Boy's surface/Proofs

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′ 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^⋆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^⋆)ⁿ = (zⁿ)^⋆, 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 g₁′ = g₁ since, for any complex number z,

Im(−z^⋆) = (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 g₂′ = g₂ since, for any complex number z,

Re(z^⋆) = 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 g₃′ = g₃. 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

Re(UV) = Re(U)Re(V) − Im(U)Im(V),

Im(UV) = Re(U)Im(V) + Im(U)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′ = ze^i2π/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^i4π = e^i2π = 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 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

$$g_1'(z) = -{3 \over 2} \mathrm{Im} \left( { z e^{i 2 \pi / 3} (1 - z^4 e^{i 8 \pi / 3}) \over z^6 e^{i 4 \pi} + \sqrt{5} z^3 e^{i 2 \pi} - 1} \right).$$ Noticing that e^i8π/3 = e^i2π/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^i4π/3 = e^−i2π/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

Re(e^iθ) = cos θ,

Im(e^iθ) = sin θ, 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

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

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