The h Bar

5:00 PM

Before we close out, I do want to mention that my plan for the next chat session in two weeks is to return to discussion of our homework policy. There will be a companion meta post sometime between now and then. Anyone who is sick of that topic is welcome to not attend, of course!

And with that, our time is up. See you all (or some) in two weeks!

Thanks everyone for attending, and now back to regularly unscheduled discussion.

Secret

Why our mobiles and computers gone screwed. Cosmic rays may be to blame

skullpetrol

Thanks @DavidZ

Slereah

"For $p\neq q$ in $\mathcal C$, $q \in J^+(p, \mathcal C) \equiv \vec{pq}$ is future pointing causal (analogously for $I^+$)"

Why is this a lemma

Isn't that true by definition, even outside of a normal neighbourhood

0celo7

5:16 PM

What's up?

@Slereah what book?

O'neill

page?

403

@Slereah Well, you need to prove that $\vec{pq}$ is future pointing causal...

but it is easy

Slereah

Isn't $J^+$ the set of points that can be reached by future pointing causal curves

0celo7

5:19 PM

yes

Slereah

I mean

I guess it's not entirely trivial

0celo7

but how do you know that $\vec{pq}$ is future pointing causal?

Slereah

But it's a one step proof

Because $q \in J^+(p)$

Hence there exists such a curve

0celo7

but why should the geodesic be such a curve?

Slereah

Ah, it's a geodesic specifically

I see

I guess that's where the normal neighbourhood comes in

Although that's still fairly trivial

OR IS IT

0celo7

5:24 PM

@Slereah The preimage of the convex set under the exponential map is convex. So just take a straight line connecting the two, this gets mapped to a geodesic in the manifold.

It's future-directed and causal because of properties of the exponential map I guess

I have to check out Lemma 5.33

Slereah

Do straight lines map to geodesic with the exponential map?

0celo7

Yes

Slereah

and vice versa

I should look up that theorem

0celo7

That's the local classification of geodesics

Works for any affine connection

Hmm. Lemma 5.33 only goes one direction.

Slereah

Hm, let's see

What theorem would that be in O'neill

I mean it makes sense since there's the whole relation between the magnitude of the vector in the tangent space and the range of the geodesic

0celo7

5:36 PM

@Slereah Take the curve you get from the definition that connects $p$ and $q$.

Pull it back via the exponential map, then Lemma 5.33 shows it lies in a timecone.

Slereah

Sounds reasonable, yeah

thx

0celo7

So you can fit a straight line inside of the timecone that connects $p$ and $q$'s preimages.

Then map that back onto the manifold, you get a geodesic.

So all you have to check is that it's a) future-directed b) causal

b) is easy because the initial velocity is causal, and goedesics have constant velocity

Slereah

Yeah

(At least if they have metric connections)

0celo7

Levi-Civita is metric

Slereah

it is famously

0celo7

5:39 PM

So how to prove a)...

What is the definition of future-directed?

Ah.

It's true at $p$

So take the vector field $X$ that defines the time orientation, let $\gamma$ be our geodesic

We want the sign of $g(X,\gamma')$ to be constant

If it's not, use continuity to find a point where $g(X,\gamma')=0$

but that's not possible for $X$ timelike and $\gamma'$ causal

@Slereah Good?

Slereah

Alright yeah

thx

0celo7

You should read all of the Lorentz algebra stuff in O'Neill and Sachs and Wu

It's very useful, for instance, no two timelike vectors are orthogonal

A timelike vector is never orthogonal to a null one

@Slereah Do you know Witt's theorem?

Slereah

i do not

0celo7

@Slereah it gives a formal way to change an arbitrary Lorentz metric to Minowski

I think Sylvester's law of inertia is a less powerful version

DanielSank

6:00 PM

@ACuriousMind to be honest, I don't really see that temptation.

Rep is rep. It's defined by the system that awards it.

I'm agreeing with you.

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