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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 $\theta$.

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 $\theta$.

(1) For the case of $M=\mathbb{R^n}$ 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 $f$.

(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 $\theta$.

(2) 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?

(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 $f$.

(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 $\theta$.

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) For the case of $M=\mathbb{R^n}$ 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.

(2) 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?

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) For the case of $M=\mathbb{R^n}$ 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 $\theta$.

(2) 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?