# Ricci and sectional curvature

The Ricci curvature tensor (AKA Ricci tensor) is formed by contracting two indices in the Riemann curvature tensor:

\displaystyle \begin{aligned}R{}_{ab} & \equiv R^{c}{}_{acb}\\ \mathrm{Ric}(v,w) & \equiv R{}_{ab}v^{a}w^{b} \end{aligned}

Using the symmetries of the Riemann tensor for a metric connection along with the first Bianchi identity with zero torsion, it is easily shown that the Ricci tensor is symmetric. A pseudo-Riemannian manifold is said to have constant Ricci curvature, or to be an Einstein manifold, if the Ricci tensor is a constant multiple of the metric tensor.

Since the Ricci tensor is symmetric for zero torsion, by the spectral theorem it can be diagonalized on a Riemannian manifold and thus is determined by

$$\displaystyle \mathrm{Ric}(v)\equiv\mathrm{Ric}(v,v),$$

which is called the Ricci curvature function (AKA Ricci function). Note that the Ricci function is not a 1-form since it is not linear in $${v}$$. Choosing a basis that diagonalizes $${R{}_{ab}}$$ is equivalent to choosing our basis vectors to line up with the directions that yield extremal values of the Ricci function on the unit vectors $${\mathrm{Ric}(\hat{v},\hat{v})}$$ (or equivalently, the principal axes of the ellipsoid / hyperboloid $${\mathrm{Ric}(v,v)=1}$$).

Finally, if we raise one of the indices of the Ricci tensor and contract we arrive at the Ricci scalar (AKA scalar curvature):

\displaystyle \begin{aligned}R & \equiv g^{ab}R_{ab}\end{aligned}

For a Riemannian manifold $${M^{n}}$$, the Ricci scalar can thus be viewed as $${n}$$ times the average of the Ricci function on the set of unit tangent vectors.

 Δ The Ricci function and Ricci scalar are sometimes defined as averages instead of contractions (sums), introducing extra factors in terms of the dimension $${n}$$ to the above definitions.

The Ricci function in terms of the curvature 2-form in an orthonormal frame $${e_{\mu}}$$ (dropping the hats to avoid clutter) on a pseudo-Riemannian manifold $${M^{n}}$$ naturally splits into terms which each also measure curvature:

$$\displaystyle \mathrm{Ric}(e_{\mu})=\sum_{i\neq\mu}g_{ii}\left\langle \check{R}(e_{i},e_{\mu})\vec{e}_{\mu},e_{i}\right\rangle$$

The term $${i=\mu}$$ vanishes due to the anti-symmetry of $${\check{R}}$$. The $${(n-1)}$$ non-zero terms are each called a sectional curvature, which in general is defined as

\displaystyle \begin{aligned}K(v,w) & \equiv\frac{\left\langle \check{R}(v,w)\vec{w},v\right\rangle }{\left\langle v,v\right\rangle \left\langle w,w\right\rangle -\left\langle v,w\right\rangle ^{2}}\\ \Rightarrow K(e_{i},e_{j}) & =g_{ii}g_{jj}\left\langle \check{R}(e_{i},e_{j})\vec{e}_{j},e_{i}\right\rangle \\ \Rightarrow\mathrm{Ric}(e_{\mu}) & =\sum_{i\neq\mu}g_{\mu\mu}K(e_{i},e_{\mu})\\ \Rightarrow R & =\sum_{j}g_{jj}\mathrm{Ric}(e_{j})\\ & =\sum_{i\neq j}K(e_{i},e_{j})\\ & =2\sum_{i<j}K(e_{i},e_{j}). \end{aligned}

Note that the sectional curvature is not a 2-form since it is not linear in its arguments; in fact it is constructed to only depend on the plane defined by them, and therefore is symmetric and defined to vanish for equal arguments. Thus for a Riemannian manifold, the Ricci function of a unit vector $${\mathrm{Ric}(\hat{v})}$$ can be viewed as $${(n-1)}$$ times the average of the sectional curvatures of the planes that include $${\hat{v}}$$, and the Ricci scalar can be viewed as $${n}$$ times the average of all the Ricci functions. For a pseudo-Riemannian manifold, the Ricci scalar is twice the sum of all sectional curvatures, or $${n(n-1)}$$ times the average of all sectional curvatures, whose count is the binomial coefficient $${n}$$ choose 2 or $${n(n-1)/2}$$.

The Cartan-Hadamard theorem states that the universal covering space of a complete Riemannian manifold $${M^{n}}$$ with non-positive sectional curvature is diffeomorphic to $${\mathbb{R}^{n}}$$. For $${M^{n}}$$ complete with constant sectional curvature $${K}$$ (sometimes called a space form), its universal covering space is isometric to $${\mathbb{R}^{n}}$$ if $${K=0}$$, $${S^{n}}$$ if $${K=1}$$, and $${H^{n}}$$ if $${K=-1}$$, where $${H^{n}}$$ is called the (real) hyperbolic space and we can generalize by noting that scaling the metric inversely scales $${K}$$. There are different ways to define $${H^{n}}$$ concretely, one being the region of $${\mathbb{R}^{n}}$$ with $${x_{0}>0}$$ and metric $${\delta_{\mu\nu}/x_{0}^{2}}$$, another being the set of points with $${x_{0}>0}$$ and $${\left\langle x,x\right\rangle =-1}$$ in $${M^{n+1}}$$ with a Lorentzian metric.

The sectional curvatures completely determine the Riemann tensor, but in general the Ricci tensor alone does not for manifolds of dimension greater than 3. However, the Riemann tensor is determined by the Ricci tensor together with the Weyl curvature tensor (AKA Weyl tensor, conformal tensor), whose definition (not reproduced here) removes all contractions of the Riemann tensor, so that it is the “trace-free part of the curvature” (i.e. all of its contractions vanish). The Weyl tensor is only defined and non-zero for dimensions $${n>3}$$.

The Einstein tensor is defined as

\displaystyle \begin{aligned}G(v,w) & \equiv\mathrm{Ric}(v,w)-\frac{R}{2}g(v,w)\\ G_{ab} & =R_{ab}-\frac{R}{2}g_{ab}. \end{aligned}

If we define $${G\equiv g^{ab}G_{ab}}$$ then we find that $${R_{ab}=G_{ab}-Gg_{ab}/(n-2)}$$, so that the Einstein tensor vanishes iff the Ricci tensor does. Now, for zero torsion the Einstein tensor is symmetric, and by the spectral theorem can be diagonalized at a given point on a Riemannian manifold with an orthonormal basis, which also diagonalizes the Ricci tensor. In terms of the sectional curvature, we have

\displaystyle \begin{aligned}G(e_{\mu},e_{\mu}) & =-g_{\mu\mu}\sum_{\begin{subarray}{c} i<j\\ i,j\neq\mu \end{subarray}}K(e_{i},e_{j}).\end{aligned}

Thus for a Riemannian manifold, the Einstein tensor $${G(\hat{v},\hat{v})}$$ applied to a unit vector twice can be viewed as $${-\left\langle \hat{v},\hat{v}\right\rangle (n-1)(n-2)/2}$$ times the average of the sectional curvatures of the planes orthogonal to $${\hat{v}}$$. For an Einstein manifold, $${R_{ab}=kg_{ab},}$$ so that $${R=nk}$$ and thus the Einstein tensor $${G_{ab}=(1-n/2)kg_{ab}}$$ is also proportional to the metric tensor. Using the second Bianchi identity with zero torsion it can be shown ( pp. 80-81) that the Einstein tensor is also “divergenceless,” i.e.

$$\displaystyle \nabla_{a}G^{ab}=0.$$

For each value of $${b}$$ in an orthonormal frame, this relation expressed in terms of the Riemann curvature tensor can be seen to be equivalent to the second Bianchi identity. Recall that unless the metric is flat, there is no conserved quantity which can be associated with this vanishing divergence for a Lorentzian metric.

 Δ Frequent references to the divergencelessness of the Einstein tensor being related to a conserved quantity for a Lorentzian metric usually refer to some kind of particular context; one simple one is that in the limit of zero curvature or infinitesimal volume, there is a set of conserved quantities due to the above equation.