Ricci curvature and volume comparison

For a tangent vector v and an orthonormal basis eᵢ, define $\operatorname{Ric}(v,v)=\sum_i g(R(e_i,v)v,e_i)$ using our curvature sign. The trace is basis independent. If v is unit, choose it as one basis vector; the remaining terms are the sectional curvatures of the planes it spans with the perpendicular basis vectors. In an n-dimensional constant-curvature space this gives Ric=(n−1)Kg, so a round sphere has positive Ricci curvature.

A Ricci lower bound constrains a sum and need not bound each sectional curvature individually in dimensions above two. Confusing these conditions can reverse an intended comparison argument. Riemannian volume also comes from the metric: in coordinates its density is the square root of the determinant of the metric matrix. A nonconstant coordinate density is not itself evidence for Ricci curvature, since a curvilinear coordinate system can give a varying density even in Euclidean space.

Let M be a complete connected n-dimensional Riemannian manifold without boundary, with n≥2 and Ric≥(n−1)kg for a real constant k. Write Vₖⁿ(r) for the radius-r ball volume in the simply connected n-dimensional constant-sectional-curvature-k model. Bishop–Gromov says the ratio of a metric ball's volume to this model volume is nonincreasing as r increases. For positive k, state this comparison for 0<r<π/√k, before the model's antipodal radius; for nonpositive k the radius range is unrestricted.

The ratio approaches one at small radius, so every allowed-radius ball has volume no greater than its model counterpart. The theorem is global despite its local-looking curvature hypothesis. Its proof compares radial Jacobi volume densities before cut points and then handles the cut locus in integration; we state the theorem rather than deriving that full measure-theoretic argument here. Replacing a Ricci lower bound by an upper bound does not preserve the asserted inequality direction.

Bonnet–Myers states that under the same complete connected setting, Ric≥(n−1)kg with k>0 forces diameter at most π/√k and makes M compact. A proof uses transverse variations of a long minimizing geodesic: summing their index forms traces curvature into Ricci and contradicts minimality past the stated length. This is a separate conclusion from volume comparison. Merely requiring Ric to be positive at every point without a uniform positive lower constant is a weaker condition and does not give this diameter estimate.

The model examples also explain scaling. Multiplying the metric by c² multiplies metric radii by c and volume elements by cⁿ, while normalized sectional curvatures scale by c⁻². A volume formula written in coordinate radius must therefore be converted before comparing intrinsic balls. In a constant rescaled Euclidean metric, these effects cancel so that the intrinsic radius-r ball still has volume ωₙrⁿ. Apparent coordinate density changes alone do not alter this intrinsic model calculation.