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Let $(M,g)$ be a compact manifold of dimension $n\in\mathbb{N}$. Is there any formula (or is it possible) to find the order of the fundamental group, $\pi_1(M)$ using the metric of the manifold in general?

Or in other words, how can we conclude if a manifold is multiply-connected i.e. $\text {order}\left[\pi_1(M)\right] \ge 2$ (see this) using the metric $g$ in general?

I know there are some theorems regarding the properties of fundamental group of a Riemannian manifold, as stated in "Fundamental group and curvature", but I'm searching for any theorem regarding the direct relation between the order of $\pi_1(M)$ and the metric.

Context: As Gauss-Bonnet theorem connects the geometry and the Euler characteristic (a topological invariant) of a compact $2$-dimensional Riemannian manifold, likewise there should be an other theorem which will connect the order of the fundamental group (again a topological invariant) and the metric (geometry) of that manifold.

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If a compact manifold has curvature $K\leq 0$, the Cartan-Hadamard theorem (https://en.wikipedia.org/wiki/Cartan%E2%80%93Hadamard_theorem) tells us that its universal cover is diffeomorphic to $\bf R ^n$, hence non compact so that its fundamental group is infinite.

For finite groups I dont think that such can exists. Consider for instance lens spaces $S^{2n-1}/C_k$ obtained by taking the quotient of the unit sphere of $C^n$ by a cyclic group of order $k$ acting diagonnaly. These spaces have all curvature equal to $+1$, and fundamental group cyclic of order $k$.

More generally a finite cover of a manifold has same curvature bounds, but a fundamental group of smaller order, whenever finite.

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    $\begingroup$ Another connection between geometry and topology is that a positive lower bound on the Ricci tensor yields compactness and finiteness of the fundamental group $\endgroup$
    – Didier
    Commented Jul 18, 2023 at 8:31
  • $\begingroup$ Just a doubt: If for $K\le0$, the fundamental group is infinite, then the manifold can never be simply-connected, right? As simply-connected manifold, by definition, needs trivial fundamental group i.e. of $\text{order}\left[\pi_1\right] = 1$. Thus any manifold with non-positive sectional curvature has to be multiply-connected. Is it the case always? $\endgroup$
    – SCh
    Commented Jul 18, 2023 at 9:13
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    $\begingroup$ Certainly, if you add the word "compact". $R^n$ is a simply connected manifold with $k=0$. it is not compact. The universal cover a a compact manifold with non positive curvature is always diffeomrophic to $R^n$. $\endgroup$
    – Thomas
    Commented Jul 18, 2023 at 12:00

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