The set of diffeomorphisms preserving some metric.

Question

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.

Answer 1

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.

Answer 2

Regarding your question (a), the set $\text{Diag(M)}$ of all diagonalizable diffeomorphisms is not always closed under composition. (So in particular it is not a group).

For example take $M=\mathbb{R}^n$ , and consider linear diffeomorphims.

It is proved here that a linear automorphism $T:V \rightarrow V$ preserves some inner product on $V$ if and only if the matrix of $T$ w.r.t an arbitrary basis is similar to an orthogonal matrix. The point is that composition of two transformation of this type is not necessarily also of that type. (Which amounts to proving that the union of conjugacy classes of $O(n)$ is not a subgroup, see here).

In particular, if you take $A = \pmatrix{0&2\\-1/2&0}$, then $A,A^T$ both preserves some inner products on $\mathbb{R}^2$ , but their product $AA^T = \pmatrix{4&0\\0&\frac14}$ does not.

While the result above holds pointwise (it is an algebraic result, true for a single vector space, which can be thought of as a tangent space), note that for linear transformations, we can identify them with their differentials (after identifying the tangent spaces canonically with $\mathbb{R}^n$).

Now arise the natural question: what about other manifolds?