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Represent the Riemann sphere by the complex projective line, with finite $z$ represented by $[z:1]$ and infinity by $[1:0]$. An invertible matrix $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ acts by $[u:v]\mapsto [au+bv:cu+dv]$, giving the stated fractional map on finite points. Applying $B$ and then $A$ is exactly matrix multiplication $AB$, including at poles and infinity. Thereforeso $\theta$ is a group homomorphism from the special linear group to the Möbius group.

Every Möbius transformation has an invertible complex matrix representative $A$. Choose $\rho \in \mathbb{C}$ with ${\rho }^2\det A=1$. Such a nonzero square root exists in $\mathbb{C}$, and $\rho A$ has determinant one while representing the same projective map. This proves surjectivity. If the map is the identity, fixing zero and infinity forces $b=c=0$, and fixing one forces $a=d$. The determinant-one condition gives $a^2=1$. Conversely both scalar matrices $I$ and $-I$ act trivially. HenceFor the normal forms, use the fact that a complex $2\times 2$ matrix has a Jordan normal form: either it is diagonalizable or it has one nontrivial Jordan block. For a determinant-one representative, distinct eigenvalues are $\lambda ,{\lambda }^{-1}$. Diagonalizing gives $\mathrm{diag}(\lambda ,{\lambda }^{-1})$, whose projective action is $z\mapsto {\lambda }^{2}z$. Distinctness means ${\lambda }^2\ne 1$, so $\mu ={\lambda }^2$ is neither zero nor one.

If the eigenvalue is repeated, the determinant condition gives $\lambda =\pm 1$. A diagonalizable representative would be $\lambda I$ and would give the identity, excluded here. A nontrivial Jordan block is similar to $\left(\begin{array}{cc}\lambda & 1\\ 0& \lambda \end{array}\right)$, giving $z\mapsto z+1/\lambda =z\pm 1$. Any invertible conjugating matrix gives a Möbius transformation; rescaling it to determinant one does not change that map. Thus these matrix similarities are genuine conjugacies in the Möbius group, proving the requested Möbius conjugacy normal forms.

Conjugacy preserves fixed points bijectively. A nonidentity scaling has exactly the two fixed points zero and infinity, while either translation has exactly one, infinity. The identity fixes every point. This proves all three possibilities and excludes any further finite number of fixed points.

Finally, suppose $P\circ T\circ P^{-1}=S$, with $S(z)=z+\epsilon$ and $\epsilon =\pm 1$. Then $S^n(z)=z+n\epsilon \to \infty$ for every finite $z$ in the topology of the Riemann sphere; infinity remains infinity. Since $P^{-1}$ is continuous on that sphere,This is iteration of a one-fixed-point Möbius transformation. The limiting statement includes the unique fixed point itself and uses spherical convergence, which also handles intermediate iterates equal to infinity.

Let $Z_t$ denote the generation-$t$ type-count vector, and let $p_i^{(t)}={\mathrm{Pr}}_i(|Z_t|>0)$ from a single type-$i$ ancestor. Every type-$j$ child independently has probability $p_j^{(t)}$ of having descendants in generation $t$. For a Poisson random variable $N$ with mean $\lambda$, $\mathbb{E}{s}^N=\exp [\lambda (s-1)]$. Independence across child types therefore givesThe events of nonempty successive generations decrease. Each individual has finitely many children almost surely, so the total tree is infinite precisely when no generation is empty. Consequently $p^{(t)}\downarrow p$, the actual survival probability of a branching process. Continuity of $T$ yieldsThe map $T$ is coordinatewise increasing on nonnegative vectors. Every nonnegative fixed point $p^{\prime }$ automatically lies in $[0,1{]}^l$, since its coordinates equal $1-e^{-z}$ with finite $z\ge 0$. Hence $p^{\prime }\le 1$ and, by induction, $p^{\prime }=T^t(p^{\prime })\le T^t(1)=p^{(t)}$. Taking limits proves $p^{\prime }\le p$ in every coordinate, so $p$ is the greatest nonnegative solution, not merely some fixed point. This argument applies without irreducibility assumptions to the multitype Poisson branching process.

The necessary and sufficient condition iswhere $\rho$ is the spectral radius of the nonnegative matrix. No irreducibility is required. We use the given nonnegative-vector characterization of its Perron eigenvalue.

For necessity, suppose $p\ne 0$ is the survival fixed point and let $S=\{i:p_i>0\}$. Since $1-e^{-z}<z$ for every $z>0$, its equation gives $(\Lambda p{)}_i>p_i$ for every $i\in S$. There are finitely many such coordinates, soOutside $S$ the inequality $\Lambda p\ge \alpha p$ still holds, because its right side is zero and its left side is nonnegative. The provided Perron–Frobenius theorem characterization therefore gives $\rho (\Lambda )\ge \alpha >1$. In particular survival is impossible at $\rho =1$ for this Poisson offspring law.

Conversely suppose $\rho >1$. The given characterization supplies a nonzero $v\ge 0$ with $\Lambda v\ge \rho v$; normalize it so ${\max }_i{v}_i=1$. For sufficiently small $\eta >0$,For example any $0<\eta \le \min (1,2(\rho -1)/{\rho }^2)$ works. Coordinates with $v_i=0$ satisfy the inequality automatically. Starting at $x_0=\eta v$, iterate $x_{t+1}=T(x_t)$. Monotonicity makes the sequence increase, and $T$ takes nonnegative vectors into $[0,1{]}^l$, so it converges to a fixed point $x_{\infty }\ge \eta v\ne 0$. Part (a) gives $p\ge x_{\infty }$, proving survival from at least one type. This proves the spectral survival criterion for multitype Poisson branching for reducible as well as irreducible reproduction matrices.

A fixed point of the left shift satisfies $x_{k+1}=x_k$ for every $k\in \mathbb{Z}$, so its entire two-sided sequence is constant. The constant sequence with value $i$ belongs to the subshift of finite type precisely when $A_{ii}=1$: this is exactly the requirement that its repeated transition be allowed by the symbolic transition matrix. There is one such fixed point for each diagonal entry equal to one. ThusThis also covers an empty subshift of finite type, for then no permitted constant sequence exists.

The equation ${\sigma }^{n}x=x$ says $x_{r+n}=x_r$ for every $r\in \mathbb{Z}$. Thus a solution is determined uniquely by the block $(x_0,\dots ,x_{n-1})$, repeated in both directions. It belongs to the subshift of finite type exactly when all transitions inside the block, and the closing transition from $x_{n-1}$ to $x_0$, are allowed. Hence these fixed points of the $n$th iteration of a map are in bijection with the closed transition chains of $n$ steps with a specified symbol at time zero. By the counting argument in part (b), the number starting at $i$ is $(A^n{)}_{ii}$. Summing over $i$ proves the trace formula for periodic points of a subshift:Every closed transition chain extends by periodic repetition, so the extension issue in part (b) does not affect this formula. We count points, not periodic orbits: a least-period-$d$ periodic orbit contributes its $d$ distinct points.

Here period $n$ means fixed by ${\sigma }^n$, so the least period may divide $n$. If $P_d$ denotes the number of points of least period exactly $d$, division with remainder shows that a point is fixed by ${\sigma }^n$ exactly when its least period divides $n$. Consequentlywhere ${\mu }_{\mathrm{M}}$ is the Möbius function and the second equality is Möbius inversion. For example, $A=(1)$ has one point fixed by every power but no point of least period two; interpreting the printed formula as an exact-period count would therefore be false.

The quadratic discrete predator-prey map equals the identity plus the vector field of part (a), so its admissible fixed points are the same $O,E,P$. Their Jacobian matrices are the identity plus the corresponding continuous-time Jacobian matrices, but map stability requires eigenvalues inside the unit circle rather than the left half-plane.

At $O$ the multipliers are $1/a,0$: it is a sink for $a>1$ and a saddle for $a<1$. At $E$ they are $2-1/a$ and $(1-a)/b$. ThereforeFor $a<1/3$, the prey multiplier is below $-1$: $E$ is a saddle when $b>1-a$ and a repeller when $b<1-a$. For $1/3<a<1$, $b<1-a$ makes it a saddle. The $+1$ crossings at $a=1$ and $a+b=1$ are the two transcritical bifurcations already encountered. The crossing at $a=1/3$ is a prey period-doubling bifurcation.

At positive coexistence,For clarity, derive the relevant Schur stability criterion directly: setting $z=(1+s)/(1-s)$ in $z^2-Tz+D$ givesThe open unit disk maps to $\mathrm{Re}s<0$. A real quadratic has both roots there precisely when all three displayed coefficients have the same positive sign. Since $1-T+D=c>0$, the coexistence fixed point is a sink precisely whenThis region exists only for $a>1/9$. In the coexistence triangle $a+b<1$, put $K=1+T+D=(3a+1-3b)/a$. If $K<0$, one root lies below $-1$ and the other in $(-1,1)$, so $P$ is a saddle. If $K>0$ but $D>1$, it is a repeller; if $K>0,D<1$, it is a sink. These signs give the complete local fixed-point partition, including real and complex multipliers.

The coexistence period-doubling bifurcation lies at $b=a+1/3$; its segment adjoining the sink region has $1/9<a<1/3$. The other multiplier there is $2-1/(3a)$ and lies strictly inside the unit circle on this segment. This coexistence flip is subcritical on that segment. To calculate its direction, normalize the $-1$ eigenvector by $q_f=(1,(2-b/a)/b{)}^T$ and the left eigenvector by $p_f{q}_f=1$. The symmetric quadratic derivative $\mathcal{B}$ is given explicitly in part (d). Eliminating the quadratic term from the one-dimensional centre manifold map gives the effective cubic flip coefficientFor a scalar reduced map $u^{\prime }=(-1+\delta )u+c_f{u}^3+\cdots$, its second iterate satisfies $u^{\prime \prime }-u=-2\delta u-2c_f{u}^3+\cdots$. Thus $u^2=-\delta /c_f>0$ on the stable-fixed-point side $\delta >0$, and the two-cycle multiplier is $1+4\delta +\cdots >1$. An unstable coexistence two-cycle meets the stable fixed point as $b$ increases to the flip curve. The other multiplier remains inside the unit circle. The complex unit-modulus crossing lies at $a+2b=1$, with $1/9<a<1$. The two curves meet at $(a,b)=(1/9,4/9)$ with a double multiplier $-1$. The prey flip and predator invasion curves meet at $(1/3,2/3)$ with multipliers $-1,+1$.

The sketches mark the sink regions and all these local thresholds. They are a local atlas, not a claim that the map has only these global attractors. Indeed the full nonnegative quadrant is not invariant: $x^{\prime }=x(1-x-ay)/a$ becomes negative when $x+ay>1$. On the prey axis the logistic map already supplies new dynamics. Its nontrivial two-cycle, obtained by factoring $F^2(x)-x$ after removing the fixed-point factors, isIts longitudinal multiplier is $4+2/a-1/a^2$, so it is stable on the axis for $1/(1+\sqrt{6})<a<1/3$. The transverse multiplier is $x_{+}{x}_{-}/b^2=a(1+a)/b^2$, making the cycle stable in the plane when additionally $b>\sqrt{a(1+a)}$. Further flips and more complicated logistic dynamics show why the differential equation's three global regions cannot be transferred unchanged to its unit-step map.

On $a+2b=1$ the coexistence fixed point hasFor $1/9<a<1$ these give a nonreal conjugate pair on the unit circle. Nearby, while the pair remains complex, its modulus is $\sqrt{D}$; increasing $a+2b$ through one changes the modulus from greater than one to less than one. This is the candidate Neimark–Sacker bifurcation, and a nonlinear coefficient is needed to decide its direction.

Here that coefficient can be calculated explicitly. Let $\rho =e^{i\theta }$, $\sigma =\sin \theta >0$, and normalize the right eigenvector and complex projection byFor deviations $(X,Y)=zq+\overline{z}\overline{q}$, the quadratic derivative isThus $f_{20}=p\mathcal{B}(q,q)/2$, $f_{11}=p\mathcal{B}(q,\overline{q})$ and $f_{02}=p\mathcal{B}(\overline{q},\overline{q})/2$ are the quadratic coefficients in $z^{\prime }$. The quadratic coordinate change hasSubstitution into the map and collection of $w^2\overline{w}$ yieldsUsing $b=(1-a)/2$ and ${\sigma }^2=(1-a)(9a-1)/(16a^2)$ simplifies the cubic radial coefficient of a quadratic planar map toConsequently, away from strong resonances of a planar map, the bifurcation is supercritical: on $a+2b>1$ coexistence is attracting, whereas on $a+2b<1$ a small attracting invariant closed curve surrounds the unstable coexistence fixed point. If $\eta =\sqrt{D}-1>0$, its leading radius is $|w{|}^2=-\eta /\ell$, and its radial multiplier is $1-2\eta +\cdots$. Motion on the curve can be quasiperiodic or frequency-locked into periodic motion; stability of the curve alone does not imply an irrational rotation.

Two interior values need qualification: $a=1/7$ is a 1:3 resonance, and $a=1/5$ is a 1:4 resonance. At the former $h_{02}$ is singular and the quadratic ${\overline{w}}^2$ term must be retained. At the latter a cubic ${\overline{w}}^3$ term remains. The resonant terms can be displayed explicitly in the same normalization. At $a=1/7$, $f_{02}=-7/6-7\sqrt{3}i/2\ne 0$. The third iterate, near the resonance and after removing nonresonant quadratics, hasHere $\delta$ is the angular detuning. Its nonzero stationary solutions have $|w|=\sqrt{{\eta }^2+{\delta }^2}/|f_{02}|$ and three phases; they constitute a saddle three-cycle. Indeed the two leading real exponents at these solutions are $3[\eta \pm \sqrt{4{\eta }^2+3{\delta }^2}]$, of opposite signs. Therefore no ordinary attracting-circle conclusion follows arbitrarily close to the exact 1:3 point.

At $a=1/5$, the cubic resonant normal form isThe extra coefficient follows from $g_{03}=f_{11}{h}_{02}+2f_{02}{\overline{h}}_{20}$ and $h=\overline{\rho }{g}_{03}$. Its radial cubic damping is uniform since $\mathrm{Re}g+|h|<0$. The slow angular equation is $\Delta \arg w=\delta +|w{|}^2\mathrm{Im}(g+h{e}^{-4i\arg w})+\cdots$, so fourfold phase locking must be considered. For example, varying $b$ at fixed $a=1/5$ gives $\delta =-\eta /2+\cdots$. The locked branches satisfy $\tan (4\arg w)=6/17$ and $|w{|}^2=-\eta /\mathrm{Re}(g+h{e}^{-4i\arg w})$. For $\eta >0$ they form one attracting and one saddle four-cycle; the two branches alternate in phase. This is a resonant alternative to quasiperiodic motion on an invariant curve. The endpoints also fail the simple-complex-pair assumptions. This distinguishes the generic nearby dynamics from unjustified claims about every point on the printed interval.

The triangular-grid form is Sperner's lemma: subdivide a triangle into small triangles meeting edge to edge, label its vertices $1,2,3$, and forbid label $i$ on the side opposite corner $i$. The corner labels are thereby forced. An odd number of small triangles, in particular at least one, carry all three labels.

Here is the parity proof. Count incidences of a small triangle with an edge whose endpoint labels are $1,2$. On the outer boundary these edges occur only on the side joining corners $1,2$; traversing that side begins with label one and ends with label two, so the number of changes of label is odd. Interior edges contribute twice to the incidence count. A small triangle with all three labels contributes once; a triangle having only labels one and two contributes twice if it has both, and all other triangles contribute zero. Reduction modulo two therefore proves the assertion.

To obtain the continuous Brouwer fixed-point theorem on the triangle, use barycentric coordinates and a continuous map $F$ from the triangle to itself. If a grid vertex is not already fixed, label it by an index $i$ with $x_i>F_i(x)$, which exists since both coordinate sums are one. A point on the side $x_i=0$ cannot receive label $i$, so the discrete lemma applies. On successively finer grids take a fully labelled small triangle. Compactness gives a subsequence converging to one point $x$; all three vertices converge to it because the mesh tends to zero. Passing to the limit in the three label inequalities gives $x_i\ge F_i(x)$ for every $i$. Equal coordinate sums force equality throughout, hence $F(x)=x$.

A closed disk is homeomorphic to a triangle: radial coordinates from an interior point, scaled by the distance to the boundary in each direction, give such a homeomorphism. Conjugation therefore gives the same fixed-point assertion on the disk. If $r$ were a continuous retraction from the unit disk to its boundary, define $F(x)=-r(x)$. A fixed point would have $|x|=1$, hence $r(x)=x$ and $x=-x$, impossible on the unit circle. Thus no continuous disk-to-boundary retraction exists. The continuous conclusion has been deduced from the grid parity argument, not used to prove it.

In the hyperbolic-ball model, the limit set of a Kleinian group is the boundary accumulation set of an orbit $Go$ of an interior point: $\Lambda (G)=\overline{Go}\cap \partial {\mathbb{H}}^3$. The choice of $o$ does not matter, since any two such orbits remain a fixed hyperbolic distance apart. It is closed and $G$-invariant. The complicated/non-elementary hypothesis excludes relatively compact groups and elementary groups with at most two limit points; in particular the limit set is nonempty. Having no finite boundary orbit alone would not exclude a compact rotation group, whose interior orbits are bounded.

If $\Lambda (G)$ were finite and nonempty, the orbit of any of its points would be a finite boundary orbit, contrary to the hypothesis. Thus it is infinite. In fact it cannot be countable. We give the essential minimality/perfectness argument. A sequence of Möbius transformations sending an interior point toward $\xi \in \Lambda$ has a subsequence converging on the boundary away from one exceptional point to the constant $\xi$. This follows by normalizing the representing $2\times 2$ matrices: a diverging sequence has a rank-one projective limit, with one kernel line and one image line. For any boundary point $\eta$, the absence of a finite orbit allows choosing $h\in G$ with $h\eta$ different from that exceptional point; then $g_{j}h\eta \to \xi$. Hence the orbit of every $\eta \in \Lambda$ is dense in $\Lambda$.

If $\xi$ were an isolated limit point, density of every orbit would make every point belong to its orbit: each orbit must hit the open singleton $\{\xi \}$. Every point would then be isolated by equivariance, making the compact set $\Lambda$ discrete and therefore finite, a contradiction. Thus $\Lambda$ is a nonempty perfect compact metric space. Repeatedly choosing two disjoint smaller neighbourhoods meeting this perfect set, with diameters tending to zero, associates distinct nested limits to all infinite binary sequences. The limit set is therefore uncountable.

Jorgensen inequality states that for a non-elementary discrete group generated by lifts $A,B\in \mathrm{SL}(2,\mathbb{C})$,For a concrete failure of discreteness, take $A=\mathrm{diag}(e^{s/2},e^{-s/2})$ and $B=\left(\begin{array}{cc}1& 1\\ -1& 0\end{array}\right)$ with $s>0$ small. The fixed points of $A$ and of $BA{B}^{-1}$ collectively include $0,\infty ,-1$, so the group is non-elementary. Direct multiplication gives both summands equal to $4{\sinh }^2(s/2)$. Their sum is less than one when $s$ is sufficiently small, proving this two-generator group is not discrete.

For a connected-limit-set example, the modular group generated by $z\mapsto z+1$ and $z\mapsto -1/z$ is discrete because its matrices have integer entries. It preserves the upper half-plane; its limit set is the extended real line, a circle. Indeed the orbit of the parabolic fixed point infinity consists of the rational points and infinity, dense in that circle. For a disconnected example, choose four pairwise disjoint round disks and two loxodromic maps pairing the exterior of each negative disk with the interior of its positive disk. The Schottky group generated by these maps is free and discrete by the disjoint-domain ping-pong construction. Successive reduced words give nested disjoint disks with shrinking diameters; their limiting points form a Cantor set. Thus its limit set is disconnected, unlike that of the modular group.

At a fixed point $y=x$, and the first equation factors. The branches areThe three branches meet at $a=0$, and the positive square-root branch meets the quadratic branch at $a=1$. The Jacobian matrix at a general fixed point isOn $x=a^2$, its trace is $-1$ and determinant $a-a^4=a(1-a^3)$: the branch is attracting for $0<a<1$ and a saddle for $a<0$ or $a>1$. On $x=\pm \sqrt{a}$, the eigenvalues are $-1$ and $2a(1\mp a^{3/2})$. The negative branch is always a saddle for $a>0$; the positive branch is a saddle for $0<a<1$ and attracting for $a>1$. At the bifurcation values the zero eigenvalue requires the nonlinear analysis below.

For $a=1+\mu$, $u=x-1$, $v=y-x$, the equations areThe extended centre manifold has $v=h(u,\mu )$ with no linear terms. Since $h_u\dot{u}$ has degree at least three, its invariance equation to degree two is $0=-h_2-(2{\mu }^2-5\mu u+2u^2)$. ThusThe reduced quadratic factors as $(u-2\mu )(2u-\mu )$. Its two roots have reduced eigenvalues $3\mu$ and $-3\mu$, respectively, and the transverse eigenvalue remains negative. They exchange stability at $\mu =0$, a transcritical bifurcation. The exact branches have $u=2\mu +{\mu }^2$ and $u=\mu /2-{\mu }^2/8+\cdots$, consistent with this second-order evolution equation.

Near $a=0$, the slow equation has leading form $\dot{x}=x^3-ax$, so this is a subcritical pitchfork bifurcation with the parameter orientation displayed. For $a<0$ there is one nearby saddle, whose unstable direction is close to $y=x$ and stable direction close to vertical. For $a>0$ the central branch is attracting and the outer two equilibria near $\pm \sqrt{a}$ are saddles; their separatrices distinguish attraction toward the central equilibrium from departure outwards. The actual central location is $(a^2,a^2)$, not exactly the origin away from $a=0$.

Adding positive $\epsilon$ breaks the pitchfork symmetry. The leading equilibrium equation near zero is $x^3-ax+\epsilon =0$: the diagram unfolds into a single branch plus a saddle-node pair, with the latter appearing for positive $a$ of order ${\epsilon }^{2/3}$. More precisely the fold is at $x=(\epsilon /2{)}^{1/3}$, $a=3(\epsilon /2{)}^{2/3}$ at leading order. Near one the reduced equation becomes $2u^2-5\mu u+2{\mu }^2+\epsilon =0$. Its discriminant is $9{\mu }^2-8\epsilon$; the crossing is replaced by two saddle-node endpoints with a no-equilibrium gap for $|\mu |<\sqrt{8\epsilon }/3$ to leading order. Outside that gap, the lower reduced root is stable and the upper one unstable. These local approximations describe the sketch; the exact perturbed curves have higher-order corrections.