/sci/ - Science and Mathematics

>>3170

To facilitate further discussion, I'll start introducing notation and terminology here and in future posts on this.

Notation: The end of proofs will be denoted with a #.

If a point P and a line m are incident, I may also say that:

P lies on m

m passes through P

or other similar terms.

The order of a projective plane is one less than the number of distinct points on a line.

Given a non-empty tuple of points, a join of those points is a line that is incident to all points in the tuple.

Given a non-empty tuple of lines, a meet of those lines is a point that is incident to all lines in the tuple.

Note that the definitions themselves do not rule out the possibility of multiple meets or joins for some set. It is our axioms that allow us to speak of 'the' meet or 'the' join when talking about tuples with more than one distinct element. Also, there is no requirement that the points specifying a join are distinct (hence the use of tuples instead of sets; the order is irrelevant), and the same goes for lines specifying a meet.

A set of points is collinear if there is a common join between all the points in the set.

A set of lines is concurrent if there is a common meet between all the lines in the set.

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For all this talk of these axioms, we still have not proven that any projective planes exist. Thus we will exhibit examples of planes.

Example 1:

A perfect difference set(PDS) modulo q is a set of (k + 1) distinct residues mod q such that for any non-zero residue arises uniquely as the difference of two distinct residues in the PDS. For any PDS, it is necessary that q = k^2 + k + 1, and PDSs always exist if k is a prime power.

Given a PDS mod q = k^2 + k + 1, k >= 2, a projective plane can be formed by assigning a distinct point and line to each residue mod q, and define a point and line incident if their residues sum to some element of the PDS. Examples of PDSs are given in pic 1, there p^n = k is the order of the resultant plane.

Axiom 1 and 2:

Given any two distinct points P_m and P_n, the difference of their indices m-n is non-zero. Thus, there are unique distinct elements j,k in the PDS such that j - k = m - n mod q. Therefore, j - m = k - n mod q. Let j - m = c. Then c + m = j and c + n = k, thus the line l_c joins P_m and P_n. The uniqueness of j and k ensures the uniqueness of l_c.

Exactly analogous reasoning shows that any two distinct lines meet in a unique point.

Axiom 3:

If we consider the set of points incident to some given line l_i, their indices are merely our given PDS shifted by -i mod q. Since this shift does not effect differences, those indices always form a PDS as well. In particular, any line's point indices cannot contain three or more distinct residues in arithmetic progression.

We now need to produce 4 four distinct points with no three collinear. Consider the residues 0, 1, and 2. Since q is at least 7, their corresponding points are distinct. Also, they are not collinear since the residues are in arithmetic progression. Consider the residue 3. Its point is not collinear with those of 1 and 2, but it may be those of 0 and 1 or 0 and 2. If neither is the case, then the points of 0, 1, 2, and 3 are the desired points.

If we assume the points of 0, 1, and 3 are collinear, then the following residues' points are collinear, from shifting.

0, 1, and 3

1, 2, and 4

-1, 0, and 2

If we now take residue 5, we can see that its point cannot lie on any of the lines above on pain of non-unique differences. Thus, the points of 0, 1, 2, and 5 are distinct with no three collinear.

If the points 0, 2, and 3 are collinear instead, then the points of -3, 0, 1, and 2 are distinct with no three collinear. In all cases Axiom 3 is satisfied.

>>3201

Another important aspect of the plane axioms is independence. Given a set of propositions, a particular proposition is independent of the others if given all the other propositions, the addition of the statement or its negation produces a consistent set of statements. We will demonstrate that the plane axioms themselves are each independent of the others, but as we consider axioms in conjunction to the plane axioms, independence of the new axioms may not hold.

We have already demonstrated the consistency of the plane axioms, at least with respect to the consistency of the constructions and their underlying theories, so what we now need is a model for each axiom in which that axiom is false, yet the other plane axioms are true. In reverse order:

Axiom 3:

There are many models for which Axioms 1 and 2 are true yet 3 is false. These are generally known as degenerate projective planes. One instructive example is if both the set of lines and the set of points are the empty. Other examples of degenerate planes include a single line with all points on it, a point with all but one line through it and one point on the non-incident line for each distinct line through the point, and so on (see pics 1-3).

Degenerate planes in general do not have the same number of distinct points on each line or the same number of distinct lines on each point, and cannot support the constructions that non-degenerate planes can. In particular, A set of four points with no three collinear, along with the six pairwise joins, is itself an important motif in future sections.

Axiom 2:

The Euclidean plane itself serves here, as although it can be made into a projective plane by the additions of a line at infinity, by itself the possibility of parallel lines means it does not satisfy Axiom 2, though Axiom 1 and 3 do apply.

The fact that parallel lines exist in the Euclidean plane also complicates the description and proof of certain Euclidean theorems, as the parallel case has to be taken into account. Conversely, a single projective theorem may lead to several different Euclidean theorems from different choices of the line at infinity.

Axiom 1:

We first emphasize that we are treating 'points' and 'lines' as undefined terms and we can take any two sets as our lines and points if we have the right incidence relation to satisfy the axioms we require. Thus, we can use the Euclidean plane as an independence example for Axiom 1 as well, if we take our points to be Euclidean lines and our lines to be Euclidean points with the standard incidence relation. Here the existence of parallel lines invalidates Axiom 1 instead of Axiom 2.

Another example comes from the sphere, where our set of points is merely the set of spherical points, but our lines are the great circles of the sphere. Here Axiom 1 fails if the distinct points chosen are antipodal.

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On a final note, from our axioms, we proved this statement as a corollary:

'For any two distinct lines, there is exactly one point incident with both lines.'

Note that it is a stronger version of Axiom 2, and so could replace it as an axiom without effect.

Additionally, this statement is the same as Axiom 1 with the words 'point' and 'line' swapped. The same is true for Axiom 3 and (1), >>3171.

Thus, if some statement about points and lines is a theorem, then the statement where the words 'point' and 'line' are swapped as well as other word pairs like 'concurrent' and 'collinear' is also a theorem.

This property of the plane axioms is called duality, and it often allows a single proof to prove two different theorems. Frequently, I will prove a statement about, say, points on a line, and immediately one can prove the dual statement of lines on a point. Upcoming sections, however, will introduce additional axioms, and the preservation of duality will need to be shown for the enlarged set of axioms.

Before we start, I should not that there are a lot of definitions in this installment. Here are the first few.

The range of a line is the set of all points incident to the line.

The pencil of a point is the set of all lines incident the point.

Both ranges and pencils are examples of forms.

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We start be proving (6.1), >>3171. We first prove a weaker version of (6.2), >3171

Theorem 1: Given a line and a point not incident to each other, the corresponding range and pencil have the same cardinality.

Proof: Let our point and line be P and l respectively. Then construct a function |f| from l's range to P's pencil so that for any point A on l, PA = |f|(A). From Theorem 1, >>3171, |f| is injective, as non-injectivity implies that a line through P is incident to two distinct points on l. From Axiom 2, |f| is surjective, because any line m though P must meet l, and then |f|(lm) = l. Thus there is a bijection between the range and the pencil, and their cardinalities must be identical.#

Lemma 1:

Given two distinct lines, there is a point which is incident to neither of them. (Proof left to reader.)

Corollary 1.1: Given any two distinct lines, their ranges have the same cardinality.

Proof: Starting with the lines a and b, from Lemma 1 we obtain a point P not on a or b. From Theorem 1, the range of a and pencil of P have the same cardinality. Again from Theorem 1, the pencil of P and the range of b have the same cardinality. Thus, the ranges of a and b have the same cardinality, as required.#

The constructions used in the proof of Theorem 1 and implied in the proof of Corollary 1.1 are the subject of this particular section. First, we need to rigorously define these constructions.

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Given a line a and a point T, the half-perspectivity from a to T is the bijection from the range of a to the pencil of T such that for any point C on a, its image is CT. (see pic 1)

The half-perspectivity from T to a also exists, and it is the bijection from the pencil to the range such that for any b on T, its image is ab. Note that this is the inverse of the half-perspectivity from a to T.

Notation: The half-perspectivity from a to T shall be called a>T, and the half-perspectivity from T to a shall be called T<a.

Given two distinct lines a, b and a point C not incident to either, the perspectivity from a to b with centre C is the function from the range of a to the range of b where for any point X on a, its image Y under the perspectivity is X' = b[XC].(see pic 2)

Given two distinct points P, Q and a line a not incident to either, the perspectivity from P to Q with axis a is the function from the pencil of P to the pencil of Q where for any line x through P, its image y under the perspectivity is x' = Q[xa]. (see pic 3)

Notation: The perspectivity from a to b with centre C can be written as a>C<b. The perspectivity from P to Q with axis a can be written as P<a>Q.

Note that both types of perspectivities can be considered a composition of two compatible half-perspectivities. We can generalize this idea as follows.

A projectivity is function between forms composed from a finite number of half-perspectivities.

Note that this definition encompasses both perspectivities and half-perspectivities. Also, the identity perspectivity, which takes a form to itself with all elements self-corresponding, is also a projectivity.

Notation: The name of a projectivity may be its presentation in half-perspectivities(e.g. a>B<c>D<e>F), a string of letters(and other notation) enclosed by vertical bars(e.g. |hKmN|, |W/(xy)Z|) or any combination of the two (e.g. P<|abc|<i>J<k, A<|x|<e). The composition of projectivities |a| and |b| is |ab|, such that for any x, |ab|(x) = |b|(|a|(x)). The inverse of |a| is |/a|. The identity projectivity is |1|, regardless of domain.

More notation: If we wish to keep track of the images of particular points, the name of a line or point in a projectivity presentation can be suffixed by a tuple of elements from the form (e.g. l(A, B, C)>|R|>P(d, e, f)). The elements in each tuple in the presentation are associated in order, first to first and so on. If the tuples are present, then the line or point names may be suppressed (e.g. abc<|T|<DEF).

Given two tuples of the same size each with elements from a single form, we say the two tuples are projective if there exists some projectivity which takes one tuple to the other tuple.

Notation: We make note of a pair of projective tuples with the ~ symbol (e.g. (U, V, W) ~ (x, y, z)).

We say that two projectivities are identical(or equal) when they have the same domain and co-domain and for each point in the domain, its image under either projectivity is the same.

When compared to a generic projectivity, perspectivities and half-perspectivities have certain special properties. One is the restriction of the type of form the domain and co-domain can be. Half-perspectivities must have domain and co-domain of different types, while for perspectivities they must be of the same type. Furthermore, the domain and the co-domain for a perspectivity must be distinct, while there is no such restriction for projectivities.

If we look carefully at the definition of a perspectivity, then the following theorem is apparent.

Theorem 2: For any perspectivity, the common point of the domain and the co-domain is self-corresponding.

For a generic projectivity, even if the domain and co-domain are of the same type yet distinct, their common element may not be self-corresponding, as shown by the theorem below.

Theorem 3: Given any two distinct lines m and n, and tuples of distinct points (A, B, C) and (A', B', C') with elements from the ranges of m and n respectively, then (A, B, C) ~ (A', B', C').

(Note: On the term “tuples of distinct points”, only distinctness within a tuple is intended, as is the case most of the time. Exceptions will be noted on occurrence.)

The following lemma will be use in the proof below and many other proofs.

Lemma 2(Collinearity Lemma(Col L)):

If the intersection of two collinear sets contains two distinct points, then the union of the two sets and any subset of it is also collinear. (Proof left to reader)

Naturally the dual Concurrency Lemma(Con L) also holds.

Proof of Theorem 3: Assume without loss of generality that neither A nor A' is the point mn. Define the points B" and C" as [[A'B][AB']] and [[A'C][AC']] respectively. If B” and C” are identical, that would imply {A, B', C'} is a collinear set, which in turn implies that A lies on n, contrary to the premise. This B” and C” are distinct and the line B”C” = o is well-defined.

If we assume the line o passes through A, then the set {A, B”, C”} is collinear. The sets {A, B”, B'} and {A, C”, C'} are collinear by construction, thus by the Col. L. the set {A, B', C'} must be as well, which again implies A lies on n, contrary to the premise. Therefore B”C” does not pass through A. An analogous argument shows that o does not pass through A'. Let A” = o[AA']. We have now shown that the following projectivity exists.

m(A, B, C)>A'<o(A”, B”, C”)>A<n(A', B', C')(see pic 1)

This projectivity shows (A, B, C) ~ (A', B', C'), as required.#

Looking closely at the proof, we that if, say, the B' = mn, then all that implies is that om = B = B”. If C = mn as well, then o = BC', C” = C' and the projectivity still exists (see pic 2). In these cases the point mn will definitely not be self-corresponding.

This result still leaves open the possibility of a converse to Theorem 2 for 'eligible' projectivities. What remains to be seen is what counts as an eligible projectivity. In the next few sections our main focus will be to introduce new axioms that bear on this question. Right now, even with just the plane axioms, we prove some consequences of a converse of Theorem 2.

Lemma 3: Given two projectivities |a| and |b|, |a| and |b| are identical iff |a/b| = |1|

Proof: Let x be an element in the common domain of |a| and |b|. First we assume |a| and |b| are identical. Then |a|(x) = |b|(x) = y. Thus, we have |/b|(y) = x, and composing gives us |a/b|(x) = y. Since x was chosen arbitrarily, |a/b| = |1| holds. Now assume |a/b| = |1|. Let |a|(x) = z. Therefore, |/b|(z) = x, and thus, |b|(x) = z = |a|(x), as required.#

Lemma 4: If |PQ| = |1|, then so does |QP|.

Proof: By the premise, for any X in the domain of |PQ|, |PQ|(X) = X.

Let Y = |P|(X). Then it must be the case that |Q|(Y) = X. Therefore, |QP|(Y ) = Y. However, since |P| is a bijection and x was chosen arbitrarily, Y is also arbitrary. Thus, |QP| = |1|.#

Theorem 4: If the projectivity a>R<b>P<c is identical to the perspectivity a>Q<c, then a>Q<c>P<b is also identical to a>R<b.

Proof: The premise implies a>R<b>P<c>Q<a = |1| by Lemma 3, which in turn implies b>P<c>Q<a>R<b = |1| by Lemma 4. Then, by Lemma 3 again, b>P<c>Q<a = b>R<a, and taking the inverse of both we have a>Q<c>P<b = a>R<b, as required.#

Theorem 5: If the projectivity a>R<b>P<c is identical to the perspectivity a>Q<c, then either {a, b, c} is concurrent, or {ac, P, R} is collinear.

Proof: Since the two projectivities are equal, then the images of one point under each are the same. In particular, the image of the point ac is the same for both. Since a>Q<c is a perspectivity, then ac is self-corresponding. If a, b, and c are concurrent, then this is the case for a>R<b>P<c anyways. If a, b, and c are not concurrent(see pic 1), then the component perspectivity a>R<b must send ac to a point not equal to ac on b. Call this point E. By construction, {E, ac, R} is collinear. Since under the full projectivity ac is self-corresponding, the second component perspectivity b>P<c must send E back to ac, thus {E, ac, P} collinear. Therefore, {P, R, ac} is collinear, by Col L.#

Theorem 6: If the projectivity a>R<b>P<c is identical to the perspectivity a>Q<c and {a, b, c} is concurrent, then {P, Q, R} is collinear.

Proof: First assume P, Q, and ac are not collinear. Let M = a[PR], M' = b[PR], and M” = c[PR](see pic 2). By assumption M, M' and M” are distinct. Note that M' is the image of M under a>R<b, and M” is the image of M' under b>P<c. By composition and the theorem premise, M” is the image of M under a>Q<c. Thus, Q lies on MM”, but MM” = PR by definition, proving the theorem. This proof breaks down if {P, R, ac} is collinear. In this case, from Theorem 4 we have b>P<c>Q<a identical to b>R<a. If {Q, P, ab} is not collinear, R can be proven to lie on QP with the argument above. If they are collinear, that implies {P, R, Q} collinear by Col L, again satisfying the theorem.#

Theorem 7: If the projectivity a>R<b>P<c is identical to the perspectivity a>Q<c and {a, b, c} is not concurrent, then Q = [R[bc]][P[ab]]

Proof: By the theorem premise, ab and bc are distinct. Note also that neither P or R can lie on b. Let F be the image of ab under b>P<c and let D be the image of cb under b>R<a(see pic 3). By composition, a>R<b>P<c, and thus a>Q<c, must take ab to F and D to bc. Thus {ab, P, Q} is collinear, as well as {Q, R, bc}. Thus, Q lies on P[ab] and R[bc]. Since P and R do not lie on b while ab and bc do, P[ab] and R[bc] are distinct lines, making Q the meet of P[ab] and R[bc].#

One last theorem of note is the existence of a particular projectivity with with identical domain and co-domain.

Theorem 8: Given any 4 distinct lines a, b, c, d in the pencil P, P(a, b, c, d)~P(b, a, d, c)

Proof: Choose points P*,P" on lines c, d respectively distinct from P. Choose X on line a distinct from P and not collinear to P* and P". Then label the following lines:

e = P*P"

a* = XP*

a" = XP"

b* = P*[a"b]

b" = P"[ab*]

(see pic 4 and 5)

Thus abcd<a">a*b*ce<a>a"b"de<b*>badc provides the required perspectivity.#

>>3335

With these definitions we can prove that:

Theorem 1: Plane sets are projective planes

Corollary 1.1: If two distinct space planes share three or more points, then those points are collinear.

The proof of these statements is rather long and somewhat tedious, so it is written separately here:(https://itmb.co/u1kfz)

Another obvious corollary is:

Corollary 1.2: Given a plane set and a line not in the plane set, there is at most one point in both the plane set and the line's range.

Proof: since the plane set is a projective plane, then if there were two distinct points of the range on the line, that would mean that the line joining those two points would be in the plane set as well, contrary to our assumption.#

The existence of such a line is guaranteed by the next theorem.

Theorem 2: For any plane set, there exists a point outside that set.

Proof: By Axiom 4, there exists two lines that do not meet. If our plane set contains one of those lines, then the other line is not, since the plane set is a projective plane. Therefore, at most one point in that line's range is in the plane set. By Axiom 3*, however, the line's range contains at least three points, so at least two points must be outside the plane set. The same reasoning applies for both of the lines when neither is in the plane set.#

We are now ready to introduce our new axiom, which is actually a theorem with respect to the space axioms:

Theorem 3(Desargues' Theorem): Given two distinct triangles with ABC and A*B*C*, (A, B, C,)><(A*, B*, C*) implies (CA, AB, BC)<>(C*A*, A*B*, B*C*).

Proof: This proof will only consider 'general' cases where all the aforementioned points and lines are distinct and there are no 'extra' collinear or concurrent sets. Verification of such 'special' cases will be left to the reader except to note that in the planar case, there is no additional difficulty if O lies on the line [[BC][B*C*]][[CA][C*A*]].(I will cover specific special cases if people ask.)

We will first consider the case when the triangles are not in the same plane set. (A, B, C)><(A*, B*, C*) implies that AA*, BB*, and CC* meet at some point O. If we consider the two lines AA* and BB*, by Axiom 2*, AB and A*B* also meet at some point C”(see first picture). Analogously, B” = [CA][C*A*] and A” = [BC][B*C*] also exist. By construction A”, B”, and C” are in both of the plane sets of triangles ABC and A*B*C*, which were assumed distinct. Thus, by Corollary 1.1, they are collinear, lying on some common line o.

In the case where the triangles are in the same plane set, this proof may fail. Note that O is in the common plane set as well. By Theorem 2, there exists some point P not in the plane set. Let p = PO, and note that by Axiom 3*, there is another point P* on p. By construction, AA* and PP* meet. By Axiom 2*, The lines PA and P*A* also meet at some point D.(see second picture) Analogously E = [PB][P*B*] and F = [PC][P*C*] also exist. Since PA, PB, and PC are not in the plane set but A, B and C are, D, E, and F are not in the plane as well (Corollary 1.2). Also, (A,B,C)><(D,E,F) and (A*,B*,C*)><(D,E,F), by centres P and P* respectively. By the non-planar case [AB][DE] = C” , [BC][EF] = A”, and [CA][FD] = B” are collinear. However, [A*B*][DE] exists as well, and since DE is not in the plane set, its range has only one distinct common point with the plane set. Thus, [AB][A*B*] = C”. Similarly [BC][B*C*] = A” and [CA][C*A*] = B” as well. This proves the theorem in the case where all triangle points are distinct and with no spurious collinear or concurrent sets.#

>>3336

If we wish to take this theorem of the space axioms as an addition to our plane axioms, there are still a few matters to consider. One of those is the principle of duality. Luckily, Desargues' Theorem can prove its own dual.

Theorem 4: If Desargues' Theorem holds, then given 2 distinct triangles ABC and A*B*C*, (A, B, C)><(A*, B*, C*) if (CA, AB, BC)<>(C*A*, A*B*, B*C*).

Proof: If the triangles share any points or lines, the theorem is trivial. Thus, assume the two triangles ABC and A*B*C* points' and lines' are all distinct. Let A” = [BC][B*C], B” = [AC][A*C*], and C” = [AB][A*B*].(see third pic) By assumption, A”, B”, and C” are collinear. Also, let O = [AA*][BB*]. Note that our theorem will be proved if {C, C*,O} can be shown collinear. To show this, consider the triangles BB*A” and AA*B”. If we take their points by pairs in the order written(AB, A*B*, and B”A”), then C” is collinear with each pair, thus (A, A*, B”)><(B, B*, A”). By Desargues Theorem, then, (B”A, AA*, A*B”)<>(A”B, BB*, B*A”). Equivalently, [B”A][A”B], [AA*][BB*], and [A*B”][B*A”] are collinear. However, these are just names for C, O, and C*, respectively. Thus, the theorem follows.#

Another consideration is consistency. It would not be of much use to adopt this new axiom if it leads to contradiction. The way we alleviate this concern is to construct a model of the space axioms using linear algebra. Thus, plane sets constructed from this space will satisfy Desargues' Theorem and our original plane axioms.

Start with a 4-dimensional vector space. Define our points to be 1-dimensional subspaces, while our lines are the 2-dimensional subspaces. We say that a point and line are incident iff the 1d-space is a subspace of the 2D-space. This is thus a model of the ED space axioms.

(Notation: Vector space names will be in bold, vector names will be underlined and italic.)

Axiom 1: Let the join of two distinct points be the direct sum of the 1D-spaces. Since the points are distinct, their basis vectors are linearly independent, and the direct sum exists.

Axiom 2*: Let X, Y, Z, and W be the 1D-subspaces in question. Since they are distinct, then their bases are pairwise linearly independent. However, since [XY][ZW] exists, then that implies that they are linearly dependent taken altogether. In particular, it implies the existence of scalars x, y, z, w (no more than one equaling zero) such that the following equation with the basis vectors holds:

xX + yY = zZ + wW

This can be rearranged to the equation xX – zZ = wW – yY. This implies the existence of [XZ][YW].

Axiom 3a: Given a 2D-space, let (P, Q) be a basis of the space. Then span(P), span(Q), and span(P+Q) are three distinct 1D subspaces.

Axiom 4: Let (A, B, C, D) be a basis of the 4D-subspace. Consider the lines span(A, B) and span(C, D). Since their dimension is 2 and the dimension of their sum is 4, the dimension of their intersection is 0. Thus the two lines have no common 1D-subspace.

We are now ready to adopt Desargues' theorem as an addition to our plane axioms.

Axiom D: Given two distinct triangles with ABC and A*B*C*, (A, B, C,)><(A*, B*, C*) implies (CA, AB, BC)<>(C*A*, A*B*, B*C*)

(Notation: Theorems and other statements with proofs using Axiom D or other statements requiring Axiom D will be marked with a (D) at the very start.

e.g. Theorem 4(D): )

Shortly we will show that Desargues' Theorem has consequences for projectivities.

The straight-line figures shown in this section are to be thought of as merely suggestive diagrams. The question of whether Axiom D applies to the Euclidean plane with a line at infinity(Example 2, >>3201) is left to a future section.

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In a previous set of posts, we introduced Axiom D by showing it was a consequence of projective plane constructed in a particular way. At the end of those posts it was implied that the new axiom had significant consequences for projectivities. At first glance, however, the statement of Axiom D has nothing to do with projectivities. To use Axiom D to prove properties of projectivities, we will need to translate it into a form that applies to projectivities. These first two theorems will do this.

Theorem 1(Stable Perspectivity Theorem(SPT))(D): Given three distinct yet concurrent lines a, b, and c and a projectivity a>P<b>Q<c, there exists a third point R such that a>P<b>Q<c is identical to a>R<c.

Proof: Note that the theorem is trivial if P and Q are not distinct, so assume they are(pic 1). Then the line PQ exists. Take any point A on a not on b or PQ. Let its image under a>P<b be B and its image under a>P<b>Q<c be C. By construction, ABC is a triangle with none of its vertices on PQ. Let R = [AC][PQ]. It is obvious then that A, ac, and a[PQ] have the same images under a>P<b>Q<c and a>R<c. Suppose we have some arbitrary point on a named X not equal to A, ab, or a[PQ]. Let Z be the image of X under a>P<b>Q<c and let Y = [PX][QZ]. Then XYZ is a distinct triangle from ABC, and (A, B, C)><(X, Y, Z) with centre ac. Note that by construction [AB][XY] = P and [BC][YZ] = Q. By Axiom D, [AC][XZ] must lie on PQ. Therefore, [AC][XZ] = R. Therefore, Z is the image of X under a>R<c. Since X was chosen arbitrarily, the images of all points on a under a>R<c and a>P<b>Q<c are equal, thus the two projectivities are identical.#

Theorem 2: From the plane axioms alone, the SPT implies Axiom D

Proof: Let ABC and XYZ be distinct triangles such that (A, B, C)><(X, Y, Z). Let a = AX, b = BY, and c = CZ(see pic 1). By assumption, a, b, and c are concurrent. Let P = [AB][XY] and let Q = [BC][YX]. If P = Q, then Axiom D is satisfied vacuously, so assume P and Q distinct. The outcome is the same if P lies on a, Q lies on c, or either P or Q lie on b, so again we assume none of these are true. Under these conditions, the projectivity a>P<b>Q<c exists. By the SPT, there exists a point R such that a>R<c is identical to a>P<b>Q<c. Since C is the image of A and Z is the image of X under a>P<b>Q<c = a>R<c, R = [AC][XZ]. Also, from Theorem 6 in >>3297, P, Q, and R are collinear. Therefore (BC, CA, AB)<>(YZ, ZX, XY), as required by Axiom D.#

The SPT has a profound consequence for all projectivities, specifically the number of (half-)perspectivities needed to construct a projectivity. Before we can prove this consequence, we need to prove a particular lemma.

Lemma 1(Substitution Lemma)(D): Assume lines i, j, and k are non-concurrent.

1a: Given a projectivity i>A<j>B<k and a line j' distinct but concurrent with i and j and not passing through B, then there exists a point A' on AB such that i>A'<j'>B<k = i>A<j>B<k

1b: Given a projectivity i>A<j>B<k and a line j* distinct but concurrent with k and j and not passing through A, then there exists a point B* on AB such that i>A<j*>B*<k = i>A<j>B<k.

Proof: The proof for 1a and 1b are analogous, so we will prove 1a here and its modification to prove 1b will be left to the reader. Starting off, since j' does not pass through B, the projectivity i>A<j>B<j'>B<k exists(see pic 2). This projectivity is self-evidently identical to i>A<j>B<k. Now consider the component projectivity i>A<j>B<j'. Since i, j, and j' are concurrent, by the SPT, there exists A' on AB such that i>A<j>B<j' = i>A'<j'. Recomposing the new perspectivity into the projectivity, we have i>A'<j'>B<k = i>A<j>B<j'>B<k = i>A<j>B<k, as required.#

Theorem 3(2-perspectivity Theorem)(D): Any projectivity between two distinct ranges is identical to a composition of at most 2 perspectivities.

Proof: The construction in Theorem 3 in >>3296 implies the theorem for a plane with 3 or 4 distinct points per range regardless of Axiom D. Thus, we can safely assume there are at least 5 distinct lines per pencil in the proof. Also note that all we have to do is to show that all projectivities composed of three perspectivities can be reduced to one composed of only two perspectivities. For larger compositions, the proof can be applied multiple times to get the desired result. Start with the projectivity h>P<i>Q<j>R<k. There are 5 relevant cases with regard to concurrency of the ranges.

Case 1: h, i, j concurrent

Case 2: i, j, k concurrent

Case 3: h, i, k concurrent

Case 4: h, j, k concurrent

Case 5: No lines concurrent

Cases 1 and 2 are immediately taken care of by the SPT. Cases 3 and 4 can be reduced to Case 5 by applying the Substitution Lemma on i or j through ij.

Focusing on case 5, first suppose that hi, jk, and Q are not collinear. Let x = [hi][jk]. By assumption, then, the perspectivity h>P<i>Q<x>Q<j>R<k exists and is identical to the original projectivity(See pic 3). Focusing on the component projectivities h>P<i>Q<x and x>Q<j>R<k, we see that the SPT applies to both by the construction of x, producing the perspectivities h>P'<x and x>R'<k. Recomposing, we have h>P'<x>R'<k = h>P<i>Q<j>R<k, as required.

In the case that hi, jk, and Q are collinear, this construction breaks down. For this situation, choose some line i' through ij not equal to i, j or passing through P, hk. By the Substitution Lemma, there exists Q' such that h>P<i>Q<j = h>P<i'>Q'<j. In particular, the image of hi on j must be jk for both projectivities. Therefore the image of hi' distinct from hi must not equal jk. Thus, hi', Q' and jk are not collinear, so we can use the reduction in the previous sub-case on the identical projectivity h>P<i'>Q'<j>R<k, satisfying the theorem.#

Finally, a couple of immediate consequences of this theorem.

Corollary 1(D): Any projectivity from a range to itself is identical to a composition of at most 3 perspectivities.

Corollary 2(D): Any projectivity between a range and a pencil is identical to a composition of at most 5 half-perspectivities.

The 2-perspectivity theorem is important in that it allows us to move from talking about projectivities in the abstract to particular configurations of lines and points. This ability will be made more clear in the next two installments, where we introduce another new axiom.

>>3503

Theorem 1: The UPC implies the SPT.

Proof: First of all, it is easily verified that the SPT is true for any projective plane with exactly three distinct points in each range. Thus, we assume there at least four distinct lines through each point throughout the proof. Also, starting with an eligible projectivity, we assume the centres of the component perspectivities are distinct, as otherwise the consequent is trivial.

Starting from a projectivity m>R<n>P<o where {m, n, o} is concurrent, note that the line RP is well-defined by assumption. We recognize three relevant cases:

Case 1: {RP, m, n, o} is concurrent.

Case 2: P lies on m and R lies on o.

Case 3: At least one of m[RP], o[RP] is distinct from any of P, R, or mo.

Case 3: Assume without loss of generality that o[RP] distinct from P, R, and mo. Choose some line n' through o[RP] distinct from o and not passing through R or P(see pic 1). Thus the projectivity m>R<n'>R<n>P<o exists and is equal to the original projectivity.

If we consider the projectivity n'>R<n>P<o, we see that by the construction of n', {n', n, o} is not concurrent yet {n'o, R, P} is collinear. Thus the UPC applies and some point P' exists such that n'>P'<o equal to n'>R<n>P<o. Therefore, m>R<n>P<o = m>R<n'>R<n>P<o = m>R<n'>P'<o.

By Theorem 7, >>3297, P' = [R[no]][P[nn']], thus {P', R, mo} is collinear. Also, by construction, {m, n', o} is not concurrent. Thus the UPC can be applied again to obtain Q such that m>Q<o is ultimately equal to the original perspectivity.

Case 2: Note that in this case RP cannot pass through mo. Choose a line n* passing through mo distinct from m, n, and o. Since P lies on m and R lies on o, the projectivity m>R<n*>R<n>P<o exists. The component projectivity n*>R<n>P<o falls under case 3 with n*[RP], so there exists a point P* such that n*>R<n>P<o = n*>P*<o. Thus, m>R<n>P<o = m>R<n*>P*<o.

If R = P*, we are done trivially. Otherwise, by Theorem 6, >>3297, RP = RP* and thus RP* does not pass through mo. If P* lies on m, then that implies P = P* and m>R<n>P<o = m>R<n*>P<o. Let X be an otherwise arbitrary point on m not equal to m[RP*] or mo. Let Y be the image of X under m>R<n, Y* be the image of X under m>R<n*, and Z be the image of X under m>R<n>P<o(see pic 2). By the component perspectivities, the following sets are collinear:

{X, R, Y}

{X, R, Y*}

{Y, P, Z}

{Y*, P, Z}

Also, since X is distinct from om, Y and Y* are distinct from each other by construction. Thus with judicious use of Col L, the union {X, R, Y, Y*, P, Z} is collinear. This implies that X lies on RP, contrary to our construction, Therefore, P* does not lie on m and m>R<n*>P*<o falls under Case 3 with m[RP], allowing for an equal perspectivity m>Q<o ultimately identical to the original projectivity m>R<n>P<o.

Case 1: If there are exactly four distinct lines in each pencil, then this case is easily taken care of without the UPC. Thus, we assume there are at least five distinct lines in each pencil.

Under that additional assumption, there exists a point B not on m, n, o, or RP. Define the following line and point.

k = [m[BR]][o[BP]]

A = [R[nk]][B[mo]]

From these constructions, the sets {B, A, mn}, {B, mk, R}, {A, nk, R}, and {B, P, ok} are collinear sets (see pic 3). Also note that k cannot pass through om and none of A, B, P, or R can lie on k. Consider the projectivity m>A<k>B<n. By construction, {m, k, n} is not concurrent but {B, A, mn} is collinear. By the UPC, there exists a point R' such that m>R'<n = m>A<k>B<n. By Theorem 7, >>3297 and the construction of A and k, the following two sets are collinear:

{A, nk, R, R'}

{B, mk, R, R'}

If R and R' were distinct, then by ColL A and B would lie on k, contrary to our construction. Therefore, R = R' and m>R<n>P<o = m>A<k>B<n>P<o. Looking at the component projectivity k>B<n>P<o, we see that {k, n, o} is not concurrent but {B, P, ok} is collinear. By the UPC, there exists a point C such that k>B<n>P<o = k>C<o, and thus m>R<n>P<o = m>A<k>C<o. By Theorem 7, >>3297, {C, B, no} is collinear. By Col L, {A, C, mo} is collinear as well. Thus, by the UPC, There exists a point Q such that m>Q<o = m>A<k>C<o = m>R<n>P<o, as required.#

Theorem 5: The Three-Fix Theorem implies Pappus's Theorem

Proof: Starting with the premise of Pappus's Theorem, let A” = [BC'][B'C], B” = [CA'][C'A], and C” = [AB'][A'B]. Our goal is to show that {A”, B”, C”} must be collinear.

Since none of A, B, C, A', B', or C' are equal to mn, the construction used in Theorem 3, >>3296 can be carried out three different ways(see pics 1-3)

|a| = m>A'<a>A<n, where a = B”C”, A* = a[AA']

|b| = m>B'<b>B<n, where b = C”A”, B* = b[BB']

|c| = m>C'<c>C<n, where c = A”B”, C* = c[CC']

Also, from the fact that A, B, C, A', B', and C' are distinct from mn, we can conclude that the also do not lie on any of a, b, or c, and that none of A”, B”, C”, A*, B*, or C* lie on m or n. From the definitions of a, b, and c, we have the following collinear sets:

{ma, B”, C”}

{mb, C”, A”}

{mc, A”, B”}

If we explicitly write out the projectivity |ab*|:

m(A, B, C, ma)>A'<a(A*, C”, B”, ma)>A<n(A', B', C', mn)>B<b(C”, B*, A”, mb)>B'<m(A, B, C, mb)

we see that A, B, and C are self-corresponding, therefore by the Three-Fix Theorem |ab*| = |1|. Therefore, since ma|ab*| = mb, ma = mb. Looking at |bc*| we can also conclude ma = mb = mc. Since mc lies on m, it is distinct from A”, B”, and C”. By ColL on the above collinear sets, we have {A”, B”, C”} collinear, as required.#

Theorem 6: Pappus's Theorem implies the UPC

Proof: Let m>R<n>P<o be a projectivity to which the UPC would apply. If P = R, then the result is trivial, so we assume otherwise. From this, the set {mo, R, P} is non-trivially collinear. Let Q = [R[no]][P[mn]]. Choose some arbitrary point X on m not equal to om, mn, or m[R[on]]. Let Y = n[RX] and Z = o[PY](see pic 4).

By construction, the image of X under m>R<n>P<o is Z and X = m[RY]. Also, Y cannot equal mn, no, or n[RP]. By the construction of Q and the premise of the proof, the images of om, mn, and m[R[on]] are the same under m>R<n>P<o and m>Q<o.

For the other points on m, we first consider the two tuples of points (R, om, P) on RP and (mn, Y, no) on n. We first assumed R and P were distinct, and if om were identical to R or P, the given projectivity would not exist. Likewise, Y is distinct from mn and no by construction, and mn cannot equal no by the premise of the proof. None of the listed points can equal mn due to either their construction or the properties of the given projectivity. Therefore, Pappus's Theorem applies and {[[om][no]][YP], [P[mn]][[no]R], [RY][[mn][om]]} is collinear. However:

[[om][no]][YP] = o[PY] = Z

[P[mn]][[no]R] = Q

[RY][[mn][om]] = m[RY] = X.

Therefore, {X, Q, Z} is collinear and Z is the image of X under m>Q<o. Since X was arbitrary, the theorem follows.#

From Theorems 2 to 6, we can conclude the following:

Theorem 7: The UPC, PerT, FTP, Three-Fix Theorem, and Pappus's Theorem are equivalent.

In the next section, we look at the immediate consequences of these statements and to incorporate these results into our other axioms.

Sorry I'm a bit late with this, I got a bit distracted.

On an unrelated topic, here is a restatement of Axiom D that does not require the 'perspective' relation:

Axiom D: Given two distinct triangles with ABC and A*B*C*, if {AA*, BB*, CC*} is concurrent then {[CA][C*A*], [AB][A*B*], [BC][B*C*]} is collinear.

———————————————

As we have seen, the assumption of the UPC in addition to the plane axioms has substantial consequences for perspectivities and projectivities in general, including implying the SPT, hence Axiom D(Theorem 2, >>3397). Thus, it would be desirable to incorporate a new axiom into our scheme equivalent to the UPC. However, we have not yet shown that the UPC is consistent with our axioms, or that the UPC has other properties we would like our axioms to have. Luckily, Theorem 7, >>3506, allows us to use any of the mentioned “theorems” to prove these properties.

First of all we show the consistency of the UPC with the plane axioms. To do this, we will use the Three-Fix Theorem.

Theorem 1: For any projective plane of order 3, the Three-Fix Theorem holds.

Proof: Recall that the order of a plane is one less than the number of distinct points on each line. Say n>|A|<n is a projectivity on a line n in the given plane such that the distinct points P, Q, R, are self-corresponding. What we require is that any other distinct point on n is self corresponding as well. However, there is only one other distinct point S on n. Since |A| is a bijection and all the other points are accounted for, S is forced to be self-corresponding, as required.#

One can generate a projective plane of order 3 with the PDS {0, 1, 3, 9} mod 13.

Another example of a projective plane where the UPC holds is the Euclidean plane with the added line at infinity. We can show this by proving Pappus's theorem in this plane. From the “planes and lines” model of this projective plane, it is clear that any line in the projective plane can act as the line at infinity. From the premise of Pappus's theorem with tuples of points (A, B, C) on m and (A', B', C') on n, if we let the line [[AB'][A'B]][[BC'][B'C]] be the line at infinity, then we can get an equivalent Euclidean version of Pappus's theorem.

Theorem 2(Euclidean Pappus Theorem): Suppose that on the Euclidean plane, there are two distinct lines m and n, and two tuples of distinct points (A, B, C) and (A', B', C') such that A, B, C lie on m, A', B', and C' lie on n, and none of the points is equal to mn if it exists. Then if AB'||A'B and BC'||B'C, then CA'||C'A also holds.

Proof: Note that this is a theorem in the Euclidean plane, thus Euclidean concepts and theorems can and must be used. Let O = mn and (OX) be the directed length from O to any point on X on m or n.(see pic 1)

By the intercept theorem, we get:

(OB)/(OA) = (OA')/(OB') = k (as AB'||A'B)

(OC)/(OB) = (OB')/(OC') = l (as BC'||B'C)

Then we have (OC)/(OA) = lk and (OA')/(OC') = kl, therefore by the converse of the intercept theorem, we have CA'||C'A.

If n and m are parallel(see pic 2), mark an arbitrary direction along the lines as positive, and if two points X and Y are both on either m or n, then (XY) is the directed length from m to n. We then have:

(BA) = (A'B') = s (as AB'||A'B)

(CB) = (B'C') = t (as BC'||B'C)

We thus have (CA) = (CB) + (BA) = t + s while (A'C') = (A'B') + (B'C') = s + t, therefore (CA) = (A'C'), and thus CA'||C'A.#

Corollary 2.1: Pappus's theorem holds in the projective plane constructed by augmenting the Euclidean plane with a line at infinity.

With Corollary 2.1, we have two distinct examples of planes where the UPC holds.

If one constructs examples of quadrangles on the Euclidean plane(+ the line at infinity), it is quickly observed that in addition to the previous facts, the diagonal points themselves are never collinear. However, unlike the above statements, a proof of this fact from our axioms is not obvious. We formally state the conjecture below.

Conjecture F: The diagonal points of a quadrangle are never collinear.

An interesting thing to note is that this conjecture is self-dual.

Theorem 1: Given Conjecture F, the diagonal lines of a quadrilateral are never concurrent.

Proof: Let the lines e, f, g, and h be the lines of the given quadrilateral(see pic 1). It can be easily shown that the set {ef, fg, gh, he} is in general linear position.

The diagonal points of the resulting quadrangle are [[ef][fg]][[gh][he]] = fh, [[ef][he]][[gh][fg]] = eg, and [[ef][gh]][[fg][he]] = D, which is the meet of two of the diagonal lines of the quadrangle. The third diagonal line is [fh][eg], which is the join of two diagonal points.

If the diagonal lines of the quadrangle are concurrent, then they must all be incident to D, including [fh][eg]. Thus {fh, D, eg} is collinear, in contradiction to conjecture F, thus the diagonal lines cannot be concurrent.#

Looking at our other examples of projective planes, it turns out at least one of them respects conjecture F.

Theorem 2: Any projective plane of order 3 must satisfy Conjecture F.

Proof: Suppose that there exists a quadrangle in the plane with collinear diagonal points. Let P, Q, R, and S be the points of the quadrangle and let D = [PQ][RS], E = [PR][QS], and F = [PS][QR].(see pic 2)

Taking together the four quadrangle points, six lines, three diagonal points, and the line on which they all lie, we have seven each of points and lines. Within this set, the lines exhaust all the joins between the points, so any line outside the set can lie incident to at most one point in the set. Additionally, each point in the set is incident to exactly three lines, one less than required by the order of the plane. Therefore, there must be at least seven lines in the plane but outside the set.

However, by (6.4), >>3171, there are 13 lines altogether in the plane, and only six lines in the plane but not in the set. Thus, our supposition must be false and all quadrangles have non-collinear diagonal points.#

While this does reassure us that Conjecture F is at least consistent with our previous axioms, the proof itself relied critically on the order of plane. If we change the order of the plane, then different results can arise.

Theorem 3: In a projective plane of order 4, the diagonal points of any quadrangle are always collinear.

Proof: Suppose that there exists a quadrangle in the plane with non-collinear diagonal points. Let P, Q, R, and S be the points of the quadrilateral and let D = [PQ][RS], E = [PR][QS], and F = [PS][QR] be the diagonal points.(see pic 1)

Let l = DE, m = QR, G = l[PS], and H = ml. From this construction we conclude the following projectivity exists:

l(DGEH)>P<m(QFRH)>S<l(EGDH)

The points D, E, G, and H on l are easily proved distinct in this situation, as are the points Q, F, R, and H on m. In particular, the points G and H are distinct as otherwise {QR, PS, l} would be concurrent and F would lie on DE, in contradiction to our assumption.

The order of the plane requires one other distinct point on both l and m. Call these points X and Y respectively. These points are distinct from each other as H is the meet of m and l. The (pre-)images of these points in the above projectivity are fixed by the (pre-)images of the other points:

l(X)>P<m(Y)>S<l(X)

Since l>P<m and m>S<l are perspectivities, then the sets {X, P, Y} and {Y, S, X} are collinear. By col L, so is {X, Y, P, S}, thus X lies on the line PS and l. Since l passes through two diagonal points and PS passes though at least one point of the 4-point, l and PS are distinct lines and thus X = l[PS] = G, in contradiction to the definition of X. Thus, such a quadrangle postulated at the start cannot exist in this plane.#

Theorem 4: In a projective plane of order 4, Axiom P holds.

Proof: We start with distinct lines m and n, with distinct points A, B, and C on m and distinct points A', B', C' on n, with none of the points equal to mn. Let A* = [BC'][B'C], B* = [CA'][C'A], and C* = [AB'][A'B]. These nine points can be proven to all be distinct mn and each other and that none of A*, B*, or C* lie on either m or n.(see pic 2)

In any case, the line o = A*C* exists, and does not pass through any of A, B, C, A', B', or C'. The line o may or may not pass through mn, but in each case we wish to prove that B* lies on o.

Case 1: o does not pass through mn.

In this case the line o must still meet m and n separately. Call these meets M and N. Additionally the line CA' must meet o at some point. It cannot be at points M or N as that would imply C is on n or A' is on m. It cannot be at points A* or C* as that would imply B' = A' or B = C respectively. Since we have ruled out four distinct points on the line, there must be one distinct point Z left on o which must be the common point with CA'.

Running through the same reasoning with C'A, we see that it must also intersect o at Z, therefore B* = Z and lies on o, as required.

Case 2: o passes through mn

In this case, the set {mn, A*, C*} is collinear. Observe that the set of points {A, B, A', B'} are in general position. The diagonal points of the quadrangle are mn, C*, and W = [AA'][BB']. The point W is easily proven distinct from mn. From Theorem 3, we know {mn, C*, W} is collinear, and {mn, W, A*, C*} is as well by Col L, therefore A*[mn] = C*[mn] = W[mn] = o and {o, AA', BB'} is concurrent.

If we look at the set of points {B, C, B', C'} they are again in general position. The diagonal points of this quadrangle are mn, A*, and [BB'][CC'] = W'. By Theorem 3, {mn, A*, W'} is collinear and thus {o, BB', CC'} is concurrent. The set {AA', BB', CC'} is as well by Con L and therefore [AA'][CC'] = W = W'.

Looking at the set of points {C, A, C', A'}, they are also in general position. This time the resultant diagonal points are mn, B*, and W. By Theorem 3, B* lies on W[mn] = o, as required.#