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′ 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 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

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

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

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

$z'=ze^{i2\pi /3}\,\!$

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

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

$g_{3}'(z)=\mathrm {Im} \left({1+z^{6}e^{i4\pi } \over z^{6}e^{i4\pi }+{\sqrt {5}}z^{i2\pi }-1}\right)-{1 \over 2}.$

Since $e^{i4\pi }=e^{i2\pi }=1,$ 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

$g_{1}'(z)=-{3 \over 2}\mathrm {Im} \left({ze^{i2\pi /3}(1-z^{4}e^{i8\pi /3}) \over z^{6}e^{i4\pi }+{\sqrt {5}}z^{3}e^{i2\pi }-1}\right).$

Noticing that $e^{i8\pi /3}=e^{i2\pi /3},$

$g_{1}'=-{3 \over 2}\mathrm {Im} \left({ze^{i2\pi /3}(1-z^{4}e^{i2\pi /3}) \over z^{6}+{\sqrt {5}}z^{3}-1}\right).$

Then, letting $e^{i4\pi /3}=e^{-i2\pi /3}$ in the denominator yields

$g_{1}'=-{3 \over 2}\mathrm {Im} \left({z(e^{i2\pi /3}-z^{4}e^{-i2\pi /3}) \over z^{6}+{\sqrt {5}}z^{3}-1}\right).$

Now, applying the complex-algebraic identity, and letting

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

we get

$g_{1}'=-{3 \over 2}\left[\mathrm {Im} (z'')\mathrm {Re} (e^{i2\pi /3}-z^{4}e^{-i2\pi /3})+\mathrm {Re} (z'')\mathrm {Im} (e^{i2\pi /3}-z^{4}e^{-i2\pi /3})\right].$

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

$g_{1}'=-{3 \over 2}\left[\mathrm {Im} (z'')\left(\cos {2\pi \over 3}-\mathrm {Re} (z^{4}e^{-i2\pi /3})\right)+\mathrm {Re} (z'')\left(\sin {2\pi \over 3}-\mathrm {Im} (z^{4}e^{-i2\pi /3})\right)\right].$

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

$g_{1}'=-{3 \over 2}\left[\mathrm {Im} (z'')\left(-{1 \over 2}-[\mathrm {Re} (z^{4})\mathrm {Re} (e^{-i2\pi /3})-\mathrm {Im} (z^{4})\mathrm {Im} (e^{-i2\pi /3})]\right)+\mathrm {Re} (z'')\left({{\sqrt {3}} \over 2}-[\mathrm {Im} (z^{4})\mathrm {Re} (e^{-i2\pi /3})+\mathrm {Re} (z^{4})\mathrm {Im} (e^{-i2\pi /3})]\right)\right].$

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

$g_{2}'=-{3 \over 2}\mathrm {Re} \left({ze^{i2\pi /3}(1+z^{4}e^{i8\pi /3}) \over z^{6}e^{i4\pi }+{\sqrt {5}}z^{3}e^{i2\pi }-1}\right).$

Simplify the exponents,

$g_{2}'=-{3 \over 2}\mathrm {Re} \left({ze^{i2\pi /3}(1+z^{4}e^{i2\pi /3}) \over z^{6}+{\sqrt {5}}z^{3}-1}\right),$

$=-{3 \over 2}\mathrm {Re} (z''(e^{i2\pi /3}+z^{4}e^{-i2\pi /3})).$

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

$g_{2}'=-{3 \over 2}\left[\mathrm {Re} (z'')\mathrm {Re} (e^{i2\pi /3}+z^{4}e^{-i2\pi /3})-\mathrm {Im} (z'')\mathrm {Im} (e^{i2\pi /3}+z^{4}e^{-i2\pi /3})\right].$

Distribute the $\mathrm {Re}$ with respect to addition, and simplify constants,

$g_{2}'=-{3 \over 2}\left[\mathrm {Re} (z'')\left(-{1 \over 2}+\mathrm {Re} (z^{4}e^{-i2\pi /3})\right)-\mathrm {Im} (z'')\left({{\sqrt {3}} \over 2}+\mathrm {Im} (z^{4}e^{-i2\pi /3})\right)\right].$

Apply the complex-algebraic identities again,

$g_{2}'=-{3 \over 2}\left[\mathrm {Re} (z'')\left(-{1 \over 2}+\mathrm {Re} (z^{4})\mathrm {Re} (e^{-i2\pi /3})-\mathrm {Im} (z^{4})\mathrm {Im} (e^{-i2\pi \over 3})\right)-\mathrm {Im} (z'')\left({{\sqrt {3}} \over 2}+\mathrm {Im} (z^{4})\mathrm {Re} (e^{-i2\pi /3})+\mathrm {Re} (z^{4})\mathrm {Im} (e^{-i2\pi /3})\right)\right].$

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.