Prove: $(\delta^\nabla\text{d}^\nabla+\text{d}^\nabla\delta^\nabla)h=\nabla^*\nabla h-\mathring{R}_gh+h\circ\text{Ricc}_g$

Let $(M,g)$ be a Riemannian manifold with Levi-Civita connection $\nabla$ and adjoint $\nabla^*$, and exterior derivative $\text{d}^\nabla$ and adjoint $\delta^\nabla$. For a symmetric 2-covariant tensor $h\in\Gamma(S^2M)\subset\Gamma(T^*M\otimes T^*M)\cong\Omega^1(T^*M)$, viewed as a $T^*M$-valued 1-form on $M$, section 12.69. in Besse Einstein Manifolds says that the following is an easy calculation: $$(\delta^\nabla\text{d}^\nabla+\text{d}^\nabla\delta^\nabla)h=\nabla^*\nabla h-\mathring{R}_gh+h\circ\text{Ricc}_g.$$ Can someone please present a proof or point to where (a detailed) one is available in the literature?

• This is a special case of more general identity called the Weitzenböck identity en.wikipedia.org/wiki/Weitzenb%C3%B6ck_identity – Branimir Ćaćić Nov 12 '14 at 23:18
• I don't have time to work this out, but you might look in Wu's book "The Bochner Technique in Differential Geometry." – Neal Nov 14 '14 at 1:40
• Petersen's notes might be a useful reference (I haven't read them though): math.ucla.edu/~petersen/BLWformulas.pdf – Phillip Andreae Nov 14 '14 at 3:14
• It's nicely done on pages 11 to 13 in Eells & Lemaire Selected topics in harmonic maps. – mdg Nov 15 '14 at 0:02

Set $\Delta=\text{d}^\nabla\delta^\nabla+\delta^\nabla\text{d}^\nabla$. From (1.34) on page 13 of Ells & Lemaire: If $h\in\Omega^1(V)\cong\Gamma(T^*M\otimes V)$ and $X\in\mathfrak{X}(M)$, then $$\Delta h_k=\nabla^*\nabla h-g^{ij}R^V(\partial_i,\partial_k)h(\partial_j)+g^{ij}h(R(\partial_i,\partial_k)\partial_j),$$ where $R^V$ is the curvature tensor the vector bundle $V\rightarrow M$.
Let $R=R^{T^*M}$ be curvature tensor on $V=T^*M$, where now $h\in\Omega^1(V)\cong\Gamma(T^*M\otimes T^*M)$ is a $T^*M$-valued 1-form, and write \begin{align*} h_k=h(\partial_k)=h(\partial_k,\cdot)&=h_{km}\text{d}x^m. \end{align*} Continue to use geodesic normal coordinates and recall that $R(\partial_i,\partial_j)\partial_k=g^{n\ell}R_{ijk\ell}\partial_n$ to find that: \begin{align*} \Delta h_k&=-g^{ij}\nabla_i\nabla_j h(\partial_k,\cdot)-g^{ij}R(\partial_i,\partial_k)h(\partial_j,\cdot)+g^{ij}h\big(R(\partial_i,\partial_k)\partial_j,\cdot\big)\\ &=-g^{ij}\nabla_i\nabla_jh_{kq}\text{d}x^q-g^{ij}h_{jq}R(\partial_i,\partial_k)\text{d}x^q+g^{ij}g^{sp}h_{pq}R_{ikjs}\text{d}x^q. \end{align*} Now evaluate at $\partial_\ell$ and use the fact that $R(\partial_i,\partial_k)\text{d}x^q(\partial_\ell)=-\text{d}x^q\big(R(\partial_i,\partial_k)\partial_\ell\big)$ to find that: \begin{align*} \Delta h_k(\partial_\ell)&=-g^{ij}\nabla_i\nabla_jh_{kq}\text{d}x^q(\partial_\ell)-g^{ij}h_{jq}R(\partial_i,\partial_k)\text{d}x^q(\partial_\ell)+g^{ij}g^{sp}h_{pq}R_{ikjs}\text{d}x^q(\partial_\ell)\\ &=-g^{ij}\nabla_i\nabla_jh_{k\ell}+g^{ij}h_{jq}\text{d}x^q\big(R(\partial_i,\partial_k)\partial_\ell\big)+g^{ij}g^{sp}h_{p\ell}R_{ikjs}\\ &=\nabla^*\nabla h_{k\ell}-g^{ij}g^{qp}R_{ki\ell p}h_{jq}+g^{sp}h_{p\ell}R_{ks}\\ &=\nabla^*\nabla h_{k\ell}-\mathring{R}_gh_{k\ell}+(h\circ\text{Ricc}_g)_{k\ell} \end{align*} as required.