Amending a differential geometry formula

Question

I'm reading my professor's notes on Differential Geometry. In the chapter regarding connections he deals with torsion and there is an observation in which he affirms that if $\alpha$ is a 1-form and $\nabla$ is a zero torsion connection then the following hols: \begin{equation} \alpha(X,Y) = (\nabla_X\alpha)(Y) - (\nabla_Y\alpha)(X) \end{equation} I'm trying to make sense of this equation but I think there is a typo of some sort. Am I missing something or should this formula be amended?

Answer 1

I guess the formula your professor intended is $$ d \alpha(X,Y) = (\nabla_X \alpha)Y - (\nabla_Y \alpha)X. $$ This is true because, you can prove that $d\alpha$ acts on a pair of vectors as $$ d \alpha(X,Y) = X(\alpha(Y)) - Y(\alpha(X)) - \alpha([X,Y]). \tag{1} $$

A connection is symmetric (or torsion-free) if $$ \nabla_X Y - \nabla_Y X = [X,Y] $$ so (1) equals to $$ d \alpha(X,Y) = X(\alpha(Y)) - Y(\alpha(X)) - \alpha(\nabla_X Y) + \alpha(\nabla_Y X). \tag 2 $$

By definition, or because of the properties of the natural dual connection, the covariant derivative of a 1-form has the following expression: $$ (\nabla_X \alpha)Y = X(\alpha(Y)) - \alpha(\nabla_X Y) $$ therefore (2) equals $$ d \alpha(X,Y) = (\nabla_X \alpha)Y - (\nabla_Y \alpha)X. $$