11
$\begingroup$

Is this simply because of marketing, hype, etc?

The bicomplex numbers (especially tessarines) look just great being commutative and all.

enter image description here

enter image description here

Images source:https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.527.356&rep=rep1&type=pdf

$\endgroup$
15
  • 23
    $\begingroup$ Quaternions are useful. In particular for representation of rotations of 3-space. $\endgroup$ – Alexandre Eremenko Feb 13 at 16:45
  • 1
    $\begingroup$ Thanks. Besides, I have written, some years ago an answer on MathStackExchange giving as well different correspondances of bi-complex numbers, quaternions, etc. with matrices here $\endgroup$ – Jean Marie Becker Feb 13 at 20:22
  • 1
    $\begingroup$ @JeanMarieBecker look at this my answer also: math.stackexchange.com/questions/1187160/… It shows advantages of bicomplex numbers over quaternions, but tessarines are even better in that its $j^2=1$ coincides with hyperbolic numbers, and we do not need to add another unity. So, we have i, j, ij, adding only complex unity and hyperbolic unity to reals. Also, they have alternating signs of squares of unities, so that we can make 16-dimentioned tessarines, which are still commutative and associative. $\endgroup$ – Anixx Feb 13 at 20:29
  • 15
    $\begingroup$ Maybe, it is a little exaggerate to say such definitive things like "a huge advantage of tessarines over all other algebras and total uselessness of quaternions"... $\endgroup$ – Jean Marie Becker Feb 13 at 22:29
  • 4
    $\begingroup$ @Anixx If a study shows the 'total uselessness' of quaternions, then that study is clearly flawed, as illustrated by the many uses of quaternions. $\endgroup$ – user13928 Feb 15 at 12:53
39
$\begingroup$

Commutativity is over-rated: in fact, it holds back bicomplex numbers:

  • It prevents your number system characterising non-commuting operations, e.g. rotations in $3$-dimensional space, Hamilton's original focus.
  • Since $0=i^2-j^2=(i-j)(i+j)$ in bicomplex numbers, you have zero divisors, so it's not a normed division algebra; no convenient conjugates, no division, which hinders the description of invertible processes - again, such as rotations. (Apart from conjugation, these are also issues with "hyperbolic" or split-complex numbers, viz. $0=1-j^2=(1-j)(1+j)$.) In this respect, your number system runs into difficulties even octonions don't face, though sedenions do. I can't offer a comprehensive overview of how this has stymied historical use, but uses of split-complex numbers have never become widespread (even when they arguably deserved to be, e.g. you can use split-complex numbers in relativity, but ironically complex numbers have proved more popular for this).
$\endgroup$
15
  • 24
    $\begingroup$ @Anixx That's just a made-up rule. You might as well say complex numbers aren't numbers because you want numbers to be totally ordered. "Numbers" are whatever we need. $\endgroup$ – J.G. Feb 13 at 23:13
  • 17
    $\begingroup$ @Anixx Do you mean because fewer rearrangements are legal? That never upset group theorists. For many applications, that division is undefined is much worse than some equations no longer being true. $\endgroup$ – J.G. Feb 13 at 23:17
  • 16
    $\begingroup$ @Anixx, the most important attribute a mathematical system can have is that it is useful. Things like "stylistic elegance" take a distant second place. $\endgroup$ – Mark Feb 13 at 23:59
  • 12
    $\begingroup$ (+1) for mentioning zero-divisors. $\endgroup$ – Franklin Pezzuti Dyer Feb 14 at 0:53
  • 17
    $\begingroup$ @Anixx You asked an HSM historical question, why have bicomplex numbers been used less than quaternions? The meaning of the word "number" is irrelevant to that question. Mathematicians, physicists etc. use whichever mathematical objects they've found use for. My answer focused on the historical reasons invertible non-commutative objects (which, yes, have a matrix representation) were more helpful than non-invertible commutative objects (which, yes, have a matrix representation). $\endgroup$ – J.G. Feb 14 at 7:29
22
$\begingroup$

Your description "total uselessness of quaternions" in a comment above is poorly chosen, and reflects more on your interests than on the real state of knowledge of mathematics. The Hamilton quaternions are the simplest nontrivial example of a quaternion algebra, which has turned out to be a really important concept in mathematics.

It is useful to think of the Hamilton quaternions $\mathbf H$ as being analogous to the ring of $2 \times 2$ real matrices ${\rm M}_2(\mathbf R)$. At first these might seem quite different: one is a division ring and the other is not. But here are analogies between them: both are 4-dimensional with center the real numbers and in both of them there is no 2-sided ideal other than $(0)$ and the whole ring. A ring that is 4-dimensional over a subfield $K$, has $K$ as its center, and has no 2-sided ideals other than $(0)$ and the whole ring is called a quaternion algebra over $K$. The matrix ring ${\rm M}_2(K)$ is a quaternion algebra over $K$, and it is the only kind that is not a division ring.

Sometimes ${\rm M}_2(K)$ is the only quaternion algebra over $K$, e.g., when $K = \mathbf C$ and when $K$ is a finite field. There are two quaternion algebras over $\mathbf R$: ${\rm M}_2(\mathbf R)$ and $\mathbf H$. Similarly, there are two quaternion algebras over the $p$-adic numbers for each prime $p$. For $p = 2$ the two examples are ${\rm M}_2(\mathbf Q_2)$ and the Hamilton quaternions with $\mathbf Q_2$-coefficients, but for odd primes $p$ you need to do something different to get the "nontrivial" example because the Hamilton quaternions with $\mathbf Q_p$ coefficients is isomorphic to ${\rm M}_2(\mathbf Q_p)$ rather than being a division ring.

Things get really interesting for fields like $K = \mathbf Q$, because there are infinitely many different quaternion algebras over $\mathbf Q$. The ring ${\rm M}_2(\mathbf Q)$ and the rational Hamilton quaternions are merely two examples out of a huge list of rational quaternion algebras. These "new" rational quaternion algebras show up in various places in mathematics: (supersingular) elliptic curves, Brauer groups, Shimura curves, and hyperbolic geometry in 2 and 3 dimensions. Take a look at John Voight's book in order to appreciate that you should not dismiss quaternions as being a "totally useless" concept. If you look at Ribet's paper that showed Fermat's Last Theorem is a consequence of modularity of elliptic curves over $\mathbf Q$, you'll see a role there for rational quaternion algebras that are division rings and are not the rational Hamilton quaternions.

$\endgroup$
11
$\begingroup$

Hamilton expected that the quaternions would be of physical interest. In this, he was right. But he was too early. He had discovered them in 1843, it was almost a century later, in 1928, when Dirac discovered his equation involving the Pauli matrices, that it was seen that the quaternions were naturally implicated in quantum field theory. (Here, the Pauli matrices are a representation of the quaternions on the space of 2 complex dimensional space).

$\endgroup$
4
  • 6
    $\begingroup$ Maxwell actually based his equations on quaterions. The whole vector calculus was in quaternions before Gibbs and Heaviside. There are other ways how to represent 3D rotations and Pauli matrices too (e.g. geometric algebra). $\endgroup$ – Vladimir F Feb 14 at 15:15
  • $\begingroup$ @Vladimir F: I'd forgotten that Maxwell had used quaternions. Coincidentally, I was just looking at geometric algebra - aka Clifford alfebras - before just coming onto the site now. $\endgroup$ – Mozibur Ullah Feb 14 at 16:29
  • 1
    $\begingroup$ The quaternions may have helped inspire the algebra of Dirac gamma matrices, but the two are different. The crucial difference is that $\gamma^0$ anticommutes with all of $\gamma^1$ through $\gamma^3$, whereas in the quaternions $1$ commutes with all of $i$, $j$, and $k$. For the Pauli matrics, we are just back to 3D rotations. $\endgroup$ – Lars H Feb 15 at 12:22
  • $\begingroup$ @Lars H: Thanks for pointing this out. I'll edit my post when I find out a bit more. I think I should be thinking about Clifford algebras rather than quaternions here. $\endgroup$ – Mozibur Ullah Feb 15 at 13:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.