Prescribing isometries

Question

Let $M$ be a smooth manifold and $f:M\to M$ a diffeomorphism. What conditions can one put on $f$ such that there is a metric $g$ for which $f$ is an isometry for $(M,g)$, i.e. $f^*g=g$?

Answer 1

If $f$ has finite order in the diffeomorphism group of $M$ (i.e. there is a $k$ such that $f^{\circ k} = \operatorname{id}_M$), then one can construct a Riemannian metric $g$ on $M$ with $f^*g = g$ as follows. Fix any Riemannian metric $g_0$ and consider the new metric

$$g = g_0 + f^*g_0 + (f^{\circ 2})^*g_0 + \dots + (f^{\circ (k-2)})^*g_0 + (f^{\circ (k-1)})^*g_0.$$

Then

\begin{align*} f^*g &= f^*g_0 + f^*f^*g_0 + f^*(f^{\circ 2})^*g_0 + \dots + f^*(f^{\circ(k-2)})^*g_0 + f^*(f^{\circ (k-1)})g_0\\ &= f^*g_0 + (f\circ f)^*g_0 + (f^{\circ 2}\circ f)^*g_0 + \dots + (f^{\circ(k-2)}\circ f)^*g_0 + (f^{\circ(k-1)}\circ f)^*g_0\\ &= f^*g_0 + (f^{\circ 2})^*g_0 + (f^{\circ 3})^*g_0 + \dots + (f^{\circ(k-1)})^*g_0 + (f^{\circ k})^*g_0\\ &=f^*g_0 + (f^{\circ 2})^*g_0 + (f^{\circ 3})^*g_0 + \dots + (f^{\circ(k-1)})^*g_0 + (\operatorname{id}_M)^*g_0\\ &= g_0 + f^*g_0 + (f^{\circ 2})^*g_0 + (f^{\circ 3})^*g_0 + \dots + (f^{\circ(k-1)})^*g_0\\ &= g. \end{align*}

Answer 2

Here's a proof of the claim I made in a comment on Michael's answer. Namely, that if $f\in G\subseteq \operatorname{Diff}(M)$ with $G$ a compact Lie group, then there is an $f$-invariant Riemannian metric. (In fact, there is a $G$-invariant metric.)

This proof I'll write below is essentially copied from Proposition 2.17 in the Introduction to Lie groups, isometric and adjoint actions and some generalizations by Alexandrino and Bettiol, so I've made it community wiki. Their proof is stated in the context of showing that compact Lie groups have bi-invariant metrics, but almost exactly the same proof establishes the much more general result.

Suppose $G$ is a compact Lie group. Let $\omega$ be a right invariant volume form on $G$, scaled so that $\int_G \omega = 1$.

Now, give $M$ an arbitrary Riemannian metric $\langle \cdot, \cdot \rangle_0$. We will define a $G$-invariant Riemannian metric $\langle \cdot, \cdot \rangle_1$ on $M$ as follows. For $v,w\in T_p M$, we set $$\langle v,w\rangle_1 = \int_G \langle g_\ast v, g_\ast w \rangle_0 \omega.$$ Here, the notation $g_\ast$ denotes the differential of the map $M\rightarrow M$ with $m\mapsto gm$.

Let's prove that this actually works.

Claim 1: $\langle \cdot, \cdot \rangle_1$ is a Riemannian metric.

Proof: Bilinearity follows because $\langle \cdot, \cdot \rangle_0$ is bilinear and both $g_\ast$ and integration are linear. Symmetry follows because $\langle \cdot, \cdot \rangle_0$ is symmetric. The fact that $\langle v,v\rangle_1 \geq 0$ follows because each $\langle g_\ast v, g_\ast v\rangle_0 \geq 0$. Obviously $\langle 0, 0\rangle_1 = 0$.

The last condition is that $\langle v,v\rangle_1 > 0$ if $v\neq 0$. Consider the function $\phi:G\rightarrow \mathbb{R}$ given by $\phi(g) = \langle g_\ast v, g_\ast v\rangle_0$. This is obviously continuous and we know that $\phi(e) > 0$. By continuity, there is a neighborhood $U$ of $e$ on which $\phi > \phi(e)/2$. Then $\langle v,v\rangle_1$ is at least $\phi(e)/2$ times the (positive) measure of $U$. $\square$

Claim 2: $\langle \cdot, \cdot \rangle_1$ is $G$-invariant.

Proof: Let $h\in G$. We need to show that $\langle h_\ast v, h_\ast w\rangle_1 = \langle v,w\rangle_1$. To that end, we will use the function $\psi:G\rightarrow \mathbb{R}$ defined by $\psi(g) = \langle g_\ast v, g_\ast w\rangle_0$.

Then \begin{align*} \langle h_\ast v, h_\ast w\rangle &= \int_G \langle g_\ast (h_\ast v), g_\ast (h_\ast w)\rangle_0 \omega\\ &= \int_G \langle (gh)_\ast v, (gh)_\ast w\rangle_0 \omega\\ &= \int_G \psi(gh) \omega.\end{align*}

Now we are going to use the change of variables formula for the diffeomorphism $R_{h}:G\rightarrow G$ given by right multiplication by $h$, $R_h(g) = gh$. That is, we'll view $\psi(gh)\omega$ as $R_h^\ast(\psi(g)\omega)$. Here, we are using the fact that $\omega$ right invariant.

So, we get \begin{align*} \int_G \psi(gh)\omega &= \int_G R_h^\ast(\psi(g)\omega)\\ &= \int_G \psi(g)\omega\\ &= \int_G \langle g_\ast v, g_\ast w\rangle_0 \omega\\ &= \langle v,w\rangle_1.\end{align*} $\square$

Answer 3

There are already some sufficient conditions here; here's a necessary one:

For each point $p$ such that $f^k(p)=p$ for some $k\ge 1$, we must have that $d_p(f^k):T_pM\to T_pM$ is orthogonal in some basis (i.e. diagonlizable over $\mathbb{C}$ with eigenvalues $|\lambda_i|=1$).

This is not a sufficient condition: the map $f:S^1\times\mathbb{R}\to S^1\times\mathbb{R}$ given by $f(z,x)=(e^{i\pi\theta}z,2x)$ with $\theta\in\mathbb{R}$ irrational has no such fixed points but is never an isometry.