The set of diffeomorphisms for a compact manifold $\operatorname{Diff}(M)$ forms a group. I was recently discussing with someone that this is a topological group, and gave the usual clever proof using the compact-open topology.
But then we realized that this can't be the right topology for $\operatorname{Diff}(M)$, since it doesn't even mention the derivatives. So after some researching, we came upon a topology built from subbasis sets that look like
$$ N_\epsilon^r(f; K, U, V) = \{g\in\operatorname{Diff}(M)\mid g(K)\subset V\text{ and } \sum_{i=0}^r\lVert\tilde{f}^{(i)}-\tilde{g}^{(i)}\rVert_K<\epsilon\} $$
where:
- $\epsilon>0$
- $r$ is a non-negative integer
- $(U,\phi)$ and $(V,\psi)$ are charts on $M$, with $K\subset U$ a compact subset, and $f(K)\subset V$
- $\widetilde{f}=\psi\circ f\circ\phi^{-1}$ and similarly for $\widetilde{g}$
- $\widetilde{f}^{(i)}$ is the $i$th derivative of $\widetilde{f}$
- $\lVert\widetilde{f}^{(i)}-\widetilde{g}^{(i)}\rVert_K=\sup_{x\in \phi(K)}\lVert\widetilde{f}^{(i)}(x)-\widetilde{g}^{(i)}(x)\rVert$
Now we're trying to prove that inversion and composition are continuous in the topology generated by this subbasis. After much trial and error, we eventually turned to looking online, and only found either sources that assume this is trivial, or others that go through several layers of indirection before getting anywhere.
So, my question:
Is there a direct proof that inversion and composition on $\operatorname{Diff}(M)$ are continuous, in the above topology?
It's pretty easy to see the topology is finer than the compact-open topology, and so a proof similar to the clever one shows that on the level of homeomorphisms, inversion and composition are continuous in this topology. What I don't know how to do is include restrictions on the derivatives.
Best Answer
Ok, I think I worked this out. The key insight, oddly enough, was that $\operatorname{Diff}(M)$ is first countable. That helps immensely, because now instead of using the "preimages of open sets are open" definition of continuity, I can use convergent sequences. This simplified things enough for me to understand what was going on.
[To see it is first countable, it's pretty easy to cover $M$ with a countable basis of precompact charts $\{U_i\}$, and then to check that the open sets $$ N_{1/m}^r(f; \overline{U_i},U_j,U_k) $$ Form a countable subbasis around $f$, where all the subscripts and superscript are positive integers.]
Now if $f_n\xrightarrow{\operatorname{Diff}}f$, then in terms of the topology described in the question, this means that for any tuple $(\epsilon, r, K, U, V)$, there is a corresponding $N$ such that for $n\ge N$,
Since the compact-open topology is coarser than this topology, convergence in the latter implies convergence in the former. So we can get rid of (1) and instead say that $f_n\xrightarrow{\operatorname{Diff}}f$ is equivalent to
[I'm writing $a\xrightarrow{L} b$ to mean uniform convergence on the compact set $L$. I'm writing $a\xrightarrow{\operatorname{Diff}}b$ to mean convergence in the topology on $\operatorname{Diff}(M)$.]
Finally, since we already know inversion and composition are continuous in the compact-open topology, we can focus on (4).
Now for inversion, suppose $f_n\xrightarrow{\operatorname{Diff}}f$. Then using the fact that $f^{-1}\circ f(x)=x$ and applying the chain rule repeatedly, we see that we can write $$ f^{-(r)}\circ f(x)=c_r(f^{(1)}(x), f^{(2)}(x), \ldots, f^{(r)}(x))$$
where $c_r:\mathbb{R}^+\times\mathbb{R}^{r-1}\rightarrow\mathbb{R}$ is continuous. By choosing $n$ large enough, we can restrict the domain of $c_r$ [essentially by (1) and/or (3) above], and we can then assume $c_r$ is uniformly continuous. Then (4) plus the above equation implies $$ f_n^{-(r)}\circ f_n\xrightarrow{K}f^{-(r)}\circ f$$
Finally, (3) implies we also have $$ f_n^{-(r)}\circ f_n\xrightarrow{K}f_n^{-(r)}\circ f$$ and the fact that $f$ is a diffeomorphism shows we reallly have $$ f_n^{-(r)}\xrightarrow{K}f^{-(r)}$$ which is enough for (4), and thus inversion is continuous.
For composition, again the chain rule applied to $g\circ f$ gives an equation like $$ (g\circ f)^{(r)}(x) = d_r(g^{(1)}\circ f(x), \ldots, g^{(r)}\circ f, f^{(1)}(x), \ldots, f^{(r)}(x))$$ where we can assume $d_r$ is uniformly continuous.
Then (4) applied to $g$ shows we have $$ (g_n\circ f)^{(r)}\xrightarrow{K}(g\circ f)^{(r)} $$ and (4) applied to $f$ then gives $$ (g_n\circ f_n)^{(r)}\xrightarrow{K}(g\circ f)^{(r)} $$
The real insight here is that using the "convergent sequence" definition of continuity really simplifies the notation and presentation, and then really it all falls on the chain rule.