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Let $M$ be a finite-dimensional, smooth manifold. Call a diffeomorphism $f : M \rightarrow M$ diagonalizable if there exists a Riemannian metric $g$ on $M$ such that $f : (M, g) \rightarrow (M, g)$ is an isometry. I have some questions regarding such objects.

a) Is the set Diag(M) of all diagonalizable diffeomorphisms a group under composition?

b) Note that, in order to be diagonalizable, a diffeomorphism must possess the following well-known property of isometries:

$$\text{If}~f(p) = p~\text{and}~df(p) = \mathrm{Id}_{T_pM}, \text{for some $p \in M$, then}~f = \mathrm{Id}_M. (*)$$

Is $(*)$ also a sufficient condition? In other words, given a diffeomorphism $f \in \mathrm{Diff}(M)$ satisfying $(*)$, is there a Riemannian metric for which $f$ is an isometry? Maybe this is too general, because any diffeomorphism not fixing any point satisfies $(*)$, but I don't know the answer.

My motivation here is to know how large is the set of diffeomorphisms that could be isometries within the set of all diffeomorphisms. I apologize if I'm missing some standard notation and/or vocabulary here. I'd appreciate, as always, any references.

Thanks in advance.

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1 Answer 1

up vote 2 down vote accepted

Here's one idea, but there are many problems in the details.

Let $f : M \to M$ be a diffeomorphism, and let $g_0$ be any Riemannian metric on $M$. For each $n$, let $g_{n+1} = f_* g_n$ be the pushforward of $g_n$ under $f$. By construction, each $f$ is an isometry $(M, g_n) \to (M, g_{n+1})$, but this is not what we want. However, it is well known that the space of Riemannian metrics is closed under positive linear combinations, so consider the following construction: set $$\bar{g}_N = \frac{1}{N} \sum_{n=0}^{N-1} g_n$$ This gives rise to a new sequence of Riemannian metrics on $M$. Observe that $$f_* \bar{g}_N = \frac{1}{N} \sum_{n=1}^{N} g_n$$ and so $$f_* \bar{g}_N - \bar{g}_N = \frac{1}{N} (g_N - g_0)$$ This suggests that if the limit $$\bar{g} = \lim_{N \to \infty} \bar{g}_N$$ exists and is a smooth Riemannian metric, and if $$\lim_{N \to \infty} \frac{1}{N} (g_N - g_0) = 0$$ then we will have $f : (M, \bar{g}) \to (M, \bar{g})$ an isometry as desired.

Of course, there is no reason to believe that any of this works. For example, let us fix a norm on the space $V = \Gamma (M, T^*M \otimes T^* M)$, and consider the linear operator $f_* : V \to V$. If $f_*$ is not a bounded linear operator on $V$ there's no reason to believe that the sequence $(\bar{g}_1, \bar{g}_2, \ldots)$ should converge. If $f_*$ is bounded but with operator norm greater than $1$ we would have to modify our construction so that $g_{N-1}$ doesn't overshadow the contributions from the earlier metrics. If it is bounded with operator norm less than $1$ we have to worry about whether $\bar{g} = 0$ or other such degeneracies. There's no reason to believe that $\bar{g}$ is smooth even when it converges. But perhaps it's somewhere to start.

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I like the idea. This is a program to be carried out assuming the hypothesis on b), right? For it would provide, ideally, a metric for which $f$ is an isometry. –  student Feb 4 '12 at 3:20

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