# 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, ${\displaystyle -{1 \over z^{\star }},}$  then the functions g1, g2, and g3 of z are left unchanged.

### Proof

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

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

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

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

Multiply both numerator and denominator by -1,

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

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

${\displaystyle (z^{\star })^{n}=(z^{n})^{\star },}$

therefore

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

therefore ${\displaystyle g_{1}'=g_{1}}$  since, for any complex number z,

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

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

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

therefore ${\displaystyle g_{2}'=g_{2}}$  since, for any complex number z,

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

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

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

therefore ${\displaystyle g_{3}'=g_{3}.}$  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

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

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

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

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

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

Since ${\displaystyle e^{i4\pi }=e^{i2\pi }=1,}$  it follows that

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

therefore ${\displaystyle g_{3}'=g_{3}.}$  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

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

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

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

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

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

Now, applying the complex-algebraic identity, and letting

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

we get

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

Both ${\displaystyle \mathrm {Re} }$  and ${\displaystyle \mathrm {Im} }$  are distributive with respect to addition, and

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

due to Euler's formula, so that

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

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

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

Simplify constants,

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

therefore

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

Applying the complex-algebraic identity to the original g1 yields

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

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

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

Simplify the exponents,

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

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

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

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

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

Apply the complex-algebraic identities again,

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

Simplify constants,

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

then distribute with respect to addition,

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

Applying the complex-algebraic identity to the original g2 yields

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

The raw coordinates of the pre-rotated point are

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

and the raw coordinates of the post-rotated point are

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

Comparing these four coordinates we can verify that

${\displaystyle g_{1}'=-{1 \over 2}g_{1}+{{\sqrt {3}} \over 2}g_{2},}$
${\displaystyle g_{2}'=-{{\sqrt {3}} \over 2}g_{1}-{1 \over 2}g_{2}.}$

In matrix form, this can be expressed as

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

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.