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.
Best Answer
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.