TITLE: Hodge Laplacian in local coordinates
QUESTION [8 upvotes]: On a Riemannian Manifold $M^n$, the Hodge Laplacian is defined on k-forms by $\Delta\omega=\operatorname{d}\operatorname{d}^*\omega+\operatorname{d}^*\operatorname{d}\omega$. For 0-forms, e.i. smooth functions, one can easily see that w.r.t. some local coordinates with inverse metric tensor $g^{ij}$ it holds:$$\Delta\omega=-\partial_j(g^{ij}\partial_i\omega)$$ This can be used to apply results of 1-dimensional elliptic regularity to $\Delta$. My goal is to understand elliptic regularity on all k-forms, $\Delta:\Omega^k(M)\rightarrow\Omega^k(M)$. My question is: How does the term of highest order (2nd order in derivatives) look like in a local trivialisation of k-forms? (e.g. $\omega=\sum_{I=(i_1,...,i_k)\\0≤i_1≤...≤i_k≤n}\omega_I \operatorname{d}x^{i_{1}}\wedge...\wedge\operatorname{d}x^{i_{k}}$ or alternatively using some local orthogonal frames). 
Is it true or wrong that its highest order is diagonal?$$(\Delta\omega)_I=A^{I,ij}\partial_i\partial_j\omega_I+\sum_JB^{I,J,i}\partial_i\omega_J+\sum_J C^{I,J}\omega_J$$
Is it maybe even true that the $A^{I,ij}=-g^{ij}$ as in the k=0 case?

REPLY [4 votes]: Your formula is true.
The Weitzenböck formula states that the Hodge Laplacian on $k$-forms satisfies
$$
\Delta\omega=(d\delta+\delta d)\omega=\nabla^*\nabla\omega +\operatorname{Ric}(\omega),
$$
where $\nabla^*\nabla$ is the Bochner Laplacian. For a proof, see Theorem 9.4.1 in: 
P. Petersen, Riemannian geometry. Third edition. Graduate Texts in Mathematics, 171. Springer, Cham, 2016.
Finally, the Bochner Laplacian in suitable local coordinates can be represented as 
$$
\nabla^*\nabla=-\sum_{k,j} \big\{\ g^{kj}\nabla_k\nabla_j+\frac{1}{\sqrt{|g|}}\partial_{x^k}\big(\sqrt{|g|}g^{kj}\big)\cdot\nabla_j\ \big\}.
$$
This is proved in Example 10.1.32 in 
L. I. Nicolaescu, Lectures on Geometry of Manifolds.
Beautiful Nicolaescu's lectures are freely available on his homepage.