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For $\mathrm{Re}z>0$,so $w=(z-1)/(z+1)$ lies in the unit disc. Its inverse is $z=(1+w)/(1-w)$, whose real part is $(1-|w{|}^2)/|1-w{|}^2>0$. Both derivatives are nonzero, proving this Möbius transformation is conformal.

Write $B=\mathrm{diag}(\lambda ,{\lambda }^{-1})$ and $q={\lambda }^2>0$, so its Möbius transformation is $z\mapsto qz$, with $q\ne 1$. For $z=u+iv$, the Hyperbolic distance in the Poincare half-plane satisfiesAt $z=i$, the final parenthesis is $1$. Since $\cosh$ is strictly increasing on nonnegative arguments, every $z$ off the imaginary axis has

The hyperbolic lines $AB$ and $CD$ are both perpendicular to $BC$. They are therefore ultraparallel hyperbolic lines, and $BC$ is their unique common perpendicular of ultraparallel hyperbolic lines.

The common perpendicular realizes the minimum hyperbolic distance between two ultraparallel lines. This follows directly in the Poincare half-plane model after a Möbius transformation sends the two lines to concentric semicircles: the supplied distance formula is minimized along their common radial geodesic. The segment $AD$ also joins the two lines, but it is different from $BC$ in a nondegenerate quadrilateral. Hence

There is an exact rational-map description of sigma-model lumps and a restricted variational rational-map description of Skyrmions. The first follows from a Bogomolny equation; the second separates angular and radial dependence in a field that generally does not saturate a Bogomolny bound.

For the exact example, take the two-dimensional O3 nonlinear sigma model with a unit field $\mathbf{n}$, energy $E=\frac{1}{2}\int |\nabla \mathbf{n}{|}^2{d}^{2}x$, and a fixed limit at infinity. Regular fields then have a one-point compactification to maps $S^2\to S^2$. Choose the orientation so that the stereographic fieldhas positive charge when it is holomorphic. Equivalently, this is the CP1 nonlinear sigma model since the target sphere is the complex projective line. Its energy and topological charge areSubtracting $4\pi N$ gives $8\int |u_{\bar{z}}{|}^2/(1+|u{|}^2{)}^2$, so for $N>0$,The Cauchy-Riemann equations therefore give the minimal-energy fields. A holomorphic map from the compactified domain Riemann sphere to the target Riemann sphere is a rational map $u=p(z)/q(z)$, with common polynomial factors cancelled. Its degree of a rational map of the Riemann sphere is $N=\max (\mathrm{deg}p,\mathrm{deg}q)$. Poles of $u$ are coordinate singularities, not singularities of $\mathbf{n}$. For instance, $u=(z-a)/\lambda$ with $\lambda \ne 0$ is an exact unit lump with energy $4\pi$, arbitrary centre $a$, and scale $|\lambda |$. Its density is $4|\lambda {|}^2/(|z-a{|}^2+|\lambda {|}^2{)}^2$, whose plane integral is $4\pi$. For negative charge use antiholomorphic maps. Scale invariance allows arbitrarily small lumps and does not by itself prevent a singular concentration limit in the time-dependent theory.

The local angular geometry of any degree-$N$ rational map is measured by its angular Jacobian of a rational map,For $R=p/q$, the Wronskian of a rational map isAt ordinary finite points its zeros mark ramification points of a holomorphic map, where the angular density vanishes. At a pole use $q/p$ as the target coordinate, and at infinity use $1/z$ as the domain coordinate. A pole of order $m$ is ramified by $m-1$. The Riemann-Hurwitz formula counts total ramification $2N-2$, including infinity, even if the affine Wronskian has smaller degree. For $R=z^N$, $W=N{z}^{N-1}$ accounts for $N-1$ zeros at zero; the reciprocal coordinate shows the other $N-1$ at infinity.

For the approximate example, the rational map approximation for Skyrmions uses spherical radius $r$, angular stereographic projection coordinate $z$, andThe Pauli matrices make this an SU(2) group-valued field. The boundary conditions give a continuous centre and the vacuum value at infinity; the profile must also make the energy finite. Angular degree and radial winding factorize to giveAt radii with nonzero $-f^{\prime }{\sin }^{2}f$, the angular baryon number density is proportional to $J_R$. Thus the Wronskian of a rational map locates zero-density directions and reveals the holes or face directions in shell-like Skyrmion configurations.

In dimensionless Skyrme model units, insert this ansatz into the quadratic and quartic derivative energies. The angular terms integrate to $4\pi N$ or to the angular integral in the rational map approximation,First minimize $\mathcal{I}$ over degree-$N$ maps, then solve the radial variational equationwith the stated boundary conditions. The Cauchy-Schwarz inequality gives $\mathcal{I}\ge N^2$ because $J_R$ has sphere average $N$. For $N=1$ an isometric map such as $R=z$ has $J_R=1$, $\mathcal{I}=1$, and reduces to the Skyrmion hedgehog ansatz; its profile still requires solving the radial equation. For general $N$, angular and radial separation restrict the allowed fields. The minimum within this class is an upper bound on the full sector minimum, not a claim that every exact Skyrmion has this separated form. The original construction is Houghton, Manton and Sutcliffe's arxiv.org/abs/hep-th/9705151.

A rotational symmetry of a rational map must satisfy $R(gz)=hR(z)$, where $g,h$ are the domain and target rotations written as Möbius transformations from SU(2) matrices. Since a target rotation is an isometry, this identity implies $J_R(gz)=J_R(z)$. For $R=z^N$, $R(e^{i\alpha }z)=e^{iN\alpha }R(z)$, giving axial spatial symmetry accompanied by an internal rotation. This explains the axial symmetry of the degree-two toroidal ansatz.

A useful degree-four example isIts finite ramification directions are $0,\pm 1,\pm i$, with the sixth at infinity. They are the six coordinate-axis directions under stereographic projection, so they are the face normals of a cube. The full map has octahedral rotational equivariance: for example $R(iz)=1/R(z)$, and the cyclic-axis generator $g(z)=(iz+1)/(1-iz)$ obeys $R(gz)=e^{2\pi i/3}R(z)$. These generate the cube's rotational group and give a cubic angular density. The associated profile then produces the familiar cubic charge-four approximation. Symmetry of the Wronskian is a useful necessary diagnostic but is not sufficient: replacing this map by $cR$, $c>0$, leaves the same ramification directions, while the identity under $z\mapsto iz$ becomes $cR(iz)=c^2/(cR(z))$, whose target transformation is a sphere rotation only when $c=1$. Equivariance of the entire rational map, rather than symmetry of its critical-point set alone, determines the physical rotational symmetry.

Suppose first that a simple continued fraction is eventually periodic. Its periodic tail $\xi$ is fixed by the Möbius transformation represented by the continued-fraction matrix of one full period:ThusThe infinite tail is not rational, since a rational simple continued fraction terminates. Hence $\xi$ is a quadratic irrational, and the finite initial block expresses the original number as another rational Möbius transform of $\xi$. Therefore every eventually periodic simple continued fraction represents a quadratic irrational.

For the purely periodic fraction $x=[7,7,7,\dots ]$,so $x^2-7x-1=0$. Positivity selects

Apply the continued-fraction algorithm to $\sqrt{23}$:The complete quotient then repeats, givingThe convergents through one term before the final $8$ areThe last of these satisfies the Pell equationso a positive solution is

Represent a Möbius transformation by a matrix, with nonzero scalar multiples representing the same map. Matrix multiplication reproduces composition, so $\theta$ is a group homomorphism. Every Möbius map has a representing matrix $A\in G{L}_2(\mathbb{C})$; multiplying $A$ by either square root of $(\det A{)}^{-1}$ gives a matrix in $S{L}_2(\mathbb{C})$ representing the same map. Thus $\theta$ is surjective. A matrix in its kernel represents the identity and is therefore scalar; determinant one leaves

If a nonidentity Möbius map $T$ has two distinct fixed points, conjugate those points to $0$ and $\infty$. The conjugated map fixes both and hence has the form $S(z)=\mu z$ with $\mu \ne 0,1$. If $T$ has one fixed point, conjugate it to $\infty$. The resulting affine map has the form $az+b$; having no finite fixed point forces $a=1$ and $b\ne 0$, and conjugating by a scaling makes it $S(z)=z+1$.

Directly, the finite fixed points satisfyIncluding $\infty$ when appropriate, a nonidentity Möbius map consequently has one or two distinct fixed points.

Finally suppose $T$ has the unique fixed point $z_0$. With the conjugacy $R$ above,Every finite point tends to $\infty$ on the Riemann sphere, while $\infty$ itself is fixed. Applying the continuous map $R^{-1}$ shows thatfor every $z\in \mathbb{C}\cup \{\infty \}$.

The Poincare disc model isThis Riemannian metric has constant Gaussian curvature $-1$.

For $a\in D$ and $|\eta |=1$, the Möbius transformationis an isometry of the disc, and every orientation-preserving disc isometry has this form. In particular, $T_{z_1,1}$ sends $z_1$ to zero and $z_2$ to a point of modulusRotations are also isometries, so it remains only to integrate the metric along a radial geodesic from $0$ to $\rho$:Thus the Hyperbolic distance in the Poincare disc is

The Möbius transformationmaps the unit disc conformally onto the upper half-plane. On the upper semicircle of its boundary, $w$ approaches the positive real axis; on the lower semicircle it approaches the negative real axis. If $0<\arg w<\pi$, the harmonic functionhas the required boundary limits.

For $z=x+iy$,Since the imaginary part is positive inside the disc,Therefore the solution of the Dirichlet problem isThe exceptional boundary endpoints are precisely the jump points of the prescribed data.

An orientation-preserving isometry fixing every point of ${\ell }_{+}$ fixes more than two points in the upper half-plane and hence, as a Möbius transformation, is the identity. The mapfixes ${\ell }_{+}$ pointwise and preserves the hyperbolic metric; it is the hyperbolic reflection in ${\ell }_{+}$. Consequently the pointwise stabilizer is exactly

For any isometry $F$, use part (a) to choose $g\in PS{L}_2(\mathbb{R})$ that agrees with $F$ on one pointed line and its tangent direction. Then $g^{-1}F$ fixes that line pointwise, so it is either the identity or its reflection. Hence the Isometry group of the Poincare half-plane consists ofandwith real coefficients, modulo multiplication of all four coefficients by a nonzero scalar.

Split the string embedding map into its constant worldsheet zero mode and orthogonal fluctuations, $X^{\mu }=x^{\mu }+X^{\prime \mu }$. The kinetic operator has no inverse on the constant mode, while its inverse on $X^{\prime }$ is the worldsheet Green function $G$. Completing the square in the Gaussian functional integral givesThe remaining ordinary integral iswhich proves (1), up to a source-independent functional determinant.

To insert $n$ tachyon vertex operators, chooseThe zero-mode integral produces target-space momentum conservation,In the nonzero-mode integral, discard the coincident self-contractions by normal ordering. With $G(z,w)=-({\alpha }^{\prime }/2)\log |z-w{|}^2$, the remaining pairwise contractions produce the Koba-Nielsen factor, so

For three closed-string tachyons on the sphere, $k_i^2=4/{\alpha }^{\prime }$ and ${\sum }_i{k}_i=0$, hence ${\alpha }^{\prime }{k}_i\cdot k_j=-2$ for $i\ne j$. The matter correlator is therefore $|z_{12}{z}_{13}{z}_{23}{|}^{-2}$. Gauge fixing the sphere's Möbius transformation group fixes three insertion points and supplies the bc ghost system correlator $|z_{12}{z}_{13}{z}_{23}{|}^2$, exactly cancelling this position dependence. ThusHere each vertex contributes one closed-string coupling $g_c$, $C_{S^2}$ contains the sphere vacuum normalization and conventions, the worldsheet zero mode gives the momentum delta function, the nonzero modes give the Koba-Nielsen factor, and the ghost determinant divides by the conformal Killing group. Since $C_{S^2}\propto g_s^{-2}$ and $g_c\propto g_s$, the net string genus expansion dependence is $g_s$.

A Möbius transformation of a Papperitz symbol changes the independent variable by a Möbius transformation. In particular,maps $a,b,c$ to $0,1,\infty$, respectively. Because a Möbius map is locally biholomorphic away from its pole, it carries each exponent pair with its singular point and givesNow define a dependent-variable rescaling of a Papperitz symbol by writing the old dependent variable asso that $v=u^{-\alpha }(1-u{)}^{-\beta }y$. This dependent-variable rescaling of a Papperitz symbol subtracts $\alpha$ from both exponents at zero, subtracts $\beta$ from both exponents at one, and adds $\alpha +\beta$ to both exponents at infinity. Hence $v$ has symbolDefineThe Fuchs relation gives ${\alpha }^{\prime }-\alpha =1-C$ and ${\beta }^{\prime }-\beta =C-A-B$. After renaming $u$ as $z$, the symbol is thereforewhich is the Gauss hypergeometric equation. Renaming $A,B,C$ as $a,b,c$ gives exactly the displayed symbol in the question.

Setand, for $n\ge 1$, defineWriting $M(a)=\left(\begin{array}{cc}a& 1\\ 1& 0\end{array}\right)$, these recurrences are equivalent to the continued-fraction matrix identityThe associated Möbius transformation sends a positive tail $\beta$ to the continued-fraction tail formula

Since $\det M(a_j)=-1$, taking determinants givesMoreover,whereasTheir signs are opposite, so ${\theta }_n$ lies between the two adjacent convergents. For $n=0$, the second endpoint is interpreted as $p_{-1}/q_{-1}=\infty$.

Write the Möbius transformation asA finite fixed point of a Möbius transformation satisfiesIf three distinct points of the Riemann sphere are fixed, this quadratic is identically zero, with the usual interpretation when one point is infinity. Hence $c=b=0$ and $d=a$, so $f(z)=z$. A nonidentity transformation's fixed-point equation is a nonzero polynomial of degree at most two on the sphere, and therefore it has one repeated fixed point or two distinct fixed points.

False. Compositions of scalings $z\mapsto \mu z$ and inversion $z\mapsto 1/z$ reduce to one of the forms $\nu z$ or $\nu /z$; they preserve the unordered pair $\{0,\infty \}$. Butis a Möbius transformation fixing $0$ and sends infinity to $1$, so it cannot be such a composition.

The boundary arcs meet at $0$ and $1$. The Möbius transformationsends these points to $0$ and infinity, sends the real boundary segment to the positive real axis, and sends the circular boundary arc to the ray of argument $-\pi /6$. Hence it maps $D\cap L$ onto the sectorThe power $\zeta =w^6$ maps this sector conformally onto the lower half-plane. Finally, the Cayley map $\zeta \mapsto (\zeta +i)/(\zeta -i)$ maps the lower half-plane onto the unit disc. Thus a requested conformal equivalence is

For a continuously differentiable curve $\gamma (t)=x(t)+iy(t)$ in the Poincare half-plane model, its hyperbolic length isThe hyperbolic lines are the geodesics: vertical Euclidean lines and Euclidean semicircles whose centres lie on the real axis.

The unique hyperbolic line through distinct $z,w\in H$ can be carried by an upper-half-plane Möbius transformation to the imaginary axis. Such transformations are isometries by hypothesis, so it is enough to consider endpoints $i{y}_0$ and $i{y}_1$. Every joining curve satisfiesEquality holds for the monotone vertical parametrizationThus the appropriately parametrized hyperbolic-line segment $[z,w]$ attains the infimum defining $\rho (z,w)$.

Now let $l,m$ have positive infimum distance. They neither intersect nor share an ideal endpoint: an intersection would give distance zero directly, while lines sharing an ideal endpoint approach one another arbitrarily closely near that endpoint. Hence they are ultraparallel. An upper-half-plane isometry puts them into the formThis is the normal form underlying the common perpendicular of ultraparallel hyperbolic lines. Put $z=e^{u+i\theta }$. Sincethe metric becomesEvery path from $l$ to $m$ therefore hasEquality requires $\theta =\pi /2$ and monotone $u$, so it is attained on the imaginary axis, which meets both semicircles orthogonally. Its intersection points $ir$ and $iR$ realize the distance

For a nonattained example, take the vertical lines $l=\{iy:y>0\}$ and $m=\{1+iy:y>0\}$. They share the ideal endpoint at infinity. The horizontal segment at height $y$ has hyperbolic length $1/y$, so their infimum distance is zero, but distinct points of $H$ always have positive distance. Hence the infimum is not attained.

Elements $g,h\in G$ are conjugate whenfor some $x\in G$. Thenfor every integer $n$, so $h^n=e$ exactly when $g^n=e$. They therefore have the same order.

The Möbius group consists of the Möbius transformationsof the Riemann sphere, under composition. If $h=xg{x}^{-1}$, thenThus $p$ is fixed by $g$ exactly when $x(p)$ is fixed by $h$. Conjugation gives a bijection between their fixed-point sets, so conjugate elements have the same number of fixed points.

A nonidentity Möbius transformation has at most two fixed points because its fixed-point equation is quadratic on the Riemann sphere. If it has only one repeated fixed point, conjugate that point to infinity; the transformation becomes a nontrivial translation $z\mapsto z+c$, which has infinite order. Consequently every nontrivial finite-order element has two distinct fixed points:This is fixed points of a finite-order Möbius transformation.

Choose a Möbius transformation of $\mathbb{H}$ that fixes $0$ and exchanges $x$ with $\infty$. By Conformal invariance of SLE, it transforms an ${\mathrm{SLE}}_6$ from $0$ to $x$ into one from $0$ to $\infty$. Until $x$ is disconnected from infinity, the discrepancy between the two target domains lies beyond the component visible from the growing tip. The Locality property of SLE therefore makes the two initial curve laws identical up to that disconnection time. Hence an ${\mathrm{SLE}}_6$ from $0$ to $x$, stopped at ${\tau }_x$, has the law of an ${\mathrm{SLE}}_6$ from $0$ to $\infty$ stopped when it disconnects $x$ from infinity.

A Möbius transformation of the Riemann sphere is a mapwith its natural values at the pole and at infinity.

For three distinct points $z_1,z_2,z_3$, defineThe usual limiting conventions cover an infinite $z_i$. This Möbius transformation sends $(z_1,z_2,z_3)$ to $(1,0,\infty )$. Defining $T_w$ similarly, the mapsends $z_i$ to $w_i$. If two Möbius transformations do so, their quotient fixes $0,1,\infty$. A Möbius transformation fixing infinity is affine, and fixing zero and one then makes it the identity. This proves uniqueness.

With this convention, the cross-ratio isSubstitution shows that translations, nonzero scalings, and inversion preserve it; since these generate the Möbius group, every Möbius transformation preserves cross-ratios.

Conversely, suppose a bijection $f$ of the Riemann sphere preserves every cross-ratio. Let $m$ be the unique Möbius transformation agreeing with $f$ at three chosen points $z_1,z_2,z_3$. For any other $z$, preservation by $f$ and $m$ givesThe last coordinate in a cross-ratio with three fixed distinct entries is injective, so $f(z)=m(z)$. The equality already holds at the three base points, hence $f=m$ everywhere and $f$ is Möbius.

Finally, the map $z\mapsto a\overline{z}+b$ is constant when $a=0$ and therefore is not Möbius. If $a\ne 0$, it is bijective and fixes infinity. Were it Möbius, it would have the affine form $\alpha z+\beta$. Equality on real $z$ forces $\alpha =a$ and $\beta =b$, whereas equality at $z=i$ would require $ai=-ai$, contradicting $a\ne 0$. Thus

The uniformization theorem says that every simply connected Riemann surface is conformally equivalent to exactly one of the Riemann sphere, the complex plane, and the unit disc. A proper simply connected plane domain is noncompact, so it is not the sphere. It cannot be uniformized by the plane: after choosing a point outside the domain, a branch of a square root and then a Möbius transformation would produce a nonconstant bounded entire function, contradicting the Liouville theorem. It is therefore conformally equivalent to the disc, which is the Riemann mapping theorem.