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

Proof

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

${g_1}^{\prime }=-\frac{3}{2}\mathrm{I}\mathrm{m}\left(\frac{-\frac{1}{z^{\star }}(1-\frac{1}{z^{\star 4}})}{\frac{1}{z^{\star 6}}-\sqrt{5}\frac{1}{z^{\star 3}}-1}\right).$

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

${g_1}^{\prime }=-\frac{3}{2}\mathrm{I}\mathrm{m}\left(\frac{-z^{\star }(z^{\star 4}-1)}{1-\sqrt{5}{z}^{\star 3}-z^{\star 6}}\right).$

Multiply both numerator and denominator by -1,

${g_1}^{\prime }=-\frac{3}{2}\mathrm{I}\mathrm{m}\left(\frac{z^{\star }(z^{\star 4}-1)}{z^{\star 6}+\sqrt{5}{z}^{\star 3}-1}\right).$

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

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

therefore

${g_1}^{\prime }=-\frac{3}{2}\mathrm{I}\mathrm{m}\left(\frac{z^{\star }(z^4-1{)}^{\star }}{(z^6+\sqrt{5}{z}^3-1{)}^{\star }}\right),$

${g_1}^{\prime }=-\frac{3}{2}\mathrm{I}\mathrm{m}(-{\left(\frac{z(1-z^4)}{z^6+\sqrt{5}{z}^3-1}\right)}^{\star })$

therefore ${g_1}^{\prime }=g_1$ since, for any complex number z,

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

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

${g_2}^{\prime }=-\frac{3}{2}\mathrm{R}\mathrm{e}\left(\frac{-\frac{1}{z^{\star }}(1+\frac{1}{z^{\star 4}})}{\frac{1}{z^{\star 6}}-\sqrt{5}\frac{1}{z^{\star 3}}-1}\right),$

$=-\frac{3}{2}\mathrm{R}\mathrm{e}\left(\frac{z^{\star }(z^{\star 4}+1)}{z^{\star 6}+\sqrt{5}{z}^{\star 3}-1}\right),$

$=-\frac{3}{2}\mathrm{R}\mathrm{e}\left(\frac{z^{\star }(z^{4\star }+1)}{z^{6\star }+\sqrt{5}{z}^{3\star }-1}\right),$

$=-\frac{3}{2}\mathrm{R}\mathrm{e}\left(\frac{z^{\star }(z^4+1{)}^{\star }}{(z^6+\sqrt{5}{z}^3-1{)}^{\star }}\right),$

$=-\frac{3}{2}\mathrm{R}\mathrm{e}\left({\left(\frac{z(z^4+1)}{z^6+\sqrt{5}{z}^3-1}\right)}^{\star }\right),$

therefore ${g_2}^{\prime }=g_2$ since, for any complex number z,

$\mathrm{R}\mathrm{e}(z^{\star })=\mathrm{R}\mathrm{e}(z).$

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

${g_3}^{\prime }=\mathrm{I}\mathrm{m}\left(\frac{1+\frac{1}{z^{\star 6}}}{\frac{1}{z^{\star 6}}-\sqrt{5}\frac{1}{z^{\star 3}}-1}\right),$

$=\mathrm{I}\mathrm{m}\left(\frac{z^{\star 6}+1}{1-\sqrt{5}{z}^{\star 3}-z^{\star 6}}\right),$

$=\mathrm{I}\mathrm{m}\left(\frac{z^{6\star }+1}{1-\sqrt{5}{z}^{3\star }-z^{6\star }}\right),$

$=\mathrm{I}\mathrm{m}(-\frac{(z^6+1{)}^{\star }}{(z^6+\sqrt{5}{z}^3-1{)}^{\star }}),$

$=\mathrm{I}\mathrm{m}(-{\left(\frac{z^6+1}{z^6+\sqrt{5}{z}^3-1}\right)}^{\star }),$

therefore ${g_3}^{\prime }=g_3.$ 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

$\mathrm{R}\mathrm{e}(UV)=\mathrm{R}\mathrm{e}(U)\mathrm{R}\mathrm{e}(V)-\mathrm{I}\mathrm{m}(U)\mathrm{I}\mathrm{m}(V),$

$\mathrm{I}\mathrm{m}(UV)=\mathrm{R}\mathrm{e}(U)\mathrm{I}\mathrm{m}(V)+\mathrm{I}\mathrm{m}(U)\mathrm{R}\mathrm{e}(V).$

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

$z^{\prime }=z{e}^{i2\pi /3}$

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

${g_3}^{\prime }(z^{\prime })=\mathrm{I}\mathrm{m}\left(\frac{1+z{\text{'}}^6}{z{\text{'}}^6+\sqrt{5}z{\text{'}}^3-1}\right)-\frac{1}{2},$

${g_3}^{\prime }(z)=\mathrm{I}\mathrm{m}\left(\frac{1+z^6{e}^{i4\pi }}{z^6{e}^{i4\pi }+\sqrt{5}{z}^{i2\pi }-1}\right)-\frac{1}{2}.$

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

${g_3}^{\prime }=\mathrm{I}\mathrm{m}\left(\frac{1+z^6}{z^6+\sqrt{5}{z}^3-1}\right)-\frac{1}{2}$

therefore ${g_3}^{\prime }=g_3.$ 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

${g_1}^{\prime }(z)=-\frac{3}{2}\mathrm{I}\mathrm{m}\left(\frac{z{e}^{i2\pi /3}(1-z^4{e}^{i8\pi /3})}{z^6{e}^{i4\pi }+\sqrt{5}{z}^3{e}^{i2\pi }-1}\right).$

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

${g_1}^{\prime }=-\frac{3}{2}\mathrm{I}\mathrm{m}\left(\frac{z{e}^{i2\pi /3}(1-z^4{e}^{i2\pi /3})}{z^6+\sqrt{5}{z}^3-1}\right).$

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

${g_1}^{\prime }=-\frac{3}{2}\mathrm{I}\mathrm{m}\left(\frac{z(e^{i2\pi /3}-z^4{e}^{-i2\pi /3})}{z^6+\sqrt{5}{z}^3-1}\right).$

Now, applying the complex-algebraic identity, and letting

$z^{″}=\frac{z}{z^6+\sqrt{5}{z}^3-1}$

we get

${g_1}^{\prime }=-\frac{3}{2}[\mathrm{I}\mathrm{m}(z^{″})\mathrm{R}\mathrm{e}(e^{i2\pi /3}-z^4{e}^{-i2\pi /3})+\mathrm{R}\mathrm{e}(z^{″})\mathrm{I}\mathrm{m}(e^{i2\pi /3}-z^4{e}^{-i2\pi /3})].$

Both $\mathrm{R}\mathrm{e}$ and $\mathrm{I}\mathrm{m}$ are distributive with respect to addition, and

$\mathrm{R}\mathrm{e}(e^{i\theta })=\cos \theta ,$

$\mathrm{I}\mathrm{m}(e^{i\theta })=\sin \theta ,$

due to Euler's formula, so that

${g_1}^{\prime }=-\frac{3}{2}[\mathrm{I}\mathrm{m}(z^{″})(\cos \frac{2\pi }{3}-\mathrm{R}\mathrm{e}(z^4{e}^{-i2\pi /3}))+\mathrm{R}\mathrm{e}(z^{″})(\sin \frac{2\pi }{3}-\mathrm{I}\mathrm{m}(z^4{e}^{-i2\pi /3}))].$

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

${g_1}^{\prime }=-\frac{3}{2}[\mathrm{I}\mathrm{m}(z^{″})(-\frac{1}{2}-[\mathrm{R}\mathrm{e}(z^4)\mathrm{R}\mathrm{e}(e^{-i2\pi /3})-\mathrm{I}\mathrm{m}(z^4)\mathrm{I}\mathrm{m}(e^{-i2\pi /3})])+\mathrm{R}\mathrm{e}(z^{″})(\frac{\sqrt{3}}{2}-[\mathrm{I}\mathrm{m}(z^4)\mathrm{R}\mathrm{e}(e^{-i2\pi /3})+\mathrm{R}\mathrm{e}(z^4)\mathrm{I}\mathrm{m}(e^{-i2\pi /3})])].$

Simplify constants,

${g_1}^{\prime }=-\frac{3}{2}[\mathrm{I}\mathrm{m}(z^{″})(-\frac{1}{2}-[-\frac{1}{2}\mathrm{R}\mathrm{e}(z^4)+\frac{\sqrt{3}}{2}\mathrm{I}\mathrm{m}(z^4)])+\mathrm{R}\mathrm{e}(z^{″})(\frac{\sqrt{3}}{2}-[-\frac{1}{2}\mathrm{I}\mathrm{m}(z^4)-\frac{\sqrt{3}}{2}\mathrm{R}\mathrm{e}(z^4)])],$

therefore

${g_1}^{\prime }=-\frac{3}{2}[-\frac{1}{2}\mathrm{I}\mathrm{m}(z^{″})+\frac{1}{2}\mathrm{I}\mathrm{m}(z^{″})\mathrm{R}\mathrm{e}(z^4)-\frac{\sqrt{3}}{2}\mathrm{I}\mathrm{m}(z^{″})\mathrm{I}\mathrm{m}(z^4)+\frac{\sqrt{3}}{2}\mathrm{R}\mathrm{e}(z^{″})+\frac{1}{2}\mathrm{R}\mathrm{e}(z^{″})\mathrm{I}\mathrm{m}(z^4)+\frac{\sqrt{3}}{2}\mathrm{R}\mathrm{e}(z^{″})\mathrm{R}\mathrm{e}(z^4)].$

Applying the complex-algebraic identity to the original g₁ yields

$g_1=-\frac{3}{2}[\mathrm{I}\mathrm{m}(z^{″})\mathrm{R}\mathrm{e}(1-z^4)+\mathrm{R}\mathrm{e}(z^{″})\mathrm{I}\mathrm{m}(1-z^4)],$

$g_1=-\frac{3}{2}[\mathrm{I}\mathrm{m}(z^{″})(1-\mathrm{R}\mathrm{e}(z^4))+\mathrm{R}\mathrm{e}(z^{″})(-\mathrm{I}\mathrm{m}(z^4))],$

$g_1=-\frac{3}{2}[\mathrm{I}\mathrm{m}(z^{″})-\mathrm{I}\mathrm{m}(z^{″})\mathrm{R}\mathrm{e}(z^4)-\mathrm{R}\mathrm{e}(z^{″})\mathrm{I}\mathrm{m}(z^4)].$

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

${g_2}^{\prime }=-\frac{3}{2}\mathrm{R}\mathrm{e}\left(\frac{z{e}^{i2\pi /3}(1+z^4{e}^{i8\pi /3})}{z^6{e}^{i4\pi }+\sqrt{5}{z}^3{e}^{i2\pi }-1}\right).$

Simplify the exponents,

${g_2}^{\prime }=-\frac{3}{2}\mathrm{R}\mathrm{e}\left(\frac{z{e}^{i2\pi /3}(1+z^4{e}^{i2\pi /3})}{z^6+\sqrt{5}{z}^3-1}\right),$

$=-\frac{3}{2}\mathrm{R}\mathrm{e}(z^{″}(e^{i2\pi /3}+z^4{e}^{-i2\pi /3})).$

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

${g_2}^{\prime }=-\frac{3}{2}[\mathrm{R}\mathrm{e}(z^{″})\mathrm{R}\mathrm{e}(e^{i2\pi /3}+z^4{e}^{-i2\pi /3})-\mathrm{I}\mathrm{m}(z^{″})\mathrm{I}\mathrm{m}(e^{i2\pi /3}+z^4{e}^{-i2\pi /3})].$

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

${g_2}^{\prime }=-\frac{3}{2}[\mathrm{R}\mathrm{e}(z^{″})(-\frac{1}{2}+\mathrm{R}\mathrm{e}(z^4{e}^{-i2\pi /3}))-\mathrm{I}\mathrm{m}(z^{″})(\frac{\sqrt{3}}{2}+\mathrm{I}\mathrm{m}(z^4{e}^{-i2\pi /3}))].$

Apply the complex-algebraic identities again,

${g_2}^{\prime }=-\frac{3}{2}[\mathrm{R}\mathrm{e}(z^{″})(-\frac{1}{2}+\mathrm{R}\mathrm{e}(z^4)\mathrm{R}\mathrm{e}(e^{-i2\pi /3})-\mathrm{I}\mathrm{m}(z^4)\mathrm{I}\mathrm{m}(e^{\frac{-i2\pi }{3}}))-\mathrm{I}\mathrm{m}(z^{″})(\frac{\sqrt{3}}{2}+\mathrm{I}\mathrm{m}(z^4)\mathrm{R}\mathrm{e}(e^{-i2\pi /3})+\mathrm{R}\mathrm{e}(z^4)\mathrm{I}\mathrm{m}(e^{-i2\pi /3}))].$

Simplify constants,

${g_2}^{\prime }=-\frac{3}{2}[\mathrm{R}\mathrm{e}(z^{″})(-\frac{1}{2}-\frac{1}{2}\mathrm{R}\mathrm{e}(z^4)+\frac{\sqrt{3}}{2}\mathrm{I}\mathrm{m}(z^4))-\mathrm{I}\mathrm{m}(z^{″})(\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}\mathrm{R}\mathrm{e}(z^4)-\frac{1}{2}\mathrm{I}\mathrm{m}(z^4))],$

then distribute with respect to addition,

${g_2}^{\prime }=-\frac{3}{2}[-\frac{1}{2}\mathrm{R}\mathrm{e}(z^{″})-\frac{1}{2}\mathrm{R}\mathrm{e}(z^{″})\mathrm{R}\mathrm{e}(z^4)+\frac{\sqrt{3}}{2}\mathrm{R}\mathrm{e}(z^{″})\mathrm{I}\mathrm{m}(z^4)-\frac{\sqrt{3}}{2}\mathrm{I}\mathrm{m}(z^{″})+\frac{\sqrt{3}}{2}\mathrm{I}\mathrm{m}(z^{″})\mathrm{R}\mathrm{e}(z^4)+\frac{1}{2}\mathrm{I}\mathrm{m}(z^{″})\mathrm{I}\mathrm{m}(z^4)].$

Applying the complex-algebraic identity to the original g₂ yields

$g_2=-\frac{3}{2}(\mathrm{R}\mathrm{e}(z^{″})\mathrm{R}\mathrm{e}(1+z^4)-\mathrm{I}\mathrm{m}(z^{″})\mathrm{I}\mathrm{m}(1+z^4)),$

$g_2=-\frac{3}{2}[\mathrm{R}\mathrm{e}(z^{″})(1+\mathrm{R}\mathrm{e}(z^4))-\mathrm{I}\mathrm{m}(z^{″})\mathrm{I}\mathrm{m}(z^4)],$

$g_2=-\frac{3}{2}[\mathrm{R}\mathrm{e}(z^{″})+\mathrm{R}\mathrm{e}(z^{″})\mathrm{R}\mathrm{e}(z^4)-\mathrm{I}\mathrm{m}(z^{″})\mathrm{I}\mathrm{m}(z^4)].$

The raw coordinates of the pre-rotated point are

$g_1=-\frac{3}{2}[\mathrm{I}\mathrm{m}(z^{″})-\mathrm{I}\mathrm{m}(z^{″})\mathrm{R}\mathrm{e}(z^4)-\mathrm{R}\mathrm{e}(z^{″})\mathrm{I}\mathrm{m}(z^4)],$

$g_2=-\frac{3}{2}[\mathrm{R}\mathrm{e}(z^{″})+\mathrm{R}\mathrm{e}(z^{″})\mathrm{R}\mathrm{e}(z^4)-\mathrm{I}\mathrm{m}(z^{″})\mathrm{I}\mathrm{m}(z^4)],$

and the raw coordinates of the post-rotated point are

${g_1}^{\prime }=-\frac{3}{2}[-\frac{1}{2}\mathrm{I}\mathrm{m}(z^{″})+\frac{1}{2}\mathrm{I}\mathrm{m}(z^{″})\mathrm{R}\mathrm{e}(z^4)-\frac{\sqrt{3}}{2}\mathrm{I}\mathrm{m}(z^{″})\mathrm{I}\mathrm{m}(z^4)+\frac{\sqrt{3}}{2}\mathrm{R}\mathrm{e}(z^{″})+\frac{1}{2}\mathrm{R}\mathrm{e}(z^{″})\mathrm{I}\mathrm{m}(z^4)+\frac{\sqrt{3}}{2}\mathrm{R}\mathrm{e}(z^{″})\mathrm{R}\mathrm{e}(z^4)],$

${g_2}^{\prime }=-\frac{3}{2}[-\frac{1}{2}\mathrm{R}\mathrm{e}(z^{″})-\frac{1}{2}\mathrm{R}\mathrm{e}(z^{″})\mathrm{R}\mathrm{e}(z^4)+\frac{\sqrt{3}}{2}\mathrm{R}\mathrm{e}(z^{″})\mathrm{I}\mathrm{m}(z^4)-\frac{\sqrt{3}}{2}\mathrm{I}\mathrm{m}(z^{″})+\frac{\sqrt{3}}{2}\mathrm{I}\mathrm{m}(z^{″})\mathrm{R}\mathrm{e}(z^4)+\frac{1}{2}\mathrm{I}\mathrm{m}(z^{″})\mathrm{I}\mathrm{m}(z^4)].$

Comparing these four coordinates we can verify that

${g_1}^{\prime }=-\frac{1}{2}{g}_1+\frac{\sqrt{3}}{2}{g}_2,$

${g_2}^{\prime }=-\frac{\sqrt{3}}{2}{g}_1-\frac{1}{2}{g}_2.$

In matrix form, this can be expressed as

$\left[\begin{array}{c}{g_1}^{\prime }\\ {g_2}^{\prime }\\ {g_3}^{\prime }\end{array}\right]=\left[\begin{array}{ccc}-\frac{1}{2}& \frac{\sqrt{3}}{2}& 0\\ -\frac{\sqrt{3}}{2}& -\frac{1}{2}& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}{g}_1\\ g_2\\ g_3\end{array}\right]=\left[\begin{array}{ccc}\cos \frac{-2\pi }{3}& -\sin \frac{-2\pi }{3}& 0\\ \sin \frac{-2\pi }{3}& \cos \frac{-2\pi }{3}& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}{g}_1\\ g_2\\ g_3\end{array}\right].$

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.