@JohnRennie It's worth it. It's only around 800 pages of "story"; the rest is indices, etc.

@JohnRennie L = life, as in "this is so horrible that it will depress you and mess up your psychological well-being".

It's no rape of Nanking

This may interest some people:

Guys

Would you agree that, for a closed geodesic to exist in a $\Bbb R^4$ manifold

At least one of the coordinate has to have an extremum

I'm pretty sure that's a correct assumption but I'm not too sure how to prove it

Well, closed curve, in general

@Slereah How exactly do you define that?

Well for a curve $x^\mu(\lambda)$

$\exists \mu, \lambda. \ \dot x^\mu(\lambda) = 0$

@Slereah I don't think that models what you want

Does it not

Take flat space and a line parallel to the 1-axis. it has $\dot{x}^\mu(\lambda) = 0$ for all $\lambda,\mu\neq 1$, but I don't see that a straight line has "extrema"

Ah yes, the second thing is

I think it should be either a minimum or a maximum

So that rules out straight lines

Also my point isn't that this condition implies that the curve will be closed

It is that, if that condition cannot be met, then a curve cannot be closed

Are you trying to very complicatedly state that a closed curve must be contained in a closed and bounded set?

Because then obviously $x^2$ has a minimum and all

I'm not sure, maybe?

Are those related

I'm trying to find a generic way to check if a space can have closed geodesics

Cannot, more to the point

For flat spacetime for instance, we have $\ddot x = 0$

@Slereah Well, see, your definition with the $\dot{x}^\mu$ is heavily coordinate-dependent and doesn't actually detect extrema. So why not actually write down the definition of extrema: $\exists c,C: \forall \lambda : c\leq x^\mu(\lambda)\leq C$

David Ige

So that no geodesic can have a minimum or maximum

Noice

I guess that could work

But would you say that sounds like a decent notion, all the same

And this then directly translates into the closed curve being contained in a bounded set, which is a coordinate-invariant and hence "good" statement

I don't mind coordinates

So for a closed curve, we'd need to have like

@Slereah Well, I think it's true, but I'm not sure that actually makes it easier to compute anything

$\exists c, C, \forall \lambda, \mu, \ x^\mu(\lambda) \in [c,C]$

Does that make sense

Every coordinate is bounded

@Slereah Yes

Alright

I'll try to work those equations

Pictured working the equations

What is the theorem called btw

that closed curves are in a bounded set

Jordan curve theorem, apparently?

@Slereah Closed curves are the continuous image of $S^1$, hence compact, and compact sets in $\mathbb{R}^n$ are closed and bounded by Heine-Borel.

Hm I wonder how that theorem would translate into my idea

I guess that since it is within two bounds, it is guaranteed to have a minima and maxima

and it can't converge to it since it is just defined on $[0,1]$

Does that make sense

What is "it"?

The coordinates of the curve

I'm pretty sure a continuous function defined on a closed interval with bounds is guaranteed to have a minimum and a maximum

Yes, because it's bounded it has to have maxima and minima - but what do you mean by "it can't converge to it"?

Well like $\tanh$ can be bounded

But not have any local minima and maxima

@Slereah That's the extreme value theorem

Good to know

I'm just discovering theorems left and right

and they have been known for 200 years

How unfortunate

Hm

I think it should be that

If a curve is closed

Then for each coordinates

Either there is both a minima and a maxima

Or there is a... the third thing

What's the $\ddot x = 0$ thing again

Saddle point?

Or in other words, the minimum is the maximum

What?

BUT

If the minimum is the maximum your function is constant!

Yes

Saddle points do not mean your function is constant!

They do if it is at every point

I say this because if there's a circle in the xy plane, for instance, then it's gonna be a constant in the z coordinate

So I can't use $\ddot x \neq 0$

I guess for this reason, maybe using the definition with bounds will be better, yeah

Otherwise there are too many cases

Although I suppose that derivatives will be useful to prove those bounds

@ACuriousMind took me a few times to parse that

"Nearly every modern citation that I have found agrees that the first correct proof is due to Veblen... In view of the heavy criticism of Jordan’s proof, I was surprised when I sat down to read his proof to find nothing objectionable about it. Since then, I have contacted a number of the authors who have criticized Jordan, and each case the author has admitted to having no direct knowledge of an error in Jordan’s proof.[4] "

@Slereah Lol

Do you know the proof

I do not

I hear it is wrong, tho

I've read that before, not sure if true

A closed interval with bounds is compact, and its image must be compact.

A closed and bounded set has a min and a max. QED.

We did that in my intro analysis course...

It's on page 5 of Bredon.

Pretty well-known thing.

Actually its true more generally for $f:X\to \Bbb R$ with $X$ a compact topological space and $f$ continuous.

@ACuriousMind probably has some sheaf-theoretic generalization

@Slereah No, that's something different.

What's up!

That's the statement that $\Bbb R^2-$closed curve has two connected components, with one homeo to a disk and one homeo to interior($\Bbb R^2-$disk).

Hm, so what would the proof be for minkowski space

@Slereah The proof of what?

@Slereah There's a really powerful generalization with properly embedded submanifolds or something

$\ddot x = 0$, so either the curve has no minima and maxima, or the function is constant

But

The proof is in Dolt and my prof invoked it in his HE proof

The only constant function is the one with initial condition $\dot x = 0$

Which is just a point

So no closed geodesic in Minkowski space

Bif baf bosh

Anyone have an idea on how to prove this?

@Slereah What? why are you now going on about saddle points?

$n$ is the dimension of Euclidean space

@0celo7 Pythagoras.

@ACuriousMind I would have never guessed that

Well I am trying to apply this theorem

To prove the lack of closed geodesics in flat space

Anyone have an idea on how to prove this using Pythagoras?

@0celo7 Consider it, planes at a time

@ACuriousMind I know that you need Pythagoras...

@NeuroFuzzy I know that you have to do that as well

@0celo7 Well, what exactly eludes you, then?

@ACuriousMind How to do it?

All of it?

I honestly don't see how you can know you need to use Pythagoras and not know how to do it.

I do not know how to apply it in $n$ dimensions.

Pythagoras is a statement about triangles, it knows no dimension

Well you need at least one dimension

$d^2=x^2+r_1^2=x^2+(y^2+r_2^2)=x^2+(y^2+(z^2+w^2))$. Suggestive enough?

@NeuroFuzzy I can do it in 3 dimensions dammit

I knew that 5 years ago

I did it in four : ^ )

@0celo7 He's done it in four :P

That doesn't help either

Also, the elegant proof would probably use induction, which you love so much, here ;)

@ACuriousMind that's an ironic nose.

Okay, I'm going to go read.

@ACuriousMind I don't see how to use induction here

@ACuriousMind This is the only part of Sard's theorem I don't understand

:(

Have fun explaining the pythagorean theorem to 0celo7 everyone!

wtf

How am I supposed to know where the triangles are in $\Bbb R^n$

@NeuroFuzzy what is $r_1$

A triangle is just a set of three points

@0celo7 The triangle made out of the n-cube's half-diagonal and the radius of the sphere (when taken downward, not upward as in the picture) has as its third side the half-diagonal of an n-1-cube.

@ACuriousMind eye twitch

what

Also isn't the pythagorean theorem basically an axiom of $R^n$

Isn't it just the metric defined on it

How do you even prove the Pythagorean theorem in $\Bbb R^n$

@ACuriousMind and what induction step!?

For the Pythagorean theorem or for the sphere

Because it's obvious that you get $\sqrt{r^2+\cdots+r^2}=\sqrt{nr^2}=\sqrt nr$.

@0celo7 What? The induction step is that if you know it for the n-1-cube, applying Pythagoras once gives you the statement for the n-cube.

@ACuriousMind ...what

I obviously do not understand induction

You can of course just say "n-fold application of Pythagoras' theorem gives" instead of the induction.

@ACuriousMind is that the same as $x^2+y^2+\cdots =r^2$?

@0celo7 Ehhh? If that's obvious, what is there left to prove?

@ACuriousMind Well is what I wrote correct?

Sure

I don't know how to prove the Pythagorean theorem :(

In $n$ dimensions

How do I do it

Rotate into the plane of the triangle and use rotational invariance of the Euclidean norm?

Sure

I don't see what you think there is to prove - the triangle lies in a plane. The rest of the space is wholly irrelevant.

Then use the definition of the norm

That's basically the norm

Proving it from Euclid's axioms might be a bit longer tho

@Slereah Oh god not that again

cut-the-knot.org/pythagoras/Proof1.shtml

This is, apparently

I don't believe there's any way for me to kick this into the review queue manually, but here's a question that was edited and has not been reviewed for reopening since. Let's get some eyes on it..

Hm

R^n with the taxicab metric does not obey Pythagoras' theorem

I wonder what Euclidian axiom it breaks

Perhaps the bit about a unique line between two points

What is even a line in the taxicab metric

I think the condition to avoid the problem that a curve with $\ddot x = 0$ everywhere should still work in my idea is to add the initial condition that $\dot x(\lambda_0) \neq 0$

Since I don't want to have a dot

Hello guys, can someone please help me with the problem which looks like basic Newtonian physics. I'm stuck at the last step here physics.stackexchange.com/questions/264148/…

should I just take a sum of vectors Fp and MJ to get Fm?

@ilich what is MJ

oh i see

Or they could bring some of the lame Superman villains

Like the toy man

Or that leprechaun guy

Oh wait

wrong window

@Slereah what

Well for three points with the taxicab metric, $a + b = c$

@Obliv yeah, internet is full of tutorials about how to find torque from force, but not vice versa. Surprisingly, StackOverflow was the only place where I found solution for Fp and derivation

now only one small step left to find force equivalent to Fp but acting along the muscle

@ilich only problem i saw with it was whether $M$ and $M'$ varied. You might need diff eq. for that.

@Obliv this is a very valid observation, I'm actually recalculating torques and forces many times per second, essentially integrating numerically. This is a simulation driven by PhysX engine which does integration for me.

oh then that's fine.

@Obliv so do you think that vector sum will work fine? like Fm = Fp + MJ

oh, actually it's not going to work because Fp and MJ are orthogonal vectors and vector sum means that Fm going to be a hypotenuse, having pi/4 angle with MJ, which is definitely not how it works in reality

@yuggib Ok my prof made me prove Sard explicitly for the $n=1$ case because he didn't like that induction.

It wasn't terrible.

arxiv.org/abs/1205.4710

It irks me that people don't know \sin is a thing

Like how incompetent can you be :/

@yuggib Basically, it came down to proving the following: let $f:\Bbb R\to\Bbb R$ be $C^1$ and let $C=\{x\mid f'(x)=0\}$. Then $f(C)$ has measure zero.

One can prove this by hand.

This is the starting point for the induction, then.

According to an exercise in Hirsch, one can even take $f$ to be merely differentiable.

But I used Taylor's theorem to prove it and I dunno if that works in that case