[proofplan] We reduce to [Non-compact Universal Cover Contains a Line](/theorems/2769) by showing the universal cover $\tilde M$ is non-compact whenever $\pi_1(M)$ is infinite. The argument is purely topological: a covering map onto a compact space, with compact total space, has only finitely many sheets, so the deck-transformation group $\Gamma \cong \pi_1(M)$ is finite. Contrapositively, $|\pi_1(M)| = \infty$ forces $\tilde M$ non-compact, and then theorem 2769 produces a line. The Ricci hypothesis $\operatorname{Ric}(g) \geq 0$ is not used in the argument; it appears in the statement only because this result is the bridge between $\pi_1(M)$ infinite and the line-splitting theory in companion results. [/proofplan] [step:Set up the universal cover, the deck-transformation group, and the action on $\tilde M$] Let $\pi: \tilde M \to M$ be the universal Riemannian cover, that is, the universal cover of $M$ as a topological space, equipped with the pullback metric $\tilde g := \pi^* g$. The deck-transformation group \begin{align*} \Gamma := \operatorname{Deck}(\pi: \tilde M \to M) \end{align*} is canonically isomorphic to $\pi_1(M, x_0)$ for any choice of basepoint $x_0 \in M$ together with a chosen lift $\tilde x_0 \in \pi^{-1}(x_0)$. The action $\Gamma \curvearrowright \tilde M$ is properly discontinuous, free, and by isometries (the latter because $\tilde g = \pi^* g$ makes deck transformations preserve the metric — see the analogous discussion in [Non-compact Universal Cover Contains a Line](/theorems/2769)). The quotient is canonically $\Gamma \backslash \tilde M = M$ as Riemannian manifolds. By hypothesis $|\Gamma| = |\pi_1(M)| = \infty$. [guided] We need to set up four objects: the universal cover, its Riemannian structure, the deck group, and the deck action. Each carries a small subtlety, so we walk through them in turn. First, the universal cover. Topologically, $\tilde M$ is the simply connected covering space of $M$, which exists because $M$ is a connected manifold (hence locally path connected and semilocally simply connected). Let $\pi: \tilde M \to M$ denote the covering map. Second, the Riemannian structure. Why pull back $g$ rather than equipping $\tilde M$ with some other metric? Because we want $\pi$ to be a local isometry — this is the only way for the geometry of $M$ to lift faithfully to $\tilde M$. There is a unique metric on $\tilde M$ making $\pi$ a local isometry, namely \begin{align*} \tilde g := \pi^* g. \end{align*} Third, the deck group. Define \begin{align*} \Gamma := \operatorname{Deck}(\pi: \tilde M \to M), \end{align*} the group of self-homeomorphisms $\varphi: \tilde M \to \tilde M$ satisfying $\pi \circ \varphi = \pi$. Why is $\Gamma$ canonically isomorphic to $\pi_1(M, x_0)$? For a regular cover (which the universal cover always is, since the trivial subgroup is normal in $\pi_1(M)$), the deck-transformation group equals the quotient $\pi_1(M)/\pi_*\pi_1(\tilde M)$. Because $\tilde M$ is simply connected, $\pi_*\pi_1(\tilde M) = 0$, so $\Gamma \cong \pi_1(M, x_0)$ once we fix a basepoint $x_0 \in M$ and a lift $\tilde x_0 \in \pi^{-1}(x_0)$. Fourth, the action $\Gamma \curvearrowright \tilde M$. This action has three properties we will use later: - *Properly discontinuous and free*: standard for the deck group of a covering map. - *By isometries*: because $\tilde g = \pi^* g$, any deck transformation $\varphi$ satisfies $\varphi^* \tilde g = \varphi^* \pi^* g = (\pi \circ \varphi)^* g = \pi^* g = \tilde g$. The same calculation appears in [Non-compact Universal Cover Contains a Line](/theorems/2769). - The quotient is canonically $\Gamma \backslash \tilde M = M$ as Riemannian manifolds. By hypothesis $|\pi_1(M)| = \infty$, hence $|\Gamma| = \infty$. This is the only fact we will carry forward into the next step. [/guided] [/step] [step:Show $\tilde M$ non-compact using the contrapositive of the finite-sheets criterion] We show: if $\tilde M$ is compact, then $|\Gamma|$ is finite. Contrapositively, if $|\Gamma| = \infty$, then $\tilde M$ is non-compact. Suppose $\tilde M$ is compact. Then the covering $\pi: \tilde M \to M$ is a *finite-sheeted* covering, in the following sense: each fiber $\pi^{-1}(p)$ for $p \in M$ is a discrete closed subset of the compact space $\tilde M$, hence finite. [claim:All fibers of a covering map have the same cardinality] [proof] The number of sheets of a covering $\pi: \tilde M \to M$ is locally constant on $M$: if $V \subseteq M$ is an evenly covered open set with $\pi^{-1}(V) = \bigsqcup_{i \in I} U_i$ and each $U_i \to V$ is a homeomorphism, then $|\pi^{-1}(p)| = |I|$ for all $p \in V$. Local constancy plus connectedness of $M$ (a hypothesis) implies global constancy. [/proof] [/claim] So all fibers have the same finite cardinality $N := |\pi^{-1}(x_0)|$. The deck group $\Gamma$ acts simply transitively on $\pi^{-1}(x_0)$ (because the universal cover is regular and the action is free), so \begin{align*} |\Gamma| = |\pi^{-1}(x_0)| = N < \infty. \end{align*} This proves: $\tilde M$ compact $\Rightarrow |\Gamma|$ finite. Contrapositively, since $|\Gamma| = \infty$ by hypothesis, $\tilde M$ is non-compact. [guided] We want to extract non-compactness of $\tilde M$ from the topological hypothesis $|\pi_1(M)| = \infty$. The natural strategy is by contrapositive: prove that compactness of $\tilde M$ would force $|\Gamma|$ finite, which contradicts our hypothesis. The chain of implications is: \begin{align*} \tilde M \text{ compact} \quad &\Rightarrow \quad \pi \text{ has finitely many sheets} \\ &\Rightarrow \quad |\pi^{-1}(x_0)| < \infty \\ &\Rightarrow \quad |\Gamma| < \infty. \end{align*} Each implication is delicate enough to spell out, and each uses a different fact about covering maps. *(a) $\tilde M$ compact $\Rightarrow$ $\pi^{-1}(x_0)$ finite.* The fiber $\pi^{-1}(x_0) \subseteq \tilde M$ is closed, since $\{x_0\}$ is closed in $M$ (a manifold is Hausdorff) and $\pi$ is continuous. The fiber is also discrete: a covering map has discrete fibers, because each point of the fiber has an open neighborhood in $\tilde M$ that maps homeomorphically to an evenly-covered neighborhood of $x_0$ in $M$, isolating it from the rest of the fiber. A discrete closed subset of a compact space is finite, so $|\pi^{-1}(x_0)| < \infty$. *(b) All fibers have the same finite cardinality $N$.* This is the local-constancy property recorded in the claim: if $V \subseteq M$ is evenly covered with $\pi^{-1}(V) = \bigsqcup_{i \in I} U_i$ and each $U_i \to V$ a homeomorphism, then every $p \in V$ satisfies $|\pi^{-1}(p)| = |I|$. So $p \mapsto |\pi^{-1}(p)|$ is locally constant on $M$; on a connected base it is globally constant. Set \begin{align*} N := |\pi^{-1}(x_0)|. \end{align*} *(c) $|\Gamma| = N$ for the universal cover.* The universal cover is a regular cover — its monodromy subgroup of $\pi_1(M)$ is the trivial subgroup, which is normal. For a regular cover, the deck group acts simply transitively on each fiber. Why? Pick any two points $\tilde y, \tilde z \in \pi^{-1}(x_0)$; because $\tilde M$ is simply connected, there is a unique deck transformation taking $\tilde y$ to $\tilde z$. Hence \begin{align*} |\Gamma| = |\pi^{-1}(x_0)| = N < \infty. \end{align*} This proves: $\tilde M$ compact $\Rightarrow |\Gamma|$ finite. The contrapositive is what we need: if $|\Gamma| = \infty$, then $\tilde M$ is non-compact. We are given $|\pi_1(M)| = \infty$, hence $|\Gamma| = \infty$, hence $\tilde M$ is non-compact. [/guided] [/step] [step:Apply [Non-compact Universal Cover Contains a Line](/theorems/2769) to conclude] We invoke [Non-compact Universal Cover Contains a Line](/theorems/2769). That theorem requires: 1. $(M, g)$ is a compact Riemannian manifold — given by hypothesis. 2. $(\tilde M, \tilde g)$ is the universal Riemannian cover of $(M, g)$ — set up in the first step. 3. $\tilde M$ is non-compact — established in the previous step. All three hypotheses hold. The conclusion of theorem 2769 is that $\tilde M$ contains a line, which is what we wanted to prove. [guided] The Ricci hypothesis $\operatorname{Ric}(g) \geq 0$ in the statement of the present theorem is *not used* in the proof. It is present because this theorem is meant as a stepping stone in the chain of results: one wants Ricci non-negative so that, downstream, the [Cheeger–Gromoll Line-Splitting Theorem](/theorems/2767) can be applied to produce a line-splitting $\tilde M = N \times \mathbb R$. That further consequence belongs to the structure theorem [Structure of Compact Manifolds with Nonneg Ricci](/theorems/2771); here we extract just the line. To see exactly what curvature plays no role in: the proof we just gave only used (i) the topological structure of the cover (compactness, fibers, deck group cardinality) and (ii) theorem 2769, which itself uses only completeness of $\tilde M$ (a consequence of compactness of $M$, *not* of curvature) and the existence of an Arzelà–Ascoli limit. The Ricci hypothesis was nowhere consumed. This is intentional and is reflected in the original statement: the result is "$\pi_1$ infinite + $M$ compact + $\operatorname{Ric} \geq 0$ $\Rightarrow$ $\tilde M$ has a line", but the actual logical content is "$\pi_1$ infinite + $M$ compact $\Rightarrow$ $\tilde M$ has a line". Stating it with the Ricci hypothesis matches how it is invoked in subsequent results. [/guided] [/step]