Suppose there are two connections over the tangent bundle of a smooth manifold with the same geodesics. What is the relation between the curvature tensors, the Ricci curvature and the scalar curvature of these two connections? Note that the question gives no address to the metric, such as Riemannian or the Levi-Civita.
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1$\begingroup$ See this question and its answer. $\endgroup$– abxCommented Aug 27, 2016 at 8:25
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4$\begingroup$ There is a subtle point: what, exactly, "the same geodesics" means. If these are the same geodesics with the same parameter, then it simply means that two connections are different by an antisymmetric tensor. $\endgroup$– Alex GavrilovCommented Aug 27, 2016 at 9:10
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$\begingroup$ @abx: That question is about flat projective connections. The question is about general metric projective connections. $\endgroup$– Ben McKayCommented Aug 27, 2016 at 10:26
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$\begingroup$ @Ben McKay: Right, I should have been more precise. I had in mind the comment by Robert Bryant to that question: "More generally, two connections are projectively equivalent if and only if they have the same (unparametrized) geodesics". $\endgroup$– abxCommented Aug 27, 2016 at 13:05
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You are looking for the theory of projective connections. See the paper of Molzon and Mortensen, The Schwarzian derivative for maps between manifolds with complex projective connections, Trans. Amer. Math. Soc., Volume 348, Number 8, August 1996 which is surely the best survey. It treats the complex case, but the details are the same as for the real case. You might also look at the paper of Kobayashi and Nagano, On projective connections.
answered Aug 27, 2016 at 8:43
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1$\begingroup$ The main examples: Euclidean space, hyperbolic space, and the sphere with one point deleted are all diffeomorphic by maps preserving unparameterized geodesics (projectively equivalent). So clearly scalar curvature changes. In fact, Ricci also changes, and I think there is very little you can say about how, but I think that the traceless part of the curvature tensor is invariant. $\endgroup$ Commented Aug 27, 2016 at 17:53