Does composition on the right by a volume-preserving diffeomorphism preserve homotopy class?

Question

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.

Answer 1

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.