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Hello;

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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  • $\begingroup$ Do you want to impose to have a "convex" cone? otherwise this seems obvious. $\endgroup$
    – Reimundo Heluani
    Commented Sep 29, 2011 at 0:30
  • $\begingroup$ Yes Reimundo, the question is whether it is convex or not. $\endgroup$
    – S.A.A
    Commented Sep 29, 2011 at 0:34
  • $\begingroup$ I added the word convex to the question. Thanks. $\endgroup$
    – S.A.A
    Commented Sep 29, 2011 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
    Commented Sep 29, 2011 at 8:26

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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.

On the other hand, since you are interested in metrics with positive sectional curvature at one point. If write the metric is written in the normal coordinates at this point, then the space of germs of Riemannian metrics (at this point) forms a cone. (By the way, the same holds for negative curvature.)

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  • $\begingroup$ What's the easiest counterexample for this? $\endgroup$
    – Deane Yang
    Commented Sep 29, 2011 at 1:27

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