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Affine Connection 📂Geometry

Affine Connection

Buildup

Given a vector field V\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 R3\mathbb{R}^{3} in the sense of differential geometry proves to be impossible as follows.

  • First Case

    Let’s consider S⊂R3S \subset \mathbb{R}^{3} as a surface and c:I→Sc : I \to S as a curve given on SS. Also, let’s say that V\mathbf{V} is a vector field following cc. Then, V(t)\mathbf{V}(t) becomes a tangent vector on c(t)c(t).

    V(t)∈Tc(t)S \mathbf{V}(t) \in T_{c(t)}S

    Then, it can be represented as a coordinate vector like this:

    V(t)=(V1(t),V2(t),V3(3)) \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:

    dVdt(t)=(V1′(t),V2′(t),V3′(3)) \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 V\mathbf{V} like above generally does not result in a tangent vector.

    dVdt(t)∉Tc(t)S \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 TSTS to treat it as a derivative. Let’s call Π:R3→TS\Pi : \mathbb{R}^{3} \to TS an orthogonal projection. Then, the derivative of the vector field is defined as follows:

    DVdt(t):=Π∘dVdt(t) \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.

    dvdt(t)=lim⁡h→0V(t+h)−V(t)h \dfrac{d \mathbf{v}}{d t}(t) = \lim\limits_{h \to 0} \dfrac{\mathbf{V}(t+h) - \mathbf{V}(t)}{h}

    However, since V(t+h)∈Tc(t+h)S\mathbf{V}(t+h) \in T_{c(t+h)}S and V(t)∈Tc(t)S\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 X(M)\mathfrak{X}(M)1 be the set of C∞C^{\infty} vector fields on a differentiable manifold MM.

X(M):={all vector fields of class C∞ on M} \mathfrak{X}(M) := \left\{ \text{all vector fields of class } C^{\infty} \text{ on } M \right\}

Let D(M)\mathcal{D}(M) be the set of C∞C^{\infty} functions defined on MM.

D(M):={all real-valued functions of class C∞ 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 MM is

∇:X(M)×X(M)→X(M)(X,Y)↦∇XY \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:

  1. ∇fX+gYZ=f∇XZ+g∇YZ\nabla_{fX + gY} Z = f \nabla _{X}Z + g\nabla_{Y}Z
  2. ∇X(Y+Z)=∇XY+∇XZ\nabla_{X}(Y + Z) = \nabla_{X}Y + \nabla_{X}Z
  3. ∇X(fX)=f∇XY+X(f)Y\nabla_{X}(fX) = f\nabla_{X}Y + X(f) Y

Explanation

In ∇XY\nabla_{X}Y, XX is the variable being differentiated, and YY is the function being differentiated. Hence, 1. ~ 3. represent the following properties of differentiation, respectively.

1.　(a∂∂x+b∂∂y)f=a∂f∂x+b∂f∂y\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.　∂∂x(f+g)=∂f∂x+∂g∂x\dfrac{\partial }{\partial x}(f+ g) = \dfrac{\partial f}{\partial x} + \dfrac{\partial g}{\partial x}

3.　∂∂x(fg)=∂f∂xg+f∂g∂x\dfrac{\partial }{\partial x}(fg) = \dfrac{\partial f}{\partial x}g + f\dfrac{\partial g}{\partial x}

Therefore, ∇X\nabla_{X} is interpreted as ∂∂x\dfrac{\partial}{\partial x}, and YY as ff.

Theorem

(∇XY)(p)(\nabla_{X}Y)(p) depends only on X(p)X(p) and Y(γ(t))Y(\gamma (t)). At this time, γ\gamma is

γ:(−ϵ,ϵ)→Mγ(0)=pγ′(0)=X(p) \gamma : (-\epsilon, \epsilon) \to M \\ \gamma (0) = p \\ \gamma^{\prime}(0) = X(p)

a curve satisfying the condition.

Proof

Choose a coordinate x:U→M\mathbf{x} : U \to M. And let’s consider X,YX, Y as a vector field.

X=∑iXi∂∂xi,Y=∑jYj∂∂xj 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,

∇XY=∇∑iXi∂∂xi∑jYj∂∂xj=∑i,j∇Xi∂∂xiYj∂∂xjby 1. and 2.=∑i,jXi∇∂∂xiYj∂∂xjby 1.=∑i,jXi(∂Yj∂xi∂∂xj+Yj∇∂∂xi∂∂xj)by 3. \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, ∇∂∂xj∂∂xj\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 Γijk\Gamma_{ij}^{k}, it can be written as follows:

∇∂∂xi∂∂xj=∑kΓijk∂∂xk \nabla_{\frac{\partial }{\partial x_{i}}}\dfrac{\partial }{\partial x_{j}} = \sum_{k} \Gamma_{ij}^{k} \dfrac{\partial }{\partial x_{k}}

Substituting this,

∇XY=∑i,jXi(∂Yj∂xi∂∂xj+Yj∇∂∂xi∂∂xj)=∑i,jXi(∂Yj∂xi∂∂xj+Yj∑kΓijk∂∂xk)=∑i,jXi∂Yj∂xi∂∂xj+∑i,j,kXiYjΓijk∂∂xk \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,ki,j,k is a dummy index, let’s change jj in the previous term to kk. Then,

∇XY=∑i,kXi∂Yk∂xi∂∂xk+∑i,j,kXiYjΓijk∂∂xk=∑i,kXi(∂Yk∂xi+∑jYjΓijk)∂∂xk \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, Γijk,∂∂xk\Gamma_{ij}^{k}, \dfrac{\partial }{\partial x_{k}} is determined by the given coordinates. Just like that, ∂Yk∂xi\dfrac{\partial Y_{k}}{\partial x_{i}} is also determined once YkY_{k} is given. Therefore, the above equation depends only on the value of X(p),Y(γ(t))X(p), Y(\gamma (t)).

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  1. It is the Lie algebra X. \mathfrak{X} ↩︎