Mathematics - 2017-05-31 (page 2 of 4)

Jorge Fernández Hidalgo

17:00

its whats used to formalize probability

bye

SBM

oh

Semiclassical

@Lozansky As for what's required...eh. I'm not great at that.

Lozansky

@Semiclassical Yeah, I agree. So is this correct $(\mathbf{A} \cdot \nabla B_i) \hat{e}_i = (\mathbf{A} \cdot \nabla) \mathbf{B}$?

Semiclassical

Not as you've written it, but that's not what you wrote before.

Lozansky

wait

Semiclassical

17:01

you missed \nabla in there, I think.

Lozansky

Yeah

Semiclassical

\cdot\nabla

There you go.

Lozansky

This is what happens when you copypaste carelessly

Semiclassical

Yeah, I think that's fine.

As I say, though, I don't know what conditions are required for this to be rigorous.

Lozansky

I'm unsure as well. They definitely need to be $\mathcal{C}^1$ on $V$ at least

Semiclassical

17:04

Sounds sensible. But I'll leave that to whoever chooses to answer the question proper.

Lozansky

Yeah, I have a feeling it's gonna get buried though :P

Semiclassical

Yeah, it happens

But at least you've got the formal side of it fine.

W.R.P.S

@JorgeFernándezHidalgo Thanks @JorgeFernándezHidalgo but is it true that ($\{(−\infty, a] : a\in \Bbb R\}$) not contains all set in $\mathscr B(\mathbb R)$?

Semiclassical

I guess what I'd focus on is that, in order for the use of the Divergence to be sound, you need $B_i \mathbf{A}$ to be sufficiently well-behaved for all of $i=1,2,3$

Which really means you need $B_i A_j$ to be well-behaved for any pair of $(i,j)$.

Lozansky

Right, that makes sense

Semiclassical

17:12

Another way to look at this, btw, is to think in terms of row vs. column vectors.

In that case, if $A,B,n$ are column vectors then we can write the above integral equivalently as $\int_S B A^T n\,dS$.

So you're effectively asking for the flux of the outer product $BA^T$, and the divergence theorem gives you a volume integral over the divergence of this outer product.

Secret

$$2\int e^{x^2}dx=\frac{e^{x^2}}{x}+\frac{\int e^{x^2}dx}{x^2}+2\frac{\iint e^{x^2}dx}{x^3}+\cdots = \sum_{n=0}^{\infty}\frac{n!\int^{(n)}e^{x^2}d^n x}{x^{n+1}}$$

Semiclassical

@secret This may be of interest: en.wikipedia.org/wiki/Cauchy_formula_for_repeated_integration

Secret

Hmm ok, let's see...

W.R.P.S

@JorgeFernándezHidalgo sorry in Def2 we have if $X^{−1}(A') \in \mathscr{A}$ for any $A' \in \mathscr{A'}$. how "for any" works in Def1 ? $A' \in \mathscr{A'}$ contian all sets in borel? . my problem is this "for any" in Def1 .

Secret

Using Cauchy repeated integral formula:
$$2\int e^{x^2}dx=\sum_{n=0}^{\infty}\frac{n!\int^{(n)}e^{x^2}d^n x}{x^{n+1}}=\sum_{n=0}^{\infty}\frac{n}{x^{n+1}}\int_a^x (x-t)^{n-1}e^{t^2}dt$$

Semiclassical

17:24

n!/(n-1)!=n.

Next question is whether or not you can resum $\sum_{n=0}^\infty \frac{n}{x^{n+1}}(x-t)^{n-1}$.

Jorge Fernández Hidalgo

@W.R.P.S $\mathscr B(\mathbb R)$ is generated by the intervals of form $(-\infty,a]$

Semiclassical

For that it helps to write it as $-\displaystyle \frac{\partial}{\partial t}\sum_{n=0}^\infty \frac{(x-t)^n}{x^{n+1}}$

Which simplifies to $-\frac{\partial}{\partial t} \frac{1/x}{1-(x-t)}=\frac{1/x}{(1-x+t)^2}$...or at least, it would if I still trusted this.

It feels pretty dubious.

Secret

uh I thought you cannot pull the $(x-t)^{n-1}$ from the integral due to the t dependence?

Semiclassical

I'm not. I'm putting the sum inside the integral

Secret

Ah I see (the two functions are bounded, thus the switching of the sum and integral is valid...)

ok let's see...

Semiclassical

17:31

(That said, it feels like something has gone wrong along the way. So, caveat emptor.)

W.R.P.S

17:45

@JorgeFernándezHidalgo @JorgeFernándezHidalgo is set of intervals in $(-\infty,a]$ form a sigma algebra? I think this class of set isn't closed under countable union , so if we want to generate $\mathscr B(\mathbb R)$ we add some sets, but how check them in $X^{-1}((−\infty, a]) \in A$ for any $a \in \Bbb R$? I can't understand it :(

Jorge Fernández Hidalgo

no no

ok

lets go step by step

the first step is to prove that the set $J$ I defined previously is a sigma algebra

s.harp

I'm looking for a word, if I have some problem and I say "well x seems like something that heurstically does the right thing, so whenever we have this problem just do x", how would one describe such a solution? ad hoc, sweeping... ?

I think "crutch" is a good word

Secret

$$\sum_{n=0}^{\infty}\frac{n}{x^{n+1}}\int_a^x (x-t)^{n-1}e^{t^2}dt=\int_a^x\sum_{n=0}^{\infty}\frac{n(x-t)^{n-1}}{x^{n+1}} e^{t^2}dt=-\int_a^x\frac{\partial}{\partial t}\sum_{n=0}^{\infty}\frac{(x-t)^n}{x^{n+1}} e^{t^2}dt=-\int_a^x\frac{\partial}{\partial t}\left(\frac{\frac{1}{x}}{1-1+\frac{t}{x}}\right) e^{t^2}dt=-\int_a^x\frac{\partial}{\partial t}\left(\frac{1}{t}\right) e^{t^2}dt=\int_a^x\frac{e^{t^2}}{t^2} dt$$

And finally:

Semiclassical

Might want to use align for that :P

Secret

$$2\int e^{x^2}dx=\int \frac{e^{t^2}}{t^2}dt$$

s.harp

17:54

what are the bounds of these integrals?

Secret

(Uh, previous long message has timed out...)

Semiclassical

heh.

In any case, I find it pretty dubious.

s.harp

were you being careful with switching sums with integrals? :)

(And derivatives with sums/integrals)

Semiclassical

The lack of bounds on these integrals is the most likely source for stuff going wrong.

s.harp

if its wrong (im not sure what it is) my bet is on a switch of sum/derivative with integral breaking the truth

Secret

17:56

s.harp: They are supposed to be indefinite integrals. I got all of this from repeatly using integration by parts (which in turn is from finding it pops up from 1D change of variables in the most general case)

That switching I am still not very comfortable with, but my fingers are not quick enough and not experienced enough to do dominant convergence proofs

s.harp

the other day I was looking $\int_{\Bbb R}e^{-x^2-x^4}\,dx$ and I thought hey, write $e^{-x^4}$ as a power series, use that $e^{-x^2}$ suppresses all polynomial integrals and use the explicit formula $\int e^{-x^2}x^n$ to find a sum for the expression!

turns out the sum you get by doing this diverges

because I wanst careful with switching the sum with the integral

lesson: even in super simple examples things go wrong if you are not careful

Secret

It seems I am still quite terrible with infinite series, I guess I need to train more

Astyx

Hi chat

s.harp

(Is the equation $\int f(x) = \sum_n \frac{n! \int^{(n)}f}{x^{n+1}}$ really true?

i havent seen it ever

Secret

(It all begins from this formula that I have been messing around with inspired from the $\int \sin (\sqrt{u})du$ problem (I have derivations of it elsewhere, will post it if needed))

Dodsy

18:02

hey @Astyx

Semiclassical

@s.harp That way be asymptotic series.

Secret

1D Change of variables
Let $u=g(v)$ and $A'(x)=a(x)$ Then
\begin{align}
\int a(f(u))du &=\int \frac{a(f(g(v)))(f(g(v)))'}{f'(g(v))}dv\\
& =\int \frac{a(f(u))f'(u)}{f'(u)}dv\\
& =\frac{A(f(u))}{f'(u)}+\int \frac{A(f(u))}{(f'(u))^2}f''(u)du
\end{align}

s.harp

@Semiclassical can you elaborate? I dont quite get what you mean with your comment :) (sorry: im stupid today)

Dodsy

I think he meant "may be"

s.harp

@Dodsy this is also confusing to me

Dodsy

18:05

Asymptotic expansion

In mathematics, an asymptotic expansion, asymptotic series or Poincaré expansion (after Henri Poincaré) is a formal series of functions which has the property that truncating the series after a finite number of terms provides an approximation to a given function as the argument of the function tends towards a particular, often infinite, point. Deep investigations by Dingle reveal that the divergent part of an asymptotic expansion is latently meaningful, i.e. contains information about the exact value of the expanded function. The most common type of asymptotic expansion is a power series in either...

Secret

Btw, if we don't switch the integral and sum, we get this instead:

\begin{align}
2\int e^{x^2}dx & =\sum_{n=0}^{\infty}\frac{n!\int^{(n)}e^{x^2}d^n x}{x^{n+1}}\\
& =\sum_{n=0}^{\infty}\frac{n}{x^{n+1}}\int_a^x (x-t)^{n-1}e^{t^2}dt\\
& =\sum_{n=0}^{\infty}\int_a^x \frac{n}{x^{n+1}}(x-t)^{n-1}e^{t^2}dt\\
& =-\sum_{n=0}^{\infty}\int_a^x \frac{\partial}{\partial t}\left(\frac{(x-t)^n}{x^{n+1}}\right)e^{t^2}dt
\end{align}

s.harp

so its like a power series expansion at $\infty$ in the riemann sphere?

Semiclassical

Asymptotic series are weird and I'd rather not try to explain in detail. Basically it amounts to: For some functions, one can obtain series expansions which are formally valid but which don't converge.

Even so, those series expansions still end up being good approximations if you truncate them after a certain number of terms.

s.harp

@Secret are you sure this is correct? when I'm looking at it right now it looks like you are missing a $g'(v)$ in the second line and I dont see how the 3rd follows from the 2nd

@Semiclassical this reminds me of listening to people talk about QFT, basically the terms of a series someitmes go to infintiy but if you cut it off cleverly (without renormalisation ? ) you get something that agrees with experiment or something like that

Semiclassical

It's the same thing, yeah.

Dodsy

18:10

It's pretty crazy.

Semiclassical

The perturbation series fails to converge and so is only useful as an asymptotic series.

You can sometimes do better than that if you do stuff like Borel summation, as a way of handling those asymptotic series in a more controlled way.

But this way be resurgence theory and a whole lot of crazy crap which I don't understand.

Secret

Change of variables 1D including derivation

Given known integrand of the form: $a(f(u))$ Find:

$$\int a (f(u))du$$

Let substitution be $u=g(v)$ such that $f(g(v))=h(v)$. Then $f'(u)du=h'(v)dv\Rightarrow du=\frac{h'(v)}{f'(u)}dv$ (If $\exists u,f'(u)=0$, then $\exists v, h(v)=C$)

Therefore: $\int a (f(u))du = \int \frac{a(f(g(v)))h'(v)}{f'(g(v))}dv= \int \frac{a(h(v))h'(v)}{f'(g(v))}dv$. Let $A(x)=\int a(x) dx$. Then using integration by parts

$$\int \frac{a(h(v))h'(v)}{f'(g(v))}dv=\frac{A(h(v))}{f'(g(v))}+\int \frac{A(h(v))}{(f'(g(v)))^2}f''(g(v))g'(v)dv=\frac{A(f(u))}{f'(g(v))}+\int \frac{A(f(u))}{(f'(u))^2}f''(u)du$$

Daminark

Hai

Semiclassical

@s.harp See these slides for what I'm alluding to: faculty.washington.edu/acherman/PCM_Resurgence.pdf

Dodsy

DAMINNNNN

Hey man how arya.

$\sigma$

Daminark

18:12

I'm alright, how about you?

s.harp

Yes, I studied physics but unfortunately as an undergraduate I loved mathemtics and all the stuff we did in the undergrad lessons I could formulate in a mathematically satisfying manner

then came QFT

and I realised I dont actually like what physicists call physics

Dodsy

Which is?

Semiclassical

QFT is weird.

Dodsy

oh

Semiclassical

I wouldn't call it the entirety of physics, though.

Dodsy

18:13

quantum field theory?

Semiclassical

Right.

That said, there's a reason why all my stuff tends to be 1D quantum mechanics.

Dodsy

You use feynmann diagrams for that

Semiclassical

Namely, so that I understand what the hell I'm doing.

Daminark

naively wonders does this have anything to do with class field theory?

Secret

Alternately, the formula can be easily derived by multiplying the integrand with $\frac{f'(u)}{f'(u)}$ and then integrate by parts

What surprised me on that derivation is how the final outcome does not depend on what change of variable $g(v)$ that is used

Semiclassical

18:13

NO

gtfo

Dodsy

D:

s.harp

@Secret it may take a while for me to read what you wrote

Semiclassical

nerd rage

Dodsy

No feynman diagrams, I understand.

Daminark

naively assumes the "no" is to Dodsy

Semiclassical

18:14

Nah, QFT is all about Feynman diagrams.

s.harp

but you made the effort so I feel compelled to do it D:

Semiclassical

@Daminark WRONG

Daminark

is surprised that Semi is contradicting himself

s.harp

it was a double valued no

Dodsy

What's happening

first no to feynman diagrams

now yes

Ted Shifrin

18:15

Hi Nate, Demonark, Semiclassic, harp. I'm taking a break from packing.

Dodsy

Oh great!

Daminark

Yeah I've got no clue what's going on, Semi seems tired

s.harp

qft has feynman diagrams, "schoolbook qm" has none of that

Daminark

Hey @Ted!

Secret

As for the $e^{x^2}$ formula I got above and discussed with Semiclassical, I basically keep integrating the e^{x^2} and differentiating the $\frac{1}{x}$ term. Repeated application of integration by parts give that infinite series

Semiclassical

18:15

Well, given that I just said that Daminark was wrong about said naive assumption... :P

Daminark

I think I got the jumble, key to success!

Dodsy

-_-

Semiclassical

I should start posting the Cryptoquip for people.

In all of its horrible punny glory.

Daminark

Nice

s.harp

but if you want you can do feynman diagrams even for stochastical dynamics! so there is no limit to where you can apply them, they really just express some combinatorical shit

Semiclassical

18:16

Yeah, virial expansion /cluster diagrams stuff

Dodsy

I like the idea of this.

@TedShifrin What time is your flight?

Semiclassical

Which is not that surprising given how much similarity there is between statistical field theory and quantum field theory

The two are very very similar in formulation.

Ted Shifrin

I hate it when early Jumblers give away the answer.

Semiclassical

The former just has the advantage of already being Euclidean from the start :P

Ted Shifrin

It leaves 7 AM tomorrow, Nate, so I have to leave the house before 5:30 AM, I guess.

s.harp

18:17

@Semiclassical im not talking about statistical field theory but the master equation on super simple systems (like for example well mixed rabbits in a cage that can fight and reproduce etc)

Daminark

@Ted sorry

Also o lawd

Ted Shifrin

no, Demonark, not you. This is hours later.

Semiclassical

I think it might still be related.

Daminark

Oh phew

Ted Shifrin

I posted it over 3 hours ago.

Semiclassical

18:18

I mean, you've already got time as a continuous variable in that case.

s.harp

well I would not disagree with that statement

Daminark

Did someone do that today?

Dodsy

QIWQ YMOLZWQLZ OILGM ZMWSSX ATSX EAZRUQAZM ELZ HX ILHM U’Y GWX IM’G WR UREMZULZ UREMZULZ YMGUTRMZ

Semiclassical

So it's probably related to the Feynman path integral, just for QM not QFT.

Daminark

"Feynman path integral"
Should I be concerned?

Alessandro Codenotti

18:19

Hi chat

s.harp

No, you should be angry!

Daminark

@Dodsy I got "Scramble" out of that

s.harp

cite that wikipedia page about no lebesgue measure on banach spaces!

Daminark

Hey @Alessandro

@s.harp wat? You mean the thing about how you can't have a finitely additive translation invariant measure on $\ell^2$ for which the measure of a ball is positive and finite?

Secret

[Random] $$\sup({}^{\omega}j|j\in\Bbb{N})=\epsilon_0$$
Proof: Let some monotonic increasing function $f$. Then:
${}^{\omega}j< f({}^{\omega}j)<{}^{\omega}(j+1)$
Take $j < \omega$

$\epsilon_0< f({}^{\omega}j)<\epsilon_0$

Therefore $f({}^{\omega}j)=\epsilon_0,j < \omega$

s.harp

18:23

yes, the path integral is formally written as $\int d\psi e^{iS[\psi]}$ where the integral is over a space of paths $C^1([0,1];\Bbb R^n)$

Ted Shifrin

hi @Alessandro

Semiclassical

@Dodsy My starting guess for that what be QIWQ = THAT, U'Y = I'D, and IM'G = HE'S

Ted Shifrin

Or I'M, Semiclassic.

Semiclassical

Not so sure about Y=D. Could be Y=M.

Dodsy

oh that's much different than where I started

I did U'Y = I'm and IM'G = It's

Ted Shifrin

18:24

That's my only contribution. I have stuffs to do.

Daminark

Weird

Semiclassical

Well, that ABCA pattern shows up a lot

@s.harp Fun fact: one way to get to semiclassical approximations is to do saddle-point analysis of the Feynman path integral

s.harp

to me this fact is not fun

Semiclassical

That way be instantons.

s.harp

I know

I did a talk once called "Instantons and Morse Theory in SUSY QFT"

Semiclassical

18:28

of course, that way also be the fluctuation determinant

s.harp

I still dont know wtf an instanton is apart from that non-sense blahblah "classical solutions of the eom"

Semiclassical

And that thing is such a pain

TheGreatDuck

hi

Semiclassical

I tend to approach semiclassical stuff from the WKB approximation. Or at least results derived from that

2

TheGreatDuck

lol

you approach all of your stuff from the WKB approximation!

Semiclassical

18:31

Bohr-Sommerfeld quantization ftw

TheGreatDuck

nvmd. bad pun.

Mike Miller

@TedShifrin I need Thom's jet bundle transversality thereom for Banach spaces. Ideas on where to find this? Golubitsky-Guillemin?

Ted Shifrin

That's a good place to try, @MikeM. I honestly don't remember their doing Banach, though.

I personally don't know what works and what makes sense.

Mike Miller

The changes should be super elementary. I could do it myself, I just don't want to.

Ted Shifrin

Is somebody finite-dimensional?

Mike Miller

18:34

No. Not really a big concern. You'll want both kernel and cokernel to be complemented.

s.harp

@Semiclassical In a mathematics class I was tutoring last semester one the questions on the exam was (an exam dressed version of) "Find the energy levels of the quantised harmonic oscillator using plancks original method"

Mike Miller

I don't know if you need eg a continuous family of complements.

Ted Shifrin

I honestly have never thought about any of this.

Mike Miller

This is the part of geometry that is almost trivial to extend.

Riemannian geometry, exponentials, those are the hard things...

Semiclassical

Not sure I know what that means exactly

I mean, doing it via Bohr-Sommerfeld is easy

Mike Miller

18:37

springer.com/us/book/9780387964997 Seems plausibly in here

Semiclassical

But I don't remember what Planck's method was

Mike Miller

I'll check our library later

Semiclassical

Something along the lines of equipartition?

s.harp

Break phase space into regions $\Lambda(a,b)=\{(x,p)\mid a≤ \text{energy of }(x,p) ≤b\}$, then define $a_1 $ so that $\Lambda(0,a_1)$ has volume hbar, define $a_n$ so that $\Lambda(a_{n-1},a_n)$ has volume hbar and then the energy $E_n$ is the expectation of the energy in $\Lambda(a_{n},a_{n+1})$

you get the right values, including the zero point energy

we thought it was pretty neat while we corrected it

W.R.P.S

@JorgeFernándezHidalgo @JorgeFernándezHidalgo $J={s : X^{-1}(s)\in A }$ I couldn't .i'm trying to prove it . but my problem is not $\mathscr {B}(\Bbb R)$ = $\sigma((−\infty, a] : a\in \Bbb R)$ i know it, but mybe I don't fully understand it, Ok i'm trying to prove J is sigma algebra Thanks

Mike Miller

18:39

@TedShifrin Yeah I think it's in there. Sorry to bug you. I'm back out.

s.harp

I dont know what Bohr-Sommerfeld is exactly, but my suspicion is that its the same

Eric Silva

Hi chat

Semiclassical

Hmm. I'm surprised that gives the zp energy right.

s.harp

@MikeMiller About this book: "The main concern in all scientific work must be the human being himself. This, one should never forget among all those diagrams and equations"

Semiclassical

To sum up Bohr-Sommerfeld: You require that the action of the classical orbit (the phase space volume enclosed by the orbit) be quantized in units of Planck's constant.

Mike Miller

18:42

@s.harp it's just a quote

s.harp

@MikeMiller I know, I thought it was a funny way to introduce the book though. Especially since its such a specialised book

Mike Miller

fair enough

Semiclassical

But to make that work you need to say that it's a half-integer multiple.

s.harp

If it were an introductory book it wouldnt be as noteworthy^

Mike Miller

it's far enough from my specialty but I'm still getting something out of it, probably :)

Waiting

18:44

@Hippalectryon o/

s.harp

@Semiclassical interesting, its so strange that these things work at all

Waiting

@Hippalectryon how is it going? I'm preparing to go jogging and return a bit later.

Semiclassical

What makes me especially suspicious is that getting the harmonic oscillator right requires that the difference between the classical minimum and the ground state is half the distance from the ground state to the first excited state

s.harp

It would be interesting to see what kind of things you can learn from doing old quantum theory like things with modern mathematics, the other day there was a talk about the connection of Riemann Hypothesis with physics, and the guy mentioned doing some old quantum theory stuff to find funky things

Semiclassical

I'm quite dubious you can reproduce that if the quantization rule is in terms of integers rather than half-integers

s.harp

18:47

what do you mean integers as opposed to half integers?

Semiclassical

0,1,2,3,... versus 1/2,3/2,5/2...

s.harp

orly

but I mean how does it connect to things said previously?

as in the volume has to be a half integer multiple of hbar?

Ted Shifrin

Hi @EricSilva. Did you ever settle that question about critical pts?

Semiclassical

Well, the classical minimum corresponds to zero energy which is action = 0 so long as the Hamiltonian is just p^2+q^2.

s.harp

ok its time for me to go home, see you around!

Semiclassical

18:49

Later, @s.harp

RaisingAgent

Hi, I got a really dumb question, but I just need confirmation (like rubberduckdebugging):
`-9x² - (-2x²) = -9x² + (2x²) = -7x²` plis :'(

Semiclassical

Yes

RaisingAgent

So - ( - ) = + ( ) ...

thx a lot

Eric Silva

@Ted I managed to settle it in the case of eigenfunctions. The idea was that critical points were all non-degenerate from calculating the Hessian. Then when you look at the function along geodesics coming out of an extremum it takes a nice form and you end up being able to show that the sets of Minima and Maxima are connected, so they only have one point

It was a fun problem

In that case the manifold had to be a sphere by Morse theory, I still wonder about if you can bound the number of critical points in a more general case

PVAL-inactive

@EricSilva What's your morse theory problem?

co(homology) gives lower bounds on the number of critical points of a given index on many manifolds.

Besides that it can get quite hard.

Astyx

19:01

Hi @Ted @EricSilva @PVAL @s.harp @everybody

Eric Silva

It isn't really a Morse theory problem, it was just trying to study the critical points of a function satisfying some equation

Hi Astyx

PVAL-inactive

It's an open question (which if true would imply a lot of the most interesting 4-manifold questions) if every simply connected smooth closed 4-manifold admits a Morse function with only minima, maxima, and critical points of index two.

Ted Shifrin

Oh, that's interesting, @EricSilva.

salut, @Astyx

@PVAL: I guess $S^1\times S^3$ would be an interesting case.

Astyx

Quoi de neuf ?

Ted Shifrin

@Astyx: Tomorrow at this time I'm on my way (almost across the US, then waiting).

Eric Silva

19:04

Yeah basically I ended up showing that a sufficient condition for some Einstein manifold to be a sphere was the existence of Eigenfunctions of a particular form

Astyx

Bientôt le grand voyage ?

PVAL-inactive

@Ted That isn't simply-connected

Ted Shifrin

Exactement, Astyx.

Oh @ simply connected.

Astyx

Then waiting ?

Ted Shifrin

Flight from Philadelphia to München, @Astyx.

Astyx

19:05

Simply connected means there is a unique path from one point to another right ?

Ted Shifrin

No, no.

Eric Silva

I want to see if one can say more but not sure

Ted Shifrin

It means every closed path is homotopic to a constant path.

There's never unique paths. Maybe unique geodesics sometimes.

PVAL-inactive

@Ted S^1 \times S^3 admits a Morse function with one critical point of index 0,1,3,4 just by summing the usual Morse functions on S^1 and S^3

Ted Shifrin

Interesting, @EricSilva. Did you talk to Neves about it at all yet?

Astyx

19:06

Constant path meaning ?

Ted Shifrin

@PVAL: Yeah, I was seeing the obvious 1 and 3.

PVAL-inactive

with homological considerations this is the minimum # of critical points of each index for this manifold.

Ted Shifrin

Meaning a path that doesn't move. So you can continuous deform any closed path to a point.

@PVAL: I wasn't seeing how to get only 0, 2, 4, was the point.

Eric Silva

Nope, although I'm gonna speak to him later today about this and some problem about conformal metrics and scalar curvature I've been thinking about

Astyx

What's a typical example of a simply connected set and a non-simply connected one ?

Ted Shifrin

19:08

Sphere and punctured plane, @Astyx.

(Any vector space is contractible — which is the ultimate "simply connected.")

This is why you have all those interesting examples of vector fields on the punctured plane who appear to be conservative but who are not.

Astyx

Oh right

Thanks !

Eric Silva

#deRham

Ted Shifrin

OK, I'm gone for now. I have to eat lunch and do serious packing. Back later, no doubt.

Eric Silva

Bye Ted

Mike Miller

@PVAL Assuming Morse.

PVAL-inactive

19:11

@MikeMiller I think any degenerate critical points either "look Morse"

or don't "look like critical points"

So the function with the minimum number of critical points should be Morse.

Astyx

Bon app' @Ted

PVAL-inactive

I guess what I said isn't quite accurate

you can have things which like x^3 on one flow line

and like x^2 on another.

I still feel that the minimum should be able to be done with Morse functions.

No actually that kind of thing doesn't look like a critical point

you can morph the function on the x^3 flow line so it's derivative is increasing along it.

Mike Miller

@PVAL You can find a function on twice punctured Sigma_g with only one critical point. The branchin locus just gets much higher.

ramification*

(Think of it as a double branched cover over the disc with one branch point at the center, and take the function to be radius

PVAL-inactive

19:27

The double branched cover over the disk with a single branch point is the disk.

Mike Miller

Yeah pretty bad argument.

PVAL-inactive

I just wanna see the gradient

That would convince me

Mike Miller

You can totally make a function with 3 critical points though but I can't draw it either

Look at the 4-author book on LS category

They have this construction where you collide critical points together and do it for the surfaces at least.

Balarka Sen

yeah this was an exercise in Hirsch

math.brown.edu/~banchoff/swork/hf.html

Mike Miller

@PVAL Do you know a reference for a Banach manifold jet bundle transversality theorem? It's easy to prove but I don't want to include a 3 page proof of a mildly relevant fact.

Balarka Sen

19:32

(The thing I linked is stronger: you can even immerse the surface in a way so that height function is the 3cp function)

anyhow i'm off now

PVAL-inactive

@Balarka They seem to prove the opposite of what you say

That there exists no immersion for $g>1$

Balarka Sen

Look at Theorem 3, which provides a polyhedral immersion

for g>1 there is no smooth immersion

PVAL-inactive

Huh

Balarka Sen

weird stuff

PVAL-inactive

I wonder if Arnold's invariants do anything here.

Dodsy

19:42

Balarka is back?

PVAL-inactive

I'm still not sure how to count critical points of these deformations of immersed curves to embedded ones.

Balarka Sen

only for this conversation

Alessandro Codenotti

is $\pi_1(X\vee Y)=\pi_1(X)\ast\pi_1(Y)$ always true? (all groups are calculated in the point where the spaces are joined) I can see it is true for locally simply connected spaces

PVAL-inactive

So this immersion should fail to be generic

Balarka Sen

I think it's false if you take wedge of two Hawaiian earrings or something. You need the $X$ and $Y$ to have nice neighborhoods in the wedge which deformation retracts back to itself

PVAL-inactive

19:44

near where the images of immersed curves are tangent to each other in their respective level sets

Hippalectryon

@Waiting I'm fine, and you ?

Alessandro Codenotti

If the spaces are locally simply connected you can just use Van Kampen taking $2$ open sets with contractible intersection

RaisingAgent

In written division, is there a difference between calculating
`12 - 24 : x ` and ` - 12 : x `

Mike Miller

@PVAL What are Arnold's invariants?

PVAL-inactive

They are invariants of immersed plane curves through regular homotopies through generic immersions.

Ted Shifrin

19:50

banishes a @Balarka for naughtiness

PVAL-inactive

Briefly, they give a signed count of how many times a regular homotopy through two immersed curves fails to be generic

Alessandro Codenotti

Rehi @Ted

Ted Shifrin

rehi @Alessandro

PVAL-inactive

(by having a triple intersection or a self-tangency)

Ted Shifrin

Salut @Hippa

Hippalectryon

19:50

@TedShifrin Salut

Ted Shifrin

@RaisingAgent: Yes. You divide before you subtract. So it's 12 - (24/x).

Dodsy

Oh hey Ted. :) Good to see you before you depart!

I wish you the best in your travels, and hope you have a safe trip.

Ted Shifrin

Thanks muchly.

Alessandro Codenotti

when are you leaving @Ted?

RaisingAgent

@TedShifrin So I divide the -24 first? Does it matter to divide one before the other?

Balarka Sen

19:54

@Alessandro I'm convinced it's def garbage for H v H wedged at bad point. You have extra elements like 'alternately loop along circles on the 1st H and the 2nd H ad infinitum' which do not exist in pi_1(H)*pi_1(H)

Ted Shifrin

First flight leaves 7 AM tomorrow.

@RaisingAgent: What do you mean? One before the other? For example, if x=6, we'll have 12-(24/6) = 12-4 = 8. There's no other way to do it.

Mike Miller

@PVAL Ah interesting.

Ted Shifrin

If you mean to subtract and then divide, you have to use parentheses and write (12-24)/x.

Balarka is never going to take his two-week break.

Balarka Sen

I am. From silly conversations.

RaisingAgent

@TedShifrin ^^ this is what I thought, the earlier parentheses confused me. Thx

Ted Shifrin

19:56

That is not what the agreement was, a Balarka.

Alessandro Codenotti

@BalarkaSen Hm, that loop you described definitely doesn't look like a member of $\pi_1(H)\ast\pi_1(H)$

PVAL-inactive

@MikeMiller Well one of the invariants is a Legendrian knot invariant

Balarka Sen

one can bend rules a little :P safe flight tomorrow, once again

Ted Shifrin

I put the earlier parentheses to make sure you understood the grouping, @RaisingAgent, but most people won't write those.

a @Balarka: My rules are rigid and not deformable.

Mike Miller

@PVAL I'm just talking feelings here, but do you see what I mean? Measuring the change as we do a self-tangency is like measuring the change under RII

PVAL-inactive

19:57

Too any immersed curve with no direct self tangencies you can associate a Legendrian in the manifold of contact elements (cooriented projective bundle of TR^2)

Mike Miller

but we have no rule for R1

Alessandro Codenotti

@TedShifrin I see, hope you have a safe travel

PVAL-inactive

Which is a contact 3-manifold

Mike Miller

Yah

Ted Shifrin

Thanks, @Alessandro. I'll wave abstractly to you when I'm traveling across Italy :)

Alessandro Codenotti

19:58

@TedShifrin First rule of the Ted club...

Mike Miller

I learned that from Bob's paper :o

PVAL-inactive

now regular homotopies with no direct self tangencies are going to correspond to Legendrian isotopies

@MikeMiller I think the setup is a little different

Doesn't he consider cusp curves with no vertical tangencies

corresponding to Legendrians in R^3

Transcript for

$Mathematics$ Mathematics