I was in trouble when I was trying to complete problem 8-7 in Lee's Introduction to Smooth Manifolds in Chapter 8. The problem is
The algebra of octonions(also called Cayley numbers) is the $8-$dimensional real vector space $\mathbb{O}=\mathbb{H}\times\mathbb{H}$ with the following bilinear product: $$ (p,q)(r,s)=(pr-sq^*,p^*+rq),\quad p,q,r,s\in \mathbb{H} $$ Show that $\mathbb{O}$ is a noncommutative, nonassociative algebra over $\mathbb{R}$, and prove that there exists a smooth global frame on $\mathbb{S}^7$ by imitating as much of Problem 8-6 as you can.
Here is my idea. Note that $\mathbb{H}=\mathbb{C}\times\mathbb{C}$, and $\mathbb{O}=\mathbb{H}\times\mathbb{H}$. Then we can think of $\mathbb{O}$ as a product manifold of $\mathbb{R}\times 8$. Then for any $Q\in \mathbb{O}$, we have the coordinate of $Q=(x^1,\cdots,x^8)$ where $x^i\in \mathbb{R}$.
Consider the next function $$ A:\mathbb{O}^*\to \mathbb{R} $$ s.t. for all $Q=(p,q)\in \mathbb{O}^*=\{(x^1,\cdots,x^8)\in\mathbb{O}:(x^1)^2+\cdots+(x^8)^2\ne 0\}$, $A(Q)=(x^1)^2+\cdots+(x^8)^2$. It's clear that $A$ is smooth and $$dA_Q=2(x^1,\cdots,x^8)$$ then $rank(dA_Q)=1$ i.e. $A$ is submersion. Note that $\mathbb{S}^7=A^{-1}(1)$ which mean that $\mathbb{S}^7$ is embedded submersion of $\mathbb{O}^*$ and $A$ is defining map of $\mathbb{S}^7$. Then $$T_Q\mathbb{S}^7=\ker dA_Q=\{P\in \mathbb{O}:Q\cdot P=0\}$$ where the $\cdot$ mean the dot product operator in $\mathbb{R}^8$. Let $$1=(1,0),\quad I=(i,0),\quad J=(j,0),\quad K=(k,0),\quad L=(0,1),\quad M=(0,i),\quad N=(0,j),\quad O=(0,k)$$ where $1, i,j,k\in \mathbb{H}$. Consider the next vector fields $$ X_1|_Q=QI,\quad X_2|_Q=QJ,\quad X_3|_Q=QK,\quad X_4|_Q=QL $$ $$ X_5|_Q=QM,\quad X_6|_Q=QN,\quad X_7|_Q=QO $$ The above vector fields are linearly independent since octonionic multiplication satisfies the distribute law. To modify the proof of problem 8-6, we need to prove that $X_1,\cdots, X_7$ tangent to $\mathbb{S}^7$.
For $X_1|_Q$, it's easy to compute that $$X_1|_Q=(-x^2,x^1,-x^4,x^3,-x^6,x^5,-x^8,x^7)$$ then $Q\cdot X_1|_Q=0$ i.e. $X_1|_Q\in T_Q\mathbb{S}^7$. So $X_1$ tangent to $\mathbb{S}^7$. But for $X_2|_Q$, we compute next $$X_2|_Q=(-x^3,x^4,x^1,-x^2,x^7,x^8,x^5,x^6)$$ and $$Q\cdot X_2|_Q=x^5x^7+x^6x^8+x^7x^5+x^8x^6$$ we can not confirm $Q\cdot X_2|_Q$ is vanish.
Where is the wrong in my computation? Could anybody help me? Thanks.
Let $J_1,\ldots,J_7$ be the right multiplication by $(i,0), (j,0), (k,0), (0,1), (0,i), (0,j)$ and $(0,k)$ respectively. Let $X=(p,q) \in \Bbb O$. Then $$ J_1X = (p,q)\cdot(i,0) = (ip, -qi), $$ and therefore $$ J_1^2 X = \big((p,q)\cdot(i,0)\big)\cdot(i,0) = (ip, -qi)\cdot(i,0) = (i^2p,qi^2) = -(p,q) = -X. $$ Similarly, $$ J_2X = (p,q)\cdot(j,0) = (jp,-qj), $$ and $$ J_2^2X = \big((p,q)\cdot(j,0)\big)\cdot(j,0) = (jp,-qj)\cdot(j,0) = (j^2p,qj^2) = -(p,q) = -X. $$ The exact same computations show that $\forall \ell \in \{1,\ldots,7\}$, $J_{\ell}^2 = - \mathrm{Id}_{\Bbb O}$.
An important property of the octonion norm is that for any $X$ and $Y$, we have $\|X \cdot Y\| = \|X\|\|Y\|$, where $\|X\| = \sqrt{X^*X}$ is the usual euclidean norm of $\Bbb O \simeq \Bbb R^8$ in the canonical coordinates. This implies that $\|J_{\ell} X\| = \|X\|$ for all $\ell \in\{1,\ldots,7\}$ and $X\in \Bbb O$. By standard euclidean theory, for $\ell\in\{1,\ldots,7\}$, $J_{\ell}$ is a linear isometry of $\Bbb O$: for all $X$ and $Y$, one has $$ \langle J_{\ell} X, J_{\ell}Y\rangle = \langle X,Y \rangle. $$ Applying this equality to $Y = J_{\ell} X$, we obtain that $J_{\ell}X \perp X$ for all $X$. In fact, if $\ell_1 \neq \ell_2 \in \{1,\ldots,7\}$, one can show that $J_{\ell_1} J_{\ell_2} = \pm J_{\ell_3}$ for some $\ell_3 \in \{1,\ldots,7\}$, and thus, we have $$ \langle J_{\ell_1}X,J_{\ell_2}X \rangle = \langle X, -J_{\ell_1}J_{\ell_2}X\rangle = \langle X, \mp J_{\ell_3}X \rangle = 0, $$ and $J_{\ell_1}X \perp J_{\ell_2}X$. It follows that if $X \neq 0$, then $\{X,J_1X,\ldots,J_7X\}$ is an orthogonal basis of $\Bbb O$. If moreover $\|X\|=1$, then it is an orthonormal basis.
Now, to go back to the original question, notice that if $X \in \Bbb S^7 =\{ Y \in \Bbb O \mid \|Y\|=1\}$, then $T_X\Bbb S^7 = \{X\}^{\perp}$. Since $\{X,J_1X,\ldots,J_7X\}$ is an orthonormal basis of $\Bbb O$, then $\{J_1X,\ldots,J_7X\}$ is an orthonormal basis of $\{X\}^{\perp} = T_X\Bbb S^7$. It follows that $T\Bbb S^7$ is trivial: a global trivialization is given by $$ (X,Y) \in T\Bbb S^7 \mapsto \big(X, (\langle Y,J_1X \rangle,\ldots,\langle Y,J_7X\rangle) \big) \in \Bbb S^7 \times \Bbb R^7. $$