In § 12.2, we considered projective lines, which are equivalence classes of points in $\KK^2$. Those equivalence classes, in turn, correspond to the (normalized) “squares” ($vv^\dagger$) of the “points” ($v$). Here, we extend the discussion to projective planes, which are described in $\KK^3$.

So what is a projective plane? Starting with $\RR$, we would like to define an equivalence relation between points, of the form \begin{equation} (b,c,r) \sim (b\chi,c\chi,r\chi) \label{equiv3} \end{equation} where $0\ne\chi\in\RR$. Again, it is more natural to consider squares of normalized columns. If we normalize $v\in\RR^3$ by \begin{equation} v^\dagger v = 1 \end{equation} where we have again written $\dagger$ instead of $T$ for transpose, then \begin{equation} (vv^\dagger)(vv^\dagger) = v(v^\dagger v)v^\dagger = vv^\dagger \label{rproj} \end{equation} Thus, normalized column vectors $v$ have matrix squares $vv^\dagger$ that are projection operators, that is, which square to themselves. Thus, we define \begin{equation} \RP^2 = \{\XXX\in\bH_3(\RR): \XXX^2 = \XXX, \tr\XXX = 1\} \label{projective3} \end{equation} where again the trace condition guarantees that $\XXX=vv^\dagger$ for some $v$.

We expect this construction to go through for the other division algebras, and it does, but there's a catch. Over $\CC$ and $\HH$, everything goes through as expected, yielding projective planes $\CP^2$ and $\HP^2$. What happens over $\OO$? The octonionic projective plane, known as the Cayley–Moufang plane, is indeed defined by \begin{equation} \OP^2 = \{\XXX\in\bH_3(\OO): \XXX^2 = \XXX, \tr\XXX = 1\} \label{points} \end{equation} but there's a catch: We used associativity in (\ref{rproj})! We must therefore ask, which matrices in $\bH_3(\OO)$ square to themselves? The answer is somewhat surprising.

Writing \begin{equation} \XXX = \begin{pmatrix} p& \bar{a}& c\\ a& m& \bar{b}\\ \bar{c}& b& n\\ \end{pmatrix} \end{equation} and squaring, we see that, for instance, the $(1,3)$ component of $\XXX^2=\XXX$ yields \begin{equation} (p+n)c +\bar{a}\,\bar{b} = c \end{equation} Since $p,n\in\RR$, $a,b,c$ must associate! Thus, the components of matrices in the Cayley–Moufang plane must lie in a quaternionic subalgebra of $\OO$! The components of different matrices can of course lie in different quaternionic subalgebras; nonetheless, this is a significant restriction on the possible matrices.

If we recall the definition of the Freudenthal product given in § 11.2, we see that \begin{equation} \XXX*\XXX = \XXX^2 - \XXX\,\tr(\XXX) + \frac12 \Big(\tr(\XXX)^2-\tr(\XXX^2)\Big) \label{FreudSq} \end{equation} so if $\XXX^2=\XXX$ and $\tr\XXX=1$, we obtain \begin{equation} \XXX*\XXX = 0 \label{OneSq} \end{equation} Conversely, if the Freudenthal square vanishes, then so does its trace \begin{equation} \tr(\XXX*\XXX) = \frac12 \Big(\tr(\XXX^2)-\tr(\XXX)^2\Big) \end{equation} so that (\ref{OneSq}) reduces to \begin{equation} \XXX^2 = \XXX\tr(\XXX) \end{equation} Thus, points in the Cayley–Moufang plane are precisely the normalized solutions of (\ref{OneSq}), that is, \begin{equation} \OP^2 = \{\XXX\in\bH_3(\OO): \XXX*\XXX=0, \tr\XXX = 1\} \end{equation}

Now that we know what the points in the projective plane are, we need to find the lines. We first briefly return to the real case. Let $v,w\in\RR^3$ be (real) vectors in three Euclidean dimensions. In § 11.2, we introduced the Jordan ($\circ$) and Freudenthal ($*$) products on the Albert algebra of octonionic Hermitian $3\times3$ matrices. We can construct such matrices from $v$ and $w$, namely $vv^T$ and $ww^T$. We than have the identities \begin{align} (vv^T) \circ (ww^T) &= (v^T w)(vw^T) + (w^T v)(wv^T) = (v^T w)(vw^T+wv^T) \label{rdot}\\ \tr(vv^T\circ ww^T) &= (v^T w)^2 \label{rdot2}\\ (vv^T) * (ww^T) &= (v\times w)(v\times w)^T \label{rcross} \end{align} where $v\times w$ denotes the usual cross product in $\RR^3$, and where of course $v^Tw=w^Tv\in\RR$. Equations (\ref{rdot})–(\ref{rcross}) justify regarding the trace of the Jordan product as a generalized dot product, and the Freudenthal product as a generalized cross product.

Over $\CC$ and $\HH$, (\ref{rdot2}) becomes \begin{equation} \tr(vv^\dagger\circ ww^\dagger) = |v^\dagger w|^2 \label{hdot2} \end{equation} since now $v^\dagger w=\bar{w^\dagger v}$.

Continuing the analogy to vector analysis, a “vector” “cross” itself must vanish. In fact, if we think of $\OP^2\otimes\RR$ as the space of (non-normalized) “vectors”, henceforth called “1-squares”, so that \begin{equation} \OP^2\otimes\RR = \{\XXX\in\bH_3(\OO): \XXX*\XXX=0\} \end{equation} then for $\AAA,\BBB\in\OP^2\otimes\RR$ we have \begin{equation} \AAA*\BBB = 0 \Longleftrightarrow \BBB=\lambda\AAA \end{equation} for some $\lambda\in\RR$, so that $\AAA$ and $\BBB$ are “parallel”.

Similar analogies can be made with the dot product. In $\HP^2$ (and hence also in $\RP^2$ and $\CP^2$), \begin{equation} \tr(vv^\dagger\circ ww^\dagger)=0 \Longrightarrow vv^\dagger\circ ww^\dagger=0 \label{ortho1} \end{equation} which follows from (\ref{hdot2}). Remarkably, (\ref{ortho1}) still holds in $\OP^2$ even though (\ref{hdot2}) does not, as can be checked by an explicit but lengthy computation. That is, \begin{equation} \tr(\XXX\circ\YYY)=0 \Longrightarrow \XXX\circ\YYY=0 \label{xtr} \end{equation} for $\XXX,\YYY\in\OP^2$.

Using our intuition about points and lines in the real projective plane, our knowledge of vector analysis in $\RR^3$, and our notions of generalized dot and cross products, we are now ready to study the properties of the octonionic projective plane.

So what are the lines in the projective plane? Recall that the points of $\OP^2$ are given by (\ref{points}). Consider by analogy the matrices \begin{equation} \Lambda = \{\XXX\in\bH_3(\OO): \XXX^2 = \XXX, \tr\XXX = 2\} \label{lines} \end{equation} and call the elements of $\Lambda$ lines.

Why is this terminology reasonable? In algebraic terms, $\Pcal$ is a primitive idempodent of $\bH_3(\OO)$. Restricting temporarily to $\HH$ to avoid associativity issues, $\Pcal$ is a projection operator on $\HH^3$ into a 1-dimensional subspace, whereas $\Lcal$ is a projection operator into a 2-dimensional subspace (and $\III$ “projects” into a 3-dimensional subspace, namely all of $\HH^3$). In the projective plane, (equivalence classes of) such 1-dimensional subspaces are “points”, and (equivalence classes of) such 2-dimensional subspaces are lines. The definitions (\ref{points}) and (\ref{lines}) therefore represent plausible generalizations of these concepts to the nonassociative case.

Using definition (\ref{lines}), if $\Qcal\in\OP^2$ is a point, then \begin{equation} \Lcal = \III-\Qcal \in\Lambda \end{equation} is a line, where $\III$ is the $3\times3$ identity matrix, which is also the unique matrix that squares to itself and has trace 3. There is thus a natural duality relationship between the lines and points of $\OP^2$. Using our intuition from $\RR^3$, where there is a unique line (through the origin) perpendicular to each plane (through the origin), and vice versa, we can use the point $\III-\Lcal$, which is “perpendicular” to $\Lcal$, to define the points $\Pcal$ that are on the line $\Lcal$ as being those points which are orthogonal to $\III-\Lcal$. What does orthogonality mean for points? Use the trace of the Jordan product, that is, define $\Pcal$ to be on $\Lcal$ if \begin{equation} \tr\Bigl( (\III-\Lcal)\circ\Pcal \Bigr) = 0 \end{equation} Since both $\Pcal,\III-\Lcal\in\OP^2$, we can use (\ref{xtr}) to drop the trace; $\Pcal$ is on $\Lcal$ if \begin{equation} (\III-\Lcal)\circ\Pcal = 0 \end{equation} or equivalently if \begin{equation} \Lcal\circ\Pcal = \Pcal \end{equation}

What is the line determined by the points $\Pcal$ and $\Qcal$? Use the cross product! If $\Pcal,\Qcal\in\OP^2$, then it does indeed follow that \begin{equation} (\Pcal*\Qcal)*(\Pcal*\Qcal) = 0 \end{equation} that is, $\Pcal*\Qcal$ is a 1-square if $\Pcal$ and $\Qcal$ are 1-squares (and is nonzero so long as $\Pcal$ and $\Qcal$ are not “parallel”). However, $\Pcal*\Qcal$ will not in general be normalized. The point “orthogonal” to $\Pcal$ and $\Qcal$ is therefore $\Pcal*\Qcal/\tr(\Pcal*\Qcal)$, so the line determined by $\Pcal$ and $\Qcal$ must be 1) \begin{equation} \Lcal_{\Pcal\Qcal} = \III - \frac{\Pcal*\Qcal}{\tr(\Pcal*\Qcal)} \label{linePQ} \end{equation} $\Pcal$ and $\Qcal$ are on this line, since \begin{equation} (\Pcal*\Qcal)\circ\Qcal = 0 \end{equation} for any 1-squares $\Pcal$ and $\Qcal$. More generally, three points $\Pcal$, $\Qcal$, and $\Scal$ are collinear if \begin{equation} (\Pcal*\Qcal)\circ\Scal = 0 \end{equation}

Recall from § 11.2–§ 11.4 that $E_6$ acts on $\bH_3(\OO)$. We briefly summarize some properties of the induced action of $E_6$ on $\OP^2$.

• The trace of the triple product is the polarization of the determinant, and hence preserved by $E_6$. Thus, $E_6$ is precisely the symmetry group which preserves the notion of collinear points in $\OP^2$.
• In fact, $E_6$ takes $p$-squares to $p$-squares, but the boosts in $E_6$ do not preserve the normalization condition on $\OP^2$. Thus, $E_6$ takes points to points, and lines to lines up to normalization (which can be corrected by slightly modifying the action).
• The action of $E_6$ on lines is not the usual action of $E_6$ on 2-squares, but rather the “squared” action induced by $\Pcal*\Qcal\longmapsto(\MMM\Pcal\MMM^\dagger)*(\MMM\Qcal\MMM^\dagger)$, with subsequent renormalization as needed.
• Since $F_4$ is the automorphism group of the Jordan product, it is also the automorphism group of the Freudenthal product, so that \begin{equation} (\MMM\Pcal\MMM^\dagger)*(\MMM\Qcal\MMM^\dagger)=\MMM(\Pcal*\Qcal)\MMM^\dagger \end{equation} for $\MMM\in F_4$.
• The remaining elements (“boosts”) in $E_6$ are generated by complex Hermitian matrices, and for such matrices direct computation shows that \begin{equation} (\MMM\Pcal\MMM^\dagger)*(\MMM\Qcal\MMM^\dagger) = (\MMM*\MMM)(\Pcal*\Qcal)(\MMM*\MMM)^\dagger \end{equation} This operation yields a “dual” action of boosts on 1-squares, given by $\MMM*\MMM$ rather than $\MMM$ — and note that $\MMM*\MMM$ is (a multiple of) $\MMM^{-1}$; these boosts go “the other way”.
• Any projective line is a 2-square with repeated eigenvalue $1$, and can therefore be written (in several ways) as $\Lcal=\Pcal+\Qcal$, where $\Pcal$, $\Qcal$ are projective points satisfying $\Pcal\circ\Qcal=0$.
• But if $\Pcal\circ\Qcal=0$, the Freudenthal product simplifies to $2\Pcal*\Qcal=\III-\Pcal-\Qcal$.
• Thus, a projective line can be written as $\Lcal=\III-2\Pcal*\Qcal$, where $\Pcal$ and $\Qcal$ are (“orthogonal”) points on the line.
• If $\Lcal\in\Lambda$ is a line in $\OP^2$, then the action of $E_6$ is given by \begin{equation} \Lcal \longmapsto \Lcal'= \begin{cases} \III-\MMM(\III-\Lcal)\MMM^\dagger & \textrm{(rotations)} \\ \III - {1\over N}(\MMM*\MMM)(\III-\Lcal)(\MMM*\MMM)^\dagger & \textrm{(boosts)} \end{cases} \end{equation} where $N$ is a normalization constant.
• Untangling these definitions, the condition that a point $\AAA$ is on $\Lcal$ is preserved by boosts, since \begin{align} (\MMM*\MMM)(\III-\Lcal)(\MMM*\MMM)^\dagger &= 2 (\MMM*\MMM)(\Pcal*\Qcal)(\MMM*\MMM)^\dagger \nonumber\\ &= 2 (\MMM\Pcal\MMM^\dagger)*(\MMM\Qcal\MMM^\dagger) \end{align} and therefore \begin{align} \Lcal'\circ\AAA'=\AAA' &\Longleftrightarrow 0 = (\MMM*\MMM)(\III-\Lcal)(\MMM*\MMM)^\dagger \circ\MMM\AAA\MMM^\dagger \nonumber\\ &\qquad\quad = 2 \Bigl( (\MMM\Pcal\MMM^\dagger)*(\MMM\Qcal\MMM^\dagger) \Bigr) \circ\MMM\AAA\MMM^\dagger \nonumber\\ &\qquad\quad = 2(\Pcal*\Qcal)\circ\AAA \nonumber\\ &\Longleftrightarrow \Lcal\circ\AAA=\AAA \end{align}
• $E_6$ is therefore the symmetry group which takes points to points and lines to lines in $\OP^2$, while preserving the incidence relation.