The h Bar

0celo7

00:00

Theorem: Spacetime is Hausdorff.

Proof: Makes life easier.

Slereah

Taub-NUT isn't Hausdorff though D:

0celo7

It's non physical

Slereah

ur non physical

(like a ghost)

0celo7

I'm a FP ghost.

user54412

Is the long line Hausdorff?

Slereah

00:04

Yes

It's not paracompact, though

Though really I'm not quite sure how a non-Hausdorff manifold would do anything for QM, really

user54412

oh right

user54412

well, the long line could be a nice warm up -- no partition of unity, but points are still rather separable

Slereah

I mean I understand the motivation behind it, but I once vaguely tried for a free particle to make sense of it

But I couldn't really find any idea for it

ACuriousMind

@ChrisWhite You already get the first problem with defining the differentiable structure for that one

Slereah

@ChrisWhite : Well spacetime itself in such a structure would be a subset of that manifold, which is itself Hausdorf :p

No worries about structures

But the behaviour of particles in that manifold, no idea

Hopefully a free quantum particle would converge to a point per timeline, and every one of those points would be different

But then how do you deal with superpositions?

There would have to be more lines where the particle is in some position than others

And I have no idea how that would work

dmckee

00:09

@0celo7 Yeah, but it found \gtrapprox ($\gtrapprox$) for me on the very first try earlier today.

Slereah

Apparently if the spacetime is orientable, there's a very limited number of compact spacetimes

Well, not exactly, but a limited number of Betti numbers

Oh, its fundamental group has to be abelian too, tho

Hm

What would be a compact manifold with a non-abelian $\pi_1$

"It is shown that, as a consequence of the no-return theorem of Tipler, a generic non-spacelike convergent compact space-time cannot admit a closed embedded edgeless spacelike hypersurface. Such a space-time cannot therefore be spatially isotropic and so cannot provide a satisfactory model of the universe."

Damn it Newman!

ACuriousMind

00:24

@Slereah Torus

Slereah

Isn't the torus's group like

$Z^2$?

ACuriousMind

That's the homology, not the fundamental group

Slereah

Oh

Hm

What is the fundamental group then

ACuriousMind

Hm

Wait

you're right, it's abelian

figure eight

Slereah

$user image$

ACuriousMind

00:27

Yep, figure eight is non-abelian, it's the free group on two generators

Slereah

Figure 8 isn't a manifold, though

ACuriousMind

Hmm...that's gonna be difficult, the example has to be at least 3D

Slereah

"It is well known ([IS]) that a noncompaet spacetime admits a Spin structure if and only if it is parallelizable. This result appears to be false in the compact case, and the necessary and sufficient conditions for a compact spacetime to admit spin structures are not wetl understood (see [41] for a detailed discussion and references)."

Whaaat

I am shocked

ACuriousMind

Aha!

Take the double torus

That one has four generators $a,b,c,d$, where $a,b$ and $c,d$ commute, but no further relations

Slereah

neat

Although

Double torus isn't a spacetime, I do believe

Since $\chi \neq 0$!

ACuriousMind

00:41

Well, then there are no 2D compact spacetimes with non-Abelian fundamental group

Slereah

heheh

There's a bunch of weird 4D ones, though

$S^1 \times S^3$ is fine

(also $S^3$)

I hate it when articles are only available in anthology books :p

Slereah

01:05

Goddamn, now that I look into it there's tons of theorems for CTCs

"An arbitrarily close return to a previous initial state of the Universe, such as is predicted by the Poincaré recurrence theorem, cannot occur in a closed universe governed by general relativity. The significance of this result for cosmology and thermodynamics is pointed out."

neat

0celo7

WHAT

Link

@Slereah LINK

I must see this :o

Slereah

It's in "Essays in general relativity"

p. 30

Well p. 21

it is a book so you must acquire it

Legally

0celo7

Legally transfer it via Skype please.

Slereah

"The concept of the eternal return—the idea that time is fundamentally cyclic—apparently played a key role in the cosmological thought of mankind as far back as 6500 B.C. [11, p. 332]."

I really hope that reference is a paper from 6500 BC

5

0celo7

Hehe

Slereah

01:13

Ug Grobarg, "Are skygods good or evil", cave painting

0celo7

lol

Slereah

Man there's like three pages on the history of cyclical time

Slereah

01:34

How does $p^a x_a$ make sense

$x$ isn't a vector

It is not part at all of the tangent bundle

Slereah

02:19

"In particular, non-Hausdorff manifolds may have any cardinality from c upwards and even in dimension 1 a non-Hausdorff manifold need no longer be orientable."

D:

0celo7

What's a one dimensional non Hausdorff manifold

Slereah

Branching real line

Line with two origins

Complete feather

stuff like that

0celo7

02:34

@Slereah isn't that R^2 with a weird quotient

Slereah

No

It's $R \sqcup R$ identified everywhere but at 0

0celo7

oh it's the union of two lines

Uh, we established that the union of two lines is the plane.

Slereah

No

Disjoint union isn't the same as cartesian product

Disjoint union is just

You have two copies of $R$

0celo7

@Slereah which is $R^2$

DanielSank

@ChrisWhite link to resume template? I dig it.

Slereah

02:40

$R^2$ is more $c$ copies of $R$

0celo7

@Slereah what

speed of light?

Slereah

Cardinality of the continuum

0celo7

what

stop making up things

Slereah

I would have said $\aleph_1$ but that's assuming the continuum hypothesis :p

user54412

03:08

@DanielSank My resume? It's just some tex I slapped together. I can email you the source.

Slereah

You know

I can't really think of a GR paper that was written in the 50's

There's a big gap in my mind between the 40's and the 60's

IIRC there wasn't much going on in the 50's

user54412

Maybe the discovery of Kerr revitalized the field?

Slereah

Maybe

Let's see if Stephani lists any 50's papers

Hm

59 paper on the Petrov classification

53 : "Certain exact solutions of the equations of general relativity with an electrostatic field"

54 : "Static magnetic fields in general relativity"

51 : "On gravitational waves"

54 : "Reciprocal static solutions of the equations of the gravitational field"

And a bunch more

Still not a lot

$\Ydown$

Aw

Not in mathjax

NeuroFuzzy

03:37

0

Q: How does quantum mechanics show forever is not tonight?

Edward Witten participated in a discussion on "The meaning of forever". He is quoted as saying, “For many years, the R&B community has posited the classic notion that forever is presumed to go on and on like our love,” and “This assertion then raises a problem of even greater complexity: how to ...

quantum-mechanics

Slereah

Damn the distinguishing but non strongly causal spacetime example is pretty elaborate

0celo7

04:13

o.o

0celo7

05:32

0

A: Why not all timelike world lines have infinite total length?

Timelike curves of finite lengths are usually obtained by considering either : Singularities, where the curve will begin and end abruptly. Consider for instance a spacetime with two points removed such that the second point removed is in the chronological future of the first point removed. Th...

@Slereah "Easy to see."

You're not a textbook author D:<

JokelaTurbine

07:37

@Danu The HSM sounds interesting. Sometime's I found it funny when the possibilities of simple trigonometry has been forgotten, like here; physics.stackexchange.com/questions/161333/… It seem's like the most of the living people have died, as defined by Erdős. I don't expect much progress in science can be made by dead people, no matter what lesson they take, or what books they have read.

Danu

07:56

@JokelaTurbine I find it funny that you are ready to pass judgment on modern-day physicists, though for all intents and purposes the previous century has easily been the most fruitful for physics (and all other sciences).

JokelaTurbine

If you are not able to think different, are you able to think at all? Of course you need to use the Math to prove your thoughts, but the history have shown, that the correct answers can be found through various approaches. So it's not even reasonable to make the same approach, which somebody else has already made. What new could possibly been found there? youtube.com/watch?v=EYPapE-3FRw&feature=youtu.be&t=450 "From all the known.." I am not in a position to pass judgement to anyone.

Danu

@JokelaTurbine I like that last sentence.

JokelaTurbine

@Danu I like this kind of answers of JR physics.stackexchange.com/questions/214459/… But I do wonder how the others are not able to calculate that little by them selfs,,,

The math really makes the judgement. I notice it my self very often. Ie here; physics.stackexchange.com/questions/216591/… if you look the edits you note the Question is very different, and found actually even a complete new approach.

The simple math gives the oxygen in Mercury's atmosphere a very obvious and logical explanation.

And though this math has been available to everyone, all the time, it seems no one has lived it live before. “There are only two ways to live your life. One is as though nothing is a miracle. The other is as though everything is a miracle.” -was once said. Erdős defined "the other" as a living dead.

DanielSank

09:03

@ChrisWhite plz do

skull petrol

We lost to the Vikings D:<

skill patrol

While the Patriots remain undefeated.

;-)

Balarka Sen

10:08

@Slereah Assuming you want a 3 dimensional example, $S^1 \times K$ where $K$ is the Klein bottle. If you want orientable, $S^1 \times M_2$ where $M_2$ is the surface of genus $2$.

yuggib

Viks first in the NFC north...who would have thought that :o

skill patrol

10:21

Ya! really :-/

Who's your team @yuggib?

Balarka Sen

Not sure if this helps because I barely know what it is: if you have a topological space $X$, you can look at the finite dimensional (real/complex) vector bundles on $X$. This admits a group structure by taking two vector bundles, and taking their direct sum by fiberwise direct sum. This group is denoted as $K^0(X)$, the $0$-th $K$-group of $X$.
Higher $K$-groups are not hard to obtain, you just need functoriality and they naturally arise from trying to get the long exact sequence axiom to work.

yuggib

@skillpatrol I don't have a favourite team; but this year I like the rams...they're nasty :-D

skill patrol

cool

yuggib

@BalarkaSen Since you talk about that, a sanity check: the cotangent bundle is the topological dual of the tangent bundle (seen as a topological manifold), right?

Even for infinite dimensional manifolds?

Slereah

10:55

@0celo7 it's just $ds^2 < 0$ for integral lines of $\phi$

Pretty easy to see

Balarka Sen

@yuggib Not sure what you mean by topological dual. It's just the dual bundle of the tangent bundle.

By dual bundle, I mean fiberwise dual.

"infinite dimensional manifold" well, first you need a set-up in which infinite-dimensional manifolds make sense and have reasonably well tangent spaces. I have never worked with them, though.

yuggib

@BalarkaSen I know it is a fiberwise dual, but I would like it to be dual in a topological sense, with respect to the topology of the tangent bundle.

Balarka Sen

What do you mean by dual as topological spaces?

yuggib

the space of continuous linear functionals on the topological vector space

Balarka Sen

there is no topological vector space here. a smooth manifold need not have a vector space structure.

yuggib

11:04

ok, yeah right...forget about the vector space structure

Balarka Sen

Then your definition does not make sense to me. How can we consider linear functionals without domain/codomain being vector spaces?

yuggib

good point

so it is something I can say only locally

for each fiber

Balarka Sen

Well, fiberwise, yes.

yuggib

that is quite annoying... but just for my purposes I guess :P

thanks anyways

Balarka Sen

no problem.

Secret

11:43

i100.independent.co.uk/article/…

Converting this paragraph taken form the link into something like a digraph, we obtained the following:

the West are indirectly supporting and countering IS so that nobody can overpower them and no worse enemy can be introduced in the system
That is, in a system level, the whole thing is a steady state

The same graph drawn without intersections
(and corrections on -2)

Numbers are calculated based on the following rules:
0. All nodes start with value 0
1. Any node receiving a green arrow get +1
2. The amount of valued gained by the node is determined by the number of arrows connecting between the node and the nonnegative node at the end of the chain
3. Any node receiving a red arrow gain -1 unless the node is negative (because it is too weak)
4. Red and green directionless edges represents hostile or friendly relationships, and don't contribute to the calculations

The question then becomes:
How can we minimise the absolute value of all the nodes in the graph if we are allowed to add only one red or green arrow

Secret

12:39

Correction on errors

0celo7

13:27

you forgot the lizard people

I hypothesize that @ACuriousMind has class on Monday around this time.

0celo7

14:06

@skillpatrol The smartest, most talented teams tend to win.

@BalarkaSen What does that do?

Balarka Sen

What does what do, @0celo7?

0celo7

@BalarkaSen K theory

I'm fairly uninterested in math for the hell of it

What can I do with K theory

Prove Atiyah-Singer?

Balarka Sen

I know a few applications in homotopy theory. I can't tell you what you can do with it in physics.

Googling tells me it's useful in high-energy physics :P

0celo7

@BalarkaSen Yes, because you can prove Atiyah-Singer

Balarka Sen

okie.

0celo7

14:14

And possibly other things

looks like there's a wiki page for it in string theory

but string theory is just math

so that doesn't actually count

Balarka Sen

fair enough.

0celo7

@BalarkaSen when I say "string theory is math"

I mean that no one has seen a string or measured a string

or anything with strings

so it's just a mathematical model at this point

Balarka Sen

Sure, but I don't really care for the stuff physicists try to do with them. Apparently string theory have a lot of mathematical structures, and they are useful in topology. That's good enough for me.

:)

0celo7

I don't think string theory increases GDP

so it's meh

Slereah

14:34

String theory is used in solid state physics

Very GDP

0celo7

@Slereah bullshit

and not all solid state physics is GDP

Danu

@HDE226868 I noted that you mention Planet Hunter on your profile

What are all the possible tags one can apply to a light curve?

I'd like to have a list overview.

@HDE226868 yay my first really nice one

Slereah

14:58

@Danu Is that the GDP

Danu

@Slereah You'd hope not.

skill patrol

It could be :P

0celo7

@Danu there is some economics graph that looks like this

maybe gdp/debt or something

dunno

booms and bursts

@Secret Missing Russia, Turkey, Kurds. Kurds are rebels, you need more granularity with the rebels. Some rebels want to kill others.

yuggib

@Danu , @ACuriousMind I have a question on smooth manifolds that you may be able to answer:

0

Q: Duality between tangent and cotangent bundles

Given a smooth manifold $M$, the cotangent bundle $T^*M$ is dual to the tangent bundle $TM$ "fiberwise", i.e. $\forall x\in M$, $T^*_x(M)=(T_x(M))^*$. Now, if the manifold is a vector space, then the tangent space is a topological vector space, and the cotangent bundle is homeomorphic to the top...

manifolds vector-bundles duality-theorems

0celo7

@yuggib what is the topological dual

yuggib

15:13

the space of continuous linear functionals

and btw, writing this question I understood that with no doubt I hate manifolds

0celo7

:o

Danu

@yuggib And I hate your functional analysis ;D

yuggib

;-D

no functional analysis in the question though....just plain old topology

for the ones interested in the physical context, here it is:

the cotangent bundle is the phase space, the tangent bundle the configuration space of a classical physical system

0celo7

@yuggib so?

yuggib

obviously, if you choose the manifold to be an open of $\mathbb{R}^d$, then the tangent bundle is homeomorphic to $\mathbb{R}^{2d}$, as it is the cotangent bundle

and they are therefore (I think) in turn "one the dual of the other"

in the sense described in the question

0celo7

15:17

(co-)tangent bundles over vector spaces are trivial

yuggib

yes exactly

and therefore everything is easy

0celo7

right, but in mechanics you don't always choose $\mathbb{R}^d$

yuggib

but is it true also in infinite dimensional smooth manifolds? is it true in general as stated?

0celo7

double pendulum is $T^2$

yuggib

and also, are there other situations where the tangent bundle is a top vec space

?

0celo7

15:19

I have no clue what a top vec space is

yuggib

If yes, then the cotangent bundle is homeomorphic to its dual?

topological vector space

a vector space with a topology

0celo7

@yuggib I know

I have no clue what that means

@yuggib meaning?

does the metric give you a topology?

yuggib

there is no metric

0celo7

if there is a metric

does it give a topology

yuggib

yes

a metric vector space is a topological vector space

0celo7

15:20

so a TVS has to be a vector space

yuggib

yes indeed

0celo7

so how many tangent bundles are vector spaces

yuggib

that is a starting question, yes

0celo7

is that what you're asking?

yuggib

not only that

0celo7

15:21

well, let's see

if it's a vector space, it must be homeo to some $\mathbb{R}^{2d}$, right?

yuggib

also if, given the tangent bundle as a vector space (and it has a topology), the cotangent bundle is the topological dual of the tangent space

@0celo7 in finite dimensions, yes

0celo7

@yuggib in finite dimensions this question is trivial!

stop being a crazy mathematician for once

yuggib

why is trivial? all the manifolds with a vector space as tangent bundle have to be vector spaces (in finite dimensions)?

0celo7

I'm pretty sure.

::waits for Danu to destroy hopes and dreams::

Apparently the tangent bundle of the torus is trivial.

Wait

Ohhhhhh

But the torus is not a vector space

yuggib

ok, and therefore also the tangent space is not one?

0celo7

15:27

@yuggib $TM\simeq \mathbb{R}^{2d}=\mathbb{R}^d\times\mathbb{R}^d\simeq\mathbb{R}^d\times M$

iff $M$ is a vector space

so

we need that $M$ is a vector space

yuggib

I am not convinced by the iff over there

0celo7

and we need the tangent bundle to be trivial

@yuggib that's for the last one

$M\simeq\mathbb{R}^d$ iff $M$ is a vector space, iirc

yuggib

ok

0celo7

and the first $\simeq$ is because you want $TM$ to be a vector space

yuggib

I agree

0celo7

15:30

so now we have to find the condition for the tangent bundle to be trivial

I shall peruse a geometry text

I bet there's some homological or homotopical conditions

apparently the tangent bundle of a Lie group is trivial, @yuggib

maybe there's something you can gleam from the proof of that for a general condition on triviality

or just wait for ACM to show up ;)

> Proposition 18.21 The tangent bundle TG of a Lie group G is trivial.
Proof: We have a global basis of sections given by the left invariant vector fields.

what

@yuggib maths.bris.ac.uk/~mazag/109c/hw4.pdf

Just solve problem 1 (a)

yuggib

ok, but that does not imply, I think, that then the tangent space could not have a vector space structure

or maybe it does, if $M$ does not have one

0celo7

if $M$ is not a vector space there is no way $TM$ can be one

I think I showed that above

or not

maybe @ACuriousMind has the answer

yuggib

well, you only showed that it can be written as a trivial bundle if $M$ is a vec space

0celo7

@yuggib right, but if $TM\simeq R^{2d}$

then $TM\simeq R^d\times R^d$

automatically

and this means that the bundle is (a) trival and (b) over a vector space

@yuggib Bam.

Covered by a single chart...

Is that not just $R^d$?

So a vector space

Although that just goes in one direction. Damn.

@yuggib You should take a look at Lee's Smooth Manifolds

He has a lot of info on the tangent bundle.

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