Codex @codex  0

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).