3
$\begingroup$

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).

Thanks.

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

$\text{Diff}^+(S^1)$ 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 $x$.

Define

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

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 $H_1(z) = z^k$.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ Yes. Being smooth is a local property, and $e$ is a local diffeomorphism. $\endgroup$ – Paul Frost Aug 2 '18 at 9:57
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.