notes/docs/mathematics/differential-geometry/curvature.md

Curvature

Let \mathrm{M} be a differential manifold with \dim \mathrm{M} = n \in \mathbb{N} used throughout the section. Let \mathrm{TM} and \mathrm{T^*M} denote the tangent and cotangent bundle, V and V^* the fiber and dual fiber bundle and \mathscr{B} the tensor fiber bundle.

Curvature operator

Definition 1: the curvature operator \Omega: \Gamma(\mathrm{TM})^3 \to \Gamma(\mathrm{TM}) is defined as

\Omega(\mathbf{v}, \mathbf{w}) \mathbf{u} = [\nabla_\mathbf{v}, \nabla_\mathbf{w}] \mathbf{u} - \nabla_{[\mathbf{v}, \mathbf{w}]}\mathbf{u},

for all \mathbf{u}, \mathbf{v}, \mathbf{w} \in \Gamma(\mathrm{TM}) with [\cdot, \cdot] denoting the Lie bracket.

It then follows from the definition that the curvature operator \Omega can be decomposed.

Proposition 1: the decomposition of the curvature operator \Omega relative to a basis \{\partial_i\}_{i=1}^n of \Gamma(\mathrm{TM}) results into

\Omega(\mathbf{v}, \mathbf{w}) \mathbf{u} = v^i w^j [D_i, D_j] u^l \partial_l,

for all \mathbf{u}, \mathbf{v}, \mathbf{w} \in \Gamma(\mathrm{TM}).

??? note "Proof:"

Will be added later.

Curvature tensor

Definition 2: the Riemann curvature tensor \mathbf{R}: \Gamma(\mathrm{T}^*\mathrm{M}) \times \Gamma(\mathrm{TM})^3 \to \mathbb{K} is defined as

\mathbf{R}(\bm{\omega}, \mathbf{u}, \mathbf{v}, \mathbf{w}) = \mathbf{k}(\bm{\omega}, \Omega(\mathbf{v}, \mathbf{w}) \mathbf{u}),

for all \bm{\omega} \in \Gamma(\mathrm{T}^*\mathrm{M}) and \mathbf{u}, \mathbf{v}, \mathbf{w} \in \Gamma(\mathrm{TM}).

The Riemann curvature defines the curvature of the differential manifold at a certain point x \in \mathrm{M}.

Proposition 2: let \mathbf{R}: \Gamma(\mathrm{T}^*\mathrm{M}) \times \Gamma(\mathrm{TM})^3 \to \mathbb{K} be the Riemann curvature tensor, with its decomposition given by

\mathbf{R} = R^i_{jkl} \partial_i \otimes dx^j \otimes dx^k \otimes dx^l,

then we have that its holor is given by

R^i_{jkl} = \partial_k \Gamma^i_{jl} + \Gamma^m_{jl} \Gamma^i_{mk} - \partial_k \Gamma^i_{jk} - \Gamma^m_{jk} \Gamma^i_{ml},

for all (i,j,k,l) \in \{1, \dots, n\}^4 with \Gamma^i_{jk} denoting the linear connection symbols.

??? note "Proof:"

Will be added later.

It may then be observed that R^i_{jkl} = - R^i_{jlk} such that

\mathbf{R} = \frac{1}{2} R^i_{jkl} \partial_i \otimes dx^j \otimes (dx^k \wedge dx^l).