# Is there a global obstruction for a diffeomorphism to be an isometry?

Let $V$ be a finite dimensional vector space. Let us call an automorphism $T:V\rightarrow V$ admissible if there exists an inner product $\langle , \rangle$ on $V$ making $T$ an isometry.

We know $T$ is admissible if and only if there exists a basis for $V$ such that the representing matrix of $T$ w.r.t to this basis is orthogonal. (see this question of mine for details).

Similarly, we call a finite sequence of linear automorphisms $T_i:V_i\rightarrow V_{i+1}$ for $i=1,...n-1$ and $T_n:V_n\rightarrow V_1$ admissible if there exists inner products $\langle , \rangle_i$ on $V_i$ making all the $T_i$ isometries.

It is easy to see that a sequence is admissible if and only if the composition $T_n \circ T_{n-1} \circ ... \circ T_1:V_1\rightarrow V_1$ is admissible.

Now assume we have a diffeomorphism $\phi:M\rightarrow M$. ($M$ is a smooth manifold). A necessary condition for the existence of a Riemannian metric on $M$ making $\phi$ an isometry is a "pointwise" condition: For every fixed point $p\in M$ of $\phi$, $d\phi_p:T_pM\rightarrow T_pM$ should be admissible. More generally, for evey finite orbit of $\phi$ the corresponding sequence of differentials must be admissible.

For infinite orbits we have no pointwise constraints, since we can just pick an inner product on one of the tangent spaces in the orbit, and take the induced (pulled-back or pushed-forward) inner products on consecutive spaces ad infinitum.

My Question: Assume $\phi$ satisfies all the poinwise conditions mentioned above. Is there a Riemannian metric making $\phi$ an isometry?

The question boils down to whether we can glue together all the different inner products into a smoothly changing inner product. (Note that we have some degrees of freedom in choosing the preserved inner product on each tangent space, since an admissible automorphism can have many different preserved products)

Update:

In the case where there are no finitie orbits (or "almost none") it is demonstrated by the answers of Matveev & Sawin below that there is a global obstruction.

Two questions arise:

1) What happens if all the orbits are finite?

Well, it turns out that any diffeomorphism which has only finite orbits is actually of finite order, and for that case the answer is always positive, since we can sum over translates of any initial metric

2) Can we find other conditions on the dynamics of the diffeomorphism which ensures a smooth metric exists?

Yes, there are global obstructions. Consider $M= S^1 \times \mathbb R$ and $\phi$ acting by an irrational rotation on $S^1$ and by multiplication by $2$ on $\mathbb R$. There are no finite orbits, but also no invariant metrics: The function that takes a point $P$ on $S^1$ to the length of a vector pointing along $\mathbb R$ at $(P,0)$ is a function on $S^1$ that, when rotated irrationally, is multiplied by $2$. No nonzero such functions exist - consider what happens to the max and min when you rotate and multiply by $2$.

The general obstruction is that if you have an orbit whose closure has positive dimension, and take a sequence of powers $n$ such that $\phi^n(x)$ converges to $x$, then the absolute values of the eigenvalues of $d\phi^n(x)$ must converge to $1$. This doesn't depend on your choice of basis for the tangent space near $x$ because the eigenvalues depend continuously on the matrix and as $\phi^n(x)$ converges to $0$ you get a good result.

• Can you please elaborate on your remark? I am not sure I understand. Why must the eigenvalues converge to one? I guess you mean the eigenvalues of the representing matrix of $(d\phi^n)_x$ w.r.t to the standard coordinate fields, since I do not know any definition of eigenvalues of transformations from a given domain to a different codomain. – Asaf Shachar Apr 20 '15 at 20:50
• In the particular example you gave, we take advantage of the special fact that the tangnet space at each points is built from two components, one of them ($\mathbb{R}$) is fixed so we can speak on eigenvalues for that part. But I do not see what you mean in a general case. (Sorry for the multiple remarks, my 5 minutes were gone...) – Asaf Shachar Apr 20 '15 at 20:56
• @asafshachar Yes, I want the representing matrix. Changing the representing matrix doesn't change the eigenvalues but it does not change this converge property. The reason is we choose a basis continuously on the space, so if we choose a new basis, the change-of-basis matrix converges to $1$ as the two points approach each other. Because $\phi^n(x)$ converges to $x$, as $n$ goes to $\infty$ in two different bases the change of basis matrix goes to $1$. – Will Sawin Apr 20 '15 at 20:57
• Rather than saying that the eigenvalues converge to $1$, let me say instead: If $x$ is a point and we have a sequence of $n$ such that $\phi^n(x)$ converge to $x$, and if $v$ is a vector in the tangent space at $x$, then $\phi^n(v)$ should be bounded. By bounded I just mean that they lie in a compact subset of $TM$. – Will Sawin Apr 20 '15 at 21:02
• If the orbit of $x$ is compact, then we could just ask that $\phi^n(v)$ be bounded and not worry about a subsequence. So if the orbit of $x$ in $M$ is compact, I want the orbit of $v$ in $TM$ to also be compact. – Will Sawin Apr 20 '15 at 21:05

The answer is no'', the pointwise condition is not enough. The example exists in dimension 1 already and can be generalized and made arbitrary weird for all dimensions.

Consider a smooth function $v(x)$ such that

$\bullet$ it is positive for $x> 0$

$\bullet\bullet$ it is odd (i.e. $v(x)=-v(-x)$). In particular it vanishes at $x=0$.

$\bullet\bullet\bullet$ all its all derivatives (first $v'$, second $v''$, third $v^{(3)}$ and so on) vanish at $x=0$.

Then, take the vector field $v(x)\frac{\partial }{\partial x}$,
its flow $F_t$, and as the diffeomorphism take $F_1$.

It can not be an isometry since it does not preserve any smooth or even continuous volume since it sends the interval $[-\varepsilon ,\varepsilon]$ to a bigger interval containing $[-\varepsilon ,\varepsilon]$.
Its only fixed or periodic point is the point $x=0$ and at $x=0$ the differential of $F_1$ is the identity endomprphism so your condition is fulfilled.

The example can be made analytic - actually one does not need that all derivatives are zero; the first two are sufficient. So $v(x)= x^3$ gives a counterexample to your statement (though of course in this case one needs to think or calculate to show that the differential of $F_1$ at $x=0$ is the identity endomorphism and if we assume that all derviatives $v^{(i)}(0)$ are zero then it is evident).

Of course one can construct an example on a closed manifold (on a circle in dimension 1) of any dimension using the same idea.

The idea of this counterexample is due to Ben McKay and is in his answer to my question Does for every vector field there always exist a volume form for which the vector field is a homothety?

• $x\mapsto x^3 \;$ is not a diffeomorphism from $\mathbf{R}$ to itself, since the inverse of that map is not differentiable at zero. $\;\;\;\;$ – user5810 Apr 19 '15 at 22:12
• @RickyDemer He means $x \to x+x^3$. – Will Sawin Apr 19 '15 at 22:16
• Actually $v(x)=x^3$ doesn't quoite work because the flow along that vector field diverges to $\infty$ in finite time. So $F_1$ is not a well defined function (it's something like $f(x) = x/ \sqrt{1-x^2}$). But just adding $x^3$ works fine, or taking a slower-growing vector field. – Will Sawin Apr 19 '15 at 22:18
• I agree that the diffeomorphism $x\mapsto x+ x^3$ is possibly the simples counterexample. The observation that the flow of $x^3 \frac{\partial}{\partial x}$ diverges does not really make any problems since it is well defined on small intervals around $0$ – Vladimir S Matveev Apr 20 '15 at 7:03
• I can but may be instead of my example you take the one which appeared in the comment of Will Sawin: as the diffeomorphism $R\to R$ you take $x\mapsto x+ x^3$. It has only one fixed or periodic point, this is the point $x=0$; its derivative at $x=0$ is $1$, and it can not be an isometry since isometry preserves volume and therefore can not map the interval [-1, 1] on the interval $[-2,2]$ as the diffeomorphism does. – Vladimir S Matveev Apr 20 '15 at 18:35

I have decided to write the details of the solution suggested by Sawin, for sake of completeness.

Let $M=\mathbb{S}^1\times\mathbb{R}$. (i.e $M$ is the infinite cylinder in $\mathbb{R}^3$).

$\phi:M \rightarrow M , \phi(p,r)=(R_{\theta}(p),2r)$. For a point $(p,x)\in \mathbb{S}^1\times\mathbb{R}$, $T_{(p,x)}(\mathbb{S}^1\times\mathbb{R})=T_p\mathbb{S}^1 \oplus T_x\mathbb{R}=T_p\mathbb{S}^1 \oplus \mathbb{R}$.

Looking at the differential, we see that $d\phi_{(p,x)}:T_p\mathbb{S}^1 \oplus \mathbb{R}\rightarrow T_{R_{\theta}(p)}\mathbb{S}^1 \oplus \mathbb{R}, d\phi_{(p,x)}(v,r)=(R_{\theta}(v),2r)$.

Now assume there is a $\phi$-invariant metric on $M$.

Now define the following function: $f:\mathbb{S}^1 \rightarrow \mathbb{R}$, $f(p)=\|(0,1)\|_{(p,0)}$ (The length of a vector which points along $\mathbb{R}$, i.e, is orthogonal to the copy of $\mathbb{S}^1$ which lies at height $0$). Now note that $\phi$-invariance of the metric implies: $f(p)=\|(0,1)\|_{(p,0)}=\|d\phi_{(p,0)}(0,1)\|_{(R_{\theta}(p),0)}=\|(0,2)\|_{(R_{\theta}(p),0)}=2\|(0,1)\|_{(R_{\theta}(p),0)}=2f(R_{\theta}(p))$.

Now, since $f$ is smooth (proof in the end), it is a real-valued continuous function on a compact domain, hence obtains a maximum. Since we can rotate the point where the maximum is achieved and the value will be multiplied (or divided) by 2, this forces the maximum of $f$ to be $0$. But this is a contradiction to the positive-definiteness of the metric.

All is left is to verify that $f$ is smooth:

Define the vector field $X:M\rightarrow TM,$ via $X(p,x)=(0,1)\in T_{(p,x)}(\mathbb{S}^1\times\mathbb{R})=T_p\mathbb{S}^1 \oplus \mathbb{R}$. $X$ is smooth since $TM=T(\mathbb{S}^1\times\mathbb{R}) \stackrel{\text{diffeomorphic}}{\cong} T\mathbb{S}^1 \times T\mathbb{R}$ and both "components" are smooth as sections of $T\mathbb{S}^1, T\mathbb{R}$ respectively.

Now consider the function $h:M \rightarrow \mathbb{R}$, $h(p,x)=\|X(p,x)\|^2_{(p,x)}=\|(0,1)\|^2_{(p,x)}$. $h$ is smooth by the definition of Riemannian metric. Since $\mathbb{S}^1=\mathbb{S}^1 \times \{0\}$ is a submanifold of $M$ the restriction $h|_{\mathbb{S}^1} = f^2$ is smooth. Since $f$ is positive (by definition of a metric) and $\sqrt{(\cdot)}:\mathbb{R}^{>0} \to \mathbb{R}^{>0}$ is smooth, it follows that $f$ is smooth as required.

An attempt to explain Sawin's remark about the eigenvalues:

Let $x\in M$. Assume $\phi^n(x)\rightarrow x$. We have $(d \phi^n) _x:T_xM \rightarrow T_{\phi^n(x)}M$. It is possible to choose an orthonormal local frame $E_1,...,E_n$ over a (small enough) neigbourhood $U$ of $x$. (since $\phi^n(x)\rightarrow x$ we can assume $\phi ^n(x) \in U$ for every $n$). Now assume $[d \phi^n]^{E_i(x)}_{E_i(\phi^n(x))}$ has eigenvalue $\lambda_n$, i.e there exists a vector $v_n\in T_xM$ such that

(*) $\lambda_n [v_n]_{E_i(x)}=[d \phi^n]^{E_i(x)}_{E_i(\phi^n(x))}[v_n]_{E_i(x)}=[(d \phi^n) _x (v_n)]_{E_i(\phi^n(x))}$.

Also, using the fact $\phi$ is an isometry implies

(**) $\|[(d \phi^n) _x (v_n)]_{E_i(\phi^n(x))}\|_{Euc} \stackrel{\text{$E_i$orthonormal}}{=}\|(d \phi^n) _x (v_n)\|_{\phi^n(x)} \\ \stackrel{\text{$\phi$isometry}}{=}\|v_n\|_x\stackrel{\text{$E_i$orthonormal}}{=}\|[v_n]_{E_i(x)}\|_{Euc}$

($\| \cdot \|_{Euc}$ is the standard Euclidean norm on $\mathbb{R}^n$)

Combining (*),(**) we get: $\| \lambda_n [v_n]_{E_i(x)}\|_{Euc}=\|[v_n]_{E_i(x)}\|_{Euc}$, hence $|\lambda_n|=1$.