Boy's surface/Proofs

Boy's surface/Proofs

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} } ight) over {1 over z^{star 6 - sqrt{5} {1 over z^{star 3 - 1} ight). 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} } ight).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} ight).

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 } ight), : g_1' = -{3 over 2} mathrm{Im} left( - left( {z (1 - z^4) over z^6 + sqrt{5} z^3 - 1} ight)^star ight) 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 ight) over {1 over z^{star 6 - sqrt{5} {1 over z^{star 3 - 1 } ight),:: = -{3 over 2} mathrm{Re} left( { z^star (z^{star 4} + 1) over z^{star 6} + sqrt{5} z^{star 3} - 1 } ight), :: = -{3 over 2} mathrm{Re} left( { z^star (z^{4 star} + 1) over z^{6 star} + sqrt{5} z^{3 star} - 1 } ight), :: = -{3 over 2} mathrm{Re} left( { z^star (z^4 + 1)^star over (z^6 + sqrt{5} z^3 - 1)^star } ight), :: = -{3 over 2} mathrm{Re} left( left( { z (z^4 + 1) over z^6 + sqrt{5} z^3 - 1} ight)^star ight), 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} ight), :: = mathrm{Im} left( { z^{star 6} + 1 over 1 - sqrt{5} z^{star 3} - z^{star 6 ight), :: = mathrm{Im} left( { z^{6 star} + 1 over 1 - sqrt{5} z^{3 star} - z^{6 star ight), :: = mathrm{Im} left( - { (z^6 + 1)^star over (z^6 + sqrt{5} z^3 - 1)^star} ight), :: = mathrm{Im} left( - left( { z^6 + 1 over z^6 + sqrt{5} z^3 - 1} ight)^star ight), therefore g_3' = g_3. Q.E.D.

ymmetry 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}(U V) = mathrm{Re}(U) mathrm{Re}(V) - mathrm{Im}(U) mathrm{Im}(V), ,!: mathrm{Im}(U V) = 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' = z e^{i 2 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} ight) - {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} ight) - {1 over 2}. Since e^{i 4 pi} = e^{i 2 pi} = 1, it follows that: g_3' = mathrm{Im}left( {1 + z^6 over z^6 + sqrt{5} z^3 - 1} ight) - {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} ight). Noticing that e^{i 8 pi / 3} = e^{i 2 pi / 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} ight). Then, letting e^{i 4 pi / 3} = e^{-i 2 pi / 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} ight).

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}) ight] . Both mathrm{Re} and mathrm{Im} are distributive with respect to addition, and: mathrm{Re}(e^{i heta}) = cos heta, ,!: mathrm{Im}(e^{i heta}) = sin heta, ,!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}) ight) + mathrm{Re}(z") left( sin {2 pi over 3} - mathrm{Im}(z^4 e^{-i 2 pi / 3}) ight) ight] . 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}) ] ight) + 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})] ight) ight] . 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) ight] ight) + mathrm{Re}(z") left( {sqrt{3} over 2} - left [ -{1 over 2} mathrm{Im}(z^4) - {sqrt{3} over 2} mathrm{Re}(z^4) ight] ight) ight] ,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) ight] .

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} ight) .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} ight), :: = -{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}) ight] . 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^{-i 2 pi / 3}) ight) - mathrm{Im}(z") left({sqrt{3} over 2} + mathrm{Im}(z^4 e^{-i 2 pi / 3}) ight) ight] . 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}) ight) - 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}) ight) ight] . 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) ight) - mathrm{Im}(z") left({sqrt{3} over 2} - {sqrt{3} over 2} mathrm{Re}(z^4) - {1 over 2} mathrm{Im}(z^4) ight) ight] , 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) ight] .

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) ight), : g_2 = -{3 over 2} left [ mathrm{Re}(z") (1 + mathrm{Re}(z^4)) - mathrm{Im}(z") mathrm{Im}(z^4) ight] , : g_2 = -{3 over 2} left [ mathrm{Re}(z") + mathrm{Re}(z") mathrm{Re}(z^4) - mathrm{Im}(z") mathrm{Im}(z^4) ight] .

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) ight] , 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) ight] , : 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) ight] . 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: egin{bmatrix} g_1' \ g_2' \ g_3' end{bmatrix} =egin{bmatrix} -{1 over 2} & {sqrt{3}over 2} & 0 \ -{sqrt{3}over 2} & -{1 over 2} & 0 \ 0 & 0 & 1 end{bmatrix} egin{bmatrix} g_1 \ g_2 \ g_3 end{bmatrix}= egin{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} egin{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".


Wikimedia Foundation. 2010.

Игры ⚽ Нужно сделать НИР?

Look at other dictionaries:

  • Surface — This article discusses surfaces from the point of view of topology. For other uses, see Differential geometry of surfaces, algebraic surface, and Surface (disambiguation). An open surface with X , Y , and Z contours shown. In mathematics,… …   Wikipedia

  • List of mathematics articles (B) — NOTOC B B spline B* algebra B* search algorithm B,C,K,W system BA model Ba space Babuška Lax Milgram theorem Baby Monster group Baby step giant step Babylonian mathematics Babylonian numerals Bach tensor Bach s algorithm Bachmann–Howard ordinal… …   Wikipedia

  • literature — /lit euhr euh cheuhr, choor , li treuh /, n. 1. writings in which expression and form, in connection with ideas of permanent and universal interest, are characteristic or essential features, as poetry, novels, history, biography, and essays. 2.… …   Universalium

  • printmaking — /print may king/, n. the art or technique of making prints, esp. as practiced in engraving, etching, drypoint, woodcut or serigraphy. [1925 30; PRINT + MAKING] * * * Art form consisting of the production of images, usually on paper but… …   Universalium

  • HEBREW GRAMMAR — The following entry is divided into two sections: an Introduction for the non specialist and (II) a detailed survey. [i] HEBREW GRAMMAR: AN INTRODUCTION There are four main phases in the history of the Hebrew language: the biblical or classical,… …   Encyclopedia of Judaism

  • epistemology — epistemological /i pis teuh meuh loj i keuhl/, adj. epistemologically, adv. epistemologist, n. /i pis teuh mol euh jee/, n. a branch of philosophy that investigates the origin, nature, methods, and limits of human knowledge. [1855 60; < Gk… …   Universalium

  • Anthropology and Archaeology — ▪ 2009 Introduction Anthropology       Among the key developments in 2008 in the field of physical anthropology was the discovery by a large interdisciplinary team of Spanish and American scientists in northern Spain of a partial mandible (lower… …   Universalium

  • Puddle — This article is about the liquid phenomenon. For other uses, see Puddle (disambiguation). A seep puddle in a forest clearing A puddle is a small accumulation of liquid, usually water, on a surface. It can form either by pooling in a depression on …   Wikipedia

  • Chromolithography — This article is about the print making method. For the Felipe Alfau novel, see Felipe Alfau. Folding Card, The Old Woman Who Lived in A Shoe, 6 April 1883 …   Wikipedia

  • BIBLE — THE CANON, TEXT, AND EDITIONS canon general titles the canon the significance of the canon the process of canonization contents and titles of the books the tripartite canon …   Encyclopedia of Judaism

Share the article and excerpts

Direct link
Do a right-click on the link above
and select “Copy Link”