An anonymous user using the tripcode !!aPRGKvtfXM wrote a series of posts about projective geometry from an axiomatic point of view in 8ch.net. I really liked his posts, but I knew that given enough time, I wouldn’t be able to find them again. I’m saving his writings here. Using Rmarkdown, I’ve upgraded the text for a more pleasant reading.
Intro to axioms
This is a bunch of words talking about a particular set of axioms, including the ones immediately below. The objects the axioms refer to are points and lines, between which a relation of incidence may be present. Despite the suggestive terminology, these points and lines are not necessarily points or lines in Euclidean space, and in fact the set of lines and points in a Euclidean plane do not obey these axioms per se. The only thing we know about points, lines, and incidence is that they obey at least our first set of axioms, and perhaps some other axioms to be introduced later. Otherwise, these objects and relation are are completely mysterious. However, this approach allows us to take anything as our points and lines, so long as we can specify an incidence relation so that at least out first set of axioms are satisfied.
These first axioms are:
Axiom 1: For any two distinct points, there is exactly one line incident with both points.
(Notation: If \(X\) and \(Y\) are the names of distinct points, \([X][Y]\) is the name of the common incident line, where the square brackets are dropped if the bracketed point name is a single letter.)
Axiom 2: For any two distinct lines, there is at least one point incident with both lines.
Axiom 3: There exist 4 distinct points such that no line is incident with 3 distinct points out of the 4.
These will be referred to as the plane axioms. There may be a few sections where the plane axioms do not apply. These exceptions will be noted then. Any set of lines and planes with the appropriate incidence relation following these axioms is a projective plane(plane for short). One simple consequence of these axioms is as follows.
Theorem 1: For any two distinct lines, there is at most one point incident with both lines.
Proof: Assume the statement is false. Then there are two distinct lines that have more than one distinct common incident point. Consider two such distinct points we call \(A\) and \(B\). By Axiom 1, there is a unique line which is incident to both. This contradicts our ‘construction’ where 2 distinct lines are incident to A and B. Thus, our assumption must have been false, and the theorem holds.
Corollary 1: For any two distinct lines, there is exactly one point incident with both lines.
(Notation: If \(x\) and \(y\) are the names of distinct lines, \([x][y]\) is the name of the common incident point, where the square brackets are dropped if the bracketed line name is a single letter.)
Other consequences of the axioms:
There exist 4 distinct lines such that no point is incident with 3 distinct lines out of the 4.
Proof: Given the 4 distinct points required by Axiom 3 are named \(A\), \(B\), \(C\), and \(D\), the lines \(AB\), \(BC\), \(CD\), \(DA\) satisfy the statement. Assume a point is incident to three of those lines. By Theorem 1, it must be one of the points mentioned previously, say \(A\). Since there are only two lines of the 4 explicitly incident to \(A\), one of the other lines, say \(BC\), is incident to \(A\). Thus there are three distinct points out of our original set of points incident to a single line, in violation of Axiom 3.
For any line there is a point not incident to it.
A direct consequence of Axiom 3.
For any point there is a line not incident to it.
Proof left to reader.
Every line is incident to at least 3 distinct points.
Proof: Starting again with the points \(A\), \(B\), \(C\), and \(D\) required by Axiom 3, assume that the line in question is incident to at most one of these points, say \(D\). Then the lines \(AB\), \(BC\), and \(CA\) all have distinct common points of incidence with our line. If the line is incident with two of the points, say \(A\) and \(B\), the third point is its common incident point with \(CD\).
Every point is incident to at least 3 distinct lines
Proof left to reader.
If there exists a line incident to exactly \(n+1\) distinct points, then:
All lines are incident to exactly \(n+1\) distinct points.
Proof left to later post.
All points are incident to exactly \(n+1\) distinct lines.
Proof left to reader.
There are \(n^2 + n + 1\) distinct points altogether.
Start with any point on the plane. By (6.2) there are only n+1 distinct lines incident to that point. By (6.1) each of those lines is incident to n distinct points aside from our starting point, constructing a set of n(n+1) points distinct from our staring point. Theorem 1 guarantees all the points in this new set are distinct, while Axiom 1 guarantees this set and our starting point are all the points there are. Thus there are n(n+1)+1 = n^2+n+1 distinct points.
There are \(n^2 + n + 1\) distinct lines altogether.
Proof left to reader.
More about notation
To facilitate further discussion, I’ll start introducing notation and terminology here and in future posts on this.
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.
Example 1:
For all this talk of these axioms, we still have not proven that any projective planes exist. Thus we will exhibit examples of planes.
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 figure 1, there \(p^n = k\) is the order of the resultant plane.
| \(p^n\) | \(q\) | \(\phi(q)/(3n)\) | perfect difference set |
|---|---|---|---|
| 2 | 7 | 2 | 0 1 3 |
| 2² | 21 | 2 | 0 1 4 14 16 |
| 2³ | 73 | 8 | 0 1 3 7 15 31 36 54 |
| 2⁴ | 273 | 12 | 0 1 3 7 15 31 63 90 116 127 136 181 194 204 233 238 255 |
| 3 | 13 | 4 | 0 1 3 9 |
| 3² | 91 | 12 | 0 1 3 9 27 49 56 61 77 81 |
| 5 | 31 | 10 | 0 1 3 8 12 18 |
| 7 | 57 | 12 | 0 1 3 13 32 36 43 52 |
| 11 | 133 | 36 | 0 1 3 12 20 34 38 81 88 94 104 109 |
| 13 | 183 | 40 | 0 1 3 16 23 28 42 76 82 86 119 137 154 175 |
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.
Example 2:
In Euclidean 3D space, pick out some arbitrary Euclidean point \(O\). We take as points in our projective plane the set of lines incident to \(O\) and our lines in the projective plane as the set of Euclidean planes incident to \(O\).
A point and line in the projective plane are incident if the Euclidean line lies on the Euclidean plane. The proof that this construction does generate a projective plane is left to the reader.
Something interesting occurs when we consider a Euclidean plane not incident to O. One can draw lines from O to any point in the Euclidean plane to obtain a point of the projective plane. The same is true of Euclidean lines and the resulting projective lines. Thus, we have an injective function from the Euclidean plane to the projective plane. Furthermore, incidence of a line and point in the Euclidean plane implies incidence of the resultant point and line in the projective plane.
This correspondence must not be perfect, however, as the Euclidean plane is not a projective plane. The flaw is the fact that the projective line parallel to the Euclidean plane, as well as any projective points on that line, do not have a Euclidean counterpart. This missing line allows for the existence of parallel lines in the Euclidean plane; the meet of the corresponding projective lines lies on the non-corresponding line.
This also suggests that the Euclidean plane can be converted to a projective plane by adding an extra line and the points on that line. First, note that it can be proven that parallelism is an equivalence relation on lines in the Euclidean plane (if lines are regarded as parallel to themselves). To each resultant equivalence class, assign a new unique point and say that each line in the class is incident to only the new point associated with the class. Then one can introduce a new line(the line at infinity) that is incident to all and only new points from the last step. It is then easy to show that the result is a projective plane.
The line at infinity in the ‘new line’ construction can be seen as equivalent to the non-corresponding line in the ‘planes and lines’ construction, so the overall constructions are equivalent as well. Note that in the projective plane constructed, the line at infinity is in no way special, as in the ‘planes and lines’ construction, any projective line can be the line at infinity depending on the choice of Euclidean plane.
Because of this near-correspondence between a projective plane and the Euclidean plane, one can use straight-line diagrams with points as diagrams for a projective plane, so long at the existence of a line at infinity is kept in mind. However, as more axioms are introduced, this may not be the case, as the projective plane constructed from the Euclidean plane has additional properties that may not be shared by other projective planes. Additionally, to prevent over-reliance on Euclidean intuitions, curved-line diagrams will be used as well.
One of a few last remarks is that while Example 1 produces planes of finite order, the plane produced from Example 2 is of infinite order. Also, Example 1 implies many distinct planes of different orders, Example 2 only produces only one plane up to isomorphism.
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).
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.
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).
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.