 Solution

Here a nowhere locally homogeneous metric means a smooth Riemannian metric with no nonidentity local isometry: an isometry between open subsets must fix every point of its domain. This is the strong local-rigidity meaning of “bumpy” used in this question, not the different convention about nondegenerate closed geodesics. Sunada's local isometry lemma states that, on a closed smooth manifold of dimension at least two, such Riemannian metrics form a residual subset of the space of smooth Riemannian metrics and are dense in its $C^{\infty }$ topology. In dimension one this conclusion is false: arclength coordinates give local translations for every Riemannian metric.

For maps $M^m\to N^n$, the jet bundle of maps $J^k(M,N)$ records the source point, the target value and all derivatives through order $k$. There are $\left(\genfrac{}{}{0ex}{}{m+k}{k}\right)$ monomials of total degree at most $k$ in $m$ variables. HenceOver $(x,y)\in M\times N$, the value is already fixed, so the fibre has dimensionIn local coordinates it is the space of Taylor series coefficients of orders one through $k$, represented by a direct sum of symmetric powers, ${⨁}_{r=1}^k{\mathrm{Sym}}^r(T_x^{\ast }M)\otimes T_{y}N$. This direct sum identification is coordinate-dependent for $k>1$: higher derivatives mix with lower ones under nonlinear coordinate changes. Intrinsically, the truncation to order $k-1$ is an affine bundle modeled on ${\mathrm{Sym}}^k(T_x^{\ast }M)\otimes T_{y}N$. For $k=1$ the fibre is canonically $\mathrm{Hom}(T_{x}M,T_{y}N)$.

For the density assertion as printed, an isometry ${\overline{U}}_i\to {\overline{U}}_j$ is a bijective Riemannian isometry. A localized volume perturbation suffices; it is not necessary to prove the stronger generic local-rigidity lemma afresh. We take the coordinate-ball basis without repetitions, and each $U_i$ as the interior of its smooth closed-ball closure. Distinct such domains have a nonempty open difference in at least one direction. Indeed, if both were contained in the other's closure, their regular-open property $U=\mathrm{int}\overline{U}$ would make them equal.

Fix any Riemannian metric $g$. If the two domains have unequal volumes, they cannot be isometric, so $g$ already lies in the desired complement. Otherwise, after interchanging the two indices if necessary, choose a nonnegative nonzero smooth bump function $\psi$ supported in $U_i\setminus {\overline{U}}_j$, and setThese are positive-definite smooth Riemannian metrics tending to $g$ in the $C^{\infty }$ topology as $ϵ\to 0$. Their Riemannian volume forms satisfy $d{V}_{g_{ϵ}}=e^{mϵ\psi }d{V}_g$. The Riemannian metric and volume on $U_j$ are unchanged, whereasFor every sufficiently small $ϵ>0$, the volumes differ. Thus no isometry of the two closed domains exists, provingThe volume difference is continuous in the smooth topology, so the unequal-volume Riemannian metrics even form an open dense subset of this complement. The regular-open assumption is also needed to justify this literal closed-domain assertion. As written, requiring only that each closure be a closed ball permits two distinct basis members with the same closure. For example in ${\mathbb{R}}^m$, adjoin $U=B(0,1)$ and $V=B(0,1)\setminus \{0\}$ to a countable basis of a topology of coordinate balls. Both are open and both closures are the same closed ball. The identity is an isometry of those closures for every Riemannian metric, so the indicated complement is empty. Thus the conclusion needs distinct closures, as in the usual basis of genuine coordinate balls used above. The no-repetitions convention is necessary as well: if the indexed basis repeats a domain, its identity map is always an isometry and the asserted complement for those two indices is empty. This volume argument concerns isometries onto the specified whole domains; the stronger assertion excluding arbitrary local isometries is the separate Sunada lemma stated above.