K-Loops from Matrix Groups over Ordered Fields I

Hubert Kiechle

Introduction

Let (L, ⊕) be a loop, i.e., L is a set with a binary operation ⊕ such that for all a, b ∈ L the equations a ⊕ x = b and y ⊕ a = b have unique solutions x, y ∈ L, and such that there exists 0 ∈ L with a ⊕ 0 = 0 ⊕ a = a. The condition a ⊕ (b ⊕ x) = (a ⊕ b) ⊕ δ_{a,b}(x) for a, b, x ∈ L then clearly defines a bijective map δ_{a,b} : L → L. Following [17] we use the phrase precession-maps to denote these maps.

L is called a K-loop if the precession-maps are automorphisms of the loop, if the automorphic inverse property is satisfied, and the Bol identity holds. To be precise, we require that for all a, b, c, d ∈ L

δ_{a,b}(c ⊕ d) = δ_{a,b}(c) ⊕ δ_{a,b}(d)
a ⊕ (b ⊕ (a ⊕ c)) = (a ⊕ (b ⊕ a)) ⊕ c (Bol identity),
⊖(a ⊕ b) = (⊖a) ⊕ (⊖b)  (automorphic inverse property)

where ⊖a is defined by a ⊕ (⊖a) = 0. The Bol identity implies the equality of left and right inverse, i.e., (⊖a) ⊕ a = 0 as well. Thus we use the term “automorphic inverse property” in the standard way. A recent result of Kreuzer [11] shows that the requirement δ_{a,b} ∈ Aut L is obsolete, it follows from the other hypotheses. It should also be noted that in the definition of K-loops the Bol identity is usually replaced by conditions on the precession-maps (including δ_{a,b} ∈ Aut L). Using [9] one easily deduces that the two definitions are equivalent.

The group Σ(L) := ⟨δ_{a,b}; a, b ∈ L⟩ ≤ Aut(L) generated by the precession-maps has been called structure group in [12].

The set of admissible velocities R_{c}^{3} := {v ∈ R^{3}; |v| < c} together with the relativistic velocity addition forms a K-loop. This has been proved by Ungar (cf. [16, 17]). Encouraged by Ungar’s discovery, many K-loops have been constructed in recent years (see, eg., [1, 2, 4, 7, 8, 9, 10, 12]). In case of Ungar’s example R_{c}^{3}, the precession-maps are the “Thomas-precessions” or “Thomas-rotations” of special relativity, hence the name.

The notion of a K-loop evolved from the study of neardomains (F, ⊕, ·). Presently, there seem to be no examples of proper neardomains known. “Proper” here means “not a nearfield”. A nearfield is a neardomain where all precession-maps are 1. See [18; V §1] for definitions and more information concerning nearfields and neardomains, and their connection with sharply 2-transitive groups.

Here we generalize a construction due to Karzel, [3], which evolved from his attempt to give an “elegant representation of [Ungar’s] results” (cf. [4; p. 339]). While Karzel used positive definite, hermitian 2 × 2-matrices over euclidian fields, we will consider symmetric and hermitian n × n-matrices over suitable ordered fields R. All of these fields happen to be pythagorean, i.e., for every a ∈ R the element 1 + a^{2} is a square in R. Hence every sum of squares is a square.

The methods used are quite different from Karzel’s, though. They employ a paper by Kreuzer and Wefelscheid [12], which makes it posible to study many classical groups. In the first section, the construction of Kreuzer and Wefelscheid is reviewed. The second section describes our construction. In the third section, we deal with the question, when the precession-maps are fixed point free. Since every automorphism fixes 0, the phrase “fixed point free” refers to the situation, where only the identity has fixed points other that 0. In fact, we require that every element of the structure group is fixed point free. This is interesting, because the precession-maps of a neardomain F generate a fixed point free group of automorphisms of the additive loop of F. Only such examples have a chance to become a proper neardomain, when they are grown up. The fourth section containes a bunch of examples, including one with fixed point free precession-maps. We also show that quite a few constructions using matrices are subsumed by our method. In particular, Karzel/Ungar’s examples, and Im’s work [1, 2].

1. Preliminaries

First, we will review the method of Kreuzer and Wefelscheid [12] to constructing K-loops from groups. Let G be a group with a subgroup Ω and a set L of representatives for the cosets of Ω in G. This means in particular G = LΩ. For A, B ∈ L let A ⊕ B be the unique element in L such that AB ∈ (A⊕B)Ω. Define d_{A,B} ∈ Ω by AB = (A⊕B)d_{A,B}. Assume that the following conditions are satisfied for all g ∈ Ω and for all A, B ∈ L:

1 ∈ L, gLg^{−1} ⊆ L, ABd_{A,B}^{−1} = d_{A,B}BA, ALA ⊆ L.  (1)

From [12; (3.8),(3.7)] we have

(1.1) (L, ⊕) is a K-loop with 0-element 1. For A, B, X ∈ L: δ_{A,B}(X) = d_{A,B}Xd_{A,B}^{−1}, thus A ⊕ (B ⊕ X) = (A ⊕ B) ⊕ δ_{A,B}(X).

Certain automorphisms of G induce automorphisms of L.

(1.2) Let σ ∈ Aut G such that σ(L) = L and σ(Ω) = Ω. Then σ L is an automorphism of (L, ⊕), and σ(d_{A,B}) = d_{σ(A),σ(B)}. In particular, for all u ∈ Ω the map û : L → L; X ↦ uXu^{−1} fulfills these conditions.

Proof. We have σ(A ⊕ B)σ(d_{A,B}) = σ(AB) = σ(A)σ(B) = (σ(A) ⊕ σ(B))d_{σ(A),σ(B)}. By uniqueness of the decomposition we must have σ(A ⊕ B) = σ(A) ⊕ σ(B) and σ(d_{A,B}) = d_{σ(A),σ(B)}.

Now, the last statement follows directly from the second condition in eq. (1).

As a direct consequence, we obtain

(1.3) The map Ω → Aut(L, ⊕); u ↦ û is a homomorphism with kernel C_{Ω}(L) = {u ∈ Ω; ∀A ∈ L : uAu^{−1} = A}.

We’ll write Ω̂ for the image of Ω under the map in the preceding lemma.

Define Ĝ := L ×_{Q} Ω̂ as follows

Ĝ := \{(A, α) : \{ \begin{matrix}L → L \\ x ↦ A ⊕ α(x) \end{matrix} ; A ∈ L, α ∈ Ω̂\}.

It turns out that Ĝ is a group, called the quasidirect product of L by Ω̂. More specific

(A, α) ∘ (B, β) = (A ⊕ α(B), δ_{A,α(B)}αβ), (A, α)^{−1} = (α^{−1}(⊖A), α^{−1})  (2)

as is easily computed, using δ_{A,⊖A} = 1 (cf. [12; (K4)]). Note also that (A, α) = 1 if and only if A = 0 and α = 1.

Remarks 1. It’s easy to see that the construction of the quasidirect product can be carried out with any subgroup of Aut(L, ⊕) which contains Σ(L) replacing Ω̂. Furthermore, in [12; (7.1)] it is shown that every automorphism of L is induced by an automorphism of Ĝ as in (1.2).

2. Our definition of the quasidirect product is only formally different from the definitions given in [6; p. 28] and [12], which use eq. (2) to define an operation on L × Ω̂. The notions are clearly equivalent. Our approach releases us from checking the associtive law, though.

3. The phrase “quasidirect product” has been used by Wilson [20] in the more general context of quasigroups. Though similar, the meaning is distinct even for loops.

(1.4) The map

π : \{ \begin{matrix}G → Ĝ \\ Au ↦ (A, û) \end{matrix} , A ∈ L, u ∈ Ω,

is an epimorphism with kernel C_{Ω}(L).

Proof. By the uniqueness of the decomposition, π is well-defined. Obviously π is surjective. Now let Au, Bv ∈ G, A, B ∈ L, u, v ∈ Ω be given. We have

AuBv = Aû(B)uv = (A ⊕ û(B))d_{A,û(B)}uv.

Together with eq. (2) this shows that π is a homomorphism, since Ω → Ω̂; u ↦ û is. The statement about the kernel is clear from the definitions.

We’ll frequently encounter diagonal matrices, and more generally block diagonal matrices. This will be indicated by diag(A_{1}, …, A_{n}), with the blocks A_{1}, …, A_{n} placed on the diagonal in that order:

diag(A_{1}, …, A_{n}) := ( \begin{matrix}A_{1} \\ & . & . & . \\ & & & & A_{n} \end{matrix} ) .

Notice that the A_{i} are either numbers or matrices. We’ll not distingish these two cases in our notation, we’d rather identify numbers with 1 × 1-matrices.

2. The construction

Let R be an ordered field, let K := R(i), where i^{2} = −1, and let n ≥ 2 be a fixed integer. Consider the set L of positive definite hermitian n × n-matrices over K, i.e.,

L = {A ∈ K^{n×n}; A = A^{∗}, ∀v ∈ K^{n} \ {0} : v^{∗}Av > 0},

where X^{∗} := \overline{X}^{T} for every X ∈ K^{m×n}, m ∈ N. Write

Ω := U(n, K) = {U ∈ K^{n×n}; UU^{∗} = I_{n}}

for the corresponding unitary group. Here I_{n} denotes the n × n identity matrix. R will be call n-real if all eigenvalues of every element of L are in K. In fact, it is well known and easy to see that in this case they are in R. Clearly, n-real implies (n − 1)-real. A field is called totally real if it is n-real for all n ∈ N. The term “totally real” is usually defined differently. From [19; Thm. 2] (see also [14]), we know that our notion is equivalent. Indeed we have

(2.1) R is totally real if and only if R is the intersection of real closed fields.

Remarks 1. In fact, both [19, 14] prove this theorem only for symmetric matrices over R. However, the proofs given in [14] apply after minor modification to the hermitian case as they stand. This is true in particular for [14; Theorem 2] and the Corollary and Remarks following.

2. The class of such fields strictly contains the class of real closed fields and the class of hereditarily euclidean fields (cf. [19]).

(2.2) R is 2-real if and only if R is pythagorean. For n ≥ 2 every n-real field is pythagorean.

The Proof is an adaption of the proof of [14; Lemma 1] to our situation.

If R is 2-real, then let a ∈ R be positive. The matrix

A := ( \begin{matrix}a + 2 & a \\ a & a \end{matrix} )

is clearly hermitian over R (it is in fact symmetric). The eigenvalues are

a + 1 ± \sqrt{1 + a^{2}},

which are to be in R, since A is clearly positive definite. Hence 1 + a^{2} is a square in R, and because a was arbitrary, we conclude that R is pythagorean.

Conversely, assume that R is a pythagorean field. A general hermitian 2 × 2-matrix A has the form

A = ( \begin{matrix}α & a \\ \overline{a} & β \end{matrix} ) ,

with α, β ∈ R and a ∈ R(i). The characteristic polynomial of A is x^{2}−(α+β)x+αβ−a\overline{a} and has discriminant (α + β)^{2} − 4(αβ − a\overline{a}) = (α − β)^{2} + 4a\overline{a}. This is a sum of squares in R, hence a square. Thus the eigenvalues of A are in R, and R is 2-real.

The last assertion follows from a previous remark.

We now return to our main concern.

(2.3) I_{n} ∈ L, L^{−1} = L and A^{∗}LA = L for all A ∈ GL(n, K).

Proof. The first statement is clear. Let v ∈ K^{n} be non zero. For A ∈ L, we have (A^{−1})^{∗} = (A^{∗})^{−1} = A^{−1}. Furthermore, 0 < (A^{−1}v)^{∗}A(A^{−1}v) = v^{∗}A^{−1}AA^{−1}v = v^{∗}A^{−1}v, hence A^{−1} is positive definite. (An alternative argument uses eigenvalues.)

For B ∈ L we have (A^{∗}BA)^{∗} = A^{∗}BA, and v^{∗}A^{∗}BAv = (Av)^{∗}B(Av) > 0, for all v ∈ K^{n} \ {0}. Hence A^{∗}BA is positive definite.

We will have use for the spectral theorem [13; XIV.12.4]. Note that our assumption for K makes sure that this theorem is true.

(2.4) Let R be n-real and let A ∈ L. The eigenvalues of A are in R, and there exists a U ∈ Ω such that UAU^{−1} is diagonal. This means there exists an orthonormal basis of K^{n}, consisting of eigenvectors of A. If A ∈ GL(n, R), then U can be taken from GL(n, R). Furthermore, it is possible to chose U such that det U = 1.

Proof of the last statement. If det U = α, then αᾱ = 1 (since det U^{∗} = \overline{det U}). Hence we can replace the first column c, say, of U by α^{−1}c to obtain a matrix U′. It is easy to see that U′ is unitary. Clearly, det U′ = 1, hence the result.

(2.5) The map κ : L → L; X ↦ X^{2} is injective.

Proof. X^{2} = X^{∗}I_{n}X ∈ L, by (2.3). Now assume that A^{2} = B^{2} for A, B ∈ L. There is no loss in generality to assume that R is n-real (or even real closed), since otherwise we pass to an extension field of R.

By the spectral theorem (2.4) we can decompose K^{n} into an orthogonal sum of eigenspaces V_{1}, …, V_{r} of A, say. Denote the corresponding (distinct!) eigenvalues by λ_{1}, …, λ_{r}. Thus Av = λ_{k}v for all v ∈ V_{k}, k ∈ {1, …, r}. Note that λ_{k} > 0 for all k ∈ {1, …, r}. We have A^{2}v = λ_{k}^{2}v = B^{2}v for all v ∈ V_{k}, k ∈ {1, …, r}. Applying the same argument to B, we conclude that the V_{k} are the eigenspaces of B. Since B is positive definite, the λ_{k} are the eigenvalues of B. This entails A = B, and the injectivity of κ.

(2.6) Let R be n-real und take A, B ∈ L, then A ∈ BΩ if and only if A = B.

Proof. Let A = diag(λ_{1}, …, λ_{n}) and C := B^{−1} = ( c_{1} … c_{n}), where c_{k} ∈ K^{n} are the columns of C. Note that C ∈ L by (2.3). We compute CA = ( λ_{1}c_{1} … λ_{n}c_{n}) ∈ Ω. Using λ_{k} ∈ R, for k ≠ j we have

0 = (λ_{k}c_{k})^{∗}(λ_{j}c_{j}) = (λ_{k}λ_{j})c_{k}^{∗}c_{j}.

Since λ_{k}λ_{j} ≠ 0 we infer that C has orthogonal columns, i.e.,

C^{2} = C^{∗}C = diag(c_{1}^{∗}c_{1}, …, c_{n}^{∗}c_{n}).

From (2.5) we conclude C = diag(μ_{1}, …, μ_{n}) for appropriate μ_{k} ∈ K. Thus CA = diag(λ_{1}μ_{1}, …, λ_{n}μ_{n}) ∈ Ω. Since λ_{k}μ_{k} > 0 we must have B^{−1}A = CA = I_{n}, as claimed.

The general case where A is not necessarily diagonal can be reduced to the case just handled by (2.4).

The following constitutes the key lemma for the further development.

(2.7) Let G be a subgroup of GL(n, K) and put L_{G} := L ∩ G, Ω_{G} := Ω ∩ G. The following are equivalent.

(I) G = L_{G}Ω_{G}; (II) AA^{∗} ∈ κ(L_{G}), for all A ∈ G.

Proof. (I) ⟹ (II): Let A = BU ∈ G, B ∈ L_{G}, U ∈ Ω_{G}. We have AA^{∗} = BUU^{∗}B^{∗} = BB^{∗} = B^{2} ∈ κ(L_{G}).

(II) ⟹ (I): For A ∈ G let B ∈ L_{G} subject to B^{2} = AA^{∗}. Now U := B^{−1}A is in Ω: UU^{∗} = B^{−1}AA^{∗}(B^{−1})^{∗} = B^{−1}B^{2}B^{−1} = I_{n}. Hence G ⊆ L_{G}Ω_{G}.

Remark. This is a generalization of [2; (1.1.e)].

We will now assume that R is n-real, and that G is a subgroup of GL(n, K) with G = L_{G}Ω_{G}. Then for A, B ∈ L_{G}, there are unique A ⊕ B ∈ L_{G} and d_{A,B} ∈ Ω_{G} such that AB = (A ⊕ B)d_{A,B}. We proceed to show that (L_{G}, ⊕) becomes a K-Loop. This will be an immediate consequence of (1.1) and the following lemma.

(2.8) For all A, B ∈ L_{G} and all U ∈ Ω_{G} we have

(1) AΩ_{G} = BΩ_{G} ⟺ A = B;

(2) UL_{G}U^{−1} = L_{G};

(3) AL_{G}A = L_{G};

(4) ABU ∈ L_{G} ⟺ U^{∗}BA ∈ L_{G}.

Proof. (1) obviously follows from (2.6). (2) and (3) are direct from (2.3), since U^{∗} = U^{−1} and A^{∗} = A.

(4) ABU ∈ L_{G} implies ABU = (ABU)^{∗} = U^{∗}BA. This entails the result.

Remark. From (2.8.4) one can deduce that A ⊕ B = κ^{−1}(AB^{2}A) (cf. the proof of [12; (3.4)]). This can be used to prove (1.2) for our special K-loops without the hypothesis “σ(Ω) = Ω”. Therefore, (1.2) implies σ(Ω_{G}) = Ω_{G}.

Tying things together, we have proved

(2.9) Theorem. Let R be n-real, and G ≤ GL(n, K) with G = L_{G}Ω_{G}, then there are A ⊕ B ∈ L_{G} and d_{A,B} ∈ Ω_{G} such that AB = (A ⊕ B)d_{A,B}.

(L, ⊕) is a K-loop with δ_{A,B} = d̂_{A,B}. Furthermore, Ω_{G}/C_{Ω_{G}}(L) is naturally isomorphic to a subgroup of Aut L.

Proof.

Remark. When R is the field of real numbers, the above theorem can be generalized to Hilbert spaces H over R or R(i). Here, G is the unit group of the Banach algebra of bounded operators H → H. In this case the polar decomposition [15; Thm. VIII.6, p. 73] serves as a substitute for (2.7.I), where L is the set of positive (self adjoint) operators in G. The details will be left to the reader.

3. Fixed points of precession-maps

We say that a K-loop L has fixed point free precession-maps if for every γ ∈ Σ(L) \ {1} and A ∈ L, the condition γ(A) = A implies A = 0. The results of this section will enable us to decide whether, or not, our examples have fixed point free precession-maps.

We begin with a (well-known?) lemma.

(3.1) Let A = diag(λ_{1}, …, λ_{n}), B ∈ K^{n×n} with |{λ_{1}, …, λ_{n}}| = n (i.e., all λ_{k} distinct), then we have

AB = BA ⟺ B = diag(α_{1}, …, α_{n}), α_{k} ∈ K.

Proof. Assume AB = BA, and let B = ( b_{k,ℓ}). Then we have for all k, ℓ ∈ {1, …, n}

λ_{k}b_{k,ℓ} = b_{k,ℓ}λ_{ℓ} hence (λ_{k} − λ_{ℓ})b_{k,ℓ} = 0.

We conclude that b_{k,ℓ} = 0 if k ≠ ℓ. Thus B is diagonal.

The converse is trivial.

(3.2) Assume that G is one of the groups SL(n, R), GL(n, R), SL(n, K), GL(n, K). If A ∈ L_{G} has a double eigenvalue, i.e., an eigenspace of A is at least 2-dimensional, then there exists U ∈ Ω_{G} such that UA U^{−1} = A, det U = 1, and Û ≠ 1.

Proof. Using the spectral theorem (2.4) there is S ∈ Ω, det S = 1 such that SA S^{−1} = diag(λ_{1}, …, λ_{n}) is diagonal with λ_{1} = λ_{2}. Put

U′ := ( \begin{matrix}0 & 1 \\ −1 & 0 \\ & & 1 \\ & & & . & . & . \\ & & & & & & 1 \end{matrix} ) ∈ SO(n, R),

then U := S^{−1}U′S has the desired properties, except possibly the last. Û ≠ 1 fol-

lows from the fact that a matrix C in the centralizer of ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) has the form

C = ( \begin{matrix}a & b \\ −b & a \end{matrix} ), which is hermitean if and only if it is diagonal.

4. Examples

We recall that R is always an ordered field and K := R(i).

(4.1) Let R be n-real. For any A ∈ GL(n, K) the eigenvalues of AA^{∗} are squares in R.

Proof. Let λ be an eigenvalue of AA^{∗} with eigenvector v, i.e., AA^{∗}v = λv. It follows that

v^{∗}AA^{∗}v = λv^{∗}v and λ = \frac{1}{(v^{∗}v)^{2}} (v^{∗}v)((A^{∗}v)^{∗}(A^{∗}v))

which is a sum of squares in R. Since R is pythagorean by (2.2), we conclude that λ is a square in R.

We’ll now give some applications of the theory presented in the preceding sections. Using the results from §3, we can also characterize those of our examples which have fixed point free precession-maps. We note that by PSO(2, R) we mean SO(2, R)/{±I_{2}}, despite the fact that SO(2, R) is commutative.

(4.2) Let R be n-real, and let G ∈ {SL(n, R), GL(n, R), SL(n, K), GL(n, K)}. Then (1) G = L_{G}Ω_{G}, hence (L_{G}, ⊕) is a K-loop;

(2) Σ(L_{G}) ≅ \{ \begin{matrix}PSO(n, R) & if G ∈ \{SL(n, R), GL(n, R)\} \\ PSU(n, K) & if G ∈ \{SL(n, K), GL(n, K)\}. \end{matrix}

(3) The K-loop (L_{G}, ⊕) has fixed point free precession-maps if and only if G = SL(2, R). This will work whenever R is a pythagorean field.

Proof. (1) By (2.7) it suffices to show that for every A ∈ G, there exists B ∈ L_{G} with B^{2} = AA^{∗}. By (2.4) there exists U ∈ Ω_{G} such that

AA^{∗} = U diag(λ_{1}, …, λ_{n}) U^{−1}.

By (4.1) the λ_{k} are squares in R. Hence \sqrt{λ_{k}} ∈ R.

For

B := U diag(\sqrt{λ_{1}} , …, \sqrt{λ_{n}}) U^{−1} ∈ L_{G}

we therefore have B^{2} = AA^{∗}, as desired.

(2) is Theorem (1.3) from [5]. Note, that [5] uses the present paper (only!) up to (1) of this theorem. Thus, there is no logical circle involved.

(3) Assume first that L_{G} has fixed point free δ_{A,B}. Then no element other than I_{n} can have a double eigenvalue, by (2) and (3.2). Hence n = 2 and G = SL(2, .), since otherwise diag(λ, λ, …) ∈ L_{G} for λ > 0. The case G = SL(2, K) must be excluded, because U := diag(i, −i) ∈ Ω_{G} and Û has fixed point diag(λ, λ^{−1}) ∈ L_{G} for every positive λ ∈ R, but Û ≠ 1, since for example

Û ( \begin{matrix}α & β \\ β & γ \end{matrix} ) = ( \begin{matrix}α & −β \\ −β & γ \end{matrix} ) .

Moreover, Û ∈ Σ(L_{G}) by (2).

Now assume G = SL(2, R), then Ω_{G} = SO(2, R). Let A ∈ L_{G}, A ≠ I_{2} and choose U ∈ Ω_{G} with UAU^{−1} = A. By (2.4) there is S ∈ SO(2, R) such that

Λ := SAS^{−1} = diag(λ, λ^{−1})

is diagonal. The second eigenvalue of Λ has to be λ^{−1}, because det Λ = 1. For V := SUS^{−1} we have V ΛV^{−1} = Λ and V ∈ Ω_{G}. Since λ > 0 and λ ≠ 1, we have λ ≠ λ^{−1}. From (3.1) we conclude that V = diag(α_{1}, α_{2}) and |α_{1}| = |α_{2}| = 1. Since det V = 1, we have α_{1} = α_{2} = ±1. Thus V̂ = 1, and so Û = 1.

Remark. For readers, who feel uneasy with the argument given in the proof of (4.2.2), we’ll sketch a proof of (4.2.3), which is independent of (4.2.2) and [5]. Let G ∈ {SL(n, R), GL(n, R), SL(n, K), GL(n, K)}, as in the setting of (4.2). Assume first n = 2. We start with a simple computation. Note that all matrices involved have determinant 1, and belong to G in all cases. Let

A := ( \begin{matrix}3 & 0 \\ 0 & 1/3 \end{matrix} ) , B := ( \begin{matrix}1 & 4 + \sqrt{6} \\ 4 + \sqrt{6} & 23 + 8\sqrt{6} \end{matrix} ) and U_{8} := \frac{\sqrt{2}}{2} ( \begin{matrix}1 & 1 \\ −1 & 1 \end{matrix} ) ,

then

AB = ( \begin{matrix}3 & 12 + 3\sqrt{6} \\ \frac{4}{3} + \frac{1}{3} \sqrt{6} & \frac{23}{3} + \frac{8}{3} \sqrt{6} \end{matrix} ) = ( \begin{matrix}\frac{15}{2} \sqrt{2} + 3\sqrt{3} & \frac{9}{2} \sqrt{2} + 3\sqrt{3} \\ \frac{9}{2} \sqrt{2} + 3\sqrt{3} & \frac{19}{6} \sqrt{2} + \frac{7}{3} \sqrt{3} \end{matrix} ) U_{8},

hence d_{A,B} = U_{8}. Moreover, U_{4} := U_{8}^{2} = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ).

For general n, it is also easy to see how the matrix U′ in the proof of (3.2) has d_{A′,B′}^{2} = U′ for suitable A′, B′ ∈ L_{G}. Now, in the proof of (4.2.3), as well as (3.2) it is enough to know that certain conjugates of U_{4}, and U′, respectively, are (d_{A,B}^{2})’s. This follows from (1.2).

For the next example, we’ll need some preparation.

(4.3) Let K be an arbitrary field and A ∈ GL(2, K) such that det A = α^{2} is a square in K, i.e., α ∈ K. Put β := tr A and assume \sqrt{2α + β} ∈ K. For

B := \frac{1}{\sqrt{2α + β}} (αI_{2} + A)

we have

(1) B^{2} = A.

(2) If K = R(i) with an ordered field R, and if A is hermitian, then B is hermitian.

(3) If, in addition to the asumptions of (2), A is positive definite, and α, \sqrt{2α + β} > 0, then B is positive definite.

(4) With assumptions as in (3), the conditions B hermitian, positive definite with B^{2} = A, uniquely determine B. In this case we write \sqrt{A} := B.

The Proof of the first part comes easily from the Cayley-Hamilton-Theorem. It is also easy to see that B is hermitian in case (2).

(3) That B is positive definite comes directly from computing v^{∗}Bv for general v ∈ R(i)^{2}.

(4) follows directly from (2.5).

For our example, we’ll assume that R is an arbitrary, ordered field. Write R^{real} for a (fixed) real closure of R.

Let K = R(i) as before, n = 2, and let GL′(2, K) := {A ∈ GL(2, K); det A ∈ R}, the invertible 2 × 2-matrices over K with “real” determinant. Consider

G := ⋃_{d>0} \sqrt{d} GL′(2, K) ⊆ GL(2, R^{real}(i)) (\sqrt{d} positive in R^{real}).

Clearly, G is a subgroup of GL(2, R^{real}(i)).

(4.4) For all A ∈ L_{G} we have AA^{∗} ∈ κ(L_{G}), hence G = L_{G}Ω_{G} and (L, ⊕) is a K-loop.

Proof. We have A = \sqrt{d}B for some d > 0 and some B ∈ GL(2, K). Thus AA^{∗} = dBB^{∗} and det(dBB^{∗}) = d^{2}(det B)^{2} is a square in R. Also AA^{∗} is hermitian, positive definite, so by (4.3) C := \sqrt{AA^{∗}} ∈ L_{G} with κ(C) = C^{2} = AA^{∗}.

The remaining assertions follow from the preceding paragraph, (2.7), and (2.9).

Remark. The above example is isomorphic to the example described by Bokhee Im [1] via the map κ. More precisely: Let H be the set of positive definite 2 × 2-matrices over K with square determinant, i.e., for all A ∈ H there is α ∈ R such that det A = α^{2}. The map κ : L_{G} → H, X ↦ X^{2} is bijective and (H, +) becomes a K-loop by A + B := κ(κ^{−1}(A) ⊕ κ^{−1}(B)). This is Im’s example! To see that κ is actually well-defined and bijective, note first that κ(X) =^{!} XX^{∗} does have square determinant. From (2.5) we know that κ is injective. Finally, for B ∈ H, having square determinant (4.3) gives κ^{−1}(B) = \sqrt{B} ∈ L_{G}.

References

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Mathematisches Institut Technische Universität München D-80290 München Germany

kiechle@mathematik.tu-muenchen.de