It's not super clear to me how a Stein structure would force curvature restrictions

Wait hahaha

You were actually talking about Stein doe

I was thinking about Einstein

'cause that's what I am interested in hahaha

oh

#fail

Interestingly, your crappy spelling in that sentence overall and lack of TeX typesetting made me think that you abbreviating Einstein to Stein was pretty plausible

yeah just to clarify my question is if $S^2 \times \Bbb R^2$ w/ the standard smooth structure admits a Stein structure (I know the answer is "no" but not why not)

ein stein v. einstein

What a misreading :P

idk anything about einstein manifolds

how to make german math writing confusing: replace all instances of Einstein with ein Stein

eine steine

I don't know anything about Stein structures. Why do you care?

It's a counterexample to Eliashberg's theorem in complex dimension 1, which says for all almost complex manifolds of complex dimension >= 2 with real topology degenerating to below or equal to complex dimension, there is a complex structure in the same homotopy class as the almost complex structure making it Stein

converse to Lefschetz hyperplane

hmm

welp

but your thing would have complex dimension... 2?

i think I've finally got working code for the split step fourier method in mathematica

albeit not implemented using FFT

@Danu Yeah I lowered the numbers by 1

Eliash is true for dim >= 3

At 2 there are counterexamples.

(I'm having mathematica do the FT and inverse FT by sampling the relevant integrands, constructing interpolating objects, and integrating those)

so far I seem to have it working as far as unitary evolution goes

now let's see what happens if I make a moving packet...

(real test will be wave-packet scattering)

not bad

Much better than anything I did with mathematica. I always just used it for solving integrals and diff eqs

well, there's definitely numerical integration under the hood here

but i'm using interpolation objects to speed it up

Well I didn't really use it for numerical integration except for a couple times. More of I was too lazy to do it by hand, but there was some symbolic solution

Too lazy may be a bit strong. I could have probably stared at some of them for hours and not gotten anywhere

learning to do weird integrals numerically, on the other hand, never gets old /s

The majority of my calc 2 class was "here's another example of a weird integral we can do by hand" and that was about enough for me lol

And that was 9am in my first semester

The other bit was "let's see if this series converges"

yeah

integrals by substitution are largely a matter of spotting the trick

series convergence is good practice for doing real analysis

but beyond that I barely care about it

(radius of convergence, by contrast, is pretty darn useful)

Yeah I did find it pretty neat when I could come up with volumes/surface areas of whatever shape I wanted

not really

if you're doing a math major, maybe

well, to some extent that's a matter of necessity

And I should probably have taken real analysis. I took complex analysis at a point that I didn't understand a ton other than seeing integrals and thinking I could do them as contour integrals

being able to understand Hilbert space stuff

Is there a good source for functional analysis? Or not so much analysis as just understanding what's going on. It seems like everything I've seen takes a different stance. Like some seem to develop calculus of functionals and some seem to just use normal derivatives by varying by something like $\delta F = F + \epsilon \eta(x)$. And I think I've seen functionals mentioned in differential geometry, but haven't gotten that far yet

I guess reading a book that does GR starting from the action may fix some of those misconceptions

yeah

i mean, i've learned more about pairwise correlation in stats since getting curious about quantum correlations

@danielunderwood GR starting from the action principle? sounds like Zee.

not sure i'd ever have bothered otherwise

@heather Yeah he does. I actually started reading that, but got a bit distracted

It's in my growing "next to read" pile lol

@danielunderwood I know the feeling. i'm a bit of a bookaholic.

used bookstores are both my nemesis and best friend. =)

@Blue I'll give them a watch. You said he had some GR lectures. Are those just part of the others?

@heather Yeah I keep buying up books on amazon as I find ones that are like $5 used

i'll ask in here, in case anyone feels like helping with an odd sort of problem:

When you say proofs, do you mean like formal proofs or derivations of why something actually works?

let's say i have three equations that represent concentric circles, along with a couple of equations for line segments that form a star. how can i take these equations and generate one equation that produces the same graph? (by same, i don't mean there can't be connecting lines, just that it looks reasonably similar)

@heather So you want a single $f(x, y)$ such that the level set $f(x, y) = 0$ is the union of all those circles and lines?

@BalarkaSen that sounds right.

If yes, multiply all those individual equations,say $f_i(x, y) = 0$, togather to $f(x, y) = f_1(x, y) \cdot f_2(x, y) \cdots f_n(x, y)$.

> Newtonian spacetime is curved!

@BalarkaSen but will that account for the fact that some of the lines are segments, i.e., have a limited domain/range?

Hmmm

@heather Ah, no, you'd have to pull a different stunt for that.

@BalarkaSen what sort of stunt? =)

Segments of lines in $\Bbb R^2$ cannot be written as $f(x, y) = 0$ for some continuous function $f$, because $\{0\}$ is a closed subset of $\Bbb R$, and if $f$ is continuous, $f^{-1}(\{0\})$ would be a closed subset of $\Bbb R^2$, which a segment of a line is not.

That's a topological problem

So you want $f$ to be discontinuous

Let's see.

Hm, well, you can demand the segment to be "closed" as in, at the endpoints.

hrm. i'm still trying with my scattering, but so far I keep coming up with nearly no reflected component

well...okay, rather stupid idea. could you approximate each line segment by a really long, thin ellipse somehow?

and I've pushed up my potential to a magnitude sufficiently large that I figured I'd definitely have reflection

Yeah (not a stupid idea), but the limit function will not be super easy to write down symbolically.

i don't even mean a limit. it just has to look similar when graphed, not be exactly the same.

$\|x-x_1\|+\|x-x_2\|=a\|x_1-x_2\|$ as $a\to 1^+$?

or perhaps this is an xy problem - the line segments are in the shape of a star, so is there a simple(r) equation for a star?

@Semiclassical I mean, think $nx^2 + y^2 = 1$ as $n \to \infty$.

though I guess you could just represent the line segment by solutions to $\|x-x_1\|+\|x-x_2\|=\|x_1-x_2\|$

The limit of the level sets is an interval segment along the y-axis, but the limit function is not that easy to write down.

if $x(t)=tx_1+(1-t)x_2$, then $\|x(t)-x_1\| = |(1-t)|\|x_1-x_2\|$ and similarly for $\|x(t)-x_2\|$

@Semiclassical no that'd not be a line segment.

so if $x=x(t)$ with $0<t<1$, then $\|x(t)-x_1\|+\|x(t)-x_2\|=[(1-t)+t]\|x_1-x_2\|=\|x_1-x_2\|$

Oh here by $x$ you mean a vector.

ya

being lazy

Yes, that works.

hmm, i never noticed this...i was messing around with the equation of a circle and if you keep increasing the powers, as long as the powers are even, you begin to approach a square.

But still just a tedious fix. Given any closed subset $A \subset \Bbb R^2$, it's the level set of $d(x, A) = 0$.

So..

of course, you can also take $x=x(t)$ for $t\in(0,1)$ as a parametric curve for the integral

but eh

it's easy to create a bunch of individual line segments and curves

Wait, does heather want the circle to be implicit or explicit?

@heather Yeah, the point is "$\lim_{n \to \infty} x^{2n}$" for $x \in [0, 1]$ is the function $f$ such that $f(1) = 1$ and $f(x) = 0$ for all $0 \leq x < 1$

@SirCumference I don't think it really matters...i just want an equation i can screw around with.

You won't get a nice symbolic expression for the equation if you want segments involved

Sigh ignore me, no idea why I thought a circle could be explicit

But you can play with piecewise stuff, or SemiC's trick

okay. thank you.

fun beans

@heather Here's something. Say $[a, b] \subset \Bbb R$ is an interval in the x-axis and you want to find a function $f : \Bbb R \to \Bbb R$ such that $f(x) = 0$ iff $x \in [a, b]$. Check that one such function is $f(x) = \text{min}_y \{|x - y| : y \in [a, b]\}$. $f$ is continuous, but not differentiable. Is it possible to have such an infinitely differentiable function?

i'm a bit confused: "$f$ is ... not differentiable" and "is it possible to have such an infinitely differentiable function?"

are you looking for a function that satisfies the requirements that $f$ does but is infinitely differentiable unlike $f$?

Correct.

well, let me see..under what conditions is a function differentiable...

in other words, you want a smooth function?

Attach some sine curves to the two ends? $f(x) = \sin(x-b)$ for $x>b$ etc...?

@heather Yeah.

my first thought then is:

$x^2 + y^2 = \sqrt{\frac{a+b}{2}}$

oh, wait, no, the function has to be 0

i was just trying to find a function that worked only in that range

What you wrote is not a function. A function is something of the form $f(x) = \text{blah in terms of x}$

@enumaris Shh. (Doesn't work, but still)

$f(x) = \sqrt{\sqrt{\frac{a+b}{2}} - x^2}$ then

but anyway, that's not 0 between a and b.

hmmm yeah, I guess that's also continuous but not differentiable

Yeah.

Yeah I think it's impossible o.O

Naw it's differentiable inside the interval.

All order derivatives must be 0 at the boundary of the intervals

Yes. So yours just work first order.

but even though all order derivatives must be 0 going to the right of $b$ you must have $f(x) \neq 0$ to the right of b...

seems impossible unless some pathologically weird function that satisfies that exists lol

Correct

zips mouth

so if you have a function that is 0 from a to b, the derivative of the function at those points is just 0, and so on, right? so a function must exist if it only needs to be infinitely differentiable in the interval.

but you want it infinitely differentiable with a domain of all of $\mathbb{R}$...

I think it might also change if you made the interval open rather than closed

If it's 0 from a to b it need not have derivative 0 at a or b

@enumaris Impossible with open for topological reasons

but not impossible with closed?

or just as impossible lol

zips mouth

I worry not about pathological functions

Everything is a pathology my man

This whole world, the planet, your face

Very pathological

(gottem, by the way)

I think there is some theorem related to this

but I only have a vague memory of it

I gather it can be done

and probably one attaches some funky function like $e^{-1/x^2}$ at the appropriate endpoints,

Pathological functions are my favorite type :D

@enumaris :)

and it don't work with $]a,b[$ cus that function is undefined at $x=0$ heh...what a funky function