2
$\begingroup$

Let $(M,g)$ be a Lorentzian manifold and let $R$ be the curvature tensor. We say $R\leq 0$ if $$ g(R(X,Y)Y,X) \leq 0\quad \forall \, X,Y \in TM.$$

My question is whether given a Lorentzian manifold $(M,g)$, it is always possible to find a metric $\hat{g}=cg$ such that the curvature of $(M,\hat{g})$ is non-positive.

$\endgroup$
1
  • 1
    $\begingroup$ I suppose you want $X,Y$ to be sections, not elements, of $TM$. What is $c$? $\endgroup$
    – S.Surace
    Jun 3, 2020 at 12:24

1 Answer 1

3
$\begingroup$

No, this is not possible if the Lorentzian manifold has a null geodesic admitting a pair of conjugate points. The notion of conjugate points for null geodesics does not depend on the conformal factor. By the Lorentzian version of the Cartan-Hadamard theorem (see e.g. the book by Beem et al. 1996 Global Lorentzian Geometry. Prop. 11.13) in a spacetime if the inequality you wrote were to hold for any pair of timelike vectors then there would not be causal geodesics with conjugate points.

$\endgroup$
1
  • $\begingroup$ It makes me wonder if this is the only obstruction, namely if the inequality can be achieved if one removes null conjugate and cut points. $\endgroup$
    – Ali
    Feb 19, 2021 at 13:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.