The h Bar

Full GR action is $$\mathcal{S}_\mathrm{EH} + \mathcal{S}_\mathrm{GHY} = \frac{1}{16 \pi} \int_\mathcal{M} \mathrm{d}^4 x \, \sqrt{-g} R + \frac{1}{8 \pi} \int_{\partial \mathcal{M}} \mathrm{d}^3 y \, \epsilon \sqrt{h}K$$

0celo7

What is $k_g$?

Slereah

$k_g$ is the geodesic curvature of the boundary

0celo7

9:17 PM

No shit

What exactly is it though

Slereah

$$k_g = \|\frac{dU^\mu}{d\lambda}\|$$

0celo7

sigh

what are these symbols

and why are you using $\mu$ in Riemannian geometry

Slereah

It's the norm of the derivative of the tangent unit vector

because I'm great

0celo7

damn physicists

Danu

@0celo7 I don't think a classification of those can be expected, no

Slereah

9:20 PM

BUT

On the other hand

0celo7

@Danu The issue resolved itself

I was being dumb

Slereah

The Hilbert part is $1/16$ and the York part is $1/8$

Why is there a factor of 2 between the two

Danu

For Riemann surfaces, the classification is very simple though

0celo7

@Danu I know

Danu

Not simple to prove, but a simple result.

Slereah

9:21 PM

Hopefully because $k_g = 2 K$

0celo7

My proof is simple: if I draw a random Riemann surface it looks like a bunch of tori :P

Danu

I'm talking about noncompact ones, too

There are only three simply connected ones!

0celo7

@Danu I was going to say "ur mum" but she's got holes ;P

what are they?

Slereah

Plane, sphere and...

Empty set???

Danu

@0celo7 Sphere, plane, disk

Slereah

9:24 PM

Well then you forgot the empty set :p

Danu

(in order of decreasing curvature)

No, I just didn't forget to exclude dumb cases

Slereah

The empty set is the best case!

It has the coarsest and the finest topology!

It's great

0celo7

@Slereah what is K

what is $\epsilon$, $\sqrt h$?

explain these symbols dude

I don't have Wald with me

Slereah

Epsilon is 1 for timelike and -1 for spacelike surfaces

0celo7

eww

Slereah

9:26 PM

$\sqrt h$ is the det of the induced metric

0celo7

is the $\epsilon$ from Stokes theorem

Slereah

$K$ is the second fundamental form

0celo7

using my correct proof

which Wald has wrong

Slereah

hm

I dunno

0celo7

@Danu I like translating rap lyrics into German

Slereah

9:27 PM

$K=\nabla_a n^a$

0celo7

they sound so silly

Slereah

Hm

0celo7

@Slereah that doesn't look like $|| U'||$ to me

Slereah

Close tho

Wait or does it

Agn

Wait

In 2D the induced metric is always flat

Since it's the metric of a 1D manifold

so $K = \partial_a n^a$

0celo7

yes

Slereah

9:30 PM

Hm

How to link $n^a$ to $U^a$

Danu

@0celo7 So for triangulizable surfaces there is a classification, at least.

0celo7

@Danu my algebraic topology sucks, what kind of surfaces cannot be triangulized (fuck that spelling)

Slereah

Weird ones

I remember getting no satisfying answer

bc fuck weird manifolds

ACuriousMind

yesterday, by ACuriousMind

@Slereah Theorem 3 here says that every surface is obtained by removing points from the sphere, then removing discs from the remnant, and then pairwise gluing the boundaries of those discs together

@Danu ^deja-vu :P

Slereah

It is a neat theorem

Danu

9:33 PM

@ACuriousMind nice

@0celo7 LOTS

Slereah

Like

Simple to understand

0celo7

Damn.

Danu

write a comma

between investor and yes

0celo7

u smart

@ACuriousMind KSP is coming to 3DS

Why won't Nintendo let me download my old DS games to my 3DS SD card

WHY NINTENDO

@Slereah hey check out the GHY wiki page

Danu

@ACuriousMind Funny thing that I thought of yesterday: $U(1)$ has exactly one subgroup with order $n$ for each $n\in \Bbb N$, and no other subgroups.

...I think.

0celo7

9:39 PM

what is the last paragraph in the intro about

Danu

9:59 PM

@0celo7 Roots of unity, yeah

ACuriousMind

@Danu What about $\{\mathrm{e}^{\mathrm{i}q}\mid q\in\mathbb{Q}\}$?

Danu

@ACuriousMind Right, there come the problems

0celo7

@Danu is that a cyclic group

Danu

The main point I had in mind is that they're discrete

0celo7

How is the group ACM defend discrete

Danu

10:02 PM

@0celo7 Yes, it's clearly isomorphic to $\Bbb Z_n$

0celo7

*defined

Danu

@0celo7 $\Bbb Q$ is discrete, you know that right

(in $\Bbb R$)

0celo7

Nope

usukidoll

hi hi

0celo7

What is discrete

Countable?

Danu

10:03 PM

Discrete topology

0celo7

What about where q is irrational

ACuriousMind

@Danu Well...is there a subset of $\mathbb{R}$ that's closed under addition and not discrete and not $\mathbb{R}$ itself?

usukidoll

anyone know metric system? that's like in base 10. what unit is only commonly found in there? is it grams?

Danu

@ACuriousMind Yeah, I didn't think about it that way. Right.

@0celo7 It will generate all of $U(1)$

(standard exercise in topology, in fact!)

0celo7

well I'm sorry that I won't have taken topology until next semester

usukidoll

10:06 PM

nevermind I found it

ACuriousMind

@0celo7 That's the whole of $\mathrm{U}(1)$ because I can add $2\pi$ to any rational to make it irrational.

Danu

@ACuriousMind But in some sense this kind of doesn't ruin the picture; one for every finite $n$, and the limit!

ACuriousMind

I'm actually not sure what kind of topology $\mathbb{R}-\mathbb{Q}$ has.

0celo7

@ACuriousMind hmm, how does that work in the other direction?

ACuriousMind

@0celo7 What other direction?

0celo7

10:08 PM

how do you get rationals out of irrationals

Danu

Indiscrete topology if you quotient out the irrational action @ACuriousMind

@0celo7 Not so easily

ACuriousMind

@0celo7 You don't

0celo7

so how do you get like $\mathrm{e}^\mathrm{i}$

ACuriousMind

You don't.

That's why it's a true subgroup

0celo7

then how is it $\mathrm{U}(1)$?

ACuriousMind

10:09 PM

@0celo7 Wait, what is "it"?

$\{\mathrm{e}^{\mathrm{i}q}\mid q\in\mathbb{Q}\}$ is not $\mathrm{U}(1)$.

0celo7

$I:=\{\mathrm{e}^{\mathrm{i}q}\mid q\in \mathbb{R}-\mathbb Q\}$

(for future reference I have given it a name)

you claim $\mathrm{U}(1)=I$?

ACuriousMind

That's $\mathrm{U}(1)$ because for any $\mathrm{e}^{\mathrm{i}\phi},\phi\in[0,2\pi)$ either $\phi$ is irrational or $\phi+2\pi$ is.

Slereah

is that related to like

the fact that an irrational rotation repeated will go through every point (almost) of a circle

0celo7

@ACuriousMind Is it not discrete as well?

proofwiki.org/wiki/Rational_Numbers_are_not_Discrete_Space

I'm confused

are they discrete or not?

ACuriousMind

I suspect Danu meant "totally disconnected", not "discrete".

0celo7

10:20 PM

oh, that's easy

@ACuriousMind I suspect $R-Q$ is also totally disconnected?

ACuriousMind

@Danu See above; We used "discrete" wrong. "Discrete" means every set is open, that's not true for the rationals.

@0celo7 That's why I said "I'm not sure what topology $\mathbb{R}-\mathbb{Q}$ has"

0celo7

@ACuriousMind well you don't even know what topology $\mathbb Q$ has

totally disconnected is not a topology, is it?

what are the open sets of $\mathbb Q$?

ACuriousMind

@0celo7 What? Sure it is, it means there are no non-singleton connected components

@0celo7 Open intervals, where the endpoints of the intervals are still allowed to be reals (subspace topology of $\mathbb{R}$).

0celo7

@ACuriousMind of course, but that's not what I was looking for

@ACuriousMind what

that doesn't tell you what the topology is...

ACuriousMind

I have a feeling we're talking apst each other :p

There can be more than one totally disconnected topology on a set, if that's what you're asking.

0celo7

10:26 PM

@ACuriousMind Well, obviously!

$\mathbb Q$ with the discrete topology is $\mathbb Z$

ACuriousMind

@0celo7 What was your question, then?

0celo7

@ACuriousMind Beats me

@ACuriousMind Have you ever seen "pseudogroup of transformations" used when defining manifolds

ACuriousMind

nope

DanielSank

10:45 PM

@BernardMeurer get enough sleep

0celo7

hmm

I've been sleeping more lately and eating less

coincidence?

Slereah

10:59 PM

Eating less is unamerican, pinko

0celo7

@Slereah I'm not American

Slereah

I knew it

0celo7

and who is pinko

Slereah

Go back to your country

Pinko is an old timey murcan word for communists

0celo7

I'm German

Bernard Meurer

11:02 PM

@DanielSank Still 20:00 here, relax :p

0celo7

@BernardMeurer HEY

Bernard Meurer

@0celo7 Hey

0celo7

Did you see the h3h3 rain florence response vid

Bernard Meurer

Yeah

Why?

0celo7

that burn victim thing

I see it in my dreams

Bernard Meurer

11:04 PM

Sounds creepy

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