# $\Delta_L(\text{im}\,\delta^*_g)\subset\text{im}\,\delta^*_g$ and $\Delta_L\big(\text{ker}\,\text{Bian}(g)\big)\subset\text{ker}\,\text{Bian}(g)$?

Let $(M,g)$ be an Einstein manifold with Levi-Civita connection $\nabla$ and whose Ricci tensor $\text{Rc}(g)=g$, in components $R_{ij}=g_{ij}$. The Lichnerowicz Laplacian of $g$ is the map \begin{align*} \Delta_L:\Gamma(S^2M)&\rightarrow\Gamma(S^2M)\\ h&\mapsto(\Delta_Lh)_{ij}=g^{st}\nabla_s\nabla_th_{ij}-g^{st}R_{is}h_{jt}-g^{st}T_{is}h_{it}-2g^{pq}g^{qt}R_{isjt}h_{pq}. \end{align*} The Bianchi operator is the map \begin{align*} \text{Bian}(g):\Gamma(S^2M)&\rightarrow\Gamma(T^*)\\ h&\mapsto\text{Bian}(g,h)=\delta_gg+\frac{1}{2}\text{d}(\text{tr}_gh) \end{align*} where the divergence is the map \begin{align*} \delta_g:\Gamma(S^2M)&\rightarrow\Gamma(T^*M)\\ h&\mapsto(\delta_gh)_k=-g^{st}\nabla_sh_{tk}. \end{align*} The formal $L^2$-adjoint of the divergence is the map \begin{align*} \delta^*_g:\Gamma(T^*M)&\rightarrow\Gamma(S^2M)\\ \omega&\mapsto(\delta^*_g\omega)_{ij}=\frac{1}{2}(\nabla_i\omega_j+\nabla_j\omega_i). \end{align*}

Question: Does it hold that \begin{align*} \Delta_L(\text{im}\,\delta^*_g)\subset\text{im}\,\delta^*_g\qquad\text{and}\qquad\Delta_L\big(\text{ker}\,\text{Bian}(g)\big)\subset\text{ker}\,\text{Bian}(g)? \end{align*} If these set inclusions do hold, how does one prove them?

## 2 Answers

Since you've done the rest of the work in your answer, I'll show that $DRc_g =$ id on im $\delta_g^*$. It ends up being just a consequence of the invariance of $Rc$ under the action of the diffeomorphism group on the metrics.

For $X$ a vector field and $\varphi_t$ the 1-parameter group of diffeomorphisms generated by $X$, this invariance gives $$Rc(\varphi_t^*g) = \varphi_t^* Rc(g).$$ Differentiating this equation, we find, since $Rc(g) = g$, $$DRc_g(\mathcal{L}_{X}g) = \mathcal{L}_X(Rc(g)) = \mathcal{L}_Xg.$$ But the Lie derivative of the metric is given by $$(\mathcal{L}_Xg)_{ij} = \frac12(g_{kj}\partial_iX^k + g_{ki}\partial_jX^k).$$ So for $\alpha$ a 1-form and $\alpha^{\#}$ it's associated dual vector field, we get $$(\mathcal{L}_{\alpha^{\#}}g)_{ij} = \frac12(\nabla_i\alpha_j + \nabla_j\alpha_i) = (\delta_g^*\alpha)_{ij}.$$ It follows that $$DRc_g(\delta_g^*\alpha) = DRc_g(\mathcal{L}_{\alpha^{\#}}g) = \mathcal{L}_{\alpha^{\#}}g = \delta_g^*\alpha.$$

First inclusion: The first inclusion $\Delta_L(\text{im}\,\delta^*_g)\subset\text{im}\,\delta^*_g$ doesn't rely on $g$ being Einstein. It relies on the fact that the linearisation $D\text{Rc}(g):\Gamma(S^2M)\rightarrow\Gamma(S^2M)$ is the identity on the image $\text{im}\,\delta^*_g$, that is, for $\delta^*_g\omega\in\text{im}\,\delta_g^*$ we have \begin{align*} D\text{Rc}(g)(\delta^*_g\omega)=\frac{1}{2}\Delta_L(\delta^*_g\omega)+\delta^*_g\big(\text{Bian}(g,\delta^*_g\omega)\big)=\delta_g^*\omega \end{align*} so we get \begin{align*} \frac{1}{2}\Delta_L(\delta^*_g\omega)&=\delta^*_g\big(\omega-\text{Bian}(g,\delta^*_g\omega)\big)\\ &=\delta_g^*\alpha \end{align*} where $\alpha=\omega-\text{Bian}(g,\delta^*_g\omega)\in\Gamma(T^*M)$. So my new question is:

Question: How does one show that $D\text{Rc}(g)=\text{id}$ on the image $\text{im}\,\delta_g^*$?

Second inclusion: The second inclusion $\Delta_L\big(\text{ker}\,\text{Bian}(g)\big)\subset\text{ker}\,\text{Bian}(g)$ makes use of the Einstein condition. The linearisation of the map $g\mapsto B(g):=\text{Bian}\big(g,\text{Rc}(g)\big)=0$ in the direction of $h\in\text{Ker}\,\text{Bian}(g)$ yields \begin{align*} 0=DB(g)h&=\partial_1\text{Bian}\big(g,\text{Rc}(g)\big)h+\text{Bian}\big(g,D\text{Rc}(g)h\big)\\ &=\partial_1\text{Bian}\big(g,\text{Rc}(g)\big)h+\frac{1}{2}\text{Bian}(g,\Delta_Lh). \end{align*} The first term is the linearisation of the map $g\mapsto\text{Bian}(g,R)$ in the direction of $h$ where we set $R=\text{Rc}(g)$. For this we calculate that \begin{align*} \big(D\text{Bian}(g,R)h\big)_m&=g^{pq}R_{pm}\text{Bian}(g,h)_q-\frac{1}{2}g^{pq}g^{st}(\nabla_pR_{sm}+\nabla_sR_{pm}-\nabla_mR_{ps})h_{qt}\\ &=0. \end{align*} The first term is $0$ since $h\in\text{Ker}\,\text{Bian}(g)$. The second term is $0$ since $g$ is Einstein so $R_{ij}=g_{ij}$.

So out of this, my only remaining question is as above: Why is $D\text{Rc}(g)=\text{id}$ on $\text{im}\,\delta^*_g$?