[seqfan] The quaternion question
Robert Munafo
mrob27 at gmail.com
Sat Nov 21 17:52:24 CET 2009
Thanks for the replies. In my case, reading the Wiki pages on formal maths
topics doesn't help me much, because they are too abstract. Like I said
before I just want to know how to calculate these d**n sequences. That's the
point of all this: we have over 200 OSIE entries that need simple
instructions on how to calculate them.
I have found a relationship to quaternions. You can set *all but four* of
the floretion components to zero, to make the definitions of X and Y like
this:
X = P + A'i + B'j + C'k, and Y = p + a'i + b'j + c'k
Then the multiplication formula becomes:
X*Y = ( Pp-Aa-Bb-Cc)
+ ( Pa+Ap+Bc-Cb)'i
+ ( Pb-Ac+Bp+Ca)'j
+ ( Pc+Ab-Ba+Cp)'k
It is now easy to verify that:
'i^2 = 'j^2 = 'k^2 = 'i * 'j * 'k = -1
'i * 'j = 'k, and 'j * 'i = -'k
'j * 'k = 'i, and 'k * 'j = -'i
'k * 'i = 'j, and 'i * 'k = -'j
These are the same multiplication rules seen in quaternions. The same thing
happens if you choose the second set of "i j k" instead.
The rest of the floretion multiplication table seems to extend from this, in
the sense that when you combine two of the same elements (e.g. when
multiplying 'i times 'ik') the two identical elements ('i in this example)
"cancel out" into a sign change. When you multiply something like 'ji' times
'i, the 'j and 'i become 'k, and the answer is 'ki' or -'ki' depending on
the order of the factors.
Perhaps this is what "containing the quaternions" means?
Joerg Arndt wrote:
> And the pdf seems to indicate that the algebra contains two copies of the
> quaternions. Progress indeed!
>
--
Robert Munafo -- mrob.com
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