Ngô Quốc Anh

November 26, 2009

R-G: Ricci curvature

Filed under: Riemannian geometry — Ngô Quốc Anh @ 19:56

In differential geometry, the Ricci curvature tensor, named after Gregorio Ricci-Curbastro, represents the amount by which the volume element of a geodesic ball in a curved Riemannian manifold deviates from that of the standard ball in Euclidean space. As such, it provides one way of measuring the degree to which the geometry determined by a given Riemannian metric might differ from that of ordinary Euclidean n-space. More generally, the Ricci tensor is defined on any pseudo-Riemannian manifold. Like the metric itself, the Ricci tensor is a symmetric bilinear form on the tangent space of the manifold.

The Ricci curvature is broadly applicable to modern Riemannian geometry and general relativity theory. In connection with the latter, it is up to an overall trace term, the portion of the Einstein field equation representing the geometry of spacetime, the other significant portion of which comes from the presence of matter and energy. In connection with the former, lower bounds on the Ricci tensor on a Riemannian manifold allow one to extract global geometric and topological information by comparison (cf. comparison theorem) with the geometry of a constant curvature space form. If the Ricci tensor satisfies the vacuum Einstein equation, then the manifold is an Einstein manifold, which have been extensively studied (cf. Besse 1987). In this connection, the Ricci flow equation governs the evolution of a given metric to an Einstein metric, the precise manner in which this occurs ultimately leads to the solution of the Poincaré conjecture.

Definition. Ricci curvature (or Ricci tensor) is an $(2,0)$-tensor field denoted by $\rm Ric$, that is ${\rm Ric} : TM \times TM \to \mathbb R$, defined by ${\rm Ric}(X,Y) = {\rm Trace}( x \to R(x, X)Y)$.

In local coordinates, $\rm Ric$ is of the form ${\rm Ric} = R_{ij} dx^i \otimes dx^j$.

We assume $\frac{\partial}{\partial x^i}$ where $i=1,2,...,n$  is an orthonormal basis for $T_pM$, then $\displaystyle R\left( {\frac{\partial }{{\partial {x^i}}},X} \right)Y = {X^j}{Y^k}R\left( {\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^j}}}} \right)\frac{\partial }{{\partial {x^k}}} = {X^j}{Y^k}R_{kij}^l\frac{\partial }{{\partial {x^l}}}$.

Thus, $\displaystyle {\rm Trace}\left( {x \mapsto R\left( {x,X} \right)Y} \right) = {X^j}{Y^k}R_{kij}^i$.

In other words, $\displaystyle {\rm Ric}\left( {\frac{\partial }{{\partial {x^j}}},\frac{\partial }{{\partial {x^k}}}} \right) = {R_{ij}}d{x^i} \otimes d{x^j}\left( {\frac{\partial }{{\partial {x^j}}},\frac{\partial }{{\partial {x^k}}}} \right) = {R_{jk}} = R_{kij}^i$.

To be exact, one should read $\displaystyle {R_{jk}} = \sum\limits_i {R_{jik}^i}$.

A simple calculation shows us that $\displaystyle {R_{jk}} = R_{jik}^i = {g^{li}}{g_{li}}R_{jik}^i = {g^{li}}{R_{ljik}}$.

Thus, Ricci tensor can be thought as the trace of curvature tensor $R_{ljil}$.

Example. For the two-dimensional spherical surface of radius $R$ whose metric is $\displaystyle{\left( {ds} \right)^2} = {R^2}\left[ {{{\left( {d\theta } \right)}^2} + {{\sin }^2}\theta {{\left( {d\phi } \right)}^2}} \right]$

we have $\displaystyle \left( {{g_{ij}}} \right) = {R^2}\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & {{{\sin }^2}\theta } \\ \end{array} } \right), \qquad \left( {{g^{ij}}} \right) = \frac{1}{{{R^2}}}\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & {\frac{1}{{{{\sin }^2}\theta }}} \\ \end{array} } \right)$.

Thus $\displaystyle\begin{gathered} {R_{11}} = R_{111}^1 + R_{121}^2 = {g^{22}}{R_{2121}} = {g^{22}}{R_{1212}} = 1, \hfill \\ {R_{12}} = R_{112}^1 + R_{122}^2 = 0, \hfill \\ {R_{22}} = R_{212}^1 + R_{222}^2 = {g^{11}}{R_{1212}} = {\sin ^2}\theta , \hfill \\ {R_{21}} = {R_{12}} = 0. \hfill \\ \end{gathered}$