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On a closed connected Riemannian manifold with nonnegative Ricci curvature, harmonic one-forms are parallel under nonnegative Ricci curvature. Evaluation at one point injects their space into the cotangent space. The Hodge decomposition theorem identifies its dimension with the first Betti number, giving $b_1(M)\le \dim M$. For example, a genus-two closed orientable surface times a circle has dimension three and first Betti number five, so no Riemannian metric on that product has nonnegative Ricci curvature.

For a complete connected $n$-dimensional Riemannian manifold with nonnegative Ricci curvature, the displayed ratio is nonincreasing, tends to one at zero, and is at most one. In particular $\mathrm{Vol}B(p,r)\le {\omega }_n{r}^n$. In geodesic polar coordinates, the radial Jacobi field determinant is at most the Euclidean density $r^{n-1}$ until the cut time. The trace of the radial Riccati equation for distance spheres, with Cauchy-Schwarz inequality, proves this comparison. Equality for every radius forces the Euclidean polar metric and absence of a cut locus.

For a complete connected $n$-dimensional Riemannian manifold with $\mathrm{Ric}\ge (n-1)kg$, $k>0$, put $D=\pi /\sqrt{k}$ andHere ${\omega }_{n-1}$ is the area of the unit $(n-1)$-sphere, and $v_k(r)$ is the Riemannian volume of a radius-$r$ ball in the round curvature-$k$ sphere. The Bishop-Gromov inequality asserts that $V_p(r)/v_k(r)$ is nonincreasing for $0<r<D$, tends to one at zero, and is at most one. In geodesic polar coordinates, the radial volume density before the cut time is at most $s_k(t{)}^{n-1}$. The bound uses trace Ricci curvature; with normalized Ricci curvature it reads $\overline{\mathrm{Ric}}\ge kg$.

On a closed Riemannian manifold, a Killing vector field $X$ satisfies ${\nabla }^{\ast }\nabla X={\mathrm{Ric}}^{♯}X$ for the nonnegative rough Laplacian. This follows by differentiating ${\nabla }_i{X}_j+{\nabla }_j{X}_i=0$, contracting and commuting covariant derivatives; the contraction also gives $\mathrm{div}X=0$. Taking the inner product with $X$ and using integration by parts gives the displayed identity. If Ricci curvature is nonpositive, the nonnegative left side must vanish, so $X$ is a parallel vector field. In particular every Killing vector field on a closed flat Riemannian manifold is a parallel vector field.

For a real one-form $\alpha$, the Hodge Laplacian differs from the rough Laplacian by the action of Ricci curvature: $(\mathrm{Ric}\cdot \alpha )(X)=\alpha ({\mathrm{Ric}}^{♯}X)$, where $g({\mathrm{Ric}}^{♯}X,Y)=\mathrm{Ric}(X,Y)$. Equivalently, with the nonnegative scalar Laplacian convention,

The graph $z=x^2+y^2$ in Euclidean three-space has the displayed induced Riemannian metric and positive Gaussian curvature. It is smooth at its vertex; the polar-coordinate singularity is only a coordinate artifact. The metric dominates the Euclidean plane metric, so upward stability of Riemannian completeness proves completeness. It has infinite diameter. In dimension two Ricci curvature equals Gaussian curvature times the metric, so this is a counterexample to compactness under merely pointwise positive Ricci curvature. Its curvature tends to zero at infinity.

The product Riemannian metric of two round unit spheres has Ricci curvature equal to its metric, since the factor Ricci tensors equal their metrics and the product connection splits. Its directional Ricci curvatures are all one. A two-plane tangent to a single sphere has sectional curvature one, while a mixed plane has sectional curvature zero. This four-dimensional Einstein manifold therefore need not have constant sectional curvature.

The product Riemannian metric of the round unit two-sphere and a circle has Ricci curvature equal to $g_{S^2}$ on the sphere directions and zero on the circle direction. Its scalar curvature is constantly two, but no single scalar multiple of the whole metric equals that Ricci tensor. In two dimensions the identity $\mathrm{Ric}=\mathrm{Scal}g/2$ does make constant scalar curvature equivalent to an Einstein metric.

On a compact connected Riemannian manifold without boundary with nonnegative Ricci curvature, the Bochner formula for one-forms makes every harmonic one-form parallel. Evaluation at any point is then injective: a parallel form that is zero there stays zero along every path. Hence $\dim {\mathcal{H}}^1(M)\le \dim M$. Flat tori attain equality. Positive-definite Ricci curvature at even one point instead forces ${\mathcal{H}}^1=0$, because the Bochner integral makes its Ricci contraction vanish everywhere.

An Einstein manifold has Ricci curvature equal to a constant multiple of its metric. In the Riemannian setting, the round unit sphere has $\Lambda =n-1$. This condition is weaker than constant sectional curvature in higher dimensions. The definition can also be applied to pseudo-Riemannian metrics, but positivity is assumed in the completeness theorems discussed here.

For a compact connected Riemannian manifold with nonnegative Ricci curvature, its finitely generated fundamental group acts freely and properly discontinuously on the complete universal cover. The preceding polynomial growth of properly discontinuous isometry groups bound applies. The degree-one Hurewicz theorem identifies first integral homology with the abelianization of the fundamental group. If its free rank is $b_1(M)$, this group maps onto ${\mathbb{Z}}^{b_1(M)}$. Choosing lifts of a basis gives at least a constant times $m^{b_1(M)}$ distinct elements in a word ball of radius a constant times $m$. The upper bound of order $m^{\dim M}$ forces the displayed inequality.

On a closed Riemannian manifold with nonnegative Ricci curvature, integration of the Bochner-Weitzenbock formula for one-forms for a harmonic one-form givesBoth terms are nonnegative, so $\nabla \alpha =0$. Riemannian volume density makes the argument valid even without orientation.

This metric pulls back the round spherical metric by $(u,v)\mapsto (\cos u\cos v,\cos u\sin v,\sin u)$. It has unit Gaussian curvature and Ricci curvature equal to its metric, yet the meridian reaches a missing boundary in finite time. Every curve between $(0,0)$ and $(0,L)$ has length at least $|L|/\sqrt{2}$, so its diameter is infinite. It shows that the positive Ricci bound alone does not replace geodesic completeness in the Bonnet-Myers theorem.

Assume the trace Ricci curvature satisfies $\mathrm{Ric}\ge (n-1)kg$, with $k>0$. If points $p,q$ have Riemannian distance $D=\pi /\sqrt{k}$, the balls $B(p,r)$ and $B(q,D-r)$ are disjoint. Bishop volume comparison with a positive-curvature model bounds their volumes below by $\mathrm{Vol}(M)v_k(r)/v_k(D)$ and $\mathrm{Vol}(M)v_k(D-r)/v_k(D)$. Since $v_k(r)+v_k(D-r)=v_k(D)$, both inequalities must be equalities. Taking $r\downarrow 0$ gives $\mathrm{Vol}M=v_k(D)$, and spherical rigidity of maximal total volume yields the round sphere.

For a Riemannian manifold of dimension $n\ge 2$, the normalized directional Ricci curvature averages, rather than sums, the $n-1$ sectional curvatures through a unit vector. Thus the Bonnet-Myers theorem gives $\mathrm{diam}M\le \pi /\sqrt{k}$ from $\overline{\mathrm{Ric}}\ge kg$. With trace Ricci curvature the corresponding hypothesis is $\mathrm{Ric}\ge (n-1)kg$. The distinction changes positive constants but not the signs of Ricci curvature.

Use the Levi-Civita connection $\nabla$ of the Riemannian metric $g$ and fix the conventionThe Leibniz rule shows that this expression is linear over smooth functions in all three arguments: the differentiated coefficient terms in the first two derivatives cancel those in the bracket term. Thus $R$ is a section of ${\Lambda }^2{T}^{\ast }M\otimes \mathrm{End}(TM)$, the Riemann curvature tensor, and $R_4$ is a covariant tensor field. This convention gives positive sectional curvature on the round sphere.

Its algebraic symmetries areThe first identity follows immediately from the definition. For the second, apply $XY-YX-[X,Y]$ to $g(Z,W)$ and use compatibility of the Levi-Civita connection with the Riemannian metric. The left side is zero and the differentiated terms combine toFor the first Bianchi identity, use normal coordinates at a point $p$, so the Christoffel symbols vanish there. Torsion-freeness gives ${\Gamma }_{ij}^{\ell }={\Gamma }_{ji}^{\ell }$, andThe cyclic sum over $i,j,k$ cancels term by term. Since this is a tensor identity, it holds for arbitrary tangent vectors everywhere.

To prove pair interchange rather than assume it, abbreviate $R_4(a,b,c,d)$ to $R_{abcd}$ and put $A=R_{abcd}$, $B=R_{cdab}$. The first Bianchi identity and the two antisymmetries giveThe first Bianchi identity applied to $(d,b,a)$ with final argument $c$ gives $R_{dbca}+R_{adcb}=R_{badc}=A$. Hence $A=2B-A$, proving $A=B$.

There is also the differential second Bianchi identity,For its proof take commuting coordinate fields, normal at $p$. The Jacobi identity for operator commutators says $[{\nabla }_X,[{\nabla }_Y,{\nabla }_Z]]+\text{cyclic}=0$. On these commuting fields, $[{\nabla }_Y,{\nabla }_Z]=R(Y,Z)$; at $p$, the covariant derivatives of the coordinate arguments vanish. The commutator identity is therefore the displayed covariant derivative identity at $p$, and tensoriality establishes it everywhere.

For a two-dimensional plane $\sigma =\mathrm{span}\{X,Y\}\subset T_{p}M$, define the sectional curvature byThe denominator is the squared area of $X,Y$. Under an invertible change of plane basis, the alternating pairs in the numerator and the area in the denominator both acquire the square of the change-of-basis determinant. Thus the quotient depends only on $\sigma$. For an orthonormal basis of $\sigma$, it is simply $g(R(X,Y)Y,X)$.

The Ricci curvature is the tracewhere $(e_i)$ is any orthonormal basis of $T_{p}M$. This is independent of the basis because it is a trace. Applying the first Bianchi identity to $(e_i,X,Y)$ and pairing with $e_i$ proves that the summand difference for $\mathrm{Ric}(X,Y)-\mathrm{Ric}(Y,X)$ vanishes; the middle term $g(R(X,Y)e_i,e_i)$ is zero by the last-pair antisymmetry. Hence Ricci curvature is a symmetric bilinear form. For a unit vector $v=e_1$ extended to an orthonormal basis,For nonzero $X$, take $v=X/|X|$ and multiply the right side by $|X{|}^2$. This determines the entire Ricci curvature by the polarization identity:Finally, the scalar curvature is the metric trace of Ricci curvature. Taking its diagonal entries in an orthonormal basis and pairing the equal contributions from $(i,j)$ and $(j,i)$ yieldsIn particular, constant sectional curvature $K_0$ gives $\mathrm{Ric}=(n-1)K_{0}g$ and $\mathrm{Scal}=n(n-1)K_0$. In dimension one the traces are zero; there are no two-dimensional tangent planes and the sums are empty.