Involuntary

A Linear Transformation of period two. Since a Linear Transformation has the form,

$$\lambda'={\alpha \lambda+\beta\over\gamma\lambda+\delta},$$ (1)

applying the transformation a second time gives

$$\lambda''={\alpha\lambda'+\beta\over\gamma\lambda'+\delta} ... ...delta)\over(\alpha+\delta)\gamma\lambda+\beta\gamma+\delta^2}.$$ (2)

For an involuntary, $\lambda''=\lambda$, so

$$\gamma(\alpha+\delta)\lambda^2+(\delta^2-\alpha^2)\lambda-(\alpha+\delta)\beta=0.$$ (3)

Since each Coefficient must vanish separately,

$$\alpha\gamma+\gamma\delta = 0$$ (4)

$$\delta^2-\alpha^2 = 0$$ (5)

$$\alpha\beta+\beta\delta = 0.$$ (6)

The first equation gives $\delta=\pm\alpha$. Taking $\delta=\alpha$ would require $\gamma=\beta=0$, giving $\lambda=\lambda'$, the identity transformation. Taking $\delta=-\alpha$ gives $\delta=-\alpha$, so

$$\lambda'={\alpha\lambda+\beta\over\gamma\lambda-\alpha},$$ (7)

the general form of an Involution.

See also Cross-Ratio, Involution (Line)

References

Woods, F. S. Higher Geometry: An Introduction to Advanced Methods in Analytic Geometry. New York: Dover, pp. 14-15, 1961.