going on at once right now

Mike Miller

7:33 PM

it 'should' be 1?

why?

Balarka Sen

the function is symmetric about $x = 0$, where it takes value $1$.

Mike Miller

ok?

Semiclassical

not exactly impeccable logic

Balarka Sen

in the case of average value over compact domains, if a function is symmetric about something, then that's where the average value lives.

Mike Miller

that's new to me

Semiclassical

7:35 PM

that don't sound right

user19405892

@anon A point is a special case because otherwise we don't need to specify that a center of symmetry lies inside the figure?

Mike Miller

so the average value of $1/(x^2+1)$ on $[-1,1]$ is $1$?

Semiclassical

the function acts symmetrically on its domain, but its range is certainly not symmetric

Balarka Sen

you're right, that was wrong.

Mike Miller

i could tell you that the actual average value there was $\pi/4$, or i could point out that because it's a continuous function and almost always less than 1, its integral is less than 2

Mambo

7:36 PM

@AkivaWeinberger you there

Semiclassical

to draw a connection between our two strands of conversation, the Gaussian will have 'average value of zero' according to Mike's prescription. But the periodized Gaussian won't.

Akiva Weinberger

@Mambo I am now

Mike Miller

i prefer mine

Semiclassical

meh

main reason i stress the role of Fourier analysis, though, is because that is precisely how things get justified in a physics textbook

Mike Miller

i want this mu

g

Semiclassical

7:38 PM

if only because Fourier analysis is easy to use

Mike Miller

thenewsharingmachine.com/products/dont-talk-to-me-mug

Mambo

@AkivaWeinberger If $\infty$ is included, how is order meaningful?

user19405892

@anon I think I was wrong. You are right that any point in the plane is a center of symmetry of a point.

Akiva Weinberger

Positive infinity. $+\infty$. Not the thing you get in the one-point compactification.

Semiclassical

i prefer this one: store.dieselsweeties.com/products/fucking-coffee-mug

Mambo

7:40 PM

Then what about $-\infty$

Balarka Sen

@MikeMiller I suppose I was sleepy. I was thinking of when the range is symmetric facepalm.

Mike Miller

i actually like everything on that site

thenewsharingmachine.com/products/yes-i-have-8x10-plaque

Mambo

Will $ \infty \in (a,b)$?

Akiva Weinberger

@Mambo I was just talking about $\Bbb R\cup\{+\infty\}$. It's not compact, but I didn't need it to be

Balarka Sen

I am being slippery about what I am saying today.

Either should stop being slippery or stop saying much.

Akiva Weinberger

7:41 PM

@Mambo Do you remember how an order topology is defined?

Mambo

In $\Bbb R$ yes

Akiva Weinberger

In any ordered set…

Semiclassical

getting back to the path integral stuff, I think the following 'joke' is relevant (no link this time)

Suppose $f(x,y)=x^2+y^2$. What is $f(r,\theta)$?

Akiva Weinberger

en.m.wikipedia.org/wiki/Order_topology @Mambo

Mike Miller

@Semiclassical nobody likes that joke when i tell them.

Semiclassical

7:43 PM

'joke' is maybe the wrong word

though it'd be more amusing with both physics people and math people in the conversation

Balarka Sen

It's $r^2 + \theta^2$.

Semiclassical

@balarkaSen and therefore you are a math guy, not a physics guy.

Mambo

Okay, with that order topology, $\Bbb R \cup \{\infty\}$ is homeomorphic to what?

Semiclassical

if you were physics, you'd have said it was $r^2$.

Balarka Sen

I know, it's old.

Akiva Weinberger

7:44 PM

@Mambo It's homeomorphic to $(0,1]$

with $\Bbb R$ being mapped to $(0,1)$

Mambo

okay

Semiclassical

the point being, of course, that the physicist would assume a certain abuse of notation, i.e. that $(x,y)$ isn't a label of inputs but rather Cartesian coordinates

so there's a certain amount of baggage that isn't being stated.

similarly, the path integral shouldn't be as a literal construction so much as a convenient way of implying a whole slew of assumptions about how such a construction should proceed.

Mambo

@Semiclassical Did you study degrees of freedom in Classical mechanics?

Semiclassical

probably

i took my last classical mechanics course five years ago

Mambo

Can you guess what theorem is being exploited out there?

Semiclassical

7:48 PM

out where?

Mambo

In calculating degrees of freedom of a system

Semiclassical

couldn't tell you

Mambo

and thats what physicists do

Forever Mozart

8:41 PM

hi everybodie

Akiva Weinberger

Remember that theorem I mentioned earlier? About how $\Bbb R^n$ can't be written as the union of two connected open sets with disconnected intersection? And there's a quick LES proof. I think it's really illuminating to see what happens at the level of cycles, because the LES proof kinda obscures what's going on.

Forever Mozart

what is a LES proof?

Akiva Weinberger

There's a proof that uses a "long exact sequence" of homology groups.

Forever Mozart

and you must assume $n>1$

Akiva Weinberger

You sure?

Forever Mozart

8:49 PM

oh I thought you said CAN

Akiva Weinberger

When I said "level of cycles" I meant "level of chains," I guess

So, first off, connectedness is the same as path-connectedness for open sets, and all the relevant sets are open

and path-connectedness is the same as saying that, for any two points $x$ and $y$, there's a 1-chain whose boundary is $x-y$

So, call the open sets $A$ and $B$; we want to show that $A\cap B$ is connected

So, choose arbitrary $x$ and $y$. Because $A$ and $B$ are connected, there's are paths (1-chains) $\alpha_1\in C_1(A)$ and $\beta_1\in C_1(B)$ whose boundaries are $x-y$

$\alpha_1-\beta_1$ is a 1-cycle in $\Bbb R^n$, so it's a boundary of some 2-chain. You can write that 2-chain as $\alpha_2+\beta_2$ for $\alpha_2\in C_2(A)$ and $\beta_2\in C_2(B)$, by subdivision

Then $\partial(-\alpha_2)+\alpha_1=\partial(\beta_2)+\beta_2$ is a chain in $A\cap B$, with boundary $x-y$

and so it's the desired 1-chain connecting $x$ and $y$. Since $x$ and $y$ were arbitrary, any two points can be connected by a 1-chain in $A\cap B$.

Thus, $A\cap B$ is connected; QED.

Forever Mozart

i will think about that

Akiva Weinberger

And that whole thing is essentially a one-line proof, when one uses (reduced) Mayer–Vietoris.

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