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4.1 $\G _2$

In 1914, Élie Cartan noted that the smallest of the exceptional Lie groups, $\G _2$, is the automorphism group of the octonions [12]. Its Lie algebra $\g _2$ is therefore $\Der (\O)$, the derivations of the octonions. Let us take these facts as definitions of $\G _2$ and its Lie algebra, and work out some of the consequences.

What are automorphisms of the octonions like? One way to analyze this involves subalgebras of the octonions. Any octonion $e_1$ whose square is $-1$ generates a subalgebra of $\O$ isomorphic to $\C$. If we then pick any octonion $e_2$ with square equal to $-1$ that anticommutes with $e_1$, the elements $e_1,e_2$ generate a subalgebra isomorphic to $\H$. Finally, if we pick any octonion $e_3$ with square equal to $-1$ that anticommutes with $e_1,e_2,$ and $e_1e_2$, the elements $e_1,e_2,e_3$ generate all of $\O$. We call such a triple of octonions a basic triple. Given any basic triple, there exists a unique way to define $e_4, \dots , e_7$ so that the whole multiplication table in Section 2 holds. In fact, this follows from the remarks on the Cayley-Dickson construction at the end of Section 2.3.

It follows that given any two basic triples, there exists a unique automorphism of $\O$ mapping the first to the second. Conversely, it is obvious that any automorphism maps basic triples to basic triples. This gives a nice description of the group $\G _2$, as follows.

Fix a basic triple $e_1,e_2,e_3$. There is a unique automorphism of the octonions mapping this to any other basic triple, say $e'_1,e'_2,e'_3$. Now our description of basic triples so far has been purely algebraic, but we can also view them more geometrically as follows: a basic triple is any triple of unit imaginary octonions (i.e.octonions of norm one) such that each is orthogonal to the algebra generated by the other two. This means that our automorphism can map $e_1$ to any point $e'_1$ on the 6-sphere of unit imaginary octonions, then map $e_2$ to any point $e'_2$ on the 5-sphere of unit imaginary octonions that are orthogonal to $e'_1$, and then map $e_3$ to any point $e'_3$ on the 3-sphere of unit imaginary octonions that are orthogonal to $e'_1,e'_2$ and $e'_1 e'_2$. It follows that

\begin{displaymath}\dim \G _2 = \dim S^6 + \dim S^5 + \dim S^3 = 14 .\end{displaymath}

The triality description of the octonions in Section 2.4 gives another picture of $\G _2$. First, recall that $\Spin (8)$ is the automorphism group of the triality $t_8 \maps V_8 \times S^+_8 \times
S^-_8 \to \R$. To construct the octonions from this triality we need to pick unit vectors in any two of these spaces, so we can think of $\G _2$ as the subgroup of $\Spin (8)$ fixing unit vectors in $V_8$ and $S^+_8$. The subgroup of $\Spin (8)$ fixing a unit vector in $V_8$ is just $\Spin (7)$, and when we restrict the representation $S^+_8$ to $\Spin (7)$, we get the spinor representation $S_7$. Thus $\G _2$ is the subgroup of $\Spin (7)$ fixing a unit vector in $S_7$. Since $\Spin (7)$ acts transitively on the unit sphere $S^7$ in this spinor representation [1], we have

\begin{displaymath}\Spin (7)/\G _2 = S^7 . \end{displaymath}

It follows that

\begin{displaymath}\dim \G _2 = \dim (\Spin (7)) - \dim S^7 = 21 - 7 = 14 .\end{displaymath}

This picture becomes a bit more vivid if we remember that after choosing unit vectors in $V_8$ and $S^+_8$, we can identify both these representations with the octonions, with both unit vectors corresponding to $1 \in \O$. Thus what we are really saying is this: the subgroup of $\Spin (8)$ that fixes $1$ in the vector representation on $\O$ is $\Spin (7)$; the subgroup that fixes $1$ in both the vector and right-handed spinor representations is $\G _2$. This subgroup also fixes the element $1$ in the left-handed spinor representation of $\Spin (8)$ on $\O$.

Now, using the vector representation of $\Spin (8)$ on $\O$, we get homomorphisms

\begin{displaymath}\G _2 \hookrightarrow \Spin (8) \to \SO (\O) \end{displaymath}

where $\SO (\O) \iso \SO (8)$ is the rotation group of the octonions, viewed as a real vector space with the inner product $\langle x,y\rangle
= \Re (x^\ast y)$. The map from $\Spin (8)$ to $\SO (\O)$ is two-to-one, but when we restrict it to $\G _2$ we get a one-to-one map

\begin{displaymath}\G _2 \hookrightarrow \SO (\O) . \end{displaymath}

At the Lie algebra level, this construction gives an inclusion

\begin{displaymath}\g _2 \hookrightarrow \so (\O) \end{displaymath}

where $\so (\O) \iso \so (8)$ is the Lie algebra of skew-adjoint real-linear transformations of the octonions. Since $\g _2$ is 14-dimensional and $\so (\O)$ is 28-dimensional, it is nice to see exactly where the extra 14 dimensions come from. In fact, they come from two copies of $\Im (\O)$, the 7-dimensional space consisting of all imaginary octonions.

More precisely, we have:

\begin{displaymath}\so (\O) = \g _2 \oplus L_{\Im (\O)} \oplus R_{\Im (\O)}

(a direct sum of vector spaces, not Lie algebras), where $L_{\Im (\O)}$ is the space of linear transformations of $\O$ given by left multiplication by imaginary octonions and $R_{\Im (\O)}$ is the space of linear transformations of $\O$ given by right multiplication by imaginary octonions [77]. To see this, we first check that left multiplication by an imaginary octonion is skew-adjoint. Using polarization, it suffices to note that

\begin{displaymath}\langle x,ax \rangle = \Re (x^*(ax)) = \Re ((x^*a)x) =
\Re ((a^*x)^*x) = -\Re ((ax)^* x) = -\langle ax,x \rangle \end{displaymath}

for all $a \in \Im (\O)$ and $x \in \O$. Note that this calculation only uses the alternative law, not the associative law, since $x, x^\ast$ and $a$ all lie in the algebra generated by the two elements $a$ and $\Im (x)$. A similar argument shows that right multiplication by an imaginary octonion is skew-adjoint. It follows that $\g _2$, $L_{\Im (\O)}$ and $R_{\Im (\O)}$ all naturally lie in $\so (8)$. Next, with some easy calculations we can check that

% latex2html id marker 1661
L_{\Im (\O)} \cap R_{\Im (\O)} = \{0\} \end{displaymath}


% latex2html id marker 1662
\g _2 \cap (L_{\Im (\O)} + R_{\Im (\O)}) = \{0\} .\end{displaymath}

Using the fact that the dimensions of the 3 pieces adds to 28, equation (4.1) follows.

We have seen that $\G _2$ sits inside $\SO (8)$, but we can do better: it actually sits inside $\SO (7)$. After all, every automorphism of the octonions fixes the identity, and thus preserves the space of octonions orthogonal to the identity. This space is just $\Im (\O)$, so we have an inclusion

\begin{displaymath}\G _2 \hookrightarrow \SO (\Im (\O)) \end{displaymath}

where $\SO (\Im (\O)) \iso \SO (7)$ is the rotation group of the imaginary octonions. At the Lie algebra level this gives an inclusion

\begin{displaymath}\g _2 \hookrightarrow \so (\Im (\O)) . \end{displaymath}

Since $\g _2$ is 14-dimensional and $\so (\Im (\O))$ is 21-dimensional, it is nice to see where the 7 extra dimensions come from. Examining equation (4.1), it is clear that these extra dimensions must come from the transformations in $ L_{\Im (\O)} \oplus R_{\Im (\O)}$ that annihilate the identity $1 \in \O$. The transformations that do this are precisely those of the form

\begin{displaymath}\ad _a = L_a - R_a \end{displaymath}

for $a \in \Im (\O)$. We thus have

\begin{displaymath}\so (\Im (\O)) \iso \Der (\O) \oplus \ad _{\Im (\O)}

where $\ad _{\Im (\O)}$ is the 7-dimensional space of such transformations.

We may summarize some of the above results as follows:

Theorem 4.   The compact real form of the Lie algebra $\g _2$ is given by
\begin{displaymath}\g _2 = \Der (\O) \subset \so (\Im (\O)) \subset \so (\O) \end{displaymath}

and we have
% latex2html id marker 1668\begin{array}{lcl}
\so (\Im (\...
... \Der (\O) \oplus L_{\Im (\O)} \oplus R_{\Im (\O)}

where the Lie brackets in $\so (\Im (\O))$ and $\so (\O)$ are built from natural bilinear operations on the summands.

As we have seen, $\G _2$ has a 7-dimensional representation $\Im (\O)$. In fact, this is the smallest nontrivial representation of $\G _2$, so it is worth understanding in as many ways as possible. The space $\Im (\O)$ has at least three natural structures that are preserved by the transformations in $\G _2$. These give more descriptions of $\G _2$ as a symmetry group, and they also shed some new light on the octonions. The first two of the structures we describe are analogous to more familiar ones that exist on the 3-dimensional space of imaginary quaternions, $\Im (\H)$. The third makes explicit use of the nonassociativity of the octonions.

First, both $\Im (\H)$ and $\Im (\O)$ are closed under the commutator. In the case of $\Im (\H)$, the commutator divided by 2 is the familiar cross product in 3 dimensions:

\begin{displaymath}a \times b = {1\over 2}[a,b] .\end{displaymath}

We can make the same definition for $\Im (\O)$, obtaining a 7-dimensional analog of the cross product. For both $\Im (\H)$ and $\Im (\O)$ the cross product is bilinear and anticommutative. The cross product makes $\Im (\H)$ into a Lie algebra, but not $\Im (\O)$. For both $\Im (\H)$ and $\Im (\O)$, the cross product has two nice geometrical properties. On the one hand, its norm is determined by the formula

% latex2html id marker 1670
\Vert a \times b\Vert^2 + \langle a,b\rangle^2 = \Vert a\Vert^2 \, \Vert b\Vert^2 , \end{displaymath}

or equivalently,

% latex2html id marker 1671
\Vert a \times b\Vert = \vert {\sin \theta}\vert \, \Vert a\Vert \, \Vert b\Vert , \end{displaymath}

where $\theta$ is the angle between $a$ and $b$. On the other hand, $a
\times b$ is orthogonal to $a$ and $b$. Both these properties follow from easy calculations. For $\Im (\H)$, these two properties are enough to determine $x \times y$ up to a sign. For $\Im (\O)$ they are not -- but they become so if we also use the fact that $x \times y$ lies inside a copy of $\Im (\H)$ that contains $x$ and $y$.

It is clear that the group of all real-linear transformations of $\Im (\H)$ preserving the cross product is just $\SO (3)$, which is also the automorphism group of the quaternions. One can similarly show that the group of real-linear transformations of $\Im (\O)$ preserving the cross product is exactly $\G _2$. To see this, start by noting that any element of $\G _2$ preserves the cross product on $\Im (\O)$, since the cross product is defined using octonion multiplication. To show that conversely any transformation preserving the cross product lies in $\G _2$, it suffices to express the multiplication of imaginary octonions in terms of their cross product. Using this identity:

\begin{displaymath}a \times b = ab + \langle a, b\rangle , \end{displaymath}

it actually suffices to express the inner product on $\Im (\O)$ in terms of the cross product. Here the following identity does the job:

% latex2html id marker 1673
\langle a,b\rangle =
-\textstyle{1\over 6}\tr (a \times (b \times \cdot\,))

where the right-hand side refers to the trace of the map

% latex2html id marker 1674
a \times (b \times \cdot\,) \maps \Im (\O) \to \Im (\O) .\end{displaymath}

Second, both $\Im (\H)$ and $\Im (\O)$ are equipped with a natural 3-form, or in other words, an alternating trilinear functional. This is given by

\begin{displaymath}\phi(x,y,z) = \langle x,yz \rangle .\end{displaymath}

In the case of $\Im (\H)$ this is just the usual volume form, and the group of real-linear transformations preserving it is $\SL (3, \R)$. In the case of $\Im (\O)$, the real-linear transformations preserving $\phi$ are exactly those in the group $\G _2$. A proof of this by Robert Bryant can be found in Reese Harvey's book [50]. The 3-form $\phi$ is important in the theory of `Joyce manifolds' [56], which are 7-dimensional Riemannian manifolds with holonomy group equal to $\G _2$.

Third, both $\Im (\H)$ and $\Im (\O)$ are closed under the associator. For $\Im (\H)$ this is boring, since the associator vanishes. On the other hand, for $\Im (\O)$ the associator is interesting. In fact, it follows from results of Harvey [50] that a real-linear transformation $T \maps \Im (\O) \to \Im (\O)$ preserves the associator if and only if $\pm T$ lies in $\G _2$. Thus the symmetry group of the associator is slightly bigger than $\G _2$: it is $\G _2 \times \Z_2$.

Now we must make an embarrassing admission: these three structures on $\Im (\O)$ are all almost the same thing! Starting with the cross product

\begin{displaymath}\times \maps \Im (\O) \times \Im (\O) \to \Im (\O) \end{displaymath}

we can recover the usual inner product on $\Im (\O)$ by equation (4.1). This inner product allows us to dualize the cross product and obtain a trilinear functional, which is, up to a constant, just the 3-form

\begin{displaymath}\phi \maps \Im (\O) \times \Im (\O) \times \Im (\O) \to \R .\end{displaymath}

The cross product also determines an orientation on $\Im (\O)$ (we leave this as an exercise for the reader). This allows us to take the Hodge dual of $\phi$, obtaining a 4-form $\psi$, i.e.alternating tetralinear functional

\begin{displaymath}\psi \maps \Im (\O) \times \Im (\O) \times \Im (\O) \times \Im (\O)
\to \R .\end{displaymath}

Dualizing yet again, this gives a ternary operation which, up to a constant multiple, is the associator:

\begin{displaymath}[\cdot, \cdot, \cdot]\maps \Im (\O) \times \Im (\O) \times \Im (\O)
\to \Im (\O) .\end{displaymath}

We conclude this section with a handy explicit formula for all the derivations of the octonions. In an associative algebra $A$, any element $x$ defines an inner derivation $\ad _x \maps A \to A$ by

\begin{displaymath}\ad _x (a) = [x,a] \end{displaymath}

where the bracket denotes the commutator $xa - ax$. In a nonassociative algebra, this formula usually does not define a derivation. However, if $A$ is alternative, any pair of elements $x,y \in A$ define a derivation $D_{x,y} \maps A \to A$ by

\begin{displaymath}D_{x,y} a = [[x,y],a] - 3[x,y,a]

where $[a,b,x]$ denotes the associator $(ab)x - a(bx)$. Moreover, when $A$ is a normed division algebra, every derivation is a linear combination of derivations of this form. Unfortunately, proving these facts seems to require some brutal calculations [77].

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© 2001 John Baez