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.

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

1 Answer 1

Reset to default
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.

Improve this answer
$\endgroup$
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
    Commented Feb 19, 2021 at 13:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .