Chemistry - Curvature tensor

In differential geometry, the Riemann curvature tensor is the most standard way to express curvature of Riemannian manifolds, or more generally, any manifold with an affine connection, torsionless or with torsion. The curvature tensor is given in terms of a Levi-Civita connection (more generally, an affine connection) $\nabla$ (or covariant differentiation) 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.

NB. Some authors define the curvature tensor with the opposite sign.

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 anticommutativity of the covariant derivative.

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

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, 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 can 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:

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

Curvature tensor

Symmetries and identities

See also