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To understand this, you should read about the Cayley-Dickson construction.
For each algebra of the Cayley-Dickson construction, the conjugate of a product is the product of the conjugates in opposite order. That is, for any two elements p and q of such an algebra, ( p q )* = q* p* . The statement for any one algebra implies the same statement for the next algebra, so it holds for all of them by induction.
Now suppose the statement is true for one step of the Cayley-Dickson construction. Then it is also true for the next step in the construction: using only the Cayley-Dickson formulas for multiplication and conjugation,
while
The statement is trivially true of real numbers, since the real numbers are commutative and a real number is its own conjugate. By induction, it is true for all steps of the construction.
That the present algebra in the Cayley-Dickson construction is associative follows from the commutativity of the previous algebra. This will imply that the quaternions are associative, but say nothing about the octonions.
The Cayley-Dickson multiplication formula gives
| [ ( a, b ) ( c, d ) ] ( e, f ) | = ( ac − bd*, bc* + da ) ( e, f ) |
| = ( [ ac − bd* ] e − [ bc* + da ] f*, [ bc* + da ] e* + f [ ac − bd* ] ) , |
while, using the fact that the conjugate of a product is the product of the conjugates in opposite order,
| ( a, b ) [ ( c, d ) ( e, f ) ] | = ( a, b ) ( ce − df*, de* + fc ) |
| = ( a [ ce − df* ] − b [ de* + fc ]*, b [ ce − df* ]* + [ de* + fc ] a ) | |
| = ( a [ ce − df* ] − b [ ed* + c*f* ], b [ e*c* − fd* ] + [ de* + fc ] a ) . |
Assuming that a, b, c, d, e and f commute, as they do if they are all complex numbers, the two expressions are seen to be the same.
To show that the octonions aren’t associative, it suffices to produce one example of three octonions p, q and r such that
Such an example is presented below.
Note that if any of the three corresponds to a real number, the order of the multiplication does not matter. If all the octonions correspond to quaternions, the order of multiplication will not matter, because the quaternions are associative. An example of three octonions that don’t associate must therefore include octonions that don’t correspond to quaternions, and must not include a quaternion corresponding to a real number.
With that in mind, it’s easy to find an example of octonions that don’t associate. Set p = (a, b), q = (c, d), r = (e, f). Then, after several applications of the multiplication formula,
A notaton for the commutator facilitates a more compact presentation: let [a, b] = ab* − b*a:
From this expression, it is very easy to find non-associative octonions. (Indeed, it appears to be rare for octonions to associate, given that it is rare for quaternions to commute). For instance, taking b = 0, f ≠ 0, any a and d that don’t commute, and any c and e at all, then the expression will not be 0.
More particularly, taking b = 0, f = 1, a = i and d = j (which don’t commute), and c = e = 0: this yields
Question: exactly when do three octonions associate? Clearly, they do if all three correspond to quaternions. Also, if any one of the three corresponds to a real number, they associate. Likewise, if any two consecutive octonions of the three correspond to complex numbers, the three associate. Are there any other cases?
Question: does the commutator reside in any particular subspace, or have any other nice properties?