Spectral Theory and Heat Trace Asymptotics

4A.8 Compact Operators and Spectra proved that the Dirichlet Laplacian on a bounded domain has an orthonormal basis of eigenfunctions, with eigenvalues $\lambda_1 \leq \lambda_2 \leq \cdots \to \infty$, and asked what they know about the domain: Weyl's law says they know its area, and Gordon, Webb and Wolpert showed they do not always know its shape. This chapter connects that question to the heat equation, and answers part of it with proofs.

The connection is the heat trace

the trace of the heat semigroup $e^{t\Delta}$. On one hand it is built from the eigenvalues alone. On the other, it is the integral of the heat kernel on the diagonal, $\int_\Omega p_t(x, x)\,dx$, and for small $t$ heat has had no time to feel the boundary or the curvature, so $p_t(x, x)$ is almost the Euclidean value $(4\pi t)^{-n/2}$. Hence $Z(t) \approx \frac{|\Omega|}{(4\pi t)^{n/2}}$: the eigenvalues determine the volume. Corrections, as $t$ grows, see the boundary (its length or area) and, on a curved manifold, the scalar curvature. A Tauberian theorem converts the small-$t$ behaviour of $Z$ into the large-$\lambda$ behaviour of the eigenvalue counting function, which is Weyl's law.

On the way, the chapter makes rigorous something 4A.8 Compact Operators and Spectra treated informally: the Laplacian is an unbounded operator, and to apply spectral theory one must say on which domain it acts and check that it is self-adjoint there. The Friedrichs extension does this through the quadratic form $\int|\nabla u|^2$, and it is the route by which every elliptic operator in this guide, including the operator $-4\Delta + R$ whose lowest eigenvalue is Perelman's $\lambda$-invariant (12A.2 Ricci Flow as a Gradient Flow), becomes a self-adjoint operator with a spectral theorem.

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

• distinguish symmetric and self-adjoint unbounded operators, and construct the Dirichlet Laplacian as a self-adjoint operator from its quadratic form;
• use the spectral theorem for operators with compact resolvent to define $e^{t\Delta}$, $\cos(t\sqrt{-\Delta})$ and other functions of the Laplacian, and write the heat kernel as an eigenfunction series;
• prove that the Dirichlet heat kernel is positive, symmetric and bounded by the Euclidean heat kernel, and that $Z(t) = \int p_t(x, x)\,dx$;
• prove $Z(t) \sim |\Omega|(4\pi t)^{-n/2}$ as $t \to 0$, and deduce Weyl's law with Karamata's Tauberian theorem;
• compute heat traces exactly for intervals, tori and spheres, and read off length, area and curvature from them.

15.1Counting the modes of a cavity

15.2Unbounded self-adjoint operators

The Laplacian is not defined on all of $L^2(\Omega)$, and on the functions where it is defined it is not bounded ($\|\Delta\sin kx\| = k^2\|\sin kx\|$). Spectral theory for such operators needs care about domains.

An unbounded operator on a Hilbert space $H$ is a linear map $A : D(A) \to H$ from a subspace $D(A)$, its domain. It is densely defined if $D(A)$ is dense, and symmetric if $\langle Au, v\rangle = \langle u, Av\rangle$ for all $u, v \in D(A)$. Its adjoint $A^*$ has domain

and $A^*v$ is the vector representing that functional (Riesz, 4A.5 Hilbert Spaces and Lax–Milgram). $A$ is self-adjoint if $A = A^*$, including equality of domains. Symmetric means $A \subseteq A^*$; self-adjoint means no room is left between them.

The distinction is not pedantic. On $L^2(0, 1)$, the operator $-\frac{d^2}{dx^2}$ on $C_c^\infty(0, 1)$ is symmetric, but its adjoint acts on all of $H^2(0, 1)$ with no boundary conditions, and it has many self-adjoint extensions: Dirichlet ($u = 0$ at both ends), Neumann ($u' = 0$), periodic, Robin, each with a different spectrum (Exercise 15.7). Physically, the boundary conditions are part of the operator. Only for a self-adjoint operator does the spectral theorem hold, and only then is $e^{-itA}$ a unitary evolution (quantum mechanics) or $e^{-tA}$ a well-defined heat flow.

The Friedrichs construction. Quadratic forms give self-adjoint operators automatically. Let $\Omega$ be bounded and open, and on $H^1_0(\Omega)$ consider the form $Q(u, v) = \int\nabla u\cdot\nabla v$.

For a smooth boundary, elliptic regularity identifies the domain: $D(-\Delta_D) = H^2(\Omega)\cap H^1_0(\Omega)$ (6A.6 Weak Solutions and Elliptic Regularity). But the construction works on any bounded open set, with no regularity of the boundary at all, which is its strength: the boundary condition "$u = 0$" is encoded in the choice of the form domain $H^1_0$. With the form domain $H^1(\Omega)$ instead, the same construction gives the Neumann Laplacian, and the condition $\partial_\nu u = 0$ appears automatically as the natural boundary condition. On a closed manifold, the form $\int|\nabla u|^2$ on $H^1(M)$ gives the Laplace–Beltrami operator, and adding $\int Ru^2$ gives Perelman's operator $-4\Delta + R$ (Exercise 15.14).

Functional calculus. With an orthonormal eigenbasis, any bounded function $F$ on $[0, \infty)$ defines a bounded operator

with $\|F(-\Delta_D)\| = \sup_k|F(\lambda_k)|$. The choices $F(\lambda) = e^{-t\lambda}$, $\frac{1}{\mu + \lambda}$, $\cos(t\sqrt\lambda)$ and $e^{-it\lambda}$ give the heat semigroup, the resolvent, the wave propagator and the Schrödinger group, all from one basis. The heat semigroup obtained this way is the one Hille–Yosida produced in 6B.13 Semigroups and Evolution Equations, and its smoothing estimate is 6B.13 Semigroups and Evolution Equations's spectral computation.

15.3The heat kernel of a domain

The heat semigroup is an integral operator, $e^{t\Delta_D}g(x) = \int_\Omega p_t(x, y)g(y)\,dy$, with the Dirichlet heat kernel

The series converges absolutely and uniformly on $\Omega\times\Omega$ for each $t > 0$: by elliptic regularity and Sobolev embedding the eigenfunctions satisfy $\sup|\phi_k| \leq C\lambda_k^{n/4}$ on interior regions (Exercise 15.13 gives a global form of this bound once the kernel is known), and Weyl's law (or the cruder bound $\lambda_k \geq ck^{2/n}$ from the heat trace below) makes $\sum_ke^{-\lambda_kt}\lambda_k^{n/2}$ converge. Physically, $p_t(x, y)$ is the temperature at $x$ at time $t$ after a unit of heat is released at $y$ at time $0$, in a body whose boundary is held at zero; probabilistically, it is the density of Brownian motion (run at rate $2$) from $y$, killed when it first hits the boundary (6D.1 Brownian Motion).

In words: a body with cold walls is everywhere cooler than the same heat in free space, because the walls only remove heat. Part (2) also gives a first bound on the heat trace, with no work: $Z(t) = \int p_t(x, x) \leq \int\Phi_t(0) = \frac{|\Omega|}{(4\pi t)^{n/2}}$.

15.4The heat trace knows the volume

The proof is short because the two halves are maximum principles: one says the walls only remove heat, the other says that in a short time heat from a point does not reach walls far away. The scale $\sqrt t$ is parabolic scaling (6A.1 What a PDE Is): in time $t$, heat explores a neighbourhood of radius about $\sqrt t$, and the boundary layer where the walls are felt has volume about $|\partial\Omega|\sqrt t$. That suggests the next term.

More terms. For a bounded domain with smooth boundary in the plane, Dirichlet conditions,

where $|\partial\Omega|$ is the perimeter and $\gamma$ the number of holes (Ake Pleijel, 1954, for the perimeter term; Henry McKean and Isadore Singer, 1967, for the constant). So one can hear the area, the perimeter and the number of holes of a drum. The perimeter term is the boundary layer: near a straight piece of wall, the Dirichlet kernel on the diagonal is $\Phi_t(0)(1 - e^{-d^2/t})$ by the method of images (6B.4 Green’s Functions, Separation of Variables and Transforms), and integrating the deficit over the distance $d$ to the wall gives $\frac{1}{4\pi t}\int_0^\infty e^{-d^2/t}dd = \frac{1}{4\pi t}\cdot\frac{\sqrt{\pi t}}{2}$ per unit length, which is $\frac{1}{8\sqrt{\pi t}}$ (Exercise 15.11).

On a closed Riemannian manifold $M$ of dimension $n$, the heat trace of the Laplace–Beltrami operator has the Minakshisundaram–Pleijel expansion (1949)

where $R$ is the scalar curvature: the spectrum hears the total scalar curvature. The higher coefficients are integrals of polynomials in the curvature and its derivatives. On the unit $2$-sphere, with eigenvalues $l(l + 1)$ of multiplicity $2l + 1$, area $4\pi$ and $R = 2$, this predicts $Z(t) = \frac1t + \frac13 + O(t)$, which a direct computation confirms (Exercise 15.10). The expansion is a statement about the small-time behaviour of the heat kernel, $p_t(x, x) = (4\pi t)^{-n/2}\big(1 + \frac t6R(x) + O(t^2)\big)$, and its derivation is the subject of 9B.7 The Heat Equation on a Manifold.

15.5From heat traces to Weyl's law

The heat trace is the Laplace transform of the eigenvalue counting function $N(\lambda) = \#\{k : \lambda_k \leq \lambda\}$:

Small $t$ in $Z$ corresponds to large $\lambda$ in $N$. If $N(\lambda) \sim C\lambda^{n/2}$, then $Z(t) \sim C\,\Gamma(\frac n2 + 1)t^{-n/2}$, an Abelian computation (Exercise 15.12). The converse, from $Z$ to $N$, is a Tauberian theorem, and it needs a positivity hypothesis, which $N$, being non-decreasing, satisfies.

Jovan Karamata gave a short proof in 1931, by approximating the indicator function of $[0, 1]$ from above and below by polynomials in $e^{-x}$ (Weierstrass approximation, 2B.5 Uniform Convergence and Arzelà–Ascoli); Feller's An Introduction to Probability Theory and Its Applications, volume 2, chapter XIII, and Taylor II, §8.3, give it. Combining it with Theorem 15.3 ($A = (4\pi)^{-n/2}|\Omega|$, $\gamma = \frac n2$):

In the plane, $N(\lambda) \sim \frac{|\Omega|}{4\pi}\lambda$, the formula of 4A.8 Compact Operators and Spectra's exercise on the square, now for every bounded domain. The constant has a semiclassical meaning: $\frac{\omega_n|\Omega|\lambda^{n/2}}{(2\pi)^n}$ is the volume of the region $\{(x, \xi) : x \in \Omega,\ |\xi|^2 \leq \lambda\}$ of phase space divided by $(2\pi)^n$, "one quantum state per cell of volume $(2\pi)^n$", which is how physicists derive it.

15.6History

Hendrik Lorentz posed the eigenvalue counting problem in 1910, and Hermann Weyl solved it in 1911–12. Richard Courant proved his min–max principle around 1920 and used Dirichlet–Neumann bracketing to give a second proof of Weyl's law. Kurt Friedrichs constructed his extension in 1934. Torsten Carleman (1934) introduced the heat-trace (Laplace transform) approach to eigenvalue asymptotics, and Jovan Karamata's Tauberian theorem dates from 1931. Subbaramiah Minakshisundaram and Åke Pleijel found the heat expansion on manifolds in 1949, Pleijel the perimeter term for planar domains in 1954, Mark Kac posed "Can one hear the shape of a drum?" in 1966, and McKean and Singer computed the constant term and the curvature terms in 1967.

15.7Exercises

Solution

(a) $\frac{\pi^2}{L^2}|k|^2 \leq \lambda$ iff $|k| \leq \frac{L\sqrt\lambda}{\pi}$; the positive octant of the ball has volume $\frac{\pi}{6}\big(\frac{L\sqrt\lambda}{\pi}\big)^3 = \frac{L^3\lambda^{3/2}}{6\pi^2}$, and the lattice count differs by a boundary term of lower order. (b) Weyl: $\frac{\omega_3L^3}{(2\pi)^3}\lambda^{3/2} = \frac{4\pi/3}{8\pi^3}L^3\lambda^{3/2} = \frac{L^3\lambda^{3/2}}{6\pi^2}$. (c) $2\cdot\frac{V}{6\pi^2}\big(\frac{2\pi\nu}{c}\big)^3 = \frac{2V\cdot8\pi^3\nu^3}{6\pi^2c^3} = \frac{8\pi V\nu^3}{3c^3}$.

Solution

(a) $k^2$, $k \geq 1$ ($\sin kx$). (b) $k^2$, $k \geq 0$ ($\cos kx$; the constant has eigenvalue $0$). (c) $4k^2$, $k \geq 0$, with multiplicity $2$ for $k \geq 1$ ($\cos2kx$, $\sin2kx$). Integrating by parts, $\int_0^\pi(-u'')v - u(-v'') = [-u'v + uv']_0^\pi$, which vanishes under each of the three conditions (for periodic, the values at $0$ and $\pi$ cancel).

Solution

$\sum_{k\in\mathbb{Z}}e^{-k^2t} = 1 + 2Z(t)$, and the right side is $\sqrt{\pi/t}(1 + O(e^{-\pi^2/t}))$. So $Z = \frac12\sqrt{\pi/t} - \frac12 + O(t^{-1/2}e^{-\pi^2/t})$. The volume term is $\frac{|\Omega|}{(4\pi t)^{1/2}} = \frac{\pi}{2\sqrt{\pi t}} = \frac12\sqrt{\frac\pi t}$, and the boundary term in dimension $1$ is $-\frac14$ per boundary point (the analogue of $-\frac{|\partial\Omega|}{4}(4\pi t)^{-(n-1)/2}$).

Solution

In one dimension, Poisson summation for $f(x) = e^{-4\pi^2x^2t}$ gives $\sum_ke^{-4\pi^2k^2t} = \sum_m\hat f(m) = \sum_m\frac{1}{\sqrt{4\pi t}}e^{-m^2/4t}$ (4A.6 The Fourier Transform, the Gaussian with $a = 4\pi t$); in $n$ dimensions the sum factorises. The volume is $1$, so the leading term is $(4\pi t)^{-n/2}$, and the corrections $e^{-|m|^2/4t}$ are the heat kernel evaluated at the translates $m$ of the origin, the paths that wind around the torus.

Solution

(a) $l(l + 1) = (l + \frac12)^2 - \frac14$, and $2l + 1 = 2(l + \frac12)$. (b) $\int_0^\infty2xe^{-x^2t}dx = \frac1t$ and $g'(0) = 2$, so the sum is $\frac1t + \frac{2}{24} + O(t) = \frac1t + \frac{1}{12} + O(t)$ (the next Euler–Maclaurin term involves $g'''(0)$, which is of order $t$). (c) $e^{t/4}\big(\frac1t + \frac{1}{12}\big) = \frac1t + \frac14 + \frac{1}{12} + O(t) = \frac1t + \frac13 + O(t)$. The prediction: $\frac{1}{4\pi t}\big(4\pi + \frac t6\cdot2\cdot4\pi\big) = \frac1t + \frac13$.

Solution

(a) $|x - \tilde x| = 2x_2$, so $\Phi_t(x - \tilde x) = \frac{1}{4\pi t}e^{-4x_2^2/4t}$. (b) $\int_0^\infty e^{-d^2/t}dd = \frac{\sqrt{\pi t}}{2}$, so the deficit is $\frac{1}{4\pi t}\cdot\frac{\sqrt{\pi t}}{2} = \frac{1}{8\sqrt{\pi t}}$.

Solution

$dN = C\gamma\lambda^{\gamma-1}d\lambda$, so the integral is $C\gamma\int_0^\infty e^{-\lambda t}\lambda^{\gamma-1}d\lambda = C\gamma t^{-\gamma}\Gamma(\gamma) = C\Gamma(\gamma + 1)t^{-\gamma}$.

Solution

Every term of the series for $p_t(x, x)$ is non-negative, so it dominates the $k$-th one. With $t = 1/\lambda_k$: $\phi_k(x)^2 \leq e\,(4\pi)^{-n/2}\lambda_k^{n/2}$.

Solution

(a) $Q(w, w) + c\int w^2 = 4\int|\nabla w|^2 + \int(R + c)w^2 \geq 4\int|\nabla w|^2 + \int w^2$ when $R + c \geq 1$. The form is then coercive on $H^1$, and the proof of Theorem 15.1 applies verbatim with $H^1(M)$ in place of $H^1_0$; $H^1(M) \hookrightarrow L^2(M)$ is compact on a closed manifold (Rellich, 4A.11 Sobolev Embeddings and Critical Exponents). (b) As in 6A.6 Weak Solutions and Elliptic Regularity, the lowest eigenvalue is the minimum of the Rayleigh quotient, and replacing $w$ by $|w|$ does not change $Q$, so a minimiser can be taken non-negative and then, by the strong maximum principle, positive. (c) $|\nabla w|^2 = \frac14|\nabla f|^2e^{-f}$ and $w^2 = e^{-f}$, so $Q(w, w) = \int(R + |\nabla f|^2)e^{-f}dV = \mathcal F(g, f)$, and the constraint $\int w^2 = 1$ is $\int e^{-f}dV = 1$.