Let $X=\text{Diff}^+(S^1)=$ set of all orientation preserving diffeomorphism of $S^1$. I want to prove that this is path connected.

I tried with the following argument:

Let us consider the natural action of $O(2,\mathbb{R})$ on $S^1$. Now define the map from $O(2)$ to $X$ defined as $A\mapsto \phi $ where $\phi$ is the action i.e. $\phi(x)=Ax$. Now, I am having problem in showing the above map is onto (that is my guess).


  • $\begingroup$ Not all diffeomorphisms of $S^1$ are rotations. $\endgroup$ – Travis Jul 17 '18 at 1:46
  • $\begingroup$ Not only is it not onto, but its codomain can't be taken to be $\text{Diff}^{+}$ because the reflections in $O(2)$ reverse orientation. You want $SO(2)$. $\endgroup$ – Qiaochu Yuan Jul 17 '18 at 1:46
  • $\begingroup$ How will we prove it is path connected? $\endgroup$ – XYZABC Jul 17 '18 at 2:52

$\text{Diff}^+(S^1)$ path connected means that for each $f \in \text{Diff}^+(S^1)$ there exists a homotopy $H : S^1 \times I \to S^1$ such that $H_t \in \text{Diff}^+(S^1)$ for all $t \in I$, where $H_t(x) = H(x,t)$, and $H_0 = f$, $H_1 = id$. Here $I = [0,1]$.

Consider the covering $e : \mathbb{R} \to S^1, e(x) = e^{2\pi i x}$. The map $f \circ e : \mathbb{R} \to S^1$ has a lift $F : \mathbb{R} \to \mathbb{R}$ (i.e. $e \circ F = f \circ e$). It is an orientation preserving diffeomorphism such that $F(x + 1) = F(x) + 1$ for all $x$.


$$\Gamma : \mathbb{R} \times I \to \mathbb{R}, \Gamma(x,t) = (1-t)F(x) + tx .$$

We have $\Gamma_t'(x) = (1-t)F'(x) + t > 0$. Therefore each $\Gamma_t$ is an orientation preserving diffeomorphisms such that $\Gamma_t(x + 1) = \Gamma_t(x) + 1$ for all $x$. Therefore $\Gamma$ induces a unique homotopy $H : S^1 \times I \to S^1$ such that $e \circ \Gamma = H \circ (e \times id_I)$. We have $H_0 = f, H_1 = id$ and all $H_t$ are orientation preserving diffeomorphisms.

Added on request:

$e : \mathbb{R} \to S^1$ is a smooth covering, $F$ is a lift of the smooth map $f \circ e$, hence $F$ is a smooth map. Consider any $x \in \mathbb{R}$. We have $e(F((x,x+1))) = f(e((x,x+1)) = f(S^1 \backslash \{ e(x) \}) = S^1 \backslash \{ f(e(x)) \}$ because $f$ is a bijection. $F((x,x+1))$ is a connected subset of $\mathbb{R}$, i.e. an interval. It is mapped by $e$ onto $S^1 \backslash \{ f(e(x)) \}$. This is only possible if $F((x,x+1)))$ is itself an open interval of length $1$. Therefore

(1) $F((x,x+1)) = (x',x'+1)$ for some $x' \in \mathbb{R}$. Note that necessarily $e(x') = f(e(x))$.

Using the charts $e_{(x,x+1)}^{-1}$ and $e_{(x',x'+1)}^{-1}$ for $S^1$, we see that

(2) $F \mid_{(x,x+1)} = e_{(x',x'+1)}^{-1} \circ f \mid_{e(x,x+1)} \circ \phantom{.} e_{(x,x+1)} : (x,x+1) \to (x',x'+1)$ is a smooth bijection having everywhere a positive derivative since $f$ is an orientation preserving diffeomorphism. In particular, $F \mid_{(x,x+1)}$ is a strictly increasing smooth bijection.

(1), (2) and continuity show us

(3) $F(x) = x', F(x+1) = x'+1$

This implies

(4) $F(x+1) = F(x) + 1$ and $F((x,x+1)) = (F(x),F(x)+1)$.

This suffices to see that $f$ is bijective. Since $F$ has a positive derivative on each open interval of length $1$, it has a positive derivative on all of $\mathbb{R}$. Therefore

(5) $F$ is an orientation preserving diffeomorphism.

Final remark:

Although not needed her, let us consider an arbitrary continuous map $f : S^1 \to S^1$. As above we find $F : \mathbb{R} \to \mathbb{R}$ such that $e \circ F = f \circ e$. We infer $e(F(x+1)) = e(F(x))$ which is equivalent to $F(x+1) - F(x) \in \mathbb{Z}$. But now the function $g(x) = F(x+1) - F(x)$ is continuous on $\mathbb{R}$ with values in $ \mathbb{Z}$, and this is possible only if $g$ is constant. Therefore there exists $k \in \mathbb{Z}$ such that $F(x+1) = F(x) + k$ for all $x$. This shows that $F$ is completey determined by $F \mid_{[0,1]}$, but that's irrelevant here. We can define a homotopy $\Gamma : \mathbb{R} \times I \times \mathbb{R}, \Gamma(x,t) = (1-t)F(x) + tkx$. Then $\Gamma_0 = F$, $\Gamma_1(x) = kx$ and each $\Gamma_t$ has the property that $\Gamma_t(x+1) = \Gamma_t(x) + k$. Hence $\Gamma$ induces a unique homotopy $H : S^1 \times I \to S^1$ such that $H_0 = f$ and $H_1(z) = z^k$.

  • $\begingroup$ Will you please tell me why $F$is an orientation preserving diffeomorphism and why $F(x+1)=F(x)+1$ ? $\endgroup$ – XYZABC Jul 19 '18 at 3:15
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    $\begingroup$ Yes. Being smooth is a local property, and $e$ is a local diffeomorphism. $\endgroup$ – Paul Frost Aug 2 '18 at 9:57
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    $\begingroup$ @SachchidanandPrasad This is impossible in the range $t \in [0,1]$. $\Gamma'_t(x)$ is a convex combination of the two numbers $F'(x) >0$ and $1$. Alternatively look at your equation. It has no solution $t$ if $F'(x) = 1$, a negative solution $t$ if $F'(x) < 1$ and a solution $t > 1$ if $F'(x) > 1$. $\endgroup$ – Paul Frost Aug 2 '18 at 10:53
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    $\begingroup$ Let $B^A$ be the set of all continuous functions $A \to B$ endowed with the compact-open topology. Then the exponential correspondence establishes a bijection $e : Z^{X \times Y} \to (Z^Y)^X $, where $f' = e(f)$ is given by $f'(x)(y) = f(x,y)$, provided $Y$ is locally compact. It is even a homeomorphism, but that is irrelevant here. In particular, for $X = I$ and locally compact $Y$ this yields a bijection between homotopies $H : I \times Y \to Z$ and paths $\omega : I \to Z^Y$. Apply this for $Y = Z = S^1$, noting that $\text{Diff}^+(S^1)$ is subspace of $(S^1)^{S^1}$. $\endgroup$ – Paul Frost Aug 4 '18 at 22:40
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    $\begingroup$ If $A$ is compact and $B$ is a metric space, then the compact-open topology on $B^A$ agrees with the topology induced by the supremum-metric on the set $B^A$. This is the topology of uniform convergence. $\endgroup$ – Paul Frost Aug 4 '18 at 22:48

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