Connections

Connections

Differentiating vector fields in \(\mathbb{R}^n\)

Given a path \(\gamma(t)\) in \(\mathbb{R}^n\), we know we can represent its velocity by \[ \dot{\gamma}(t) = \dot{\gamma}^i(t) \partial_i. \] Furthermore, in the case of \(\mathbb{R}^n\), we understand the acceleration of the curve \(\gamma(t)\) to be given by \[ \ddot{\gamma}(t) = \frac{d }{d t}\left(\dot{\gamma}^i(t) \partial_i\right) = \ddot{\gamma}^i(t) \partial_i. \] In computing the acceleration, we are differentiating a vector field. More generally in \(\mathbb{R}^n\), we understand the derivative of some vector field \(X\) in the direction \(v \in T_p\mathbb{R}^n\) at point \(p\), which we denote \(\nabla_v X\), to be \[ \nabla_v X = \left.\frac{d }{d t}\right|_{t=0}X(\gamma(t)) = \left.\frac{d }{d t}\right|_{t=0}X^i(\gamma(t))\partial_i = v(X^i)\partial_i. \] where \(\gamma(t)\) is a smooth curve such that \(\dot{\gamma}(0) = v\). This directional derivative enjoys the following nice properties:

  1. First, the directional derivative is linear with respect to the direction. Indeed, for any vectors \(v,w\) based at point \(p \in \mathbb{R}^n\), real numbers \(a,b\), and vector field \(X\) we find \[ \nabla_{av + bw}X = (av + bw)X^i \partial_i = avX^i\partial_i + bwX^i\partial_i = a\nabla_vX + b\nabla_wX. \]
  2. Second, this directional derivative is linear with respect to the vector fields. Indeed, for any vector \(v\) based at \(p \in \mathbb{R}^n\), any real numbers \(a,b\), and any vector fields \(X,Y\) that \[ \nabla_v(aX + bY) = v(aX^i + bY^i)\partial_i = avX^i\partial_i + bvY^i\partial_i = a\nabla_vX + b\nabla_vY \]
  3. Finally, we have a product rule. For any vector \(v\) at point \(p \in \mathbb{R}^n\), vector field \(X\), and smooth function \(f\) we find \[ \nabla_v(fX) = v(fX^i)\partial_i = (vf)X^i\partial_i + f(vX^i)\partial_i = (vf)X + f\nabla_vX \]

The problem with differentiating vector fields on manifolds

However, the definition of the directional derivative of a vector field \(X\) in direction \(v\) \[ \left.\frac{d }{d t}\right|_{t=0} X(\gamma(t)) = \lim_{t \to 0}\frac{X(\gamma(t)) - X(\gamma(0))}{t} \] does not make sense on a general smooth manifold! The vectors \(X(\gamma(t))\) and \(X(\gamma(0))\) belong to the two different tangent spaces \(T_{\gamma(t)}M\) and \(T_{\gamma(0)}M\), so it does not make sense to subtract them. Note that in the case of \(\mathbb{R}^n\), there is a natural identification between different tangent spaces \(T_p\mathbb{R}^n\) and \(T_q\mathbb{R}^n\) by simply translating the vectors. However, this a consequence of having a nice coordinate frame \((\partial_i)\), for we are naturally identifying \(v^i\partial_i \in T_p\mathbb{R}^n\) with \(v^i\partial_i \in T_q\mathbb{R}^n\). In order to differentiate vector fields on manifolds, we need more structure on the manifold.

Connections

We are looking to define a directional derivative operation \(\nabla_v X\) on a general smooth manifold. Formally, we are looking for a map \(\nabla: T_pM \times \Gamma(TM) \to T_pM\) varying smoothly with \(p\). If this map is denoted \(\nabla: (v, X) \mapsto \nabla_vX\), we want \(\nabla\) to satisfy the following three properties we expect from a directional derivative. In the following, \(a,b \in \mathbb{R}\) are real numbers, \(v,w \in T_pM\) are vectors based at \(p\), and \(X,Y \in \Gamma(TM)\) are vector fields on the manifold.

  1. Linearity with respect to the direction: \[ \nabla_{av + bw}X = a\nabla_vX + b\nabla_wX. \]
  2. Linearity with respect to the vector field: \[ \nabla_v(aX + bY) = a\nabla_vX + b\nabla_vY. \]
  3. Product rule: \[ \nabla_v(fX) = (vf)X + f\nabla_vX. \]

Such a map \(\nabla: T_pM \times \Gamma(TM) \to T_pM\) varying smoothly with \(p\) that satisfies the three properties above is called a connection. Equivalently, we can consider two vector fields \(X,Y\) and say \(\nabla_X Y\) is vector field such that \((\nabla_X Y)(p) = \nabla_{X(p)} Y\) so that we now have a map \(\nabla: \Gamma(TM) \times \Gamma(TM) \to \Gamma(TM)\). The directional derivative \((\nabla_X Y)\) corresponding to a connection is called the covariant derivative of \(Y\) in the direction of \(X\). The main question, however, is how do we choose which connection to use? In general, there are many connections on a smooth manifold: choose any coordinate frame \((\partial_i)\) and decide the derivative of each coordinate frame in the direction of all the other coordinate frames at every point. That is, choose functions \(A_{ij}^k\) such that \[ \nabla_{\partial_i}\partial_j = A_{ij}^k \partial_k. \] Then given any arbitrary vector fields \(X = X^i\partial_i\) and \(Y = Y^j\partial_j\), we can determine what the covariant derivative \(\nabla_XY\) should be by \[ \nabla_X Y = \nabla_{X^i \partial_i}(Y^j \partial_j) = X^i((\partial_iY^j)\partial_j + Y^j \nabla_{\partial_i}\partial_j) = X^i(\partial_iY^k + Y^jA_{ij}^k)\partial_k. \label{eq:As} \tag{A} \] For any smooth functions \(A_{ij}^k\) that we choose, the above formula in fact defines a connection.

Exercise. Verify that for any smooth functions \(A_{ij}^k\), the formula (\ref{eq:As}) defines a connection by checking the three necessary properties are satisfied.

Thus there are many possible connections on a general smooth manifold. For some intuition for connections, note that choosing a connection gives us a sense of acceleration on our manifold. Given a curve \(\gamma(t)\), it’s velocity at each point along the curve is given by the tangent vectors \(\dot{\gamma}(t)\). Then the acceleration should be the change of \(\dot{\gamma}(t)\) in direction \(\dot{\gamma}(t)\). That is, the acceleration of \(\gamma(t)\) is defined to be \(\ddot{\gamma}(t) = \nabla_{\dot{\gamma}(t)}\dot{\gamma}(t)\).