3.4 $\OP^2$ and the Exceptional Jordan Algebra

The octonions are fascinating in themselves, but the magic really starts when we use them to construct the exceptional Jordan algebra $\h _3(\O)$ and its associated projective space, the octonionic projective plane. The symmetry groups of these structures turn out to be exceptional Lie groups, and triality gains an eerie pervasive influence over the proceedings, since an element of $\h _3(\O)$ consists of 3 octonions and 3 real numbers. Using the relation between normed division algebras and trialities, we get an isomorphism

$$% latex2html id marker 1622 \begin{array}{ccc} \h _3(\O)... ... \right) & \mapsto & ((\alpha,\beta,\gamma),x,y,z) \end{array}$$

where $\alpha,\beta,\gamma \in \R$ and $x,y,z \in \O$. Examining the Jordan product in $\h _3(\O)$ then reveals a wonderful fact: while superficially this product is defined using the $\ast$-algebra structure of $\O$, it can actually be defined using only the natural maps

$$V_8 \times S_8^+ \to S_8^- , \qquad V_8 \times S_8^- \to S_8^+ , \qquad S_8^+ \times S_8^- \to V_8$$

together with the inner products on these 3 spaces. All this information is contained in the normed triality

$$t_8 \maps V_8 \times S_8^+ \times S_8^- \to \R ,$$

so any automorphism of this triality gives an automorphism of $\h _3(\O)$. In Section 2.4 we saw that $\Aut (t_8) \iso \Spin (8)$. With a little thought, it follows that

$$\Spin (8) \subseteq \Aut (\h _3(\O)) .$$

However, this picture of $\h _3(\O)$ in terms of 8-dimensional Euclidean geometry is just part of a bigger picture — a picture set in 10-dimensional Minkowski spacetime! If we regard $\h _2(\O)$ as sitting in the lower right-hand corner of $\h _3(\O)$, we get an isomorphism

$$% latex2html id marker 1626 \begin{array}{ccc} \h _3(\O)... ...\\ \end{array} \right) & \mapsto & (\alpha,a,\psi) \end{array}$$

We saw in Section 3.3 that $a \in \h _2(\O)$ and $\psi \in \O^2$ can be identified with a vector and a spinor in 10-dimensional Minkowski spacetime, respectively. Similarly, $\alpha$ is a scalar.

This picture gives a representation of $\Spin (9,1)$ as linear transformations of $\h _3(\O)$. Unfortunately, most of these transformations do not preserve the Jordan product on $\h _3(\O)$. As we shall see, they only preserve a lesser structure on $\h _3(\O)$: the determinant. However, the transformations coming from the subgroup $\Spin (9) \subset \Spin (9,1)$ do preserve the Jordan product. We can see this as follows. As a representation of $\Spin (9)$, $\h _2(\O)$ splits into 'space' and 'time':

$$\h _2(\O) \iso V_9 \oplus \R$$

with the two pieces corresponding to the traceless elements of $\h _2(\O)$ and the real multiples of the identity, respectively. On the other hand, the spinor representation of $\so (9)$ splits as $S_8^+ \oplus S_8^-$ when we restrict it to $\so (8)$, so we have

$$\O^2 \iso S_9 .$$

We thus obtain an isomorphism

$$% latex2html id marker 1629 \begin{array}{ccc} \h _3(\O)... ...array} \right) & \mapsto & ((\alpha,\beta),a,\psi) \end{array}$$

where $a \in \h _2(\O)$ has vanishing trace and $\beta$ is a real multiple of the identity. In these terms, one can easily check that the Jordan product in $\h _3(\O)$ is built from invariant operations on scalars, vectors and spinors in 9 dimensions. It follows that

$$\Spin (9) \subseteq \Aut (\h _3(\O)) .$$

For more details on this, see Harvey's book [50].

This does not exhaust all the symmetries of $\h _3(\O)$, since there are other automorphisms coming from the permutation group on 3 letters, which acts on $(\alpha,\beta,\gamma) \in \R^3$ and $(x,y,z) \in \O^3$ in an obvious way. Also, any matrix $g \in \OO (3)$ acts by conjugation as an automorphism of $\h _3(\O)$; since the entries of $g$ are real, there is no problem with nonassociativity here. The group $\Spin (9)$ is 36-dimensional, but the full automorphism group $\h _3(\O)$ is much bigger: it is 52-dimensional. As we explain in Section 4.2, it goes by the name of $\F _4$.

However, we can already do something interesting with the automorphisms we have: we can use them to diagonalize any element of $\h _3(\O)$. To see this, first note that the rotation group, and thus $\Spin (9)$, acts transitively on the unit sphere in $V_9$. This means we can use an automorphism in our $\Spin (9)$ subgroup to bring any element of $\h _3(\O)$ to the form

$$% latex2html id marker 1631 \left( \begin{array}{ccc} \a... ...^* \\ z & \beta & x \\ y & x^* & \gamma \end{array} \right)$$

where $x$ is real. The next step is to apply an automorphism that makes $y$ and $z$ real while leaving $x$ alone. To do this, note that the subgroup of $\Spin (9)$ fixing any nonzero vector in $V_9$ is isomorphic to $\Spin (8)$. When we restrict the representation $S_9$ to this subgroup, it splits as $S_8^+ \oplus S_8^-$, and with some work [50] one can show that $\Spin (8)$ acts on $S_8^+ \oplus S_8^- \iso \O^2$ in such a way that any element $(y,z) \in \O^2$ can be carried to an element with both components real. The final step is to take our element of $\h _3(\O)$ with all real entries and use an automorphism to diagonalize it. We can do this by conjugating it with a suitable matrix in $\OO (3)$.

To understand $\OP^2$, we need to understand projections in $\h _3(\O)$. Here is where our ability to diagonalize matrices in $\h _3(\O)$ via automorphisms comes in handy. Up to automorphism, every projection in $\h _3(\O)$ looks like one of these four:

$$% latex2html id marker 1596 p_0 = \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right) ,$$

$$% latex2html id marker 1597 p_1 = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right) ,$$

$$% latex2html id marker 1598 p_2 = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right) ,$$

$$% latex2html id marker 1599 p_3 = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) .$$

Now, the trace of a matrix in $\h _3(\O)$ is invariant under automorphisms, because we can define it using only the Jordan algebra structure:

$$\tr (a) = {1\over 9} \tr (L_a) , \qquad \qquad a \in \h _3(\O)$$

where $L_a$ is left multiplication by $a$. It follows that the trace of any projection in $\h _3(\O)$ is 0,1,2, or 3. Furthermore, the rank of any projection $p \in \h _3(\O)$ equals its trace. To see this, first note that $\tr (p) \ge \rank (p)$, since $p < q$ implies $\tr (p) < \tr (q)$, and the trace goes up by integer steps. Thus we only need show $\tr (p) \le \rank (p)$. For this it suffices to consider the four projections shown above, as both trace and rank are invariant under automorphisms. Since $p_0 < p_1 < p_2 < p_3$, it is clear that for these projections we indeed have $\tr (p) \le \rank (p)$.

It follows that the points of the octonionic projective plane are projections with trace 1 in $\h _3(\O)$, while the lines are projections with trace 2. A calculation [50] shows that any projection with trace 1 has the form

$$% latex2html id marker 1634 p = \left( \begin{array}{c} x ... ...& y y^* & y z^* \\ z x^* & z y^* & z z^* \end{array} \right)$$

where $(x,y,z) \in \O^3$ has

$$% latex2html id marker 1635 (xy)z = x(yz), \qquad \qquad \Vert x\Vert^2 + \Vert y\Vert^2 + \Vert z\Vert^2 = 1 .$$

On the other hand, any projection with trace 2 is of the form $1 - p$ where $p$ has trace 1. This sets up a one-to-one correspondence between points and lines in the octonionic projective plane. If we use this correspondence to think of both as trace-1 projections, the point $p$ lies on the line $p'$ if and only if $p < 1 - p'$. Of course, $p < 1 - p'$ iff $p' < 1 - p$. The symmetry of this relation means the octonionic projective plane is self-dual! This is also true of the real, complex and quaternionic projective planes. In all cases, the operation that switches points and lines corresponds in quantum logic to the 'negation' of propositions [91].

We use $\OP^2$ to stand for the set of points in the octonionic projective plane. Given any nonzero element $(x,y,z) \in \O^3$ with $(xy)z = x(yz)$, we can normalize it and then use equation (3.4) to obtain a point $[(x,y,z)] \in \OP^2$. Copying the strategy that worked for $\OP^1$, we can make $\OP^2$ into a smooth manifold by covering it with three coordinate charts:

• one chart containing all points of the form $[(x,y,1)]$,
• one chart containing all points of the form $[(x,1,z)]$,
• one chart containing all points of the form $[(1,y,z)]$.

Checking that this works is a simple calculation. The only interesting part is to make sure that whenever the associative law might appear necessary, we can either use the alternativity of the octonions or the fact that only triples with $(xy)z = x(yz)$ give points $[(x,y,z)] \in \OP^2$.

We thus obtain the following picture of the octonionic projective plane. As a manifold, $\OP^2$ is 16-dimensional. The lines in $\OP^2$ are copies of $\OP^1$, and thus 8-spheres. For any two distinct points in $\OP^2$, there is a unique line on which they both lie. For any two distinct lines, there is a unique point lying on both of them. There is a 'duality' transformation that maps points to lines and vice versa while preserving this incidence relation. In particular, since the space of all points lying on any given line is a copy of $\OP^1$, so is the space of all lines containing a given point!

To dig more deeply into the geometry of $\OP^2$ one needs another important structure on the exceptional Jordan algebra: the determinant. We saw in Section 3.3 that despite noncommutativity and nonassociativity, the determinant of a matrix in $\h _2(\O)$ is a well-defined and useful concept. The same holds for $\h _3(\O)$! We can define the determinant of a matrix in $\h _3(\O)$ by

$$% latex2html id marker 1636 \det \left( \begin{array}{ccc}... ... \beta \Vert y\Vert^2 + \gamma \Vert z\Vert^2) + 2 \Re (xyz) .$$

We can express this in terms of the trace and product via

$$\det(a) = {1\over 3} \tr (a^3) - {1\over 2} \tr (a^2) \tr (a) + {1\over 6} {\tr (a)}^3 .$$

This shows that the determinant is invariant under all automorphisms of $\h _3(\O)$. However, the determinant is invariant under an even bigger group of linear transformations. As we shall see in Section 4.4, this group is 78-dimensional: it is a noncompact real form of the exceptional Lie group $\E _6$. This extra symmetry makes it worth seeing how much geometry we can do starting with just the determinant and the vector space structure of $\h _3(\O)$.

The determinant is a cubic form on $\h _3(\O)$, so there is a unique symmetric trilinear form

$$(\cdot,\cdot,\cdot) \maps \h _3(\O) \times \h _3(\O) \times \h _3(\O) \to \R$$

such that

$$(a,a,a) = \det(a) .$$

By dualizing this, we obtain the so-called cross product

$$\times \maps \h _3(\O) \times \h _3(\O) \to \h _3(\O)^*.$$

Explicitly, this is given by

$$(a \times b)(c) = (a,b,c) .$$

Despite its name, this product is commutative.

We have already seen that points of $\OP^2$ correspond to trace-1 projections in $\h _3(\O)$. Freudenthal [36] noticed that these are the same as elements $p \in \h _3(\O)$ with $\tr (p) = 1$ and $p \times p = 0$. Even better, we can drop the equation $\tr (p) = 1$ as long as we promise to work with equivalence classes of nonzero elements satisfying $p \times p = 0$, where two such elements are equivalent when one is a nonzero real multiple of the other. Each such equivalence class $[p]$ corresponds to a unique point of $\OP^2$, and we get all the points this way.

Given two points $[p]$ and $[q]$, their cross product $p \times q$ is well-defined up to a nonzero real multiple. This suggests that we define a 'line' to be an equivalence class of elements $p \times q \in \h _3(\O)^*$, where again two such elements are deemed equivalent if one is a nonzero real multiple of the other. Freudenthal showed that we get a projective plane isomorphic to $\OP^2$ if we take these as our definitions of points and lines and decree that the point $[p]$ lies on the line $[L]$ if and only if $L(p) = 0$. Note that this equation makes sense even though $L$ and $p$ are only well-defined up to nonzero real multiples.

One consequence of all this is that one can recover the structure of $\OP^2$ as a projective plane starting from just the determinant on $\h _3(\O)$: we did not need the Jordan algebra structure! However, to get a 'duality' map switching points and lines while preserving the incidence relation, we need a bit more: we need the nondegenerate pairing

$$\langle a,b \rangle = \tr (ab)$$

on $\h _3(\O)$. This sets up an isomorphism

$$\h _3(\O) \iso \h _3(\O)^* .$$

This isomorphism turns out to map points to lines, and in fact, it sets up a one-to-one correspondence between points and lines. We can use this correspondence to think of both points and lines in $\OP^2$ as equivalence classes of elements of $\h _3(\O)$. In these terms, the point $p$ lies on the line $\ell$ iff $\langle \ell,p \rangle = 0$. This relationship is symmetrical! It follows that if we switch points and lines using this correspondence, the incidence relation is preserved.

We thus obtain a very pretty setup for working with $\OP^2$. If we use the isomorphism between $\h _3(\O)$ and its dual to reinterpret the cross product as a map

$$\times \maps \h _3(\O) \times \h _3(\O) \to \h _3(\O) ,$$

then not only is the line through distinct points $[p]$ and $[q]$ given by $[p \times q]$, but also the point in which two distinct lines $[\ell]$ and $[m]$ meet is given by $[\ell \times m]$. A triple of points $[p], [q]$ and $[r]$ is collinear iff $(p,q,r) = 0$, and a triple of lines $[\ell]$, $[m]$, $[n]$ meets at a point iff $(\ell,m,n) = 0$. In addition, there is a delightful bunch of identities relating the Jordan product, the determinant, the cross product and the inner product in $\h _3(\O)$.

For more on octonionic geometry, the reader is urged to consult the original papers of Freudenthal [35,36,37,38], as well as those of Jacques Tits [87,88] and Tonny Springer [80,81,82]. Unfortunately, we must now bid goodbye to this subject and begin our trip through the exceptional groups. However, we shall return to study the symmetries of $\OP^2$ and the exceptional Jordan algebra in Sections 4.2 and 4.4.

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