The Heat Equation on a Manifold

Book 6A studied the heat equation in $\mathbb{R}^n$. Book 11A studies the Ricci flow, a heat equation for a metric. This chapter is the bridge: the heat equation $\partial_tu = \Delta u$ on a closed Riemannian manifold, where there is no boundary and no infinity, and where curvature enters the estimates. Everything is in place. The spectral theorem gives the solution as an eigenfunction expansion (9A.6 The Laplacian and the Bochner Formula, 4A.7 Compact Operators and Spectra). The maximum principle works as in 6A.4 Maximum Principles, with nothing to check at a boundary. The Bochner formula turns a Ricci lower bound into the Li–Yau gradient estimate, the model for Hamilton's Harnack inequality. The chapter ends with time-dependent metrics, and derives the conjugate heat operator $-\partial_t - \Delta + R$ that sits at the centre of Perelman's work.

By the end of this chapter you will be able to:

• construct the heat kernel of a closed manifold from eigenfunctions, and list its basic properties;
• prove the maximum principle and conservation of total heat on a closed manifold, and use Hamilton's trick;
• state the short-time asymptotics (Minakshisundaram–Pleijel, Varadhan) and read geometry from them;
• state and check the Li–Yau gradient estimate and its Harnack inequality;
• derive the adjoint of the heat operator for a time-dependent metric, and recognise the conjugate heat equation of the Ricci flow.

7.1Distances from heat

7.2The heat kernel of a closed manifold

Let $M$ be closed and connected, and let $\phi_0, \phi_1, \dots$ be an orthonormal basis of $L^2(M)$ of eigenfunctions, $-\Delta\phi_k = \lambda_k\phi_k$, $0 = \lambda_0 < \lambda_1 \leq \lambda_2 \leq \dots$ (9A.6 The Laplacian and the Bochner Formula). The solution of $\partial_tu = \Delta u$ with $u(\cdot, 0) = u_0$ is

with the heat kernel

The series converges smoothly for $t > 0$, by Weyl's law for the growth of $\lambda_k$ and elliptic estimates. The heat kernel has the properties of the Gaussian of 6A.3 The Heat Equation on ℝⁿ:

• it is symmetric in $x, y$, solves the heat equation in each variable, and tends to the delta function at $y$ as $t \to 0$;
• it is positive, and $\int_Mp(x, y, t)\,dV(y) = 1$, since constants are preserved;
• it satisfies the semigroup law $p(x, z, t + s) = \int p(x, y, t)p(y, z, s)\,dV(y)$;
• as $t \to \infty$, $p \to \frac{1}{\operatorname{Vol}M}$, exponentially fast at rate $\lambda_1$: the spectral gap is the rate of equilibration (Exercise 7.3).

On the unit sphere $S^2$ the eigenfunctions are spherical harmonics, and the kernel is a Legendre series, $p = \sum_l\frac{2l + 1}{4\pi}e^{-l(l + 1)t}P_l(\cos\theta)$, with $\theta$ the angle between $x$ and $y$ (Figure 7.1).

7.3The maximum principle without boundary

The same argument handles equations with lower-order terms, systems, and tensors: it is the template of 6A.4 Maximum Principles with the boundary case deleted. Hamilton's trick packages it. If $u$ is smooth on $M\times[0, T]$, the function $\mu(t) = \max_Mu(\cdot, t)$ is Lipschitz, and at each $t$ its upper right derivative is at most $\max\{\partial_tu(x, t) : u(x, t) = \mu(t)\}$ (Exercise 7.5). So pointwise information at a maximum, where $\nabla u = 0$ and $\Delta u \leq 0$, becomes an ODE inequality for $\mu$. That is how the evolution of the minimum of scalar curvature was handled in the last exercise of 9B.6 Scalar Curvature and Topology.

Conservation. $\frac{d}{dt}\int_Mu\,dV = \int_M\Delta u\,dV = 0$: total heat is conserved on a fixed closed manifold. The next sections ask what replaces this when the metric moves.

7.4Short times: what heat knows about geometry

As $t \to 0$, the heat kernel looks like the Euclidean one, corrected by curvature. Subbaramiah Minakshisundaram and Åke Pleijel (1949) proved an asymptotic expansion

with higher coefficients given by curvature invariants. Integrating over $M$, the heat trace satisfies

So the eigenvalues determine the dimension, the volume and the total scalar curvature: part of the answer to Mark Kac's 1966 question "Can one hear the shape of a drum?" For a surface, $\int R = 4\pi\chi$ by Gauss–Bonnet, so the spectrum determines the Euler characteristic (Exercise 7.6).

Off the diagonal, Varadhan's formula (S. R. S. Varadhan, 1967) recovers the distance:

The normalisation matches the Euclidean kernel $(4\pi t)^{-n/2}e^{-|x - y|^2/4t}$ and the analyst's Laplacian of this guide (Figure 7.2).

7.5Li–Yau

Integrating the estimate along a path from $(x, t_1)$ to $(y, t_2)$ gives the Harnack inequality

which compares temperatures at different places and times: heat cannot be very concentrated now if it will be spread out later. Hamilton's Harnack inequality for the Ricci flow (11B.2 Ancient Solutions and the Harnack Inequality) is the analogue for the curvature itself, and Perelman's reduced distance (12A.5 Reduced Distance and Reduced Volume) is the length functional that this integration suggests.

7.6Moving metrics and the conjugate heat equation

Now let the metric depend on time, $\partial_tg = h$. The volume form changes (Exercise 7.8):

For the heat operator $\square = \partial_t - \Delta_{g(t)}$, integrate by parts in space and time for functions $u, v$ on $M\times[a, b]$:

So the formal adjoint of $\square$ is

Under the Ricci flow, $h = -2\operatorname{Ric}$ and $\frac12\operatorname{tr}h = -R$, so

the conjugate heat operator. If $\square u = 0$ and $\square^*v = 0$, then $\int_Muv\,dV$ is constant in time; with $u \equiv 1$, a solution of $\square^*v = 0$ keeps $\int_Mv\,dV$ constant. A solution of the conjugate heat equation runs backwards in time, like heat flowing from the future into the past, and conserves total mass.

7.7History

Fourier's Théorie analytique de la chaleur (1822) founded the subject. Minakshisundaram and Pleijel's expansion appeared in 1949, and Kac's drum question in 1966, followed by McKean and Singer's study of the heat trace in 1967. Varadhan's formula dates from 1967. Peter Li and Shing-Tung Yau published their estimate in 1986, Hamilton his matrix Harnack inequality for the Ricci flow in 1993, and Perelman the conjugate heat equation's central role in 2002.

7.8Exercises

Solution

$\bar u$ is the coefficient of $\phi_0$, which is not damped. In the expansion, $u - \bar u = \sum_{k \geq 1}e^{-\lambda_kt}c_k\phi_k$, so $\|u - \bar u\|^2 = \sum_{k \geq 1}e^{-2\lambda_kt}c_k^2 \leq e^{-2\lambda_1t}\sum c_k^2$. Directly: $\frac{d}{dt}\int(u - \bar u)^2 = 2\int(u - \bar u)\Delta u = -2\int|\nabla u|^2 \leq -2\lambda_1\int(u - \bar u)^2$, and Gronwall (2B.10 Ordinary Differential Equations).

Solution

The normalised eigenfunctions are $\frac{e^{ikx}}{\sqrt{2\pi}}$ with eigenvalues $k^2$. The wrapped Gaussian solves the heat equation, is periodic, and tends to the periodic delta function, so it is the same kernel. For $|x - y| < \pi$, the $m = 0$ term dominates as $t \to 0$, the others being smaller by factors $e^{-c/t}$, so $-4t\log p = (x - y)^2 + 2t\log(4\pi t) + o(1) \to (x - y)^2$.

Solution

$\mu(t + s) - \mu(t) \leq u(x_s, t + s) - u(x_s, t) = s\,\partial_tu(x_s, t + \sigma_s)$ for some $\sigma_s \in (0, s)$. Along a subsequence $x_s \to x_*$, and by continuity $u(x_*, t) = \lim u(x_s, t + s) = \lim\mu(t + s) = \mu(t)$ ($\mu$ is continuous), so $x_*$ is a maximum point at time $t$ and the difference quotients converge to $\partial_tu(x_*, t)$.

Solution

$\sum(2l + 1)e^{-l(l + 1)t} = e^{t/4}\sum f(l + \frac12)$, and $\int_0^\infty2xe^{-tx^2}dx = \frac1t$, $f'(0) = 2$. So the trace is $(1 + \frac t4 + \dots)(\frac1t + \frac{1}{12} + O(t)) = \frac1t + \frac14 + \frac{1}{12} + O(t) = \frac1t + \frac13 + O(t)$. The prediction: $\frac{1}{4\pi t}(4\pi + \frac t6\cdot8\pi) = \frac1t + \frac13$. (Numerically, the trace at $t = 0.01$ is $100.334$.)

Solution

$\log u = -\frac n2\log(4\pi t) - \frac{|x|^2}{4t}$, so $|\nabla\log u|^2 = \frac{|x|^2}{4t^2}$ and $\partial_t\log u = -\frac{n}{2t} + \frac{|x|^2}{4t^2}$; the difference is $\frac{n}{2t}$. Equality in the gradient estimate along the path that realises it gives equality in the integrated Harnack inequality, so the constants cannot be improved.

Solution

(a) As stated, since $dV = \sqrt{\det g}\,dx$. (b) $\frac{d}{dt}\int v\,dV = \int(\partial_tv + \frac12(\operatorname{tr}h)v)\,dV = \int(-\Delta v)\,dV = 0$. (c) $-\partial_tv - \Delta v + Rv = 0$. For constant $v$: $v' = Rv$. On the shrinking sphere $\frac{d}{dt}\operatorname{Vol} = -\int R\,dV = -R\operatorname{Vol}$ ($R$ is constant in space), so $\frac{d}{dt}\frac{c}{\operatorname{Vol}} = \frac{cR\operatorname{Vol}}{\operatorname{Vol}^2} = R\frac{c}{\operatorname{Vol}}$. Its integral $v\operatorname{Vol} = c$ is constant, as (b) predicts.