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For $f_c(z)=z^2+c$, the Mandelbrot set is the set of parameters for which the finite critical orbit $0,f_c(0),f_c^2(0),\dots$ remains bounded. It is a compact connected parameter set. Its main cardioid consists of parameters with an attracting fixed point, with a period-two bulb attached on the left, smaller bulbs attached recursively and fine filamentary structure. It is symmetric about the real axis and meets that axis in $[-2,1/4]$. The quadratic filled Julia set is connected exactly when $c$ belongs to this parameter set; outside it, the rational critical point escapes and the Julia set is a Cantor set. Thus it organizes the contrasting phase-space dynamics of the quadratic family, rather than itself being a Julia set in the dynamical $z$-plane.

The affine map $h(z)=1/2-z$, its own inverse, givesThis value is unique even among Möbius conjugacies. The finite fixed point of $P$ is zero and its periodic-orbit multiplier is one. Holomorphic conjugacy preserves that periodic-orbit multiplier. Infinity has periodic-orbit multiplier zero for a quadratic polynomial, so a corresponding multiplier-one fixed point of $w^2+c$ must be finite; it obeys $2w=1$ and $w^2+c=w$, forcing $c=1/4$. To locate it explicitly, a quadratic fixed point of periodic-orbit multiplier $\lambda$ gives $c=\lambda /2-{\lambda }^2/4$. The boundary parametrization $|\lambda |=1$ has its cusp at $\lambda =1$, so $c=1/4$ is the right-hand cusp on the Mandelbrot boundary. It is in the set: the critical orbit starts at zero, stays in $[0,1/2]$ and increases to the fixed point $1/2$. For real $c>1/4$, the increment $f_c(x)-x=(x-1/2{)}^2+c-1/4$ is strictly positive; a bounded critical orbit would have a limit solving $x^2+c=x$, which has no real solution. These nearby real parameters escape, proving that the cusp belongs to the boundary of the full Mandelbrot set as well.

The fixed points of $P$ on the sphere are zero and infinity. There is no neighborhood of zero on which all iterates converge to zero, even pointwise: a negative real initial point satisfies $P(x)=x-x^2<x<0$, so its iterates decrease without a finite limit and escape to infinity. Every neighborhood of zero contains such points. At infinity there is a uniformly attracting neighborhood in the chordal metric. In the reciprocal coordinate $w=1/z$,For $|w|\le 1/4$, its modulus is at most $|w|/3$. Thus the reciprocal iterates tend uniformly to zero there, proving uniform convergence to infinity on the corresponding sphere neighborhood.

For the given disc, $|z-1/2|<1/2$ is equivalent to $|z{|}^2<\Re z$, and hence to $\Re (1/z)>1$. Under $w=1/z$ the iteration becomesIf $\Re w>1$, then $\Re (1/(w-1))>0$, so $\Re T(w)>\Re w+1$. Consequently the half-plane is forward invariant andThis proves uniform convergence to zero on the disc and therefore $\Delta \subset F(P)$. It is the explicit parabolic disk for z minus z squared.

Let $U=F_{\Delta }$. The iterate family is normal on the connected open set $U$. Every subsequential limit along $n_j\to \infty$ equals zero on $\Delta$; by the identity theorem it is zero throughout $U$. If the full sequence failed to converge uniformly on some compact subset, a further normal subsequence would contradict that conclusion. Hence $P^n\to 0$ locally uniformly throughout $U$. Complete invariance of the Fatou set puts $P(U)$ in a Fatou component. Since $P(\Delta )\subset \Delta$, that component is $U$, so $P(U)\subseteq U$.

We prove the reverse-image assertion, rather than assuming that a basin has only one component. Each connected component $V$ of $P^{-1}(U)$ maps properly to $U$: for compact $K\subset U$, the sphere preimage $P^{-1}(K)$ is compact, and its intersection with $V$ is closed in it because $V$ is a relatively closed component of $P^{-1}(U)$. Its image is open by holomorphic openness. To prove it is closed in $U$, lift a convergent sequence of image points to preimages in $V$. Their images eventually lie in a compact neighborhood inside $U$; properness gives a convergent preimage subsequence with limit in $V$ and the desired limiting image. Thus the image is also closed, and equals the connected target $U$. Moreover $U$ itself is one such component: it lies in $P^{-1}(U)$ by forward invariance, and any connected inverse-image component containing it lies in $F(P)$ and hence in $U$ by maximality.

Now $1/4\in \Delta$ and its entire preimage is $\{1/2\}$, counted twice. Every component $V$ of $P^{-1}(U)$ must contain a preimage of $1/4$, so every one must be $U$. ThereforeThis is the Fatou component containing an entire fiber is completely invariant argument. The convergence on this component is locally uniform: it cannot be uniform on all of $U$, because $P^n(U)=U$ for every $n$ and $U$ contains nonzero points.

Finally suppose zero belonged to the Fatou set. A normal subsequence on a disc around zero would have limit zero, since that disc meets $\Delta$ in an open set where the entire sequence tends to zero. The identity theorem forces this limit to be zero on the whole disc. Uniform convergence in a smaller finite coordinate disc would then imply derivative convergence at zero by the Cauchy integral formula. Butfor every $n$, whereas the derivative of the zero limit is zero. This contradiction proves

Use the normalized chordal metricUnder stereographic projection it is half the Euclidean chord distance on the unit sphere, which proves the metric axioms and the agreement with the usual sphere topology. If $\rho$ is spherical geodesic distance, then $\rho =2\arcsin \chi$, so $2\chi \le \rho \le \pi \chi$.

A rational map is holomorphic, hence smooth, as a map of compact spheres. At a pole use $1/R$ as target coordinate, and at infinity use $1/z$ as source coordinate; neither is a singularity of the sphere map. Its differential in the round metric has a finite maximum norm $L$. Integrating its norm along a minimizing spherical geodesic gives $\rho (Rz,Rw)\le L\rho (z,w)$. The preceding comparisons implyThis proves that the rational map is Lipschitz in the chordal metric. For a constant map take constant zero; for a nonconstant map the bound is finite by compactness. In finite coordinates the differential norm is $R^{\#}(z)=|R^{\prime }(z)|(1+|z{|}^2)/(1+|R(z){|}^2)$, with its continuous sphere extension.

Now suppose $R,S$ are commuting maps. We first prove $S(F(R))\subseteq F(R)$. The family $\{R^n\}$ is locally equicontinuous on $F(R)$, by the normality equivalence in question 1. Commutation and the Lipschitz estimate give, for all $n$,For $x_0\in F(R)$ choose a source ball within that open set, small enough to make the right side less than any prescribed $ϵ$ uniformly in $n$. Holomorphic openness makes its image contain a ball about $S(x_0)$. Every point $w$ of that target ball has some preimage $x$ in the chosen source ball, so the same inequality proves equicontinuity of the iterates of $R$ at $S(x_0)$. Repeating at each point of the image gives local normality on that image. Thus the stated inclusion holds even where $S$ has a rational critical point.

Carefully state the permitted Montel theorem: a family of meromorphic maps on an open domain, all omitting the same three distinct values of the Riemann sphere, is normal for locally uniform spherical convergence. Since $\mathrm{deg}R\ge 2$, question 1 shows that $J(R)$ contains at least three distinct points. The forward inclusion just proved means every iterate of $S$ maps $F(R)$ into itself and therefore omits those three points there. Montel gives $F(R)\subseteq F(S)$. Exchanging the roles, using $\mathrm{deg}S\ge 2$, gives the opposite inclusion. ConsequentlyThis proves that commuting rational maps of degree at least two have the same Julia set without assuming either is an iterate of the other.

For commuting maps of degree one with different Julia sets, takeTheir two compositions are the identity, and the Julia sets follow from the expanding and contracting Möbius cases in question 1.

For equal Julia sets without commutation, takeBoth affine maps contract toward their finite fixed points and have a repelling periodic point at infinity. But $R\circ S=z/4+1/2$, while $S\circ R=z/4+1$, so they do not commute.

Finally the distinct Chebyshev polynomialshave degrees two and three and neither is a pure power of $z$. Direct expansion gives both compositions equal to $32z^6-48z^4+18z^2-1$. Equivalently $T_m(T_n(z))=T_{mn}(z)$, first seen at $z=\cos \theta$ and then as a polynomial identity. This provides the requested higher-degree pair.

Write $n=km$, with positive integers $k,m$, and put $Q=P^m$. Polynomial substitution respects congruence, so $Q(z)\equiv z(\mathrm{mod}Q(z)-z)$ implies inductively $Q^j(z)\equiv z$ for every $j\ge 1$. Since $Q^k=P^n$,Equivalently one can factor each $Q^{j+1}(z)-Q^j(z)$ by $Q(z)-z$ and telescope. If $Q(z)-z$ is identically zero, then $Q$ is the identity and $P^n-z$ is also zero, so the assertion still holds in the sense that the latter is a polynomial multiple of the former.

For the holomorphic cycle, the chain rule giveswith indices cyclically understood. These are the same product at every point. If its modulus is less than one at one point, it is less than one at every point. Taylor expansion of each return map gives a contracting neighborhood of its fixed point, exactly as in question 1. Hence every point of the holomorphic cycle is attracting for $P^n$ whenever one is; a zero factor makes every return periodic-orbit multiplier zero.

For the quadratic family, factor the second-iterate fixed-point equation:The first factor gives fixed points. A root of the second is also fixed only if subtracting the two factors gives $2z+1=0$, so $z=-1/2$ and $c=-3/4$. If $c\ne -3/4$, the second factor has discriminant $-4c-3\ne 0$ and has two distinct roots, neither fixed. They are exchanged by $P$, since each is fixed by $P^2$ but not by $P$, and form an exact two-cycle. Therefore the unique parameter with no exact two-cycle isAt this parameter,Thus the four finite solutions counted with multiplicity are $3/2$ once and $-1/2$ three times: there are only two distinct finite solutions, both already fixed by $P$. The periodic-orbit multiplier at $3/2$ is three (nine for $P^2$), so it is repelling. At $-1/2$ the periodic-orbit multiplier for $P$ is minus one; for $P^2$ it is one. Indeed, with $u=z+1/2$, the local first-return expression for $P$ is $-u+u^2$, and its square is $u-2u^3+u^4$. This is a parabolic return with a triple fixed-point root. The two-cycle has merged into this fixed point; the triple root is not three distinct periodic points. In the Mandelbrot set picture this is the period-two bulb's attachment to the main cardioid.

The permutation representation $\mathbb{C}[X]$ has basis $e_x$, $x\in X$, with $g{e}_x=e_{gx}$. Its matrix is a permutation matrix, whose diagonal entry is one exactly at fixed points. Therefore

The Brouwer fixed-point theorem says that every continuous map from a closed Euclidean ball to itself has a fixed point. For the closed disc $D\subset {\mathbb{R}}^2$ its equivalent no-retraction theorem says that there is no continuous $r:D\to \partial D$ satisfying $r(z)=z$ on $\partial D$.

Assume the fixed-point theorem and suppose such a retraction exists. Then $F(x)=-r(x)$ maps $D$ continuously into itself. A fixed point must lie on the boundary, where $r(x)=x$, giving $x=-x$, impossible on the unit circle.

Conversely suppose $F:D\to D$ is continuous and has no fixed point. From $F(x)$ draw the ray through $x$ and let $r(x)$ be its exit point on the circle. To check continuity rather than merely relying on the picture, put $v=x-F(x)\ne 0$. The exit point is $F(x)+t(x)v$, whereThis is continuous, lies on the circle and has $t(x)\ge 1$. On the boundary the exit point is $x$, so $r$ is a retraction, contradicting the assumed theorem. Hence the two disc formulations are equivalent; the same argument works for a ball in any dimension.

A horseshoe for an interval map consists of an open interval $J$ containing two disjoint open subintervals each mapped onto $J$. Glendinning chaos means that some positive iterate has a horseshoe.

Solving on the appropriate branches of the tent map givesHere $F(x_{-2})=x_{-1}$ and $F(x_{-1})=x_0$. On $J=[x_{-1},x_0]$ the second iterate isBoth endpoints map to $x_0$, and the minimum is $v=\mu -{\mu }^2/2$. Both monotone branches cover $J$ when $v\le x_{-1}$, equivalently $(\mu -1)({\mu }^2-2)\ge 0$. Restricting them to preimages of the interior proves a horseshoe for $F^2$ when $\mu \ge \sqrt{2}$.

For $1<\mu \le \sqrt{2}$, $G(J)\subseteq J$. Set $d=(\mu -1)/(\mu +1)$ and $h(x)=(x_0-x)/d$. This maps $J$ onto $[0,1]$; substitution in each linear piece gives $hG{h}^{-1}=F_{{\mu }^2}$. The renormalization of the tent map near its fixed point therefore makes $F^4$ conjugate on $J$ to $F_{{\mu }^2}^2$. Applying the preceding threshold proves its horseshoe when $2^{1/4}\le \mu <\sqrt{2}$.

For $\mu \ge \sqrt{2}$, the inverse branch of $G$ fixing $x_0$ contracts distance to $x_0$ by ${\mu }^{-2}$. Compose $m$ copies of it with the other inverse branch. This contraction has a fixed point with a periodic itinerary, distinct from $x_0$ and at distance $O({\mu }^{-2m})$. Hence nontrivial periodic points accumulate at $x_0$.

For $\mu <\sqrt{2}$, the conjugate tent parameter $\nu ={\mu }^2<2$ has every nonzero periodic point in $[\nu -{\nu }^2/2,\nu /2]$. This interval is forward invariant; below it, a nonzero point increases until entering it. The positive lower bound separates those periodic points from zero, which corresponds to $x_0$. Thus no nontrivial periodic points are sufficiently close to $x_0$ from the left. A point sufficiently close on the right maps to such a left-hand point; periodicity would pass to its image and is impossible. Thus $x_0$ is isolated among periodic points, even though a fourth-iterate horseshoe exists elsewhere.

For a nonidentity element of ${\mathrm{PSL}}_2(\mathbb{R})$ represented by a determinant-one matrix, the classification is elliptic, parabolic or hyperbolic according as the squared trace is less than, equal to or greater than four. This follows either from the eigenvalues or from the discriminant of its fixed-point equation on the boundary of the upper half-plane.

The first matrix has determinant one and trace two, and is not scalar. It is therefore parabolic. Its fixed-point equation isThus its unique fixed point is $z=-2$ on the real boundary, of multiplicity two; infinity is not fixed and there is no fixed point in the upper half-plane. The second matrix has determinant one and trace $-4$, so it is hyperbolic. Its two boundary fixed points are $(-1\pm \sqrt{3})/2$, consistently with this classification.

A strict Lyapunov function is a continuously differentiable function $V$ with $V(0)=0$, $V(x)>0$ off the origin, and $\nabla V\cdot f<0$ off the origin in the domain. Its compact sublevel sets inside the domain are forward invariant and force convergence to the equilibrium. The domain of stability, or basin of attraction, consists of initial points whose forward solutions exist for all positive time and converge to the fixed point.

For $V=(x^2+y^2)/2$, direct differentiation and factorization giveIf $R^2=x^2+y^2<1/2$, then $x^2+4y^2\le 4R^2<2$, so $\dot{V}<0$ off zero. For any initial point in that disc, choose a slightly larger closed disc still inside it. The vector field points inward on its boundary; the trajectory is bounded, exists globally, and the strict Lyapunov decrease forces its limit to be zero. Thus the origin is asymptotically stable and the whole open disc lies in its basin.

If $R^2>2$, then $x^2+4y^2\ge R^2>2$, so $\dot{V}>0$. A solution starting there cannot cross inward through $R^2=2$ and cannot approach zero. Therefore

A plane crystallographic group is a discrete subgroup of the Euclidean plane's isometry group with compact quotient; equivalently a compact set has translates covering the plane. We prove the crystallographic restriction theorem without assuming its conclusion. The orientation-preserving subgroup $G^{+}$ has index at most two and still has a cocompact group action. Its subgroup $T$ of translations is discrete as an additive subgroup of ${\mathbb{R}}^2$.

Suppose $T$ were trivial. Linear parts of orientation-preserving plane isometries commute, so every group commutator in $G^{+}$ is a translation. Thus $G^{+}$ would be an abelian group. Every member would then commute with the given nontrivial rotation $g$ and preserve its unique fixed point. A discrete subgroup of the compact rotation group about that point is finite. Then $G$ would be finite, contradicting cocompactness on the unbounded plane. Hence $T$ contains a nonzero vector $v$.

If $g$ is a half-turn its order is already two. Otherwise its linear part $R$ rotates $v$ to a linearly independent vector $Rv$, and conjugating translations shows that $Rv\in T$. Thus $T$ spans the plane and is a euclidean lattice. A discrete additive subgroup meets every bounded closed set in finitely many points: an accumulating sequence would give nonzero differences tending to zero, contradicting isolation of the identity. To see the euclidean lattice assertion directly, choose a shortest nonzero translation $v$. Translations on its line are integer multiples of $v$. Reducing all parallel components modulo $v$ shows that nonzero perpendicular components are bounded away from zero, since otherwise infinitely many translation vectors would accumulate in a compact rectangle. Choose a vector $w$ with the smallest positive perpendicular component; reducing by $w$ and then $v$ proves $T=\mathbb{Z}v+\mathbb{Z}w$.

Conjugation by $g$ preserves this euclidean lattice. In its euclidean lattice basis, $R$ is therefore an invertible integer matrix with determinant one. Its trace is an integer and is also $2\cos \theta$, where $\theta$ is the rotation angle. Since $g$ is nontrivial,These values give angles of order $2,3,4,6$ respectively. Hence

The nonzero fixed point of the logistic map is $x_{\ast }=(\mu -1)/\mu$, with multiplier $F^{\prime }(x_{\ast })=2-\mu$. It is stable for $1<\mu <3$, and its multiplier crosses $-1$ at $\mu =3$. To identify the emerging orbit explicitly, factorThe remaining two roots, interchanged by $F$, areThey coincide with $x_{\ast }$ at three and are distinct for $\mu >3$. Their cycle multiplier isIt lies between $-1$ and $1$ for $3<\mu <1+\sqrt{6}$. This establishes the first period-doubling bifurcation and the stable bifurcating two-cycle.

At ${\mu }_2=1+\sqrt{6}$, ${\Lambda }_2=-1$ and $d{\Lambda }_2/d\mu =-2\sqrt{6}\ne 0$. The second flip is nondegenerate: for $H=F^2$ at either cycle point, the cubic flip coefficient is $c=(H^{\prime \prime \prime }/6)+(H^{\prime \prime }/2{)}^2=-\mathcal{S}H/6$ when $H^{\prime }=-1$. Here $\mathcal{S}$ denotes the Schwarzian derivative. Since $\mathcal{S}F=-6/(1-2x{)}^2<0$ and $\mathcal{S}(F^2)=(\mathcal{S}F\circ F)(F^{\prime }{)}^2+\mathcal{S}F$, this coefficient is positive. Locally, after following the fixed point of $H$, its second iterate has the form $H^2(y)-y=2a(\mu -{\mu }_2)y-2c{y}^3+\cdots$, with $a=2\sqrt{6}>0$. Nonzero roots therefore appear with $y^2\sim a(\mu -{\mu }_2)/c$ on the increasing-parameter side. These are a four-cycle of $F$, proving the second period-doubling bifurcation occurs at $\mu =1+\sqrt{6}$.