📂Geometry

Coordinate Representation of the Riemann Curvature Tensor

Explanation¹

Given a Riemannian manifold $(M, g)$, let’s say the coordinate system at $p$ is referred to as $(U, \mathbf{x})$. And let the tangent vector be denoted as follows.

$$\dfrac{\partial }{\partial x_{i}} \overset{\text{denote}}{=} X_{i}$$

Now, consider $R(X_{i}, X_{j})X_{k}$. By the definition of the Riemann curvature tensor $R$, it is also a vector field. Therefore, it can be expressed as follows.

$$R(X_{i}, X_{j})X_{k} = \sum_{l}R_{ijk}^{l}X_{l}$$

The coefficient of the above vector is determined by $X_{i}, X_{j}, X_{k}$, so it is denoted as $R_{ijk}^{l}$. Now let’s say the vector field $X, Y, Z$ is as follows.

$$X = \sum u^{i}X_{i},\quad Y = \sum v^{j}X_{j},\quad Z = \sum w^{k}X_{k}$$

Then, by the linearity of $R$, we get the following.

$$R(X, Y)Z = \sum_{i, j, k, l}R_{ijk}^{l}u^{i}v^{j}w^{k}X_{l}$$

Looking back at $R(X_{i}, X_{j})X_{k}$ to express $R_{ijk}^{l}$ in terms of the Christoffel symbols $\Gamma_{ij}^{k}$, since $[X_{i}, X_{j}] = 0$, and by definition since $\nabla_{X_{i}}X_{j} = \Gamma_{ij}^{k}X_{k}$,

$$\begin{align*} R(X_{i}, X_{j})X_{k} &= \nabla_{X_{j}}\nabla_{X_{i}}X_{k} - \nabla_{X_{i}}\nabla_{X_{j}}X_{k} \\ &= \nabla_{X_{j}}\left( \sum_{l}\Gamma_{ik}^{l}X_{l} \right) - \nabla_{X_{i}}\left( \sum_{l}\Gamma_{jk}^{l}X_{l} \right) \\ &= \sum_{l}\Gamma_{ik}^{l} \nabla_{X_{j}} X_{l} + \sum_{l}\dfrac{\partial \Gamma_{ik}^{l}}{\partial x_{j}}X_{l} - \sum_{l}\Gamma_{jk}^{l}\nabla_{X_{i}}X_{l} - \sum_{l}\dfrac{\partial \Gamma_{jk}^{l}}{\partial x_{i}}X_{l}\\ &= \sum_{l}\Gamma_{ik}^{l} \sum_{s}\Gamma_{jl}^{s}X_{s} - \sum_{l}\Gamma_{jk}^{l} \sum_{s} \Gamma_{il}^{s}X_{s} + \sum_{l}\dfrac{\partial \Gamma_{ik}^{l}}{\partial x_{j}}X_{l} - \sum_{l}\dfrac{\partial \Gamma_{jk}^{l}}{\partial x_{i}}X_{l} \\ &= \sum_{s} \sum_{l} \left( \Gamma_{ik}^{l}\Gamma_{jl}^{s} - \Gamma_{jk}^{l} \Gamma_{il}^{s} \right)X_{s} + \sum_{l}\dfrac{\partial \Gamma_{ik}^{l}}{\partial x_{j}}X_{l} - \sum_{l}\dfrac{\partial \Gamma_{jk}^{l}}{\partial x_{i}}X_{l} \\ &= \sum_{s} \sum_{l} \left( \Gamma_{ik}^{l}\Gamma_{jl}^{s} - \Gamma_{jk}^{l} \Gamma_{il}^{s} \right)X_{s} + \sum_{s}\dfrac{\partial \Gamma_{ik}^{s}}{\partial x_{j}}X_{s} - \sum_{s}\dfrac{\partial \Gamma_{jk}^{s}}{\partial x_{i}}X_{s} \\ &= \sum_{s} \left( \sum_{l}\Gamma_{ik}^{l}\Gamma_{jl}^{s} - \sum_{l}\Gamma_{jk}^{l} \Gamma_{il}^{s} + \dfrac{\partial \Gamma_{ik}^{s}}{\partial x_{j}} - \dfrac{\partial \Gamma_{jk}^{s}}{\partial x_{i}} \right)X_{s} \\ &= \sum_{s} R_{ijk}^{s}X_{s} \\ \end{align*}$$

Therefore,

$$R_{ijk}^{s} = \sum_{l}\Gamma_{ik}^{l}\Gamma_{jl}^{s} - \sum_{l}\Gamma_{jk}^{l} \Gamma_{il}^{s} + \dfrac{\partial \Gamma_{ik}^{s}}{\partial x_{j}} - \dfrac{\partial \Gamma_{jk}^{s}}{\partial x_{i}}$$

In differential geometry at $\mathbb{R}^{3}$, conversely, the above equation is defined as the coefficients of the Riemann curvature tensor. Hence, $R(X_{i}, X_{j}, X_{k}, X_{l})$ is as follows.

$$\begin{align*} R(X_{i}, X_{j}, X_{k}, X_{l}) &= g( R(X_{i}, X_{j})X_{k}, X_{l} ) = \left\langle R(X_{i}, X_{j})X_{k}, X_{l} \right\rangle \\ &= \sum_{s}g(R_{ijk}^{s}X_{s}, X_{l}) \\ &= \sum_{s}R_{ijk}^{s}g(X_{s}, X_{l}) \\ &= \sum_{s}R_{ijk}^{s}g_{sl} \\ &\overset{\text{denote}}{=} R_{ijkl} \end{align*}$$

Looking at the above equation, whether it is $R_{ijk}^{s}$ or $R_{ijkl}$, the only difference is multiplying by the metric $g_{ks}$. For this reason, the notation $R(X, Y)Z$ and $R(X,Y,Z,W)$ is redundantly used, and essentially considered the same.

Symmetry

By the Bianchi identity, the following holds.

$$R_{ijk}^{s} + R_{jki}^{s} + R_{kij}^{s} = 0,\quad \forall s$$

By the symmetry of Riemann curvature, the following holds.

$$\begin{align*} R_{ijkl} + R_{jkil} + R_{kijl} &= 0 \\ R_{ijkl} &= -R_{jikl} \\ R_{ijkl} &= -R_{ijlk} \\ R_{ijkl} &= R_{klij} \end{align*}$$