Can Quaternions Be Generalized Beyond Four Dimensions?

[lpetrich]

Working with rotation matrices led me to consider quaternions, and that led me to consider the question of generalizing the quaternion group over different numbers of components of the quaternion vector. I've attempted to work out what possibilities there are, though someone else may have anticipated me in the professional literature. That someone may even have come up with a simpler proof than what I'd come up with.

The quaternion operation I define for elements q1 and q2 as (q1*q2)_i = c_ijkq1_jq2_k

The identity will be qid, and the inverse for element q will be inv(q) = (Mi.q)/(q.M.q) for matrices M and Mi. Matrix M is symmetric, and it must have nonzero eigenvalues. It can be diagonalized by by redefining elements q: q -> D.q for some appropriate D. So without loss of generality, we can set
inv(q) = (Mi.q)/(q.q)

Turning to the question of the identity, one can rotate the q's, so that the identity has the form (id,0,0,0,...,0). Rescaling the q's gives it the value (1,0,0,0,...,0). This results in the familiar scalar-vector decomposition of the quaternions. The scalar index of the q's I will call s and the vector part v:
qid_s = 1, qid_v = 0

In what follows, indices i,j,k, ... will run over the vector part.

The constraint q*qid = qid*q = q leads to
c_sss = 1
c_iss = c_sis = c_ssi = 0
c_ijs = c_isj = d_ij
Mi_ss = 1
Mi_is = 0
with d_ij being a shorthand for delta_ij.
Likewise, the constraint q*inv(q) = inv(q)*q = qid leads to
c_sij + c_sji = - 2*d_ij
c_ijk + c_ikj = 0
Mi_si = 0
Mi_ij = - d_ij
thus making the inverse
inv((q_s,q_v)) = (q_s,-q_v)/(q_s² + q_v.q_v)
It's easy to show that inv(inv(q)) = q, as expected.
In this notation, qid = (1,0)

The remaining constraint is associativity. It yields
c_sij = c_sji = - d_ij
c_kij = c_ijk = c_jki = - c_kji = - c_ikj = - c_jik
- d_ijd_kl + d_ild_jk + c_lmkc_mij - c_limc_mjk = 0

Taking +(ijk)+(jki)-(kij) yields
c_mijc_mkl = d_ikd_jl - d_ild_jk

Multiplying by c_lpq and renaming indices yields
d_ikc_jlm - d_jkc_ilm = d_kmc_lij - d_klc_mij

Contract over kl, and for n vector dimensions, (n - 3) * c_ijm = 0

So the only possible solutions are:

n = 0 -- trivial
n = 1 -- construction of complex extensions of the real numbers and its subfields
n = 3 -- quaternions: c₁₂₃ = +1 or -1

For all other n >= 2, one gets a contradiction -- the c's must either be all zero or some of them must be nonzero.

So it looks like it's not possible to generalize quaternions any further without sacrificing some group property.

 

[D H]

There are biquaternions, octonions, sedonions, etc. You might want to look into the Cayley-Dickson construction.

Stepping up in the progression (reals, complex numbers, quaternions, ...) results in a loss of something. Complex numbers can't be compared (asking whether i is less than 1 doesn't make sense). Multiplication is associative but not commutative in the quaternions, alternative but not associative in the octonions. Beyond that, there isn't much structure left at all.

 

[lpetrich]

Like John Baez's Octonions?

The Cayley-Dickson construction of the next level of 2^n-ion is to create pairs of members of the current level, with component-by-component addition and scalar multiplication, and conjugation and multiplication defined by

conjg((a,b)) = (conjg(a),-b)
conjg(real number) = itself

(a1,b1)*(a2,b2) = (a1*a2 - b2*conjg(b1), conjg(a1)*b2 + b1*a2)
* on real numbers: multiplication

 

[D H]

Baez did not invent the octonions. They were invented long before he was born.

 

[lpetrich]

True, but that seemed to me a good document on them.

For the 3D case, I found 4 solutions that satisfy associativity and having an identity, thus making the system a monoid. Though I could find inverses, inv(q) was either (quadratic in q)/(cubic in q), with a factorable denominator, or both. None of these solutions could be turned into form
(Mi.q)/(q.q)

From Cramer's rule, the n-D case has inverses with the general form
(degree-(n-1) polynomial in q)/(degree-n polynomial in q)

I think that what I'd posted shows that they can be reduced to form (Mi.q)/(q.q) only for 1D, 2D, and 4D -- the first three members of the Cayley-Dickson series.

 

[Deveno]

"Über lineare Substituionen und bilineare Formen" Journal für die Reine und Angewandte mathematik, Vol. 84 (1877)

if a division ring is finite-dimensional over the field R, it is algebraic over R. it is a consequence of the Fundamental Theorem of Algebra that a division ring which is algebraic over C is in fact, C itself.

now if D is a division ring that is also a vector space over R, then the subspace R1 = {r1, r in R, 1 the multiplicative identity of D} lies in the center of D. if D is commutative, D = C, if D is not commutative, then the center of D = R.

the gist of the theorem of Frobenius is that if D is a division ring algebraic over R, with the center of D = R, then D is isomorphic to the quaternions.

in other words, the complex numbers are "the best possible algebraic extension field of the reals", and the quaternions are "the best possible extension division ring of the reals".

it is possible to construct several interesting "hyper-complex" number systems, but they all suffer from various kinds of algebraic defects.

loss of associativity for the octonions is particularly hard-hitting, because we would like to be able to view a member of a division algebra (a dvision ring that is also a vector space over a field F) as a function that somehow "operates" on other elements of the division algebra, and functions have associativity built-in by the way of associativity of composition.

 

[lpetrich]

A division ring is sometimes called a division algebra or a skew field, an algebraic field where multiplication need not be commutative. A finite skew field is always a true field (Wedderburn's theorem), while that need not be the case for an infinite one. Wedderburn's theorem relies on member-counting and divisibility arguments, arguments that require finiteness.

For constructing a division ring D from a field F, one can express the elements of D as
a₀ + a₁*e₁ + a₂*e₂ + ...

where the a's are members of F, and where 1 and the e's form a basis set. There has to be at least 2 of the e's for D to be noncommuting, and
e_i*e_j = f_ij0 + f_ijke[_k

where the f's are in F.


One can construct a multiplicative inverse of an element a in terms of its a_i's and the f values, though that is no guarantee that every element is invertible. For n e_i's, the inverse's numerator has degree n in the a's, and its denominator has degree (n+1) in the a's, though they may share common factors. That's from Cramer's rule for systems of linear equations.

For noncommuting n = 2 / vector-space dimension 3, the denominator has three linear factors, one of which cancels out with the numerators. This leaves two, and that means that some nonzero elements can lack multiplicative inverses, violating the division-ring condition for D.

So for being noncommuting, one needs at least 3 e's, yielding a vector-space dimension of at least 4.

The quaternion solution,
e_i*e_j = - delta_ij + epsilon_ijke_k

is one solution, but I've been unable to show that there are no other noncommuting ones, to within isomorphism.

That is no violation of Wedderburn's theorem, because a quaternion's inverse requires dividing by its absolute square, and because it is always possible to choose finite-field members that make that absolute square zero. From Lagrange's four-square theorem, any nonnegative integer can be expressed as the sums of squares of 4 nonnegative integers, and taking modulo some value p, it's possible to choose 4 nonzero values such that their sum of squares is 0 mod p.

 

[lpetrich]

With only one of the e's, one gets a quadratic extension of the original field, like the complex numbers from real numbers. If all nonzero-element inverses exist, then for GF(pⁿ), one gets GF(p²ⁿ).

Life becomes *much* easier if one uses some other constraints, like having a norm:

|a| = N_ija_ia_j
|a*b| = |a|*|b|

for a, b, and N having components in F, while relaxing associativity for *. Let the norm have the constraint that if |a+b| = |a| + |b| for all b, then a = 0 (N has no nullspace).

For F being GF(2ⁿ) (characteristic 2), |a+b| = |a| + |b|. Otherwise, one can define an inner product from the norm:
<a,b> = (1/2)*(|a+b| - |a| - |b|)
<a,b> = N_ija_ib_j

It is symmetric and bilinear, and one can prove various theorems, like
<a*c,b*c> = <c*a,c*b> = <a,b>*|c|
<a*c,b*d> + <a*d,b*c> = 2*<a,b>*<c,d>
If <a,b> = 0 for all b, then a = 0
Suppose this to be true for GF(2ⁿ) also. In that case, <a*c,b*d> = <a*d,b*c>.

Setting d = 1, one can define a conjugate, cjg(a) = 2*<a,1> - a
and we find <a*c,b> = <a,b*cjg(c)> and <c*a,b> = <a,cjg(c)*b>
This leads to several results, like
cjg(cjg(a)) = a, cjg(a*b) = cjg(b)*cjg(a), |cjg(a)| = |a|, a*cjg(a) = |a|, and a*b + b*a = 2(a*<b,1>+b*<a,1>-<a,b>).

Also, inv(a) = cjg(a)/|a|. So a will have an inverse if |a| is nonzero.

For GF(2ⁿ), cjg(a) = a, a*b = b*a -- commutativity, and <a*b,c> = <a,b*c>. One can also prove associativity. But inverses need not exist, and they need not exist for other finite fields.

Back to the general case, though * need not be associative, one can show that * is alternating (2-element associative): (a*a)*b = a*(a*b), (a*b)*a = a*(b*a), (b*a)*a = b*(a*a).


We can address the question of what extension algebras are possible by imagining extending subalgebras of it with single elements. Take a subalgebra S and an element not in it, i, to make a superalgebra of it: Si. That extra element is chosen to satisfy <i,a> = 0 for all a in S, which is always possible here. One gets the Cayley-Dickson construction:
(a + b*i)*(c + d*i) = (a*c - |i|*d*cjg(b)) + (b*cjg(c) + d*a)*i
cjg(a + b*i) = cjg(a) - b*i
|a + b*i| = |a| + |i|*|b|
One can prove several things:
* is associative in S
* is commutative in S <-> * is associative in Si
All elements of S are self-conjugate: a = cjg(a) <-> * is commutative in Si

So one finds Hurwitz's theorem, which states that the only normed division-algebra extensions of a field are that field itself, a 2D complex-like extension, a 4D quaternion-like one, and an 8D octonion-like one. Attempting to continue with the CD construction yields a 16D sedenion-like one, but it has zero divisors, meaning that |a*b| need not equal |a|*|b|.

 

[lpetrich]

For the complex extension, we get pairs (a,b) of members of F satisfying

(a,b) * (c,d) = (a*c - |i|*b*d, a*d + b*c)
cjg((a,b)) = (a,-b)
|(a,b)| = a² + |i|*b²
inv(x) = cjg(x)/|x|

|i| is a member of F, of course, and it need not be 1.

This will be a field as long as sqrt(-|i|) is not in the field. That is the case for the real numbers and its subfields, but not for the complex numbers and many of its subfields.

For finite fields, GF(pⁿ) has a complex extension GF(p²ⁿ) if p is odd, but has none of it is odd. That is because its multiplicative group is a cyclic group with order pⁿ-1, and if that quantity is even, then a member with order pⁿ-1 is the square of no other member.

Furthermore, (-1) has order 2 in the multiplicative group for p odd, though it equals 1 for p = 2. We can thus set sqrt(-|i|) = sqrt(-1)*sqrt(|i|).

One gets the quaternion extension by using two i's, i₁, then i₂. It can be shown that a necessary but not sufficient condition for invertibility is that none of sqrt(-|i₁|), sqrt(-|i₂|), and sqrt(-|i₁|*|i₂|) may be in F. Since we are already limited to p odd, we can set |i₁| = g, a multiplicative-group generator, and |i₂| = g^m, where m is relatively prime to the multiplicative group's order. Thus, |i₁|*|i₂| = g^m+1, and since m must be odd, m+1 must be even, meaning that sqrt(|i₁|*|i₂|) is in F.

Thus, the quaternion extension of a finite field cannot be a division ring / (skew) field, since some of its nonzero elements are not invertible. But for F being the real numbers, we can set |i| = 1 for each i.

 

[lpetrich]

Returning to GF(2ⁿ) (characteristic 2), one can deduce from the Cayley-Dickson construction that all CD extension algebras have these properties:

All elements are self-conjugate -- mathematical induction
Multiplication is always commutative -- mathematical induction
Some elements do not have inverses, and the algebra has nonzero divisors of zero -- from every element of the base field having a square root in that field

Back to lost properties when repeating the CD construction, we have:

1. Complex: self-conjugacy lost
2. Quaternion: multiplication loses commutativity
3. Octonion: multiplication loses associativity (no matrix representations possible), though still alternating
4. Sedenion: multiplication no longer alternating, it also loses norm. Divisors of zero appear

No further losses of properties.

For the real numbers and its subfields, multiplication is always power-associative; it does not matter what order one multiplies to make a power of some element. Is that true in general?

I've found it hard to find a proof of power-associativity. I've tried to find one myself, using mathematical induction (if it's true for one level of C-D construction, it ought to be true for the next one, and if it's true for one power, it ought to be true for the next one), but I've had no success in doing so.

ETA: It ought to suffice to prove aⁿ = a^m*a^n-m for all a and all m such that 0 < m < n.

 

[lpetrich]

I've demonstrated power-associativity of Cayley-Dickson extension algebras, by explicit construction. It took me some experimenting with Mathematica, but I was pleasantly surprised at how simple the resulting formula is.

[itex]P(n,a,b) = \sum_k {n \choose 2k} a^{n-2k} (-b)^k[/itex]
[itex]Q(n,a,b) = \sum_k {n \choose 2k+1} a^{n-2k-1} (-b)^k[/itex]
Re(a) = (a + cjg(a))/2 = (all but the first member set to zero)
Im(a) = (a - cjg(a))/2 = (the first member set to zero)

aⁿ = P(n,Re(a),|Im(a)|) + Im(a)*Q(n,Re(a),|Im(a)|)

This formula will be true for any starting field, since the binomial coefficients are integers.


I've verified that result by using the Cayley-Dickson construction, and the identities
P(m+n,a,b) = P(m,a,b)*P(n,a,b) - b*Q(m,a,b)*Q(n,a,b)
Q(m+n,a,b) = P(m,a,b)*Q(n,a,b) + Q(m,a,b)*P(n,a,b)

which can be derived from
(a + sqrt(-b))ⁿ = P(n,a,b) + sqrt(-b)*Q(n,a,b)

An alternate approach to that formula starts with
aⁿ = (Re(a) + Im(a))ⁿ = binomial expansion

This gets the above formulas with the help of
Re(a) being proportional to the identity
Im(a) commuting with Re(a)
Im(a)² = - Im(a)*cjg(Im(a)) = - |Im(a)|

I hope that my reasoning is correct; John Baez in The Octonions made no mention of that formula or its derivation.

In my efforts to derive that formula, I had discovered that power-associativity can be expressed in generating-function fashion:
E((t+u)*a) = E(t*a)*E(u*a)
where
[itex]E(a) = \sum_{k=0}^{\infty} \frac{a^k}{k!}[/itex]
a generalization of the exponential function.

 

[lpetrich]

Power - associativity can be extended to negative powers, for elements that have multiplicative inverses:
a = Re(a) + Im(a), cjg(a) = Re(a) - Im(a), inv(a) = cjg(a)/|a|

|a| = Re(a)² + |Im(a)|
Invertibility : |a| != 0

a ^{- n} = (P (n,Re(a),|Im (a)|) - Im(a)*Q (n,Re (a),|Im (a)|))/ |a|ⁿ

Also, |aⁿ| = |a|ⁿTurning back to the quaternion - extension problem of a few posts back, where
e _i*e _j = f_ij + g_ija k _k

I've made some progress. Associativity is

g_ijaf_ak - g_jkaf_ia = 0
f_ijd_kl - f_jkd_il + g_ijag_akl - g_jkag_ial = 0

One can also redefine the e's: e_i' = T_iae_a + D_i
f_ij' = T_iaT_jbf_ab + D_iD_j - g_ija'D_a
g_ija'T_ak = T_iaT_jbg_abk + T_ikD_j + T_jkD_i

If one can set D_i = T_iaD_a' then
f_ij' = T_iaT_jb(f_ab - D_a'D_b' - g_abcD_c')
g_ija'T_ak = T_iaT_jb(g_abk + d_akD_b' + d_bkD_a')

With such transformations, it's possible to diagonalize f in algebraically-closed fields like the algebraic numbers.

In the quaternionic case (3 e's), I've tried finding a general solution, but even Mathematica's equation-crunching fails to give me any simple forms.