The Cross-Ratio Function
Created | Updated Dec 13, 2008
Let x be a complex number. The formal values of the cross-ratio function are x, 1-x, (1-x)-1, -x(1-x)-1, -x-1(1-x), and x-1 (where x≠0 or 1). The values equal x, f(x), gf(x), fgf(x), gfgf(x), fgfgf(x) where f(x)=1-x and g(x)=1/x. Let h0(x)=x, h1(x)=f(x), h2(x)=gf(x), h3(x)=fgf(x), .... Then h0=h6 and hence hi=hi+6, i=0, 1, 2, .... In this paper, the equality of the functions h0 and h6 is considered to be the defining concept of the cross-ratio. This allows generalization. For example, if f(x)=2-x and g(x)=2/x where x≠0, 1, or 2, then h0=h8, and if f(x)=3-x and g(x)=3/x where x≠0, 1, 2, 3/2, or 3, then h0=h12. However, as will be shown, if f(x)=k-x and g(x)=k/x where k is a natural number other than 1, 2, or 3, then h0≠hi for any i>0. Let x1 and x2 be the roots of x2-x+1=0. Then x1+x2=1 and x1x2=1 (the symmetric equations), and hence x2=1-x1 and x1=1/x2. Therefore if x is a root of x2-x+1, then h0(x)=h2(x)=h4(x)=... and h1(x)=h3(x)=h5(x)=.... So there is a relationship (in some sense) between the equation x2-x+1=0 and the cross-ratio. This is the motivation for seeking "cross-ratios" associated with the general quadratic equation x2-ax+b=0.
(1) Ui=(x1i-x2i)/(x1-x2) where x1 and x2 are the roots of x2-ax+b=0.
(2) If i divides j, then Ui divides Uj if Ui≠0, or Uj=0 if Ui=0.
(3) Ui2-Ui-1Ui+1=bi-1.
Let h0(x)=x, h1(x)=f(x), h2(x)=gf(x), h3(x)=fgf(x), ....
(4) h1(x), h2(x), h3(x), ..., hk(x) are defined only if Uix≠Ui+1, i=1, 2, 3, ..., k. If i is odd and 0<i≤k, then hi(x)=(Uj+1-Ujx)/(Uj-Uj-1x) where j=(i+1)/2. If i is even and 0<i≤k, then hi(x)=b(Uj-Uj-1x)/(Uj+1-Ujx) where j=i/2.
Proof: The proof is by induction. Suppose hi(x)=b(Uj-Uj-1x)/(Uj+1-Ujx) where i is even, i≥0, and j=i/2. Then hi+1(x)=f(hi(x))=a-b(Uj-Uj-1x)/(Uj+1-Ujx)=[(aUj+1-bUj)-(aUj-bUj-1)x]/(Uj+1-Ujx)=(Uj+2-Uj+1x)/(Uj+1-Ujx). Therefore hi+1(x)=(U[(i+1)+1]/2+1-U[(i+1)+1]/2)/(U[(i+1)+1]/2-U[(i+1)+1]/2-1) where i+1 is odd. Now suppose hi(x)=(Uj+1-Ujx)/(Uj-Uj-1x) where i is odd, i>0, and j=(i+1)/2. Then hi+1(x)=g(hi(x))=b(Uj-Uj-1x)/(Uj+1-Ujx). Therefore hi+1(x)=b(U(i+1)/2-U(i+1)/2-1x)/(U(i+1)/2+1-U(i+1)/2x) where i+1 is even. Also, h1(x)=f(x)=a-x=(U2-U1x)/(U1-U0x) and h2(x)=b/(a-x)=b(U1-U0x)/(U2-U1x).
Proof: Suppose h0(x)=hi(x) where i is odd. Then x=(Uj+1-Ujx)/(Uj-Uj-1x) where j=(i+1)/2 and hence -Uj-1x2+2Ujx-Uj+1=0. Therefore x=[-2Uj±√(4Uj2-4Uj-1Uj+1)]/(-2Uj-1) if Uj-1≠0, or x=Uj+1/(2Uj) if Uj-1=0. Then since Uj2-Uj-1Uj+1=bj-1, x=(Uj±b(j-1)/2/Uj-1 if Uj-1≠0. Conversely, if x=(Uj±b(j-1)/2)/Uj-1 then h0(x)=hi(x), and if x=Uj+1/(2Uj), Uj-1=0, then h0(x)=hi(x).
Therefore h0≠hi where i is odd (since h0(x)=hi(x), i is odd, if and only if x has a certain value). (Some U values corresponding to the cross-ratio are U0=0, U1=1, U2=1, U3=0, U4=-1, U5=-1, U6=0. Therefore h0(x)=h1(x) if and only if x=U2/(2U1)=1/2, h0(x)=h3(x) if and only if x=(U2±1)/U1=0, 2, and h0(x)=h5(x) if and only if x=(U3±1)/U2=1, -1. Therefore h0(x)=hi(x) where i is odd and a=b=1 if and only if x=1/2, 2, or -1 [since x=0, 1 are not considered to be in the domain of f and g].)
(6) h0(x)=hi(x) where i>0, i is even, if and only if Ui/2(x2-ax+b)=0.
Proof: Suppose h0(x)=hi(x) where i>0, i is even. Then x=b(Uj-Uj-1x)/(Uj+1-Ujx) where j=i/2 and hence Ujx2-(Uj+1+bUj-1)x+bUj=0, that is, Uj(x2-ax+b)=0. Conversely, if Ui/2(x2-ax+b)=0, then h0(x)=hi(x).
There can be up to i/2+1 x values for which hi(x) is not defined if h0(x)=hi(x), i>0, i is even. These x values and the two x values satisfying x2-ax+b=0 are not considered to be in the domain of f and g. Therefore h0=hi where i>0, i is even, if and only if Ui/2=0. If a=0, then U2=0, h1(x)=-x, h2(x)=-b/x, h3(x)=b/x, and h4(x)=x=h0(x). This is a trivial "cross-ratio", so it is stipulated that a≠0 in the following.
Proof: First, some lemmas will be proved. Let c(i,j) denote i "choose" j (a binomial coefficient).
(8) If i is even, Ui=ai-1-c(i-2,1)ai-3b+c(i-3,2)ai-5b2-c(i-4,3)ai-7b3+...+(-1)i/2-2c(i/2+1,i/2-2)a3bi/2-2+(-1)i/2-1c(i/2,i/2-1)abi/2-1. If i is odd, Ui=ai-1-c(i-2,1)ai-3b+c(i-3,2)ai-5b2-c(i-4,3)ai-7b3+...+(-1)(i-1)/2-1c((i-1)/2+1,(i-1)/2-1)a2b(i-1)/2-1+(-1)(i-1)/2c((i-1)/2,(i-1)/2)b(i-1)/2.
Proof: The proof is by induction. Suppose Ui=ai-1-c(i-2,1)ai-3b+c(i-3,2)ai-5b2-...+(-1)i/2-1c(i/2,i/2-1)abi/2-1 and Ui-1=ai-2-c(i-3,1)ai-4b+c(i-4,2)ai-6b2-...+(-1)(i-2)/2c((i-2)/2,(i-2)/2)b(i-2)/2 where i is even. Then using the relations Ui+1=aUi-bUi-1, c(i,j-1)+c(i,j)=c(i+1),j), j=1, 2, 3, ..., i, and collecting terms gives Ui+1=ai-c(i-1,1)ai-2b+c(i-2,2)ai-4b2-...+(-1)i/2c(i/2,i/2)bi/2, that is, Ui+1=a(i+1)-1-c((i+1)-2,1)a(i+1)-3b+c((i+1)-3,2)a(i+1)-5b2-...+(-1)[(i+1)-1]/2c([(i+1)-1)]/2,[(i+1)-1]/2)b[(i+1)-1]/2 where i+1 is odd. Now suppose Ui=ai-1-c(i-2,1)ai-3b+c(i-3,2)ai-5b2-...+(-1)(i-1)/2c((i-1)/2,(i-1)/2)b(i-1)/2 and Ui-1=ai-2-c(i-3,1)ai-4b+c(i-4,2)ai-6b2-...+(-1)(i-1)/2-1c((i-1)/2,(i-1)/2-1)ab(i-1)/2-1 where i is odd. Then using the relations and collecting terms as before gives Ui+1=a(i+1)-1-c((i+1)-2,1)a(i+1)-3b+c((i+1)-3,2)a(i+1)-5b2-...+(-1)(i+1)/2-1c((i+1)/2,(i+1)/2-1)ab(i+1)/2-1 where i+1 is even. Finally, if i=1, Ui=ai-1=(-1)(i-1)/2c((i-1)/2,(i-1)/2)b(i-1)/2=1 and if i=2, Ui=ai-1=(-1)i/2-1c(i/2,i/2-1)abi/2-1=a.
(9) Ui=0, i>0, only if √(a2-4b) is imaginary.
Proof: Ui=0 only if (x1i-x2i)/(x1-x2)=0 where x1=[a+√(a2-4b)]/2 and x2=[a-√(a2-4b)]/2. Now {[a+√(a2-4b)]/2}i-{[a-√(a2-4b)]/2}i=(1/2i)ad(c(i,1)ai-2+c(i,3)ai-4d2+...+c(i,i-1)di-2) if i is even, or (1/2i)d(c(i,1)ai-1+c(i,3)ai-3d2+...+c(i,i)di-1) if i is odd, where d=√(a2-4b), d=x1-x2, therefore Ui=0 only if d is imaginary and hence Ui=0 only if b>0.
(10) Ui=0 where i>0, i is even, only if b=a2/q where q is a positive divisor of i/2. Ui=0 where i is odd only if b=a2.
Proof: Suppose Ui=0 where i>0, i is even. (Note that U2≠0.) Then by Lemma (8), b divides ai-1 and a2 divides c(i/2,i/2-1)bi/2-1. Let k=g.c.d.(a, b) and k1=g.c.d.(a, b/k). Then a{(a/k1)i/2-1(a/k)i/2-1-c(i-2,1)(a/k1)i/2-2(a/k)i/2-2[b/(kk1)]+c(i-3,2)(a/k1)i/2-3(a/k)i/2-3[b/(kk1)]2-...+(-1)i/2-2c(i/2+1,i/2-2)(a/k1)/(a/k)[b/(kk1)]i/2-2+(-1)i/2-1c(i/2,i/2-1)[b/(kk1)]i/2-1}=0. Now g.c.d.(a/k, b/k)=1 and g.c.d.(a/k1, b/(kk1))=1, therefore (a/k1)/(a/k) divides c(i/2,i/2-1), that is, a2/(kk1) divides i/2. Also, b/(kk1) divides 1. Therefore kk1=±b and hence b=a2/±q where q is a positive or negative divisor of i/2. By Lemma (9), b>0, therefore b=a2/q where q is a positive divisor of i/2. Now suppose Ui=0 where i is odd. Then (a/k1)(i-1)/2(a/k)(i-1)/2-c(i-2,1)(a/k1)(i-1)/2-1(a/k)(i-1)/2-1[b/(kk1)]+c(i-3,2)(a/k1)(i-1)/2-2(a/k)(i-1)/2-2[b/(kk1)]2-...+(-1)(i-1)/2-1c((i-1)/2+1,(i-1)/2-1)(a/k1)(a/k)[b/(kk1)](i-1)/2-1+(-1)(i-1)/2c((i-1)/2,(i-1)/2)[b/(kk1)](i-1)/2=0. Therefore (a/k1)/(a/k) divides c((i-1)/2,(i-1)/2), that is, a2/(kk1) divides 1. Also, b/(kk1) divides 1. Therefore kk1=±b and hence b=a2.
√(a2-4b) is imaginary, b=a2/q, only if q=1, 2, or 3. Therefore Ui=0, i>0, i is even, only if q=1, 2, or 3. Now U3=a2-b, U4=a(a2-2b), and U6=a5-4a3b+3ab2=a(a2-3b)(a2-b), therefore Ui=0, i>0, i is even, if q=1, 2, or 3.
Comptes Rendus Paris, 82, 1876, pp. 1303-5.
Relationship of "h" Functions to a Lucas Series
Let f(x)=a-x and g(x)=b/x where a and b are integers, b≠0. Let Ui be defined by the recurrence relation Ui-aUi-1+bUi-2=0, i=2, 3, 4, ..., U1=1, U0=0 (a Lucas series). Some properties of this series (commonly cited in the mathematical literature) are;(1) Ui=(x1i-x2i)/(x1-x2) where x1 and x2 are the roots of x2-ax+b=0.
(2) If i divides j, then Ui divides Uj if Ui≠0, or Uj=0 if Ui=0.
(3) Ui2-Ui-1Ui+1=bi-1.
Let h0(x)=x, h1(x)=f(x), h2(x)=gf(x), h3(x)=fgf(x), ....
(4) h1(x), h2(x), h3(x), ..., hk(x) are defined only if Uix≠Ui+1, i=1, 2, 3, ..., k. If i is odd and 0<i≤k, then hi(x)=(Uj+1-Ujx)/(Uj-Uj-1x) where j=(i+1)/2. If i is even and 0<i≤k, then hi(x)=b(Uj-Uj-1x)/(Uj+1-Ujx) where j=i/2.
Proof: The proof is by induction. Suppose hi(x)=b(Uj-Uj-1x)/(Uj+1-Ujx) where i is even, i≥0, and j=i/2. Then hi+1(x)=f(hi(x))=a-b(Uj-Uj-1x)/(Uj+1-Ujx)=[(aUj+1-bUj)-(aUj-bUj-1)x]/(Uj+1-Ujx)=(Uj+2-Uj+1x)/(Uj+1-Ujx). Therefore hi+1(x)=(U[(i+1)+1]/2+1-U[(i+1)+1]/2)/(U[(i+1)+1]/2-U[(i+1)+1]/2-1) where i+1 is odd. Now suppose hi(x)=(Uj+1-Ujx)/(Uj-Uj-1x) where i is odd, i>0, and j=(i+1)/2. Then hi+1(x)=g(hi(x))=b(Uj-Uj-1x)/(Uj+1-Ujx). Therefore hi+1(x)=b(U(i+1)/2-U(i+1)/2-1x)/(U(i+1)/2+1-U(i+1)/2x) where i+1 is even. Also, h1(x)=f(x)=a-x=(U2-U1x)/(U1-U0x) and h2(x)=b/(a-x)=b(U1-U0x)/(U2-U1x).
Necessary and Sufficient Conditions for Equal "h" Functions
(5) h0(x)=hi(x) where i is odd if and only if x=(Uj±b(j-1)/2)/Uj-1 where j=(i+1)/2 and Uj-1≠0, or x=Uj+1/(2Uj) where Uj-1=0.Proof: Suppose h0(x)=hi(x) where i is odd. Then x=(Uj+1-Ujx)/(Uj-Uj-1x) where j=(i+1)/2 and hence -Uj-1x2+2Ujx-Uj+1=0. Therefore x=[-2Uj±√(4Uj2-4Uj-1Uj+1)]/(-2Uj-1) if Uj-1≠0, or x=Uj+1/(2Uj) if Uj-1=0. Then since Uj2-Uj-1Uj+1=bj-1, x=(Uj±b(j-1)/2/Uj-1 if Uj-1≠0. Conversely, if x=(Uj±b(j-1)/2)/Uj-1 then h0(x)=hi(x), and if x=Uj+1/(2Uj), Uj-1=0, then h0(x)=hi(x).
Therefore h0≠hi where i is odd (since h0(x)=hi(x), i is odd, if and only if x has a certain value). (Some U values corresponding to the cross-ratio are U0=0, U1=1, U2=1, U3=0, U4=-1, U5=-1, U6=0. Therefore h0(x)=h1(x) if and only if x=U2/(2U1)=1/2, h0(x)=h3(x) if and only if x=(U2±1)/U1=0, 2, and h0(x)=h5(x) if and only if x=(U3±1)/U2=1, -1. Therefore h0(x)=hi(x) where i is odd and a=b=1 if and only if x=1/2, 2, or -1 [since x=0, 1 are not considered to be in the domain of f and g].)
(6) h0(x)=hi(x) where i>0, i is even, if and only if Ui/2(x2-ax+b)=0.
Proof: Suppose h0(x)=hi(x) where i>0, i is even. Then x=b(Uj-Uj-1x)/(Uj+1-Ujx) where j=i/2 and hence Ujx2-(Uj+1+bUj-1)x+bUj=0, that is, Uj(x2-ax+b)=0. Conversely, if Ui/2(x2-ax+b)=0, then h0(x)=hi(x).
There can be up to i/2+1 x values for which hi(x) is not defined if h0(x)=hi(x), i>0, i is even. These x values and the two x values satisfying x2-ax+b=0 are not considered to be in the domain of f and g. Therefore h0=hi where i>0, i is even, if and only if Ui/2=0. If a=0, then U2=0, h1(x)=-x, h2(x)=-b/x, h3(x)=b/x, and h4(x)=x=h0(x). This is a trivial "cross-ratio", so it is stipulated that a≠0 in the following.
Necessary and Sufficient Conditions for Lucas Series Values of Zero
(7) Ui=0, i>0, if and only if b=a2/k where k=1, 2, or 3 and k divides a. If b=a2, then Ui=0, i>0, if and only if 3 divides i. If b=a2/2, then Ui=0, i>0, if and only if 4 divides i. If b=a2/3, then Ui=0, i>0, if and only if 6 divides i.Proof: First, some lemmas will be proved. Let c(i,j) denote i "choose" j (a binomial coefficient).
(8) If i is even, Ui=ai-1-c(i-2,1)ai-3b+c(i-3,2)ai-5b2-c(i-4,3)ai-7b3+...+(-1)i/2-2c(i/2+1,i/2-2)a3bi/2-2+(-1)i/2-1c(i/2,i/2-1)abi/2-1. If i is odd, Ui=ai-1-c(i-2,1)ai-3b+c(i-3,2)ai-5b2-c(i-4,3)ai-7b3+...+(-1)(i-1)/2-1c((i-1)/2+1,(i-1)/2-1)a2b(i-1)/2-1+(-1)(i-1)/2c((i-1)/2,(i-1)/2)b(i-1)/2.
Proof: The proof is by induction. Suppose Ui=ai-1-c(i-2,1)ai-3b+c(i-3,2)ai-5b2-...+(-1)i/2-1c(i/2,i/2-1)abi/2-1 and Ui-1=ai-2-c(i-3,1)ai-4b+c(i-4,2)ai-6b2-...+(-1)(i-2)/2c((i-2)/2,(i-2)/2)b(i-2)/2 where i is even. Then using the relations Ui+1=aUi-bUi-1, c(i,j-1)+c(i,j)=c(i+1),j), j=1, 2, 3, ..., i, and collecting terms gives Ui+1=ai-c(i-1,1)ai-2b+c(i-2,2)ai-4b2-...+(-1)i/2c(i/2,i/2)bi/2, that is, Ui+1=a(i+1)-1-c((i+1)-2,1)a(i+1)-3b+c((i+1)-3,2)a(i+1)-5b2-...+(-1)[(i+1)-1]/2c([(i+1)-1)]/2,[(i+1)-1]/2)b[(i+1)-1]/2 where i+1 is odd. Now suppose Ui=ai-1-c(i-2,1)ai-3b+c(i-3,2)ai-5b2-...+(-1)(i-1)/2c((i-1)/2,(i-1)/2)b(i-1)/2 and Ui-1=ai-2-c(i-3,1)ai-4b+c(i-4,2)ai-6b2-...+(-1)(i-1)/2-1c((i-1)/2,(i-1)/2-1)ab(i-1)/2-1 where i is odd. Then using the relations and collecting terms as before gives Ui+1=a(i+1)-1-c((i+1)-2,1)a(i+1)-3b+c((i+1)-3,2)a(i+1)-5b2-...+(-1)(i+1)/2-1c((i+1)/2,(i+1)/2-1)ab(i+1)/2-1 where i+1 is even. Finally, if i=1, Ui=ai-1=(-1)(i-1)/2c((i-1)/2,(i-1)/2)b(i-1)/2=1 and if i=2, Ui=ai-1=(-1)i/2-1c(i/2,i/2-1)abi/2-1=a.
(9) Ui=0, i>0, only if √(a2-4b) is imaginary.
Proof: Ui=0 only if (x1i-x2i)/(x1-x2)=0 where x1=[a+√(a2-4b)]/2 and x2=[a-√(a2-4b)]/2. Now {[a+√(a2-4b)]/2}i-{[a-√(a2-4b)]/2}i=(1/2i)ad(c(i,1)ai-2+c(i,3)ai-4d2+...+c(i,i-1)di-2) if i is even, or (1/2i)d(c(i,1)ai-1+c(i,3)ai-3d2+...+c(i,i)di-1) if i is odd, where d=√(a2-4b), d=x1-x2, therefore Ui=0 only if d is imaginary and hence Ui=0 only if b>0.
(10) Ui=0 where i>0, i is even, only if b=a2/q where q is a positive divisor of i/2. Ui=0 where i is odd only if b=a2.
Proof: Suppose Ui=0 where i>0, i is even. (Note that U2≠0.) Then by Lemma (8), b divides ai-1 and a2 divides c(i/2,i/2-1)bi/2-1. Let k=g.c.d.(a, b) and k1=g.c.d.(a, b/k). Then a{(a/k1)i/2-1(a/k)i/2-1-c(i-2,1)(a/k1)i/2-2(a/k)i/2-2[b/(kk1)]+c(i-3,2)(a/k1)i/2-3(a/k)i/2-3[b/(kk1)]2-...+(-1)i/2-2c(i/2+1,i/2-2)(a/k1)/(a/k)[b/(kk1)]i/2-2+(-1)i/2-1c(i/2,i/2-1)[b/(kk1)]i/2-1}=0. Now g.c.d.(a/k, b/k)=1 and g.c.d.(a/k1, b/(kk1))=1, therefore (a/k1)/(a/k) divides c(i/2,i/2-1), that is, a2/(kk1) divides i/2. Also, b/(kk1) divides 1. Therefore kk1=±b and hence b=a2/±q where q is a positive or negative divisor of i/2. By Lemma (9), b>0, therefore b=a2/q where q is a positive divisor of i/2. Now suppose Ui=0 where i is odd. Then (a/k1)(i-1)/2(a/k)(i-1)/2-c(i-2,1)(a/k1)(i-1)/2-1(a/k)(i-1)/2-1[b/(kk1)]+c(i-3,2)(a/k1)(i-1)/2-2(a/k)(i-1)/2-2[b/(kk1)]2-...+(-1)(i-1)/2-1c((i-1)/2+1,(i-1)/2-1)(a/k1)(a/k)[b/(kk1)](i-1)/2-1+(-1)(i-1)/2c((i-1)/2,(i-1)/2)[b/(kk1)](i-1)/2=0. Therefore (a/k1)/(a/k) divides c((i-1)/2,(i-1)/2), that is, a2/(kk1) divides 1. Also, b/(kk1) divides 1. Therefore kk1=±b and hence b=a2.
√(a2-4b) is imaginary, b=a2/q, only if q=1, 2, or 3. Therefore Ui=0, i>0, i is even, only if q=1, 2, or 3. Now U3=a2-b, U4=a(a2-2b), and U6=a5-4a3b+3ab2=a(a2-3b)(a2-b), therefore Ui=0, i>0, i is even, if q=1, 2, or 3.
References
E. LucasComptes Rendus Paris, 82, 1876, pp. 1303-5.
