Affine Connection
Buildup
Given a vector field $\mathbf{V}$ on a differentiable manifold, we can differentiate functions defined on the manifold using the vector field. Naturally, one might also want to differentiate the vector field itself. However, approaching the differentiation of the vector field $\mathbb{R}^{3}$ in the sense of differential geometry proves to be impossible as follows.
First Case
Let’s consider $S \subset \mathbb{R}^{3}$ as a surface and $c : I \to S$ as a curve given on $S$. Also, let’s say that $\mathbf{V}$ is a vector field following $c$. Then, $\mathbf{V}(t)$ becomes a tangent vector on $c(t)$.
$$ \mathbf{V}(t) \in T_{c(t)}S $$
Then, it can be represented as a coordinate vector like this:
$$ \mathbf{V}(t) = \left( V_{1}(t), V_{2}(t), V_{3}(3) \right) $$
Therefore, one would desperately want to differentiate the vector like this:
$$ \dfrac{d \mathbf{V}}{d t}(t) = \left( V_{1}^{\prime}(t), V_{2}^{\prime}(t), V_{3}^{\prime}(3) \right) $$
However, defining the derivative of $\mathbf{V}$ like above generally does not result in a tangent vector.
$$ \dfrac{d \mathbf{V}}{d t}(t) \notin T_{c(t)}S $$
Differential geometry is interested in objects with intrinsic properties, but such a definition makes the derivative of the vector field not intrinsic. Therefore, the vector field is instead projected onto the tangent bundle $TS$ to treat it as a derivative. Let’s call $\Pi : \mathbb{R}^{3} \to TS$ an orthogonal projection. Then, the derivative of the vector field is defined as follows:
$$ \dfrac{D \mathbf{V}}{d t}(t) := \Pi \circ \dfrac{d \mathbf{V}}{d t}(t) $$
This is called the covariant derivative and is intrinsic.
Second Case
Consider the differentiation of the function defined by the limit as follows.
$$ \dfrac{d \mathbf{v}}{d t}(t) = \lim\limits_{h \to 0} \dfrac{\mathbf{V}(t+h) - \mathbf{V}(t)}{h} $$
However, since $\mathbf{V}(t+h) \in T_{c(t+h)}S$ and $\mathbf{V}(t) \in T_{c(t)}S$, the two terms in the numerator are elements of different spaces. Therefore, addition operation is impossible.
For these reasons, the differentiation of the vector field is defined as an abstract concept that satisfies the formal conditions that differentiation must have.
Definition
Let $\mathfrak{X}(M)$1 be the set of $C^{\infty}$ vector fields on a differentiable manifold $M$.
$$ \mathfrak{X}(M) := \left\{ \text{all vector fields of class } C^{\infty} \text{ on } M \right\} $$
Let $\mathcal{D}(M)$ be the set of $C^{\infty}$ functions defined on $M$.
$$ \mathcal{D}(M) := \left\{ \text{all real-valued functions of class } C^{\infty} \text{ defined on } M \right\} $$
Then, an affine connection $\nabla$ on the differentiable manifold $M$ is
$$ \begin{align*} \nabla : \mathfrak{X}(M) \times \mathfrak{X}(M)& \to \mathfrak{X}(M) \\ (X, Y) &\mapsto \nabla_{X}Y \end{align*} $$
defined as such a mapping, satisfying the following properties:
- $\nabla_{fX + gY} Z = f \nabla _{X}Z + g\nabla_{Y}Z$
- $\nabla_{X}(Y + Z) = \nabla_{X}Y + \nabla_{X}Z$
- $\nabla_{X}(fX) = f\nabla_{X}Y + X(f) Y$
Explanation
In $\nabla_{X}Y$, $X$ is the variable being differentiated, and $Y$ is the function being differentiated. Hence, 1. ~ 3. represent the following properties of differentiation, respectively.
1. $\left( a\dfrac{\partial }{\partial x} + b\dfrac{\partial }{\partial y} \right)f = a\dfrac{\partial f}{\partial x} + b\dfrac{\partial f}{\partial y}$
2. $\dfrac{\partial }{\partial x}(f+ g) = \dfrac{\partial f}{\partial x} + \dfrac{\partial g}{\partial x}$
3. $\dfrac{\partial }{\partial x}(fg) = \dfrac{\partial f}{\partial x}g + f\dfrac{\partial g}{\partial x}$
Therefore, $\nabla_{X}$ is interpreted as $\dfrac{\partial}{\partial x}$, and $Y$ as $f$.
Theorem
$(\nabla_{X}Y)(p)$ depends only on $X(p)$ and $Y(\gamma (t))$. At this time, $\gamma$ is
$$ \gamma : (-\epsilon, \epsilon) \to M \\ \gamma (0) = p \\ \gamma^{\prime}(0) = X(p) $$
a curve satisfying the condition.
Proof
Choose a coordinate $\mathbf{x} : U \to M$. And let’s consider $X, Y$ as a vector field.
$$ X = \sum_{i} X_{i} \dfrac{\partial }{\partial x_{i}},\quad Y = \sum_{j} Y_{j}\dfrac{\partial }{\partial x_{j}} $$
Then, due to the properties of $\nabla$,
$$ \begin{align*} \nabla_{X}Y =& \nabla_{\sum_{i} X_{i}\frac{\partial}{\partial x_{i}}}\sum_{j}Y_{j}\dfrac{\partial }{\partial x_{j}} \\ =& \sum_{i,j} \nabla_{X_{i}\frac{\partial}{\partial x_{i}}}Y_{j}\dfrac{\partial }{\partial x_{j}} &\text{by 1. and 2.}\\ =& \sum_{i,j} X_{i}\nabla_{\frac{\partial}{\partial x_{i}}}Y_{j}\dfrac{\partial }{\partial x_{j}} &\text{by 1.} \\ =& \sum_{i,j} X_{i} \left( \dfrac{\partial Y_{j}}{\partial x_{i}}\dfrac{\partial }{\partial x_{j}} + Y_{j}\nabla_{\frac{\partial}{\partial x_{i}}}\dfrac{\partial }{\partial x_{j}} \right) &\text{by 3.} \end{align*} $$
At this point, $\nabla_{\frac{\partial}{\partial x_{j}}}\dfrac{\partial }{\partial x_{j}}$ is a value that depends only on the choice of coordinates, independent of the vector field. Since this is also a vector field according to the definition of affine connection, if the coefficients are said to be $\Gamma_{ij}^{k}$, it can be written as follows:
$$ \nabla_{\frac{\partial }{\partial x_{i}}}\dfrac{\partial }{\partial x_{j}} = \sum_{k} \Gamma_{ij}^{k} \dfrac{\partial }{\partial x_{k}} $$
Substituting this,
$$ \begin{align*} \nabla_{X}Y =& \sum_{i,j} X_{i} \left( \dfrac{\partial Y_{j}}{\partial x_{i}}\dfrac{\partial }{\partial x_{j}} + Y_{j}\nabla_{\frac{\partial}{\partial x_{i}}}\dfrac{\partial }{\partial x_{j}} \right) \\ =& \sum_{i,j} X_{i} \left( \dfrac{\partial Y_{j}}{\partial x_{i}}\dfrac{\partial }{\partial x_{j}} + Y_{j}\sum_{k} \Gamma_{ij}^{k} \dfrac{\partial }{\partial x_{k}} \right) \\ =& \sum_{i,j} X_{i} \dfrac{\partial Y_{j}}{\partial x_{i}}\dfrac{\partial }{\partial x_{j}} + \sum_{i,j,k} X_{i}Y_{j}\Gamma_{ij}^{k} \dfrac{\partial }{\partial x_{k}} \end{align*} $$
Here, since $i,j,k$ is a dummy index, let’s change $j$ in the previous term to $k$. Then,
$$ \begin{align*} \nabla_{X}Y =& \sum_{i,k} X_{i} \dfrac{\partial Y_{k}}{\partial x_{i}}\dfrac{\partial }{\partial x_{k}} + \sum_{i,j,k} X_{i}Y_{j}\Gamma_{ij}^{k} \dfrac{\partial }{\partial x_{k}} \\ =& \sum_{i,k} X_{i} \left( \dfrac{\partial Y_{k}}{\partial x_{i}} + \sum_{j} Y_{j}\Gamma_{ij}^{k}\right) \dfrac{\partial }{\partial x_{k}} \\ \end{align*} $$
Here, $\Gamma_{ij}^{k}, \dfrac{\partial }{\partial x_{k}}$ is determined by the given coordinates. Just like that, $\dfrac{\partial Y_{k}}{\partial x_{i}}$ is also determined once $Y_{k}$ is given. Therefore, the above equation depends only on the value of $X(p), Y(\gamma (t))$.
■
It is the Lie algebra X.
\mathfrak{X}↩︎