[proofplan] The proof is an induction on the number of splittings. At each stage the Cheeger-Gromoll Splitting Theorem supplies one new $\mathbb{R}$ factor from a line in the current residual manifold, while the residual factor remains complete and has nonnegative Ricci curvature because it is a totally geodesic product factor. Composing the successive product isometries gives a single isometry from $M$ to the final residual factor times $\mathbb{R}^k$. The maximality hypothesis then identifies the final residual factor as the desired line-free manifold $N$. [/proofplan]

[step:Record the residual splittings supplied by the hypothesis] Set $(M_0,g_0):=(M,g)$. For each integer $j$ with $1 \leq j \leq k$, the hypothesis gives a connected Riemannian manifold $(M_j,g_j)$ and an isometry \begin{align*} \Phi_j : (M_{j-1},g_{j-1}) \longrightarrow (M_j \times \mathbb{R},\, g_j + d t_j^2), \end{align*} where $t_j:\mathbb{R}\to\mathbb{R}$ denotes the standard coordinate on the $j$th extracted Euclidean factor. This isometry is obtained by applying the Cheeger-Gromoll Splitting Theorem (citing a result not yet in the wiki: Cheeger-Gromoll Splitting Theorem) to a line in $(M_{j-1},g_{j-1})$. The condition that the chosen line lies in the residual factor means that the next application of the splitting theorem is made to $(M_j,g_j)$, not to one of the already produced Euclidean factors. Thus the data form an iterated residual decomposition rather than repeated rediscovery of the same Euclidean direction. [/step]

[step:Show that each residual factor remains complete and has nonnegative Ricci curvature]We prove by induction on $j$ that each residual factor $(M_j,g_j)$ is complete and satisfies $\operatorname{Ric}_{g_j}\geq 0$. For $j=0$, this is exactly the hypothesis on $(M,g)$. Assume that $(M_{j-1},g_{j-1})$ is complete and satisfies $\operatorname{Ric}_{g_{j-1}}\geq 0$ for some $1\leq j\leq k$. The isometry \begin{align*} \Phi_j : (M_{j-1},g_{j-1}) \longrightarrow (M_j \times \mathbb{R},\, g_j+d t_j^2) \end{align*} shows that the product $(M_j\times\mathbb{R},g_j+d t_j^2)$ is complete and has nonnegative Ricci curvature. For a point $a\in\mathbb{R}$, define the slice inclusion \begin{align*} \iota_a : M_j &\longrightarrow M_j\times\mathbb{R} \\ p &\longmapsto (p,a). \end{align*} The image $\iota_a(M_j)=M_j\times\{a\}$ is closed because $\{a\}$ is closed in $\mathbb{R}$. With the distance induced by the product metric, this slice is isometric to $(M_j,g_j)$; since a closed subspace of a [complete metric space](/page/Complete%20Metric%20Space) is complete, $(M_j,g_j)$ is complete. It remains to identify the Ricci tensor on the product factor. Let $p\in M_j$, let $a\in\mathbb{R}$, and let $v\in T_pM_j$. Under the splitting \begin{align*} T_{(p,a)}(M_j\times\mathbb{R}) \cong T_pM_j \oplus T_a\mathbb{R}, \end{align*} the product connection is the direct-sum connection, so the curvature of the product restricts on vectors tangent to $M_j$ to the curvature of $(M_j,g_j)$. Therefore \begin{align*} \operatorname{Ric}_{g_j}(v,v) = \operatorname{Ric}_{g_j+d t_j^2}((v,0),(v,0)). \end{align*} The right-hand side is nonnegative because the product is isometric to $(M_{j-1},g_{j-1})$, whose Ricci curvature is nonnegative by the induction hypothesis. Hence $\operatorname{Ric}_{g_j}\geq 0$. By induction, $(M_j,g_j)$ is complete and has nonnegative Ricci curvature for every $0\leq j\leq k$.[/step]

[guided]We need to justify that after splitting off one line, the remaining factor is still eligible for the next splitting. That requires two properties: completeness and nonnegative Ricci curvature. The base case is $(M_0,g_0)=(M,g)$, and the theorem assumes that $(M,g)$ is complete with $\operatorname{Ric}_{g}\geq 0$. Now fix an integer $j$ with $1\leq j\leq k$ and assume that $(M_{j-1},g_{j-1})$ is complete with $\operatorname{Ric}_{g_{j-1}}\geq 0$. The splitting hypothesis gives an isometry \begin{align*} \Phi_j : (M_{j-1},g_{j-1}) \longrightarrow (M_j \times \mathbb{R},\, g_j+d t_j^2). \end{align*} Because isometries preserve metric completeness and Ricci curvature, the product manifold $(M_j\times\mathbb{R},g_j+d t_j^2)$ is complete and has nonnegative Ricci curvature. First consider completeness of the residual factor. Choose a point $a\in\mathbb{R}$ and define the slice inclusion \begin{align*} \iota_a : M_j &\longrightarrow M_j\times\mathbb{R} \\ p &\longmapsto (p,a). \end{align*} The subset $M_j\times\{a\}$ is closed in $M_j\times\mathbb{R}$ because $\{a\}$ is closed in $\mathbb{R}$. The map $\iota_a$ is an isometry from $(M_j,g_j)$ onto this slice with the induced product metric. Since a closed subset of a complete [metric space](/page/Metric%20Space) is complete, the slice is complete, and therefore $(M_j,g_j)$ is complete. Now consider Ricci curvature. At a point $(p,a)\in M_j\times\mathbb{R}$, the tangent space splits as \begin{align*} T_{(p,a)}(M_j\times\mathbb{R}) \cong T_pM_j \oplus T_a\mathbb{R}. \end{align*} This identification is the differential of the product coordinate projections. Let $v\in T_pM_j$. The corresponding tangent vector in the product is $(v,0)$. For the product metric $g_j+d t_j^2$, the Levi-Civita connection is the direct-sum connection, so the curvature tensor on vectors tangent to the $M_j$ factor is exactly the curvature tensor of $(M_j,g_j)$. Taking the trace over an [orthonormal basis](/page/Orthonormal%20Basis) of $T_pM_j$ gives \begin{align*} \operatorname{Ric}_{g_j}(v,v) = \operatorname{Ric}_{g_j+d t_j^2}((v,0),(v,0)). \end{align*} The product Ricci curvature is nonnegative because the product is isometric to $(M_{j-1},g_{j-1})$, and the induction hypothesis gives $\operatorname{Ric}_{g_{j-1}}\geq 0$. Thus $\operatorname{Ric}_{g_j}(v,v)\geq 0$ for every $p\in M_j$ and every $v\in T_pM_j$, which means $\operatorname{Ric}_{g_j}\geq 0$. This proves the induction step. Therefore every residual factor $(M_j,g_j)$ remains complete and has nonnegative Ricci curvature.[/guided]

[step:Compose the product isometries into one global splitting] For each integer $r$ with $0\leq r\leq k$, define the Euclidean metric \begin{align*} g_{\mathrm{Euc},r}:=d t_1^2+\cdots+d t_r^2 \end{align*} on $\mathbb{R}^r$, with the convention that $\mathbb{R}^0$ is a point and $g_{\mathrm{Euc},0}$ is the zero metric. We claim that for every $r$ with $0\leq r\leq k$ there is an isometry \begin{align*} \Psi_r : (M,g) \longrightarrow (M_r\times\mathbb{R}^r,\, g_r+g_{\mathrm{Euc},r}). \end{align*} For $r=0$, take $\Psi_0=\operatorname{id}_M$. Assume $\Psi_{r-1}$ has been constructed for some $1\leq r\leq k$. Define \begin{align*} \widehat{\Phi}_r : M_{r-1}\times\mathbb{R}^{r-1} &\longrightarrow M_r\times\mathbb{R}\times\mathbb{R}^{r-1} \\ (q,s) &\longmapsto (\Phi_r(q),s), \end{align*} where $q\in M_{r-1}$ and $s\in\mathbb{R}^{r-1}$. Since $\Phi_r$ is an isometry and the identity map on $\mathbb{R}^{r-1}$ is an isometry, $\widehat{\Phi}_r$ is an isometry from \begin{align*} (M_{r-1}\times\mathbb{R}^{r-1},\, g_{r-1}+g_{\mathrm{Euc},r-1}) \end{align*} onto \begin{align*} (M_r\times\mathbb{R}\times\mathbb{R}^{r-1},\, g_r+d t_r^2+g_{\mathrm{Euc},r-1}). \end{align*} Composing with the coordinate permutation \begin{align*} \Pi_r : M_r\times\mathbb{R}\times\mathbb{R}^{r-1} &\longrightarrow M_r\times\mathbb{R}^r \\ (p,t,s_1,\dots,s_{r-1}) &\longmapsto (p,s_1,\dots,s_{r-1},t), \end{align*} which is an isometry of Euclidean factors, we define \begin{align*} \Psi_r:=\Pi_r\circ \widehat{\Phi}_r\circ \Psi_{r-1}. \end{align*} Thus $\Psi_r$ is an isometry. Induction gives an isometry \begin{align*} \Psi_k : (M,g) \longrightarrow (M_k\times\mathbb{R}^k,\, g_k+g_{\mathrm{Euc},k}). \end{align*} [/step]

[step:Use maximality to identify the final residual factor] Define \begin{align*} (N,h):=(M_k,g_k). \end{align*} The preceding step gives an isometry \begin{align*} (M,g)\cong (N\times\mathbb{R}^k,\, h+g_{\mathrm{Euc},k}). \end{align*} The second step proves that $(N,h)$ is complete and satisfies $\operatorname{Ric}_h\geq 0$. The maximality hypothesis says precisely that the final residual factor $(M_k,g_k)$ contains no line, and therefore $(N,h)$ contains no line. This is the asserted decomposition. [/step]