Talk:Versor

Versors in Clifford Algebra
According to the classic book Clifford Algebra to Geometric Calculus (CAGC), a versor is a product of vectors, and those vectors need not be invertible. Some texts, such as Geometric Algebra for Computer Science (GACS) require a versor to be invertible. However, the non-invertible concept is more general, and more useful, since many results hold for these 'k-products' of vectors, rather than just for invertible ones. A case in point is the formula for the contraction over a geometric product of vectors. Unit-versors are exactly the pinors, even unit-versors are exactly the spinors, and rotors are exactly the special spinors (GAGC has an erroneous claim here, which GACS propagates). I was hoping to get the non-invertible definition through as the most useful definition. What do you think?

--Kaba3 (talk) 01:41, 8 April 2013 (UTC)

Added detail
I heard this mentioned in a mechanics class today in reference to a unit vector. I've added what I've found online, but it needs an expert's eyes. Still, it looks like it's a real thing &mdash;BenFrantzDale 02:51, September 9, 2005 (UTC)

This objection is a work in progress
I think I remember when the article on versors was practically empty! I want to complement you on the outstanding job you are doing. I like the way you have noted hamilton's terminology!

This is just a draft objecting, I want to come back later and refine it.

Nit picking a little on finer points of classical theory
In classical notation as your outstanding article points out there is something called the tensor of a versor.

If the versor is written in quadranomial form it is just

$$ Tq = \sqrt{w^2 + x^2 + y^2 + z^2} \,$$

The above operation always equals one for a versor.

What hamilton called the common norm of a versor is could be written as

common norm = Nq = qKq

In other words the common norm is quaternion times its conjugate. The 'common norm' and the 'tensor' of a versor were two completely different operations. The common norm of a quaternion is equal to the square of the tensor of the same quaternion.

The common norm of a versor is also equal to one.

One approach I considered, but don't like very well was that since there is more than one 'norm' if you called the tensor of a quaternion the Euclidian-Hamilton norm, that would distinguish it from the 'common norm', as well as the corresponding norm operations in $$R^4$$

Along these lines terms used by both Hamilton in his classical theory and in other branches of math could be made less ambigious by affixing the word Hamilton in front of them.

For example of this nomenclature consider the section that you link to the Euclidian norm which you might consider writing  a formula identical to that of the 'hamiltonian tensor' of a quaternion.

By affixing the term hamiltonian to the word tensor I mean the an abreviation of what hamilton called a tensor.

Your treatment of modern quaternions
Your problem is confounded by the fact that they keep changing the names of the operations on quaternions on the main page. Thus leading to some major ambiguity.

While the classical quaternion point of view is very well defined, and can be well documented, and its definitions of every word are very precise, the same is not true about modern thinking on the subject.

Sometimes they are calling the classical tensor, the modulus, and sometimes the absolute value.

Eucledian norm is an operation on real vector spaces and the vector space associated with H is three complex numbers C3. So it is an operation on a different space, and therefore not identical to any operation on H

On the other hand the word tensor has been taken over these days by a whole branch of math. So using the word tensor, for the euclidian norm would confuse your readers, in your introductory paragraph.

Hobojaks (talk) 03:51, 21 July 2008 (UTC)
 * You are so right about tensor now being a part of multilinear algebra and the term is hardly suitable for modern quaternion study. The article mentions norm as appropriate for those developing the metric space concept.

The whole idea of rethinking the quaternion as some type of normed vector space is a point of view that needs to be documented in terms of the versor concept.

However some of these view are very much non-classical views.

I can provide you with some good references to material by writers who held this view, of which one of the important early ones is Gibbs. Gibbs book vector analysis devotes considerable time to his view of what a versor is, but it is really really different from hamiltons. What makes it interesting is that it is written by a writer well versed in classical quaternion thinking.

In my view the most interesting candidate corresponding to the idea of a norm in a normed vector spaces when it comes to quaternions is a version of the lorentz invariant. In other words take the product of each individual element of quaternion with itself, keeping the i j and k. You get

$$w^2 -x^2 - y^2 - z^2$$

Which as Rgdboer points out is the scalar part of the square of a quaternion.

The thing is that the whole idea of a normed vector space to my way of thinking is pretty much quaternion negitive nomenclature.

A classical quaternion has only one product, and since it is a quaternion and not a scaler, it is excluded from the definition of a normed vector space.

It seems to me that the motivation behind normed vector spaces is to incorperate the idea of an angle into them, but every quaternion already has a built in angle characteristic, so elevating one characteristic of a quaternion above the others and bestowing the name norm on it so that you can define an angle, or for that matter define what you mean by distance seems a bit contrived.

In my way of thinking, and of course this is just my opinion and probably could not be included on wikipedia, a quaternion has a number of quantities that could be thought of as norms.

(1)The distance idea to my way of thinking is the tensor of the vector of the quaternion. This is the true eucledian norm.

(2)The lorenz invariant, as I am pleased to discover, in other words the four dimensional distance in space time is the scaler of the product of a quaternion with itself.

(3)The tensor of a quaternion, looks a lot like the norm of a real four space, so I can understand why you guys coming from the non-classical point of view that advocates remaking a quaternion into a normed vector space like it so much.

Once you start to extend the classical quaternion idea, depending on how far you go, you can't really be sure any more that the common norm is still equal to the square of the tensor.

Hamilton proved this was true for a classical quaternion, however I was thinking that if you wanted to avoid quaternion negative nomenclature and felt that the term tensor was to obscure for most of your readers, may I suggest simply in the classical section of your article."Square root of the common norm".

In another approach I was thinking of including the sentence:  The square root of the common norm of a quaternion is a positive definite tensor of rank zero. This characteristic of a quaternion written $$Tq$$ has been found so useful that in contexts where no other kinds of tensors are being discussed it is simply called "the tensor". The term Hamiltonian tensor is preferable when discussing this characteristic in a mixed context to distinguish it from other uses of the word tensor. A versor rotates any vector in its plane through its angle, but preserves its length. A positive definite zero ordered tensor preforms an act of tension either streaching or shrinking a vector. If you pick two arbitrary vectors, you can multiply the first by a quaternion consisting of the proper choice tensor and versor and the result will be the second.

These two operations are commutative. In symbols.

qβ = Tq.Vqβ = Vq.Tqβ

Hobojaks (talk) 21:20, 25 July 2008 (UTC)

Woops, the text below somehow got moved around, sorry for that.

Similarly the real part of the square of an ordinary complex number is the modulus of the corresponding split-complex number.Rgdboer (talk) 21:51, 24 July 2008 (UTC)

Would the term 'evil tensor' be less confusing?
The term evil tensor for the useful quantity we are discussing would not only be consistent with quaternion negative nomenclature, but helpful for remembering things because it starts with the letter e, helping people to remember that it corresponds with the idea of what is called in quaternion negative nomenclature the euclidean norm of real space. If the first basis vector of quaternion space is thought of as an 'imaginary' vector, then you might explain to your non technical readers that tensors are terrifying numbers. It would help them to rememberTq. This might also be reflective of a commonly held view of students of multi-linear algebra when first encountering the tensor of a quaternion concept.

It would unify the concept basis vectors and square roots for the non-technical readers since it is consistent with calling the square roots of minus one purely imaginary numbers, and with the naming conventions that generate expressions like real number.

It would give an introduction to the article a little less technical tone, than using the convention of naming concepts after the people who invented them, as in Hamiltonian Tensor Norm, or Euclidean norm.

I find 'the norm' very confusing because this could also mean the Lorentz invariant or scaler of a quaternion. Perhaps affixing some term with a negative connotation like satanic scaler would be in keeping with the traditional naming conventions of multi-linear algebra? A more neutral term starting with the letter for Sq might be special realitivistic norm. —Preceding unsigned comment added by Hobojaks (talk • contribs) 12:10, 27 July 2008 (UTC)

Some ideas I have been having on the subject.
A great need exists to explain the basic properties of a quaternion in a way people with out a phd in math can understand.

Below is an attempt that I started working on, but I am not really sure that I like it very much or if it would be allowed on wikipedia.

The main difficulty is finding words that are at the same time non-technical and non-ambigious.

Also this text is incomplete, because I quit working on it for today at the point were it took up the subject of the angle of a versor. I notice that no one has commented on the technique. —Preceding unsigned comment added by Hobojaks (talk • contribs) 02:10, 28 July 2008 (UTC)

Hi I cut out the non-technical introduction that I was working on here and paisted it into my talk page. I don't think it could ever be acceptable for wikipedia, because it contains some made up words. In other words unless I could find a book were someone was already calling tensors terrible numbers, I don't think it would be appropriate.

One of these days if I ever write my own book on quaternions I think it is a good idea, but not for here and I don't think that there is any way to fix it.

Here is an important conclusion that I have reached. There are a lot of words that quaternion theory have in common, and it makes writing about linear algebra and quaternions in the same article hard because words like vector and tensor have meaning in each.

My conclusion is that when possible it may be better to have a section about one subject or the other, so as to not make the language so confusing. An entire article written using one nomenclature, for example a whole article on the vectors of linear algebra or on the vectors of quaternion calculus can use the word vector with out confusion.

Hobojaks (talk) 17:40, 5 August 2008 (UTC)

SO(3)?
Is a versor just another word for a member of SO(3)? —Ben FrantzDale (talk) 02:42, 4 December 2008 (UTC)
 * Short answer: Yes. Caveat: versor is an antiquated term from a time before continuous groups were classified. The versor concept represents a seed-idea of Lie theory: group elements can be described as evolving out of the identity (1). Thus the close association with a great circle arc. The term drags behind it the Hamilton-algebra with its controversial allusion to four dimensions. One who considers using the term versor might do well to clear with the audience the admissablility of the meta-physical 4-space.Rgdboer (talk) 00:49, 7 December 2008 (UTC)
 * No. SO(3) is not a 3-sphere. You both probably need to read the Spin group article. Incnis Mrsi (talk) 19:34, 23 January 2014 (UTC)

Geometric algebra?
In geometric algebra, it appears that "versor" has a broader meaning than just rotation. One paper I'm reading says versors are multivectors that "have the property that they are a geometric product of vectors." Is that right? Should this page be expanded to reflect this notion? —Ben FrantzDale (talk) 21:44, 24 January 2009 (UTC)
 * This article uses versor in the way it has historically been used. The Geometric Algebra people have appropriated the term for their own use apparently in your reading about multivectors, the material of multilinear algebra. Take a look at Algebra of physical space, one of the Geometric Algebra pages. In the references you see work by W.E. Baylis, one of the main proponents. There you have a link to his 2002 paper in Canadian Journal of Physics, in the ArXiv 0406158. He says on page 3, the "algebraic approach advocated here for introductory courses ... is isomorphic (equivalent in structure) to complex quaternions,...". In WP we call complex quaternions by the name biquaternions. When you go to page 16 explaining "Boosts as spacetime rotations" you see that he writes the equation for a hyperbolic versor as found in this article. In general, the Geometric Algebra or Clifford Algebra proponents write linear algebra and multilinear algebra, but the formalism doesn't automatically generate meaning. The references to Baylis show superficiality in Geometric Algebra.Rgdboer (talk) 20:46, 1 September 2009 (UTC)
 * To shorten the hunt, the reference and link are now here:


 * Baylis, William (2002). Relativity in Introductory Physics, Can. J. Phys. 82 (11), 853--873 (2004). (ArXiv:physics/0406158)
 * Interestingly, Baylis cites biquaternion work by Conway and Silberstein, but doesn't weave their efforts into the physical story leading up to GA.Rgdboer (talk) 21:35, 2 September 2009 (UTC)

Spherical Triangle?
There seems to be something seriously misleading going on in the "Presentation on the sphere" section! I'm talking about the part I have copied below. It makes it look like the arcs on the sphere form a spherical triangle, and the text sounds that way too.


 * Multiplication of quaternions of norm one corresponds to the "addition" of great circle arcs on the 2-sphere. Hamilton writes


 * $$q = \beta: \alpha = OB:OA \ \ $$ and
 * $$q' = \gamma:\beta = OC:OB \,$$


 * imply


 * $$ q' q = \gamma:\alpha = OC:OA . \, $$

The problem with this is that the edge CA can't really be a great circle arc! The product q'q is the rotation that you get by first rotating by q and then rotating by q'. Doing that rotation to point A will bring it to point C, but there are many rotations that have that effect and only one of them takes A along the great circle it shares with C. In fact, unless A, B, and C are all on the same great circle, AC is sure to be a small circle arc. Just imagine what happens to point B when rotated by q'q: rotating it by q and then by q' is very different from the result of the rotation that would take A to C along the great circle.

I'm having a hard time figuring out what this section is actually trying to say. Lilwik (talk) 04:09, 1 May 2012 (UTC)


 * Your argument that many of the potential rotations that move A to C are not on great arcs does not imply that the one formed by the product q′q is not the special one that does so on a great circle. Roger Penrose in his book The Road to Reality §11.4 provides a very similar construction for a spherical triangle, and explicitly mentions that all the arcs are on great circles.  He does however point out a catch: the triangle's sides are exactly half the arc length of the rotations concerned. I am not familiar enough with the detail to make an authoritative comment.  Take as a simple example: i, j and k represent rotations of $\pi$ (not π/2) radians respectively around the x, y and z axes.  The composition of the i and then j rotations is ji = −k, which you can check by rotating a book 180° for each, though this corresponds somewhat counterintuitively to an equilateral right spherical triangle addition.  A rotation of 90° around the x-axis would be represented by q=(1+i)/√2, with a corresponding triangle side of π/4.  This halving seems to be hidden by the treatment here due to it dealing only with the spherical triangle representing the summation and not with the rotations represented.  I see this as an omission in the article.  — Quondum☏ 10:50, 1 May 2012 (UTC)


 * I didn't prove that q' q is not a great arc from A to C, but it's a fact. I find a spherical triangle with 180° edges hard to picture, but it's very easy to picture one with 90° edges and that's pretty close to the triangle in the image, so just use that. Take any two of the edges and make the rotation of their arcs into quaternions q and q' which will both be 90° rotations and the axis of q is perpendicular to the axis of q. Now notice that q' q is a quaternion with a 120° rotation, and rightly so because it needs to rotate all points to their correct positions, not just A. But there's no way that a 120° rotation represents the third side of the triangle; A is being rotated along a small circle arc instead of taking the direct route like it seems to show in the image. — Lilwik (talk) 12:49, 1 May 2012 (UTC)


 * You seem to have missed my point. Rotation of a solid object in space by an angle of θ around the x-axis is represented by a quaternion q = exp(−iθ/2), and the corresponding rotation of any vector r is calculated as qrq*, where q* is the conjugate of q (negate the vector part only).  This will rotate any vector (or an object described by a number of individual vectors) by the correct amount around the x axis.  In contrast, your construction rotates only a vector oriented to the corner of the spherical triangle correctly, and the great-circle "spherical vector sum" only works for that vector, but you seem to be wanting to interpret it as a rotation for any vector.  To take your example of two 90° rotations, say around the x-axis and then around the y-axis, we'd put q=exp(−iπ/4)=(1−i)/√2, q′=exp(−jπ/4)=(1−j)/√2, giving us q′q=(1−j)(1−i)/2=(1−i−j−k)/2=exp(−((i+j+k)/√3)π/3), which corresponds to a rotation of 2π/3=120° about an axis equiangular from the three +x, +y and +z semi-axes.  And as you've said, the original vector in the actual rotation does not move on a great circle in the composed rotation, but the semi-angle vectors describing the corners of the spherical triangle picture of the rotations are joined by three great circle arcs of 45°, 45°, making a third side of 60°.  You must keep in mind that the triangle picture does not happen in the same space as the rotation being described; they're related by a halving of angles around the origin.
 * By way of contrast, if you take the 90° spherical triangle (like in the picture, with each corner being 90°), if you want to interpret that as representing a rotation of an object, the angles of rotation are double, thus giving 180° actual rotations being represented. Take a book, put it on the desk in front of you.  Rotate it 180° around the x-axis pointing to your right, leaving the book face down and its top towards you.  Next rotate it 180° around the y-axis parallel pointing away from you along the desk, leaving the book face up but top towards you.  This combination is equivalent to a 180° rotation around the z-axis upwards out of the desk.  This interpretation works exactly, for all angles and axes of rotation.  If you do not use the halving/doubling of angles, everything falls apart.  — Quondum☏ 14:28, 1 May 2012 (UTC)


 * Now I see your point. If I had been paying attention I would have noticed that this whole versor page is missing the θ/2 that is used when the quaternions are representing rotations. I just had an unshakable assumption that a quaternion represented one axis and one angle, and that it would always be the same angle no matter how the quaternion was being used. Now I see that you can use quaternions to calculate great triangles, but if you want to use the axis and angle θ of one of those quaternions for a rotation, you need to construct a new quaternion using w = cos(θ/2) as the scalar. And now that I see your point, I see that you were right that it is a shame that this issue isn't expressed more forcefully in this article than a mere failing to divide the angle by two. I just can't think how that should be done. — Lilwik (talk) 21:24, 1 May 2012 (UTC)
 * Another poor quaternion expert who never heard of the Spin group. Incnis Mrsi (talk) 08:34, 24 January 2014 (UTC)

Very Prominent Unbalanced Parenthesis
Why is there is a very bold unbalanced closing parenthesis in the first formula of this article? It bothers me if this is an example of someone's real notation for something. Please tell me it was a mistake and clean it up or enlighten me about why it could possibly make sense. — Preceding unsigned comment added by 209.66.90.5 (talk) 23:55, 10 May 2012 (UTC)


 * This is a standard notation, explained in Interval (mathematics). — Quondum☏ 07:25, 11 May 2012 (UTC)

Versor in geometry and physics
I created a separate article for the meanings of the term versor in geometry and physics. A disambiguation page already existed, and at the very beginning of the article we say this article is only about an "algebraic rotation representation". In geometry and physics a versor does not represent a rotation, but the direction of an axis or vector. Paolo.dL (talk) 10:58, 16 August 2013 (UTC)


 * I'm happy for the separation: I'm not a fan of putting multiple uses of a term into one article simply because they share a name: an article should be about a topic, not a name. I'm not convinced that "(physics)" is the appropriate disambiguating parenthesis for the new article – "(geometry)" might have been better – but that can be debated there. — Quondum 11:59, 16 August 2013 (UTC)

Introductory phrase
have no opinion on merits of a separate article versor (physics) mentioned above, but obscured the fact that versors as a rotational formalism do not need to belong to $H$. The comment “It is customary… to specify the context… For other meaning of versor, see disambiguation” is indeed hypocritical. There is no special article about hyperbolic versors. One article should explain evolution of the term from “a norm-1 element of $H$ expressed as blah-blah” in Hamilton’s jargon to “such thing $q$ that $q v q^{−1}$ in certain associative algebra specifies a rotation of $v$” in modern algebra. tried to achieve it in the lead. Paolo.dL discarded “an algebraic representation of rotations” and left only “in classical quaternion theory”. By the way, how abstract algebra is related to versors? They rely upon associative algebras (with some additional structure such as *-algebra or a quadratic norm form) over the field of scalars. An associative algebra is already an algebra of certain signature, hence we are out of the domain of abstract algebra. Such things as ideal (algebra) or simple (algebra) are abstract. Versors are not especially so. Incnis Mrsi (talk) 20:57, 22 January 2014 (UTC)
 * Note that Paolo.dL, after discarding “an algebraic representation of rotations”, invented his own
 * LoL: can a rotation (mathematics) exist around a non-fixed axis? See also criticism of other Paolo.dLisms in the section below. Incnis Mrsi (talk) 19:34, 23 January 2014 (UTC)

Description in lede as a "directed arc"
The sentence
 * A versor may be viewed as a directed arc of a circle with radius 1 connecting the tips of quaternions 1 and q, where a is the angle (in radians) subtended by the arc.

does not seem to make sense, for several reasons: Because the intent of the statement is so obscure that I cannot even repair it, and a versor has already been described as a rotation, I am simply removing it. —Quondum 02:31, 23 January 2014 (UTC)
 * Quaternions, especially those that have nonzero scalar part, do not have a natural interpretation as vectors, and hence the inference by the word "tips" does not work. If intended as a reference to a four dimensional vector space and four-dimensional rotations or arcs, if valid (which I doubt), this is considerably too obscure for the lede.
 * If they are considered to represent a 3-d rotation, the angle is 2a, not a.
 * If the arc is considered to represent the trajectory of a vector that is rotated, it is not unique, as the vector is not specified. The same quaternion would represent multiple arcs.


 * Two of the references, Hardy and Molenbroeck, interpret versors as arcs on the sphere. The versor has a vector part which is the axis of the great circle containing the arc. The length can be obtained from inverse cosine of the real part, or inverse sine of the norm of the vector part. However, since the arc is an equivalence class of such, there is some complexity that is better served in the section on the arc interpretation. Nevertheless, a versor is not a rotation of the 2-sphere until the adjoint map is applied.Rgdboer (talk) 02:59, 23 January 2014 (UTC)
 * You're going beyond my understanding with your "not ... until". Nevertheless, it does sound like I was not entirely out of line in removing the statement, which got somewhat morphed from a more interpretable form.  The arc concept (strictly speaking not only great circle arcs, and also not only of unit radius) in the form of an equivalence class can be developed in the body of the article if appropriate.  Since there is no 1-to-1 mapping between quaternions and geometric arcs, the connection seems too tenuous to put in the lede.  And indeed the path of the arc only becomes defined in terms of the exponential interpolation, which is not necessary to the fundamental concept of quaternions as versors, references notwithstanding, which suggests they would be using the concept for illustration, not as a mathematically equivalent concept. The interpretation as representing a rotation (adjoint map or not) seems appropriate though, as a primary interpretation of what quaternion versor may be used to represent.  So I think we're good. —Quondum 06:07, 23 January 2014 (UTC)


 * do you copy? How do you refer to rotations around an axis while the action on 3-vectors is not yet defined? already said once that there are no rotations of a quaternion (1 in our case) around an axis, because the space of quaternions is 4-dimensional! Incnis Mrsi (talk) 15:14, 23 January 2014 (UTC)
 * Indeed, look at the text:
 * It is a confusion over another confusion. First of all, angle of which rotation and where? If a (single) multiplication in $H$, then true. If a sandwiched product, then false: its angle is 2a.
 * Second, the motion of a point where? If in $H$, then see above. Otherwise, the same as which? r is a 3-vector. The arc lies in $S^{3}$. think, everybody (but, possibly, Paolo.dL) now understood the situation. If there will not be objections in next 10 hours, then  proceed with wiping the gibberish out. Incnis Mrsi (talk) 19:34, 23 January 2014 (UTC)

Directed arcs: continuation
A versor is a point on the 3-sphere, it is represented by an equivalence class of arcs on the 2-sphere, namely those with the same axis and the same arc length.Rgdboer (talk) 20:35, 23 January 2014 (UTC)
 * Rgdboer, see you understand the subject very well. Though, you miss one point: your “[families of] directed arcs on $S^{2}$” are barely more illustrative than “double cover of the rotation group”. It’s me who understand what are you speaking about, and it’s me who sees that you really understand what are you speaking about . An average reader wouldn’t feel the difference between your families of arcs (note not all of them have the radius 1) and Paolo’s “motion of a point rotating about r along a directed arc of a circle with radius 1”. Incnis Mrsi (talk) 06:30, 24 January 2014 (UTC)
 * did not understand myself the $S^{2}$ arc feature at the moment: see below. Incnis Mrsi (talk) 10:13, 26 January 2014 (UTC)
 * No, you don’t it well. Not the same arc length, but the same subtended angle. Try to think better what is a rotation of $S^{2}$. Incnis Mrsi (talk) 07:59, 24 January 2014 (UTC)
 * Rgdboer might assume great circle “arcs” again (in $S^{2}$ they are merely line segments, so this terminology is confusing). do not know what Hardy and Molenbroeck said about the formalism of directed segments on $S^{2}$ (it was already discussed at ) and its relation to the Spin group, but it can’t explain the double cover feature. First thought is: directed segments are great, they distinguish the rotation by $&alpha;$ in one direction and the rotation by $360° − &alpha;$ in the opposite direction! But actually this distinction is not preserved under composition in any meaningful way (hint: think about directed segments interpretation of the composition of a 90° rotation and the rotation by a variable $&alpha;$ in some perpendicular direction). It appears that user:Quondum was right from the beginning – entire $S^{2}$ segments stuff is irrelevant to versors. Indeed, are my “paths from 1 to $q$ in $S^{3}$” an original research? Incnis Mrsi (talk) 07:36, 25 January 2014 (UTC)


 * This final question is tricky to answer, but is interesting to try to understand. It is fairly straightforward to see that there is a 1-to-1 mapping from unit quaternions to S3, albeit not natural (there is no "best" point in S3 to map 1 to). I don't think that the concept of "path" is helpful in the sense of a trajectory since there are many paths that join a given pair of endpoints. It might be useful to illustrate the double cover aspect, because antipodal points in S3 correspond to two distinct versors that rotate all vectors identically (i.e. that have the same group action). This is topologically significant: S3 is a double cover of the set of rotations, the latter being topologically the projective space P3(R), and while the space of versors (topologically S3) is simply connected, the space of rotations (topologically P3(R)) is not.
 * On the directed arcs in S2, while I do not have access to the references already mentioned, Penrose's The Road to Reality §11.4 does give a construction (illustrated quite nicely) of great-circle arcs in S2, and in particular how they can be used to compose versors. Care with interpretation is needed. In particular, certain caveats apply:
 * The arcs have half the angle of the rotation that they describe, but have the same axis and orientation.
 * The arcs are directed great-circle arcs, but like with vector addition, they are not considered to have any particular starting point: they are freely rotated on the great circle to get the tail to coincide with the head of the previous arc. to which they are to be "added" (corresponding to the rotation with which its rotation is to be composed). This requires each arc to be rotated so that the correct end lies on the great circle of the other arc, before the construction will work.
 * A consequence of the previous point is that a sequence of "arc addition" operations requires the previous result to be rotated into the correct position to fit the next addition operation.
 * The final result is again essentially an "axis and angle" output: a directed arc of a given (half) angle on a great circle, but has no given start or end point as it is still free to be moved around its axis. It is the angle (and axis) that counts, remembering that this angle is half that of the axis–angle representation.
 * These arcs must not be confused with the paths of the points of vectors that they rotate.
 * Considering the caveats, this should be treated as an illustrative geometric construction. Properly presented, it may be considered to have a place in the body of the article, especially as its geometric nature is an aid to visualization. I think that it may be regarded as notable, especially if the references (Hardy and Molenbroeck) given by user:Rgdboer are actually giving the same interpretation. I did not notice this connection earlier in this thread. —Quondum 18:32, 25 January 2014 (UTC)
 * Certainly, arcs in $S^{2}$ do not qualify as a good rotation formalism.

spent considerable time attempting to understand what these directed line (great circle) segments in $S^{2}$ can do for $Spin(3)$ or $SO(3)$. concluded that they can represent the conjugacy, but not a simple composition of rotations (essentially due to arguments already presented by user:Lilwik above). my new picture that shows a conjugacy in $SO(3)$. It is an interesting observation that a simple geometric interpretation fails for the composition of geometric rotations (unfortunately haven’t a picture for a composition), but works for the group conjugacy; versors also act by conjugacy.


 * So, we hope that the action in $S^{3}$ can help, but it is not a trivial thing. For a non-identity versor it does not have poles (fixed points) because it is constructed as a Hopf fibration. Incnis Mrsi (talk) 20:31, 25 January 2014 (UTC)

Indeed, we need an equivalence relation that shifts beginning of the segment to any point of $S^{3}$. We wouldn’t be satisfied with your “free rotation on the great circle”: two great circles in the general position do not intersect. A versor $q_{1}$ rotates 1 along certain great circle. You always can start from certain $v_{0} : q_{1} v_{0} = 1$ and then say that for the product $q_{2} q_{1}$ of versors $v_{2} := q_{2} 1$ is the endpoint ( use this superfluous notation to emphasize that we have a group action). So, the product is represented by the shortest great circle route from $v_{0}$ to $v_{2}$. The problem is that this route does not pass, in general position, through 1 and you have to shift it transversally (via right quaternion multiplication, using associativity) to put its beginning point to 1 again. Incnis Mrsi (talk) 20:31, 25 January 2014 (UTC)

Yesterday was reluctant to think on Lilwik’s arguments thoroughly; that’s why  didn’t understand Quondum’s counter-arguments, and also stuffed my mind with operations in SO(3) and their geometric interpretation. Now was reluctant to read attentively Quondum’s text – because of this, inventing the same construction by my own mind (after Roger Penrose and after Quondum’s explication) was prescribed. What is now in the article is virtually my original research, but hope it doesn’t deviate much from the road of Sir Roger. evaluated the stuff is the article mistakenly: it wasn’t actually an “incompetent gibberish” ( am sorry, Paolo.dL). There was only one mistake: explaining the directed segments on $S^{2}$ as SO(3) rotations. It does not add anything to understanding. Sorry for this poor text: I have to sleep in this time. Incnis Mrsi (talk) 23:57, 25 January 2014 (UTC)
 * The arc-representation of versors goes back to Hamilton: see his Lectures. There is also the article rotations in 4-dimensional Euclidean space which takes a general view of the 3-sphere.Rgdboer (talk) 00:39, 26 January 2014 (UTC)

S3, Spin(3), and the group of unit quaternions: subtle differences
A comment to It is fairly straightforward to see that there is a 1-to-1 mapping from unit quaternions to S3, albeit not natural (there is no "best" point in S3 to map 1 to). No so simple, indeed. When you get $S^{2}$ as a Riemannian manifold, fix the 0 point there and the orientation, then you obtain the complex projective line up to U(1) symmetry only. You can multiply points to complex numbers, and you know their complex absolute values, but you do not know their exact values as $a + bi$. To establish an isomorphism you have to fix the 1 point somewhere halfway between 0 and ∞.

With versors we have an analogous (but not identical) fable: when you get $S^{3}$ as a Riemannian manifold, fix the 1 point in it and the orientation, then you have a Spin(3) structure. You instantly define the scalar part as the cosine of the distance from 1, hence have the plane (unit sphere) of unit vectors, and the arc structure described above, where the orientation specifies the direction of arcs, hence the order of multiplication of versors. But the entire picture is still SO(3)-invariant. Unlike the situation with $CP^{1}$ you can multiply $S^{3}$ points, but not a point to a unit quaternion (specified with 4 components), nor can you get/put quaternionic values. You have to fix a correctly oriented orthonormal basis i,j,k in the space of 3-vectors for an isomorphism between a geometric picture and numeric components. Incnis Mrsi (talk) 10:13, 26 January 2014 (UTC)


 * I was not intending to reference any structure other than topological and metric (i.e. the mapping to which I refer is an isometry and a diffeomorphism). We know that the multiplicative group of the unit quaternions is isomorphic to SU(2)=Spin(3) (lede of Special unitary group), but it is not clear how to me how to depict the multiplication on S3 even though we have a mapping, so your introduction of arcs in S3 does not help me understand how these relate to composing them (corresponding to quaternion multiplication or composition of Spin(3) group elements). —Quondum 03:11, 27 January 2014 (UTC)
 * Metric space and Riemannian manifold structures on a $n$-sphere imply each other. Each of them virtually describes an embedding of the $n$-sphere into Euclidean $(n + 1)$-space. But none of them specifies an orthonormal basis there. If you fix the 1 point on $S^{3}$, then you choose only one its element of four. Then you have the scalar part function and its zero set, the $S^{2}$ of unit 3-vectors. Then you can encode points of $S^{3}$, except 1 and −1, as "directed great arcs" $S^{2}$ (up to known equivalence) using geometric means: find a great circle on $S^{2}$ orthogonal to the versor (i.e. lying at π/2 distance), choose the direction according to the orientation structure, and its length should be equal to the distance of the versor from 1. Now you can multiply versors although you miss 3/4 of the basis. Then you can observe that multiplication of a versor $q$ and any right versor orthogonal to $q$ gives another right versor orthogonal to $q$: it just demonstrates that the great circle is multiplication-invariant. Incnis Mrsi (talk) 09:29, 27 January 2014 (UTC)

Illustration of a versor’s action on S3
attempted to make a picture that shows how one-sided multiplication action is related to directed great arcs on $S^{2}$ (because partially used a stereographic projection, it is shown as the (yellow) plane). How is it related to the discussion? Look how the action $v ↦ 1 + i + j + k⁄2v$ is arranged near the 3-vectors’ sphere. Inside the red roundabout it pulls elements down (towards negative scalar part). Outside the red roundabout it pulls elements up (towards positive scalar part). On the red circle itself (note the geometry in the current version is a crap) it effects a 60° rotation, conserving the entire circle. When we compose this left multiplication with the right multiplication to the inverse versor, then up/down movements cancel each other on the whole $S^{3}$, but the rotation will remain and become a 120° rotation.

The picture is unfinished (the region around −1 is almost empty, and a net of the octahedron of $±i, ±j, ±k$ is not visible yet), there are visualization glitches and some positions are incorrect, but do you have an timely advice about adding or altering something? Incnis Mrsi (talk) 22:49, 26 January 2014 (UTC)
 * Finished in about two days thereafter. Incnis Mrsi (talk) 15:21, 9 February 2014 (UTC)


 * You start with "directed great arcs on $S^{2}$". Do you mean "$S^{3}$"?  I have difficulty following the diagram at this point, and so have difficulty commenting. —Quondum 03:16, 27 January 2014 (UTC)
 * ’m sorry the current version is unclear about it. The great circle on $S^{2}$ is one of invariant lines (great circles) under the left multiplication action. The latter has many invariant lines (see above, and parts of several such lines are depicted: one with white arrows downwards, and yet three in different hues (blue, magenta, orange). But the red circle is the only invariant line that entirely lies on the $S^{2}$ of unit 3-vectors. Directed arcs on $S^{2}$ are a consequence of the left multiplication action, and vice versa (see above). Incnis Mrsi (talk) 09:29, 27 January 2014 (UTC)
 * I too have great difficulty following the diagram, though even that is too generous. It makes no sense to me at all, even at full size, and at the size it appears in articles it's impossible to see any part of it. I don't know if the problem's so much with the quality of it but with what I think it's trying to do. Trying to represent four dimensions in two is very problematical, and usually only works where the viewer has a clear understanding before they view the image, or where they can extrapolate from a familiar three dimensional image into four. Unless that can be done here – a clear analogy can be made with something in 3D that's familiar or easily represented – I don't think this is suitable for using in articles.
 * (I've boldly adjusted the size to make it clear what I mean: at 200px it's even less clear. I also moved it right as on the left it was hiding the indentation of messages and replies)-- JohnBlackburne wordsdeeds 13:36, 9 February 2014 (UTC)
 * $S^{3}$ has three dimensions and one point at infinity, not four dimensions. The picture shows all 24 unit Hurwitz quaternions (a.k.a. elements of the binary tetrahedral group), as well as the “plane” of unit 3-vectors. Which objects are represented poorly? What should be changed in the picture? If you want to understand the action, then you can do it with the current version. realized some details (such as the rhombic dodecahedron on the “plane”) while drawing it. Incnis Mrsi (talk) 15:21, 9 February 2014 (UTC)
 * $S^{3}$ is three dimensional but it can't be drawn in three Euclidian dimensions. Unit 3-vectors don't lie in a plane in three dimensions. The main problem is it's unclear how these 4D objects are represented as 3D objects. I don't see 24 arrows (I assume those are the quaternions). But if you were representing the binary tetrahedral group in 4D you'd have a 24-cell, but none of the representations of that look anything like this. If anything this looks like a set of rotations in 3D, not 4D. Again, mostly the problem I think is is representing anything four dimensional is very difficult. You write earlier you 'partially' use a stereographic projection but I don't know what that means - a projection is either stereographic or it's not. It's difficult for me to say what should be changed because I don't know what it's showing.-- JohnBlackburne wordsdeeds 16:36, 9 February 2014 (UTC)
 * Do you see 24 balls? These are Hurwitz quaternions. Sorry missed two interconnections (they would clutter the picture too) and another six arrow heads for straight coloured segments. Do you know such thing as “elliptic geometry”? Really do you not? FYI: a great 2-sphere on the 3-sphere is a plane. The projection is not exactly stereographic. It is stereographic on the “plane of vectors” and is extended upwards and downwards in a convenient way. Surely  could draw this in a purely stereographic projection, but the region of interest around $i + j + k⁄√3$ would become too small, whereas not especially interesting elements $1 − i − j − k⁄2$ and $−1 − i − j − k⁄2$ would protrude too far upwards and downwards respectively. Incnis Mrsi (talk) 17:40, 9 February 2014 (UTC)
 * Incnis, the tone is not helpful. Elucidation/interpretation of the representation and explanation is what is failing here; anything worth putting into in WP should be understandable to someone with enough grounding to be interested in the detail.  I can understand the $S^{3}$ is a three-dimensional manifold, and that a conformal map $S^{3} → E^{3}$ exists; looking at n-sphere, and that this would be a stereographic projection.  What you appear to be doing, as far as I can tell, is showing a (nearly stereographic) map of specific points in the unit 3-sphere (namely the Hurwitz quaternions), which are primarily relevant in that they give a reference in which all the vertices, angles etc. are uniform, since these naturally become distorted by the projection.  You then (and I'm guessing here), appear to be illustrating trajectories of specific points under one-sided multiplication by some specific quaternion.  Any one-sided quaternion multiplication corresponds to an isoclinic rotation in E4.  In short, it looks to me like a geometric illustration of isoclinic rotation by concentrating on the points at radius one and a mapping that discards no information, but it sheds no real light on either the quaternions (other than as specific vectors being rotated), not in how such rotations compose, and not on how quaternions relate to the rotations.  If I have misinterpreted this, I apologize.  However, if you are to get agreement, the description must be enhanced so that it is understandable on a first reading to most mathematicians without making guesses. —Quondum 19:00, 9 February 2014 (UTC)
 * Yes, it’s an isoclinic 60° rotation (or, in terms of elliptic geometry, a Clifford’s translation). Do you really not see a distinctive Hopf link structure? It would be difficult to demonstrate the second rotation, the right multiplication to the reciprocal (or conjugate) versor, without cluttering the picture too much. In words, the second rotation will rotate the plane of vectors by the same 60°, but will cancel the “vertical” part of the first one. The main point of my picture is not the relation between the one-sided multiplication and 3D rotation, but the relation between the one-sided multiplication and the arcs’ formalism, do you remember? My red roundabout shows six arcs. Exactly those we need in the arcs’ formalism. The red circle is, repeat, the unique invariant line (a.k.a. great circle) on the plane (a.k.a. 2-sphere) of vectors. It is a substantiation for arcs, not an attempt to show how the 3D rotation is composed (although with a sufficient geometric imagination about the plane reflection in 3D the picture will help you to think about it also). Incnis Mrsi (talk) 20:04, 9 February 2014 (UTC)
 * I will need time to review (and learn to understand). And yes, I see how every orbit of the action of the 1-d subgroup constituting a continuous rotation seems to form a Hopf link with every other such orbit within this 3-d space, but have not discerned the insight to be gained from this, other than to show that the red circle is the only orbit contained in the plane in the illustration.  Since you refer to a substantiation of the formalism using arcs, I see the plane that represents the 2-sphere, and the invariant circle. What I get from this is that there is only one 2-d subspace of vectors in 3-d for which one-sided multiplication of a quaternion having zero scalar component corresponds to rotation.  I have yet to connect this to the formulation using arcs in 3-d, though. —Quondum 02:16, 10 February 2014 (UTC)

Etymology and meaning
T.Y. Lam, in his paper Hamilton's Quaternions p. 22 says: "Any quaternion $q ∈ H^{∗}$ can be written in the form $ρ ⋅ (cos θ + u sin θ)$, where $ρ = N(q)^{1/2}$, and $θ ∈ [0, π]$ is called the polar angle of $q$.  (Hamilton, who had a penchant for inventing strange names, called $cos θ + u sin θ$ the versor of $q$.)"  He makes no reference to the idea of versor as being used as in this article. If the term versor has come to mean a quaternion of norm 1 (which is not the same thing at all), a clearer picture of the origin of this use of the word would be appropriate. The two linked online dictionaries define the term versor compatibly with Hamilton's use (in a sense as part of a quaternion), and not in the main sense as used in the article. As the distinction is subtle, I'll labour the point: Hamilton's use appears to have described an operation applied to any quaternion and its result, not to an arbitrary quaternion that may be the result of such an operation (just as it sounds strange to refer to an arbitrary real number $x ∈ [0, π]$ as a polar angle), nor to a quaternion representing a rotation of a vector. Under his usage it would be proper to say that $p$ is the versor of some quaternion, but it would not be proper to say that $p$ is a versor. —Quondum 17:23, 17 February 2014 (UTC)

Complicated distraction
This image was removed as unhelpful in its detail. Rgdboer (talk) 02:01, 1 June 2017 (UTC)

Representation theory more advanced than a general reader might assimilate:
 * In the quaternion algebra a versor $$q = \exp(a \mathbf{r})$$ will rotate any quaternion v through the sandwiching product map $$ v \mapsto q v q^{-1} $$ such that the scalar part of v is preserved. If this scalar part (the fourth dimension of the quaternion space) is zero, i.e. v is a Euclidean vector in three dimensions, then the formula above defines the rotation through the angle 2a around the unit vector r. For this case, this formula expresses the adjoint representation of the Spin(3) Lie group in its respective Lie algebra of 3-dimensional Euclidean vectors, and the factor "2" is due to the double covering of Spin(3) over the rotation group SO(3). In other words, q v  q−1 rotates the vector part of v around the vector r. See quaternions and spatial rotation for details.


 * A quaternionic versor expressed in the complex 2 ×  2 matrix representation is an element of the special unitary group SU(2). Spin(3) and SU(2) are the same group. Left multiplication qv of a quaternion v to a versor q is another kind of quaternion rotation as a 4-dimensional real vector space, identical to the SU(2) action on the 2-dimensional complex space identical to quaternions (v = A + Bj). Angles of rotation in this λ = 1/2 representation are equal to a; there is no "2" factor in angles unlike the λ = 1 adjoint representation mentioned above; see representation theory of SU(2) for details.

New subsection Representation of SO(3) was substituted.Rgdboer (talk) 21:52, 4 June 2017 (UTC)

Pretty sure 'a' should be between 0 and 2pi in the introduction
Otherwise versors aren't closed under multiplication. I'll make the change, and if anyone can find a source that disagrees, they can roll it back. — Preceding unsigned comment added by 62.172.100.253 (talk) 14:00, 16 August 2018 (UTC)

[edit]

I've thought about it, and the original definition does work. Why? Because if $$a$$ is increased by $$\pi$$, then you can negate $$r$$, and subtract $$\pi$$ from $$a$$ again. One could even argue that it's more "efficient" to define it that way, because there are fewer representations of the same versor. But if a versor has $$r=0$$ then there are still multiple representations under the old definition. I don't see any advantage in it.

[edit]

To be conservative, I've rolled back the edit. — Preceding unsigned comment added by 62.172.100.253 (talk) 14:12, 16 August 2018 (UTC)


 * Hello London. Yes, a half-turn is enough given the choice of r and –r. Compare elliptic geometry with the geometry of a sphere. Also right about ambiguity of angle when r = 0. BTW, why not register as a WP:User and get a WP:watchlist as well as a user page? — Rgdboer (talk) 23:14, 16 August 2018 (UTC)