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I am given to understand that the group of diffeomorphisms of the unit circle, $\operatorname{Diff}(\mathbb{S}^1)$, has two connected components, $\operatorname{Diff}^+(\mathbb{S}^1)$ and $\operatorname{Diff}^-(\mathbb{S}^1)$, the diffeomorphisms that preserve or reverse the canonical (counterclockwise) orientation, respectively.

Question 1: How does one prove that?

Question 2: Given $\Phi, \Psi \in \operatorname{Diff}^+(\mathbb{S^1})$, can one construct an explicit path joining them?

I'd be satisfied already with a proof that we can join $\Psi, \Phi \in \operatorname{Diff}^+ (\mathbb{S}^1)$, without giving the path explicitly. I tried the obvious path, $t \mapsto \dfrac{t\Phi + (1-t)\Psi}{|t\Phi + (1-t)\Psi|}$, but that doesn't seem to work.

Thanks.

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I don't know how relevant it is, but in case anyone is wondering, consider the compact-open topology on $Diff(\mathbb{S}^1)$. –  student Dec 4 '11 at 23:08
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It is enough to connect everything in the positive part to the identity. To do that, pick a diffeo and lift it to a map $\mathbb R\to\mathbb R$. Then interpolate with a map given by a translation. Then descend back to the circle. –  Mariano Suárez-Alvarez Dec 4 '11 at 23:12
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@Leandro It seems pretty relevant! –  Dylan Moreland Dec 5 '11 at 0:48

1 Answer 1

up vote 13 down vote accepted

Question 2:

As Mariano points out, we can lift $\Phi \in \operatorname{Diff}^+{(\mathbb S^1)}$ uniquely to a diffeomorphism $\phi : \mathbb{R} \to \mathbb{R}$ such that $\phi(0) \in [0,1)$ and $\phi(x+1) = \phi(x) + 1$ for all $x \in \mathbb{R}$. Lift $\Psi$ similarly to $\psi$. For $t \in [0,1]$ the map $\gamma_t = (1-t) \phi + t\psi$ is a diffeomorphism $\mathbb{R} \to \mathbb{R}$ (it is strictly monotonically increasing because $ \gamma_{t}^\prime(x) \gt 0$ for all $x \in \mathbb{R}$ and all $t \in [0,1]$) such that $\gamma_t(0) \in [0,1)$ and $\gamma_t(x+1) = \gamma_t(x)+1$. Thus $\gamma_t$ descends to a diffeomorphism $\Gamma_t \in \operatorname{Diff}^+(\mathbb{S}^1)$, $\Gamma_0 = \Phi$ and $\Gamma_1=\Psi$. It is straightforwad to check that $t \mapsto \Gamma_t$ is a continuous path.

What we exploited here is that we have a short exact sequence (in fact a central extension) $$ 0 \to \mathbb{Z} \to \operatorname{Diff}_{\mathbb{Z}}{(\mathbb{R})} \to \operatorname{Diff}^+{(\mathbb{S}^1)} \to 1 $$ where $\operatorname{Diff}_{\mathbb{Z}}{(\mathbb{R})}$ denotes the group of diffeomorphisms $\mathbb{R} \to \mathbb{R}$ commuting with the shift $\tau(x) = x+1$ and $\mathbb{Z} = \langle \tau \rangle$ and the lift $\Phi \mapsto \phi$ is a section of this central extension.

Question 1:

It is clear that a diffeomorphism $\mathbb{S}^1 \to \mathbb{S}^1$ either preserves or reverses orientation and that the orientation-preserving diffeomorphisms $\operatorname{Diff}^+{(\mathbb{S}^1)}$ form a normal subgroup of $\operatorname{Diff}{(\mathbb{S}^1)}$. Now simply use the conjugation diffeomorphism $z \mapsto \bar{z}$ to see that $\operatorname{Diff}^+{(\mathbb{S}^1)}$ has index 2.


For an excellent introduction to $\operatorname{Homeo}^+{(\mathbb{S}^1)}$ and $\operatorname{Diff}^+{(\mathbb{S}^1)}$ and their subgroups I recommend Étienne Ghys's article Groups acting on the circle, Enseign. Math. (2) 47 (2001), no. 3–4, 329–407, MR2111644. The short exact sequence mentioned above plays a central rôle in the theory.

For more on the diffeomorphism group of the circle, I recommend consulting the work of Andrés Navas, and, of course, the classic article:

Michael R. Herman, Sur la conjugaison différentiable des difféomorphismes du cercle à des rotations, Publications Mathématiques de l'IHÉS, 49 (1979), p. 5–233.

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Excellent answer, and I'll certainly look up those references. Thanks a lot. –  student Dec 5 '11 at 0:59
    
What does $0$ mean in your exact sequence? –  Alexei Averchenko Dec 6 '11 at 5:20
    
@Alexei: $0$ as $0 \in \mathbb{Z}$, which is the neutral element and can be identified with the trivial group. –  t.b. Dec 6 '11 at 5:22

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