# Diffeomorphism group of a manifold is never a Lie group?

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).

• 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... – user98602 Aug 22 '15 at 21:33
• 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. – user98602 Aug 22 '15 at 21:34
• 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.) – user98602 Aug 23 '15 at 2:31
• @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). – Asaf Shachar Aug 23 '15 at 15:07
• 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 $G$. – user98602 Aug 23 '15 at 15:10

## 1 Answer

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 $(\tau,A)$.

Proof:

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)