Talk:Quaternion

regarding basis shown in Matrix representations

Even though the section says that there are at least two ways, should'nt it be explicitly said that the basis made up of four 4x4 matrices shown in the example are not unique and that other matrices which have the same properties can be used to represent i,j and k. also how many such bases can be possible? a trivial case is a basis which is made of the transpose (equivalent to choosing a basis of -i, -j and -k) or basis where matrices corresponding to i, j and k are cyclicaly shifted. does another basis which cannot be made up by doing these two operations exist? Does the basis have to be made up of 0, 1 and -1? Cplusplusboy (talk) 13:22, 20 January 2012 (UTC)

These questions make decent research projects, but they will not be appropriate for the article (unless there is some very nice citable result). (Ordered) bases of the type you described will correspond naturally to the ring automorphisms of H. Rschwieb (talk) 13:58, 20 January 2012 (UTC)

Arbitrary 4 × 4 real matrix without Jordan blocks with same eigenvalues (namely, {i, i, −i, −i} ) is eligible to represent the quaternion i. You may construct real 4-dimensional quaternions' representations by algebraic conjugation: X → U⁻¹ X U where X is a canonical representation and U is an arbitrary reversible 4 × 4 real matrix chosen for this particular representation. This is actually nothing but a (two-side) intertwiner, or simply a change of basis, and is considered the same in the representation theory. Incnis Mrsi (talk) 14:30, 20 January 2012 (UTC)

(Note to OP: the conjugation described here produces an automorphism of H. Rschwieb (talk) 15:12, 20 January 2012 (UTC)

It is an automorphism of ℍ only if U belongs to SO(4). I guess that it is also sufficient (the 3-sphere of unit quaternions in the canonical representation seems to be the same as left-isoclinic rotations), but am not completely sure. Moreover, as we discuss representations by arbitrary matrices, U does not even have to be orthogonal, this means that U⁻¹ X U not necessary is a canonical representation of any quaternion. Incnis Mrsi (talk) 16:20, 20 January 2012 (UTC)

Oh. I've never heard of a reversible matrix, so I was guessing it meant special orthogonal. Rschwieb (talk) 20:25, 20 January 2012 (UTC)

Having considered the group of matrices that may be U, this does not directly say the obvious things about the resulting representation. For example, the first matrix always remains the identity matrix. Next, it would seem to me that the remainder of the basis matrices obey a linear transformation law, which, unlike U, has only three dimensions: the symmetry group of S²? — Quondum^☏✎ 05:18, 21 January 2012 (UTC)

Ahem. Perhaps we can keep this to language accessible to those who do not already know the answer to the original question? Cplusplusboy may have a point that "There are at least two ways of representing quaternions as matrices" may be so weak a statement as to be misleading, and should at least be rephrased. There are an infinite number of ways (for example derivable from each of those representations via 3-dimensional rotations and reflections of the (i,j,k) basis on a 4×4 real matrix representation alone (the ring automorphism group being isomorphic to O(3), I guess). So perhaps it would be reasonable to change this to "There are many ways of representing quaternions as matrices" – even without citations. Those given just happen to be two of the "neat" ways. — Quondum^☏✎ 14:50, 20 January 2012 (UTC)

Hehe, I'm not very familiar in this math and so didn't want to edit the article myself. I was just comparing an example given in a book on quaternions and found that the bases it showed differed from wikipedia's. Since I was under the impression that the basis was unique, I tried to do the check of ijk=-1 property on both bases and found that both were right and wanted to confirm the fact. Should this talk be removed as the confusion is resolved? I didn't see anything about that in the guidelines. Cplusplusboy (talk) 16:36, 21 January 2012 (UTC)

I've edited the article in an attempt to address the initial problem; we'll see what others make of it. No, we leave the discussion as is; there are tight constraints on any editing of prior comments; it'll be removed in due course by the archiving process. See Wikipedia:Talk page guidelines#Editing comments. — Quondum^☏✎ 06:35, 22 January 2012 (UTC)

Please, in an article on mathematics, be more precise. For instance: The sentence ,Using 4x4 real matrices ...' is clear, since a,b,c,d must be real numbers, but this should be stated there, even if this is tedious. But in the 2-dimensional matrix representation some lines above, nothing is clear: Are the a,b,c,d real numbers as well or complex numbers? Obviously complex!? And why there are different letters for a,b,c,d in the text and in the matrix representation? And the same for i. Is this the same complex unit as in the text line before. The same question some blocks before in the determinant - please state whether these a,b,c,d are real or complex numbers. — Preceding unsigned comment added by 130.133.155.70 (talk) 13:17, 18 July 2014 (UTC)

Error?

I've never made an edit (except for the occasional spelling fix) so not sure of protocol.

The section "Three-dimensional and four-dimensional rotation groups" refers to the 3-sphere as a three dimensional sphere, it isn't, the 3-sphere is four dimensional (its hypersurface has 3 dimensions)

Ds1392 (talk) 14:25, 20 October 2013 (UTC)

The 3-sphere or sphere of dimension three is a manifold of dimension 3 that may be embedded as an hypersurface in the Euclidean space of dimension 4. This embedding is realized by defining the 3-sphere as the zero set of the equation $x^{2}+y^{2}+z^{2}+t^{2}-1=0.$ Thus the article is correct, although somehow too technical.

There is no protocol for editing. You have just to edit. However, if your edit is wrong or does not follows Wikipedia rules and policies, it is likely that it will be quickly reverted. D.Lazard (talk) 14:48, 20 October 2013 (UTC)

Point taken :D Perhaps the wording could be adjusted a smidge to make that clearer? I guess it's difficult to satisfy both the requirement that wikipedia be readable by a general audience (where intuitively, an n-dimensional object is one that can be embedded in Rⁿ) and the requirement that the information be accurate (an n sphere is an n-dimensional manifold.) If I can think of a way to improve the phrasing that isn't too wordy, I'll make the edit. Ds1392 (talk) 01:19, 21 October 2013 (UTC)

I don't think it can be clarified without properly distinguishing what "dimensions" are being discussed. It has (geometric) dimension 4 when embedded in R⁴, but has (topological) dimension 3. Rschwieb (talk) 13:23, 21 October 2013 (UTC)

No, an n-sphere always has dimension n. It can be embedded in a larger dimensional space, but that does not change its dimension. See manifold. Ozob (talk) 14:07, 21 October 2013 (UTC)

I agree with Ozob. The problem may come that for many people the distinction between a sphere and a ball is unclear: The sphere of dimension n is the boundary of the ball of dimension n+1. The surface of Earth is roughly a 2-sphere, while Earth in the whole is roughly a 3-ball. The lead of Sphere deserve to be edited to emphasize this distinction. D.Lazard (talk) 14:28, 21 October 2013 (UTC)

@Ozob (cc @D.Lazard): I'm saying that there are two subsets of humans: those who think of dimension in the topological way and those thinking of it in the geometric way. I'm pretty sure most laypeople carry the geometric dimension learned in grade school through 2-d an 3-d geometry. So, for example, they will report that the 2-sphere is a "three dimensional object," even if it is just the surface of a ball. I've seen this misconception cleared up a handful of times in graduate and undergraduate setting, so it is even common among non-laypersons.

Anyhow in summary, you and I know it has an intrinsic dimension that doesn't depend on where it's embedded, but stubbornly pretending that everyone else will understand it that way if we say so is an invitation for misunderstanding. Rschwieb (talk) 15:00, 21 October 2013 (UTC)

Nobody would say that a line is anything more than one dimensional or that a plane is anything more than two dimensional. I agree that some people are confused about the precise meaning of dimension, but the standard definition is both not too surprising and used universally within mathematics. I don't see any reason why this article should equivocate on this point. Ozob (talk) 18:44, 21 October 2013 (UTC)

I've changed it to "3-sphere S³". In this instance, using a less familiar term might lead to less confusion, for the reason that it does not as readily trigger an unintended interpretation. — Quondum 00:41, 22 October 2013 (UTC)

This change works for me. I was tripped up even though I should know better. Quondum's point re "less familiar" terms is quite valid; It's probably a good idea to avoid phrases with a natural language interpretation as much as possible because it's hard to avoid the reflexive interpretation. In everyday speech I refer to the 3-ball/2-sphere-in-R³ as a "three dimensional sphere" (formally correct or not this is how natural language is, and natural language "got there first" so to speak.) If I'm referring to the manifold I'll explicitly use the term "3-sphere" to avoid ambiguity. My 2c anyway :) Ds1392 (talk) 12:13, 23 October 2013 (UTC)

Matrix vector product

The following text was removed from Matrix representation:

The multiplication of two quaternions

$ab=c$

can be represented by matrix vector multiplication:

${\begin{bmatrix}a_{0}&-a_{1}&-a_{2}&-a_{3}\\a_{1}&a_{0}&-a_{3}&a_{2}\\a_{2}&a_{3}&a_{0}&-a_{1}\\a_{3}&-a_{2}&a_{1}&a_{0}\end{bmatrix}}{\begin{bmatrix}b_{0}\\b_{1}\\b_{2}\\b_{3}\end{bmatrix}}:={\begin{bmatrix}c_{0}\\c_{1}\\c_{2}\\c_{3}\end{bmatrix}}$

If we define

$B(a)\equiv {\begin{bmatrix}a_{0}&-a_{1}&-a_{2}&-a_{3}\\a_{1}&a_{0}&-a_{3}&a_{2}\\a_{2}&a_{3}&a_{0}&-a_{1}\\a_{3}&-a_{2}&a_{1}&a_{0}\end{bmatrix}}$

and a, b, and c are real column vectors constructed from quaternions, we can rewrite the multiplication as

$B(a)b=c$

or

$B(a)B(b)=B(c)$.

We can also define

$A(a)\equiv {\begin{bmatrix}a_{0}&a_{1}&a_{2}&a_{3}\\-a_{1}&a_{0}&-a_{3}&a_{2}\\-a_{2}&a_{3}&a_{0}&-a_{1}\\-a_{3}&-a_{2}&a_{1}&a_{0}\end{bmatrix}}$.

The A and B matrix constructions have the following basic properties.

$A(a)b=B(b)a^{*}$

$A(a)B(b)=B(b)A(a)$

Two matrices must be multiplied to represent the quaternion product. The text removed today was unreferenced and made a false assertion.Rgdboer (talk) 20:51, 18 July 2014 (UTC)

While it would be easy enough to correct this, the matrix representations as they stand in the article are sufficient. Also not being referenced makes it look like the OR it probably is. I agree with the removal. —Quondum 21:28, 18 July 2014 (UTC)

But this looks very useful. Why would you multiply the whole matrix; that's four times the amount of work???

Think it was just transposed incorrectly:

${\begin{bmatrix}a_{0}&a_{1}&a_{2}&a_{3}\\-a_{1}&a_{0}&-a_{3}&a_{2}\\-a_{2}&a_{3}&a_{0}&-a_{1}\\-a_{3}&-a_{2}&a_{1}&a_{0}\end{bmatrix}}{\begin{bmatrix}b_{0}&b_{1}&b_{2}&b_{3}\\-b_{1}&b_{0}&-b_{3}&b_{2}\\-b_{2}&b_{3}&b_{0}&-b_{1}\\-b_{3}&-b_{2}&b_{1}&b_{0}\end{bmatrix}}:={\begin{bmatrix}c_{0}&c_{1}&c_{2}&c_{3}\\-c_{1}&c_{0}&-c_{3}&c_{2}\\-c_{2}&c_{3}&c_{0}&-c_{1}\\-c_{3}&-c_{2}&c_{1}&c_{0}\end{bmatrix}}$

${\begin{bmatrix}a_{0}&a_{1}&a_{2}&a_{3}\end{bmatrix}}{\begin{bmatrix}b_{0}&b_{1}&b_{2}&b_{3}\\-b_{1}&b_{0}&-b_{3}&b_{2}\\-b_{2}&b_{3}&b_{0}&-b_{1}\\-b_{3}&-b_{2}&b_{1}&b_{0}\end{bmatrix}}:={\begin{bmatrix}c_{0}&c_{1}&c_{2}&c_{3}\end{bmatrix}}$

${\begin{bmatrix}a_{0}&-a_{1}&-a_{2}&-a_{3}\\a_{1}&a_{0}&a_{3}&-a_{2}\\a_{2}&-a_{3}&a_{0}&a_{1}\\a_{3}&a_{2}&-a_{1}&a_{0}\end{bmatrix}}{\begin{bmatrix}b_{0}\\b_{1}\\b_{2}\\b_{3}\end{bmatrix}}:={\begin{bmatrix}c_{0}\\c_{1}\\c_{2}\\c_{3}\end{bmatrix}}$

213.205.240.129 (talk) 13:06, 12 September 2014 (UTC)

Type of isomorphism is unclear

In the section Quaternion#Matrix_representations, the following sentence occurs:

• Restricted to unit quaternions, this representation provides an isomorphism between S³ and SU(2).

This sentence has the problem that technically it is ill-defined, or more correctly, since both these objects are only in the same category as sets, this only says that they have the same cardinality. I expect that most people will find a natural interpretation as an isomorphism of topologies and/or as a congruence of geometric objects in Euclidean 4-space. Given that the representation is given as the basis of the isomorphism, the geometric interpretation may be intended, but is inappropriate (we would not normally call a linear mapping between representations an isomorphism in the algebraic context). However, S³ regarded as a topological object or a geometric object, H regarded as a ring and SU(2) regarded as a group leaves room for confusion of what isomorphism is meant. Could someone with more knowledge in the area please qualify this to clarify what is meant? Perhaps leave S³ out of it altogether, and simply state that there is a group isomorphism between the multiplicative group of unit quaternions and SU(2)? —Quondum 13:56, 19 July 2014 (UTC)

Where to even begin.... there's lots of confusion all around, it should be clarified.

• (lower-case) su(2) is an algebra, and specifically a Lie algebra.
• the quaternions are an algebra, too. (an algebra is a vector space endowed with multiplication, usually a non-commutative multiplication)
• the structure constants of su(2) are equal to those of H, except that they are multiplied by an extra factor of ${\sqrt {-1}}$. Thus, the generators of su(2) when squared, give you +1, instead of -1 when you square the generators of H.

• (upper case) SU(2) is a Lie group it corresponds to the fundamental representation of the algebra. Give a point ${\vec {\theta }}$ in the lie algebra su(2), you get the corresponding group element $U=exp(i{\vec {\theta }}\cdot {\vec {\sigma }})$, which is a 2x2 unitary matrix. Here exp is the exponential map used to convert dirction vectors into geodesics. The ${\vec {\sigma }}$ are the Pauli matricies.

• You can do exactly the same thing as above, using +1, i,j,k instead of using the identity matrix plus the pauli matricies. You get exactly the same thing (except for an extra confusing factor of ${\sqrt {-1}}$ that floats around and makes thing randomly confusing.

• The 3x3 matrix group representation of su(2) is call the rotation group SO(3). The explicit mapping is this. Let ${\vec {v}}$ be a 3D vector. Let R be a 3x3 rotation matrix. Then, $R\cdot {\vec {v}}=U^{\dagger }{\vec {v}}\cdot {\vec {\sigma }}U$ where U is same as above. By contrast, R is given by $exp({\vec {L}}\cdot {\vec {\theta }}/2)$ where ${\vec {L}}$ are the generators of angular momentum i.e. the purely real 3x3 matrixes that generate SO(3) rotations. anyway, its the same theta in U and R.

• (upper case) SU(2) is a manifold that is topologically isomorphic to S_3

• (upper case) SO(3) is a topological manifold that is covered twice over (double cover) by SU(2).

• The complex projective plane CP(2) is topologically isomorphics to SU(2)

• The last bit is made use of in quantum mechaics, where a spinor is a 2D complex vector of unit length (thus its projective) and is spun around by elements of SU(2).

• The resulting manifold is called the Bloch sphere.

• The metric on SU(2) aka S_3 aka CP(2) is called the ... crap, I don't remember the name. Oh right Fubini-Study metric. Its basically just the standard metrix on the sphere, but it looks interesting when you write it out for the typcal notation used in QM and in algebraic geometry which each have thier unique notiations (alg. geom studies complex projective spaces).

• One of the things that confuses people is the relationship between SU(2) and su(2) because both use 2x2 complex matrices. They're not the same tho, because SU(2) is a group, su(2) is an algebra. Likewise, there is a similar confusion for quaternions: There is the algebra H and there is the group H and they both use 1, i, j, k to understand how these differ, it helps to keep su(2) vs SU(2) firmly rooted in mind.

• Wait -- there's more... if you allow the vector ${\vec {v}}$ to be complex, then you get representations of the group SL(2,C) which have SO(3,1) as a representation -- this is the group of special relativity. which is why the outer product of two relativistic spinors is a spin-1 boson. Add the sqrt(-1) and you can say the same with quaternions, if you wanted to. You could write out Einstein's equations for general relativity using quaternions, if you wanted to. This is because the quaternions are $sqrt{-1}$ times the usual generators of sl(2,C). People have actually done this: its vaguely instructive to see those eqns as SL(2,C) instead of text-book standard SO(3,1).

• The outer product of two quaternions gives the Runge-Lenz vector: it describes the orbital mechanics of planetary systems (planets orbiting a sun) and the conserved qauantities are given by SO(4) (not just SO(3)).

• The restriction of SL(2,C) to real numbers gives SL(2,R) which is a hyperbolic manifold important in number theory and seems to have somethhing to do with the Riemann hypothesis. This is one of the many Siren's calls that string theorists are unable to resist. (well, string world sheets are Riemann surfaces which leads to conformal field theory and AdS/CFT correspondance which gives monstrous moonshine. Heh. So there.

• In short, when you really start fucking with it, you find all of these low-dimensional concepts are isomorphic or homomorphic to each other, which makes for a very rich playground of things related to each other.

Anyway, I clearly had too much fun writing the above. Thanks for posing the question. 67.198.37.16 (talk) 01:39, 13 February 2015 (UTC)

I see the question served its purpose: Ozob addressed it with this edit. Yes, the connections are varied and deep, and fun if you live with this stuff. —Quondum 03:52, 13 February 2015 (UTC)

Terminology: "scalar part" and "vector part"

I find this terminology inept and ugly. Why not stick with real and imaginary parts?

Is that terminology standard in the literature (the maths literature, not the physics one)? Not if I trust the few papers I've read today, but I could be wrong. If it is standard, then I guess some could argue that Wikipedia should continue to spread this ugliness. Otherwise it would be a shame that Wikipedia helps set up or propagate an ill-suited standard for no reason.--Seub (talk) 05:59, 11 March 2015 (UTC)

Real and imaginary parts is used in complex numbers where the parts are equal in dimension. scalar and vector parts emphasises that they are not equally sized, but a 1-dimensional scalar part and a 3-dimensional vector part. And they are vectors in a very real sense too; vector operations such as the cross product, dot product scalar product arise from quaternion product by considering how the 'vectors' in them multiply.--JohnBlackburne^words_deeds 12:41, 11 March 2015 (UTC)

I think that there are arguments both ways. Doesn't the scalar/vector terminology date to Hamilton, whereas real/imaginary and other variants tend to be more contemporary? I am not particularly a fan of the use of "vector part", because it does not seem to generalize directly to hypercomplex numbers or Clifford algebras, and would be destined to fade. John is correct that it does relate directly to vector algebra, but only in exactly 3 dimensions. IMO, it would be more appropriate to speak of the "scalar" and "nonscalar" parts. At least "real" can "imaginary" do fit as a generalization of the use of the concepts from complex numbers, and I've seen the term "imaginary" used to refer to any nonreal component that squares to a real number, such as in split-complex numbers. For the purposes of this article, perhaps we could switch to "real" and "nonreal" (just a suggestion)? —Quondum 17:54, 11 March 2015 (UTC)

At the level of this article vector algebra is done in 3D. E.g. you have the cross product, defined only in three dimensions, as a key operation. This and the dot product show how it is a 'vector' part as both operations arise directly from the quaternion product if restricted to products only of the vector parts. At least that's how I learned it. I later learned how to generalise it in various ways, which leads to other ways to think of the non-scalar parts, as imaginaries, as bivectors . But vectors I think is most usual at a less advanced level.--JohnBlackburne^words_deeds 00:21, 12 March 2015 (UTC)

Just because quaternions can be related to vectors and rotations in Euclidean 3-space doesn't mean that it is their nature or purpose, it's just a use you can make of them. So I disagree with the argument that the imaginary part of a quaternion is a vector in a "very real sense". And the "scalar part" term is no more clever imho, why would you want to think of it as a scalar? It is just a quaternion that is identified as a real. I recall that scalars are what you name the elements of the base field of a vector space, when you think of them acting on vectors by multiplication. If you work with vector spaces over the quaternions, then the quaternions themselves are all scalars. I also recall that in (not really) modern mathematics, vectors are not just vectors in the plane and the 3-space. If you define quaternions as a 4-dimensional algebra over the reals, then all quaternions are "vectors". Anyway, my recommendation is "real" and "imaginary", I see no drawback in that terminology. I also think it's the more common usage in modern mathematics (we could survey the arXiv a little), but I'm not certain of that.--Seub (talk) 10:17, 12 March 2015 (UTC)