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Next: E6 Up: Exceptional Lie Algebras Previous: F4

4.3 The Magic Square

Around 1956, Boris Rosenfeld [74] had the remarkable idea that just as $\F _4$ is the isometry group of the projective plane over the octonions, the exceptional Lie groups $\E _6$, $\E _7$ and $\E _8$ are the isometry groups of projective planes over the following three algebras, respectively:

There is definitely something right about this idea, because one would expect these projective planes to have dimensions 32, 64, and 128, and there indeed do exist compact Riemannian manifolds with these dimensions having $\E _6$, $\E _7$ and $\E _8$ as their isometry groups. The problem is that the bioctonions, quateroctonions and and octooctonions are not division algebras, so it is a nontrivial matter to define projective planes over them!

The situation is not so bad for the bioctonions: $\h _3(\C \tensor \O)$ is a simple Jordan algebra, though not a formally real one, and one can use this to define $(\C \tensor \O)\P^2$ in a manner modeled after one of the constructions of $\OP^2$. Rosenfeld claimed that a similar construction worked for the quateroctonions and octooctonions, but this appears to be false. Among other problems, $\h _3(\H \tensor \O)$ and $\h _3(\O \tensor \O)$ do not become Jordan algebras under the product $a \circ b = {1\over 2}(ab + ba)$. Scattered throughout the literature [5,38,39] one can find frustrated comments about the lack of a really nice construction of $(\H \tensor \O)\P^2$ and $(\O \tensor \O)\P^2$. One problem is that these spaces do not satisfy the usual axioms for a projective plane. Tits addressed this problem in his theory of `buildings', which allows one to construct a geometry having any desired algebraic group as symmetries [90]. But alas, it still seems that the quickest way to get our hands on the quateroctonionic and octooctonionic `projective planes' is by starting with the Lie groups $\E _7$ and $\E _8$ and then taking quotients by suitable subgroups.

In short, more work must be done before we can claim to fully understand the geometrical meaning of the Lie groups $\E _6, \E _7$ and $\E _8$. Luckily, Rosenfeld's ideas can be used to motivate a nice construction of their Lie algebras. This goes by the name of the `magic square'. Tits [89] and Freudenthal [37] found two very different versions of this construction in about 1958, but we shall start by presenting a simplified version published by E. Vinberg [92] in 1966.

First consider the projective plane $\KP^2$ when $\K$ is a normed division algebra $\K$. The points of this plane are the rank-1 projections in the Jordan algebra $\h _3(\K)$, and this plane admits a Riemannian metric such that

\begin{displaymath}\isom (\KP^2) \iso \Der (\h _3(\K)). \end{displaymath}

Moreover, we have seen in equation (4.2) that

\begin{displaymath}\Der (\h _3(\K)) \iso \Der (\K) \oplus \sa _3(\K) . \end{displaymath}

Combined with Rosenfeld's observations, these facts might lead one to hope that whenever we have a pair of normed division algebras $\K$ and $\K'$, there is a Riemannian manifold $(\K \tensor \K')\P^2$ with

\begin{displaymath}\isom ((\K\tensor \K')\P^2) \iso
\Der (\K) \oplus \Der (\K') \oplus \sa _3(\K \tensor \K') \end{displaymath}

where for any $\ast$-algebra $A$ we define

% latex2html id marker 1703\begin{array}{lcl}
\sh _n(A) &...
...{ x \in A[n] \colon \; x^* = -x, \; \tr (x) = 0\}.

This motivated Vinberg's definition of the magic square Lie algebras:

\M (\K,\K') = \Der (\K) \oplus \Der (\K') \oplus \sa _3(\K \tensor \K').

Now, when $\K \tensor \K'$ is commutative and associative, $\sa _3(\K
\tensor \K')$ is a Lie algebra with the commutator as its Lie bracket, but in the really interesting cases it is not. Thus to make $\M (\K,\K')$ into a Lie algebra we must give it a rather subtle bracket. We have already seen the special case $\K' = \R$ in equation (4.2). In general, the Lie bracket in $\M (\K,\K')$ is given as follows:
  1. $\Der (\K)$ and $\Der (\K')$ are commuting Lie subalgebras of $\M (\K,\K')$.
  2. The bracket of $D \in \Der (K) \oplus \Der (\K')$ with $x \in \sa _3(\K \tensor \K')$ is given by applying $D$ to every entry of the matrix $x$, using the natural action of $\Der (K) \oplus \Der (\K')$ as derivations of $\K \tensor \K'$.
  3. Given $X,Y \in \sa _3(\K \tensor \K')$,

    \begin{displaymath}[X,Y]= [X,Y]_0 +
{1\over 3}\displaystyle{\sum_{i,j = 1}^3 D_{X_{ij},Y_{ij}} } .\end{displaymath}

    Here $[X,Y]_0$ is the traceless part of the $3 \times 3$ matrix $[X,Y]$, and given $x,y \in \K \tensor \K'$ we define $D_{x,y} \in
\Der (\K) \oplus \Der (\K')$ in the following way: $D_{x,y}$ is real-bilinear in $x$ and $y$, and

    \begin{displaymath}D_{a \tensor a',b \tensor b'} = \langle a',b' \rangle D_{a,b} +
\langle a,b \rangle D_{a',b'} \end{displaymath}

    where $a,b \in \K$, $a',b' \in \K'$, and $D_{a,b},D_{a',b'}$ are defined as in equation (4.1).
With this construction we magically obtain the following square of Lie algebras:

  $\K' = \R$ $\K' = \C$ $\K' = \H$ $\K' = \O$
$\K = \R$ $\so (3)$ $\su (3)$ $\symp (3)$ $\f _4$
$\K =
\C$ $\su (3)$ $\su (3) \oplus \su (3)$ $\su (6)$ $e_6$
$\K = \H$ $\symp (3)$ $\su (6)$ $\so (12)$ $e_7$
$\K = \O$ $\f _4$ $e_6$ $e_7$ $\e _8$
Table 5 — Magic Square Lie Algebras $\M (\K,\K')$

We will mainly be interested in the last row (or column), which is the one involving the octonions. In this case we can take the magic square construction as defining the Lie algebras $\f _4$, $e_6$, $e_7$ and $\e _8$. This definition turns out to be consistent with our earlier definition of $\f _4$.

Starting from Vinberg's definition of the magic square Lie algebras, we can easily recover Tits' original definition. To do so, we need two facts. First,

% latex2html id marker 1707
\sa _3(\K \otimes \K')
\iso \sa _3(\K') \, \oplus \, (\Im (\K) \! \tensor \! \sh _3(\K')).

This is easily seen by direct examination of the relevant matrices. Second,

\begin{displaymath}\Der (\h _3(\K)) \iso \Der (\K) \oplus \sa _3(\K)

as vector spaces. This is just equation (4.2). Starting with Vinberg's definition and applying these two facts, we obtain
% latex2html id marker 1709\begin{array}{lcl}
\M (\K,\K') ...
...(\K')) \, \oplus \, (\Im (\K) \tensor \sh _3(\K'))

The last line is Tits' definition of the magic square Lie algebras. Unlike Vinberg's, it is not manifestly symmetrical in $\K$ and $\K'$. This unhappy feature is somewhat made up for by the fact that $\Der (\K)
\oplus \Der (\h _3(\K'))$ is a nice big Lie subalgebra. This subalgebra acts on $\Im (\K) \tensor \sh _3(\K')$ in an obvious way, using the fact that any derivation of $\K$ maps $\Im (\K)$ to itself, and any derivation of $\h _3(\K')$ maps $\sh _3(\K')$ to itself. However, the bracket of two elements of $(\Im (\K) \tensor \sh _3(\K'))$ is a bit of a mess.

Yet another description of the magic square was recently given by Barton and Sudbery [4]. This one emphasizes the role of trialities. Let $\Tri (\K)$ be the Lie algebra of the group $\Aut (t)$, where $t$ is the normed triality giving the normed division algebra $\K$. From equation (2) we have

% latex2html id marker 925
\Tri (\R) &\iso & \{0\} \\  \Tr...
...)^2 \\  \Tri (\H) &\iso & \symp (1)^3 \\  \Tri (\O) &\iso & \so (8)

To express the magic square in terms of these Lie algebras, we need three facts. First, it is easy to see that

\begin{displaymath}\sh _3(\K) \iso \K^3 \oplus \R^2 .\end{displaymath}

Second, Barton and Sudbery show that as vector spaces,

\begin{displaymath}\Der (\h _3(\K)) \iso \Tri (\K) \oplus \K^3 . \end{displaymath}

This follows in a case-by-case way from equation (3), but they give a unified proof that covers all cases. Third, they show that as vector spaces,

\begin{displaymath}\Tri (\K) \iso \Der (\K) \oplus \Im (\K)^2 . \end{displaymath}

Now starting with Tits' definition of the magic square, applying the first two facts, regrouping terms, and applying the third fact, we obtain Barton and Sudbery's version of the magic square:
% latex2html id marker 1713\begin{array}{lcl}
\M (\K,\K') ...
...i (\K) \oplus \Tri (\K') \oplus (\K \tensor \K')^3

In the next three sections we use all these different versions of the magic square to give lots of octonionic descriptions of $e_6$, $e_7$ and $\e _8$. To save space, we usually omit the formulas for the Lie bracket in these descriptions. However, the patient reader can reconstruct these with the help of Barton and Sudbery's paper, which is packed with useful formulas.

As we continue our tour through the exceptional Lie algebras, we shall make contact with Adams' work [1] constructing $\f _4,\e _6,\e _7,$ and $\e _8$ by means of spinors and rotation group Lie algebras:

% latex2html id marker 1714\begin{array}{lcl}
\f _4 &\iso...
...S_{12}^+ \\ \e _8 &\iso & \so (16) \oplus S_{16}^+

as vector spaces. Note that the numbers 9, 10, 12 and 16 are 8 more than the dimensions of $\R,\C,\H$ and $\O$. As usual, this is no coincidence! In terms of the octonions, Bott periodicity implies that

\begin{displaymath}S_{n+8} \iso S_n \tensor \O^2 .\end{displaymath}

This gives the following description of spinors in dimensions $\le 16$:

$S_1 = \R$ $S_9 = \O^2$
$S_2 = \C$ $S_{10} = (\C \tensor \O)^2$
$S_3 = \H$ $S_{11} = (\H \tensor \O)^2$
$S_4^\pm = \H$ $S_{12}^\pm = (\H \tensor \O)^2$
$S_5 = \H^2$ $S_{13} = (\H^2 \tensor \O)^2$
$S_6 = \C^4$ $S_{14} = (\C^4 \tensor \O)^2$
$S_7 = \O$ $S_{15} = (\O \tensor \O)^2$
$S_8^\pm = \O$ $S_{16}^\pm = (\O \tensor \O)^2$

Table 6 -- Spinor Representations Revisited

Since spinors in dimensions 1,2,4 and 8 are isomorphic to the division algebras $\R,\C,\H$ and $\O$, spinors in dimensions 8 higher are isomorphic to the `planes' $\O^2, (\C \tensor \O)^2, (\H \tensor \O)^2$ and $(\O \tensor \O)^2$ — and are thus closely linked to $\f _4$, $e_6$, $e_7$ and $\e _8$, thanks to the magic square.

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Next: E6 Up: Exceptional Lie Algebras Previous: F4

© 2001 John Baez