Let $\phi : M \to M$ be a diffeomorphism. Is there a metric $g$ on $M$ and a diffeomorphism $\psi$ isotopic to $\phi$ so that $\psi$ is an isometry with respect to $g$? I'm guessing the answer is no, since there are manifolds such that all metrics admit no nontrivial isometries and maybe some of these manifolds have a nontrivial mapping class group.


I'm actually particularly interested in the case of the 3-torus which has mapping class group $GL\_3(\mathbb{Z})$. Which of these mapping classes can be realized as isometries for some metric?



Realizing mapping classes as isometries?
As you suspect, in general, no.


For example, if $M$ is compact, and $\psi:M\to M$ fixes a metric $g$ on $M$, then the closure of $\{\psi^k\ |\ k\in\mathbb{Z}\ \}$ is a compact abelian subgroup of $\mathrm{Isom}(M,g)$, and, hence, its identity component is a torus, so $\psi$ must be isotopic in $\mathrm{Isom}(M,g)$ to a $g$-isometry of finite order. In particular, the action of $\psi$ on the deRham cohomology of $M$ must have finite order, as must any diffeomorphism $\phi:M\to M$ homotopic to $\psi$.


Conversely if the closure of $\{\psi^k | k\in\mathbb{Z}\ \}$ in the compact open topology is a compact group, then, yes $\psi$ fixes a metric $g$.


