Ricci curvature

Let ${\displaystyle U\subset \mathbb {R} ^{n}}$ be an open set, and for each ${\displaystyle i,j=1,\ldots ,n}$ let ${\displaystyle g_{ij}:U\to \mathbb {R} }$ be a smooth function, subject to the restriction that for each ${\displaystyle p\in U}$ the matrix

${\displaystyle {\begin{pmatrix}g_{11}(p)&\cdots &g_{1n}(p)\\\vdots &\ddots &\vdots \\g_{n1}(p)&\cdots &g_{nn}(p)\end{pmatrix}}}$

is symmetric and invertible. Let ${\displaystyle g^{ij}:U\to \mathbb {R} }$ denote the components of the inverse matrices, so that for each ${\displaystyle p\in U}$ and each ${\displaystyle i,j=1,\ldots ,n}$ one has

${\displaystyle {\begin{pmatrix}g_{11}(p)&\cdots &g_{1n}(p)\\\vdots &\ddots &\vdots \\g_{n1}(p)&\cdots &g_{nn}(p)\end{pmatrix}}{\begin{pmatrix}g^{11}(p)&\cdots &g^{1n}(p)\\\vdots &\ddots &\vdots \\g^{n1}(p)&\cdots &g^{nn}(p)\end{pmatrix}}={\begin{pmatrix}1&\cdots &0\\\vdots &\ddots &\vdots \\0&\cdots &1\end{pmatrix}}.}$

Now define for each ${\displaystyle i,j=1,\ldots ,n}$ the function ${\displaystyle R_{ij}:U\to \mathbb {R} }$ by

${\displaystyle {\begin{aligned}R_{ij}&=-{\frac {1}{2}}\sum _{a,b=1}^{n}{\Big (}{\frac {\partial ^{2}g_{ij}}{\partial x^{a}\partial x^{b}}}+{\frac {\partial ^{2}g_{ab}}{\partial x^{i}\partial x^{j}}}-{\frac {\partial ^{2}g_{ib}}{\partial x^{j}\partial x^{a}}}-{\frac {\partial ^{2}g_{jb}}{\partial x^{i}\partial x^{a}}}{\Big )}g^{ab}\\&\qquad +{\frac {1}{2}}\sum _{a,b,c,d=1}^{n}{\Big (}{\frac {1}{2}}{\frac {\partial g_{ac}}{\partial x^{i}}}{\frac {\partial g_{bd}}{\partial x^{j}}}+{\frac {\partial g_{ic}}{\partial x^{a}}}{\frac {\partial g_{jd}}{\partial x^{b}}}-{\frac {\partial g_{ic}}{\partial x^{a}}}{\frac {\partial g_{jb}}{\partial x^{d}}}{\Big )}g^{ab}g^{cd}\\&\qquad -{\frac {1}{4}}\sum _{a,b,c,d=1}^{n}{\Big (}{\frac {\partial g_{jc}}{\partial x^{i}}}+{\frac {\partial g_{ic}}{\partial x^{j}}}-{\frac {\partial g_{ij}}{\partial x^{c}}}{\Big )}{\Big (}2{\frac {\partial g_{bd}}{\partial x^{a}}}-{\frac {\partial g_{ab}}{\partial x^{d}}}{\Big )}g^{ab}g^{cd}.\end{aligned}}}$

As presented, there is nothing intuitive about this definition. However, it satisfies the following remarkable property. Let ${\displaystyle V\subset \mathbb {R} ^{n}}$ be an open set and let ${\displaystyle y:V\to U}$ be a smooth map whose matrix of first derivatives

${\displaystyle J(q)={\begin{pmatrix}D_{1}y^{1}(q)&\cdots &D_{n}y^{1}(q)\\\vdots &\ddots &\vdots \\D_{1}y^{n}(q)&\cdots &D_{n}y^{n}(q)\end{pmatrix}}}$

is invertible for all ${\displaystyle q\in V.}$ Define ${\displaystyle {\widetilde {g}}_{ij}:V\to \mathbb {R} }$ by the matrix product

${\displaystyle {\begin{pmatrix}{\widetilde {g}}_{11}(q)&\cdots &{\widetilde {g}}_{1n}(q)\\\vdots &\ddots &\vdots \\{\widetilde {g}}_{n1}(q)&\cdots &{\widetilde {g}}_{nn}(q)\end{pmatrix}}=J(q)^{T}{\begin{pmatrix}g_{11}(y(q))&\cdots &g_{1n}(y(q))\\\vdots &\ddots &\vdots \\g_{n1}(y(q))&\cdots &g_{nn}(y(q))\end{pmatrix}}J(q)}$

Compute the functions ${\displaystyle {\widetilde {R}}_{ij}:V\to \mathbb {R} }$ using the same formula as above but replacing ${\displaystyle g}$ by ${\displaystyle {\widetilde {g}}}$ everywhere that it appears. Then one could check, using the product rule and the chain rule, that

${\displaystyle {\begin{pmatrix}{\widetilde {R}}_{11}(q)&\cdots &{\widetilde {R}}_{1n}(q)\\\vdots &\ddots &\vdots \\{\widetilde {R}}_{n1}(q)&\cdots &{\widetilde {R}}_{nn}(q)\end{pmatrix}}=J(q)^{T}{\begin{pmatrix}R_{11}(y(q))&\cdots &R_{1n}(y(q))\\\vdots &\ddots &\vdots \\R_{n1}(y(q))&\cdots &R_{nn}(y(q))\end{pmatrix}}J(q)}$

for all ${\displaystyle q\in V.}$ This is quite unexpected since, directly plugging the formula which defines ${\displaystyle {\widetilde {g}}_{ij}}$ into the formula defining ${\displaystyle {\widetilde {R}}_{ij},}$ one sees that one will have to consider up to third derivatives of ${\displaystyle x,}$ arising when the second derivatives in the first four terms of ${\displaystyle {\widetilde {R}}_{ij}}$ act upon the components of ${\displaystyle J.}$ The "miracle" is that the unwieldy collection of first derivatives, second derivatives, and inverses comprising the definition of ${\displaystyle R_{ij}}$ is perfectly set up so that all of these higher derivatives of ${\displaystyle y}$ cancel out, and one is left with the remarkable clean matrix formula above which relates ${\displaystyle R_{ij}}$ and ${\displaystyle {\widetilde {R}}_{ij}.}$ Although it is a somewhat small matter next to this cancellation, note also that, as can be seen from the grouping of the terms in the definition of ${\displaystyle R_{ij},}$ one has the symmetry ${\displaystyle R_{ij}(p)=R_{ji}(p)}$ for all ${\displaystyle i,j=1,\ldots ,n}$ and all ${\displaystyle p\in U.}$ Importantly, there is (very loosely speaking) no simpler way of using ${\displaystyle g}$ to prescribe ${\displaystyle R}$ which will make the same sort of cancellations work out.

In physical terms, this above property is a manifestation of "general covariance" and is a primary reason that Einstein made use of the formula defining ${\displaystyle R_{ij}}$ when formulating general relativity. In mathematical terms, this shows that a (sufficiently differentiable) pseudo-Riemmannian metric defines descendant tensor fields. This is discussed from the perspective of differentiable manifolds in the next subsection, although the underlying content is virtually identical to that of this subsection.

Let ${\displaystyle (M,g)}$ be a smooth Riemannian or pseudo-Riemannian n-manifold. Given a smooth chart ${\displaystyle (U,\varphi )}$ one then has a function ${\displaystyle g_{ij}:\varphi (U)\to \mathbb {R} }$ for each ${\displaystyle i,j=1,\ldots ,n,}$ and functions ${\displaystyle g^{ij}:\varphi (U)\to \mathbb {R} }$ for each ${\displaystyle i,j=1,\ldots ,n}$ such that

${\displaystyle \sum _{k=1}^{n}g^{ik}(x)g_{kj}(x)={\begin{cases}1&i=j\\0&i\neq j\end{cases}}}$

for all ${\displaystyle x\in \varphi (U).}$ The functions ${\displaystyle g_{ij}}$ are defined by evaluating ${\displaystyle g}$ on coordinate vector fields, while the functions ${\displaystyle g^{ij}}$ are defined so that, as a matrix-valued function, they provide an inverse to the matrix-valued function ${\displaystyle x\mapsto [g_{ij}(x)].}$

Now for each ${\displaystyle i,j,k=1,\ldots ,n}$ define ${\displaystyle \Gamma _{ij}^{k}:\varphi (U)\to \mathbb {R} }$ by

${\displaystyle \Gamma _{ij}^{k}={\frac {1}{2}}\sum _{c=1}^{n}{\Big (}{\frac {\partial g_{jc}}{\partial x^{i}}}+{\frac {\partial g_{ic}}{\partial x^{j}}}-{\frac {\partial g_{ij}}{\partial x^{c}}}{\Big )}g^{kc},}$

and then, for each ${\displaystyle i,j=1,\ldots ,n,}$ define ${\displaystyle R_{ij}^{(U,\varphi )}:\varphi (U)\to \mathbb {R} }$ by

${\displaystyle R_{ij}^{(U,\varphi )}=\sum _{a=1}^{n}{\frac {\partial \Gamma _{ij}^{a}}{\partial x^{a}}}-\sum _{a=1}^{n}{\frac {\partial \Gamma _{aj}^{a}}{\partial x^{i}}}+\sum _{a=1}^{n}\sum _{b=1}^{n}{\Big (}\Gamma _{ab}^{a}\Gamma _{ij}^{b}-\Gamma _{ib}^{a}\Gamma _{aj}^{b}{\Big )}.}$

If ${\displaystyle (U,\varphi )}$ and ${\displaystyle (V,\psi )}$ are smooth charts with ${\displaystyle U\cap V\neq \emptyset ,}$ then one has, as can be checked by a calculation with the chain rule and the product rule,

${\displaystyle R_{ij}^{(U,\varphi )}(x)=\sum _{k,l=1}^{n}R_{kl}^{(V,\psi )}{\Big (}\psi \circ \varphi ^{-1}(x){\Big )}D_{i}{\Big |}_{x}(\psi \circ \varphi ^{-1})^{k}D_{j}{\Big |}_{x}(\psi \circ \varphi ^{-1})^{l}.}$

Now, for any ${\displaystyle p\in U,}$ one can define a bilinear map ${\displaystyle \operatorname {Ric} _{p}:T_{p}M\times T_{p}M\to \mathbb {R} }$ by

${\displaystyle \operatorname {Ric} _{p}(X,Y)=\sum _{i,j=1}^{n}R_{ij}^{(U,\varphi )}(\varphi (x))X^{i}(p)Y^{j}(p),}$

where ${\displaystyle X^{1},\ldots ,X^{n}}$ and ${\displaystyle Y^{1},\ldots ,Y^{n}}$ are the components of ${\displaystyle X}$ and ${\displaystyle Y}$ relative to the coordinate vector fields. According to the relation between ${\displaystyle R^{(U,\varphi )}}$ and ${\displaystyle R^{(V,\psi )}}$ above, the definition of ${\displaystyle \operatorname {Ric} _{p}}$ does not depend on the choice of ${\displaystyle (U,\varphi ).}$

The above expressions and notation, although very formal, is quite cumbersome in practice. Many expositions in local coordinates would say the same thing in the following language:

Let ${\displaystyle M}$ be a smooth manifold, and let ${\displaystyle g}$ be a Riemannian or pseudo-Riemannian metric. In local coordinates, define the Christoffel symbols

${\displaystyle \Gamma _{ij}^{k}={\frac {1}{2}}g^{kl}(\partial _{i}g_{jl}+\partial _{j}g_{il}-\partial _{l}g_{ij})}$

and

${\displaystyle R_{jk}=\partial _{i}\Gamma _{jk}^{i}-\partial _{j}\Gamma _{ik}^{i}+\Gamma _{ip}^{i}\Gamma _{jk}^{p}-\Gamma _{jp}^{i}\Gamma _{ik}^{p}.}$

Since ${\displaystyle R_{jk}={\widetilde {R}}_{ab}{\frac {\partial {\widetilde {x}}^{a}}{\partial x^{i}}}{\frac {\partial {\widetilde {x}}^{b}}{\partial x^{j}}},}$ this defines a (0,2)-tensor field on ${\displaystyle M,}$ called the Ricci tensor of ${\displaystyle g.}$

Suppose that (M, g) is an n-dimensional Riemannian or pseudo-Riemannian manifold, equipped with its Levi-Civita connection ∇. The Riemann curvature of M is a map which takes smooth vector fields ${\displaystyle X,Y,Z}$ and returns a vector field

${\displaystyle R(X,Y)Z=\nabla _{X}\nabla _{Y}Z-\nabla _{Y}\nabla _{X}Z-\nabla _{[X,Y]}Z}$

on vector fields X, Y, Z. The crucial property of ${\displaystyle R}$ is that if ${\displaystyle X,Y,Z}$ and ${\displaystyle X',Y',Z'}$ are smooth vector fields such that ${\displaystyle X_{p}=X_{p}',}$ ${\displaystyle Y_{p}=Y_{p}',}$ and ${\displaystyle Z_{p}=Z_{p}',}$ then

${\displaystyle {\big (}R(X,Y)Z{\big )}_{p}={\big (}R(X',Y')Z'{\big )}_{p}.}$

So the Riemann curvature actually defines, for each ${\displaystyle p\in M,}$ a (multilinear) map

${\displaystyle \operatorname {Rm} _{p}:T_{p}M\times T_{p}M\times T_{p}M\to T_{p}M}$

such that ${\displaystyle \operatorname {Rm} _{p}(u,v,w)={\big (}R(X,Y)Z{\big )}_{p}}$, where ${\displaystyle X,Y,Z}$ are smooth vector field extensions of the vectors ${\displaystyle u,v,w.}$ Define ${\displaystyle \operatorname {Ric} _{p}:T_{p}M\times T_{p}M\to \mathbb {R} }$ by

${\displaystyle \operatorname {Ric} _{p}(Y,Z)=\operatorname {tr} {\big (}X\mapsto \operatorname {Rm} _{p}(X,Y,Z){\big )}.}$

Note that some sources define ${\displaystyle R(X,Y)Z}$ to be what would here be called ${\displaystyle -R(X,Y)Z;}$ they would then define ${\displaystyle \operatorname {Ric} }$ as ${\displaystyle -\operatorname {tr} (X\mapsto \operatorname {Rm} _{p}(X,Y,Z)).}$ Although sign conventions differ about the Riemann tensor, they do not differ about the Ricci tensor.

The two above definitions are identical. The formulas defining ${\displaystyle \Gamma _{ij}^{k}}$ and ${\displaystyle R_{ij}}$ in the coordinate approach have an exact parallel in the formulas defining the Levi-Civita connection, and the Riemann curvature via the Levi-Civita connection. Arguably, the definitions directly using local coordinates are preferable, since the "crucial property" of the Riemann tensor mentioned above requires ${\displaystyle M}$ to be Hausdorff in order to hold. By contrast, the local coordinate approach only requires a smooth atlas. It is also somewhat easier to connect the "invariance" philosophy underlying the local approach with the methods of constructing more exotic geometric objects, such as spinor fields.

Note also that the complicated formula defining ${\displaystyle R_{ij}}$ in the introductory section is the same as that in the following section. The only difference is that terms have been grouped so that it is easy to see that ${\displaystyle R_{ij}=R_{ji}.}$

The Ricci curvature is sometimes thought of as (a negative multiple of) the Laplacian of the metric tensor (Chow & Knopf 2004, Lemma 3.32) harv error: no target: CITEREFChowKnopf2004 (help). Specifically, in harmonic local coordinates the components satisfy

${\displaystyle R_{ij}=-{\tfrac {1}{2}}\Delta \left(g_{ij}\right)+{\text{lower-order terms}},}$

where ${\displaystyle \Delta =\nabla \cdot \nabla }$ is the Laplace–Beltrami operator, here regarded as acting on the locally-defined functions g_ij. This fact motivates, for instance, the introduction of the Ricci flow equation as a natural extension of the heat equation for the metric. Alternatively, in a normal coordinate system based at p, at the point p

${\displaystyle R_{ij}=-{\tfrac {2}{3}}\Delta \left(g_{ij}\right).}$

Trivially, one has

${\displaystyle \operatorname {Ric} =Z+{\frac {1}{n}}Rg.}$

It is less immediately obvious that the two terms on the right hand side are orthogonal to each other:

${\displaystyle {\Big \langle }Z,{\frac {1}{n}}Rg{\Big \rangle }_{g}\equiv g^{ab}{\Big (}R_{ab}-{\frac {1}{n}}Rg_{ab}{\Big )}=0.}$

An identity which is intimately connected with this (but which could be proved directly) is that

${\displaystyle |\operatorname {Ric} |_{g}^{2}=|Z|_{g}^{2}+{\frac {1}{n}}R^{2}.}$

By taking a divergence, and using the contracted Bianchi identity, one sees that ${\displaystyle Z=0}$ implies ${\displaystyle \textstyle {\frac {1}{2}}dR-{\frac {1}{n}}dR=0.}$ So, provided that n ≥ 3 and ${\displaystyle M}$ is connected, the vanishing of ${\displaystyle Z}$ implies that the scalar curvature is constant. One can then see that the following are equivalent:

• ${\displaystyle Z=0}$
• ${\displaystyle \operatorname {Ric} =\lambda g}$ for some number ${\displaystyle \lambda }$
• ${\displaystyle \operatorname {Ric} ={\frac {1}{n}}Rg}$

In the Riemannian setting, the above orthogonal decomposition shows that ${\displaystyle R^{2}=n|\operatorname {Ric} |^{2}}$ is also equivalent to these conditions. In the pseudo-Riemmannian setting, by contrast, the condition ${\displaystyle |Z|_{g}^{2}=0}$ does not necessarily imply ${\displaystyle Z=0,}$ so the most that one can say is that these conditions imply ${\displaystyle R^{2}=n|\operatorname {Ric} |_{g}^{2}.}$

In particular, the vanishing of trace-free Ricci tensor characterizes Einstein manifolds, as defined by the condition ${\displaystyle \operatorname {Ric} =\lambda g}$ for a number ${\displaystyle \lambda .}$ In general relativity, this equation states that (M,g) is a solution of Einstein's vacuum field equations with cosmological constant.