I was just reading about the Laplace Beltrami operator, which is a linear, second order, self-adjoint operator on a general Riemannian metric $g_\mu\nu$ space.
$$\Delta A_{\mu}=\nabla^{\alpha}\nabla_{\alpha}A_{\mu}$$
Where $A_\alpha$ is some arbitrary vector field component on M that the operator is acting on. Suppose We're looking at some integral over the manifold $M$ such as:
$$\intop_{M}B^{\alpha}\Delta A_{\alpha}dvol_{M}$$
Where $B^\alpha$ is another vector field component. Invoking the self-adjointness of $\Delta$ One may write this as:
$$\intop_{M}B^{\alpha}\Delta A_{\alpha}dvol_{M}=\intop_{M}<dB^{\alpha},dA_{\alpha}>dvol_{M}=\intop_{M}A^{\alpha}\Delta B_{\alpha}dvol_{M}$$
This might be a dumb question, but is the $d$ in the middle equation the exterior derivative of $A,B$ (respectively)? The paper I was reading skipped the derivation, and didn't explain the notation (same with the wiki page).
If the connection is the Levi-Civita of the metric then the exterior derivative is obtained antisymmetrizing the covariant derivative of the form. If your manifold is orientable and closed then you have in general that $\int_M \langle B, \nabla^* \nabla A\rangle = \int_M \langle \nabla B, \nabla A \rangle$. Then alternating and using the inner product on forms (at this point using the metric we regard $B$ and $A$ as forms) you should get that $\int_M \langle \nabla B, \nabla A \rangle = \int_M \langle d B^\sharp, d A^\sharp \rangle$.