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0Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2014, Article ID 690806, 11 pages http://dx.doi.org/10.1155/2014/690806
Research Article
Quasi-Jordan Banach Algebras
Reem K. Alhefthi, Akhlaq A. Siddiqui, and Fatmah B. Jamjoom
Department of Mathematics, College of Science, KingSaud University, P.O. Box 2455-5, Riyadh 11451, Saudi Arabia Correspondence should be addressed to Fatmah B. Jamjoom; fatmahj@yahoo.com Received 16 November 2013; Accepted 14 January 2014; Published 25 March 2014 Academic Editor: Marco Sabatini
Copyright © 2014 Reem K. Alhefthi et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We initiate a study of quasi-Jordan normed algebras. It is demonstrated that any quasi-Jordan Banach algebra with a norm 1 unit can be given an equivalent norm making the algebra isometrically isomorphic to a closed right ideal of a unital split quasi-Jordan Banach algebra; the set of invertible elements may not be open; the spectrum of any element is nonempty, but it may be neither bounded nor closed and hence not compact. Some characterizations of the unbounded spectrum of an element in a split quasiJordan Banach algebra with certain examples are given in the end.
1. Introduction
Lo oking b ack at the development of mo dern mathematics, we see that early formal study of algebra was mostly commutative and associative. With an abstract study of functions and matrices, it became noncommutative but still associative; then introduction of nonassociative structures, such as Lie structures due to Sophus Lie [1] and Jordan structures due to Jordan [2], has led us to the mathematics which at present is noncommutative as well as nonassociative.
There is a strong relationship between Lie algebras and Jordan algebras [3]. Jordan structures have been extensively studied by a large number of mathematicians: P. Jordan, von Neumann, E. Wigner, N. Jacobson, K. McCrimmon, R. Braun, M. Koecher, E. Neher, and O. Loos, to name but a few. A vast literature containing many important results on Jordan algebras has been developed (cf. [3, 4]). After the mid-1960s, people began studying Jordan structures from the point of view of functional analysis. Interesting theories of Jordan Banach algebras, /B-algebras, JB*-algebras, and JB*-triples have been developed, which closely resemble that of C*-algebras and have found surprisingly important applications in a wide range of mathematical disciplines including analysis, physics, and biology (cf. [5-7]).
A Jordan algebra is a nonassociative algebra J with the product x o y satisfying x ° y = y ° x and the Jordan identity: (x2 o y) o x = x2 o (y o x), where x2 = x o x. Any associative algebra A becomes a Jordan algebra A+ with the same linear
spacestructureandtheJordanproductxoy := (1/2)(xy+yx); it becomes a Lie algebra under the skew-symmetric product [x, y] := xy - yx, so called the Lie bracket (cf. [4]). For any Jordan algebra J, there is a Lie algebra L(J) such that J is a linear subspace of L(J) and the product of J can be expressed in terms of the Lie bracket in L(J). Moreover, the universal enveloping algebra of a Lie algebra has the structure of an associative algebra; see the original work due to Kantor, Koecher, and Tits appearing in [8-10].
Loday introduced a generalization of Lie algebras, called the Leibniz algebras [11, 12], and successfully demonstrated that the relationship between Lie algebras and associative algebras can be translated into an analogous relationship between Leibniz algebras and the so-called dialgebras (cf. [13]): a dialgebra over a field K is a K-module D equipped with two bilinear associative maps H, h: D x D ^ D satisfying x H (y h z) = x H (y H z); (x h y) H z = x h (y H z); and (x H y)hz = (x hy) hz. The maps H and h are called the left product and the right product, respectively. Any dialgebra (D, H, h) becomes a Leibniz algebra under the Leibniz bracket [x, y] := x H y - yhx, and the universal enveloping algebra of a Leibniz algebra has the structure of a dialgebra; for details, see [12, 13].
Recently in [14], Velasquez and Felipe introduced the notion of quasi-Jordan algebras, which have relation with the Leibniz algebras similar to the existing relationship between the Jordan algebras and the Lie algebras [15]. The quasiJordan algebras are a generalization of Jordan algebras where
the commutative law is replaced by a quasicommutative identity and a special form of the Jordan identity is retained. These facts indicate the significance of studying the quasi-Jordan algebras; within a few years time, many mathematicians, including M. K. Kinyon, M. R. Bremner, L. A. Peresi, J. Sanchez-Ortega, and V. Voronin, have got their interests in this new area. In [16], Felipe made an attempt to study dialgebras from the functional analytic point of view.
A Jordan Banach algebra is a real or complex Jordan algebra with a complete norm || • || satisfying \\xy\\ < ||x||||y||; basics of Jordan Banach algebras maybe seen in [7]. In this paper, we initiate a study of the quasi-Jordan normed algebras. The class of complete quasi-Jordan normed algebras, called quasi-Jordan Banach algebras, properly includes all Jordan Banach algebras and hence all C*-algebras (cf. [7]). This study may provide a better mathematical foundation for some important areas such as quantum mechanics. We are interested specially in extending, as much as possible, the theory of Jordan Banach algebras to the general setting of quasi-Jordan Banach algebras. We investigate the notions of invertibility and spectrum of elements in the setting of unital quasi-Jordan Banach algebras. Among other results, we demonstrate that any quasi-Jordan Banach algebra I with a norm 1 unit can be given an equivalent norm that makes the algebra I isometrically isomorphic to a norm closed right ideal of a unital split quasi-Jordan Banach algebra. We show that the set of invertible elements in a unital quasi-Jordan Banach algebra generally is not open and that the spectrum of any element is nonempty but it may be neither bounded nor closed, hence not compact. Moreover, if the spectrum of an element in a complex unital split quasi-Jordan Banach algebra (see below) is unbounded then it coincides with the whole complex plane and vice versa. Some examples are also given in the end.
2. Quasi-Jordan Banach Algebras
We begin by recalling some basics of the quasi-Jordan algebra theory from [14,17]. A quasi-Jordan algebra is a vector space I over a field K of characteristic = 2 equipped with a bilinear map <: I x I ^ I, called the quasi-Jordan product, satisfying x < (y < z) = x < (z < y) (right commutativity) and (y < x) < x1 = (y < x2) < x (right Jordan identity), for all x,y,z e I. Here, x := x < x and x" := xn~ < x for n > 2; in the sequel, we will see that x2 < x may not coincide with x < x . Every quasi-Jordan algebra I includes two important sets: Iann := the linear span of the elements x < y - y < x with x,y e I and Z(I) := [z e I : x < z = 0,Vx e I}, respectively, called the annihilator and the zero part of I. It follows from the right commutativity that Iann c Z(I) and that the quasi-Jordan algebra I is a Jordan algebra if and only if Iann = [0}.
Example 1. Let (D, H, h) bea dialgebra over a field K of characteristic = 2. One can define another product <1: DxD ^ D by x < y := (\/2)(x H y + y h x) for all x,y e D, which satisfies the identities x < (y < z) = x < (z < y); (y < x) < x2 = (y < x2) < x; and x2 < (x < y) = x <
(x2 < y); however, this quasi-Jordan product generally is not commutative. Therefore, (D, <) is a quasi-Jordan algebra, which may not be a Jordan algebra (cf. [14]). This quasiJordan algebra is denoted by D+, called a plus quasi-Jordan algebra. Any quasi-Jordan algebra which is a homomorphic image of a plus quasi-Jordan algebra is called special.
We define a quasi-Jordan normed algebra as a quasi-Jordan algebra (I, <) over the field C of complex numbers endowed with a norm || • || satisfying \\x < y\\ < ||x||||y||, for all x e I. Thus,thequasi-Jordanproduct"<"inanormedquasi-Jordan algebra I is continuous. A quasi-Jordan normed algebra is called a quasi-Jordan Banach algebra if it is complete as a normed space.
If there is an element e in a quasi-Jordan algebra I satisfying e < x = x for all x e I, called the left unit, then I becomes commutative and hence a Jordan algebra. Due to this fact, we will consider only the right unit elements; henceforth, unit in a quasi-Jordan algebra would mean a right unit. An element e in a quasi-Jordan algebra I is called a unit if x < e = x, for all x e I.A quasi-Jordan algebra may have many (right) units (cf. [15,17]). If the dialgebra D has a bar-unit e (i.e., x H e = x = ehx) then x < e = (\/2)(x H e + e h x) = x, for all x e D, and hence e is a unit of the plus quasi-Jordan algebra D+.
For any quasi-Jordan algebra I with unit e, we know from [17]that Iann and Z(I) are two-sided ideals of I, Iann = [x e I : e < x = 0} = Z(I), and U(I) = [x + e : x e Z(I)}, where U(I) denotes the set of all (right) units in I.
One can always attach a (two-sided) unit to any Jordan algebra by following the standard unitization process. This unitization process no longer works for quasi-Jordan algebras (cf. [17, pages 210-211]). Adding a unit to a quasi-Jordan algebra is yet an open problem. In giving a partial solution to this unitization problem, Velasquez andFelipe [17]introducedthe following special class of quasi-Jordan algebras, called split quasi-Jordan algebras: let I bea quasi-Jordan algebra and let I be an ideal in I such that Iann c I c Z(3).Wesaythat I is a split quasi-Jordan algebra (more precisely, I is split over I) if there exists a subalgebra J of I such that I = /©/, the direct sum of J and I. It is easily seen that such a subalgebra J is a Jordan algebra. One can attach a unit to any split quasi-Jordan algebra; details of such a unitization process are given in [17]. If the algebra I has a unit then Iann = Z(I). Thus, a quasiJordan algebra I with unit is a split quasi-Jordan algebra if and only if I = J ® Z(I) for some subalgebra J of I; the algebra J is called the Jordan part of I. In such a case, each element x e I has a unique representation x = Xj + xz with Xj e J and e Z(3),respectively, calledthe Jordan part and the zero part of x. Moreover, if I is split quasi-Jordan algebra with unit e, then there exists a unique element ej e J which acts as a unit of the quasi-Jordan algebra I and at the same time a unit of the Jordan algebra /.
Proposition 2. Let I := J ® Z(I) be a unital split quasiJordan Banach algebra with unit e e J. Then the Jordan part J is a norm closed subalgebra of I and hence a unital Jordan Banach algebra.
Proof. From the above discussion, it is clear that the Jordan part J is a unital Jordan normed algebra. Next, let {xn} be any fixed Cauchy sequence in /. Then, the same {xn} is Cauchy sequence also in the quasi-Jordan Banach algebra I since J is a normed subspace of I. Hence, xn ^ x for some x e I.
Now, since each xn e J and since e is the unit of the Jordan algebra J under the (restricted) product "<," we get e < xn = xn for all n. So, by the continuity of the product, xn = e < xn ^ e < x. Hence, by the uniqueness of the limit of any convergent sequence in a normed space, e < x = x. However, x = e < (xj + xz) = e < Xj + e < xz = e < Xj + 0 = Xj e J. Thus, the required result follows. □
We know from [17] that for any quasi-Jordan algebra I, I := {(x, Ry) : x,y e I} equipped with the sum (x, Ry) + (a,Rh) := (x + a, Ry+b), scalar multiplication X(x,Ry) := (\x,RXy), and product (x,Ry) < (a,Rh) := (x < b,Ry<b) is a split quasi-Jordan algebra and the map x ^ <p(x) := (x,Rx) is an embedding of I into 3j, where Rz stands for the usual right multiplication operator on I. It is easily seen that {0} x R(I) is the Jordan part of I and the zero part Z(11) = I x {0}. Moreover, the embedding f preserves the units: clearly, (a,Rh) < (e,Re) = (a < e,Rb<e) = (a,Rh) for all (a, Rb) e I1 so that (e, Re) is a unit in I1 whenever e is a unit in I. In fact, (x, Re) is a unit in I1, for all x e I;it may be noted here that (0, Re) is the only unit in the Jordan part of
From [17], we also know that R(I) := {Rx : x e I}, with product "•" defined by Rx • Ry = Rx<y, for all x,y e I, is a quasi-Jordan algebra. Moreover, 11 is the direct product of the quasi-Jordan algebras I and R(I).
Indeed, the split quasi-Jordan algebra 11 is a quasi-Jordan Banach algebra with unit (e, Re) whenever I is a quasi-Jordan Banach algebra with a norm 1 unit e, and that f(I) = {(x,Rx) : x e I} is a closed unital quasi-Jordan normed subalgebra of 11. To justify this claim, we need the following result.
Proposition 3. Suppose I is a quasi-Jordan Banach algebra with a norm 1 unit e. Then the algebra R(I) as above is a quasiJordan Banach algebra with unit Re of norm 1.
Proof. Clearly, each Rx is a bounded linear operator with \\RXW < IM|. Hence, the usual operator norm is a norm on the quasi-Jordan algebra R(I) with the quasi-Jordan product Rx • Ry = Rx<y. Further, we observe that
\\Rx • Ry\\ - ll-^yll
= sup \\Rx<y (z)\\
ze3,||z||=1
- sup \\z < (x < y)\\
ze3,||z||=1
- sup \\z < ((e < x) < y)\
ze3,||z||=1
- sup \\z<(Ry (Rx (e))}\\
ze3,|z|=1
< sup \\z\\ (Rx (e))\ \
ze3,||z||=1
-K (Rxe)\\ < \\«,\\ \k\\ \\e\\
- \\^*\\ \\Ry\\ •
Alternately, by exploiting the right commutativity of the quasi-Jordan product in I, we get
Rx '^yW- sup
ze3,||z||=1
ze3,||z||=1
ze3,|z|=1
ze3,|z|=1
ze3,|z|=1
ze3,||z||=1
z63,|z|=1
z < (x < y)\\ z < ((x < e) < (y < e z < ((x < e) < (e < z < ((e < y) < (x < e z < ((e < y) < (e < x z<(Ry (e)<Rx (e)}\
z\\ \\Ry\\ \ \ Rx\\ \\er - \\ Rx\\ \ \ Ry
Thus, R(I) together with the operator norm is a quasiJordan normed algebra. Moreover, for any x e R(I), we have Rx • Re = Rx<e = Rx and 1 = \Ie\I > =
supo = zeg(||ae(z)||/||z||) > IIRe(e)II/IIe\I = 1; that is, Re is a norm 1 (right) unit in R(I).
Suppose {Rx } is any fixed Cauchy sequence in R(I). Then, for any fixed a e I, Ha < xm - a < xnH = (a) -
Rx„(*)II = II(RXn - Rx„)(«)I < IIRXn - Rx„IMI ^ 0 as
m,n ^ >x>, and so {a < xn} is a Cauchy sequence in I. But, I is complete. Hence, the sequence {a < xn} for any a e I is convergent in I. In particular, the sequence {e < xn} converges to some ye I. Moreover, the Cauchy sequence {Rx } converges to Ry e R(I) in the operator norm because II(RiXn - Ry)(z)II = Hz < (x„ - y)II = Hz<xn -z< yH = Hz < (xn < e) - z < yH = Hz < (e < xn) - z < yH = Hz < (e < xn - y)H < HzIIHe < xn - yH ^ 0 as n ^ >x>, for all z e I. Thus, the quasi-Jordan normed algebra R(I) is complete.
Now, we show that the corresponding algebra 11 is a unital split quasi-Jordan Banach algebra.
Proposition 4. Let I be a quasi-Jordan Banach algebra with a norm 1 unit e and let 11 be as above. Then 11 is the direct product of the quasi-Jordan algebras I and R(I). Moreover, 11 equipped with the product norm ||(x, Ry)H := ||x|| + ||.Ry|| is a split quasi-Jordan Banach algebra with unit (e,Re), and f(I) = {(x, Rx) : x e I} is a closed right ideal in 11 in the norm topology.
Proof. For the first part, see [17]. Clearly, ||(x, .Ry)|| := ||x|| + ||-Ry|| is a norm, and it satisfies
||(x,.Ry)<(z,.Rj| = |(x<w,.Ry<m)|
= ^ (*)|| + |k, • KJ
<(|M| + |K||)|K|| <(M + ||aJ)(||z|| + M = ||M,)|| ||MJ||,
for all x, y, z e 3. Keeping in view Proposition 3, we deduce that 31 being the product of complete spaces 3 and £(3) is complete in the product norm. Thus, 31 is a split quasi-Jordan Banach algebra with (right) unit (e, £e).
Clearly, <p(3) is a subspace of 31 with (x, < (y, £z) = (% < z, £*<z) e <p(3) for all (%, e <p(3), (y, £z) e 3^so that <p(3) is a quasi-Jordan normed subalgebra with the unit (e, £e) included. Further, let |(xn, )} be any fixed Cauchy sequence in <p(3). Then ||%m - xj| < ||%m - xj| + ||.R* -
|| = ||(*m - - )| = ) - (*».**„
0 as m, n ^ œ, so that |xn} is a Cauchy sequence in (the complete space) 3 and hence it converges to some x e 3. Now, by using the fact || .Rz ||<|| z || for all z e 3 (since e is a norm 1 unit in 3; see above), we get the convergence of arbitrarily fixed Cauchy sequence |(xn, .R* )} to (x, e <p(3) since ||(x„,.R*n)-(%,.Rj|| = ||(%„-x,.R*n_"j|| = ||%„-%|| + ||-R* _*|| < 2||%„-%|| ^ 0 as n ^ œ. Thus, <p(3) being complete is a closed right ideal in 31 in the norm topology. □
Then, x has an inverse y with respect to e ^ y < x = e+e<(x) and y < x2 = x + e<(%) + e<(x2).
We know from the above discussion that the embedding x ^ <p(x) := (%, of a quasi-Jordan algebra I into the split quasi-Jordan algebra 3j preserves the units. The embedding <p also preserves the corresponding invertible elements: if y is an inverse of x with respect to a unit e in 3, then y < x = e + (e < x - x) and y < x2 = x + (e < x - x) + (e < x2 - x2). Hence,
= (e + (e < x - %), + £e<x -+ ((e<x2,^e<x2 ))
= (*.-R*) + (Me)<(*.-R«)-(*.-Rj)
+ (Me)<(x,^)2 -MJ2).
Next, we observe the isometry between 3 and <p(3).
Proposition 5. Let 3 foe a quasi-Jordan Banach algebra with norm 1 unit. Then there exists an equivalent norm that makes 3 isometrically isomorphic to a quasi-Jordan closed subalgebra <p(3) of the split quasi-Jordan Banach algebra 3j.
Proof. Clearly, ||%||0 := ||%|| + ||.R,J defines a norm on the quasi-Jordan algebra 3. It follows that (3, || • ||0) is a quasiJordan Banach algebra. Moreover, 3 is isomorphic to the subalgebra |(x, : x e 3} of 3i under the isomorphism <p : x ^ (x, ), as seen above. Further, we observe that <p is an isometry since ||%||0 = ||%|| + ||-RJ = ||(^,-RJH = ||^(*)||. Finally, we note that the two norms || • || and || • ||0 on 3 are equivalent since ||%|| < ||%|| + ||-RJ = |ML = H*H + H-RJ < 2||%||. □
3. Invertible Elements
As in [17], an element x in a quasi-Jordan algebra 3 is called invertible with respect to a unit e e 3 if there exists ye 3 such that y<x = e+(e<x-x) and y < x2 = x + (e < x - x) + (e < x2 - x2); such an element y is called an inverse ofx with respect to e. Let e<(%) denote the element e < x - x.
Thus, (y, is an inverse of (%, RJ, with respect to the unit (e, .Re),in 3j.
Let 3 = 7 ® Z(3) be a unital split quasi-Jordan algebra. Then, (y) = y<x = y<(xj + = y<Xj + =
y < Xj + 0 = R (y), for all 3. Thus, = for all x£ 3.
In this section, we demonstrate that the set of invertible elements, with respect to a fixed unit, in a quasi-Jordan Banach algebra, may not be open. For this, we proceed as follows.
Proposition 6. Let 3 := 7 ® 2(3) be a unital split quasiJordan algebra with unit e e J. If x e 3 is invertible with respect to e, then so is Ax for all A = 0.
Proof. Let y be an inverse of x in 3 with respect to e. We show that y' := (1/A)^j + is an inverse of Ax with respect to e.
Observe that
< * + < ^ = (j/ + < *
= y < x = e + e< (x) = e + e< (x/ + %z) = e + (-xz),
2 2 2 yj < X + yz < X = y < X
= x + e< (Xj + xz) + e< ((.Xj + XZ)2) = Xj + (-xz < x).
Hence, by the uniqueness of the representation as sum of Jordan and zero parts, we get yj < x = e, yz < x = -xz, yj < x2 = Xj, and yz < x2 = -xz < x. Therefore, y < Xx = X(((!/X)yj + yz) < x) = yj < x + Xyz < x = e - Xxz = e + e<(Xx) and y' < (Xx)2 = X2((l/X)yf < x2 + yz < x2) = Xxj - X2xz < x = Xx + e4(Xx) + e4((Xx)2) because e4(Xx) = e < (Xx) - Xx = -Xxz and e4((Xx)2) = e < X2x2 - X2x2 = X2(e<(x2)) = -X2xz < x. □
Proposition 7. Let I be a quasi-Jordan normed algebra with a unite.LetGe(1) := {x e I: x is invertible with respect to e} be an open set and x e Ge(1). Then x + z e Ge(1) for all z e Z(I).
Proof. Suppose x e Ge(I) and z e Z(I). Ifz = 0 then x+z = x e Ge(1). Next, suppose z = 0. Since Ge(1) is an open set, there exists e > 0 such that a e Ge(1) whenever \\x - a\\ < e. Hence, x + z0 e Ge(1) with z0 = (e/2\\z\\)z.
Let y and y0 be inverses of x and x + z0 with respect to e, respectively. Then
<(x + z„) = e + e< (x + z„) = e + e< (x)- z„ = y < x - z0,
y0<(x + = x + za+e< (x + za) + e< ((x + z0)2) ^ ) = x + e< (x) + e< (x2) - z„ < x
= y < x2 - z0 < x.
Hence, by setting y1 = y + <x(y„ - y) with a = 2\\z\\/e, we see that
y1 < (x + z) = (y + a (y0 - y)) < x
= y < x + a (y0 < (x + z0) - y < x) = e + e< (x) + a (-z„) = e + e< (x) - z = e + e< (x + z),
y1 <(x + z)2 = (y + a (y„- y)) <
= y < x2 + a (y0 < (x + z„)2 - y < x2) = x + e< (x) + e< (x2) - a.(z„ < x) = x + e< (x) + e< (x2) - z < x = (x + z) + e< (x + z) + e< ((x + z)2) ,
since z e Z(I). Thus, y1 is an inverse of x + z with respect to the unit e. □
Corollary 8. Under the hypothesis of Proposition 7, Xe - x e Ge(1) implies Xe-(x + z) e Ge(I) for all z e Z(I).
Proposition 9. Let I = J ® Z(I) be a split quasi-Jordan normed algebra and let e e J beaunitin I such that the set Ge(1) is open. Then x < x2 = x2 < x for all x e I.
Proof. Of course, the element e is the unit of the Jordan algebra /. Then, for any fixed element a e J, there exists X e C such that Xe - a is invertible in J; otherwise, we would get the negation of the well-known fact that spectrum of an element of a unital Jordan algebra is bounded. That is, there exists y e J such that y < (Xe - a) = e and y < (Xe-a) = (Xe-a).However,e<(a) = 0 = e<(a2).Hence, y is an inverse of a in the quasi-Jordan algebra I with respect to the unit e. By Corollary 8, Xe - (a + z) is also invertible for any z e Z(I). This in turn gives the existence of be I satisfying b < (Xe -(a + z)) = e + z and b < (Xe - (a + z)) = (Xe -a)+z < (Xe - a). Multiplying the first equation from the right by (Xe - (a + z)) , the second equationby (Xe-(a + z)), and using the right Jordan identity, we get
(e + z) < (Xe - (a + z))
= ((Xe - a) + z < (Xe - a)) < (Xe - (a + z)).
Hence,
(e + z) < (Xe - a) = ((Xe - a) + z < (Xe - a)) < (Xe - a),
I that
e < (Xe - a)2 + }?z - 2X(z < a) + z < a = (Xe - a)2 + }?z - 2Xz < a + (z < a) < a.
This last equation reduces to e < (Xe - a)2 = (Xe - a)2 since (Xe - a) e J and e is the unit of /. Hence, z < a = (z < a) < a for all a e J and z e Z(I). Now, for any x e I, the last equation with a = Xj and z = xz gives xz < Xj = (xz < Xj) < Xj. Thus, X < X2 = Xj < X2 +xz < x2 = x2 < Xj + (xz < Xj) < Xj = x2 < x for all x e I. □
Corollary 10. If a unital split quasi-Jordan algebra has an element x with x < x = x < x then the set of invertible elements, with respect to the unit of the Jordan part, is not open.
In the sequel, we will show the existence of a unital split quasi-Jordan Banach algebra with elements x such that x < x2 = x2< x. Thus, the above result establishes that the set of invertible elements, with respect to a fixed unit, in a quasiJordan Banach algebra may not be open.
4. The Spectrum of Elements in a Unital Quasi-Jordan Algebra
As usual, we define the spectrum of an element x in a unital quasi-Jordan algebra (I,e), denoted by 0(g>e)(x), to be the collection of all complex numbers A for which Ae -x is not invertible. Thus, (x) := {Я e C : Ae -x is not invertible}. Here, the subscript e indicates that the invariability depends on the choice of unit e, which generally is not unique.
Proposition 11. Let I be a unital quasi-Jordan algebra. Then a(3,e)(e') = {1} for all e, e' e U(I).
Proof. Let e,e' e U(3). Then, for any A = 1, у := (1/(A- 1))e' is an inverse of Ae - e' with respect to the unit e because у < (Ae - e') = (1/(A - 1))e' < (Ae - e') = e + e<(Ae - e') and
у < (Ae - e')2 = (Ae - e') + e<(Ae - e') + e<((Ae - e')2). □
Proposition 12. Let I be a quasi-Jordan algebra with unit e. Then <T(gje)(z) = {0} for all z e Z(3).
Proof. For any fixed z e Z(3) and nonzero scalar A, the vector у := (1/A)(e + z) satisfies у < (Ae -z) = Ay = e + z = e + e< (Ae - z) and у < (Ae - z) = A2y = Ae + Az = (Ae - z) + e<(Ae - z) + e<((Ae - z)2). So (1/A)(e + z) is an inverse of Ae - z with respect to e. This means A £ CT(ge)(z) for all A = 0. However, the zero vector is not invertible. Thus, CT(3e)(z) = {0}. ' □
Proposition 13. Let I = J® Z(3) be a split quasi-Jordan algebra with unit e e /. Then 0(g>e)(p) С {0,1} for all idempo-tents p e I (i.e., p2 = p).
Proof. Let p be any fixed idempotent in I. Since p e I, p has a unique representation p = pj + p2 with pj e / and p2 e
Z(3). Clearly, p^ + Pz < Р/ = p2 = P = Р/ + Pz. Then, by uniqueness of the representation in the split quasi-Jordan algebra I, p2 < pj = p2 and p^ = pj; this means pj is an idempotent in the Jordan algebra /. Hence, <Гд>е) (pj) с {0,1}. Thus, рл := Ae - pj is invertible in / with the unique inverse p-1, for all A £ {0,1}.
We show that у := p-1 + (1/(A - 1))p2 is an inverse of Ae - p in I with respect to the unit e; for this, we note that
Ae - P = Рл - Pz, (Рл - Pz)2 = Рл - Pz < Pл, e<(Ae -P) = е<.(Рл - Pz) = e < (Рл - Pz) - (Рл - Pz) = Pz, and e<((Ae - p)2) = ед((рл - p2)2) = - p2 < рл) = p2 < Рл = Pz < (¿e-p;) = Ap2-p2 < p; = Ap2-p2 = (A- 1)pz. Hence, у < (Ae - p) = у < (рл - pz) = у < рл = (рл1 +
(1/(A - 1))p2) < рл = p-1 < Рл + (1/(А- 1))pz < Рл = е + (1/(А - 1))р2 < (Ae - р;) = е + (1/(А - 1))(Ар2 - р2 < р;) = е + (1/(А - 1))(Ар2 - pz) = е + Pz = е + е<(Ае - р)
and у < (Ае - р)2 = у < (р| - р2 < Рл) = У < Рл = (р-1 +(1/(А-1))р2) < р| = рл + (1/(А-1))р2 < (Ae-pj)2 = рл + (1/(А-1))р2 < (А2е-2Ар; + р;2) = рл + (1/(А-1))р2 < (А2е - 2Apj + р;) = рл + (1/(А - 1))(А2р2 < е - 2Ар2 < Р/ + Pz < Р/) = Рл + (1/(А - 1))(А2р2 - 2Ар2 + р2) =
pA + (1/(A-1))(A-1)2p2 = Ae - pj + (A-1)p2 = (Ae-p) + e<(Ae-p) + e<((Ae-p)2). □
As mentioned in Section 2, if I is a quasi-Jordan algebra with a unit e then the set je + z : z e Z(3)| coincides with the set U(3) of all units in I.
Proposition 14. Let I = / ® Z(3) be a «rata/ split quasiJordan algebra with unit e e / and x e I invertible with respect to some e' e U(I). Then Xj is invertible, with respect to the unit e, in the Jordan algebra J.
Proof. Clearly, e <1 x - x = -x2 and e <1 x2 - x2 = -x2 < Xj. Since e' e U(I), we have e' := e + z for some z e Z(3). Hence, the invertibility of x in I, with respect to the unit e', gives the existence of ye I such that yj <1 Xj + y2 <1 Xj = = y<x = e + z+(e + z)<lx-x = e + z + z<lx-x2 and yj <1x2 + y2 < ^2 = y < = y <1 x2 = (e + z) < x + (e + z) <1 x2 - x2 = Xj + z<x + z<x2 - x2 <1 Xj.So, by the uniqueness of the representations in the split algebra I, we get < Xj = e and <1x2 = Xj. Thus, is the inverse of Xj in the Jordan algebra /. □
Next, we observe that the spectrum of x in a unital split quasi-Jordan algebra with respect to any unit includes the spectrum of Xj in the Jordan part /.
Corollary 15. Let I = /®Z(3) be a unital split quasi-Jordan algebra with unit e e /, and let x e I. Then CT(je)(xj) c <T(3,e')(x) for all e' e U(I).
Proof. Let e' := e + z with z e Z(3), and let A £ a(g e/)(x). Then, Ae' - x is invertible in I with respect to the unit e'. Hence, its Jordan part Ae-Xj is invertible in the Jordan algebra / by Proposition 14. Thus, A £ Oy^Xj). □
It is well known that the spectrum of any element in a unital Jordan Banach algebra is nonempty (cf. [7]). This together with Proposition 2 and Corollary 15 gives the following result.
Corollary 16. The spectrum of any element in a unital split quasi-Jordan Banach algebra is nonempty.
The next result extends Corollary 16 to any quasi-Jordan Banach algebra with a norm 1 unit.
Proposition 17. The spectrum of any element in a quasiJordan Banach algebra with a norm 1 unit is nonempty.
Proof. Let I be a unital quasi-Jordan Banach algebra with a norm 1 unit e. From Section 2, we know that the map <p : x ^ <p(x) = (x, .R.J embeds I into the unital split quasi-Jordan Banach algebra := j(x, .Ry) : x, y e I}, equipped with the sum (x, .Ry) + (a, .Rfc) := (x lar multiplication A(x, .Ry) := (Ax, .RAy), and product (x, .Ry) < (a, .Rfc) := (x < fc, .Ry<lfc), and the image <p(3) is a norm closed right ideal isomorphic to I with norm 1 unit (e, .Re). Moreover, it is seen in Section 3 that the embedding
f also preserves the corresponding invertible elements; that is, (y,Ry) is an inverse of (x,Rx), with respect to the unit (e,Re), in I whenever y is an inverse of x, with respect to a unit e, in 3. Hence, by Corollary 16, it follows that
$ = a(3 1,(e,Re)) (x,Rx) C a(f(S),(e,Re)) (x'RX) = a(1,e) ^ for aU
X 6 3. □
Proposition 18. Let 3 = J ® Z(3) be a unital split quasiJordan algebra with unit e 6 J and let x 6 J. Then 0(Je) (x) = V(Ie) (x).
Proof. By Corollary 15, 0(Je)(x) c a(3e)(x). For the reverse inclusion, let X i a^je)(x), then Xe - x is invertible in J; that is, there exists y 6 J such that y A (Xe - x) = e and y A (Xe - x)2 = Xe - x. However, e<(Xe - x) = 0 = e<((Xe - x)2). Hence, y is an inverse of Xe - x in 3 with respect to the unit e; that is, X i d(3e)(x). Thus, a(3,e)(x) c 0(Je)(x). □
Proposition 19. Let 3 be a unital quasi-Jordan normed algebra, and let e be a unit in 3 for which Ge(3) is open. Then a(3,e) (X) = a(3,e)(X + z),for all Z 6 Z(3).
Proof. By Corollary 8, Xe - x is invertible if and only if Xe -(x + z) is invertible, for all z 6 Z(3). Thus, X i a(3,e)(x) if and only if X i a(3 e)(x + z) for all z 6 Z(3). □
Corollary 20. Let 3 = J®Z(3) be a unital split quasi-Jordan normed algebra, and let e be a unit in 3 such that Ge(3) is open. Then &(3,e)(x) = 0(3 e)(xj) for all x = Xj + xz 6 3. Further, if the unit e 6 J then <J(3e)(x) = 0(3e)(xj) = Oge)(xj).
Lemma 21. Let 3 = J ® Z(3) be a unital split quasi-Jordan algebra with a unit e 6 J, and let x 6 3 be invertible with respect to e. Then xz < x = (xz < x) < x.
Proof. As x is invertible, there exists y 6 3 such that y < x = e + (e < x - x) = e - xz and y < x2 = x + (e < x - x) + (e < x - x ) = Xj - xz < x. Hence, by the uniqueness of the representation in a split quasi-Jordan algebra, we obtain
Proof. (1) Let X £ U(3,e)(x). Then Xe - x = (Xe - Xj) + (-xz) is invertible with respect to e. By Lemma 21, we have
< x = -xz>
(11) (12)
yz < X = -xz < X. Thus,
xz < x2 = - (yz < x) < x2 (by (11))
= - (yz ^ x2) < x (by the right Jordan identity) = (xz < x) < x (by (12)).
Proposition 22. Let I = J ® Z(I) be a unital split quasiJordan algebra with a unit e e J; x = Xj + xz e I satisfies a(1,e) (x) = C Then,
(1) xz < x2 = (xz < x) < X,
(2) x2 < x = x < x2.
xz < (Xe - xf = (xz < (Xe - x)) < (Xe - x), since the zero part of Xe - x is -xz. However,
< (Xe - x)
= xz < (X2e -Xe<x-Xx<e + x2) = X2 (xz < e) - X (xz < (e < x))
- X (xz < (x < e)) + xz Ax2 = X2xz - 2Xxz A x
+ xz Ax2 (by the right commutativity of a) , (xz A (Xe - x)) A (Xe - x)
= (X (xz A e) - xz A x) A (Xe - x) = (Xxz - xz A x) A (Xe - x) = X2 (xz A e) - X (xz A x) A e
- Xxz A X + (xz A x) A X
= X2xz - 2Xxz A x + (xz A x) A X.
Therefore, (14) becomes
X2xz - 2Xxz A x + xz Ax2
= X2xz - 2Xxz A x + (xz A x) A X,
which after simplification reduces to the required equation xz Ax2 = (xz A x) A X.
(2) Since x = (xj + xz) = Xj + xz A Xj, we have
x1 A X = (xj + xz A x) A X
= xj A X + (xz A x) A X = xj A Xj + (xz A x) A X
2 2 = Xj A Xj + Xz AX
by the part (1). But, xj A Xj = Xj A xj since Xj is in the Jordan algebra J. Therefore, x2 A x = xj A Xj + xz Ax2 = Xj A
x2 + xz Ax2 = X A x2.
Remark 23. In any quasi-Jordan algebra, if an element x satisfies x = x A x = x A x , then xn Ax = xn+ for all positive integers n. For this, suppose x satisfies x2 A x = x A x2 and xm A x2 = xm+2 for any fixed m> 1. Then xm+1 A x2 = (xm A x) A x2 = (xm A x2) A x (by the right Jordan identity) = xm+2 A x = xm+3.
Proposition 24. Let I be a unital quasi-Jordan Banach algebra with unit e, and let x e I satisfy (e - x) A (e - x) = (e - x) A (e - x). Then x e Ge(1) whenever \\e - x\\ < 1.
Proof. First note that ||e-%|| < 1 gives \\(e - x)"\\ < \\e - x\\n < Taking the limit as n ^ >x>, we get 1 for all n = 1,2,3,____Hence, the infinite geometric series
e + J^=l(e-x)n converges absolutely to some element ye I. y < x2 = e + w-2e<w-w2 + e<w2 We show that the geometric series sum y is an inverse of x, with respect to the unit e. For any fixed positive integer n, let yn := e + ^n=1(e-x) . Then, the sequence [yn] of partial sums converges to y. By setting w = e - x,we get
= e + (e - x) - 2e < (e - x) - (e - x)2 + e < (e - x)2
yn <x=(e+^wk)<(e-w)
n / n+1
= e + Xwk - ( e < w + Jwk k=l \ k=2
=e+w-e<w-w
= e + (e - x) - e < (e - x) - (e - x)n
= e+(e-x-e + e<x)-(e- x)n+
= e + (e < x - x) - (e - x)n+1.
Thus, by allowing n ^ >x>, we obtain x<y = e+(e< x - x) = e + e4(x) since \\e - x\\ < 1. Next, by Remark 23, we have
Jn <x2 = Jn <(e- w)2
= yn <(e-e<w-w<e + w2) = yn < (e -2w + w2)
= yn <e - 2y„ <w + yn <w'
= 7n -2ï„ <w + yn <w
= I e + Xwk ) - 2 ( e + I < w
k=i J V k=l
= 2e-x-2e + 2e<x-e + e<x + x- x
+ e-2e<x + e<x = x + (e < x - x) + (e < x2 - x2) .
Proposition 25. Let I be a unital split quasi-Jordan Banach algebra with unit e. If x e I with (e - x) < (e - x) = (e - x) < (e - x), then |A| < ||%|| for all A e a^^x).
Proof. If X e a(3,e)(x) with A = 0, then, the noninvertibility of Xe - x means the noninvertibility of e - (1/X)x, with respect to the unit e. However, by Proposition 24, e - (1/X)x must be invertible with respect to the unit e,whenever (l/|A|)||x|| < 1. It follows that |A| < ||%|| for all X e açi^(x). □
5. Unbounded and Nonclosed Spectrum
In this section, we show that the spectrum of an element in a split quasi-Jordan Banach algebra may be neither bounded nor closed, and hence not compact. The following result gives a couple of characterizations of the unbounded spectrum of an element in a split quasi-Jordan Banach algebra.
Proposition 26. Let x be an element of a unital split quasiJordan Banach algebra I = J ® Z(I) with a unit e e J. Then, the following statements are equivalent.
(1) 0(ie)(x) = C.
(2) x2 < x = x < x2.
(3) |A| < HxH, for all X e açi^(x).
Proof. (1^2): See Proposition 22.
(2^3): Suppose x2 < x = x < x2. Then,
(e - x)2 <(e-x) = (e-e<x-x + x2) < (e - x)
+ I e + ^wk ) < w2 k=l
n \ / n+1
= I e + Xwk ) - 2 ( e < w + ^wk
k=l J \ k=2
= e-2e<x-x + 2x 2
+ (e < x) < x - x < x = e - 2xj - x + 2x2 + e < x2 - x3, (e - x) < (e - x)2 = (e - x) < (e - 2x + x2)
+ I e <w2 + Xwk
= e- 2(e<x) + e<x - x
22 + 2x - x < x
~ 2 „2 n+1 n+2
= e + w-2e<w-w + e <w - w + w .
= e - 2xj + e < x2 - x + 2x2 - x3.
From (21), we get
(e - x)2 < (e - x) = (e - x) < (e - x)2.
Hence, |A| < ||x||, for all A e (x) by Proposition 25.
(3^1): Immediate. □
Remark 27. There do exist unital split quasi-Jordan algebras containing elements that have the spectrum, with respect to the unit of the Jordan part, equal to the whole of C, and hence unbounded. To justify this claim, we proceed as follows.
Let A be a unital associative algebra and let M be an Abimodule. Let f : M ^ A be an A-bimodule map (i.e., an additive map satisfying f(ax) = af(x) and f(xa) = f(x)a, for all a e A, x e M). Then, one can put a dialgebra structure on M as follows: x H y := xf(y) and xhy := f(x)y (cf. [13, Example 2.2(d)]). Hence, M+ is a quasi-Jordan algebra under the quasi-Jordan product "<"givenbyx < y := (\/2)(xf(y) + f(y)x). Further, for any x e M, we observe that
Hence, f is an A-bimodule map. Thus, by Remark 27, M+ is a quasi-Jordan algebra with the quasi-Jordan product as below:
fe11 b12 fe21 fe22
(\a11 au\ f(\b11 ^12 ])
([a21 a22] J ([fe21 b22 \J
Ai fe12 ) aii a12
h2\ b22. ) a2i a22.
a11b11 b11 + b22
b11 + b22
a22b22
Indeed, M+ = J® Z(M+), where J = {[ a00b ]: a,b e C} is a subalgebra of M+ and Z(M+) = {[ h ]: a, b e C}. Any matrix of the form [ ihal ] with a,b e C is a (right) unit in M+. Thus, M+ is a unital split quasi-Jordan algebra with the matrix I = [ 11 ] as the unit of its Jordan part /.
Further, a natural norm is defined on M+ as follows:
X2 < X = 2 (xf (x) + f (x) x) < X = 1((xf(x) + f(x)x)f(x) +f(x) (xf(x)+f(x)x))
a11 a12 At
:= \a11\ + \U12 \ + \a-
21 22 This norm also satisfies
a11 a12
b11 b12 b21 b22
a11 a12
bn 0 0 fe,.
= 4 (xf (x) f (x) + 2f (x) xf (x) +f(x)f (x) x),
bn 0 0 h,.
a11 a12
X < X2 = 2X < (xf (x) + f (x) x)
= 1 (xf (xf (x) + f(x)x) + f (xf (x) + f (x)x)x)
= -(2xf(x)f(x)+2f(x)f(x)x).
a11 a12
a11 a12
bu 0 0 h,.
fe11 fe12
Next, for any x = [ ^ ] e M+, we observe that
However, the right hand sides of the above equations (23) may not be equal; see the following example (Example 28). For such elements x, we have x2 < x = x < x2. Hence, by Proposition 26, the spectrum of x is unbounded whenever M+ is a unital split quasi-Jordan Banach algebra.
Example 28. Let M be the collection of 2x2 matrices with entries from the field C, and let A be the algebra of all matrices of the form ["p ] with a, p e C. Then, it is easily seen that M is an A-bimodule. Next, we define f : M ^ A
by f[ «21 «22 ] := [ "a1 a22 ]. Of course, f is an additive map satisfying
a 0 0 ß
a11 a12 a21 a22
a21 a22.
a 0 0 ß.
a 0 0 ß
( ( a11 a12
a21 a22_
a12 ) a 0
a22. ) .0 ß
x2 = a11 a12 < a11 a12
a21 a22. a21 a22.
a, i + fl
x < x =
X < X =
21 [ 4
(a11 + a22)2
au + a-
12 2 — (a11 +a22)
fl12 ( 2 2 N
fl21 ( 2 2 N
So, for any x = [ % ] e M+, x2 < x = x < x2 & a\2 = a2\ = 0 or an = a22. Thus, by Proposition 26,
a(M+j)([ a! ]) = C whenever au = a22 and |fl12, a21| = {0}. In particular, for x := [ J 2 ] 6 we have
X < X =
4(1+2)2
X < X =
-(1 + 2)2
(12 + 22)
2(12 + 22 ) 23
1 5i 2 0 8
Concerning the inequality between the right hand sides of (23) in Remark 27, we observe for x as above that
4 (*/ (*) / (x) + 2/ (x) x/ (x) + /(*)/ (x) x)
1 0 0 2
1 0 0 2
1 0 0 2
1 0 0 2
1 0 0 2
1 0 0 2
4(2x/(x)/(x)+2/(x)/(x)x)
1 5 2 0 8
1 0 0 2
1 0 0 2
1 0 0 2
1 0 0 2
Hence, x2 < x = x < x2. Thus, 0(m+j)([o2]) = C by Proposition 26.
Further, suppose the matrix [ aa\ ^ ] 6 M+ is invertible with respect to the unit 7; that is, [ ^ ^ ] 6 G7(M+). Then there exists [ ^ ^ ] 6 M+ such that
~k fe2 h h
a i a -)
a i a -)
a, a-)
a i a -)
a i a -)
ai + a.
a, a-)
«i + «4 2
However,
h3 fe4
fe1 fe2 fe3 fe4
fl fl9
fl fl9
fe1fl1
fl + fl,.
fe1fl1
22 fl, + fl.
fl1 + fl4 fe4fl4
22 fl, + fl.
fe4fl4t
It follows that = 0, fl4 = 0, ^ = 1/fl,, fc4 = 1/fl4, fc2((fl! + fl4)/2) = -fl2, fc3((fl1 +fl4)/2) = -fl3, fc2((fl2 +a4)/2) = -fl2((fl1 + fl4)/2), and fc3((«2 + fl,)/2) = -fl3((fl1 + fl4)/2). From these equations, we get fc2(a1 - fl4)2 = 0 and fc3(fl1 - fl4)2 = 0,sothat (fc2-fc3)(fl1 - fl4) = 0. Then,forfl1 = fl4,weobtainfc2 = fc3 = 0 and hence fl2 = fl3 = 0. Therefore,
G, (M+) =
fl fe a 0"
C fl , X = 0 ß.
6 M+ : fl = 0,
a = 0,ß = 0,a = ß
The set G:(M+) is not open: clearly [ J 1 ] e G:(M+); for any e > 0,
1 1 0 1
= - < e 2
but [ 1-0/4 1+U ]iG,(M+).
Now, if A = [ -q1 1 ], then A/ - A = [ A+1 ^ ] £ G:(M+), and so A e CT(M+j)(A) for all A e C. Thus, CT(M+j)(A) = C, an unbounded spectrum.
Next, we observe that the spectrum of an element with respect to a unit is closed whenever the corresponding set of invertibles is open.
Proposition 29. Let I be a quasi-Jordan normed algebra with aunite suchthat Ge(3) isopen. Then 0(g>e)(x) is closed, for all x e I.
Proof. Define / : C ^ I by /(A) = Ae - x. Since / is continuous, the inverse image of the open set Ge(3) is open in C and so its complement 0(g>e)(x) is closed. □
We conclude this paper with the following example of a nonclosed spectrum.
Example 30. Let M and M+ be as in Example 28. Let E = [ ¿1 ] and A = [ 0 1 ] with fl = b both different from 1. Then E is a unit in M+. We show that CT(M+>E)(A) = C \ {1}. For this, let us first investigate when can an element of the form B = [ " ^ ] be invertible? Assuming that B is invertible, we get the existence of an element CeD+ such that C < B = E + (£),
C < £2 = B + (S) + (S2) .
From these equations, we get a = 0, y = 0, and
if. ô I
(38) [6 [7
where ß. satisfies the following two equations: ß. {a-+1) = 1-ß+a + Y
a2 + y2 \ a + v a2 + v2 „a + y 1 - +-—--ß-
2/2 2 '2
Multiplying the last equation by 2(a + y)/(a2 + y2) and then using the other equation, we get
2(l-ß) + (a + y) = (l-ß)(A+^ + (« + r),
or equivalently
this equation is satisfied for p = 1 or a = y. Hence, the matrix B is invertible if and only if B = [ 0 1 ] or B = [ ] for a, y e C \ {0} and p e C.
We conclude that AA := XE - A = [¿-b ] is invertible with respect to E if and only if X - b = 0, X - a = 0, and X = 1; that is, Ax is invertible if a = X and b = X and X = 1. Hence, Ax is invertible only if X = 1 as we assumed that a = b and both are not 1. So, for all X = 1, Aa i GE(M+). Thus, o(M+ E)(A) = C \ {1}, which is neither bounded nor closed.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
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Acknowledgment
This project was supported by King Saud University, Dean-ship of Scientific Research, College of Science Research Center.
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