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 $T\_xA$



Normal tubular neighborhood theorem for semi(or pseudo)-riemannian manifolds
For a pseudo-Riemannian manifold, the local version of the exponential slice theorem is the Luna slice theorem,


* MR0342523 (49 #7269) 
Luna, Domingo
Slices étales. (French) Sur les groupes algébriques, pp. 81–105. Bull. Soc. Math. France, Paris, Memoire 33 Soc. Math. France, Paris, 1973.


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 (\*).


