# Linearisation of Einstein operator

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*}

• $\delta_g^*$ is not the formal adjoint of the divergence of a vector field. The divergence of vector field is a scalar and is $L^2$ paired against scalars, and the formal adjoint of that operator acts on scalar fields as the exterior derivative. It looks like that you are writing is the formal adjoint of the "divergence of a symmetric two tensor" operator, now acting on vector fields. Note that by definition it is equal to $$\delta_g^* X = \frac12 L_X g$$ Feb 17, 2015 at 8:25

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.