Riemann Curvature Tensor, Gauss Equation, and Codazzi-Mainardi Equation in Differential Geometry 📂Geometry

Riemann Curvature Tensor, Gauss Equation, and Codazzi-Mainardi Equation in Differential Geometry

Definition¹

The coefficients of the Riemannian curvature tensor $R_{ijk}^{l}$ are defined as follows.

$$ R_{ijk}^{l} = \dfrac{\partial \Gamma_{ik}^{l}}{\partial u^{j}} - \dfrac{\partial \Gamma_{ij}^{l}}{\partial u^{k}} + \sum_{p} \left( \Gamma_{ik}^{p} \Gamma_{pj}^{l} - \Gamma_{ij}^{p}\Gamma_{pk}^{l}\right) \text{ for } 1 \le i,j,k,l \le 2 $$

Here, $\Gamma_{ij}^{k}$ is the Christoffel symbol.

Explanation

Since Christoffel symbols are intrinsic, the Riemann curvature tensor is also intrinsic.

The so-called coefficients that appear in differential geometry do not depend on the coordinate system. We call these entities tensors.

The Gauss equation provides an extrinsic expression of $R_{ijk}^{l}$ from the perspective of the second fundamental form and the Weingarten map.

Theorems

Gauss’s Equations

$$ R_{ijk}^{l} = L_{ik}L_{j}^{l} - L_{ij}L_{k}^{l} $$

Codazzi-Mainardi Equations

$$ \dfrac{\partial L_{ij}}{\partial u^{k}_{}} - \dfrac{\partial L_{ik}}{\partial u^{j}} = \sum\limits_{l} \left( \Gamma_{ik}^{l}L_{lj} - \Gamma_{ij}^{l}L_{lk} \right) $$

Here, $L_{j}^{i}$ is a component of the matrix representation of the Weingarten map.

Proof

Both equations are proved simultaneously. Let $\mathbf{x} : U \to \R^{3}$ be a coordinate patch mapping. Let $(u^{1}, u^{2})$ be the coordinates of $U$.

Gauss’s Formula
$$ \mathbf{x}_{ij} = L_{ij} \mathbf{n} + \sum \limits_{l=1}^{2} \Gamma_{ij}^{l} \mathbf{x}_{l} $$

First, by Gauss’s formula, we obtain the following.

$$ \begin{align*} \mathbf{x}_{i j k} &= \dfrac{\partial}{\partial u^{k}}\left( L_{ij} \mathbf{n} + \sum \limits_{l=1}^{2} \Gamma_{ij}^{l} \mathbf{x}_{l} \right) \\ &= \frac{\partial L_{ij}}{\partial u^{k}}\mathbf{n} + L_{i j} \mathbf{n}_{k}+\sum\limits_{l} \frac{\partial \Gamma_{i j}^{l}}{\partial u^{k}} \mathbf{x}_{l}+\sum\limits_{l}\Gamma_{i j}^{l} \mathbf{x}_{l k} \end{align*} $$

Here, since $\mathbf{n}_{k} = \mathbf{x}_{k}\mathbf{n} = - L(\mathbf{x}_{k}) = -\sum\limits_{l}L_{k}^{l}\mathbf{x}_{l}$, the second term is $L_{ij}\mathbf{n}_{k} = -\sum\limits_{l} L_{i j} L_{k}^{l} \mathbf{x}_{l}$. Also, applying Gauss’s formula to the fourth term again,

$$ \sum\limits_{l} \Gamma_{ij}^{l} \mathbf{x}_{l k} = \sum\limits_{l} \Gamma_{i j}^{l}L_{lk} \mathbf{n} + \sum\limits_{l,m}\Gamma_{i j}^{l}\Gamma_{lk}^{m} \mathbf{x}_{m} $$

Substituting this, we obtain the following.

$$ \begin{align*} \mathbf{x}_{i j k} &= \frac{\partial L_{ij}}{\partial u^{k}}\mathbf{n} -\sum\limits_{l} L_{i j} L_{k}^{l} \mathbf{x}_{l} + \sum\limits_{l} \frac{\partial \Gamma_{i j}^{l}}{\partial u^{k}} \mathbf{x}_{l} + \sum\limits_{l} \Gamma_{i j}^{l}L_{lk} \mathbf{n} + \sum\limits_{l,m}\Gamma_{i j}^{l}\Gamma_{lk}^{m} \mathbf{x}_{m} \\ &= \frac{\partial L_{ij}}{\partial u^{k}}\mathbf{n} -\sum\limits_{l} L_{i j} L_{k}^{l} \mathbf{x}_{l} + \sum\limits_{l} \frac{\partial \Gamma_{i j}^{l}}{\partial u^{k}} \mathbf{x}_{l} + \sum\limits_{l} \Gamma_{i j}^{l}L_{lk} \mathbf{n} + \sum\limits_{p,l}\Gamma_{i j}^{p}\Gamma_{pk}^{l} \mathbf{x}_{l} \\ &= \left( \frac{\partial L_{ij}}{\partial u^{k}} + \sum\limits_{l} \Gamma_{i j}^{l}L_{lk} \right)\mathbf{n} + \sum\limits_{l} \left(\frac{\partial \Gamma_{i j}^{l}}{\partial u^{k}} - L_{i j} L_{k}^{l} + \sum\limits_{p}\Gamma_{i j}^{p}\Gamma_{pk}^{l} \right)\mathbf{x}_{l} \end{align*} $$

$l,m$ is a dummy index, so we change the index of the last term to $(l,m) \to (p,l)$ and grouped the terms. Similarly, we obtain the following.

$$ \mathbf{x}_{ikj} = \left( \frac{\partial L_{ik}}{\partial u^{j}} + \sum\limits_{l} \Gamma_{i k}^{l}L_{lj} \right)\mathbf{n} + \sum\limits_{l} \left(\frac{\partial \Gamma_{i k}^{l}}{\partial u^{j}} - L_{i k} L_{j}^{l} + \sum\limits_{p}\Gamma_{i k}^{p}\Gamma_{pj}^{l} \right)\mathbf{x}_{l} $$

Assuming the coordinate patch mapping $\mathbf{x}$ is sufficiently differentiable,

$$ \mathbf{x}_{i j k}=\frac{\partial^{3} \mathbf{x}}{\partial u^{k} \partial u^{j} \partial u^{i}}=\frac{\partial^{3} \mathbf{x}}{\partial u^{j} \partial u^{k} \partial u^{i}}=\mathbf{x}_{i k j} $$

$\left\{ \mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{n} \right\}$ is the basis of $\mathbb{R}^{3}$, so each component of $\mathbf{x}_{ijk}$ and $\mathbf{x}_{ikj}$ must be the same. Therefore, we obtain the following.

$$ \begin{align*} && \frac{\partial L_{ij}}{\partial u^{k}} + \sum\limits_{l} \Gamma_{i j}^{l}L_{lk} &= \frac{\partial L_{ik}}{\partial u^{j}} + \sum\limits_{l} \Gamma_{i k}^{l}L_{lj} \\ \implies && \frac{\partial L_{ij}}{\partial u^{k}} - \frac{\partial L_{ik}}{\partial u^{j}} &= \sum\limits_{l} \left( \Gamma_{i k}^{l}L_{lj} - \Gamma_{i j}^{l}L_{lk} \right) \end{align*} $$

The Codazzi-Mainardi equations are proved. Continuing with the same logic, the following equality holds.

$$ \frac{\partial \Gamma_{i j}^{l}}{\partial u^{k}} - L_{i j} L_{k}^{l} + \sum\limits_{p}\Gamma_{i j}^{p}\Gamma_{pk}^{l} = \frac{\partial \Gamma_{i k}^{l}}{\partial u^{j}} - L_{i k} L_{j}^{l} + \sum\limits_{p}\Gamma_{i k}^{p}\Gamma_{pj}^{l} $$

Well organized, we get the following.

$$ L_{i k} L_{j}^{l} - L_{i j} L_{k}^{l} = \frac{\partial \Gamma_{i k}^{l}}{\partial u^{j}} - \frac{\partial \Gamma_{i j}^{l}}{\partial u^{k}} + \sum\limits_{p} \left( \Gamma_{i k}^{p}\Gamma_{pj}^{l} - \Gamma_{i j}^{p}\Gamma_{pk}^{l} \right) = R_{ijk}^{l} $$

The Gauss’s equations are proved.

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