Editing Polar decomposition (section)

==Quaternion polar decomposition==
The polar decomposition of [[quaternion]]s <math>\mathbb{H}</math> with [[orthonormal basis]] quaternions <math>1, \hat\imath, \hat\jmath, \hat k</math> depends on the unit 2-dimensional sphere <math>\hat r \in \{x \hat \imath + y \hat \jmath + z \hat k \in \mathbb{H} \setminus \mathbb{R} : x^2 + y^2 + z^2 = 1\}</math> of [[quaternion#Square roots of −1|square roots of minus one]], known as ''[[right versor]]s''. Given any <math>\hat r</math>  on this sphere and an angle {{nobr|{{math|−{{pi}} < ''a'' ≤ {{pi}}}},}} the [[versor]] <math>e^{a \hat r} = \cos a + \hat r \sin a</math> is on the unit [[3-sphere]] of <math>\mathbb{H}.</math> For {{nobr|{{math|''a'' {{=}} 0}}}} and {{nobr|{{math|''a'' {{=}} {{pi}}}},}} the versor is 1 or −1, regardless of which {{mvar|r}} is selected. The [[norm (mathematics)|norm]] {{mvar|t}} of a quaternion {{mvar|q}} is the [[Euclidean distance]] from the origin to {{mvar|q}}. When a quaternion is not just a real number, then there is a ''unique'' polar decomposition:
<math display="block">
 q = t \exp(a \hat r).
</math>
Here {{mvar|r}}, {{mvar|a}}, {{mvar|t}} are all uniquely determined such that {{mvar|r}} is a right versor {{nobr|({{math|''r''<sup>2</sup> {{=}} –1}}),}} {{mvar|a}} satisfies {{nobr|{{math|0 < ''a'' < {{pi}}}},}} and {{nobr|{{math|''t'' > 0}}.}}