Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Why is it that for quaternions, $u*v = \mathrm{cross}(u,v)-\mathrm{dot}(u,v)$?

I wonder for what reason they are equal to each others.

share|cite|improve this question
@Jason: Here $u$ and $v$ are pure quaternions, which we identify with $\mathbb R^3$ and the scalar part of a quaternion we identify with scalars. – Eric O. Korman Feb 20 '11 at 4:14
up vote 6 down vote accepted

Assume that $u = bi + cj + dk$ and $v = xi + yj + zk$ are imaginary quaternions (no real part). Then a straightforward computation using the identities $i^2 = j^2 = k^2 = ijk = -1$ gives \[ uv = -(bx + cy + dz) + (cz - dy)i + (dx - bz)j + (by - cx)k = - \langle u,v \rangle + u \times v \] with the usual identification $\mathbb{R}^{3} = \operatorname{Im}\mathbb{H}$.

share|cite|improve this answer
In view of sweetser's answer below let me add that there is no difficulty in getting his "more general formula" from yours by recalling that the real quaternions are in the center of $\mathbb{H}$ (in fact they are equal to the center of $\mathbb{H}$). Namely, for real $a',b'$ we have $$(a' + u)(b'+v) = a'b' + a'v + b'u + uv = (a'b'-\langle u,v \rangle) + (a'v + b'u + u \times v),$$ as desired, where the two parentheses in the rightmost formula indicate real and imaginary parts, respectively. – t.b. Jun 16 '11 at 11:38

The quaternions written are called "pure quaternions", meaning the scalar value is zero. Let me write a quaternion as a scalar and a 3-vector, where the 3-vector has an arrow. Then: $(0, \vec{u})(0, \vec{v}) = (-u \cdot v, \vec{u} \times \vec{v})$ This is not very general because for a different inertial observer, the scalar will no longer be zero. In that case: $(a', \vec{u'})(b', \vec{v'}) = (a' b'- \vec{u'} \cdot \vec{v'}, a' \vec{v'} + b' \vec{u'} + \vec{u'} \times \vec{v'})$ If the 3-vectors represent a position in space, then the scalars are time. If the 3-vectors are 3-momentums, then the scalars are energy.

share|cite|improve this answer
There were some typos and inconsistencies in notation in your original answer. I tried to correct that and hope I didn't mess up your intended meaning. – t.b. Jun 16 '11 at 11:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.