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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)?

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4  
Change the plus sign to a minus sign. –  Will Jagy Oct 27 '11 at 7:12
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More is true: $[X, Y] = \nabla_Y X - \nabla_X Y$, for arbitrary vector fields $X$ and $Y$ and any torsion-free connection $\nabla$. –  Zhen Lin Oct 27 '11 at 7:18
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What's your definition $\nabla_XY$? (Also, @ZhenLin, do you mean $[Y,X]$?) –  Jesse Madnick Oct 27 '11 at 15:07
    
@Jesse: Oops, yes, of course. –  Zhen Lin Oct 27 '11 at 16:29

1 Answer 1

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, $df$ comes from the smooth structure, the gradient needs more.

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