# The Gottschalk-Hedlund theorem, cocycles, and small divisors

Jordan Bell
July 23, 2014

## 1 Introduction

This note consists of my working through details in the paper Resonances and small divisors by Étienne Ghys.1 Aside from containing mathematics, Ghys makes thoughtful remarks about the history of physics, unlike the typically thoughtless statements people make about the Ptolemaic system. He insightfully states “Kepler’s zeroth law”: “If the orbit of a planet is bounded, then it is periodic.” I can certainly draw a three dimensional bounded curve that is not closed, but that curve is not the orbit of a planet. It is also intellectually lazy to scorn Kepler’s correspondence between orbits and the Platonic solids (“Kepler’s fourth law”).

## 2 Almost periodic functions

Suppose that $f:\mathbb{R}\to\mathbb{C}$ is continuous. For $\epsilon>0$, we call $T\in\mathbb{R}$ an $\epsilon$-period of $f$ if

 $|f(t+T)-f(t)|<\epsilon,\qquad t\in\mathbb{R}.$

$T$ is a period of $f$ if and only if it is an $\epsilon$-period for all $\epsilon>0$.

We say that $f$ is almost periodic if for every $\epsilon>0$ there is some $M_{\epsilon}>0$ such that if $I$ is an interval of length $>M_{\epsilon}$ then there is an $\epsilon$-period in $I$.

If $f$ is periodic, then there is some $M>0$ such that if $I$ an interval of length $>M$ then at least one multiple $T$ of $M$ lies in $I$, and hence for any $t\in\mathbb{R}$ we have $f(t+T)-f(t)=f(t)-f(t)=0$. Thus, for every $\epsilon>0$, if $I$ is an interval of length $>M$ then there is an $\epsilon$-period in $I$. Therefore, with a periodic function, the length of the intervals $I$ need not depend on $\epsilon$, while for an almost periodic function they may.

## 3 The Gottschalk-Hedlund theorem

The Gottschalk-Hedlund theorem is stated and proved in Katok and Hasselblatt.22 2 Anatole Katok and Boris Hasselblat, Introduction to the Modern Theory of Dynamical Systems, p. 102, Theorem 2.9.4. The following case of the Gottschalk-Hedlund theorem is from Ghys. We denote by

 $\pi_{1}:\mathbb{R}/\mathbb{Z}\times\mathbb{R}\to\mathbb{R}/\mathbb{Z},\qquad% \pi_{2}:\mathbb{R}/\mathbb{Z}\times\mathbb{R}\to\mathbb{R}$

the projection maps.

###### Theorem 1 (Gottschalk-Hedlund theorem).

Suppose that $u:\mathbb{R}/\mathbb{Z}\to\mathbb{R}$ is continuous, that

 $\int_{0}^{1}u(x)dx=0,$

that $x_{0}\in\mathbb{R}/\mathbb{Z}$, and that $\alpha$ is irrational. If there is some $C$ such that

 $\left|\sum_{k=0}^{n}u(x_{0}+k\alpha)\right|\leq C,\qquad n\geq 0,$ (1)

then there is a continuous function $v:\mathbb{R}/\mathbb{Z}\to\mathbb{R}$ such that

 $u(x)=v(x+\alpha)-v(x),\qquad x\in\mathbb{R}/\mathbb{Z}.$
###### Proof.

Say there is some $C>0$ satisfying (1). Define $g:\mathbb{R}/\mathbb{Z}\times\mathbb{R}\to\mathbb{R}/\mathbb{Z}\times\mathbb{R}$ by

 $g(x,y)=(x+\alpha,y+u(x)),\qquad x\in\mathbb{R}/\mathbb{Z}.$

For $n\geq 0$,

 $g^{n}(x_{0},0)=\Big{(}x_{0}+n\alpha,\sum_{k=0}^{n}u(x_{0}+k\alpha)\Big{)}$

The set $\{g^{n}(x_{0},0):n\geq 0\}$, namely the orbit of $(x_{0},0)$ under $g$, is contained in $\mathbb{R}/\mathbb{Z}\times[-C,C]$. Let $K$ be the closure of this orbit. Because $K$ is a metrizable topological space, for $(x,y)\in K$ there is a sequence $a(n)$ such that $g^{a(n)}(x_{0},0)\to(x,y)$. As $g$ is continuous we get $g^{a(n)+1}(x_{0},0)\to g(x,y)$, which implies that $g(x,y)\in K$. This shows that $K$ is invariant under $g$. Let $\mathscr{K}$ be the collection of nonempty compact sets contained in $K$ and invariant under $g$. Thus $K\in\mathscr{K}$, so $\mathscr{K}$ is nonempty. We order $\mathscr{K}$ by $A\prec B$ when $A\subset B$. If $\mathscr{C}\subset\mathscr{K}$ is a chain, let $C_{0}=\bigcap_{C\in\mathscr{C}}C$. It follows from $K$ being compact that $C_{0}$ is nonempty, hence $C_{0}\in\mathscr{K}$ and is a lower bound for the chain $\mathscr{C}$. Since every chain in $\mathscr{K}$ has a lower bound in $\mathscr{K}$, by Zorn’s lemma there exists a minimal element $M$ in $\mathscr{K}$: for every $A\in\mathscr{K}$ we have $M\prec A$, i.e. $M\subset A$. To say that $M$ is invariant under $g$ means that $g(M)\subset M$, and $M$ being a nonempty compact set contained in $K$ implies that $g(M)$ is a nonempty compact set contained in $K$, hence by the minimality of $M$ we obtain $g(M)=M$.

The set $M$ is nonempty, so take $(x,y)\in M$. Because $M$ is invariant under $g$, $\{g^{n}(x,y):n\geq 0\}\subset M$. The set

 $\pi_{1}\{g^{n}(x,y):n\geq 0\}=\{x+n\alpha:n\geq 0\}$

is dense in $\mathbb{R}/\mathbb{Z}$, hence $\pi_{1}(M)$ is dense in $\mathbb{R}/\mathbb{Z}$. Moreover, $M$ being compact implies that $\pi_{1}(M)$ is closed, so $\pi_{1}(M)=\mathbb{R}/\mathbb{Z}$.

For $t\in\mathbb{R}$, define $\tau_{t}:\mathbb{R}/\mathbb{Z}\times\mathbb{R}\to\mathbb{R}/\mathbb{Z}\times% \mathbb{R}$ by $\tau_{t}(x,y)=(x,y+t)$. For any $t$,

 $\tau_{t}\circ g(x,y)=\tau_{t}(x+\alpha,y+u(x))=(x+\alpha,y+u(x)+t)=g(x,y+t)=g% \circ\tau_{t}(x,y),$

so $\tau_{t}\circ g=g\circ\tau_{t}$. Hence, if $A\subset\mathbb{R}/\mathbb{Z}\times\mathbb{R}$ and $g(A)\subset A$, then $g(\tau_{t}(A))=\tau_{t}\circ g(A)\subset\tau_{t}(A)$, namely, if $A$ is invariant under $g$ then $\tau_{t}(A)$ is invariant under $g$. Therefore $\tau_{t}(M)$ is invariant under $g$, and so $M\cap\tau_{t}(M)$ is invariant under $g$. This intersection is compact and is contained in $K$, so either $M\cap\tau_{t}(M)=\emptyset$ or by the minimality of $M$, $M\cap\tau_{t}(M)=M$. Suppose by contradiction that for some nonzero $t$, $M\cap\tau_{t}(M)=M$. Then using $g(M)=M$ we get $\tau_{t}(M)=M$, and hence for any positive integer $k$ we have $\tau_{kt}(M)=\tau_{t}^{k}(M)=M$. But because $M$ is compact, $\pi_{2}(M)$ is contained in some compact interval $I$, and then there is some positive integer $k$ such that $\pi_{2}(\tau_{kt}(M))$ is not contained in $I$, a contradiction. Therefore, when $t\neq 0$ we have $M\cap\tau_{t}(M)=\emptyset$. Let $x\in\mathbb{R}/\mathbb{Z}$. If there were distinct $y_{1},y_{2}\in\mathbb{R}$ such that $(x,y_{1}),(x,y_{2})\in M$, then with $t=y_{2}-y_{1}\neq 0$ we get $\tau_{t}(x,y_{1})=(x,y_{2})\in M$, contradicting $M\cap\tau_{t}(M)=\emptyset$. This shows that for each $x\in\mathbb{R}/\mathbb{Z}$ there is a unique $y\in\mathbb{R}$ such that $(x,y)\in M$, and we denote this $y$ by $v(x)$, thus defining a function $v:\mathbb{R}/\mathbb{Z}\to\mathbb{R}$. Then $M$ is the graph of $v$, and because $M$ is compact, it follows that the function $v$ is continuous. Let $(x,v(x))\in M$. As $M$ is invariant under $g$,

 $(x+\alpha,v(x)+u(x))=g(x,v(x))\in M,$

and as $M$ is the graph of $v$ we get $v(x)+u(x)=v(x+\alpha)$ and hence $v(x+\alpha)-v(x)=u(x)$, completing the proof. ∎

## 4 Cohomology

In this section I am following Tao.33 3 Terence Tao, Cohomology for dynamical systems, http://terrytao.wordpress.com/2008/12/21/cohomology-for-dynamical-systems/ Suppose that a group $(G,\cdot)$ acts on a set $X$ and that $(A,+)$ is an abelian group. A cocycle is a function $\rho:G\times X\to A$ such that

 $\rho(gh,x)=\rho(h,x)+\rho(g,hx),\qquad g,h\in G,\quad x\in X.$ (2)

If $F:X\to A$ is a function, we call the function $\rho(g,x)=F(gx)-F(x)$ a coboundary. This satisfies

 $\rho(gh,x)-\rho(g,hx)=F((gh)x)-F(x)-F(g(hx))+F(hx)=F(hx)-F(x)=\rho(h,x),$

showing that a coboundary is a cocycle. We now show how to fit the notions of cocycle and coboundary into a general sitting of cohomology. We show that they correspond respectively to a $1$-cocycle and a $1$-coboundary.

For $n\geq 0$, an $n$-simplex is an element of $G^{n}\times X$, i.e., a thing of the form $(g_{1},\ldots,g_{n},x)$, for $g_{1},\ldots,g_{n}\in G$ and $x\in X$. We denote by $C_{n}(G,X)$ the free abelian group generated by the collection of all $n$-simplices, and an element of $C_{n}(G,X)$ is called an $n$-chain. In particular, the elements of $C_{0}(G,X)$ are formal $\mathbb{Z}$-linear combinations of elements of $X$. For $n<0$, we define $C_{n}(G,X)$ to be the trivial group.

For $n>0$, we define the boundary map $\partial:C_{n}(G,X)\to C_{n-1}(G,X)$ by

 $\displaystyle\partial(g_{1},\ldots,g_{n},x)$ $\displaystyle=$ $\displaystyle(g_{1},\ldots,g_{n-1},g_{n}x)$ $\displaystyle+\sum_{k=1}^{n-1}(-1)^{n-k}(g_{1},\ldots,g_{k-1},g_{k}g_{k+1},g_{% k+2},\ldots,g_{n},x)$ $\displaystyle+(-1)^{n}(g_{2},\ldots,g_{n},x).$

For $n\leq 0$ we define $\partial:C_{n}(G,X)\to C_{n-1}(G,X)$ to be the trivial map. If $n\leq 1$ then of course $\partial^{2}=0$. If $n\geq 2$, one writes out $\partial^{2}(g_{1},\ldots,g_{n},x)$ and checks that it is equal to $0$, and hence that $\partial^{2}=0$. Thus the sequence of abelian groups $C_{n}(G,X)$ and the boundary maps $\partial:C_{n}(G,X)\to C_{n-1}(G,X)$ are a chain complex.

We denote the kernel of $\partial:C_{n}(G,X)\to C_{n-1}(G,X)$ by $Z_{n}(G,X)$, and elements of $Z_{n}(G,X)$ are called $n$-cycles. We denote the image of $\partial:C_{n+1}(G,X)\to C_{n}(G,X)$ by $B_{n}(G,X)$, and elements of $B_{n}(G,X)$ are called $n$-boundaries. Because $\partial^{2}=0$, an $n$-boundary is an $n$-cycle. $Z_{n}(G,X)$ and $B_{n}(G,X)$ are abelian groups and $B_{n}(G,X)$ is contained in $Z_{n}(G,X)$, and we write

 $H_{n}(G,X)=Z_{n}(G,X)/B_{n}(G,X),$

and call $H_{n}(G,X)$ the $n$th homology group.

We define $C^{n}(G,X,A)=\mathrm{Hom}(C_{n}(G,X),A)$, which is an abelian group. Elements of $C^{n}(G,X,A)$ are called $n$-cochains. That is, an $n$-cochain is a group homomorphism $C_{n}(G,X)\to A$. Because $C_{n}(G,X)$ is a free abelian group generated by the collection of all $n$-simplices, an $n$-cochain is determined by the values it assigns to $n$-simplices. We thus identity $n$-cochains with functions $G^{n}\times X\to A$.

We define the coboundary map $\delta:C^{n-1}(G,X,A)\to C^{n}(G,X,A)$ by

 $(\delta F)(c)=F(\partial c),\qquad F\in C^{n-1}(G,X,A),c\in C_{n}(G,X).$

Explicitly, for $F\in C^{n-1}(G,X,A)$ and for an $n$-simplex $(g_{1},\ldots,g_{n},x)$,

 $\displaystyle(\delta F)(g_{1},\ldots,g_{n},x)$ $\displaystyle=$ $\displaystyle F(\partial(g_{1},\ldots,g_{n},x))$ $\displaystyle=$ $\displaystyle F(g_{1},\ldots,g_{n-1},g_{n}x)$ $\displaystyle+\sum_{k=1}^{n-1}(-1)^{n-k}F(g_{1},\ldots,g_{k-1},g_{k}g_{k+1},g_% {k+2},\ldots,g_{n},x)$ $\displaystyle+(-1)^{n}F(g_{2},\ldots,g_{n},x).$

For $F\in C^{n-2}(G,X,A)$, write $G=\delta F$ and take and $c\in C_{n}(G,X)$. Then,

 $(\delta^{2}F)(c)=(\delta G)(c)=G(\partial c)=(\delta F)(\partial c)=F(\partial% ^{2}c)=F(0)=0,$

showing that $\delta^{2}=0$. Thus the sequence of abelian groups $C^{n}(G,X,A)$ and the coboundary maps $\delta:C^{n-1}(G,X,A)\to C^{n}(G,X,A)$ are a cochain complex.

We denote the kernel of $\delta:C^{n}(G,X,A)\to C^{n+1}(G,X,A)$ by $Z^{n}(G,X,A)$, and elements of $Z^{n}(G,X,A)$ are called $n$-cocycles. We denote the image of $\delta:C^{n-1}(G,X,A)\to C^{n}(G,X,A)$ by $B^{n}(G,X,A)$, and elements of $B^{n}(G,X,A)$ are called $n$-coboundaries. Because $\delta^{2}=0$, an $n$-coboundary is an $n$-cocycle. $Z^{n}(G,X,A)$ and $B^{n}(G,X,A)$ are abelian groups and $B^{n}(G,X,A)$ is contained in $Z^{n}(G,X,A)$, and we write

 $H^{n}(G,X,A)=Z^{n}(G,X,A)/B^{n}(G,X,A),$

which we call the $n$th cohomology group.

Take $n=1$. We identify $C^{1}(G,X,A)$, the group of $1$-chains, with functions $G\times X\to A$. For $\rho\in C^{1}(G,X,A)$, to say that $\rho$ is a $1$-cocycle is equivalent to saying that for any $(g,h,x)\in G^{2}\times X$, $(\delta\rho)(g,h,x)=0$, i.e. $\rho(g,hx)-\rho(gh,x)+\rho(h,x)=0$, i.e.

 $\rho(gh,x)=\rho(h,x)+\rho(g,hx).$

To say that $\rho$ is a $1$-coboundary is equivalent to saying that there is a $0$-chain $F$ (a function $X\to A$) such that $\rho=\delta F$, i.e., for any $(g,x)\in G\times X$,

 $\rho(g,x)=(\delta F)(g,x)=F(gx)-F(x).$

## 5 Small divisors

Suppose that $u:\mathbb{R}/\mathbb{Z}\to\mathbb{R}$ be $C^{\infty}$ and satisfies

 $\int_{0}^{1}u(x)dx=0.$

For each $n\in\mathbb{Z}$, let

 $\widehat{u}(n)=\int_{0}^{1}e^{-2\pi inx}u(x)dx.$

We have $\widehat{u}(0)=0$. For any $x\in\mathbb{R}/\mathbb{Z}$,

 $u(x)=\sum_{n\in\mathbb{Z}}\widehat{u}(n)e^{2\pi inx},$

and $\sum_{n\in\mathbb{Z}}|\widehat{u}(n)|<\infty$; for these statements to be true it suffices merely that $u$ be $C^{\beta}$ for some $\beta>\frac{1}{2}$.

Let $\alpha$ be irrational. We shall find conditions under which there exists a continuous function $v:\mathbb{R}/\mathbb{Z}\to\mathbb{R}$ such that

 $u(x)=v(x+\alpha)-v(x),\qquad x\in\mathbb{R}/\mathbb{Z}.$ (3)

Supposing that for each $x$, $v(x)$ is equal to its Fourier series evaluated at $x$ and that its Fourier series converges absolutely,

 $v(x)=\sum_{n\in\mathbb{Z}}\widehat{v}(n)e^{2\pi inx},$

then for each $x\in\mathbb{R}/\mathbb{Z}$,

 $v(x+\alpha)-v(x)=\sum_{n\in\mathbb{Z}}\widehat{v}(n)\left(e^{2\pi in(x+\alpha)% }-e^{2\pi inx}\right)=\sum_{n\in\mathbb{Z}}\widehat{v}(n)(e^{2\pi in\alpha}-1)% e^{2\pi inx}.$

Then using $u(x)=v(x+\alpha)-v(x)$ we obtain

 $\widehat{u}(n)=\widehat{v}(n)(e^{2\pi in\alpha}-1),\qquad n\in\mathbb{Z},$

or,

 $\widehat{v}(n)=\frac{\widehat{u}(n)}{e^{2\pi in\alpha}-1},\qquad n\neq 0;$ (4)

because $\alpha$ is irrational, the denominator of the right-hand side is indeed nonzero for $n\neq 0$. The value of $\widehat{v}(0)$ is not determined so far. We shall find conditions under which the continuous function $v$ we desire can be defined using (4).

A real number $\beta$ is said to be Diophantine if there is some $r\geq 2$ and some $C>0$ such that for all $q>0$ and $p\in\mathbb{Z}$,

 $\left|\beta-\frac{p}{q}\right|>Cq^{-r}.$ (5)

It is immediate that a Diophantine number is irrational. Suppose that $\alpha$ satisfies (5). Let $n\neq 0$ and let $p_{n}$ be the integer nearest $n\alpha$. Then

 $\displaystyle|e^{2\pi in\alpha}-1|$ $\displaystyle=$ $\displaystyle|e^{2\pi i(n\alpha-p_{n})}-1|$ $\displaystyle\geq$ $\displaystyle\frac{2}{\pi}|2\pi(n\alpha-p_{n})|$ $\displaystyle=$ $\displaystyle 4|n\alpha-p_{n}|$ $\displaystyle=$ $\displaystyle 4|n|\left|\alpha-\frac{p_{n}}{n}\right|$ $\displaystyle>$ $\displaystyle 4|n|\cdot C|n|^{-r}$ $\displaystyle=$ $\displaystyle 4C|n|^{-r+1}.$

Because $u\in C^{\infty}$, it is straightforward to prove that for each nonnegative integer $k$ there is some $C_{k}>0$ such that

 $|\widehat{u}(n)|\leq C_{k}|n|^{-k},\qquad n\neq 0.$

Therefore, for each nonnegative integer $k$, using (4) we have

 $|\widehat{v}(n)|=\frac{|\widehat{u}(n)|}{|e^{2\pi in\alpha}-1|} (6)

One can prove that if $h_{n}$ are complex numbers satisfying (6) then the function defined by

 $h(x)=\sum_{n\in\mathbb{Z}}h_{n}e^{2\pi inx},\qquad x\in\mathbb{R}/\mathbb{Z}$

is $C^{\infty}$. Therefore, we have established that if $\alpha$ is Diophantine then there is some $v:\mathbb{R}/\mathbb{Z}\to\mathbb{R}$ that is $C^{\infty}$ and that satisfies (3).

On the other hand, for $\alpha=\sum_{n=1}^{\infty}10^{-n!}$, Ghys constructs a $C^{\infty}$ function $u:\mathbb{R}/\mathbb{Z}\to\mathbb{R}$ such that there is no continuous function $v:\mathbb{R}/\mathbb{Z}\to\mathbb{R}$ satisfying $u(x)=v(x+\alpha)-v(x)$ for all $x\in\mathbb{R}/\mathbb{Z}$.