Adam Cave
Geometry 300
Classifying Models of Incidence Geometry up to Isomorphism
This paper will focus on classifying all the possibilities of four and five point models of incidence geometry up to isomorphism. The idea for this paper arose while working through a problem concerning whether or not two models of incidence geometry each with exactly four points must be isomorphic. The answer, which we will soon find out to be no, posed more complex issues. For example, although not all of these models are isomorphic, there certainly exist some four-point models that are isomorphic. To break these models down and classify them requires proving which various four-point models exist, and which of these are isomorphic. Taking this idea to the next level deals with the same ideas in relation to five-point models. In more general terms, this idea encompasses what it means for two models to be isomorphic.
By definition, isomorphisms of models of incidence geometry are “situations” where “there exists a one-to-one correspondence… between the points of the models and a one-to one correspondence… between the lines of the models such that P lies on l if and only if P’ lies on l’ ” (Greenberg 56). In other words, for two models to be isomorphic, everything about points, lines, and incidence that is true for one must also hold true for the other. Using this, we will first prove exactly which four-point models of incidence geometry exist. Next we will show whether two random four-point models are isomorphic or not. Once accomplished, we will repeat the procedure for models with exactly five points. We will conclude with how these classifications can be applied and possibly expanded.
**Note: For the purpose of clarity in this paper, the standard notation of a line will be altered slightly. The notation of a line will include all points that line is incident with.
Example: Line AC will still denote a line incident with points A and C, but it will have the
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added meaning that line AC is incident only with points A and C. Likewise, line PQRS is incident with only points P, Q, R, and S.
Classifying Four-Point Models of Incidence Geometry
The first step is to discover and classify four-point models of incidence geometry.
We begin with M, a model of incidence geometry with exactly four points. As stated before, since M is a model, we know that all the incidence axioms must hold in M. By the law of excluded middle we know that each line in M must have exactly zero, one, two, three, or four points incident with it. Incidence axiom-1 tells us that a line with zero points or one point can not exist in a model. Similarly, there can be no line in M incident with all four points. This would result in all points being collinear, a violation of incidence axiom-3. This leaves us with the possibility of lines having either two or three points incident with them. These possibilities will be broken down into cases:
Case 1: All lines have exactly two points incident on them (no three points are collinear). This produces six lines.
A D
B C
Case 2: Exactly one line has three points incident on it (exactly three points A,
B, and C are collinear). This produces four lines.
A B C
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D
Case 3: Two distinct lines have three points incident with them. (A, B, C are collinear and B, C, D are collinear)
A B C
D
Cases 1 and 2 hold, but Case 3 does not. The picture for Case 3 should immediately look wrong. To prove that case 3 does not hold, we take the lines ABC and
BCD. By the definition of a model , we know that all the propositions for incidence geometry must hold. Proposition 2.1 tells us that these lines have a unique point in common. By the notation previously defined, ABC and BCD share two points in common: B and C, which contradicts Proposition 2.1. Incidence axiom-one tells us that a unique line exists between these two points. Thus ABC and BCD are the same line.
Using the same proof, it can be argued that any additional lines with three points will overlap with some other line on two points. This would simply be a re-labeling of an existing line.
Since all the other possibilities for models have been proven impossible, we know that Cases 1 and 2 are the only four-point models of incidence geometry. Now, given any unknown four-point model, we can distinguish which type it is if we can discover whether it has any three collinear points or not. Now we are left to prove which four-point models are isomorphic.
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4-Point Model Isomorphisms
Two Models from Case 1:
**Note: These first proofs will be in some detail but once the basic technique has been laid, subsequent proofs of isomorphism will be abbreviated.**
Let M and N be four-point models as described in Case 1 from above. Thus both
M and N have no three points collinear. M contains points A, B, C, and D; N contains points P, Q, R, and S. Incidence axiom-one tells us:
M has six unique lines: AB, AC, AD, BC, BD, CD
N has six unique lines: PQ, PR, PS, QR, QS, RS
In order to prove isomorphism, we must exhibit a one-to-one correspondence between points and lines. To do this we must prove that there are no more lines possible in either
M or N. We begin with an RAA hypothesis where we suppose line l in M distinct from the six lines already defined by IA-1. IA-2 tells us that x and y are distinct points on l .
We know, by definition, that this model has only four points, so x and y must be some two of A, B, C, D. But every combination of these letters was taken care of in the six lines from above. Therefore line l must be equal to one of the six lines from above.
This is a contradiction of our RAA hypothesis which said that l was distinct. This proof allows us to say that there are exactly six lines in M. Using similar arguments, we know that N similarly has exactly six lines. Now we are ready to set up the isomorphism between these two models.
Point Correspondence
A
B
C
D
P
Q
R
S
Line Correspondence
AB
AC
AD
BC
PQ
PR
PS
QR
Incidence normal incidence between lines and points
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BD
CD
QS
RS
This allows us to say that for any two four-point models of incidence geometry where each model has no three points are collinear, the two models must be isomorphic.
Two Models from Case 2:
Let M and N be four-point models of incidence geometry as described in Case 2 from above. M and N each have one line incident with three points. M contains A, B,
C, D; N contains P, Q, R, S. In M, we relabel so that A, B, C are collinear and in N, we relabel so that P, Q, R are collinear. Therefore we know:
M has distinct lines: ABC, AD, BD, CD
N has distinct lines: PQR, PS, QS, RS
Using a similar proof from case A, we know that there exists no line l in either M or N distinct from these lines. So we know that each model has exactly four lines. This again allows us to set up an isomorphism.
Point Correspondence
A P
B
C
Q
R
Line Correspondence
ABC PQR
AD
BD
PS
QS
Incidence the normal incidence between lines and points
D S CD RS
This relationship allows us to say that for any two models of four-point geometry where three points are collinear, the two models must be isomorphic.
A Model from Case 1 and a Model from Case 2:
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Let M and N be four-point models of incidence geometry. M, like those models in case 1, has no three collinear points. N, like those models in case 2, has three collinear points: P, Q, and R. From Case A, we know that M has exactly 6 lines; from Case B we know that N has exactly 4 lines. Setting up another isomorphism table allows us to see the discrepancies between these two four-point models.
Point Correspondence Line Correspondence Incidence
D
A
B
C
S
P
Q
R
AB
AC
AD
PQR
PS
QS
BC RS
BD
CD normal incidence
As shown by the graph, although M and N have a one-to-one correspondence between points, the one-to-one correspondence between lines and incidence is not preserved. For example, A and B correspond to P and Q respectively. Since line AB exists in M, for isomorphism to hold, there must be a line PQ in N. There is a line PQR that has points P and Q, but that line is also incident with point R. The only way for isomorphism to hold is if a line ABC exists in M. From the chart it clearly does not. This difference in point collinearity is enough to assure that these models are not isomorphic. Thus we know that any two four-point models of incidence geometry with differences in point collinearity are not isomorphic.
Classifying Five-point Models of Incidence Geometry:
We now move on to finding every unique five-point model of incidence geometry.
We begin with a model of incidence geometry, M , that contains five points: A, B, C, D,
E. As in the four-point models, we know that lines in M can have a minimum of two points incident on them. Similarly, we know that all five points can not be collinear because then IA-3 would not hold. So we know that lines must either have two, three, or
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four points incident on them. In order to deduce all possible five-point models, it is best to look at individual cases and narrow them down using the law of excluded middle in regards to collinear points.
Case 1 : All lines have exactly two points. (No three points are collinear.)
A D
E
B C
This model satisfies all the incidence axioms and produces 10 distinct lines.
Case 2 : Exactly one line has three points incident on it. All other lines have exactly two with
A
points incident on them. (In this graph, ABC has been chosen as the line
three points)
B C
D E
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A D
B
E
C
This model satisfies all the necessary requirements using only six lines.
Case 4 : At least three distinct lines with three points incident on each.
A
This model likewise satisfies all the incidence axioms but produces eight distinct lines.
Case 3: Exactly two distinct lines have exactly three points incident on them. In this case; AEC and BED.
B C
D
E
This case fails because in five-point geometry, three distinct lines with three points incident on each can not exist. If we suppose they do exist, we could relabel to call them
ABC, BCD, and CDE. By the definition of notation set up earlier in the paper, we know that ABC contains A, B, C; BCD contains B, C, D; and CDE contains C, D, E. If ABC and BCD are distinct lines, then proposition 2.1, which must hold by the definition of a
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model, says that they meet at a distinct point X. From notation, however, we know that B and C are both incident on ABC and BCD. Thus, X=B=C, contradicting the given statement that B and C are distinct. Therefore, we can say that ABC and BCD are not distinct lines. We could go on to prove that these three lines are, in fact, the same line, but what we have now is enough to tell us that in an arbitrary five-point model, there can be no three distinct lines with three points incident on each.
Case 5 : Four points are collinear. (Exactly one line with four points incident with it;
ABCD)
A B
E
C D
This case brings up a possibility unavailable in four-point geometry: four collinear points.
In this five-point model, the axioms are satisfied and five lines are created.
Case 6: At least two distinct lines with four points incident on each. (ABCD, BCDE)
A B C D
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E
It should be apparent from previous examples that each of these lines contains the points
B, C, and D. Thus these lines cannot be distinct. This tells us that in five-point models no two distinct lines can exist where four points are incident on each.
Case 4 tells us that no more than two lines with three points can exist and Case 6 tells us that no more than one line with four points can exist. Combining these, we can prove that no other combinations (i.e. a model containing one line with four points and one line with three points) occur in five-point geometry. So, with Cases 1, 2, 3, and 5, we have found all the possible models of five-point geometry.
Isomorphism in Five-Point Models
Let us refer back to our arbitrary models of incidence geometry: M and N . For our purposes, let M have the points A, B, C, D, E; let N have the points P, Q, R, S, T.
Using the previous proofs from the four-point models as examples, we can assume that the lines formed from IA-1 and the definition of the individual model are all the lines in that model. Now it is relatively easy for us to set up one-to-one correspondences between models from similar cases.
Two Models from Case 1:
Let M and N both be five-point models of incidence geometry with no three points collinear. Each model will have the same number of points (5), the same number of lines
(10), and preserve incidence.
Point Correspondence
A
B
C
P
Q
R
Line Correspondence
AB
AC
AD
PQ
PR
PS
Incidence normal incidence between points and lines
10

D
E
S
T
Two Models from Case 2:
AE
BC
BD
BE
CD
CE
DE
PT
QR
QS
QT
RS
RT
ST
Let M and N both be five-point models where exactly one line has three points incident on it. Since each model has eight lines and preserves incidence, these models are isomorphic.
Point Corr.
E
A
B
C
D
T
P
Q
R
S
Two Models from Case 3:
Line Corr.
BE
CD
CE
DE
ABC PQR
AD PS
AE
BD
PT
QS
QT
RS
RT
ST
Incidence normal incidence between points and lines
Let M and N both be five-point models where exactly two lines each have three points incident on them. These models each contain six lines and are isomorphic as well.
Point Corr.
A
B
C
D
E
P
Q
R
S
T
Line Corr.
AEC
BED
AB
BC
CD
AD
PTR
QTS
PQ
QR
RS
PS
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Incidence normal incidence

Models from Case 5:
Let M and N be five-point models of incidence geometry both containing exactly one line with four points incident on it. From before, we know that each model will produce five lines. Since incidence remains the same as well, these models are isomorphic.
Point Corr.
A
B
C
D
E
P
Q
R
S
T
Line Corr.
ABCD PQRS
AE PT
BE
CE
DE
QT
RT
ST
Incidence normal incidence
With this information in hand and the results from the four-point models, we can now deduce that models from different cases are not isomorphic. Despite the fact that each of the models has the same number of points, lines are not preserved in a one-to-one correspondence so incidence is not preserved. Different numbers of lines, therefore, preclude the one-to-one correspondence.
Conclusions:
Now that we have classified four and five-point models of incidence geometry, it would be possible for us to prove whether two arbitrary n-point models are isomorphic or not. If we have a known 16-point model of incidence geometry, M, and want to find out if it is isomorphic to an unknown model, N, a series of questions allows us to quickly
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answer. First, we need to know if N has 16 points. If so, we then need to known if N has the same number of lines as M. If M and N do indeed have the same number of points and lines, the final step is to determine whether the incidence of points and lines in
N matches with the incidence of points and lines in M. For example, if M has exactly three lines with four points incident on each of them and N has exactly four lines with four points incident on each of them, we know that M and N are not isomorphic. This can be a very powerful tool because determining that two models are isomorphic gives you the power to assume that the models are “essentially the same” (Greenberg 57). The basic tools laid out in this paper for finding isomorphisms in four and five-point models are valuable and give us the power to find similarities that would otherwise be left hidden.
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Works Cited
Greenberg, Marvin J., ed. Euclidean and Non-Euclidean Geometries:
Development and History . 3 rd ed. New York: W. H. Freeman and Company, 1993.
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