The h Bar

Geroch's toolkit for spacetime topology

"smooth the corners"

does he actually tell you how to do that

I don't recall

I can get Hawking & Israel for $16.11

This seem like a good meta post?

6:52 PM

It's a nice book

It's one of those odds and ends book

Who cares about pixelating smh

@0celo7 Computer graphic nerds

Let $X$ be a topological vector space, and let $x\in X$. Let $x_n=x/n$. Is $\lim x_n=0$?

Prolly?

Has to be.

Ah.

Multiplication by $1/n$ is continuous.

So it's just $=x\lim 1/n=0$.

Q: Shouldn't we get a vector logo?

6:56 PM

0

Right now, we use a raster logo instead of a vector logo. For those who don't know, a raster is an image made of pixels. If you zoom enough, you'll see the image become pixelated (as is the case with our logo). Vector graphics don't become pixelated when you zoom into them. Almost every SE site...

How do u define limits in a vector space

@0celo7 Eddington books.google.com/books/… has the nicest way to work out Schwarzschild, uses simplified Christoffels and Ricci and is nicer than Zee, Straumann etc!

Q: Shouldn't we get a vector logo?

Don't you need a norm

yuggib

@0celo7 probably you need local convexity

Jaime Gallego

I see the text like this

Physics Meta

6:58 PM

0

Right now, we use a raster logo instead of a vector logo. For those who don't know, a raster is an image made of pixels. If you zoom enough, you'll see the image become pixelated (as is the case with our logo). Vector graphics don't become pixelated when you zoom into them. Almost every SE site...

discussion design

On the other hand, suppose $\lim x_n=y\ne 0$. Maybe we can go for a contradiction.

@JaimeGallego Wtf

Can define limits using filters

Just define a limit like you do in any topological space

I s'ppose

6:58 PM

How do you define a limit in any topological space?

$x_n\to x$ if $(x_n)$ is eventually in any neighborhood of $x$.

Question 26 on that quizz I know the counterexample!

It's in Sanchez

Well, for the spacetime with only one closed null curve, anyway

Limit Point, hmm

I guess proving that there's no spacetime with only one closed timelike curve is just

Deforming a CTC slightly

@JaimeGallego It fixed now?

7:01 PM

Just wiggle it in a normal neighbourhood or something

Jaime Gallego

@SirCumference Yes!

@JaimeGallego Wooo, thanks

2017

@SirCumference I think you should add the feature-request tag

@2017 Done :)

So we know that $f:\Bbb R\times X\to X,(a,x)\mapsto ax$ is continuous.

7:06 PM

proof?

Definition of linear topological space.

Now, it should be true in any setting that if $f:A\to B$ is continuous, and $a_n\to a$ in $X$, then $f(a_n)\to f(a)$.

I think this was an exam question in Topology 1.

So, we consider the sequence $k_n=(1/n,x)$ in $\Bbb R\times X$.

This has the limit $(0,x)$.

So we have: $\lim x/n=\lim f(1/n,x)=f(0,x)=0\cdot x=0$. @yuggib Good?

So back to the earlier question, I have $U$ a nbhd of $0$. I want a nbhd $W$ of $0$ s.t. $W=-W$, and $W+W\subset U$.

"No compact spacetime admits an achronal slice" is false

Which seems odd

Huh

I mean I know that it's not mandatory that there's a CTC through every point

but still

I think I want to construct $W$ for the second condition initially.

Because once I have that I just take $W'=W\cap (-W)$.

Well, we know that $+:X\times X\to X$ is continuous.

So we can take $+^{-1}U\subset X\times X$.

That's open, and we can take a product neighborhood $V\times V\subset +^{-1}U$.

I think that's good enough

7:16 PM

"In a geodesically complete spacetime, every maximally extended timelike curve has infinite length" is false D:

what is the counterexample

Oh wait, do you count the length of CTCs for only one loop?

Because torus I guess then

@Slereah A loop is a map $[a,b]\to \mathscr M$

The integral only goes from $a$ to $b$

7:32 PM

@BernardoMeurer youtube.com/watch?v=x0fFzqX_Xgo

Uzi is an otaku lol

Did you ever find out how to prove that spacetimes were globally hyperbolic or not

@Slereah I guess $\Bbb R^2$ minus $\Bbb Z$ works.

Z as in the integers in the real axis

@Slereah Not really, no

You just isotope integers without 0 to the integers and nothing happens

@BalarkaSen That's what I said...

7:37 PM

Oh, I didn't read the transcript. Sorry.

@Slereah In lots of cases you can check directly that there's a Cauchy surface

For Schwarzschild you can check the Penrose diagram

I guess checking the causal diamond is doable for Minkowski space

Since $J^\pm$ is always closed for Minkowski space I guess the intersection will also always be closed

Mostafa

U.S. Nuclear Weapons Tests Come to YouTube

Though I suppose you also need to prove that it's causal

Mostafa

Thousands of films of U.S. atmospheric nuclear tests between 1945 and 1962 are made declassified ^

7:41 PM

Hm

what's a simple proof that Minkowski space is causal

And how often do you have to write Minkowski to forget that there's "cow" in the middle

what?

MinKOWski

Very distracting

@Slereah what's the definition of causal again?

$\forall p,\ p \not \ll p$

Wait

$p \not \leq p$

Otherwise that would be chronological

So no CCCs?

7:45 PM

yes

what the?

You can probably use the intermediate value theorem or something.

i just got +100 reputation on every site

how?

Show that if you have a CCC then it's spacelike somewhere.

"because we trust you" ?

what does that mean?

7:46 PM

You get +100 reps on other sites if you're trusted on one site

Or maybe something like Borsuk Ulam

Ah yeah that makes sense

how do you guys know if im trusted?

if you gain enough rep

@Slereah actually you can prove that at some point, it has to be past-directed

7:47 PM

is it like 6 months complete or something?

then use the fact that you can't go from one side of the light cone to another without being spacelike or having a cusp

for that you just use calculus

if $T$ is the time component of the curve, then you have $\int_a^b T'=0$

so it has to be both positive and negative, i.e. switches time cones

Good?

yeah sounds reasonable

Someday i should read the proof about the fact that minkowski space is $\approx$ its own tangent space

I think it's in O'neill

I assume it a lot

Do you want me to prove it

It's true for any smooth manifold with a linear structure

Is that anything other than Minkowski space

The best proof is in, uh

I don't remember the book name

Metric Structures in Differential Geometry

@Slereah minkowski space refers to the metric...

7:58 PM

metric + manifold

I wouldn't call $(T^2, \eta)$ Minkowski space

Yes, I know

O'neill calls the twin in the twin paradox Jean and Evelyn

DROPPED

dropped?

Obviously a terrible book

The twins are called Alice and Bob

@Slereah Do you want the proof that $V\approx T_vV$

it's really just parallel translation

there's a canonical parallelism

8:12 PM

Sho'

Just check page 9 of Walschap

I'm busy

It's also in Sachs and Wu, which I know you have

the hell is walschap

vzn

in Mos Eisley, 8 hours ago, by Himarm

↑ for 0celo7... almost as much fun as topology? :)

8:59 PM

@vzn she's a communist

AccidentalFourierTransform

lol

we are ALL communists on this blessed day :)

"avada kedavra!" "salute, comandante"

9:15 PM

Let's see, if we have $\Bbb R^2 \setminus (0, k)$ for $k > 0$, then we can map it to $\Bbb R^2 \setminus (0, k)$ for $k \geq 0$ by a simple translation, no?

Just move all charts by $(0,-1)$

vzn

@0celo7 what?