Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 20 / 3 / Solution

Here “bumpy” means a nowhere locally homogeneous metric, not the different convention concerning nondegenerate closed geodesics. A smooth Riemannian metric has this property if every local isometry between connected open subsets is the identity on its domain; in particular, no two distinct open subsets are isometric. The Sunada local isometry lemma states that, on a compact smooth manifold without boundary of dimension at least two, such Riemannian metrics form a residual set and are dense in the $C^{\infty }$ topology. The dimension restriction matters: in dimension one, arclength coordinates always give nontrivial local isometries.

An element $j_x^{k}f$ of the jet bundle of maps $J^k(M,N)$ records the source point $x$, target value $f(x)$ and all derivatives through order $k$. If $\dim M=m$ and $\dim N=n$, there are $\left(\genfrac{}{}{0ex}{}{m+r-1}{r}\right)$ derivative multi-indices of order $r$, and $n$ components for each one. ThusFor the projection $j_x^{k}f\mapsto (x,f(x))$ onto $M\times N$, the fiber has dimension $n((\genfrac{}{}{0ex}{}{m+k}{k})-1)$. In source and target coordinates its Taylor coefficients identify it withThis is a coordinate-dependent identification for $k>1$, since a change of coordinates mixes different derivative orders. For $k=1$ the fiber is canonically $\mathrm{Hom}(T_{x}M,T_{y}N)$; for higher orders, truncation to the previous jet bundle of maps is an affine bundle modeled on ${\mathrm{Sym}}^k(T_x^{\ast }M)\otimes T_{y}N$. For $k=0$ the fiber is a point.

The density assertion in the printed question requires distinct closed domains. Literally, for $i=k$, the identity is an isometry for every Riemannian metric, so ${\mathcal{S}}_{ii}$ is the whole metric space and its complement is empty. Merely indexing the same closed ball twice has the same defect. The intended assertion is for distinct embedded closed balls $K_i={\overline{U}}_i$ and $K_k={\overline{U}}_k$, with an isometry meaning an onto diffeomorphism preserving the metric. The proof below establishes precisely this corrected assertion.

Distinct embedded closed balls are regular closed sets, so at least one of $\mathrm{int}{K}_i\setminus K_k$ and $\mathrm{int}{K}_k\setminus K_i$ contains a nonempty open set. Indeed, if both were empty, taking closures would give both inclusions and hence equality. Interchange the labels if necessary and choose a nonzero nonnegative smooth bump function $\eta$ supported in $\mathrm{int}{K}_i\setminus K_k$. For any given Riemannian metric $g$, make the localized volume perturbationThe Riemannian volume form satisfies $d{V}_{g_t}=e^{mt\eta }d{V}_g$. ThereforeIf the original volumes agree, every sufficiently small $t>0$ makes them unequal. If they already disagree, continuity preserves that inequality for sufficiently small $t$. An onto isometry must preserve volume, so each of these perturbed Riemannian metrics belongs to $\mathcal{C}{\mathcal{S}}_{ik}$. Finally $g_t\to g$ in $C^{\infty }$: all derivatives of $e^{2t\eta }-1$ tend to zero uniformly on compact sets. This proves density of $\mathcal{C}{\mathcal{S}}_{ik}$ for distinct closed balls, in fact in every positive dimension. The localized volume perturbation settles this particular density claim directly. It also does not imply the stronger Sunada local isometry lemma by itself, since a general local isometry need not send one preselected basis ball onto another.