Let $\mathcal{G}$ be the set of Riemannian metrics on a compact, smooth manifold $M$. Let $g \in \mathcal{G}$ and $[g]$ be its conformal class.
I recently read that since $[g]$ is the range of $C^{\infty}(M,\mathbb{R})$ under the map \begin{equation} f \longrightarrow e^{2f}g, \end{equation} it is a connected subset of $\mathcal{G}$.
I'm not sure how to see that this fact is true. Does it have something to do with the fact that $\mathcal{G}$ is a Frechet space with the $C^{\infty}$-topology? Or is there some simple, clear explanation which I am failing to see?
Thanks in advance for any help!
In fact $[g]$ is path-connected with respect to the $C^k$-, $C^\infty$-, or $C^{k,\alpha}$-topologies.
$C^\infty(M,\mathbb{R})$ is path-connected: if $f,h\in C^\infty(M,\mathbb{R})$, then $\gamma(t)=th+(1-t)f$ is a continuous path connecting them. The map $$ C^\infty(M,\mathbb{R}) \ni f \mapsto e^{2f}g \in \mathcal{G} $$ is continuous. Hence the image of this map, namely $[g]$, is path-connected.
I usually just think of this argument as the observation that $t\mapsto e^{2th + 2(1-t)f}g$ is a continuous path from $e^{2f}g$ to $e^{2h}g$.