# What is the oriented Fano plane?

One way to remember the multiplication table of the octonions is to use the following diagram (which I got from John Baez's online paper): 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 $\mathbb O$.

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, &c.

What kind of object is this oriented Fano plane?

NB1: Naive googling informs of the concept of Mendelsohn triple systems and of transitive triple systems, 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.

NB2: 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.

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Let me add one more object which it isn't. Viewing the Fano plane as a matroid, this oriented Fano plane is not an oriented matroid either. See here. – Tony Huynh Apr 7 '10 at 2:25
I hope this does not evolve into a big-list of things it isn't! :) – Mariano Suárez-Alvarez Apr 7 '10 at 2:27

Here is one answer: It is an oriented line over $\mathbb{F}_7$.

An affine line over $\mathbb{F}_7$ 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.

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 $\mathbb{F}_7^\times$, since each of the three translation actions yield the same answer. These two data let us reconstruct the diagram from the oriented line structure.

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 $B^+(\mathbb{F}_7)/\mathbb{F}_7^\times$, where $B^+(\mathbb{F}_7)$ is the group of upper triangular matrices with entries in $\mathbb{F}_7$ and invertible square determinant, and $\mathbb{F}_7^\times$ 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 $\mathbb{P}^1(\mathbb{F}_7)$. In this sense, we can view the simple group as the automorphism group of an oriented projective line, since it is the subgroup of $PGL_2(\mathbb{F}_7)$ whose matrices have square determinant.

Unfortunately, I do not know a natural notion of orientation on an $\mathbb{F}_2$-structure. I tried something involving torsors over $\mathbb{F}_8^\times$ and the Frobenius, but it became a mess.

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Indeed. This construction proceeds from what's called a $\lambda$-difference set, a subset $D$ of $\mathbb Z_n$ such that each non-zero element of $\mathbb Z_n$ can be written in exactly $\lambda$ ways as a difference of elements of $D$; in the case of the Fano plane, $n=7$, $D={1,2,4}$ and $\lambda=1$. When $\lambda$ is one, the resulting hypergraph is a projective plane. – Mariano Suárez-Alvarez Apr 8 '10 at 20:38
I am confused. The number of $F_7$-affine structures is 6!=720. The number of orientations on each line is $2^7=128$. Clearly, some affine structures are no good, for instance, those that have 3 collinear points follow one another. How do you account for this? – Bugs Bunny Apr 9 '10 at 10:37
@Bugs: Admissible sets of line orientations need to admit 3 independent Hamiltonian cycles, taken to each other by a transitive symmetry (or, see NB2 in the question). Admissible triples of affine structures have to obey the collinearity condition I gave above. There are 21 of each. – S. Carnahan Apr 9 '10 at 18:58

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.

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