Categories: Surfaces | Geometric topology

Boy's surface

In geometry, Boy's surface is an immersion of the real projective plane in 3-dimensional space found by Werner Boy in 1901. Unlike the Roman surface and the cross-cap, it has no singularities (pinch points ), but it does self-intersect.

To make a Boy's surface:

Start with a sphere. Remove a cap.
Attach one end of each of three strips to alternate sixths of the edge left by removing the cap.
Bend each strip and attach the other end of each strip to the sixth opposite the first end, so that the inside of the sphere at one end is connected to the outside at the other. Make the strips skirt the middle rather than go through it.
Join the loose edges of the strips. The joins intersect the strips.

Boy's surface is discussed (and illustrated) in Jean-Pierre Petit's Le Topologicon.

Boy's surface was first parametrized correctly by Bernard Morin in 1978.

Contents

1 Parametrization of Boy's surface

2 Property of R. Bryant's parametrization

2.1 Proof
2.2 Relating the Boy's surface to the real projective plane

3 Symmetry of the Boy's surface

3.1 Proof

4 Structure of the Boy's surface

4.1 Sections of the Boy's surface
4.2 Pathways on a Boy's surface

5 External links

Parametrization of Boy's surface

Boy's surface can be parametrized in several ways. One parametrization, discovered by R. Bryant, is the following: given a complex number z whose magnitude is less than or equal to one, let

$g_1 = -{3 \over 2} \mathrm{Im} \left( {z (1 - z^4) \over z^6 + \sqrt{5} z^3 - 1} \right),$

$g_2 = -{3 \over 2} \mathrm{Re} \left( {z (1 + z^4) \over z^6 + \sqrt{5} z^3 - 1} \right),$

$g_3 = \mathrm{Im} \left( {1 + z^6 \over z^6 + \sqrt{5} z^3 - 1} \right) - {1 \over 2},$

$g = g_1^2 + g_2^2 + g_3^2,$

so that

$X = {g_1 \over g},$

$Y = {g_2 \over g},$

$Z = {g_3 \over g},$

where X, Y, and Z are the desired Cartesian coordinates of a point on the 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,

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

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.

Relating the Boy's surface to the real projective plane

Let $P (z) = (X (z), Y (z), Z (z))$ denote a point on Boy's surface, where $\| z \| \le 1$ . Then

$P(z) = P\left( -{1 \over z^\star} \right)$

but only if $\| z \| = \sqrt{z z^\star} = 1$ . What if $\| z \| < 1 ?$ Then

$\left\| - {1 \over z^\star} \right\| > 1$

because

$- {1 \over z^\star} = {- z \over z^\star z} = {-z \over \| z \|^2}$

whose magnitude is

${\| z \| \over \| z \|^2} = {1 \over \| z \|}$ ,

but $\| z \| < 1$ , so that

${1 \over \| z \|} > 1.$

Since P(z) belongs to the Boy's surface only when $\|z\| \le 1$ , this means that $P\left( - {1 \over z^\star} \right)$ belongs to Boy's surface only if $\| z \| = 1$ . Thus $P (z) = P ( - z)$ if $\| z \| = 1$ , but all other points on the Boy's surface are unique. The Boy's surface has been parametrized by a unit disk such that pairs of diametrically opposite points on the perimeter of the disk are equivalent identically. Therefore the Boy's surface is homeomorphic to the real projective plane, RP².

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(U V) = Re(U)Re(V) - Im(U)Im(V),

Im(U V) = 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' = z e i 2π / 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 $e i 4π = e i 2π = 1$ , 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

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

Noticing that $e i 8π / 3 = e i 2π / 3$ ,

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

Then, letting $e i 4π / 3 = e - i 2π / 3$ 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 i θ) = cosθ,

Im(e i θ) = sinθ,

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

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.

Structure of the Boy's surface

Figure 1.

Figure 1 shows a Boy's surface seen from the bottom. The nearly triangular portion is the outside surface of its "inner chamber". A Boy's surface is roughly like a bottle with three openings. Three "passageways" lead out of the inner chamber.

Figure 2.

Figure 2 shows the Boy's surface seen from near the top. The continuations of the three passageways are visible. The passageways end up intersecting into each other, forming a closed curve of double points.

Figure 3.

Figure 3 shows the Boy's surface seen from the top. About a half of the curve of double points can be seen in this figure, with the other half hidden under the folds. At the center of Figure 3 the three passageways intersect at a single triple point, which is also the point where the curve of double points intersects itself three times.

The three points of the triple point are not identically equivalent because they have different tangent planes.

Figure 4.

Figure 4 depicts the Boy's surface seen from the side. One of the surface's three "cave entrances" is shown, with a passageway coiling on top of it.

Rotate Figure 4 a slight angle counterclockwise about the Z-axis and the result is Figure 5.

Figure 5.

The passageway on the left in Figure 5 leads to the one at the center which arches over the cave entrance on the right, showing a passageway at the center growing up out of the inner chamber at the bottom and slanting towards the right to form an arch.

Figure 6.

Another slight counterclockwise rotation about the Z-axis produces Figure 6. This image shows the passageway on the left pointing towards the cave entrance to the right. Also, another cave entrance is revealed: a hole on the left, under the left passageway which arches above it -under the arch of the central passageway- which feeds into the passageway which grows on the left side and arches above another cave entrance through which one may peek directly into the interior surface of the inner chamber.

Sections of the Boy's surface

The Boy's surface can be cut into six sections. Let them be called A, B, C, D, E, and F. Then sections A, C, and E are mutually congruent, and sections B, D, and F are mutually congruent.

Section A	Section B	Section C
Section D	Section E	Section F

These six sections arrange themselves into a circle, or rather a hexagon: each section corresponding to one side of the hexagon. The sections are arranged in this order: A, B, C, D, E, F -- counterclockwise around the hexagon. Each section has three sides which have been shown as orange, green, and blue. Each section can be converted through a homotopy into a triangle. The colors of the edges show how the sections are supposed to fit together. Only sides of the same color are allowed to coincide.

Section A′s green edge matches section D′s, B′s green edge with E, C′s green edge with F. Thus, opposite sides of the hexagon match through the green sides.

Notice that if A is rotated counterclockwise by 120°, it looks the same as C, and if it is rotated further another 120° then it looks the same as E. A similar case holds for B, D, and F.

Figure 7.

Opposite sections B and E are shown in Figure 7 joined together along their common green edge. The outside of section B becomes the inside of section E. Section B has the cave entrance under the top portion of section B which resembles an archway. The cave entrance of B leads to section E′s inner "passageway" (inner side of outer "tentacle") which eventually makes a 180° turn becoming part of the "inner chamber".

Figure 8.

Figure 8 shows another combination of opposite sections, this time sections A and D, joined along their common green edge. Section D has the cave entrance and section A has the cave's passageway into the inner chamber. There are three cave entrances in total: sections D, F, and B -- three ways to move from the outside of the Boy's surface to the inside (including the inner chamber).

Figure 9.

Figure 9 shows sections A, B, and E. Sections A and E are joined along their common green edge, and sections A and B are joined along their common blue edge. B has the cave entrance and A′s tentacle frames the top of B′s cave entrance. Every cave entrance which leads into one passageway is framed on top (like an arch) by another passageway belonging to another cave entrance. Section A intersects section E's continuation of section B′s cave entrance, such that section A becomes a "passage barrier" to section B′s cave entrance. This passage barrier can be considered to separate the cave entrance from its passageway.

Figure 10.

Figure 10 shows another view of sections A, B, and E, rotated 30° in the +z direction according to the right hand rule. The green edge of section B coincides with the green border of section E, but E seems invisible at the bottom portion of the green edge of B. This is because E curls up tightly behind B in this view.

Figure 11.

In Figure 11, sections A, B and E of the previous figure have been complemented with sections F, C and D, completing the Boy's surface. The cave entrance of section B remains in place, showing up as a hole pointed out by a pale blue circle surrounding it, through which one may see the interior surface of section A. This shows that the cave entrance opens directly (visually) into the "inner chamber" without being obstructed (visually) by the surface of another section. However, only a small portion of the inner chamber can be seen through this hole: this small portion is the passage barrier, which is circumscribed (and defined by) a loop of double points.

Pathways on a Boy's surface

Let a "topological ant" start out walking from the bottom of the Boy's surface (shown in Figure 12), on the outside. Let this ant walk along the green path into a cave entrance. This cave entrance is located under an archway which is like one of the tentacles of an octopus.

Figure 12.

The ant goes under this archway along the dotted green path. Then the ant passes through a surface belonging to the same archway (a "passage barrier"), like a ghost passing through a wall, then walks along the inside surface of another tentacle -- a "passage" -- which feeds into the cave entrance which the ant previously walked through. Now the ant walks along the yellow path (inner surface) towards the root of the "tentacle" which leads directly into the inner chamber. The ant walks in the inner chamber towards the bottom and lands in the same point where it started, but oriented inwards. Therefore the Boy's surface, as a whole (globally), is non-orientable.

Figure 13.

Figure 13 shows the reverse case of the path shown in Figure 12. An ant starts out from the blue X on the outside of a "tentacle" then walks lengthwise along the tentacle -- along the green path -- towards another tentacle. The ant passes the other tentacle like a ghost through a wall, then finds itself in an interior surface: the inner side of a cave entrance. So it walks along the dotted yellow line until it reaches the surface of the inner chamber and the yellow path turns solid. Then the ant walks towards the orange O near the center of the inner chamber. (The ant moves from section C through its green edge into section F′s inner surface, then it walks up through section F′s blue edge into section E, and walks down section E′s inner side towards the bottom of the inner chamber.)

What if the ant were to walk along the width of a tentacle in an outward direction (away from the triple point)? See Figure 14.

Figure 14.

Then the ant, starting from section C, will cross the blue edge of section C into section D′s outer side. Then it can keep moving into section D′s cave entrance, cross section D′s green edge, move on to the top, inner side of section A (yellow path). Keep walking along the passageway (inside of tentacle) towards inner chamber at the bottom of section A.

What if the ant, starting at the blue X on the outside of the tentacle on the right (see Figure 15), were to move widthwise inbound towards the triple point?

Figure 15.

The ant starts on the outer surface of section E and walks towards its orange edge. The ant knows nothing about double or triple points; it goes right through them, like a ghost through a wall. As it crosses the orange edge, the ant walks into the outside surface of the cave entrance of section D. It can choose to walk into the cave, but let us suppose it walks the other way. It is on the higher side of D so it walks out of the archway on top of D′s cave entrance (green), crosses D′s blue edge and walks on the outer surface of section C. The ant ends on the blue O on top of another tentacle.

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Categories: Surfaces | Geometric topology

Last updated: 05-24-2005 23:11:39

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Boy's surface

Parametrization of Boy's surface

Property of R. Bryant's parametrization

Proof

Relating the Boy's surface to the real projective plane

Symmetry of the Boy's surface

Proof

Structure of the Boy's surface

Sections of the Boy's surface

Pathways on a Boy's surface

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Encyclopedia

Dictionary

Quotes

Boy's surface

Parametrization of Boy's surface

Property of R. Bryant's parametrization

Proof

Relating the Boy's surface to the real projective plane

Symmetry of the Boy's surface

Proof

Structure of the Boy's surface

Sections of the Boy's surface

Pathways on a Boy's surface

External links