Boy's surface/Proofs

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This mathematics article is devoted entirely to providing mathematical proofs and support for claims and statements made in the article Boy's surface. This article is currently an experimental vehicle to see how well we can provide proofs and details for a math article without cluttering up the main article itself. See Wikipedia:WikiProject Mathematics/Proofs for some current discussion. This article is "experimental" in the sense that it is a test of one way we may be able to incorporate more detailed proofs in Wikipedia.

1 Property of R. Bryant's parametrization
- 1.1 Proof
2 Symmetry of the Boy's surface
- 2.1 Proof

[edit] Property of R. Bryant's parametrization

If z is replaced by the negative reciprocal of its complex conjugate, $- {1 \over z^\star},$ then the functions g₁, g₂, and g₃ of z are left unchanged.

[edit] Proof

Let g₁′ be obtained from g₁ by substituting z with $- {1 \over z^\star}.$ Then we obtain

$g_1' = -{3 \over 2} \mathrm{Im} \left( {- {1 \over z^\star} \left( 1 - {1 \over z^{\star 4} } \right) \over {1 \over z^{\star 6}} - \sqrt{5} {1 \over z^{\star 3}} - 1} \right).$

Multiply both numerator and denominator by $z^{\star 6},$

$g_1' = -{3 \over 2} \mathrm{Im} \left( {-z^\star (z^{\star 4} - 1) \over 1 - \sqrt{5} z^{\star 3} - z^{\star 6} } \right).$

Multiply both numerator and denominator by -1,

$g_1' = -{3 \over 2} \mathrm{Im} \left( {z^\star (z^{\star 4} - 1) \over z^{\star 6} + \sqrt{5} z^{\star 3} - 1} \right).$

It is generally true for any complex number z and any integral power n that

$(z^\star)^n = (z^n)^\star,$

therefore

$g_1' = -{3 \over 2} \mathrm{Im} \left( { z^\star (z^4 - 1)^\star \over (z^6 + \sqrt{5} z^3 - 1)^\star } \right),$

$g_1' = -{3 \over 2} \mathrm{Im} \left( - \left( {z (1 - z^4) \over z^6 + \sqrt{5} z^3 - 1} \right)^\star \right)$

therefore $g 1' = g 1$ since, for any complex number z,

$\mathrm{Im} (-z^\star) = \mathrm(z).$

Let g₂′ be obtained from g₂ by substituting z with $- {1 \over z^\star}.$ Then we obtain

$g_2' = -{3 \over 2} \mathrm{Re} \left( { - {1 \over z^\star} \left( 1 + {1 \over z^{\star 4}} \right) \over {1 \over z^{\star 6}} - \sqrt{5} {1 \over z^{\star 3}} - 1 } \right),$

$= -{3 \over 2} \mathrm{Re} \left( { z^\star (z^{\star 4} + 1) \over z^{\star 6} + \sqrt{5} z^{\star 3} - 1 } \right),$

$= -{3 \over 2} \mathrm{Re} \left( { z^\star (z^{4 \star} + 1) \over z^{6 \star} + \sqrt{5} z^{3 \star} - 1 } \right),$

$= -{3 \over 2} \mathrm{Re} \left( { z^\star (z^4 + 1)^\star \over (z^6 + \sqrt{5} z^3 - 1)^\star } \right),$

$= -{3 \over 2} \mathrm{Re} \left( \left( { z (z^4 + 1) \over z^6 + \sqrt{5} z^3 - 1} \right)^\star \right),$

therefore $g 2' = g 2$ since, for any complex number z,

$\mathrm{Re} (z^\star) = \mathrm{Re} (z).$

Let g₃′ be obtained from g₃ by substituting z with $- {1 \over z^\star}.$ Then we obtain

$g_3' = \mathrm{Im} \left( { 1 + {1 \over z^{\star 6}} \over {1 \over z^{\star 6}} - \sqrt{5} {1 \over z^{\star 3}} - 1} \right),$

$= \mathrm{Im} \left( { z^{\star 6} + 1 \over 1 - \sqrt{5} z^{\star 3} - z^{\star 6}} \right),$

$= \mathrm{Im} \left( { z^{6 \star} + 1 \over 1 - \sqrt{5} z^{3 \star} - z^{6 \star}} \right),$

$= \mathrm{Im} \left( - { (z^6 + 1)^\star \over (z^6 + \sqrt{5} z^3 - 1)^\star} \right),$

$= \mathrm{Im} \left( - \left( { z^6 + 1 \over z^6 + \sqrt{5} z^3 - 1} \right)^\star \right),$

therefore $g 3' = g 3 .$ 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

$\mathrm{Re}(U V) = \mathrm{Re}(U) \mathrm{Re}(V) - \mathrm{Im}(U) \mathrm{Im}(V), \,\!$

$\mathrm{Im}(U V) = \mathrm{Re}(U) \mathrm{Im}(V) + \mathrm{Im}(U) \mathrm{Re}(V). \,\!$

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

$z' = z e^{i 2 \pi / 3} \,\!$

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

$g_3'(z') = \mathrm{Im} \left( {1 + z'^6 \over z'^6 + \sqrt{5} z'^3 - 1} \right) - {1 \over 2},$

$g_3'(z) = \mathrm{Im} \left( {1 + z^6 e^{i 4 \pi} \over z^6 e^{i 4 \pi} + \sqrt{5} z^{i 2 \pi} - 1} \right) - {1 \over 2}.$

Since $e i 4π = e i 2π = 1,$ it follows that

$g_3' = \mathrm{Im}\left( {1 + z^6 \over z^6 + \sqrt{5} z^3 - 1} \right) - {1 \over 2}$

therefore $g 3' = g 3 .$ 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

$g_1'(z) = -{3 \over 2} \mathrm{Im} \left( { z e^{i 2 \pi / 3} (1 - z^4 e^{i 8 \pi / 3}) \over z^6 e^{i 4 \pi} + \sqrt{5} z^3 e^{i 2 \pi} - 1} \right).$

Noticing that $e i 8π / 3 = e i 2π / 3,$

$g_1' = -{3 \over 2} \mathrm{Im} \left( {z e^{i 2 \pi / 3} (1 - z^4 e^{i 2 \pi / 3}) \over z^6 + \sqrt{5} z^3 - 1} \right).$

Then, letting $e i 4π / 3 = e - i 2π / 3$ in the denominator yields

$g_1' = -{3 \over 2} \mathrm{Im} \left( { z (e^{i 2 \pi / 3} - z^4 e^{-i 2 \pi / 3}) \over z^6 + \sqrt{5} z^3 - 1} \right).$

Now, applying the complex-algebraic identity, and letting

$z'' = {z \over z^6 + \sqrt{5} z^3 - 1}$

we get

$g_1' = -{3 \over 2} \left[ \mathrm{Im}(z'') \mathrm{Re}(e^{i 2 \pi / 3} - z^4 e^{-i 2 \pi / 3}) + \mathrm{Re}(z'') \mathrm{Im}(e^{i 2 \pi / 3} - z^4 e^{-i 2 \pi / 3}) \right].$

Both $Re$ and $Im$ are distributive with respect to addition, and

$\mathrm{Re}(e^{i \theta}) = \cos \theta, \,\!$

$\mathrm{Im}(e^{i \theta}) = \sin \theta, \,\!$

due to Euler's formula, so that

$g_1' = -{3 \over 2} \left[ \mathrm{Im}(z'') \left( \cos {2 \pi \over 3} - \mathrm{Re}(z^4 e^{-i 2 \pi / 3}) \right) + \mathrm{Re}(z'') \left( \sin {2 \pi \over 3} - \mathrm{Im}(z^4 e^{-i 2 \pi / 3}) \right) \right].$

Applying the complex-algebraic identities again, and simplifying $\cos {2 \pi \over 3}$ to -1/2 and $\sin {2 \pi \over 3}$ to $\sqrt{3} / 2,$ produces

$g_1' = -{3 \over 2} \left[ \mathrm{Im}(z'') \left( -{1 \over 2} - [ \mathrm{Re}(z^4) \mathrm{Re}(e^{-i 2 \pi / 3}) - \mathrm{Im}(z^4) \mathrm{Im}(e^{-i 2 \pi / 3}) ] \right) + \mathrm{Re}(z'') \left( {\sqrt{3} \over 2} - [ \mathrm{Im}(z^4) \mathrm{Re}(e^{-i 2 \pi / 3}) + \mathrm{Re}(z^4) \mathrm{Im} (e^{-i 2 \pi / 3})] \right) \right].$

Simplify constants,

$g_1' = -{3 \over 2} \left[ \mathrm{Im}(z'') \left( -{1 \over 2} - \left[ -{1 \over 2} \mathrm{Re}(z^4) + {\sqrt{3} \over 2} \mathrm{Im}(z^4) \right] \right) + \mathrm{Re}(z'') \left( {\sqrt{3} \over 2} - \left[ -{1 \over 2} \mathrm{Im}(z^4) - {\sqrt{3} \over 2} \mathrm{Re}(z^4) \right] \right) \right],$

therefore

$g_1' = -{3 \over 2} \left[ -{1 \over 2} \mathrm{Im}(z'') + {1 \over 2} \mathrm{Im}(z'') \mathrm{Re}(z^4) - {\sqrt{3} \over 2} \mathrm{Im}(z'') \mathrm{Im}(z^4) + {\sqrt{3} \over 2} \mathrm{Re}(z'') + {1 \over 2} \mathrm{Re}(z'') \mathrm{Im}(z^4) + {\sqrt{3} \over 2} \mathrm{Re}(z'') \mathrm{Re}(z^4) \right].$

Applying the complex-algebraic identity to the original g₁ yields

$g_1 = -{3 \over 2} [ \mathrm{Im}(z'') \mathrm{Re}(1 - z^4) + \mathrm{Re}(z'') \mathrm{Im}(1 - z^4) ],$

$g_1 = -{3 \over 2} [ \mathrm{Im}(z'') (1 - \mathrm{Re}(z^4)) + \mathrm{Re}(z'') (-\mathrm{Im}(z^4))],$

$g_1 = -{3 \over 2} [ \mathrm{Im}(z'') - \mathrm{Im}(z'') \mathrm{Re}(z^4) - \mathrm{Re}(z'') \mathrm{Im}(z^4) ].$

Plug in z′ for z in g₂(z), resulting in

$g_2' = -{3 \over 2} \mathrm{Re} \left( {z e^{i 2 \pi / 3} (1 + z^4 e^{i 8 \pi / 3}) \over z^6 e^{i 4 \pi} + \sqrt{5} z^3 e^{i 2 \pi} - 1} \right) .$

Simplify the exponents,

$g_2' = -{3 \over 2} \mathrm{Re} \left( {z e^{i 2 \pi / 3} (1 + z^4 e^{i 2 \pi / 3}) \over z^6 + \sqrt{5} z^3 - 1} \right),$

$= -{3 \over 2} \mathrm{Re} ( z'' (e^{i 2 \pi / 3} + z^4 e^{-i 2 \pi / 3})).$

Now apply the complex-algebraic identity to g′₂, obtaining

$g_2' = -{3 \over 2} \left[ \mathrm{Re}(z'') \mathrm{Re}(e^{i 2 \pi / 3} + z^4 e^{-i 2 \pi / 3}) - \mathrm{Im}(z'') \mathrm{Im}(e^{i 2 \pi / 3} + z^4 e^{-i 2 \pi / 3}) \right].$

Distribute the $Re$ with respect to addition, and simplify constants,

$g_2' = -{3 \over 2} \left[ \mathrm{Re}(z'') \left(-{1 \over 2} + \mathrm{Re}(z^4 e^{-i 2 \pi / 3}) \right) - \mathrm{Im}(z'') \left({\sqrt{3} \over 2} + \mathrm{Im}(z^4 e^{-i 2 \pi / 3}) \right) \right].$

Apply the complex-algebraic identities again,

$g_2' = -{3 \over 2} \left[ \mathrm{Re}(z'') \left(-{1 \over 2} + \mathrm{Re}(z^4) \mathrm{Re}(e^{-i 2 \pi / 3}) - \mathrm{Im}(z^4) \mathrm{Im}(e^{-i 2 \pi \over 3}) \right) - \mathrm{Im}(z'') \left({\sqrt{3} \over 2} + \mathrm{Im}(z^4) \mathrm{Re}(e^{-i 2 \pi / 3}) + \mathrm{Re}(z^4) \mathrm{Im}(e^{-i 2 \pi / 3}) \right) \right].$

Simplify constants,

$g_2' = -{3 \over 2} \left[ \mathrm{Re}(z'') \left( -{1 \over 2} - {1 \over 2} \mathrm{Re}(z^4) + {\sqrt{3} \over 2} \mathrm{Im}(z^4) \right) - \mathrm{Im}(z'') \left({\sqrt{3} \over 2} - {\sqrt{3} \over 2} \mathrm{Re}(z^4) - {1 \over 2} \mathrm{Im}(z^4) \right) \right],$

then distribute with respect to addition,

$g_2' = -{3 \over 2} \left[ -{1 \over 2} \mathrm{Re}(z'') - {1 \over 2}\mathrm{Re}(z'') \mathrm{Re}(z^4) + {\sqrt{3}\over 2} \mathrm{Re}(z'') \mathrm{Im}(z^4) - {\sqrt{3}\over 2} \mathrm{Im}(z'') + {\sqrt{3}\over 2} \mathrm{Im}(z'') \mathrm{Re}(z^4) + {1 \over 2} \mathrm{Im}(z'') \mathrm{Im}(z^4) \right].$

Applying the complex-algebraic identity to the original g₂ yields

$g_2 = -{3 \over 2} \left( \mathrm{Re}(z'') \mathrm{Re}(1 + z^4) - \mathrm{Im}(z'') \mathrm{Im}(1 + z^4) \right),$

$g_2 = -{3 \over 2} \left[ \mathrm{Re}(z'') (1 + \mathrm{Re}(z^4)) - \mathrm{Im}(z'') \mathrm{Im}(z^4) \right],$

$g_2 = -{3 \over 2} \left[ \mathrm{Re}(z'') + \mathrm{Re}(z'') \mathrm{Re}(z^4) - \mathrm{Im}(z'') \mathrm{Im}(z^4) \right].$

The raw coordinates of the pre-rotated point are

$g_1 = -{3 \over 2} [ \mathrm{Im}(z'') - \mathrm{Im}(z'') \mathrm{Re}(z^4) - \mathrm{Re}(z'') \mathrm{Im}(z^4) ],$

$g_2 = -{3 \over 2} \left[ \mathrm{Re}(z'') + \mathrm{Re}(z'') \mathrm{Re}(z^4) - \mathrm{Im}(z'') \mathrm{Im}(z^4) \right],$

and the raw coordinates of the post-rotated point are

Comparing these four coordinates we can verify that

$g_1' = -{1 \over 2} g_1 + {\sqrt{3} \over 2} g_2,$

$g_2' = -{\sqrt{3} \over 2} g_1 -{1 \over 2} g_2.$

In matrix form, this can be expressed as

$\begin{bmatrix} g_1' \\ g_2' \\ g_3' \end{bmatrix} = \begin{bmatrix} -{1 \over 2} & {\sqrt{3}\over 2} & 0 \\ -{\sqrt{3}\over 2} & -{1 \over 2} & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} g_1 \\ g_2 \\ g_3 \end{bmatrix} = \begin{bmatrix} \cos {-2 \pi \over 3} & -\sin {-2 \pi \over 3} & 0 \\ \sin {-2 \pi \over 3} & \cos {-2 \pi \over 3} & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} g_1 \\ g_2 \\ g_3 \end{bmatrix}.$

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.

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