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

\[ \langle \nabla^3 u, \nabla^3 v\rangle =-\langle \nabla^2 u, \div (\nabla^3 v)\rangle. \]

{#eq:eq1}

In the next step we investigate \(\div (\nabla^3 v)\). In local coordinates this term reads \({v_{ijp}}^p\), where we use the convention discussed in (0x67dcf74c) .

Applying Ricci’s identity twice yields

\begin{align} v_{ijp}{}^p&=v_{ipj}{}^p+ R_{jpi}{}^m {v_m}^p \\ &=v_{ip}{}^p{}_j + R_j{}^p{}_i{}^m v_{mp} + R_j{}^p{}_p{}^m v_{im} + R_{jpi}{}^m {v_m}^p. \end{align}

We used that the Riemann curvature endomorphism of the sphere is covariantly constant, i.e. \(R_{ijk}{}^l{}_{;m}=0\).

The second and the fourth terms are equal. Identifying the third term with the Ricci curvature , as in \(k=2\) , and applying the identity for \(v_{ip}{}^p\), as also calculated in that case, gives

\[ v_{ijp}{}^p=v_p{}^p{}_{ij} + R_i{}^m v_{mj} + R_j{}^m v_{im} + 2 R_{jpi}{}^m v_m{}^p. \]

As mentioned in the previous case , we have \(R_i{}^m=\frac{d-1}{R^2}g_i{}^m\).

We proceed with the last term on the right hand side. The curvature endomorphism on the sphere is given by \(R_{ijkl} =\frac{1}{R^2}(g_{il}g_{jk} - g_{ik}g_{jl})\). This gives

\[ R_{jpi}{}^m v_m{}^p = \frac{1}{R^2} (g_j{}^m g_{pi} - g_{ji}g_p{}^m)v_m{}^p = \frac{1}{R^2}(v_{ji}-g_{ij}v_p{}^p). \]

{#eq:eq3}

We apply \(\langle \nabla^2 u, \cdot\rangle_g\) on [@eq:eq3] and we obtain in local coordinates

\[ 2R_{jpi}{}^m v_m{}^p u^{ij} = \frac{2}{R^2} (v_{ji}u^{ij} - v_p{}^pu_q{}^q).\]

Using integration by parts and Ricci’s identity we obtain for the second term

\begin{align} \int_{\mathbb{S}^d_R} v_p{}^p u_q{}^q &= - \int_{\mathbb{S}^d_R} v_p{}^{pq} u_q\\ &= - \int_{\mathbb{S}^d_R} v_p{}^{qp} u_q - \int_{\mathbb{S}^d_R} R^{pq}{}_p{}^m v_m u_q \\ &=\int_{\mathbb{S}^d_R} v_p{}^{q} u_q{}^p + \frac{d-1}{R^2} \int_{\mathbb{S}^d_R} v^q u_q . \end{align}

At this point it should be mentioned that according to our convention, we actually apply \(\langle \cdot, \nabla^2 u\rangle\). Since the inner product on tensor fields is symmetrical, the expressions are identical.

Thus we have

\[ \int_{\mathbb{S}^d_R} 2R_{jpi}{}^m v_m{}^pu^{ij} = \frac{2}{R^2}\int_{\mathbb{S}^d_R} (v_{ji} u^{ij} - v_p{}^p u_q{}^q) = -\frac{2(d-1)}{R^4}\langle \nabla u, \nabla v\rangle. \]

Combining everything and using the result for \(k=2\), we get for the right hand side in [@eq:eq1]

\begin{align} -\langle \nabla^2 u, \div(\nabla^3 v)\rangle&= \langle \nabla^2 u, \nabla^2 (-\Delta v)\rangle - \frac{2(d-1)}{R^2} \langle \nabla^2 u, \nabla^2 v\rangle + \frac{2(d-1)}{R^4}\langle \nabla u, \nabla v\rangle \\ &= \langle u, P(-\Delta )v\rangle \end{align}

with

\[ P(t)=\biggl(t^2-\frac{d-1}{R^2}t\biggr)\biggl(t-\frac{2(d-1)}{R^2}\biggr)+\frac{2(d-1)}{R^4}t. \]