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I was reading about the Fano plane, the smallest possible projective plane. After playing around with it, it seems that any projective plane of 7 points will be isomorphic to the Fano plane.

However, I've always been troubled with showing isomorphisms between sets of points and lines, because it seems like any function between two sets is really just an assignment of points to other points that preserves lines, but there is no nice fixed function that maps elements to others based on some hard and fast rule, like say $x\mapsto x^2$ or $(x,y)\mapsto x-y$, etc.

Is there an elegant way to show that any projective plane of 7 points is necessarily isomorphic to the Fano plane? I could only think of exhausting all permutations of points and lines and saying "Look, if these points $A$, $B$, $C$ are on a line $l$ here, then $f(A), f(B), f(C)$ are on a line $f(l)$ here." But this seems like it's brute forcing the matter, and not very efficient at all, considering that there are many ways to connect the points. What is the best way to go about something like this? Thanks.

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Perhaps I'm missing something, but if you are defining projective plane as a quotient of $\mathbb{F}^3 -(0,0,0)$ by $\mathbb{F}^\times$. Then only finite fields can give finite projective planes. The order of all finite fields are known. So isn't it just a counting game as this point? –  solbap Feb 8 '11 at 17:53
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@solbap: there is a combinatorial definition of what a projective plane is, which you will find in Wikipedia. In particular, not all such projective planes come from fields in the way you mention. –  Mariano Suárez-Alvarez Feb 8 '11 at 18:10
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2 Answers

up vote 3 down vote accepted

It seems you can use the enormous symmetry of the Fano plane for this. You can pick a line ABC and a point off the line D and pick f(A) and f(B) arbitrarily, f(C) is the third point on the line AB, and pick f(D) arbitrarily different from the first three. Then if the lines through D are ADE, BDF, and CDG you can define f(E), f(F) and f(G) in the obvious way. Now you just need to prove that f(CEF), f(AGF), and f(BEG) are lines. But there has to be a meet between the lines f(CE) and f(AG) and it can't be f(B) or f(D), so must be f(F).

Added: support for this comes from that fact that any isomorphism should be able to be composed with all the automorphisms of the destination plane to make another. This isomorphism has 7*6*4 possibilities as the first, second, and fourth points can be chosen at random and 7*6*4=168 is the number of automorphisms of the Fano plane.

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Thanks for your answer Ross, do you mind explaining why the intersection of $f(CE)$ and $f(AG)$ can't be $f(B)$ or $f(D)$? Also, how does that prove $f(BEG)$ is a line, I only see it for $f(CEF)$ and $f(AGF)$? –  Hobbie Feb 10 '11 at 9:55
    
f(CE) can't be f(B) because ABC is a line, so BCE can't be. Similarly CDG is a line, so F(CDG) is, so f(ADG) can't be. As all pairs of lines have to meet in a point, CE and AG must meet in one. We just ruled out f(B) and f(D), so it is f(F). So f(CEF) and f(AGF) are lines. We now have six lines and need seven. All the points are on three lines except f(B), f(E), and f(G), so they have to be on one. –  Ross Millikan Feb 10 '11 at 13:33
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Let $P$ be a projective plane on $7$ points. Fix an ordered $4$-tuple of distinct points $(p_1,p_2,p_3,p_4)$ in $P$ such that no line is incident with more than two of them. If $\pi=\{\{u,v\},\{w,t\}\}$ is a partition of $\{1,2,3,4\}$ in two parts of size $2$, let $q_\pi$ be the point of intersection of the line joining $p_u$ and $p_v$ with the line joining $p_w$ and $p_t$.

This provides a labeling $p_1$, $p_2$, $p_3$, $p_4$, $q_{\{\{1,2\},\{3,4\}\}}$, $q_{\{\{1,3\},\{2,4\}\}}$, $q_{\{\{1,4\},\{2,3\}\}}$ of the $7$ points.

Now show from the labeling alone you can reconstruct the lines.

This proves uniqueness.

NB: Notice that there are the $4$-element set $\{p_1,p_2,p_3,p_4\}$ is determined by its complement, which is a line. It follows that there are $7$ possible sets, one for each line, and each such set can be turned into an ordered $4$-tuple in $4!$ ways. The above reasoning shows then that the automorphism group of $P$ has order $4!\cdot 7=168$.

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+1 Very elegant! –  Ben Blum-Smith Nov 12 '11 at 14:16
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