Prove: Gravitation operator is invertible. Let $(M,g)$ be a Riemannian manifold and $\Gamma(S^2M)$ the space of symmetric 2-covariant tensors.  Define the gravitation operator as the map
\begin{align*}
G:\Gamma(S^2M)&\rightarrow\Gamma(S^2M)\\
h&\mapsto Gh:=h-\frac{1}{2}(\text{tr}_gh)g,
\end{align*}
where the metric trace $\text{tr}_gh=g^{pq}h_{pq}$.  
Question:  Is the gravitation operator invertible, and if so how does one show this?
 A: It is not invertiable at least when the dimension $n$ of $M$ is two. Take $h = g$. Then 
$$(Gh)_{ij} = (Gg)_{ij} = g_{ij} - \frac{1}{2} g^{kl}g_{kl} g_{ij} = g_{ij} - \frac{n}{2} g_{ij} = 0 . $$
A: Setting $k=Gh$, see if you can write $\mathrm{tr}_gk$ in terms of $\mathrm{tr}_gh$ (and thus vice versa).  You should find a relatively clean expression, and from there it's just simple algebra: suppose $\mathrm{tr}_gh = f(\mathrm{tr}_gk)$ for some $f$.  Then $k_{ij}=h_{ij}-\frac12f(\mathrm{tr}_gk)g_{ij}$, so $h_{ij}=k_{ij}+\frac12f(\mathrm{tr}_gk)g_{ij}$.
A: For dimension $n>2$ the inverse of 
$
y=Gh=h-\frac{1}{2}(\text{tr}_gh)g
$ 
is
$$
x=G^{-1}y=y-\frac{1}{n-2}(\text{tr}_gy)g
$$
since
\begin{align}
G^{-1}y&=G^{-1}\Big(h-\frac{1}{2}(\text{tr}_gh)g\Big)\\
&=h-\frac{1}{2}(\text{tr}_gh)g-\frac{1}{n-2}\Big(\text{tr}_gh-\frac{n}{2}\text{tr}_gh\Big)g\\
&=h
\end{align}
and
\begin{align*}
Gx&=G\Big(y-\frac{1}{n-2}(\text{tr}_gy)g\Big)\\
&=y-\frac{1}{n-2}(\text{tr}_gy)g-\frac{1}{2}\Big(\text{tr}_gy-\frac{n}{n-2}\text{tr}_gy\Big)g\\
&=y.
\end{align*}
