On the diagonalizability of two-dimensional metrics

Assume that $M$ is a two dimensional manifold with (pseudo-)Riemannian metric $g$. In some local chart $(\tau,\sigma)$ we have $$g=A(\tau,\sigma)d\tau^2+B(\tau,\sigma)d\sigma^2+C(\tau,\sigma)(d\tau\otimes d\sigma+d\sigma\otimes d\tau ).$$

I am aware that this metric is always diagonalizable (in fact, all two dimensional metrics are conformally flat), however if we make the additional condition that we must keep the $\sigma$ coordinate fixed, I'm stuck.

It is clear that the metric is diagonized, if we find an exact form $dt$ which is orthogonal to $d\sigma$.

This orthogonality may be expressed in local coordinates as $$g^{ij}\partial_i t\partial_j \sigma=g^{i \sigma}\partial_i t=0.$$ Renaming $g^{i \sigma}=V^i$, this is just $V^i\partial_i t=0$. The functions $A,B,C$ are arbitrary (as long as they form a metric), so the inverse metric functions $g^{\tau\sigma},g^{\sigma\sigma}$ are essentially arbitrary.

So the problem is equivalent to asking whether an arbitrary smooth vector field (in 2 dimensions) has a (locally) exact annulator.

I can feel this is very easy to prove, but I am unable to. I have no idea how to show that the equation $$V^\tau(\tau,\sigma)\frac{\partial t}{\partial\tau}+V^\sigma(\tau,\sigma)\frac{\partial t}{\partial \sigma}=0$$ is soluble.

Any other method is also welcome. I tried to look for situations in which Poincaré's lemma or Frobenius' theorem can be applied to help out - to no avail, but as I said, I think I'm missing something trivial and obvious.

• Isn't $\tau$ and r are too much similar? It really makes it harder to follow. – Lucky Nov 7 '17 at 18:56
• @Rookie Renamed $r$ to $\sigma$ and tensor component indices to latin letters. – Bence Racskó Nov 7 '17 at 19:04
• So please update the text. There's still at least one $r$. – Ted Shifrin Nov 7 '17 at 19:46
• You don't need anything as sophisticated as Frobenius - you can just choose $t$ to be constant on the integral curves of $V$. In this case $V$ is non-vanishing, so it's locally $\partial/\partial x$ for some chart $(x,y)$ and thus you can just take any smooth $t(y).$ – Anthony Carapetis Nov 8 '17 at 1:14
• @AnthonyCarapetis This helped a lot, thanks! I knew this was simple, but for some reason it eluded me. If you post this as an answer I'll upvote/accept. – Bence Racskó Nov 8 '17 at 15:55