[proofplan] We iterate the [Cheeger–Gromoll Line-Splitting Theorem](/theorems/2767). At each stage, if the current factor contains a line, we split off an $\mathbb R$ factor; the remaining factor still has $\operatorname{Ric} \geq 0$ because Ricci on a Riemannian product is the sum of factor Riccis. The dimension drops by one with each split, so iteration terminates after at most $\dim M$ steps in a factor $X$ with $\operatorname{Ric}(g_X) \geq 0$ containing no line. The total Euclidean factor is $\mathbb R^q$ for $q$ equal to the number of splits performed. Uniqueness of $X$ and $q$ comes from the maximality: any further $\mathbb R$ factor in $X$ would correspond to a parallel unit vector field on $X$, which by Cheeger–Gromoll would arise from a line in $X$ — but $X$ has none. [/proofplan] [step:Set up the iteration: split off one $\mathbb R$ factor whenever a line exists] Let $(M_0, g_0) := (M, g)$. We define a sequence \begin{align*} (M_0, g_0), (M_1, g_1), (M_2, g_2), \ldots \end{align*} of complete connected Riemannian manifolds with $\operatorname{Ric}(g_k) \geq 0$ at every stage, by the following recursion. **Base step:** Set $(M_0, g_0) := (M, g)$. Note $\operatorname{Ric}(g_0) \geq 0$ by hypothesis. **Inductive step:** Suppose $(M_k, g_k)$ has been constructed and is a complete connected Riemannian manifold with $\operatorname{Ric}(g_k) \geq 0$. - *Case A:* $M_k$ contains no line. Stop the iteration: $(X, g_X) := (M_k, g_k)$ and $q := k$. - *Case B:* $M_k$ contains a line $\ell_k : \mathbb R \to M_k$. Apply [Cheeger–Gromoll Line-Splitting Theorem](/theorems/2767) to $(M_k, g_k)$ with the line $\ell_k$, producing an isometry \begin{align*} (M_k, g_k) \cong (M_{k+1} \times \mathbb R, g_{k+1} + dt^2) \end{align*} for some complete connected Riemannian manifold $(M_{k+1}, g_{k+1})$. Continue with $(M_{k+1}, g_{k+1})$ in the next stage. [claim:At every stage, $\operatorname{Ric}(g_k) \geq 0$] [proof] By induction on $k$. The base case $k = 0$ is the hypothesis $\operatorname{Ric}(g) \geq 0$. For the inductive step, suppose $\operatorname{Ric}(g_k) \geq 0$ and $(M_k, g_k) \cong (M_{k+1} \times \mathbb R, g_{k+1} + dt^2)$. The Ricci tensor of a Riemannian product decomposes block-diagonally on the product tangent space $T_{(p, t)}(M_{k+1} \times \mathbb R) = T_p M_{k+1} \oplus T_t \mathbb R$: \begin{align*} \operatorname{Ric}_{(p, t)}(g_{k+1} + dt^2) = \operatorname{Ric}_p(g_{k+1}) \oplus \operatorname{Ric}_t(dt^2) = \operatorname{Ric}_p(g_{k+1}) \oplus 0, \end{align*} since the Euclidean line $(\mathbb R, dt^2)$ is flat with vanishing curvature. Non-negativity on the LHS therefore forces non-negativity of the $M_{k+1}$ block: $\operatorname{Ric}(g_{k+1}) \geq 0$. [/proof] [/claim] [claim:At every stage, $(M_k, g_k)$ is complete] [proof] A Riemannian product $(N \times \mathbb R, g_N + dt^2)$ is complete if and only if both factors are complete. Hence completeness of $(M_k, g_k) \cong (M_{k+1} \times \mathbb R, g_{k+1} + dt^2)$ implies completeness of $(M_{k+1}, g_{k+1})$. [/proof] [/claim] So both hypotheses for applying [Cheeger–Gromoll Line-Splitting Theorem](/theorems/2767) — completeness and $\operatorname{Ric} \geq 0$ — are preserved at every stage, and the iteration is well-defined. [guided] We want to extract Euclidean factors from $M$ one at a time, using [Cheeger–Gromoll Line-Splitting Theorem](/theorems/2767) as the engine. Each time the current factor contains a line, the theorem trades that line for an $\mathbb R$ summand, leaving a smaller factor behind. The natural way to organise this is a recursion. Why a recursion rather than a single application? Because Cheeger–Gromoll only gives one $\mathbb R$ factor per use, and the remaining factor may itself contain a line — so we have to feed the leftover back into the theorem until no line remains. To make the recursion precise, we build a sequence \begin{align*} (M_0, g_0), (M_1, g_1), (M_2, g_2), \ldots \end{align*} of complete connected Riemannian manifolds with $\operatorname{Ric}(g_k) \geq 0$ at every stage. The key invariant being preserved is the hypothesis class of Cheeger–Gromoll itself: "complete, connected, $\operatorname{Ric} \geq 0$". As long as this invariant survives, we can keep applying the theorem. **Base step.** Set $(M_0, g_0) := (M, g)$. The hypothesis $\operatorname{Ric}(g) \geq 0$ is given, so the invariant holds at $k = 0$. **Inductive step.** Suppose $(M_k, g_k)$ is constructed and satisfies the invariant. We branch on whether $M_k$ contains a line. - *Case A:* $M_k$ contains no line. Then we cannot (and need not) apply Cheeger–Gromoll again. Stop the iteration and set $(X, g_X) := (M_k, g_k)$ and $q := k$. This $X$ is what we want: a complete connected Riemannian manifold with $\operatorname{Ric} \geq 0$ and no line. - *Case B:* $M_k$ contains a line $\ell_k : \mathbb R \to M_k$. Both hypotheses of Cheeger–Gromoll are met (completeness and $\operatorname{Ric}(g_k) \geq 0$, by the invariant; existence of a line is what makes the theorem applicable in the first place). Applying [Cheeger–Gromoll Line-Splitting Theorem](/theorems/2767) yields an isometry \begin{align*} (M_k, g_k) \cong (M_{k+1} \times \mathbb R, g_{k+1} + dt^2) \end{align*} for some complete connected Riemannian manifold $(M_{k+1}, g_{k+1})$. We then continue the recursion at stage $k+1$. For this construction to make sense, we must verify that $(M_{k+1}, g_{k+1})$ inherits the invariant. There are two facts to check. *Completeness is inherited.* A standard fact about Riemannian products: $(N \times \mathbb R, g_N + dt^2)$ is complete if and only if both factors are complete. Since $(M_k, g_k)$ is complete and isometric to the product $(M_{k+1} \times \mathbb R, g_{k+1} + dt^2)$, both factors must be complete; in particular, $(M_{k+1}, g_{k+1})$ is complete. *Non-negative Ricci is inherited.* This is where the Ricci-splitting on a product becomes essential. The Levi-Civita connection of a Riemannian product is the direct sum of the factor Levi-Civita connections; consequently the Riemann tensor is the direct sum of the factor Riemann tensors, and contracting gives that the Ricci tensor is the direct sum of the factor Riccis. Block-diagonally on $T_{(p, t)}(M_{k+1} \times \mathbb R) = T_p M_{k+1} \oplus T_t \mathbb R$, \begin{align*} \operatorname{Ric}_{(p, t)}(g_{k+1} + dt^2) = \operatorname{Ric}_p(g_{k+1}) \oplus \operatorname{Ric}_t(dt^2) = \operatorname{Ric}_p(g_{k+1}) \oplus 0, \end{align*} where the second equality uses that Euclidean space $(\mathbb R, dt^2)$ is flat (its Riemann tensor, hence its Ricci, vanishes). Non-negativity of the LHS — which holds by the inductive hypothesis $\operatorname{Ric}(g_k) \geq 0$ — therefore forces non-negativity of each diagonal block, in particular $\operatorname{Ric}(g_{k+1}) \geq 0$. Both facts together show the invariant "complete, connected, $\operatorname{Ric} \geq 0$" passes from stage $k$ to stage $k+1$. By induction, the iteration is well-defined: at every stage, the manifold $(M_k, g_k)$ satisfies the hypotheses of Cheeger–Gromoll, so Case B is actually applicable whenever a line is present. Why this pattern works in general: any time you have a class $\mathcal C$ of objects closed under taking factors, together with a structure theorem of the form "object in $\mathcal C$ with substructure $\Rightarrow$ object in $\mathcal C$ as a product", you can iterate to peel off substructures one by one. Here $\mathcal C$ = "complete connected Riemannian, $\operatorname{Ric} \geq 0$", the substructure is "a line", and Cheeger–Gromoll is the structure theorem. The closure-under-factors property is exactly the two facts we just verified. [/guided] [/step] [step:Show the iteration terminates after at most $\dim M$ steps] We show the iteration cannot continue indefinitely. The dimensions satisfy \begin{align*} \dim M_{k+1} = \dim M_k - 1 \end{align*} whenever the splitting is performed at stage $k$ (because $M_k \cong M_{k+1} \times \mathbb R$ and $\mathbb R$ has dimension $1$). Starting from $\dim M_0 = \dim M$ and decreasing by $1$ at each step, the dimensions form the strictly decreasing sequence $\dim M, \dim M - 1, \ldots, 1, 0$. If we ever reach $\dim M_k = 0$, then $M_k$ is a single point, which contains no line for the dimensional reason that a line requires a non-constant curve $\mathbb R \to M_k$, impossible for a $0$-dimensional space. Even before reaching $0$, at $\dim M_k = 1$, $M_k$ is either $\mathbb R$ or a circle $S^1$. The circle has a smallest unbounded universal cover that is $\mathbb R$, but $S^1$ itself contains no line (geodesics on $S^1$ are not globally minimizing because they wrap around). So $M_k = S^1$ also halts the iteration. Only $M_k = \mathbb R$ as a $1$-dimensional case continues, and then one further Cheeger–Gromoll step produces $M_{k+1} = \{*\}$, which terminates. In any case, the iteration must terminate at some finite $q \leq \dim M$ with $(M_q, g_q) = (X, g_X)$ a complete connected Riemannian manifold with $\operatorname{Ric}(g_X) \geq 0$ and *no line*. Composing the splittings, we obtain \begin{align*} (M_0, g_0) &\cong (M_1, g_1) \times \mathbb R \\ &\cong (M_2, g_2) \times \mathbb R^2 \\ &\quad \vdots \\ &\cong (M_q, g_q) \times \mathbb R^q = (X, g_X) \times \mathbb R^q, \end{align*} where the product metric on $(X, g_X) \times \mathbb R^q$ is $g_X + g_{\mathrm{Eucl}}$ ($g_{\mathrm{Eucl}}$ the standard Euclidean metric on $\mathbb R^q$). This gives the existence of the splitting in the theorem. [guided] We need to show two things in this step: that the iteration set up in Step 1 actually halts at some finite stage, and that what comes out at the end is the splitting $(M, g) \cong (X, g_X) \times \mathbb R^q$ promised by the theorem. **Why the iteration terminates.** The mechanism is a strictly decreasing dimension count. Whenever Case B fires at stage $k$, Cheeger–Gromoll produces $(M_k, g_k) \cong (M_{k+1} \times \mathbb R, g_{k+1} + dt^2)$, and dimensions add under products: $\dim M_k = \dim M_{k+1} + \dim \mathbb R = \dim M_{k+1} + 1$. Rearranging, \begin{align*} \dim M_{k+1} = \dim M_k - 1. \end{align*} Starting from $\dim M_0 = \dim M$ and dropping by $1$ at every step where Case B fires, the dimensions form the strictly decreasing sequence $\dim M, \dim M - 1, \dim M - 2, \ldots$ of non-negative integers. A strictly decreasing sequence of non-negative integers must eventually reach $0$, so Case B can fire at most $\dim M$ times. We do not need to know in advance how many steps it takes — termination is automatic from this dimension count. What if we worry about edge cases at low dimension? At $\dim M_k = 0$, $M_k$ is a single point, which trivially contains no line (a line requires a non-constant geodesic $\mathbb R \to M_k$, impossible on a $0$-dimensional manifold). At $\dim M_k = 1$, the complete connected possibilities are $\mathbb R$ and $S^1$; the circle $S^1$ has no line because its geodesics wrap around and fail to be globally minimising, so the iteration halts there too via Case A. Only $M_k = \mathbb R$ at dimension $1$ admits one further Cheeger–Gromoll application, after which $\dim M_{k+1} = 0$ and the iteration halts. So no matter the low-dimensional behaviour, termination occurs at some finite stage $q \leq \dim M$. What does termination *mean* qualitatively? It means the final factor $(X, g_X) := (M_q, g_q)$ falls under Case A: it is a complete connected Riemannian manifold with $\operatorname{Ric}(g_X) \geq 0$ (preserved at every stage by Step 1's invariant) that contains *no line*. The "no line" condition is the essential qualitative property — it is what distinguishes $X$ from the Euclidean factors and what will give uniqueness in Step 3. (Such "no-line" manifolds need not be compact and need not be simply connected; their classification is an active research topic. The present theorem is content-free as a classification — it just isolates the Euclidean factor.) **Assembling the splitting.** Each Case-B application at stage $k$ gives $(M_k, g_k) \cong (M_{k+1}, g_{k+1}) \times \mathbb R$. Composing these isometries inductively, after $q$ stages we get \begin{align*} (M_0, g_0) &\cong (M_1, g_1) \times \mathbb R \\ &\cong (M_2, g_2) \times \mathbb R^2 \\ &\quad \vdots \\ &\cong (M_q, g_q) \times \mathbb R^q = (X, g_X) \times \mathbb R^q, \end{align*} where the chain of isometries uses associativity of the Riemannian product: if $A \cong B \times \mathbb R$ and $B \cong C \times \mathbb R$, then $A \cong (C \times \mathbb R) \times \mathbb R \cong C \times \mathbb R^2$, and so on. The product metric on $(X, g_X) \times \mathbb R^q$ is $g_X + g_{\mathrm{Eucl}}$, where $g_{\mathrm{Eucl}}$ is the standard Euclidean metric on $\mathbb R^q$ (it is $dt_1^2 + \cdots + dt_q^2$, which is exactly the iterated sum of the $dt^2$ metrics from the individual splittings). This establishes the *existence* part of the theorem: $(M, g) \cong (X, g_X) \times \mathbb R^q$ with $X$ complete, connected, $\operatorname{Ric}(g_X) \geq 0$, and line-free. Uniqueness of $X$ and $q$ — that they do not depend on the choices made during the iteration (which line to pick at each stage, in particular) — is the content of Step 3. [/guided] [/step] [step:Show uniqueness of $X$ and $q$] We show that the integer $q$ and the manifold $(X, g_X)$ in the splitting $(M, g) \cong (X, g_X) \times \mathbb R^q$ are determined up to isometry by $(M, g)$. The strategy is to characterize $q$ as the dimension of an intrinsic vector space attached to $(M, g)$ — namely the space of parallel vector fields — and to recover $X$ as the leaf of an intrinsically defined parallel distribution. Define \begin{align*} \mathcal P(M, g) &:= \{\xi \in \mathfrak X(M) : \nabla \xi = 0\}, \end{align*} the real vector space of $\nabla$-parallel vector fields on $M$, where $\nabla$ is the Levi-Civita connection of $g$. The set $\mathcal P(M, g)$ depends only on the isometry class of $(M, g)$ (since isometries intertwine Levi-Civita connections). [claim:$\dim \mathcal P(M, g) = q$] [proof] Using the splitting $(M, g) \cong (X \times \mathbb R^q, g_X + g_{\mathrm{Eucl}})$, we compute $\dim \mathcal P(X \times \mathbb R^q, g_X + g_{\mathrm{Eucl}})$ in three substeps. *Substep 1: parallel fields on a product split as direct sums of parallel fields on the factors.* The Levi-Civita connection of a Riemannian product is the direct sum of the factor Levi-Civita connections: under the canonical identification $T_{(p,t)}(X \times \mathbb R^q) = T_p X \oplus T_t \mathbb R^q$, every vector field $\xi$ on $X \times \mathbb R^q$ decomposes uniquely as $\xi = (\xi_X, \xi_{\mathbb R^q})$ where $\xi_X$ is the $TX$-component and $\xi_{\mathbb R^q}$ is the $T\mathbb R^q$-component. The product-connection identity gives, for any vector field $Z$ on the product, \begin{align*} \nabla_Z \xi &= \nabla^X_{Z_X} \xi_X + \nabla^{\mathbb R^q}_{Z_{\mathbb R^q}} \xi_X + \nabla^X_{Z_X} \xi_{\mathbb R^q} + \nabla^{\mathbb R^q}_{Z_{\mathbb R^q}} \xi_{\mathbb R^q}, \end{align*} where we have used that the off-diagonal connection terms vanish by the product structure (mixed covariant derivatives across factors are zero). Hence $\nabla \xi = 0$ if and only if $\nabla^X \xi_X = 0$ on $X$ (the $X$-component is parallel for the connection on $X$, with no dependence on the $\mathbb R^q$-coordinate) and $\nabla^{\mathbb R^q} \xi_{\mathbb R^q} = 0$ on $\mathbb R^q$ (likewise). This gives a direct-sum decomposition \begin{align*} \mathcal P(X \times \mathbb R^q, g_X + g_{\mathrm{Eucl}}) &\cong \mathcal P(X, g_X) \oplus \mathcal P(\mathbb R^q, g_{\mathrm{Eucl}}). \end{align*} *Substep 2: $\dim \mathcal P(\mathbb R^q, g_{\mathrm{Eucl}}) = q$.* The Levi-Civita connection on flat Euclidean space is the standard directional derivative: for $\xi = \sum_{i=1}^q a_i(t) \partial_{t_i}$ with $a_i \in C^\infty(\mathbb R^q)$, $\nabla \xi = 0$ is equivalent to $\partial_{t_j} a_i = 0$ for all $i, j$, i.e., each $a_i$ is constant. Hence $\mathcal P(\mathbb R^q, g_{\mathrm{Eucl}}) = \operatorname{span}_{\mathbb R}\{\partial_{t_1}, \ldots, \partial_{t_q}\}$, which has dimension $q$. *Substep 3: $\dim \mathcal P(X, g_X) = 0$.* Suppose for contradiction that $\xi_X \in \mathcal P(X, g_X)$ is non-zero at some point $p \in X$. Then $\xi_X$ is non-zero everywhere on $X$ (since parallel transport is an isometry, $|\xi_X|_{g_X}$ is constant along every curve, and by connectedness of $X$ it is a positive constant on $X$). Rescale by $|\xi_X|_{g_X}^{-1}$ to assume $|\xi_X|_{g_X} \equiv 1$ (rescaling by a constant preserves parallelism). The vector field $\xi_X$ is autoparallel, $\nabla^X_{\xi_X} \xi_X = 0$ (this is the parallel condition $\nabla^X \xi_X = 0$ contracted against $\xi_X$), so its integral curves are unit-speed geodesics of $(X, g_X)$. By completeness of $X$ (Step 2 preserved completeness across the iteration, and $X = M_q$ is complete), each integral curve $\gamma : \mathbb R \to X$ of $\xi_X$ is defined on all of $\mathbb R$. Now apply the converse direction of [Cheeger–Gromoll Line-Splitting Theorem](/theorems/2767): the existence of a parallel unit vector field $\xi_X$ on $(X, g_X)$ produces a Riemannian splitting $(X, g_X) \cong (N, g_N) \times (\mathbb R, dt^2)$ with the integral curves of $\xi_X$ corresponding to the $\mathbb R$-factor lines. (This converse is a direct consequence of the parallel-vector-field-flow being a Riemannian submersion onto its orbit space: parallel transport along orbits is an isometry between the orthogonal complements of $\xi_X$, which exhibits $X$ as a metric product.) Each integral curve of $\xi_X$ is therefore a line in $X$ — a unit-speed geodesic $\gamma : \mathbb R \to X$ that is globally minimizing. But $X$ contains no line, by construction in Step 2. Contradiction. Hence $\xi_X = 0$, so $\mathcal P(X, g_X) = \{0\}$, with $\dim \mathcal P(X, g_X) = 0$. *Combining the substeps:* $\dim \mathcal P(M, g) = \dim \mathcal P(X, g_X) + \dim \mathcal P(\mathbb R^q, g_{\mathrm{Eucl}}) = 0 + q = q$. Since $\mathcal P(M, g)$ depends only on $(M, g)$, the integer $q$ is uniquely determined. [/proof] [/claim] [claim:The manifold $(X, g_X)$ is uniquely determined up to isometry] [proof] We construct $X$ intrinsically as a leaf of a parallel distribution. *The parallel sub-bundle $\mathcal P \subset TM$.* Evaluation at a point gives, for each $x \in M$, a linear map $\mathrm{ev}_x : \mathcal P(M, g) \to T_x M$, $\xi \mapsto \xi(x)$. This map is injective: if $\xi(x) = 0$ then by parallel transport $\xi \equiv 0$ along any curve through $x$, and by connectedness $\xi = 0$ on $M$. Let $\mathcal P_x := \mathrm{ev}_x(\mathcal P(M, g)) \subseteq T_x M$, a $q$-dimensional subspace varying smoothly in $x$ to give a rank-$q$ sub-bundle $\mathcal P \subset TM$. The bundle $\mathcal P$ is parallel: $\nabla_Z \eta \in \Gamma(\mathcal P)$ for any vector field $Z$ and any section $\eta$ of $\mathcal P$ that arises as $\mathrm{ev}(\xi)$ for $\xi \in \mathcal P(M, g)$, since $\nabla \xi = 0$ then gives $\nabla_Z \eta = 0 \in \Gamma(\mathcal P)$. *The orthogonal distribution $\mathcal D := \mathcal P^\perp \subset TM$.* Define $\mathcal D_x := (\mathcal P_x)^\perp \subseteq T_x M$ with respect to $g_x$. This is a rank-$(\dim M - q)$ smooth distribution. We claim $\mathcal D$ is parallel as a sub-bundle of $TM$. Indeed, for any vector field $Z$ on $M$ and any section $\eta$ of $\mathcal D$, and any $\xi \in \mathcal P(M, g)$, metric compatibility of $\nabla$ gives \begin{align*} 0 &= Z\, g(\eta, \xi) = g(\nabla_Z \eta, \xi) + g(\eta, \nabla_Z \xi) = g(\nabla_Z \eta, \xi) + 0, \end{align*} where $g(\eta, \xi) = 0$ identically (since $\eta \in \mathcal D = \mathcal P^\perp$ and $\xi \in \mathcal P$) and $\nabla_Z \xi = 0$ (since $\xi$ is parallel). Hence $g(\nabla_Z \eta, \xi) = 0$ for all $\xi$, so $\nabla_Z \eta \in \mathcal P^\perp = \mathcal D$. So $\mathcal D$ is parallel. *$\mathcal D$ is integrable.* For sections $\eta_1, \eta_2 \in \Gamma(\mathcal D)$, the Lie bracket satisfies $[\eta_1, \eta_2] = \nabla_{\eta_1} \eta_2 - \nabla_{\eta_2} \eta_1$ (because the Levi-Civita connection is torsion-free). Both terms on the right lie in $\Gamma(\mathcal D)$ by the previous paragraph. Hence $[\Gamma(\mathcal D), \Gamma(\mathcal D)] \subseteq \Gamma(\mathcal D)$, i.e., $\mathcal D$ is involutive. By Frobenius' theorem, $\mathcal D$ is integrable: through every $x \in M$ there passes a unique maximal connected integral submanifold $L_x \subset M$ with $T_y L_x = \mathcal D_y$ for every $y \in L_x$. *Construction of $X$.* Pick any basepoint $x_0 \in M$ and set $X := L_{x_0}$, equipped with the induced Riemannian metric $g_X := g|_X$. *All leaves are isometric.* Let $L_{x_0}$ and $L_{y_0}$ be two leaves. Pick any $\xi \in \mathcal P(M, g)$ whose flow $\Phi_t^\xi$ moves $x_0$ to $y_0$ for some $t$ (such $\xi$ exists because the orbits of $\mathcal P(M, g)$-flows fill out the leaves of the foliation orthogonal to $\mathcal D$; this is precisely the dual foliation, whose leaves are the orbits of the parallel-vector-field flows, and these are exactly the $\mathbb R^q$-fibers of the splitting). The flow $\Phi_t^\xi$ is an isometry of $(M, g)$ (because $\xi$ is parallel implies $\nabla \xi = 0$, which implies $\xi$ is a Killing field — for a parallel field, $g(\nabla_Y \xi, Z) + g(Y, \nabla_Z \xi) = 0$ since both terms vanish individually — so its flow preserves $g$). Moreover, $\Phi_t^\xi$ maps $\mathcal D$ to $\mathcal D$ because $d\Phi_t^\xi$ commutes with parallel transport along the orbit. Hence $\Phi_t^\xi(L_{x_0}) = L_{y_0}$, and the restriction $\Phi_t^\xi|_{L_{x_0}} : L_{x_0} \to L_{y_0}$ is a Riemannian isometry. *Uniqueness of the splitting.* Suppose $(M, g) \cong (X', g_{X'}) \times \mathbb R^{q'}$ is another splitting with $X'$ line-free. By the previous claim $q' = q$. The construction of the parallel sub-bundle $\mathcal P$ depends only on $(M, g)$, so it agrees in the two splittings; in particular, the orthogonal distribution $\mathcal D$ is the same in both splittings. Under the second splitting, the $X' \times \{t_0\}$-slices for $t_0 \in \mathbb R^{q'}$ are leaves of $\mathcal D$ (by the same computation as in Substep 1: the tangent bundle of $X'$ inside $T(X' \times \mathbb R^{q'})$ is exactly the orthogonal complement of $\mathcal P(\mathbb R^{q'})$, hence equals $\mathcal D$). Hence $X' \cong L_{x_0} = X$ as Riemannian manifolds (taking any leaf as the representative). This shows the $X$-factor is uniquely determined up to isometry. [/proof] [/claim] This completes the proof. [guided] The goal of this step is to prove that the splitting $(M, g) \cong (X, g_X) \times \mathbb R^q$ from Steps 1–2 is unique up to isometry: the integer $q$ and the Riemannian manifold $(X, g_X)$ are invariants of $(M, g)$, not artifacts of the iteration. The strategy is to attach to $(M, g)$ two intrinsic objects — a vector space and a distribution — that recover $q$ and $X$ respectively. **The intrinsic vector space.** Define \begin{align*} \mathcal P(M, g) &:= \{\xi \in \mathfrak X(M) : \nabla \xi = 0\}, \end{align*} the space of $\nabla$-parallel vector fields on $M$. This is intrinsic: it is defined purely from $(M, g)$ via the Levi-Civita connection $\nabla$, with no reference to any chosen splitting. We will show $\dim \mathcal P(M, g) = q$, which forces $q$ to be an isometry-invariant of $(M, g)$. *Why parallel vector fields?* A parallel vector field has constant length (parallel transport is an isometry, so $|\xi|_g$ is constant along curves, hence constant on the connected manifold $M$). Its integral curves are autoparallel — they are geodesics — and on a complete manifold these geodesics are defined on all of $\mathbb R$. The Cheeger–Gromoll splitting says: a line yields a parallel unit vector field. The converse is also true (and is what we will use): a parallel unit vector field yields a line, hence a splitting. So parallel vector fields are exactly the *infinitesimal generators* of $\mathbb R$-factors in a splitting. *Computation on the model.* If we were given the splitting, we would compute $\dim \mathcal P(X \times \mathbb R^q)$ as follows. The Levi-Civita connection of a Riemannian product is the direct sum of the factor connections (this is a standard fact about products: under the canonical splitting $T(X \times \mathbb R^q) = TX \oplus T\mathbb R^q$, the connection acts diagonally and there are no cross-terms). So a parallel vector field on the product decomposes as $(\xi_X, \xi_{\mathbb R^q})$ with $\nabla^X \xi_X = 0$ on $X$ and $\nabla^{\mathbb R^q} \xi_{\mathbb R^q} = 0$ on $\mathbb R^q$. Packaging these two factor-wise parallel conditions into a single statement gives a clean direct-sum decomposition of the parallel-fields space: \begin{align*} \mathcal P(X \times \mathbb R^q, g_X + g_{\mathrm{Eucl}}) &\cong \mathcal P(X, g_X) \oplus \mathcal P(\mathbb R^q, g_{\mathrm{Eucl}}). \end{align*} This reduces the dimension count to two independent calculations, one on each factor — exactly what we want, since the two factors have very different qualitative character ($X$ is line-free; $\mathbb R^q$ is flat). We compute each factor separately. *Parallel fields on $\mathbb R^q$.* Euclidean space has the flat standard connection $\nabla = d$, the directional derivative. Parallel means each component is locally constant, hence (by connectedness) constant. So $\mathcal P(\mathbb R^q, g_{\mathrm{Eucl}}) = \operatorname{span}\{\partial_{t_1}, \ldots, \partial_{t_q}\}$ has dimension $q$. *Parallel fields on $X$.* Here is the crux. We claim $\mathcal P(X, g_X) = \{0\}$, using the line-free hypothesis on $X$. Suppose $\xi_X \neq 0$ is parallel. After rescaling we can assume $|\xi_X|_{g_X} \equiv 1$. The integral curves of $\xi_X$ are unit-speed geodesics (because $\nabla_{\xi_X} \xi_X = 0$, contracting the parallel condition against $\xi_X$), and by completeness they are defined on all of $\mathbb R$. We now want to conclude that these geodesics are *lines* — globally minimizing — but we cannot conclude this just from being parallel and unit-speed. The bridge is the converse direction of Cheeger–Gromoll: the existence of a parallel unit vector field on a complete Riemannian manifold produces a Riemannian splitting $X \cong N \times \mathbb R$, and the integral curves of $\xi_X$ correspond to the $\mathbb R$-factor curves $\{n\} \times \mathbb R$, which are lines (the $\mathbb R$-factor in a Riemannian product is always a line: distance in a product equals the Pythagorean combination of factor distances, so in particular a curve confined to one factor realizes its endpoint distance). So $X$ contains a line, contradicting the construction of $X$ in Step 2 as the line-free factor at the end of the iteration. Hence $\xi_X = 0$, and $\dim \mathcal P(X, g_X) = 0$. *Conclusion for $q$.* Combining: $\dim \mathcal P(M, g) = 0 + q = q$. Since the left-hand side depends only on $(M, g)$, so does $q$. This is the uniqueness of $q$. **The intrinsic distribution and the construction of $X$.** Now we recover $X$. Two pieces of structure are intrinsic: the $q$-dimensional sub-bundle $\mathcal P \subset TM$ obtained by evaluating elements of $\mathcal P(M, g)$ at each point (this map is injective on each fiber, by parallel transport, so $\dim \mathcal P_x = q$ everywhere), and its orthogonal complement $\mathcal D = \mathcal P^\perp$, which has rank $\dim M - q$. *$\mathcal D$ is parallel.* This is a general fact: the orthogonal complement of a parallel sub-bundle, in a metric-compatible connection, is parallel. The argument: for $\eta \in \Gamma(\mathcal D)$ and $\xi \in \mathcal P(M, g)$, $g(\eta, \xi) = 0$ identically. Differentiating along any $Z$, \begin{align*} 0 = Z g(\eta, \xi) = g(\nabla_Z \eta, \xi) + g(\eta, \nabla_Z \xi) = g(\nabla_Z \eta, \xi), \end{align*} using metric compatibility and $\nabla_Z \xi = 0$ (parallel). Since $\xi$ ranges over a basis of $\mathcal P_x$, this forces $\nabla_Z \eta \in \mathcal P^\perp = \mathcal D$. *$\mathcal D$ is integrable.* For sections $\eta_1, \eta_2$ of $\mathcal D$, the torsion-free property of $\nabla$ (Levi-Civita is torsion-free) gives \begin{align*} [\eta_1, \eta_2] = \nabla_{\eta_1} \eta_2 - \nabla_{\eta_2} \eta_1 \in \Gamma(\mathcal D), \end{align*} because both summands lie in $\Gamma(\mathcal D)$ (by the parallel-distribution property just established). So $\mathcal D$ is involutive. By Frobenius' theorem, $\mathcal D$ is integrable: $M$ is foliated by maximal connected integral submanifolds $\{L_x\}_{x \in M}$, with $T_y L_x = \mathcal D_y$ at every $y \in L_x$. *Define $X := L_{x_0}$* for any choice of basepoint $x_0$, with the induced metric $g_X = g|_X$. Different choices of basepoint give isometric leaves: parallel transport along any flow $\Phi_t^\xi$ of a parallel vector field $\xi \in \mathcal P(M, g)$ is an isometry of $(M, g)$ (a parallel field is a Killing field, since $\nabla \xi = 0$ directly gives $g(\nabla_Y \xi, Z) + g(Y, \nabla_Z \xi) = 0$), and the flow maps leaves of $\mathcal D$ to leaves of $\mathcal D$. The two leaves are connected by such a flow because the parallel fields generate the orthogonal foliation (whose leaves are the $\mathbb R^q$-fibers in a splitting) and the foliation is transverse to $\mathcal D$. *Uniqueness of $X$.* If $(M, g) \cong (X', g_{X'}) \times \mathbb R^{q'}$ is another splitting with $X'$ line-free, then $q' = q$ by the dimension argument above, and the parallel sub-bundle $\mathcal P$ — being intrinsic to $(M, g)$ — is the same in both splittings. Hence so is $\mathcal D$. Under the second splitting, the slice $X' \times \{t_0\}$ is tangent to $\mathcal D$ everywhere (because $T(X' \times \{t_0\}) = TX' = (T\mathbb R^{q'})^\perp = \mathcal P^\perp = \mathcal D$, where we have used that the parallel-vector-field bundle for $X' \times \mathbb R^{q'}$ is $T\mathbb R^{q'}$, by the same Substep-2 argument applied to the second splitting). So $X' \times \{t_0\}$ is a leaf of $\mathcal D$, hence isometric to $X = L_{x_0}$ by the previous paragraph. **Why the maximality matters.** The hypothesis "$X$ contains no line" was used at exactly one place: to conclude $\mathcal P(X, g_X) = 0$, which forced $\dim \mathcal P(M, g) = q$. Without maximality, $X$ might still admit further parallel vector fields, and the splitting would not be uniquely determined — for instance, $\mathbb R^q \cong \mathbb R^k \times \mathbb R^{q-k}$ for any $0 \le k \le q$, and there is no preferred $k$ unless we demand $X$ to be line-free. The maximality clause is what fixes the splitting. **Standard facts used.** We have used: (i) the Levi-Civita connection of a Riemannian product is the direct sum of the factor Levi-Civita connections; (ii) a Riemannian product is complete iff both factors are complete; (iii) the Ricci tensor of a Riemannian product is the direct sum of the factor Ricci tensors. Facts (ii) and (iii) were already cited in Steps 1–2; fact (i) is the underlying reason for both of them and for the parallel-field decomposition above. [/guided] [/step]