Your formula is true.

The Weitzenböck formula states that the Hodge Laplacian on $k$-forma 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 a 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 Manidolfds.

Beautiful Nicolaescu's lectures are freely available on his homepage.

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.

I'm on shaky ground here, I believe your formula is true.

The Weitzenböck formula states that the Hodge Laplacian on $k$-forma 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 a 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 Manidolfds.

Beautiful Nicolaescu's lectures are freely available on his homepage.

Your formula is true.

The Weitzenböck formula states that the Hodge Laplacian on $k$-forma 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 a 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 Manidolfds.

Beautiful Nicolaescu's lectures are freely available on his homepage.