I realize that my question will sound silly, but I am reading Petersen's book on Riemannian geometry (third edition) and I can't quite figure out one step in his proof of the Weitzenböck identity for forms (Theorem 9.4.1, page 347). My notation is as follows. Let $T$ be a tensor on a Riemannian manifold $M$, with Levi-Civita connection $\nabla$ and let $R$ denote the curvature tensor with the following sign convention $$R(X,Y)T=\nabla^2_{X,Y}T-\nabla^2_{Y,X}T=\nabla_X(\nabla_Y T)-\nabla_Y(\nabla_X T)-\nabla_{[X,Y]}T$$ where $\nabla^2_{X,Y}T=\nabla_X(\nabla_YT)-\nabla_{\nabla_X Y}T$. Let $\{E_1,\dots,E_n\}$ be a local orthonormal frame. Then, for a $k$-form $\omega$ on $M$ and local vector fields $X_0,X_1,\dots,X_k$ we have that its exterior derivative can be expressed as $$(d\omega)(X_0,\dots,X_k)=\sum_{i=0}^n(-1)^i(\nabla_{X_i}\omega)(X_0,\dots,\widehat{X}_i\dots,X_k)$$ and the $L^2$-adjoint $\delta$ of $d$ is given by $$(\delta\omega)(X_2,\dots,X_k)=-\sum_{j=0}^n(\nabla_{E_j}\omega)(E_j,X_2,\dots, X_k)$$ Finally, let us define the tensor $\text{Ric}(\omega)$ by $$(\text{Ric}(\omega))(X_1,\dots,X_k)=\sum_{i,j=1}^n (R(E_j,X_i)\omega)(X_1,\dots,E_j,\dots,X_k)$$ where $E_j$ is in the $i$-th slot, instead of $X_i$. Let the "rough Laplacian" $\Delta=-\nabla^*\nabla$ be defined as $\sum_j \nabla^2_{E_j,E_j}$, so that $$-(\Delta\omega)(X_1,\dots,X_k)=(\nabla^*\nabla\omega)(X_1,\dots,X_k)=-\sum_j(\nabla^2_{E_j,E_j}\omega)(X_1,\dots,X_k)$$ and let the De Rham Laplacian $\Delta_H$ be defined as usual by $\Delta_H=d\delta+\delta d$, so that $$(\Delta_H\omega)(X_1,\dots,X_k)=(d(\delta\omega)+\delta(d\omega))(X_1,\dots,X_k)$$
The Weitzenböck identity in question is $$\Delta_H\omega=\nabla^*\nabla\omega+\text{Ric}(\omega)$$
I have no problems in showing that $$(\delta d\omega)(X_1,\dots,X_k)=(\nabla^*\nabla\omega)(X_1,\dots,X_k)+\sum(\nabla^2_{E_j,X_i}\omega)(X_1,\dots,E_j,\dots,X_k)$$ but I can't seem to get the other half of the formula, i.e. $$(d\delta\omega)(X_1,\dots,X_k)=-\sum(\nabla^2_{X_i,E_j}\omega)(X_1,\dots,E_j,\dots,X_k)$$
I'll explain the problem in the basic case where $k=1$, i.e. $\omega$ is a 1-form and $\delta\omega$ is a smooth function. Let $X$ be a local vector field. Then $$(d\delta\omega)(X)=X(\delta\omega)=-\sum X((\nabla_{E_j}\omega)(E_j))=-\sum\left((\nabla_X(\nabla_{E_j}\omega))(E_j)+(\nabla_{E_j}\omega)(\nabla_X E_j)\right)$$ and it is not clear to me that the term inside the sum is the second covariant derivative $(\nabla^2_{X,E_j}\omega)(E_j)$. I know I could fix the problem by adapting the frame so that $\nabla_{E_i}E_j=0$ at a point and use tensoriality, but I was trying to see if I could get the full calculation in a general orthonormal frame to work out.
What am I missing? Where is my mistake?
P.S. I was super pedantic with brackets in order to avoid any notation misunderstandings. Sorry if the formulae are not very legible.