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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).


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?


(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 $\theta$.

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  • 4
    $\begingroup$ Suppose $f$ is a non-trivial involution (so $f^2=f\circ f= Id$ and $f\neq Id$ then for any metric $g$ the "balanced metric" $g_f=g+f^*g$ is preserved by $f$. In particular, $Met(f)$ is infinite dimensional. More generally, it seems that if $G_f=\lbrace f^n: n\in \mathbb{Z}\rbrace$ which is a subgroup of $Diff(M)$ acts ``nicely" enough" (e.g., freely and properly discontinuously), then one could pull back any metric on the quotient to get a element of $Met(f)$. $\endgroup$ – foliations Aug 23 '15 at 17:30
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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.

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  • $\begingroup$ Very nice!A few points: 1) Why the ratio $\frac{α_1}{α_2}$ needs to be irrational? 2) In your example the differential of the the diffeomorphism was trivial (the identity, after canonically identifying all the tangent spaces). So preserving the metric amounted to the metric being invariant under translates. In general, even if we are guaranteed that there is no dense orbit, how can we deduce the preserved cone will always be infinitely-dimensional? 3) Ho can we make it one dimensional? Don't we always have freedom to choose the metric on one tangent space?(See my added answer below). $\endgroup$ – Asaf Shachar Aug 28 '15 at 23:33
  • $\begingroup$ 1. The ratio should be irrational because otherwise the orbit of any point will be discrete and therefore not dense. Like in my first example. $\endgroup$ – Vladimir S Matveev Aug 29 '15 at 21:54
  • $\begingroup$ 2. Take any orbit and its closure. Denote by $d$ the distance to the orbit (in a background metric for which your mapping is an isometry). The function $d$ is preserved by isometries. Choose any positive function $f$ and conformally change the metric by multiplying it by $f(d)$. You will obtain a metric which is preserved by isometries, and you have a functional freedom, namely a choice of a function $f$. If the function $f$ is such that it is nonconstant only for small values, the resulting metric will be a smooth one. $\endgroup$ – Vladimir S Matveev Aug 29 '15 at 22:12
  • $\begingroup$ 3. Let me give an example, again on the torus, such that the set of the metrics is two-dimensional. It is similar to my example 2, but instead of shifting w.r.t vector $(\alpha_1, \alpha_2)$ take the sliding symmetry with respect to this vector. Since the square of the sliding symmetry is translation, the metric must have constant components in our coordinate system. The condition that it should be preserved by the sliding symmetry is an additional restriction on the metric that kills one from thee freedoms in changing the metric. $\endgroup$ – Vladimir S Matveev Aug 29 '15 at 22:22
  • $\begingroup$ 3 continued: I think you now see the phenomena: In order to make the set of metrics smaller, the orbit should be dense and should return close to a point with different differentials. I do not think that one can construct a two-dimensional example with one-dimensional set of metrics but I think I know an example, though complicated one, in dimension 4 $\endgroup$ – Vladimir S Matveev Aug 29 '15 at 22:25
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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?


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 $\epsilon_i \neq 0)$

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