Weitzenböck identity for $TM$-valued differential forms

Let $$M$$ be a Riemannian manifold, and let $$\nabla$$ denote its Levi-Civita connection. We have two second order differential operators $$\Gamma(T^*M \otimes TM) \to \Gamma(T^*M \otimes TM)$$:

The Bochner Laplacian $$\Delta_B=\nabla^* \nabla$$, and the "Hodge" Laplacian $$\Delta_H=\delta_{\nabla} d_{\nabla}+d_{\nabla}\delta_{\nabla}$$, where here $$d_{\nabla}: \Omega^1(M,TM) \to \Omega^2(M,TM)$$ is the covariant exterior derivative associated with the Levi-Civita connection $$\nabla$$, and $$\delta_{\nabla}$$ is its adjoint.

Is there some kind of Weitzenböck formula connecting these two Laplacians? something like $$\Delta_H-\Delta_B=C(R^{\nabla})$$, where $$R^{\nabla}$$ is the curvature tensor of $$\nabla$$?

I know that for real-valued one-forms, the classical Weitzenböck formula is $$\Delta_H-\Delta_B=\text{Ric}$$, where "$$\text{Ric}$$" denotes the Ricci curvature of the manifold. Can we do something similar here?

• Please look at this question and the discussion in the comments there. In particular, please pay attention to P.Peterson's note where this topic is covered in a greater generality. – Yuri Vyatkin Apr 20 at 23:58
• A proof of a similar identity for the case of vector-bundle-valued 1-forms one can find on p.18 of my thesis. Perhaps, specializing that calculation for the case of $TM$-valued 1-forms would be a partial answer to your question. – Yuri Vyatkin Apr 21 at 0:30
• Thank you, I will take a look. – Asaf Shachar Apr 21 at 14:41