# Calculating the Covariant Derivatives on Latitudes of a Sphere

I am currently trying to solve a question, but am $\textit{really}$ struggling with the idea of Covariant Derivative, and would love for someone to help me out. The question reads:

On a sphere of Radius $R$, parametrized by: $$\sigma(\theta, \phi) = (R\cos(\phi)\sin(\theta), R\sin(\phi)\sin(\theta), R\cos(\theta))$$ Consider the curves $\theta=c\in\mathbb{R}$. Parametrize the Curves by arc-length $s$. Then, consider the southward pointing Vector Field $X_{\theta}$. Compute the Covariant Derivative: $$\frac{D }{\partial s}(X_{\theta})$$ Further, what is the Covariant Derivative $$\frac{D}{\partial s}(\sin(s)\cdot X_{\theta})$$

Now, I have calculated the curves by arc-length. However, I am confused by how to compute the Covariant Derivative. In $\textit{Do Carmo's Differential Geometry of Curves and Surfaces}$, the definition of the Covariant Derivative is given by: \begin{align*} \frac{DW}{\partial t} &= (a'+\Gamma_{11}^1au'+\Gamma_{12}^1av'+\Gamma_{12}^1 bu'+\Gamma_{22}^1 bv')x_u+(b'+\Gamma_{11}^2au'+\Gamma_{12}^2av'+\Gamma_{12}^2bu'+\Gamma_{22}^2bv')x_v \end{align*} Where the surface is parametrized as $x(u(t),v(t))$ and $w(t)$ is given by: $$w(t)=a(t)x_u+b(t)x_v$$ Now, in my case, $X_{\theta}$ is (I'm assuming) just the partial derivative w.r.t $\theta$ of my parametrization? But how do I compute $a(t)$ and $b(t)$? Is there an easier way to solve this question? I know of the way of finding the Christoffel Symbols and finding $a(t)$ and $b(t)$ through ODE's, but am I missing something?

• I think I've mentioned this to you before, but you might find my differential geometry text a useful reference to supplement doCarmo with more concrete examples. In particular, I have this problem completely worked out in there. Apr 27, 2017 at 5:05

I am going to try and solve this, but I'm not sure if I'm doing it right. So, since $X_{\theta}=\sigma_{\theta}$, we require that if we write: $$X_{\theta}=a(s)\cdot \sigma_{\theta}+b(s)\cdot\sigma_{\phi}$$ Then $a(s)=1$ and $b(s)=0$. Plugging this into the equation in the question:
$$\frac{D}{\partial s}(X_{\theta}) = (\Gamma_{11}^1\theta'+\Gamma_{12}^1\phi')\cdot\sigma_{\theta}+(\Gamma_{11}^2\theta'+\Gamma_{12}^2\phi')\cdot\sigma_{\phi}$$ Where the Prime denotes $\frac{\partial}{\partial s}$. Now, having found that: $$\theta'=0, \>\>\phi' = \frac{1}{\rho\sin(c)}$$ Also, noting that: $$\Gamma_{11}^1=0, \>\>\Gamma_{12}^2=0, \>\>\Gamma_{11}^2 = 0, \>\>\Gamma_{12}^2 = \cot(\theta)=\cot(c)$$Plugging these values in above yields: $$\frac{D}{\partial s}(X_{\theta}) = \frac{\cot(c)}{\rho\sin(c)}\cdot\sigma_{\phi}$$