# stability of two-sided sectional curvature bounds in Lorentzian geometry

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 $$K_2$$?

• Non-degerate space-like two plane has positive curvature by definition. How can it be bounded above by $K_1<0$? Mar 4, 2021 at 22:55

• 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 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).

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.

Your condition is somewhat restrictive, and might not be what you want. Let $$Y$$ be a future directed lightlike vector and let $$X$$ be a spacelike vector orthogonal to it. Consider $$Y'=Y+tU$$ where $$U$$ is a unit future directed timelike vector. Then for $$t<0$$ with $$\vert t\vert$$ sufficiently small, $$\textrm{Span}(X,Y')$$ is spacelike, that is, $$g(X,X)g(Y',Y')-g(X,Y')^2$$ is positive actually going to zero for $$t\to 0$$. For $$t>0$$, $$\textrm{Span}(X,Y')$$ is timelike and $$g(X,X)g(Y',Y')-g(X,Y')^2$$ is negative actually going to zero for $$t\to 0$$. Thus if $$g(R(X,Y)X,Y)> 0$$ the 'spacelike' sectional curvature for $$t\to 0-$$ goes to plus infinity while in the 'timelike' sectional curvature for $$t\to 0+$$ goes to minus infinity, so the two bounds cannot be satisfied. We conclude that $$g(R(X,Y)X,Y)\le 0$$ for a generic pair as above. Maybe from here one can obtain further constraints on the Riemann tensor.