The same map in different coordinates

The tent map is two straight lines. The logistic map at $r = 4$ is a parabola. One is not differentiable at its peak and the other is smooth everywhere; one has constant slope and the other has slope varying from $4$ to $-4$; and drawn side by side they have almost nothing in common.

They are the same map. There is a change of coordinate carrying each to the other, and once it is written down every dynamical statement about one becomes a statement about the other with no work at all.

The identity #

Write $T$ for the tent map at slope two, $x \mapsto 2\min(x, 1-x)$, and $L$ for the logistic map at four, $x \mapsto 4x(1-x)$. Set

$$h(y) = \sin^2\!\left(\frac{\pi y}{2}\right).$$

Then $L(h(y)) = h(T(y))$ for every $y$ in the interval. That is a trigonometric identity and it can be checked directly: for $y \le 1/2$,

$$L(h(y)) = 4\sin^2\tfrac{\pi y}{2}\cos^2\tfrac{\pi y}{2} = \sin^2(\pi y) = h(2y) = h(T(y)),$$

using the double-angle formula, and the case $y > 1/2$ follows from the symmetry of both sides.

Three properties of $h$ make it a change of coordinate rather than merely a function: it carries the interval onto itself, it is strictly increasing, and it is continuous with a continuous inverse. So it renames the points of the interval without losing or merging any, and the identity says that renaming turns one map into the other.

The figure checks all four claims — the identity at every sampled point, the two endpoints, and that $h$ is increasing across the whole interval — because a conjugacy that fails any of them is not one.

What the identity does #

Once $L \circ h = h \circ T$, applying it twice gives $L^2 \circ h = h \circ T^2$, and by induction $L^n \circ h = h \circ T^n$ for every $n$. That single fact carries every dynamical property across.

Orbits correspond. The orbit of $y$ under $T$ is carried point by point to the orbit of $h(y)$ under $L$. The figure asserts exactly that on a drawn orbit.

Cycles correspond. A point of period $k$ for $T$ goes to a point of period $k$ for $L$, and the correspondence is a bijection. So the two maps have the same number of cycles of every length — which the figure counts, on both sides, by bisection on $f^k(x) - x$, and finds $2^k$ each time.

Chaos corresponds. Sensitive dependence, density of periodic points, and topological transitivity are all preserved, because all three are statements about orbits and neighbourhoods and $h$ respects both — so a difference too small to draw on one side is one on the other.

That is a great deal for one identity, and it is worth appreciating the economy. The tent map’s dynamics can be worked out by hand — it is a shift of binary places, and an orbit is literally a word — and the conjugacy hands all of it to the logistic map, whose parabola makes none of it visible.

What it does not preserve #

A conjugacy is only required to be continuous with a continuous inverse, so anything depending on more than that can be destroyed — and here two things are.

The multiplier at a fixed point. $T$ has slope $\pm 2$ everywhere; $L$ has slope $4$ at $0$ and $-2$ at its non-zero fixed point. The two maps’ fixed points correspond and their derivatives do not, because $h$ is not differentiable in a useful way at the ends — $h'(0) = 0$, and the chain rule’s factor is degenerate exactly there.

Any statement about distance. Nearby points of the interval in one coordinate may not be nearby in the other; $h$ compresses near the ends and stretches in the middle, by a factor that goes to zero at $0$ and $1$.

What is preserved is what a homeomorphism preserves: the topology, and therefore everything defined from open sets and continuity. What is lost is everything defined from a metric or a derivative.

A conjugacy that is smooth with a smooth inverse would preserve the multipliers too, and there is no such conjugacy here — the multipliers differ, which proves it. So the maps are topologically the same and not smoothly the same, and that distinction is the reason the word topological is in the name.

The one quantity that survives anyway #

There is an exception worth stating because it looks like a counterexample and is not.

The Lyapunov exponent — the average rate at which nearby orbits separate — is $\log 2$ for the tent map, since its slope is $\pm2$ everywhere. For the logistic map at four the slope varies wildly, and the exponent is an average of $\log|4 - 8x|$ against the map’s invariant density. The answer is also $\log 2$.

That is not the conjugacy preserving it, because the conjugacy does not preserve derivatives — how fast two orbits part is a metric quantity and metric quantities are exactly what is lost. It is the two computations happening to agree, and the agreement has a reason: the exponent is a time average along an orbit, and the conjugacy carries orbits to orbits, so the two averages are averages of different functions over corresponding orbits. The equality comes from the change-of-variable formula for the integral rather than from topology.

The rule to carry is that a quantity survives conjugacy when it is defined from the orbit structure and not otherwise — and where a metric quantity survives anyway, there is a separate reason and it is worth finding.

How a conjugacy is found #

The identity above looks like it was pulled out of the air, and it is worth saying how such a thing is arrived at, because the method is general.

Both maps are unimodal: continuous on the interval, increasing then decreasing, with a single maximum, and mapping the interval onto itself. For unimodal maps whose critical orbit behaves the same way, a theorem says a conjugacy exists — and the conjugacy is essentially determined, because it must carry the maximum to the maximum, the fixed points to the fixed points, and the preimages of those to the preimages.

That determines $h$ on a dense set, which determines it everywhere by continuity. So the construction is: match the critical points, match their orbits, match all the preimages, and take the limit.

The closed form $\sin^2(\pi y/2)$ is then a piece of luck rather than a piece of method. It exists because the logistic map at exactly $r = 4$ is the Chebyshev polynomial $T_2$ in disguise, and Chebyshev polynomials are cosines in disguise — so the conjugating map is trigonometric for the same reason a Chebyshev polynomial is. At any other parameter there is a conjugacy to some tent map and no formula for it.

Two maps that are not conjugate, and how one tells #

A relation is only useful when it can fail, and it is worth seeing the test that separates two maps as well as the identity that joins two.

Compare the logistic map at $r = 4$ with the same map at $r = 3.2$. The second has a fixed point and a two-cycle and nothing else; the first has $2^k$ points of period dividing $k$ for every $k$. Since a conjugacy is a bijection on orbits, the counts would have to agree — and they do not, at $k = 3$ already, where one map has eight periodic points and the other has two.

Counting cycles is the cheapest invariant and it settles most cases. Where it does not, the standard tool is topological entropy — the growth rate of the number of distinguishable orbit segments — which is $\log 2$ for both maps in this essay and zero for anything that settles onto a cycle.

That gives a working procedure. To show two maps are the same, produce the change of coordinate; to show they are different, produce a quantity a conjugacy would preserve and evaluate it on both. The second is usually much easier, which is the ordinary situation with equivalence relations and is why the invariants get invented first.

Slopes other than two, and the family #

The conjugacy above is between two particular maps. The general picture is a family on each side and a matching between them.

The tent maps $x \mapsto s\min(x, 1-x)$ for $1 < s \le 2$ and the logistic maps $x \mapsto rx(1-x)$ for $3.57\ldots < r \le 4$ are both one-parameter families of chaotic unimodal maps, and there is a correspondence: for most $r$ in the chaotic range the logistic map is conjugate to the tent map of slope $e^{\lambda(r)}$, where $\lambda$ is the Lyapunov exponent.

So the tent family is a normal form for the whole chaotic range, with the exponent as the coordinate — one number identifying which tent map a given parabola is. That is a considerable reduction: a two-parameter question becomes a one-parameter one, and the parameter is a quantity that can be measured from an orbit.

The correspondence is not perfect and the gaps are the interesting part. The chaotic range is interrupted by windows in which the map settles onto a cycle, and inside a window there is no tent map to be conjugate to — the entropy is zero and the tent family has no member with zero entropy above slope one. The windows are visible in the bifurcation diagram as the vertical white stripes.

Why anybody wants one #

Conjugacy is the dynamical version of a habit that runs through mathematics: solve the problem in the coordinates where it is easy, and transport the answer.

The pattern is everywhere on this site. A map rewritten in its own invariant directions becomes two independent stretches, and its powers become cheap; the same idea moves an exponential of a matrix into a basis where it is a list of scalars. A quadratic completed into a square becomes a shifted square, and its roots become obvious. A signal written in frequencies becomes a list of numbers that differentiation multiplies.

Here the transport is exact and the destination is the tent map, whose dynamics is binary arithmetic. Every hard statement about the parabola — that its periodic points are dense, that it has orbits of every period, that it is sensitive to initial conditions everywhere — becomes an observation about shifting bits.

The one thing to keep in view is what the transport costs, and this rung’s answer is unusually clean: it costs every metric statement and keeps every topological one. A change of coordinate that preserved everything would be a relabelling of nothing.

When only half of it holds #

A conjugacy needs the change of coordinate to be reversible. Dropping that gives a weaker relation which is often all that is available, and it is worth having a name for.

A map $\pi$ that is continuous and onto, with $g \circ \pi = \pi \circ f$, is a semi-conjugacy, and it says that $g$ is a factor of $f$ — a coarser view of the same dynamics. Orbits of $f$ project to orbits of $g$, so everything $g$ does, $f$ does; but $g$ may have forgotten distinctions $f$ makes.

The standing example is next door. The doubling map $x \mapsto 2x \bmod 1$ on the circle is semi-conjugate to the tent map by $\pi(x) = 1 - |2x - 1|$, which folds the circle onto the interval two to one. Every tent orbit lifts to a doubling orbit and each lifts in two ways, so the tent map is the doubling map with the direction of travel forgotten.

That is why the orbit written as a word works for both, and why the symbolic description is really a description of the doubling map with the tent map inheriting it.

The distinction matters for what can be concluded. A conjugacy transports statements in both directions; a semi-conjugacy transports them one way only. Knowing that a complicated map has a simple factor bounds its entropy from below and says nothing about how much more it does — which is usually the situation with a real system, and is why the word appears more often in practice than the stronger one.

What the picture cannot show #

The figure draws one orbit on each side and the correspondence between its points. The claim is about every orbit, and about the maps rather than about any orbit; the drawing is an instance.

It cannot show that $h$ has no smooth inverse, which is the reason the maps are only topologically the same. The curve looks perfectly smooth on the page and is: what fails is that its derivative vanishes at the endpoints, so the inverse has an infinite derivative there, and infinite derivatives are not visible.

And it cannot show the cycle counts, which are the figure’s most convincing assertion. Both maps have $2^k$ points of period dividing $k$, found on both sides by bisection over six thousand cells and required to agree; that is a computation reported in the caption, and a picture of two curves cannot display it.

Where the ladder goes next #

Above: what a long orbit leaves behind, which is where the conjugacy’s failure to preserve distances starts to matter — the two maps have different invariant densities, and the change of variable is exactly what carries one to the other.

Two debts. The theorem that unimodal maps with matching critical orbits are conjugate is quoted and not proved; it is the beginning of kneading theory and is a real subject. And the claim that the logistic map at four is a disguised Chebyshev polynomial is stated in a sentence — the substitution $x = \sin^2\theta$ is the whole of it, and it deserves the two lines it takes.

What a change of coordinate is worth #

Two maps are the same map when a relabelling of the points carries one to the other, and every question about orbits then has one answer for both.

The tent map and the parabola are the standing example because they look so unlike. What the example teaches is that looking unlike is a fact about the coordinates: the slope, the smoothness, the shape of the graph are all properties of the labelling, and the dynamics — which points go where, and what returns — is what is left when the labelling is stripped away.