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12:00 AM
@Balarka: We use one-dimensional normal bundle and suppress that.
We should let DogAteMy catch up.
 
True, true
 
Right I'm way behind
So the assumption is that $x_u$ and $x_v$ are eigenvectors?
 
NOOOO.
We're just doing the computation for any parametrization of the surface.
 
We're doing the computation that'd show $x_u \cdot S_p(x_v) = x_v \cdot S_p(x_u)$
 
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).
 
@BalarkaSen Oh, OK, that's 'cause $x_{vu}=x_{uv}$.
 
That's the only interesting part, as clearly $x_u\cdot S_P(x_u) = x_u\cdot S_P(x_u)$, etc.
 
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 $\overline{AE}$?
 
@AkivaWeinberger Exactly.
 
I've gotten everything but the last part, finding $||\overline{CD}|| / ||\overline{AE}||$
 
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.
 
OK hold on but
 
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 $\{x_u, x_v\}$
 
Yeah, I let $\overrightarrow{AB} = \vec{x}$ and $\overrightarrow{AC} = \vec{y}$
 
This would prove $S_p$ is self-adjoint.
 
Cool. But the question is to compare the two pieces of $\overline{AE}$. Not what you said.
 
12:05 AM
@BalarkaSen So if they are eigenvectors, with eigenvalues $\lambda_v$ and $\lambda_w$
 
Oh...Not the magnitudes?
 
Mhm, then it's a dot product fact
 
That explains all the difficulties haha
 
$v \cdot w = w \cdot v$
 
then we have $v\cdot(\lambda_w w)=(\lambda_v v)\cdot w$
 
12:06 AM
Right, Cookie.
 
and because the, uh, lambdae are unequal, the dot has to be zero
 
DogAteMy: That was the general linear algebra fact I asked you to show before.
 
Yep
 
Of course, they could be equal. But then ... the shape operator is a scalar multiple of the identity.
 
Right, and then everyone's an eigenvector.
And our shape's a sphere (up to the second degree approximation).
 
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 $T_p M$
 
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
 
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.
 
I was wondering why the first parts went so smoothly if I was having so much trouble with this one lol
Anyways, thank you!
 
@Cookie: As I told my students frequently, reading is a prerequisite for mathematics :P
 
what is the motivation behind excluding the whole ring as a prime ideal?
 
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?
 
12:10 AM
"umbillic points are like your belly buttons"
 
No idea
 
How many do you have, Balarka?
 
not a lot
i am not a differential geometry
9
 
To be clear, these are the eigenvectors.
They give birth to the curvature?… I have no idea
 
Equally curving in all directions, DogAteMy ... like a belly button (probably an outie).
No, eigenvalues.
 
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
 
Right. Back to work now.
 
call me when you see a saddle only
hyperbolic geometry intensifies
 
@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 $s + t = 1$.
 
Ah, yes?
 
By describe, do you mean give all possible vectors?
 
12:16 AM
I need to peace out
 
Well, yes, but describe them geometrically in words.
 
I stayed up way too long
 
Night, @Balarka.
I'll finish the forms stuff with DogAteMy if he hasn't lost interest.
You've been through it before anyhow.
 
@AkivaWeinberger Come to the dark side and learn symplectic geometry with me
@Ted Yeah... I might end up learning extrinsic geometry again
 
Ok, thank you @Ted :)
 
12:16 AM
Let's see what happens
 
Night, Balarka.
 
I wanted to know about Willmore theory
 
That's Eric's game.
 
omg ted you really want me to be gone don't you
yeah
 
how many prime ideals does Z_p have? I’m thinking one, ie pZ_p
 
12:18 AM
it's a field....
 
You need sleep, boy.
 
@BalarkaSen p-adics
 
then make it clear you strawhead
 
sorry
 
lol
 
12:19 AM
Does anyone have an intuitive way to think about adjunctions?
 
context?
 
@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)
 
ah
 
@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)
 
im going to slither off to my opium den now
y'all smoke weed without me
 
12:24 AM
Bye @Balarka
 
keeps mum
 
neon meat dream of the octopus, remember that
 
@Semiclassical these do look like they're versions of each other
 
yeah
I'm not convinced they're entirely identical yet
 
@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.
 
12:27 AM
but it seems veeeery suggestive
 
Okay, I'll look at examples then. Thanks
 
Main reason I say that, I should note, is because $\vec{E}=-\nabla V$ where $V(x,y,z)$ is the scalar potential
 
DogAteMy, are we still working?
 
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
 
The reciprocal of the principal radii gives you the principal curvatures whose sum is mean curvature up to a scalar
 
12:30 AM
@EricSilva right
 
that -2H
 
so I could just as well write the RHS as $-2HE$
 
mhm
 
of course, the variation formula I linked is for $dA$
 
by $E_{n}$ do you mean $\langle E, n \rangle$
 
12:31 AM
right
that's implicit in what Purcell has
 
got it
interesting
 
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?
 
i am wondering what this really means other than being a happy coincidence
 
Why is $de_1=\omega_{12}+\omega_{13}$?
 
12:35 AM
Did you finish understanding it for the classical proof, DogAteMy? I assume so.
 
@EricSilva maybe it's an upset confidence
 
OK
So $de_1\cdot e_1 = 0$ (same argument we've used a dozen times already today.)
 
@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
 
Therefore $de_1 = \omega_{12}e_2 + \omega_{13}e_3$ for some diff forms $\omega_{1j}$.
 
12:36 AM
Oh, OK. 'Cause it has no $e_1$ part.
 
Right.
 
So if we know how $dA$ changes as we move away from the surface, then we can deduce how $E_n$ changes.
 
Wait, $\cdot$ instead of $\wedge$?
 
@Semiclassical aha
 
No, neither.
Officially, you could put tensor symbols in there.
$de_1(v) = \omega_{12}(v)e_2 + \omega_{13}(v)e_3$.
 
12:38 AM
The tricky thing here is that Gauss's law, expressed using differential forms, is $d(\mathbf{E}\cdot d\mathbf{A})=\frac{1}{\epsilon_0}\rho dV$
 
How am I interpreting $de_1\cdot e_1$?
 
(I think that's the right statement)
 
As a function evaluated at an input?
$de_1(e_1)$?
 
No, THAT is a dot.
You're taking the $e_1$ component of the vector-valued differential form $de_1$.
 
you can use this to give electrostatics proofs of some diff geo things i guess @Semi
which is cute
 
12:39 AM
right
 
That is $[de_1\cdot e_1](v) = de_1(v)\cdot e_1$.
 
or vice versa
 
yup
 
Okay, okay. And because $\|e_1\|=1$, that's $0$.
 
@TedShifrin this is referencing a problem that I know we talked about at one point
 
12:40 AM
Similarly for $e_2$ and $e_3$. I'm back on the same page again.
 
I'm working on taxes and talking to DogAteMy, @Semiclassic, so I'm not paying detention.
 
fair enough
 
OK, DogAteMy. We're almost there.
 
tax forms
 
smacks Eric :P
 
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
 
DogAteMy: So we're trying to see that the shape operator (i.e., $-dn$) is self-adjoint.
 
Meaning that it's a symmetric matrix?
 
maybe you can use this to give a physical proof of the fact that there are no closed minimal surfaces
 
Or that it fits into your favorite formula the right way
 
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]
 
12:43 AM
Like $u\cdot d_n(v)=d_n(u)\cdot v$
 
tbf this sounds a lot like how people did stuff when they thought Dirichlet's principle was obvious
 
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.
 
@TedShifrin I see.
 
So here's the punchline.
 
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
 
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
 
that's basically just the normal proof
 
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
 
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.
 
2
Q: Verifying an integration method for step function integrals.

The Great DuckI 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
 
this is all extrinsic geometry is the thing
 
12:46 AM
please take advantage of it!
 
So why is that $2\times 2$ matrix symmetric, you ask, DogAteMy?
 
@TedShifrin Wait, what are $\omega_1$ and $\omega_2$ again?
 
you can do the heat kernel stuff without any ambient space
 
The basis $1$-forms dual to $e_1,e_2$.
 
12:47 AM
Oh, dual basis for $V^*$ to a basis for $V$ means ... $\phi_i(e_j) = \delta_{ij}$.
 
Like mapping $a\hat\imath+b\hat\jmath+c\hat k$ to $adx+bdy+cdz$?
 
Right.
 
@TedShifrin any ideas on how to fix my proof?
 
$\omega_i(e_j) = \delta_{ij}$.
I'm busy, Duck.
 
12:48 AM
ah sorry.
 
And then we need $h_{12}=h_{21}$.
 
500 rep bounty
fyi
 
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
 
OK, DogAteMy, so compute $\omega_{13}\wedge\omega_1 + \omega_{23}\wedge\omega_2$.
 
mhmm
 
12:49 AM
@Semiclassical the heat equation is even spreading correct?
 
In that case I suspect the analogy to electrostatics is fairly precise, since steady state heat equation = Laplace's equation
 
like a smoothing operation?
 
@TedShifrin So if that's $0$ then we're done?
 
Right.
 
hmhmhmhmmm
 
12:49 AM
yeah, heat which is localized to one spot initially spreads out over time
 
my mental picture for heat equation is $S^{1}$ cause it's so nice there
 
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?
 
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
 
(Note I'm not asking $de_3$ !!!).
 
12:50 AM
if so
it is never uniform
i got to go
 
on the real line, it can spread out indefinitely
 
When did you go back to being a duck?
 
Here's something to think about, DogAteMy. $\omega_i(v) = v\cdot e_i$ for $v$ in the tangent space. Right?
 
but on the sphere it can't, so it'll eventually reach some steady state temperature
 
the heat kernel on noncompact guys is wacky
 
12:51 AM
@TedShifrin Right
 
@Semiclassical my version can do that. It is discreet as that is the operation of smoothing an image.
 
So what is $\omega_3(v)$?
 
discrete bounded grid doesnt either
 
For $v$ in $\Bbb R^3$, no? Why just the tangent space? @TedShifrin
 
so unless by symmetry it does... no
 
12:52 AM
Because we're doing all this on a surface in $\Bbb R^3$.
 
@TedShifrin Oh, for $v$ in the tangent space, it'd be $0$.
 
@AkivaWeinberger when I finally completed a valid construction of the implied derivative.
 
the problem with the sphere or real line example is that there's no variation of curvature
 
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?
 
Dec. 25th of last year
 
12:53 AM
and I suspect that's what will make the heat kernel business interesting
 
Well it won't be $de_3$?
 
@Semiclassical google image smoothing
it is equivalent
look at image smoothing on the surface of a 3D model
it is equivalent
 
@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
 
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).
 
We'll get $0$, clearly
 
12:55 AM
you're right about curvature showing up
 
@TheGreatDuck yes, the heat equation is used to do smoothing
 
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
 
OK. So, here's what's called the structure equations: $d\omega_i = \sum_j\omega_{ij}\wedge \omega_j$
 
not sure what you're objecting to, though. i'm making a claim about the asymptotics
 
Hello the great mathematicians ,
 
12:56 AM
and while it'll never become uniform in finite time, it can certainly be asymptotic to a uniform distribution
 
How are you ?
 
the interesting thing is that the asymptotics of the heat kernel have nothing to do with curvature
 
I have one question
 
(Not surprisingly, just like your other proof used equality of mixed partials, this one will use $d\circ d = 0$.)
 
Can you please help me ?
 
12:56 AM
@EricSilva huh
 
only has to do with topology
 
@IccheGuri Good
 
0
Q: In what position , the dogs will reside?

Standard EquationFour 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...

I have this question .
 
short time behavior is dominated by curvature effects and long time behavior gets rid of them
 
12:57 AM
that's the reason you can prove Gauss-Bonnet with heat equation stuff
 
Can you please explain me , in where the dogs will reside after infinite time ?
 
I suspect what's not clear to me is what curvature exactly means in this context
 
Yippee Gauss-Bonnet ...
 
@Semiclassical so for a surface Gaussian curvature
 
12:58 AM
if dogs passes distance $ \frac{1}{k}$ in every time run
 
not mean curvature?
 
$\frac{1}{R_{1}R_{2}}$ from our previous discussion i guess
 
disappoint
 
Please help me
 
nah because mean curvature depends how you live in space
 
12:59 AM
hmm
 
@IccheGuri: Most of us are in the middle of things here. You can't expect us to drop our discussions just in an instant.
 
ok. I am waiting
 
LOL
 
Thank you for your explaining :p
 

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