Mathematics - 2018-01-18 (page 1 of 8)

Ted Shifrin

12:00 AM

@Balarka: We use one-dimensional normal bundle and suppress that.

We should let DogAteMy catch up.

Balarka Sen

True, true

Akiva Weinberger

Right I'm way behind

So the assumption is that $x_u$ and $x_v$ are eigenvectors?

Ted Shifrin

NOOOO.

We're just doing the computation for any parametrization of the surface.

Akiva Weinberger

OK

Balarka Sen

We're doing the computation that'd show $x_u \cdot S_p(x_v) = x_v \cdot S_p(x_u)$

Ted Shifrin

12:02 AM

Away from certain kinds of points, you can always do it to make it what you said, but that requires a real differential equations theorem (proved in appendix of my notes if you care).

Akiva Weinberger

@BalarkaSen Oh, OK, that's 'cause $x_{vu}=x_{uv}$.

Ted Shifrin

That's the only interesting part, as clearly $x_u\cdot S_P(x_u) = x_u\cdot S_P(x_u)$, etc.

Cookie Toast

Hey @Ted, could you give me a hint on this problem? Given the triangle $ABC$, with a point $D$ $\frac{2}{3}$ down $\overrightarrow{AB}$, and point $E$ $\frac{2}{5}$ down $\overrightarrow{CB}$, where the point $Q$ bisects $\overline{CD}$. Show that $||\overrightarrow{AQ}|| = c||\overrightarrow{AE}||$. Find the value $||\overrightarrow{AQ}|| / ||\overrightarrow{AE}||$. What is the value for which $\overline{CD}$ divides $\overline{AE}$?

Balarka Sen

@AkivaWeinberger Exactly.

Cookie Toast

I've gotten everything but the last part, finding $||\overline{CD}|| / ||\overline{AE}||$

Ted Shifrin

12:04 AM

With all problems like that, @Cookie, as I said in the lectures, write $\overrightarrow{AB} =\vec x$ and $\overrightarrow{AC} = \vec y$ and write everything under the sun in terms of $\vec x$ and $\vec y$.

No, you're misreading my question, @Cookie.

Akiva Weinberger

OK hold on but

Balarka Sen

So once you do that, like Ted said, you can say $v \cdot S_p(w) = w \cdot S_p(v)$ for any pair $v,w \in T_p M$, by using the basis $\{x_u, x_v\}$

Cookie Toast

Yeah, I let $\overrightarrow{AB} = \vec{x}$ and $\overrightarrow{AC} = \vec{y}$

Balarka Sen

This would prove $S_p$ is self-adjoint.

Ted Shifrin

Cool. But the question is to compare the two pieces of $\overline{AE}$. Not what you said.

Akiva Weinberger

12:05 AM

@BalarkaSen So if they are eigenvectors, with eigenvalues $\lambda_v$ and $\lambda_w$

Cookie Toast

Oh...Not the magnitudes?

Balarka Sen

Mhm, then it's a dot product fact

Cookie Toast

That explains all the difficulties haha

Balarka Sen

$v \cdot w = w \cdot v$

Akiva Weinberger

then we have $v\cdot(\lambda_w w)=(\lambda_v v)\cdot w$

Ted Shifrin

12:06 AM

Right, Cookie.

Akiva Weinberger

and because the, uh, lambdae are unequal, the dot has to be zero

Ted Shifrin

DogAteMy: That was the general linear algebra fact I asked you to show before.

Balarka Sen

Yep

Ted Shifrin

Of course, they could be equal. But then ... the shape operator is a scalar multiple of the identity.

Akiva Weinberger

Right, and then everyone's an eigenvector.

And our shape's a sphere (up to the second degree approximation).

Balarka Sen

12:08 AM

The big man that comes up from this computation is the 2nd fundamental form $\Bbb{II}(v, w) = v \cdot S_p(w)$. This theorem says it's a billinear form on $T_p M$

Cookie Toast

I spent literally a week on that problem @Ted, and all it's asking is how would one rewrite $\frac{5}{6}$ as a ratio :P I am an idiot

Ted Shifrin

Those points are called umbilics, DogAteMy (you can explain why). Those are the bad points to which I referred earlier where you cannot construct such a parametrization.

Cookie Toast

I was wondering why the first parts went so smoothly if I was having so much trouble with this one lol

Anyways, thank you!

Ted Shifrin

@Cookie: As I told my students frequently, reading is a prerequisite for mathematics :P

Leaky Nun

what is the motivation behind excluding the whole ring as a prime ideal?

Akiva Weinberger

12:08 AM

@TedShifrin Can I?

Latin, DogAteMy.

Shadows?

No, that's umbra.

...Umbilical cord?

I was just typing that hint.

So why are these called umbilic points?

Balarka Sen

12:10 AM

"umbillic points are like your belly buttons"

Akiva Weinberger

No idea

Ted Shifrin

How many do you have, Balarka?

Balarka Sen

not a lot

i am not a differential geometry

9

Akiva Weinberger

To be clear, these are the eigenvectors.

They give birth to the curvature?… I have no idea

Ted Shifrin

Equally curving in all directions, DogAteMy ... like a belly button (probably an outie).

No, eigenvalues.

Akiva Weinberger

12:12 AM

Oh, I misunderstood

So the umbilics are the points on the surface where it looks like a sphere

Not the directions in which you have the maximum curvature

OK, that makes a lot more sense. Yeah, they look like bumps or dips

Ted Shifrin

Right. Back to work now.

Balarka Sen

call me when you see a saddle only

hyperbolic geometry intensifies

Cookie Toast

@Ted can I get one more clarification? Let $\vec{u},\vec{v}$ $\in$ $\mathbb{R}^{2}$. Describe the vectors $\vec{x} = s\vec{u} + t\vec{v}$, where $s + t = 1$.

Ted Shifrin

Ah, yes?

Cookie Toast

By describe, do you mean give all possible vectors?

Balarka Sen

12:16 AM

I need to peace out

Ted Shifrin

Well, yes, but describe them geometrically in words.

Balarka Sen

I stayed up way too long

Ted Shifrin

Night, @Balarka.

I'll finish the forms stuff with DogAteMy if he hasn't lost interest.

You've been through it before anyhow.

Balarka Sen

@AkivaWeinberger Come to the dark side and learn symplectic geometry with me

@Ted Yeah... I might end up learning extrinsic geometry again

Cookie Toast

Ok, thank you @Ted :)

Balarka Sen

12:16 AM

Let's see what happens

Ted Shifrin

Night, Balarka.

Balarka Sen

I wanted to know about Willmore theory

Ted Shifrin

That's Eric's game.

Balarka Sen

omg ted you really want me to be gone don't you

yeah

Leaky Nun

how many prime ideals does Z_p have? I’m thinking one, ie pZ_p

Balarka Sen

12:18 AM

it's a field....

Ted Shifrin

You need sleep, boy.

Leaky Nun

@BalarkaSen p-adics

Balarka Sen

then make it clear you strawhead

Leaky Nun

sorry

Balarka Sen

lol

gian

12:19 AM

Does anyone have an intuitive way to think about adjunctions?

Leaky Nun

context?

MatheinBoulomenos

@LeakyNun don't forget the zero ideal, it's prime in every integral domain (in fact it's prime iff the ring is an integral domain)

Leaky Nun

ah

MatheinBoulomenos

@gian this is a vague analogy, but ithere are some formal analogies to adjoint operators in linear algebra if you think about $\operatorname{Hom}(-,-)$ in analogy to the canonical pairing $V^* \times V \to K$ or as a "scalar product" (I think that's where the name comes from)

Balarka Sen

im going to slither off to my opium den now

y'all smoke weed without me

MatheinBoulomenos

12:24 AM

Bye @Balarka

Ted Shifrin

keeps mum

Balarka Sen

neon meat dream of the octopus, remember that

Eric Silva

@Semiclassical these do look like they're versions of each other

Semiclassical

yeah

I'm not convinced they're entirely identical yet

MatheinBoulomenos

@gian I'd say that I have an intuition about adjunctions, but it's difficult to put it in words. My intuition comes from seeing/recognizing lots of examples and perceiving some sense of analogy/similarity between them.

Semiclassical

12:27 AM

but it seems veeeery suggestive

gian

Okay, I'll look at examples then. Thanks

Semiclassical

Main reason I say that, I should note, is because $\vec{E}=-\nabla V$ where $V(x,y,z)$ is the scalar potential

Ted Shifrin

DogAteMy, are we still working?

Semiclassical

So the formula $\frac{dE_n}{dx}=-\left(\frac{1}{R_1}+\frac{1}{R_2}\right)E_n$ definitely seems like it should boil down to a statement of geometry alone

Eric Silva

The reciprocal of the principal radii gives you the principal curvatures whose sum is mean curvature up to a scalar

Semiclassical

12:30 AM

@EricSilva right

Eric Silva

that -2H

Semiclassical

so I could just as well write the RHS as $-2HE$

Eric Silva

mhm

Semiclassical

of course, the variation formula I linked is for $dA$

Eric Silva

by $E_{n}$ do you mean $\langle E, n \rangle$

Semiclassical

12:31 AM

right

that's implicit in what Purcell has

Eric Silva

got it

interesting

Semiclassical

yeah

Akiva Weinberger

44 mins ago, by Ted Shifrin

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.

42 mins ago, by Ted Shifrin

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 $de_i\cdot e_i = 0$?)

@TedShifrin That's where we were?

Eric Silva

i am wondering what this really means other than being a happy coincidence

Akiva Weinberger

Why is $de_1=\omega_{12}+\omega_{13}$?

Ted Shifrin

12:35 AM

Did you finish understanding it for the classical proof, DogAteMy? I assume so.

Akiva Weinberger

Yes

Daminark

@EricSilva maybe it's an upset confidence

Ted Shifrin

OK

So $de_1\cdot e_1 = 0$ (same argument we've used a dozen times already today.)

Semiclassical

@EricSilva I want to say it's something like: If I move away from the surface of the conductor, $\mathbf{E}\cdot d\mathbf{A}=E_n dA$ should remain the same since the charge is confined to the conductor's surface

Ted Shifrin

Therefore $de_1 = \omega_{12}e_2 + \omega_{13}e_3$ for some diff forms $\omega_{1j}$.

Akiva Weinberger

12:36 AM

Oh, OK. 'Cause it has no $e_1$ part.

Ted Shifrin

Right.

Semiclassical

So if we know how $dA$ changes as we move away from the surface, then we can deduce how $E_n$ changes.

Akiva Weinberger

Wait, $\cdot$ instead of $\wedge$?

Eric Silva

@Semiclassical aha

Ted Shifrin

No, neither.

Officially, you could put tensor symbols in there.

$de_1(v) = \omega_{12}(v)e_2 + \omega_{13}(v)e_3$.

Semiclassical

12:38 AM

The tricky thing here is that Gauss's law, expressed using differential forms, is $d(\mathbf{E}\cdot d\mathbf{A})=\frac{1}{\epsilon_0}\rho dV$

Akiva Weinberger

How am I interpreting $de_1\cdot e_1$?

Semiclassical

(I think that's the right statement)

Akiva Weinberger

As a function evaluated at an input?

$de_1(e_1)$?

Ted Shifrin

No, THAT is a dot.

You're taking the $e_1$ component of the vector-valued differential form $de_1$.

Eric Silva

you can use this to give electrostatics proofs of some diff geo things i guess @Semi

which is cute

Semiclassical

12:39 AM

right

Ted Shifrin

That is $[de_1\cdot e_1](v) = de_1(v)\cdot e_1$.

Semiclassical

or vice versa

Eric Silva

yup

Akiva Weinberger

Okay, okay. And because $\|e_1\|=1$, that's $0$.

Semiclassical

@TedShifrin this is referencing a problem that I know we talked about at one point

Akiva Weinberger

12:40 AM

Similarly for $e_2$ and $e_3$. I'm back on the same page again.

Ted Shifrin

I'm working on taxes and talking to DogAteMy, @Semiclassic, so I'm not paying detention.

Semiclassical

fair enough

Ted Shifrin

OK, DogAteMy. We're almost there.

Eric Silva

tax forms

Ted Shifrin

smacks Eric :P

Semiclassical

12:41 AM

basically how the principal radii of curvature at some point on the surface of a conductor influence the resulting electric field at that point

Ted Shifrin

DogAteMy: So we're trying to see that the shape operator (i.e., $-dn$) is self-adjoint.

Akiva Weinberger

Meaning that it's a symmetric matrix?

Eric Silva

maybe you can use this to give a physical proof of the fact that there are no closed minimal surfaces

Akiva Weinberger

Or that it fits into your favorite formula the right way

Ted Shifrin

From what you reprinted up there, we have $-dn = \omega_{13}e_1 + \omega_{23}e_2$ [note this gives a linear map from tangent space to tangent space]

Akiva Weinberger

12:43 AM

Like $u\cdot d_n(v)=d_n(u)\cdot v$

Semiclassical

tbf this sounds a lot like how people did stuff when they thought Dirichlet's principle was obvious

Ted Shifrin

Right, DogAteMy. Since we're working with respect to an orthonormal basis $e_1,e_2$, it will literally be a symmetric matrix in a moment.

Akiva Weinberger

@TedShifrin I see.

Ted Shifrin

So here's the punchline.

Eric Silva

if there were a closed minimal surface you do this stuff on it and whatnot and you would have $\partial E/ \partial n = 0$ so the normal components of these dudes are constant but closed surfaces dont work like that cause height functions so baddabing baddaboom impossible

Akiva Weinberger

12:44 AM

"Note that $dn\cdot e_i=−de_i\cdot n$" — for $i\ne3$, right?

Oh wait

There as well 'cause they're both 0 there

Eric Silva

that's basically just the normal proof

Semiclassical

cool beans

but this was originally coming from me being curious how much physical content there was in heat kernel stuff

so i was probably being silly :P

Ted Shifrin

Right, DogAteMy. Writing the forms in terms of basis $1$-forms, we have $\omega_{13} = h_{11}\omega_1 + h_{12}\omega_2$ and $\omega_{23} = h_{21}\omega_1 + h_{22}\omega_2$ for some functions $h_{ij}$.

There's your $2\times 2$ matrix.

The Great Duck

2

Q: Verifying an integration method for step function integrals.

I should note that this is an integration algorithm and therefore intermediate steps DO appear to be unjustified. The purpose of this question is to justify or reject this algorithm as always giving correct solutions. Suppose we have a function $f(x)$ that has no vertical asymptotes and can be w...

my bounty expires soon

Eric Silva

this is all extrinsic geometry is the thing

The Great Duck

12:46 AM

please take advantage of it!

Ted Shifrin

So why is that $2\times 2$ matrix symmetric, you ask, DogAteMy?

Akiva Weinberger

@TedShifrin Wait, what are $\omega_1$ and $\omega_2$ again?

Eric Silva

you can do the heat kernel stuff without any ambient space

Ted Shifrin

The basis $1$-forms dual to $e_1,e_2$.

Akiva Weinberger

Dual?

Ted Shifrin

12:47 AM

Oh, dual basis for $V^*$ to a basis for $V$ means ... $\phi_i(e_j) = \delta_{ij}$.

Akiva Weinberger

Like mapping $a\hat\imath+b\hat\jmath+c\hat k$ to $adx+bdy+cdz$?

Ted Shifrin

Right.

The Great Duck

@TedShifrin any ideas on how to fix my proof?

Akiva Weinberger

OK.

Ted Shifrin

$\omega_i(e_j) = \delta_{ij}$.

I'm busy, Duck.

The Great Duck

12:48 AM

ah sorry.

Akiva Weinberger

And then we need $h_{12}=h_{21}$.

The Great Duck

500 rep bounty

fyi

Semiclassical

i more mean something along the lines of: If I transfer a finite amount of heat to some point on the surface of a sphere and allow it to diffuse according to the heat equation, I expect it to eventually look uniform

Ted Shifrin

OK, DogAteMy, so compute $\omega_{13}\wedge\omega_1 + \omega_{23}\wedge\omega_2$.

Eric Silva

mhmm

The Great Duck

12:49 AM

@Semiclassical the heat equation is even spreading correct?

Semiclassical

In that case I suspect the analogy to electrostatics is fairly precise, since steady state heat equation = Laplace's equation

The Great Duck

like a smoothing operation?

Akiva Weinberger

@TedShifrin So if that's $0$ then we're done?

Ted Shifrin

Right.

Eric Silva

hmhmhmhmmm

Semiclassical

12:49 AM

yeah, heat which is localized to one spot initially spreads out over time

Eric Silva

my mental picture for heat equation is $S^{1}$ cause it's so nice there

Ted Shifrin

DogAteMy, so we really had a complete basis $e_1,e_2,e_3$, with $e_3 = n$. What is the differential form $\omega_3$ on the surface?

The Great Duck

no i meant like it is equivalent to the operation of replacing every element in a grid with the average of the element and the elements around it

Ted Shifrin

(Note I'm not asking $de_3$ !!!).

Akiva Weinberger

$n$?

The Great Duck

12:50 AM

if so

it is never uniform

i got to go

Semiclassical

on the real line, it can spread out indefinitely

Akiva Weinberger

When did you go back to being a duck?

Ted Shifrin

Here's something to think about, DogAteMy. $\omega_i(v) = v\cdot e_i$ for $v$ in the tangent space. Right?

Semiclassical

but on the sphere it can't, so it'll eventually reach some steady state temperature

Eric Silva

the heat kernel on noncompact guys is wacky

Akiva Weinberger

12:51 AM

@TedShifrin Right

The Great Duck

@Semiclassical my version can do that. It is discreet as that is the operation of smoothing an image.

Ted Shifrin

So what is $\omega_3(v)$?

The Great Duck

discrete bounded grid doesnt either

Akiva Weinberger

For $v$ in $\Bbb R^3$, no? Why just the tangent space? @TedShifrin

The Great Duck

so unless by symmetry it does... no

Ted Shifrin

12:52 AM

Because we're doing all this on a surface in $\Bbb R^3$.

Akiva Weinberger

@TedShifrin Oh, for $v$ in the tangent space, it'd be $0$.

The Great Duck

@AkivaWeinberger when I finally completed a valid construction of the implied derivative.

Semiclassical

the problem with the sphere or real line example is that there's no variation of curvature

Ted Shifrin

Right. So $\omega_3 = 0$ as a differential form on the surface. So let's take the derivative of that $1$-form. What will it be?

The Great Duck

Dec. 25th of last year

Semiclassical

12:53 AM

and I suspect that's what will make the heat kernel business interesting

Akiva Weinberger

Well it won't be $de_3$?

The Great Duck

@Semiclassical google image smoothing

it is equivalent

look at image smoothing on the surface of a 3D model

it is equivalent

Semiclassical

@ericsilva I could be talking nonsense, though, since the fact that you were talking about cohomology makes me think I'm taking the analogy too literally

Ted Shifrin

Nooo, DogAteMy. We're differentiating the $1$-form that is identically $0$. We get a $2$-form (we'll compute what it is in a moment).

Akiva Weinberger

We'll get $0$, clearly

Eric Silva

12:55 AM

you're right about curvature showing up

Semiclassical

@TheGreatDuck yes, the heat equation is used to do smoothing

Eric Silva

like formally it's hard to work with $\Delta$ on a manifold so you want to look at the spectral decomposition (if you live in a compact world that is, closed surfaces and whatnot) and as it turns out that spectrum is loaded with curvature

Ted Shifrin

OK. So, here's what's called the structure equations: $d\omega_i = \sum_j\omega_{ij}\wedge \omega_j$

Semiclassical

not sure what you're objecting to, though. i'm making a claim about the asymptotics

Icche Guri

Hello the great mathematicians ,

Semiclassical

12:56 AM

and while it'll never become uniform in finite time, it can certainly be asymptotic to a uniform distribution

Icche Guri

How are you ?

Eric Silva

the interesting thing is that the asymptotics of the heat kernel have nothing to do with curvature

Icche Guri

I have one question

Ted Shifrin

(Not surprisingly, just like your other proof used equality of mixed partials, this one will use $d\circ d = 0$.)

Icche Guri

Can you please help me ?

Semiclassical

12:56 AM

@EricSilva huh

Eric Silva

only has to do with topology

Akiva Weinberger

@IccheGuri Good

Icche Guri

0

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Four dogs stand in four corners of a square . The side of the square is $1$ km . Now closing eyes, each dog runs at the same velocity to the dog residing to the right . By this, they cover half distance . After opening eyes , each dog runs at the same velocity to dog in the right and covers the s...