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Ricci lower bound does not control every sectional curvature in dimension at least three

Statement refuted

False claim: if a Riemannian manifold of dimension $n \geq 3$ satisfies a Ricci lower bound $Ric \geq (n - 1) k g$ , then every sectional curvature satisfies $K \geq k$ ; in particular the Ricci lower bound controls the individual sectional curvatures.

The product $M : = H^{2} (- 1) \times R^{n - 2}$ of the hyperbolic plane $H^{2} (- 1)$ of constant sectional curvature $- 1$ with the flat Euclidean factor, with the product metric, is a counterexample for every $n \geq 3$ :

$M$ is complete;
$Ric \geq - g = (n - 1) k g$ for $k : = - \frac{1}{n - 1}$ ;
a two-plane tangent to the $H^{2}$ factor at any point has $K = - 1 < k$ .

So a Ricci lower bound with $k < 0$ does not force the sectional curvature bound $K \geq k$ once $n \geq 3$ . Both factors are needed: on a surface Ricci is the sectional curvature, so no such failure occurs when $n = 2$ .

Facts & Assumptions

Given: The hyperbolic plane $H^{2} (- 1)$ realized as the upper half-plane $U^{2} = {y > 0}$ with $g_{1} = y^{- 2} (d x^{2} + d y^{2})$ , the Euclidean space $R^{n - 2}$ with the Euclidean metric and $n \geq 3$ , the product $M = H^{2} (- 1) \times R^{n - 2}$ with the product metric, and the inherited ${A C}_{ω}$ of [A1].

[A1]

The countable-choice premise is the inherited ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )), carried by the hyperbolic-completeness and product-completeness suppliers used below; the curvature and trace computations add no selection.

[F1]

Product manifolds and metrics (Products of smooth manifolds have a canonical product smooth structure, Canonical tangent and cotangent splittings for products, Coordinate criterion for a riemannian metric): the product of smooth manifolds carries its canonical product smooth structure, its tangent spaces split canonically as $T_{(p_{1}, p_{2})} (M_{1} \times M_{2}) ≅ T_{p_{1}} M_{1} \oplus T_{p_{2}} M_{2}$ , and a smooth symmetric positive-definite $(0, 2)$ -tensor field is a Riemannian metric; in a product chart $(x^{1}, \dots, x^{n_{1}}, y^{1}, \dots, y^{n_{2}})$ the product metric has the block-diagonal matrix $diag ((g_{1, i j} (x)), (g_{2, α β} (y)))$ , the first block depending only on $x$ and the second only on $y$ .

[F2]

Levi-Civita symbols and coordinate curvature (Fundamental theorem of riemannian geometry, Christoffel formula for the levi civita connection, Connection laws in directional form, Coordinate formula for the curvature tensor, Curvature is a type (1,3) tensor): the Levi-Civita connection is unique; in a chart its Christoffel symbols are $Γ^{a}_{b c} = \frac{1}{2} g^{a d} (\partial_{b} g_{c d} + \partial_{c} g_{b d} - \partial_{d} g_{b c})$ , and the curvature components are $R^{a}_{b c d} = \partial_{c} Γ^{a}_{d b} - \partial_{d} Γ^{a}_{c b} + Γ^{a}_{c e} Γ^{e}_{d b} - Γ^{a}_{d e} Γ^{e}_{c b}$ ; the curvature is $C^{\infty}$ -linear in all three argument fields, so a formula verified on a coordinate frame holds for all tangent vectors.

[F3]

The two factors (Upper half-space model geometry, Euclidean space has zero curvature, $R$ and $R^{n}$ for $n \geq 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $R$ ): the upper half-plane $U^{2} = {y > 0}$ with $g_{1} = y^{- 2} (d x^{2} + d y^{2})$ is the case $a = 1$ , $n = 2$ of the half-space model, hence a complete Riemannian surface of constant sectional curvature $- 1$ ; Euclidean $R^{n - 2}$ has identically zero Riemann curvature and is metrically complete.

[F4]

Ricci is the trace $Ric (X, Y) = tr (Z \mapsto R (Z, X) Y)$ and, in an orthonormal basis $(e_{1}, \dots, e_{n})$ , $Ric (X, Y) = \sum_{i} Rm (e_{i}, X, Y, e_{i})$ , independently of the basis (Ricci curvature, Ricci curvature is symmetric and basis independent). For orthonormal $e_{i}, e_{j}$ , $Rm (e_{i}, e_{j}, e_{j}, e_{i}) = K (span (e_{i}, e_{j}))$ (Sectional curvature).

[F5]

Product completeness (A Riemannian product is complete iff each factor is complete): a finite Riemannian product is metrically complete exactly when each factor is metrically complete.

Counterexample

1.1F1F2given

The product connection splits. [F1, F2, given] In a product chart of [F1] the metric matrix is block diagonal, the first block a function of $x$ alone and the second a function of $y$ alone, and its inverse has the same block structure and dependence. In the Christoffel formula of [F2], if the upper index and the two lower indices are not all contained in the same block, then each of the three derivative terms either differentiates a mixed component, which vanishes identically, or differentiates a component of one block with respect to a coordinate of the other block, and each term $g^{a d}$ with $a$ in one block and $d$ in the other vanishes; hence $Γ^{a}_{b c} = 0$ whenever the three indices do not all lie in one block. When all three do lie in one block the formula is literally the Christoffel formula of that factor's metric.

2.1F1F2step 1.1

The curvature splits accordingly. [F1, F2, step 1.1] Insert the symbols of step 1.1 into the coordinate curvature formula of [F2]. If all four indices lie in the first block, then the symbols with all indices in that block are exactly those of $g_{1}$ and the first block of the inverse depends only on $x$ , so the formula reproduces the coordinate formula for the curvature of $g_{1}$ ; the same holds in the second block. If the indices meet both blocks, then every derivative term differentiates either an identically zero symbol or a block symbol with respect to a coordinate of the other block, and every quadratic term contains a symbol whose indices meet both blocks; so the component vanishes. By the tensoriality in [F2] and the canonical splitting of [F1], for tangent vectors decomposed accordingly, $R^{M_{1} \times M_{2}} ((u_{1}, u_{2}), (v_{1}, v_{2})) (w_{1}, w_{2}) = (R^{M_{1}} (u_{1}, v_{1}) w_{1}, R^{M_{2}} (u_{2}, v_{2}) w_{2}) .$ Consequently a two-plane inside the first factor has the sectional curvature computed from $g_{1}$ with the same Gram determinant, and a plane inside the second factor has the sectional curvature computed from $g_{2}$ ; in particular the planes inside the Euclidean factor are flat.

3.1F3step 2.1given

The curvature table of the product. [F3, step 2.1, given] Fix a point of $M$ and an orthonormal frame $e_{1}, \dots, e_{n}$ with $e_{1}, e_{2}$ tangent to the $H^{2}$ factor and $e_{3}, \dots, e_{n}$ tangent to the $R^{n - 2}$ factor. By [F3] the hyperbolic factor has constant sectional curvature $- 1$ and the Euclidean factor has identically zero curvature, so step 2.1 gives: $K (span (e_{1}, e_{2})) = - 1, K (span (e_{i}, e_{j})) = 0 otherwise,$ because every other pair of frame vectors either lies inside the flat factor, where [F3] gives vanishing curvature, or is mixed, in which case the curvature endomorphism of step 2.1 annihilates the pair.

4.1F4step 3.1

The Ricci tensor of the product. [F4, step 3.1] By the orthonormal-basis formula of [F4], the diagonal Ricci entries are $Ric (e_{j}, e_{j}) = \sum_{i \neq j} K (span (e_{i}, e_{j}))$ . Step 3.1 gives $Ric (e_{1}, e_{1}) = Ric (e_{2}, e_{2}) = - 1$ and $Ric (e_{j}, e_{j}) = 0$ for $j \geq 3$ . The curvature splitting of step 2.1 makes the mixed Ricci entries zero; on the two-dimensional hyperbolic factor the Ricci tensor is $- g_{1}$ , so its off-diagonal entry in the chosen orthonormal frame is also zero. Hence for $X = \sum_{j} x^{j} e_{j}$ , $Ric (X, X) = - (x^{1})^{2} - (x^{2})^{2} \geq - \sum_{j = 1}^{n} (x^{j})^{2} = - ∣ X ∣^{2},$ with equality exactly on $span (e_{1}, e_{2})$ . Thus $Ric \geq - g$ , and since $k = - 1 / (n - 1)$ gives $(n - 1) k = - 1$ , this is exactly $Ric \geq (n - 1) k g$ .

5.1F3F5step 4.1∎

The sectional bound fails and $M$ is complete. [F3, F5, step 4.1] The two-plane tangent to the hyperbolic factor has $K = - 1 < - \frac{1}{n - 1} = k$ , because $n \geq 3$ makes $\frac{1}{n - 1} < 1$ . So the Ricci lower bound does not imply $K \geq k$ . For completeness: by [F5] the product is metrically complete exactly when both factors are, and the hyperbolic plane and Euclidean space are metrically complete by [F3]; hence $M$ is metrically complete. In $n = 2$ the same construction degenerates to $H^{2} (- 1)$ alone, and there Ricci is the sectional curvature, so the failure genuinely needs dimension at least three. The product, its factors and the two-plane are explicit, so the inherited ${A C}_{ω}$ of [A1] is not drawn on beyond its declaration.

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Sources

Ved Datar, Lectures on Riemannian Geometry (2025) (standard reference, not scraped)
J.-H. Eschenburg, Comparison Theorems in Riemannian Geometry (standard reference, not scraped)