LECTURES ON R-ALGEBRAS

by Kurt Meyberg ^{1}

1 Introduction and basic notations

1.1 Introduction.

The theory of nonassociative algebras deals mostly with algebras defined by certain identities, like Lie algebras, alternative algebras, Jordan algebras, power-associative algebras - and others. Most extensions, generalizations of these types, use ”weaker” versions of the defining identities. But there are other ways to characterize certain types of algebras, and the one which leads to R-algebras – which may be viewed as a generalization of power-associative algebras – has its roots in analysis: It was observed that every ordinary differential equation ẏ = p(y) in a finite dimensional real or complex vector space V, with homogeneous quadratic right hand side p(y) ^{2} may be viewed as a homogeneous Riccati equation

ẏ = y^{2}

in the commutative, nonassociative algebra A = (V, ·), the product (x, y) ↦ x · y being given by

x · y = \frac{1}{2} [p(x + y) − p(x) − p(y)] .

1These lecture notes were written while I was visiting the University of Sao Paulo, Brasil. I thank the IME for their kind hospitality and Prof. I. Shestakov for his interest in this work and stimulating discussions. Support of FAPESP (Proc. 2001/02074-4) is gratefully acknowledged.

2Many important systems of differential equations in physics, chemistry, biology and other areas are quadratic, others can be reduced to quadratic systems. The importance of these systems is reflected in the following statement of a famous mathematician: ”The world is quadratic.”

Thus the differential equation gives rise to an algebra and properties of its solutions are reflected in the structure of the corresponding algebra. And conversely, given a finite dimensional real or complex algebra A, it is of interest to study the initial value problem

ẏ = y^{2}, y(0) = x

for any x ∈ A. This interplay between differential equations and algebras, first studied by L. Markus [9] in the 60’s of the last century has since then successfully been employed to prove results concerning the existence of certain special solutions, first and higher integrals, symmetries and other features (see the bibliography at the end of this report). If one recalls that linear systems are successfully treated using linear algebra, this new algebraic approach to quadratic systems seems to be very natural. But – other than the linear case – arbitrary algebras still remain too unmanageable as to expect a similar clear and complete picture. One has to impose further restrictions, either on the algebra or on the differential equation. R-algebras were introduced by S. Walcher in [17], they are characterized by the property that the solution y(x, t) of the initial value problem ẏ = y^{2}, y(0) = x in A is rational in x and t. Since power-associative algebras share this property they may be viewed as a generalization of power-associative algebras. This report is an extended version of a minicourse given in the algebra seminar (I. Shestakov et al.) at the IME, University of Sao Paulo, Brasil, in Sept./Oct. 2001. Most of the material is taken from [2], [10], [11], [13], [14], [17], [19].

1.2 Some notations.

Let 𝔽 be the field of real or complex numbers ^{3} and A be a finite dimensional commutative 𝔽-algebra. Thus A is an 𝔽-vectorspace equipped with a bilinear multiplication A×A → A, (x, y) ↦ x·y, such that x·y = y·x for all x, y ∈ A. Note that, in general, we do not assume associativity. But, as is customary, we omit very often the multiplication dot and write xy for the product. The ordinary powers x^{k} of an element x ∈ A are defined recursively

3We freely want to use differentiation and other tools from analysis, therefore we restrict to these fields. By some general principles this will be enough for R-algebras over fields of characteristic 0.

x^{1} = x, x^{k+1} = x^{k}x (k = 1, 2, 3, …)

and x^{0} = e, if A is unital with unit element e. The multiplication operator L(x) : A → A (left multiplication) is defined by L(x)y = xy. Let V, W be 𝔽-vectorspaces. The derivative of a differentiable function f : U → W (U ⊆ V an open subset) at x ∈ U is denoted by Df(x). The derivative of f(x) in direction y is Df(x)y - this is the linear mapping Df(x) applied to y ∈ V and we often write Df(x) · y instead.

Df(x) · y = \frac{1}{t} [f(x + ty) − f(x)]_{t→0}

Observe, that this expression makes sense for arbitrary fields, if f is polynomial. In this case the differentiation can be extended by means of the quotient rule to rational functions. The higher derivatives (if they exist) are denoted by D^{k}f(x). In particular, D^{2}f(x) : V × V → W is bilinear,

D^{2}f(x)(y, z) = D(Df(x) · y) · z .

D^{k}f(x) is k-linear. The mapping f : U → W is called polynomial, if with respect to bases of V and W all components of f(x) are polynomials in the coordinates x_{1}, x_{2}, … of x. This is of course independent of any choice of bases. And f is called rational, if all component functions are rational. If f : V → W is a homogeneous polynomial map of degree r > 0, i.e. f polynomial and f(tx) = t^{r}f(x) for all x ∈ V, t ∈ 𝔽, then there is a unique symmetric r-linear map f̂ : V^{r} → W such that f(x) = f̂(x, …, x) for all x ∈ V, namely f̂ = \frac{1}{r!}D^{r}f(0). As is customary, we will not distinguish f and f̂ typographically from now on. Thus for a homogeneous polynomial map f : V → W of degree k > 0 we write f(x) = f(x, …, x) (k times x). In this case the derivative Df(x)y is easily obtained from the coefficient of t in the expansion of f(x + ty) = f(x + ty, …, x + ty) = f(x) + kf(x, …, x, y)t + terms of higher t-order. Thus

(1.1)  Df(x) · y = kf(x, …, x, y) .

(Replace, like in the product rule, all the x, one after another, by y and then use symmetry: f(y, x, …, x) + f(x, y, x, …, x) + ⋯ + f(x, …, x, y) = kf(x, …, x, y).)

1.3 The homogeneous polynomials g_{k}(x).

Let A be an arbitrary commutative 𝔽-algebra. We define homogeneous polynomial maps g_{k} : A → A recursively by

(1.2) g_{0}(x) = x,  (k + 1)g_{k+1}(x) = Dg_{k}(x) · x^{2} (k = 1, 2, 3, …)

Then (by definition)

(1.2′)  \begin{aligned}g_{k}(x) & = & g_{k}(x, x, …, x) (k + 1−linear), \\ g_{k+1}(x) & = & g_{k}(x, …, x, x^{2}), \end{aligned}

and for small k we have

g_{0}(x) = x,  g_{1}(x) = x^{2},  g_{2}(x) = x^{3},

g_{3}(x) = \frac{1}{3}(2x^{4} + x^{2}x^{2}),

g_{4}(x) = \frac{1}{2·3}(2x^{5} + 3x^{2}x^{3} + x(x^{2}x^{2})),

g_{5}(x) = \frac{1}{5·6}(4x^{6} + 8x^{2}x^{4} + 6x^{3}x^{3} + 4x^{2}(x^{2}x^{2}) + 6x(x^{2}x^{3}) + 2x(x(x^{2}x^{2}))).

An easy induction argument shows

(1.3)  (k + 1)g_{k+1}(x) = ∑_{j=0}^{k} g_{j}(x)g_{k−j}(x) .

Proof : For k = 0 the formula reduces to g_{1}(x) = g_{0}(x)g_{0}(x) = x · x = x^{2}. Assume (1.3) holds for k. Then by differentiating (1.3) in direction x^{2} we get (using the product rule and the definition (1.2)):

(k + 2)(k + 1)g_{k+2}(x) = ∑_{j=0}^{k} (j + 1)g_{j+1}(x)g_{k−j}(x)+

+ ∑_{j=0}^{k} g_{j}(x)(k − j + 1)g_{k−j+1}(x)

= ∑_{j=1}^{k+1} jg_{j}(x)g_{k+1−j}(x) + ∑_{j=0}^{k} (k + 1 − j)g_{j}(x)g_{k+1−j}(x)
= (k + 1)g_{k+1}(x)g_{0}(x) + ∑_{j=1}^{k} (k + 1)g_{j}(x)g_{k+1−j}(x)

+(k + 1)g_{0}(x)g_{k+1}(x)

= (k + 1) ∑_{j=0}^{k+1} g_{j}(x)g_{k+1−j}(x) .
□

The algebra A is called power-associative, if for all x ∈ A and all m, n ∈ ℕ we have x^{n}x^{m} = x^{m+n} (that is, if the powers of an element associate as one is used to) equivalently, if the subalgebras generated by one element are associative. Due to a theorem of A.A.Albert the algebra A is powerassociative, if and only if x^{4} = x^{2}x^{2} for all x ∈ A. In power-associative algebras, as can be seen from (1.3), we have g_{k}(x) = x^{k+1} for all k. Thus in this case the polynomials g_{k}(x) are the ”ordinary” powers. But in the more general situation the polynomials g_{k}(x) (which are, in general, different from x^{k+1}) will take over the rôle of powers.

1.4 The initial value problem.

The polynomials g_{k}(x) in A (A an arbitrary commutative 𝔽-algebra) enter the scene in a natural way when studying the initial value problem

(1.4)  ẏ = y^{2},  y(0) = x

for any x ∈ A. Thus we want to determine the curve y = y(t) which satisfies \frac{d}{dt}y(t) = y(t)^{2} and passes through the point x at time t = 0. Due to the Cauchy-Kowalewski-theorem this initial value problem has a unique solution y(x, t) which can be extended in a power series y(x, t) = ∑_{k≥0} G_{k}(x)t^{k}. Comparing the coefficients of t^{k} in ẏ = y^{2} we obtain the recursion (k + 1) G_{k+1}(x) = ∑_{j=0}^{k} G_{j}(x)G_{k−j}(x) and therefore G_{k}(x) = g_{k}(x) since the initial values are the same, x = y(x, 0) = G_{0}(x) = g_{0}(x). Thus (1.4) has the unique solution

(1.5)  y(x, t) = ∑_{k≥0} g_{k}(x)t^{k}

(for t from an interval containing t = 0). Since g_{k}(x) is homogeneous of degree k + 1 we have

(1.6)  ty(x, t) = y(tx, 1) .

1.5 The definition of R-algebras.

If we want to employ more algebra (and considering important examples, see par.2) it appears natural to study those algebras, in which the solution y(x, t) of our initial value problem (1.4) is rational (rational in x and t). These algebras are called R-algebras. From (1.6) we see that y(x, t) is rational, if and only if y(x, 1) is rational in x (whenever defined, resp. for x from an open neighbourhood of 0 ∈ A). Assume y(x, 1) to be rational. Then it can be written as

y(x, 1) = \frac{1}{q(x)}P (x)

where P (x) is a polynomial in A and the scalar polynomial q : A → 𝔽 a common denominator for all components of y(x, 1) (with respect to some basis). Thus y(x, 1) = ∑_{k≥0} g_{k}(x) is rational (x from an open subset U ⊆ A containing the zero element) if and only if

(1.7)  q(x) (∑_{k≥0} g_{k}(x)) = P (x) .

Moreover, we may assume that no prime divisor of q(x) is a factor of all components of P (x) - otherwise it will be cancelled - and q(0) ≠ 0, since y(x, t) has no singularity at x = 0. We decompose q(x) as sum of its homogeneous parts q(x) = q_{0}(x)+q_{1}(x)+⋯+q_{m}(x) where q_{k}(x) is homogeneous of degree k, q_{m}(x) ≠ 0. Since q_{0} ≠ 0 we may normalize and assume q_{0} = 1,

q(x) = 1 + q_{1}(x) + ⋯ + q_{m}(x) .

In this normalized and reduced (cancelled) form we call q the exact denominator of y(x, 1).

Lemma 1.1. Let s(x) = s_{0}(x) + ⋯ + s_{k}(x) be a scalar polynomial function on A of degree k, s_{j} homogeneous of degree j, then s(x)y(x, 1) is polynomial if and only if there exists an integer r_{0} ≥ k such that

∑_{j=0}^{k} s_{j}(x)g_{r−j}(x) = 0

for all r ≥ r_{0} and all x ∈ A.

Proof : In (s_{0}(x)+s_{1}(x)+⋯+s_{k}(x))·∑_{k≥0} g_{k}(x) we collect all homogeneous terms of equal degree r + 1 ≥ k : s_{o}(x)g_{r}(x) + s_{1}(x)g_{r−1}(x) + ⋯ + s_{k}(x)g_{r−k}(x). Then s(x)y(x, 1) is a polynomial, if and only if there is a degree r_{0} such that these terms vanish for all r ≥ r_{0}. □

In particular

Corollary. A is an R-algebra, if and only if there is a polynomial q(x) = 1+q_{1}(x)+⋯+q_{m}(x), q_{j}(x) homogeneous of degree j, and an integer r_{0} ≥ m such that

(1.8)  g_{r}(x) + q_{1}(x)g_{r−1}(x) + ⋯ + q_{m}(x)g_{r−m}(x) = 0

for all x ∈ A and all r ≥ r_{0}. □

Examples will be given in the next paragraph.

1.6 The invariant chain of an R-algebra.

Let A be an R-algebra, y(x, 1) = \frac{1}{q(x)}P (x) as in (1.5) and q(x) = 1 + q_{1}(x) + ⋯ + q_{m}(x) the exact denominator of y(x, 1). If q_{1}(x) = ⋯ = q_{m}(x) = 0 then y(x, 1) = P (x) is a polynomial. In this case A is called a P-algebra (R for ”rational”, P for ”polynomial”) and we have g_{r}(x) = 0 for all x ∈ A, r ≥ r_{0} (see 1.8). Thus P-algebras behave like nilalgebras (even worse). Therefore, from now on, we will always assume that the exact denominator q(x) = 1 + q_{1}(x) + ⋯ + q_{m}(x), q_{m}(x) ≠ 0 has degree m ≥ 1.

Theorem 1.2. The homogeneous polynomials q_{j}(x) satisfy the following recursion relation

(1.9) \begin{aligned}jq_{j}(x, …, x, x^{2}) + q_{1}(x)q_{j}(x) & = & (j + 1)q_{j+1}(x), 1 ≤ j < m, \\ mq_{m}(x, …, x, x^{2}) + q_{1}(x)q_{m}(x) & = & 0. \end{aligned}

Proof : We differentiate (1.8) in direction x^{2}, use Dg_{k}(x)·x^{2} = (k+1)g_{k+1}(x) and obtain

∑_{j=1}^{m} jq_{j}(x, …, x, x^{2})g_{r−j}(x) + ∑_{j=0}^{m} q_{j}(x)(r − j + 1)g_{r−j+1}(x) = 0.

Rearranging yields

∑_{j=2}^{m} [(j − 1)q_{j−1}(x, …, x, x^{2}) + (r − j + 1)q_{j}(x)] g_{r−j+1}(x)

+mq_{m}(x, …, x, x^{2})g_{r−m}(x) + (r + 1)g_{r+1}(x) + rq_{1}(x)g_{r}(x) = 0.

Using the (r + 1)-fold of (1.8) (with r + 1 instead of r) we can substitute

(r + 1)g_{r+1} + (r + 1)q_{1}(x)g_{r}(x) = − ∑_{j=2}^{m}(r + 1)q_{j}(x)g_{r+1−j}(x)

in the foregoing equation to obtain

∑_{j=2}^{m} [(j − 1)q_{j−1}(x, …, x, x^{2}) − jq_{j}(x)] g_{r−j+1}(x)−

−q_{1}(x)g_{r}(x) + mq_{m}(x, …, x, x^{2})g_{r−m}(x) = 0.

Here we replace −q_{1}(x)g_{r}(x) by

−q_{1}(x)g_{r}(x) = ∑_{j=1}^{m} q_{1}(x)q_{j}(x)g_{r−j}(x) = ∑_{j=2}^{m+1} q_{1}(x)q_{j−1}(x)g_{r−j+1}(x)

(which again comes from (1.8)) to end up with

∑_{j=2}^{m} [jq_{j}(x) − (j − 1)q_{j−1}(x, …, x, x^{2}) − q_{1}(x)q_{j−1}(x)] g_{r−j+1}(x)

+ [−mq_{m}(x, …, x, x^{2}) − q_{1}(x)q_{m}(x)] g_{r−m}(x) = 0,

or in abbreviated form

(1.10)  ∑_{j=2}^{m+1} s_{j}(x)g_{r+1−j}(x) = 0

for all r ≥ r_{0}. This in turn shows that

s(x) ∑_{k≥0} g_{k}(x) =: P̂(x)

is a polynomial, where s(x) = s_{2}(x) + ⋯ + s_{m+1}(x) (see lemma 1.1). We want to show s(x) = 0. Assume s ≠ 0. Then

\frac{1}{s(x)}P̂(x) = ∑_{k≥0} g_{k}(x) = \frac{1}{q(x)}P (x)

on a Zariski open nonempty set. Componentwise

s(x)p_{i}(x) = q(x)p̂_{i}(x)

for all x ∈ A, 1 ≤ i ≤ dim A, where p_{i}, p̂_{i} denote the components of P, resp. P̂ (with respect to a basis). Since no prime divisor of q divides all p_{i}, q(x) must divide s(x), say s(x) = q(x)h(x). We now compare degrees. From the definition of s(x) we have deg s(x) ≤ m + 1. deg s(x) = m = deg q(x) would imply that s(x) is a scalar multiple of q(x) (because of s = qh) which contradicts s(0) = 0, q(0) = 1. Thus m + 1 = deg s(x) = deg q(x)+ deg h(x) and therefore deg h(x) = 1. Again, comparing the homogeneous parts of the same degree in

s = s_{2} + ⋯ + s_{m+1} = h + hq_{1} + ⋯ + hq_{m} (q_{0} = 1),

we get h = 0, thus s = qh = 0, a contradiction. □

Definition. The m-tuple of polynomials (q_{1}, …, q_{m}) is called the invariant chain of A. The degree m of the exact denominator q(x) = 1 + q_{1}(x) + ⋯ + q_{m}(x) of y(x, 1) is called the degree of A. □

NOTE: We are always assuming deg A = m ≥ 1. The recursion (1.9) shows that all members of the invariant chain of R-algebra can be computed from the linear form q_{1}(x):

2q_{2}(x) = q_{1}(x^{2}) + q_{1}(x)^{2}
3q_{3}(x) = q_{1}(x^{3}) + \frac{3}{2}q_{1}(x)q_{1}(x^{2}) + \frac{1}{2}q_{1}(x)^{3}
4q_{4}(x) = \frac{1}{3}q_{1}(2x^{4} + x^{2}x^{2}) + \frac{4}{3}q_{1}(x)q_{1}(x^{3}) + \frac{1}{2}q_{1}(x^{2})q_{1}(x^{2})+
+q_{1}(x)^{2}q_{1}(x^{2}) + \frac{1}{6}q_{1}(x)^{4}
etc.

We will show later that in unital R-algebras we also have ”going-down” formulas which allow the computation of q_{j}(x) (j ≤ m) from q_{m}(x).

1.7 Arbitrary invariant chains.

Let A be an arbitrary commutative 𝔽-algebra. An m-tuple (s_{1}, …, s_{m}) of homogeneous polynomials s_{j} : A → 𝔽 of degree j is called an invariant chain in A, if it satisfies the recursion (1.9), i.e. if

(1.11) js_{j}(x, …, x, x^{2}) + s_{1}(x)s_{j}(x) = (j + 1)s_{j+1}(x) (1 ≤ j < m)
ms_{m}(x, …, x, x^{2}) + s_{1}(x)s_{m}(x) = 0

for all x ∈ A.

Example 1. We consider the 2-dimensional algebra A = (ℝ^{2}, ·) with product

x · y = (\begin{matrix}−x_{1}y_{1} + x_{1}y_{2} + x_{2}y_{1} \\ −2x_{2}y_{2} \end{matrix})

It is easy to verify that the initial value problem ẏ = y^{2}, y(0) = x in A has unique solution

y_{1}(x, t) = \frac{x_{1} + 2x_{1}x_{2}t}{1 + x_{1}t + x_{1}x_{2}t^{2}}
y_{2}(x, t) = \frac{x_{2}}{1 + 2x_{2}t}

The exact denominator is

\begin{aligned}q(x) & = & (1 + x_{1}t + x_{1}x_{2}t^{2})(1 + 2x_{2}t) \\ & = & 1 + (x_{1} + 2x_{2})t + 3x_{1}x_{2}t^{2} + 2x_{1}x_{2}^{2}t^{3}. \end{aligned}

Thus A is an R-algebra. The invariant chain of A is (q_{1}, q_{2}, q_{3}), where

q_{1}(x) = x_{1} + 2x_{2},  q_{2}(x) = 3x_{1}x_{2},  q_{3}(x) = 2x_{1}x_{2}^{2}.

We will show that (s_{1}, s_{2}) with

s_{1}(x) = x_{1}, s_{2}(x) = x_{1}x_{2}

is another invariant chain in A. For this we have to verify (1.11):

s_{1}(x) = x_{1} is linear.
s_{1}(x^{2}) + s_{1}(x)^{2} = (−x_{1}^{2} + 2x_{1}x_{2}) + x_{1}^{2} = 2x_{1}x_{2} = 2s_{2}(x).

And using 2s_{2}(x, y) = x_{1}y_{2} + y_{1}x_{2} we get

2s_{2}(x, x^{2}) + s_{1}(x)s_{2}(x) = x_{1}(−2x_{2}^{2}) + x_{2}(−x_{1}^{2} + 2x_{1}x_{2}) + x_{1}^{2}x_{2} = 0
□

Theorem 1.3. Let (s_{1}, …, s_{m}) be an invariant chain in the commutative algebra A and y(t) a solution of ẏ = y^{2} in A. Then

z(t) := (s_{1}(y(t)), s_{2}(y(t)), …, s_{m}(y(t)))^{T}

is a solution of the following system of differential equations in 𝔽^{m}

(1.12)  ż_{1} = −z_{1}z_{1} + 2z_{2}^{2} ; ż_{2} = −z_{1}z_{2} + 3z_{3} ; ⋮ ; ż_{m−1} = −z_{1}z_{m−1} + mz_{m} ; ż_{m} = −z_{1}z_{m} ; briefly ż = S(z).

Proof : Using the recursion relations (1.11), ẏ = y^{2} and the differentiation rule \frac{d}{dt}s_{j}(y) = \frac{d}{dt}s_{j}(y, …, y) = js_{j}(y, …, y, ẏ) = js_{j}(y, …, y, y^{2}) we obtain for z_{j}(t) = s_{j}(y(t))

ż_{j} = js_{j}(y, …, y, y^{2}) = −s_{1}(y)s_{j}(y) + (j + 1)s_{j+1}(y)
= −z_{1}z_{j} + (j + 1)z_{j+1}, if 1 ≤ j < m

and

ż_{m} = ms_{m}(y, …, y, y^{2}) = −s_{1}(y)s_{m}(y) = −z_{1}z_{m}
□

The value of this theorem comes from the fact that the solutions of the system (1.12) are explicitely known: The initial value problem

ż = S(z), z(0) = (a_{1}, …, a_{m})^{T}

in 𝔽^{m} has the solution

(1.13)  z_{k}(t) = \frac{1}{k!} \frac{p^{(k)}(t)}{p(t)}

with p(t) = 1 + a_{1}t + ⋯ + a_{m}t^{m}.

Proof : By direct verification. □

We combine (1.13) with theorem 1.2: Let (s_{1}, …, s_{m}) be an invariant chain in A and y(x, t) the solution of the initial value problem ẏ = y^{2}, y(0) = x in A, then z(t) = (z_{1}(t), …, z_{m}(t))^{T} with z_{k}(t) = s_{k}(y(x, t)) is the solution of the corresponding initial value problem ż = S(z), z(0) = (s_{1}(x), …, s_{m}(x))^{T}. Hence - by (1.13) -

(1.14)  s_{k}(y(x, t)) = \frac{1}{k!} \frac{p^{(k)}(t)}{p(t)}  (1 ≤ k ≤ m)

with p(t) = p(x, t) = 1 + s_{1}(x)t + ⋯ + s_{m}(x)t^{m}. In particular

(1.15)  s_{m}(y(x, t)) = \frac{s_{m}(x)}{1 + s_{1}(x)t + ⋯ + s_{m}(x)t^{m}}

As a consequence we get

Corollary. Let (s_{1}, …, s_{m}) be an invariant chain in A and y(x, t) the solution of the standard initial value problem ẏ = y^{2}, y(0) = x. Then the algebraic varieties

V_{k} = {x ∈ A | s_{k}(x) = s_{k+1}(x) = ⋯ = s_{m}(x) = 0}

are invariant for x ↦ y(x, t).

Proof : If s_{j}(x) = 0 for j ≥ k then p(x) = 1 + s_{1}(x)t + ⋯ + s_{k−1}(x)t^{k−1} and p^{(j)}(x) = 0, and consequently s_{j}(y(x, t)) = 0 for k ≤ j ≤ m by (1.14).

□

These results - of course - apply in particular to the invariant chain (q_{1}, …, q_{m}) of an R-algebra. They are also useful to construct the solution of ẏ = y^{2} from an invariant chain. If the chain is given and (1.14) can be solved for y, then we get the solution y = y(t) explicitely.

Example 2. The algebra A we considered in example 1 has an invariant chain s_{1}(x) = x_{1}, s_{2}(x) = x_{1}x_{2}. Then, with p(t) = 1 + x_{1}t + x_{1}x_{2}t^{2}, (1.14) gives

y_{1}(x, t) = \frac{x_{1} + 2x_{1}x_{2}t}{1 + x_{1}t + x_{1}x_{2}t^{2}}

and

y_{1}(x, t)y_{2}(x, t) = \frac{x_{1}x_{2}}{1 + x_{1}t + x_{1}x_{2}t^{2}} ,

which leads to the solution given before. □

1.8 The AutA-invariance of q_{j}.

In this section we generalize an important result from the theory of powerassociative algebras to R-algebras.

Theorem 1.4. Let A be an R-algebra with invariant chain q_{1}, …, q_{m}. Then

q_{j}(T x) = q_{j}(x) (1 ≤ j ≤ m)

for all x ∈ A and every automorphism T of A.

Proof : It is immediate from the definition that g_{k}(T x) = T g_{k}(x) for all k ≥ 0 (T ∈ Aut A). Then using (1.8) we get

T (∑_{j=0}^{m} q_{j}(T x)g_{r−j}(x)) = ∑_{j=0}^{m} q_{j}(T x)g_{r−j}(T x) = 0, thus

∑_{j=0}^{m} q_{j}(T x)g_{r−j}(x) = 0

for all x ∈ A, r ≥ r_{0}, T ∈ Aut A. Then by lemma 1.1, (1 + q_{1}(T x) + ⋯ + q_{m}(T x))y(x, 1) = P̂(x) is polynomial. Comparing components (relative to a basis) of y(x, 1) = \frac{1}{q(x)}P (x) = \frac{1}{q(T x)} P̂(x) we get q(T x)p_{i}(x) = q(x)p̂_{i}(x). And again, same reasoning as in the proof of theorem 1.2, q(x) divides q(T x). But then q(x) = q(T x). □

A linear map D : A → A is a derivation of A, if D(x · y) = (Dx) · y + x · (Dy) for all x, y ∈ A. The algebra DerA of all derivations of A with product given by the commutator [D_{1}, D_{2}] = D_{1}D_{2} − D_{2}D_{1} is a Lie algebra and one is always interested getting information about DerA.

Corollary.

q_{j}(x, …, x, Dx) = 0 (1 ≤ j ≤ m)

for all x ∈ A, D ∈ DerA.

Proof : For D ∈ DerA the exponential e^{tD} is an automorphism of A (for all t ∈ R). This is a standard result and can easily be verified. Thus

q_{j}(e^{tD}x, e^{tD}x, …, e^{tD}x) = q_{j}(x)

by the theorem. Expansion of the left hand side gives a power series

q_{j}(x) + jq_{j}(x, …, x, Dx)t + terms of higher order

whose non constant coefficients are all zero. □

2 Examples

Before going on with the theory it’s time to present some examples.

2.1 Example 1. Power-associative algebras.

If A is power-associative (x^{m}x^{n} = x^{m+n}), then g_{k}(x) = x^{k+1} and the initial value problem ẏ = y^{2}, y(0) = x has the explicit solution (→ 1.5 and geometric series)

y(x, t) = ∑_{k≥0} x^{k+1}t^{k} = ∑_{k≥0} L(tx)^{k}x = [Id − tL(x)]^{−1}x

where L(x) denotes (as usual) the left multiplication, L(x)y = xy. The inverse exists, if t is small enough. This solution is rational in x and t, thus A is an R-algebra. This solution formula is characteristic for power-associative algebras:

Theorem 2.1. A commutative algebra A is power-associative if and only if the initial value problem ẏ = y^{2}, y(0) = x has solution

y(x, t) = [Id − tL(x)]^{−1}x.

Proof : We still have to prove the ”only if” part. Assume [Id − L(x)]^{−1}x = y(x, 1) = ∑_{k≥0} g_{k}(x), then x = (Id − L(x)) ∑_{k≥0} g_{k}(x) = x + (g_{3}(x) − x^{4}) + terms of higher degree on an open subset of A. Thus (comparing homogeneous parts) g_{3}(x) = x^{4}, equivalently x^{2}x^{2} = x^{4}, from which follows power associativity by Albert’s theorem. □

We continue to look at the special case A = 𝔽^{n×n}, the algebra of all n × n - matrices over 𝔽. Using the formula for the inverse of an n × n - matrix, in this case the solution Y (X, t) = (E − tX)^{−1}X can be written as

Y (X, t) = \frac{1}{det(E − tX)}(E − tX)^{∗}X

where E denotes the identity matrix, det the determinant and (E − tX)^{∗} the adjoint of E − tX (which is polynomial). Since the determinant is irreducible

q(X) = det(E − tX) = 1 − (traceX)t + ⋯ + (−1)^{n}(detX)t^{n}

is the exact denominator of the solution and

q_{1}(X) = −traceX, …, q_{n}(X) = (−1)^{n}detX

the invariant chain.

2.2 Example 2. Bernstein algebras.

A commutative (nonassociative) 𝔽-Algebra A is called a Bernstein algebra, if there exists a nontrivial homomorphism ω : A → 𝔽 such that

(2.1)  x^{2}x^{2} = ω(x)^{2}x^{2} for all x ∈ A.

ω is called the weight homomorphism of A. Since it is a homomorphism its kernel

N = { x ∈ A | ω(x) = 0 }

is an ideal in A. If e = e^{2} ≠ 0 is an idempotent, then ω(e) = 1, and conversely, with any element x ∈ A with weight ω(x) ≠ 0 we get an idempotent e := \frac{1}{ω(x)^{2}} x^{2}. Thus idempotents are abundant. Now let e ≠ 0 be an idempotent in A. Then A has a Peirce decomposition

A = 𝔽e ⊕ U ⊕ V
where U = \{ x ∈ N | ex = \frac{1}{2}x \}
and V = { x ∈ N | ex = 0 }.

Moreover one has a series of ”Peirce relations”. Most secrets of the structure of a Bernstein algebra are hidden in the Hentzel-Peresi-ideal

L = { x ∈ U | xU = 0 }.

In any Bernstein algebra A we have

x^{3} − ω(x)x^{2} ∈ L for all x ∈ A.

(see [5])

A Bernstein algebra A is called nuclear, if A^{2} = A. From the Peirce relations it is easily seen that this is equivalent to U = V^{2}. Moreover, in this case (A nuclear) the Peirce relations yield LN = 0 and

(2.2)  \begin{aligned}f_{4}(x) & = (L(x) − \frac{1}{2}ω(x) Id) (x^{3} − ω(x)x^{2}) \\ & = x^{4} − \frac{3}{2}ω(x)x^{3} + \frac{1}{2}ω(x)^{2}x^{2} = 0 \end{aligned}

Proof of (2.2): Since x^{3} − ω(x)x^{2} ∈ L ⊆ U, x − ω(x)e ∈ N and LN = 0, we get 0 = (x^{3} − ω(x)x^{2})(x − ω(x)e) = x^{4} − ω(x)x^{3} − \frac{1}{2}ω(x) [x^{3} − ω(x)x^{2}].

□

Theorem 2.2. The initial value problem

ẏ = y^{2}, y(0) = x

in a nuclear Bernstein algebra (A, ω) over 𝔽 has the unique solution

y(x, t) = x + tx^{2} + \frac{t^{2}}{1 − tω(x)}x^{3}.

In particular, nuclear Bernstein algebras are R-Algebras.

Proof : Since ω(x)^{2}x^{2} = x^{2}x^{2} (see (2.1) ) formula (2.2) can be written as

\frac{1}{3}[2x^{4} + x^{2}x^{2}] = ω(x)x^{3}

or, using g_{k}-notation, as

g_{3}(x) = ω(x)g_{2}(x).

Then by induction, using ω(x^{k})ω(x) = ω(x^{k+1}) and (k + 1)g_{k+1}(x) = Dg_{k}(x) · x^{2} we get

g_{k}(x) = ω(x)^{k−2}x^{3} for all k ≥ 2.

Thus

\begin{aligned}y(x, 1) & = ∑_{k≥0} g_{k}(x) = x + x^{2} + x^{3} + ω(x)x^{3} + ⋯ + ω(x)^{k−2}x^{3} + … \\ & = x + x^{2} + \frac{1}{1−ω(x)}x^{3} \end{aligned}

(by the geometric series, |ω(x)| < 1). The solution formula now follows from y(x, t) = \frac{1}{t}y(tx, 1). □

This example (nuclear Bernstein algebras) is a special (but very interesting) case of a wider class of Bernstein algebras which are R-algebras. For arbitrary Bernstein algebras (A, ω) one can prove the following identity

g_{k}(x) − ω(x)g_{k−1}(x) = \frac{2^{k−1}}{k!} [L(x) − \frac{1}{2}ω(x) Id]^{k−2} (x^{3} − ω(x)x^{2}) =: f_{k}(x)

for all x ∈ A, k ≥ 2 (see [4]). Thus, if f_{m}(x) = 0 for some m ∈ N, then f_{k}(x) = 0 for all k ≥ m and g_{k}(x) = ω(x)g_{k−1}(x) for all k ≥ m. Again an easy induction argument shows g_{m+j}(x) = ω(x)^{j+1}g_{m−1}(x) and as in the case m=3 (nuclear Bernstein algebras) we then get a rational solution formula of our initial value problem:

y(x, t) = ∑_{k=0}^{m−2} g_{k}(x)t^{k} + \frac{t^{m−1}}{1 − tω(x)}g_{m−1}(x).

In particular, these Bernstein algebras (containing the class of Bernstein algebras which are train algebras, see [4]) are R-Algebras. The degree is m = 1, the invariant chain has length 1, q_{1}(x) = −ω(x) and q_{1}(x^{2})+q_{1}(x)q_{1}(x) = 0.

2.3 Example 3.

The initial value problem

\begin{aligned}ẏ_{1} & = y_{1}^{2} \\ ẏ_{2} & = y_{2}^{2} + 2μy_{1}y_{2} \end{aligned}

y_{1}(0) = x_{1}, y_{2}(0) = x_{2}, μ ∈ \frac{1}{2}ℤ \ \{\frac{1}{2}\}

can be solved successively:

y_{1}(t) = \frac{x_{1}}{1 − tx_{1}}

y_{2}(t) = \frac{(1 − 2μ)x_{1}x_{2}}{(1 − tx_{1})x_{2} + [(1 − 2μ)x_{1} − x_{2}](1 − tx_{1})^{2μ}}

Since this solution is rational, the corresponding 2-dimensional algebra with product

( \begin{matrix}x_{1} \\ x_{2} \end{matrix} ) ( \begin{matrix}y_{1} \\ y_{2} \end{matrix} ) = ( \begin{matrix}x_{1}y_{1} \\ x_{2}y_{2} + μ(x_{1}y_{2} + x_{2}y_{1}) \end{matrix} )

is an R-Algebra. Obviously the unit vectors e_{1} = ( \begin{matrix}1 \\ 0 \end{matrix} ), e_{2} = ( \begin{matrix}0 \\ 1 \end{matrix} ) are

idempotents, e_{1}e_{2} = μe_{2}. Thus the left multiplication L(e_{1}) has eigenvalue μ and therefore the algebra is not power-associative, if we choose μ ≠ 0, 1, \frac{1}{2}. (In power-associative algebras the eigenvalues of L(c), c^{2} = c, are always among 0,1, \frac{1}{2}.) A general construction principle for this type of R-Algebras will be given in 3.4.

2.4 R-Algebras constructed from an invariant chain.

We start with a chain (s_{1}, …, s_{m}) of polynomials on a vectorspace V, s_{j} homogenous of degree j, and try to solve

(2.3) \begin{aligned}js_{j}(x, …, x, y) & = (j + 1)s_{j+1}(x) − s_{1}(x)s_{j}(x),  1 ≤ j < m \\ ms_{m}(x, …, x, y) & = −s_{1}(x)s_{m}(x) \end{aligned}

for y. In general the solution will be rational. But if the solution y = p(x) is a homogenous quadratic polynomial, then x^{2} := p(x) defines an algebra A = (V, ·), x · y = \frac{1}{2}[(x + y)^{2} − x^{2} − y^{2}] and (s_{1}, …, s_{m}) is an invariant chain in this algebra and a solution of (1.14) will give the solution y(x,t) of our standard initial value problem. As an example we consider on V = 𝔽^{4} the chain s_{1} = x_{1}, s_{2} = x_{1}x_{2}, s_{3} = x_{1}x_{2}x_{3} and s_{4} = x_{1}x_{2}x_{3}x_{4}. Then js_{j}(x, …, x, y) = y_{1}x_{2} … x_{j} + x_{1}y_{2} … x_{j} + x_{1} … x_{j−1}y_{j} and (2.3) becomes

\begin{aligned}y_{1} & = 2x_{1}x_{2} − x_{1}^{2} \\ y_{1}x_{2} + x_{1}y_{2} & = 3x_{1}x_{2}x_{3} − x_{1}^{2}x_{2} \\ y_{1}x_{2}x_{3} + x_{1}y_{2}x_{3} + x_{1}x_{2}y_{3} & = 4x_{1}x_{2}x_{3}x_{4} − x_{1}^{2}x_{2}x_{3} \\ y_{1}x_{2}x_{3}x_{4} + x_{1}y_{2}x_{3}x_{4} + x_{1}x_{2}y_{3}x_{4} + x_{1}x_{2}x_{3}y_{4} & = −x_{1}^{2}x_{2}x_{3}x_{4} \end{aligned}

This system has the solution

y = ( \begin{matrix}2x_{1}x_{2} − x_{1}^{2} \\ 3x_{2}x_{3} − 2x_{2}^{2} \\ 4x_{3}x_{4} − 3x_{3}^{2} \\ −4x_{4}^{2} \end{matrix} ) =: x^{2}

Thus (s_{1}, s_{2}, s_{3}, s_{4}) is an invariant chain of the algebra A = (𝔽^{4}, ·), where the square x^{2} of x ∈ A is given by this solution. Moreover in this case, the system (1.14) of equations has a ”triangular” form. It can be successively solved for y_{1}(t), then y_{2}(t), … to give a rational solution y(x, t). Thus A is an R-Algebra. (s_{1}, s_{2}, s_{3}, s_{4}) is an invariant chain of A but it is not THE invariant chain of the R-Algebra. This example is easily extended to 𝔽^{n}. But all the essentials are better seen from this special case n = 4.

3 Unital R-algebras

3.1 A going-down-formula for g_{k}(x).

We will see in the next section that adjoining a unit element preserves Ralgebras. Therefore we will henceforth assume that the R-algebra A has a unit element e. In this case we have some simplifications and more information. We first observe

(3.1)  Dg_{k}(x) · e = (k + 1)g_{k−1}(x)

for all x ∈ A and k ≥ 0 (g_{−1}(x) := e). Or, if we employ the corresponding (k + 1)-linear map g_{k}(x) = g_{k}(x, …, x), (3.1) appears as

(3.1′)  g_{k}(x, …, x, e) = g_{k−1}(x).

Proof : The formula is immediately verified, if k = 0, 1 : g_{0}(e) = e, g_{1}(x, e) = x · e = x = g_{0}(x). We proceed by induction. For k > 1 we differentiate

g_{k+1}(x) = g_{k}(x, …, x, x^{2})

(see (1.2’)) in e-direction:

(k + 2)g_{k+1}(x, …, x, e) = kg_{k}(x, …, x^{2}, e) + 2g_{k}(x, …, x, xe)
= Dg_{k}(x, …, x, e) · x^{2} + 2g_{k}(x)

= Dg_{k−1}(x) · x^{2} + 2g_{k}(x) by induction hypothesis

= kg_{k}(x) + 2g_{k}(x) (see (1.2))
= (k + 2)g_{k}(x)
□

Using the definition (1.2) we can compute g_{k}(x) from g_{l}(x), l < k, by differentiating in x^{2}-direction. These are the ”going-up-formulas”. With a unit e at hand we get ”going-down-formulas” which allow the computation of g_{k}(x) from the higher degree polynomials g_{l}(x), k < l.

3.2 Going down formulas for the invariant chain

Theorem 3.1. Let (q_{1}, …, q_{m}) be the invariant chain of an R-Algebra with unit element e. Then

(a ) q_{1}(e) = −m, q_{j}(e) = (−1)^{j} ( \begin{matrix}m \\ j \end{matrix} )

(b ) jq_{j}(x, …, x, e) = −(m − j + 1)q_{j−1}(x) (2 ≤ j ≤ m)

(c ) q_{j}(x) = (−1)^{m+j} ( \begin{matrix}m \\ j \end{matrix} ) q_{m}(x, …, x, e, …, e) (j-times x)

(d ) \begin{aligned}q(x) & = (−1)^{m}q_{m}(e − x) = q_{m}(x − e) \\ q(e − x) & = (−1)^{m}q_{m}(x) \end{aligned}
where q(x) = 1 + q_{1}(x) + ⋯ + q_{m}(x).

Proof : We differentiate the defining identity (1.8)

∑_{j=0}^{m} q_{j}(x)g_{r−j}(x) = 0

for r > m in e-direction. Using the going down formula (3.1) and (r + 1) ∑_{j=0}^{m} q_{j}(x)g_{r−j−1}(x) = 0 ((3.1) with r − 1 instead of r) we get

0 = ∑_{j=0}^{m}[jq_{j}(x, …, x, e)g_{r−j}(x) + (r − j + 1)q_{j}(x)g_{r−j−1}(x)]
= ∑_{j=1}^{m}[jq_{j}(x, …, x, e)g_{r−j}(x) − jq_{j}(x)g_{r−j−1}(x)]
= q_{1}(e)g_{r−1}(x) + ∑_{j=1}^{m−1}[(j + 1)q_{j+1}(x, …, x, e) − jq_{j}(x)]g_{r−j−1}(x)
−mq_{m}(x)g_{r−1−m}(x)
=: ∑_{j=0}^{m} s_{j}(x)g_{r−1−j}(x)

Then, by lemma (1.1), s(x)y(x, 1) =: P̂(x) is polynomial, where s(x) = s_{0}(x) + s_{1}(x) + ⋯ + s_{m}(x), s_{0}(x) = q_{1}(e), s_{m}(x) = −mq_{m}(x), in particular degree s = degree q = m. Again (compare with the proof of theorem (1.2))

\frac{1}{s(x)}P̂(x) = y(x, 1) = \frac{1}{q(x)}P (x)

and then componentwise (with respect to a basis)

s(x)p_{i}(x) = q(x)p̂_{i}(x).

Since q(x) is the exact denominator of y(x, 1), no prime divisor of q divides all p_{i}, thus q divides s and therefore s(x) = αq(x) for some α ∈ 𝔽 since both have the same degree. The homogenous parts are equal, s_{j}(x) = αq_{j}(x); and s_{m}(x) = −mq_{m}(x) (by definition) gives α = −m. Thus

s_{j}(x) = −mq_{j}(x) (0 ≤ j ≤ m),

and from the definition of the s_{j},

(3.2)  \begin{aligned}q_{1}(e) = s_{0}(x) = −mq_{0}(e) & = −m \\ (j + 1)q_{j+1}(x, …, x, e) − jq_{j}(x) & = s_{j}(x) = −mq_{j}(x) \\ & (1 ≤ j ≤ m − 1). \end{aligned}

This proves b) and the first equation in a). Now we differentiate b) (s − 1)-times in e-direction (1 ≤ s ≤ j) to obtain

jq_{j}(x, … x, \underbrace{e, …, e}_{s terms}) = −(m − j + 1)q_{j−1}(x, … x, \underbrace{e, …, e}_{s−1 terms}).

Starting with j = m and applying this formula repeatedly we end at

mq_{m}(x, …, x, \underbrace{e, …, e}_{k}) = −q_{m−1}(x, …, x, \underbrace{e, …, e}_{k−1})

= \frac{(−1)^{2}}{m−1} 2q_{m−2}(x, …, x, \underbrace{e, …, e}_{k−2})

= \frac{(−1)^{3}3!}{(m−1)(m−2)}q_{m−3}(x, …, x, e, …, e)

⋮

= (−1)^{k} \frac{k!}{(m−1)(m−2)...(m−k+1)}q_{m−k}(x)

This is c) with j = m − k. Part d) of the theorem is obtained by expanding q_{m}(x − e) - one may think of the binomial formula – and using c):

\begin{aligned}q_{m}(x − e) & = ∑_{j=0}^{m} (−1)^{m+j} ( \begin{matrix}m \\ j \end{matrix} ) q_{m}(\underbrace{x, …, x}_{j}, e, …, e) \\ & = ∑_{j=0}^{m} q_{j}(x) = q(x). \end{aligned}

This is d). The choice x = 0 gives (−1)^{m}q_{m}(e) = q(0) = 1 and then (see c))

q_{j}(c) = (−1)^{j} ( \begin{matrix}m \\ j \end{matrix} ) for all j. This completes the proof.

□

A characterization of R-algebras in terms of identities was given in (1.8). A is an R-algebra if and only if

(3.3)  ∑_{j=0}^{m} q_{j}(x)g_{r−j}(x) = 0

for all r ≥ r_{0}. We will use the foregoing formulas to give another characterization:

Theorem 3.2. A unital 𝔽-algebra is an R-algebra if and only if it has an invariant chain (q_{1}, …, q_{m}) (see (1.11) for the definition) such that

(3.4)  g_{m−1}(x) + q_{1}(x)g_{m−2}(x) + ⋯ + q_{m}(x)e = 0

for all x ∈ A.

Proof : We use the foregoing ”going-down-formulas” and differentiation in e-direction to show that ∑_{j=0}^{m} q_{j}(x)g_{r−j}(x) = 0 for some r ≥ m leads to (3.4). And conversely, going up, repeated differentiation in x^{2}-direction leads to the identity for r > m. □

NOTE: The descent (differentiation in e-direction) stops at (3.4). The next step will give zero on both sides.

As a consequence of this theorem we get a clear picture of the solution y(x, t) of the standard initial value problem ẏ = y^{2}, y(0) = x in A. Since ty(x, t) = y(tx, 1) we only need to know y(x, 1).

(3.3) and (3.4) show with q = 1 + q_{1} + ⋯ + q_{m} that q(x)y(x, 1) = P (x) is polynomial of degree ≤ m (see lemma (1.1)). If P_{1}, …, P_{m} denote the homogenous parts (of corresponding degree), then P (x) = ∑_{j=1}^{m} P_{j}(x) and P_{k}(x) = g_{k−1}(x) + q_{1}(x)g_{k−2}(x) + ⋯ + q_{k−1}(x)g_{0}(x) (1 ≤ k ≤ m), in particular P_{m}(x) = g_{m−1}(x) + q_{1}(x)g_{m−2}(x) + ⋯ + q_{m−1}(x)x = −q_{m}(x)e by (3.3). Thus degree P (x) = m on a dense subset of A.

3.3 Small degrees.

We pause to look at unital R-algebras of small degree:

m=1: x + q(x)e = 0; thus A = 𝔽e.
m=2: x^{2} + q_{1}(x)x + q_{2}(x)e = 0;

then the left multiplication L(x^{2}) is a linearcombination of L(x) and L(e) = Id. Thus L(x^{2}) and L(x) commute and A is a Jordan algebra.

m=3: In this case we have

\begin{aligned}x^{3} + q_{1}(x)x^{2} + q_{2}(x)x + q_{3}(x)e & = 0 \\ \frac{1}{3}(2x^{4} + x^{2}x^{2}) + q_{1}(x)x^{3} + q_{2}(x)x^{2} + q_{3}(x)x & = 0, \end{aligned}

and if we multiply the first equation by x and then subtract from the second we get x^{2}x^{2} = x^{4}. Thus (due to Albert’s theorem) A is power-associative.

We can single out the Jordan algebras among the power-associative algebras (degree ≤ 3).

Theorem 3.3. A unital R-algebra of degree ≤ 3 is Jordan, if and only if the bilinear form φ(x, y) := q_{1}(xy) is associative; φ(xy, z) = φ(x, yz).

Proof : In Jordan algebras all L(x)L(y) − L(y)L(x), where L(x) denotes left multiplication, are derivations. Then, by the corollary of theorem (1.3), we have q_{j}(z, …, z, x(yz) − y(xz)) = 0, in particular q_{1}(x(yz)) = q_{1}((xy)z). Conversely (for m=3), linearizing x^{3} + q_{1}(x)x^{2} + q_{2}(x)x + q_{3}(x)e = 0 yields 2x(xy) + x^{2}y + 2q_{1}(x)y + q_{1}(y)x^{2} + q_{2}(x)y + 2q_{2}(x, y)x + 3q(x, x, y)e = 0. On the one hand we multiply this equation by x, on the other hand we replace y by xy and then we substract the results: x(x^{2}y) − x^{2}(xy) + [3q_{3}(x, x, y) − 2q_{2}(x, xy) − q_{1}(y)q_{2}(x)]x − [3q_{3}(x, x, xy) + q_{1}(y)q_{3}(x)]e = 0. It is now a routine calculation, using the recursion formulas for q_{2} and q_{3} (see (1.3)), to verify that the coefficients of x and e vanish – if q_{1} is associative. Hence x^{2}(xy) = x(x^{2}y) for all x, y ∈ A, and A is a Jordan algebra. □

The difficulties arise for higher degree.

3.4 Adjoining idempotents.

Since unital R-algebras behave much nicer than the others we employ a standard trick: if there is no unit element, adjoin it! We do it more generally, we adjoin an idempotent. Let A be an arbitrary algebra over the field 𝔽 and c an element (a symbol) not in A. On the direct sum

A(c, μ) := 𝔽c ⊕ A

we define for any μ ∈ 𝔽 a product by

(αc + x) · (βc + y) := αβc + μ(αy + βx) + x · y

where x·y is the product in A. This composition is obviously bilinear (hence a multiplication). In particular c · c = c, c · y = μy for all y ∈ A. We call A(c, μ) an algebra obtained from A by adjoining the idempotent c. (If μ = 1, then c is a unit element in A(c, 1).)

Theorem 3.4. If A is an R-algebra, c ∉ A and μ ∈ \frac{1}{2}ℤ \ \{\frac{1}{2}\}, then the algebra A(c, μ) obtained from A by adjoining an idempotent c is again an R-algebra.

Proof : The differential equation in A(c, μ) is

\frac{d}{dt}(ηc + y) = (ηc + y)^{2} = η^{2}c + 2μηy + y^{2},

equivalently

(3.5)  \begin{aligned}η̇ & = η^{2} \\ ẏ & = 2μηy + y^{2} \end{aligned}

Let the initial value be η(0) = ξ, y(0) = x (with x ∈ A and ξ ≠ 0). Then η(t) = \frac{ξ}{1−ξt} and substitution yields

(3.6)  ẏ = y^{2} + 2μ\frac{ξ}{1 − ξt}y.

To obtain a pure quadratic equation we substitute y = αz , α(0) = 1, z(0) = x: α̇z + αż = ẏ = y^{2} + 2μ\frac{ξ}{1−ξt}y = α^{2}z^{2} + 2μ\frac{ξ}{1−ξt}αz. Thus

ż = αz^{2}, z(0) = x

if we choose α to be the solution of

α̇ = 2μ\frac{ξ}{1 − ξt}α, α(0) = 1,

that is, if

α = (1 − ξt)^{−2μ}.

Next we observe, if (as always in these notes) y(x, t) denotes the solution of ẏ = y^{2} then y(x, φ(t)) solves ż = αz^{2}, if

φ̇(t) = α(t), φ(0) = 0,

thus , if

φ(t) = ∫_{0}^{t} α(s)ds = \frac{1}{ξ(2μ − 1)}[(1 − ξt)^{−2μ+1} − 1].

Putting things together we obtain the solution in A(c, μ): (3.7)

y(ξc + x, t) = \frac{ξ}{1 − tξ} c + \frac{1}{(1 − tξ)^{2μ}} y (x, \frac{1}{ξ(2μ − 1)}[(1 − ξt)^{−2μ+1} − 1]) .

If μ ≠ \frac{1}{2} and 2μ an integer, and if y(x, t) is rational in x and t, then this solution is rational, too. □

Remark: It still seems to be quite complicated to compute the invariant chain of A(c, μ) from the invariant chain in A.

Corollary. Adjoining a unit element preserves R-algebras. In this case (μ = 1) the initial value problem η̇c+ẏ = (ηc+y)^{2}, η(0) = ξ, y(0) = x has the solution

(3.8)  y(ξc + x, t) = \frac{ξ}{1 − ξt}c + \frac{1}{(1 − ξt)^{2}} y(x, \frac{t}{1 − ξt}).
□

Example 1. In any power-associative algebra A the solution is y(x, t) = (Id − tL(x))^{−1}x (see (2.1)). Then the solution in A(c, 1) obtained from A by adjoining the unit element c is

y(ξc + x, t) = \frac{ξ}{1 − ξt}c + \frac{1}{(1 − ξt)^{2}} (Id − \frac{t}{1 − ξt}L(x))^{−1} x.

On the other hand, adjoining a unit element preserves power-associative algebras. Thus, in A(c, 1) the solution formula

y(ξc + x, t) = [c − t(ξc + x)]^{−1}(ξc + x)

applies as well. One may check that the right hand sides are indeed equal.

Example 2. In a nuclear Bernstein algebra (A, ω) the solution of ẏ = y^{2}, y(0) = x is

y(x, t) = x + tx^{2} + \frac{t^{2}}{1 − tω(x)}t^{3}

(see (2.2)). In A(c, 1), the algebra obtained from A by adjoining the unit c (which is no longer a Bernstein algebra ) the solution of ż = z^{2}, z(0) = ξc+x is

z(ξc+x, t) = \frac{ξ}{1 − tξ} c+\frac{1}{(1 − tξ)^{2}} [x + \frac{t}{1 − tξ} x^{2} + \frac{t^{2}}{(1 − ξt)(1 − ξt − ω(x)t)}x^{3}]

The denominator of z(ξc + x, 1) is

q(ξc + x) = (1 − ξ)^{3}(1 − ξ − ω(x))

Example 3. Starting from a one dimensional algebra A_{1} = 𝔽b_{1}, b_{1}^{2} = b_{1} (A ≅ 𝔽) and adjoining repeatedly idempotents we construct the Ralgebra A with a basis (b_{1}, …, b_{n}) of idempotents b_{i}^{2} = b_{i} and the multiplication given by

b_{i}b_{j} = μ_{j}b_{i} (1 ≤ i < j ≤ n)

where 2μ_{i} are integers and μ_{i} different from \frac{1}{2}. The solution of our standard initial value problem in A can be constructed successively from the one dimensional solution \frac{ξ_{1}}{1−ξ_{1}t}. □

4 The inversion j

4.1 j(x) replaces x^{−1}.

In a unital associative algebra A an element x ∈ A is said to be invertible, if there is an element y ∈ A such that xy = yx = e. In this case y is uniquely determined (xy = yx = e ⇒ y = ye = y(xy) = (yx)y = ey = y) and one usually writes y = x^{−1}. x^{−1} is called THE inverse of x ( if it exists). Observe, to show uniqueness one needs associativity. Since in powerassociative algebras all subalgebras generated by one element are associative, the concept of an inverse can be carried over to power-associative algebras. In this case x ∈ A is called invertible, if there is a y in the subalgebra generated by x, such that xy = yx = e. Again x^{−1} := y is unique. For arbitrary nonassociative algebras this concept cannot be carried over. But if, in a power-associative unital algebra A, we replace x by e − x in y(x, 1) = (e − x)^{−1}x (see (2.1)) we get a formula for this inverse

x^{−1} = y(e − x, 1) + e

which can immediately be generalized to arbitrary commutative unital algebras.

Definition. Let A be a unital (with unit element e ∈ A) commutative algebra over the field 𝔽 and y(x, t) = ∑_{k≥0} g_{k}(x)t^{k} the unique solution of the initial value problem ẏ = y^{2}, y(0) = x in A. Then the inversion j of A is defined by

(4.1)  j(x) := y(e − x, 1) + e
□

j is defined on the open subset of A where y(e − x, 1) is defined. We observe that we can recover y(x, t) from j:

(4.2)  \begin{aligned}j(e − x) & = y(x, 1) + e & ⇒ \\ y(x, 1) & = j(e − x) − e & ⇒ \\ y(x, t) & = \frac{1}{t}y(tx, 1) = \frac{1}{t}[j(e − tx) − e]. \end{aligned}

We use this formula to characterize those algebras, for which j(x) is the usual inverse:

Theorem 4.1. Let A be a unital algebra and the inversion j in A be defined by (4.1) . Then xj(x) = e (whenever defined) if and only if A is power-associative.

Proof : If A is power-associative, then (as we have noticed before) x^{−1} = y(e − x, 1) + e = j(x). Conversely, if xj(x) = e (whenever defined), then we get from (4.2)

\begin{aligned}L(e − tx)y(x, t) & = \frac{1}{t}[(e − tx)j(e − tx) − (e − tx)] \\ & = \frac{1}{t}[e − (e − tx)] = x. \end{aligned}

Thus y(x, t) = L(e − tx)^{−1}x and A is power-associative , by theorem 2.1. □

Example 1. j(x) = x^{−1} in power-associative algebras. □

Example 2. Let A(c, 1) be the algebra obtained by adjoining a unit element c to a nuclear Bernstein algebra (A, ω). Then (see (3.4) , example 2)

j(ξc + x) = z(c − (ξc + x), 1) + c
= z((1 − ξ)c − x, 1) + c
= \frac{1}{ξ}c + \frac{1}{ξ^{2}} [−x + \frac{1}{ξ}x^{2} − \frac{1}{ξ(ξ+ω(x))}x^{3}]
□

4.2 More about j.

Although we don’t have x · j(x) = e in general, the inversion j has some rather nice strong properties. First of all, we notice that j(x) is rational in R-algebras. In arbitrary commutative algebras this may not be the case.

Theorem 4.2. Let A be a unital R-algebra. Then the map j defined on an open subset of A by (4.1) is involutory and homogenous of degree -1, hence an inversion on A:

(a ) j(j(x)) = x

(b ) j(tx) = \frac{1}{t}j(x)

Proof : In tj(tx) = ty(e−tx, 1)+te we use the series expansion y(e−tx, 1) = ∑_{k≥0} g_{k}(e − tx) (which converges and may be differentiated termwise near t = 1, x = e) to compute the derivative:

\frac{d}{dt}tj(tx) = ∑_{k≥0} g_{k}(e − tx) + e + t ∑_{k≥0} Dg_{k}(e − tx)(−x).

The second term on the right hand side equals

∑_{k≥0} Dg_{k}(e − tx) · (e − tx) − ∑_{k≥0} Dg_{k}(e − tx) · e
= ∑_{k≥0}(k + 1)g_{k}(e − tx) − ∑_{k≥0}(k + 1)g_{k−1}(e − tx)
= − ∑_{k≥0} g_{k}(e − tx) − e

where we used Dg_{k}(y) · y = (k + 1)g_{k}(y) – g_{k} is homogenous of degree k + 1 – and the going down formula Dg_{k}(y) · e = (k + 1)g_{k−1}(y) (see (3.1)). Thus \frac{d}{dt}tj(tx) = 0 and tj(tx) is locally constant (near t = 1). Since this is a rational function it is constant everywhere: tj(tx) = j(x), this is b). a): From the theory of differential equation (uniqueness theorem) it is well known

y(y(x, t), s) = y(x, t + s)

(the solution curve which passes at time s = 0 through y(x, t) is given by y(x, t + s)). In particular we have

y(y(x, 1), −1) = y(x, 0) = x.

Since j(x) − e = y(e − x, 1) (by definition) this equation implies

y(j(x) − e, −1) = y(y(e − x, 1), −1) = e − x.

Then we get – using ty(x, t) = y(tx, 1) for t = −1 –

x = e − y(j(x) − e, −1) = e + y(e − j(x), 1) = j(j(x)).
□

Although j(x) is not the inverse x^{−1} in general, it maintains some important properties of inverses:

(a) j(e) = e, Dj(e) = −Id
(4.3)  (b) Dj(x) · x^{2} = −e
(c) j(e − j(x)) = e − j(e − x).

Proof : We use the series expansion

j(x) = y(e − x, 1) + e = ∑_{k≥0} g_{k}(e − x) + e

to compute the derivative in y-direction

(4.4)  \begin{aligned}Dj(x)y & = − ∑_{k≥0} Dg_{k}(e − x) · y \\ & = − ∑_{k≥0}(k + 1)g_{k}(e − x, …, e − x, y) \\ & = −y − ∑_{k≥0}(k + 1)g_{k}(e − x, …, e − x, y) \end{aligned}

This reduces with x = e to Dj(e) · y = −y. This proves a). (j(e) = e is immediate from the definition.) b): The standard procedure, substituting x by e − x in (4.4), gives

Dj(e − x) = − ∑_{k≥0} Dg_{k}(x),

and then

Dj(e − x) · (e − x)^{2} = − ∑_{k≥0} Dg_{k}(x) · e + 2 ∑_{k≥0} Dg_{k}(x) · x−

− ∑_{k≥0} Dg_{k}(x) · x^{2}

= −e − ∑_{k≥0}(k + 1)g_{k−1}(x) + 2 ∑_{k≥0}(k + 1)g_{k}(x)

− ∑_{k≥0}(k + 1)g_{k+1}(x)

(use the going up and going down formulas (1.2), (3.1) and
Dg_{k}(x) · x = (k + 1)g_{k}(x) )
… = −e − ∑_{k≥0}(k + 2)g_{k}(x) + 2 ∑_{k≥0}(k + 1)g_{k}(x)−

− ∑_{k≥0} kg_{k}(x)

= e.

This shows Dj(x) · x^{2} = −e.

We pause to give two interpretations of this formula.

1^{st} With Dj(x) · x^{2} = −e, or equivalently, with x^{2} = −Dj(x)^{−1} · e we can reconstruct the squaring on A on an open subset of A. But this is enough to reconstruct the product x · y on A.

2^{nd} If y(t) is a solution of ẏ = y^{2}, then

\frac{d}{dt}j(y(t)) = Dj(y(t))ẏ(t) = Dj(y(t))ẏ(t)^{2} = −e.

Thus j maps parametrized solutions of ẏ = y^{2} onto parametrized solutions of ż = −e, which are straight lines (z(t) = −te + a). For this reason j is called a straightening function for ẏ = y^{2}. Taking y(t) = y(x, t) we get

j(y(x, t)) = −te + j(x),

equivalently (using j(j(x)) = x)

(4.5)  y(x, t) = j(j(x) − te).

Now we continue to prove (4.3) c): in (4.2) we constructed y(x, t) from j,

y(x, t) = \frac{1}{t}[j(e − tx) − e]

and comparing this with (4.5) we obtain

tj(j(x) − te) = j(e − tx) − e.

The obvious choice t = 1 gives the desired result.

□

Remark: Applying j on both sides of (4.3 c) and then substituting x by j(x) gives

(4.4)  x = e − j(e − j(e − j(x)))

which is equivalent to (4.3 c), the foregoing manipulations are reversible. In the theory of associative algebras a very similar formula is known as Hua’s identity:

aba = a − (a^{−1} − (a − b^{−1})^{−1})^{−1}.

For this reason (4.4) – or the equivalent formula (4.3 c) – is called HUA’s identity at the unit element.

4.3 R-inversions.

It is well known that the inverse x ↦ x^{−1} plays a substantial role in the theory of Jordan algebras [1]. An axiomatic approach to the inverse was achieved by T.A. Springer and K. McCrimmon ([15],[8]). They introduced and investigated j-structures on vectorspaces and they showed, roughly speaking, that an involutory unitary inversion j on a vector space which satisfies a Hua identity is precisely the inverse j(x) = x^{−1} on a unital Jordan algebra. The inversion j in unital R-algebras defined by (4.1) has all the properties which are required in Springers’s and McCrimmon’s work, with the exception that for our j we have only a weak Hua identity, the Hua identity at e (see (4.4)). But these properties completely characterize unital R-algebras.

Theorem 4.3. Let j be a rational map on an 𝔽-vectorspace V with the following properties: j(j(x)) = x, j(tx) = \frac{1}{t}j(x), j has an isolated fixed point e ∈ V . Then j is the inversion of an R-algebra A = (V, ·) with unit element e, j(x) = y(e − x, 1) + e in A, if and only if Hua’s identity at e

(4.5)  x = e − j(e − j(e − j(x)))

holds.

Proof : If A is an R-algebra with unit e, then j defined by (4.1) has all these properties and satisfies Hua’s identity at e. Conversely, using j with the given properties we have to define a multiplication on V leading to an R-algebra A = (V, ·) such that j(x) = y(e − x, 1) + e in this algebra. We start by replacing x by j(tx) in (4.5) to obtain j(e − j(e − tx)) = e − j(tx). Then (apply j on both sides)

(4.6)  t^{−1}(j(e − tx) − e) = j(j(x) − te)

Taylor expansion about t = 0 on both sides leads to

t^{−1}[j(e − tx) − e] = t^{−1}[−tDj(e)x + \frac{t^{2}}{2} D^{2}j(e)(x, x) + ⋯]

= j(j(x)) + tDj(j(x))(−e) + ⋯

Equating coefficients gives

(4.7) Dj(e) = −Id,  +\frac{1}{2}D^{2}j(e)(x, x) = Dj(j(x))(−e) = Dj(x)^{−1}(−e).

(The last equality in (4.7) comes from j(j(x)) = x, thus Dj(j(x))Dj(x) = Id.) Since the second derivative is bilinear, there is a product on V defined by

(4.8)  x · y = \frac{1}{2}D^{2}j(e)(x, y)

Next we observe that Dj(x) is homogeneous of degree −2 (differentiate tj(tx) = j(x) in x-direction), use Euler’s differentiation rule for homogeneous mappings and obtain

D^{2}j(x)(y, x) = D(Dj(x)y)x = −2Dj(x)y,

in particular (x = e, Dj(e) = Id, see(4.7))

ey = \frac{1}{2}D^{2}j(e)(e, y) = \frac{1}{2}(−2Dj(e)y) = y.

Thus e is the unit element of A = (V, ·). Now we show that (V, ·), the product given by (4.8) is an R-algebra. Let y(t) = y(x, t) be the solution of the corresponding initial value problem ẏ = y^{2}, y(0) = x. Then by (4.7)

\frac{d}{dt}j(y(t)) = Dj(y(t))ẏ(t) = Dj(y(t))y(t)^{2}
= Dj(y(t))(\frac{1}{2}D^{2}j(e)(y(t), y(t))
= Dj(y(t))Dj(y(t))^{−1}(−e) = −e

and therefore

j(y(x, t)) = −te + j(x)

This implies (apply j on both sides and use (4.6))

y(x, t) = j(j(x) − te) = \frac{1}{t}(j(e − tx) − e)

which shows that the solution y(x, t) is rational (what we wanted to show) and moreover (for t = 1 and x replaced by e − x)

j(x) = y(e − x, 1) + e.
□

Remark: of course, if we start with an R-algebra A and the inversion j in A defined by (4.1) then the above construction brings nothing new. The product (4.8) is the ’old’ product in A. To distinguish j from any other inversion on A we call it the R-inversion of the R-algebra A. Although in general j(x) is not the inverse of x, nevertheless it behaves like an inverse, in particular inside the invariant chain.

Theorem 4.4. Let A be a unital R-algebra with invariant chain q_{1}, q_{2}, …, q_{m} and R-inversion j, then the following identities hold (whenever defined):

a) q_{m}(x)q_{m}(e − j(x)) = q_{m}(e − x),

b) q_{m}(x)q_{k}(j(x)) = q_{m−k}(x) (1 ≤ k ≤ m, q_{0}(x) := 1).

Proof : (1.15) with s_{m} = q_{m}, q(x) = 1 + q_{1}(x) + ⋯ + q_{m}(x) and t = 1 gives

q_{m}(y(x, 1)) = \frac{q_{m}(x)}{q(x)},

equivalently (x ↔ e − x)

q(e − x)q_{m}(y(e − x, 1)) = q_{m}(e − x).

Since y(e − x, 1) = j(x) − e (by definition) and q(e − x) = (−1)^{m}q_{m}(x) (by Theorem 3.1), the left hand side equals (−1)^{m}q_{m}(x)q_{m}(j(x) − e) = q_{m}(x)q_{m}(e − j(x)). This shows a). Again Theorem 3.1 d) allows us to replace q_{m}(e − y) by (−1)^{m}q(y) (for y = x, resp. y = j(x)) in a) to obtain q_{m}(x)q(j(x)) = q(x). Now the second identity follows by comparing the homogeneous summands of the same degree. □

4.4 A formula for j(x).

Let A be a unital R-algebra of degree m, q(x) = 1 + q_{1}(x) + ⋯ + q_{m}(x) the exact denominator of y(x, 1). We already know that j(x) = y(e − x, 1) + e is rational and can be represented as

j(x) = \frac{1}{h(x)}H(x)

with a scalar polynomial h(x) and a vector valued polynomial H(x). We call h(x) the exact denominator, if it does not have a common prime factor with all components of H(x) (with respect to a basis). We are going to determine h(x) and H(x). From the two identities

j(x) = y(e − x, 1) + e = \frac{1}{h(x)}H(x)

y(x, 1) = j(e − x, 1) − e = \frac{1}{q(x)}P (x)

we get

\frac{1}{q(e − x)}P (e − x) + e = \frac{1}{h(x)}H(x) and

\frac{1}{h(e − x)}H(e − x) − e = \frac{1}{q(x)}P (x).

Since both, h and q are exact denominators we obtain h(x) = aq(e − x) for some constant a ∈ F. Normalizing, we may assume a = 1. Thus h(x) = q(e − x) = (−1)^{m}q_{m}(x) and

(4.9)  j(x) = \frac{1}{q(e − x)}H(x)

with a homogeneous polynomial H(x) of degree m − 1:

H(tx) = (−1)^{m}q_{m}(tx)j(tx) = (−1)^{m}t^{m}q_{m}(x)t^{−1}j(x) = t^{m−1}H(x)

Next, we expand

(4.10)  q(x)j(e − x) = H(e − x) = ∑_{k=0}^{m−1} H_{k}(x)

as a sum of homogeneous polynomials H_{k} of degree k and observe at first H_{m−1}(x) = (−1)^{m−1}H(x). This is a general fact concerning homogeneous maps: remember, there is a corresponding (m − 1)-linear map (also denoted by H) such that H(x) = H(x, x, …, x). Then H(e − x) = H(e − x, e − x, …, e − x) = H(e) − (m − 1)H(e, …, e, x) + ⋯ + (−1)^{m−1}H(x, …, x).

On the left hand side of (4.10) we use j(e − x) = y(x, e) + e = e + ∑_{k≥0} g_{k}(x) (which may be called the geometric series in A) and q(x) = ∑_{k=0}^{m} q_{k}(x), and then compare homogeneous terms of equal degree:

H_{k}(x) = g_{k−1}(x) + q_{1}(x)g_{k−2} + ⋯ + q_{k}(x)e, 1 ≤ k ≤ m − 1.

In particular

H(x) = (−1)^{m−1}H_{m−1}(x) = (−1)^{m−1}[g_{m−2}(x)+q_{1}(x)g_{m−3}(x)+⋯+q_{m−1}(x)e].

Replacing this expression in (4.9) and using q(e − x) = (−1)^{m}q_{m}(x) we find a nice formula for j(x):

(4.11)  j(x) = \frac{1}{q_{m}(x)}[g_{m−2}(x) + q_{1}(x)g_{m−3}(x) + ⋯ + q_{m−1}(x)e]

Remark: This formula corresponds to the formula for x^{−1} in power-associative algebras

x^{−1} = \frac{1}{q_{m}(x)}[x^{m−1} + q_{1}(x)x^{m−2} + ⋯ + q_{m−1}(x)e],

which is obtained from the generic minimum polynomial

x^{m}+q_{1}(x)x^{m−1}+⋯+q_{m}(x)e = 0 ⇒ x[x^{m−1}+q_{1}(x)x^{m−2}+q_{m−1}(x)e] = −q_{m}(x)e.

So there is a nice correspondence of the generic minimum polynomial, geometric series and the formula for the inverse in power-associative algebras and R-algebras – one only has to replace x^{k+1} by g_{k}(x) and x^{−1} by j(x).

4.5 The quasi-inverse.

In the theory of associative rings as well as in Jordan theory there is the useful notion of quasi-invertibility. It may, for example, be used to define the Jacobson radical. We recall that an element x in an associative ring R is called quasi-invertible, if c − x is invertible in the Ring R̂ = Rc ⊕ R obtained from R by adjoining a unit element c ∉ R. (Observe, the unit element of R̂ is c, no matter whether R has a unit or not.) If c − x is invertible with inverse αc + y, then c = (c − x)(αc + y) = αc − αx + y − xy and therefore α = 1 and xy = x − y. The inverse of c − x has the form c + y; y is called the quasi-inverse of x. Let A be an R-algebra and Â = A(c, 1) = 𝔽c ⊕ A the algebra obtained from A by adjoining a unit element c ∉ A. (see 3.4) Definition. x ∈ A is called quasi-invertible with quasi-inverse y ∈ A, if j(c − x) exists in Â and has the form j(c − x) = c + y. □

Note, by definition j(z) = y(c − z, 1) + c in Â and then

j(c − x) = y(x, 1) + c (x ∈ A).

This sheds new light on quasi-invertibility: x ∈ A is quasi-invertible, if and only if y(x, 1) exists. In this case y = y(x, 1) is the quasi-inverse of x.

5 Nilpotents, P-elements

5.1 P-elements

Let A be an R-algebra over 𝔽 with invariant chain q_{1}(x), …, q_{m}(x), m ≥ 1. As always in this report y(x, t) denotes the solution of ẏ = y^{2}, y(0) = x in A, which has a power series expansion

y(x, t) = ∑_{k≥0} g_{k}(x)t^{k}

where the homogeneous polynomials g_{k}(x) of degree k + 1 replace the ’ordinary’ powers x^{k+1}. The computation of the g_{k}(x) is step by step, always differentiating in x^{2}-direction

g_{k+1}(x) = g_{k}(x, …, x, x^{2}),

or using the recursion

g_{k+1}(x) = \frac{1}{k + 1} ∑_{j=0}^{k} g_{j}(x)g_{k−j}(x).

Since ty(x, t) = y(tx, 1), the solution is completely determined by

y(x, 1) = ∑_{k≥0} g_{k}(x) = \frac{1}{q(x)}P (x)

where P (x) is a vectorial polynomial and q(x) = 1 + q_{1}(x) + ⋯ + q_{m}(x) the exact denominator of y(x, 1).

We are used to call an element u ∈ A nilpotent, if there is a k such that x^{k} = 0. In this case x^{l} = 0 for all l ≥ k. But nothing can be said about arbitrary monomials of degree l ≥ k, unless A is power-associative. One may, of course, introduce the notion of strong nilpotence and define an element u ∈ A to be strongly nilpotent, if all monomials in x of degree l ≥ k are zero. This seems to be an unneccessary strong condition. In R-algebras we use P -elements, instead.

Definition. An element v ∈ A is called a P -element if there is an index k_{0} such that g_{k}(v) = 0 for all k ≥ k_{0}. □

Obviously, v ∈ A is a P -element, if and only if the solution y(v, t) = ∑_{k≥0} g_{k}(v)t^{k} is a polynomial in t.

Lemma 5.1. An element v ∈ A is a P -element, if and only if q_{1}(v) = q_{2}(v) = ⋯ = q_{m}(v) = 0.

Proof : If q_{i}(v) = 0 then q(tv) = 1 + q_{1}(v)t + ⋯ + q_{m}(v)t^{m} and ty(v, t) = y(tv, 1) = P (tv). Since P (0) = 0 the factor t can be cancelled, thus y(v, t) is a polynomial. Conversely, assume that v is a P -element and q_{l}(v) ≠ 0 for some l, 1 ≤ l ≤ m. Let’s take the maximal index l with this property. Then, if l < m we have q_{l+1}(v) = ⋯ = q_{m}(v) = 0 and also q_{l+1}(y(v, t)) = ⋯ = q_{m}(y(v, t)) = 0 by the corollary of theorem 1.2. From the basic recursion formula (1.9) for the invariant chain we obtain for y = y(v, t)

lq_{l}(y, …, y, y^{2}) + q_{1}(y)q_{l}(y) = (l + 1)q_{l+1}(y) = 0,

equivalently (use ẏ = y^{2})

(5.1)  \frac{d}{dt}q_{l}(y(v, t)) = −q_{1}(y(v, t))q_{l}(y(v, t))

We notice that we get the same equation from the recursion (1.9) also for case l = m.

Since v is a P -element – y(v, t) a polynomial – the scalar function q_{l}(y(v, t)) is a polynomial. It is nonzero, since at t = 0 we have q_{l}(y(v, 0)) = q_{l}(v) ≠ 0 by assumption. But then q_{1}(y(v, t)) = 0, since the differentiation on the left hand side in (5.1) reduces the degree, an effect which could not be achieved by multiplication with a factor ≠ 0 on the right hand side. Thus

q_{1}(y(v, t)) = 0.

But then – use the recursion (1.9) again, y = y(v, t) –

\frac{d}{dt}q_{j}(y) = jq_{j}(y, …, y, y^{2}) + q_{1}(y)q_{j}(y) = (j + 1)q_{j+1}(y)

and consequently (j = 1, 2, …) q_{2}(y(v, t)) = 0, …, q_{l}(y(v, t)) = 0, in particular q_{l}(v) = q_{l}(y(v, 0)) = 0 which contradicts our assumption q_{l}(v) ≠ 0.

□

If v ∈ A is a P -element, then (by definition) the solution of ẏ = y^{2}, y(0) = v has the form y(v, t) = v + tv^{2} + ⋯ + t^{k}g_{k}(v). In particular, this solution is defined at all times t, y(v, 1) = v + v^{2} + ⋯ + g_{k}(v) exist, so does the quasiinvers (see 4.5). In the algebra Â = 𝔽e ⊕ A obtained from A by adjoining a unit element e (e ∉ A) the ”inverse” j(e − v) has form

j(e − v) = e + v + ⋯ + g_{k}(v).

This formula generalizes the well known formula

(E − N)^{−1} = E + N + N^{2} + ⋯ + N^{k}

for nilpotent matrices (N^{k+1} = 0, E the identity matrix).

5.2 2-nilpotents

An element u ∈ A is called a 2-nilpotent, if its square u^{2} vanishes, u^{2} = 0. These elements are the most ”strongest” nilpotents in A. Any monomial in u of degree ≥ 1 vanishes, in particular g_{k}(u) = 0 for all k ≥ 1 (if u^{2} = 0). Since the 2-nilpotents are stationary points of the differential equation (y(v, t) = v = const for all t) they are of particular interest from the analytic point of view. Lemma 5.2. If the R-algebra A contains a nonzero P -element, then it contains a nonzero 2-nilpotent. Proof : Let v ≠ 0 be a P -element, y(v, t) = v + tv^{2} + ⋯ + t^{k}g_{k}(v) the polynomial solution of ẏ = y^{2}, y(0) = v with highest coefficient g_{k}(x) ≠ 0. Then, comparing the coefficients of t^{2k} in \frac{d}{dt}y(v, t) = y(v, t)^{2} immediately shows g_{k}(v)^{2} = 0. □

In associative or alternative algebras the left multiplication satisfies L(x^{k}) = L(x)^{k} for all k. Thus, if x is nilpotent, then L(x) is nilpotent, too. This does not hold in general. Take, for example, a twodimensional algebra with basis x, y and product being defined by x · x = 0, x · y = y. But this is no R-algebra.

Theorem 5.3. If A is an R-algebra and v ∈ A a 2-nilpotent (v^{2} = 0) then the left multiplication L(v) is nilpotent.

Proof : We may assume A unital and 𝔽 = ℂ. In this case a linear map on A is nilpotent, if and only if all its eigenvalues are zero. Let λ ∈ ℂ be an eigenvalue of L(v) and w ∈ A a corresponding eigenvector, vw = λw. We linearize (differentiate) the recursion 1.9 in w-direction to obtain

k(k − 1)q_{k}(v, …, v, w, v^{2}) + 2kq_{k}(v, …, v, vw)
+q_{1}(w)q_{k}(v) + kq_{1}(v)q_{k}(v, …, v, w)
= (k + 1)^{2}q_{k+1}(v, …, v, w)

and

m(m − 1)q_{m}(v, …, v, w, v^{2}) + 2mq_{m}(v, …, v, vw)
+q_{1}(w)q_{m}(v) + mq_{1}(v)q_{m}(v, …, v, w) = 0.

Using v^{2} = 0, vw = λw and q_{1}(v) = ⋯ = q_{m}(v) = 0 (see lemma 5.1) these equations reduce to

2kλq_{k}(v, …, v, w) = (k + 1)^{2}q_{k+1}(v, …, v, w)
mλq_{m}(v, …, v, w) = 0.

Assume λ ≠ 0, then these two equations show

q_{k}(v, …, v, w) = 0 for all k ≥ 1.

Now we differentiate the basic identity ∑_{i=0}^{m} q_{i}(x)g_{m−i}(x) = 0 in w-direction,

substitute x = v and use q_{i}(v) = q_{i}(v, …, v, w) = 0 for all i ≥ 1 to obtain

(5.2)  g_{m}(v, …, v, w) = 0.

Linearizing g_{m}(x) = g_{m−1}(x, …, x, x^{2}) gives

(m + 1)g_{m}(x, …, x, y) = 2g_{m−1}(x, …, x, xy) + (m − 1)g_{m−1}(x, …, x, y, x^{2})

and substituting x = v, y = w leads to g_{m−1}(v, …, v, w) = 0 (use v^{2} = 0, vw = λw and λ ≠ 0). We repeat this argument (induction) to descend to g_{1}(v, w) = vw = λw = 0, which contradicts λ ≠ 0, w ≠ 0. □

5.3 P -ideals

Next, we want to characterize ideals in A which consists entirely of P - elements.

Definition. An ideal U ⊆ A is called a P -ideal, if all elements u ∈ U are P -elements. □

If U ⊆ A is a P -ideal, then q_{i}(u) = 0 for all u ∈ U, 1 ≤ i ≤ m (see lemma 5.1). Since U is an ideal, these and even much stronger properties follow from q_{1}(u) = 0 for all u ∈ U.

Lemma 5.4. If U ⊆ A is an ideal such that q_{1}(u) = 0 for all u ∈ U, then U is a P -ideal. Moreover q(x + u) = q(x) for all x ∈ A, u ∈ U.

Proof : At first we show by induction on i

(5.3)  q_{i}(x, …, x, u) = 0 for all x ∈ A, u ∈ U.

For i = 1 this is true by assumption. Suppose (5.3) holds for k ≤ i. Then all terms on the right hand side of

(i + 1)^{2}q_{i+1}(x, …, x, u) = i(i − 1)q_{i}(x, …, x^{2}, u) + 2iq_{i}(x, …, x, xu)
+q_{1}(u)q_{i}(x) + iq_{1}(x)q_{i}(x, …, x, u)

(which is a linearization of (1.9)) vanish for u ∈ U. Thus (5.3) is true for i + 1 and the assertion follows. In particular, put x = u, we have q_{i}(u) = 0, 1 ≤ i ≤ m, for all u ∈ U, thus U is a P -ideal by lemma 5.1. Moreover, in the expansion q_{i}(x + u, …, x + u) = q_{i}(x) + iq_{i}(x, …, x, u) + ⋯ + q_{i}(u) all terms on the right hand side containing at least one argument u vanish (see (5.3) and its linearization). Thus

q(x + u) − q(x) = ∑ [q_{i}(x + u) − q_{i}(x)] = 0
□

As a consequence we have a characterization of P -ideals by an equation:

Theorem 5.5. An ideal U ⊆ A is a P -ideal if and only if q(x + u) = q(x) for all x ∈ A, u ∈ U.

Proof : The ”only if”-part of this theorem is in lemma 5.4; we still have to prove the ”if”-part. Let u be an element from U. Then αu ∈ U for all α ∈ 𝔽 and q(x) = q(x + αu) for all x ∈ A (by assumption); in particular (x = 0)

1 = q(0) = q(αu) = 1 + q_{1}(u)α + ⋯ + q_{m}(u)α^{m}

for all α ∈ 𝔽. Thus q_{1}(u) = ⋯ = q_{m}(u) = 0 and u is a P -element (again lemma 5.1). □

As an application of this characterization we show

Lemma 5.6. The sum U_{1} + U_{2} of two P -ideals U_{i} ⊆ A is again a P -ideal.

Proof : The assertion follows from

q(x + (u_{1} + u_{2})) = q((x + u_{1}) + u_{2}) = q(x + u_{1}) = q(x)

for all x ∈ A, u_{i} ∈ U_{i}, and theorem 5.5. □

It is now natural to consider the sum of all P -ideals in A.

P -radA := ∑ {U ⊂ A | U is a P -ideal}

is called the P -radical of A.

Any P -ideal in A is (by definition) contained in P -radA and P -radA itself is a P -ideal (see lemma 5.6), thus it is the unique maximal P -ideal in A. Moreover, by lemma 5.4,

P -radA ⊆ {u ∈ A | q(x + u) = q(x) for all x ∈ A}.

It is not clear, whether the set on the right hand side is always an ideal, we can only show that it is a subalgebra.

Lemma 5.7. For u ∈ A the following conditions are equivalent

(5.4) a) q(x + u) = q(x) for all x ∈ A, b) q_{i}(x, …, x, u) = 0 for all x ∈ A, 1 ≤ i ≤ m.

Proof : ”b) ⇒ a)” is contained in the proof of lemma 5.4. Let’s assume a). Then (by induction) q(ku + x) = q(x) for all x ∈ A and all k ≥ 1, and the polynomial h(t) := q(tu + x) − q(x) (for fixed u, x) has infinitely many zeros. Thus

q(tu + x) = q(x) for all x ∈ A, t ∈ 𝔽.

By taking the derivative (with respect to t) at t = 0 we find ∑_{i} iq_{i}(x, …, x, u) =

0 for all x ∈ A, which is obviously equivalent to b) (consider the homogeneuos parts). □

Theorem 5.8. Let A be an R-algebra with invariant chain q_{1}, …, q_{m}, q = 1 + q_{1} + ⋯ + q_{m}, then the set

N = {u ∈ A | q(u + x) = q(x) for all x ∈ A}

is a subalgebra.

Proof : Obviously N is a subspace (see (5.4 b)). Next we note that for u ∈ N not only q_{i}(x, …, x, u) = 0 for all x ∈ A, all linearizations vanish, too: q_{i}(x_{1}, …, x_{i−1}, u) = 0 for all x_{1}, …, x_{i−1} ∈ A. Then from a linearization of (1.9) we also get

(5.5)  q_{i}(x, …, x, xu) = 0 for all x ∈ A, u ∈ N.

Once again, linearization gives for u, v ∈ N

0 = q_{i}(x, …, x, vu) + q_{i}(x, …, x, xu, v) = q_{i}(x, …, x, vu),

thus uv ∈ N. □

5.4 Addendum

We include in these notes a very general result on arbitrary real nonassociative algebras which motivated much of subsequent work on algebras and differential equations.

Theorem 5.9. Let A be a finite dimensional nonassociative algebra over ℝ without 2-nilpotents ≠ 0 then A contains at least one idempotent (c^{2} = c).

Proof : If there is a nonzero element u ∈ A such that u^{2} = αu, α ≠ 0, then c := \frac{1}{α}u is an idempotent. Assume that there is no idempotent and no 2-nilpotent ≠ 0 in A. Then, by the foregoing remark, u^{2} ≠ αu for all α ∈ ℝ and we have a homotopy

S(u, λ) := \frac{λu^{2} − (1 − λ)u}{|λu^{2} − (1 − λ)u|}, 0 ≤ λ ≤ 1

on the sphere which connects g(u) = u^{2} with the antipodal map j(u) = −u. But then, using the right notion of a degree, which is invariant under homotopy, we come to a contradiction. The degree of g is even and the degree of j is odd. □

This theorem implies a basic result on quadratic differential equations.

Corollary. Any homogeneous quadratic system of differential equations has a solution ray consisting entirely of stationary points or carrying the orbit of an unbounded solution.

Proof : The quadratic system may be written as ẏ = y^{2}. If u ≠ 0 is a 2nilpotent, then all y(t) = αu, α ∈ ℝ, are stationary points. And if c ≠ 0 is an idempotent, then y(t) = \frac{α}{1−αt}c (α ≠ 0 ) is an unbounded solution of ẏ = y^{2}.

□

6 Idempotents

6.1 Basic properties.

Now we concentrate on idempotents. An element c ∈ A is called an idempotent, if c^{2} = c. Idempotents always play a special role in the structure theory because they usually lead to a decomposition of the algebra, which reveals parts of its structure. If A is finite dimensional power-associative and x ∈ A not nilpotent (x^{k} ≠ 0 for all k ≥ 1) then the subalgebra ⟨x⟩ generated by x contains an idempotent. This basic theorem (due to A.A.Albert) guarantees the existence of many idempotents in power-associative algebras which are not nil. In general we only know that there are idempotents if A has no 2-nilpotents (see theorem 5.9). Below we will prove another existence theorem for idempotents, but at first we evaluate the polynomials g_{k}(x) and q_{j}(x) at an idempotent x = c, c^{2} = c.

Lemma 6.1. Let A be an R-algebra with unit element e and c = c^{2} ∈ A an idempotent, 0 ≠ c ≠ e, then

a) g_{k}(c) = c for all k ≥ 0,

b) there is a k, 1 ≤ k < m, such that

q_{1}(c) = −k, q_{j}(c) = (−1)^{j} ( \begin{matrix}k \\ j \end{matrix} ) if 1 ≤ j ≤ k and

q_{j}(c) = 0 if j > k.

Proof : An easy induction using

g_{k+1}(x) = g_{k}(x, …, x, x^{2}) (see(1.2′))

proves the first assertion.

To prove b) we evaluate the basic identity (3.4)

g_{m−1}(x) + q_{1}(x)g_{m−2}(x) + ⋯ + q_{m}(x)e = 0

at x = c and obtain

[1 + q_{1}(c) + ⋯ + q_{m−1}(c)]c + q_{m}(c)e = 0.

Thus

1 + q_{1}(c) + ⋯ + q_{m−1}(c) = 0 and q_{m}(c) = 0

since c and e are linearly independent. Next, we use the recursion relation (1.9) for the invariant chain at x = c to obtain

(6.1)  (j + q_{1}(c)) q_{j}(c) = (j + 1)q_{j+1}(c), 1 ≤ j < m.

Thus, if q_{1}(c) = 0, then q_{j}(c) = 0 for all j, contradicting the fact that c is not a P -element (see 5.1). Hence q_{1}(c) ≠ 0. Let k be the maximum of all j such that q_{j}(c) ≠ 0, then 1 ≤ k < m and

(k + q_{1}(c)) q_{k}(c) = 0

by (6.1), and consequently q_{1}(c) = −k. Finally, using the recursion (6.1) repeatedly we get

q_{j}(c) = \frac{j − k}{j} q_{j−1}(c) = \frac{(j − k)(j − 1 − k)}{j(j − 1)}q_{j−2}(c) = ⋯ = (−1)^{j} ( \begin{matrix}k \\ j \end{matrix} )

□

Definition. A idempotent c ∈ A is called primary (resp. secondary, resp. ternary, etc) if q_{1}(c) = −1 (resp. q_{1}(c) = −2, resp. q_{1}(c) = −3, etc).

□

Obviously, primary idempotents are indecomposable,they cannot be decomposed as sum c_{1} + c_{2} of nontrivial idempotents. Primary idempotents c satisfy (by definition and lemma 6.1)

q_{1}(c) = −1, q_{2}(c) = ⋯ = q_{m}(c) = 0.

Conversely, if the system of equations

q_{1}(x) + 1 = q_{2}(x) = ⋯ = q_{m}(x) = 0

has a solution in A then A contains a primary idempotent. Theorem 6.2. If c ∈ A satisfies

q_{1}(c) = −1, q_{2}(c) = ⋯ = q_{m}(c) = 0,

then g_{r}(c) is a primary idempotent for some r ≥ 0 . Proof : For k ≥ 0, let c_{k} := g_{k}(c). From the defining identities (1.8)

g_{k}(x) + q_{1}(x)g_{k−1}(x) + ⋯ + q_{m}(x)g_{k−m}(x) = 0 (k ≥ k_{0} ≥ m) we get with x = c

c_{k} − c_{k−1} = 0 for all sufficiently large k.

Thus there is an integer r such that

(6.2)  c_{r+j} = c_{r} for all j ≥ 0.

Now we take a closer look at the terms in the recursion formula (1.3)

c_{2r} = \frac{1}{2r} ∑_{j=0}^{2r−1} c_{j}c_{2r−1−j}.

If j ≥ r then c_{j} = c_{r} (by 6.2) and if j < r then 2r−1−j ≥ r and c_{2r−1−j} = c_{r}. Therefore

c_{2r} = \frac{1}{2r} ∑_{j=0}^{r−1} c_{j}c_{2r−1−j} + \frac{1}{2r} ∑_{j=r}^{2r−1} c_{j}c_{2r−1−j}
= \frac{1}{2r} ∑_{j=0}^{r−1} c_{j}c_{r} + \frac{1}{2r} ∑_{j=r}^{2r−1} c_{r}c_{2r−1−j}
= \frac{2}{2r} (∑_{j=0}^{r−1} c_{j}) c_{r}.

The same argument, starting with 2r + 2 instead of 2r gives

c_{2r+2} = \frac{2}{2r + 2} (∑_{j=0}^{r} c_{j}) c_{r} = \frac{2}{2r + 2} (∑_{j=0}^{r−1} c_{j}) c_{r} + \frac{2}{2r + 2}c_{r}^{2}

From these two relations we find (2r + 2)c_{2r+2} = 2rc_{2r} + 2c_{r}^{2}, and recalling c_{2r+2} = c_{2r} = c_{r} (see (6.2)), this equation shows that c_{r} is idempotent, c_{r}^{2} = c_{r}.

To prove that c_{r} is primary (q_{1}(c_{r}) = −1) we employ (1.14) only for k = 1, s_{k} = q_{k}, and x = c the element in A satisfying q_{1}(c) = −1, q_{j}(c) = 0, if j > 1:

q_{1}(y(c, t)) = \frac{−1}{1 − t}.

With the power series expansion y(c, t) = ∑_{k≥0} g_{k}(c)t^{k} = ∑_{k≥0} c_{k}t^{k} on the left

hand side and the geometric series on the right hand side we obtain

∑_{k≥0} q_{1}(c_{k})t^{k} = − ∑_{k≥0} t^{k},

thus q_{1}(c_{k}) = −1 for all k ≥ 0. □

6.2 The eigenvalues of L(c), c^{2} = c.

In order to decompose the algebra with respect to an idempotent c we need all eigenvalues of the left multiplication L(c) and their multiplicities. The determination of these eigenvalues is achieved in three steps. Let A be an arbitrary nonassociative algebra, c = c^{2} an idempotent and – as usual – the homogeneous polynomial g_{k}(x) defined recursively by g_{0}(x) = x, g_{k+1}(x) = g_{k}(x, …, x, x^{2}).

Lemma 6.3. For all x ∈ A and k ≥ 0 we have

g_{k}(c, …, c, x) = \frac{1}{k + 1}p_{k}(L(c))x

with the polynomial

p_{k}(t) = \frac{1}{k!} ∏_{j=0}^{k−1}(2t + j).

Proof : (by induction). The assertion is obviously true for k = 0, 1 (g_{0}(c) = c, g_{1}(c, x) = cx = L(c)x). Next, for the step from k to k + 1 we differentiate g_{k+1}(y) = g_{k}(y, …, y, y^{2}) in x-direction, replace y by c and use the induction hypothesis to obtain

(k + 2)g_{k+1}(c, …, c, x) = kg_{k}(c, …, c, x) + 2g_{k}(c, …, c, cx)
= \frac{k}{k + 1}p_{k}(L(c))x + \frac{2}{k + 1}p_{k}(L(c))L(c)x
= \frac{1}{k + 1}p_{k}(L(c))[2L(c) + k Id]x
= p_{k+1}(L(c))x.

This completes the induction. □

In a second step we prove an identity for the polynomials p_{k}(t):

(6.3)  ∑_{j=0}^{k}(−1)^{j} ( \begin{matrix}k \\ j \end{matrix} ) p_{m−j}(t) = \frac{1}{m!} ∏_{j=−k}^{m−k−1}(2t + j)

for all m and all k < m.

Proof : (by induction). For k = 1 the left hand side of (6.3) equals

p_{m}(t) − p_{m−1}(t) = \frac{1}{m!} ∏_{j=0}^{m−1}(2t + j) − \frac{1}{(m − 1)!} ∏_{j=0}^{m−2}(2t + j)

= \frac{1}{m!} ∏_{j=0}^{m−2}(2t + j)[2t + m − 1 − m]

= \frac{1}{m!} ∏_{j=−1}^{m−2}(2t + j)

Thus (6.3) is true for k = 1 and all m > 1. Next we assume that (6.3) holds for a fixed m = l − 1 and we want to show that (6.3) also holds for m = l. We achieve this by induction on k. If k = 1 the formula is true for all m, in particular for m = l. Let’s clearly state again our assumptions

i) (6.3) holds for m = l − 1 and all k, 1 ≤ k < l − 1 ii) (6.3) holds for m = l and some fixed k.

Since ( \begin{matrix}k + 1 \\ j \end{matrix} ) = ( \begin{matrix}k \\ j \end{matrix} ) + ( \begin{matrix}k \\ j − 1 \end{matrix} ) we get

∑_{j=0}^{k+1}(−1)^{j} ( \begin{matrix}k + 1 \\ j \end{matrix} ) p_{l−j}(t) =

= ∑_{j=0}^{k}(−1)^{j} ( \begin{matrix}k \\ j \end{matrix} ) p_{l−j}(t) − ∑_{j=0}^{k}(−1)^{j} ( \begin{matrix}k \\ j \end{matrix} ) p_{l−1−j}(t).

Using assumption i) in the second term of the right hand side and ii) in the first term, we find that the left hand side equals

\frac{1}{l!} ∏_{j=−k}^{l−k−1}(2t + j) − \frac{1}{(l − 1)!} ∏_{j=−k}^{l−k−2}(2t + j) = \frac{1}{l!} ∏_{j=−k−1}^{l−k−2}(2t + j)

what we wanted to prove. □

From now on let A be an R-algebra, without loss of generality with unit element e, of degree m ≥ 4 (see theorem 3.3), c = c^{2} a nontrivial idempotent in A (c ≠ 0, c ≠ e) and q_{1}(c) = −k. We may assume 1 ≤ k ≤ \frac{m}{2}, otherwise we replace c by the idempotent e − c, q_{1}(e − c) = −m + k = −(m − k). And moreover, we may assume that all eigenvalues of L(c) are in 𝔽 (or even 𝔽 = ℂ). We recall from lemma 6.1

q_{j}(c) = (−1)^{j} ( \begin{matrix}k \\ j \end{matrix} ) if 1 ≤ j ≤ k and q_{j}(c) = 0 if j > k.

Now let λ ∈ 𝔽 be an eigenvalue of L(c) with eigenvector y. Since the eigenvector c with eigenvalue 1 is already known we may assume λ ≠ 1, c and y linearly independent. We linearize the basic identity

g_{m}(x) + q_{1}(x)g_{m−1}(x) + ⋯ + q_{m}(x)x = 0

to obtain

(6.4) 0 = ∑_{j=0}^{m}(m+1−j)q_{j}(x)g_{m−j}(x, …, x, y)+∑_{j=0}^{m} jq_{j}(x, …, x, y)g_{m−j}(x).

With x = c, g_{k}(c) = c, q_{j}(c) = (−1)^{j} ( \begin{matrix}k \\ j \end{matrix} ) (if j ≤ k)

q_{j}(c) = 0 (if j > k) and (see lemma 6.3)

g_{k}(c, …, c, y) = \frac{1}{k + 1}p_{k}(L(c))y = \frac{1}{k + 1}p_{k}(λ)y

this reduces to

(6.5)  0 = (∑_{j=0}^{k}(−1)^{j} ( \begin{matrix}k \\ j \end{matrix} ) p_{m−j}(λ)) y + (∑_{j=1}^{m} jq_{j}(c, …, c, y)) c.

Since c, y are linearly independent, both coefficients are zero and λ is a zero of (see (6.3))

∑_{j=0}^{k}(−1)^{j} ( \begin{matrix}k \\ j \end{matrix} ) p_{m−j}(t) = \frac{1}{m!} ∏_{j=−k}^{m−k−1}(2t + j).

In particular 2λ ∈ ℤ. This rather strong result on the possible eigenvalues of L(c) can be sharpened:

Theorem 6.4. Let A be a unital R-algebra of degree m and c ≠ e an idempotent of A, q_{1}(c) = −k. Then L(c) is semisimple, the possible eigenvalues are

\frac{−m + k + 2}{2}, \frac{−m + k + 3}{2}, …, \frac{k}{2} and 0, 1

Proof : We recall two basic identities

g_{m−1}(x) + q_{1}(x)g_{m−2}(x) + ⋯ + q_{m−1}(x)x + q_{m}(x)e = 0
g_{m}(x) + q_{1}(x)g_{m−1}(x) + ⋯ + q_{m}(x)x = 0

(see theorem 3.2; the second is obtained from the first by differentiation in x^{2}-direction). We multiply the first identity by x and substract from the second to find

0 = ∑_{j=0}^{m−1} q_{j}(x) (g_{m−j}(x) − xg_{m−j−1}(x)) .

We linearize in y-direction:

0 = ∑_{j=0}^{m−1} jq_{j}(x, …, x, y)(g_{m−j}(x) − xg_{m−j−1}(x))

+ ∑_{j=0}^{m−1} q_{j}(x)((m − j + 1)g_{m−j}(x, …, x, y) − yg_{m−j−1}(x) −
−x(m − j)g_{m−j−1}(x, …, x, y))

With x = c, lemma 6.3, lemma 6.1 we obtain

0 = ∑_{j=0}^{k}(−1)^{j} ( \begin{matrix}k \\ j \end{matrix} ) (p_{m−j}(L(c)) − L(c) − L(c)p_{m−j−1}(L(c))) y

for all y ∈ A. Since ∑_{j=0}^{k}(−1)^{j} ( \begin{matrix}k \\ j \end{matrix} ) = 0 the term in the middle can be

neglected and we see that L(c) is annihilated by the polynomial

f_{k,m}(t) := ∑_{j=0}^{k}(−1)^{j} ( \begin{matrix}k \\ j \end{matrix} ) (p_{m−j}(t) − tp_{m−j−1}(t)) .

From (6.3) we obtain

f_{k,m} = \frac{1}{m!} ∏_{j=−k}^{m−k−1}(2t + j) − \frac{t}{(m − 1)!} ∏_{j=−k}^{m−k−2}(2t + j)

= \frac{1}{m!} ∏_{j=−k}^{m−k−2}(2t + j) [(2 − m)t + m − k − 1]

If m = 2, k = 1 then f_{k,m}(t) = 0. But in this case we already know that A is power-associative and the result holds. We may assume m ≥ 4. If (2 − m)λ + m − k − 1 ≠ 0 for every eigenvalue λ of L(c), then L(c) is

already annihilated by ∏_{j=−k}^{m−k−2}(2t + j) and the assertion follows.

We still have to discuss the case that the zero λ = \frac{m−k−1}{m−2} is among the (possible) eigenvalues of L(c). In this case 2λ ∈ ℤ. For 1 ≤ k < m we find that 2\frac{m−k−1}{m−2} is integral only in the following three cases

i) k = m − 1, (2 − m)t + m − k − 1 = (2 − m)t ii) k = 1, (2 − m)t + m − k − 1 = (2 − m)(t − 1) iii) m = 2k, (2 − m)t + m − k − 1 = (1 − k)(2t − 1) In the cases i) and ii) f_{k,m} has only simple roots and L(c) is semisimple. The only multiple root which is still possible is λ = \frac{1}{2} with multiplicity 2 (case iii). Then there would exist nonzero elements x, y in A such that

(2L(c) − Id)x = y and (2L(c) − Id)y = 0.

On the other hand, from (6.4), (6.3) and the computation following (6.4) we obtain

0 = ∑_{j=0}^{k}(−1)^{j} ( \begin{matrix}k \\ j \end{matrix} ) p_{m−j}(L(c))x + (∑_{j=1}^{m} jq_{j}(c, …, c, x)) c

= \frac{1}{m!} [ ∏_{j=−k}^{m−k−1}(2L(c) + j Id)] x + (∑_{j=1}^{m} jq_{j}(c, …, c, x)) c

= \frac{1}{m!} [∏_{j≠1}(2L(c) − j Id)] y + (∑_{j=1}^{m} jq_{j}(c, …, c, x)) c

= \frac{1}{m!} [∏_{j≠1}(1 − j)] y + (∑_{j=1}^{m} jq_{j}(c, …, c, x)) c

which is a contradiction, since c, y are linearly independent. Hence in each case, L(c) has only simple eigenvalues which are among the roots of f_{k,m}. This completes the proof. □

This result on the eigenvalues of L(c) is best possible. There are R-algebras having idempotents with the full range of possible eigenvalues.

Example. Let A be a commutative algebra on an n + 2 dimensional vector space with basis (b_{1}, …, b_{n}, c_{1}, c_{2}), the products being defined by

c_{i}^{2} = c_{i}, c_{1}c_{2} = 0, c_{1}b_{j} = μ_{j}b_{j}, c_{2}b_{j} = (1 − μ_{j})b_{j}

with μ_{1}, …, μ_{n} ∈ \frac{1}{2}ℤ, all other products b_{i}b_{j} = 0. In this algebra the initial value problem ẏ = y^{2}, y(0) = x is (with y = y_{1}b_{1} +⋯+y_{n}b_{n} +y_{n+1}c_{1} +y_{n+2}c_{2} and similar representation for x) equivalent to the system

\begin{aligned}ẏ_{i} & = 2(1 − μ_{i})y_{i}y_{n+1} + 2μ_{i}y_{i}y_{n+2} & , i = 1, …, n \\ ẏ_{i} & = y_{i}^{2} & , i = n + 1, n + 2 \end{aligned}

with initial value y_{i}(0) = x_{i} (i = 1, …, n + 2). This system can be solved successively, the solution is

y_{i}(t) = \frac{x_{i}}{1 − x_{i}t} , i = n + 1, n + 2
y_{i}(x) = x_{i}(1 − x_{n+1}t)^{−2(1−μ_{i})}(1 − x_{n+2}t)^{−2μ_{i}} , i = 1, …, n

Let m_{a} := max{2(1 − μ_{1}), …, 2(1 − μ_{n}), 1}
m_{b} := max{2μ_{1}, …, 2μ_{n}, 1}

then this solution has the form

y(x, t) = \frac{1}{q(x, t)}P (x, t)

with (exact) denominator

q(x, t) = (1 − x_{n+1}t)^{m_{a}}(1 − x_{n+2}t)^{m_{b}}

and a vectorial polynomial P (x, t). Thus A is a unital R-algebra (unit element is e = c_{1} + c_{2}). Expanding q(x, t) = 1 + q_{1}(x)t + ⋯ + q_{m}(x)t^{m} we obtain the degree m = m_{a} + m_{b} and q_{1}(x) = −m_{a}x_{n+1} − m_{b}x_{n+2} q_{1}(c) = −m_{b} = −k.

7 R-algebras of degree 4

Let A be a unital R-algebra of degree m with invariant chain q_{1}(x), …, q_{m}(x) and idempotent c. Since the left multiplication L(c) is semisimple, the algebra decomposes as direct sum of its eigenspaces:

A = ⨁_{μ∈Λ} A_{μ}

where Λ is the set of eigenvalues and A_{μ} = {x ∈ A | cx = μx}. Much information about the structure of A comes from multiplication rules for x_{μ}y_{λ} where x_{μ} ∈ A_{μ}, y_{λ} ∈ A_{λ}. The computations are rather elaborate and will only partly be carried out in the next section. In this section we discuss in some more detail the case m = 4. Since unital R-algebras of degree ≤ 3 are power-associative (see 3.3) this is the smallest degree where the structure cannot be understood from the theory of power-associative algebras. We start by giving an example of a non power-associative unital R-algebra of degree m = 4.

7.1 Example.

In ℝ^{3} (written as row-vectors) we define a product x·y = \frac{1}{2}[(x+y)^{2} −x^{2} −y^{2}] by

x^{2} = (x_{1}^{2} + αx_{2}^{2}, 2x_{1}x_{2}, −3αx_{2}^{2} + x_{3}^{2}).

Then e = (1, 0, 1) is a unit element and c = (0, 0, 1) an idempotent. An easy (but lengthy) computation shows that for any α ∈ ℝ

q_{1}(x) = −3x_{1} − x_{3}
q_{2}(x) = 3x_{1}^{2} + 3x_{1}x_{3}
q_{3}(x) = −x_{1}^{3} − 3x_{1}^{2}x_{3} − αx_{1}x_{2}^{2} + αx_{2}^{2}x_{3}
q_{4}(x) = x_{1}^{3}x_{3} + αx_{1}^{2}x_{2}^{2} − αx_{1}x_{2}^{2}x_{3} − α^{2}x_{2}^{2}

is an invariant chain in this algebra and

\frac{1}{3}(2x^{4} + x^{2}x^{2}) + q_{1}(x)x^{3} + q_{2}(x)x^{2} + q_{3}(x)x + q_{4}(x)e = 0.

This shows that A = (ℝ^{3}, ·) is an R-algebra of degree ≤ 4. If α ≠ 0 then a short computation gives c · z = 0, c · z^{2} ≠ 0 for z = (0, 1, 0), showing that (in this case) the eigenspace A_{0} is not a subalgebra. Hence A cannot be power-associative.

7.2 Basic identities.

If A is a unital R-algebra of degree m = 4 then there is an invariant chain (q_{1}, q_{2}, q_{3}, q_{4}) in A such that

(7.1)  \frac{1}{3}(2x^{4} + x^{2}x^{2}) + q_{1}(x)x^{3} + q_{2}(x)x^{2} + q_{3}(x)x + q_{4}(x)e = 0

Since the recursion formula (1.9) for invariant chains allows the computation of q_{j}(x) from q_{1}, we obtain (in our case)

(7.2)  \begin{aligned}2q_{2}(x) & = q_{1}(x^{2}) + q_{1}(x)^{2} \\ 3q_{3}(x) & = q_{1}(x^{3}) + \frac{3}{2}q_{1}(x)q_{1}(x^{2}) + \frac{1}{2}q_{1}(x)^{3} \\ 4q_{4}(x) & = \frac{1}{3}q_{1}(2x^{4} + x^{2}x^{2}) + \frac{4}{3}q_{1}(x)q_{1}(x^{3}) + \frac{1}{2}q_{1}(x^{2})^{2}+ \\ & +q_{1}(x)^{2}q_{1}(x^{2}) + \frac{1}{6}q_{1}(x)^{4} \end{aligned}

and the linearization of q_{4}(x) in direction x^{2} satisfies

(7.3)  q_{4}(x, x, x, x^{2}) = −q_{1}(x)q_{4}(x).

The linearization of (7.1) in direction x^{2} is

(7.4)  \begin{matrix}\frac{1}{6}(2x^{5} + 3x^{2}x^{3} + x(x^{2}x^{2})) + \frac{1}{3}q_{1}(x)(2x^{4} + x^{2}x^{2})+ \\ +q_{2}(x)x^{3} + q_{3}(x)x^{2} + q_{4}(x)x = 0. \end{matrix}

Multiplying (7.1) by x, subtracting from (7.4), and simplifying, we obtain

(7.5)  2x^{5} − 3x^{2}x^{3} + x(x^{2}x^{2}) + 2q_{1}(x)(x^{4} − x^{2}x^{2}) = 0.

We also know from the general theory that

g_{r}(x) + q_{1}(x)g_{r−1}(x) + ⋯ + q_{4}(x)g_{r−4}(x) = 0

for all r ≥ 5 (not only for r = 5 as stated in (7.4)). Let, as always in this text, y(x, t) denote the solution of ẏ = y^{2}, y(0) = x in A and j(x) = y(e − x, 1) + e the corresponding inversion. We also recall (1.14) and (4.11) for m = 4

(7.6)  q_{j}(y(x, t)) = \frac{1}{j!} \frac{q^{(j)}(t)}{q(t)}

with q(t) = 1 + q_{1}(x)t + q_{2}(x)t^{2} + q_{3}(x)t^{3} + q_{4}(x)t^{4}

(7.7)  j(x) = −\frac{1}{q_{4}(x)}[x^{3} + q_{1}(x)x^{2} + q_{2}(x)x + q_{3}(x)e]

Comparing (7.7) and (7.1) we obtain

(7.8)  q_{4}(x)(j(x)x − e) = \frac{1}{3}[x^{2}x^{2} − x^{4}]

showing again the known fact (but more directly) that j(x)x = e if and only if x^{2}x^{2} − x^{4} (provided j(x) exists, q_{4}(x) ≠ 0). Without giving a definite answer, we still study

h(x) := x^{2}x^{2} − x^{4}

which measures the deviation of A from being power-associative, and which causes all the trouble.

Formula (7.5) implies

Dh(x) · x^{2} = 2x^{5} − 3x^{2}x^{3} + x(x^{2}x^{2}) = −2q_{1}(x)h(x)

And then, if we consider h(x) along solution curves H(t) := h(y(x, t)), we find (with y = y(x, t))

\frac{d}{dt}H(t) = Dh(y)ẏ = Dh(y)y^{2} = −2q_{1}(y)h(y) = −2λ(t)H(t)

with λ(t) := q_{1}(y(x, t)). Recalling with (1.14)

q_{1}(y(x, t)) = \frac{q̇(t)}{q(t)}, q(t) = 1 + q_{1}(x)t + ⋯ + q_{4}(x)t^{4},

we see that H(t) satisfies the differential equation

Ḣ = −2\frac{q̇(t)}{q(t)}H, H(0) = h(x) = x^{2}x^{2} − x^{4}

with solution

H(t) = \frac{1}{q(t)^{2}} H(0).

Observing q(t) = −q_{4}(e − tx) (see theorem 3.1) we proved

(7.9)  x^{2}x^{2} − x^{4} = q_{4}(e − tx)[y(x, t)^{2}y(x, t)^{2} − y(x, t)^{4}]

showing in particular, that the set

N := {x ∈ A | x^{2}x^{2} − x^{4} = 0}

is invariant. If the solution y(x, t) starts in N it remains in N for all times (whenever defined). With t = 1 and replacing x by e − x we get from (7.9)

– observe h(x) = h(e − x) –

(7.10)  x^{2}x^{2} − x^{4} = q_{4}(x)[j(x)^{2}j(x)^{2} − j(x)^{4}]

From (7.8) and y(x, t) = \frac{1}{t}[j(e − tx) − e] we derive

(e − tx)y(x, t) = \frac{1}{t}[(e − tx)j(e − tx) − (e − tx)]
= x + \frac{t^{3}}{3q(t)}[x^{2}x^{2} − x^{4}],

thus

(7.11)  y(x, t) = [Id − tL(x)]^{−1}x + \frac{t^{3}}{3q(t)}[Id − tL(x)]^{−1}(x^{2}x^{2} − x^{4})

which reduces to

y(x, t) = [Id − tL(x)]^{−1}x,

if the initial value satisfies x^{2}x^{2} = x^{4}.

7.3 Peirce decomposition

If c = c^{2} is an idempotent of A, we know (see lemma 6.1) q_{1}(c) = −k where k is an integer with 1 ≤ k ≤ 4 and q_{1}(c) = −4 only if c = e. Furthermore,

q_{j}(c) = (−1)^{j} ( \begin{matrix}k \\ j \end{matrix} ), if 1 ≤ j ≤ k and q_{j}(c) = 0, if j > k (assuming c ≠ e,

k < 4). The idempotent c is called primary, if q_{1}(c) = −1, secondary, if q_{1}(c) = −2. Thus

c primary ⇒ q_{1}(c) = −1, q_{2}(c) = q_{3}(c) = q_{4}(c) = 0
c secondary ⇒ q_{1}(c) = −2, q_{2}(c) = 1, q_{3}(c) = q_{4}(c) = 0

If c is primary, the possible eigenvalues are

−\frac{1}{2}, 0, \frac{1}{2}, 1

and A is the direct sum of the corresponding eigenspaces (Peirce spaces).

A = A_{− \frac{1}{2}} ⊕ A_{0} ⊕ A_{\frac{1}{2}} ⊕ A_{1}

If c is secondary, the possible eigenvalues are 0, \frac{1}{2}, 1, then A = A_{0} ⊕ A_{\frac{1}{2}} ⊕ A_{1} and in case k = 3 we have the possible eigenvalues 0, \frac{1}{2}, 1, \frac{3}{2} and A = A_{0} ⊕ A_{\frac{1}{2}} ⊕ A_{1} ⊕ A_{\frac{3}{2}} . But if q_{1}(c) = 3 then c′ := e − c is primary, A_{\frac{3}{2}} (c) = A_{− \frac{1}{2}} (c′). Therefore it is (in degree 4 case) enough to study the cases k = 1, 2. Primary idempotents play in a certain sense the role of absolutely primitive idempotents, the Peirce-1-space is trivial:

Theorem 7.1. If A is a unital R-algebra and c a primary idempotent in A then A_{1} = {x ∈ A | cx = x} = 𝔽c.

Proof : Formula (6.5) combined with (6.3) gives for k = 1 and y ∈ A_{1} the relation

(∑_{j=1}^{m} jq_{j}(c, …, c, y)) c + (\frac{1}{m!} ∏_{j=−1}^{m−2}(2 + j)) y = 0

showing y ∈ 𝔽c. □

To obtain more information about the Peirce decomposition we linearize (7.5) twice, in direction u and in direction v. With x = c, cu = λu, cv = μv, q_{1}(c) = −k we then get

4c(c(c(uv))) + [4μ + 4λ − 2 − 4k]c(c(uv)) +
+ [4μ^{2} − 4μ + 8μλ − 4λ + 4λ^{2} − 6 − 2k(2μ + 2λ − 4)]c(uv) +
+ [4μ^{3} + 6μ^{2} + 2μ − 12μ^{2}λ − 12μλ − 12μλ^{2} + 2λ + 6λ^{2} + 4λ^{3} −
− 2k(2μ^{2} + μ − 8μλ + λ + 2λ^{2})]uv + 2q_{1}(u)(2μ^{3} − 3μ^{2} + μ)v +
+ 2q_{1}(v)(2λ^{3} − 3λ^{2} + λ)u = 0

We sum up the Peirce information which can be obtained by discussing this formula for k = 1 and k = 2.

Theorem 7.2. Let A be a unital R-algebra of degree m = 4, c an idempotent in A and A_{λ} = {x ∈ A | cx = λx}.

a) If c is primary, q_{1}(c) = −1, then

i) A = 𝔽c ⊕ A_{\frac{1}{2}} ⊕ A_{1} ⊕ A_{− \frac{1}{2}} , A_{1} = 𝔽c

ii) (A_{\frac{1}{2}} )^{2} ⊆ 𝔽c ⊕ A_{0} ⊕ A_{− \frac{1}{2}} A_{0}^{2} ⊆ A_{0} ⊕ A_{\frac{1}{2}} ⊕ 𝔽c A_{0}A_{\frac{1}{2}} ⊆ A_{− \frac{1}{2}} ⊕ A_{\frac{1}{2}} ⊕ 𝔽c

iii) y_{− \frac{1}{2}} z_{0} + \frac{1}{2}q(z_{0})y_{− \frac{1}{2}} ∈ A_{0}; (y_{− \frac{1}{2}} )^{2} = −q(y_{− \frac{1}{2}} )y_{− \frac{1}{2}} y_{− \frac{1}{2}} z_{\frac{1}{2}} + q(z_{\frac{1}{2}} )y_{− \frac{1}{2}} ∈ A_{0} ⊕ A_{\frac{1}{2}} ⊕ 𝔽c

b) If c is secondary, q_{1}(c) = −2, then

i) A = A_{1} ⊕ A_{\frac{1}{2}} ⊕ A_{0}

ii) A_{1}^{2} ⊆ A_{1}, A_{0}^{2} ⊆ A_{0} A_{1}A_{\frac{1}{2}} ⊆ A_{0} + A_{\frac{1}{2}} , A_{0}A_{\frac{1}{2}} ⊆ A_{1} + A_{\frac{1}{2}}

Proof : If m = 4, k = 1 the only possible eigenvalues are −\frac{1}{2}, 0, \frac{1}{2}, 1; L(c) is semisimple and A is the direct sum of eigenspaces. Moreover A_{1} = 𝔽c by theorem 7.1. If k = 1, λ = μ = 0 the above equation reduces to

2c(c(c(uv))) − 3c(c(uv)) + c(uv) = 0.

Decomposing uv = αc + (uv)_{0} + (uv)_{\frac{1}{2}} + (uv)_{− \frac{1}{2}} into a sum of Peirce components this equation reduces further to (uv)_{0} = 0. And, as yet, we don’t get more information about uv:

uv ∈ 𝔽c ⊕ A_{\frac{1}{2}} ⊕ A_{1}

The same procedure (putting k = 1 (or k = 2, if c is secondary), taking different combinations λ, μ from the possible eigenvalues and decomposing the product uv into its Peirce components) gives all other results. We omit these (lengthy, but elementary) computations. □

We derived the Peirce relations from formula (7.5), which was a consequence of (7.1). So, if in addition we start the same procedure with (7.1) – linearizing twice in u- and v-direction, putting x = c, u ∈ A_{λ}, v ∈ A_{μ} and discussing the different cases, we obtain as the only essentially new information

q(A_{\frac{1}{2}} ) = 0

for an arbitrary idempotent.

8 Peirce decomposition with a primary idempotent

8.1 The general case.

Let A be a unital R-algebra of degree m with invariant chain q_{1}, …, q_{m} and c = c^{2} a primary idempotent, q_{1}(c) = −1, q_{2}(c) = ⋯ = q_{m}(c) = 0. The possible simple eigenvalues of L(c) are

Λ = \{\frac{−m + 3}{2}, \frac{−m + 4}{2}, …, \frac{1}{2}, 0, 1\}

and A decomposes as direct sum

A = ⨁_{λ∈Λ} A_{λ}

of the Peirce spaces (eigenspaces) A_{λ} = {x ∈ A | cx = λx}. We already know (see theorem 7.1) A_{1} = 𝔽c. To obtain information about A_{λ}A_{μ} for λ, μ ∈ Λ (the socalled Peirce-2-relations) we start with the two basic identities

g_{m−1}(x) + q_{1}(x)g_{m−2}(x)+ ⋯ +q_{m−1}(x)x + q_{m}(x)e = 0
g_{m}(x) + q_{1}(x)g_{m−1}(x)+ ⋯ +q_{m−1}(x)x^{2} + q_{m}(x)x = 0

multiply the first equation by x and subtract from the second:

(8.1)  ∑_{j=0}^{m−1} q_{j}(x)[g_{m−j}(x) − xg_{m−j−1}(x)] = 0

We linearize twice, in direction y and z, and obtain

∑_{j=2}^{m−1}(j − 1)jq_{j}(x, …, x, y, z)[g_{m−j}(x) − xg_{m−j−1}(x)] +

+ ∑_{j=1}^{m−1} jq_{j}(x, …, x, y)[(m + 1 − j)g_{m−j}(x, …, x, z) −

−(m − j)xg_{m−j−1}(x, …, x, z) − zg_{m−j−1}(x)] +

+ ∑_{j=1}^{m−1} jq_{j}(x, …, x, z)[(m + 1 − j)g_{m−j}(x, …, x, y) −

−(m − j)xg_{m−j−1}(x, …, x, y) − yg_{m−j−1}(x)] +

+ ∑_{j=0}^{m−2} g_{j}(x)[(m − j)(m + 1 − j)g_{m−j}(x, …, x, y, z) −

−(m − j)yg_{m−j−1}(x, …, x, z) − (m − j)zg_{m−j−1}(x, …, x, y) − −(m − j − 1)(m − j)xg_{m−j−1}(x, …, x, y, z)] = 0.

Putting x = c and choosing y ∈ A_{μ}, z ∈ A_{λ} we get

(8.2)  \begin{aligned}& ∑_{j=1}^{m−1} jq_{j}(c, …, c, y)[p_{m−j}(λ) − λp_{m−j−1}(λ) − λ]z+ \\ + & ∑_{j=1}^{m−1} jq_{j}(c, …, c, z)[p_{m−j}(μ) − μp_{m−j−1}(μ) − μ]y− \\ − & ∑_{j=0}^{1}(−1)^{j}[g_{m−j−1}(λ) + g_{m−j−1}(μ)]yz+ \\ + & ∑_{j=0}^{1}(−1)^{j}[g_{m−j−1}^{(λ,μ)}(L(c)) − L(c)g_{m−j−2}^{(λ,μ)}(L(c))]yz = 0 \end{aligned}

where (as in section 6)

(j + 1)g_{j}(c, …, c, x) = p_{j}(L(c))x

with the polynomial

p_{j}(t) = \frac{1}{j!} ∏_{i=0}^{j−1}(2t + i)

and where g_{l}^{(μ,λ)}(t) denotes the polynomial of degree l defined by

g_{l−1}^{(μ,λ)}(L(c))yz := l(l + 1)g_{l}(c, …, c, y, z).

We decompose yz = ∑_{α∈Λ}(yz)_{α} (with (yz)_{α} ∈ A_{α}) and obtain from (8.2)

a linear combination of eigenvectors y, z, (yz)_{α} (α ∈ Λ). And now – in principle – after explicitely computing the coefficients we derive the Peirce-2-relations by comparing components from the same eigenspace.

(i) The coefficient ∑_{j=1}^{m−1} jq_{j}(c, …, c, y)[p_{m−j}(λ) − λp_{m−j−1}(λ) − λ]

Linearizing the invariant chain condition

(j + 1)q_{j}(x) = jq_{j}(x, …, x, x^{2}) + q_{1}(x)q_{j}(x)

in direction y ∈ A_{μ} yields with x = c the recursion

(j + 1)q_{j+1}(c, …, c, y) = \frac{j − 1 + 2μ − k}{j + 1} jq_{j}(c, …, c, y) + \frac{1}{j + 1}q_{j}(c)q_{1}(y)

leading to

jq_{j}(c, …, c, y) = α_{j}(μ)q_{1}(y)

with some α_{j}(μ). The coefficients of z and y in (8.2) then are of the form

∑_{j=1}^{m−1} α_{j}(μ)[g_{m−j}(λ) − λg_{m−j−1}(λ) − λ]q_{1}(y).

The further discussion of (8.2) in its full generality (even for primary idempotents) is an arduous task and is not recommended to do ”by hand”.

9 Homotopes and the Jordan kernel

9.1 Background

Let V be an 𝔽-vector space and U ⊆ V be an open nonempty subset. On the vector space A(U) of all analytic vector fields on U (with respect to a basis of V every component of f ∈ A(U) is analytic) the Lie bracket [ , ] is defined by

[f, g](x) = Dg(x)f(x) − Df(x)g(x)

(the derivative of g(x) in direction f(x) minus the derivative of f(x) in direction g(x)). Then (A(U), [ , ]) is a Lie algebra; the Lie algebra of analytic vector fields on U. We often write

[f, g](x) = [f(x), g(x)]

Let h : U → U be an analytic map such that the derivative Dh(x) is invertible for all x ∈ U.

Theorem 9.1. If f_{i}, g_{i} ∈ A(U) satisfy

(9.1)  Dh(x)g_{i}(x) = f_{i}(h(x)) (i = 1, 2),

then

(9.2)  Dh(x)[g_{1}, g_{2}](x) = [f_{1}, f_{2}](h(x)).

Proof : We start with the left hand side of (9.2), use the standard rules for derivatives, in particular the symmetry of the second derivative

D^{2}h(x)(u, v) := D(Dh(x)u)v = D(Dh(x)v)u,

and obtain

Dh(x)[g_{1}, g_{2}](x) =
= Dh(x)Dg_{2}(x)g_{1}(x) − Dh(x)Dg_{1}(x)g_{2}(x)  (by definition of [ , ])
+ D^{2}h(x)(g_{2}(x), g_{1}(x)) − D^{2}h(x)(g_{1}(x), g_{2}(x))  (add D^{2} − D^{2})
= D(Dh(x)g_{2}(x))g_{1}(x) − D(Dh(x)g_{1}(x))g_{2}(x) (chain rule)
= D[f_{2}(h(x))]g_{1}(x) − D[f_{1}(h(x))]g_{2}(x)  (by assumption)
= Df_{2}(h(x))Dh(x)g_{1}(x) − Df_{1}(h(x))Dh(x)g_{2}(x) (chain rule)
= Df_{2}(h(x))f_{1}(h(x)) − Df_{1}(h(x))f_{2}(h(x))  (again (9.1))
= [f_{1}, f_{2}](h(x)).
□

For f ∈ A(U) and h as above, and in addition Dh(x) invertible, we define

(9.3)  f^{h}(x) := [Dh(x)]^{−1}f(h(x)).

Then, with g_{i}(x) = f_{i}^{h}(x), (9.2) is equivalent to

(9.4)  [f_{1}^{h}, f_{2}^{h}](x) = [f_{1}, f_{2}]^{h}(x),

showing the important fact, that f ↦ f^{h} is an automorphism of the Lie algebra of analytic vector fields on U.

Example 1. If p(x) is a homogeneous polynomial of degree k then Dp(x)·x = kp(x) and

[Id, p](x) = Dp(x)x − (Dx)p(x)
= kp(x) − p(x) = (k − 1)p(x)
Id^{h}(x) = Dh(x)^{−1}h(x), p^{h}(x) = Dh(x)^{−1}p(h(x))
□

Example 2. With f_{i}(x) = a_{i} ∈ V (constant) we get

[Dh(x)^{−1}a_{1}, Dh(x)^{−1}a_{2}] = Dh(x)^{−1}[a_{1}, a_{2}] = 0
□

For any f ∈ A(U) the centralizer

C(f) := {g ∈ A(U) | [g, f] = 0}

is of particular interest. To understand its importance we make a short detour to the theory of differential equations. For any f, g ∈ A(U) ^{4} we denote by F (x, t) resp. G(x, t) the (unique, maximal) solution of the initial value problem in U

ẏ = f(y), y(0) = x, resp. ẏ = g(y), y(0) = x

A map h : U → U is called solution preserving from ẏ = f(y) into ẏ = g(y), if h maps F (x, t) onto G(h(x), t), that is, if

(9.5)  h(F (x, t)) = G(h(x), t)

4Since we only want to enlighten the background we restrict to this special case. For the more general situation f ∈ A(U_{f}), g ∈ A(U_{g}) we refer to the article [21].

whenever defined. There is a criterion which allows the discussion of solution preserving maps without knowing the general solutions.

Criterion. h : U → U is solution preserving from ẏ = f(y) onto ẏ = g(y) if and only if

(9.6)  Dh(x)f(x) = g(h(x)).

Proof : If h(F (x, t)) = G(h(x), t), then the derivative with respect to t is

Dh(F (x, t))f(F (x, t)) = g(G(h(x), t)).

At time t = 0 this reduces to (9.6). Conversely, if (9.6) holds for all x ∈ U, then

\frac{d}{dt}h(F (x, t)) = Dh(F (x, t))f(F (x, t))
= g(h(F (x, t)))

showing that h(F (x, t)) solves ẏ = g(y), y(0) = h(x). Thus h(F (x, t)) = G(h(x), t) by uniqueness. □

NOTE: Any analytic h : U → U with invertible derivative Dh(x) is (by definition and (9.6)) solution preserving from ẏ = f^{h}(y) onto ẏ = f(y).

Any analytic invertible solution preserving map h : U → U from ẏ = f(y) onto itself is called symmetry of this differential equation. Thus an invertible h is a symmetry of ẏ = f(y) if and only if Dh(x)f(x) = f(h(x)); equivalently

(9.7)  h is a symmetry of ẏ = f(y) ⇔ f^{h} = f.

Given f, g ∈ A(U). We consider (for s near 0) the maps G_{s}(x) = G(x, s). Due to the basic flow relation G(G(x, s), t) = G(x, s+t) the maps G_{s} : x → G_{s}(x) and the derivative DG_{s}(x) (where D denotes the derivative with respect to x) are invertible.

We have now all the ingredients for the proof of a basic theorem from the theory of differential equations, due to Sophus Lie:

Theorem 9.2. All G_{s} are symmetries of ẏ = f(y) if and only if [g, f] = 0 in the Lie algebra A(U).

Proof : The crucial step in the proof is the following differentiation rule

(9.8)  \frac{∂}{∂s}f^{G_{s}}(x) = [g, f]^{G_{s}(x)}(x).

This formula will be proved as follows: By definition we have \frac{∂}{∂s}G_{s}(x) = g(G_{s}(x)), and then (where again D denotes the derivative with respect to x)

\frac{∂}{∂s}DG_{s}(x) = D\frac{∂}{∂s}G_{s}(x) = D[g(G_{s}(x)] = Dg(G_{s}(x))DG_{s}(x)

(by the chain rule). Employing these two equations and standard differentiation rules, including the derivative of the inverse

\frac{∂}{∂s}X(s)^{−1} = −X(s)^{−1}[\frac{∂}{∂s}X(s)]X(s)^{−1},

we obtain:

\frac{∂}{∂s}f^{G_{s}}(x) = \frac{∂}{∂s}[DG_{s}(x)^{−1}f(G_{s}(x))]
= −DG_{s}(x)^{−1}[\frac{∂}{∂s}DG_{s}(x)]DG_{s}(x)^{−1}f(G_{s}(x))

+DG_{s}(x)^{−1}Df(G_{s}(x))\frac{∂}{∂s}G_{s}(x)

= DG_{s}(x)^{−1}[Df(G_{s}(x))g(G_{s}(x)) − Dg(G_{s}(x))f(G_{s}(x))
= [g, f]^{G_{s}}(x).

Now the rest is easy: If [g, f] = 0, then \frac{∂}{∂s}f^{G_{s}}(x) = 0 (see (9.8)) thus f^{G_{s}}(x) = f^{G_{0}}(x) = f(x), and G_{s} is a symmetry (see (9.7)). Conversely, if G_{s} is a symmetry, then f^{G_{s}}(x) = f(x) and

[g, f]^{G_{s}}(x) = \frac{∂}{∂s}f^{G_{s}}(x) = \frac{∂}{∂s}f(x) = 0

for all s. Taking s = 0 we obtain [g, f] = 0. □

9.2 The centralizer of p(x) = x^{2}

Since we are dealing mainly with the differential equation ẏ = y^{2} we are mostly interested in the centralizer

C(p) = {g ∈ A(U) | [p, g] = 0}

of the quadratic polynomial p(x) = x^{2} (the square in a commutative algebra).

Theorem 9.3. If A is a unital commutative algebra, then

C(p) = DerA ⊕ C_{2}

where DerA is the Lie algebra of derivations of A and C_{2} = {q_{a} | a ∈ J (A)} with q_{a}(x) = 2x(xa) − x^{2}a and

J (A) = {a ∈ A | 2x(x(xa)) − 2x(x^{2}a) − x^{2}(xa) + x^{3}a = 0 for all x ∈ A}

Proof : If g(x) = ∑ g_{i}(x) is the Taylor series expansion about 0, g_{i}(x) homogeneous of degree i, then obviously g ∈ C(p) if and only if all g_{i} ∈ C(p). Degree 0: If g_{0} = c ∈ C(p) then 0 = [x^{2}, c] = Dc · x^{2} − Dx^{2} · c = −2xc for all x ∈ A. Taking x = e (the unit element) we get c = 0. Degree 1: The homogeneous linear term Bx is in C(p) iff 0 = [x^{2}, Bx] = B(x^{2}) − 2xBx, thus iff B is a derivation. Degree 2: This is the most interesting part. The homogeneous quadratic polynomial q(x) = q(x, x) is in C(p) iff 0 = [x^{2}, q(x, x)] = 2q(x, x^{2})−2xq(x, x). We linearize and obtain

q(y, x^{2}) + 2q(x, xy) = yq(x, x) + 2xq(x, y).

The special choice x = e, resp. y = e, leads immediately to

q(y, e) = yq(e) = ya (with a := q(e)) and
q(x) = 2xq(x, e) − q(e, x^{2}) = 2x(xa) − x^{2}a.

Thus q ∈ C(p) has the form

(9.9)  q(x) = q_{a}(x) = 2x(xa) − x^{2}a, a := q(e)

and such a polynomial is in C(p) iff

0 = [x^{2}, 2x(xa) − x^{2}a] = 2x^{2}(xa) + 2x(x^{2}a) − 2x^{3}a −
−2x(2x(xa) − x^{2}a)

Degree k ≥ 3: 0 = [x^{2}, q(x, …, x)] = kq(x, …, x, x^{2}) − 2xq(x), thus

(9.10)  2xq(x) = kq(x, …, x, x^{2}).

With x = e, k ≠ 2 we find q(e) = 0.

We linearize (9.10): (9.11) 2yq(x) + 2kxq(x, …, x, y) = k(k − 1)q(y, x, …, x, x^{2}) + 2kq(x, …, x, xy)

With x = e, q(e) = 0 we get q(e, …, e, x) = 0 for all x ∈ A. The same procedure, linearizing (9.11) again and again and putting x = e gives q(e, …, e, x, x) = 0, …, q(e, x, …, x) = 0. Finally (9.11) with y = e then shows q(x) = 0. □

9.3 The Jordan kernel J (A)

Let A be an arbitrary commutative unital algebra with product xy = L(x)y. From theorem 9.3 we know that the centralizer of p(x) = x^{2} has only a linear part DerA and a quadratic part C_{2}, where q ∈ C_{2} is homogeneous of degree 2, uniquely determined by a := q(e),

q(x) = q_{a}(x) = 2x(xa) − x^{2}a.

Moreover, we saw in the proof of theorem 9.3 that q_{a} ∈ C(p) if and only if

x^{2}(xa) + 2x(x^{2}a) − 2x(x(xa)) − x^{3}a = 0 for all x ∈ A.

It appears that the subspace (sometimes called the Jordan kernel of A)

J (A) := {q(e) | q ∈ C_{2}} = {a ∈ A | x^{2}(xa) + 2x(x^{2}a) − 2x(x(xa)) − x^{3}a = 0 for all x ∈ A}

plays an important role for the structure of A.

The following observation will be rather useful

(9.12)  q_{a}(x) = [ax, x^{2}]

Next, we want to show that J (A) is a subalgebra of A. Let a, b ∈ J (A), then (by definition) [x^{2}, q_{a}(x)] = [x^{2}, q_{b}(x)] = 0 and using the Jacobi identity we find

[[q_{a}(x), q_{b}(x)], x^{2}] = [[q_{a}(x), x^{2}], q_{b}(x)] + [q_{a}(x), [q_{b}(x), x^{2}]] = 0

Thus [q_{a}, q_{b}] is a centralizer element of degree 3, so it is zero (by theorem 9.3). Then, again using Jacobi’s identity we get

0 = [[bx, q_{a}(x)], x^{2}] + [[q_{a}(x), x^{2}], bx] + [[x^{2}, bx], q_{a}(x)]
= [[bx, q_{a}(x)], x^{2}] (observe [x^{2}, bx] = −q_{b}(x)).

showing that q(x) := [bx, q_{a}(x)] is a centralizer element. Since it is homogeneous of degree 2 it has the form q(x) = q_{c}(x) with c = [bx, q_{a}(x)] |_{x=e}= 2q_{a}(x, bx) − bq_{a}(x) |_{x=e}= ab.

We proved: If a, b ∈ J (A) then

(9.13)  [bx, q_{a}(x)] = q_{ab}(x) = [ax, q_{b}(x)].

(Since A is commutative we have symmetry in a, b). (9.13) shows in particular, that J (A) is a subalgebra.

Theorem 9.4. Let A be a commutative algebra and

J (A) = {a ∈ A | x^{2}(xa) + 2x(x^{2}a) − 2x(x(xa)) − x^{3}a = 0 for all x ∈ A}

then

a) a^{2}(ax) = a(a^{2}x) for all x ∈ A, a ∈ J (A),

b) J (A) is a Jordan subalgebra of A,

c) All L(a)L(b) − L(b)L(a), a, b ∈ J (A), are derivations of A .

Proof : Adjoining a unit element e, Â = 𝔽e ⊕ A, we get J (Â) = 𝔽e ⊕ J (A). Therefore we may assume that A is unital. a): A special case of (9.13) is q_{a^{2}}(x) = [ax, q_{a}(x)], explicitly

2x(xa^{2}) − x^{2}a^{2} = 2(ax)(ax) + 2x((ax)a) − 2(x(ax))a − 2a(x(xa)) + a(x^{2}a)

Linearizing in direction a yields 6a(a^{2}x) = 6a^{2}(ax) for all x. b): J (A) is a subspace (clear from the definition), closed under multiplication (see (9.13)) and all a, b ∈ J (A) satisfy the Jordan identity a^{2}(ab) = a(a^{2}b) (part a) of this theorem). c):

[a(bx) − b(ax), x^{2}] = [[bx, ax], x^{2}] (def. of [ , ])
= [[bx, x^{2}], ax] + [bx, [ax, x^{2}]]  (Jacobi identity)
= [q_{b}(x), ax] + [bx, q_{a}(x)]  ( → (9.12))
= 0  ( → (9.13)).

Thus [L(a), L(b)] is a linear map in the centralizer of p(x) = x^{2} and then a derivation by theorem 9.3. □

Remark: Not only L(a), L(a^{2}) commute. One can show ([13])

a^{k}(a^{m}x) = a^{m}(a^{k}x) for all a ∈ J (A), k, m nonnegative integers and all x ∈ A.

Example. If A is power-associative, then a ∈ J (A) if and only if x^{2}(xa) = x(x^{2}a) for all x ∈ A.

Proof : The linearized version of x^{2}x^{2} = x^{4} is

4x^{2}(xy) = yx^{3} + x(x^{2}y) + 2x(x(xy)).

On the other hand, a ∈ J (A) iff

x^{2}(xa) = x^{3}a − 2x(x^{2}a) + 2x(x(xa)).

And now compare both identities. □

9.4 Homotopes

Let’s return to R-algebras. Let A be a unital R-algebra, y(x, t) = ∑_{k≥0} g_{k}(x)t^{k}

the solution of our standard initial value problem ẏ = y^{2}, y(0) = x in A and the inversion j defined by

j(x) = y(e − x, 1) + e

So far we know (see section 4)

j(tx) = \frac{1}{t}j(x), j(j(x)) = x, j(e) = e

Dj(e) = −Id, Dj(x) · x^{2} = −e, x^{2} = −(Dj(x))^{−1} · e j(e + x) + j(e + j(x)) = e (Hua at e)

Moreover (”geometric” series)

j(e − x) = e + x + x^{2} + x^{3} + g_{3}(x) + g_{4}(x) + ⋯
j(x) = −\frac{1}{q_{m}(x)}[g_{m−2}(x) + q_{1}(x)g_{m−3}(x) + ⋯ + q_{m−1}(x)e]

Using Dj(x) · x^{2} = −e our basic differential equation in A may be written as

ẏ = −[Dj(y)]^{−1}e

(whenever j(y) is defined). This motivates the study of the initial value problem

(9.14)  ẏ = −[Dj(y)]^{−1}a, y(0) = x

for arbitrary a ∈ A. The right hand side is not always polynomial, but we can easily determine the solution:

(9.14) ⇔ Dj(y)ẏ = −a ⇔ \frac{d}{dt}j(y(t)) = −a
⇔ j(y(t)) = −at + j(x)
⇔ y(t) = j(j(x) − at)

Since j is rational, this solution is rational, too!

Let’s pause to look at the situation when A is Jordan. In this case j(y) = y^{−1} and −[Dj(y)]^{−1}a = 2y(ya) − y^{2}a for all a ∈ A. The right hand side defines a new squaring ⊥_{a} (resp. a new commutative product) on A

y ⊥_{a} y := 2y(ya) − y^{2}a

and it turns out that A together with this new product is again a Jordan algebra, the socalled a-homotope A_{a}. Since the homotopes play an important role in the structure theory of Jordan algebras we are naturally led to the question for which a (in an R-algebra) is −[Dj(x)]^{−1}a a homogeneous quadratic polynomial. A complete answer will be given in the next section.

But in any case, for arbitrary a ∈ A (A a unital R-algebra) it still makes sense to study the initial value problem ẏ = −[Dj(y)]^{−1}a, y(0) = x with solution

(9.15)  y_{a}(x, t) = j(j(x) − at).

Generalizing the notion of quasi-invertibility (→ 4.5) we say that (x, y) ∈ A × A is quasi-invertibel with quasi-inverse

(9.16)  q(x, y) := j(j(x) − y),

if the right hand side (the solution at t = 1) exists. If y = e then

q(x, e) = j(j(x) − e) = j(e − x) − e (Hua’s identity)
= y(x, 1),

which is the quasi-inverse of x in A (see 4.5).

We still mention the following addition formula for the quasi-inverse:

(9.17)  q(q(x, y), z) = q(x, y + z)

(whenever defined).

Proof :

q(q(x, y), z) = j(−z + j(q(x, y)) =
= j(−z + j(j(j(x) − y))) = j(−z + j(x) − y)
= j(j(x) − (y + z)) = q(x, y + z).
□

9.5 J (A) in R-algebras.

The key for a better understanding of the Jordan kernel J (A) (see 9.3) in a unital R-algebra can be found in the structure algebra of our inversion j. The structure group of j is defined as the set

G(j) := {T ∈ Gl(A) | There is a T^{#} ∈ Gl(A) such that
j(T x) = T^{# −1}j(x)}

Obviously G(j) is a group and for T, T_{1}, T_{2} ∈ G(j) we have T^{##} = T, (T_{1}T_{2})^{#} = T_{2}^{#}T_{1}^{#}.

By a standard construction one associates to G(j) the Lie algebra g(j) ⊆ EndA

B ∈ g(j) ⇔ e^{tB} ∈ G(j) for all t ∈ ℝ

and obtains

g(j) = {B ∈ EndA | There is a B^{#} ∈ EndA such that

Dj(x)Bx = −B^{#}j(x) (whenever defined) }

Theorem 9.5. Let A be a unital R-algebra with inverse j(x) = y(e−x, 1)+e. For a ∈ A the following are equivalent:

a) a ∈ J (A);

b) The left multiplication L(a) is in the structure algebra g(j); c) [Dj(x)]^{−1}a = −q_{a}(x) = x^{2}a − 2x(xa);

d) x ↦ [Dj(x)]^{−1}a is a polynomial map;

e) a ∈ g(j)e = {Be | B ∈ g(j)}.

Proof : For a ∈ J (A) we have (by definition, see 9.3) [x^{2}, q_{a}(x)] = 0, thus every y(x, s) – in particular y(x, 1) – is a symmetry of the differential equation ẏ = q_{a}(y) (see theorem 9.2) and therefore (see the criterion in 9.1)

Dy(x, 1) · q_{a}(x) = q_{a}(y(x, 1)).

Replacing x by e − x we obtain

Dy(e − x, 1)q_{a}(e − x) = q_{a}(y(e − x, 1)),

and then, using the definition of q_{a} and j

−Dj(x) · q_{a}(e − x) = q_{a}(j(x) − e).

Evaluation yields (with q_{a}(e − x) = a − 2ax + q_{a}(x))

−Dj(x)a + 2Dj(x)(xa) − Dj(x)q_{a}(x) = q_{a}(j(x)) − 2aj(x) + a.

We compare the homogeneous terms of degree 0, −1 on both sides (remember that Dj(x) is homogeneous of degree −2, see 4.3) and obtain

−Dj(x) · q_{a}(x) = a,
−Dj(x)(ax) = aj(x),

equivalently

q_{a}(x) = −[Dj(x)]^{−1}a and L(a) ∈ g(j) (L(a)^{#} = L(a)).

So far we proved a) ⇒ b) and a) ⇒ c). The implications c) ⇒ d) and b) ⇒ e) are obvious. In the next step we prove e) ⇒ d): Let B be an element from the structure algebra g(j) and a := Be, then

(9.18)  Dj(x)Bx = −B^{#}j(x) for some B^{#}.

With x = e, Dj(e) = −Id and j(e) = e we get B^{#}e = a. We differentiate(9.18) in x^{2}-direction and also Dj(x)·x^{2} = −e in Bx-direction to obtain

D^{2}j(x)(x^{2}, Bx) + Dj(x)Bx^{2} = −B^{#}Dj(x) · x^{2} = B^{#}e = a and
D^{2}j(x)(x^{2}, Bx) + 2Dj(x)(xBx) = 0.

Combining these two identities we find

Dj(x)[2x(Bx) − Bx^{2}] = −a,

showing that −(Dj(x))^{−1}a = 2x(Bx) − Bx^{2} is polynomial. Next, as a final step, we prove d) ⇒ a): Let r_{a}(x) := −Dj(x)^{−1}a be polynomial. Since Dj(x) is homogeneous of degree −2 (see 4.3), r_{a}(x) is homogeneous of degree 2. To show that r_{a} is in the centralizer of x^{2}, [x^{2}, r_{a}(x)] = 0, we apply the automorphism f ↦ f^{j}, f^{j}(x) = Dj(x)^{−1}f(j(x)) (see 9.1) to the trivial relation [e, a] = 0 and obtain

0 = Dj(x)^{−1}[e, a] = [Dj(x)^{−1}e, Dj(x)^{−1}a]
= [x^{2}, r_{a}(x)].

According to theorem 9.3 the homogeneous degree 2 polynomial r_{a}(x) in the centralizer of p(x) = x^{2} has the form r_{a}(x) = q_{b}(x) with b = q_{b}(e) = r_{a}(e) = −[Dj(e)]^{−1}a = a, thus r_{a}(x) = q_{a}(x) and a ∈ J (A). This completes the proof. □

10 Nondegenerate R-algebras

10.1 Basics.

It is known that in various types of algebras a ”nice” bilinear form is a useful tool to explore further properties of the algebra. Let A be an arbitrary finite dimensional 𝔽-algebra. A bilinear form φ : A × A → 𝔽 is called symmetric, if φ(x, y) = φ(y, x) for all x, y ∈ A and associative, if φ(xy, z) = φ(x, yz) for all x, y, z ∈ A. Moreover, φ is called nondegenerate, if φ(x, y) = 0 for all x ∈ A implies y = 0. We recall the following basic facts from linear algebra: Assume (for simplicity) that φ is symmetric.

Lemma 10.1.

a) If φ is nondegenerate and λ : A → 𝔽 a linear form on A then there is a unique vector l ∈ A such that λ(x) = φ(l, x) for all x ∈ A. b)If φ is nondegenerate and f : A → A linear, then there is a unique linear map f^{∗} : A → A (the adjoint) such that φ(f^{∗}x, y) = φ(x, fy) for all x, y ∈ A. f → f^{∗} satisfies f^{∗∗} = f, (fg)^{∗} = g^{∗}f^{∗}.

If φ is symmetric, associative and nondegenerate and if L(x), R(x) denote left- and right multiplication in A, L(x)y = xy = R(y)x, then the associativity of φ is equivalent to L(x)^{∗} = R(x) for x ∈ A.

A linear form λ : A → 𝔽 is called a trace form on A, if the associated bilinear form φ(x, y) := λ(xy) is symmetric, associative and nondegenerate. We demonstrate the influence of a nice bilinear form or a trace form on A with the following (well known) results:

Theorem 10.2.

a) A (finite dimensional) algebra with a trace form is unital (has a unit element).

b) If A has a symmetric, associative and nondegenerate bilinear form and if A has no ideal B ≠ 0 with B^{2} = 0, then A is semisimple (A is a direct sum of simple ideals; theorem of Dieudonné).

c) A commutative (finite dimensional) power-associative algebra with a nondegenerate symmetric and associative bilinear form is a Jordan algebra (satisfies xy = yx and x^{2}(xy) = x(x^{2}y) for all x, y ∈ A).

Proof : a) If λ is a trace form on A then the associated bilinear form (x, y) ↦ λ(xy) is symmetric, associative and nondegenerate. By lemma 10.1 a) there is a unique vector e ∈ A such that λ(x) = λ(ex) for all x. Then λ((ex)y) = λ(e(xy)) = λ(xy) for all y ∈ A implies ex = x for all x. Thus e is unit element.

b) If φ is associative and nondegenerate and U ⊆ A an ideal, then

U^{⊥} = {z ∈ A | φ(U, z) = 0}

is also an ideal of dimension dim A − dim U. The ideal L := U ∩ U^{⊥} satisfies φ(L^{2}, A) = φ(L, LA) = 0 since L ⊆ U and LA ⊆ U^{⊥}. Thus L^{2} = 0 and then L = 0, by assumption. This shows A = U ⊕ U^{⊥} as direct sum of ideals.

c) If A is power-associative, then x^{2}x^{2} = x^{4} holds for all x ∈ A. We linearize to obtain

4x^{2}(xy) = 2x(x(xy)) + x^{3}y

which is equivalent to

4L(x^{2})L(x) = 2L(x)^{3} + L(x^{3})

Taking the adjoint on both sides gives

4L(x)L(x^{2}) = 2L(x)^{3} + L(x^{3})

since L(x)^{∗} = L(x) for all x (by assumption). Thus L(x^{2})L(x) = L(x)L(x^{2}), which is equivalent to x^{2}(xy) = x(x^{2}y). □

These results motivate the study of R-algebras with a nice bilinear form or trace form. A natural candidate is λ(x) = q_{1}(x), the linear form from the invariant chain.

10.2 Nondegenerate R-algebras

. Let A be an R-algebra with invariant chain (q_{1}, q_{2}, …, q_{m}), q = 1+q_{1} +⋯+ q_{m}. We call q_{1} associative, if the associated bilinear form (x, y) ↦ q_{1}(xy) is associative. In 5.3 we introduced the P -radical of A and showed that P -radA is contained in the subalgebra

U = {u ∈ A | q(x + u) = q(x) for all x ∈ A}
= {u ∈ A | q_{j}(x, …, x, u) = 0 for all x ∈ A and 1 ≤ j ≤ m}.

Lemma 9.3. Let A be a unital R-algebra. If q_{1} is associative, then

P -radA = U = {y ∈ A | q_{1}(xy) = 0 for all x ∈ A}

Proof : Let y ∈ U. Then (by definition) q_{1}(y) = q_{2}(x, y) = 0 for all x ∈ A. Using the linearization of 2q_{2}(x) = q_{1}(x^{2}) + q_{1}(x)^{2} we get 0 = 2q_{2}(x, y) = q_{1}(xy) + q_{1}(x)q_{1}(y) = q_{1}(xy) for all x ∈ A. Thus

U ⊆ Ũ := {y | q_{1}(xy) = 0 for all x}.

If, conversely, y ∈ Ũ then q_{1}(y) = 0 (take x = e) and q_{2}(x, y) = \frac{1}{2}[q_{1}(xy) + q_{1}(x)q_{1}(y)] = 0. We proceeed by induction on j to show that q_{j}(x, …, x, y) = 0. Using the linearization of (1.9) ,

(j + 1)^{2}q_{j+1}(x, …, x, y) = j(j − 1)q_{j}(x, …, x, x^{2}, y) + 2jq_{j}(x, …, x, xy) + q_{1}(y)q_{j}(x) + jq_{1}(x)q_{j}(x, …, x, y)

we see that every term on the right hand side vanishes (observe xy ∈ Ũ). Thus Ũ ⊆ U consequently U = Ũ. In particular, U is a P -ideal and therefore P -radA = U. □

Definition. Let A be an R-algebra with invariant chain (q_{1}, …, q_{m}). Then we call A nondegenerate, if q_{1} is a trace form on A (symmetric, associative and nondegenerate). □

We know from lemma 10.1 that a nondegenerate R-algebra has a unit element e. As a matter of notation, we denote the associated bilinear form by φ, φ(x, y) = q_{1}(xy). Then φ is symmetric, associative and nondegenerate. Whenever we take the adjoint f^{∗} of a linear map f it is with respect to φ.

Theorem 10.4. Every nondegenerate R-algebra is semisimple (direct sum of simple ideals).

Proof : We only have to show that B^{2} = 0 for an ideal B of A implies B = 0. The assertion then follows from Dieudonné’s theorem (theorem 10.2 b). The recursion

(j + 1)q_{j+1}(x) = jq_{j}(x, …, x, x^{2}) + q_{1}(x)q_{j}(x)
and 0 = mq_{m}(x, …, x, x^{2}) + q_{1}(x)q_{m}(x)

shows for x = b ∈ B, B^{2} = 0,

(j + 1)q_{j+1}(b) = q_{1}(b)q_{j}(b), q_{1}(b)q_{m}(b) = 0,

which implies (by induction) q_{1}(b)^{m+1} = 0 thus q_{1}(b) = 0 for all b ∈ B. Since B is an ideal we then get q_{1}(xb) = 0 for all x ∈ A and consequently b = 0, since q_{1} is nondegenerate. □

10.3 The inversion ψ

Let A be a nondegenerate R-algebra, φ(x, y) = q_{1}(xy). We first want to establish a different description of φ starting with q_{m}.

For every x in the open set {x ∈ A | q_{m}(x) ≠ 0} we have the linear form

(10.1)  λ_{x}(y) := −\frac{Dq_{m}(x)y}{q_{m}(x)}

(logarithmic derivative of q_{m}) and the symmetric bilinear form

(10.2) \begin{aligned}φ_{x}(y, z) & := −(Dλ_{x}(y))z \\ & = \frac{1}{q_{m}(x)^{2}} [q_{m}(x)D^{2}q_{m}(x)(y, z) − Dq_{m}(x)y · Dq_{m}(x)z] \end{aligned}

This construction leads back to φ:

Lemma 10.5. φ = φ_{e}

Proof : Differentiation of

(−1)^{m}q_{m}(e − x) = q(x) = 1 + q_{1}(x) + ⋯ + q_{m}(x)

(see theorem 3.1) yields

(−1)^{m+1}Dq_{m}(e − x) = Dq_{1}(x) + ⋯ + Dq_{m}(x)

and

(−1)^{m+2}D^{2}q_{m}(e − x) = D^{2}q_{2}(x) + ⋯ + D^{2}q_{m}(x).

For x = 0 we find (−1)^{m} = q_{m}(e), Dq_{m}(e)y = (−1)^{m+1}Dq_{1}(0)y = (−1)^{m+1}q_{1}(y), and D^{2}q_{m}(e)(y, z) = (−1)^{m}D^{2}q_{2}(0)(y, z) = (−1)^{m}2q_{2}(y, z). This shows

φ_{e}(y, z) = 2q_{2}(y, z) − q_{1}(y)q_{1}(z) = q_{1}(yz).
□

Since (by assumption) φ = φ_{e} is nondegenerate there is a unique vector ψ(x) such that

(10.3)  φ(ψ(x), y) = λ_{x}(y) = −\frac{Dq_{m}(x)y}{q_{m}(x)}

(This is a general principle to construct inversions [3]). From the definition it is clear that ψ(x) is rational (in x) and homogeneous of degree −1. Moreover, differentiating (10.3) we obtain from the definitions (10.1), (10.2)

(10.4)  φ(Dψ(x)z, y) = D(λ_{x}(y))z = −φ_{x}(y, z)

from which we conclude

(10.5)  Dψ(e) = −Id

and

(10.6)  Dψ(x)^{∗} = Dψ(x)

(the right hand side of (10.4) is symmetric in y and z, so is the left hand side, showing that Dψ(x) is selfadjoint with respect to φ).

We will also need the following identities.

(10.7)  φ(ψ(x), x^{2}) = q_{1}(x)
(10.8)  Dψ(x)x^{2} + 2x · ψ(x) = e

Proof : The ”norm” q_{m}, the last member of the invariant chain, satisfies Dq_{m}(x)x^{2} + q_{1}(x)q_{m}(x) = 0. Therefore (if q(x) ≠ 0)

φ(ψ(x), x^{2}) = −\frac{Dq_{m}(x)x^{2}}{q_{m}(x)} = q_{1}(x).

To obtain (10.8) we differentiate (10.7),

φ(Dψ(x)y, x^{2}) + 2φ(ψ(x), xy) = q_{1}(y) = φ(e, y),

use selfadjointness of Dψ(x) and of L(x) to obtain

φ(Dψ(x)x^{2}, y) + 2φ(ψ(x) · x, y) = φ(e, y)

which is equivalent to (10.8) – since φ is nondegenerate. □

10.4 New identities.

Now let H(x) := ψ(e − x) and H(x) = ∑_{k≥0} h_{k}(x) its Taylor expansion about

the origin 0 with h_{k}(x) homogeneous of degree k. By definition, h_{k}(y) = \frac{1}{k!}D^{k}H(0)(y, …, y) = \frac{(−1)^{k}}{k!} D^{k}ψ(e)(y, …, y) for all y ∈ A and all k. This, in particular, shows that the h_{k}(y) are homogeneous of degree k. But furthermore, since ψ(x) is homogeneous of degree −1, D^{k−1}ψ(x) homogeneous of degree −k, Euler’s equation for homogeneous maps gives

D^{k}ψ(x)(x, y, …, y) = −kD^{k−1}ψ(x)(y, …, y).

Putting x = e we obtain

Dh_{k}(y)e = \frac{(−1)^{k}}{(k − 1)!}D^{k}ψ(e)(e, y, …, y)
= k \frac{(−1)^{k−1}}{(k − 1)!} D^{k−1}ψ(e)(y, …, y) = kh_{k−1}(y);

again:

(10.9)  Dh_{k}(x) · e = kh_{k−1}(x) for all x ∈ A.

The h_{k}(x)’s are actually homogeneous polynomials. In the general situation (unless A is power-associative) we don’t know how the ordinary powers x, x^{2}, …, x^{n}, … are related. So far we only have identities for the polynomials g_{k}(x) (see (1.8)). Now the new set of polynomials h_{k}(x) enters the scene in a natural way.

The h_{k}(x) satisfy the following recursion: h_{0}(x) = e and

(10.10)  (k + 2)h_{k+1}(x) = Dh_{k}(x) · x^{2} + 2xh_{k}(x)

Proof : We compute DH(x)(e − x)^{2} in two different ways. On the one hand

DH(x)(e − x)^{2} = ∑_{k≥0} Dh_{k}(x)(e − 2x + x^{2}) =
= ∑_{k≥1} kh_{k−1}(x) − 2 ∑_{k≥1} kh_{k}(x) + ∑_{k≥1} Dh_{k}(x)x^{2}

by (10.9) and homogeneity. On the other hand,

DH(x)(e − x)^{2} = −Dψ(e − x) · (e − x)^{2}
= 2(e − x)ψ(e − x) − e
= 2(e − x) ∑ h_{k}(x) − e

by virtue of (10.8). Comparing terms of equal degree shows the recursion (10.10). □

For 0 ≤ k ≤ 5 we have

h_{0}(x) = e, h_{i}(x) = x^{i} (i = 1, 2, 3)
h_{4}(x) = \frac{1}{5}(4x^{4} + x^{2}x^{2})
h_{5}(x) = \frac{1}{30}(16x^{5} + 8x^{3}x^{2} + 6(x^{2}x^{2})x)

Remark: D^{2}ψ(e)(y, z) = 2h_{2}(y, z) = 2y · z shows that the algebra structure (the multiplication) can be recovered from the inversion ψ. The going down formulas in 3.2 show that q_{1} is also determined by q_{m}. Thus ψ and therefore the algebra structure is completely determined by q_{m}.

Theorem 10.5. Notations as above. Then

(10.11) h_{r}(x) + q_{1}(x)h_{r−1}(x) + q_{2}(x)h_{r−2}(x) + ⋯ + q_{m}(x)h_{r−m}(x) = 0

for all x ∈ A and all r ≥ m.

Proof : From the definition

φ(ψ(e − x), y) = −\frac{Dq_{m}(e − x)y}{q_{m}(e − x)}

we read off that ψ(e − x) = H(x) = ∑ h_{k}(x) is rational with denominator q_{m}(e − x) and numerator of degree < m. And now, exactly as in lemma 1.1 we see that q_{m}(e − x) ∑ h_{k}(x) is a polynomial of degree < m if and only if (10.11) holds. □

Remark: The polynomials h_{k}(x) are the ”adjoints” of the g_{k}(x) in the following sense: If we linearize the recursion

g_{k+1}(x) = g_{k}(x, …, x, x^{2})

we obtain

(k + 2)g_{k+1}(x, …, x, y) = kg_{k}(x, …, x, x^{2}, y) + 2g_{k}(x, …, x, xy).

With M_{k}(x)y := g_{k}(x, …, x, y) this reads as

(k + 2)M_{k+1}(x) = kM_{k}(x, …, x, x^{2}) + 2M_{k}(x)L(x).

Define H_{k}(x) := M_{k}(x)^{∗}e, then the foregoing equation gives the recursion

(k + 2)H_{k+1}(x) = DH_{k}(x) · x^{2} + 2x · H_{k}(x), H_{0}(x) = e.

We compare with (10.10) and see H_{k}(x) = h_{k}(x). Although this approach seems to be more elementary, our approach via the inversion ψ(x) contains from the very beginning a lot of more information.

Now we have two inversions on A, the R-inversion

j(x) = y(e − x, 1) + e

(see (4.1)) and the LD-inversion (logarithmic derivative inversion) ψ(x)

φ(ψ(x), y)) = −\frac{Dq_{m}(x)y}{q_{m}(x)} ,

and it is natural to ask how they are related. It is not difficult to show

ψ(x) = −Dj(x)^{∗}ψ(j(x)),

but as yet this relation does not seem to be very useful, so we restrict to the question when these two inversions are equal. The answer is given, together with other characterisations, in

Theorem 10.6. Let A be a nondegenerate R-algebra. Then the following properties are equivalent:

a) A is a Jordan algebra,

b) x · j(x) = e wherever j(x) is defined,

c) x · ψ(x) = e wherever ψ is defined,

d) j = ψ,

e) Dj(x)^{∗} = Dj(x) wherever defined.

Proof : A is Jordan if and only if A is power-associative (theorem 10.2.c). Therefore theorem 4.1 shows the equivalence of a) and b). In the same manner one shows a) ⇔ c). If A is power-associative, then g_{k}(x) = x^{k+1}, h_{k}(x) = x^{k} and j(e − x) = ∑_{k≥0} x^{k} = ψ(e − x), thus j = ψ. And since Dψ is selfadjoint (see (10.6)) we

get d) ⇒ e). To prove ”e) ⇒ a)” we note that our assumption implies that all Dg_{k} in Dj(e − x) = − ∑ Dg_{k}(x) are selfadjoint. An easy computation gives

3Dg_{3}(x) = 4L^{3}(x) + 2L(x)L(x^{2}) + 2L(x^{3}) + 4L(x^{2})L(x).

Forming the adjoint mappings on both sides and comparing the resulting equation with the original we obtain L(x)L(x^{2}) = L(x^{2})L(x). □

10.5 Idempotents

One might expect that the presence of the new identity (10.11) will have an impact on the eigenvalues of idempotents. This is indeed the case.

Theorem 10.7. Let A be a nondegenerate R-algebra of degree m and c ≠ e an idempotent of A, q_{1}(c) = −k. Then all eigenvalues ≠ 0, 1 of L(c) are in

\{\frac{−m + k + 2}{4}, \frac{−m + k + 3}{4}, …, \frac{k + 2}{4}\} ∩ \frac{1}{2}ℤ

Proof : We only give an outline, the computations are similar to those in chapter 6. We already know (theorem 6.4) that L(c) has only simple eigenvalues and that they are all in \{\frac{−m+k+2}{2}, \frac{−m+k+3}{2}, …, \frac{k}{2}, 0, 1\}. Let λ ∈ ℂ be an eigenvalue of L(c) with eigenvector y, cy = λy and e, c, y linearly independent. Define polynomials p_{j} of degree j − 1 by

jh_{j}(c, …, c, y) = p_{j}(λ)y.

Then the linearization of (10.11)

h_{m}(x) + q_{1}(x)h_{m−1}(x) + ⋯ + q_{m}(x)e = 0

in y-direction leads with x = c, q_{j}(c) = (−1)^{j} ( \begin{matrix}k \\ j \end{matrix} ) if 1 ≤ j ≤ k, and

q_{j}(c) = 0 if j > k to the equation

∑_{j=0}^{k}(−1)^{j} ( \begin{matrix}k \\ j \end{matrix} ) p_{m−j}(λ) = 0.

In a final step one uses the recursion (10.10) for the h_{j}(x) to derive the identity

∑_{j=0}^{k}(−1)^{j} ( \begin{matrix}k \\ j \end{matrix} ) p_{m−j}(t) = \frac{3!}{(m + 1)!}(2t − 1) ∏_{j=0}^{m−k−2}(4t + 2) ∏_{j=4}^{k+2}(4t − j)

from which the result follows. □

10.6 Nondegenerate R-algebra of degree ≤ 5.

We already know that unital R-algebras of degree ≤ 3 are Jordan, if q_{1} is associative (theorem 3.3). Nondegeneracy of q_{1} allows us to make a stronger statement.

Theorem 10.8. A nondegenerate R-algebra of degree ≤ 5 is Jordan.

Proof : We have to show the assertion only for m = 4 and 5. For m = 4 we have the identities

g_{3}(x) + q_{1}(x)x^{3} + q_{2}(x)x^{2} + q_{3}(x)x + q_{4}(x)e = 0

and

h_{4}(x) + q_{1}(x)x^{3} + q_{2}(x)x^{2} + q_{3}(x)x + q_{4}(x)e = 0

((3.4) and (10.11) with m = 4). We immediately find g_{3}(x) = h_{4}(x) and consequently x^{4} = x^{2}x^{2}. Thus A is power-associative (and then Jordan, theorem 10.2). The case m = 5 needs more work. We start by taking the difference of

g_{4}(x) + q_{1}(x)g_{3}(x) + ⋯ + q_{5}(x)e = 0

and

h_{5}(x) + q_{1}(x)h_{4}(x) + ⋯ + q_{5}(x)e = 0

((3.4) and (10.11) for m = 5) to arrive at g_{4}(x) − h_{5}(x) + q_{1}(x)(g_{3}(x) − h_{4}(x)) = 0; explicitly

(10.12)  6x^{5} − 7x^{2}x^{3} + x(x^{2}x^{2}) + 4q_{1}(x)(x^{4} − x^{2}x^{2}) = 0.

Applying φ(·, x) on both sides and using associativity, we obtain

φ(x^{5}, x) = φ(x^{2}x^{2}, x^{2}).

The linearization hereof is

φ(x^{5}, y) + φ(2x(x(x(xy))) + x(x(x^{2}y)) + x(x^{3}y) + x^{4}y, x)
= 4φ(x^{2}(xy), x^{2}) + 2φ(x^{2}x^{2}, xy),

and associativity and nondegeneracy imply

(10.13)  2x^{5} + x^{2}x^{3} − 3x(x^{2}x^{2}) = 0.

Eliminating the x^{2}x^{3}-term in (10.12) with the help of (10.13), we get

(10.14)  (x + \frac{1}{5}q_{1}(x)e)(x^{4} − x^{2}x^{2}) = 0.

From (10.12) and (10.13) we also get

D(x^{4} − x^{2}x^{2}) · x^{2} = −\frac{8}{5}q_{1}(x)(x^{4} − x^{2}x^{2}),

and then repeated differentiation of (10.14) in x^{2}-direction gives

(10.15)  (x^{l} + \frac{1}{5}q_{1}(x^{l})e)(x^{4} − x^{2}x^{2}) = 0

for l = 1, 2, 3. We now linearize (10.14)

y(x^{4} − x^{2}x^{2}) + x(2x(x(xy)) + x(x^{2}y) + x^{3}y − 4x^{2}(xy)) =
−\frac{1}{5}[q_{1}(y)(x^{4} − x^{2}x^{2}) − q_{1}(x)(2x(x(xy)) + x(x^{2}y) + x^{3}y − 4x^{2}(xy))]

and put y = x^{4} − x^{2}x^{2}. Due to (10.15) and associativity of q_{1} we obtain

(x^{4} − x^{2}x^{2})^{2} = 0,

thus φ(x^{4} − x^{2}x^{2}, x^{4} − x^{2}x^{2}) = 0. Linearization shows

φ(2x(x(xy)) + x(x^{2}y) + x^{3}y − x^{2}(xy), x^{4} − x^{2}x^{2}) = 0

and then (by associativity and nondegeneracy)

2x(x(xh(x))) + x^{2}(xh(x)) + x^{3}h(x) − 4x(x^{2}h(x)) = 0

with h(x) := x^{4} − x^{2}x^{2}. Using (10.15) on each summand we obtain

(10.16)  [2q_{1}(x)^{3} + 15q_{1}(x)q_{1}(x^{2}) + 25q_{1}(x^{3})](x^{4} − x^{2}x^{2}) = 0.

Now assume that A is not power-associative. Then x^{4} − x^{2}x^{2} ≠ 0 on a nonempty Zariski-open subset of A, thus

2q_{1}(x)^{3} + 15q_{1}(x)q_{1}(x^{2}) + 25q_{1}(x^{3}) = 0

for all x ∈ A. Linearizing this identity and again using associativity and nondegeneracy of q_{1} yields

75x^{2} + 30q_{1}(x)x + (15q_{1}(x^{2}) + 6q_{1}(x)^{2})e = 0

for all x. But then (obviously) x^{4} = x^{2}x^{2} for all x, which contradicts our assumption. Thus A is power-associative and then Jordan. □

10.7 Example.

We will give an example of a nondegenerate R-algebra of degree 6 which is not Jordan; showing that theorem 10.8 cannot be improved. Let A = 𝔽^{3} with commutative product defined by

x^{2} = ( \begin{matrix}x_{1}^{2} + x_{2}^{2} + x_{3}^{2} \\ 2x_{1}x_{2} + 2αx_{2}x_{3} \\ 2x_{1}x_{3} + α(x_{2}^{2} − x_{3}^{2}) \end{matrix} ) with α^{2} = \frac{16}{5}, x = (x_{1}, x_{2}, x_{3})^{t}.

Then it is easily seen that e = (1, 0, 0)^{t} is a unit element of A and that the linear form q_{1}(x) := −6x_{1} is symmetric, associative and nondegenerate. It needs some work (which has actually been done by MAPLE) to show that (q_{1}, q_{2}, …, q_{6}) with

q_{j+1}(x) := \frac{1}{j + 1}(Dq_{j}(x) · x^{2} + q_{1}(x)q_{j}(x))

is an invariant chain (q_{7}(x) = 0) and that

g_{5}(x) + q_{1}(x)g_{4}(x) + ⋯ + q_{6}(x)e = 0

for all x ∈ A. Thus A is a nondegenerate R-algebra of degree 6. That A is not Jordan can be seen, for example, from a^{2}a^{2} ≠ a^{4} for a = (0, 1, 0)^{t}. It is easily seen that this algebra is simple. The classification of simple Ralgebras, or even the restricted problem, the classification of nondegenerate R-algebras of degree 6 still seems to be out of reach.

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