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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?

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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 $h_{ij}=k_{ij}+\frac12f(\mathrm{tr}_gk)g_{ij}$.

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It is not invertiable at least when the dimension $n$ of $M$ is two. Take $h = g$. Then

$$(Gh)_{ij} = (Gg)_{ij} = g_{ij} - \frac{1}{2} g^{kl}g_{kl} g_{ij} = g_{ij} - \frac{n}{2} g_{ij} = 0 . $$

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    $\begingroup$ Conceptually, this fails in two dimensions exactly because $G$ is (in that dimension) the operator that gives exactly the tracefree part of a tensor in $\Gamma(S^2 M)$. In particular, its kernel is exactly the trace elements, that is, multiples of $g_{ij}$. $\endgroup$ – Travis Willse Dec 2 '14 at 2:32
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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*}

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