Famous Theorems of Mathematics/Boy's surface

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Property of R. Bryant's parametrizationEdit

If z is replaced by the negative reciprocal of its complex conjugate, then the functions g1, g2, and g3 of z are left unchanged.

ProofEdit

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 surfaceEdit

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.

ProofEdit

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.