Trace of $\nabla \alpha$ where $\alpha$ is a 1-form. Let $X$ be a Riemannian manifold, $\nabla$ be a connexion and $\alpha$ be a 1-form. How do I show that $\text{Tr}(\nabla\alpha)=\sum e_{i}\alpha(e_{i})-\alpha(\nabla_{e_{i}}e_{i})$ (where $e_{i}$'s are orthonormal frame)? I have tried using the formula
$(\nabla\alpha)(X,Y)=\nabla_{X}\alpha(Y)-\nabla_{Y}\alpha(X)-\alpha([X,Y])$
but taking the trace takes me to zero.... where have I gone wrong? Or how should I proceed? Thanks!
 A: Your formula is pretty much the definition of the trace. Observe that ($n$=dim $M$)
$$
\operatorname{Tr}\nabla \alpha := \sum_{i=1}^n \nabla_{e_i}\alpha(e_i)=\sum_{i=1}^n\left[e_i \alpha (e_i)-\alpha\left(\nabla_{e_i}e_i\right)\right]
$$
The second equality is simply a definition of how to extend the covariant derivative on general tensor fields. Roughly speaking you want the extension of the covariant derivative to commute with contractions. So for a $(1,1)$-Tensor $\omega\otimes Y$ whose contraction $C(\omega\otimes Y)$ is just the function $\omega(Y)$ one requires that
$$
\nabla_X(C(\omega\otimes Y))=C(\nabla_X(\omega\otimes Y))=C((\nabla_X\omega)\otimes Y)+C(\omega\otimes( \nabla_X Y)),
$$
where you use the Leibnizrule. Rewriting gives
$$
X(\omega(Y))=(\nabla_X\omega)(Y)+\omega (\nabla_X Y)
$$
For a $(0,1)$--Tensor resp. a one-form one has
$$
(\nabla_X\omega)(Y)=X(\omega(Y))-\omega (\nabla_X Y).
$$
Here $\omega$ denotes a one-form and $Y$ a vector field.
EDIT: The formula you wrote looks suspicious to me (don't confuse $\nabla$ with the exterior derivative d).
A: Let me give some remarks to the accepted answer which are too long to be placed in the comments.
So, we have a $1$-form $\alpha$, that is $\alpha \in \Gamma(T^* M)$. The covariant derivative of $\alpha$ is a $(0,2)$-tensor $\nabla \alpha$ defined as
$$
(\nabla \alpha)(X,Y) = X \alpha (Y) - \alpha (\nabla_X Y)
$$
User @frog has correctly explained why this definition is used.
The Riemannian structure on $M$ provides a notion of (local) orthonormal frames in the tangent bundle, so we can pick one, say $\{e_i\},\,i=1,\dots,n$, and construct the following expression:
$$
\sum_{i=1}^n (\nabla \alpha)(e_i,e_i) = \sum_{i=1}^n \left[ e_i \alpha (e_i) - \alpha (\nabla_{e_i} e_i) \right]
$$
It turns out that this expression does not depend on a particular orthonormal frame, so at each point we obtain a well-defined function. This is why we can give it a name and call it the trace of tensor $\nabla \alpha$. Therefore, being careful, we must write
$$
\mathrm{Tr} \nabla \alpha := \sum_{i=1}^n \left[ e_i \alpha (e_i) - \alpha (\nabla_{e_i} e_i) \right]
$$
The (classical) notation $\nabla_{e_i}\alpha(e_i)$ is extremely misleading. Of course, it means $(\nabla \alpha)(e_i,e_i)$ really. This is what I wanted to point out.
The OP's identity $(\nabla\alpha)(X,Y)=\nabla_{X}\alpha(Y)-\nabla_{Y}\alpha(X)-\alpha([X,Y])$ is the definition (or an expression, depending on the approach) of the exterior derivative $\mathrm{d} \alpha$ of the $1$-form $\alpha$ in terms of a torsion-free connection $\nabla$ (and this is our case, because we mean the Levi-Civita connection $\nabla$ of the Riemannian metric on $M$). The purpose of the exterior derivative is that it gives a $2$-form $\mathrm{d}\alpha$ out of a $1$-form $\alpha$, that is $\mathrm{d}\alpha(X,Y) = - \mathrm{d}\alpha(Y,X)$.
Certainly, $\nabla$ and $\mathrm{d}$ should not be confused.
