Suppose that $(M,g)$ is a Lorentzian manifold of signature $(-,+,\ldots,+)$. Given a two plane $\Pi=\textrm{Span}\{X,Y\}$ with $X,Y \in T\_pM$, we say that $\Pi$ is non-degenerate if
$$ g(X,X)g(Y,Y)-g(X,Y)^2 \neq 0.$$
Moreover, given a non-degenerate two-plane we say that it is timelike or spacelike if the above quantity is negative or positive respectively. Finally, the sectional curvature $\textrm{Sec}(\Pi)$ for a non-degenerate two plane $\Pi$ as above is defined by
$$\textrm{Sec}(\Pi)=\frac{g(R(X,Y)X,Y)}{ g(X,X)g(Y,Y)-g(X,Y)^2 }.$$


**Question.** Suppose that given any non-degenerate space-like two plane in $(M,g)$ its sectional curvature is bounded from above by a fixed $K\_1<0$ and that given any non-degenerate time-like two plane in $(M,g)$ its sectional curvature is bounded from below by a fixed $K\_2>K\_1$. Suppose that $\widetilde{g}$ is another Lorentzian metric on $M$ that is obtained by adding a sufficiently small smooth compactly supported tensor $h$ to $g$. Is it true that there exists constants $K\_1'$ and $K\_2'$ such that similar sectional curvature bounds hold for the perturbed metric $\widetilde g$ with the new constants $K\_1'$ and $K\_2'$ in place of $K\_1$ and $K\_2$?



stability of two-sided sectional curvature bounds in Lorentzian geometry
To add on to [Ettore Minguzzi's answer](https://mathoverflow.net/a/385541/3948):


* If $g(R(X,Y)X,Y) < 0$ for all pairs $X,Y$ as he defined, then since this is an open condition stability is automatic for small smooth compactly supported perturbations.
* If $g(R(X,Y)X,Y) = 0$ for some pair, then stability can be violated.


This can be seen from taking $\tilde{g} = e^{2\phi} g$ a small conformal change, where $\phi\in C^\infty\_0$ is small. We will choose further that $\phi$ to have a critical point at the base of $X,Y$


Since $\tilde{g}$ is conformal, $X,Y$ are still a pair with the same properties. Our goal is to try to make the corresponding curvature tensor
$ \tilde{g}(\tilde{R}(X,Y)X,Y) > 0$. From [well-known formulas](https://en.wikipedia.org/wiki/List_of_formulas_in_Riemannian_geometry#(4,0)_Riemann_curvature_tensor_2) since we chose $\nabla\phi = 0$ at the point of interest, we have
$$ \tilde{g}(\tilde{R}(X,Y)X,Y) = (-\tilde{g}\odot \nabla^2\phi)(X,Y,X,Y)$$
where $\odot$ is the Kulkarni-Nomizu product.
The orthogonality properties of $X$ and $Y$ further implies (I'm too lazy to figure out the currect sign)
$$ \tilde{g}(\tilde{R}(X,Y)X,Y) = \pm\tilde{g}(X,X) \nabla^2\_{Y,Y}\phi $$


So choosing $\phi$ suitably convex/concave in the direction of $Y$ you can make the quantity positive.


In particular, applying this construction to Minkowski space gives you a counterexample to stability (of Minguzzi's condition).




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Returning to your actual question: it turns out that the case $g(R(X,Y)X,Y)=0$ is automatically ruled out for smooth manifolds.


Using Minguzzi's analysis, consider
$$ \rho(t) = g(R(X, Y+tU)X, Y+tU)$$
and
$$ \sigma(t) = g(X,X)g(Y+tU,Y+tU) - g(X, Y+tU)^2$$
Observe that $\sigma(t)$ vanishes at $t = 0$ to only *first order*.


The requirement that $\rho(t) / \sigma(t) > K\_2$ when $t > 0$ and $\rho(t) / \sigma(t) < -K\_1$ for $t < 0$ requires that $\rho$ is strictly negative when $t \neq 0$. So if $\rho(0) = 0$ this would make it a critical point and hence vanish to second order.


But this would mean that $\lim\_{t\to 0} \rho(t)/\sigma(t) = 0$ which is a contradiction.


Therefore in the end, we find that you must be in the situation where stability holds.


