每日彙整: 2017-09-15

樹莓派、樹莓派之學習、樹莓派之教育

GoPiGo 小汽車︰格點圖像算術《投影幾何》【五‧線性代數】《導引七‧變換組合 VIII‧⚂ 》

Timothy Peil 教授引的妙

4.6 Investigation into the Uniqueness of a Projectivity between Pencils of Points
 Teach to the problems, not to the text.
 —Kim Nebeuts, Return to Mathematical Circles (1988)

Theorem 4.10. If A, B, C and A’, B’, C’ are distinct elements in pencils of points with distinct axes p and p’, respectively, then there exists a projectivity such that ABC is projectively related to A’B’C’.

The theorem and its constructive proof, gave a procedure to determine a corresponding point D’ on axis p’ by following the perspectivities when a fourth point D on axis p was given. That is, let D be an element of axis p, then let D1 = DP · A’C and D’ = D1 Q · p’. Is the point D’ unique? Or, does the point D’ depend on the choice of the point P?

Construct Projectivity Between Two Pencils of Points

Select the steps for the construction.
Drag the points A, B, and C on axes p.
Drag the points A’, B’, and C’ on axes p’.
Select the steps for constructing D and D’.
Drag point D.

Timothy Peil, 6 February 2013, Created with GeoGebra

 

文章寫的好︰

Further, the theorem and its constructive proof give a procedure to determine a corresponding point D’ on axis p’ by following the perspectivities when a fourth point D on axis p is given. That is, let D be an element of axis p. First, find D1 on the pencil of points with A’, B1, and C by mapping D through center P; that is, let D1 = DP · A’C. Next, map D1 to D’ by mapping D1 through the center Q to p’, i.e., D’ = D1 Q · p’. Then  and . Hence, .

In the above theorem, we have shown a projectivity exists between two pencils of points with three elements. But, is the projectivity unique? Since an arbitrary point was chosen in the construction, a different point would give different perspectivities. That is, would a different point  be determined, if the perspectivities were different? Click here to investigate the uniqueness of the perspectivity GeoGebra or JavaSketchpad constructed in the proof of the previous theorem. Also, to help answer these questions consider Axiom 6.

Axiom 6. If a projectivity on a pencil of points leaves three distinct points of the pencil invariant, it leaves every point of the pencil invariant.

Axiom 6 implies that a projectivity on a pencil that leaves three elements of the pencil invariant is the identity mapping. What implications does this axiom have for distinct pencils of points? What implications does this axiom have for a projectivity on a pencil of points where no group of three points are mapped to themselves? Can this axiom extend the above theorem for constructing a projectivity between two pencils of points to more than three points? All of these questions are answered by the Fundamental Theorem of Projective Geometry, which has the surprising result that only three pairs of points are needed to determine a unique projectivity between two pencils of points.

 

直逼『投影幾何』之『基本定理』現形乎︰

Theorem 4.11. (Fundamental Theorem of Projective Geometry)  A projectivity between two pencils of points is uniquely determined by three pairs of corresponding points.

In other words, if A, B, C, D are in a pencil of points with axis p and A’, B’, C’ are in a pencil of points with axis p’, then there exists a unique point D’  on p’ such that .

Proof.  Assume A, B, C, D are in a pencil of points with axis p and A’, B’, C’ are in a pencil of points with axis p’. We have shown that there exists a point D’ on p’ such that . Suppose there is a projectivity and a point D” such that . Since  and , we have . Therefore, by Axiom 6, D’ = D”. //

 

現在要問的『問題』是『投影代數』早已預先假設『投影』可用『莫比烏斯變換』 z^{'} = \frac{a \cdot z +b}{c \cdot z +d} 來描述耶？故對當於『公設六』，才講相異三點決定『唯一』之『莫比烏斯變換 』勒！

Specifying a transformation by three points

Given a set of three distinct points z1, z2, z3 on the Riemann sphere and a second set of distinct points w1, w2, w3, there exists precisely one Möbius transformation f(z) with f(zi) = wi for i = 1,2,3. (In other words: the action of the Möbius group on the Riemann sphere is sharply 3-transitive.) There are several ways to determine f(z) from the given sets of points.

Mapping first to 0, 1, ∞

It is easy to check that the Möbius transformation

f_{1}(z)={\frac {(z-z_{1})(z_{2}-z_{3})}{(z-z_{3})(z_{2}-z_{1})}}

with matrix

{\mathfrak {H}}_{1}={\begin{pmatrix}z_{2}-z_{3}&-z_{1}(z_{2}-z_{3})\\z_{2}-z_{1}&-z_{3}(z_{2}-z_{1})\end{pmatrix}}

maps z1, z2, z3 to 0, 1, ∞, respectively. If one of the zi is ∞, then the proper formula for  {\mathfrak {H}}_{1} is obtained from the above one by first dividing all entries by zi and then taking the limit zi → ∞.

If  {\mathfrak {H}}_{2} is similarly defined to map w1, w2, w3 to 0, 1, ∞, then the matrix  {\mathfrak {H}} which maps z1,2,3 to w1,2,3 becomes

  {\mathfrak {H}}={\mathfrak {H}}_{2}^{{-1}}{\mathfrak {H}}_{1}.

The stabilizer of {0, 1, ∞} (as an unordered set) is a subgroup known as the anharmonic group.

Explicit determinant formula

The equation

w={\frac {az+b}{cz+d}}

is equivalent to the equation of a standard hyperbola

\,cwz-az+dw-b=0

in the (z,w)-plane. The problem of constructing a Möbius transformation  {\mathfrak {H}}(z) mapping a triple  (z_{1},z_{2},z_{3}) to another triple (w_{1},w_{2},w_{3}) is thus equivalent to finding the coefficients a, b, c, d of the hyperbola passing through the points  (z_{i},w_{i}). An explicit equation can be found by evaluating the determinant

\det {\begin{pmatrix}zw&z&w&1\\z_{1}w_{1}&z_{1}&w_{1}&1\\z_{2}w_{2}&z_{2}&w_{2}&1\\z_{3}w_{3}&z_{3}&w_{3}&1\end{pmatrix}}\,

by means of a Laplace expansion along the first row. This results in the determinant formulae

a=\det {\begin{pmatrix}z_{1}w_{1}&w_{1}&1\\z_{2}w_{2}&w_{2}&1\\z_{3}w_{3}&w_{3}&1\end{pmatrix}}\,
b=\det {\begin{pmatrix}z_{1}w_{1}&z_{1}&w_{1}\\z_{2}w_{2}&z_{2}&w_{2}\\z_{3}w_{3}&z_{3}&w_{3}\end{pmatrix}}\,
  c=\det {\begin{pmatrix}z_{1}&w_{1}&1\\z_{2}&w_{2}&1\\z_{3}&w_{3}&1\end{pmatrix}}\,
d=\det {\begin{pmatrix}z_{1}w_{1}&z_{1}&1\\z_{2}w_{2}&z_{2}&1\\z_{3}w_{3}&z_{3}&1\end{pmatrix}}

for the coefficients a,b,c,d of the representing matrix  \,{\mathfrak {H}}={\begin{pmatrix}a&b\\c&d\end{pmatrix}}. The constructed matrix  {\mathfrak {H}} has determinant equal to (z_{1}-z_{2})(z_{1}-z_{3})(z_{2}-z_{3})(w_{1}-w_{2})(w_{1}-w_{3})(w_{2}-w_{3}) which does not vanish if the zi resp. wi are pairwise different thus the Möbius transformation is well-defined. If one of the points zi or wi is ∞, then we first divide all four determinants by this variable and then take the limit as the variable approaches ∞.