# Torsion-free covariant derivative

Let $$\nabla$$ be an affine connection on a smooth manifold $$\mathcal M$$. We say $$\nabla$$ is torsion-free if for any vector fields $$X, Y$$ on $$\mathcal M$$, we have $$$$\nabla_X Y - \nabla_Y X = [X, Y],$$$$ where $$[X, Y] = X \circ Y - Y \circ X$$ is the Lie bracket of $$X$$ and $$Y$$. Is it equivalent to state the torsion-free property as $$$$\nabla_X Y = X \circ Y$$$$ for all vector fields $$X, Y$$? Why or why not?

$$\nabla_XY$$ is a vector field and hence acts as a derivation on $$C^\infty(M)$$, i.e. $$\nabla_XY(fg)=f\nabla_XY(g)+g\nabla_XY(f)$$ for any $$f,g\in C^\infty(M)$$. The object $$X\circ Y$$ is a second order differential operator \begin{align}X\circ Y(fg)&=X(fY(g)+gY(f))\\ &=X(f)Y(g)+f(X\circ Y)(g)+X(g)Y(f)+gX\circ Y(g)\\ &\neq f(X\circ Y)(g)+g(X\circ Y)(f) \end{align} Since $$X\circ Y$$ is not a derivation for all $$f,g\in C^\infty(M)$$ it does not give a well defined vector field so $$\nabla_XY\neq X\circ Y$$.
• +1. OP should also note that the computation done here is exactly the reason why the Lie bracket $[X,Y]$ is defined the way it is. Originally, one would like $X\circ Y$ to be a vector field, but the presence of the second-order terms is an obstruction. How to get rid of it? Subtract $Y\circ X$ and note that all the second-order terms go away, since second-order mixed partial derivatives commute. Feb 28 at 20:24