Do pp-wave spacetimes have a well-defined signature/index?

Question

A pp-wave spacetime in Brinkmann coordinates has metric $$ ds^2 = H(u,x,y) \, du^2 + 2 \, du \, dv + dx^2 + dy^2 $$ and is asserted to be a Lorentzian manifold, i.e. has index 1.

This is indeed true assuming assuming that $H(u, x, y) \ne 0$. However, when $H(u, x, y) = 0$, then the metric looks like $$ ds^2 = 2 \, du \, dv + dx^2 + dy^2 $$ which has index 0, since this can be transformed to $$ ds^2 = du'^2 + dv'^2 + dx^2 + dy^2\text{.} $$

It's stated that $H$ can be any smooth function, and the Brinkmann coordinates article specifically mentions that "The coordinate vector field $\partial_u$ can be spacelike, null, or timelike at a given event in the spacetime, depending upon the sign of $H(u, x, y)$ at that event."

But doesn't this conflict with the definition of a semi-Riemannian manifold, which requires that the metric have constant index everywhere? If so, what kind of manifold exactly is a pp-wave spacetime?

Answer 1

I think assuming that $H \neq 0$, then there is no problem with $\partial_u$ changing causal character. Clearly $\partial_x$ and $\partial_y$ are both unit fields, and orthogonal to $\partial_u$ and $\partial_v$. But $\partial_v$ is lightlike, and $\partial_u$ and $\partial_v$ are not orthogonal, in general. It is analogous to the situation where, say, in Lorentz-Minkowski space you can take a basis of a timelike plane consisting of two spacelike vectors (which are necessarily non-orthogonal).

In the case where $H = 0$, I think you have a mistake in the calculation. Can't we transform $2\,du\,dv$ into $du'^2-dv'^2$ instead? Think of lightlike coordinates in $\Bbb R^2_1$.

Either way, the index will be $1$.