# covariant derivative of covectors

How do you think that the covariant derivative in $\mathbb{R}^3$ is extended over covector fields defined over a surface $\Phi:{\mathbb{R}}^2\hookrightarrow\Sigma\subset{\mathbb{R}}^3$?

We use the Riesz Representation’s Lemma, so if

$dx^k(\quad)=\langle\partial^k,\quad\rangle=\langle g^{sk}\partial_s,\quad\rangle$

then

$\nabla_{\partial_i}dx^k(\quad)=\langle \nabla_{\partial_i}\partial^k,\quad\rangle=\langle -{\Gamma^k}_{is}\partial^s,\quad\rangle$

This implies that we have:

$\nabla_{\partial_i}dx^k=-{\Gamma^k}_{is}dx^s$

This contrast nicely with $\nabla_{\partial_i}\partial_k={\Gamma^s}_{ik}\partial_s$

For a general $w=w_sdx^s$, we use the Leibniz’s rule to get

$\nabla_{\partial_i}w=({w_s}_{,i}-{w_t\Gamma^t}_{si})dx^s$

and

$\nabla_{\partial_i}(w\otimes\theta)=(\nabla_{\partial_i}w)\otimes\theta+w\otimes\nabla_{\partial_i}\theta$

The proof that $\nabla_{\partial_i}\partial^k=\nabla_{\partial_i}(g^{sk}\partial_s)=-{\Gamma^k}_{is}\partial^s$ is very fun!

You gotta remember firmly that the $\partial^k=g^{sk}\partial_s$ form the reciprocal coordinated basis, still tangent vectors but representing (à la Riesz) the coordinated covectors $dx^k$.