$\DeclareMathOperator\ddiv{div}\DeclareMathOperator\tr{tr}\newcommand{\conf}{\mathrm{conf}}$Consider this PDE on a symmetric tensor $h$ on $S^2$:


$$\Delta \text{tr}(h) - \ddiv(\ddiv(h)) + \tr(h) = f$$
where $f \in L^2(S^2)$ and $\Delta$, $\ddiv$ and $\tr$ are with respect to the round metric on $S^2$.


I wish to show that there exists at least one solution to this.


If we assume for simplicity that $h = \frac{1}{2} \tr(h) g\_{S^2}$, then the PDE becomes
$$\Delta \tr(h) + 2\tr(h) = 2f$$
which doesn't have a solution for every $f$ since $-2$ is an eigenvalue of $\Delta$ (I am assuming that $\Delta + \lambda$ is not surjective if $-\lambda$ is an eigenvalue; is this correct?). I am not sure how to approach this.


One approach is decomposing $h$ into its trace part and a conformal Lie derivative of a vector field $X$: $h = \frac{1}{2} \tr(h) g\_{S^2} + \mathcal{L}\_{\conf}X$. Then the PDE becomes:


$$\frac{1}{2}\Delta \tr(h) - \ddiv(\Delta\_{\conf}X) + \text{tr}(h) = f$$
where $\Delta\_{\conf}$ is the conformal laplacian on vector fields.


I am not able to continue.
Any help is appreciated.



Solving $\Delta \text{tr}(h) - \mathrm{div}(\mathrm{div}(h)) + \text{tr}(h) = f$ on $S^2$
Your second-order differential operator appears when one takes the variation of the scalar curvature of the sphere by a symmetric tensor $h\mapsto \frac{d}{dt}|\_{t=0}\mathrm{Sc}\_{g+th}$. In constant sectional curvature, such operators were studied for instance by Calabi (60'), and then in constant scalar curvature by Ebin (69').


Edit: I have taken another look at Ebin's paper and figured out a similar, yet simpler argument than my original one (you can still find it below). I think this argument provides an answer to your question in any geometry: let $(M,g)$ be an arbitrary closed Riemmanian manfiold. Define a linear second order operator $\gamma:S^{2}(T^{\*}M)\rightarrow C^{\infty}(M)$ by $\gamma h=-\Delta\_{g}tr\_{g}h+\mathrm{div}\mathrm{div}h-(h,Ric\_{g})\_{g}$. Consider its $L^{2}$-dual $\gamma^{\*}:C^{\infty}(M)\rightarrow S^{2}(T^{\*}M)$. It can be checked to assume the form $\gamma^{\*}f=\mathrm{Hess}\_{g}f-fRic\_{g}-\Delta\_{g}fg$. The symbol of $\gamma$ is $|\xi|^{2}tr+i\_{\xi}i\_{\xi}$ while the symbol of $\gamma^{\*}$ is $\xi\otimes\xi+|\xi|^{2}I$. Thus the symbol of $\gamma\gamma^{\*}$ is a constant multiplied by $|\xi|^{4}$, so $\gamma\gamma^{\*}$ is an elliptic fourth order operator (a "bilaplacian"). Therefore, if you want to solve $\gamma h=f$ then replace $h=\gamma^{\*}g$ and solve $\gamma\gamma^{\*} g=f$. This has a solution if and only if $f$ satisfies a compatibility condition: it must be $L^{2}$-orthogonal to the finite dimensional kernel of $\gamma\gamma^{\*}$, which in this case is equal to the kernel of $\gamma^{\*}$.


In your original case of the sphere, $Ric\_{g}=g$ and so $(h,Ric\_{g})\_{g}=tr\_{g}h$.


My original answer is below:


Here is what I suggest to solve your question in the special case of $S^{2}$.


If you have a Riemmanian manifold with constant sectional curvature, $(M,g)$, consider the second order differential operator $H\_{g}=\frac{1}{2}{(d^{\nabla^{g}}d^{\nabla^{g}}\_{V}+d^{\nabla^{g}}\_{V}d^{\nabla^{g}}})-\frac{1}{2}\kappa g\wedge$.
$\kappa$ here is the sectional curvature of the space, so in your case it can be taken to be $\kappa=1$. The wedge product here is known as the Kulkarni-Numizu product of symmetric tensor fields (there is a more generlized version of this wedge product, operating on "double forms", also adressed by Kulkarni, but this notion will do). $d^{\nabla^{g}}$ and $d^{\nabla^{g}}\_{V}$ here are the exterior covariant derviatvie of the first and second index of a symmetric tensor, respectively, where we think of symmetric tensors as examples of vector-valued differential forms $\Omega^{1}(M ; T^{\*}M)$. The image of $H\_{g}$ lies in $\Omega^{2}(M;\Lambda^{2}T^{\*}M)$, and in the case where $M$ is two-dimensional every element in this space is fully determined by its "scalar curvature", namely if $\sigma\in\Omega^{2}(M;\Lambda^{2}T^{\*}M)$ then $\sigma=\frac{1}{4}(tr\_{g}tr\_{g}\sigma) g\wedge g$.


In the case where $\kappa=1$, a direct calcultion shows that taking the trace twice from $H\_{g}$ yields the operator $tr\_{g}tr\_{g}H\_{g}=-\Delta\_{g}tr\_{g}+\mathrm{div}\mathrm{div}-tr\_{g}$, so solving your equation is equivelent of solving $H\_{g}h=\frac{1}{4} f g\wedge g$, where $g$ is the metric of the sphere. Replacing $h=H^{\*}\_{g}\psi$ for $\psi\in\Omega^2(M;\Lambda^{2}T^{\*}M)$ where $H\_{g}^{\*}$ is the formal $L^{2}$ dual of $H\_{g}$ yields the equation $H\_{g}H^{\*}\_{g}\psi=\frac{1}{4} f g\wedge g$. Note how since $H\_{g}$ operates on symemtric tensors, the image of $H^{\*}\_{g}$ is a symmetric tensor in $\Omega^{1}(M;T^{\*}M)$.


Another calculation then shows that the principle symbol of $H\_{g}H^{\*}\_{g}$ in the case where $M$ is two dimensional is $|\xi|^{4}$. Thus this is an elliptic fourth order differential opertor (a "bilaplcian"), and so the equation $H\_{g}H^{\*}\_{g}\psi=\frac{1}{4} f g\wedge g$ is solvable for $\psi$ if and only if $ \frac{1}{4} f g\wedge g$ is orthogonal to the kernel of $H\_{g}H^{\*}\_{g}$. By duality, this kernel is equal to $\mathrm{ker} H^{\*}\_{g}$. I am not sure if this kernel is trivial when $M$ is simply connected. If not, then this orthogonality yields a compatability condition which $f$ must satisfy in order for the equation to have a solution.


