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.

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Filed under differential geometry, multilinear algebra

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