How can I prove triviality of tangent bundle of $S^3$ using "Quaternion group"? My professor said "You can check triviality of tangent bundle of  $S^3$ using Quaternion group. In fact $S^3 \subset \mathbb{R}^4 = \mathbb{H} = \{a+bi+cj+dk|ij = k, jk = i, ji = -k, ik = -j,i^2 = j^2 = k^2 = -1\}$ and multiplication in $\mathbb{H}$ preserve being tangent, this implies the fact of $TS^3$ is trivial". I can't find out how to use this structure to prove the triviality of $TS^3$. 
 A: Consider the quaternion multiplication $\mu : \mathbb H \times \mathbb H \to \mathbb H,\mu(x,y) = xy$, and let $1,i,j,k$ be the standard basis of $\mathbb H = \mathbb R^4$.
For each $x \in S^3$ the vectors $x 1, x  i, x  j, x  k$ are linearly independent in the $\mathbb R$-vector space $\mathbb H$: For $a_r \in \mathbb R$ we have $$0 = a_1(x1) + a_2(xi) + a_3(xj) +a_4(xk) = (a_1x)1 + (a_2x)i + (a_3x) +(a_4x)k \\= (x1)a_1 +(xa_2)i + (xa_3)j + (xa_4)k = x(a_1 1) + x(a_2i) + x(a_3j) + x(a_4k) \\= x(a_1 1 + a_2 i + a_3 j + a_4 k)$$ iff $$0 = a_1 1 + a_2 i + a_3 j + a_4 k .$$ Note that real numbers commute with all quaternions.
Write $x = a + bi +cj + dk$ with $a,b,c,d  \in \mathbb R$. Then $xi = -b + ai + dj -ck$, thus $\langle x, xi \rangle = -ab +ab + cd -cd = 0$. Similarly $\langle x, xj \rangle = 0$ and $\langle x, xk \rangle = 0$. Therefore $xi, xj, xk$ span the orthogonal complement of $x$ which is $T_xS^3$. The map
$$\phi : S^3 \times \mathbb H \to S^3 \times \mathbb H, \phi(x,y) = (x,xy)$$
has the property $\phi(\{x\} \times \text{span}(i,j,k)) = \{x\} \times T_x S^3$ and thus gives a trivialization of $TS^3$.
