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

\[ \langle \nabla^4 u, \nabla^4 v\rangle =-\langle \nabla^3 u, \div (\nabla^4 v)\rangle. \]

{#eq:eq1}

In the next step we investigate \(\div (\nabla^4 v)\). In local coordinates this term reads \({v_{ijkp}}^p\), where we use the convention discussed in (0x67dcf74c) .

Applying Ricci’s identity twice yields

\begin{align} v_{ijkp}{}^p&=v_{ijpk}{}^p+ R_{kpi}{}^m {v_{mj}}^p + R_{kpj}{}^m {v_{im}}^p\\ &=v_{ijp}{}^p{}_k + R_{kpi}{}^m {v_{mj}}^p + R_{kpj}{}^m {v_{im}}^p +\\ &\quad+ R_k{}^p{}_i{}^m v_{mjp} + R_k{}^p{}_j{}^m v_{imp} + R_k{}^p{}_p{}^m v_{ijm}. \end{align}

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

As for \(k=3\) we sum up identical terms, identify the last one with the Ricci curvature and use an identity for \(v_{ijp}{}^p\) calculated in the previous case. This gives us

\begin{align} v_{ijkp}{}^p&= v_p{}^p{}_{ijk} + R_i{}^m v_{mjk} + R_j{}^m v_{imk} + R_k{}^m v_{ijm} + \\ & \quad + 2R_{kpj}{}^m v_{im}{}^p + 2R_{kpi}{}^m v_{mj}{}^p + 2 R_{jpi}{}^m v_m{}^p{}_k. \end{align}

Terms with mixed derivatives Link to heading

We proceed with the last terms of the above equality. Using the identity

\[ R_{ijkl} =\frac{1}{R^2}(g_{il}g_{jk} - g_{ik}g_{jl}) \]

{#eq:R_sphere} for the curvature endomorphism on the sphere yields

1. \(2R_{kpj}{}^m v_{im}{}^p= \frac{2}{R^2} (v_{ikj} - g_{jk} v_{ip}{}^p)\).
2. \(2R_{kpi}{}^m v_{mj}{}^p = \frac{2}{R^2} (v_{kji} - g_{ik} v_{pj}{}^p)\),
3. \(2R_{jpi}{}^m v_m{}^p{}_k = \frac{2}{R^2} (v_{jik} - g_{ij} v_p{}^p{}_k)\),

We proceed by applying \(\langle \nabla^3 u, \cdot\rangle\) to each term separately and we obtain in local coordinates

1. \(\int_{\mathbb{S}^d_R} 2R_{kpj}{}^m v_{im}{}^p u^{ijk} = \int_{\mathbb{S}^d_R} \frac{2}{R^2} (v_{ikj} u^{ijk} - v_{ip}{}^p u^{iq}{}_q)=:\mathcal{R}_1\),
2. \(\int_{\mathbb{S}^d_R} 2R_{kpi}{}^m v_{mj}{}^p u^{ijk} = \int_{\mathbb{S}^d_R} \frac{2}{R^2} (v_{kji} u^{ijk} - v_{pj}{}^p u^{qj}{}_q)=:\mathcal{R}_2\),
3. \(\int_{\mathbb{S}^d_R} 2R_{jpi}{}^m v_m{}^p{}_k u^{ijk} = \int_{\mathbb{S}^d_R} \frac{2}{R^2} (v_{jik}u^{ijk} - v_p{}^p{}_k u^q{}_q{}^k) =: \mathcal{R}_3\).

At this point it should be mentioned that according to our convention, we actually apply \(\langle \cdot, \nabla^3 u\rangle\). Since the inner product on tensor fields is symmetrical, the expressions are identical.

We follow the idea of identifying both, minuend and subtrahend, with each other in each expression. To do this, we shift indices and use partial integration.

Term \(\mathcal{R}_1\) Link to heading

Applying Ricci’s identity and the identity [@eq:R_sphere] for the curvature tensor on the sphere gives us

\[ v_{ikj}=v_{ijk}+R_{kji}{}^m v_m = v_{ijk} + \frac{1}{R^2}(g_{ij}v_k-g_{ik}v_j).\]

From this follows that

\begin{align} v_{ikj}u^{ijk}&=v_{ijk}u^{ijk}+\frac{1}{R^2}(v_k u_p{}^{pk} - v_j u_p{}^{jp})\\ &=v_{ijk}u^{ijk}+\frac{1}{R^2}(v_k u_p{}^{pk} - v_j u_p{}^{pj} - v_jR^{jp}{}_p{}^m u_m)\\ &= v_{ijk}u^{ijk} - \frac{(d-1)}{R^4} v_iu^i. \end{align}

On the other hand we have

\begin{align*} \int_{\mathbb{S}^d_R} v_{ip}{}^p u^{iq}{}_q &= - \int_{\mathbb{S}^d_R} v_{ip}{}^p{}_q u^{iq}\\ &= - \Bigl[\int_{\mathbb{S}^d_R} v_{ipq}{}^p u^{iq} + R^p{}_{qi}{}^m v_{mp} u^{iq} + R^p{}_{qp}{}^m v_{im} u^{iq}\Bigr]\\ &= - \Bigl[\int_{\mathbb{S}^d_R} v_{iqp}{}^p u^{iq} - \frac{d-1}{R^2} v_{iq} u^{iq} + 2R^p{}_{qi}{}^m v_{mp} u^{iq} \Bigr]\\ &= \int_{\mathbb{S}^d_R} v_{iqp} u^{iqp} + \frac{d-1}{R^2} \int_{\mathbb{S}^d_R} v_{iq} u^{iq} + \int_{\mathbb{S}^d_R} (-2) R^p{}_{qi}{}^m v_{mp} u^{iq} \\ \end{align*}

We have

\[ -2R^p{}_{qi}{}^m v_{mp} u^{iq} = -\frac{2}{R^2}(g^{pm}g_{qi}-g^p{}_ig_q{}^m)v_{mp}u^{iq}=\frac{2}{R^2}(v_{qi} u^{iq} - v_p{}^p u^q{}_q ). \]

We discussed this expression in the case \(k=3\) and thus

\[\int_{\mathbb{S}^d_R} (-2) R^p{}_{qi}{}^m v_{mp} u^{iq} = - \frac{2(d-1)}{R^4}\langle \nabla u, \nabla v\rangle. \]

In summary,

\begin{align} \mathcal{R}_1 &= \int_{\mathbb{S}^d_R} \frac{2}{R^2} \Bigl( - \frac{d-1}{R^4} v_iu^i - \frac{d-1}{R^2} v_{ij}u^{ij} + \frac{2(d-1)}{R^4} v_iu^i\Bigr) \\ &= - \frac{2(d-1)}{R^4} \langle \nabla^2 u, \nabla^2 v\rangle + \frac{2(d-1)}{R^6} \langle \nabla u, \nabla v\rangle. \end{align}

Term \(\mathcal{R}_2\). Link to heading

Using again Ricci’s identities we transform \(\mathcal{R}_2\) into \(\mathcal{R}_1\) with some additional terms. That is

\begin{align} \mathcal{R}_2&=\int_{\mathbb{S}^d_R} \frac{2}{R^2} (v_{kji} u^{ijk} - v_{pj}{}^p u^{qj}{}_q)\\ &= \frac{2}{R^2}\int_{\mathbb{S}^d_R} v_{kij} u^{ijk} + R_{jik}{}^m v_m u^{ijk} - v_{ip}{}^pu^{iq}{}_q\\ &= \mathcal{R}_1 + \frac{2}{R^2}\int_{\mathbb{S}^d_R} \frac{1}{R^2} (v_j u^{pj}{}_p - v_i u^{ip}{}_p) \\ &= \mathcal{R}_1, \end{align}

since the last integral vanishes.

Term \(\mathcal{R}_3\) Link to heading

We transform \(\mathcal{R}_3\) into \(\mathcal{R}_1\) with some additional terms \begin{align} \mathcal{R}3&=\int{\mathbb{S}^d_R} \frac{2}{R^2} (v_{jik}u^{ijk} - v_p{}^p{}k u^q{}q{}^k)\ &= \frac{2}{R^2} \int{\mathbb{S}^d_R} v{jki}u^{ijk} + R_{ikj}{}^m v_m u^{ijk}

• (v_{pk}{}^p u^{qk}{}q + v{pk}{}^p R_q{}^{kq}{}m u^m + R^p{}{kp}{}^m v_m u^{qk}{}q + R^p{}{kp}{}^m R_q{}^{kq}{}_l v_m u^l ) \ &= \mathcal{R}1 + \frac{2}{R^2} \int{\mathbb{S}^d_R} \frac{1}{R^2}(v_i u^{ip}{}_p - v_k u^p{}p{}^k) + \frac{d-1}{R^2} v{pk}{}^p u^k + \frac{d-1}{R^2} v_k u^{qk}{}_q - \frac{(d-1)^2}{R^4} v_k u^k \ &= \mathcal{R}1 + \frac{2}{R^2} \int{\mathbb{S}^d_R} \frac{1}{R^2}(v_i u^{p}{}p{}^i + v_i R^i{}{ppm} u^m - v_k u^p{}_p{}^k)
  • \frac{2(d-1)}{R^4} \int_{\mathbb{S}^d_R} v_{kp} u^{kp} + v_{kq} u^{qk}- \frac{2(d-1)^2}{R^6} \int_{\mathbb{S}^d_R} v_k u^k \ &= \mathcal{R}_1 + \frac{2(d-1)}{R^6} \langle \nabla u, \nabla v\rangle - \frac{4(d-1)}{R^4} \langle \nabla^2 u, \nabla^2 v\rangle - \frac{2(d-1)^2}{R^6} \langle \nabla u, \nabla v\rangle \ &= \mathcal{R}_1 - \frac{4(d-1)}{R^4} \langle \nabla^2 u, \nabla^2 v\rangle - \frac{(2(d-1)-2)(d-1)}{R^6} \langle \nabla u, \nabla v\rangle \ &= \mathcal{R}_1 - \frac{4(d-1)}{R^4} \langle \nabla^2 u, \nabla^2 v\rangle - \frac{2(d-2)(d-1)}{R^6} \langle \nabla u, \nabla v\rangle. \end{align}

Summary Link to heading

Combining everything and using the results for previous cases, we get for the right hand side in [@eq:eq1]

\begin{align} \langle \nabla^4 u, \nabla^4 v\rangle &= \langle \nabla^3 u , \nabla^3 (-\Delta v)\rangle - \frac{3(d-1)}{R^2} \langle \nabla^3 u, \nabla^3 v\rangle - (\mathcal{R}_1 + \mathcal{R}_2 + \mathcal{R}_3)\\ &= \langle \nabla^3 u , \nabla^3 (-\Delta v)\rangle - \frac{3(d-1)}{R^2} \langle \nabla^3 u, \nabla^3 v\rangle - \Bigl(3\mathcal{R}_1 - \frac{4(d-1)}{R^4} \langle \nabla^2 u, \nabla^2 v\rangle - \frac{2(d-2)(d-1)}{R^6} \langle \nabla u, \nabla v\rangle\Bigr) \\ &= \langle \nabla^3 u , \nabla^3 (-\Delta v)\rangle - \frac{3(d-1)}{R^2} \langle \nabla^3 u, \nabla^3 v\rangle +\frac{10(d-1)}{R^4} \langle \nabla^2 u, \nabla^2 v\rangle + \frac{2(d-1)(d-5)}{R^6} \langle \nabla u, \nabla v\rangle\\ &= \langle u, P_4(-\Delta) v\rangle, \end{align}

with

\begin{align} P_4(t)&=P_3(t)\Bigl(t-\frac{3(d-1)}{R^2}\Bigr) + \frac{10(d-1)}{R^4} P_2(t) + \frac{2(d-1)(d-5)}{R^6} P_1(t), \end{align}

where \(P_1\), \(P_2\) and \(P_3\) are defined in the previous cases.