Decomposition of $S^7=\operatorname{Spin}(7)/G_2$

Question

$\DeclareMathOperator\Spin{Spin}$The seven-sphere can be written as the reductive space $S^7=\Spin(7)/G_2$. Has the decomposition $\Spin(7)=G_2\times S^7$ been calculated somewhere; maybe in terms of Cayley numbers?

Answer 1

This is too long for a comment. (But I have now added an answer to the original question.)

As André and Ryan have hinted at in the comments, what you have is a principal fibre bundle $G_2 \to \operatorname{Spin}(7) \stackrel{\pi}{\to} S^7$. Fibre bundles are locally trivial, so of course around every point of the 7-sphere there is a neighbourhood $U$ so that $\pi^{-1}U \cong U \times G_2$. The diffeomorphism depends on a choice of section: a way to assign to every point in $U$ a unique element of $\operatorname{Spin}(7)$.

What I know how to describe in terms of octonions (i.e., Cayley numbers) is the above fibre bundle, and maybe this helps you.

Let $\mathbb{O}$ denote the Cayley numbers. They form an 8-dimensional vector space with basis $e_1,\dots,e_8$, where $e_1,\dots,e_7$ are imaginary units and $e_8 = 1$. Let $L_i$ denote left multiplication by the imaginary unit $e_i$, for $i=1,\dots, 7$. The $L_i$ are endomorphisms of $\mathbb{O}$ which obey the Clifford relations $$ L_i \circ L_j + L_j \circ L_i = \begin{cases} - 2\operatorname{id} & i=j \cr 0 & i\neq j \end{cases}$$ whence they define a linear representation of the Clifford algebra $C\ell(7)$ on $\mathbb{O}$. (You could also use right multiplication and this would give the other inequivalent Clifford module of $C\ell(7)$. Both Clifford modules turn out to be equivalent under the spin group.)

The spin group $\operatorname{Spin}(7)$ naturally lives inside $C\ell(7)$, whence we also have a linear representation of $\operatorname{Spin}(7)$ on $\mathbb{O}$. This is nothing but the spinor representation $\operatorname{Spin}(7) \to \operatorname{SO}(\mathbb{O})$. The orbit of $1 \in \mathbb{O}$ under $\operatorname{Spin}(7)$ is the sphere of unit octonions, which we can identify with $S^7$. The stabiliser of $1$ in $\operatorname{Spin}(7)$ is precisely a $G_2$ subgroup. This then gives the principal bundle $$G_2 \to \operatorname{Spin}(7) \stackrel{\pi}{\to} S^7$$ with $\pi(g) = g \cdot 1$.

Added

Here is how to construct a local section of $S^7 \to \operatorname{Spin}(7)$ near $1 \in S^7 \subset\mathbb{O}$ using octonions.

Under the identification of $S^7$ with the unit-norm octonions, the tangent space $T_1 S^7$ is identified with the imaginary octonions $\operatorname{Im}(\mathbb{O})$. Let $\varphi$ denote the $G_2$-invariant 3-form on $\operatorname{Im}(\mathbb{O})$. The image of the map $\operatorname{Im}(\mathbb{O}) \to \Lambda^2\operatorname{Im}(\mathbb{O})$ which sends $\xi$ to $\iota_\xi \varphi$ is a $G_2$-invariant complement of $\mathfrak{g}_2 \subset \mathfrak{spin}(7)$, and we can exponentiate them to elements of $\operatorname{Spin}(7)$ near the identity. Acting on $1 \in S^7$ we coordinatise a neighbourhood of $1$.

More explicitly, in case this is useful, the components $\varphi_{ijk}$ of $\varphi$ relative to the basis $e_i$ above are given by octonion multiplication as follows: $$ e_i e_j = -\delta_{ij} 1 + \sum_{k=1}^7 \varphi_{ijk} e_k $$ Then the point of $S^7$ with coordinates $\xi = \sum_{i=1}^7 \xi_i e_i \in \operatorname{Im}(\mathbb{O})$, is given by $$ \exp\left(\tfrac12 \sum_{i,j,k=1}^7\xi_i \varphi_{ijk} L_j\circ L_k\right) \cdot 1 $$ where the $L_j$ endomorphisms were defined above.

Answer 2

The bundle $G_2 \to Spin(7) \to S^7$ is nontrivial, indeed. The formula above $$g_{S^7} = \exp\left\{ \frac 12 \xi_j \phi_{jkl} \Gamma_k \Gamma_l \right \} \qquad \qquad (1)$$ ($\Gamma_j$ are the matrices $8 \times 8$ satisfying the Clifford algebra) is a local section, but not a global section of this bundle. It can be proven in quite explicit "physical" way. I do so in 2102.07415 [hep-th]. The digest of the proof is the following.

(1) is a section because

(i) Seven generators $T_j = \phi_{jkl} \Gamma_k \Gamma_l$ are orthogonal to 14 generators of $G_2$ and

(ii) (1) parameterize the whole $S^7$: the point $\xi_j = 0$ is its north pole with $g= \mathbb{1}$ and, if $\|\xi_j\| = \pi$, one obtains $g = - \mathbb{1}$, and this is the south pole.

However, (1) is not a global section by the following not quite trivial reason:

A subgroup $G_2 \subset Spin(7)$ are the matrices that do not transform a particular 8-component real spinor $\psi_0$. Consider the action of the matrices (1) on this spinor. One can show that $g \psi_0 = \cos(3\alpha) \psi_0 + \sin(3\alpha) \psi_1$, where $\alpha = \|\xi_j\|$ and $\psi_1$ is some other spinor. If $\alpha = 0, \pm 2\pi/3$, $\psi_0$ is left intact. Thus, many elements of (1) belong to one and the same fiber of the bundle and hence (1) cannot be a section.