# Prove Levi-Civita connection is torsion free

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

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

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

## 1 Answer

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 $$\nabla_XY- \nabla_YX - [X,Y] = 0.$$