Ricci curvature ($\mathrm{Ric}$)

Ricci curvature is the trace of the Riemann curvature tensor in its first and third arguments. For a unit tangent vector $X$ and an orthonormal basis $X,e_2,\dots ,e_n$,

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 $\overline{\mathrm{Ric}}\ge kg$.

Under the hypotheses of Bishop volume comparison with a positive-curvature model, the Bonnet-Myers theorem gives diameter at most $D$. If total Riemannian volume equals $v_k(D)$, every comparison ball ratio is one. In a small normal ball the pointwise radial density bound therefore becomes equality. Writing $h={\partial }_r\log J$, the traced radial Riccati equation for distance spheres and Cauchy-Schwarz inequality force the radial operator to equal $(s_k^{\prime }/s_k)I$. The angular metric is consequently $s_k(r{)}^2{g}_{S^{n-1}}$, so the metric is locally round. The classification of complete positive constant-curvature manifolds gives a quotient $S_k^n/\Gamma$, whose volume is $v_k(D)/|\Gamma |$. Equality forces $\Gamma$ trivial.

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.

The Bishop-Gromov inequality forces every volume ratio to equal one if the asymptotic ratio is one. Equality in the radial Jacobian comparison forces the radial operator $A(X)={\nabla }_X{\partial }_r$ of each distance sphere to be $r^{-1}I$ (it is the negative of the outward shape operator in the convention $S=-\nabla \nu$), by the traced radial Riccati equation for distance spheres and equality in Cauchy-Schwarz inequality. Any finite cut time would remove a positive cone of radial volume at a later radius, contradicting equality. Thus the Riemannian exponential map is global and one-to-one, and the polar metric is $d{r}^2+r^2{g}_{S^{n-1}}$, the Euclidean metric.

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.

At a point of a Riemannian manifold, Ricci curvature is isotropic when $\mathrm{Ric}(v,v)/g(v,v)$ is independent of nonzero tangent direction. Polarization makes this equivalent to the displayed tensor identity. In dimension three, sectional curvatures from Ricci curvature in dimension three gives every sectional curvature at that point the value $c_p/2$. In higher dimensions isotropic Ricci curvature does not determine every two-plane curvature.

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.

A Riemannian or pseudo-Riemannian metric is Einstein when its Ricci tensor is a constant scalar multiple of the metric. Constant-sectional-curvature metrics satisfy this condition with $\Lambda =(d-1)k$. The Killing-form metric on a real semisimple Lie group is another example, with $\Lambda =-1/4$ for $g=B$ in the curvature convention $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$.

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.

For an orthonormal triple and distinct $i,j,k$, tracing sectional curvature gives ${\mathrm{Ric}}_{ii}=K_{ij}+K_{ik}$. Solving the three equations gives the displayed formula. Equivalently $K_{ij}={\mathrm{Ric}}_{ii}+{\mathrm{Ric}}_{jj}-\mathrm{Scal}/2$. This pointwise assertion does not require the triple to diagonalize the Ricci curvature.

Scalar curvature is the complete trace of the Riemann curvature tensor:An $n$-manifold of constant sectional curvature $K$ has $\mathrm{Scal}=n(n-1)K$.

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.

This is the positive-definite metric case of a Ricci-flat manifold.

A Ricci-flat Riemannian manifold has identically zero Ricci tensor. Every flat Riemannian manifold is Ricci-flat, while the converse can fail in dimension at least four.

For a connected Riemannian manifold of dimension $n\ge 2$, assume geodesic completeness and $\mathrm{Ric}\ge (n-1)\kappa g$ for a constant $\kappa >0$. The Riemannian distance then has diameter at most $\pi /\sqrt{\kappa }$; the manifold is compact and its fundamental group is finite. The dimension restriction excludes the vacuous one-dimensional Ricci bound.

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.

If a connected complete Riemannian manifold has $\mathrm{Ric}\ge (n-1)\kappa g$ with $\kappa >0$, its universal cover with the lifted metric is complete and satisfies the same bound. The Bonnet-Myers theorem makes that cover compact. A covering fiber is closed and discrete and hence finite; the number of its elements equals the order of the fundamental group. Applying the diameter theorem only to the base would not prove this conclusion: the positive lower bound must also be used on the cover.

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.

This is a Riemannian splitting theorem whose hypotheses include a globally minimizing geodesic line.

A complete connected Riemannian manifold with nonnegative Ricci curvature that contains a line splits isometrically as $N\times \mathbb{R}$.

If a connected closed manifold $M^3$ has a noncontractible universal cover, then $M\times \mathbb{R}$ has no complete metric making it a Ricci-flat Riemannian manifold. The product has two ends and hence contains a line in a Riemannian manifold for any complete metric. The Cheeger-Gromoll splitting theorem would split it isometrically as $N^3\times \mathbb{R}$. Its three-dimensional factor is Ricci-flat and therefore flat by three-dimensional curvature from the Ricci tensor. The complete simply connected cover of the product would be Euclidean four-space. But that cover is $\tilde{M}\times \mathbb{R}$, which retracts onto $\tilde{M}$, contradicting noncontractibility.