Do eigenfunctions determine the geometry of a manifold? If so, do finitely many suffice? Let $X$ be a smooth, Riemannian manifold. It is known that the geometry of $X$ can be recovered from its heat kernel $k_{t}(x,y)$, using Varadhan's Lemma: $\displaystyle\lim_{t \to 0} t \log k_{t}(x,y) = -\frac{1}{4}d^{2}(x,y)$. Since the heat kernel $k_{t}(x,y)$ can be expressed in terms of eigenfunctions and eigenvalues,
$$k_{t}(x,y) = \sum_{i=0}^{\infty}e^{-\lambda_{i}t}\phi_{i}(x)\phi_{i}(y)$$
we can say that the knowledge of the eigenvalues and eigenfunctions of $X$ determines its geometry.
Now, the following result of Bates: https://arxiv.org/pdf/1605.01643.pdf tells us that there is a constant $d(X)$, depending on the dimension and geometry of $X$, such that the map $X \to \mathbb{R}^{d(X)}$ sending $x \in X$ to $\langle \phi_{0}(x), \cdots, \phi_{d(X)}(x)\rangle$ is injective.
My question therefore is: we see that a map to Euclidean space using finitely many eigenfunctions is enough to recover the topological type of $X$. Can we also recover its metric (up to some simple transformation)? What if we used infinitely many eigenfunctions? In either case we do not have access to the eigenvalues. 
 A: Here is a sketch of an idea of how to show that the set $\mathcal{E}(g)\subset C^\infty(M)$ of all the eigenfunctions of the metric $g$ on a compact manifold $M$ determines $g$ up to a constant multiple.  (Note that I do not assume that the corresponding eigenvalues are given, which would be easier.)  Clearly, this is the best one can hope for, since for any constant $c>0$, $\mathcal{E}(cg) = \mathcal{E}(g)$.  In fact, I think it's very likely that knowing a sufficiently 'large' finite subset of $\mathcal{E}(g)$ should be sufficient, but that remains to be seen.
The basic idea is this:  Let $J^k(M)\to M$ denote the vector bundle consisting of the $k$-jets of smooth functions on $M$.  When $M$ has dimension $n$, the bundle $J^k$ has rank ${n+k}\choose n$ over $M$. Given $g$, it is easy to show that there is a closed quadratic cone bundle $Q(g)\subset J^3(M)$ of codimension $n$ such that the $3$-jets of all the local eigenfunctions of $g$ lie in $Q(g)$.  In fact, the $3$-jets of local eigenfunctions fill out $Q(g)$.
Note that $Q(g)$ is not a linear subbundle of $J^3(M)$ precisely because we are not specifying the eigenvalues of the eigenfunctions. Of course, the $3$-jets of local eigenfunctions with eigenvalue $\lambda$, fill out a linear subbundle $Q(g,\lambda)\subset J^3$, but the union of the $Q(g,\lambda)$ as $\lambda$ varies is not a linear subbundle.  However, it is easy to show that, for any $x\in M$, the subset $Q_x(g)\subset J^3_x$ is the zero locus of an ideal generated by $n$ polynomials homogeneous of degree $2$ on the vector space $J^3_x$. ($Q_x(g)$ is not a smooth variety in $J^3_x$ but the singular locus is quite small.)
I claim that the subbundle $Q(g)\subset J^3$ determines $g$ up to a constant factor (assuming that $M$ is connected).  Here is why:  Let $Q^0(g)\subset Q(g)$ denote the subset consisting of those $3$-jets in $Q(g)$ whose $0$-jet vanishes.  The projection $\pi^2_3(Q^0(g)_x)$ of $Q^0(g)_x$ into $J^2_x$ has codimension $n{+}1$ in $J^2_x$, cut out by the linear equation that says that the $0$-jet vanishes and the $n$ quadratic equations that then turn out to all be multiples of a single linear equation on the space of $2$-jets whose $0$-jet vanishes, as is easy to verify in local coordinates.
Consequently, it follows that there exist second order, elliptic differential operator of the form
$$
L u = a^{ij}\,\frac{\partial^2u}{\partial x^i\,\partial x^j} + b^i\,\frac{\partial u}{\partial x^i} + c\,u
$$
(with $(a^{ij})$ positive definite) such that every local eigenfunction of $g$ satisfies an equation of the form $Lu = \phi(u)u$ where $\phi(u)$ is a smooth function (that could depend on $u$), and, moreover, $L$ is unique up to scalar multiplication and the addition of a $0$-th order term.
Finally, the requirement that there exist a function $f>0$ such that $fL$ can be expressed in the divergence form
$$
(fL)(u) = |h|^{-1/2}\frac{\partial}{\partial x^i}\left(|h|^{1/2}h^{ij}\,\frac{\partial u}{\partial x^j}\right)
$$
is easily seen to imply that $h = cg$ for some constant $c\not=0$.  Thus, $g$ can be recovered, up to a constant multiple, from knowledge of the subbundle $Q(g)\subset J^3$, as claimed.
Finally, what one would expect is that, when $M$ is compact, if we now look at the $3$-jets of the elements of $\mathcal{E}(g)$, i.e., the global eigenfunctions of $g$, that a sufficiently large subset will determine sufficiently many points in each $J^3_x$ that they will determine $Q(g)_x$, which, after all, is known to be cut out by $n$ homogeneous quadratic polynomials for each $x$.  (Of course, the number of elements of $\mathcal{E}(g)$ needed to do this at each point could be large, even for $n=2$, but it will be finite.)  Assuming such a 'density' result, $\mathcal{E}(g)$ will determine $Q(g)\subset J^3$, which, as we have seen, will determine $g$ up to a constant multiple.
