Significant differences between Riemannian and pseudo-Riemannian submanifolds In Chapter 8 of his Introduction to Riemannian Manifolds, Jack Lee discusses (Riemannian) submanifolds. In particular, in the first section, which is titled The Second Fundamental Form and presents the fundamental equations of submanifold geometry (Weingarten, Gauss, Codazzi), he writes:

The results in the first section of this chapter apply virtually without modification
to Riemannian submanifolds of pseudo-Riemannian manifolds (ones on which the
induced metric is positive definite), so we state most of our theorems in that case. $\ldots$ Some of the results can also be extended to pseudo-Riemannian submanifolds of mixed signature, but there are various pitfalls to watch out for in that case; so for simplicity we restrict to the case of Riemannian submanifolds.

This makes me wonder where exactly these "pitfalls" are. In fact, by comparing with Chapter 4 in O'Neill's Semi-Riemannian Geometry With Applications to Relativity, I fail to see immediate differences between the two theories.
So my question is, what significant differences are there (both in terms of results and proofs thereof) between Riemannian and pseudo-Riemannian submanifolds?
EDIT: This is not a duplicate of What significant differences are there between a Riemannian manifold and a pseudo-Riemannian manifold?. I am asking about differences between a submanifold of a pseudo-Riemannian manifold whose induced metric is positive definite and one whose induced metric is merely nondegenerate. I am not interested in comparing them at the level of manifolds, but specifically as submanifolds. To reiterate, I want to understand why the material presented in (the first section of) Chapter 8 of Lee's book cannot be extended more or less trivially to pseudo-Riemannian submanifolds.
 A: Here is a result that is probably closer to what you are asking, instead of the links in the comments (which are mostly only geodesics and thus the previous chapter in Lee's book):
Recall the sectional curvature $K$ is only defined on non-degenerate tangent $2$-planes in pseudo-Riemannian case, as we need to divide by $g(v,v)g(w,w)-g(v,w)^2$.  This means the domain of $K_p$ is a noncompact open subset of $\operatorname{Gr}_2(T_pM)$, and hence as a result we can't expect it to be bounded in general (unlike Riemannian case $\operatorname{dom}K_p=\operatorname{Gr}_2(T_pM)$ is compact).  In fact:
Theorem (Kulkani 1979, Nomizu 1983): Let $M$ be a pseudo-Riemannian manifold of dimension $\geq 3$ and index $>0$ (i.e., $g$ is indefinite).  Then at each point $p\in M$, TFAE:

*

*$K_p$ is constant

*$K_p$ is bounded below

*$K_p$ is bounded above

*$K_p$ is bounded on indefinite planes

*$K_p$ is bounded on definite planes

Corollary: The sectional curvature of such $M$ is unbounded from above or below, unless the manifold has constant sectional curvature.
So a lot of naive analogue of comparison theorems involving sectional curvature become completely trivial in pseudo-Riemannian case (e.g., $\frac14$-pinching theorem).  In particular, relating back to the question of submanifolds:  there are results (collectively called Chen's inequalities) for Riemannian case bounding intrinsic and extrinsic curvature invariants that allow you to say something more for, e.g., minimal submanifolds, but the same is much harder, if not downright impossible, in pseudo-Riemannian case.
A: Here's one difference that comes to mind. Let $(M,g)$ be a pseudo-Riemannian manifold, and $S\subseteq M$ be a non-degenerate hypersurface. Assume that one has a normal vector field $N$ along $S$. Then, we can always talk about the shape operator at each point $x\in S$, that is, $A_x\colon T_xS \to T_xS$ given by $A_x(v) = -(\nabla_vN)^\top$, where $\nabla$ is the Levi-Civita connection of $(M,g)$. On the level of vector fields, this is characterized by the relation $$g(A(X),Y) = g(\sigma(X,Y),N),$$where $\sigma$ is the (vector-valued) second fundamental form of $S$ relative to $M$. As $\sigma$ is symmetric, $A$ is self-adjoint relative to $g$.
Now, you would love to say that at each point, $A_x$ is now diagonalizable relative to a $g_x$-orthonormal basis of $T_xS$, right? But... no way José, this is generally false. If $S$ is a spacelike hypersurface (i.e., $g$ restricted to $S$ is Riemannian), sure, but this can go horribly wrong if $S$ is not spacelike.
Milnor's version of the spectral theorem kicks in: if $(V,g)$ is a pseudo-Euclidean vector space with dimension greater or equal to $3$, and $T\colon V\to V$ is a self-adjoint operator, such that $g(Tv,v) \neq 0$ for all lightlike $v$, then $T$ is diagonalizable relative to a $g$-orthonormal basis.
Morally, the possibility of the operator having lightlike eigenvectors is what makes things go wrong. This means that one can no longer talk about principal directions and principal curvatures, in general. For surfaces, this version of the spectral theorem cannot be applied, but there's still something we can say about it.
Consider Lorentz-Minkowski space $\mathbb{L}^3 = (\mathbb{R}^3, {\rm d}x^2+{\rm d}y^2 - {\rm d}z^2)$. For a non-degenerate surface $S\subseteq \mathbb{L}^3$, write $\varepsilon$ for the indicator of the normal direction to $S$, i.e., $\varepsilon = -1$ if $S$ is Riemannian, and $\varepsilon = 1$ if $S$ is Lorentzian. Then:
Theorem: If $S\subseteq\mathbb{L}^3$ is a non-degenerate surface equipped with a unit normal field $N$, $p\in S$ is any point, $K(p)$ and $H(p)$ stand for the Gaussian and mean curvatures of $S$ at $p$, and we let $D(p) = H(p)^2 -\varepsilon K(p)$, then we have that:

*

*if $D(p)>0$, then $-{\rm d}N_p$ is diagonalizable.


*if $D(p) < 0$, then $-{\rm d}N_p$ is not diagonalizable.


*if $D(p)=0$ and $S$ is Riemannian, then $p$ is umbilic (hence $-{\rm d}N_p$ is diagonalizable).


*if $D(p) = 0$ and $S$ is Lorentzian, then $-{\rm d}N_p$ may be diagonalizable, or it may not be diagonalizable.


*if $D(p)= 0$, $S$ is Lorentzian, and $-{\rm d}N_p$ has no lightlike eigenvectors, then $p$ is umbilic and $-{\rm d}N_p = -H(p){\rm Id}_{T_pS}$.
I highly suspect that this must be true replacing $\mathbb{L}^3$ with any $3$-dimensional Lorentzian manifold, but I never bothered to check.
