The aims of this paper is to prove that every semisimple Jordan algebra bundle is locally trivial and establish the decomposition theorem for locally trivial Jordan algebra bundles using the decomposition theorem of Lie algebra bundles.
Using the decomposition theorem of Lie algebra bundles, this paper proves the decomposition theorem for locally trivial Jordan algebra bundles.
Findings of this paper establish the decomposition theorem for locally trivial Jordan algebra bundles.
To the best of the author’s knowledge, all the results are new and interesting to the field of Mathematics and Theoretical Physics community.
1. Introduction
In modern mathematics, an important notion is that of non-associative algebra. In Ref. [1], we gave a relationship between two important classes of non-associative algebras, namely, Lie algebras (introduced in 1870 by the Norwegian mathematician Sophus Lie in his study of the groups of transformations) and Jordan algebras (introduced in 1932–1933 by the German physicist Pasqual Jordan (1902–1980) in his algebraic formulation of quantum mechanics [2–4]). These two algebras are interconnected, as was remarked for instance by Kevin McCrimmon [5, p. 622]:
We are saying that if you open up a Lie algebra and look inside, 9 times out of 10 there is a Jordan algebra (of pair) which makes it work.
Here, we recall some connections between Jordan algebra bundles and Lie algebra bundles [1, 6]. If ξ is a locally trivial Jordan algebra bundle in which each fibre ξx has a unit element then
is a Lie algebra bundle, where Derξx is the vector space of all derivation defined on ξx, L(ξx) is the vector space of all left translations of ξx and an isomorphic copy of ξx.
[1] Let ξ be a locally trivial Jordan algebra bundle over X, with each fibre ξx having a unit element ex, x ∈ X. Then
[1] If ξ is a locally trivial Jordan algebra bundle in which each fibre has a unit element, then , is a Lie algebra bundle, where is an isomorphic copy of the Jordan algebra bundle ξ. Further, ξ can be imbedded in K(ξ) such that the Jordan multiplication on ξ can be given in terms of the Lie multiplication on K(ξ).
Given a locally trivial Jordan algebra bundle ξ in which each fibre ξx has a unit element, consider the Lie algebra bundle . Let h(ξ) = ∪x∈Xh(ξx), where each h(ξx) = L(ξx) ⊕ [L(ξx), L(ξx)] is an ideal of g(ξx) = Derξx + L(ξx) [7]. Let ϕ : U × J → p−1(U) be the local triviality of the Jordan algebra bundle ξ, where J is a Jordan algebra. Set g(ϕ) = Endϕ|U×g(J), where the vector bundle morphism Endϕ : U ×EndJ → ∪x∈U Endξx is given by . Further, h(ξ) is a vector bundle since g(ϕ)|U×h(J) maps U × h(J) onto ∪x∈Uh(ξx). Hence h(ξ) is an ideal subbundle of g(ξ). We denote by L(ξ), the ideal subbundle of K(ξ).□
In this paper, we prove that any semisimple Jordan algebra bundle is locally trivial and we supply an example to show that the converse need not be true. Further, we prove that a semisimple Jordan algebra bundle can be written as the direct sum of simple ideal bundles.
1.1 Notations and terminology
All Jordan algebra bundles ξ = (ξ, p, X, θ) are over the arbitrary topological space X unless otherwise mentioned.
2. Preliminaries
A Jordan algebra bundle is a vector bundle ξ = (ξ, p, X) together with a vector bundle morphism θ : ξ ⊗ ξ → ξ inducing a Jordan algebra structure on each fibre ξx, x ∈ X.
By a trivial Jordan algebra bundle, we mean a trivial vector bundle (X × J, p, X), where J is a Jordan algebra.
A morphism f : ξ → ζ of Jordan bundles ξ and ζ is a morphism of the underlying vector bundles such that for every x ∈ X, fx : ξx → ζx is a Jordan algebra homomorphism. If f is bijective and f−1 is continuous, then f is called an isomorphism.
By a subalgebra (ideal) bundle of a Jordan algebra bundle ξ = (ξ, p, X), we mean a vector sub-bundle ξ′ = (ξ′, p, X) of ξ such that each fibre (ξ′)x is a subalgebra (ideal) of ξx, ∀ x ∈ X.
By a semisimple Jordan algebra bundle, we mean a Jordan algebra bundle in which each fibre is a semisimple Jordan algebra.
If ξ is a Jordan algebra bundle with a nontrivial multiplication θ : ξ ⊗ ξ → ξ inducing the Jordan algebra bundle structure and if ξ has no ideal bundles except itself and the zero bundle, then we call ξ a simple Jordan algebra bundle.
A Jordan algebra bundle ξ is said to be the direct sum of the ideal bundles ξ1, ξ2, …, ξn provided, ξ = ξ1 ⊕ ξ2 ⊕⋯ ⊕ ξn
3. Semisimple Jordan algebra bundles
A Jordan algebra bundle ξ = (ξ, p, X) is locally trivial if, for every open set U ⊆ X, there exists a Jordan algebra J, together with a (vector bundle) trivialization φ : ξ|U → U × J, which is fibrewise a homeomorphism of Jordan algebras.
Not every Jordan algebra bundle is locally trivial, as the following example shows.
Let J be a nonzero real Jordan algebra and the field of real numbers. Then is a trivial vector bundle. Let θ : ξ ⊗ ξ → ξ be defined by, θ(t, (u, v)) = t(uv) for all , u, v ∈ J. Then, θ is continuous and induces a Jordan algebra structure on each fibre ξx = x × J. Hence, ξ is a Jordan algebra bundle. But ξ is not locally trivial because ξt for t ≠ 0 is a nonzero algebra whereas ξ0 is the zero algebra.
From Atiyah [8, p. 4], we reproduce the following things without any changes. Suppose that V and W are vector spaces and that E = X × V and F = X × W are the corresponding product bundles. Then ϕ : E → F determines a map Φ : X → Hom(V, W) by formula Φ(x)(v) = ϕ(v). Moreover, if we give Hom(V, W) its usual topology, then Φ is continuous; conversely, any such continuous map Φ : X → Hom(V, W) determines a homomorphism ϕ : E → F.
Every semisimple Jordan algebra bundle is locally trivial.
Proof. Let ξ be a semisimple Jordan algebra bundle. The local triviality of ξ as a vector bundle is given by the vector bundle isomorphism α : U × V → p−1(U). Then θ : ξ ⊗ ξ → ξ induces the morphism given by,
Then, from above Remark (3.3), defines a continuous mapping from U to M ⊆ Hom(V × V, V), the space of all Jordan multiplications defined on V endowed with the subspace topology. Since the Jordan algebra is semisimple, J0 is rigid by [9, Corollary 1.3]. That is the orbit
with respect to the Lie group G = Aut(V) is open in M.
Let . Then, for each x ∈ U′, there exists a gx in G such that
Further, since G and G(x0) satisfy the hypothesis of Aren’s theorem [10] G/G0 is homeomorphic to G(x0), where G0 is the stability subgroup corresponding to . Also G → G/G0 is a principal bundle [11, p. 33] together with a local cross section given by gG0 → g [12, p. 126]. Hence, the map x↦gx becomes the composition of the continuous maps
Therefore, we can define the vector bundle isomorphism ϕ : U × J0 → p−1(U) by ϕ(x, v) = αx(gx(v)). The map ϕ preserves the Jordan multiplication
Hence, ϕ gives the required local triviality of the Jordan algebra bundle ξ.□
4. Decomposition theorem for Jordan algebra bundle
Let (X × V, q, X) be a trivial vector bundle and (X × J, p, X) a trivial semisimple Jordan bundle. Suppose ϕ : X × V → X × J is a vector bundle monomorphism such that for each x ∈ X, ϕ(x, V) is an ideal in J. Then, there exists a finite open partition ∪iXi = X such that ϕx(V) = ϕy(V) for x, y ∈ Xi. In particular, if X is connected, for all x, y ∈ X, ϕx(V) = ϕy(V).
Proof. The map ϕ : X × V → X × J being a vector bundle morphism, x↦ϕx is a continuous map from X to Hom(V, J), the vector space of all linear transformations from V to J. If denotes the collection of all distinct ideals of J whose dimension is equal to that of V, then by the semisimplicity of J, is a finite set [13, Corollary.4.6, p. 98]. Let and let Xi = {x ∈ X| ϕx(V) = Ii} for i = 1, 2, …, n. Let x ∈ X. Since ϕx(V) is an ideal in J, and dim ϕx(V) = dim V, we have ϕx(V) = Ii for some i. Thus, x ∈ Xi, and consequently X = ∪iXi.
It is enough to prove that each Xi is open in X. Let be the vector subspace . Then is a closed subset of Hom(V, J) being a linear subspace of the vector space Hom(V, J). Since x↦ϕx is continuous and is closed in Hom(V, J), is closed in X.
To prove:
. Now if , . Then, ϕx(V) ⊆ Ii. But since dim ϕx(V) = dim Ii, ϕx(V) = Ii. So x ∈ Xi. Thus, .
Consider and let i ≠ j
Therefore,
Thus, are all disjoint collection of closed sets. Hence, Xj is the complement of ∪i≠jXi. So Xj is open in X. We here note that if X is connected, then for all x, y ∈ X, ϕx(V) = ϕy(V). □
Let ξ′ be an ideal bundle of a semisimple Jordan algebra bundle ξ. Then, is an ideal bundle of h(ξ) where
Proof. Each is an ideal in h(ξx). It is enough to prove that h1(ξ′) is a vector bundle. Since the bundle ξ is semisimple and ξ′ is semisimple being its ideal bundle, ξ and ξ′ are locally trivial by Theorem (3.4). So we have Jordan algebras J and J′ and Jordan bundle isomorphisms Φ : U × J → ∪x∈Xξx and . Then, Φ−1Ψ : U × J′ → U × J is a vector bundle monomorphism satisfying the hypotheses of Lemma (4.1). Hence, there exists a finite open cover {Ui} for U such that (Φ−1Ψ)(x, J′) = (Φ−1Ψ)(x′, J′) for all x, x′ ∈ Ui. Consequently, the neighbourhood U can be shrunk so that there exists an ideal I of J and such that Φ maps U × I onto . Let h1(I) = {T ∈ h(J)| T(J) ⊆ I}. Given T ∈ h1(I), consider .
Hence, . That is, g(Φ) maps U × h1(I) onto . That is, h1(ξ′) is an ideal sub-bundle of h(ξ).□
Every semisimple Jordan algebra bundle can be uniquely written as the direct sum of simple ideal bundles.
Proof. Let ξ be a semisimple Jordan algebra bundle. Each fibre ξx has a unit element being a semisimple Jordan algebra [13, Theorem 4.7, p. 99]. Hence, the corresponding Lie algebra bundle L(ξ) exists [1]. The semisimplicity of ξx implies that of L(ξx) [7, p. 805] and so L(ξ) is a semisimple Lie algebra bundle. Then, L(ξ) can be written as follows:
where each Li is a simple ideal bundle of L(ξ) [14]. Also each Li is of the form , [7, Lemma 1, p. 789]. Each is an ideal bundle of ξ and each ξi is simple since Li is simple. Hence, ξ = ξ1 ⊕ ξ2 ⊕⋯ ⊕ ξn.
Let us prove the uniqueness of the decomposition. Let, ξ be expressed as follows:
where each ξi and are simple ideal bundles. Consider , where . We have . Let L(a) ∈ h(ξx), then a can be written as . Since is an ideal in ξx, L(ai) maps ξx into . So
Let [L(a), L(b)] ∈ h(ξx), where and , . Since for i ≠ j, we obtain that
Consequently, . Therefore, h(ξ) = h1(ξ1) ⊕ h1(ξ2) ⊕⋯ ⊕ h1(ξn). Similarly, . Hence
where L1(ξi) = h1(ξi) ⊕ ξi ⊕ ξi and are ideal bundles of L(ξ) by Lemma (4.2). Then by [14, Theorem 2.8], we obtain that m = n and L1(ξi) coincides with one of the . Then, obviously coincides with ξi except for the order.□
The author would like to thank the referee for constructive remarks that improve the presentation of the paper and for spotting several errors in the previous version of the paper. Also, the author would like to thank REVA University for its continuous support and encouragement.
