# Normal tubular neighborhood theorem for semi(or pseudo)-riemannian manifolds

Suppose you have a manifold $M$ and a closed sub-manifold $A$, and let $g$ be a semi-riemannian metric,ie, $g_x$ defines a quadratic form on $T_xM$ such that $g_x(v,v)\ge0$, but $g_x(v,v)=0$ not necessarily imply $v=0$.
Also, and additional hypothesis (*) is: if $x\in A$, $g_x$ defines an inner product on $T_xM$.

I have the following question:

is it possible to prove a tubular neighborhood theorem is this case? I mean, find an open neighborhood of the zero section in the normal bundle of $A$ in $M$, and a diffeomorphism to an open neighborhood of $A$ in $M$. For example, I was thinking on using the exponential map, but I really don't know much about semi-riemannian metrics, and if the geodesics and the exponential are well defined, and if they could be used to prove what i want in this case.

Notice that: the additional hyp (*) allow us to define the normal bundle as the vectors of $T_xM$ wich are orthogonal to $T_xA$

• Yes, there is a tubular neighbourhood theorem. You do not technically need any kind of (pseudo) Riemann metric to prove the tubular neighbourhood theorem. In this context it's simply called the "topological tubular neighbourhood theorem". In particular you define the topological normal bundle to be the fibrewise quotient of the pull-back of the ambient tangent bundle by the submanifold's tangent bundle. – Ryan Budney Jan 16 '14 at 4:15
• Moreover, the different definitions of the normal bundle when you have any kind of extra structure like a Riemann metric are all isomorphic. – Ryan Budney Jan 16 '14 at 4:16

There are singular points in the resulting tubular non-neighborhood. A good example for honing your understanding is the adjoint action of $SL(2,\mathbb R)$.
In your case, the exponential slice theorem works okay, since your semi Riemannian metric is a Riemannian metric on an open neighborhood of the submanifold $A$, by your condition (*).