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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.

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Do you want to impose to have a "convex" cone? otherwise this seems obvious. – Reimundo Heluani Sep 29 '11 at 0:30
Yes Reimundo, the question is whether it is convex or not. – S.A.A Sep 29 '11 at 0:34
I added the word convex to the question. Thanks. – S.A.A Sep 29 '11 at 0:35
The following reference does not answer your question, but seems relevant: 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. – Benoît Kloeckner Sep 29 '11 at 8:26
up vote 16 down vote accepted

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.

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What's the easiest counterexample for this? – Deane Yang Sep 29 '11 at 1:27
@Deane, I include a source for such examples. – Anton Petrunin Sep 29 '11 at 1:45
Thanks! I know impressively little about these things. – Deane Yang Sep 29 '11 at 1:58

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