Let $M, N$ be smooth manifolds with $M$ orientable and compact. Let $\sigma$ be some volume form on $M$ and consider the set $\mathcal{M}$ of smooth maps from $M$ to $N$ in a fixed homotopy class. Now consider the group $G$ of volume-preserving diffeomorphisms on $M$. This group naturally acts on the whole space of smooth maps $C^{\infty}(M, N)$ by composition on the right.


My question is: can this action be restricted to $\mathcal{M}$? Clearly, it cannot if composition on the right does not preserve homotopy classes. If every volume-preserving diffeomorphism were homotopic to the identity, then the action would in fact preserve homotopy classes. However, I haven't been able to come up with a proof or a counterexample of this.



Does composition on the right by a volume-preserving diffeomorphism preserve homotopy class?
Let $X$ be a smooth, compact, orientable manifold and let $\omega$ be a choice of volume form. On $X\times X$, we have the natural volume form $\sigma = \pi\_1^\*\omega \wedge \pi\_2^\*\omega$ where $\pi\_i : X\times X \to X$ is projection onto the $i^{\text{th}}$ factor. The diffeomorphism $s : X\times X \to X\times X$ given by $s(x\_1, x\_2) = (x\_2, x\_1)$ satisfies $\pi\_1\circ s = \pi\_2$ and $\pi\_2\circ s = \pi\_1$, so


\begin{align\*}
s^\*\sigma &= s^\*(\pi\_1^\*\omega \wedge \pi\_2^\*\omega)\\ 
&= (s^\*\pi\_1^\*\omega)\wedge(s^\*\pi\_2^\*\omega)\\ 
&= (\pi\_1\circ s)^\*\omega\wedge(\pi\_2\circ s)^\*\omega\\
&= \pi\_2^\*\omega\wedge\pi\_1^\*\omega\\ 
&= (-1)^{\dim X}\pi\_1^\*\omega\wedge\pi\_2^\*\omega\\ 
&= (-1)^{\dim X}\sigma.
\end{align\*}


Therefore $s$ is a volume-preserving diffeomorphism if and only if $X$ is even-dimensional.


Consider the homotopy class of maps $X\times X \to X$ containing $\pi\_1$. If the action of volume-preserving diffeomorphisms preserves this class, then $\pi\_1\circ s = \pi\_2$ must be homotopic to $\pi\_1$. Fix $x\_0 \in X$ and define $i : X \to X\times X$ by $i(x) = (x, x\_0)$. If $\pi\_1$ and $\pi\_2$ were homotopic, then $\pi\_1\circ i = \operatorname{id}\_X$ would be homotopic to $\pi\_2\circ i = c\_{x\_0}$, the constant map with value $x\_0$, and hence $X$ would be contractible. So for any even-dimensional, non-contractible choice of $X$, we obtain an example which demonstrates that the question has a negative answer.


