A question on the DeTurck trick I am probably being obtuse here, but there is something in the DeTurck trick that I do not understand precisely. I was reading from Andrews Hopper, and they (on page 91) say that the equation $\frac{\partial g}{\partial t} = Q$ has a solution from the existence of parabolic equations. They have to prove that $Q$ is elliptic first. They are calculating that the linearisation of $Q$ is $DQ (h) = \Delta h + A$, from which the principal symbol is $\hat{\sigma}\[DQ\](\xi)h = |\xi|^2 h$. My question is: is it obvious that the principal symbol of $Q$ is as given above? My suspicion comes from the fact that the $\Delta$ is front of $h$ is still time dependent. 
Please forgive me if this is too obvious.
 A: The operator $g\mapsto-2\operatorname{Ric}$ is not strictly elliptic since by
the diffeomorphism invariance of the Ricci tensor, if $\frac{\partial
}{\partial s}g=\mathcal{L}_{X}g$, then $\frac{\partial}{\partial s}\left(
-2\operatorname{Ric}\right)  =-2\mathcal{L}_{X}\operatorname{Ric}$. I think
DeTurck's trick is the parabolic version of considering the Ricci tensor in
harmonic coordinates (see Hamilton's Formation of Singularities paper). Here
are some related calculations, although not complete nor tied together well.
The linearization of the map $g\mapsto-2\operatorname{Ric}$ is given by: if
$\frac{\partial}{\partial s}g=v$, then
$$
D_{v}\left(  -2\operatorname{Ric}\right)  =\frac{\partial}{\partial s}\left(
-2\operatorname{Ric}\right)  =\Delta v-2\operatorname{Sym}\left(
\operatorname{tr}{}_{1,4}(\nabla^{2}v)\right)  +\nabla^{2}V,
$$
where $V=\operatorname{tr}v$. By commuting covariant derivatives, we may
rewrite this as $\frac{\partial}{\partial s}\left(  -2\operatorname{Ric}
\right)  =\Delta_{L}v-\mathcal{L}_{W}g$, where $W=\operatorname{div}
(v-\frac{V}{2}g)$ and $\Delta_{L}$ is the Lichnerowicz Laplacian. (Digression:
e.g., taking $v=\operatorname{Ric}$, we obtain under Ricci flow that
$\frac{\partial}{\partial t}\operatorname{Ric}=D_{-2v}\left(
\operatorname{Ric}\right)  =D_{v}\left(  -2\operatorname{Ric}\right)
=\Delta_{L}\operatorname{Ric}$ since $W=0$ by the contracted second Bianchi identity.)
The variation of the Levi-Civita connection $\nabla:\Gamma(TM)\times
\Gamma(TM)\rightarrow\Gamma(TM)$ is the tensor (use $g$ to identify $TM$ and
$T^{\ast}M$)
$$
(\frac{\partial}{\partial s}\nabla)(X,Y,Z)=\frac{1}{2}(\nabla_{X}
v(Y,Z)+\nabla_{Y}v(X,Z)-\nabla_{Z}v(X,Y)),
$$
so that $\operatorname{tr}{}_{1,2}(\frac{\partial}{\partial s}\nabla
)=\operatorname{div}v-\frac{1}{2}\nabla V=W$. Since $\Delta=\operatorname{tr}
_{g}\nabla^{2}$, we have $\frac{\partial}{\partial s}\Delta=-v\cdot\nabla
^{2}-\mathcal{L}_{W}$, where the highest order term in $v$ on the right side
is $-\mathcal{L}_{W}$. The same is essentially true for the map Laplacian if
we linearize with respect to the domain metric.
Fix a connection $\tilde{\nabla}$ and define $U=\operatorname{tr}{}
_{1,2}(\nabla-\tilde{\nabla})$. The Ricci-DeTurck flow is $\frac{\partial
}{\partial t}g=-2\operatorname{Ric}+\mathcal{L}_{U}g\doteqdot Q(g)$. We have
$\frac{\partial U}{\partial s}=W$ up to lower (zeroth) order terms in $v$. So
$\frac{\partial}{\partial s}Q(g)=\Delta v+P\left(  v\right)  $, where $P$ is
lower order in $v$. Replacing $\nabla_{i}$ by $\xi_{i}$ in $\Delta
=g^{ij}\nabla_{i}\nabla_{j}$, we obtain the symbol $\left\vert \xi\right\vert
^{2}$ for $\xi\in T^{\ast}M$ as Robert Haslhofer wrote.
