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  1. Section 2
  2.  
  3. Post 1(>>8317513)
  4.  
  5. To continue, we need tools to examine the projective plane we work with, like circles in ancient Greek geometry. We introduce the most fundamental of those tools here, but before we define it, we need a few definitions.
  6.  
  7. The set of all points incident to a given line is the [math]\textbf{range}[/math] of that line.
  8.  
  9. The set of all lines incident to a given point is the [math]\textbf{pencil}[/math] of that line.
  10.  
  11. In a projective plane, let ranges and pencils be [math]\textbf{incidence sets}[/math].
  12.  
  13. The incidence sets of the point p and line L can be written as In(,L) and In(p,) respectively.
  14.  
  15. Before we explicitly define this tool, we demonstrate it with the proof of the following theorem:
  16.  
  17. Theorem 2.1: Any two distinct ranges from the same plane have the same cardinality.
  18.  
  19. Proof: We observe that every line has a unique range and vice versa, thus the two resultant lines, which we call A and B, are also distinct. By Axiom 5*, we have a point o not incident to either A or B. Now, consider the following function.
  20.  
  21. [eqn]\begin{align*}
  22. Pe : In(,\textup{A}) &\to In(,\textup{B})\\
  23. \textup{a} &\mapsto \textup{B[oa]}.
  24. \end{align*}[/eqn]
  25.  
  26. The construction of the points B[o?] for points on A is shown in the pic.
  27.  
  28. If d and e are any distinct points on A, then the lines od and oe are distinct lines which meet at o. Since o does not lie on B, then by Axiom 1, the points B[od] and B[oe] are distinct. Since d and e were arbitrary, then Pe is injective.
  29.  
  30. Now take an arbitrary point x on B. Let z = A[ox]. The point z exists and lies on A, so Pe(z) exists. {z, o, x} is collinear and x lies on B, so Pe(z) = B[oz] = x, thus Pe is surjective.
  31.  
  32. Since Pe is surjective and injective, it is a bijection between In(,A) and In(,B), thus they have the same cardinality. #
  33.  
  34. The function used in the proof above is the tool we seek, which we will formally define in the next post.
  35.  
  36. Post 1.5
  37.  
  38. Section 0 text: http://pastebin.com/50w9MYjD
  39. Section 1 text: http://pastebin.com/fMdRzhTf
  40.  
  41. I mis-numbered some theorems in the Section 1 text. They have been properly numbered in the 'cheat sheets' below. I also added and cleaned up some definitions.
  42.  
  43. Definitions and notations: https://www.dropbox.com/s/ia4m7s0fy59z9jv
  44. Axioms, Theorems, and other statements: https://www.dropbox.com/s/6ch0btu3adiz2ob
  45.  
  46. Post 2:
  47.  
  48. A [math]\textbf{perspectivity}[/math] from line L to a distinct line M with [math]\textbf{centre}[/math] c not incident to L or M is the function that, for any point x on L, returns the point M[ox].
  49.  
  50. A [math]\textbf{perspectivity}[/math] from point p to a distinct point q with [math]\textbf{axis}[/math] C not incident to p or q is the function that, for any point X on p, returns the point q[AX].
  51.  
  52. A [math]\textbf{projectivity}[/math] is a finite composition of perspectivities.
  53.  
  54. A [math]\textbf{half-perspectivity}[/math] from line L to a non-incident point p is the function that, for any point x on L, returns the line xp. The half-perspectivity from point p to a non-incident line L is the inverse of the function above.
  55.  
  56. A [math]\textbf{half-projectivity}[/math] is a composition of an odd number of half-perspectivities.
  57.  
  58. Given two ordered n-tuples of objects from a single incidence set each (X_1, ... , X_n) and (y_1, ... , y_n), we say these tuples are [math]\textbf{projectvely related}[/math] if there is some (half)-projectivity Pr such that Pr(X_k) = y_k for all k such that [math]1\leq k\leq n[/math].
  59.  
  60. Since we will use (half-) perspectivities and projectivities so often, we need a lot of new notation. We use two notations for (half-) perspectivities and projectivities. The two systems describe the perspectivity from L to M with centre p as follows:
  61.  
  62. L>p<M
  63.  
  64. [eqn]L\stackrel{p}{\barwedge}M[/eqn]
  65.  
  66. The perspectivity from p to q with axis L is noted as follows:
  67.  
  68. p<L>q
  69.  
  70. [eqn]p\stackrel{L}{\barwedge}q[/eqn]
  71.  
  72. Projectivities can be notated by concatenating perspectivities.
  73.  
  74. p<N>q<L>r
  75.  
  76. [eqn]A\stackrel{x}{\barwedge}B\stackrel{y}{\barwedge}C\stackrel{z}{\barwedge}D[/eqn]
  77.  
  78. The LaTeX-based notation does not deal well with half-perspectivities and half-projectivities, but the text-based notation does.
  79.  
  80. p<L
  81.  
  82. X>a<Y>b
  83.  
  84. The next post will contain some more notation, and some examples.
  85.  
  86. Post 3:
  87. If two tuples are (X_1, ... , X_n) and (y_1, ... , y_n) projectively related, we can note this in our two systems:
  88.  
  89. (X_1, ... , X_n) ~ (y_1, ... , y_n)
  90.  
  91. [eqn](X_{1},..., X_{n}) \doublebarwedge (y_{1},..., y_{n})[/eqn]
  92.  
  93. If the relating (half-)projectivity is known explicitly, we can combine the notations. Note the tuples are prefixed by the object its elements are all incident to.
  94.  
  95. p(A,B,C)<L>q<M(a,b,c)
  96.  
  97. [eqn]A(p, q, r, s)\stackrel{x}{\barwedge}B(p', q', r', s')\stackrel{y}{\barwedge}C(p”, q”, r”, s”).[/eqn]
  98.  
  99. With the notation taken care of, we can now look at a few examples of perspectivities.
  100.  
  101. Recall that the real projective plane is essentially the Euclidean plane with a line at 'infinity' whose points are uniquely associated with particular slopes. Now consider the function that takes points from the x-axis to the line y=1 such that (k,0) gets mapped to (k+1, 1). This is a perspectivity from the x-axis to the line y=1 with the centre. Note that since the two given lines are parallel in the Euclidean point, they meet at an 'ideal' point in the real projective plane.
  102.  
  103. Another example is the perspectivity from the x-axis to the y-axis with centre (1,1), as shown in the posted pic. It turns out that this perspectivity maps points (k,0) to points (0, 1 + 1/(k-1)). Note that in this case the meet of the lines is a 'real' point and unlike the first example, 'real' points get mapped to 'ideal' points and vice versa. In this case it is the points (1,0) and (0, 1) that are involved.
  104.  
  105. We can also look at the perspectivities in finite projective planes. In the projective plane generated by the PDS {0, 1, 3, 9} mod 13, let (k, 0) = p_k and (k, 1) = L_k. The point p_0 has pencil {L_0, L_1, L_3, L_9} and the point p_9 has the pencil {L_0, L_4, L_5, L_7}.
  106.  
  107. The line L_2 is not incident to either point, so the projectivity from p_0 to p_9 with axis L_2 exists. Working it out, it is apparent that p_0(L_0, L_1, L_3, L_9)<L_2>p_9(L_0, L_4, L_5, L_7).
  108.  
  109. Post 4:
  110.  
  111. With those examples done, we can see what the more general (half-)projectivities are like. Since in general, projectivities are compositions of perspectivities, explicitly working them out could be quite messy. Instead, what we will do is show that certain incidence set tuples are projectively related, so demonstrating what certain projectivities can do.
  112.  
  113. Theorem 2.2: Given three tuples of distinct points (a, b, c) and (a', b', c') with elements taken from I(,L) and I(,M) respectively, then (a, b, c)~(a', b', c').
  114.  
  115. Proof: If a = a' = LM, then by Axiom 1 b, c, b', and c' are all distinct, and thus [bb'][cc'] = o exists, o does not lie on L or M, and L(a, b, c)>o<M(a', b', c') holds. (see http://i.imgur.com/WedhlEe.png)
  116.  
  117. If c = b' = LM, then and a != a and c' != b, thus N = bc' exists. Also, a lies on none of M, N and a' lies on none of L, N. Thus the projectivity L>a'<N>a<M exists. It is then apparent that L(a, b, LM)>a'<N(N[aa'], b, c')>a<M(a', b', c'). (See http://i.imgur.com/EkBlhm5.png)
  118.  
  119. This second case can be generalized in steps to yield the general case by modifying the projectivity L>a'<N>a<M. If b' != LM, then replace point a in the projectivity with [bb'][aa'] while if c != LM, replace a' with [cc'][aa']. (See http://i.imgur.com/pkuBUSW.png)#
  120.  
  121. Theorem 2.3: Given four distinct yet concurrent lines A, B, C, and D, (A, B, C, D) ~ (B, A, D, C).
  122.  
  123. Proof: Choose points p*, p' on lines C, D respectively distinct from p. Choose x on line B distinct from p and not collinear to p* and p'. Then label the following lines:
  124.  
  125. E = p'p*
  126. A* = xp*
  127. A' = xp'
  128. B* = p*[AA']
  129. B' = p'[AA*]
  130.  
  131. This is shown in the posted pic and at http://i.imgur.com/lDEFXD5.png. Of the points and lines p, p', p*, A, A', and A*, only the pairs (p, A), (p', A'), and (p*, A*) are incident. Therefore, the projectivity p<A'>p*<A>p'<A*>p exists. From there one can conclude the following:
  132.  
  133. p(A, B, C, D)<A'>p*(B*, A*, C, E)<A>p'(A', B', D, E)<A*>p(B, A, D, C)
  134.  
  135. as required by the theorem.#
  136.  
  137. Post 5:
  138.  
  139. From the definitions of projectivity and perspectivity, it is apparent that perspectivities are a subcategory of projectivities. It would seem that to determine whether a projectivity is a perspectivity, one would just look at the construction of it. However, by convention (half-)projectivities are considered extensional. That means two (half-)projectivities are equal if two have the same domain and co-domain, and if for each element in the common domain, its image for each function is the same.
  140.  
  141. The upshot of that is you can't just look at a projectivity's construction and tell whether it's a perspectivity, as even if its known construction is very long, there might be a much shorter one, even up to only a single perspectivity.
  142.  
  143. Still, from looking at certain properties of a projectivity, we can often immediately rule it out as a perspectivity. First of all, it is impossible for the domain and co-domain of a perspectivity to be equal. Theorem 2.3 demonstrated that this is not the case for general projectivities.
  144.  
  145. Given that condition, the domain and co-domain have just one common element. Another condition for a perspectivity is that this common point be its own image. This is not true for projectivities in general. Consider the points a, b, c, and d which fulfil Axiom 4+. Then, by Theorem 2.2, (a, [ab][cd], b) ~ (c, d, [ab][cd]).
  146.  
  147. Now down to projectivities between distinct objects with self-corresponding meet/join, we cannot narrow our candidates any further. We know that at least some perspectivities are there, so the simplest possibility is that they are all perspectivities.
  148.  
  149. Conjecture 2.1(Perspectivity Conjecture): Every projectivity with distinct domain and co-domain such that the common element of the domain and co-domain is its own image is a perspectivity.
  150.  
  151. This conjecture will be our focus(if indirectly) for most of the next few threads. For now, the next large post will focus on some specific consequences/conditions.
  152.  
  153. Post 5:
  154.  
  155. For this whole post we assume that L>r<M>p<N is equal to L>q<N for some q.
  156.  
  157. Lemma 2.4: L>q<N>p<M = L>r<M
  158.  
  159. The proof of this statement is left to the reader.
  160.  
  161. Theorem 2.5: Either {L, M, N} is concurrent or {p, r, LN} collinear.
  162.  
  163. Proof: We assume that {L, M, N} is non-concurrent. The image of LN under the component perspectivity L>r<M is not itself. Call the image a. Then we have {r, a, LN} collinear. Since L>r<M>p<N is a perspectivity, then it must be the case that LN is the image of a under M>p<N, so we have {p, a, LN} collinear. Then our conclusion follows from Col. L.#
  164.  
  165. Theorem 2.6: If {L, M, N} is concurrent, then {p, q, r} is collinear.
  166.  
  167. Proof: We first assume that {r, p, LN} is not collinear, so L[pr], M[pr], N[pr],are all distinct and the set {p, r, L[pr], M[pr], N[pr]} is collinear. The image of L[pr] under L>r<M>p<N is N[pr]. Therefore, the image of L[pr] under L>q<N is N[pr], so {q, L[pr], N[pr]} is collinear, so {p, q, r} is collinear by Col. L.
  168.  
  169. If {r, p, LN} is collinear, you can try L>q<N>p<M and L>r<M instead by Lemma 2.4. If {p, q, LM} is collinear too, since LM = LN, the conclusion again follows Col. L.#
  170.  
  171. Lemma 2.7: If {L, M, N} is non-concurrent, then none of the pairs (p, L), (q, M), or (r, N) is incident.
  172.  
  173. Proof: By Theorem 2.5, {p, r, Ln} is collinear, so if p lies on L, so does r, which is not possible since L>r<M exists. Similar reasoning carries for the pair (r, N). Looking at L>q<N>p<M = L>r<M with this reasoning, then (q, M) non-incident also follows.#
  174.  
  175. Theorem 2.8: If {L, M, N} is non-concurrent, then q = [p[LM]][r[MN]].
  176.  
  177. Proof: From Lemma 2.4 and Theorem 2.5, {p, q, LM} is collinear. The inverses N>q<L and N>p<M>r<L are equal as well, and applying 2.4 and 2.5 to that we get {r, q, MN} collinear. By Lemma 2.7, the lines p[LM] and r[MN] exist. If they were equal, then {p, r, q, MN, LM} is collinear and thus a subset of In(,M), contradicting Lemma 2.7, so p[LM] and r[MN] are distinct. The point q lies on both, so q = [p[LM]][r[MN]].#
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