Riemannian metrics preserved by diffeomorphisms

Question

Let $f \neq Id$ be a diffeomorphism (of a smooth manifold $M$) which admits some Riemannain metric on $M$ making it an isometry. How many different metrics are preserved by $f$?

Note that $Met(f)=\{g|f^*g=g\}$ is a convex cone.

Question:

Is $Met(f)$ necessarily a finite dimensional manifold? (The set of all Riemannian metrics is an infinite dimensional manifold)

What happens if we assume $f$ has no fixed points?

Edit: It turns out that even if $f$ has no fixed points, $Met(f)$ can be infinite dimensional. (For example take the antipodal map on the sphere, see remark by foliations).

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Partial Results and further Questions:

(1) "Galois" Correspondence:

Fix some Riemannian metric $g$ on $M$. A natural point of interest is the correspondence (something analogous to Galois Correspondence) between subgroups $K \subseteq Iso(M,g)$ and $Met(K)= \{h|f^*h=h , \forall f \in K\}$. Of course this is of interest only when $Iso(M,g)$ is "rich" (if $Iso(M,g)= {Id}$ for instance this is clearly uninteresting). Suppose $Iso(M,g)$ is a positive dimensional Lie group. Is it true that for every subgroup ${Id} \neq K \subseteq Iso(M,g)$ , $Met(K)$ is a finite dimensional manifold? Is there a connection between the dimensions of $K$ and $Met(K)$?

For example we can think on the sphere $\mathbb{S}^n$ with the round metric $g_0$. It is well known that it is the only $O(n)$-invatiant metric on $\mathbb{S}^n$ (up to scalar multiple), and that $Iso(\mathbb{S}^n,g_0)=O(n)$. So in this case, for $K=Iso(\mathbb{S}^n,g_0)$, $Met(K)=\{\lambda g_0|\lambda > 0\}$ is a one dimensional cone. What happens for smaller $K$? (Do we really need $K$ to be the whole isometry group in order to uniquely determine the metric up to scalar multiple?)

Has these kind of questions been investigated before?

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(2) The case $M=\mathbb{R^n}$ and $f\in GL(\mathbb{R^n})$ :

Since we can assume $f \in O(n)$ (w.r.t to a suitable basis, see here for details), we can obtain results based on the particular form of $f$.

It is proved here that for a vector space $V$, and $T \in GL(V)$ the inner product on $V$ is uniquely determined (up to scalar multiple) on each two-dimensional subspace where the restriction of $T$ is a proper rotation (by angle $\theta \neq 0,\pi$).

In particular, if we take $n=2$ $(M=\mathbb{R^2})$ , $f \in Diff(\mathbb{R^2})$ to be a proper rotation, and $g_0 \in Met(f)$ to be the Euclidean metric, then we can multiply $g_0$ by a positive radial function $h(r)$ and still get a preserved metric.

In particular $Met(f)$ is infinite dimensional. (Since it contains a copy of the infinite dimensional real cone of positive smooth functions $M \rightarrow \mathbb{R}$).

Actually if $g \in Met(f)$ and $f=R_{\theta}$ is a rotation by angle $\theta$ , then by the result stated above $g_{(r,\alpha )} = h(r,\alpha)\cdot g_0$ where $h$ must satisfy: $h(r,\alpha)= h(r,\alpha+\theta)$ for every $r\in (0,\infty), \theta \in [0,2\pi)$.

There are two cases:

(a) $f$ is of infinite order. ($\frac{\theta}{2\pi} \notin \mathbb{Q})$ Fix $r$. Then we get the function $\alpha \rightarrow h(r,\alpha)$ which is continuous and has arbitrarily small peroids. (This follows from Dirichlet's Theorem, since $n\cdot \theta$ can be arbitrarily close to $0$ modulo $2\pi$). This forces the function to be independet of $\alpha$. So the only freedom alowed is indeed multiplication by a radial function.

(b) $f$ is of finite order. ($\frac{\theta}{2\pi} \in \mathbb{Q})$. Then $h(r,\alpha)$ must be periodic in the angle coordinate with period which divides $\theta$.

Answer 1

The answer depends on the diffeomorphism.

Let me give two examples, both on the standard torus $\mathbb{R}^2/_{\mathbb{Z}^2}$ with coordinates $x,y$.

(Example 1:) $$\phi(x,y)= (x+ 1/2,y).$$

For this example the cone of metrics which is preserved by this $\phi$ is infinitely dimensional, since any metric $g_{ij}(x,y)$ such that the entries depend on $x$ periodically with period $1/2$ and on $y$ periodically with period $1$ is preserved by it.

(Example 2:) Take irrational $\alpha_1, \alpha_2$ such that the ratio $\alpha_1/\alpha_2$ is also irrational and consider the diffeomorphism
$$\phi(x,y)= (x+\alpha_1, y+ \alpha_2).$$ Then every metric that is preserved by this diffeomorphism has constant entries in the coordinates $x,y$, so the space of metrics is finitely dimensional.

These two examples show the phenomena, and actually give an answer:

If there exists a point such that the orbit w.r.t. the iterations of the diffeomorphism (i.e., the set $\lbrace x, \phi(x), \phi(\phi(x)),...\rbrace$ is dense on the manifold, then the cone of metrics preserved by the diffeomorphism is finite-dimensional.

(One can construct examples such that it has dimension $ 1$)

Otherwise it is infinitely-dimensional.

Answer 2

I am just adding a few details to Vladimir's answer:

Lemma: Assume there exists a point $x$ such that the orbit w.r.t. the iterations of the $\phi$ (i.e., the set $\{x,ϕ(x),ϕ(ϕ(x)),...\}$ is dense in $M$. Then any $g \in Met(\phi)$ is completely determined by its restriction $g_x$ to $T_xM$.

Corollary:
The cone of metrics preserved by $ϕ$ is finite-dimensional. (In fact its dimension is bounded above by $\frac{n(n+1)}{2}$ which is the dimension of the manifold of all inner products on an $n$-dimensional vector space).

Proof of lemma:
Let $g \in Met(\phi)$. Take $y \in M$. By the density assumption, it follows that the there exist a sequence $n_k \in \mathbb{N}$, such that $\phi^{n_k}(x)$ converges to $y$. Take a coordinate neighbourhood around $y$. Then continuity of the metric implies: $g_{ij}\big(\phi^{n_k}(x)\big) \rightarrow g_{ij}(y)$. However, $\phi^{n_k} \in \text{Iso(M,g)} \Rightarrow g_x$ determines $g_{\phi^{n_k}(x)}$ so we are done.

Question: Can we choose the metric $g_x$ arbitrarily?

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Some details concerning the example on the torus:
Let $N \in \mathbb{N}$. By Dirichlet's approximation theorem, there exist integer numbers $1 \le q \le N^2,p$ such that: $|q\alpha_1-p|\le \frac{1}{N^2+1}$. Now, applying the theorem again (now for $q\alpha_2$) we can obtain $1 \le q' \le N,p'$ such that: $|q'(q\alpha_2)-p'|\le \frac{1}{N+1}$. So,$|q'q\alpha_1-q'p|\le N \cdot \frac{1}{N^2+1} \le \frac{1}{N}$. Denoting $qq'=m,q'p=n_1,p'=n_2$ we get: $|m\alpha_1-n_1|\le \frac{1}{N},|m\alpha_2-n_2|\le \frac{1}{N}$, so putting $\epsilon_1 = m\alpha_1-n_1,\epsilon_2 = m\alpha_2-n_2$ we get $g_{ij}(x,y)=g_{ij}(x+\epsilon_1,y+\epsilon_2)$ for any $(x,y)$. Hence $g_{ij}$ is constant. (Since it's continuous and has arbitrarily small peroids).

(Where we used the irrationality of $\alpha_i$ in the fact that the obtained periods $\epsilon_i \neq 0)$