Is $\textrm{Spin}(8)$ a direct product of $\textrm{Spin}(7)$ and $S^7$?
I met this statement in the literature, but without a reference. If it is true, where is it explicitly written?
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2$\begingroup$ $S^7$ is not a group, what do you mean by "direct product"? $\endgroup$– YCorCommented Jul 5, 2021 at 14:54
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2$\begingroup$ Direct product as of topological spaces. Meaning that the bundle Spin(8) -> S^7 with the layer Spin(7) is trivial. Is it? I know a simple explicit formula for a local section of this bundle, but it is not global. Sorry for the sloppiness in terminology, I am a physicist ;) $\endgroup$– Andrei SmilgaCommented Jul 5, 2021 at 14:59
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7$\begingroup$ Yes. This is true because $S^7$ is parallelizable. $\endgroup$– Robert BryantCommented Jul 5, 2021 at 15:07
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17$\begingroup$ Specifically: Regard $S^7$ as the unit octonions. We have a map $L:S^7\to\mathrm{SO}(8)$ given by letting $L(u)$ satisfy $L(u)v = uv$ for $v$ an octonion. For $g\in\mathrm{SO}(8)$, we have a smooth factorization $g = L\bigl(g(1)\bigr)\cdot \bigl( L\bigl(g(1)\bigr)^{-1} g\bigr)$, and the second factor clearly belongs to $\mathrm{SO}(7)$. This establishes the diffeomorphism $\mathrm{SO}(8)\simeq S^7\times\mathrm{SO}(7)$. Now pass to the double cover on both sides. $\endgroup$– Robert BryantCommented Jul 5, 2021 at 15:15
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4$\begingroup$ Direct product of topological spaces, isn't it just "product"? "direct" refers to other group "semidirect" structures on the set-wise product, in the group context. $\endgroup$– YCorCommented Jul 5, 2021 at 15:34
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2 Answers
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As I suspected, the statement that the bundle $\mathrm{SO}(8)\to S^7$ is a product bundle, i.e., that $$ \mathrm{SO}(8)\simeq S^7\times\mathrm{SO}(7)\tag1 $$ as bundles over $S^7$ is in N. Steenrod's The topology of fibre bundles (as part of Theorem 8.6).
Note that Steenrod uses the notation $R_n$ for what is more commonly notated $\mathrm{SO}(n)$, these days. (The '$R$' is for 'rotation group'. He uses $O_n$ for the orthogonal group.) His proof, based on the octonions (which he calls 'the Cayley numbers'), is exactly what I outlined in my comment above. He does not give a reference to an earlier statement of the result, but refers to L. E. Dickson's famous Linear Algebra for the properties of the Cayley numbers that are used in the proof.
He does not discuss the spin groups, but, obviously, the equivalence of bundles $$ \mathrm{Spin}(8)\simeq S^7\times\mathrm{Spin}(7)\tag2 $$ follows from (1) by passing to the respective double covers.
answered Jul 10, 2021 at 14:56
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$\begingroup$ Thank you, Robert. I've posted a comment this morning, but it was not correct and I deleted it. I've understood the construction with octonion multiplication. However, octonions do not have a matrix representation - their multiplication is not associative. I tried before to realize this bundle and find the section using the 8-dimensional matrix language. And this allows one to construct only local sections, but not a global one. $\endgroup$ Commented Jul 12, 2021 at 9:15
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$\begingroup$ @AndreiSmilga: A section of the bundle $\mathrm{SO}(8)\to S^7$ is given by $L:S^7\to \mathrm{SO}(8)$, where $L(u)$ is left-multiplication by $u\in S^7\subset\mathbb{O}$. Giving an explicit section of $\mathrm{Spin}(8)\to S^7$ depends on having an explicit representation of $\mathrm{Spin}(8)$. One such representation is as the group of matrices in $\mathrm{SO}(8)^3$ generated by $C(u) = \bigl(L(u),L(u){\circ}R(u),R(u)\bigr)$ for $u\in S^7$. Then $C:S^7\to\mathrm{Spin}(8)$ provides an explicit section of the spin bundle defined by $\mathrm{Spin}(8)\ni g = (g_1,g_2,g_3)\mapsto g_1(1)\in S^7$. $\endgroup$ Commented Jul 12, 2021 at 9:36
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Robert Bryant answered this question by interpreting an element of SO(8) as an octonion multiplication. But I've understood now (after exchanging messages with Robert) that there is a more direct construction. Let $\Gamma_{j=1,\ldots,7}$ be antisymmetric purely imaginary $8 \times 8$ matrices that satisfy the Clifford algebra $$ \Gamma_j \Gamma_k + \Gamma_k \Gamma_j = 2\delta_{jk} \,.$$ Then 28 mutually orthogonal Hermitian generators of SO(8) are $\Gamma_j$ and $\Sigma_{jk} = i \Gamma_j \Gamma_k$. $\Sigma_{jk}$ are also the generators of Spin(7), and it is a bit more convenient to consider instead of $SO(8)$ the so-called Semispin(8) group (see e.g. hep-th/9906059), which is isomorphic to $SO(8)$, but rotates spinors rather than vectors. Then any element of Semispin(8) can be represented as
$$ g_8 \ =\ \exp(i\alpha_j \Gamma_j) \exp(\beta_{kl} \Gamma_k \Gamma_l) \ =\ g_{S^7} g_7, $$ where $g_7$ is an element of Spin(7) and $g_{S^7}$ is an embedding of $S^7$ into Semispin(8) if restricting $\|\alpha_j\| \leq \pi$.
This already looks as a direct product searched for and the only nuissance is that the representation above is not unique: $\mathbb{1}$ of Semispin(8) may be represented either as $\mathbb{1} \times \mathbb{1}$ or as $(-\mathbb{1})\times (-\mathbb{1})$. This nuissance disappears if one goes from Semispin(8) to its double cover Spin(8). The latter is represented by two distinct matrices $\exp(i\alpha_j \Gamma_j) $ and $g_7$, and this representation is unique.