# Handling degenerate planes in pseudo-Riemannian geometry: impact on sectional curvature and comparison theorems

I've been studying Riemannian and pseudo-Riemannian manifolds and came across an intriguing point regarding the definition of sectional curvature in both geometries.

In pseudo-Riemannian geometry, for a manifold $$(M, g)$$ with Riemann curvature tensor $$R$$ and a nondegenerate plane $$\Pi \subset T_p M$$ spanned by independent vectors $$v, w \in T_p M$$, the sectional curvature of $$\Pi$$ is defined as:

$$\begin{equation*} K(\Pi):=K(v, w):=\frac{g(R(v, w) w, v)}{g(v, v) g(w, w)-g(v, w)^2} . \end{equation*}$$

Conversely, in Riemannian geometry, for a 2-plane $$\sigma$$ in the tangent space $$T_{p} \mathcal{M}$$ at a point $$p$$ in a manifold $$\mathcal{M}$$, spanned by vectors $$X_{p}$$ and $$Y_{p}$$, sectional curvature is defined as:

$$$$K_{p}(\sigma)=\frac{\left\langle R\left(X_{p}, Y_{p}\right) Y_{p}, X_{p}\right\rangle}{\left|X_{p}\right|^{2}\left|Y_{p}\right|^{2}-\left\langle X_{p}, Y_{p}\right\rangle^2}.$$$$

It's clear that in Riemannian geometry, the positive-definite nature of the metric ensures that the denominator in the sectional curvature definition is never zero. However, in pseudo-Riemannian geometry, situations where the sectional curvature's denominator becomes zero seem possible, especially in degenerate planes.

My question is: In pseudo-Riemannian geometry, when dealing with a degenerate plane (where the sectional curvature formula's denominator is zero), how should this situation be interpreted? Does this degeneracy impact the analysis of sectional curvature and the application of comparison theorems? Are there any established methods or theoretical frameworks to address such scenarios in the context of pseudo-Riemannian manifolds? Any insights or references to relevant literature would be greatly appreciated.

How should this situation be interpreted?

Why should there be any interpretation? The geometric intuition behind sectional curvature in the Riemannian setting is this: given a plane $$\Pi$$ in $$T_pM$$, consider the surface $$\exp_p \Pi$$, where $$\exp_p$$ is the exponential map based at $$p$$. The intrinsic scalar curvature of $$\exp_p \Pi$$ at $$p$$ is the value of the sectional curvature of the plane $$\Pi$$.

In the case where $$\Pi$$ is non-degenerate, then we know that near $$p$$ the surface $$\exp_p \Pi$$ is a pseudo-Riemannian manifold with the same signature as the signature of $$\Pi$$, and hence has a well-defined notion of scalar curvature.

In the case where $$\Pi$$ is degenerate, you would need to define a curvature for a 2 manifold with degenerate metric. But you run into a few problems:

• when the metric is degenerate, there may be multiple compatible torsion free linear connections (Koszul's formula cannot be used to solve for the connection from the metric).
• when the metric is degenerate, there is also no longer a canonical pairing between the tangent and cotangent spaces, and hence no canonical inverse metric.

So let's throw out the geometric intuition here, and think instead about the curvature operator.

The Riemann curvature on a pseudo-Riemannian manifold defines a quadratic form $$\mathfrak{R}$$ on the space of two-forms. In general, let $$V$$ be a vector space, and let $$\Omega$$ be any star-shaped set containing the origin. Then we can specify a two-homogeneous function $$f$$ on $$V$$ by specifying its value on $$\partial\Omega$$. In the case of Riemannian geometry, it happens that the relevant vector space $$V = \Lambda^2M$$ is equipped with an inner product (induced from the Riemannian one). So we can preferentially choose $$\partial\Omega$$ to be the unit sphere with respect to the inner product. The resulting value along $$\partial\Omega$$ is the sectional curvature.

In the non-Riemannian setting, the "unit sphere" does not cover all possible directions (the degenerate planes are missing). On the other hand, given any star-shaped $$\Omega$$ in $$\Lambda^2_pM$$ you can still define a scalar function corresponding to the value of the quadratic form $$\mathfrak{R}$$ on $$\partial\Omega$$.

Perhaps an interpretation is this: the Riemann curvature tensor has unit "length squared". To convert it to a scalar you need to evaluate it on something that has with the inverse units. Degenerate planes have no invariant notion of size (or rather, they have all size $$0$$), so there is no natural way to select a representative "unit" degenerate plane. But this doesn't mean that you cannot think about the quadratic form $$\mathfrak{R}$$, just that you cannot think about a normalized version.

Does this degeneracy impact the analysis of sectional curvature and the application of comparison theorems?

For general pseudo-Riemannian manifolds, I am not aware of anything that looks remotely like a comparison theorem.

Part of the difficulty is simply algebraic. Given a vector space $$V$$ equipped with a positive definite inner product $$I$$. Given any quadratic form $$\omega$$ on $$V$$, there exists $$\lambda, \Lambda\in \mathbb{R}$$ such that $$\lambda I \leq \omega \leq \Lambda I$$ where for quadratic forms $$A \leq B$$ means $$B-A$$ is positive definite.

Combine this with the fact that model spaces (hyperbolic space, plane, and sphere) have curvature of the form $$k I$$ you now have the beginning of comparison theorems.

In the pseudo-Riemannian case, the scalar product $$I$$ induced by the metric on $$V = \Lambda^2 M$$ is indefinite, and model spaces (plane, de Sitter, anti de Sitter) also have curvature of the form $$k I$$. But now it is not the case that every quadratic form $$\omega$$ on $$V$$ can be bounded between $$\lambda I$$ and $$\Lambda I$$ (well, in fact, if $$\lambda_1\neq \lambda_2$$, then $$\lambda_1 I$$ and $$\lambda_2 I$$ are not comparable....)

So I don't think the reason that we don't have general comparison theorems in pseudo-Riemannian geometry can be solely attributed to sectional curvature not being well-defined for degenerate planes.

Are there any established methods or theoretical frameworks to address such scenarios in the context of pseudo-Riemannian manifolds?

If you are interested in only Lorentzian cases, however, there is a general strategy available, which has seen some success. The basic idea is this: given a fixed time-like vector $$v$$, then its orthogonal complement has a positive definite inner product. So, for example, if instead of looking at all two-forms $$\omega$$, you look only at those two-forms of the form $$\tau \wedge \eta$$ where $$\tau$$ is a fixed time-like co-vector and $$\eta$$ is arbitrary, then this subspace has a positive-definite inner product and you can deal with "sectional curvature" in this setting.

One major success of this line of ideas is the Lorentzian splitting theorem; there are also plenty of developments in the study of timelike geodesics using this idea, based on the fact that variation fields of time-like geodesics are space-like vector fields and we can regain the positive definite structure here. There are also various comparison theorems established for the proper-time function (actually a partial function, only defined for points that can be joined by a time-like curve).

Of course, the previous idea basically just gives up on thinking about degenerate things, instead studying a case where the degenerate things don't bother us.

For looking at degenerate planes, notice that while we can no longer sensibly compare the sectional curvature to that of a "model geometry", due to us not having a correct sense of "scale" in the degenerate direction, we can still form comparisons that are invariant under scaling. Namely, while you cannot say what the value of the sectional curvature is for a degenerate $$\Pi$$, it still makes sense to talk about whether $$\mathfrak{R}(\Pi,\Pi)$$ is positive, negative, or zero. This can, for example, be used to study the geometry of null geodesics, which leads to, for example, the Penrose singularity theorem.

• Thank you for your insightful explanation! It greatly clarified the topic for me. Much appreciated! Commented Mar 19 at 15:02