For convenience, we omit the labeling of the scalar products and norms that appear, as we assume that it is understandable from the context.

Applying (0x67dbf0ac) implies

\[ \langle \nabla^2 u, \nabla^2 v\rangle =-\langle \nabla u, \div (\nabla^2 v)\rangle. \]

{#eq:eq1}

In the next step we investigate \(\div (\nabla^2 v)\). In local coordinates this term is expressed by \({v_{ip}}^p\), where we use the convention discussed in (0x67dcf74c) .

Then

\[ v_{ip}{}^p=v_{pi}{}^p={v_p}{}^p{}_i + R_i{}^p{}_p{}^m v_m, \]

where we used the symmetry of the covariant Hessian in the former equality and the Ricci identity in the latter. The tensor \(R_{ijk}{}^l\) is the Riemann curvature endomorphism and we obtain \(R_i{}^j{}_k{}^l\) by raising the second index .

Using the symmetry properties of the curvature endomorphism we simplify the second expression to \(R_i{}^m\), where \(R_{ij}\) is the local expression of the Ricci curvature . On the sphere we have \(\Rc=\frac{d-1}{R^2}g\) on the sphere (see (0x67dd126e) ) and thus [@eq:eq1] simplifies to

\[ \langle \nabla^2 u, \nabla^2 v\rangle = \langle \nabla u, \nabla (-\Delta v)\rangle - \frac{d-1}{R^2}\langle \nabla u, \nabla v\rangle =\biggl\langle u, \Bigl[(-\Delta )^2 - \frac{d-1}{R^2} (-\Delta )\Bigr]v\biggr\rangle. \]

In the last equality we used the result of the case \(k=1\) .