A rotation in different coordinates

The staircase that is flat almost everywhere names Denjoy’s theorem in one line and leaves it there: “a smooth enough map with irrational rotation number is conjugate to the rigid rotation — so an irrational rotation number really does mean nothing but a rotation, in different coordinates.”

That is an unusually strong statement. It says the unlocked parameters are not merely well behaved; the map at them is the rigid rotation, relabelled.

What a conjugacy is, and how one is built #

Two maps $f$ and $g$ of the circle are conjugate when there is an increasing continuous map $h$ of the circle onto itself with $h \circ f = g \circ h$. Then $h$ relabels the points and $f$ becomes $g$: every orbit of $f$ is carried to an orbit of $g$, every periodic point to a periodic point, every rotation number preserved.

The construction of $h$ is the whole content and it is one idea. Take an orbit of $f$ — the points $x, f(x), f^2(x), \ldots$ — and take an orbit of the rotation $g$, which is $0, w, 2w, \ldots$. Both are dense when $w$ is irrational. Now define $h$ by sending the $k$-th point of $f$’s orbit to the $k$-th point of $g$’s, and extend to the rest of the circle by continuity.

That is well defined because the two orbits are in the same cyclic order. The order in which a circle map’s orbit visits the circle depends only on the rotation number, not on the map: if $f^i(x)$ comes before $f^j(x)$ going round, then $iw$ comes before $jw$. So the relabelling is increasing, and an increasing map defined on a dense set extends to an increasing continuous map — provided the dense set really is dense.

The hero figure is that construction carried out. Sorting the orbit round the circle gives each point a rank; plotting rank against position is $h$; and the check is that consecutive orbit points’ ranks differ by the rotation number times the orbit length, which is what “the map becomes a rotation” means.

Where the hypothesis is spent #

The construction needs the orbit to be dense, and that is what the smoothness hypothesis buys. An orbit with an irrational rotation number need not be dense.

If it is not, its closure is a closed set which is not the whole circle and has no isolated points — so it is a Cantor set — and the intervals it leaves out are wandering: each is carried by the map to a disjoint interval, never returning. A map with a wandering interval is not conjugate to a rotation, because a rotation has no wandering interval: every interval under an irrational rotation returns arbitrarily close to itself.

So the theorem has two halves. The easy one is the construction above, which works whenever the orbit is dense. The hard one is Denjoy’s: a circle map that is continuously differentiable with derivative of bounded variation, and has irrational rotation number, has dense orbits. The hypothesis is exactly strong enough, and the reason it is interesting is that the plain $C^1$ version is false.

The counterexample, as arithmetic #

Denjoy’s map is built rather than found, and the construction is worth following because every condition in the theorem appears in it as something that has to be arranged.

Start with the rigid rotation by an irrational $w$ and pick one orbit, indexed by the integers. Now replace each orbit point $nw$ by an interval of length $\ell_n$, and define a new map sending the interval at $n$ onto the interval at $n+1$. The intervals must fit in a circle, so $\sum \ell_n$ must be finite; normalise it to one.

The result is a circle map with irrational rotation number $w$ whose orbit of the interval at zero never returns — the images are disjoint by construction. So no conjugacy to the rotation exists, and the Cantor set left over has measure nought because the intervals take up all the length.

Whether the map is differentiable depends on the $\ell_n$. The map takes an interval of length $\ell_n$ affinely onto one of length $\ell_{n+1}$, so its derivative there is $\ell_{n+1}/\ell_n$; for the map to be $C^1$ those ratios must tend to one as $|n|$ grows, so that the derivative extends continuously to the Cantor set with value one.

Both conditions can be met at once. Take $\ell_n$ proportional to $1/(|n|+2)^2$: the sum converges and the ratios tend to one. The figure computes both — the sum, by checking that doubling the range barely moves it, and the ratio, by evaluating it far out.

So there is a $C^1$ circle map with irrational rotation number that is not conjugate to a rotation. What fails is the bounded variation of the derivative: the derivatives $\ell_{n+1}/\ell_n$ are each close to one and they wobble either side of it infinitely often, and the total of those wobbles diverges.

The family matters because it shows the counterexample is not delicate. Any exponent above one gives a convergent sum; any exponent at all gives ratios tending to one. So there is a whole family of $C^1$ maps with irrational rotation number and a wandering interval, and no amount of adjusting the construction removes the phenomenon.

What does remove it is the extra hypothesis. Every map in the family has a derivative whose variation is infinite, and the reason is visible in the lengths: the derivative on the interval at $n$ is $\ell_{n+1}/\ell_n$, which for the inverse-square choice is $((|n|+2)/(|n|+3))^2$ — below one on one side of the orbit and above it on the other, with the deviations falling like $1/|n|$ and their total diverging like the harmonic series.

Why bounded variation is the right hypothesis #

That last sentence is the point at which this essay meets another part of the collection, and the meeting is not a coincidence of vocabulary.

Bounded variation is the condition that a function’s total rise and fall is finite, and it is exactly what fails for the derivative of Denjoy’s map. The derivative visits values above and below one infinitely often, each visit contributing a little to the total variation, and the contributions add up like a harmonic series.

Denjoy’s proof uses the variation directly. The obstruction to density is a wandering interval, and a wandering interval’s images have lengths summing to at most one — so most of them are tiny. The proof estimates the distortion of the map along a long orbit, which is a product of derivatives, and controls that product by the variation of the logarithm of the derivative. Bounded variation makes the product bounded; a divergent variation lets it run away, and a running-away distortion is precisely what shrinks the interval’s images fast enough for them to fit.

So the hypothesis is not a technical convenience: it is a bound on a total, and the counterexample is what happens when the total diverges — the same divergence that separates a graph with a length from one without. The same quantity decides whether a graph has a length and whether a circle map is a rotation in disguise, which is an unexpected connection and is the reason both are in this collection.

What the conjugacy does not promise #

The conjugacy is continuous and need not be smooth. Denjoy’s theorem produces a homeomorphism, and whether it can be taken differentiable is a much harder question with a different answer. The answer depends on the arithmetic of the rotation number: Herman and Yoccoz showed that for rotation numbers badly approximated by rationals — a condition of the kind the golden fraction satisfies — a smooth map is smoothly conjugate to the rotation, and for rotation numbers approximated too well it need not be.

So there are three levels and they come apart. Every irrational rotation number gives a topological conjugacy, for a smooth enough map. Only the badly approximable ones give a smooth conjugacy. And the boundary between them is a condition on the continued fraction expansion, which is a statement about number theory appearing inside a statement about maps.

That is where the golden case’s special status comes from. An earlier essay records that the last invariant curve to break as the nonlinearity rises is the golden one, and this is the reason: the smoothness of the conjugacy degrades as the rotation number is better approximated by rationals, and the golden fraction is the worst-approximated number there is.

The two regimes, and what is left over #

The picture an earlier essay draws is now complete in the following sense. At a locked parameter the map has an attracting periodic orbit, the average is a rational, and the orbit is destroyed by a collision at the edge of the plateau. At an unlocked parameter the map is a rigid rotation in different coordinates, every orbit is dense, and nothing attracts anything.

Those two descriptions are exhaustive for an invertible smooth circle map, and that is worth stating as a theorem rather than an impression: the rotation number is either rational, in which case a periodic orbit exists, or irrational, in which case the map is conjugate to the rotation. There is no third behaviour — no chaos, no strange attractor, no sensitive dependence.

Which is why the family stops being interesting above $K = 1$ and starts being interesting in a different way. The dichotomy needs invertibility; lose it and the third behaviours arrive, and the subject becomes the one the interval maps belong to, where an attracting orbit loses stability by splitting rather than by colliding.

The unlocked orbit, for comparison #

The object the theorem conjugates to is worth having on the page, because it is where every property of the unlocked case comes from.

A rigid rotation by an irrational amount has one behaviour and it is completely understood: every orbit is dense, no orbit repeats, the gaps between the first $n$ points take exactly three lengths — which is what three gaps and no more is about — and the orbit distributes itself evenly in the limit.

Denjoy’s theorem says each of those transfers. The nonlinear map’s orbits are dense because the rotation’s are and the conjugacy is a homeomorphism. Its orbits never repeat for the same reason. The three-gap theorem does not transfer, and the failure is instructive: three gap lengths is a statement about distances, the conjugacy distorts distances, and only the properties invariant under relabelling come across.

That is the standing limitation on every conjugacy result. What transfers is the topology and the combinatorics — which point comes before which, how many orbits there are, whether a set is dense. What does not is anything metric: lengths, measures, rates. So an unlocked circle map has the rotation’s orbit structure and its own distances, and questions about the second are what the smooth conjugacy theory is for.

What the pictures cannot show #

The conjugacy is built from nine hundred orbit points, so the plotted staircase has nine hundred steps and a genuine conjugacy is continuous. What the figure shows is an approximation whose resolution is stated, and the check it runs — that consecutive ranks differ by the rotation number — is accurate to the same resolution and no better.

The Cantor set in the counterexample is not drawn, and could not be: it has measure nought and is nowhere dense, so at any resolution it is invisible. What the figure draws instead is the arithmetic of the interval lengths, which is where the construction’s two conditions live.

And no picture distinguishes a dense orbit from one that merely looks dense. An orbit of nine hundred points on a circle drawn a few hundred pixels wide fills it at any resolution, and the figure’s claim — that the largest gap between consecutive orbit points is below a stated bound — is the honest version of the claim.

Still open: how smooth the conjugacy can be made #

The topological conjugacy is settled and the smooth one is not, for the cases between the extremes. For rotation numbers satisfying a Diophantine condition the conjugacy is as smooth as the map, minus a little, and the precise loss of smoothness is known in some ranges and not in others; for Liouville rotation numbers there are maps with non-smooth conjugacies, and the exact boundary between the two classes of rotation number is not drawn.

The question with the clearest statement concerns the critical case. At $K = 1$ the map has a point of vanishing derivative, so it is no longer a diffeomorphism and Denjoy’s hypothesis fails at that point; the conjugacy to a rotation then exists but is singular, and its regularity is described by exponents that are conjectured universal and computed numerically. That the exponents are universal is the renormalisation statement an earlier essay quotes, and it is proved in some cases and not in general.

And the higher-dimensional analogue is largely open. The two-dimensional version — a map of a torus with an irrational rotation vector — has no Denjoy theorem in the same strength, the counterexamples are more varied, and which rotation vectors force conjugacy is a subject rather than a theorem.

What the ordering argument is doing #

The construction of the conjugacy rests on a claim made in one sentence above and it is the load-bearing one: two orbits with the same irrational rotation number visit the circle in the same cyclic order. It is worth proving, because it is the reason a single number determines the whole map.

Suppose $f$ has rotation number $w$ and consider two of its orbit points, $f^i(x)$ and $f^j(x)$. The claim is that the order in which they and a third appear round the circle matches the order of $iw$, $jw$ and the third multiple.

The reason is that the lift is increasing and commutes with adding one. Whether $f^i(x)$ comes before $f^j(x)$ going round from $x$ is decided by comparing the lifts $F^i$ and $F^j$ at the lifted point, modulo one; and because $F$ is increasing and $F(y+1) = F(y) + 1$, the comparison is decided by whether $F^{i-j}$ moves a point past an integer — which is decided by the rotation number, since $F^n(y) - y$ is within one of $nw$ for every $y$ and every $n$.

So the combinatorics of the orbit is a function of the rotation number alone, and the conjugacy is that function’s inverse read as a map. Once stated this way the theorem is less surprising: an irrational rotation number pins down every ordering question about the orbit, and a homeomorphism of the circle is nothing more than a consistent set of answers to ordering questions.

What the argument does not give is density. It says the orbit is ordered like the rotation’s; whether it is dense, which is what the extension of $h$ needs, is the separate question the smoothness hypothesis answers — and the separation between the two is exactly the gap the counterexample lives in.

What the relabelling was worth #

A rotation number is a single number extracted from a map by averaging, and it looks like a summary. The theorem says that in the irrational case it is the whole map: everything about the dynamics is determined by it, up to relabelling the circle.

That is as strong a classification as a subject ever gets, and it is bought with a hypothesis that can be measured. Continuously differentiable with derivative of bounded variation is a condition somebody can check, the condition is sharp, and the map that fails it fails by an amount that is a divergent sum.

The habit to carry is the one the counterexample teaches. When a theorem carries a hypothesis that looks like a technicality — smooth enough, of bounded variation, sufficiently regular — the useful question is what breaks without it, and the answer is usually a specific construction rather than a vague pathology. Here it is one orbit blown up into intervals whose lengths were chosen to satisfy two arithmetic conditions at once, and the construction is what tells anybody why the hypothesis says what it says.