Dear Beanz:
The answer is provably “Yes!” if the algebra is further assumed to possess a multiplicative identity element. “Yes” also appears to be the answer under several different sets of weaker hypotheses, although proofs have not been found. We will discuss each of these variations on a common theme shortly, but first briefly sketch out necessary definitions.
Let R denote the field of real numbers. Let A denote an R-algebra, that is, A is a vector space over R together with a bilinear map
A × A ---------------> A
(x, y) --------------> x · y
known as vector multiplication, which satisfies r(x · y) = (rx) · y = x · (ry) for all r ∈ R and all x, y ∈ A. Let
Z = {x ∈ A : x · y = 0 for some nonzero y ∈ A}
be the set of left zero divisors of A. Note that we consider 0 ∈ Z which is uncustomary. Define A to be
* m-associative if there exists an m-dimensional subspace S of A with the property that (y · x) · z = y ·(x · z) for all y, z ∈ A, for all x ∈ S
* tame if Z is a finite union of subspaces of A.
It clearly follows that
* a 2-dimensional, 0-associative algebra is necessarily tame
whereas
* a 4-dimensional, 4-associative algebra need not be tame (consider the algebra of real 2-by-2 matrices).
What can be said about the tameness of 3-dimensional, m-associative algebras, 0 ≤ m ≤ 3?
3-dimensional, m-associative algebras
On one hand, a 3-dimensional, 1-associative algebra need not be tame, as exemplified by R³ together with the following vector multiplication:
Zero Divisor Structure in Real Algebras, 1
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Left zero divisors (a, b, c) are precisely those points for which the matrix
Zero Divisor Structure in Real Algebras, 2
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is singular, i.e., for which the determinant a · (a² + b · c) = 0. The solution set of this cubic equation is obviously not a finite union of planes and lines in R³. This example is new, as far as I know.
On the other hand, we conjecture that a 3-dimensional, 2-associative algebra is necessarily tame. A proof is known only for the special case in which the algebra is 3-associative and possesses a multiplicative identity element. The proof goes as follows. An algebra as such must be isomorphic to one of the following:
R × R × R, C × R, R × R[X]/(X²), R[X]/(X³),
R[X,Y]/(X², X · Y, Y²) or T
where C denotes the field of complex numbers and T is the algebra of upper-triangular real 2-by-2 matrices. There is an elementary proof of this classification theorem. The left zero divisors constitute a finite union of subspaces for each of the six representative R-algebras. Hence the original R-algebra via isomorphism must be tame.
Extensive computer work confirms the correctness of this conjecture. A more thorough report will come at a later time. For now, we describe what lies beneath the calculations. A 3-dimensional, 2-associative algebra gives rise to a natural linear transformation
μ : R³ ---------------> R3 × 3
satisfying
μ(μ(y) (x)) = μ(y) μ(x) for all y ∈ A, for all x ∈ S
Without loss of generality, assume
S = {s · e1 + t · e2 : s, t ∈ R}
where {e1, e2, e3} is the standard basis for A. Taking x = e1 above and matching polynomial coefficients in y-components, one obtains 27 homogeneous quadratic equations in 27 unknowns. The unknowns here determine the value of multiplication in A. The same happens when taking x = e2 and there exist dependencies among the net 54 equations which can be removed. Here are the quadratic equations (in original form). Random sampling of equation solutions is equivalent to random sampling of 2-associative algebras. Clearly
Z = {x ∈ A : det(μ(x)) = 0}
hence the components of a zero divisor x must satisfy a certain homogeneous cubic equation. So for a fixed random 2-associative algebra A, one can readily sample four independent zero divisors at a time and determine whether they are coplanar. Conversely, one can fix the form of the expression det(μ(x)) and attempt to find a corresponding 2-associative algebra A. These methods suggest that 3-dimensional, 2-associative algebras are always tame.
3-dimensional, alternate algebras
An R-algebra A is alternate if
(x · y) · y = x · (y · y) and (x · x) · y = x · (x · y)
for all x, y ∈ A.
(Reference: R. D. Schafer, An Introduction to Non-Associative Algebras, Academic Press, 1966; MR 96j:17001.)
Methods similar to the above suggest that 3-dimensional, alternate algebras are always tame as well.
Finite-dimensional, commutative, associative algebras
An associative R-algebra A is commutative if
x · y = y · x for all x, y ∈ A
For a finite-dimensional, commutative, associative algebra A with identity, the set Z is the union of the maximal ideals, of which there are finitely many. (Reference: M. F. Atiyah and I. G. MacDonald, Introduction to Commutative Algebra, Addison-Wesley, 1969; MR 39 #4129.) This is much more than is needed to deduce that A is tame.
Is the existence of the multiplicative identity optional for tameness to hold? Can the assumption of commutativity be weakened? No computer work has been done here for n>3.
It is always puzzling in mathematics when several different sets of hypotheses imply the same conclusion. One feels that there ought to be a general framework of which the above are only special cases.
Observe that the zero divisor structure proof for 3-dimensional, 3-associative algebras with identity rested on a classification theorem. This, in my mind, seems inelegant. There should be a way to demonstrate tameness without the full strength of a list of isomorphism classes. Indeed, it may be essential to find such a way in order to attack the 3-dimensional, 2-associative or alternate cases.
― -rainbow bum- (-rainbow bum-), Thursday, 3 February 2005 18:13 (twenty years ago)
seven years pass...