Projective linear group

The projective linear group is $PG{L}_n(F)=G{L}_n(F)/F^{\times }$ and acts on projective space.

The Möbius group is the group of all Möbius transformations of the Riemann sphere, under function composition. An invertible $2\times 2$ complex matrix represents $z\mapsto (az+b)/(cz+d)$, and nonzero scalar multiples represent the same transformation. Rescaling a representative to determinant one gives a surjective group homomorphism $S{L}_2(\mathbb{C})\to \mathcal{M}$ with kernel of a group homomorphism $\{I,-I\}$. Thus it is also the projective linear group $PG{L}_2(\mathbb{C})$ and the quotient $PS{L}_2(\mathbb{C})=S{L}_2(\mathbb{C})/\{\pm I\}$.

For ordered triples of distinct points of the Riemann sphere, a unique Möbius transformation sends the first triple to the second. Normalize each triple to $0,1,\infty$ using $T_z(z)=(z-z_1)(z_2-z_3)/((z-z_3)(z_2-z_1))$, with the usual limits at infinity, and compose $T_w^{-1}\circ T_z$. Any two choices differ by a Möbius transformation fixing those three standard points. Fixing infinity makes it affine; fixing zero and one then makes it the identity. Thus the Möbius group action is sharply three-transitive, a stronger property than mere transitivity on triples.

For two distinct points of the Riemann sphere, conjugate by a Möbius transformation sending them to zero and infinity. A transformation fixing those two points has the form $z\mapsto \lambda z$, $\lambda \ne 0$. Composition multiplies the parameter, so the pointwise stabilizer is isomorphic to the complex multiplicative group. For zero and one, conjugation by $z/(1-z)$ gives $z\mapsto \lambda z/[1+(\lambda -1)z]$.

The Möbius transformations preserving a three-point set act faithfully on it: one fixing all three is the identity. Every permutation is realized because a Möbius transformation can send any ordered distinct triple to any other, by composing the two normalizations to $0,1,\infty$. For that normalized set the six maps are $z$, $1-z$, $1/z$, $1/(1-z)$, $z/(z-1)$ and $(z-1)/z$. Thus its setwise stabilizer is the symmetric group $S_3$. Normalizing any other three-point set conjugates its stabilizer to this group.

The Möbius transformations generated by $D_{\lambda }(z)=\lambda z$ with $\lambda \ne 0$ and $I(z)=1/z$ have exactly the two displayed forms. The relations $I^2=1$ and $I{D}_{\lambda }=D_{{\lambda }^{-1}}I$ reduce every word to $D_{\lambda }{I}^{ϵ}$, $ϵ\in \{0,1\}$. This is a semidirect product of ${\mathbb{C}}^{\times }$ with a cyclic group of order two acting by multiplicative inversion. It preserves the unordered pair $\{0,\infty \}$, and only its scaling elements fix zero.

A finite abelian subgroup of the Möbius group is cyclic or isomorphic to $C_2\times C_2$. If it contains an element of order greater than two, conjugate that element's two fixed points to $0,\infty$. Every commuting map must preserve these points individually: an interchange would conjugate the element to its inverse. All maps are then scalings, forming a cyclic finite subgroup of ${\mathbb{C}}^{\times }$. If every nonidentity element has order two, choose one with fixed points $0,\infty$; the subgroup acts on this pair with image of size at most two and kernel contained in $\{z,-z\}$. Its size is therefore at most four, giving the asserted possibilities.

A Möbius transformation of the Riemann sphere has the formThe group is generated by translations, nonzero complex scalings, and inversion $z\mapsto 1/z$.

A useful way to map a circular lens is to send its two corner points to zero and infinity with a Möbius transformation. The two bounding circles then become straight rays, and a suitable power opens their sector into a half-plane.

For the lens $|z+1|<\sqrt{2}$, $|z-1|<\sqrt{2}$, the corners are $\pm i$. Put $w=(z-i)/(z+i)$, with inverse $z=i(1+w)/(1-w)$. For $w=u+iv$, substitution givesThus the image is precisely $u<-|v|$, or $3\pi /4<\arg w<5\pi /4$. Rotation through $-3\pi /4$ followed by squaring gives the displayed conformal map onto the upper half-plane. The open sector contains no opposite pair, so squaring is injective; it has nonzero derivative away from its excluded vertex. These facts also prove bijectivity and conformality, rather than merely identifying boundary images.

Let $A=\{z:|z-i|<\sqrt{2},|z+i|<\sqrt{2}\}$. Writing $t=(1+z)/(1-z)=u+iv$, its inverse gives $1-|z{|}^2=4u/|t+1{|}^2$ and $\mathrm{Im}z=2v/|t+1{|}^2$. The lens condition is therefore exactly $u>|v|$. Multiplication by $1+i$ converts that to positive real and imaginary parts. Thus the displayed Möbius transformation maps the lens bijectively to the first quadrant. Its two boundary arcs meet at $-1,1$, which map to zero and infinity. The centre zero maps to $1+i$. The derivative never vanishes inside the lens.

For nonzero real $a,b$, the two conditions describe circles $(x-1/(2a){)}^2+y^2=1/(4a^2)$ and $x^2+(y+1/(2b){)}^2=1/(4b^2)$, respectively, with the origin removed because $1/z$ is undefined there. Their unique actual intersection is $z=(a-ib)/(a^2+b^2)$. The map $z\mapsto 1/z$ is conformal away from zero and maps these circles to perpendicular coordinate lines, so they intersect orthogonally. The excluded origin is an intersection of their closures, not of the original loci. For $a=b=1$, both lie inside the unit circle and touch it at $1$ and $-i$.

For a nonzero integer $n$, the equation $((z+1)/(z-1){)}^n=i$ has $|n|$ distinct roots given by the displayed formula for $0\le k<|n|$. Put $w=(z+1)/(z-1)$; its roots have unit modulus and angles $(\pi /2+2\pi k)/n$. The inverse map is $(w+1)/(w-1)=-i\cot (\arg (w)/2)$, proving that all these roots are purely imaginary. The value $w=1$ never occurs. For $n=0$ there are no solutions; for negative $n$ the original map's zero at $z=-1$ is excluded as well as its pole at $z=1$.

Let $T_a(z)=z+a$, $S_k(z)=kz$ with $k\ne 0$, and $I(z)=1/z$ on the Riemann sphere. For $F(z)=(az+b)/(cz+d)$ and $\Delta =ad-bc\ne 0$, $c=0$ gives $F=T_{b/d}\circ S_{a/d}$. If $c\ne 0$, division gives $F=T_{a/c}\circ S_{-\Delta /c^2}\circ I\circ T_{d/c}$. The expressions agree at poles and at infinity by the sphere conventions. These elementary maps therefore generate the Möbius transformation group.

For a nonidentity Möbius transformation whose two fixed points are antipodal under stereographic projection, conjugate by a sphere rotation sending them to $0,\infty$. The conjugated map fixes both, hence is $z\mapsto az$ with $a\ne 0$. If $|a|=1$ it is an axial rotation; if $|a|<1$ every finite orbit tends to zero; if $|a|>1$ every nonzero orbit tends to infinity. Undoing the rotation gives either a sphere rotation or one attracting fixed point with the other as the sole exceptional initial point.

A nonidentity Möbius transformation has one or two distinct fixed points on the Riemann sphere. If there are two, send them to zero and infinity; the conjugated map fixes both and is a nonzero scaling. If there is one, send it to infinity; the resulting affine map cannot have slope different from one, since that would give another fixed point. It is a nonzero translation, which a scaling conjugates to $z+1$. The identity is the scaling $\mu =1$. Fixed-point sets are carried bijectively under conjugacy, so a translation is not conjugate to a scaling.

A nonidentity Möbius transformation with only one fixed point is conjugate to a nonzero translation of the Riemann sphere. Its translation iterates are $z+nb$, tending to infinity in the topology of the Riemann sphere for finite $z$, with infinity itself fixed. Conjugating back proves that every point tends to the unique fixed point. This is convergence on the sphere; if the fixed point is finite, intermediate iterates can still pass through infinity.

A nonidentity Möbius transformation of finite order $n$ is conjugate to multiplication by a primitive $n$th root of unity. Zero and infinity are fixed; for any other point, a return after $k$ iterates requires ${\mu }^k=1$, whose least positive solution is $n$. Thus its only orbit sizes are one and $n$. For the identity, every orbit has size one. This is a direct orbit computation, independent of an appeal to the orbit-stabilizer theorem.

This Möbius transformation maps the upper half-plane onto the unit disk. For $\mathrm{Im}z>0$, the identity $1-|C(z){|}^2=4\mathrm{Im}z/|z+i{|}^2$ puts the image inside the unit disk. Its inverse is $z=i(1+w)/(1-w)$, so the map is a conformal bijection. On the extended real boundary it maps onto the unit circle. It is the scalar complex-analytic version of a Cayley map, rather than the matrix parametrization used in Cayley transform.

A nonconstant Möbius transformation mapping the real line into the unit circle has the displayed form with $|\lambda |=1$ and $k$ nonreal. Indeed $|ax+b{|}^2=|cx+d{|}^2$ for real $x$ yields $|a|=|c|\ne 0$, $|b/a|=|d/c|$ and $\mathrm{Re}(b/a)=\mathrm{Re}(d/c)$. Thus $d/c$ is $b/a$ or its complex conjugate; the first possibility makes the map constant, so the second is necessary. Conjugate numerator and denominator have equal modulus on the real line, proving the converse as well.

Let $T_b(z)=z+b$ and $I(z)=1/z$ on the Riemann sphere. For $s\ne 0$, direct composition gives the displayed identity. Every nonzero complex scaling $z\mapsto \lambda z$ is obtained by choosing $s^2=-\lambda$. Since translations, scalings and reciprocal inversion generate the Möbius group, translations and reciprocal inversion alone also generate it.

For $|w|<R$, this Möbius transformation maps the radius-$R$ disc to the unit disc, vanishes at $w$ and has modulus one on the boundary. Expanding gives $|R^2-\overline{w}z{|}^2=R^2|z-w{|}^2$ when $|z|=R$; its pole lies outside the disc. Dividing a holomorphic function by factors corresponding to its interior zeros removes those zeros without changing its boundary modulus. This gives a direct factorization proof of Jensen's formula.

The Möbius transformation $M(z)=(3z-1)/(3-z)$ fixes the antipodal pair $1,-1$ under stereographic projection. But $M^{\prime }(1)=2$, so it multiplies infinitesimal spherical lengths at that fixed point by two, as follows from the round line element $2|dz|/(1+|z{|}^2)$. It is not an isometry and hence not a sphere rotation. Fixing an antipodal pair is therefore necessary for a nontrivial rotational Möbius transformation, but is not sufficient.

A Möbius transformation with two distinct fixed points is conjugate to $z\mapsto az$. It is loxodromic when $|a|\ne 1$; on three-dimensional hyperbolic space it translates along the axis joining its fixed points and rotates around that axis by $\arg a$. The positive-real multiplier case has no rotation and is also called hyperbolic. When listing elliptic, parabolic, hyperbolic and loxodromic as separate types, loxodromic refers to the remaining case with nonzero rotation. A determinant-one matrix with nonreal trace has reciprocal eigenvalues that are neither real nor of modulus one, so it supplies this latter case.

The special linear group $S{L}_2(\mathbb{R})$ acting by Möbius transformations has exactly three group orbits on the Riemann sphere: the extended real line and the two open half-planes. The identity $\mathrm{Im}((az+b)/(cz+d))=\mathrm{Im}z/|cz+d{|}^2$ preserves these sets. Upper triangular determinant-one matrices send $i$ to any prescribed point of the complex upper half-plane, and similarly send $-i$ through the lower half-plane. The stabilizer subgroups of $i$ and $-i$ are both $SO(2)$.

Two disjoint ordinary circle boundaries can be taken to distinct concentric circles by a Möbius transformation. After their centres are put at $0,d$ with radii $a,b$, the map $(z-s)/(z-t)$ works when $st=a^2$ and $s+t=(d^2+a^2-b^2)/d$. Disjointness makes the quadratic have distinct real roots. Already concentric circles require no transformation.

For four distinct points of the Riemann sphere, their cross-ratio iswith the evident limiting conventions at infinity. Every Möbius transformation preserves it.

For the convention in cross-ratio normalization of four projective points, permuting four distinct projective points produces the six displayed rational expressions. Interchanging the first pair sends $\lambda$ to $1/\lambda$; interchanging the middle pair sends it to $1-\lambda$. The four permutations consisting of the identity and the three double transpositions form a Klein four-group preserving the cross-ratio. The resulting six-element action gives every displayed expression.

For real $\lambda \notin \{0,1\}$, repeated values occur exactly for $\lambda \in \{-1,1/2,2\}$, and then the orbit has size three. To prove this, a repetition gives a nonidentity stabilizer after composing one operation with the other's inverse. Its fixed-point equations give $-1$, $1/2$, $2$, or ${\lambda }^2-\lambda +1=0$, with the last case nonreal. With one point at infinity, $\lambda =(x_1-x_3)/(x_1-x_2)$; the three real exceptional values mean that one finite point is the midpoint of the other two. Thus these harmonic conjugate points configurations have three collinear equally spaced finite points in that affine coordinate.

For four distinct projective points on the complex projective line, use the conventionwhere $[p_i,p_j]$ is the determinant of chosen representative vectors. Rescaling any representative cancels, and applying an invertible matrix multiplies every bracket by its determinant, which also cancels. Thus the expression is a projective invariant. The projective general linear group action on the projective line has a unique element $T$ taking the first three points to $\infty ,0,1$. In those coordinates the displayed expression is exactly $T(P_3)$. Consequently the cross-ratio completely determines the orbit of the ordered quadruple under the projective linear group. Its value lies in $\mathbb{C}\setminus \{0,1\}$.

An injective self-map of the complex projective line preserving all cross-ratios is a projective transformation; neither continuity nor prior surjectivity is needed. Choose the unique projectivity agreeing with the map on three distinct projective points and compose with its inverse. The resulting map fixes those three points. Normalize them to $\infty ,0,1$. Preservation of the cross-ratio $(\infty 0:1z)=z$ now fixes every other point as well, proving that the normalized map is the identity map. Undoing the composition proves the assertion.

Distinct projective points $C,D$ on the projective line are harmonic conjugates relative to distinct $A,B$ when $(AB:CD)=-1$. Use the cross-ratio convention $(AB:CD)=[a,c][b,d]/([a,d][b,c])$, where the brackets are two-by-two determinants. In an affine coordinate with $A=0$, $B=\infty$, the condition is $D=-C$. A complete quadrangle constructs the fourth point using joins and intersections only. Characteristic two is excluded because then $-1=1$, and the four points could not be distinct.

Normalize three distinct points on a generalized circle by a Möbius transformation $f$ taking them to $1,0,\infty$. The inverse image of the extended real line is their generalized circle, and $J=f^{-1}\circ c\circ f$, with $c(z)=\overline{z}$, fixes that circle pointwise and squares to the identity. Changing the three normalizing points replaces $f$ by $M\circ f$ with a real-coefficient Möbius map $M$. Since $M$ commutes with conjugation, $J$ is unchanged. Thus this construction depends only on the circle. It is inversion in a circle for an ordinary circle and Euclidean reflection for a straight line, each extended to the Riemann sphere.

For four vertices in cyclic order on a circle, the product of the diagonal lengths is the sum of the products of opposite side lengths. One proof uses the cross-ratio with second point mapped to zero: its negative sign gives $|r|+1=|1-r|$, which becomes the displayed length relation.

Every Möbius transformation preserves the cross-ratio. The identity $M(u)-M(v)=(ad-bc)(u-v)/[(cu+d)(cv+d)]$ makes all extra factors cancel in the ratio; limiting conventions handle points at infinity.

This cross-ratio convention is the value of $z_3$ under the Möbius transformation sending $z_1,z_2,z_4$ to $1,0,\infty$. It is one minus the convention $(z_1-z_3)(z_2-z_4)/[(z_1-z_2)(z_3-z_4)]$. Exchanging the first two points replaces $r$ by $1-r$. For four points in cyclic order on a circle, this convention gives $r<0$ and $1-r>1$.

Normalize three distinct points of the Riemann sphere by a Möbius transformation $f$ sending them to $\infty ,0,1$. A fourth point lies on their generalized circle exactly when $f$ takes a real value there: the image of that generalized circle is the extended real line. Other standard cross-ratio ordering conventions differ by real fractional-linear transformations and preserve this reality criterion.

A generalized circle is either a Euclidean circle or a straight line together with the point at infinity. Every Möbius transformation maps generalized circles to generalized circles. A Euclidean circle becomes a line exactly when it contains the pole, the preimage of infinity.

Write a generalized circle equation as $a|z{|}^2+b\bar{z}+\bar{b}z+c=0$, where $a,c$ are real. Under the reciprocal Möbius transformation $z=1/w$, with $w\ne 0$, multiply by $|w{|}^2$ to obtainThus the coefficient triple transforms as $(a,b,c)\mapsto (c,\bar{b},a)$. A nondegenerate circle through zero has $c=0$ and becomes a straight line; a straight line missing zero becomes a circle through zero on the Riemann sphere. When working only with finite nonzero coordinates, retain the excluded point $w=0$ even if it satisfies the multiplied equation.

This reciprocal transformation is different from inversion in a circle, whose unit-circle formula is $z\mapsto 1/\bar{z}$. The latter reverses orientation; the reciprocal transformation is holomorphic away from its pole.

Inversion in a circle of centre $a$ and radius $R$ is an orientation-reversing involution of the Riemann sphere fixing that circle pointwise and exchanging its interior and exterior. It is an anti-Möbius transformation. Reflection in a line is the corresponding inversion for a generalized circle through infinity. A composition of two such inversions is a Möbius transformation.

The transformations that take every Euclidean circle to a Euclidean circle are preciselyThey form the stabilizer of infinity in the Möbius group and are not a normal subgroup.

A fixed point of a Möbius transformation satisfies a quadratic equation on the Riemann sphere.

For $f(z)=(az+b)/(cz+d)$ with distinct finite fixed points of a Möbius transformation $\alpha ,\beta$, the fixed-point equation gives $f(z)-\alpha =(a-c\alpha )(z-\alpha )/(cz+d)$. Dividing the two identities gives $k=(a-c\alpha )/(a-c\beta )$, which is nonzero because $(a-c\alpha )(a-c\beta )=ad-bc\ne 0$. The identities extend over the Riemann sphere. This conjugates the transformation to a dilation.

For $f(z)=(az+b)/(z+1)$, the finite fixed points are the roots ofThus a repeated fixed point or two prescribed fixed points determine $a$ and $b$ by the usual sum and product of roots.

Every nonidentity finite-order Möbius transformation has exactly two fixed points on the Riemann sphere. A transformation with one repeated fixed point is conjugate to a nontrivial translation and therefore has infinite order.

A condition $\arg ((z-z_1)/(z-z_2))=\theta$ fixes the oriented angle subtended by $z_1,z_2$. Its locus is an arc of a circle through those two points, with the endpoints excluded.

For determinant-one real representatives, trace magnitude below, equal to, or above two gives elliptic, parabolic, or hyperbolic type.

The Möbius transformations commuting with $J(z)=1/\overline{z}$ are exactlyThey preserve the open unit disc exactly when $|a|>|b|$. The identityproves the latter assertion.

A Möbius transformation is represented by an invertible two-by-two matrix, with nonzero scalar multiples representing the same map.