# An unusual metric reconstruction problem


For which smooth functions $f$ is the correspondence

$$g\mapsto \nabla^g f$$

injective, i.e., from the knowledge that $\nabla^{g_0} f=\nabla^{g_1} f$ on $U$ we can conclude that $g_0=g_1$ on some neighborhood $V$ of $0$ in $\bR^n$, $V\subset U$.

Clearly, the above correspondence is not injective if $f\equiv const$, or $f(x)$ is linear in $x$.

The next interesting case is when $q$ is quadratic. Suppose that

$$f(x)=q(x):=\frac{1}{2}(x_1^2+\cdots +x_n^2).$$

Denote by $g_0$ the canonical Euclidean metric on $\bR^n$ so that

$$\nabla^{g_0}q =\sum_i x_i\pa_{x_i}.$$

The special case of the question goes as follows.

If $g$ is a real analytic Riemann metric defined on a neighborhood $U$ of $0$ and $\nabla^g q(x)= \nabla^{g_0} q(x)$, $\forall x\in U$, can we conclude that $g=g_0$ in a neighborhood of $0$?

I was not able to track results of this kind in the literature, and I would appreciate any pointers.

If $V$ is a vector field such that $Vf=0$ at every point, then the 2-tensor $$h^{ij}=g^{ij}+V^iV^j$$ is also an inverse metric and satisfies $\nabla^gf=\nabla^hf$. In a coordinate chart this amounts to finding a vector field orthogonal (in the Euclidean sense on the chart) pointwise orthogonal to the gradient of $f$. In the case of your second example such a vector field has to vanish at the origin but need not vanish elsewhere.
Examples of such vector fields are not hard to find. Just take $V=(\partial_2f,-\partial_1f,0,\dots,0)$ and change the indices if needed. If this simple idea doesn't give a nonzero vector field, then $\nabla f\equiv0$ and any $V$ will do. In any neighborhood one can find a nonzero vector field and therefore a distinct metric with the same gradient of $f$.
Therefore one function $f$ is never enough. If you have a collection of functions, you need any vector field orthogonal to every one of them to vanish identically. You need at least $n$ functions in $n$ dimensions.
If you take a non-analytic function $f$ and ask the question for analytic metrics, the answer may be different, but that's another story…
• It's probably worth pointing out, though, that there are constraints: In $n$ dimensions, if you specify $n$ functions $f^1,\ldots, f^n$ and $n$ vector fields $X^1,\ldots, X^n$ and you want to know whether there is a metric $g$ such that $X^i=\nabla^g f^i$, then you will need it to be true that the functions $h^{ij} = \mathrm{d}f^i(X^j)$ satisfy $h^{ij} = h^{ji}$, since, by definition, $$h^{ij} = \mathrm{d}f^i(X^j) = g(\nabla^gf^i,\nabla^gf^j).$$ Of course, you also need the matrix $h = (h^{ij})$ to be positive definite as well as symmetric. With these assumptions, $g$ exists and is unique. – Robert Bryant Apr 26 '15 at 10:33