The rotations around axes appear to be commutative. If I perform a $180^{\circ}$ rotation around the $x$-axis and then another rotation around the $y$-axis, I obtain:
Rotation around the $x$-axis: $$x,\, y,\, z \Rightarrow x,\, -y,\, -z$$
Rotation around the $y$-axis: $$x,\, -y,\, -z \Rightarrow -x,\, -y,\, z$$
However, if I reverse the order and rotate $180^{\circ}$ around the $y$-axis first and then around the $x$-axis, I still get the same result:
Rotation around the $y$-axis: $$x,\, y,\, z \Rightarrow -x,\, y,\, -z$$
Rotation around the $x$-axis: $$-x,\, y,\, -z \Rightarrow -x,\, -y,\, z$$
As you can see, the composition of rotations is commutative. In quaternion algebra, quaternion multiplication is defined as the composition of rotations. But based on what we have seen above, it is not necessary for this multiplication to be non-commutative, as the composition of rotations in practice is not.
However, I have read that in order to maintain the consistent algebraic structure of quaternions, Hamilton formulated:
$i^{2} = j^{2} = k^{2} = i j k = -1$
Where $i j k = -1$ precisely because multiplication should not be commutative. It is not clear to me why the non-commutativity of quaternion multiplication arises from this definition ($i j k = -1$), nor why this non-commutativity is necessary despite not being observed in practice. Is there something that I am missing or have not understood well? What led Hamilton to understand that $i j k = -1$?
Best Answer
The reason why $ijk=−1$ and not $ijk=1$ is a consequence of the non-commutativity of quaternion multiplication. If we were to assume $ijk=1$, then this would imply that quaternion multiplication is commutative, which is not the case. the multiplication of $i, j, k$ is defined as: $ij=k$, $jk=i$, $ki=j$, $ji=−k$, $kj=−i$, $ik=−j$. From these multiplication rules, we can see that $ijk=i(jk)=i(i)=−1$.