Simultaneous diagonalization of two Riemmanian metrics

Question

Let $M$ be a smooth manifold and $g,h$ two Riemannian metrics on $M.$ Can one find a local frame $(e_i)$ orthonormal with respect to $g$ such that $$ h(e_i,e_j)=0$$ $\forall i \neq j?$

Answer 1

It's always possible to find such a frame at a single point, as a consequence of the finite-dimensional spectral theorem applied to the self-adjoint operator $A$ that satisfies $h(X,Y) = g(X,AY)$. But it might not be possible to find a continuous local frame in a neighborhood of each point. Here's a simple counterexample: Let $g$ be the Euclidean metric on $\mathbb R^2$ and let $h$ be the metric $$ h = (1+ x^2 - y^2) dx^2 + 4 x y\, dx\,dy + (1 + y^2 - x^2) dy^2. $$ You can check that away from the origin, the following $g$-orthonormal frame diagonalizes $h$: $$ e_1 = \frac{(x,y)}{\sqrt{x^2 + y^2}}, \quad e_2 = \frac{(y,-x)}{\sqrt{x^2 + y^2}}. $$ Because the self-adjoint operator associated to $h$ has distinct eigenvalues on $\mathbb R^2\smallsetminus \{0\}$, this orthonormal frame is unique up to $\pm$ signs and reordering. But it has no continuous extension to the origin.