2.5 KiB
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).