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.