The h Bar

Der Wixxer (English: The Trixxer; refers to German "Wichser", wanker) is a 2004 German parody of likewise German crime films based on works by Edgar Wallace, especially the film Der Hexer (1964) - a German adaption of The Ringer. The film was directed by Tobi Baumann and written by Oliver Kalkofe, Oliver Welke and Bastian Pastewka. It is about two policemen, Inspector Very Long (Pastewka) and Chief Inspector Even Longer (Kalkofe) who must find the Wixxer, a gangster who wants to take over London's crime world. == Plot == The film begins at BlackWhite castle, one of the last castles in black & white...

ACuriousMind

5:21 PM

@JohnRennie Not that obsure for a fellow German ;)

0celo7

@ACuriousMind Does $\Bbb Z_2*\Bbb Z_2$ have any index 3 normal subgroups?

ACuriousMind

@0celo7 not the slightest idea.

0celo7

@JohnRennie Ah, the Legendary Master Baiter.

Yashas Samaga

-3

Q: The Theory of Relativity when looked at in-reverse?

Please read my following comments and advise on whether General Relativity makes sense as well when viewed in reverse? Doing so it seems makes sense with Quantum Mechanics then, considering the two would then share a common reference of time. To greatly simplify, Einstein’s solution of Relati...

general-relativity

that is a personal theory question, right?

He is talking about varying speed of light with fixed time

0celo7

@Slereah did you read that paper?

Slereah

5:23 PM

Not in detail

John Rennie

@0celo7 Is that in the film as well? I remember that joke being made when I was at school in the early 70s so it's at least 45 years old! :-)

Slereah

I think I've read it before

0celo7

@JohnRennie I recall the joke from an episode of Bones.

I think it had to do with geocaching.

(I don't know if they have that on the old continent. Probably not.)

@Slereah there are no details

he says that three things are wrong, but never explains why

ACuriousMind

@YashasSamaga yup.

Slereah

Those things are similar to what JD says

JD is very much against Kruskal coordinates

Though he does not say why

0celo7

5:27 PM

> If you do not like ham, you may prefer to call it the tofu sandwich theorem.

lol

Slereah

what about

Hamburger moment problem

In mathematics, the Hamburger moment problem, named after Hans Ludwig Hamburger, is formulated as follows: given a sequence { mn : n = 1, 2, 3, ... }, does there exist a positive Borel measure μ on the real line such that m n = ∫ − ∞ ∞ x n d μ ( x ) ? {\displaystyle...

So anyway are there any proposed solutions to solve the triviality of interacting fields?

0celo7

What's with all of the censorship above?

Slereah

Or do we just kinda hope that string theory will do that

0celo7

@ACuriousMind Suppose I have some connected space $X$. I want to find a connected triple cover of $X\vee X$. How about taking $Y=X\sqcup X\sqcup X$ and then gluing $Y$ to another $Y$ in some way?

ACuriousMind

@0celo7 How do you know there is a triple cover of $X\vee X$?

0celo7

5:33 PM

The only issue is what happens to the basepoints for the gluing

Slereah

Help topology has invaded the chat

Speaking of

I should get Munkres

0celo7

@ACuriousMind let's assume it's locally connected with infinite $\pi_1$

Slereah

Since I've been on a big book buying bender

0celo7

so the universal cover is infinite-sheeted

ACuriousMind

@0celo7 Leaving replies to deleted messages in the chat history is pointless, because they make little sense.

0celo7

5:34 PM

I think there should be a triple cover in that case

Slereah

Quite the aliteration

0celo7

might not be regular, but still

ACuriousMind

@0celo7 Why? $\mathbb{Z}$ is infinite but does not have subgroups of order three.

Slereah

amazon.fr/Topology-James-Munkres/dp/0131816292/…

Is it that one

0celo7

@ACuriousMind if the cover is not regular, then the fund grp of the cover doesn't have to be related to the original space at all

so drop that condition altogether

is it not possible to always find a triple cover?

I don't care how bad it is

@Slereah yes

ACuriousMind

5:36 PM

@0celo7 Sure, take the threefold disjoint union of the space, done. Non-regular covers are not interesting.

0celo7

I also recommend Lee

Slereah

Let's get topological baby

0celo7

@ACuriousMind connected

I'm supposed to find a non-regular cover for my homework for some reason

"Find a connected three-sheeted covering space of B that is not regular, and use it to show that
the fundamental group of B is not commutative."

I don't know how to do the second thing once I have the first...a non-regular cover says nothing about the fundamental group, no?

Slereah

Munkres ordered

Maybe I'll understand your gobbledigook soon

0celo7

what if I just do $\vee ^6 X$?

ACuriousMind

5:40 PM

@0celo7 That's a very odd exercise in my eyes

I've never cared about non-regular covers

0celo7

@ACuriousMind Maybe they're important in knot theory

Slereah

Why can't you find the universal covering of $\Bbb R^2 \# T^2$ instead

0celo7

We found a non-regular triple cover of $S^1\vee S^1$ in class

@ACuriousMind Is $X\vee X$ a double cover of $X$?

I doubt it. Finding an admissible neighborhood around the basepoint seems hard.

ACuriousMind

@0celo7 No, it isn't, exactly for that reason.

Yashas Samaga

Are questions which can be answered with one line good questions?

Slereah

5:44 PM

What is even $\vee$ here

Yashas Samaga

0

Q: Equation of motion

I'm not sure this equation is the right one if we invert the direction of the positive y-axis. Am I right if we assume the positive direction of the $y-axis$ downwards, that the correct equation of motion would be $m\ddot{y}=mg-{\frac{\rho A C}{2}\dot{y}^2}$ as the spacecraft tries to land do...

homework-and-exercises newtonian-mechanics drag free-body-diagram

like that one

is it a homework without effort question?

ACuriousMind

@YashasSamaga You tell me :P

Yashas Samaga

lol

ACuriousMind

@Slereah The "smash product", i.e. just gluing together two pointed spaces at their basepoint.

Slereah

What

0celo7

5:45 PM

@ACuriousMind Ok, so we found that $\vee^4 S^1$ is a non-regular three-fold cover of $S^1\vee S^1$

Slereah

I'm no big city mathematician but $S^1$ seems pretty not pointy to me

0celo7

@Slereah "pointed" means with a special point identified

a "basepoint"

Slereah

Oh

So not a manifold at all

ACuriousMind

@0celo7 I'm pretty sure that your $\vee^4$ there is not gluing together four spheres at the same point, is it?

Slereah

Oh wait

Is it a non-hausdorff manifold

0celo7

5:46 PM

@ACuriousMind oops

ACuriousMind

@Slereah A pointed space is just a topological space $X$ together with a chosen $x_0\in X$ as the "basepoint".

0celo7

I meant $(S^1\vee S^1)\vee (S^1\vee S^1)$ with all gluings at different points

@ACuriousMind I just said that!

So...does $(X\vee X)\vee (X\vee X)$ work in general?

(as a 3-cover of $X\vee X$)

ACuriousMind

I think it does

0celo7

god, to show it's not regular I need to find the lifts of generators of $\pi_1(X)$

I hate topology!

@Slereah Munkres is good because he talks about function spaces a bit

Slereah

I could use some topology

Like I know the basics for physics

But I can't really follow topology that well

0celo7

5:52 PM

@ACuriousMind So...what exactly would the map $p:\vee ^2(\vee ^2 X)\to \vee^2 X$ be?

I didn't write it down in my notes, stupidly.

oh, no

ACuriousMind

I can draw it for the $S^1$s, I cannot for general $X$ and I am not entirely sure this works in general.

0celo7

We used the fact that $S^1\to S^1$ is a double cover, namely $z\mapsto z^2$

So I need to unwrap $X$ once

(in my case $X=RP^2$, so that's ok)

So we'd have $(RP^2\vee S^2)\vee (S^2\vee RP^2)$

@ACuriousMind You mean wedge.

ACuriousMind

Yeah, I do

0celo7

Smash is $X\times Y/X\vee Y$.

Ah, $S^2$ is simply connected, so the fundamental group of this is not horrible!

It's just $\Bbb Z_2*\Bbb Z_2$ again I think.

Moses

There should really be some better policy (or implementation) on people downgrading your posts because they are guessing that it is homework...

2

0celo7

6:01 PM

@ACuriousMind Does this make sense?

It's a three-fold covering because the preimage of a point lands up twice in one $S^2$ and once in an $RP^2$

ACuriousMind

@Moses What do you mean?

If by "downgrading" you mean downvoting, then voting will never be restricted - it is anonymous and completely up to the users by design.

If you mean closing, then accountability is already given by five users having to agree to close a question and the possibiity to reopen.

@0celo7 I think so, yes

0celo7

@ACuriousMind I think one has to use the homotopy lifting theorem now. If the lifts don't commute, then the original things don't commute either, right?

things = homotopy classes of curves

Slereah

I should learn more about homotopies

Apparently there are neat theorems relating timelike homotopies and causal structures

0celo7

Ah! One has to show that the lifts of two products end up at different points!

ACuriousMind

@0celo7 Yes, a covering $X\to Y$ induces a monomorphism $\pi_1(X)\to \pi_1(Y)$ regardless of regularity.

0celo7

6:20 PM

@ACuriousMind what does that mean?

ACuriousMind

@0celo7 That means if you find non-commutative loops in the cover, the fundamental group of anything covered by it can't be commutative.

monomorphism=injection, if that's the problem :P

0celo7

@ACuriousMind yep

@ACuriousMind why not just say injection?

ACuriousMind

Because I'm more used to mono/epi than injective/surjective in algebraic contexts.

0celo7

@ACuriousMind but I recall there being something wrong with that from the categorical viewpoint

ACuriousMind

@0celo7 Nope, a group monomorphism and an injective group homomorphism are the same thing. You're probably thinking of a ring epimorphism not having to be surjective.

0celo7

6:33 PM

Ah, yes

Confusing terminology!!!

Moses

@ACuriousMind So the two dimensional complex vector space $\mathbb{C}^2$ you refer to is the vector space spanned by $|0 \rangle$ and $|1\rangle$ with complex coefficients. How do we get from that to $| 0 \rangle = \hat{z}$ and $|1 \rangle= - \hat{z}$ since surely we don't have $|0\rangle = - |1 \rangle$? since $|0 \rangle$ and $|1 \rangle$ are linearly independent.

ACuriousMind

@Moses Writing $\lvert 0\rangle = \hat{z}$ and $\lvert 1\rangle = -\hat{z}$ is very confusing - the equality is not an equality as vectors, it is just an equality as points on the sphere.

You get to writing this by looking at your general form for $\lvert \psi\rangle$ and defining the map $\mathbb{C}P^2\to S^2, \lvert \psi\rangle\mapsto (\phi,\theta)$ using polar coordinates for the $S^2$, then by embedding the $S^2$ into an $\mathbb{R}^3$ and then observing that the two poles get mapped to the $\hat{z}$ and $-\hat{z}$ in the $\mathbb{R}^3$.

None of these maps respects the vector space structure, so talking about the $\hat{z}$ is rather pointless instead of helpful, in my opinion.

Moses

@ACuriousMind Okay that makes a bit of sense. Is $\mathbb{C}P^2$ the set of equivalence classes $v \sim w$ if and only if $v = \lambda w$?

ACuriousMind

@Moses Yep, exactly

0celo7

@Moses lambda nonzero of course

Slereah

6:49 PM

Also the vectors have to be non-zero

Moses

@ACuriousMind Okay and we define the mapping as $(\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta)$?

ACuriousMind

@Moses The embedding of the sphere? Yep.

Moses

@ACuriousMind Is there a particular reason why we can do this? I know it has to do with the normalization $|a|^2 + |b|^2 = 1$ requirement.

ACuriousMind

@Moses I'm not sure what you mean by a "reason". It's a mathematical fact that $\mathbb{C}P^2$ is isomorphic to $S^2$. This is because every vector in $\mathbb{C}^2$ is related to one of the form of $\lvert \psi\rangle$ by multiplying with a complex scalar, and I just wrote down how to map the $\lvert \psi\rangle$s to the $S^2$.

DanielSank

@heather I've organized the AMA questions from meta into what I think is a reasonable set. When it's time, I'll get the show started etc. and ask the first question. I'm planning to stay out of the way as much as possible and only nudge the conversation back on track if it starts to get spaghetti-like.

If the conversation stalls etc. then I'll go down the list I made and ask a new question.

If any of this is not to your liking, please let me know. I'm at your service.

0celo7

7:07 PM

@DanielSank when is the AMA?

DanielSank

@0celo7 It's in the room scheduled events.

0celo7

I'm on mobile, how do I access that?

DanielSank

chat.stackexchange.com/rooms/info/71/the-h-bar?tab=schedule

click on that ^

0celo7

Thanks.

Looks like I can make it

Slereah

7:34 PM

The proof of Taylor's expansion theorem is simpler than I thought

0celo7

@Slereah are you reading Jost!?

Slereah

Why not

Is it FORBIDDEN?

anonymous

Aaaarrgghh, I have to wake up at midnight to attend the AMA :P I wish the earth was flat :)

2

DanielSank

@Kaumudi.H Please provide recipe.

Mostafa

@anonymous I've asked 14 questions and yet can't attend because of its time :(
it's so unfair

anonymous

7:45 PM

@Mostafa Yeah same here. I think I will go to sleep now and set an alarm for 2 hours later. Let's see if I can make it! Otherwise I'll have to be happy with the transcript :)

Lenz

buenas a todos

Slereah

Wait I'm not actually sure I get the proof

At some point there's supposed to be $$f(x) = \sum_{\nu = 0}^n \frac{1}{\nu!}f^{(\nu)}(x_0) (x - x_0)^\nu + \frac{1}{(n+1)!} (x - x_0)^{n+1} z$$

which is apparently true by some choice of $z$

But I'm not sure why this is true

At least if $z$ doesn't depend on $x$

dmckee

8:01 PM

@anonymous I've participated in two particle physics projects with collaborators in the Americas, Europe and east Asia.

There is no time when you can hold a phone conference in that situation when someone who needs to attend ought to be sleeping.

Most times no only should someone be sleeping but another group ought to have left the office hours ago.

Makes for tense scheduling sessions.

0celo7

@Slereah page?

Slereah

41

0celo7

@Slereah Just solve the equation for $z$.

$z$ does depend on $x$ in theory

but you're trying to show that it doesn't

Slereah

But then he differenciates $\varphi$ without $z'$

Moses

@ACuriousMind Just to confirm, is $S^2$ a three dimensional unit sphere?

0celo7

8:14 PM

because it's just $f^{(n+1)}(\xi)$

@Slereah with respect to $y$

$y$ is a new variable

Slereah

Ah yes

I see

might be good form to write it $z(x)$

ACuriousMind

@Moses It's the unit sphere in $\mathbb{R}^3$. Whether you call it 3-dimensional (because it sits in $\mathbb{R}^3 naturally) or 2-dimensional (because it's a 2d manifold) is a matter of convention.

0celo7

@ACuriousMind You won't find anyone calling it 3-dimensional except for the wiki article...

ACuriousMind

@0celo7 ...Moses just called it 3-dimensional :P

Moses

@ACuriousMind I was influenced by wiki :/

0celo7

8:17 PM

See!

@Moses 99% of mathematicians call $\{x\in\Bbb R^{n+1}:||x||=1\}$ the $n$-sphere.

Sir Cumference

Did we get mchem btw?

0celo7

@ACuriousMind I'm trying to learn some algebra

Danu

o/

what kind of algebra @0celo7?

0celo7

@Danu \o

@Danu Constant curvature manifolds

Danu

@ACuriousMind Also, nobody calls it 3-dimensional.

0celo7

8:23 PM

so representation theory

Danu

@0celo7 Einstein?

Or constant Riemann curvature

0celo7

constant sectional curvature

the problem is that the geometry is unintelligible because it's written by an algebraist

Danu

Sectional contains all the info of the Riemann tensor right? Is there a difference between R constant and K constant?

0celo7

@Danu What does it mean for R to be constant?

it's a tensor

But yes, sectional curvature does determine the Riemann tensor

Moses

Valid point but people say all kinds of crazy stuff...

Danu

8:25 PM

@0celo7 So is K?

@0celo7 Constant components

0celo7

@Danu it's just a function

@Danu That's always a bad definition!

Danu

@0celo7 No it isn't---unless you're working with 2-manifolds

It takes a tangent plane as input

0celo7

The metric of flat space is wildly varying in polar coordinates

Danu

Yeah, good point

0celo7

@Danu that's still a function

Danu

8:26 PM

@0celo7 But not on $M$

0celo7

@Danu yeah

but "K is constant" make sense because the codomain is just R

that's what I meant

Danu

@0celo7 Then you can also define $R$ as a function on $\mathfrak X(M)$ x 4

by $g(R(X,Y)Z,W)$

0celo7

@Danu your vectors there could vary wildly

Danu

"wildly" being "in $\Bbb R^n$"? :P

0celo7

@Danu wildly means "more that nothing" :P

Danu

8:28 PM

But so can they for $K$

There are still many planes inside each tangent space

0celo7

In any case, "constant curvature" means $K(p,E)$ is constant for all $p\in M$ and plane sections $E$ of $T_pM$

Danu

unless you have some action taking all of them into each other I don't see why that's so much nicer than R

Okay

AccidentalFourierTransform

Did countto10 leave again?

Danu

Why are you interested in constant sectional curvature?

0celo7

because I would like to learn more representation theory and this is a nice excuse to do so

and because "Einstein Manifolds" says the book is good

Danu

8:31 PM

Which book is it? :)

0celo7

I'm kind of burnt out on analysis for the time being

Danu

I'm also a little interested in representation theoretic Riemannian geometry stuff---though I don't actually want to get into the representation theory much. I like quoting the results.

0celo7

@Danu "Manifolds of Constant Curvature" I think

By Wolf

Danu

Ah, Wolf

I'm using some Wolf spaces in my thesis :P

Spaces of CC is the name

Tell me when you get to chapters 8-9

I'm interested in knowing a little about symmetric spaces

0celo7

There's a surprising number of typos and flat-out wrong things for a book in its 6th edition.

Moses

8:34 PM

@ACuriousMind To summarize the Bloch sphere would this be a reasonable thing to state to summarize it:

$\mathbb{C}^2 \to \mathbb{C}P^2 \to S^2 \hookrightarrow \mathbb{R}^3$ given by $| \psi \rangle \to [| \psi \rangle] \mapsto (\phi, \theta) \mapsto (\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta)

0celo7

@Danu Will do.

ACuriousMind

@Moses Yep :)

Slereah

It is our friend John Duffield

Hello friend

Danu

?

0celo7

@JohnDuffield hi

Moses

8:38 PM

@ACuriousMind Great thanks.

0celo7

@Danu He has a bad habit of not stating hypotheses. I'm trying to figure out what some "appropriate vector" is whose identity is a mystery to me.

Danu

"a vector that does the job at hand"

Presumably

then one just needs to see that such a thing exists

(??)

ACuriousMind

@Moses You're welcome

0celo7

@Danu indeed. Only I don't know what the job is. I think I need to recheck a previous section for some lemmas

(Thermo class right now)

He might just mean "nonzero"

Danu

@0celo7 Let's hope those exist :P

ACuriousMind

8:47 PM

lol, my first question on rpg.SE went hot.

Danu

@ACuriousMind lol

That site is such hot-network garbage

ACuriousMind

It doesn't even have a clickbaity title!

And only one answer

Yet I just saw it in the HNQ sidebar

Danu

I don't see it.

Slereah

People have stopped answering me on the science fiction SE chat

Danu

You clickbaiting me bro?

Slereah

8:48 PM

I am talking too much about star trek even for nerds

ACuriousMind

@Slereah lol

@Danu Always!

Danu

@ACuriousMind You're coming across as extremely pedantic German mode in that question lol

"BUT THE RULE BOOK SAYS [...]" :D

I'm pretty sure you also came up with the answer given by yourself---I'm wondering what you were looking for in an ideal answer

0celo7

What? Alicia is American, not German.

ACuriousMind

@Danu All the other discussion of DW always emphasises that the GM must always follow the DM rules instead of falling back to the usual improvisational free-form. I find it incoherent that apparently there is a large area of play not governed by those rules that no one ever cares to mention. The answer to every other question about DW GMing is always "follow the GM rules"...except in this case, where I'm expected to just wing it.

Sir Cumference

Found this in my old files

ACuriousMind

8:55 PM

Not that I have a problem with that, but it's a flaw in the rules-as-written.

Sir Cumference