Boy's surface/Proofs
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[edit] 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.
[edit] 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 g1' = g1 since, for any complex number z,
Let g2′ be obtained from g2 by substituting z with Then we obtain
therefore g2' = g2 since, for any complex number z,
Let g3′ be obtained from g3 by substituting z with Then we obtain
therefore g3' = g3. Q.E.D.
[edit] 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.
[edit] 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 ei4π = ei2π = 1, it follows that
therefore g3' = g3. 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 ei8π / 3 = ei2π / 3,
Then, letting ei4π / 3 = e − i2π / 3 in the denominator yields
Now, applying the complex-algebraic identity, and letting
we get
Both Re and Im 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 Re 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.