Talk:Dual quaternion

Limits to anti-commutativity

It seems, when multiplication associates, that ε = e can only anti-commute with two of the i,j,k. For example, assume ei + ie = 0 and ej + je = 0. Then

ek = e(ij) = (ei)j = −(ie)j = −i(ej) = ije = ke.

The section refering to the skew theory algebra is marred by this problem. A separate article about a (revamped) appropriate algebra for the skew transformation is in order. Should associativity be dropped? Maybe only two non-commutativities are necessary. I realize that the dual quaternions are common fare for the mechanical engineers in some places. Nevertheless all mathematics must be consistent and there is a reason that the dual quaternion literature has not made it into standard mathematical texts. Expectations are too high.Rgdboer (talk) 22:56, 13 August 2008 (UTC)

Checking the Martin Baker reference shows there is no claim of associativity; non-associativity is acknowleged with an example. For the time being this alternative dual quaternion idea is under consideration.Rgdboer (talk) 23:40, 22 September 2008 (UTC)

Revisions to this article

I would like to make some changes to this article to make it easier to read and perhaps more useful. Prof McCarthy (talk) 03:45, 30 June 2011 (UTC)

I commented out the section on dual quaternions for rigid motions because is claims that in this case the dual unit ε anti-commutes with the quaternion units i, j, k. This is not correct. It propagates a confusion introduced by Baker. The correct description is found in O. Bottema and B. Roth, Theoretical Kinematics, North Holland Publ. Co., 1979 Prof McCarthy (talk) 17:47, 30 June 2011 (UTC)

I changed the notation to eliminate some of the unnecessary math formatting so it reads easier, and otherwise keep what was already there. However, I found some errors that needed to be corrected. Prof McCarthy (talk) 21:16, 30 June 2011 (UTC)

I added the section on dual quaternions and spatial displacements, including formulas for the composition of spatial displacements. Prof McCarthy (talk) 00:11, 2 July 2011 (UTC)

I am not sure it is proper, but I increased the quality and importance ratings a little. Prof McCarthy (talk) 22:39, 2 July 2011 (UTC)

Conjugate of a quaternion

I hope the issue of anti-commutation of the dual unit ε is resolved. There may be eight dimensional algebras constructed from ε, i, j, and k, in which anti-commutivity exists and associativity is lost, however, it is not correct to call them dual quaternions. It is possible to show using Clifford Algebra theory that the construction of dual quaternions yields a dual unit multivector that commutes with the quaternion unit multi vectors---they are constructed as the even Clifford algebra on four dimensional space with a degenerate (3,0) metric. The reason that I bring this up is that I would like to remove the specialized conjugates that have been introduced and focus on the usual and useful dual quaternion conjugate. Prof McCarthy (talk) 14:40, 12 August 2011 (UTC)

Alternate Usage of word "conjugate"

I have encountered the term "conjugate" used for the quantity A* - ε B* (quaternions are conjugated and dual part is also negated), in papers on kinematics. This initially caused me hours of confusion. I do not know if this is considered "correct", or if there is a better term. This seems to be so widespread that I think the reader should be informed/warned about this usage, with an explanation of why people use it, and a reference. If there is a better, standard, term for this quantity, I think it should be mentioned, since it seems to be widely used.

Also, the "conjugate" section seems to define two kinds of conjugates (in addition to the one I mention above), without naming them. Do these have names, such as "quaternion conjugate" and "dual conjuguate"? Gsspradlin (talk) 20:28, 24 August 2013 (UTC)

The section about "Dual quaternions and 4×4 homogeneous transforms" indicates this as the way to apply a dual transformation to a vector in $R^{3}$

${\vec {v}}'={\hat {q}}\cdot {\hat {v}}\cdot {\hat {q}}^{-1}$

This is not correct. It should be using the other conjugate, the one defined as

${\hat {A}}^{*}=(A^{*},-B^{*})=A^{*}-\epsilon B^{*}.\!$

which is not the inverse.

An easy example is if ${\vec {v}}=(0,0,0)$, then ${\hat {v}}=1$ and ${\vec {v}}'={\hat {q}}\cdot {\hat {v}}\cdot {\hat {q}}^{-1}={\hat {q}}\cdot 1\cdot {\hat {q}}^{-1}={\hat {q}}\cdot {\hat {q}}^{-1}=1$, while the correct answer should be the translation embedded in the double quaternion. Audetto (talk) 20:22, 27 April 2017 (UTC)

Norm or Seminorm

In the article, a norm for dual-quaternions is defined. However, I was wondering whether this is a norm or a seminorm, because dual quaternions with a zero non-dual part would have the norm zero no matter what their dual part is, right? If I'm correct it might be better to write seminorm instead of norm to avoid confusion. — Preceding unsigned comment added by 2A00:1398:200:200:2677:3FF:FE1F:6278 (talk) 11:10, 19 February 2014 (UTC)

No, it is not zero, but a (non-invertible) dual number (there are three “conjugations” in this algebra, see the section above). The article does not claim this “norm” makes a normed vector space. Dual quaternions are the quaternion algebra over the dual number ring, and this “norm” reflects such structure, as well as convenient properties of the square root operation on dual numbers. Incnis Mrsi (talk) 18:10, 19 February 2014 (UTC)

Is rotation or displacement applied first?

The sentence

The displacement can be written as $dr^{*}=0+{\frac {\Delta x}{2}}i+{\frac {\Delta y}{2}}j+{\frac {\Delta z}{2}}k$.

is not meaningful unless the article specifies whether its convention for the dual Quaternion's rigid transform is T*R or R*T.--2A02:8084:2562:5E80:9ED0:F6E5:50EF:B820 (talk) 11:39, 21 October 2018 (UTC)