# Order of the fundamental group from the metric of a Riemannian manifold

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.

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 $$k$$.
• 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?
• 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 $R^n$. Commented Jul 18, 2023 at 12:00