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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 $\mathbb{C}\setminus \{0,1\}$.

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 $PS{L}_2(\mathbb{C})=S{L}_2(\mathbb{C})/\{\pm I\}$.

Nonzero scalar matrices form a subgroup $N$ of the general linear group, since products and inverses remain nonzero scalar matrices. Moreover, for every invertible $g$,Thus $N$ lies in the center of a group and is a normal subgroup. The quotient group is the complex projective linear group ${\mathrm{PGL}}_n(\mathbb{C})$, whose elements identify invertible matrices differing by a nonzero scalar.

The canonical group homomorphism $\pi :g\mapsto [g]$ obeys $[g][h]=[gh]$. Composing it with the given group representation givesConsequentlyThis is the projective representation associated with $R$: a nonzero scalar acts trivially on each line in the projective space of $V$. Its kernel of a group homomorphism consists of those $g$ for which $R(g)$ is scalar, which can be larger than the kernel of a group homomorphism of $R$ itself.

Use the matrices of solution. Their projective classes satisfy $[-M]=[M]$, since $-I_2\in N$. The group homomorphism is therefore explicitlyThese four classes are distinct: the two diagonal representatives are not scalar multiples of one another, the two off-diagonal representatives are not scalar multiples of one another, and a diagonal representative cannot be a nonzero scalar multiple of an off-diagonal one.

Although $A$ and $B$ anticommute as matrices, their projective classes commute:Furthermore $[A{]}^2=[B{]}^2=[I_2]$ and $[C{]}^2=[-I_2]=[I_2]$. Thus every nonidentity element of the image has order two. An explicit group isomorphism to the Klein four-group isIndeed reordering products introduces only a sign, which disappears in the projective linear group, so this map turns multiplication into addition modulo two in each coordinate. Equivalently, $\ker \tilde{R}=G\cap N=\{\pm I_2\}$ and the first isomorphism theorem givesThis projective image of the square symmetry group illustrates how a noncommutative group can have an Abelian projective image: its scalar commutator becomes the identity class.

Use the cross-ratio convention compatible with the permutation identities in this paper. Represent a projective point $P_i\in {\mathbb{CP}}^1$ by a nonzero vector $p_i\in {\mathbb{C}}^2$, and write $[p_i,p_j]=\det (p_i,p_j)$. DefineThe denominator is nonzero because the four projective points are distinct. Rescaling any one representative multiplies numerator and denominator by the same nonzero scalar, so the definition is independent of the chosen homogeneous coordinates. In finite affine coordinates it readswith the point at infinity of the complex projective line handled by the homogeneous formula.

A projective transformation is represented by an invertible matrix $A\in {\mathrm{GL}}_2(\mathbb{C})$. Since $[A{p}_i,A{p}_j]=\det (A)[p_i,p_j]$, the two factors of $\det (A)$ in the numerator cancel the two in the denominator. Therefore every projective transformation preserves this cross-ratio. This also shows that the formula descends from matrices to the projective linear group, whose representatives differ by a nonzero common scalar.

The needed general-position theorem is sharp triple transitivity of the projective linear group on the projective line: for any two ordered triples of distinct projective points, exactly one projective transformation takes the first triple to the second.

In particular, let $T_P$ take $(P_0,P_1,P_2)$ to $(\infty ,0,1)$. In these coordinates,so cross-ratio invariance gives $T_P(P_3)=\lambda$, where $\lambda =(P_0{P}_1:P_2{P}_3)$. Distinctness implies $\lambda \ne 0,1$ and $\lambda$ is finite. The entire ordered quadruple is thus normalized toFor a second quadruple $(Q_0,Q_1,Q_2,Q_3)$ with the same cross-ratio, its normalizing map $T_Q$ yields the same quadruple. ConsequentlyThis proves the equivalence. Uniqueness on the first three points also shows that the projectivity carrying one ordered quadruple to the other is unique. The cross-ratio normalization of four projective points therefore gives a complete invariant for this ordered classification.

A projective transformation of $\mathbb{P}({\mathbb{F}}^{n+1})$ is induced by an invertible linear map $A:{\mathbb{F}}^{n+1}\to {\mathbb{F}}^{n+1}$:This is well defined because replacing $v$ by a nonzero scalar multiple does not change $[Av]$. Multiplying $A$ by a nonzero scalar also leaves $\tau$ unchanged. Conversely these are precisely the transformations in the projective linear group ${\mathrm{PGL}}_{n+1}(\mathbb{F})$. No field automorphism is included in this definition.

Let $A=\mathrm{diag}(1,-1)$ and $B=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. They generate the dihedral group of a square, realized by $\{\pm I,\pm A,\pm B,\pm AB\}$. Its image in the projective linear group has four elements, since opposite matrices become equal. The projective classes of $A,B$ commute because $AB=-BA$, and each squares to the identity. Sending $[A^a{B}^b]$ to $(a,b)$ gives the Klein four-group. This is an example in which a nontrivial scalar commutator disappears after passing to a projective representation.

A scalar matrix is a scalar multiple of the identity matrix, equivalently a diagonal matrix with all diagonal entries equal. It commutes with every square matrix of the same size. In $G{L}_n(F)$, the nonzero scalar matrices form the central subgroup removed to define the projective linear group.

For any field $F$, a projective linear group element is uniquely determined by its images of three distinct points of the projective line, and any ordered distinct target triple is attainable. Given representatives $v_z,v_x$ forming a basis and $v_y=A{v}_z+B{v}_x$, the matrix with columns $A{v}_z,B{v}_x$ sends $(0,1,\infty )$ to $(x,y,z)$. Distinctness forces $A,B\ne 0$. The matrices realizing one projective map differ by a nonzero scalar; over ${\mathbb{F}}_p$ there are $p-1$ such matrices.