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Open problems in “Finite Geometry”: the Prime Power Conjecture

February 26, 2010

I would like to begin a series of posts on what I and others might regard as the most important problems in “Finite Geometry”. I won’t attempt to give a complete survey, but I will attempt to highlight the most significant steps towards the problem. To begin, who could go past…

The Prime Power Conjecture

A projective plane is a finite incidence structure of points and lines satisfying the following axioms:

  • every two points lie on a unique line,
  • every two lines meet in a unique point.

We may also require a “non-degeneracy” condition, such as, “there exist four points with no three lying on a line”. The simplest example of a projective plane is the Fano plane:

Notice that the Fano plane has seven points and seven lines (the circle counts as a line). Now suppose we have two distinct lines \ell_1 and \ell_2. By the axioms above, we know that \ell_1 and \ell_2 meet in a point P. Consider a point Q which is not on both lines. Then every point X on \ell_1 gives us a unique line on both X and Q; we can call this line XQ. This line must meet \ell_2 in a unique point, and this point cannot be P (exercise: why not?). So we have a bijection between the points of \ell_1 minus P, and the points of \ell_2 minus P; specifically, this map is X\mapsto XQ\cap \ell_2. So we see that every line of a projective plane contains the same number of points. We define the order of a projective plane to be one less than the number of points on any line. So we see that the Fano plane has order 2. In fact, if we take the one-dimensional subspaces of a 3-dimensional vector space over a finite field GF(q) to be the points and if we take the two-dimensional subspaces to be the lines, we have an example of a projective plane, which we call PG(2, q). So we have for every prime power q a projective plane of order q. In fact, there are four projective planes of order 9, three of which are not equivalent to PG(2,9), and these are examples of non-Desarguesian planes.

Now we are ready to state the conjecture:

Conjecture: The order of any finite projective plane is the power of a prime number.

The best result we have so far is the Bruck-Ryser-Chowla Theorem (1949-1950):

If a finite projective plane of order q exists and q is congruent to 1 or 2 (mod 4), then q must be the sum of two squares.

So there are no projective planes of orders 6 or 14. Tarry (1901) proved in his paper “Le problème des 36 officiers” that there is no finite projective plane of order 6, and a controversial computer-aided proof was given by Lam (1988) for the non-existence of a plane of order 10. There is much much more on the prime power conjecture, so we will leave you now to discover more; there are plenty of articles about and they are easy to find on the web…

Now for a survey (I can’t resist!):

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3 Comments leave one →
  1. Ania permalink
    December 1, 2013 9:24 pm

    So, I can say, that if it is order s in projektive plane, for example Fano Plane, it is s + 1 points of every line, and every line has s +1 points. I found formula, that when you have order, you can know, how many lines and points is there: |P| = s^2 + s + 1 = |L|. Do you know why? Just because it is (s+1)(s+1) (lines times vertices) and we have subtract s…but why? Can you explain this formula?

    • December 2, 2013 9:49 am

      Sure, it is very simple. Take one point P. Then there are s+1 lines on P. On each of these lines, there are s points incident with the line that are not equal to P. Since every point must be on one of these lines (since two distinct points are co-incident with a unique line), the number of points not equal to P is s(s+1). So in total we have s(s+1)+1 points. By the Principle of Duality, we have the same formula for lines.

      • Ania permalink
        December 10, 2013 10:21 pm

        Thank you very much! Good luck in your work and greetings from Poland.

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