What is the value of the 2-form $X(u(Y))-Y(u(X))-u([X,Y])$, is it $2du(X,Y)$? Let $X,Y$ be two vector fields on a Riemannian manifold $(M,g,\nabla)$ where $\nabla $ is the symmetric metric connection on $M$ and let $u$ be a 1-form. I want to find the value or a simple form or interpretation of the following
\begin{equation}
(1)\hspace{.51cm} X(u(Y))-Y(u(X))-u([X,Y])\\
(2)\hspace{.51cm} X(u(Y))-u(\nabla_XY)-u(X)u(Y)
\end{equation}
John show that the first expression is a 2-form, I have been trying to prove that it is $2du$ as he claimed. We can take $X=\Sigma X^i \frac{\partial}{\partial x_i}$ , $Y=\Sigma Y^i \frac{\partial}{\partial x_i}$ and $u(\frac{\partial}{\partial x_i})=u_i$
Thanks in advance
 A: For the first one, note that if we consider $X\mapsto fX$ for some function $f$, then 
$$(fX)\big(u(Y)\big) - Y\big(u(fX)\big) - u([fX, Y]) = f\big(X\big(u(Y)\big) - Y\big(u(X)\big) - u([X, Y])\big).$$
Together with the fact that the expression is antisymmetric in $X$ and $Y$, we see that the first expression is a two form. Putting $X = \frac{\partial}{\partial x_i}$ and $Y=\frac{\partial}{\partial x_j}$ shows that the first expression is $2du$. 
For the second one, the first two terms can be written $\nabla u(X, Y)$. Not sure what to do with the last term. 
A: Expression  (1) is not in general identically equal to zero.  Indeed we have, for any $1$-form $u$, 
$du(X, Y) = X(u(Y)) - Y(u(X)) - u([X, Y]), \tag{1}$
where $X$ and $Y$ are vector fields on $M$.  (1) is a well-known result and is the $p = 1$ case of a broader formula expressing $d\omega$ for any $p$-form $\omega$.  For more information about (1) and related matters see  http://en.m.wikipedia.org/wiki/Exterior_derivative.  
As for the second expression,
$X(u(Y)) - u(\nabla_X Y) - u(X) u(Y), \tag{2}$
the best I can do is note that
$X(u(Y)) = \nabla_X (u(Y)) = (\nabla_X u)(Y) + u(\nabla_X Y), \tag{3}$
which holds since $u(Y)$ is a function and the differential operators $X$ and $\nabla_X$ agree on functions.  In light of (3), (2) becomes
$X(u(Y)) - u(\nabla_X Y) - u(X) u(Y) = (\nabla_X u)(Y) + u(\nabla_X Y) - u(\nabla_X Y) - u(X) u(Y) = (\nabla_X u)(Y) - u(X) u(Y); \tag{4}$
I don't think there is any obvious, simple way to simplify (4) further since $u(X) u(Y)$ is "quadratic", that is, of second "degree" in $u$; since $\nabla_X$ is a linear operator, most of the standard identities don't address products such as $u(X) u(Y)$.  So that's about as fat as I can take things.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
