Famous Theorems of Mathematics/Boy's surface

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.

Proof
Let g1&prime; 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&prime; 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&prime; 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 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.

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&prime;1, g&prime;2, and g&prime;3.

Substitute z&prime; 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&prime; 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&prime; 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&prime;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&prime; on the complex plane is equivalent to rotating P(z) by -120° about the Z-axis to P(z&prime;). This means that the Boy's surface has 3-fold symmetry, quod erat demonstrandum.