The surface area is $2\theta r^2$, which is nice because $2\theta$ is the "angle excess"

If $M$ is a 2-submanifold of $\Bbb R^3$ with corners, $$\int_{\text{int} M} K dA + \int_{\partial M} k_g d\ell + \sum \theta_i = 2\pi \chi(M)$$

(A digon on the plane has total angle 0, so a lune with angle $\theta$ has $2\theta$ more angle than it should)

(and in general a spherical polygon has area $r^2$ times its angle excess)

Where $\theta_i$ are angles at the corners

@BalarkaSen Interior angles or exterior angles

Interior I am sure

Like, if its almost straight at a corner, is it near $0$ or near $2\pi$

Near 0

Right, that'd be exterior

Is it? Okay

So yeah, that should be the same as considering the curvature to be a Dirac delta thingy

I just mean the angle drawn from inside the manifold

So for an equilateral triangle in the plane its $120^\circ+120^\circ+120^\circ$ and not $60^\circ+60^\circ+60^\circ$

I think the version of G-B you use for a flat plant polygon has an application in physics

@Akiva It should be the latter, no?

Nah, the total curvature needs to be $2\pi$ for a simple curve in the plane

Oops right

That's peculiar

A better example would be, a square would be $90^\circ+$oh wait no that's a bad example

But I'll take it

A hexagon would be $6(60^\circ)$ and not $6(120^\circ)$, I guess. Which is the same value as the triangle, which is nice

Can we go back to the world of closed manifolds please?

@Akiva Yeah

Closed PL 1-manifolds in the plane, right?

(Nah jk)

:P

I still don't quite know how Gaussian curvature is defined. I feel like you take a tiny triangle and divide its its angle excess by its area—in accordance with how spherical triangles work—but that's probably not the usual way

(And it kinda makes Gauss–Bonnay seem a bit too easy)

how on the earth did Atiyah and Macdonald generate their multitude of notoriously difficult problems...

RNG

Er, RWG

@AkivaWeinberger So say $\Sigma \subset \Bbb R^3$ is a 2-manifold

'Kay

Then it has a Gauss map $f : \Sigma \to S^2$ given by sending each point $p$ to the unit normal at $p$ to $S$.

@LeakyNun but they're fun! I remember having a great time working through them

@MatheinBoulomenos yes, they are fun

Mrmhrm.

Think about $df$

What is this, a 3x3 matrix?

2x2?

2x2, yeah. It's a map $T_p \Sigma \to T_{f(p)} S^2$

Both are 2D vector spaces

OK

hi @TedShifrin

@LeakyNun interesting fact: if the ring is Noetherian, then the degree of nilpotency is bounded

hi @Leaky

Hi @Ted

und @Mathein

ich bin noch in Chapter 1

des AM

Chapter = Kapitel

@Leaky: You got my note re $1+x$ earlier?

ya, that's my "constructive" proof

Sounds like capítulo

as in Das Kapitel, nicht wahr?

:P

not quite

Oh.

(1+x)(1-x+x^2-...) = 1+(-x)^n = 1

Kapital $\neq$ Kapitel

Oh, OK, @Leaky. I missed that. Well, that's the right proof :P

well, that's the constructive proof :P

are you a constructivist?

Turns out they both come from Latin (capitulum)

@Mathein, Du haßt mein :P bemerkt? :)

@AkivaWeinberger I am not surprised

@Ted Ah, sorry. I'm German, I don't have humour

Oh, right.

As @Balarka knows, it's tough for those of us with developing Alzheimers to remember.

@TedShifrin Das hab ich auch nicht gesehen, hab's aber trotzdem lustig gefunden hehe

It's German humor, it's no laughing matter

Almost as infamous as Yiddish humor, DogAteMy.

@Akiva Now define the "shape operator" $S_p : T_p M \to T_p M$ to be $S_p(v) = df_p(v)$, the directional derivative of the Gauss map at $p$ in the direction of $v$. So you're moving the unit normal at $p \in M$ slightly in the direction of $v$ and thinking about how much that changes. It's a formal fact that $S_p$ does indeed land in $T_p M$ and not elsewhere in $\Bbb R^3$.

Tthe hell are you talking about, Yiddish humor is hilarious

In this case, the constructive proof is simple enough, but for other stuff, I sometimes prefer the nonconstructive proofs because explicitely writing everything down gets messy. For example, prove that $a_0+a_1T+\dots a_nT^n \in A[T]^\times$ iff $a_0 \in A^\times$ and $a_i$ nilpotent for $i>0$. The nonconstructive proof is simple and elegant, but the constructive proof is non-intuitive and messy imo

@BalarkaSen Er, $M$?

$M$ is $\Sigma$?

Fuck me.

$M = \Sigma$

One direction follows from the $1+x$ thing

Is DogAteMy learning differential geometry now?

@MatheinBoulomenos :o I only know the constructive proof

In the same way $\omega=3$, seems like @BalarkaSen

I'm a constructivist :P

@TedShifrin I'm telling him about the Gaussian curvature

@Akiva lmao

Formal fact? Nonsense.

Unless you're punning.

@LeakyNun you can't really use Atiyah-Macdonald, sorry. Zorn's lemma everywhere

but go on with your nonconstructive proof

$n\cdot n = 1 \implies n(p)\cdot dn_p = 0$.

It's not visually obvious to me. You have to draw the curve and compute ($\gamma \cdot \gamma' = 0$ if $\gamma$ is on the unit sphere)

Yes, I'd call that formal. There's no visual logic to that, it's a calculation

@MatheinBoulomenos it's more of the style of proving that bothers me

No, it's the physics that to stay on the unit sphere, your velocity must be orthogonal to your position.

Totally visual/physical.

they proved that $x \in J(A) \iff (\forall y, 1-xy \in A^*)$ using double contrapositive

Wait, but it's not on the unit sphere, it's on $T_pM$?

To compute a directional derivative, you follow a curve on the surface. Following the unit normal along that curve gives, in turn, a curve on the sphere.

@TedShifrin Maybe. I don't see it :)

Like a lot of differential geometry in general...

I know how $\gamma\cdot\gamma'=0$ implies it's on a sphere, I'm trying to see how that connects

If you have a particle moving on a sphere, if its velocity had any normal component, the particle would move off the sphere (its distance from the center would increase/decrease).

Right

That tells you the displayed equation I wrote 12 lines up, DogAteMy.

Hi chat

@AkivaWeinberger The Gauss map lands on the unit sphere

hi demonic @Alessandro

Yippee ... My diff geo notes are now available on the AMS Open Notes webpage.

@BalarkaSen And then $S_p$ is the directional derivative of the gauss map. OK. So it's orthogonal to the unit sphere

Shouldn't that make it land on $T_{f(p)}S^2$, then?

That's right

First we prove that $A[x]^\times=A^\times$ if $A$ is an integral domain. But that's obvious, because $\operatorname{deg}(fg)=\operatorname{deg}(f)+\operatorname{deg}(g)$ as there can't be any cancelation, so all units have degree 0.

@Akiva $T_{f(p)} S^2$ is tangent to the unit sphere!

I know! But you said it lands on $T_pM$!

But $T_{f(p)} S^2 = T_p M$.

Think what $f$ is.

Now if $a_0+a_1T+\dots+a_nT^n$ is a unit in $A[T]$, then for every prime ideal $\mathfrak{p} \in \operatorname{Spec}A$, we have that $\overline{a_0+a_1T+\dots+a_nT^n}$ is a unit in $A/\mathfrak{p}[T]$, but that's an integral domain, so $\overline{a_i}= 0$ in $A/\mathfrak{p}[T]$, thus $a_i \in \mathfrak{p}$ for $i>0$. As $\mathfrak{p}$ was arbitrary, $a_i$ is in every prime ideal of $A$ for $i>0$, thus $a_i$ is nilpotent for $i>0$

...oh

Shouldn't we just tell DogAteMy Gauss's definition of Gaussian curvature?

OK, I get it, continue

@Ted I thought I would disclose that after the formality

OK.

You can mumble Radon-Nikodym derivative to impress him.

Isn't that a type of synthetic fabric

@MatheinBoulomenos genius

No, it's an element. What am I thinking of?

Only if we have a rayon of hope, DogAteMy.

Nylon.

Or rayon.

@AkivaWeinberger So the shape map $S_p : T_p M \to T_p M$ is given by taking the rate of change of the normal at $S$ if you move it a little bit in the direction of $v \in T_p M$, just to summarize.

sniped

two months ago

'Kay, sure.

Now the Gaussian curvature is defined to be the determinant of $S_p$.

@Balarka: I'm curious to see your "conceptual" proof of self-adjointness of the map. For me, that's easiest with differential forms.

Here's an explanation

Mhm.

I'm gonna want to see an example

It'd be easier once I explain it

You can see the saddle, eg

@BalarkaSen OK

Think about the eigenspaces of $S_p$

On a saddle?

Well, generally, for now

Say you change of basis $T_p M$ to have the $x$ and $y$-axes as the eigenspaces of $S_p$ (so you appropriately parameterize $M$ near $p$ on the first place)

For a saddle I guess it's the bit that goes straight up and the bit that goes straight down

That's right

If you move the point down the down-bit, the normal goes the same way

Of course you're not discussing what I said.

If you move the point up the up-bit, it goes the wrong way

The orthogonality of the eigenspaces is super important.

@Akiva Yeah. And det = product of eigenvalues

And I imagine somewhere in the middle you gotta have it stay the same

Which is something... negative here because +ve * -ve

@TedShifrin Ah ok that

@BalarkaSen That's true regardless of orthogonality of eigenspaces, isn't it?

That $\det=\prod\lambda$?

Yep.

Er, what letter is eigenvalues

For a non-orthogonal saddle ($z=(x+y)^2-y^2$ rather than $x^2-y^2$) this is more surprising, DogAteMy.

Oh, yeah, that's always true.

That's cuz eigenvalues are roots of the characteristic polynomial.

Graphy thingy

BTW, there are pictures of this non-orthogonal saddle in my diff geo notes.

@TedShifrin By self-adjointness you mean $u \cdot S_p(v) = v \cdot S_p(u)$, yeah?

I.e., the IInd fundamental form is symmetric

I think the only definition of Gaussian curvature I saw is by contracting the Riemann curvature tensor

Hi @Daminark

Garbage definition, the Riemann curvature tensor is usually understood by the Gaussian curvature

It's a nonsense tensor. In Gromov's words, nobody understands it

Hello!

So for a sphere

The reason big spheres have small curvature is because the normal doesn't move a lot on a big sphere

@BalarkaSen Yeah.

So why is orthogonality important? And how do you prove it?

Spectral theorem, DogAteMy.

And can this be related to the area-over-angle-excess thingy?

@TedShifrin Er, remind me

It follows from that self-adjointness

Symmetric matrices are flipm flop

@BalarkaSen Oh, so you don't have me on ignore anymore. I want to apologize for what I said, that was wrong and I shouldn't have said it. Sorry

So, just as Jacobian determinants give the fudge factor for area/volume for mappings of Euclidean space, the determinant of the Gauss map gives you the factor by which the Gauss map distorts area. That was Gauss's original interpretation. This is the Radon-Nikodym derivative to which I earlier referred facetiously.

@Balarka: As I warned in my notes, the matrix of the shape operator need not be a symmetric matrix.

@MatheinBoulomenos No need to be. I probably got unnecessarily annoyed, but actually was a little sad because you and Narcissus misinterpret my jokes as algebra-hate.

The Rayon–Nylon derivative.

DogAteMy: Do you remember Ted's favorite formula? :)

$Ax\cdot y=x\cdot A^\top y$?

Right.

I'd joke about Germans having no humour, but yeah I maybe took your jokes too seriously

So if $A=A^\top$, use that to prove the eigenspaces for different eigenvalues are orthogonal.

@TedShifrin I have yet to learn the Radon-Nikodym derivative

That's in graduate measure theory/integration theory, Balarka.

It's the change of variables theorem for measures.

But yeah, I was about to tell Akiva that's it's basically, if you draw a small coordinate xy-square on the surface with one vertex at p, the amount of stretching the Gauss map does to the square is K

And note the sign reflects orientation (pun intended).

x and y being the eigendirections

True

@TedShifrin Why is $A=A^\top$?

@MatheinBoulomenos I should clarify that I meant no harm. I like algebra.

Ah, that's the self-adjointness of the Gauss map to which I referred.

Can I use differential forms or do I need to do a classical proof?

"It's a calculation" as differential geometers say

#stereotypingdifferentialgeometers

I imagine if it's true for quadratic surfaces it's true for all surfaces

'cause you probably only need the series expansion out to degree 2 in general

Not even obvious there.

(i.e. everything is locally a quadratic surface)

True @TedShifrin

It's an exercise in Ted's book

But, yeah, for a graph, you can read off the curvature from the 2nd order Taylor polynomial, of course.

The Diff Geo one?

Yes

What's an exercise? Hell no. I proved it carefully.

By "it" I meant that the IInd fundamental form determines the surface upto second order

@TedShifrin And everything's a graph with respect to one of the axes

'cause of the implicit value theorem

DogAteMy: The idea of the proof is to work in coordinates (if I don't get to use differential forms) and use the product rule.

What's the forms proof?

If the surface is parametrized by $x(u,v)$, $S_P(x_u) = -Dn_P(x_u) = -n_u(P)$.

$n$ is the Gauss map?

Dot with $x_v$ and you get $-n_u\cdot x_v = n\cdot x_{vu}$. Now you can figure out the rest.

Yup.

What's $x_u$

$\frac{\partial x}{\partial u}$?

Unit tangent in the u-direction

Yes.

$\partial/\partial u$

Differential forms proof. Let $e_1,e_2,e_3$ be an orthonormal frame, with $e_3=n$. Let $\omega_1,\omega_2$ be the dual basis of $1$-forms.

Note that $dn\cdot e_i = - de_i\cdot n$ (product rule again). Write $de_1 = \omega_{12}e_2+\omega_{13}e_3$, and similarly $de_2 = \omega_{21}e_1+\omega_{23}e_3$. (Why is $de_i\cdot e_i = 0$?)

Sorry I had to go momentarily

@TedShifrin I don't understand that second equality

Directional derivative in direction of $x_u$ is (by abuse of notation) $\partial n/\partial u$ (thinking of $n = n\circ x$).

OK

I probably shouldn't keep typing the differential forms stuff until you catch up :P

@TedShifrin $x_{vu}$? Second derivative?

Yup. Diff geo is so full of derivatives that subscripts are way less cumbersome than usual $\partial$ notation.

@TedShifrin Two of those four forms is $\Bbb{II}(e_i, -)$, for $i = 1, 2$ IIRC.

There's too much stuff going on in moving frames

Just $\omega_{13}$ and $\omega_{23}$ I think

I was getting to that, but stalled. Yes, so $-dn = \omega_{13}e_1 + \omega_{23}e_2$ (tensor notation deleted).

@TedShifrin So that comes from $\partial u(n\cdot x_v)=0$? 'Cause $x_v$ is in the tangent plane and thus normal to $n$

The normal components, 'cuz II lives in the normal bundle

@Balarka: That's cuz $-dn\cdot e_i = -de_3\cdot e_i = de_i\cdot e_3$.

Right, DogAteMy.

Mmm yes