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Given a quaternion of the form,

$$q= a + bi + cj + dk$$

Which is the norm of $q$?

(1) $\sqrt{a^2+b^2+c^2+d^2}$

(2) $a^2+b^2+c^2+d^2$

This page from MathWorks says (1) but another page says (2). Wikipedia says (1). My lecture slides say:

$$|q|^2=a^2+b^2+c^2+d^2$$

but it is not clear whether $|q|$ is the norm or $|q|^2$ is the norm. It is not clear whether $||^2$ is just the L2 norm notation or it is the power of 2 notation.

I am trying to figure this out because I am trying to normalize and compute inverses of quaternions. Both of these operations depend on the norm.

takfuruya
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    Conventions vary. But which one of these does normalise $q$? How do you express the inverse in terms of each? You can easily work this out yourself. – ancient mathematician Apr 04 '19 at 06:37
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    You wouldn't use $|q|^2$ as a norm, for if $c\in\mathbb{R}$ we would have $|cq|^2=|c|^2|q|^2$ were $|c|$ is just absolute value. We want to have absolute scalability, so $|q|^2$ refers to the norm squared. – Melody Apr 04 '19 at 06:38
  • Agree with @ancientmathematician. Conventions vary. In pure algebra (ignoring all real analysis), the norm of $q$ would be the determinant of left multiplication by $q$ which comes out as $N(q)=(a^2+b^2+c^2+d^2)^2$. A degree four polynomial on the coefficients reflecting the fact that the quaternions form a 4-dimensional space. But, when studying division algebras, this would be replaced with the reduced norm $nr(q)=a^2+b^2+c^2+d^2$. Of course, if your context dictates that the norm should have something to do with the length of a vector, then you would take the square root one more time. – Jyrki Lahtonen Apr 04 '19 at 06:49
  • From the point of view calculating inverses I would use the reduced norm, for then we have $$q^{-1}=\frac{\overline{q}}{nr(q)}.$$ If you want to scale them to unit quaternions, then you normalize as $q\mapsto q/\sqrt{a^2+b^2+c^2+d^2}$ (handy when using quaternions to represent rotations as in 3D-graphics programming). But, really, the choice is very much context dependent. Pick one, and stick to it. – Jyrki Lahtonen Apr 04 '19 at 06:50
  • The reason for this variety is that norm is one of the more overloaded terms in math. Here there's a bit of a clash between various uses. As long as the readers are aware of the different possibilities, and thus know to check the local convention, we'll do fine :-) – Jyrki Lahtonen Apr 04 '19 at 06:58

1 Answers1

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From a practitioner point of view, you have two cases:

  1. Your quaternion is not normalized, i.e., $Q Q^* \neq 1$, then the rotation is:

$$v' = Q v Q^{-1}$$

Where $Q^{-1} = Q^* / \|Q\|^2$ and $\|Q\|^2 = a^2 + b^2 + c^2 + d^2$.

As you see the nornalization term must be squared since Q appears twice in the quaternion ("sandwitch") product.

  1. Your quaternion is unit quaternion, i.e., $Q Q^* = 1$, then the rotation is just:

$$v' = Q v Q^*$$

Now, a unit quaternion is just $Q / \|Q\|$ where $\|Q\| = \sqrt { a^2 + b^2 + c^2 + d^2}$.

Mauricio Cele Lopez Belon
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