📂Geometry

Levi-Civita Connection, Riemannian Connection, Coefficients of Connection, Christoffel Symbols

Theorem¹

Let $(M,g)$ be a Riemannian manifold. Then, there uniquely exists an affine connection $\nabla$ on $M$ satisfying the following:

• $\nabla$ is symmetric.
• $\nabla$ is compatible with $g$.

Such $\nabla$ specifically satisfies the following equation:

$$\begin{align*} g(Z, \nabla_{Y}X) =&\ \dfrac{1}{2}\Big( X g(Y, Z) + Y g(Z, X) - Z g(X, Y) \\ &\ - g([X, Z], Y) - g([Y, Z], X) - g([X, Y], Z) \Big) \tag{1} \end{align*}$$

Description

Such a connection $\nabla$ is called the Levi-Civita (or Riemannian) connection.

Let’s denote the basis of the tangent space as $\left\{ \dfrac{\partial }{\partial x_{i}} \right\} \overset{\text{denote}}{=} \left\{ X_{i} \right\}$. By the definition of connection, $\nabla_{X_{i}}X_{j}$ is also a vector field. Thus, it can be represented as a linear combination of $X_{k}$. By Einstein notation,

$$\nabla_{X_{i}}X_{j} = \sum_{k}\Gamma_{ij}^{k}X_{k} = \Gamma_{ij}^{k}X_{k}$$

Since the vector field is determined by $X_{i}, X_{j}$, let’s denote the coefficients by $\Gamma_{ij}^{k}$. These are called the coefficients of the connection $\nabla$ or the Christoffel symbols of the connection. In differential geometry, the Christoffel symbols are defined as the coefficients of the second-order derivatives $\mathbf{x}_{ij}$ of coordinate mappings $\mathbf{x}$, and it can be shown that they are the same. By substituting $X_{i}, X_{j}, X_{k}$ into the left side of $(1)$,

$$\begin{align*} g(\nabla_{X_{j}}X_{i}, X_{k}) = g\left( \Gamma_{ji}^{l}X_{l}, X_{k} \right) = \Gamma_{ji}^{l}g_{lk} \end{align*}$$

Calculating the right side yields $[X_{i}, X_{j}] = 0$, hence,

$$\begin{align*} & \dfrac{1}{2}\left( X_{i}g(X_{j}, X_{k}) + X_{j}g(X_{i}, X_{k}) - X_{k}g(X_{i}, X_{j}) \right) \\ =& \dfrac{1}{2}\left( X_{i}g_{jk} + X_{j}g_{ik} - X_{k}g_{ij} \right) \\ \end{align*}$$

Therefore,

$$\begin{align*} && \Gamma_{ji}^{l}g_{lk} &= \dfrac{1}{2}\left( X_{i}g_{jk} + X_{j}g_{ik} - X_{k}g_{ij} \right) \\ \implies && \sum_{k}\Gamma_{ji}^{l}g_{lk}g^{ks} &= \sum_{k}\dfrac{1}{2}g^{ks}\left( X_{i}g_{jk} + X_{j}g_{ik} - X_{k}g_{ij} \right) \\ \implies && \Gamma_{ji}^{l}\delta_{l}^{s} &= \sum_{k}\dfrac{1}{2}g^{ks}\left( X_{i}g_{jk} + X_{j}g_{ik} - X_{k}g_{ij} \right) \\ \implies && \Gamma_{ji}^{s} &= \sum_{k}\dfrac{1}{2}g^{ks}\left( X_{i}g_{jk} + X_{j}g_{ik} - X_{k}g_{ij} \right) \\ \end{align*}$$

Summarizing, the following can be obtained:

$$\Gamma_{ij}^{k} = \dfrac{1}{2}g^{mk}\left( \dfrac{\partial }{\partial x_{i}}g_{jm} + \dfrac{\partial }{\partial x_{j}}g_{im} - \dfrac{\partial }{\partial x_{m}}g_{ij} \right)$$

This is the same as the equation obtained for surfaces on $\mathbb{R}^{3}$ in differential geometry. Particularly, in Euclidean space $\mathbb{R}^{n}$, since the metric is constant as $g_{ij} = \delta_{ij}$, it follows that $\Gamma_{ij}^{k} = 0$.

When initially defining affine connections, $\nabla_{X}Y$ was not explicitly given and was defined only as an abstract concept satisfying certain properties. However, when the Riemannian metric $g$ is given to such a connection $\nabla$, it is clear that $\nabla_{X}Y$ is determined by the coefficients of the metric $g_{ij}$. If we denote this as $X = u^{i}X_{i}, Y= v^{j}X_{j}$, then

$$\begin{align*} \nabla_{X}Y = \nabla_{u^{i}X_{i}}v^{j}X_{j} &= u^{i}X_{i}(v^{j})X_{j} + u^{i}v^{j}\nabla_{X_{i}}X_{j} \\ &= u^{i}X_{i}(v^{j})X_{j} + u^{i}v^{j}\Gamma_{ij}^{k}X_{k} \\ &= u^{i}X_{i}(v^{k})X_{k} + u^{i}v^{j}\Gamma_{ij}^{k}X_{k} \\ &= \left( u^{i}X_{i}(v^{k}) + u^{i}v^{j}\Gamma_{ij}^{k}\right)X_{k} \\ &= \left( u^{i}X_{i}(v^{k}) + u^{i}v^{j}\Gamma_{ij}^{k}\right)X_{k} \\ \end{align*}$$

If we denote this as $X = X^{i}\dfrac{\partial }{\partial x_{i}}, Y = Y^{i}\dfrac{\partial }{\partial x_{j}}$, then

$$\begin{align*} \nabla_{X}Y &= \left( X^{i}\dfrac{\partial Y^{k}}{\partial x_{i}} + X^{i}Y^{j}\Gamma_{ij}^{k}\right)\dfrac{\partial }{\partial x_{k}} \\ &= \sum_{i,k}\left( X^{i}\dfrac{\partial Y^{k}}{\partial x_{i}} + \sum_{j}X^{i}Y^{j}\Gamma_{ij}^{k}\right)\dfrac{\partial }{\partial x_{k}} \end{align*}$$

Moreover, the covariant derivative of the vector field $V = v^{j}X_{j}$ is as follows.

$$\dfrac{DV}{dt} = \sum_{k} \left( \dfrac{d v^{k}}{dt} + \sum_{i,j} v^{j}\frac{dc_{i}}{dt} \Gamma_{ij}^{k} \right) X_{k}$$

Proof

• Part 1. Uniqueness

  Assume that a connection $\nabla$ satisfying the conditions of the theorem exists. Since $\nabla$ is compatible, for vector fields $X,Y,Z \in$ $\mathfrak{X}(M)$, the following holds:

  $$\begin{align*} X g(Y, Z) =&\ g(\nabla_{X}Y, Z) + g(Y, \nabla_{X}Z) \\ Y g(Z, X) =&\ g(\nabla_{Y}Z, X) + g(Z, \nabla_{Y}X) \\ Z g(X, Y) =&\ g(\nabla_{Z}X, Y) + g(X, \nabla_{Z}Y) \\ \end{align*}$$

  By adding the first equation and the second, and subtracting the third, as $\nabla$ is symmetric, we obtain the following:

  $$\begin{align*} &\ X g(Y, Z) + Y g(Z, X) - Z g(X, Y) \\ =&\ g(\nabla_{X}Y, Z) + {\color{red}g(Y, \nabla_{X}Z)} + {\color{blue}g(\nabla_{Y}Z, X)} + g(Z, \nabla_{Y}X) - {\color{red}g(\nabla_{Z}X, Y)} - {\color{blue}g(X, \nabla_{Z}Y)} \\ =&\ {\color{red}g(\nabla_{X}Z-\nabla_{Z}X, Y)} + {\color{blue}g(\nabla_{Y}Z - \nabla_{Z}Y, X)} + g(\nabla_{X}Y, Z) + g(Z, \nabla_{Y}X) \\ =&\ g([X, Z], Y) - g([Y, Z], X) - g(\nabla_{X}Y, Z) + g(Z, \nabla_{Y}X) \end{align*}$$

  Arranging this with $0=g(\nabla_{Y}X, Z)-g(\nabla_{Y}X, Z)$ added yields:

  $$X g(Y, Z) + Y g(Z, X) - Z g(X, Y) \\ = g([X, Z], Y) + g([Y, Z], X) + g([X, Y], Z) + 2g(Z, \nabla_{Y}X)$$

  Arranging the last term of the right side yields:

  $$\begin{align*} g(Z, \nabla_{Y}X) =&\ \dfrac{1}{2}\Big( X g(Y, Z) + Y g(Z, X) - Z g(X, Y) \\ &\ - g([X, Z], Y) - g([Y, Z], X) - g([X, Y], Z) \Big) \tag{1} \end{align*}$$

  Now, let’s assume that another connection $\nabla^{\prime}$ exists.

  $$\begin{align*} g(Z, \nabla^{\prime}_{Y}X) =&\ \dfrac{1}{2}\Big( X g(Y, Z) + Y g(Z, X) - Z g(X, Y) \\ &\ - g([X, Z], Y) - g([Y, Z], X) - g([X, Y], Z) \Big) \end{align*}$$

  Subtracting these two equations,

  $$g(Z, \nabla_{Y}X)-g(Z, \nabla^{\prime}_{Y}X) = g(Z, \nabla_{Y}X - \nabla^{\prime}_{Y}X) = 0$$

  According to the properties of inner product, for the above equation to hold for all $Z$, it must be $\nabla_{Y}X - \nabla^{\prime}_{Y}X=0$. Therefore, such a connection $\nabla$ is unique.

  $$\nabla_{Y}X = \nabla^{\prime}_{Y}X$$

• Part 2. Existence

  If $\nabla$ is defined as in $(1)$, it is well-defined and satisfies the conditions of the theorem well.

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