Dynamical Systems · Chapter VI, §5

The Poincaré–Birkhoff annulus theorem

Original

5. The geometric theorem of Poincaré.*

Poincaré showed that the existence of an infinite number of periodic orbits in the restricted problem of three bodies and other dynamical problems would follow at once from a certain geometric theorem to which the lemma of section 1 is intimately related.

For convenience we shall first state:

Poincaré's Theorem. Given a ring $0<a\leq r\leq b$ in the $r,\theta$ plane ($r,\theta$ being polar coordinates), and a one-to-one, continuous, area-preserving transformation $T$ of the ring into itself, which advances points on $r=a$ and regresses points on $r=b$. Then there will exist at least two points of the ring invariant under $T$.

This section is essentially the same as section 34 of my paper, Dynamical Systems with Two Degrees of Freedom, Trans. Amer. Math. Soc., vol. 18 (1917).

We will indicate briefly the proof of this theorem.

Let us take $x=\theta,y=r^2$ as the rectangular coordinates of a point in the $x,y$ plane. The ring then appears as a strip $a^2\leq y\leq b^2$. The transformation $T$ of this strip advances points of the boundary $y=a^2$ to the right, and moves points on $y=b^2$ to the left. Moreover $T$ is area-preserving in the $x,y$ plane (for we have $2r\,dr\,d\theta=dx\,dy$), and displaces any two points which have the same ordinate and whose abscissas differ by a multiple of $2\pi$ in the same way.

Let us combine $T$ with a further transformation $T_\varepsilon$ which effects a translation of the $x,y$ plane in the direction of the $y$ axis through a distance $\varepsilon>0$. The transformation $T$ followed by $T_\varepsilon$ yields an area-preserving transformation $TT_\varepsilon$ which shifts the given strip into the strip $a^2+\varepsilon\leq y\leq b^2+\varepsilon$.

Suppose if possible that there exists no invariant point of $T$. There exists then a positive quantity $d$ such that all points are displaced at least a distance $d$ by the transformation $T$. Choose $\varepsilon$ less than $d$.

Consider now the narrow strip $a^2\leq y\leq a^2+\varepsilon$. By the transformation $TT_\varepsilon$ the lower edge of this strip is carried into the upper edge and the strip is carried into a second strip lying wholly above the first one save along the common edge. By a repetition of the transformation $TT_\varepsilon$, the second strip goes into a third, and so on.

By a continuation of this process, a series of strips is obtained forming consecutive strata. Each of these strata is unaltered by a shift of $2\pi$ to the right. This follows from the fact that $T$ and $T_\varepsilon$ are single-valued over the ring.

The images of these strata on the ring are a set of closed strata about the ring, all having equal area of course since $TT_\varepsilon$ is an area-preserving transformation in the $r,\theta$ as well as in the $x,y$ plane. Consequently some one of the strata on the infinite strip, say the $k$-th, must overlap the upper edge $y=b^2$.

In the $x,y$ plane let $Q$ be a point of the upper edge of the $k$-th stratum for which $y$ is a maximum. Let $P$ be the point of $y=a^2$ from which $Q$ is derived by $k$-fold repetition of $TT_\varepsilon$, and let $P',P'',\cdots,P^{(k)}=Q$ denote the successive images of $P$ under the iteration of $TT_\varepsilon$. Draw the straight line $PP'$ which will obviously lie on the first stratum. The successive images of this line, $PP',P'P'',\cdots,P^{(k-1)}P^{(k)}$ will lie in the successive strata, and will have no points in common except that successive arcs have an end-point in common. Thus we get a single arc $PQ$, made up of all these lines, which is without double points.

Consider now a vector $LL'$ drawn from a point $L$ to its image $L'$ under $TT_\varepsilon$, of which the initial point moves from $P$ to $P^{(k-1)}$ along the line $PQ$.

Historical diagram of the successive arcs and their endpoint images.

The angle which this vector makes with the positive direction of the $x$ axis at the outset may be taken to be a positive acute angle, since the image $P'$ of $P$ lies to the right of and above $P$. When $L$ has varied to its final position $P^{(k-1)}$, the same angle lies in the second or third quadrant, since $P^{(k)}$ lies to the left of $P^{(k-1)}$, by the hypothesis of the theorem.

Our construction of the successive arcs $PP',P'P'',\cdots$ renders it apparent that as $L$ moves from $P$ to $P^{(k-1)}$, its image $L'$ moves along the same curve from $P'$ to $Q$. Therefore we see at once from the figure that $LL'$ has rotated through the least positive angle from the first direction to the second. If $L'$ be moved further in a vertical direction from $Q$ up to $y=b^2+\varepsilon$ the same statement will still hold of the standard curve described by $L$ provided that $\varepsilon$ is small, since $Q$ lies at most $\varepsilon$ above $y=b^2$.

Suppose now that $L$ moves in any manner from a point of $y=a^2$ to a point of $y=b^2$ in the given strip. The transformation $TT_\varepsilon$ leaves no points of this region invariant, so that the point $L$ will never coincide with $L'$. In the initial position for $L$ on $y=a^2$ the angle made by $LL'$ lies in the first quadrant. In the final position it lies in the second or third quadrant. But the total variation of angle during the variation of $L$ has been seen to be through the least positive angle in a special case. Since any one path of $L$ from $y=a^2$ to $y=b^2$ can be varied continuously into any other, the same must be true always.

Now let $\varepsilon$ approach zero. As $\varepsilon$ becomes smaller the vector $LL'$ continues to have a definite direction, since no invariant points under $TT_\varepsilon$ are present. By a limiting process we infer that, for the transformation $T$, the angular variation of $LL'$ is through the least positive angle consistent with its initial and final directions. It should be observed that for $L$ on $y=a^2$ the direction of $LL'$ is that of the positive $x$ axis, while on $y=b^2$ the direction is that of the negative $x$ axis.

Consider now the inverse transformation $T^{-1}$ which is of the same type as $T$ although it moves points on $y=a^2$ to the left, and points on $y=b^2$ to the right. By an entirely analogous argument to that given above we are led to infer that if a vector $LL^{(-1)}$ with end-point $L^{(-1)}=T^{-1}(L)$ has its initial point $L$ varied from a point of $y=a^2$ to a point of $y=b^2$, the total angular variation will be the least negative angle consistent with its initial and final positions.

But the total rotation of $LL^{(-1)}$ is precisely the same as that of the oppositely directed vector $L^{(-1)}L$ which joins a point $L^{(-1)}$ of $y=a^2$ to its image $L$ under $T$.

Hence by our earlier result the total angular variation of $L^{(-1)}L$ must also be the least positive angle consistent with the two positions. Thus we have been led to a contradiction, so that there must exist at least one invariant point.

To prove that there are at least two invariant points we may adopt the method used by Poincaré.

Let the point $L$ describe the fundamental rectangle $$0\leq x\leq2\pi,\qquad a^2\leq y\leq b^2,$$ in the $x,y$ plane, in a positive sense. It is obvious that the total rotation of the vector $LL'$ is $0$ over this circuit, since there is no rotation along $y=a^2$ or $y=b^2$, and the rotations along $x=0$ and $x=2\pi$ are the negatives of one another. But around a simple invariant point the rotation is $\pm2\pi$. Hence, according to obvious reasoning, there will either be at least two simple invariant points with rotations $+2\pi$ and $-2\pi$, or there will be at least one multiple invariant point. As a matter of fact, there will always be at least two geometrically distinct invariant points.*

See my paper, An Extension of Poincaré's Last Geometric Theorem, Acta Mathematica, vol. 47 (1926).

Modernized

The annulus fixed-point theorem

A fixed point is a point $z$ with $T(z)=z$. Boundary advance and regression are measured in one continuous lift of the angle. Composition is written in the source’s chronological convention: $TT_\varepsilon$ means apply $T$, then the vertical translation. The source’s topological construction is retained as a sketch.

For convenience we shall first state:

Poincaré's Theorem. Given an annulus $0<a\leq r\leq b$ in the $r,\theta$ plane ($r,\theta$ being polar coordinates), and a continuous injective, area-preserving transformation $T$ of the annulus into itself, which advances points on $r=a$ and regresses points on $r=b$. Then there will exist at least two points of the annulus invariant under $T$.

This section is essentially the same as section 34 of Birkhoff’s paper, Dynamical Systems with Two Degrees of Freedom, Trans. Amer. Math. Soc., vol. 18 (1917).

Lift to a strip and translate

The source gives the following proof sketch.

Let us take $x=\theta,y=r^2$ as the rectangular coordinates of a point in the $x,y$ plane. The annulus then appears as a strip $a^2\leq y\leq b^2$. The transformation $T$ of this strip advances points of the boundary $y=a^2$ to the right, and moves points on $y=b^2$ to the left. Moreover $T$ is area-preserving in the $x,y$ plane (for we have $2r\,dr\,d\theta=dx\,dy$), and displaces any two points which have the same ordinate and whose abscissas differ by a multiple of $2\pi$ in the same way.

Suppose if possible that there exists no fixed point of $T$. There exists then a positive quantity $d$ such that all points are displaced at least a distance $d$ by the transformation $T$. Choose $\varepsilon$ less than $d$.

The images of these strata on the annulus are a set of closed strata about the annulus, all having equal area of course since $TT_\varepsilon$ is an area-preserving transformation in the $r,\theta$ as well as in the $x,y$ plane. Consequently some one of the strata on the infinite strip, say the $k$-th, must overlap the upper edge $y=b^2$.

The displacement-vector contradiction

Consider now a vector $LL'$ drawn from a point $L$ to its image $L'$ under $TT_\varepsilon$, of which the initial point moves from $P$ to $P^{(k-1)}$ along the line $PQ$.

Historical diagram of the successive arcs and their endpoint images.

Now let $\varepsilon$ approach zero. As $\varepsilon$ becomes smaller the vector $LL'$ continues to have a definite direction, since no fixed points under $TT_\varepsilon$ are present. By a limiting process we infer that, for the transformation $T$, the angular variation of $LL'$ is through the least positive angle consistent with its initial and final directions. It should be observed that for $L$ on $y=a^2$ the direction of $LL'$ is that of the positive $x$ axis, while on $y=b^2$ the direction is that of the negative $x$ axis.

But the total rotation of $LL^{(-1)}$ is precisely the same as that of the oppositely directed vector $L^{(-1)}L$ which joins a point $L^{(-1)}$ of $y=a^2$ to its image $L$ under $T$.

Counting fixed points by rotation

To prove that there are at least two fixed points we may adopt the method used by Poincaré.

Let the point $L$ describe the fundamental rectangle $$0\leq x\leq2\pi,\qquad a^2\leq y\leq b^2,$$ in the $x,y$ plane, in a positive sense. It is obvious that the total rotation of the vector $LL'$ is $0$ over this circuit, since there is no rotation along $y=a^2$ or $y=b^2$, and the rotations along $x=0$ and $x=2\pi$ are the negatives of one another. But around a simple fixed point the rotation is $\pm2\pi$. Hence, according to obvious reasoning, there will either be at least two simple fixed points with rotations $+2\pi$ and $-2\pi$, or there will be at least one multiple fixed point. As a matter of fact, there will always be at least two geometrically distinct fixed points.*

See Birkhoff’s paper, An Extension of Poincaré's Last Geometric Theorem, Acta Mathematica, vol. 47 (1926).

Reinterpreted

Annulus, lift and boundary twist

Let $A=(\mathbb R/2\pi\mathbb Z)\times[a,b]$ be a closed annulus, and let $F:A\to A$ be an orientation-preserving, area-preserving homeomorphism preserving each boundary component. Choose a lift $\widetilde F=(X,Y):\mathbb R\times[a,b]\to\mathbb R\times[a,b]$ with

$$\widetilde F(x+2\pi,y)=\widetilde F(x,y)+(2\pi,0).$$

Assume, for this single chosen lift,

$$X(x,a)-x>0,\qquad X(x,b)-x<0\qquad\text{for every }x.$$

Poincaré–Birkhoff theorem (external theorem). Under these hypotheses there are at least two distinct points modulo deck translation with $\widetilde F(x,y)=(x,y)$. Their projections are distinct fixed points of $F$.

The boundary signs refer to the same lift; choosing different full turns at the two boundaries would destroy the meaning of the condition. A fixed point of $F$ need not be fixed by an arbitrary lift. No monotonicity of $X$ in the interior radial variable is assumed in this theorem.

What area contributes to Birkhoff’s construction

For a plane annulus with polar radius $r$, the change $x=\theta$, $y=r^2$ converts positive area $r\,dr\,d\theta$ to half the rectangular area density. Birkhoff perturbs the lifted map by $S_\varepsilon(x,y)=(x,y+\varepsilon)$ and iterates a narrow strip next to the lower boundary. As long as the construction remains in the domain, successive strata have equal positive area in a fundamental period. They cannot all remain disjoint inside the finite-area annulus.

That finite-area obstruction forces the constructed arc to reach the upper edge. The delicate part of the source proof is then topological: it controls the rotation of the nonzero displacement vector along this arc and transfers the result to other boundary-to-boundary paths. Applying the same construction to the inverse produces an incompatible rotation. The figure and this argument are retained in Modernized.

The brief historical account does not supply every extension, separation and limiting argument needed for a full proof for arbitrary homeomorphisms. We therefore invoke the theorem as stated, rather than treating a few area inequalities as its complete proof. Birkhoff’s cited 1926 paper supplies the extended treatment.

The displacement index explains the second point

For a differentiable map and an isolated zero $z$ of $\widetilde F-\mathrm{id}$, the fixed-point index is the winding number of

$$\frac{\widetilde F(w)-w}{|\widetilde F(w)-w|}$$

as $w$ traverses a small positively oriented circle around $z$. If $\det(D\widetilde F(z)-I)\ne0$, linearization gives

$$\operatorname{ind}_z(F)=\operatorname{sign}\det(D\widetilde F(z)-I).$$

Choose a fundamental rectangle with no zero on its vertical edges. On its lower horizontal edge the displacement points right; on its upper edge it points left. The two vertical edges contribute opposite winding because of periodicity. The total boundary degree is consequently zero. Additivity of degree implies

$$\sum_z\operatorname{ind}_z(F)=0$$

for finitely many interior fixed points in this lift class. If a first point exists and all points are nondegenerate, there must be at least one of each sign. This proves the two-point conclusion in that case once existence is known.

A degenerate isolated point can have index zero. Thus index cancellation alone does not prove two distinct points in the general case. That remaining conclusion belongs to the full Poincaré–Birkhoff theorem, exactly where the source refers to the longer treatment.

Rational rotation gives periodic points

Suppose a lift has constant boundary displacements $2\pi\alpha$ and $2\pi\beta$, with $\alpha<\beta$. For any rational $k/n$ strictly between them, apply the theorem, with reversed boundary signs if needed, to

$$G=\widetilde F^n-(2\pi k,0).$$

A zero of $G-\mathrm{id}$ projects to an $n$-periodic point with winding $k$. If $\gcd(k,n)=1$, its least period is $n$: a smaller return period $d$ would divide $n$, and the total winding would be $(n/d)$ times an integer, contradicting coprimality. This is the mechanism used for the billiard and geodesic applications. Boundary behavior must be verified for those particular maps.

For the integrable shear $F(x,y)=(x+\tau(y),y)$, with strictly increasing $\tau$, a rational target is attained on a whole circle. The theorem guarantees at least two points, not exactly two and not necessarily isolated points. The normal-form argument of §1 extracts periodic points without assuming invariant circles in the perturbed problem.