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Is there any paper about the $\pi_1$ group and curvature ?

Because how close a curve depends on the curvature near the curve . I think there must have some condition which decide whether there is closed curve on manifold. Besides, curvature decide the genus.So, I guess there should be some connection between fundamental group and curvature.

If there are any unclear , sorry for my poor English.

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    $\begingroup$ The following sentence is a bit reductionist, but fun to write: there is a whole field on the relation between the topology of a space and its curvature, known as differential geometry. $\endgroup$ – user98602 Jun 13 '16 at 2:27
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Here are four theorems from Riemannian geometry that might interest you. All of them can be found (with proofs) in Chapters 9 and 12 of Manfredo do Carmo's book "Riemannian Geometry."

Bonnet-Myers Theorem: Let $(M,g)$ be a complete Riemannian manifold with $\text{Ric}(g) \geq k > 0$ for some constant $k > 0$. Then $M$ is compact and $\pi_1(M)$ is finite.

Synge Theorem: Let $(M^{2n},g)$ be a compact Riemannian manifold that is orientable, even-dimensional, and having positive sectional curvature. Then $\pi_1(M) = 0$.

Theorem (Preissman): Let $(M,g)$ be a compact Riemannian manifold with negative sectional curvature. Then $\pi_1(M)$ is not abelian.

Theorem (Byers): Let $(M,g)$ be a compact Riemannian manifold with negative sectional curvature. Then every solvable subgroup of $\pi_1(M)$ is either the identity itself or is infinite cyclic. Moreover, $\pi_1(M)$ has no cyclic subgroup of finite index.

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    $\begingroup$ I read too little. $\endgroup$ – lanse7pty Jun 13 '16 at 2:57

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