3
$\begingroup$

Recently i read the article "Hamilton, Rodrigues, Gauss, Quaternions and Rotations: A Historical Reassessment", which can be found freely on the internet. This article is by far the most comprehensive paper i found on the early history of quaternions: it discusses Euler's glimpses of a quaternions algebra, and Gauss's anticipations of it in his posthomously published 1819 fragment "mutations des raumes" (english: "rotations of space").

The article gives a quite complete assessement of this fragment of Gauss: it lists (on p. 11-12) several aspects of Gauss's "algebra of rotations", which correspond almost exactly to the eight subsections of part I of Gauss's fragment. However, it doesn't comment about subsection 5; in it Gauss writes down, as far as i understand it, several congruences involving the elements (i.e, the quaternion coefficients $a,b,c,d$) of a composition of two spatial rotations (two quaternions) modulo the norm of one quaternion.

More specifically, Gauss defines the following:

$$ A = a\alpha - b\beta - c\gamma - d\delta$$ $$B = a\beta + b\alpha - c\delta + d\gamma $$ $$C = a\gamma + b\delta + c\alpha - d\beta$$ $$D = a\delta - b\gamma + c\beta + d\alpha$$ and the norms: $m = a^2 + b^2 + c^2 + d^2, \mu = \alpha^2 + \beta^2 + \gamma^2 + \delta^2$, and then observes that, for example:

$$\frac {{C+Di}}{{A + Bi}} \equiv \frac {{c + di}}{{a + bi}} \pmod m$$

One thing that immediately came to my mind is what is "$i$"? he didn't define it previously, and my second question is: what's the meaning of this congruence? in fact Gauss writes six such congruences (anyone interested can find it in Gauss's werke, volume 8, p.359)

$\endgroup$
  • $\begingroup$ I just want to acknowledge and encourage the groundwork you are doing with tracing lesser known facts through the original sources. I wish I could be of more help but unfortunately I do not have as much time as I used to. $\endgroup$ – Conifold Aug 8 '18 at 5:15
  • $\begingroup$ Thanks sincerely Conifold! It's very encouraging to know that people here don't only view me as a snooze user. $\endgroup$ – user2554 Aug 9 '18 at 8:01
1
$\begingroup$

I asked a mathematician who is an expert to abstract algebra and he showed me that Gauss's congruence was actually correct. To prove Gauss congruence let's introduce the following notation: $$x = a+bi,y = c+ di, u = \alpha + \beta i, v = \gamma + \delta i, X = A + Bi, Y = C + Di$$.

First of all, one has to understand that Gauss's notation is different from modern convention in two aspects:

  • When Gauss designates a quaternion by a collection of four coefficients $(a,b,c,d)$, he means $a+ib+jc+kd$ (not $a+bi+cj+dk$).
  • Secondly, Gauss defines quaternions multiplication in such way that the product of two fundamental quaternions gives the third one with positive sign if the multiplication operation is done counterclockwise (not clockwise like in modern convention); that is, $ij=-k$ and $jk = -i$. That he defines quaternions multiplication in this way is evident from the bilinear expressions he give for the four coefficients $(A,B,C,D)$ of the product.

Therefore each of the quaternions $q_1= a + ib + jc +kd $ and $q_2 = \alpha + i\beta +j\gamma + k\delta $ can be written as: $$q_1 = x + jy$$ $$q_2 = u + jv$$

and:

$$q_3 = q_1q_2 = X+jY = (x + jy)(u+jv) = (xu - \bar y v) + j(yu + \bar x v)$$

The complex conjugate symbols appear because of the non-commutitivity of quaternions algebra; that is, $jy = \bar yj$. Finally, to prove Gauss's congruence, let's make the following step (multiplying both sides of the congruence by $(A+iB)(a+ib)$):

$$(C+Di)(a + bi)-(A+Bi)(c+di) = Yx - Xy = (yu + \bar x v)x - (xu - \bar y v)y = (x\bar x + y\bar y)v + (xy - xy)u =\rVert x + jy \rVert ^2 v$$

Since $\rVert x + yj \rVert ^2 = m$ one gets that that the result equals the product of m and an integral complex number.

$\endgroup$
  • $\begingroup$ The mathematician who gave this reconstruction also explained Gauss's motivation: he said something about orders of quaternion algebras and the structure of their quotient rings modulo the norm ideal (so Gauss's result can be interpreted as a research in this direction). $\endgroup$ – user2554 Jan 15 at 16:29

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.