Projective Geometry Part 1

not an argument.

Looks like we're done here

>saying "but that's ok because it is an axiom" doesn't make it less false.
TIL all of modern math is false

>Wrong. Give the point which is incident to both the line x = 5 and the line x = 6 (in r^2)
The incidence of two parallel lines is at infinity. As an example, if you were to look down an infinite railroad track, the rails would appear to cross at the vanishing point.

The simplest projective geometry consists of just seven points, say {a,b,c,d,e,f,g} (Known as the Fano Plane)

Each "line" can then be defined as a set of three points:
{a,b,c}
{a,d,e}
{a,f,g}
{b,d,f}
{b,e,g}
{c,d,g}
{c,e,f}

Any two points can be seen to lie on exactly one line, and any two lines intersect at exactly one point.

"infinity" isn't a number, dumbass

not an argument :^)

Actually to extend R^2 to a projective geometry, you have to add a whole line "at infinity" AND a point "at infinity".

Each line in the extended space then includes a point on the line at infinity with value equal to it's gradient, so parallel lines then intersect at that point.
Vertical lines include the point at infinity (as does the line at infinity)

yes it is, you're just too stupid to realise it. I asked for a point in R^2 (that is incident to bot x=5 and x= 6) and you did not give one because infinity is not a number.

There doesn't exist a point in R^2 where they intersect.

What you're too stupid to realise is that this doesn't contradict anything stated in this thread.

>claiming that infinity is an element of R

You're done, kid.

>doesn't contradict anything stated in this thread, except for
fixed ;^)

Again, R^2 is not a projective geometry. Anyone asserting otherwise is wrong.

>yes it is, you're just too stupid to realise it. I asked for a point in R^2 (that is incident to bot x=5 and x= 6) and you did not give one because infinity is not a number.
Just because I replied doesn't mean I attempted to answer your question. Your question is nonsensical in projective geometry. There is no metric in projective geometry.

>I replied to you when you were asking a question b-b-but I w-w-wasn't trying to answer your question
>back peddling this hard

ok champ.
So you admit that there is no point where the lines x= 5 and x= 6 intersect and thus the axiom that any two lines have at least one point of incidence is wrong.

check and mate, retard

I'll respond to the comments in general and specifically before posting the notation I intend to use from now on.

-'Points' and 'lines' are primitive terms here, which means that while these terms have 'outside' meaning, in this context right here, they have no meaning outside of the axioms we have them obey.

-The axioms posted so far are not complete in that they do not exclude uninteresting cases, such as the case where there are no lines or points at all. In addition, even now these aren't the most general axioms for projective geometry.

Before we explain our notation, we should emphasize that while the terminology of 'points' and 'lines' is suggestive of Euclidean geometry, they can be anything that satisfies the the axioms that we are using. To reinforce this idea, when 'points' and 'lines' are given single letter names, we will go against common convention and give 'points' lowercase names and 'lines' uppercase names.

The full naming system is thus defined as follows:

A valid point name may be:

- A single lowercase letter with optional punctuation (not [] or any other reserved characters) and subscripts.
Examples: c, a', g”, r_1, e*

- A string of the form [Λ][Δ], where Λ and Δ are names of distinct lines which are incident at the named point. Square brackets are omitted for line names which are single letters.

A valid line name may be:

- A single uppercase letter with optional punctuation (not [] or any other reserved characters) and subscripts.
Examples: F, L', G”, B_2, A**

- A string of the form [λ][δ], where λ and δ are names of distinct points which are incident to the named line. Square brackets are omitted for point names which are single letters.

Examples of point names: A*B, C_2[de], [fg][hi], O[p[QR]], [a[[wx][yz]]][mn], [a[KL]][b[MN]]

Examples of line names: c”d, c_3[PQ], [AB][A*B*], [Q[ef]]x, [PQ][R[[EF][GH]]], [A'[bc]][A[b'c']]

Thanks to Theorems 1.2 and 1.3, this system not only names points and lines uniquely, but also describes their construction. If Axiom 2 or 3 do not apply, however, names can be generated that do not name any object. In this case, other axioms must be used to guarantee the existence of named objects.

When a point and a line are incident, it can be said that the point lies on the line, the line passes through the point, or other similar terms, to simplify discussion. To that end, we also define some terms.

Given a set of lines, a [b]meet[/b] of that set is a point which lies on all of the lines of the set.

Given a set of points, a [b]join[/b] of that set is a line which passes through all of the points of the set.

A set of lines is [b]concurrent[/b] if there exists a meet of that set.

A set of points is [b]collinear[/b] if there exists a join of that set.

A [b]configuration[/b] is a finite set of points and lines which obeys Axiom 1 in itself.

A configuration where every line passes though at least three distinct points of the configuration and every point lies on at least three distinct lines of the configuration is [b]confined[/b]. A configuration which is not confined is [b]open[/b].

We may use figures of configurations to illustrate proofs by drawing 'points' as points and 'lines' as lines. Since typical screens are Euclidean, such figures may not represent the intended geometry accurately. Even if they do, the diagram may suggest arguments and theorems not supported by the axioms used. To guard against this, curves may often represent 'lines' and apparent intersections ignored unless explicitly marked.

Three lemmas that will be useful later will also be stated here; proofs are left to the reader.

Lemma 1.4 (Collinearity Lemma): Given two collinear sets of points whose intersection has at least size two, their union and any subset thereof is also collinear.

Lemma 1.5 (Concurrency Lemma): Given two concurrent sets of lines whose intersection has at least size two, their union and any subset thereof is also concurrent.

Lemma 1.6 (Cross Lemma): Given distinct lines L, M and distinct points p, q, {p, LM, q} collinear iff {L, pq, M} concurrent.

The next post will show some deficiencies with our axiom system, and how to fix them.

While these axioms can be shown to have many different models, some of these are uninteresting mathematically. Firstly, since our system contains no unconditional existence axiom, an empty set of points and lines fulfills our axioms. The way eliminate such uninteresting models, such as the one posted, is to add more axioms to our system.

Axiom 4: Every line is incident to at least three distinct points.

Axiom 5: For any line, there exists a point not incident to it.

Axiom 6: There exists a line.

These axioms aren't our only options. If we only wanted to add one axiom, we use do the following:

Axiom 4+: There exist four points such that no line is incident with any three.

Another set of axioms can be more useful in some circumstances.

Axiom 4*: Given any two points, there exists a line not incident to either point.

Axiom 5*: Given any two lines, there exists a point not incident to either line.

Axiom 6*: There exist two distinct points.

All of these axiom sets can be shown to be equivalent, as these theorems will show:

Theorem 1.4a: Given axioms 1-3, Axiom 4-6 imply Axioms 4*-6*

Proof: Given two distinct points a and b, Theorem 1.2 guarantees the existence of a unique line ab. Axiom 4 guarantees the existence of a distinct point c on the line ab, while Axiom 5 guarantees some point d not on line ab(see i.imgur.com/p9ZjX7I.png). The line cd is the line required by Axiom 4*.

Given two distinct lines A and B, Theorem 1.1 and Axiom 4 implies that there exist points a lying on A but not on B and b lying on B but not on A. The line ab, by Axiom 4, passes through a third distinct point c, as shown at i.imgur.com/hqzkaCU.png. Point c fulfills the requirement for Axiom 5*.

Axiom 6 asserts that some line L exists. Axiom 4 implies that distinct points p and q lie on L. These two points are what is needed for Axiom 6*. #

Post 5

Theorem 1.4b: Given Axioms 1-3, Axioms 4*-6* imply Axiom 4+.

Proof: We start with the two given points p and q granted by Axiom 6*. In addition to line pq, we also get a line L not passing through p or q from Axiom 4*. Axiom 5* gets us a point r not on either L or pq, as shown in the posted pic.

The points p, r, L[pq], and L[rq] are a set of points that satisfy Axiom 4+, as if three of these points were collinear it would imply that p or r lie on L either directly or by cross lemma, which cannot be true by construction. #

Theorem 1.4c: Given Axioms 1-3, Axiom 4+ implies Axioms 4-6.

Proof: Given some line L, first assume it passes through at most one of 4 points given by Axiom 4+. Then take three of the points not on L. Call these points a, b, and c. Since they are not collinear, the lines ab, bc, and ca are distinct. By Axiom 3, each of these lines must meet L(i.imgur.com/d7bCKrd.png). These meets must be distinct, as otherwise some of the points a, b, c would lie on L. Thus, these meets are three distinct points lying on L, as required by Axiom 4.

Now, consider the case where L passes through two of the four given points. Call these points p and q, and call the other points r and s. The line rs must meet L at L[rs](i.imgur.com/WOfG0s0.png). The point L[rs] must be distinct from both p and q, as otherwise three of the given four points are collinear. Thus, the points p, q, and L[rs] are three distinct points on L, again as required by Axiom 4.

Axiom 4+ implies Axiom 5 as no line can pass through all four points by construction, while it implies Axiom 6 as you can simply take the join of two of the four given points.#

These sub-theorems culminate in the following theorem:

Theorem 1.4: Given Axioms 1-3, Axioms 4-6, Axiom 4+, and Axioms 4*-6* are all equivalent.

We say that any system of lines and points which satisfies Axioms 1-6 is a '''projective plane'''.

We define the following:

A '''morphism''' between two projective planes is function that takes the points of first plane to the points of the second and the lines of the first plane to the lines of the second such that if a line and a point are incident in the first plane, then their images in the second plane are incident as well.

A similar definition can be made for configurations as well. Once the notion of morphism between such objects defined, other terms naturally arise (isomorphic systems, automorphisms, etc.). We generally treat isomorphic (or nearly isomorphic, in the case of configurations) systems as the same.

We still need to ensure that there are models for these new axioms. The one defined below is of particular interest:

Let the point set, the line set, and the incidence relation be defined as follows:

Pr = Point set: The set of all Euclidean lines through the origin in R^3.

Lr = Line set: The set of all Euclidean planes through the origin in R^3.

Ir = Incidence: A line and point are incident if the Euclidean plane contains the Euclidean line.

Theorem 1.5a: {Pr, Lr, Ir} is a projective plane.

Proof: First we check if Axioms 1-3 hold.

Axiom 1: In R^3, distinct planes have at most one common contained line. If Axiom 1 were violated, two planes would exist that have two common lines between then, in violation of this fact.

Axiom 2: Take two non-zero vectors such that each given line contains the tip of one vector. Their cross-product is guaranteed to be non-zero and orthogonal to both vectors. The equation that defines orthogonality to that cross product is a single linear equation with three unknowns, so it defines a plane which clearly contains the two given lines.

Axiom 3: In fact, either two distinct planes contain a common line, or they share no common points at all. Since the two given plane each contain the origin, they must have a common line.

To finish, the x, y, and z axes, along with the line x=y=z satisfy Axiom 4+. #

If you could teach all that to a random stranger privately that would do their best to absorb it all. Would you? Hmm.

>The simplest projective geometry consists of just seven points, say {a,b,c,d,e,f,g} (Known as the Fano Plane)

Pretty sure the simplest one is the empty set.

I went over this.

Well, desu. you're technically correct, but only in the same sense that the empty set = R^-1 is a Euclidean space.

Given the projective plane constructed above, consider a plane not passing through the origin. For those 'points' and 'lines' which do intersect the off-origin plane make intersections of points and lines on that plane.

The only 'line' which does not intersect is the plane parallel to the off-origin plane. Furthermore, parallel lines on the off-origin plane correspond to 'lines' which are concurrent to the parallel plane, exactly analogous to the ideal line and points discussed in the last thread. In this way the idea of a Euclidean plane + a line at 'infinity' is made rigorous.

Note in particular, with this construction, it is abundantly clear that the ideal 'line' is exactly like any of the other 'lines' of the projective plane. Any 'line' can be made into an ideal 'line' with the right off-origin plane.

The most important consequence of this is that a single theorem of projective planes can imply several distinct Euclidean plane theorems, depending on the choice of the line at infinity. Going the other way, if you want to see if a condition applies to this particular projective plane, with a judicious choice of ideal line, you can translate the condition into an equivalent Euclidean one, then bring the techniques of Euclidean Geometry to bear.

This consequence is important enough to give this plane a name. We call the plane isomorphic to the Euclidean plane with an added line at 'infinity' the [b]real projective plane[/b].

Another model of projective plane axioms involves the use of perfect difference sets(PDS) modulo n. These are subsets of Z/nZ such that for any non-zero residue, there exist a unique ordered pair of residues in the PDS such that the difference of the pair is equal to the given residue. Note that to satisfy Axioms 4-6, n must be greater than three.

The first formula line in the posted pic says that [math]R_{diff}[/math] is our PDS mod n, and the next two show how to obtain a the point set [math]P_{diff}[/math], line set [math]L_{diff}[/math], and incidence relation [math]I_{diff}[/math], from [math]R_{diff}[/math].

Theorem 1.5b: The system [math]\{P_{diff},\, L_{diff},\, I_{diff}[\math] is a projective plane.

Proof: The proof that the system complies with Axioms 1-3 follows very easily from the definition of a PDS. We now note the following facts:

- The set of first residues of points which are incident to a particular line always form a PDS mod n. In particular, it is a shifted version of the generating PDS.
- No PDS can contain the same non-zero difference between two distinct pairs of residues. In particular, no PDS can contain any arithmetic progression of residues.
- [math]n = k^{2}+k+1[/math] where k is is one less than the size of the PDS. Therefore, n is at least 7.

Now, consider the set of points {(0,0), (1,0), (2,0)}. These three points are non-collinear for the reasons stated above. If the point (3,0) is collinear with {(0,0), (2,0)}, then let p = (4,0). If the point (3,0) is collinear with {(0,0), (1,0)}, let p = (5,0). Otherwise, let p = (3,0). Thus, it can be proven with the facts above that the set {(0,0), (1,0), (2,0), p} satisfies Axiom 4+.#

Let's try to repost that
_____________________

Another model of projective plane axioms involves the use of perfect difference sets(PDS) modulo n. These are subsets of Z/nZ such that for any non-zero residue, there exist a unique ordered pair of residues in the PDS such that the difference of the pair is equal to the given residue. Note that to satisfy Axioms 4-6, n must be greater than three.

The first formula line in the posted pic says that [math]R_{diff}[/math] is our PDS mod n, and the next two show how to obtain a the point set [math]P_{diff}[/math], line set [math]L_{diff}[/math], and incidence relation [math]I_{diff}[/math], from [math]R_{diff}[/math].

Theorem 1.5b: The system [math]\{P_{diff},\, L_{diff},\, I_{diff}[/math] is a projective plane.

Proof: The proof that the system complies with Axioms 1-3 follows very easily from the definition of a PDS. We now note the following facts:

- The set of first residues of points which are incident to a particular line always form a PDS mod n. In particular, it is a shifted version of the generating PDS.
- No PDS can contain the same non-zero difference between two distinct pairs of residues. In particular, no PDS can contain any arithmetic progression of residues.
- [math]n = k^{2}+k+1[/math] where k is is one less than the size of the PDS. Therefore, n is at least 7.

Now, consider the set of points {(0,0), (1,0), (2,0)}. These three points are non-collinear for the reasons stated above. If the point (3,0) is collinear with {(0,0), (2,0)}, then let p = (4,0). If the point (3,0) is collinear with {(0,0), (1,0)}, let p = (5,0). Otherwise, let p = (3,0). Thus, it can be proven with the facts above that the set {(0,0), (1,0), (2,0), p} satisfies Axiom 4+.#

P.S.:
More detail in the proof:

Case 1: {(0,0), (1,0), (3,0)} is collinear
Then so are {(1,0), (2,0), (4,0)} and {(-1,0), (0,0), (2,0)}
Thus (5, 0) cannot lie on any of their joins on pain of non-unique differences. (5 - 3 = 3 - 1, 2 - 5 = -1 - 2, 2-1 = 5-4).

Case 2: {(0,0), (2,0), (3,0)} is collinear
Then so are {(-1,0), (1,0), (2,0)} and {(-2,0), (0,0), (1,0)}
Thus (4, 0) cannot lie on any of their joins on pain of non-unique differences. (4 - 2 = 2 - 0, 1 - 4 = -2 - 1, -1 - 1 = 2 - 4).

Examples of PDSs are {0, 1, 4, 14, 16} mod 21 and {0, 1, 3, 8, 12, 18} mod 31. It has been shown that for any prime power k, there exists a PDS mod [math]k^{2}+k+1[/math], where each point lies on k+1 lines. Thus we have an infinite number of non-isomorphic projective planes, though each plane has only finitely many points and lines.

We finish off this series of posts by looking at a feature that our full axiom set has. Before we do that, we need to prove a certain theorem.

Theorem 1.6: There exist two distinct lines.

Proof: We are given two distinct points p and q from Axiom 6*. In addition to line pq, we also get a line L not incident to either p or q from Axiom 4*. As pq passes though both points, these lines must be distinct. #

Now consider the pairs of propositions below:

Axiom 1 – Theorem 1.1
Axiom 2 – Axiom 3
Axiom 4* - Axiom 5*
Axiom 6* - Theorem 1.6

Note that the statements within each pair are the same but with the words 'point' and 'line' swapped, plus some insignificant wording changes. Also, each of these propositions is true for any projective plane, and these pairs encompass the full set of axioms so far. This has some important consequences:

- Any theorem proved with the plane axioms in any of their equivalent forms has a dual theorem obtained by swapping the words 'point' and 'line' and similar term pairs (join-meet, collinear-concurrent, etc.). While the dual theorem may just be equivalent to the original, it is always provable if the original theorem is.

-For any projective plane, the [math]\textbf{dual plane}[/math] is defined as follows.
- The points of the dual plane are precisely the lines of the staring plane.
- The lines of the dual plane are precisely the points of the staring plane.
- Incidence in the dual plane is inherited from the starting plane.
The property of the axioms discussed above guarantee that the dual plane is a projective plane in its own right. They do not imply that the dual plane is isomorphic to the original.

This property of the plane axioms is called [math]\textbf{duality}[/math]. It often allows us to prove two distinct theorems with one proof. If we wish to extend the plane axioms, we need to take steps if we wish duality to still hold.

The next thread will introduce the major 'tool' we use to examine projective planes.

>The next thread will introduce the major 'tool' we use to examine projective planes.
To lead into this, you can try to prove the following projective plane theorems.

-The set of all points on a given line has the same cardinality for any two lines.

-The set of all points on a given line has the same cardinality as the set of all lines passing through a given point for any point and line.