Whether preserving inner products follows from preserving intrinsic distances I know that if a map F between two surfaces preserves inner products of tangent vectors and is 1-1 and onto, then it must preserve intrinsic distances, but I’m not sure whether the inverse is true. Namely, is it true that any map F between two surfaces that is 1-1, onto, and preserves intrinsic distances will automatically preserve inner products of tangent vectors?
Thanks a lot!
 A: Suppose $$F : (S, g) \to (S', g')$$ is a metric isometry (homeomorphism that preserves the distance respective functions $d_g, d_{g'}$ of $g, g'$, that is, that satisfies $d_g = F^* d_{g'}$). Then, yes, $F$ is actually a (Riemannian) isometry, i.e., $F^* g' = g$. 
To do this, it's certainly enough to give an explicit formula for $g$ in terms of $d_g$ (or even just $g(Z, Z)$, since we can recover $g$ by the Polarization Formula).
Hint Compare $g$ with the Euclidean metric in normal coordinates to establish that for any $p$ and $\epsilon > 0$ there is a constant $C > 0$ for which the comparison
$$(1 - C|t|) |t X - t Y|_g \leq d_g(\exp t X, \exp t Y) \leq (1 + C|t|)|t X - t Y|_g$$ holds when $|t| \leq 1$ for all $X, Y \in B_{\epsilon}(p)$.
Conclude that $$|X - Y|_g^2 = \lim_{t \to 0} \frac{d_g(\exp tX, \exp tY)^2}{t^2} .$$
For a more detailed hint, see the erratum for p. 112, Problem 6-2 in the errata to Lee's Riemannian Manifolds.
A: Yes, because $4a\cdot b=(a+b)^2-(a-b)^2$ or, in the complex case, $4a\cdot b=\sum_{n=0}^3 i^{-n}\Vert a+i^n b\Vert^2$.
