# Famous Theorems of Mathematics/Boy's surface

## Property of R. Bryant's parametrizationEdit

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.

### ProofEdit

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,

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

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.

## 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

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

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 $e^{i 4 \pi} = e^{i 2 \pi} = 1,$ 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 $e^{i 8 \pi / 3} = e^{i 2 \pi / 3},$

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

Then, letting $e^{i 4 \pi / 3} = e^{-i 2 \pi / 3}$ 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 $\mathrm{Re}$ and $\mathrm{Im}$ 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 $\mathrm{Re}$ 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

$g_1' = -{3 \over 2} \left[ -{1 \over 2} \mathrm{Im}(z'') + {1 \over 2} \mathrm{Im}(z'') \mathrm{Re}(z^4) - {\sqrt{3} \over 2} \mathrm{Im}(z'') \mathrm{Im}(z^4) + {\sqrt{3} \over 2} \mathrm{Re}(z'') + {1 \over 2} \mathrm{Re}(z'') \mathrm{Im}(z^4) + {\sqrt{3} \over 2} \mathrm{Re}(z'') \mathrm{Re}(z^4) \right],$
$g_2' = -{3 \over 2} \left[ -{1 \over 2} \mathrm{Re}(z'') - {1 \over 2}\mathrm{Re}(z'') \mathrm{Re}(z^4) + {\sqrt{3}\over 2} \mathrm{Re}(z'') \mathrm{Im}(z^4) - {\sqrt{3}\over 2} \mathrm{Im}(z'') + {\sqrt{3}\over 2} \mathrm{Im}(z'') \mathrm{Re}(z^4) + {1 \over 2} \mathrm{Im}(z'') \mathrm{Im}(z^4) \right].$

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.