# 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 $\mathbf{u}, \mathbf{v}, \mathbf{w} \in \Gamma(\mathrm{TM})$.