Curvature of Riemannian manifolds

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In mathematics, specifically differential geometry, the infinitesimal geometry of Riemannian manifolds with dimension at least 3 is too complicated to be described by a single number at a given point. Riemann introduced a way to describe it as a "little monster tensor". Similar notions have found applications everywhere in differential geometry.

For a more elementary discussion see the article on curvature which discusses the curvature of curves and surfaces in 2 and 3 dimensions.

The curvature of a Pseudo-Riemannian manifold can be expressed in the same way with only slight modifications.

1 Ways to express the curvature of a Riemannian manifold
2 Further curvature tensors
3 Calculation of curvature

[edit] Ways to express the curvature of a Riemannian manifold

[edit] The curvature tensor

The curvature of Riemannian manifold can be described in various ways; the most standard one is the curvature tensor, given in terms of a Levi-Civita connection (or covariant differentiation) $\nabla$ and Lie bracket $[ * , * ]$ by the following formula:

$R(u,v)w=\nabla_u\nabla_v w - \nabla_v \nabla_u w -\nabla_{[u,v]} w .$

Here $R (u, v)$ is a linear transformation of the tangent space of the manifold; it is linear in each argument. If $u=\partial/\partial x_i$ and $v=\partial/\partial x_j$ are coordinate vector fields then $[u, v] = 0$ and therefore the formula simplifies to

$R(u,v)w=\nabla_u\nabla_v w - \nabla_v \nabla_u w$

i.e. the curvature tensor measures noncommutativity of the covariant derivative.

The linear transformation $w\mapsto R(u,v)w$ is also called the curvature transformation or endomorphism.

NB. There are a few books where the curvature tensor is defined with opposite sign.

If you change the metric by a factor $e 2 f$ , the curvature tensor changes to (seen as a (0,4)-Tensor): $e^{2f}(R+(Hess(f)-df\otimes df+\frac{1}{2}\|grad(f)\|^2)\circ g)$

where $\circ$ denotes the Kulkarni-Nomizu product.

[edit] Symmetries and identities

The curvature tensor has the following symmetries:

$R(u,v)=-R(v,u)^{}_{}$

$\langle R(u,v)w,z \rangle=-\langle R(u,v)z,w \rangle^{}_{}$

$R(u,v)w+R(v,w)u+R(w,u)v=0 ^{}_{}$

The last identity was discovered by Ricci, but is often called the first Bianchi identity, just because it looks similar to the Bianchi identity below. These three identities form a complete list of symmetries of the curvature tensor, i.e. given any tensor which satisfies the identities above, one could find a Riemannian manifold with such a curvature tensor at some point. Simple calculations show that such a tensor has $n 2 (n 2 - 1) / 12$ independent components. Yet another useful identity follows from these three:

$\langle R(u,v)w,z \rangle=\langle R(w,z)u,v \rangle^{}_{}$

The Bianchi identity (often the second Bianchi identity) involves the covariant derivatives:

$\nabla_uR(v,w)+\nabla_vR(w,u)+\nabla_w R(u,v)=0$

[edit] Sectional curvature

Sectional curvature is a further, equivalent but more geometrical, description of the curvature of Riemannian manifolds. It is a function $K (σ)$ which depends on a section $σ$ (i.e. a 2-plane in the tangent spaces). It is the Gauss curvature of the $σ$ -section at p; here $σ$ -section is a locally-defined piece of surface which has the plane $σ$ as a tangent plane at p, obtained from geodesics which start at p in the directions of the image of $σ$ under the exponential map at p.

If $v, u$ are two linearly independent vectors in $σ$ then

$K(\sigma)= K(u,v)/|u\wedge v|^2\ \ \mbox{where}\ \ K(u,v)=\langle R(u,v)v,u \rangle$

The following complex formula indicates that sectional curvature describes the curvature tensor completely:

$6\langle R(u,v)w,z \rangle =^{}_{}$

$[K(u+z,v+w)-K(u+z,v)-K(u+z,w)-K(u,v+w)-K(z,v+w)-K(v+z,u)+K(u,w)+K(v,z)]-^{}_{}$

$[K(u+w,v+z)-K(u+w,v)-K(u+w,z)-K(u,v+z)-K(w,v+z)-K(u+w,v)+K(v,w)+K(u,z)].^{}_{}$

[edit] Curvature form

The Cartan formalism gives a very elegant way to describe curvature. It is used more for general vector bundles, and for principal bundles, but it works just as well for the tangent bundle with the Levi-Civita connection. The curvature of n-dimensional Riemannian manifold is given by an antisymmetric n×n matrix $\Omega^{}_{}=\Omega^i_{\ j}$ of 2-forms (or equivalently a 2-form with values in $s o (n)$ , the Lie algebra of the orthogonal group $O (n)$ , which is the structure group of the tangent bundle of a Riemannian manifold).

Let $e i$ be a local section of orthonormal bases. Then one can define the connection form, an antisymmetric matrix of 1-forms $\omega=\omega^i_{\ j}$ which satisfy from the following identity

$\omega^k_{\ j}(e_i)=\langle \nabla_{e_i}e_j,e_k\rangle$

Then the curvature form $\Omega=\Omega^i_{\ j}$ is defined by

$\Omega=d\omega +\omega\wedge\omega$

The following describes relation between curvature form and curvature tensor:

$R(u,v)w=\Omega(u\wedge v)w.$

This approach builds in all symmetries of curvature tensor except the first Bianchi identity, which takes form

$\Omega\wedge\theta=0$

where $θ = θ i$ is an n-vector of 1-forms defined by $\theta^i(v)=\langle e_i,v\rangle$ . The second Bianchi identity takes form

D Ω = 0

D denotes the exterior covariant derivative

[edit] The curvature operator

It is sometimes convenient to think about curvature as an operator $Q$ on tangent bivectors (elements of $Λ 2 (T)$ ), which is uniquely defined by the following identity:

$\langle Q (u\wedge v),w\wedge z\rangle=\langle R(u,v)w,z \rangle.$

It is possible to do this precisely because of the symmetries of the curvature tensor (namely antisymmetry in the first and last pairs of indices, and block-symmetry of those pairs).

[edit] Further curvature tensors

In general the following tensors and functions do not describe the curvature tensor completely, however they play an important role.

[edit] Scalar curvature

Scalar curvature is a function on any Riemannian manifold, usually denoted by Sc. It is the full trace of the curvature tensor; given an orthonormal basis ${e i}$ in the tangent space at p we have

$S\! c=\sum_{i,j}\langle R(e_i,e_j)e_j,e_i\rangle=\sum_{i}\langle Ric(e_i),e_i\rangle,$

where Ric denotes Ricci tensor. The result does not depend on the choice of orthonormal basis. Starting with dimension 3, scalar curvature does not describe the curvature tensor completely.

[edit] Ricci curvature

Ricci curvature is a linear operator on tangent space at a point, usually denoted by Ric. Given an orthonormal basis ${e i}$ in the tangent space at p we have

$Ric(u)=\sum_{i} R(u,e_i)e_i.^{}_{}$

The result does not depend on the choice of orthonormal basis. Starting with dimension 4, Ricci curvature does not describe the curvature tensor completely.

Explicit expressions for the Ricci tensor in terms of the Levi-Civita connection is given in the article on Christoffel symbols.

[edit] Weyl curvature tensor

The Weyl curvature tensor has the same symmetries as the curvature tensor, plus one extra: its Ricci curvature must vanish. In dimensions 2 and 3 Weyl curvature vanishes, but if the dimension n > 3 then the second part can be non-zero.

The curvature tensor can be decomposed into the part which depends on the Ricci curvature, and the Weyl tensor.
If g′=fg for some positive scalar function f — a conformal change of metric — then W ′ = W.
For a manifold of constant curvature, the Weyl tensor is zero.
- Moreover, W=0 if and only if the metric is locally conformal to the standard Euclidean metric (equal to fg, where g is the standard metric in some coordinate frame and f is some scalar function).

[edit] Calculation of curvature

For calculation of curvature

of hypersurfaces and submanifolds see second fundamental form,
in coordinates see the list of formulas in Riemannian geometry or covariant derivative,
by moving frames see Cartan connection and curvature form.
the Jacobi equation can help if one knows something about the behavior of geodesics.

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Category: Riemannian geometry