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On the unit sphere $S^n\subseteq {\mathbb{R}}^{n+1}$ choose the outward unit normal $N(x)=x$. For tangent vector fields $X,Y$,soIf $X,Y$ are orthonormal, the Gauss equation givesThus the round unit sphere has sectional curvature one. Tracing over an orthonormal basis gives its scalar curvature

Give $S^2\times \mathbb{R}$ the Riemannian product of the unit round metric and the Euclidean metric. It is complete and has infinite diameter. The round sphere has scalar curvature $2$, while the line has scalar curvature $0$; scalar curvature is additive under Riemannian products, soThus this manifold has a strictly positive uniform lower bound on scalar curvature but violates the conclusion of the Bonnet-Myers theorem. Its Ricci curvature vanishes in the $\mathbb{R}$ direction, showing precisely why a scalar-curvature bound is insufficient.

The Penrose inequality compares asymptotic energy and an appropriate horizon or enclosing area in geometrized units. Its physical motivation uses censorship, positive radiated energy, settling to a Kerr black hole, and Hawking's area theorem. The area variable requires care: the area of an arbitrary apparent horizon on non-time-symmetric data does not give a universally valid inequality. The Riemannian Penrose inequality uses the relevant outermost minimal surface under nonnegative scalar curvature.

For complete asymptotically flat three-dimensional Riemannian initial data with nonnegative scalar curvature and an outermost compact minimal surface boundary of area $A$, the Riemannian Penrose inequality gives the displayed bound. Equality is realized by the spatial exterior of Schwarzschild spacetime. Time-symmetric gravitational initial data have this form when the Einstein field equations and energy hypothesis give the required nonnegative curvature. This theorem does not justify substituting arbitrary non-time-symmetric apparent horizon area into the same formula.

For dimension $n\ge 3$, the Schouten tensor packages the Ricci tensor and scalar curvature in the trace part of the Riemann curvature tensor. In dimension three it is $S_{ij}=R_{ij}-R{g}_{ij}/4$, and the full Riemann curvature tensor is reconstructed from it.