Talk:Quaternion

Discovery or invention?

To me, it seems that some things in mathematics are discoveries, and some are inventions. I consider $pi$ and $e$ to be discoveries, since they are fundamental to so much. Matrices I consider to be an invention, since, despite their flexibility and utility value, I've always regarded them as being rather arbitrary (full disclosure: I never did like matrices :). Quaternions also seem to fall into the invention category (more full disclosure: I love quaternions). Complex numbers are harder to so categorize; while the term "imaginary part" may argue for "invention", they are so closely tied to fundamentals (e.g., two-dimensional Euclidean space) that "discovery" also seems accurate. BMJ-pdx (talk) 22:39, 5 June 2023 (UTC)

Maybe make a blog or social media post out of this instead of chitchatting about it here. Cf. WP:NOTFORUM. –jacobolus (t) 00:46, 6 June 2023 (UTC)

See Philosophy of Mathematics. --50.47.155.64 (talk) 15:51, 15 August 2023 (UTC)

Many quantum and particle physicists would say that hypercomplex numbers fully exist in the natural world. LagrangianFox (talk) 18:58, 11 October 2024 (UTC)

Polynomial equations

This link studies a certain family of quadratic equations. In general, a homogeneous quaternionic polynomial of degree of n is of the form

$\sum _{i}a_{i,1}xa_{i,2}x\cdots a_{i,n}xa_{i,n+1},$

and there can be arbitrarily many terms in the summation.

Example of a linear polynomial: $ax+xb=ax1+1xb$;

Example of a quatratic polynomial: $ax^{2}+xbx+x^{2}c=ax1x1+1xbx1+1x1xc$. 129.104.244.74 (talk) 11:42, 26 May 2025 (UTC)

Power series for quaternions

It seems to me that any function of real or complex numbers that can be equated to a power series can be seamlessly extended to a function of quaternions, via the power series. This would apply to trigonometric and hyperbolic functions, etc. If this is true and notable, should we add mention of it to the article?

Furthermore, in the case of complex numbers, such a power series can often be extended beyond its radius of convergence via analytic continuation. If there is a similar concept for quaternions, should we mention that too?

Thoughts? —Quantling (talk | contribs) 16:24, 16 June 2025 (UTC)

I think the power series of a quaternion corresponds fairly trivially to that of a complex number with its imaginary part equal to the norm of the vector part of the quaternion. Any geometric interpretations, of the complex number, should have a corresponding geometric interpretation in the plane generated by the quaternion and its powers. This makes sense as the quaternions can be derived by the Cayley–Dickson construction from complex numbers so contains multiple copies of them.

I don't see any quaternion specific applications though. If you try to generalise it to quaternions not in the same plane then things stop working like complex numbers. Even simple formula like e^(a+b) = e^a * a^b stop working as multiplication is no longer commutative. --2A04:4A43:900F:FA65:253D:1E14:39A3:1BCC (talk) 14:47, 17 June 2025 (UTC)

I'd have to put some effort into understanding your first paragraph, but if you would summarize the topic for the article itself, that sounds like an improvement to me. Yes, non-commutativity would mean that the exponential function so defined would not behave as we might first have expected, and similarly for other functions. I'd like to see mention of something along these lines in the article too. So long as this isn't original research but an actual reflection of what mathematicians have done ... I'd like to see it in the article. —Quantling (talk | contribs) 15:15, 17 June 2025 (UTC)

It definitely is original research as I did not use any sources writing that. But again I think it's a trivial consequence of how quaternions contain the complex numbers, so the properties of complex numbers apply to such a subset. Much like the complex numbers contain the real numbers so you can do real number math within them. It's not very interesting using complex numbers for that though. --2A04:4A43:900F:FA65:253D:1E14:39A3:1BCC (talk) 16:18, 17 June 2025 (UTC)

Find a reliable source for your conjecture. Hawkeye7 (discuss) 19:20, 17 June 2025 (UTC)

Which "conjecture" are we talking about here? You are asking whether the exponential function of a quaternion can be defined as a power series (or defined some other way and proven to be equal to a power series)? Yes, that is standard and easy to find references for.

The functions sinh and cosh are just the even and odd parts of the exponential function, which can be, trivially, the even and odd terms of the power series for exp. Defining quaternion-valued sine or cosine isn't really very insightful in my opinion, but you can probably find someone doing it if you look around. What you can do is take the sinh or cosh of an "imaginary" quaternion (more generally, bivector) and pull out the "imaginary" direction (unit bivector) out front which leaves the sine or cosine of a real quantity.

The lack of commutativity of multiplication means that, as the IP editor mentioned, exponential identities must be treated carefully. There are certainly sources about this, both for quaternions per se and for more general kinds of multivectors. –jacobolus (t) 20:08, 17 June 2025 (UTC)

Here's a source from the 1870s, JSTOR 25138496. –jacobolus (t) 20:18, 17 June 2025 (UTC)

So, the question I have is whether this is important enough for the article. I think a brief mention is called for. For example, we could change the lead sentence of the Exponential, logarithm, and power functions section from Given a quaternion, to A function of a quaternion can be defined from a power series. For example, given a quaternion,. That touches on the topic about as lightly as I can imagine, excepting that it is the lead sentence of that section. We could do more. —Quantling (talk | contribs) 20:23, 17 June 2025 (UTC)

Apologies. I was put on the defensive by "it's a trivial consequence of how quaternions contain the complex numbers, so the properties of complex numbers apply to such a subset." The quaternions are not a subset, so some properties of complex numbers, like commutativity, do not apply to quaternions.

Returning to your point: Can "any function of real or complex numbers that can be equated to a power series can be seamlessly extended to a function of quaternions, via the power series"? Hawkeye7 (discuss) 00:56, 18 June 2025 (UTC)

Firstly, a power series defines a function only if the variable commutes with the coefficients. So, the question makes sense only for series with real coefficients. As mentioned in the article, the quaternion form a Banach algebra over the reals. Series on Banach algebas have been widely studied, and all general results on Banach algebras apply to quaternions. I am not sure whether there are results specific to quaternions. D.Lazard (talk) 11:10, 18 June 2025 (UTC)

Good to know about the coefficients being real numbers. Would it be okay if I change the lead sentence of the Exponential, logarithm, and power functions section from Given a quaternion, to A function of a quaternion can be defined from a power series with real coefficients. For example, given a quaternion,. —Quantling (talk | contribs) 13:40, 18 June 2025 (UTC)

Section: square roots of given arbitrary Quaternions: FOUND Varying results! What is correct?

I am working square Roots of Quaternion Numbers. ANY MATH SPCIALIST Out HERE

There is almost nothing on the internet.

I get different results and need to approve the imaginary, the vector part

for the real part of the square-root, I get the same matching results. following My own calculus: q = [ s, v ] √q = [ x, y ]

√q = [(√( s + ||q||) , ??? v *√2*√( s + ||q|| ) ]

I looked up some resurces. The result for the imaginary part varies over different resources.

Google AI gives [...] Microsoft BING-Copilot states: QUERY: "What is the square root of an arbitrary quaternion number ?"

√q = [ √(|q|+s)/2) , v / |v| * √(|q|-s}/2)

x = √(|q|+s)/2)

y = v / |v| * √(|q|-s}/2)

QUESTION: a) The last (MS-Copilot matches with our Wikipedia solution.

b) Is the Google solution correct, too

c) Are both results correct?

d) where can I find the related resources to look up the calculus??

- Frank.Haferkorn

P.S. the variation is in

• - the sign +s or -s
• . the factor sqrt(2)
• and use of |q| vs. |v|

2A02:3100:9CBA:B800:840C:DC8D:E863:E44E (talk) 23:41, 4 August 2025 (UTC)

This is not really the right venue for this kind of question. You should try Wikipedia:Reference desk/Mathematics or some other forum. But anyway, AI agents are a poor tool to use for factual questions. This article already covers this topic at § Square roots of arbitrary quaternions. If you have one or more formulas for a square root, you can verify them by squaring and checking that the original quaternion is recovered, which just takes a bit of basic algebra. If you want reliable sources to cite, since this article doesn't give a source, I recommend you search the academic literature. Quite a lot has been written about this topic, dating back over a century. A google scholar search for the exact phrase "square root of a quaternion" returns quite a few results. (There is also quite a lot of information about this available on the internet, if you try web searches instead. But Google is sadly a lot less effective at finding it than it used to be 20 years ago.) –jacobolus (t) 04:59, 5 August 2025 (UTC)

By the way, if you want to derive an expression for yourself, the concept here is to take a quaternion ⁠$\mathbf {q}$⁠, add a scalar of the same magnitude, which results in a quaternion pointing in the correct direction of the square root, and then re-normalize that to the desired length. If we let $r=|\mathbf {q} |$ then we want: $${\sqrt {\mathbf {q} }}={\sqrt {r}}{\frac {\mathbf {q} +r}{{\bigl |}\mathbf {q} +r{\bigr |}}}.$$ In the case you started with a unit quaternion, then ⁠$r=1$⁠, and you get: $${\sqrt {\mathbf {q} }}={\frac {\mathbf {q} +1}{{\bigl |}\mathbf {q} +1{\bigr |}}}.$$ –jacobolus (t) 05:17, 5 August 2025 (UTC)

Hallo Jacobulus,

The Square Root of a quaternion seems not to be common knowledge.

I am looking for resources in the internet for a some time, now.

The older resources you mention are from the 19th century) are from Hamilton himself and contain no roots at all. Can you show me ANY foreign resource, that has the given RESULT?

LET ME SUMMARIZE in order to repeat ...

For the imaginary part, the related chapter "§Square roots of arbitrary quaternions" says:

y = v / ||v|| * √( ||q||-r)/√2

The square of a quaternion q² is following chapter Scalar and vector parts

says:

p² = [ x*x-y*y , 2*x*y + ( y cross y) ) //with (y cross y) = 0

an the norm ||q|| = sqrt(r^2+y*y)

NOTATION:

q = [ r, v ]

√q = [ x, y ]

q = √q * √q

so: [r, v] = [ x, y ]* [ x, y ] = [ x*x-y*y , 2*x*y ]

so we need to solve by means of x and y.

in order to express in properties of q likewise r and v and maybe using ||q||. 2A02:3100:A121:A00:6C4C:D60F:24D9:B87D (talk) 16:10, 5 August 2025 (UTC)

A research on Scholar Google of "square root of a quaternion" gives (among many other results) two articles of I. Niven in The American Monthly entitled The roots of a quaternion (1942) and Equations in quaternions (1941). I did not read these articles, but their title suggest strongly that they contain the results summarized in the last paragraph of § Square roots of arbitrary quaternions, and that these results are common knowledge.

By the way, I checked Jacobolus formula, which is correct for quaternions that are not negative real numbers (a "⁠$\pm$⁠" must be added before Jacobolus's formula). For negative real numbers, this formula gives 0/0 and does not provide a square root. D.Lazard (talk) 17:01, 5 August 2025 (UTC)

That's true, but negative real numbers are a degenerate case with infinitely many square roots. Thus "the square root" is not well defined for them. –jacobolus (t) 17:44, 5 August 2025 (UTC)

For the REAL-part, I get an additional Plus-Minus before and within the term.

What I did is right hand side multiply the imaginary part

v = 2*x*y by y

and inserting y*y in the real-equation

For the real-part I solve the 4-th order equation: r = x² - v²/4x²

to

x_{A.B} = sqrt( 1/2*(r \pm\ ||q|| ) )

These have an extra plus/minus before x_A, x_B

Putting that directly into v = 2 *x *y

y = v / (\pm sqrt(2) * sqrt( r \pm ||q|| ))

as ||v|| = sqrt( ||q||² - r² ) = sqrt (||q||- r) sqrt (||q|| + r)

the imaginary part is then

y = m v / ||v|| * [ \pm sqrt(2) * sqrt( - (r \mp ||q|| ) ) ]

and has two separated plus/minus switces.

AM I MAKING a mistake?

Greetings,

Frank.Haferkorn 2A02:3100:A121:A00:6C4C:D60F:24D9:B87D (talk) 19:37, 5 August 2025 (UTC)

Can you try to edit all of your comments here to use LaTeX for the mathematical notation? (See Help:Displaying a formula.) –jacobolus (t) 19:49, 5 August 2025 (UTC)

For example:

For the REAL-part, I get an additional Plus-Minus before and within the term. What I did is right hand side multiply the imaginary part ⁠$v=2xy$⁠ by ⁠$y$⁠ and inserting ⁠$y\cdot y$⁠ in the real-equation. For the real-part I solve the 4th order equation: $$r=x^{2}-{\frac {v^{2}}{4x^{2}}}$$ to $$x_{A.B}={\sqrt {{\tfrac {1}{2}}{\bigl (}r\pm \|q\|{\bigr )}}}.$$ These have an extra plus/minus before ⁠$x_{A}$⁠, ⁠$x_{B}$⁠. Putting that directly into ⁠$v=2xy$⁠, $$y={\frac {v}{\pm {\sqrt {2}}{\sqrt {r\pm \|q\|}}}}$$ as $$\|v\|={\sqrt {\|q\|^{2}-r^{2}}}={\sqrt {\|q\|-r}}{\sqrt {\|q\|+r}}.$$ The imaginary part is then $$y={\frac {mv}{\|v\|\cdot \left(\pm {\sqrt {2}}{\sqrt {-(r\mp ||q||)}}\right)}}$$ and has two separated plus/minus switces.

I must admit I still am quite confused about what you are trying to say in the above. In any event, if you want both square roots of an arbitrary quaternion (not a negative real scalar), the plus-or-minus should go on the whole expression; the two quaternions are additive inverses of each-other. –jacobolus (t) 20:13, 5 August 2025 (UTC)

Related to your way derive an expression for yourself,

I do not get the point at all.

You add the scalar part r to q in order to afterwards normalize it ? 2A02:3100:A121:A00:6C4C:D60F:24D9:B87D (talk) 16:18, 5 August 2025 (UTC)

Yes, if you plot the quaternion ⁠$\mathbf {q}$⁠ and the scalar $r=|\mathbf {q} |$ in the plane containing ⁠$\mathbf {q}$⁠ and scalars, they are the two sides of a rhombus. When you take the sum (see vector addition) it is the diagonal of that rhombus, which bisects the angle, and therefore points in the same direction as the square root. Then you can normalize this diagonal to the desired length, which should be the square root of the magnitude of your original quaternion ⁠${\sqrt {r}}$⁠, so you multiply by that number and divide by the magnitude of the sum. –jacobolus (t) 17:45, 5 August 2025 (UTC)

The general idea is exactly the same for quaternions as it is for the square root of a complex number ⁠$z=x+iy=re^{i\theta }$⁠. Here, we can derive the whole thing step by step: $${\begin{aligned}{\sqrt {z}}={\sqrt {r}}{\frac {z+r}{|z+r|}}&={\frac {{\sqrt {r}}(x+r+iy)}{\sqrt {(x+r+iy)(x+r-iy)}}}\\[6mu]&={\frac {{\sqrt {r}}(x+r+iy)}{\sqrt {2r^{2}+2xr}}}\\[6mu]&={\frac {x+r}{{\sqrt {2}}{\sqrt {x+r}}}}+{\frac {iy}{{\sqrt {2}}{\sqrt {x+r}}}}\\[6mu]&={\frac {\sqrt {x+r}}{\sqrt {2}}}+{\frac {iy{\sqrt {r-x}}}{{\sqrt {2}}{\sqrt {y^{2}}}}}\\[6mu]&={\sqrt {\textstyle {\tfrac {1}{2}}(r+x)}}+{\frac {iy}{|iy|}}{\sqrt {\textstyle {\tfrac {1}{2}}(r-x)}}.\end{aligned}}$$ It's just the same with a quaternion, except that ⁠$iy$⁠ must be replaced by an arbitrary imaginary quaternion. –jacobolus (t) 18:27, 5 August 2025 (UTC)

I hadn't seen this formula for the square root of a complex number ... nice! It looks like the tangent-half angle formula, yes? If y / x is the tangent of the angle θ of z with respect to the positive real axis then the square root of z will have angle θ / 2 which has tangent equal to sin θ / (1 + cos θ), which is (y/r) / (1 + x/r) = y / (r + x), where indeed y and r + x are, respectively, the imaginary and real parts of z + r. The rest is scaling that to a length of ⁠${\sqrt {r}}$⁠. —Quantling (talk | contribs) 21:37, 5 August 2025 (UTC)

Yes, the stereographic projection also invoves taking a square root, but then the length is normalized a different way. –jacobolus (t) 22:30, 5 August 2025 (UTC)

Cool. So we can also use the formula

$$\tan {\frac {\alpha +\beta }{2}}={\frac {\sin \alpha +\sin \beta }{\cos \alpha +\cos \beta }}$$

to calculate the geometric mean between two complex numbers. That is, for z and w as complex numbers with unit magnitude we get ${\sqrt {zw}}={\frac {z+w}{|z+w|}}$. When z and w don't necessarily have unit magnitude then it is a little hairier, but not too bad: $${\sqrt {zw}}={\sqrt {|z||w|}}{\frac {|w|z+|z|w}{|(|w|z+|z|w)|}}\,.$$ And, of course w = 1 and |z| = r gets us back to ⁠${\sqrt {z}}$⁠, and we can have fun with the likes of w = −1 or w = ±i too! Thank you for pointing me in this direction. —Quantling (talk | contribs) 01:36, 6 August 2025 (UTC)

Yes, that's true. You can think of the expression for square root as a special case of the version for two arbitrary quaternions (or complex numbers). If you want you can simplify $\textstyle {\sqrt {|z|\cdot |w|}}$ to $\textstyle {\sqrt {|zw|}}$. –jacobolus (t) 05:30, 6 August 2025 (UTC)

Maybe this all generalizes to a quadratic field too? That is, with an expression like ⁠$z=a+b{\sqrt {d}}$⁠, we can think of a as ⁠$|z|\cosh \eta$⁠ and ⁠$b{\sqrt {d}}$⁠ as ⁠$|z|\sinh \eta$⁠ and then compute a square root with a tanh η/2 formula somehow. Or maybe only for those cases where the field already contains the square root that we are aiming to compute? Do you happen to know? —Quantling (talk | contribs) 18:53, 7 August 2025 (UTC)

I don't quite understand your question, but I'm the wrong person to ask and this is the wrong venue. –jacobolus (t) 23:22, 7 August 2025 (UTC)

For $q=t\exp(\theta r),\quad t>0,\quad r^{2}=-1$, the square root of q is given by $t^{1/2}\exp({\frac {\theta }{2}}r)$. This method uses the concepts of versor and Polar_decomposition#Quaternion_polar_decomposition. — Rgdboer (talk) 21:22, 5 August 2025 (UTC)

Real quaternions and scalars

Rgdboer has recently changed the terminology for scalar quaternions by removing the alternative name of "real quaternion", because this phrase is also used in the contex of biquaternions for distinguishing quaternions from complex quaternions (in the linked article, "real" is often written between parentheses). Nevertheless, the phrase "real quaternion" is commonly used for scalar quaternions and, therefore, must be mentioned in the definitions.

Rgdboer renamed scalar quaternions simply as "scalars". This conflicts with the usual terminology: in ⁠$a+b\mathbf {i} +c\mathbf {j} +d\mathbf {k}$⁠, all ⁠$a,b,c,d$⁠ are scalars, not only ⁠$a$⁠.

Therefore, I edited the article for replacing everywhere (I hope) "scalar" alone with "scalar quaternion". That is, for not using "scalar" as a noun.

By the way the article contains multiple repetitions, and I have not tryed to fix them. D.Lazard (talk) 11:16, 23 September 2025 (UTC)

I just edited the article before reading this ... apologies. I would have discussed it first here, but I hope it is okay nonetheless. I added that "scalar quaternions" are sometimes referred to simply as "scalars" though I am happy that we don't do that in this article. —Quantling (talk | contribs) 13:35, 23 September 2025 (UTC)

This is fine. Thanks. D.Lazard (talk) 14:11, 23 September 2025 (UTC)

Axis of rotation - making things clearer for learners in the opening

A rotation in space entails a vector perpendicular to the plane of rotation

This article needs a small blurb in the opening that dumbs things down for people who get here from a google search after encountering a quaternion in a game engine or something similar, which is probably most of the people arriving here from google. Basically, emphasize that when used to describe a rotation, a quaternion is essentially a rotation amount, and a vector denoting the axis of rotation. This is mentioned in the article but it's quite buried and I doubt many people get there after encountering the word salad in the opening.

I understand there are other articles specifically for quaternions and rotations, but that doesn't matter; people are going to come to THIS page when they google "what is a quaternion". And many, if not the majority, of people encountering quaternions for the first time are encountering it in the context of rotations. And upon arrival, the article is very impenetrable.

To reiterate: I think we need at least one sentence in the opening that explains the rotation aspect in crystal clear language, because a large portion of people visiting this page are doing it because they don't know what a quaternion is yet. "When used to describe a rotation, a quaternion has two components; a rotation amount, and a vector in 3D space that represents the axis upon which rotation occurs." Something like that.Binglederry (talk) 22:45, 12 October 2025 (UTC)

I disagree. Since quaternions are an advanced algebraic concept and rotations are only one application of them, the word salad in the lead is necessary. Rotations are already mentioned in the lead, and if someone visiting the page wants to know more about the connection between quaternions and rotations, they can follow the links or read the other sections of the article.—Anita5192 (talk) 23:10, 12 October 2025 (UTC)

Agreed that the promise of quaternions is explication of rotation geometry. As Euler's formula expresses the unit circle and the implicit group action upon itself as a planar rotation, so in quaternion expression with versors the geometry of three-dimensional rotations is given. Hamilton wrote when spherical distances and spherical trigonometry were current practice. Today the 3-sphere with versor products (quaternion multiplication) exhibits a compact Lie group. — Rgdboer (talk) 01:01, 13 October 2025 (UTC)

Please correct typesetting error "{{{1}}}"

The text under the Dublin bridge's photo has a typesetting error.

It's second-last line says "{{{1}}}"

Someone familiar with the formatting please correct this.

As of 5-Jan-2026 it appears, in various browsers, as

"Quaternion plaque on Brougham (Broom) Bridge, Dublin, which reads: Here as he walked by on the 16th of October 1843 Sir William Rowan Hamilton in a flash of genius discovered the fundamental formula for quaternion multiplication

{{{1}}}

& cut it on a stone of this bridge "

Thank you. -kho ~2026-11603-2 (talk) 15:42, 6 January 2026 (UTC)

 Done Thank you for pointing this out.—Anita5192 (talk) 16:32, 6 January 2026 (UTC)

Quaternion square root

There are new results on how to calculate the quaternion square root in Lemma 3.2 in https://arxiv.org/abs/2510.04629. I also posted it here:

https://math.stackexchange.com/questions/382431/square-roots-of-quaternions

What do you think, could you consider this for your article? ~2026-24970-47 (talk) 12:54, 23 April 2026 (UTC)

Per WP:OR, an ArXiV preprint is not not suitable for Wikipedia, and, for being mentioned in WIkipedia, a new mathematical result requires nonly to be reliably published, but also to be discussed in other reliably published articles or books. D.Lazard (talk) 13:59, 23 April 2026 (UTC)

It is not my intention to link to the ArXiv preprint mentioned here. I only cited the source to refer to the relevant proof. The article (currently only an ArXiv preprint) has been submitted to a mathematics journal and is still under review, so it has not yet been published.

While preparing this article, we searched in vain through the literature and online for a consistent representation of the quaternion square root, because this would also ensure that the main results of the article—the singular solutions to the homogeneous and inhomogeneous Sylvester equations—are presented in a consistent manner. These, in turn, serve as an important intermediate results for a subsequent article in which we provide a closed-form solution to Wahba’s problem in the field of quaternions. Had we been unable to use a closed-form representation of the quaternion square root, many special cases would have arisen, each of which would have entailed a kind of unesthetic presentation of the solutions and increased computational effort. ~2026-25619-69 (talk) 17:11, 26 April 2026 (UTC)

With a topic as old and established as quaternions, any result claiming to be new has a high hurdle to overcome. If it makes it to several textbooks or, possibly, if it gets attention in multiple non-academic sources then it could be appropriate to cover. But even then we'd cite the textbooks or an earlier primary source. On the other hand, without those textbooks, chances are good that we won't be able to establish the result as sufficiently noteworthy. —Quantling (talk | contribs) 15:40, 23 April 2026 (UTC)

Which part is "new"? This looks like material that has been known since the 19th century. This topic seems to be covered fine in § Square roots, and no further proofs seem necessary. Calculating the square root of a quaternion is essentially the same as calculating the square root of a complex number. The basic idea to find the square root of a unit-norm quaternion is to form a parallelogram with the quaternion and 1 as two of the sides, and then intersect the diagonal with the unit sphere. To find the square root of a quaternion of arbitrary norm, you can decompose the quaternion into a unit-norm quaternion times a scalar; the square root of the product is the product of square roots. –jacobolus (t) 16:32, 23 April 2026 (UTC)

The following should be noted regarding the proposed representation of the square root of a quaternion. First, let’s revisit the actual statement:

The quaternion square root of a nonzero quaternion $a$ is given as

${\sqrt {a}}=\pm {\sqrt {|a|}}\,{\frac {p}{|p|}},$

where $p$ is either an arbitrary pure quaternion if $a$ is a negative real number, or $p=a+|a|$ if $a$ is a nonreal quaternion or a positive real number.

1. This presentation provides a common, easy-to-understand format for all cases, specifying the necessary conditions. It also allows one to immediately determine the number of possible solutions for each case.

• If $a$ is a negative real number, there are infinitely many solutions, since $p$ is an arbitrary quaternion.
• If $a$ is a non-real quaternion or a positive real number, there are only two solutions, taking into account the change in sign, since $p=a+|a|$.

2. The representation is explicit, even for negative real numbers. Furthermore, it does not require any decomposition into the scalar and vector components of quaternion $a$.

3. The proof of this statement is direct, constructive, and does not rely on any geometric interpretations. This can be particularly useful for mathematical work based on it, as it provides ideas for constructing proofs. ~2026-25619-69 (talk) 17:29, 26 April 2026 (UTC)

Fix in 1. first item:

If $a$ is a negative real number, there are infinitely many solutions, since $p$ is an arbitrary pure quaternion. ~2026-25619-69 (talk) 17:46, 26 April 2026 (UTC)

This result is essentially the same as what is shown in the article (§ Square roots of arbitrary quaternions), except that the article's version separates it by grade. Neither expression is at all novel, I'm sure you can find something extremely similar in plenty of 19th century sources (and also later sources). It's also the same as the square root of a complex number (with the caveat that the square root of ⁠$-1$⁠ is trickier), for which you should also be able to find many sources. –jacobolus (t) 18:09, 26 April 2026 (UTC)

Can you cite a reference for calculating the square root of a quaternion as suggested here? ~2026-25619-69 (talk) 21:11, 26 April 2026 (UTC)

How about Niven, Ivan (December 1941). "Equations in Quaternions". The American Mathematical Monthly. 48 (10): 654–661. JSTOR 2303304. and Niven, Ivan (June–July 1942). "The Roots of a Quaternion". The American Mathematical Monthly. 49 (6): 386–388. JSTOR 2303134.? Hawkeye7 (discuss) 23:53, 26 April 2026 (UTC)

A search for «complex "square root"» turns up among its first couple results doi:10.1007/BF01206319.

An elementary geometric construction for the complex square root of ⁠${\sqrt {z}}$⁠ in the open upper half plane would proceed as follows:

1. Normalization: Compute $\textstyle a=|z|^{-1}\cdot z$.
2. Bisection of angle: Compute $b=a+1$.
3. Normalization: Compute $\textstyle c=|b|^{-1}\cdot b$.
4. Correction of length: Compute $\textstyle d={\sqrt {|z|}}\cdot c$.

The quaternion process is identical (except for negative scalars, where you get a choice among infinitely many possible square roots). This is the same as your version except you skip the first step above and use the alternative second step $b=z+|z|$. The result is the same either way. I have seen multiple sources of exactly this in the context of quaternions, as well as similar for other kinds of multivector-like objects etc., including versions that normalize twice and versions that skip one of the normalizations, but I didn't find one in 1 minute of searching and I have other things to do than more thorough literature review. But this is the well known basic method of computing a square root. For complex numbers you can probably even find more geometrically phrased sources from the 18th century or before. If you split the result into separate scalar and vector components, you effectively get the half-angle formulas for sine and cosine, which are centuries older still. –jacobolus (t) 00:48, 27 April 2026 (UTC)

A google web search for quaternion square root turns up http://squoze.net/math/quatrot.pdf –jacobolus (t) 01:06, 27 April 2026 (UTC)

Yes, of course, I’m familiar with Niven’s papers. We’ve also linked to them in our article. However, none of the publications mentioned contain the general formulation for the square root of quaternions discussed here. It’s also not helpful to refer to complex numbers, since these are entirely different mathematical objects, and statements about complex numbers cannot simply be applied to quaternions. Of course, such statements about complex numbers can suggest whether something similar might also hold true for quaternions. Ultimately, however, only a proof can confirm such a conjecture. We have provided such a proof.

The statement made here can even be extended to the trivial case of the square root of the zero quaternion:

The quaternion square root of an arbitrary quaternion $a$ is given as

${\sqrt {a}}=\pm {\sqrt {|a|}}\,{\frac {p}{|p|}},$

where $p$ is either an arbitrary pure quaternion if $a$ is a non-positive real number, or $p=a+|a|$ if $a$ is a nonreal quaternion or a positive real number. ~2026-25619-69 (talk) 12:56, 29 April 2026 (UTC)

This is the same as what is described above. The $p/|p|$ factor is just the normalized bisection of the chord between 1 and a/|a|. Sławomir Biały (talk) 13:18, 29 April 2026 (UTC)

And what if the quaternion is a non-positive real number? ~2026-25619-69 (talk) 14:00, 29 April 2026 (UTC)

Then it's also completely standard (it is solved in any etale subalgebra). Sławomir Biały (talk) 14:06, 29 April 2026 (UTC)

If it's 0, then the square root is 0. If it's a negative scalar, then the result is degenerate and you can pick among the infinite choices or throw an error, depending on your context. –jacobolus (t) 17:03, 29 April 2026 (UTC)

If these individual results are already known, why isn’t a consistent presentation of the quaternion square root in the wiki article desirable? When I first asked, I was told that only established knowledge can be included in wiki articles. Now I’m being told that everything is already known. So which is it?

Isn't the point to do everything possible to present well-known information to Wikipedia users in the simplest way possible, so that they can easily reuse it for their own purposes?

The current Wikipedia article does not provide a consistent representation of the square root of a quaternion. In the case of negative real quaternions, there is no explicit representation of the square root of a quaternion at all. There is merely a description of the solution space. This is unsatisfactory for Wikipedia users, myself included. It can be done much better, and I would like to contribute to and help improve it.

So, to repeat my question: Wouldn't it make sense to use a consistent notation for the square root of a quaternion in the wiki article, one that applies to arbitrary quaternions? ~2026-25619-69 (talk) 09:50, 1 May 2026 (UTC)

It seems to me that everything that you ask for is aleady in section § Square roots of arbitrary quaternions. Please, be clearer on which points are lacking. D.Lazard (talk) 14:52, 1 May 2026 (UTC)

I don't have a problem with also mentioning that the square root of a quaternion can be found by a process equivalent to (1) bisecting the angle, (2) normalizing, and (3) multiplying by the square root of the norm. But the version currently in the article, which is equivalent but split into separate real/imaginary components, would also be fine to leave alone. Your claim that it is not explicit or consistent doesn't seem justified. –jacobolus (t) 14:53, 1 May 2026 (UTC)

Commutativity of multiplication of H-holomorphic functions

My recent edit of the article "Quaternion" was rejected by D. Lazard. Therefore, I propose the following for discussion. The gist of my editing of the quaternion paper is as follows. The non-commutativity of quaternion products is not an absolute truth. There are quaternion functions, called holomorphic, whose quaternion product is commutative. The commutative behavior of the product of H-holomorphic functions is a property of quaternion holomorphic functions established in 2020 (see Theorem 3.5 in https://pubs.sciepub.com/ajma/8/1/3/index.html). This property is familiar to many readers of the Research Gate website and can therefore hardly be considered original research at this time. This has been 100% theoretically proven and repeatedly confirmed by dozens of example functions (see Example 3.6 in https://pubs.sciepub.com/ajma/8/1/3/index.html, Section 4.3 in https://arxiv.org/abs/2402.08487). To avoid time-consuming computations, a special tool for checking commutativity is available in the Wolfram Mathematica language (see the Appendix at http://pubs.sciepub.com/ajma/8/1/3). H-holomorphic analogs of any two (complex) holomorphic functions can be obtained by replacing the complex variable as a single whole with a quaternion variable in the expressions for the original (complex) holomorphic functions. You can then represent the resulting H-holomorphic functions in the Cayley-Dixon construction and check their commutativity manually or using a computer-based commutativity tool. I propose the following experiment: give me any two (complex) holomorphic functions you like. I will construct their H-holomorphic analogues and show computationally that their quaternion product behaves commutatively. I believe that without this important property of quaternionic (holomorphic) functions, the Wikipedia article "Quaternion" does not provide a complete understanding of quaternions, and I ask you to reconsider your assessment. neronchaga nerch (talk) 15:44, 29 April 2026 (UTC)

I removed the section that you added, because it does not respect the fundamental policies of Wikipedia, especially WP:No original research. The references that you provide are not published in a peer reviewed journal. Moreover, for been acceptable, a content must have been discussed in a reliable WP:secondary source, which seems not be the case here. D.Lazard (talk) 18:45, 29 April 2026 (UTC)

Deriving the multiplication table

The article does not go into any depth (yet) on how the multiplication table is derived, so I will. The secret is in the loss of commutativity, and the subsequent splitting of multiplication into left and right-hand variants.

Zundamon's theorem does a wonderful job iterating through explaining this, so I'll replicate it here, but from a different starting point.

We'll start from the core identity, $ijk=-1$. We can either left-multiply by $i$, or right-multiply by $k$.

For brevity, I'll only do the left multiplication path, which gives us the following:

$ijk=-1$

Left-multiply both sides by $i$ to get $iijk=-i$, which cleans up to $i^{2}jk=-i$

$i^{2}$ turns to -1, so we get $-jk=-i$, and now we can invert on both sides to get $jk=i$.

This process can then be iterated, using either left or right-side multiplication, to build out the identities for positive and negative $i$, $j$, and $k$. ~2026-38641-42 (talk) 17:34, 6 July 2026 (UTC)

I've made a stab at editing the article accordingly. What do you think? —Quantling (talk | contribs) 20:06, 6 July 2026 (UTC)

I think the previous one was fine (and probably slightly better even) but it doesn't really make too much difference. –jacobolus (t) 23:03, 6 July 2026 (UTC)

It's close, but doesn't quite bootstrap the consequences of what losing the commutative property does to the process, and how you get left and right multiplication as a result.

Perhaps I've been spoiled by things like the video above being a full self-contained primer? ~2026-38641-42 (talk) 17:03, 7 July 2026 (UTC)

I don't think the video's argument is really appropriate for an encyclopedia article. (I also don't personally find it too illuminating, YMMV.) 20:46, 7 July 2026 (UTC) –jacobolus (t) 20:46, 7 July 2026 (UTC)