[proofplan] We pass to the universal Riemannian covering space and equip it with the lifted metric. Because the covering projection is a local isometry, the Ricci lower bound and completeness lift to the universal cover. The Bonnet-Myers theorem then makes the universal cover compact. Finally, $\pi_1(M)$ acts on the universal cover as the deck transformation group; an infinite deck group would give an infinite properly discontinuous orbit in a [compact space](/page/Compact%20Space), which is impossible. [/proofplan] [step:Lift the Riemannian structure and Ricci lower bound to the universal cover] Let \begin{align*} \pi:\widetilde M \to M \end{align*} be the universal covering map of the connected manifold $M$. Define the lifted Riemannian metric $\widetilde g$ on $\widetilde M$ by \begin{align*} \widetilde g_{\widetilde p}(\widetilde v,\widetilde w) := g_{\pi(\widetilde p)}\bigl(d\pi_{\widetilde p}(\widetilde v),d\pi_{\widetilde p}(\widetilde w)\bigr) \end{align*} for every $\widetilde p \in \widetilde M$ and every $\widetilde v,\widetilde w \in T_{\widetilde p}\widetilde M$. Since $\pi$ is a covering map, the differential \begin{align*} d\pi_{\widetilde p}:T_{\widetilde p}\widetilde M \to T_{\pi(\widetilde p)}M \end{align*} is a linear isomorphism in local coordinates, so $\widetilde g$ is a Riemannian metric and $\pi:(\widetilde M,\widetilde g)\to(M,g)$ is a local isometry. The Ricci tensor is invariant under local isometries. Therefore, for every $\widetilde p \in \widetilde M$ and every $\widetilde v \in T_{\widetilde p}\widetilde M$, \begin{align*} \operatorname{Ric}_{\widetilde g}(\widetilde v,\widetilde v) &= \operatorname{Ric}_{g}\bigl(d\pi_{\widetilde p}(\widetilde v),d\pi_{\widetilde p}(\widetilde v)\bigr) \\ &\geq (n-1)K\, g_{\pi(\widetilde p)}\bigl(d\pi_{\widetilde p}(\widetilde v),d\pi_{\widetilde p}(\widetilde v)\bigr) \\ &= (n-1)K\, \widetilde g_{\widetilde p}(\widetilde v,\widetilde v). \end{align*} Thus $(\widetilde M,\widetilde g)$ satisfies the same positive Ricci lower bound. [guided] We first move the problem from $M$ to its universal cover because the fundamental group appears naturally as the deck transformation group of that cover. Let \begin{align*} \pi:\widetilde M \to M \end{align*} be the universal covering map. We define a Riemannian metric $\widetilde g$ on $\widetilde M$ by pulling back $g$ through the differential of $\pi$: \begin{align*} \widetilde g_{\widetilde p}(\widetilde v,\widetilde w) := g_{\pi(\widetilde p)}\bigl(d\pi_{\widetilde p}(\widetilde v),d\pi_{\widetilde p}(\widetilde w)\bigr), \end{align*} where $\widetilde p \in \widetilde M$ and $\widetilde v,\widetilde w \in T_{\widetilde p}\widetilde M$. This formula is meaningful because the differential \begin{align*} d\pi_{\widetilde p}:T_{\widetilde p}\widetilde M \to T_{\pi(\widetilde p)}M \end{align*} is a linear isomorphism: locally, a covering map is a diffeomorphism onto an evenly covered neighbourhood. Hence $\widetilde g$ is positive definite, symmetric, and smooth. By construction, $\pi:(\widetilde M,\widetilde g)\to(M,g)$ is a local isometry. The reason this definition is useful is that curvature tensors are preserved by local isometries. In particular, Ricci curvature is transported exactly through $d\pi_{\widetilde p}$. Therefore, for every point $\widetilde p \in \widetilde M$ and tangent vector $\widetilde v \in T_{\widetilde p}\widetilde M$, \begin{align*} \operatorname{Ric}_{\widetilde g}(\widetilde v,\widetilde v) &= \operatorname{Ric}_{g}\bigl(d\pi_{\widetilde p}(\widetilde v),d\pi_{\widetilde p}(\widetilde v)\bigr). \end{align*} Now we use the hypothesis on $(M,g)$, applied to the tangent vector $d\pi_{\widetilde p}(\widetilde v) \in T_{\pi(\widetilde p)}M$: \begin{align*} \operatorname{Ric}_{g}\bigl(d\pi_{\widetilde p}(\widetilde v),d\pi_{\widetilde p}(\widetilde v)\bigr) &\geq (n-1)K\, g_{\pi(\widetilde p)}\bigl(d\pi_{\widetilde p}(\widetilde v),d\pi_{\widetilde p}(\widetilde v)\bigr). \end{align*} Using the definition of $\widetilde g$, the right-hand side is \begin{align*} (n-1)K\, \widetilde g_{\widetilde p}(\widetilde v,\widetilde v). \end{align*} Thus \begin{align*} \operatorname{Ric}_{\widetilde g}(\widetilde v,\widetilde v) \geq (n-1)K\, \widetilde g_{\widetilde p}(\widetilde v,\widetilde v), \end{align*} so the positive Ricci lower bound lifts unchanged to the universal cover. [/guided] [/step] [step:Show that the lifted metric is complete] Let \begin{align*} \widetilde\gamma:[0,b) \to \widetilde M \end{align*} be a unit-speed geodesic for $\widetilde g$, where $0<b<\infty$. The composed curve \begin{align*} \gamma:[0,b) \to M,\qquad \gamma(t):=\pi(\widetilde\gamma(t)) \end{align*} is a unit-speed geodesic for $g$, because $\pi$ is a local isometry. Since $(M,g)$ is complete, $\gamma$ extends to a geodesic \begin{align*} \bar\gamma:[0,b+\varepsilon) \to M \end{align*} for some $\varepsilon>0$. The path-lifting property for the covering map $\pi$ gives a unique lift \begin{align*} \overline{\widetilde\gamma}:[0,b+\varepsilon) \to \widetilde M \end{align*} of $\bar\gamma$ with $\overline{\widetilde\gamma}(0)=\widetilde\gamma(0)$. By uniqueness of lifted paths, $\overline{\widetilde\gamma}|_{[0,b)}=\widetilde\gamma$. Hence every finite-time geodesic in $(\widetilde M,\widetilde g)$ extends past its endpoint. Therefore $(\widetilde M,\widetilde g)$ is geodesically complete, and hence complete. [/step] [step:Apply Bonnet-Myers to make the universal cover compact] The manifold $(\widetilde M,\widetilde g)$ is connected, complete, and $n$-dimensional, and it satisfies \begin{align*} \operatorname{Ric}_{\widetilde g} \geq (n-1)K\,\widetilde g \end{align*} with $K>0$. By the Bonnet-Myers Theorem (citing a result not yet in the wiki: Bonnet-Myers Theorem), $\widetilde M$ has finite diameter, with \begin{align*} \operatorname{diam}(\widetilde M,\widetilde g) \leq \frac{\pi}{\sqrt K}. \end{align*} In particular, the complete Riemannian manifold $(\widetilde M,\widetilde g)$ is compact. [/step] [step:Identify the fundamental group with the deck transformation group] Fix a basepoint $p_0 \in M$ and choose a point $\widetilde p_0 \in \pi^{-1}(\{p_0\})$. Let $\operatorname{Deck}(\widetilde M/M)$ denote the group of diffeomorphisms \begin{align*} \Phi:\widetilde M \to \widetilde M \end{align*} such that $\pi\circ\Phi=\pi$. Since $\widetilde M$ is the universal cover of the connected manifold $M$, the standard monodromy action identifies $\pi_1(M,p_0)$ with $\operatorname{Deck}(\widetilde M/M)$. The action of $\operatorname{Deck}(\widetilde M/M)$ on $\widetilde M$ is free. Indeed, if $\Phi \in \operatorname{Deck}(\widetilde M/M)$ fixes some point $\widetilde p \in \widetilde M$, then $\Phi$ and the identity deck transformation agree at $\widetilde p$; by [uniqueness of lifts](/theorems/1885) on the connected covering space, $\Phi=\operatorname{id}_{\widetilde M}$. [/step] [step:Use compactness to rule out infinitely many deck transformations] Suppose, for contradiction, that $\operatorname{Deck}(\widetilde M/M)$ is infinite. The orbit \begin{align*} \mathcal O := \{\Phi(\widetilde p_0):\Phi \in \operatorname{Deck}(\widetilde M/M)\} \subset \widetilde M \end{align*} is infinite because the deck action is free. Since $\widetilde M$ is compact, the infinite subset $\mathcal O$ has an accumulation point $\widetilde q \in \widetilde M$. On the other hand, the deck action of a covering space is properly discontinuous: there exists an open neighbourhood $U \subset \widetilde M$ of $\widetilde q$ such that \begin{align*} \{\Phi \in \operatorname{Deck}(\widetilde M/M): \Phi(U)\cap U\neq \varnothing\} \end{align*} is finite. Since $\widetilde q$ is an accumulation point of $\mathcal O$, infinitely many distinct points of $\mathcal O$ lie in $U$. Choose distinct deck transformations $\Phi_j$ with $\Phi_j(\widetilde p_0)\in U$. For any two such transformations $\Phi_i$ and $\Phi_j$, the deck transformation $\Phi_j\circ\Phi_i^{-1}$ satisfies \begin{align*} (\Phi_j\circ\Phi_i^{-1})\bigl(\Phi_i(\widetilde p_0)\bigr)=\Phi_j(\widetilde p_0)\in U, \end{align*} while $\Phi_i(\widetilde p_0)\in U$. Hence \begin{align*} (\Phi_j\circ\Phi_i^{-1})(U)\cap U\neq \varnothing. \end{align*} As $j$ varies with $i$ fixed, the transformations $\Phi_j\circ\Phi_i^{-1}$ are distinct, contradicting the finiteness required by proper discontinuity. Therefore $\operatorname{Deck}(\widetilde M/M)$ is finite. Since \begin{align*} \pi_1(M,p_0)\cong \operatorname{Deck}(\widetilde M/M), \end{align*} the group $\pi_1(M,p_0)$ is finite. The finiteness of the fundamental group is independent of the chosen basepoint because $M$ is connected. Thus $\pi_1(M)$ is finite. [/step]