I am following some lecture notes on Ricci flow and reached the section where we linearize the Ricci tensor and obtain the principal symbol for the resulting operator. We have $T \in \: \Gamma(Sym^2 T^{\*}M)$ smooth, fixed and positive definite and then compute the time derivative for the divergence of $G(T)$:


$\bigg(\frac{\partial}{\partial t}\delta G(T) \bigg)Z = -T \bigg((\delta G(h))^{\#},Z\bigg) + \bigg<h,\nabla T(.,.,Z) - \frac{1}{2}\nabla\_{Z}T \bigg>, $


where $\frac{\partial g}{\partial t}=h$, $Z$ an arbitrary vector field and $G(T)=T-\frac{1}{2}(tr \: T)h$. 


The notes then say that this implies 


$\frac{\partial}{\partial t}T^{-1}\delta G(T) = -\delta G(h)\: + \:\: ...$


where the dots indicate terms which don't have derivatives of $h$. I see that you can apply $T^{-1}$ as it doesn't depend on $t$ and then lose the $Z$ as it is arbitrary, but not sure exactly what happens to the right-hand side such that we end up with $-\delta G(h)$ or where the sharp goes.



Principal Symbol for the Ricci-DeTurck Flow
It looks like this is from the *Lecture notes on Ricci flow* from Peter Topping. He mentions that he uses the symbol $T$ as the symmetric, positive definite bilinear form and also for the map $\Gamma(TM^\ast) \to \Gamma(TM^\ast)$ induced by $T$ and the metric $g$ in the following way
$$
T(\alpha)(Z)=T(\alpha^\#,Z)
$$
where $\alpha^\#$ is the dual vector field to the $1$-form $\alpha$ with respect to $g$. Thus looking at 
\begin{eqnarray\*}
\bigg(\frac{\partial}{\partial t}\delta G(T) \bigg)Z &=& -T \bigg((\delta G(h))^{\#},Z\bigg) + \bigg<h,\nabla T(.,.,Z) - \frac{1}{2}\nabla\_{Z}T \bigg> \\
&=& -T(\delta G(h))(Z) + \ldots
\end{eqnarray\*}
 we just apply to the equation the inverse of $T$ to obtain your desired expression (note that $\delta G(T)$ is a $1$-form)


