Heat kernel on a manifold

MatheinBoulomenos

Jun 30, 2019 00:54

can someone explain to me the Laplace-Beltrami operator and why its spectrum is that important? Apparantly the Selberg trace formula is important for NT, but I never worked with the spectrum of a differential operator before

Ted Shifrin

@Mathein: It starts with hearing the shape of a vibrating drum :)

Ryan Unger

@MatheinBoulomenos you can hear the euler characteristic of a closed surface

anakhro

Sounds like a "kill your father" thing.

Ted Shifrin

cleans out his ears

anakhro

holds a plate to his ear and hears "... 1 1 1 1 1 1 1 1 1 1 ..."

Darn it's not closed.

emits a final cry before being devoured by geometers.

Ryan Unger

Jun 30, 2019 00:58

I think the same is only true when you have a boundary if you know that the boundary is totally geodesic

Ted Shifrin

Hmmm, this sounds like a good thing for you to teach us, @Ryan.

anakhro

@TedShifrin are you fond of Eliashberg?

Ryan Unger

right so the basic result is that on a surface $$\sum_{j=0}^\infty e^{-\lambda_jt}=\frac{V(M^2,g)}{4\pi t}+\frac{\chi(M)}{6}+O(t)$$

Ted Shifrin

I don't know him personally, nor most of his work.

Ryan Unger

to no-one's surprise, the chi appears because of Gauss-Bonnet

MatheinBoulomenos

Jun 30, 2019 01:01

what is $V(M^2,g)$?

Ryan Unger

volume

Ted Shifrin

This sounds like classic heat equation stuff for Gauss-Bonnet.

Ryan Unger

yes the thing on the left is the trace of the heat kernel

Ted Shifrin

@Mathein: $g$ is the metric and your Laplace-Beltrami is defined dependent upon $g$.

Ryan Unger

@MatheinBoulomenos do you know what the heat kernel is

MatheinBoulomenos

Jun 30, 2019 01:03

it's the kernel of the heat operator

no, I have no idea

Ted Shifrin

kernel ≠ kernel

anakhro

equality is't reflexive anymore?

Ryan Unger

ok you probably know functional analysis so if we want to solve some abstract problem like $x'(t)=Ax(t)$ where $x(t)$ is a curve in some Banach space and $A$ is a linear operator then we can write $x(t)=e^{At}x_0$, where $x_0$ is the "initial condition"

MatheinBoulomenos

yeah I know functional analysis

Ted Shifrin

LOL, that's freshman linear algebra :P

Ryan Unger

Jun 30, 2019 01:05

to make this rigorous you need some conditions on $A$ of course

anakhro

@TedShifrin tbh never learned this in linear algebra

Ryan Unger

so we want to study the solutions of $(\partial_t-\Delta)u=0$ (heat equation) on a Riemannian manifold

anakhro

I had to read it in Hirsch & Smale

Ryan Unger

I might call such a $u$ a caloric function

anakhro

should have been in our ODE course, but it is a WASTE.

Ted Shifrin

Jun 30, 2019 01:06

I taught it my linear algebra courses and, of course, it's in my books, @anakhro.

Ryan Unger

so we want to solve this for $u(x,t)$ where $(x,t)\in M\times[0,\infty)$ and $\Delta$ is taken wrt a fixed metric $g$ on $M$

and we have some initial condition $u_0\in C^\infty(M)$, which means we want $u(\cdot,0)=u_0$

anakhro

@TedShifrin one day I will be like you and write a book I can reference.

Ted Shifrin

I hope so, @anakhro.

anakhro

See Anakhro [2].

Ryan Unger

now using the functional calculus we can write $u(t)=e^{t\Delta}u_0$

Ted Shifrin

Jun 30, 2019 01:08

Now, that's more obnoxious than necessary :D

Ryan Unger

this can be made rigorous because $\Delta$ is essentially self-adjoint on $C^\infty(M)$

the "heat kernel" is the operator $e^{t\Delta}$

Ted Shifrin

notes that Ryan is doing a splendid job so far

5

I was expecting a fundamental solution type of kernel, @Ryan.

Ryan Unger

i was going to write an integral kernel in a second

now $e^{t\Delta}:L^2\to L^2$ is actually a trace-class operator for $t>0$

and we have the nice formula $$\mathrm{tr}\,e^{t\Delta}=\sum_{j=0}^\infty e^{-t\lambda_j}$$

now the problem is that actually computing the RHS is kind of impossible from what I just said, so we have to consider another representation of the heat kernel

To fix notation let $\{\phi_j,\lambda_j\}_{j=0}^\infty$ be the spectral resolution for the Laplacian (on a closed Riemannian manifold)

this means that $\Delta \phi_j+\lambda_j\phi_j=0$, $\phi_j\in C^\infty$, $\{\phi_j\}$ is an orthonormal basis for $L^2$, and $0=\lambda_0<\lambda_1\le \lambda_2\le\cdots\to \infty$

MatheinBoulomenos

this means that $\phi_j$ are eigenvectors and $\lambda_j$ are eigenvalues, right?

sniped

Ryan Unger

@TedShifrin If you're familiar with quantum mechanics you'll see that I've taken the following from my QM course

we have the identity $$\sum_{j=0}^\infty\phi_j\otimes \phi_j=id_{L^2}$$

the tensor means the following: if $f,g,h\in L^2$, then $$(f\otimes g)h=(g,h)_{L^2}f$$

Ted Shifrin

Jun 30, 2019 01:17

(using $L^2 \cong (L^2)^*$?)

Ryan Unger

oh yeah

MatheinBoulomenos

yeah, I assumed as much

Ryan Unger

this all works for a Laplacian on a complex vector bundle in which case one needs to add some transposes and whatnot

here it's simple

ok so we take $u(t)=e^{t\Delta}u_0$ and expand $u_0=\sum a_j\phi_j$

now we can just compute that $e^{t\Delta}\phi_j=e^{-\lambda_jt}\phi_j$

so by "linearity" (this part needs to be checked), $u(t)=\sum e^{-\lambda_jt}a_j\phi_j$

but we can rewrite this as $u(t)=(\sum_{j}e^{-\lambda_jt}\phi_j\otimes\phi_j)u_0$

so, at least formally, $e^{t\Delta}=\sum e^{-\lambda_jt}\phi_j\otimes \phi_j$

good so far?

MatheinBoulomenos

yeah, I'm following

thanks a lot for the great exposition

Ryan Unger

Let's go back to the tensor product. With everything written out we have $$(f\otimes g)h(x)=\int_M f(x)g(y)h(y)\,dy$$

so we can represent the tensor product of two functions as an integral operator

so the key to bringing PDE back into the fold is writing: $$u(x,t)=e^{t\Delta}u_0(x)=\int_MH(x,y,t)u_0(y)\,dy$$

where $$H(x,y,t)=\sum_{j=0}^\infty e^{-\lambda_jt}\phi_j(x)\phi_j(y)$$

MatheinBoulomenos

Jun 30, 2019 01:27

So is $H(x,y,t)$ the heat kernel?

Ryan Unger

yeah

it turns out that $H$ is smooth on $M\times M\times (0,\infty)$

and it blows up along the diagonal as $t\searrow 0$

so on R^n we can actually write down the heat kernel explicitly

I know that's not a compact manifold but the same ideas work

The answer is $$e(x,y,t)=(4\pi t)^{-n/2}e^{-|x-y|^2/4t}$$

Ted Shifrin

And that's the classic fundamental solution.

If you compute Laplacian of that, you get $\delta_y$.

Ryan Unger

right so this thing satisfies $$(\partial_t-\Delta_x)e(x,y,t)=\delta_y(x)$$

Ted Shifrin

So ... think convolution and what Ryan's been doing.

Ryan Unger

where $\delta_y$ is the Dirac mass at $y$

MatheinBoulomenos

Jun 30, 2019 01:31

this is a weak derivative, right?

Ryan Unger

er sorry that's slightly wrong

it satisfies the heat equation, i.e. $(\partial_t-\Delta_x)e(x,y,t)$ for $t>0$

and $e(\cdot,y,t)\to \delta_y$ as $t\searrow 0$

MatheinBoulomenos

ah, I see

Ted Shifrin

=0

Ryan Unger

in various senses, let's keeps this informal haha

Ted Shifrin

You're doing great.

Ryan Unger

Jun 30, 2019 01:34

Ted and I both mixed this up with the Laplace equation

Ted Shifrin

well, not with $t$ in there :P

Ryan Unger

@MatheinBoulomenos So the derivatives here are classical ones

Ted Shifrin

Yeah, when I said "compute Laplacian" I was an idiot.

Ryan Unger

$e$ is a smooth function for $t>0$

ok so the story continues with Minakshisundaram and Pleijel

Ted Shifrin

Wow ... what a name to have to type!

Ryan Unger

Jun 30, 2019 01:36

I think I did it correctly

Ted Shifrin

I don't know it/him/her.

Ryan Unger

They asked to what extent does $H(x,y,t)$ look like $e(x,y,t)$ as $x\to y$ and $t\searrow 0$

Recall that on a Riemannian manifold we have a distance function $d$. So we consider the "naive heat kernel" $$\mathcal E(x,y,t)=(4\pi t)^{-n/2}e^{-d(x,y)^2/4t}$$

their idea is that we should be able to add correction terms to this and obtain the real heat kernel

since this thing obviously is like $e$ for short distances and times, and if the correction terms are small, then $H$ looks like $e$ too

it turns out the right thing to look at is $$H_k(x,y,t)=\mathcal E(x,y,t)\sum_{j=0}^kt^ju_j(x,y)$$

and we want to compute $u_j$ so that they satisfy $$(\partial_t-\Delta_x)H_k=-\mathcal Et^k\Delta_x u_k$$

I don't have a great reason for why this equation, but it works out in the end

Ted Shifrin

This looks like classic perturbation theory, I guess.

Ryan Unger

it's definitely motivated by that

there's probably rules for doing this stuff for ODEs and just just made it work in this setting

so you can compute the $u_j$'s from this, at least in principle.

it gives a recursion relation and you start with $u_0= 0$ on the diagonal

this turns out to be enough to determine all the others, but actually computing them is very hard

it involves the metric determinant in polar coordinates in a neighborhood of the diagonal, kind of nasty

now what is particularly interesting are the numbers $$a_j=\int u_j(x,x)\,dx$$

because these give the coefficients in the $t$-expansion of $$\mathrm{tr}\,e^{t\Delta}=\int H(x,x,t)\,dx$$

it turns out that $u_1(x,x)=R(x)/6$, where $R$ is the scalar curvature

so by the Gauss--Bonnet formula, $$a_1=\frac 46\pi\chi$$ when $n=2$

Ted Shifrin

I hate to leave in the middle of this lecture, but I have to cook dinner. I'll catch up later :) Thanks, Ryan.

Ryan Unger

Jun 30, 2019 01:49

cheers

it's basically done anyway

one can show that for $k$ sufficiently large, $H_k$ approximates $H$ very well and we have $$\sum e^{-\lambda_jt}=(4\pi t)^{-n/2}\left(\sum_{j=0}^k t^j\int u_j(x,x)\,dx\right)+O(t^\alpha),$$

where $\alpha$ is some power

I actually wanted $u_0(x,x)=1$ haha

so the first term gives the volume, always

and when $n=2$ you get the Euler characteristic in the second term

MatheinBoulomenos

wow, really cool

thanks a lot

Ryan Unger

np

@BalarkaSen there might be something interesting to you above

the heat kernel is one of the main tools in geometric analysis

the relationship to the Poincare conjecture is particularly mysterious

$Mathematics$ Mathematics

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