Background: Recently I've read this post of one person trying to prove that the pullback $F^*g$ of a riemannian metric $g$ is a Riemannian metric iff $F: N \to (M,g)$ is a smooth immersion.
Question: Now assuming that $M$ and $N$ are both non-compact and that $M$ is a Lorentzian manifold instead of a Riemannian one, I'm interested if there is any result that gives conditions that make $F^* g$ non-degenerate only . Does it change something if $N$ becomes a non-compact (smooth) manifold with boundary?
The main problem to me is that in the Riemannian case the non-generacy case is equivalent to being positive definite. Positivity of the form drops down in the Lorentzian case, so non-degeneracy is a bit more general. In particular I can't prove non-degeneracy by only requiring that the differential $dF$ is injective at every point of $M$.
Best Answer
In general you cannot expect such a result for a non-positive form. This is really due to linear algebra:
Sylvesters theorem of inertia tells us that without loss of generality, we may assume that any bilinear form on a $d$ dimensional real vector space can be taken to be $(\mathbb{R}^d, Q_{q,p,n-p-q})$ where $Q_{q,p,r}$ is the bilinear form with $q$, $p$, and $r$, $1$'s, $-1$'s and $0$'s along the diagonal respectively. The pointwise version of your question is: if $V\subset \mathbb{R}^d$ a subspace, under what conditions can we show that $Q_{d-1, 1,0}\vert_{V}$ is of a particular type (positive definite, Lorentzian, etc.)? By the decomposition, it is easy to see that a subspace may be taken in the form $(\mathbb{R}^{d}, Q_{q,p,0})$ for $p\leq 1$ and $q+p\leq d$ by taking $V$ to be spanned by the first $q$ columns and the final column if $p=1$. We can even make the form non-definite by taking $V=\mathrm{span}(e_i+e_d)$ for $i\leq d-1$. The local version of Ivo Terek's comment says that if $V$ is codimension $1$ and its orthogonal complement is a negative definite subspace, $V$ is Lorentzian, and if $V$'s compliment is negative definite, $V$ is positive definite.
In the nondegenerate cases, the preceding argument guarantees that one can find nonproper submanifolds with induced quadratic form of a desired type. A natural way to investigate this type of question further may be through the lens of distributions. The problem then becomes: Given a semi-Riemannian manifold $(M,g)$, does there exist a distribution $\xi\subset TM$ of a desired type, which is integrable. Answering this question in the positive means that the entirety of $M$ is foliated by submanifolds of a given type (think maximal spacelike hyperplanes in $\mathbb{R}^{d-1,1}$)