Covariant Derivative

Covariant Derivative

Let $f$ be a differentiable function on a manifold $M$ defined in a neighborhood of a point $p_0\in M$. The derivative of $f$ along a tangent vector $v\in T_pM$ is defined as

$$D_v f = \frac{d}{dt}f(p(t))|_{t=0}=\lim_{t\to 0}\frac{1}{t}\left(f(p(t)) - f(p_0)\right)$$

However, this directional derivative cannot be defined on arbitrary tensor field on $M$, as one cannot subtract tensors at different point on $M$. We have to define a new operation for directional derivative along a tangent vector for a tensor field. We call that operation, the covariant derivative.

A covariant derivative $\nabla$ on a manifold $M$ is an operation satisfying the following axioms:

$$\begin{align} &\nabla_{(u+v)}X = (\nabla_{u}X) + (\nabla_{v}X)\\ &\nabla_{av}X = a\nabla_vX\\ &\nabla_v(X+Y) = (\nabla_vX) + (\nabla_uY)\\ &\nabla_v(fX) = (D_vf)X + f(p)(\nabla_vX)\\ &\text{If \(X\) and \(Y\) are smooth, so is \(\nabla_XY\)} \end{align}$$

where $u$, $v$ are vectors on the tangent space $T_pM$, $X$ and $Y$ are smooth vector fields. Let $x^1,...,x^n$ be a coordinate system on $M$. Let $\partial_j = \partial/\partial x^j$. Since $\partial_j$ span the tangent space $T_pM$, $\nabla X//\partial x^j$ is the covariant derivative of $X$ along $\partial_j$, and vector fields $X$, $Y$ can be written as $X = \sum_j X^j\partial_j$ and $Y = \sum_j Y^j\partial_j$. The covariant derivative of $X$ along a vector of $Y$ is $$\begin{align} \nabla_YX &= \nabla_{Y^j\partial_j}(X^i\partial_i)\\ &= Y^j(\partial_j X^i)\partial_i + Y^jX^i(\nabla_j\partial_i)\text{(by axiom 2 and 4)}.\\ \end{align}$$ By axiom 5, $\nabla_j\partial_i$ is also a smooth vector field, which can be written as $$\nabla_j\partial_i = \sum_k \Gamma^k_{ij}\partial_k.$$ So, $$\nabla_YX = (Y^j\partial_j X^k + \Gamma^k_{ij} Y^j X^i)\partial_k.$$ Let $p(t)$ be a differentiable curve on $M$. The covariant derivative of $X$ along the tangent vector $\dot{p}(t)$ of the curve $p(t)$ is given by $$\begin{align} \nabla_{\dot{p}}X &= \left(\partial_jX^k\frac{dx^j}{dt}+\Gamma^k_{ij}\frac{dx^j}{dt}X^i\right)\partial_k\\ &= \left(\frac{dX^k}{dt} + \Gamma^k_{ij}\frac{dx^j}{dt}X^i\right)\partial_k. \end{align}$$ A covariant derivative on a manifold is also called an affine connection, or so sometimes just connection, as it gives a connection for tangent vectors at different points on a curve. $\Gamma^k_{ij}$ is called the connection coefficients.

Covariant Derivative of the metric tensor

Since a manifold is locally flat, , i.e. the Christoffel symbols are zero locally. At a point $P$, $$V^\alpha_{;\beta}=V^\alpha_{,\beta}.$$ This is also true for a metric tensor $$g_{\alpha\beta;\gamma}=g_{\alpha\beta,\gamma}.$$ Since the metric tensor derivative is $$g_{\alpha\beta,\gamma}=0$$ locally, we conclude $$\bbox[5px,border:2px solid #666] { g_{\alpha\beta;\gamma}=0 }$$ for all basis at different point.

Parallel Transport

Let $\lambda$ be the parameter of the curve. The tangent vector of the curve is given by $$\vec{U}=\frac{d\vec{x}}{d\lambda}.$$ Let $\vec{V}$ be a vector that is constant along curve, i.e. $$\frac{dV^\alpha}{d\lambda}=0$$ at a point $P$. Then, $$\begin{align} \frac{dV^\alpha}{d\lambda} &= \frac{\partial V^\alpha}{\partial x^\beta}\frac{dx^\beta}{d\lambda}\\ &= U^\beta V^\alpha_{,\beta}.\\ \end{align}$$ Since it is locally flat, the Christoffel symbol is 0, $$U^\beta V^\alpha_{,\beta} = U^\beta V^\alpha_{;\beta} = 0.$$ Since covariant derivative is true for all basis, along the curve with different $\lambda$ value, we have $$U^\beta V^\alpha_{;\beta}=0.$$ Then, $\vec{V}$ is said to be parallel-transported along the curve.

Commutator of Covariant Derivative

For the second derivative of the covariant derivative locally where the Christoffel symbols are $$\begin{align} \nabla_\alpha\nabla_\beta V^\mu &= \nabla_\alpha(V^\mu_{;\beta}) \\ &= (V^\mu_{;\beta})_{,\alpha} + \Gamma^\mu_{\sigma\alpha}V^\sigma_{;\beta}-\Gamma^\sigma_{\beta\alpha}V^\mu_{;\sigma} \\ &= V^\mu_{,\beta\alpha}+\Gamma^\mu_{\nu\beta,\alpha}V^\nu. \end{align}$$ Relabelling the dummy variables, $$\nabla_\beta\nabla_\alpha V^\mu = V^\mu_{,\alpha\beta} + \Gamma^\mu_{\nu\alpha,\beta}V^\nu.$$ Then the commutator of the second derivative of covariant derivative is $$\begin{align} [\nabla_\alpha,\nabla_\beta]V^\mu & := \nabla_\alpha\nabla_\beta V^\mu - \nabla_\beta\nabla_\alpha V^\mu \\ &= \left(\Gamma^\mu_{\nu\beta,\alpha}-\Gamma^\mu_{\nu\alpha,\beta}\right)V^\nu \\ \end{align}$$