Let $M$ be a smooth manifold.

I heard there is a way to introduce a topology and a structure of infinite dimensional manifold (something like a Banach or a Frechet manifold) on $\text{diff}(M)$ making it a group whose multiplication (composition) has some degree of smoothness with respect to the above structure.

My question: Is there a way to prove there aren't any topology and smooth structure making $\text{diff}(M)$ a finite dimensional Lie group w.r.t composition?

A possible strategy:

Finding topological obstructions.

I guess both topologies which are usually used (weak and strong) are not locally-compact, hence cannot serve as a ground for a finite dimensional smooth manifold.

But what about other possible topologies?

As noted by Matt it is possible to endow $\text{diff}(M)$ with a locally-compact, locally-connected and Hausdorff topology making it a topological group (The discrete topology).

Is it possible to endow it with locally-compact and locally-connected Hausdorff second countable topology making it a topological group whose action on $M$ is continuous?

(All these properties are necessary for the topology of a Lie group, hence if we show this is impossible, we are done).

  • 1
    $\begingroup$ It is a result of Kallman (see here) that $\text{Homeo}(M)$ has precisely one completely metrizable, separable group topology. (I learned this from Katie Mann.) As far as I know, this is still open for $\text{Diff}(M)$ - it would follow from an automatic continuity principle, which some people are trying to prove. If you make some very simple assumptions (the map $tX \mapsto \text{exp}(tX)$ is a Lie group homomorphism $\Bbb R \to \text{Diff}(M)$) you can prove that your hope is false... $\endgroup$ – user98602 Aug 22 '15 at 21:33
  • 1
    $\begingroup$ at least for $M$ compact. Probably for $M$ noncompact, too. (Actually, for noncompact $M$, I don't even know what the right topology is on $\text{Diff}(M)$. People interested in the topology often study eg $\text{Diff}_c(M)$ here, where it's obvious.) As a separate note, $\text{Diff}(M)$ is not a(n $\infty$-dim) Banach group - not even $\text{Diff}^k(M)$. There is a theorem along the lines that if a Banach group acts transitively on a finite dimensional manifold, then the group is finite-dim. So you need $\text{Diff}$ to be Frechet. $\endgroup$ – user98602 Aug 22 '15 at 21:34
  • 1
    $\begingroup$ The vector field thing was a red herring, sorry. The real point here is that if $\text{Diff}(M)$ acts continuously on $M$, then it's infinite-dimensional. This is because it's $n$-transitive for all $M$, so there is a surjective continuous map $\text{Diff}(M) \to \text{Sym}^n(M)$ given by acting on some specific set of points in $M$. By invariance of domain, $\text{Diff}(M)$ cannot be finite dimensional. (This uses the assumption that $\text{Diff}(M)$ is second-countable, so that it has only countably many connected components.) $\endgroup$ – user98602 Aug 23 '15 at 2:31
  • 1
    $\begingroup$ @MikeMiller: OK, so your map is obtained by fixing some $n$ different points $p_1,...p_n$ and sending $\phi \in Diff(M)$ to $(\phi(p_1),...,\phi(p_n))$. I agree it is surjective and continuous. But why assuming $Diff(M)$ is finite dimensional contradicts invariance of domain? To what version of it do you refer? (There are continuous surjections from $\mathbb{R}^n $ to $\mathbb{R}^m $ for $m > n$, space filling curves). $\endgroup$ – Asaf Shachar Aug 23 '15 at 15:07
  • 1
    $\begingroup$ Oops! You're right. Anyway, that was overkill. Given any transitive action of a topological group $G$ on a space $X$, $G/\text{Stab}_p \cong X$. Stabilizers are closed subgroups, so if $G$ is a Lie group, $G/\text{Stab}_p$ is a manifold of at most the same dimension of $G$. $\endgroup$ – user98602 Aug 23 '15 at 15:10

For sake of completeness I am writing a solution based on the suggestion of Mike Miller:

Theorem: There are no topology $\tau$ and a compatible (finite-dimensional) smooth structure $A$ on $\text{Diff(M)}$ satisfying:

(1) The action of $\text{Diff(M)}$ on $M$ is continuous w.r.t $\tau$.
(2) $\text{Diff(M)}$ is a Lie group w.r.t $(\tau,A)$.


We assume by contradiction there exist such a pair $(\tau,A)$.

Define $X_n = \{(p_1,...p_n)\in M^n |p_i \neq p_j \forall i \neq j \}$, and look at the map $\psi : \text{Diff(M)} \times X_n \rightarrow X_n$ , $\psi\left(\phi,(p_1,...p_n)\right)=(\phi(p_1),...\phi(p_n))$.

(1) It is easy to see that $X_n$ is an $n$-dimensional manifold. (It's an open subset of $M^n$).
(2) Continuity of the action of $\text{Diff(M)}$ on $M$ implies $\psi$ is continuous.
(3) n-transitivity of $\text{Diff(M)}$ implies the action $\psi$ (of $\text{Diff(M)}$ on $X_n$) is transitive.

Now fix some point $q=(q_1,...q_n)\in X_n$, and denote by $G_q = \{\phi \in \text{Diff(M)} | \phi(q)=q \} $ the stabilizer group. It is closed, hence by the closed subgroup theorem $G_q$ is an embedded Lie subgroup of $\text{Diff(M)}$.

So, The left coset space $\text{Diff(M)} / G_q$ is a topological manifold* of dimension $dim(\text{Diff(M)})-dim(G_q)$.
(It can also be given a unique smooth structure making the quotient map $\pi: \text{Diff(M)} \rightarrow \text{Diff(M)} / G_q$ a smooth submersion, but that is irrelevant for our discussion**).

Now , by easy proposition on continuous homogenous $G$-spaces (i.e topological spaces with a continuous transitive action of a topological group $G$) it follows that $\text{Diff(M)} / G_q$ and $X_n$ are homeomorphic.

In particular their dimension as topological manifolds are equal, hence: $dim(\text{Diff(M)}) \ge dim(\text{Diff(M)})-dim(G_q) = dim(X_n)=n$ for every $n$ which is a contradiction.

*See Lee's book (Intro to smooth manifolds) Thm 21.17.

** If we assume the action of $\text{Diff(M)}$ on $M$ is smooth (not merely continuous), then we get that $\text{Diff(M)} / G_q$ and $X_n$ are diffeomorphic. (See Thm 21.18)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.