I am trying to understand the proof that the Levi-Civita connection is torsion free. In the notes in theorem 6.8 it is written that

$g(\nabla_XY, Z) - g(\nabla_YX, Z) = g(Z, [X,Y])$

proves that connection is torsion free. My question is how do we show that the above relation satisfies the 0 torsion definition

$\nabla_XY - \nabla_YX = [X, Y]$?

  • 2
    $\begingroup$ If the relation is true for any vector field $Z$ then this follows from the fact that $g$ is bilinear and positive definite. $\endgroup$ – Thomas Jun 25 '17 at 11:23

Swap the two arguments in the $g$ on the right-hand side, and move it to the left. Now you have $$ g(\nabla_XY, Z) - g(\nabla_YX, Z) - g( [X,Y], Z) = 0 $$ Use the bilinearity of $g$ to change that into $$ g(\nabla_XY- \nabla_YX - [X,Y], Z) = 0. $$

Since this holds for every $Z$, and $g$ is nondegenerate, you get that $$ \nabla_XY- \nabla_YX - [X,Y] = 0. $$


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