Discussion Groups / English Usage / November 2006

They end up being isomorphic, of course.

Most people use it the other way: i*j = k, j*k = i, k*i=j. That way,
the first equation is in "alphabetical order".

The "deal" is that it doesn't matter, so long as you are consistent.

Let us assume you want to define quaternions "your" way, with i*k = j,
j*i=k, and k*j=i.

Every quaternion can be written uniquely as a + b*i + c*j + d*k.

Okay: I'm going to define "J" to be shorthand for "-j". So

3 + 4*i -5*J + k

means

3 + 4*i -5(-j) + k = 3 + 4*i + 5j + k.

Now, clearly, every quaternion can be written uniquely in terms of i,
J, and k, just as it can in terms of i, j, and k.

And multiplication will now obey the following rules:

i*J = i*(-j) = -(i*j) = -(-k) = k.
J*i = (-j)*i = -(j*i) = -(k) = -k.
J*k = (-j)*k = -(j*k) = -(-i) = i.
k*J = k*(-j) = -(k*j) = -(i) = -i.
i*k = j = -J.
k*i = -j = J.

so the multiplication table for i, J, k is just the "other" possible
definition. It is also pretty easy to see that if you let x be any
quaternion written in terms of i, j, and k; and you let f(x) be any
quaternion written in terms of i, J, and k, then

f(x+y) = f(x) + f(y)
f(xy) = f(x)f(y)

so that anything you do using i, j, and k, you get the same thing
(with a flipped sign) if you use i, J, and k instead.

So this means that you can define the multiplication either way, and
everything will be "essentially" the same. It is just a matter of
convenience. And, as I mentioned, as far as I know most people find
the definition i*j = k more convenient.

(You could do the same thing replacing i with I=-i, or k with K=-k,
instead of replacing j).

Consider that the basis vectors 1, i, j, k have in themselves no meaning
apart from the quaternion division algebra.  The form I learned
[Herstein, _Topics in Algebra_, 1964] with the multiplication table

  x  1  i  j  k
  1  1  i  j  k
  i  i -1  k -j
  j  j -k -1  i
  k  k  j -i -1

can be written with J=k and K=j as

  x  1  i  J  K
  1  1  i  J  K
  i  i -1 -K  J
  J  J  K -1 -i
  K  K -J  i -1

which corresponds to your form.

In either form,
1) ii = jj = kk = ijk = -1
2) ij = -ji
3) jk = -kj
4) ki = -ik

The more common form has the advantage that (2-4) are all positive, and
the cycle ij (=k), jk (=i), ki (-j) seems intuitive.
In your form, the positive products are ik (=j), kj (=i), and ji (=k),
an order which doesn't have alphabetization to speak for it. In any
case, there is little reason other than convention to lead one to prefer
one of these equivalent forms over the other.

The conventions "i*j = k etc" and "j*i = k etc" are transformed into
each other by quaternion conjugation.

The issue reminds me of a joke my teacher of Latin and Greek told us in
the classroom: "The Iliad and Odyssey were not written by Homer, but by
another poet also named Homer.".

Johan E. Mebius

Iam really surprised to see the convention you have used.but as most of
them have discussed it is all with how we assign the three axis.let me
just give a rule to clearly come to conclusion of the sign.
let us consider three mutually perpendicular unit vecors namely i,,j,k.
now let me define,what is i* j.
first you look from the direction of 'k' vector,then you try to rotate
'i' vector to reach 'j' vector. if your direction of rotation is
counter-clockwise then your result is '+k',if your direction of
rotation is clockwise,then the result is '-k'.
this helps you to easily find the sign and reduce our confusion.

As I learned it, in the US, i × j = k, but in the UK, i × j = -k.  It
goes along with driving on opposite sides of the street.

It's a convention, not something you can determine, like whether
electrical current flows from positive to negative or negative to
positive.  It flows in the direction your instructor tells you that it
flows.