2
$\begingroup$

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)?

  • 4
    Change the plus sign to a minus sign.2011-10-27
  • 7
    More is true: $[X, Y] = \nabla_Y X - \nabla_X Y$, for arbitrary vector fields $X$ and $Y$ and _any_ torsion-free connection $\nabla$.2011-10-27
  • 2
    What's your definition $\nabla_XY$? (Also, @ZhenLin, do you mean $[Y,X]$?)2011-10-27
  • 1
    @Jesse: Oops, yes, of course.2011-10-27

1 Answers 1