Famous Theorems of Mathematics/Boy's surface

Property of R. Bryant's parametrization If z is replaced by the negative reciprocal of its complex conjugate, then the functions g1, g2, and g3 of z are left unchanged. Proof Let g1′ be obtained from g1 by substituting z with Then we obtain Multiply both numerator and denominator by Multiply both numerator and denominator by -1, It is generally true for any complex number z and any integral power n that therefore therefore since, for any complex number z, Let g2′ be obtained from g2 by substituting z with Then we obtain therefore since, for any complex number z, Let g3′ be obtained from g3 by substituting z with Then we obtain therefore Q.E.D. Symmetry of the Boy's surface Boy's surface has 3-fold symmetry. This means that it has an axis of discrete rotational symmetry: any 120° turn about this axis will leave the surface looking exactly the same. The Boy's surface can be cut into three mutually congruent pieces. Proof Two complex-algebraic identities will be used in this proof: let U and V be complex numbers, then Given a point P(z) on the Boy's surface with complex parameter z inside the unit disk in the complex plane, we will show that rotating the parameter z 120° about the origin of the complex plane is equivalent to rotating the Boy's surface 120° about the Z-axis (still using R. Bryant's parametric equations given above). Let be the rotation of parameter z. Then the "raw" (unscaled) coordinates g1, g2, and g3 will be converted, respectively, to g′1, g′2, and g′3. Substitute z′ for z in g3(z), resulting in Since it follows that therefore This means that the axis of rotational symmetry will be parallel to the Z-axis. Plug in z′ for z in g1(z), resulting in Noticing that Then, letting in the denominator yields Now, applying the complex-algebraic identity, and letting we get Both and are distributive with respect to addition, and due to Euler's formula, so that Applying the complex-algebraic identities again, and simplifying to -1/2 and to produces Simplify constants, therefore Applying the complex-algebraic identity to the original g1 yields Plug in z′ for z in g2(z), resulting in Simplify the exponents, Now apply the complex-algebraic identity to g′2, obtaining Distribute the with respect to addition, and simplify constants, Apply the complex-algebraic identities again, Simplify constants, then distribute with respect to addition, Applying the complex-algebraic identity to the original g2 yields The raw coordinates of the pre-rotated point are and the raw coordinates of the post-rotated point are Comparing these four coordinates we can verify that In matrix form, this can be expressed as Therefore rotating z by 120° to z′ on the complex plane is equivalent to rotating P(z) by -120° about the Z-axis to P(z′). This means that the Boy's surface has 3-fold symmetry, quod erat demonstrandum.