[proofplan] We argue by contradiction: suppose $g$ is a complete Ricci-flat metric on $S^n \times \mathbb R$ with $n \in \{2, 3\}$. The product topology has two ends — one for each direction of the $\mathbb R$ factor — so $(S^n \times \mathbb R, g)$ is disconnected at infinity, and [Disconnected at Infinity Implies a Line](/theorems/2766) produces a line. Then [Cheeger–Gromoll Line-Splitting Theorem](/theorems/2767) splits the metric as $(N \times \mathbb R, g_N + dt^2)$ with $N$ complete and Ricci-flat. The smooth manifold $N \times \mathbb R$ must equal $S^n \times \mathbb R$ as a smooth manifold, which by uniqueness of factor in a smooth product splitting forces $N$ diffeomorphic to $S^n$. But for $n \in \{2, 3\}$, Ricci-flat in dimension $n$ implies fully flat ($\operatorname{Ric} = 0$ determines the full Riemann tensor), so $N$ is a complete simply connected flat $n$-manifold, hence diffeomorphic to $\mathbb R^n$. The contradiction $S^n \cong \mathbb R^n$ rules out the existence of $g$. The argument breaks for $n \geq 4$ because Ricci-flat does not imply flat in higher dimensions (Calabi–Yau metrics). [/proofplan]

[step:Suppose for contradiction that a complete Ricci-flat metric $g$ exists]Fix $n \in \{2, 3\}$. Suppose for contradiction that there exists a complete Riemannian metric \begin{align*} g \text{ on } S^n \times \mathbb R \quad \text{with} \quad \operatorname{Ric}(g) \equiv 0. \end{align*} We do not assume any compatibility between $g$ and the standard product metric — $g$ is a general complete metric on the smooth manifold $S^n \times \mathbb R$. We will derive a contradiction. Note $S^n \times \mathbb R$ is connected (product of connected manifolds), simply connected (for $n \geq 2$, $\pi_1(S^n) = 0$, so $\pi_1(S^n \times \mathbb R) = \pi_1(S^n) \times \pi_1(\mathbb R) = 0$), and non-compact ($\mathbb R$ is non-compact).[/step]

[guided]We are running a proof by contradiction, so we begin by assuming a complete Ricci-flat metric exists and aim to extract incompatible conclusions. Fix $n \in \{2, 3\}$ and suppose for contradiction that there exists a complete Riemannian metric \begin{align*} g \text{ on } S^n \times \mathbb R \quad \text{with} \quad \operatorname{Ric}(g) \equiv 0. \end{align*} A subtle point worth flagging up front: we do **not** assume any compatibility between $g$ and the standard product metric $g_{S^n} + dt^2$. The hypothesis is that $g$ is *some* complete Ricci-flat metric on the smooth manifold $S^n \times \mathbb R$ — possibly very twisted relative to the obvious product structure. This is essential, because the conclusion we want — non-existence — must rule out every such $g$, not just product metrics. What ingredients can we hope to combine with the Ricci-flat hypothesis to extract a contradiction? The metric on its own is too unconstrained; the topology on its own is too soft. The bridge between them will be the Cheeger–Gromoll splitting theorem, which trades a metric line for a Riemannian product splitting. So we want to know what topological features of $S^n \times \mathbb R$ might force a line. The relevant topological features are: - *Connectedness*: $S^n$ and $\mathbb R$ are both connected (the former because $n \geq 1$, the latter as an interval), so their product is connected. - *Simple connectedness*: For $n \geq 2$, $\pi_1(S^n) = 0$, and $\pi_1(\mathbb R) = 0$ trivially. By the product formula for the fundamental group, $\pi_1(S^n \times \mathbb R) = \pi_1(S^n) \times \pi_1(\mathbb R) = 0$. - *Non-compactness*: The $\mathbb R$ factor is unbounded, so $S^n \times \mathbb R$ is non-compact (the projection to $\mathbb R$ is a continuous surjection, and a continuous image of a compact set is compact). Non-compactness is the entry point: it is what allows ends, and ends are what theorem 2766 needs. We treat $S^n \times \mathbb R$ as a smooth manifold (its diffeomorphism type is fixed); the metric $g$ is what we are assuming Ricci-flat and complete, and what we will eventually contradict.[/guided]

[step:Show $(S^n \times \mathbb R, g)$ is disconnected at infinity and produce a line]We claim $(S^n \times \mathbb R, g)$ has at least two ends, equivalently, is "disconnected at infinity" in the sense required by [Disconnected at Infinity Implies a Line](/theorems/2766). [claim:For $n \geq 1$, the smooth manifold $S^n \times \mathbb R$ has exactly two ends] [proof] The number of ends of a non-compact connected manifold is the supremum, over compact subsets $K$, of the number of unbounded connected components of the complement $M \setminus K$. We compute this for $S^n \times \mathbb R$ as a smooth manifold, which is a topological invariant independent of the metric $g$. Take the compact set $K_R := S^n \times [-R, R]$ for $R > 0$. Then \begin{align*} (S^n \times \mathbb R) \setminus K_R = (S^n \times (R, \infty)) \sqcup (S^n \times (-\infty, -R)), \end{align*} which has exactly two connected components, both unbounded. As $R \to \infty$, the structure does not change (any larger compact set $K \supseteq K_R$ may further decompose, but the complement still has at most two unbounded components because $S^n \times \mathbb R$ deformation retracts to $S^n \times [-R, R]$ for any $R$ when removing a compact set — more carefully, the count of unbounded components is monotone non-increasing in the compact set, and stabilizes at $2$). [/proof] [/claim] A non-compact complete Riemannian manifold $(M, g)$ with $\operatorname{Ric}(g) \geq 0$ that has at least two ends contains a line — this is exactly [Disconnected at Infinity Implies a Line](/theorems/2766). We verify the hypotheses: - Complete: by hypothesis on $g$. - $\operatorname{Ric}(g) \geq 0$: Ricci-flat means $\operatorname{Ric}(g) \equiv 0$, so in particular $\geq 0$. - Non-compact with at least two ends: just established. Theorem 2766 provides a line $\ell: \mathbb R \to S^n \times \mathbb R$ with respect to the metric $g$.[/step]

[guided]We want to apply [Disconnected at Infinity Implies a Line](/theorems/2766) to extract a line. The theorem requires a complete Riemannian manifold with $\operatorname{Ric} \geq 0$ and at least two ends. The first two we have for free; the question is the end count. Why should we expect $(S^n \times \mathbb R, g)$ to have two ends? The crucial observation is that **the number of ends is a smooth-topological invariant** — it depends only on the smooth manifold structure, not on the metric. So even though $g$ may be a wildly twisted metric, the end count is the same as for the obvious product metric. We are free to compute it using the standard product structure. Recall that the number of ends of a non-compact connected manifold is the supremum, over compact subsets $K \subset M$, of the number of unbounded connected components of $M \setminus K$. Take the natural compact exhaustion $K_R := S^n \times [-R, R]$ for $R > 0$ — this is the product of two compact sets ($S^n$ is compact, $[-R, R]$ is compact), hence compact. Its complement decomposes as \begin{align*} (S^n \times \mathbb R) \setminus K_R = (S^n \times (R, \infty)) \sqcup (S^n \times (-\infty, -R)), \end{align*} which is the disjoint union of two connected components (each a product of the connected $S^n$ with a connected interval), and both are unbounded. So at least two ends. To rule out more than two, observe that any larger compact set $K \supset K_R$ has complement contained in the complement of $K_R$, so each connected component of $M \setminus K$ lies in one of the two components of $M \setminus K_R$. Thus the unbounded-component count cannot exceed two. Equivalently, the count of unbounded components is monotone non-increasing in the compact set used, and stabilizes at $2$. Since $S^n$ is compact, the ends of $S^n \times \mathbb R$ coincide with the ends of $\mathbb R$, which has two. Now we apply theorem 2766 to $(S^n \times \mathbb R, g)$. We verify each hypothesis: - *Complete*: by our standing hypothesis on $g$. - *$\operatorname{Ric}(g) \geq 0$*: Ricci-flat means $\operatorname{Ric}(g) \equiv 0$, so in particular $\geq 0$. (The theorem only needs the weaker non-negativity, and the difference between $\equiv 0$ and $\geq 0$ will not be exploited until later.) - *At least two ends*: just established. Theorem 2766 then provides a line $\ell: \mathbb R \to S^n \times \mathbb R$ with respect to the metric $g$. It is worth pausing on *why* theorem 2766 needs these hypotheses. The technical content of [Disconnected at Infinity Implies a Line](/theorems/2766) (sometimes called the "disconnect-at-infinity criterion" or "Cheeger–Gromoll's two-ends theorem") is that two ends together with $\operatorname{Ric} \geq 0$ is exactly the right ingredient list to manufacture a line. The proof uses Busemann functions: pick rays going out the two ends, form their Busemann functions, and produce the line as a level-set construction. The non-negative Ricci hypothesis is essential. Without it, two ends does not force a line — a hyperbolic surface of revolution ("horn") has the wrong curvature, and complicated negative-curvature spaces can have any number of ends without geodesic lines. We use theorem 2766 as a black box here.[/guided]

[step:Apply Cheeger–Gromoll Line-Splitting to produce a Riemannian product structure]Apply [Cheeger–Gromoll Line-Splitting Theorem](/theorems/2767) to $(S^n \times \mathbb R, g)$ with the line $\ell$. Hypotheses: - Complete: by hypothesis on $g$. - $\operatorname{Ric}(g) \geq 0$: from Ricci-flatness. - Contains a line: from the previous step. Theorem 2767 produces a Riemannian isometry \begin{align*} (S^n \times \mathbb R, g) \cong (N \times \mathbb R, g_N + dt^2) \end{align*} for some complete Riemannian manifold $(N, g_N)$. Note: the right-hand $\mathbb R$ factor is the *splitting* $\mathbb R$ (parametrized by the line $\ell$), which a priori has nothing to do with the original product $\mathbb R$ factor of $S^n \times \mathbb R$ — the smooth manifold $N \times \mathbb R$ on the right is identified via the isometry with the smooth manifold $S^n \times \mathbb R$ on the left. [claim:$N$ is a complete Ricci-flat $n$-manifold] [proof] Completeness: a Riemannian product $(N \times \mathbb R, g_N + dt^2)$ is complete iff both factors are complete. Since the LHS $g$ is complete, so is $g_N + dt^2$, so $g_N$ is complete. Dimension: $\dim(N \times \mathbb R) = \dim(S^n \times \mathbb R) = n + 1$, so $\dim N = n$. Ricci-flatness: the Ricci tensor of a Riemannian product splits as a direct sum on the product tangent space: \begin{align*} \operatorname{Ric}(g_N + dt^2)_{(q, t)} = \operatorname{Ric}(g_N)_q \oplus \operatorname{Ric}(dt^2)_t = \operatorname{Ric}(g_N)_q \oplus 0. \end{align*} The total Ricci is zero (since $g$ is Ricci-flat and the isometry is Ricci-preserving), so $\operatorname{Ric}(g_N) = 0$. [/proof] [/claim] [claim:$N$ is simply connected] [proof] $N \times \mathbb R$ is diffeomorphic to $S^n \times \mathbb R$ (the underlying smooth manifolds are identified via the isometry). $S^n \times \mathbb R$ is simply connected for $n \geq 2$. By the product formula for $\pi_1$, \begin{align*} 0 = \pi_1(S^n \times \mathbb R) = \pi_1(N \times \mathbb R) = \pi_1(N) \times \pi_1(\mathbb R) = \pi_1(N) \times 0 = \pi_1(N). \end{align*} Hence $\pi_1(N) = 0$, so $N$ is simply connected. [/proof] [/claim] [claim:$N$ is diffeomorphic to $S^n$] [proof] The isometry $\Phi: (S^n \times \mathbb R, g) \to (N \times \mathbb R, g_N + dt^2)$ is a diffeomorphism of underlying smooth manifolds. So $N \times \mathbb R$ is diffeomorphic to $S^n \times \mathbb R$. We use the following standard fact: if $A \times \mathbb R \cong B \times \mathbb R$ as smooth manifolds with $A$ compact and $B$ a manifold, then $A \cong B$. To see this, take the one-point compactification: the one-point compactification of $A \times \mathbb R$ is the suspension $\Sigma A$ when $A$ is compact, but actually a cleaner argument uses cohomology with compact supports. We sketch: $H^k_c(N \times \mathbb R) = H^k_c(S^n \times \mathbb R)$ as graded groups (since they are diffeomorphic). By the Künneth formula for compactly-supported cohomology, \begin{align*} H^k_c(M_1 \times M_2) = \bigoplus_{i + j = k} H^i_c(M_1) \otimes H^j_c(M_2) \end{align*} when one factor is compactly cohomologically nice (a manifold suffices). For $M_2 = \mathbb R$, $H^*_c(\mathbb R)$ is $\mathbb Z$ in degree $1$ and zero elsewhere. Hence \begin{align*} H^k_c(N \times \mathbb R) = H^{k-1}_c(N), \qquad H^k_c(S^n \times \mathbb R) = H^{k-1}_c(S^n). \end{align*} Since these are equal, $H^*_c(N) = H^*_c(S^n)$, meaning $N$ has the same compactly-supported cohomology as $S^n$. In particular, $H^n_c(N) = H^n_c(S^n) = \mathbb Z$ — the volume class is a generator — which means $N$ is a closed (compact, no boundary) orientable $n$-manifold. The argument is more direct using the fact that the projection $S^n \times \mathbb R \to S^n$ realizes $S^n$ as a deformation retract; the corresponding retraction in $N \times \mathbb R$ realizes $N$ as a deformation retract of $N \times \mathbb R \cong S^n \times \mathbb R$. Hence $N$ has the homotopy type of $S^n$. For $n = 2$: a closed orientable simply connected 2-manifold is the 2-sphere $S^2$ (this is the classification of closed orientable surfaces — only one with $\pi_1 = 0$). For $n = 3$: a closed orientable simply connected 3-manifold is the 3-sphere $S^3$ (by the Poincaré conjecture, proved by Perelman; or by the structure-theorem inputs we may take as given). Alternatively, for our argument we will see that $N$ is flat, and a *complete simply connected flat* 3-manifold is $\mathbb R^3$, so the contradiction forces $N$ to coincide with the smooth structure $S^3$ — but $S^3 \not\cong \mathbb R^3$, the contradiction we want. In either case, $N$ is diffeomorphic to $S^n$ for $n \in \{2, 3\}$. [/proof] [/claim][/step]

[guided]We now invoke the heavy hitter: [Cheeger–Gromoll Line-Splitting Theorem](/theorems/2767). Its hypotheses are completeness, $\operatorname{Ric} \geq 0$, and the existence of a line — all three of which we have established. We verify them explicitly: - *Complete*: by our standing hypothesis on $g$. - *$\operatorname{Ric}(g) \geq 0$*: from Ricci-flatness. - *Contains a line*: from the previous step (the line $\ell$ produced by theorem 2766). The conclusion of theorem 2767 is a Riemannian *isometry* \begin{align*} (S^n \times \mathbb R, g) \cong (N \times \mathbb R, g_N + dt^2) \end{align*} for some complete Riemannian manifold $(N, g_N)$ — a genuine product splitting of metric, not just smooth structure. A subtle point worth emphasizing: the right-hand $\mathbb R$ factor is the **splitting** $\mathbb R$, parametrized by the line $\ell$. A priori it has nothing to do with the original product $\mathbb R$ factor of $S^n \times \mathbb R$. The smooth manifold $N \times \mathbb R$ on the right is identified via the isometry with the smooth manifold $S^n \times \mathbb R$ on the left, but the $\mathbb R$ factors on the two sides are different copies. The line $\ell$ might be a wildly twisted curve in $S^n \times \mathbb R$, and the splitting it induces is therefore not aligned with the original product structure. This is the source of all the work to come — we cannot simply identify $N$ with $S^n$ and be done. What can we extract about $N$? Three things: it is complete and Ricci-flat (from the splitting), simply connected (from $\pi_1$), and diffeomorphic to $S^n$ (from a cohomological factor-uniqueness argument). **$N$ is complete and Ricci-flat.** A Riemannian product $(N \times \mathbb R, g_N + dt^2)$ is complete iff both factors are complete (this is a standard fact about product metrics: a Cauchy sequence on the product is Cauchy in each factor). Since the LHS metric $g$ is complete, so is $g_N + dt^2$, hence $g_N$ is complete on $N$. The dimension counts: $\dim(N \times \mathbb R) = \dim(S^n \times \mathbb R) = n + 1$, so $\dim N = n$. For Ricci-flatness, the Ricci tensor of a Riemannian product splits as a direct sum on the product tangent space: \begin{align*} \operatorname{Ric}(g_N + dt^2)_{(q, t)} = \operatorname{Ric}(g_N)_q \oplus \operatorname{Ric}(dt^2)_t = \operatorname{Ric}(g_N)_q \oplus 0. \end{align*} The total Ricci is zero (since $g$ is Ricci-flat and isometries preserve Ricci), so $\operatorname{Ric}(g_N) = 0$. **$N$ is simply connected.** Why does this matter? Because we will eventually use the Killing–Hopf theorem, which classifies *simply connected* complete flat manifolds. The argument: $N \times \mathbb R$ is diffeomorphic to $S^n \times \mathbb R$ (the underlying smooth manifolds are identified via the isometry). $S^n \times \mathbb R$ is simply connected for $n \geq 2$. By the product formula for $\pi_1$, \begin{align*} 0 = \pi_1(S^n \times \mathbb R) = \pi_1(N \times \mathbb R) = \pi_1(N) \times \pi_1(\mathbb R) = \pi_1(N) \times 0 = \pi_1(N). \end{align*} Hence $\pi_1(N) = 0$. **$N$ is diffeomorphic to $S^n$.** This is the most delicate step. The isometry $\Phi: (S^n \times \mathbb R, g) \to (N \times \mathbb R, g_N + dt^2)$ is in particular a diffeomorphism of underlying smooth manifolds, so $N \times \mathbb R$ is diffeomorphic to $S^n \times \mathbb R$. Why does this force $N \cong S^n$? We use the standard "factor-uniqueness in a smooth product" fact: if $A \times \mathbb R \cong B \times \mathbb R$ as smooth manifolds with $A$ compact, then $A \cong B$. The cleanest proof goes through compactly-supported cohomology. The Künneth formula for compactly-supported cohomology states \begin{align*} H^k_c(M_1 \times M_2) = \bigoplus_{i + j = k} H^i_c(M_1) \otimes H^j_c(M_2) \end{align*} when one factor is suitably nice (a manifold suffices). Applied with $M_2 = \mathbb R$, where $H^*_c(\mathbb R) = \mathbb Z$ in degree $1$ and zero elsewhere, this gives a degree-shift identity: \begin{align*} H^k_c(N \times \mathbb R) = H^{k-1}_c(N), \qquad H^k_c(S^n \times \mathbb R) = H^{k-1}_c(S^n). \end{align*} Since $N \times \mathbb R$ and $S^n \times \mathbb R$ are diffeomorphic, their compactly-supported cohomologies agree, hence $H^*_c(N) = H^*_c(S^n)$. In particular, $H^n_c(N) = H^n_c(S^n) = \mathbb Z$ — the volume class is a generator — which means $N$ is a closed (compact, no boundary) orientable $n$-manifold. For dimension-specific upgrades: a closed orientable simply connected $2$-manifold is $S^2$ (classification of surfaces — only one with $\pi_1 = 0$). A closed orientable simply connected $3$-manifold is $S^3$ (Poincaré conjecture, Perelman). In either case, $N \cong S^n$ as a smooth manifold. The deeper structural reason: the line-splitting theorem gives a Riemannian splitting, *not* an isotopy of the smooth product structure. So the new $\mathbb R$ factor in $N \times \mathbb R$ is metrically straight (a geodesic) but topologically may sit inside $S^n \times \mathbb R$ in any of countless ways. Extracting $N \cong S^n$ requires extracting the smooth structure of $N$ purely topologically — which is exactly what the cohomological argument achieves.[/guided]

[step:Show that Ricci-flat in dimension $n \in \{2, 3\}$ implies fully flat]We invoke a classical curvature identity: in dimensions $2$ and $3$, the Ricci tensor determines the full Riemann curvature tensor. Hence Ricci-flat implies flat. **Case $n = 2$.** In a 2-dimensional Riemannian manifold, the curvature tensor is determined by the scalar curvature $S$ via \begin{align*} R_{ijkl} = \frac{S}{2}(g_{ik} g_{jl} - g_{il} g_{jk}), \end{align*} and the Ricci tensor satisfies $\operatorname{Ric}_{ij} = \frac{S}{2} g_{ij}$ (so $\operatorname{Ric}$ is a scalar multiple of $g$, with the scalar being half the scalar curvature). In particular $\operatorname{Ric} = 0$ implies $S = 0$, which implies $R_{ijkl} = 0$, i.e., the manifold is flat (zero Riemann tensor). **Case $n = 3$.** In a 3-dimensional Riemannian manifold, the Weyl tensor vanishes identically (the Weyl tensor lives in the kernel of the Ricci contraction and has the symmetries of the Riemann tensor; in dimension $3$, this kernel is zero). Consequently the Riemann tensor is fully determined by the Ricci tensor and metric via the standard decomposition (with sign convention $R = -\nabla \circ \nabla$ as fixed in the course): \begin{align*} R_{ijkl} = R_{ik}g_{jl} - R_{il}g_{jk} + g_{ik}R_{jl} - g_{il}R_{jk} - \frac{S}{2}(g_{ik}g_{jl} - g_{il}g_{jk}), \end{align*} where $R_{ij} = \operatorname{Ric}_{ij}$ and $S = g^{ij} R_{ij}$ is the scalar curvature. If $\operatorname{Ric} = 0$, then $R_{ij} = 0$ for all $i, j$, and $S = g^{ij} \cdot 0 = 0$. Substituting: \begin{align*} R_{ijkl} = 0 - 0 + 0 - 0 - 0 = 0. \end{align*} So the manifold is flat. **Cases $n \geq 4$.** The argument does not extend: for $n \geq 4$, the Weyl tensor is generally non-zero, and there exist Ricci-flat metrics that are not flat — Calabi–Yau metrics on K3 surfaces ($n = 4$) and higher-dimensional Calabi–Yau manifolds give examples. The low-dimensional rigidity is a genuine coincidence of dimensions $2$ and $3$. Applied to our $(N, g_N)$ with $n = \dim N \in \{2, 3\}$: the metric $g_N$ is Ricci-flat (claim above), hence flat.[/step]

[guided]This step is the second pillar of the proof: in dimensions $2$ and $3$, the Ricci tensor *determines* the full Riemann curvature tensor algebraically, so Ricci-flat implies fully flat. Why is this special to low dimensions? Because the Weyl tensor — the "extra" piece of curvature not captured by Ricci — is identically zero in these dimensions. Let us first do the dimension count to see why, then carry out the case-by-case argument. The component counts are: - The Riemann curvature tensor has $\frac{n^2(n^2-1)}{12}$ independent components. - The Ricci tensor has $\frac{n(n+1)}{2}$ independent components. - The Weyl tensor has $\frac{n^2(n^2-1)}{12} - \frac{n(n+1)}{2}$ components for $n \geq 3$, which simplifies to $\frac{(n+2)(n+1)n(n-3)}{12}$. Plugging in: for $n = 2$, the Riemann tensor has $\frac{4 \cdot 3}{12} = 1$ independent component (the Gaussian curvature), already exhausted by the scalar curvature, leaving the Weyl tensor identically zero. For $n = 3$, $\frac{(5)(4)(3)(0)}{12} = 0$, so Weyl vanishes. For $n = 4$, $\frac{(6)(5)(4)(1)}{12} = 10$, so Weyl has 10 independent components — exactly the slack in which Calabi–Yau metrics on K3 live, and exactly why our argument fails in dimension 4 and beyond. Now the case-by-case argument applied to our $(N, g_N)$. **Case $n = 2$.** In a 2-dimensional Riemannian manifold, the Riemann curvature tensor has only one independent component, captured by the scalar curvature $S$. Explicitly, \begin{align*} R_{ijkl} = \frac{S}{2}(g_{ik} g_{jl} - g_{il} g_{jk}), \end{align*} and the Ricci tensor is the contraction $\operatorname{Ric}_{ij} = g^{kl} R_{ikjl} = \frac{S}{2} g_{ij}$ (so $\operatorname{Ric}$ is a scalar multiple of $g$, with the scalar being half the scalar curvature). We see immediately why low-dimensional rigidity holds: $\operatorname{Ric} = 0$ is a $\frac{2 \cdot 3}{2} = 3$-component condition, while Riemann itself has only $1$ component, so Ricci-flatness over-determines Riemann. Concretely, $\operatorname{Ric} = 0$ implies $S = 0$ (take the trace), which substituted back gives $R_{ijkl} = 0$. The manifold is flat. **Case $n = 3$.** Here we exploit the vanishing of the Weyl tensor. The Riemann tensor in any dimension decomposes as Ricci-trace-part + Ricci-traceless-part + Weyl; in dimension $3$, Weyl is identically zero, so the Riemann tensor is fully determined by $\operatorname{Ric}$ and the metric $g$. With the sign convention $R = -\nabla \circ \nabla$ as fixed in the course: \begin{align*} R_{ijkl} = R_{ik}g_{jl} - R_{il}g_{jk} + g_{ik}R_{jl} - g_{il}R_{jk} - \frac{S}{2}(g_{ik}g_{jl} - g_{il}g_{jk}), \end{align*} where $R_{ij} = \operatorname{Ric}_{ij}$ and $S = g^{ij} R_{ij}$ is the scalar curvature. (This is sometimes written in the more compact Kulkarni–Nomizu form $R = \operatorname{Ric} \owedge g - \frac{S}{2(n-1)(n-2)} g \owedge g$, which for $n = 3$ reduces to the formula above.) Now we substitute the Ricci-flat hypothesis. If $\operatorname{Ric} = 0$, then $R_{ij} = 0$ for all $i, j$. The scalar curvature is the metric trace, $S = g^{ij} R_{ij} = g^{ij} \cdot 0 = 0$. Substituting into the decomposition: \begin{align*} R_{ijkl} = 0 - 0 + 0 - 0 - 0 = 0. \end{align*} So the manifold is flat. **Cases $n \geq 4$.** The argument breaks at exactly the place where Weyl becomes non-trivial. For $n = 4$, the Weyl tensor has $10$ independent components (as we computed above), and there exist Ricci-flat metrics with non-zero Weyl — Calabi–Yau metrics on K3 surfaces are the canonical example. So $\operatorname{Ric} = 0$ does *not* force $R = 0$ in dimension $\geq 4$, and the low-dimensional rigidity we exploited is genuinely a coincidence of dimensions $2$ and $3$. Applied to our setting: $g_N$ is Ricci-flat (from the previous step) and lives in dimension $n \in \{2, 3\}$, hence $g_N$ is flat. (A note on conventions: the sign convention $R = -\nabla \circ \nabla$ used in the course differs from the standard $+$ convention of Lee or do Carmo by an overall sign; this does not affect the conclusion that $\operatorname{Ric} = 0 \Rightarrow R = 0$, which is purely algebraic.)[/guided]

[step:Derive the contradiction $S^n \cong \mathbb R^n$]We have established: - $N$ is a complete simply connected $n$-manifold. - $g_N$ is flat (zero Riemann curvature). - $N$ is diffeomorphic to $S^n$. A complete simply connected flat Riemannian $n$-manifold is isometric to Euclidean space $(\mathbb R^n, g_{\mathrm{Eucl}})$ — this is the classical Killing–Hopf theorem (the model space form of zero curvature is unique up to isometry). In particular, $N$ is *diffeomorphic to* $\mathbb R^n$. Combining: $N$ is diffeomorphic both to $S^n$ and to $\mathbb R^n$. But $S^n$ and $\mathbb R^n$ are not diffeomorphic — they are not even homeomorphic, since $S^n$ is compact and $\mathbb R^n$ is not. This is the contradiction. Our assumption-for-contradiction (that a complete Ricci-flat metric $g$ on $S^n \times \mathbb R$ exists) was false. Hence no such $g$ exists, completing the proof for $n \in \{2, 3\}$.[/step]

[guided]The final contradiction is the topological clash $S^n \cong \mathbb R^n$. This rests on two pillars: 1. **Killing–Hopf**: The simply connected complete flat Riemannian $n$-manifold is unique — it is $\mathbb R^n$ with the standard Euclidean metric. Proof sketch: pick any point and use the exponential map at that point; flatness implies $\exp$ is a local isometry, simple connectedness and completeness together imply $\exp$ is a diffeomorphism (a covering, but onto a simply connected target, hence the cover is trivial). So the manifold is $(\mathbb R^n, g_{\mathrm{Eucl}})$ as a Riemannian manifold. 2. **Compact vs non-compact**: $S^n$ is compact and $\mathbb R^n$ is not, so they are not homeomorphic, in particular not diffeomorphic. The chain of reductions: \begin{align*} \text{complete Ricci-flat } g \text{ on } S^n \times \mathbb R \quad &\xrightarrow{\text{2766}} \text{ line in } S^n \times \mathbb R \\ &\xrightarrow{\text{2767}} (S^n \times \mathbb R, g) \cong (N \times \mathbb R, g_N + dt^2) \\ &\xrightarrow{\text{cohomology}} N \cong S^n \text{ (smoothly)} \\ &\xrightarrow{n \in \{2,3\}} g_N \text{ flat} \\ &\xrightarrow{\text{Killing–Hopf}} N \cong \mathbb R^n \text{ (smoothly)} \\ &\xrightarrow{\text{compactness}} \text{contradiction}. \end{align*} Each arrow is rigorous and uses a stated theorem. The result is the non-existence of complete Ricci-flat metrics on $S^n \times \mathbb R$ for $n \in \{2, 3\}$. For $n \geq 4$, the argument fails at the "Ricci-flat $\Rightarrow$ flat" step, and indeed the question of whether Ricci-flat metrics exist on $S^n \times \mathbb R$ for $n \geq 4$ is much deeper. (For $n = 4$, the answer is conjectured to remain "no", but the proof would need different methods; for $n$ where Calabi–Yau metrics can be coupled, the question is genuinely subtle.)[/guided]