Lie bracket and covariant derivatives I came across the following equality
$[\text{grad} f, X] = \nabla_{\text{grad} f} X + \nabla_X \text{grad} f$
Is this true, and how can I prove this (without coordinates)?
 A: No. Replace all three occurrences of the gradient by any vector field, call it $W,$ but then replace the plus sign on the right hand side by a minus sign, and you have the definition of a torsion-free connection, $$ \nabla_W X - \nabla_X W = [W,X].$$  If, in addition, there is a positive definite metric, the Levi-Civita connection is defined to be torsion-free and satisfy the usual product rule for derivatives, in the guise of $$ X \, \langle V,W \rangle = \langle \nabla_X V, W \,\rangle + \langle V, \, \nabla_X W \, \rangle.  $$ Here $\langle V,W \rangle$ is a smooth function, writing $X$ in front of it means taking the derivative in the $X$ direction. Once you have such a connection, it is possible to define the gradient of a function, for any smooth vector field $W$ demand $$ W(f) = df(W) =  \langle \, W, \, \mbox{grad} \, f \, \rangle   $$ Note that physicists routinely find use for connections with torsion. Also, $df$ (the gradient) comes from the smooth structure, the connection needs more.
