Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)?

share|cite|improve this question
Change the plus sign to a minus sign. – Will Jagy Oct 27 '11 at 7:12
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
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

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.