Classify the vector spaces with an associative product and a nonzero derivation d(ab) = da b + a db.
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Problem
Classify the vector spaces with an associative product and a nonzero
derivation d(ab) = da b + a db.
Let V be a vector space with a bilinear function p: V x V -> V (with
p(a,b) written as ab), and a linear function d: V -> V not identically
0, (with d(a) written as da), satisfying the properties:
(V1) (ab)c = a(bc) (so V is a ring)
(V2) d(ab) = da b + a db
and possibly one or more of the following properties
(V3) ab = 0 -> a = 0 or b = 0
(V4) there exists a unit, 1, such that a1 = a = 1a
(it follows that d1 = 0)
(V5) ab = ba
This space is meant to generalize on the concept of a differential
manifold.
I know the following:
(A) There are only 2 finite dimensional vector spaces satisfying V3,
V4, V5 (but not V2): the real numbers and the complex numbers
(i.e. the 2-dimensional vector space spanned by 1 and i)
(B) There are no finite dimensional vector spaces satisfying all of
V1-V5. But there are infinite dimensional ones, e.g. the Fourier
Space, U(1), spanned by set (take s(0) = 0 out to make it a basis):
{ c(0), c(1), c(2), c(3), ..., s(0), s(1), s(2), s(3), ... }
with c(0) = 1, s(0) = 0, with the usual multiplication rules for
sines and cosines (c(a) c(b) = 1/2 c(a+b) + 1/2 c(|a-b|), etc), and
with:
d c(a) = -a s(a), d s(a) = a c(a)
(C) There are finite dimensional solutions satisfying V1-V4, but not
V5. For example, the vector space Q, spanned by the Quaternion
units { 1, i, j, k } with the Quaternion multiplication rule (ij =
-ji = k, jk = -kj = i, ki = -ik = j, i^2 = j^2 = k^2 = -1 [Editor:
"= 1" in the original]), and with:
dv = 1/2 (hv - vh) = [h, v], for all v in Q
for some fixed non-zero quaternion, h.
V1, V2, and V4 are satisfied, and V2 [Editor: "V3" in the original]
is simply the statement:
[h, vw] = [h, v] w + v [h, w]
(or: (hvw - vwh)/2 = (hv - vh) w/2 + v(hw - wh)/2)
(D) All (C-infinity) differential manifolds generate sets satisfying
V1, V2, V3, and V5, where V is the space of C-infinity functions
over the manifold, and df is the Lie derivative taken with respect
to an arbitrary non-zero vector field.
[ Editor: V3 fails here: two smooth functions with disjoint
supports are both nonzero but have product 0. ]
For the U(1) space, it can also be proven that the d operator presented
above is essentially unique, given the multiplication rule, and for the
Q space the form of the d operator (that dv is the commutator [h, v])
is uniquely determined by the multiplication rule.
Define the d-dimension of the space V, with respect to the product
operator p, to be the maximum number of linearly independent d
operators that can be defined with respect to p satisfying V2. The set
U(1) has a d-dimension of 1, the set Q has a d-dimension of 3, and any
of the sets described in case (D) generated by an N-dimensional
manifold I believe will have d-dimension N.
OK. Now the tough problem. Note the suspicious similarity to Lie
Algebras. Classify all the possible spaces satisfying V1 and V2, and
all the different combinations of V3, V4, and V5. Of particular
interest, are the finite dimensional spaces that satisfy V1-V4. Also
of interest: prove the d-dimension hypothesis.
(Note: this is not a homework problem).
-- Mark <markh@csd4.csd.uwm.edu> (sci.math, April 1993)
Show solution
A full classification of all spaces with V1 and V2 is out of reach
(see the end), but the case singled out as "of particular interest"
has a short complete answer, and the "d-dimension" hypothesis needs
correcting before it can be proved. Throughout, the scalars are the
real numbers. In standard terms, V with V1 is an associative
algebra, and a d with V2 is a derivation.
1. Finite-dimensional, V1-V4 (with or without V5).
Answer: the only one is the quaternions H, and its derivations are
exactly d(v) = hv - vh with h a nonzero pure quaternion. There are
no examples with V5.
Proof. Let V be finite-dimensional, associative, nonzero and without
zero divisors (V1, V3). For a nonzero a, the linear map x -> ax is
injective, hence onto, so V is a division algebra; V4 even follows
from V1 and V3 (choose e with ae = a; then a(ex - x) = 0 gives
ex = x, and (xe - x)e = 0 gives xe = x). By Frobenius' theorem
(1878) a finite-dimensional associative real division algebra is R,
C or H. R has no nonzero derivation (d1 = 0 forces d = 0), nor has
C: from i^2 = -1, 2i d(i) = d(-1) = 0, so d(i) = 0. Every derivation
of H is inner, d(v) = hv - vh (true for any finite-dimensional
central simple algebra, by the Skolem-Noether theorem), and h can be
taken pure imaginary since real h give 0. So the derivations of H
form a 3-dimensional space, isomorphic to the Lie algebra of
rotations in 3-space, as stated in (C). This also proves (A) and
(B) of the posting (in (A), for commutative V the possibilities are
R and C).
2. The "d-dimension". For any V the derivations form a vector space
Der(V); it is also a Lie algebra under [d1, d2] = d1 d2 - d2 d1, which
explains the "suspicious similarity to Lie algebras". But if V is
commutative (V5), Der(V) is much bigger than the posting assumes: if
d is a derivation and f any element of V, then f d, defined by
(f d)(a) = f (da), is again a derivation:
(f d)(ab) = f (da b + a db) = (f da) b + a (f db).
So in the U(1) example d, c(1) d, s(1) d, c(2) d, ... are linearly
independent derivations, and the d-dimension is infinite, not 1.
What is true is that d is unique up to such a factor: every
derivation D of U(1) is g d with g = c(1) D(s(1)) - s(1) D(c(1)).
(Proof: write c = c(1), s = s(1). From c^2 + s^2 = 1,
c D(c) + s D(s) = 0. Then g d(c) = -s g = s^2 D(c) - s c D(s) =
D(c) - c (c D(c) + s D(s)) = D(c), and in the same way
g d(s) = c g = D(s) - s (c D(c) + s D(s)) = D(s). A derivation is
determined by its values on the generators c and s.)
The right notion of dimension for commutative V is therefore the rank
of Der(V) as a module over V. With that change the hypothesis in (D)
is a standard theorem: every derivation of the algebra of smooth
functions on an N-dimensional manifold is a smooth vector field (the
Lie derivative along it), and these form a module of rank N (locally
free; globally free exactly when the manifold is parallelizable).
For the noncommutative H the plain dimension, 3, is the right count.
3. Dropping V3 or V4. Here the variety is enormous:
- If all products are 0 (ab = 0), V1 holds and every linear map is
a derivation.
- Any graded algebra V = V0 + V1 + V2 + ... has the derivation
d(a) = n a for a in Vn; e.g. the dual numbers R[e]/(e^2), with
d(e) = e, are finite-dimensional and satisfy V1, V2, V4 and V5.
- Every matrix algebra M_n(R) satisfies V1, V2 and V4 with
d(v) = hv - vh, and its derivation space has dimension n^2 - 1.
A classification of all algebras with a nonzero derivation would
include, for instance, all nilpotent associative algebras, for which
no classification is known, so the general question is open and not
expected to have a tidy answer. Infinite-dimensional examples with
V1-V5 are plentiful (polynomial rings with d/dx, the U(1) example,
which is the coordinate ring of the circle).
-- Claude (rec.puzzles Oracle, 2026-11-02)