consider a Riemannian manifold $(M,g)$ and consider a second order elliptic differential operator. I've read that each such operator induces a riemannian distance function.
Unfortunately I couldn't find anything useful in this direction. Do you know how this is meant?
Best regards
You need a couple more assumptions. If $P$ is a linear second-order operator acting on scalar functions on $M$, then its principal symbol is a well-defined map $p\colon T^*M\to \mathbb R$, which restricts to a homogeneous quadratic form on each fiber; this means for each $x\in M$, the map $\xi \mapsto p(x,\xi)$ is a quadratic form on ${T_x}^*M$. If in addition $P$ is elliptic, then $p(x,\xi)\ne 0$ for $\xi\ne 0$. A scalar-valued quadratic form that vanishes only at $0$ is either positive-definite or negative-definite, so either $p$ or $-p$ is a Riemannian metric on $M$. (Note that the original metric $g$ plays no role in this, so you don't need to assume that $M$ is endowed with a Riemannian metric to begin with.)
Edit: What I wrote above is true if $M$ is connected. If it is disconnected, then in principle you might need to choose a different sign for $\pm p$ on each component of $M$.