Linearisation of Einstein operator

Question

Let $(M,g)$ be a $(m+1)$-dimensional Riemannian manifold with Levi-Civita connection $\nabla$.

The Ricci curvature can be viewed as a differential operator $\text{Rc}:\Gamma(S^2_+M)\rightarrow\Gamma(S^2M)$ from the space of metrics on $M$ to the space of symmetric tensors on $M$ given by $g\mapsto\text{Rc}(g)$.

The formal adjoint of the divergence is the map ${\delta}_g^*:\Gamma(TM)\rightarrow\Gamma(S^2M)$ given locally by \begin{align*} ({\delta}_g^*X)_{ij}&=\frac{1}{2}\big[g(\nabla_iX,\partial_j)+g(\partial_i,\nabla_jX)\big]\\ &=\frac{1}{2}(L_Xg)_{ij} \end{align*} where we identify vector fields and covector fields via the musical isomorphisms. The Lie derivative with respect to a vector field $X\in\Gamma(TM)$ of a symmetric tensor is the map $L_X:\Gamma(S^2M)\rightarrow\Gamma(S^2M)$ given locally by \begin{align*} (L_Xr)_{ij}&=\nabla_X(r_{ij})-r\big([X,\partial_i],\partial_j\big)-r\big(\partial_i,[X,\partial_j]\big)\\ &=(\nabla_Xr)_{ij}+(\nabla_iX)^kr_{kj}+(\nabla_jX)^kr_{ki} \end{align*} where the Lie bracket $[X,Y]=\nabla_XY-\nabla_YX$. Now set \begin{align*} r(g)=\text{Rc}_g+mg\in\Gamma(S^2M). \end{align*} We get a map $r:\Gamma(S^2_+M)\rightarrow\Gamma(S^2M)$ given by $g\mapsto r(g)=\text{Rc}(g)+mg$, sometimes called the Einstein operator since if $r(g)=0$ then $\text{Rc}(g)=-mg$

Question: Does the linearisation $2Dr_g({\delta}^*_gX)$ of $r$ at $g$ in the direction of $2{\delta}^*_gX$ equal \begin{align*} 2Dr_g({\delta}^*_gX)=Dr_g(L_Xg)=L_Xr(g)? \end{align*}

Answer 1

Let $\Phi_s$ denote the one parameter family of diffeomorphisms generated by $X$ and let $g(s) = \Phi_s^* g$. Note that $r(\Phi_s^* g) = \Phi_s^* r(g)$ by definition, so you have that the right hand side of your equation is exactly $$ L_X r(g) = \lim_{s\to 0} \frac{\Phi_s^* r(g) - r(g)}{s} .$$

On the other hand, the left hand side of your expression is $$ \lim_{s \to 0} \frac{r(g(s)) - r(g)}{s} = D r_g (\dot{g}) $$ by definition of "linearisation". And we have that $\dot{g} = \frac{d}{ds} g(s) | _{s = 0} = L_X g$ by construction. So yes, the two sides are equal.