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I am curious how people picture geometrically how the covariant derivative acts with regards to vector fields. I understand that it is merely an object which has been defined for some use and obeys certain conditions but how would you picture such a thing geometrically? In other words, when someone asks you what the covariant derivative is, how do you explain it in absence of a formal definition?

J.Main
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  • Related: https://math.stackexchange.com/questions/62602/understanding-covariant-derivative-connexion?rq=1 – Watson Jun 15 '17 at 21:13

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Suppose that the manifold $M$ is embedded in $\mathbb R^n$. Then a vector $v \in T_p M$ at point $p \in M$ can be considered a vector in $\mathbb R^n$.

The covariant derivative $\nabla_u v$ for some $u \in T_p M$ then is the ordinary derivative of $v$ in the direction $u$ projected on $T_p M$.

md2perpe
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    Note that you can't take the covariant derivative of a single vector, only of a vector field which is defined in a neighborhood of $p$ or along a curve whose velocity at $p$ is $u$. – levap Jun 15 '17 at 19:01
  • I know. Unfortunately I didn't have time to make it more precise. – md2perpe Jun 15 '17 at 19:06