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Clarification: My question concerns the homotopy type of the space of $C^k$ diffeomorphisms with the compact-open $C^k$ topology, where $0< k \leq\infty$. I have stated my question below with $k=1$ for definiteness and simplicity. I am not particularly interested in any specific value of $k$.$\newcommand{\Diff}{\operatorname{Diff}}$$\newcommand{\RR}{\mathbb{R}}$

It is fairly well-known that the space $\Diff(M)$ of $C^1$ diffeomorphisms of a closed smooth manifold $M$ is homotopy equivalent to a CW-complex. Here are the relevant facts:

  • $\Diff(M)$ is a Banach manifold modelled locally on the space of $C^1$ vector fields on $M$.

  • Metrizable Banach manifolds have the homotopy type of CW-complexes, as shown by Palais.

[Remark: When $M$ is closed, the spaces of $C^k$ diffeomorphisms of $M$ (for $0 < k \leq\infty$) are all homotopy equivalent to each other via the natural inclusions. This can be shown by embedding $M$ smoothly in $\RR^N$, and then using smoothing operators defined by taking convolution with a mollifier.]

When we allow $M$ to not be compact, there are several common topologies on $\Diff(M)$. I am interested in the compact-open (or weak) $C^1$ topology.

Questions: Is it known whether the space $\Diff(M)$ of $C^1$ diffeomorphisms with the compact-open $C^1$-topology is homotopy equivalent to a CW-complex when $M$ is a smooth manifold without boundary? Is there a known counter-example? Are there particular cases where the answer is known, for example if $M$ is the interior of a compact manifold?

Feel free to use instead $C^k$ diffeomorphisms and/or the compact-open $C^k$ topology for any $0 < k \leq\infty$.

I would also be interested in hearing about any known results related to this question: e.g. for spaces of embeddings of manifolds in the compact-open/weak topology when the source is not the interior of a compact manifold.

Edit: Allen Hatcher gave a very nice answer to my question. Afterwards, I also posted an answer very similar to Allen's, which I was writing when Allen posted his. A pertinent question still remains: (1) Does the result hold for the interior of a compact manifold? Here is a perhaps less pertinent question: (2) For $M$ without boundary, do the path components of $\Diff(M)$ have the homotopy type of a CW-complex?

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4 Answers 4

up vote 18 down vote accepted

Here is an example where ${\rm Diff}(M)$ with the compact-open topology is not homotopy equivalent to a CW complex. Take $M$ to be a surface of infinite genus, say the simplest one with just one noncompact end. I will describe an infinite sequence of diffeomorphisms $f_n:M\to M$ converging to the identity in the compact-open topology and all lying in different path-components of ${\rm Diff}(M)$. Assuming this, suppose $\phi:{\rm Diff}(M) \to X$ is a homotopy equivalence with $X$ a CW complex. The infinite sequence $f_n$ together with its limit forms a compact set in ${\rm Diff}(M)$, so its image under $\phi$ would be compact and hence would lie in a finite subcomplex of $X$, meeting only finitely many components of $X$. Thus $\phi$ would not induce a bijection on path-components, a contradiction.

To construct $f_n$, start with an infinite sequence of disjoint simple closed curves $c_n$ in $M$ marching out to infinity, and let $f_n$ be a Dehn twist along $c_n$. The $f_n$'s converge to the identity in the compact-open topology since the $c_n$'s approach infinity. We can choose the $c_n$'s so that they represent distinct elements in a basis for $H_1(M)$ and then the $f_n$'s will induce distinct automorphisms of $H_1(M)$. If two different $f_n$'s were in the same path-component of ${\rm Diff}(M)$ they would have to induce the same automorphism of $H_1(M)$ since any path joining them would restrict to an isotopy of any simple closed curve in $M$ (see the next paragraph below) and a basis for $H_1(M)$ is represented by simple closed curves.

If $g_t$ is a path in ${\rm Diff}(M)$ then the images $g_t(c)$ of any simple closed curve $c$ vary by isotopy since this is true as $t$ varies over a small neighborhood of a given $t_0$, so since the $t$-interval $[0,1]$ is compact, a finite number of these neighborhoods cover $I$ and the claim follows.

Remark: The $f_n$'s were chosen to be Dehn twists just for convenience. Many other choices of diffeomorphisms would work just as well. One can easily see how to generalize to higher dimensions.

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Thank you very much for your answer, Allen. I actually wrote an answer using the same idea, and you seem to have posted yours while I was writing mine. :) By the way, a continuous path in the compact-open $C^1$ topology on $\operatorname{Diff}(M)$ also gives a continuous path in the compact-open $C^0$-topology, and such a path is exactly a continuous homotopy $M\times I\to M$. That is another way to prove the claim you require at the end: different $f_n$'s are not in the same component. –  Ricardo Andrade Apr 8 '13 at 23:30

Check the paper Antonelli, P. L.; Burghelea, D.; Kahn, P. J. The non-finite homotopy type of some diffeomorphism groups. Topology 11 (1972), 1–49.

There they mention an old result of Palais (Homotopy theory of infinite dimensional manifolds. Topology 5 (1966), 1-16) which states that the identity component of ${\rm Diff}_0(M)$ has the homotopy type of a countable $CW$-complex.

In their paper, Antonelli, Burghelea and Kahn prove that for many smooth manifolds (including spheres of dimension $\geq 7$) the group ${\rm Diff}_0(M)$ does not have the homotopy type of a finite $CW$-complex. (This is highly nontrivial.)

Above, by diffeomorphisms they mean smooth diffeomorphisms and the topology is the $C^\infty$-topology.

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Regarding the finite CW case, there's the simpler example of Hatcher that $Diff(S^1 \times S^2)$ does not have the homotopy-type of a finite CW complex. –  Ryan Budney Apr 8 '13 at 17:12

$\newcommand{\Diff}{\operatorname{Diff}}$ First let $M$ be closed. Then $\Diff^\infty(M)$ is locally modelled on the space of smooth vector fields in $M$ which is a Frechet space. All Frechet spaces are homeomorphic. So this is reduced to the case you described. The same is true for $\Diff^k(M)$.

Let $M$ be open (non-compact without boundary). Then $\Diff^\infty(M)$ is open in $C^\infty(M,M)$ in the Whitney $C^\infty$ topology. But it is not locally contractible in this topology, and the arc connected component of the identity is contained in the group $\Diff^\infty_c(M)$ of diffeomorphisms which differ from the identity only on a compact set. The group $\Diff^\infty_c(M)$ is a regular Lie group locally modeled on the space of $\mathfrak X_c(M)$ smooth vector fields with compact support which is a nuclear (LF)-space. Similar for $\Diff^k(M)$. I do not remember whether $\mathfrak X_c(M)$ is a CW-complex.

See [Peter W. Michor: Manifolds of differentiable mappings. Shiva Mathematics Series 3, Shiva Publ., Orpington, (1980), iv+158 pp., MR 83g:58009, ZM 433.58001]
for lots of details on this, including the case of manifolds with boundary and with corners. See also section 41 of [here].

Edit: Vidit Nanda pointed out that the compact open $C^1$ or $C^\infty$ topology was asked for. This is not a good topology: In general it is not locally contractible, so no manifold modelled on topological vector spaces. Also $\Diff^k(M)$ is not open in $C^k(M,M)$ for any $k\ge 1$ if $M$ is not compact.

I once tried to describe a setting where $\Diff^\infty(M)$ with the compact $C^\infty$ topology would be a Lie group: smooth manifolds based on smooth curves instead of charts; but you need a lot of other structures like a geodesic structure, locally convex spaces as tangent spaces, and parallel transport. The resulting category of manifolds is monodially closed, and the the manifolds with finite dimensional (or even Banach) tangent spaces are exactly the usual ones. The topology is the final topology with respect to the smooth curves, and for $\Diff(M)$ it is indeed the compact $C^\infty$-topology. The theory is horrendibly complicated, and nobody ever used it. I have no idea whether this helps for the quest for CW-complexes. See:

  • Peter W. Michor: A convenient setting for differential geometry and global analysis, I, II. Cahiers Topologie Geometrie Differentielle 25 (1984), 63--109, 113--178. (pdf of part 1) (pdf of part 2)

Liviu Nicolaescu's answer seems to be the best for your question.

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Peter, isn't the Whitney topology very different from the compact-open topology which Ricardo is asking about? –  Vidit Nanda Apr 8 '13 at 14:32
    
@Peter: Thank you for the interesting results and references. I had actually taken a look at your book "Manifolds of differentiable mappings" when thinking about this question. Since I could not find the answer there, I posted my question here. –  Ricardo Andrade Apr 8 '13 at 16:29
    
Peter: thank you for addressing the compact open topology. Also, +1 for "horrendibly". –  Vidit Nanda Apr 8 '13 at 19:10

[Edit: Allen Hatcher posted an answer while I was writing this one. Both answers seem to use similar ideas. I will leave my answer here anyway.]

$\newcommand{\Diff}{\operatorname{Diff}}$$\newcommand{\ZZ}{\mathbb{Z}}$$\newcommand{\RR}{\mathbb{R}}$$\newcommand{\CC}{\mathbb{C}}$$\newcommand{\connsum}{\mathbin{\#}}$$\newcommand{\id}{\mathrm{id}}$$\newcommand{\set}[1]{\lbrace #1 \rbrace}$Reading the answer by Peter Michor gave me an idea for a counter-example, which I explain below. The question still remains as to whether the result is valid for the interior of a compact manifold.

Claim: The path components of a space homotopy equivalent to a CW-complex are all open.

A homotopy equivalence induces a bijection on path components. Consequently, if a space $X$ is homotopy equivalent to a space $Y$, and the path components of $Y$ are open in $Y$, then the path components of $X$ are open in $X$. Finally, observe that any CW-complex is locally path connected, and thus its path components are open.

Construction of the counter-example

It thus suffices to present a smooth manifold $M$ without boundary such that the path components of $\Diff(M)$ are not open. Define $M$ to be the open submanifold of $\RR \times S^1$ given by $$ M = (\RR \times S^1) \setminus (\ZZ \times \set{1}) $$ where $1\in S^1 \subset\CC$. One can also view $M$ as a connected sum of infinitely many punctured spheres $P=S^2\setminus\set{(1,0,0)}$: $$ M \cong \, \cdots \connsum P \connsum P \connsum P \connsum \cdots $$

Proof that the path component of $\id_M$ is not open in $\Diff(M)$

Pick a neighbourhood $U$ of $\id_M$ in $\Diff(M)$. I will describe a diffeomorphism $\varphi\in U$ which is not in the path component of $\id_M$ in $\Diff(M)$, thus concluding the proof. By definition of the compact-open topology on $\Diff(M)$, there exists a compact subspace $K$ of $M$ such that a given diffeomorphism of $M$ is in $U$ if it is the identity on $K$. Let $n$ be a positive integer large enough so that $K\subset [-n,n]\times S^1$. Now we define the required diffeomorphism $\varphi$ of $M$: $$ \varphi(t,x) = \bigl( t , e^{i\cdot\theta(t-n)} x \bigr) $$ where $\theta:\RR\to\RR$ is any smooth function which is identically zero on $(-\infty,0]$, and equals $2\pi$ on $[1,+\infty)$. Viewing $M$ as the connected sum of infinitely many punctured spheres, the diffeomorphism $\varphi$ is the result of applying a Dehn twist at the junction cylinder joining the spheres numbered $n$ and $n+1$.

It is easy to see that $\varphi:M\to M$ is not even homotopic to the identity: in fact, the homomorphism induced by $\varphi$ on $\pi_1(M)$ (based at the point $(n,-1)$) is not conjugate to the identity. Here is a simple way to see this:

  1. The group $\pi_1 M$ is free on infinitely many generators.

  2. The homomorphism $\pi_1 \varphi$ coincides with the identity on all loops in $M$ contained in $(-\infty,n]\times S^1$. In particular, $\pi_1 \varphi$ fixes two distinct (actually, infinitely many) free generators of $\pi_1 M$.

  3. The homomorphism $\pi_1 \varphi$ is not the identity homomorphism on $\pi_1 M$: a loop going once around the puncture $(n+1,1)\in\RR\times S^1$ is not fixed by $\pi_1 \varphi$.

  4. In a free group, conjugation by a non-identity element can fix at most one of the free generators.

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