We know that the space of riemannian metrics on a compact manifold is an open cone in the space of symmetric 2-tensors. Is it reasonable to think that metrics with positive sectional curvature (even positive at a specific point $x \in M$) also form a convex cone?

This is a question about the local behaviour of metrics, so I am not imposing the condition that the sectional curvature be positive everywhere.

Also, due to certain corollary of this statement for a class of metrics, I am quite certain that this cannot hold for metrics of negative curvature.

  • $\begingroup$ Do you want to impose to have a "convex" cone? otherwise this seems obvious. $\endgroup$ – Reimundo Heluani Sep 29 '11 at 0:30
  • $\begingroup$ Yes Reimundo, the question is whether it is convex or not. $\endgroup$ – S.A.A Sep 29 '11 at 0:34
  • $\begingroup$ I added the word convex to the question. Thanks. $\endgroup$ – S.A.A Sep 29 '11 at 0:35
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    $\begingroup$ The following reference does not answer your question, but seems relevant: front.math.ucdavis.edu/0907.2444 There Fernando Coda Marques proves that the space of positive scalar curvature metrics on a given orientable 3-dimensional manifold is path-connected. It makes essential use of the Ricci flow and the work of Perelman. $\endgroup$ – Benoît Kloeckner Sep 29 '11 at 8:26

No, the formula for curvature is nonlinear with respect to metric tensor in a very essential way.

In particular, a convex combination of two positively curved metrics can have negative curvature. In fact, arbitrary large negative sectional curvature.

For example, the induced metric on any embedding $\mathbb{S}^2\hookrightarrow\mathbb{S}^2\times \mathbb{S}^2$, such that both projections are diffeomorphisms is a convex combination of metrics with constant curvature 1/2. But it is not hard to make such metric arbitrary bad.

  • $\begingroup$ What's the easiest counterexample for this? $\endgroup$ – Deane Yang Sep 29 '11 at 1:27

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