Do diffeomorphisms preserving a parallelisms act properly?

Question

Let $M$ be a connected $n$-dimensional manifold and $\omega:TM\to \Bbb R^n$ a parallelism, that is a $\mathbb R^n$ valued $1$-form that is an isomorphism at each tangent space, alternatively described by a frame $(b_1,...,b_n)$ of vector fields $\mathfrak X(M)$ so that the restriction to any tangent space is a basis.

Let $f_n:M \to M$ is a sequence of diffeomorphisms preserving the parallelism, that is $f_n^*(\omega)=\omega$ or $D_xf_n(b_{i,x})=b_{i,f_n(x)}$ for all $x\in M$. My question is:

If there exists a point $x\in M$ so that $f_n(x)$ converges, does there exist a diffeomorphism $f$ and subsequence $f_{n_k}$ so that $f_{n_k}\to f$?

My question can be reformulated more concisely:

Let $\mathrm{Aut}(\omega,M)$ be the group of diffeomorphisms preserving a parallelism $\omega$, does $\mathrm{Aut}(\omega,M)$ act properly on $M$?

The topology of $\mathrm{Aut}(\omega,M)$ is the inherited topology of $\mathrm{Diffeo}(M)$, which should mean that the notion of convergence is that of uniform convergence on compacta.

Answer 1

Notice that we can endow $M$ with a Riemannian metric using the frame of vector fields $b_1,\ldots,b_n$. Then your question is true if $M$ is compact:

A diffeomorphism $f$ preserves $\omega$ iff it is an isometry.

Let $K$ be a compact nhood containing $\{f_n(x)\}_{n\in\mathbb N}$. Consider the family $\mathcal F$ of all isometries $g:M\rightarrow M$ such that $g(x)\in K$. Then in the compact-open topology of $C(M,M)$, $\mathcal F$ is equicontinuous as $\mathcal F$ consists of isometries. $\mathcal F$ is pointwise relative compact, i.e., for any $z\in M$ we have that $\{g(z):g\in \mathcal F\}$ is relatively compact since $M$ is compact.

If $\{f_n\}_{n\in\mathbb N}$ is a sequence of metric isometries $M\rightarrow M$ which converge to a function $f$, then $f$ is a metric isometry, and thus by the Myers-Steenrod theorem we get that $f$ is an isometry of Riemann manifolds. Therefore $\mathcal F$ is closed in $C(M,M)$.

Thus $\mathcal F$ is compact in $C(M,M)$ by the generalized Arzelà-Ascoli theorem, and hence in particular your question is true.

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As Moishe Kohan indicates in the comments the Arzelà-Ascoli holds using a diagonal argument if the metric given to $M$ turns out to be complete, as $M$ by definition is second countable.

In this case we define $$\mathcal F:=\{M\xrightarrow gM:g\text{ is an isometry and }g(x)\in K\},$$

where $K$ is some compact subset of $M$ containing the sequence $\{f_n(x)\}_{n\in\mathbb N}$.

Equicontinuity of $\mathcal F$ holds as again $\mathcal F$ consists of isometries of $M$. $\mathcal F$ is closed using again the Myers-Steenrod theorem and that $K$ is closed. Finally, $\mathcal F$ is pointwise relatively compact since for any $z\in M$ the set $\{g(z):g\in\mathcal F\}$ has bounded distance from $K$ as $g(x)\in K$ for all $g\in\mathcal F$, and thus the completeness of $M$ implies $\overline{\{g(z):g\in\mathcal F\}}$ is compact.

Answer 2

I figured I'd write up an answer to the general case.

Yes, the automorphism group of a parallelism always acts properly if the manifold is connected (not necessarily compact!).

Most of the steps are taken care of in Lemma 2.7 from these lecture notes by Werner Ballmann. From this we can find that if $f_n(x)$ converges for a point $x$ and $f_n\in\mathrm{Aut}(M,\omega)$ then there is a pointwise limit of $f_n$ that is smooth and an automorphism. I'll summarise the steps briefly:

Consider the constant fields $\omega^{-1}(v)$, $v\in\Bbb R^n$. These fields span the tangent space at any point, and an automorphism of $\omega$ necessarily commutes with the flow of these fields (that is $f(\varphi^t_v(x))=\varphi_v^t(f(x))$) whenever the flow is defined. By dimensional reasons these flows explore a neighbourhood of any point and by connectedness of the manifold if we start at a point $x$ we can reach any other point $y$ by iterating flows along these contant fields.

Now if $f_n\in\mathrm{Aut}(M,\omega)$ and $x$ so that $f_n(x)$ converges we have that $f_n(y)$ converges for every $y$. For let $\varphi=\varphi^{t_1}_{v_1}\circ ...\circ \varphi^{t_n}_{v_n}$ be a composition of flows taking us from $x$ to $y$, then $f_n(y)=f_n(\varphi(x))=\varphi(f_n(x))$ which converges. So we may define a function $f:M\to M$ as the pointwise limit of $f_n$.

We note that $f$ also commutes with the flows of the constant fields, which will be used to show that $f$ is smooth. For a point $x\in M$ let $U_x\subset \Bbb R^n$ open around $0$ so that the flow of constant fields at $x$ is a diffeomorphism from $U_x$ to some neighbourhood of $x$, do the same for $f(x)$ and assume that $U_x=U_{f(x)}$ by making both smaller if necessary. Then the flow of constant fields are two coordinate neighbourhoods around these points, in these coordinates the function $f$ is the identity transformation, which is smooth, hence $f$ is smooth.

Finally $f$ is an automorphism as it preserves the flow of the constant fields, meaning that $\mathcal{L}_{\omega^{-1}(v)}(f)=0$ for any constant field $\omega^{-1}(v)$, which implies that $f$ is an automorphism.

Next one can show that $f_n\to f$ uniformly on compacta. This is implies the first formulation of my question. The second formulation, that $\mathrm{Aut}(M,\omega)$ acts properly, seems a bit stronger, so I will prove that also.

To do this take a basis of $\Bbb R^n$ and as in the other answer pull back the usual inner product of this basis back to get a Riemannian metric on $M$, denote with $d:M\times M\to \Bbb R$ the induced distance function. Automorphisms of $\omega$ automatically preserve this metric and thus also the distance function.

Now we can appeal to two simple general results:

1. If $(X,d)$ is a locally compact metric space and $f_n$ a sequence of isometries that converge pointwise to an isometry $f$ then there is a sub-sequence that converges uniformly on compacta.

2. Let a group $G$ act on metric space $(X,d)$ by isometries, then (if $f_n\in G$, $x\in X$ so that $f_n(x)$ converges $\implies$ $\exists$ a subsequence $f_{n_k}$ and an $f\in G$ so that $f_{n_k}\to f$) is equivalent to ($G$ acts properly on $X$).

The first step follows from considering $\sup_{z\in K}d(f_n(z), f(z))$ for a compactum $K$. If that doesn't converge to $0$ there must be a sequence $z_n\in K$ so that the above is $>\epsilon$, by compactness of $K$ we may assume $z_n$ converges to some $z\in K$. Then $$\epsilon<d(f_n(z_n),f(z_n))≤ d(f_n(z_n),f_n(z)) + d(f_n(z),f(z)) + d(f(z), f(z_n)).$$ However since $f$ and $f_n$ are isometries the first and the last summand may be replaced by $d(z_n, z)$ which converges to $0$. The middle summand also converge s to $0$ since $f_n$ converges pointwise to $f$. This contradiction means that $f_n$ has a sub-sequence that converges uniformly on $K$. By doing a diagonalisation argument and making use of an exhausting sequence of compacta (use second-countability of metric spaces and local compactness), we find a sub-sequence $f_{n_k}$ convering uniformly on any compactum.

Next, acting properly means that whenever $f_n(x_n)$ converges with $x_n$ convergent it follows that $f_n$ admits a convergent sub-sequence. Thus for the equivalence desired the direction "$\impliedby$" is trivial and we show "$\implies$".

Now suppose the left-hand side of the equivalence holds and we have sequences $f_n(x_n)\to y, x_n\to x$, then: $$d(f_n^{-1}(y),x) ≤ d(f_n^{-1}(y),x_n)+ d(x_n,x) = d(y, f_n(x_n)) + d(x_n,x)$$ hence $f_n^{-1}(y)$ converges to $x$. The left-hand side implies that $f_n^{-1}$ then converges to some $f^{-1}$, thus $f_n$ converges to $(f^{-1})^{-1})$.