The h Bar

John Duffield

12:17

@Slereah : no, on GR he's not fine.

Balarka Sen

wow, Casson handles are really messed objects.

Bass

@JohnDuffield can you explain your diagram in Simple English?

ACuriousMind

@BalarkaSen I thought you don't like the smooth world?

Danu

@ChrisWhite Interesting!

Balarka Sen

@ACuriousMind I just don't know much about the smooth category (working on it). Besides, Casson handles are topological constructs.

They are used to classify simply connected 4-folds in TOP, or so I have heard.

It's hard to do topology without liking DIFF.

Danu

12:29

Would you agree that topology is essentially a type of geometry?

Balarka Sen

No.

Danu

Would you agree that it fits the description of geometry given by Klein in his famous Erlangen program address?

Balarka Sen

Geometry takes account of a lot of extra structures. (e.g., Riemannian metric, Lie group action)

What fits the description of Erlangen program, @Danu?

It's just a systematic way to study groups as isometry groups of metric spaces.

Danu

That's not the interpretation I got from reading it.

Balarka Sen

What is your interpretation?

Danu

12:31

Are you sure that it's about just metric spaces?

My interpretation is that the statement is essentially the following:

Do do geometry, you (1) pick a group of transformations (2) study invariants under this group.

Balarka Sen

As far as I know, yes. You study symmetry groups of metric spaces.

John Duffield

@Bass : it isn't my diagram, it's a Penrose diagram. And no, I can't explain it in simple English. Or things like the prallel antiverse. But see Wikipedia and note that "the precursors to the Penrose diagrams were Kruskal–Szekeres diagrams". I can explain the problem with Kruskal-Szekeres coordinates.

Danu

In the case of topology that'd be the homeomorphisms.

Secret

(((http://www.mathworks.com/matlabcentral/newsreader/view_thread/284171
Now I finally understand why I can never label stem3 stems

You need to do this row by row)))

Danu

@BalarkaSen My interpretation was much more broad

Balarka Sen

12:32

@Danu Transformations = isometries of metric spaces.

Danu

@BalarkaSen Does it really have to be?

Balarka Sen

No, Klein did not have in mind topology when he said that.

@Danu Well, that's what Klein's program is :P

Danu

@BalarkaSen Okay, if you say so. I don't know much about Klein himself---I just read the lecture.

yuggib

@Danu That is a swear T_T

Danu

But it's been a while so I don't recall the details.

Balarka Sen

12:35

Studying homeomorphisms of topological manifolds, even, is hard stuff.

Danu

In any case, my interpretation expounded above was heavily influenced by the view taken by one of my professors---who admits that, at heart, he's a geometer. I wouldn't be surprised if his position is heavily biased :P

I think it's a conceptually nice picture, though.

It is not so far off to say that topology is the study of homeomorphism-invariants, is it?

Balarka Sen

@Danu No, I agree that it isn't.

I am just saying it's not a part of Erlangen program.

Danu

But it is a natural generalization of the idea behind the Erlangen program, perhaps.

ACuriousMind

@Danu However, many topological invariants are weaker in that they are only homotopy-invariant.

Balarka Sen

I am going to fight with anyone who labels homotopy theory as topology :P

I am so convinced homotopy theory is essentially combinatorics, lol.

Danu

12:39

hahahaha

Balarka Sen

Something not very related, but fun: Erlangen programme studies geometric objects by studying groups which act on it. There is a way to do this in the opposite : studying groups by looking at the geometric objects (in particular, graphs) it act on. Turns out the geometry of the objects a certain group act on is specific to the groups nature, in a sense (so, there are things like hyperbolic groups).

This is called geometric group theory.

Danu

Nice

That's very interesting.

My basic topology course is doing a lot of stuff on group actions right now

Actions of finite, compact, discrete groups

ACuriousMind

@Danu Then you'll probably soon see universal coverings and fundamental groups.

Danu

Exactly, that's the way we're headed.

It will be algebraic topology from next week on, I think

Balarka Sen

So, for example, if you look at the Cayley graph of the free group on 2 generators, you'll see surprising similarities with the Poincare disk model of the hyperbolic plane.

@Danu that's nice.

Essentially, if $X$ is a simply connected topological space, $G$ is a group which acts on $X$ by homeomorphisms freely and properly discontinuously (for any pt $x \in X$, there is an nbhd $U$ of $x$ such that orbits of $U$ by $G$-action are all disjoint) then $G$ is called the fundamental group of $X/G$ :P

And $X$ is called the universal cover of $X/G$. I am lying slightly, because that's not how it's defined, but you can think of it like this.

ACuriousMind

12:45

You're missing simply-connectedness of $X$ there.

Balarka Sen

Ah, thanks, @ACuriousMind

Danu

Right, we did proper discont. last lecture.

Also that's the crappiest piece of terminology ever.

A properly discontinuous continuous action

yeah, right...

Balarka Sen

:) Hatcher uses covering space action.

Danu

mhm

Balarka Sen

For free + properly disc.

Danu

12:48

By the way

Does any one of you have any idea about the following:

If you type up the notes to a lecture, writing down pretty much only what the lecturer writes on the board

(type up as in TeX it)

how many pages do you end up with for a one-semester course?

(this is for mathematics lectures---physicists never write jackshit on the board :P)

ACuriousMind

@Danu Whaaaat?

Danu

@ACuriousMind Have you ever seen a physicist write out full sentences on the board?

ACuriousMind

@Danu Yes.

Danu

Not here :P

Balarka Sen

That's because physicists don't speak full sentences either. :P /joke

ACuriousMind

12:52

But you are right that physicists have a tendency to have a more abbreviated style

Huy

I'm at 50 pages now and there's still a month to go.

ACuriousMind

@Danu I can't tell because I write down what's on the board+what is said, not just what's on the board :P

Also, I'm a lazy fuck and generally don't TeX my notes

Bass

hi guys, the Dirac gamma matrices are no tensors, are they?

Danu

@ACuriousMind Yes, sure, but for the mathematics lecturers here at Munich that tends to be <10%.

Mostly more elaborate explanations for proofs

(I do write them out afterwards, lest I forget :P)

I'm asking because I feel Leeb is covering extraordinarily much.

@Huy How many months have passed? :P

Huy

the semester started mid Sept

Danu

12:56

Right.

Huy

only 3 lectures a week though

Danu

I'm at 40 and my semester started mid-October :P

2 lectures a week

haha

Leeb-speed confirmed

I think we'll end up with ~100 pages

Also @BalarkaSen

@Minhyong Kim: So Klein's notion of geometry is neither more nor less restrictive than the Riemannian one. Klein's includes examples like topology, which have no system of measurement, so in that sense it's more general than Riemannian geometry, not less. — Ben Crowell Jan 28 '13 at 15:56

Balarka Sen

Might be, but I am going to fight with you that Klein didn't really work with Homeo(X) :P

Danu

Of course he didn't---at the time topology hadn't even been made precise!

Balarka Sen

That's my point.

Danu

13:01

But the spirit was there :P

Balarka Sen

Sure.

John Duffield

13:23

@0celo7 : I know this. It's fairly straightforward, see my answer. Note that the curvature relates to the tidal force, which in general increases towards the centre. This is why people talk about the strong curvature regime in the context of black holes. For the Earth it increases then reduces then increases then reduces.

Maybe I should scrap that last "reduces".

Secret

14:06

RpfnoR 27
and before anyone shout, no, illuminati is not confirmed

It doesn't matter, because NOC will win

0celo7

never seen anyone write out full sentences on the board

@JohnDuffield If you can't explain it in simply English maybe you're not so good at GR...

Bass

turns out the gamma matrices are a tensor, with two spinor indices (one upper, one lower) and one contravariant "normal" index.

0celo7

@Bass uh, what?

having a "spinor" index does not make it a tensor

it makes it a spinor ":P

Bass

@0celo7 a spinory tensor then :P

or spinning tensor

whatever

0celo7

well

it's not a spinor either

it has two indices

David Z

14:22

@ChrisWhite about that flag: ironically we didn't see it because the community was "too efficient" at getting the post automatically deleted :-P we should figure out some way to deal with that

Bass

@0celo7 seems to be something like a higher representation of a spin group.. like a normal tensor is a higher representation of the orthogonal group. right?

0celo7

@Bass why can't someone write a book called

SPINORS

and have 800 pages of representation theory

and proofs of every gamma matrix thing

then we would have all of the answers

@Bass a tensor is a rep of the general linear group

it transforms under invertible matrices

Bass

@0celo7 and a spinory tensor, of the Spin group?

0celo7

@Bass dunno, maybe?

ACuriousMind

@Bass: Tip: Linear maps from $V\to V$ are automatically elements in $V\otimes V^\ast$.

Bass

14:26

spinory tensor, I thereby baptize thee in the name of Clifford, Dirac and Pauli to the name Spinsor

ACuriousMind

I.e. every matrix is a tensor. That shouldn't come as a surprise

0celo7

@ACuriousMind i'm shocked and appalled

Bass

@ACuriousMind maybe I'm mixing stuff, but the christoffel symbols are kinda matrices but they are no tensors? If every matrix is a tensor, why all these exercises "prove that xyz transforms like a tensor"?

0celo7

rant incoming?

ACuriousMind

@Bass The Christoffel symbols are indeed not tensors. What do you mean when you say they're "kinda matrices".

Secret

14:29

32

Q: Differences between a matrix and a tensor

What is the difference between a matrix and a tensor? Or, what makes a tensor, a tensor? I know that a matrix is a table of values, right? But, a tensor?

Bass

@ACuriousMind okay they are "three-dimensional" matrices (because they have three indices)

ACuriousMind

Perhaps I should clarify that when I say "every matrix is a tensor" I mean "every matrix that is meant to represent a linear map (as matrices are wont to do) is a tensor".

Secret

All rank 2 tensors have matrix representations, but not all matrices are tensors

ACuriousMind

@Bass See my statement above, to me, a mere collection of numbers doesn't make a matrix

Bass

ok, makes sense

Secret

14:31

for example, if I wrote down a matrix that count the number of apples in a 2x2 stack of cubicles, this thing obviosuly does not depend on which coordinate system we are choosing

ACuriousMind

And the three indices on the Christoffels are actually "of different type" - it is actually a one-form taking values in $n\times n$ matrices, see my answer here

Bass

but if you just look at a matrix, you don't know its transformation behaviour, right?

because you don't know which indices are upper/lower and which ones are spinor

Secret

Yes, thus you need to see what happens when you change your basis

ACuriousMind

@Bass Yes. A "matrix" needs to be accompanied by the information whether it is to be interpreted as a (0,2),(1,1) or (2,0) tensor

Which are interpretations as quadratic form, linear map and quadratic form on the dual, respectively

Bass

@Secret and then you're in represenation theory of some Lie group, right? so, is it correct to say that a Spinsor (tensor with spinor indices) is composed of irreps of the Spin group?

Secret

14:33

$$\begin{pmatrix}1 & 2 \\ 3 & 4\end{pmatrix}$$

The numbers you saw here are only valid for a certain basis, or may not be related to the basis at all if it is just something like the "apple matrix" I mentioned above

@Bass I am still not terribly good at spinors to comment anything on that, Acuriousmind might be a better person to ask

0celo7

@Secret not really

Jun 9 at 18:30, by ACuriousMind

God, I don't know anything about spinors, it seems

Secret

seriously???

ok nvm then

Bass

he had more than five months to learn them since then :P

0celo7

he's been playing far too much Witcher 3 to do any of that

and he is currently playing Fallout 4

Bass

@ACuriousMind is it correct to say that a Spinsor (tensor with spinor indices) is composed of irreps of the Spin group? like a normal tensor is composed of irreps of the general linear group (or orthogonal?)

Secret

14:37

well, everyone are being sucked into fallout 4 currently, I have at least 4 friends playing it all day long

ACuriousMind

@Bass Each gamma matrix is a mapping $S\to S$ for $S$ the spinor representation (since you use it to multiply a spinor and get again a spinor). Hence each $\gamma^\mu$ transforms in $S\otimes S^\ast$. This is intertwined with the transformation of the vector index as $$ \rho_S(\Lambda)^{-1}\gamma^\mu\rho_S(\Lambda) = (\rho_V(\Lambda)\gamma)^\mu$$

0celo7

wtf

ACuriousMind

cf. this answer by joshphysics to show I'm not making this up

Secret

2

RpfnoR 27b

ACuriousMind

@Secret What exactly do you want to tell us by posting that image?

Secret

14:42

Nothing is tagged there, thus it is nothing important

@Acuriousmind anyway, what is $\rho_S$ and $\rho_V$?

Bass

@ACuriousMind $S\otimes S^*$ meaning that each gamma matrix has one upper and one lower spinor index?

ACuriousMind

@Secret The representation of the Lorentz group on the spinor space and the four-vector space, respectively.

Secret

@ACuriousMind is 4 vector space minkowski space?

or just $\mathbb{R}^4$

ACuriousMind

@Bass ...I suppose so? I don't really think about it with indices. For any vector space $V$, a linear map $V\to V$ is equivalently a bilinear map $V\times V^\ast \to \mathbb{R}$ since $V\cong \mathrm{Hom}(V^\ast,\mathbb{R})$, and such bilinear maps are precisely the elements of the tensor product $V\otimes V^\ast$

0celo7

@ACuriousMind I think it's a cry for help

he's been kidnapped and needs our help

@ACuriousMind what are the indices of the gamma matrix in dot-undotted notation

ACuriousMind

14:46

@0celo7 I have no fucking clue what dotted/undotted indices do

Secret

@0celo7 nah, it's more like a "honours hangover" thing, imagine spending 7 weeks just to get a program work!

Bass

@ACuriousMind $\rho_V(\Lambda)$ is usually denoted as just $\Lambda$, right? for example in $S\gamma^\lambda S^{-1}={\Lambda^\gamma}_\mu\gamma^\mu$

ACuriousMind

@Bass Exactly, the representation on the four-vectors is just the fundamental

0celo7

@ACuriousMind oh, I knew that

your notation is just strange :P

ACuriousMind

@Bass Your indices are off, though :P

Secret

14:47

@ACuriousMind Ok I see, so $\rho_V$ is basically representation of minkowski space...

ACuriousMind

@Secret Well, not exactly, because a representation space need not be an inner product space

But you may think of it as Minkowski space, yes

Bass

@ACuriousMind are they? I copied this equation 1:1 from Zee.. (not that this would mean anything)

ACuriousMind

@Bass Look at it! There's a free $\lambda$ on the left, and a free $\gamma$ on the right. Should both be $\lambda$, imo.

@0celo7 Perfectly standard among actual group theorists :P

0celo7

@ACuriousMind typo

Bass

@ACuriousMind oh, shame on me. Didn't see the upper $\lambda$ even when proof-looking at it.. ::goes hiding in corner::

0celo7

14:51

@ACuriousMind I know

ACuriousMind

@0celo7 Well, the damn thing with indices is that you never know if the mismatches are typos or actual errors :P

0celo7

but I've never seen it in a QFT book loke that

@ACuriousMind is index free better 100% of the time

Secret

(This message was supposed to be posted 30 mins ago): Finally some nice discussion going on in the chat, I am so bored to death by the recent lack of activity!

ACuriousMind

@0celo7 Is that a question or a statement? ;)

0celo7

because the Bianchi identity is pretty damn ugly in index free

Balarka Sen

14:52

You call this nice discussion?

0celo7

it's like

$\nabla R(V,W,X,Y,Z)+otherstuff=0$

Bass

@ACuriousMind just to be sure, in your equation above, everything is to be read as operators or matrices, acting on something on their right, is that correct?

0celo7

hardly nicer than $R_{ijkl;m}+otherstuff=0$

Secret

@BalarkaSen Far better than some of the random plus lack of activity approx 2 hours ago, as right now, there's some academic discussion going on, thus a chance to learnt more stuff!

Balarka Sen

I see the usual stuff physicists discuss over tea, so it looks thoroughly un-nice to me.

:P

0celo7

14:54

@BalarkaSen what do you want us to discuss

tbh I dunno why you come here

you just complain every time

Balarka Sen

$\infty$-categories of course.

@0celo7 It was a joke, don't take it seriously. I just come here because I am curious about what physicists do.

And I find that most of the time they either play video games or curse...

ACuriousMind

@Bass Well...on the l.h.s. there are three matrices just being multiplied, on the r.h.s., there is a matrix acting on a 4-vector.

Balarka Sen

That last thing was a joke too :P

Secret

@0celo7 From my 3 days of observations, he often come with questions to ask, quite similar when at approx. 3:00 in Syd time, you and Acuriousmind often have nice discussion about topology, group theory and abstract algebra

0celo7

@BalarkaSen that's not actually that wrong

Secret

14:57

@BalarkaSen The recent fallout 4 stole a lot of opporunity of otherwise interesting discussions, because too mnya people are playign on it and I was stuck with a malfunctioning program in my honours project

0celo7

note that no one in here is a physicist though

ACuriousMind

@BalarkaSen What physicists do you know that drink tea?! ;)

Secret

I thought dmckee is a particle physicist?

0celo7

@Secret he's not here right now

ACuriousMind

@0celo7 dmckee and DavidZ might disagree :P

Secret

14:58

ok fair enough

0celo7

@ACuriousMind neither is here right now

Balarka Sen

@0celo7 ah, glad to hear.

0celo7

@BalarkaSen although it's probably more true of physics students

I can't see Witten playing video games

maybe some tetris

or pong

Balarka Sen

lol.

ACuriousMind

@0celo7 Also, at least I am a physicist-in-training and already have a Bachelor's degree in physics :P

Secret

14:59

same here

Balarka Sen

I just had a mental image of Ed Witten playing tetris.

Secret

even though I am also a chemist

in training*

Balarka Sen

@ACuriousMind Haha, what do they drink, then?

0celo7

@ACuriousMind you're a mathematician-in-training who has been tricked

ACuriousMind

@BalarkaSen Coffee, beer or liquor, depending on the hour and the difficulty of the problem.

Mostly coffee, though

Balarka Sen

15:00

Cool. You need to try some ffee.

0celo7

irish coffee

@ACuriousMind you drink coffee o.o

ACuriousMind

@BalarkaSen Yeah, and a coconut is just a nut, right?

0celo7

wtf is a coconut, anyway

Balarka Sen

@ACuriousMind The nut needs to be finite dimensional for that to hold.

0celo7

@BalarkaSen what is the dimension of a nut

Balarka Sen

15:01

Depends on the nut.

ACuriousMind

Well, infinite-dimensional nuts would just be totally nuts

0celo7

that hurts

stop

Vi Esr

Hi, I wanted to ask you people something. In general, we all learn everything someday. It can be from a source or a thought itself. Would you like to be able to share it somewhere, as in kind of a social network where people share their learning experiences(with source) and you can follow people you like. Does this sound interesting?

something everyday*

ACuriousMind

Best typo I've seen in a while :D

Vi Esr

Lol

0celo7

15:04

@ACuriousMind what level are you in Fallout?

Secret

My recent "sharing" due to the honours hangover has caused quite a lot of flooding issues in this chat room recently...

0celo7

at least you don't have 1/6 of the posts

Balarka Sen

What kind of video game has the name "fallout"??

0celo7

postcount++

Secret

lol true

0celo7

15:05

@BalarkaSen wtf kind of rock do you live under D:

Balarka Sen

How is that relevant?

I haven't classified the rock I live under, I am not a geologist.

4

Slereah

@0celo7 Sometimes I do!

Vi Esr

Any views on the question I asked. They can be on the negative side too. I don't mind.

:)

Secret

are you promoting something?

Vi Esr

No, just wanted to ask if something like this sounds interesting to other people

ACuriousMind

15:08

@ViEsr Why would you need a new network for that. People can already share that on the existing social networks, no?

Secret

just twitter and facebook alreayd does that job, no?

ACuriousMind

@BalarkaSen One that is set after a nuclear war. Makes sense, I think.

0celo7

@ACuriousMind have you been to the glowing sea yet?

Vi Esr

Yes, but its kind of too noisy. Don't you think?

Secret

so what's the thing you know of that is better than them?

0celo7

15:11

actually, the most radiation in the game is in the mass fusion building

you get lik 80+ rads

it's nuts

Slereah

don't you have a hazmat suit

Vi Esr

Well, it'd be nothing special as such. Just more focused on this one aspect.

John Duffield

@0celo7 : I can explain a lot of things in GR. But I can't explain the parallel universe, the antiverse, the parallel antiverse, the new universe, or the new parallel universe. How on Earth anybody believes in that stuff beats me.

0celo7

@JohnDuffield We believe it because people like Dr. Penrose tell us.

Secret

The reason of the hangover: what I have been working on in the past 7 weeks

ACuriousMind

15:15

@0celo7 no

Vi Esr

Mostly it can help evaluate the learning outcomes and efficiency of a particular learning source and help you meet and connect to new like minded people.

0celo7

@ACuriousMind well that's where the bomb that destroyed the commonwealth got dropped

lots of fallout there

Slereah

There are

4 of them

B)

Though really, if you have the hazmat suit you basically don't notice it

0celo7

4 of what

Slereah

Fallout 4

0celo7

15:19

oh ffs

did you ever finish the main quest?

I need to know what happens

Slereah

No

I am kinda like eh

Fallout 4 you have disappoint me

0celo7

was new vegas better?

Slereah

yeah

I think really the problem is the lack of settlements

John Duffield

@0celo7 : you shouldn't, because the precursors to the Penrose diagrams were Kruskal–Szekeres diagrams. See Wikipedia. And Kruskal-Szekeres coordinates are akin to Gullstrand-Painleve coordinates, which Einstein dismissed for good reason.

Slereah

There's like two big settlements, three if you add the institute

That doesn't go very far

0celo7

15:25

@JohnDuffield He has a PhD and you don't, so guess who I'm gonna believe...

John Duffield

@0celo7 : Einstein.

0celo7

@JohnDuffield I thought the Kruskal stuff was in the 60s, after Einstein's death

@JohnDuffield Why? He died a long time ago.

John Duffield

@0celo7 : The Gullstrand-Painleve stuff wasn't.

@0celo7 : Because he was the originator of GR.

0celo7

@JohnDuffield So? We don't quote Euler on mechanics.

@JohnDuffield What is Wikipedia supposed to tell me?

How do you even dismiss coordinates?

John Duffield

@0celo7 : mechanics is a whole field, GR is a particular theory. There's a big difference. And when you read up on GR in the Einstein digital papers you appreciate that Penrose diagrams are borne of misunderstanding.

0celo7

15:38

@JohnDuffield What misunderstanding is that?

John Duffield

@0celo7 : it's a failure to understand that the parrot is dead.

0celo7

@JohnDuffield Lol, what?

John Duffield

@0celo7 : check out the dead parrot sketch. The parrot is dead, and nothing you do is ever going to change that.

0celo7

To be clear, you're not going to explain why Kruskal coordinates are wrong and why Penrose diagrams are built on a misunderstanding?

@JohnDuffield I don't see what that has to do with anything.

John Duffield

@0celo7 : I will explain why KS coordinates are wrong and why Penrose diagrams are built on a misunderstanding. It's to do with the coordinate speed of light. See this:

"For example, at the event horizon of a black hole the coordinate speed of light is zero, while the proper speed is c.[1] The coordinate speed of light (both instantaneous and average) is slowed in the presence of gravitational fields. The local instantaneous proper speed of light is always c."

The parrot is dead, and nothing you do is ever going to change that.

0celo7

15:53

@JohnDuffield It is zero in which coordinates?

John Duffield

@0celo7 : the coordinates preferred by the distant observer.

I'm the distant observer, and you're at the event horizon in a bubble of artistic licence. You are at a location where the coordinate speed of light is zero. The parrot is dead.

0celo7

@JohnDuffield Please show this.

@JohnDuffield No reference to Einstein or diagrams please.

John Duffield

@0celo7 : you have an optical clock. But because you're at a place where the coordinate speed of light is zero, this clock doesn't tick. Not ever. Never ever ever. Gravitational time dilation is infinite. But Kruskal-Szekeres replace the t coordinate with a T coordinate and claim that the clock somehow starts ticking "in your frame". They effectively claim that a stopped clock ticks from the point of view of a stopped observer.

Slereah

Hm

On a more serious matter

How to show that every spacetime that isn't time orientable has CTCs going through it

0celo7

@JohnDuffield Yes, there is no issue with this.

Slereah

16:07

It is hinted at but I've never seen a proof

0celo7

@Slereah Where?

Slereah

In the few examples I can find

"The light-like geodesic in $D^2$ lying in the 2-plane $x - t = 0, y = 1$ projects to a light-like geodesic in $M^2$ which is self-intersecting (since the points $t = - 1, x = - 1, y = 1$ and $t = 1, x = 1, y = 1$ of $D^2$ become identified in $M^2$). Time-like geodesics in $D^2$ near the above light-like geodesic also project to self-intersecting geodesics in $M^2$."

0celo7

@JohnDuffield Anyway, since it's clear that you don't want to talk physics, I think it would be best for us not to converse in the future. It never ends well.

Slereah

Also I recall a paper mentionning that non-time orientable always have non-contractible CTCs that disappear in the double cover

0celo7

Proof?

John Duffield

16:12

@0celo7 : the stopped clock doesn't tick. Sticking a stopped observer doesn't make it start ticking again. So KS coordinates represent a misunderstanding of GR. And "the precursors to the Penrose diagrams were Kruskal–Szekeres diagrams".

Slereah

Trying to find the paper

John Duffield

@0celo7 : I'm talking physics. What's the problem?

0celo7

@JohnDuffield You're not, that's the problem.

And now his picture is small.

We have achieved peace in h Bar.

John Duffield

@0celo7 : I've just explained why Penrose diagrams are borne of a misunderstanding. That's what you wanted.

Bass

@JohnDuffield can you explain this in Simple English?

John Duffield

16:21

@Bass : yes. If you're at a place where light doesn't move, some people claim that you don't see anything unusual. But the truth of the matter is that you're at a place where light doesn't move. So you don't see anything.

Slereah

"Some spacetimes, such as G¨odel spacetime, do not admit any global time slices. This is a consequence of three features: it is time orientable; a CTC passes through each point; and it is simply connected. The edge of an achronal surface S is the set of points p such that every open neighborhood O 3 p includes points in I+(p) and I−(p) that can be connected by a timelike curve that does not cross S."

:O

Bass

@JohnDuffield have you ever been at such a place?

John Duffield

@Bass : no. But we're fairly confident that black holes exist.

Bass

@JohnDuffield so you're saying light doesn't move at black holes?

Slereah

I guess it's the whole argument of HE

But more general

But why do people keep making statements about proofs they don't write down or give a reference to

lawl

Wait why does he even mention that the spacetime is simply connected and time orientable

All simply connected spacetimes are time orientable

Hey @Qmechanic

Apparently the standard term for a non-time orientable de Sitter space is the "universe from nothing"

"For a large class of these spacetimes, one can always choose metrics without CTCs; time nonorientability is then their only causal pathology."

Hm

I guess not all such spacetimes have CTCs then

0celo7

16:49

@Slereah proof?

Slereah

@0celo7 The double cover is time orientable

ACuriousMind

@0celo7 You're like a broken record :P

Slereah

There's no cover for simply connected spaces

0celo7

@Slereah proof?

@ACuriousMind Your face is like a broken record

Slereah

Orientability

In mathematics, orientability is a property of surfaces in Euclidean space that measures whether it is possible to make a consistent choice of surface normal vector at every point. A choice of surface normal allows one to use the right-hand rule to define a "clockwise" direction of loops in the surface, as needed by Stokes' theorem for instance. More generally, orientability of an abstract surface, or manifold, measures whether one can consistently choose a "clockwise" orientation for all loops in the manifold. Equivalently, a surface is orientable if a two-dimensional figure such as in the space...

just need a double cover to detwist that line element into a proper vector field :p

ACuriousMind

16:53

@Slereah Strictly speaking not true, but the covers are just disjoint unions of themselves.

0celo7

@Slereah proof?

@ACuriousMind proof?

@Slereah proof?

stop saying things without proof >:(

ACuriousMind

@0celo7 Elementary covering space theory.

0celo7

@ACuriousMind your face is elementary

ACuriousMind

Yes, it is the very element of handsomeness :)

Slereah

\newcommand{\h}{\frac{1}{2}}

The most important latex command to add

ACuriousMind

16:56

Hm, I prefer to use one that just takes one argument and then writes $\frac{1}{\text{argument}}$ instead of hardcoding the 2 there

Slereah

Yeah but there are so many instances of a half

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