Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 7 / 5

The Gauss lemma says that, for $v,w\in T_{p}M$ for which the exponential map is defined,In particular radial and angular directions are orthogonal, and the radial direction retains its length.

Take the variation through geodesics $\alpha (t,s)={\exp }_p(t(v+sw))$, and put $T={\partial }_t\alpha$, $J={\partial }_s\alpha$. The Levi-Civita connection is metric compatible and torsion-free, so ${\nabla }_{t}J={\nabla }_{s}T$. Each $t$-curve is a geodesic, hence ${\nabla }_{t}T=0$, and its constant squared speed is $|v+sw{|}_p^2$. ThereforeAt $t=0$ the variation starts at the fixed point $p$, so $J(0,s)=0$. Integrating from $0$ to $1$ and setting $s=0$ gives the displayed identity, because $T(1,0)=(d{\exp }_p{)}_{v}v$ and $J(1,0)=(d{\exp }_p{)}_{v}w$.

In geodesic normal coordinates based on an orthonormal basis at $p$, choose $v=x$ and $w$ to be the $j$th coordinate vector. The lemma givesThese identities also hold at $x=0$ by continuity. They are the radial identities required in part (b).

Use $a^{ij}=g^{ij}$, $r^2=|x{|}^2$, and $E=x\cdot \nabla$. The Gauss lemma from part (a) givesFor the Gaussian $q_t(x)=(4\pi t{)}^{-n/2}{e}^{-r^2/(4t)}$, we have ${\partial }_i{q}_t=-x_i{q}_t/(2t)$. Applying precisely the divergence-form operator in the paper yieldsSince ${\partial }_t{q}_t=(-n/(2t)+r^2/(4t^2))q_t$, the Gaussian itself already satisfies $({\partial }_t+L)q_t=0$.

For completeness, the product rule gives for an arbitrary smooth $G$,Indeed the cross term in $L(q_{t}G)$ is $-2a^{ij}({\partial }_i{q}_t)({\partial }_{j}G)=q_{t}EG/t$. Substituting the formal power series $G={\sum }_{k\ge 0}{t}^k{B}_k$, with $B_0=1$, gives the radial transport equationsAlong a radial segment their unique smooth solution isThe integral is smooth including at the centre, since each derivative adds a nonnegative power of $s$ to the integrable factor $s^{k-1}$. Uniqueness follows because a homogeneous solution $h$ has $d(s^{k}h(sx))/ds=0$; smoothness makes its limit at $s=0$ zero, so $h(x)=0$. This also excludes singular ray-dependent homogeneous solutions.

Here $L1=0$, so the recursion successively gives $\boxed{B_k=0(k\ge 1),G(x,t)=1}$. Thus both existence and uniqueness hold, with no convergence issue because the formal solution is the exact Gaussian. This is the normal-coordinate divergence-form heat parametrix. The positive Laplace-Beltrami operator $-{\rho }^{-1}{\partial }_j(\rho g^{ij}{\partial }_i)$, where $\rho =\sqrt{\det g}$, has an additional first-order term and a different leading amplitude; it is essential to use the operator actually specified here.

We use a closed manifold of positive dimension $n$. The nonnegative positive Laplace-Beltrami operator on functions isIt satisfies $⟨\Delta f,f⟩={\int }_M|df{|}_g^{2}d{\mathrm{vol}}_g$. Its domain as a self-adjoint operator on $L^2(M)$ is $H^2(M)$. The analytic argument in Question 4(d), with $q=0$, gives a smooth orthonormal eigenbasis and ordered eigenvalues $0={\lambda }_0\le {\lambda }_1\le \cdots \to \infty$.

We prove a quantitative counting bound from local Fourier series, rather than assume a spectral asymptotic formula. Choose a finite coordinate cover and smooth functions ${\chi }_{\alpha }$ compactly supported in its charts with ${\sum }_{\alpha }{\chi }_{\alpha }^2=1$. This can be obtained by normalizing an ordinary nonnegative partition of unity by the square root of the sum of its squares. Embed each chart support in a fixed coordinate torus, and extend $v_{\alpha }={\chi }_{\alpha }f$ by zero there. Uniform comparison of the coordinate and metric norms, and differentiation of the cutoffs, give constants independent of $f$ such thatLet $P_R$ retain the Fourier modes $|k|\le R$. Parseval identity givesThe map $f\mapsto (P_R{v}_{\alpha }{)}_{\alpha }$ has target dimension at most $C_0{R}^n$, since there are at most $C_n{R}^n$ lattice points in each cutoff ball.

Let $E_{\Lambda }$ span all eigenfunctions with eigenvalues at most $\Lambda$. For $f\in E_{\Lambda }$, $\Vert df{\Vert }_2^2\le \Lambda \Vert f{\Vert }_2^2$. If every retained local coefficient vanishes, the preceding estimates implyChoose $R=b\sqrt{1+\Lambda }$ with fixed $b$ sufficiently large. Then this is impossible unless $f=0$, so the finite-dimensional coefficient map is injective on $E_{\Lambda }$. Therefore the eigenvalue counting function satisfiesTaking $\Lambda ={\lambda }_k$ gives $k+1\le C_1(1+{\lambda }_k{)}^{n/2}$, so ${\lambda }_k\ge c_1(k+1{)}^{2/n}-1$. For all sufficiently large $k$, this implies ${\lambda }_k\ge A_0{k}^{2/n}$.

If $M$ is connected, the zero eigenvalue is simple: $\Delta f=0$ implies $df=0$, hence $f$ is constant. Thus every ${\lambda }_k$ for $k\ge 1$ is positive. Taking the minimum of $A_0$ and the finitely many positive ratios ${\lambda }_k/k^{2/n}$ below the large-$k$ threshold provesConnectedness is implicit in the printed claim. If $M$ has $b>1$ connected components, zero has multiplicity $b$, and the claim fails already at $k=1$. The same proof gives ${\lambda }_k\ge A\max \{k-b+1,0{\}}^{2/n}$ for all $k$, as well as the universal counting bound above. These finite initial zero modes do not affect part (d).

Let $({\varphi }_j)$ be a smooth orthonormal eigenbasis of the positive Laplace-Beltrami operator, with eigenvalues $({\lambda }_j)$. The orthonormal eigenbasis definesThe large-$j$ lower bound of part (c), valid also after the finitely many zero modes of a disconnected compact manifold, gives ${\lambda }_j\ge a{j}^{2/n}$ for sufficiently large $j$. Henceby comparison with ${\int }_0^{\infty }{e}^{-ta{x}^{2/n}}dx$. The operator is positive and diagonal with these summable coefficients, so it is a trace-class operator and its operator trace is their sum.

Its proposed heat kernel isWe verify smoothness and the kernel action. The elliptic regularity estimate established in Question 4(d) gives, by induction,The Sobolev embedding theorem therefore gives, for every derivative order $a$, constants $C_a$, $r_a$ with $\Vert {\varphi }_j{\Vert }_{C^a}\le C_a(1+{\lambda }_j{)}^{r_a}$. Every differentiated term of the kernel series is bounded by $C_{a,b}{e}^{-t{\lambda }_j}(1+{\lambda }_j{)}^{r_a+r_b}$. An exponential absorbs any fixed power: for $t\ge \tau >0$ this is at most $C_{a,b,\tau }{e}^{-\tau {\lambda }_j/2}$, a summable bound. Time derivatives merely introduce more powers of ${\lambda }_j$. Thus the series converges uniformly with all spatial and time derivatives on $[\tau ,\infty )\times M\times M$. In particular $K_t$ is smooth for every $t>0$.

Integrating its uniformly convergent series against $f\in L^2(M)$ is legitimate since the compact manifold has finite volume and $L^2(M)\subseteq L^1(M)$. It gives exactly the spectral expression above, soFinally $K_t(x,x)={\sum }_j{e}^{-t{\lambda }_j}|{\varphi }_j(x){|}^2$ is nonnegative. Integrating term by term, either by uniform convergence or by Tonelli theorem, and using $\Vert {\varphi }_j{\Vert }_2=1$ proves the heat kernel trace formula