What is the oriented Fano plane? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T17:24:32Z http://mathoverflow.net/feeds/question/20567 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20567/what-is-the-oriented-fano-plane What is the oriented Fano plane? Mariano Suárez-Alvarez 2010-04-06T23:52:24Z 2010-04-09T09:16:30Z <p>One way to remember the multiplication table of the octonions is to use the following diagram (which I got from John Baez's <a href="http://math.ucr.edu/home/baez/octonions/node4.html" rel="nofollow">online paper</a>): if $(e_i,e_j,e_k)$ is one of the lines listed according to the cyclic order indicated in the diagram, then $e_ie_j=e_k$ and $e_je_i=-e_k$ in $\mathbb O$.</p> <p><img src="http://img191.imageshack.us/img191/4401/fano.jpg" alt="alt text"></p> <p>If we forget the cyclic orientation of the lines, this is of course a well-known depiction of the Fano plane $P^2(\mathbb F_2)$, which is an example of many different structures: it is a Steiner triple system, a quasigroup, &amp;c.</p> <p>What kind of object is this <em>oriented</em> Fano plane?</p> <p><strong>NB1:</strong> Naive googling informs of the concept of <em>Mendelsohn triple systems</em> and of <em>transitive triple systems</em>, both of which are enrichments of the notion of Steiner triple systems with orderings on the blocks. The oriented Fano plane above is not an example of these concepts, though.</p> <p><strong>NB2:</strong> One way to reconstruct the orientation is as follows: it is (up to projective linear automorphisms) the unique way to cyclically orient the lines in the plane in such a way that for each point $x$, the set of three points which follow $x$ in the three lines that go through it is itself a line. In fact, it is the only Steiner triple system which can be oriented with this property.</p> http://mathoverflow.net/questions/20567/what-is-the-oriented-fano-plane/20733#20733 Answer by Bugs Bunny for What is the oriented Fano plane? Bugs Bunny 2010-04-08T14:33:10Z 2010-04-08T14:33:10Z <p>No clue how to answer your question but one way to choose an orientation is to choose a basis of the vector space ${F_2}^3$. By basis I mean a totally ordered basis. For instance, the orientation on the picture corresponds to the basis $e_1$, $e_2$, $e_3$ where I take the liberty to identify a line with its non-zero element. If you think of the vector space with a basis as "the standard vector space" then you can think of the oriented Fano plane as the standard projective space.</p> http://mathoverflow.net/questions/20567/what-is-the-oriented-fano-plane/20766#20766 Answer by S. Carnahan for What is the oriented Fano plane? S. Carnahan 2010-04-08T17:53:25Z 2010-04-08T23:42:09Z <p>Here is one answer: It is an oriented line over <code>$\mathbb{F}_7$</code>.</p> <p>An affine line over <code>$\mathbb{F}_7$</code> is a set of 7 points with a simply transitive action of $\mathbb{Z}/7\mathbb{Z}$, but no distinguished origin. Here, we don't have a distinguished origin and we also don't remember the precise translation action, but we have a distinguished notion of addition by a square (think of what this would mean for real numbers). In other words, it is a set with seven elements, equipped with an unordered triple of simply transitive actions of $\mathbb{Z}/7\mathbb{Z}$, such that translation by 1 under one of the actions is equivalent to translation by the square classes $2$ and $4$ under the other two actions.</p> <p>If you take any pair of points $(x,y)$ in the above picture and subtract their indices, the orientation of the arrow between them is $x \to y$ if and only if $y-x$ is a square mod 7. Furthermore, a triple of points $(x,y,z)$ with directed arrows $x \to y \to z$ is collinear if and only if $\frac{z-y}{y-x} = 2$. Even though the numerator and denominator are only well-defined up to multiplication by squares, the quotient is a well-defined element of <code>$\mathbb{F}_7^\times$</code>, since each of the three translation actions yield the same answer. These two data let us reconstruct the diagram from the oriented line structure.</p> <p>There is a group-theoretic interpretation of this object. The oriented hypergraph you've given has automorphism group of order 21, generated by the permutations $(1234567)$ (one of the translation actions) and $(235)(476)$ (changes translation action by conjugating). This can be identified with the quotient <code>$B^+(\mathbb{F}_7)/\mathbb{F}_7^\times$</code>, where <code>$B^+(\mathbb{F}_7)$</code> is the group of upper triangular matrices with entries in $\mathbb{F}_7$ and invertible square determinant, and <code>$\mathbb{F}_7^\times$</code> is the subgroup of scalar multiples of the identity. This group is the stabilizer of infinity under the transitive action of the simple group of order 168 on the projective line <code>$\mathbb{P}^1(\mathbb{F}_7)$</code>. In this sense, we can view the simple group as the automorphism group of an oriented projective line, since it is the subgroup of <code>$PGL_2(\mathbb{F}_7)$</code> whose matrices have square determinant.</p> <p>Unfortunately, I do not know a natural notion of orientation on an <code>$\mathbb{F}_2$</code>-structure. I tried something involving torsors over <code>$\mathbb{F}_8^\times$</code> and the Frobenius, but it became a mess.</p>