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It is known for $f$ a smooth function and vector fields $X$, the covariant derivative obeys product rule

$$\nabla_Y(fX) = (Yf)\nabla_YX+f\nabla_Y X$$

I just did this calculation and i keep getting $\nabla_Y(fX) = X \nabla_Yf +f\nabla_Y X$.

Basically, $\nabla_Y(fx) = (Y(fX^1), \dots, Y(fX^n))$.

Now $Y(fX^1) = <grad(fX^1), Y> = <X^1gradf,Y>+ <fgrad X^1, Y>$.

So putting it back I should get $\nabla_Y(fX) = X \nabla_Yf +f\nabla_Y X$.

But my first term does not match

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Lemon
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  • You are correct. Your formula is NOT correct. –  Apr 06 '15 at 07:59
  • So how do I correct the erroneous part? – Lemon Apr 06 '15 at 08:22
  • Your formula $\nabla_Y(fX) = X \nabla_Yf +f\nabla_Y X$ is the correct one. –  Apr 06 '15 at 08:54
  • @John, I know, but why? – Lemon Apr 06 '15 at 08:55
  • Don't quite understand your question. To me the formula IS included in the definition of a connection. (I mean, the one that you derived IS the correct one, the one you stated at the beginning is the wrong one) –  Apr 06 '15 at 09:03
  • @John, oh so you are saying my calculations are right. I got the formula from online notes: http://people.math.gatech.edu/~ghomi/LectureNotes/LectureNotes14G.pdf – Lemon Apr 06 '15 at 09:13
  • Yes, you are correct and the note wrong. –  Apr 06 '15 at 09:14
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    Thanks and there goes 2 hours of my life... – Lemon Apr 06 '15 at 09:15

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