Exercise 7, § 4

(a) Let $(\lambda, x) \mapsto \lambda.x$ be a law of continuous left operation of $\mathbf{R}$ on $\mathbf{T}$. Then one of the two following holds: (1) either there exists a fixed point and the stabilizer of every point is either $\mathbf{R}$ or $\{0\}$; (2) or there exists a point whose stabilizer is an infinite discrete subgroup of $\mathbf{R}$ and $\mathbf{R}$ operates transitively on $\mathbf{T}$. (If $\{\lambda \in \mathbf{R} | \lambda.x = x\} = \{0\}$, the orbit of $x$ is homeomorphic to $\mathbf{R}$ and has frontier points $\lim_{\lambda \to +\infty} \lambda.x$ which are fixed. If $\{\lambda \in \mathbf{R} | \lambda.x = x\} = \mathbf{Z}a$ with $a \neq 0$, the orbit of $x$ is homeomorphic to $\mathbf{T}$ and $\mathbf{R}$ is transitive.)

(b) If $f$ is a homeomorphism of $\mathbf{T}$ onto itself and $k$ an integer $\geqslant 1$, a point $x \in \mathbf{T}$ is called periodic of period $k$ for $f$ if $f^k(x) = x$ and $f^h(x) \neq x$ for $1 \leqslant h < k$. In the notation of (a), let $f_\lambda$ be the homeomorphism $x \mapsto \lambda.x$. Then the set of periodic points of period $k > 1$ for $f_\lambda$ is empty or equal to $\mathbf{T}$. (If there exists a periodic point of period $k > 1$ for $f_\lambda$, its stabilizer is distinct from $\{0\}$ and $\mathbf{R}$, hence $\mathbf{R}$ operates transitively on $\mathbf{T}$ and all the points of $\mathbf{T}$ are periodic of period $k$ for $f_\lambda$.)

(c) Let $\Omega$ be a neighbourhood of $e$ in $\mathrm{Diff}^\infty(\mathbf{T})$ (Differentiable and Analytic Manifolds, R, 15.3.8, Note (1)). There exist an integer $k > 1$ and $f \in \Omega$ without fixed point such that the set of periodic points of period $k$ for $f$ is neither empty nor equal to $\mathbf{T}$. (Let $p : \mathbf{R} \to \mathbf{T}$ be the canonical mapping. Let $\rho$ be the mapping $p(x) \mapsto p\left(x + \frac{2\pi}{k}\right)$ of $\mathbf{T}$ into $\mathbf{T}$. Let $\Omega'$ be a neighbourhood of $e$ in $\mathrm{Diff}^\infty(\mathbf{T})$ such that ${\Omega'}^2 \subset \Omega$. For sufficiently large $k$, $\rho \in \Omega'$. On the other hand, let $\sigma$ be a mapping of $\mathbf{T}$ into $\mathbf{T}$ such that $\sigma(y) = y$ for

and $\sigma(p(x)) = p(x + \lambda(x))$ for $x \in ]0, \frac{2\pi}{k}[$, where can be chosen such that $\sigma \in \Omega'$. Then $f = \rho \circ \sigma$ has the required properties.)

Such an $f$ cannot, by (b), be in the image of a continuous morphism of $\mathbf{R}$ into $\mathrm{Diff}^\infty(\mathbf{T})$.

(d) Let $Y$ be a compact differentiable manifold of dimension $\geqslant 1$. Every neighbourhood of $e$ in $\mathrm{Diff}^\infty(Y)$ contains an element $h$ with the following property: $h$ belongs to the image of no continuous morphism of $\mathbf{R}$ into $\mathrm{Diff}^0(Y)$. (The manifold $Y$ contains an open submanifold of the form $\mathbf{T} \times D$, where $D$ is the open Euclidean ball of radius 1 and centre 0 in $\mathbf{R}^n$ ($n = \dim Y - 1$). Let $g \in \mathrm{Diff}^\infty(D)$ be such that $g(0) = 0$, $\|g(x)\| > \|x\|$ for $0 < \|x\| < \frac{1}{2}$ and $g(x) = x$ for $\|x\| \geqslant \frac{1}{2}$ and $g$ very close to $e$ in $\mathrm{Diff}^\infty(D)$. Let $f$ be as in (c), and, for $0 < t < 1$, let $f_t : \mathbf{T} \to \mathbf{T}$ be defined as follows: for $x \in \mathbf{T}$, $y \in p^{-1}(x)$, $z \in p^{-1}(f(x))$ and $|z - y| < \pi$, we write $f_t(x) = p(ty + (1 - t)z)$. Then there exists $h \in \mathrm{Diff}^\infty(Y)$ such that $h(x, y) = (f_{2\|y\|}(x), g(y))$ for $x \in \mathbf{T}$, $y \in D$ and $h(u) = u$ for $u \in Y - (\mathbf{T} \times D)$; moreover, we can choose $f$ and $g$ such that $h$ is arbitrarily close to $e$ in $\mathrm{Diff}^\infty(Y)$. The points of $\mathbf{T} \times \{0\}$ are the $y \in Y$ such that when $n$ tends to $+\infty$, $h^{kn}(x)$ tends to a point not fixed under $h$ (in fact it is periodic of period $k$). Therefore, every homeomorphism of $Y$ onto $Y$ which commutes with $h$ leaves $\mathbf{T} \times \{0\}$ invariant. Then apply (c).)

(e) Let $Y$ be a compact differentiable manifold of dimension $\geqslant 1$. There do not exist a Lie group $G$ and continuous morphisms $\mathrm{Diff}^\infty(Y) \to G$, $G \to \mathrm{Diff}^0(Y)$ whose composition is the canonical injection of $\mathrm{Diff}^\infty(Y)$ into $\mathrm{Diff}^0(Y)$. (Use (d).)