The Fano plane is the projective plane over the field $\mathbf Z/2$. It can be used to remember octonion multiplication, as nicely explianed in John Baez's article on octonions (see http://math.ucr.edu/home/baez/octonions/).

The picture (taken from Baez's website) is as follows:

The Fano plane

It indicates for example, using the cyclic orderings on the lines, that $e_6 \cdot e_1= e_5$ but that $e_6 \cdot e_4= -e_3$.

Two natural questions arise for me:

  1. Why did Baez label the circles in such a weird order. There is probably a good reason which is implicit. (A priori, if I permute the labels arbitrarily, I'll get something isomorphic).

  2. Also, one could decide to choose other orientations for the arrows. So again, why did Baez choose these orientations? And what could we get with other orientations?

  • 2
    $\begingroup$ related: What is the oriented Fano plane? @ MO $\endgroup$
    – Grigory M
    Nov 27, 2015 at 10:20
  • $\begingroup$ The order initially looked weirder to me than it did 30 seconds later. If you go from $1$ to $2$, then to $3$, then to $4$, you're following arrows, and I thought if you go from there to $5$, you're not. But then I notice you said "cyclic orderings", which would seem to mean there's an arrow, not shown ,from $4$ to $5$ (and similarly from $6$ to $3$, etc. $\endgroup$ Jul 23, 2018 at 23:41

1 Answer 1


The circles are labeled in such a way that the lines are given by $(i,i+1,i+3)$ modulo $7$. (Interpreted in the interval $1,...,7$). This is with good reason, see below.

The orientations are more difficult to explain in a sentence, but two remarks are in order:

  • They are drawn in such a way that the automorphisms group $G\cong \mathsf{PSL}(3,2)\cong\mathsf{PSL}(2,7)$ of the Fano plane will induce automorphisms of the octonions. For instance, invariance under rotations (of oder $3$) is immediately visible from the picture; an element of order $7$ in $G$ is given by mapping $e_i\to e_{i+1}$ and since the orientation is always $(i,i+1,i+3)$ the arrows are invariant under these automorphisms too. (The automorphisms of order $2$ correspond to reflections the Fano plane, but you also have to introduce some signs like $e_i\mapsto -e_j$ so this is less obvious.) This is somewhat useful since it will immediately provide you with a bunch of automorphisms of the octonions.
  • You could in principle draw arbitrary arrows and study the resulting algebra generated by the same mechanism. But it is a fact that if you want the outcome to be a composition algebra, then the quadratic form determines the octonion algebra entirely. (This quadratic form is a so called Pfister form, and I think this theorem is usually attributed to Pfister. [The theorem is actually due to Jacobson, see the comment by Mariano Suárez-Alvarez here. ]) So in some sense, if you specify what $e_i^2$ is (usually one studies $e_i^2=-1$ for all $i$, this corresponds to the so called "compact real octonions") then you have specified the quadratic form entirely and there is essentially only one way to orient the edges to obtain a composition algebra.
  • It's very well possible that drawing different orientations on the edges gives rise to other algebra's with interesting properties. I would be very interested if anyone knew of results in that direction.
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    $\begingroup$ Thanks. It definitely looks like an answer to my question. But I have to admit that I don't understand your second remark on orientations. Is the theorem of Pfister mentionned the same as this one: math.uconn.edu/~kconrad/blurbs/linmultialg/pfister.pdf ? $\endgroup$
    – Oblomov
    Nov 27, 2015 at 10:49
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    $\begingroup$ Hm that link doesn't open right now but I'm guessing it will refer to Pfisters theorem on quadratic forms which is more general. The theorem which I mentioned says that "octonion algebras are determined by their norm"; for instance check out the accepted anser to this MO question who attributes the theorem to Jacobson or Albert. You will definitely find a proof in the "book of involutions". $\endgroup$
    – Myself
    Nov 27, 2015 at 11:50
  • $\begingroup$ Ok a comment from Mariano Suárez-Alvarez say it is a theorem of Jacobson, I'll edit my answer. $\endgroup$
    – Myself
    Nov 27, 2015 at 11:51
  • $\begingroup$ This is an especially nice answer. $\endgroup$ Nov 27, 2015 at 16:32

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