Uniqueness under Spectral Variation in the Socle of a Banach Algebra
aa r X i v : . [ m a t h . F A ] A ug UNIQUENESS UNDER SPECTRAL VARIATION IN THE SOCLEOF A BANACH ALGEBRA
F. SCHULZ AND R. BRITS
Abstract.
Let A be a complex semisimple Banach algebra with identity, anddenote by σ ′ ( x ) and ρ ( x ) the nonzero spectrum and spectral radius of anelement x ∈ A , respectively. We explore the relationship between elements a, b ∈ A that satisfy one of the following conditions: (1) σ ′ ( ax ) ⊆ σ ′ ( bx ) for all x ∈ A , (2) ρ ( ax ) ≤ ρ ( bx ) for all x ∈ A . The latter problem was identified byBreˇsar and ˇSpenko in [7]. In particular, we use these conditions to spectrallycharacterize prime Banach algebras amongst the class of Banach algebras withnonzero socles, as well as to obtain spectral characterizations of socles whichare minimal two-sided ideals. Introduction By A we denote a complex Banach algebra with identity element and invert-ible group G ( A ). Moreover, it will be assumed throughout that A is semisimple(i.e. the Jacobson radical of A , denoted Rad A , only contains 0). We will write Z ( A ) for the center of A , that is, for the set of all x ∈ A such that xy = yx for all y ∈ A . For x ∈ A we denote by σ A ( x ) = { λ ∈ C : λ − x / ∈ G ( A ) } , ρ A ( x ) = sup {| λ | : λ ∈ σ A ( x ) } and σ ′ A ( x ) = σ A ( x ) − { } the spectrum, spectralradius and nonzero spectrum of x , respectively. If the underlying algebra is clearfrom the context, then we shall agree to omit the subscript A in the notation σ A ( x ), ρ A ( x ) and σ ′ A ( x ). This convention will also be followed in some of the forthcomingdefinitions. We shall also agree to reserve the notation ∼ = exclusively for algebraisomorphisms. Moreover, we recall that an element x of A is called quasinilpotentif σ ( x ) = { } .In [7] M. Breˇsar and ˇS. ˇSpenko consider two interesting problems which resultedfrom certain questions centered around Kaplansky’s problem on spectrum preserv-ing maps [10]: Problem
1. Suppose that a, b ∈ A satisfy σ ( ax ) = σ ( bx ) for all x ∈ A . Does thisimply a = b ? Problem
2. Suppose that a, b ∈ A satisfy(1.1) ρ ( ax ) ≤ ρ ( bx ) for all x ∈ A. What is the relation between a and b ?The first problem has been settled by G. Braatvedt and R. Brits in [5]: Mathematics Subject Classification.
Key words and phrases. rank, socle, trace, spectrum, spectral radius.
Theorem 1.1. [5, Theorem 2.1, Theorem 2.6]
Let a, b ∈ A . Then the followingare equivalent: (i) a = b . (ii) σ ( ax ) = σ ( bx ) for all x ∈ A such that ρ ( x − ) < . (iii) σ ( a + x ) = σ ( b + x ) for all x in some open neighbourhood of − b . Problem 2, as to be expected, is slightly more intricate. Evidence such as [7, Ex-ample 3.3] suggests that the answer to this question may depend on the algebra oron the elements under consideration. Indeed, in the special situation where b = it was found in [6] that a must then belong to Z ( A ). Moreover, in [7] Breˇsar andˇSpenko investigated the special case where A is a prime C ∗ -algebra. The conclu-sion in this case is that the elements a and b satisfying (1.1) are necessarily linearlydependent. We recall that A is a prime algebra if all nonzero two-sided ideals I and J of A satisfy IJ = { } . In particular, we will see that the linear dependenceobtained in the prime C ∗ -algebra case extends to the case where A is assumed to beprime with a nonzero socle. Furthermore, the consideration of Problem 2 leads tospectral characterizations of socles which are minimal two-sided ideals. Other char-acterizations of such socles were recently obtained by the authors and G. Braatvedt(cf. [12, Theorem 3.8, Theorem 3.9] and [11, Theorem 4.4]).The notions of rank, trace and determinant are well-established for operator the-ory. Moreover, in a more general setting, these notions provide an analytic meansto investigate the socle of a semisimple Banach algebra. This latter idea was madeprecise by B. Aupetit and H. Du. T. Mouton in [3] where they managed to showthat these notions can be developed, without the use of operators, in a purely spec-tral and analytic manner. This paper is fundamental to our discussion here, so asin [12] we briefly summarize some of the theory in [3] before we proceed.For each nonnegative integer m , let F m = { a ∈ A : σ ′ ( xa ) ≤ m for all x ∈ A } , where the symbol K denotes the number of distinct elements in a set K ⊆ C .Following Aupetit and Mouton in [3], we define the rank of an element a of A asthe smallest integer m such that a ∈ F m , if it exists; otherwise the rank is infinite.In other words, rank ( a ) = sup x ∈ A σ ′ ( xa ) . If a ∈ A is a finite-rank element, then E ( a ) = { x ∈ A : σ ′ ( xa ) = rank ( a ) } is a dense open subset of A [3, Theorem 2.2]. A finite-rank element a of A issaid to be a maximal finite-rank element if rank ( a ) = σ ′ ( a ). With respect torank it is useful to know results such as Jacobson’s Lemma [1, Lemma 3.1.2], theSpectral Mapping Theorem [1, Theorem 3.3.3]) and the Scarcity Theorem [1, The-orem 3.4.25]. It can be shown [3, Corollary 2.9] that the socle , written Soc A ,of a semisimple Banach algebra A coincides with the collection S ∞ m =0 F m of fi-nite rank elements. We mention a few elementary properties of the rank of anelement [3, p. 117]. Firstly, σ ′ ( a ) ≤ rank ( a ) for all a ∈ A . Furthermore,rank ( xa ) ≤ rank ( a ) and rank ( ax ) ≤ rank ( a ) for all x, a ∈ A , with equality if NIQUENESS UNDER SPECTRAL VARIATION IN THE SOCLE 3 x ∈ G ( A ). Moreover, the rank is lower semicontinuous on Soc A . It is also sub-additive, i.e. rank ( a + b ) ≤ rank ( a ) + rank ( b ) for all a, b ∈ A [3, Theorem 2.14].Finally, if p is a projection of A , then p has rank one if and only if p is a minimalprojection, that is, if pAp = C p [3, p. 117]. It is also worth mentioning here thata projection p is minimal if and only if Ap is a nontrivial left ideal which doesnot contain any left ideals other than { } and itself, that is, if and only if Ap is anontrivial minimal left ideal [4, Lemma 30.2]. A similar result holds true for theright ideal pA . We will also define a minimal two-sided ideal in this manner, thatis, as a two-sided ideal which does not contain any two-sided ideals other than { } and itself.The following result is fundamental to the theory developed in [3] and is mentionedhere for convenient referencing later on: Diagonalization Theorem [3, Theorem 2.8]: Let a ∈ A be a nonzero maximalfinite-rank element and denote by λ , . . . , λ n its nonzero distinct spectral values.Then there exists n orthogonal minimal projections p , . . . , p n ∈ Aa ∩ aA such that a = λ p + · · · + λ n p n . In particular, the Diagonalization Theorem easily implies the well-known resultthat every element of the socle is
Von Neumann regular , that is, for each a ∈ Soc A ,there exists an x ∈ Soc A ⊆ A such that a = axa [3, Corollary 2.10].If a ∈ Soc A we define the trace of a as in [3] byTr ( a ) = X λ ∈ σ ( a ) λm ( λ, a ) , where m ( λ, a ) is the multiplicity of a at λ . A brief description of the notion of multi-plicity in the abstract case goes as follows (for particular details one should consult[3]): Let a ∈ Soc A , λ ∈ σ ( a ) and let B ( λ, r ) be an open disk centered at λ suchthat B ( λ, r ) contains no other points of σ ( a ). It can be shown [3, Theorem 2.4] thatthere exists an open ball, say U ⊆ A , centered at such that σ ( xa ) ∩ B ( λ, r )] isconstant as x runs through E ( a ) ∩ U . This constant integer is the multiplicity of a at λ . It can also be shown that m ( λ, a ) ≥ X α ∈ σ ( a ) m ( α, a ) = (cid:26) a ) if 0 ∈ σ ( a )rank ( a ) if 0 / ∈ σ ( a ) . Furthermore, we note that the trace has the following useful properties:(i) Tr is a linear functional on Soc A ([3, Theorem 3.3] and [12, Lemma 2.1]).(ii) Tr ( ab ) = Tr ( ba ) for each a ∈ Soc A and b ∈ A [12, Corollary 2.5].(iii) For any a ∈ A , if Tr ( ax ) = 0 for each x ∈ Soc A , then a Soc A = { } .Moreover, if a ∈ Soc A , then a = 0 [3, Corollary 3.6].(iv) If f is an analytic function from a domain D of C into Soc A , then λ Tr ( f ( λ )) is holomorphic on D [3, Theorem 3.1].Let λ ∈ σ ( a ) and suppose that B ( λ, r ) separates λ from the remaining spectrumof a . Let f λ be the holomorphic function which takes the value 1 on B ( λ, r ) andthe value 0 on C − B ( λ, r ). If we now let Γ be a smooth contour which surrounds F. SCHULZ AND R. BRITS σ ( a ) and is contained in the domain of f λ , then p ( λ, a ) = f λ ( a ) = 12 πi Z Γ f λ ( α ) ( α − a ) − dα is referred to as the Riesz projection associated with a and λ . By the HolomorphicFunctional Calculus, Riesz projections associated with a and distinct spectral valuesare orthogonal, all commute with a and for λ = 0(1.3) p ( λ, a ) = a πi Z Γ f λ ( α ) α ( α − a ) − dα ∈ Aa ∩ aA. It is also worth mentioning that the orthogonal minimal projections obtained inthe conclusion of the Diagonalization Theorem are in fact the Riesz projections ofthe maximal finite-rank element associated with each of its corresponding nonzerospectral values.In the operator case, A = B ( X ) (bounded linear operators on a Banach space X ),the “spectral” rank and trace both coincide with the respective classical operatordefinitions.2. Uniqueness under Spectral Variation in the Socle
Let a ∈ A . J. Zem´anek has shown that ρ ( a + x ) = 0 for all quasinilpotent x in A if and only if a ∈ Rad A [1, Theorem 5.3.1]. In order to get some feeling forthe subject matter, we start by utilizing the aforementioned result to show thatcondition (iii) in Theorem 1.1 can be substantially relaxed: Theorem 2.1.
Let a, b ∈ A . Then the following are equivalent: (i) a = b . (ii) ρ ( a + x ) ≤ ρ ( b + x ) for all x in some open neighbourhood of − b .Proof. Certainly, (i) ⇒ (ii). We therefore proceed to show that (ii) ⇒ (i). Weclaim that ρ ( a − b + q ) = 0 for all quasinilpotent elements q in A : Let q be anyquasinilpotent element in A . Consider the analytic function f : C → A defined by f ( λ ) = a − b + λq . By hypothesis and the Spectral Mapping Theorem, there exists areal number k > ρ ( a − b + λq ) ≤ ρ ( λq ) = 0 whenever | λ | < k . Hence, σ ( f ( λ )) = { } whenever | λ | < k . By the Scarcity Theorem we may thereforeconclude that σ ( f ( λ )) = { α ( λ ) } for all λ ∈ C , where α is a mapping from C into C . By [1, Corollary 3.4.18], α is an entire function. However, α ( λ ) = 0 whenever | λ | < k , and so, from basic Complex Analysis it must be the case that α ( λ ) = 0for all λ ∈ C . This proves our claim. Consequently, a − b ∈ Rad A by [1, Theorem5.3.1]. Thus, by semisimplicity we have the result. (cid:3) Theorem 2.2.
Let a, b ∈ Soc A . Then a = b if and only if any one of the followingholds true: (i) σ ( ax ) = σ ( bx ) for all rank one elements x ∈ A . (ii) σ ( a + x ) = σ ( b + x ) for all rank one elements x ∈ A .Proof. Obviously, if a = b then conditions (i) and (ii) both hold. So let a, b ∈ Soc A and assume that condition (i) holds. Then Tr ( ax ) = Tr ( bx ) for all rank oneelements x ∈ A . Let y ∈ Soc A be arbitrary. Clearly, Tr ( ay ) = Tr ( by ) if y = 0.So assume that y = 0. By the Diagonalization Theorem and the density of E ( y )there exist rank one projections p , . . . , p n , α , . . . , α n ∈ C and a u ∈ G ( A ) such NIQUENESS UNDER SPECTRAL VARIATION IN THE SOCLE 5 that y = α up + · · · + α n up n . Thus, by the linearity of the trace we readilyobtain Tr ( ay ) = Tr ( by ) for all y ∈ Soc A . Consequently, Tr (( a − b ) y ) = 0 forall y ∈ Soc A . Thus, since a − b ∈ Soc A , it follows from [3, Corollary 3.6] that a − b = 0. Next take a, b ∈ Soc A and assume that condition (ii) holds. Fix any λ / ∈ σ ( a ) ∪ σ ( b ) and 0 = α ∈ C . If x ∈ A has rank one, then we have λ − (cid:0) a + α − x (cid:1) ∈ G ( A ) ⇔ λ − (cid:0) b + α − x (cid:1) ∈ G ( A ) . Consequently,( λ − a ) (cid:16) + ( λ − a ) − α − x (cid:17) ∈ G ( A ) ⇔ ( λ − b ) (cid:16) + ( λ − b ) − α − x (cid:17) ∈ G ( A ) . Since the first term on the left of each expression is invertible, it follows that α ∈ σ (cid:16) ( λ − a ) − x (cid:17) ⇔ α ∈ σ (cid:16) ( λ − b ) − x (cid:17) . Hence, σ ′ (cid:16) ( λ − a ) − x (cid:17) = σ ′ (cid:16) ( λ − b ) − x (cid:17) for all rank one elements x ∈ A .Thus, Tr (cid:16) ( λ − a ) − x (cid:17) = Tr (cid:16) ( λ − b ) − x (cid:17) for all rank one elements x ∈ A .Moreover, since( λ − a ) − − ( λ − b ) − = ( λ − a ) − ( a − b ) ( λ − b ) − ∈ Soc A, it follows as before from [3, Corollary 3.6] that ( λ − a ) − − ( λ − b ) − = 0. Hence, a = b , which establishes the result. (cid:3) Let J be a two-sided ideal of A . Denote by l ( J ) the left-annihilator of J , that is, l ( J ) := { x ∈ A : xJ = { }} . Similarly, we define the right-annihilator of J by r ( J ) := { x ∈ A : Jx = { }} . Theorem 2.3.
Suppose that
Soc A = { } . Then l (Soc A ) = { } if and only if thefollowing are equivalent for any a, b ∈ A : (i) a = b . (ii) σ ( ax ) = σ ( bx ) for all rank one elements x ∈ A . (iii) σ ( a + x ) = σ ( b + x ) for all rank one elements x ∈ A .Proof. Suppose first that l (Soc A ) = { } and let a, b ∈ A . Certainly, (i) ⇒ (ii) and(i) ⇒ (iii). Using the argument in the proof of Theorem 2.2 we see that (ii) impliesTr (( a − b ) y ) = 0 for all y ∈ Soc A . Hence, by [3, Corollary 3.6] it follows that( a − b ) Soc A = { } . Hence, a − b ∈ l (Soc A ) = { } , so (ii) ⇒ (i). Similarly, theargument in the proof of Theorem 2.2 can also be used to show that (iii) impliesTr (cid:16)(cid:16) ( λ − a ) − − ( λ − b ) − (cid:17) y (cid:17) = 0 for all y ∈ Soc A , where λ / ∈ σ ( a ) ∪ σ ( b ) isfixed. Hence, by [3, Corollary 3.6]( λ − a ) − − ( λ − b ) − ∈ l (Soc A ) = { } . Thus, (iii) ⇒ (i). This proves the forward implication. For the converse, we arguecontrapositively. Suppose that l (Soc A ) = { } . Let 0 = a ∈ l (Soc A ) be fixed.Moreover, pick y ∈ Soc A . Since a = 0, a + y = y . However, since a Soc A = { } , itfollows that σ (( a + y ) x ) = σ ( yx ) for all rank one elements x ∈ A . Hence, (ii) (i). This completes the proof. (cid:3) F. SCHULZ AND R. BRITS
In [5, Theorem 2.5] it was shown that properties (i) to (iii) are equivalent for anytwo bounded linear operators on a Banach space X . Consequently, Theorem 2.3implies the well-known fact that l (Soc B ( X )) = { } . Lemma 2.4.
Suppose that
Soc A is a minimal two-sided ideal. Let a, b ∈ A andsuppose that b = pt , where p = p ∈ Soc A and t ∈ G ( A ) . If ρ ( ax ) ≤ ρ ( bx ) for all x ∈ A , then a = λb for some λ ∈ C with | λ | ≤ .Proof. If p = 0, then by semisimplicity a = 0 and we are done. So assume that p = 0. By hypothesis, ρ ( a ′ x ) ≤ ρ ( px ) for all x ∈ A , where a ′ = at − . It will sufficeto show that a ′ = λp for some λ ∈ C , since of course the assumption in conjunctionwith the Spectral Mapping Theorem automatically yields | λ | ≤
1. Replacing x by( − p ) x , we get ρ ( a ′ ( − p ) x ) = 0 for all x ∈ A . Thus, a ′ ( − p ) ∈ Rad A = { } ,and so, a ′ = a ′ p . Moreover, if we replace x by x ( − p ), then by Jacobson’s Lemmawe have ρ (( − p ) a ′ x ) = 0 for all x ∈ A . Hence, as before, the semisimplicity of A yields a ′ = pa ′ . Consequently, a ′ = pa ′ p . Now, by [2, Lemma 2.5] pAp is a closedsemisimple subalgebra of A with identity p . Moreover, σ ′ pAp ( pxp ) = σ ′ A ( pxp ) for all x ∈ A . Hence, by hypothesis, we have ρ pAp (( pa ′ p ) ( pxp )) ≤ ρ pAp ( pxp ) for all x ∈ A. Hence, by the result in [6] it follows that a ′ ∈ Z ( pAp ). However, since Soc A isa minimal two-sided ideal, by [12, Theorem 3.8, Theorem 3.9] we may infer that pAp ∼ = M n ( C ). Consequently, Z ( pAp ) = C p . So, a ′ = λp for some λ ∈ C . Theresult now follows. (cid:3) Let p be a projection of A with rank ( p ) ≤
1. By J p we denote the two-sided idealgenerated by p , that is, we let J p := n X j =1 x j py j : x j , y j ∈ A, n ≥ . By [11, Lemma 2.2] these J p are minimal two-sided ideals. Moreover, by [12, Lemma3.5] there exists a collection of pairwise orthogonal two-sided ideals { J p : p ∈ P} such that every element of Soc A can be written as a finite sum of members of the J p . In particular, this implies that Soc A is a minimal two-sided ideal whenever A is prime. Theorem 2.5.
Soc A is a minimal two-sided ideal if and only if the following areequivalent for any a ∈ A and b ∈ Soc A : (i) ρ ( ax ) ≤ ρ ( bx ) for all x ∈ A (ii) a = λb for some λ ∈ C with | λ | ≤ .Proof. Suppose first that Soc A is a minimal two-sided ideal and let a ∈ A and b ∈ Soc A . Obviously, (ii) ⇒ (i), so assume that condition (i) holds. If b = 0, thenby hypothesis and the semisimplicity of A we have a = 0. So assume b = 0. By theDiagonalization Theorem and the density of E ( b ) we can find mutually orthogonalrank one projections p , . . . , p n , α , . . . , α n ∈ C − { } and a u ∈ G ( A ) such that b = α p u + · · · + α n p n u . Observe firstly that if we set p := p + · · · + p n , then p = p and pb = b . Consequently, by hypothesis and Jacobson’s Lemma it followsthat ρ (( − p ) ax ) = 0 for all x ∈ A . Hence, ( − p ) a ∈ Rad A = { } , and so, NIQUENESS UNDER SPECTRAL VARIATION IN THE SOCLE 7 a = pa . By orthogonality it follows that (cid:0) α − p + · · · + α − n p n (cid:1) b = pu . Thus, byhypothesis and Jacobson’s Lemma it follows that ρ (cid:0)(cid:0) α − p + · · · + α − n p n (cid:1) ax (cid:1) ≤ ρ ( pux ) for all x ∈ A. Thus, by Lemma 2.4 we may infer that (cid:0) α − p + · · · + α − n p n (cid:1) a = λpu for some λ ∈ C . Hence, a = pa = ( α p + · · · + α n p n ) (cid:0) α − p + · · · + α − n p n (cid:1) a = ( α p + · · · + α n p n ) ( λpu )= λ ( α p u + · · · + α n p n u ) = λb. This proves the forward implication. For the reverse implication we argue con-trapositively. Suppose that Soc A is not a minimal two-sided ideal. Then by [12,Lemma 3.5] we may infer the existence of two rank one projections, say p and q ,such that J p J q = J q J p = { } . In particular, p = λ ( p + q ) for all λ ∈ C . Let x ∈ A be arbitrary. Then ( px )( qx ) = ( qx )( px ) = 0. Hence, by [1, Chapter 3, Exercise 9]it follows that σ ′ (( p + q ) x ) = σ ′ ( px ) ∪ σ ′ ( qx ). So, ρ ( px ) ≤ ρ (( p + q ) x ). Since x ∈ A was arbitrary, this shows that (i) (ii), which completes the proof. (cid:3) Lemma 2.6.
Suppose that for any a, b ∈ A we have that the following are equiva-lent: (i) ρ ( ax ) ≤ ρ ( bx ) for all x ∈ A (ii) a = λb for some λ ∈ C with | λ | ≤ .Then A is a prime algebra.Proof. We shall argue contrapositively. If A is not prime, then we can find twononzero two-sided ideals I and J such that IJ = { } . Let 0 = a ∈ I . If a ∈ J ,then aAa = { } . But then, by semisimplicity, it follows that a = 0 which is absurd.Hence, a / ∈ J . Pick 0 = b ∈ J . In particular then, a = λb for all λ ∈ C . Wefirstly claim that I ⊆ r ( J ). Let x ∈ l ( J ) and let y ∈ J be arbitrary. By Jacobson’sLemma and the fact that J is a two-sided ideal, it follows that ρ ( yxw ) = 0 forall w ∈ A . Hence, yx ∈ Rad A = { } . Since y ∈ J was arbitrary, it follows that I ⊆ r ( J ) as claimed. Since a = λb for all λ ∈ C and b = 0, we may infer that a = λ ( b + a ) for all λ ∈ C . Let x ∈ A be arbitrary. Then ax ∈ l ( J ) ∩ r ( J ).Consequently, ( ax )( bx ) = ( bx )( ax ) = 0. Thus, by [1, Chapter 3, Exercise 9] itfollows that σ ′ (( a + b ) x ) = σ ′ ( ax ) ∪ σ ′ ( bx ). Hence, ρ ( ax ) ≤ ρ (( a + b ) x ). Since x ∈ A was arbitrary, this gives the result. (cid:3) Theorem 2.7.
Let A be a C ∗ -algebra. Then A is prime if and only if for any a, b ∈ A we have that the following are equivalent: (i) ρ ( ax ) ≤ ρ ( bx ) for all x ∈ A (ii) a = λb for some λ ∈ C with | λ | ≤ .Proof. This is immediate from [7, Theorem 3.7] and Lemma 2.6. (cid:3)
Lemma 2.8.
Suppose that
Soc A is a minimal two-sided ideal and that l (Soc A ) = { } . Then for any a, b ∈ A we have that the following are equivalent: (i) ρ ( ax ) ≤ ρ ( bx ) for all x ∈ A (ii) a = λb for some λ ∈ C with | λ | ≤ . F. SCHULZ AND R. BRITS
Proof.
Let a, b ∈ A . If a = 0, then we are done. So assume a = 0. It sufficesto show that (i) ⇒ (ii). Let y ∈ Soc A be arbitrary but fixed. By hypothesis, ρ ( ayx ) ≤ ρ ( byx ) for all x ∈ A . Hence, by Theorem 2.5 there exists a λ y ∈ C suchthat ay = λ y by . Let f b : Soc A → C and f a : Soc A → C be defined as follows: f b ( y ) = Tr ( by ) and f a ( y ) = Tr ( ay ) for y ∈ Soc A . Then f b and f a are nonzero linearfunctionals on the linear space Soc A . Moreover, by our first observation it followsthat Ker f b ⊆ Ker f a . Hence, from linear algebra (see [9, p. 10]), it follows that f a = λf b for some λ ∈ C . Thus, by the linearity of the trace, Tr (( a − λb ) y ) = 0 forall y ∈ Soc A . Hence, by [3, Corollary 3.6] it follows that a − λb ∈ l (Soc A ) = { } which gives the result. (cid:3) Theorem 2.9.
Suppose that
Soc A = { } . Then A is prime if and only if for any a, b ∈ A we have that the following are equivalent: (i) ρ ( ax ) ≤ ρ ( bx ) for all x ∈ A (ii) a = λb for some λ ∈ C with | λ | ≤ .Proof. The reverse implication follows immediately from Lemma 2.6. So assumethat A is prime. Since Soc A = { } , we may infer that l (Soc A ) = { } . Moreover,since A is prime, it readily follows from the remark preceding Theorem 2.5 thatSoc A is a minimal two-sided ideal. The forward implication therefore follows fromLemma 2.8. (cid:3) Corollary 2.10.
Suppose that
Soc A = { } . Then A is prime if and only if Soc A is a minimal two-sided ideal and l (Soc A ) = { } .Proof. This is a direct consequence of Lemma 2.8 and Theorem 2.9. (cid:3)
Let 0 = a ∈ A and 0 = b ∈ Soc A . It turns out that the condition σ ′ ( ax ) ⊆ σ ′ ( bx ) for all x ∈ A ⇒ a = b can also be used to characterize socles which are minimal two-sided ideals. Firstly,however, we will prove some related results.The next result was obtained by G. Braatvedt and R. Brits in [5]. We state ittogether with a short new proof based on the spectral trace: Theorem 2.11. [5, Corollary 2.3]
Let N be an arbitrary nonempty open subset of A and let a, b ∈ A . If σ ( ax ) and σ ( bx ) are finite and equal for all x ∈ N , then a = b .Proof. Let y ∈ A . A standard argument using Baire’s Category Theorem andthe Scarcity Theorem can be used to show that if σ ( yx ) is finite for all x in somenonempty open set N of A , then y has finite rank. We may therefore infer that both a and b have finite rank. Furthermore, since E ( a ) and E ( b ) are both open densesubsets of A , it readily follows that E ( a ) ∩ E ( b ) is a dense subset of A . Consequently,we can find an x ∈ N such that ax and bx are both maximal finite-rank elements.Let y ∈ A be arbitrary but fixed, and define analytic functions from C into Soc A as follows: f ( λ ) = a [(1 − λ ) x + λy ] and g ( λ ) = b [(1 − λ ) x + λy ] ( λ ∈ C ) . Since ( E ( a ) ∩ E ( b )) ∩ N is a nonempty open set and x belongs to this set, thereexists a real number ǫ > λ ∈ B (0 , ǫ ) we have that f ( λ ) and g ( λ ) NIQUENESS UNDER SPECTRAL VARIATION IN THE SOCLE 9 are maximal finite-rank elements and σ ( f ( λ )) = σ ( g ( λ )). By the DiagonalizationTheorem the functions λ Tr ( f ( λ )) and λ Tr ( g ( λ ))agree on B (0 , ǫ ). Thus, since these functions are entire by [3, Theorem 3.1], it mustbe the case that they agree on all of C . With the particular value λ = 1 we getTr ( ay ) = Tr ( by ). Since y ∈ A was arbitrary we conclude by [3, Corollary 3.6] that a = b . (cid:3) From Theorem 2.11 it is clear that if σ ( ax ) and σ ( bx ) are finite and equal for all x in some nonempty open set N , then σ ( ax ) = σ ( bx ) for all x ∈ A . In fact, we havethe following result: Lemma 2.12.
Let N be an arbitrary nonempty open subset of A and let a, b ∈ A .If σ ( bx ) is finite and σ ′ ( ax ) ⊆ σ ′ ( bx ) for all x ∈ N , then σ ′ ( ax ) ⊆ σ ′ ( bx ) for all x ∈ A .Proof. The hypotheses allows us to infer that both a and b are finite-rank elements.Recall that E ( a ) and E ( b ) are open and dense in A . Hence, E ( a ) ∩ E ( b ) is openand dense in A . Fix any x ∈ ( E ( a ) ∩ E ( b )) ∩ N and let x ∈ A be arbitrary. Definethe following analytic functions from C into Soc A : f ( λ ) = a [(1 − λ ) x + λx ] and g ( λ ) = b [(1 − λ ) x + λx ] ( λ ∈ C ) . Let rank ( a ) = k and rank ( b ) = n and note that k ≤ n (since ( E ( a ) ∩ E ( b )) ∩ N = ∅ ). By the Scarcity Theorem there exist two closed and discrete subsets of C , say F a and F b , such that σ ′ ( f ( λ )) = k for all λ ∈ C − F a and σ ′ ( g ( λ )) = n forall λ ∈ C − F b . Moreover, by the Scarcity Theorem, our choice of x , and thedefinitions of f and g , there exists a real number ǫ > λ ∈ B (0 , ǫ ), σ ′ ( f ( λ )) = { α ( λ ) , . . . , α k ( λ ) } , σ ′ ( g ( λ )) = { γ ( λ ) , . . . , γ n ( λ ) } ,σ ′ ( f ( λ )) ⊆ σ ′ ( g ( λ )), and the α i ’s and γ i ’s are all holomorphic on B (0 , ǫ ). Let i ∈ { , . . . , k } be arbitrary but fixed. We claim that α i = γ j for some j ∈ { , . . . , n } :Fix any β in B (0 , ǫ ) and let ( λ m ) be any sequence in B (0 , ǫ ) − { β } which con-verges to β . Since σ ′ ( f ( λ )) ⊆ σ ′ ( g ( λ )) for each λ ∈ B (0 , ǫ ), it follows that α i ( λ m ) = γ j m ( λ m ) for some j m ∈ { , . . . , n } . However, by the Pigeon Hole Prin-ciple we may infer the existence of a subsequence, denoted by ( λ m ) for conve-nience, and a j ∈ { , . . . , n } such that α i ( λ m ) = γ j ( λ m ). However, then theset { λ ∈ B (0 , ǫ ) : α i ( λ ) − γ j ( λ ) = 0 } contains a limit point. So, from elementaryComplex Analysis we conclude that α i = γ j . This proves our claim. Withoutloss of generality we may therefore assume that σ ′ ( f ( λ )) = { γ ( λ ) , . . . , γ k ( λ ) } for each λ ∈ B (0 , ǫ ). Pick any λ ∈ ∂B (0 , ǫ ) ∩ [ C − ( F a ∪ F b )] (which exists since F a and F b are discrete), and let z ∈ C − ( F a ∪ F b ) be arbitrary. We claim that σ ′ ( f ( z )) ⊆ σ ′ ( g ( z )): Since F a and F b are discrete, we can find a path Γ in C − ( F a ∪ F b ) which connects λ and z . Now, for each λ ∈ Γ, there exists anonempty open disk B λ := B ( λ, r λ ) such that for β ∈ B λ , σ ′ ( f ( β )) = n α ( λ )1 ( β ) , . . . , α ( λ ) k ( β ) o and σ ′ ( g ( β )) = n γ ( λ )1 ( β ) , . . . , γ ( λ ) n ( β ) o , where the α ( λ ) i ’s and γ ( λ ) i ’s are all holomorphic on B λ . By compactness we can find λ , . . . , λ m ∈ Γ such that B λ i ∩ B λ i +1 = ∅ for i ∈ { , . . . , m − } and B λ m ∩ B z = ∅ . Now, observe that σ ′ ( f ( β )) = { γ ( β ) , . . . , γ k ( β ) } and σ ′ ( g ( β )) = { γ ( β ) , . . . , γ n ( β ) } for each β ∈ B (0 , ǫ ) ∩ B λ . Since B (0 , ǫ ) ∩ B λ is a nonempty open set, it followsin a similar way as before that σ ′ ( f ( β )) = n γ ( λ )1 ( β ) , . . . , γ ( λ ) k ( β ) o for each β ∈ B λ . Hence, σ ′ ( f ( β )) ⊆ σ ′ ( g ( β )) for each β ∈ B λ . Repeatingthis argument with the chain of intersecting open disks we may conclude that σ ′ ( f ( β )) ⊆ σ ′ ( g ( β )) for each β ∈ B z . This proves our claim. Since z ∈ C − ( F a ∪ F b )was arbitrary, σ ′ ( f ( z )) ⊆ σ ′ ( g ( z )) for all z ∈ C − ( F a ∪ F b ). Thus, by a straight-forward argument, using the upper semicontinuity of the spectrum and Newburgh’sTheorem [1, Theorem 3.4.4], we may conclude that the spectral containment ex-tends to all of C . Hence, σ ′ ( ax ) = σ ′ ( f (1)) ⊆ σ ′ ( g (1)) = σ ′ ( bx ) . Since x ∈ A was arbitrary, this establishes the result. (cid:3) The Jacobson radical formula is really only a particular case of a more general typeof spectral calculus: Suppose σ ( bx ) is finite for all x ∈ A . If for each x ∈ A we havethat σ ′ ( ax ) is a portion of σ ′ ( bx ), then “ a is a portion of b ” in the following sense: Theorem 2.13.
Let N be an arbitrary nonempty open subset of A and let a, b ∈ A .If σ ( bx ) is finite for each x ∈ N , and σ ′ ( ax ) ⊆ σ ′ ( bx ) for each x ∈ N then a commutes with b and, either a = 0 , or there exist rank one elements a , . . . , a n , and k ≤ n such that a = a + · · · + a k and b = a + · · · + a n . Moreover, a is orthogonal to b − a .Proof. As before, by the hypotheses above, it follows that both a and b have finiterank. Moreover, by Lemma 2.12 it follows that the spectral containment assumption“for all x ∈ N ” may be replaced by “for all x ∈ A ”. Now, if σ ( ax ) = { } for all x ∈ A , then by the semisimplicity of A we may infer that a = 0. We may thereforeassume that a = 0 and conclude that rank ( b ) = n ≥
1. Recall that E ( a ) ∩ E ( b ) is anopen dense subset of A since E ( a ) and E ( b ) are both open and dense. Further, since σ ′ ( ax ) ⊆ σ ′ ( bx ) for each x ∈ A , it follows in particular that rank ( a ) ≤ rank ( b ).Since G ( A ) is open and E ( a ) ∩ E ( b ) is dense, we can fix an x ∈ ( E ( a ) ∩ E ( b )) ∩ G ( A ).By the Diagonalization Theorem and our hypothesis on the spectrums of ax and bx , we can find n mutually orthogonal rank one projections p , . . . , p n , k mutuallyorthogonal rank one projections q , . . . , q k (with k ≤ n ), and nonzero complexnumbers α , . . . , α n such that(2.1) bx = α p + · · · + α n p n and ax = α q + · · · + α k q k . Set b ′ = bx and a ′ = ax . Let p be any rank one projection such that a ′ p = 0.Then a ′ p has rank one. Moreover, by the containment above and the fact that E ( a ′ p ) is dense, it follows that σ ′ ( a ′ py ) = σ ′ ( b ′ py ) for all y in a dense subsetof A . Thus, Tr ( a ′ py ) = Tr ( b ′ py ) for all y in a dense subset of A . However, by[12, Lemma 2.3] the trace is continuous on the set of rank one elements. Hence,Tr ( a ′ py ) = Tr ( b ′ py ) for all y ∈ A , and so, a ′ p = b ′ p by [3, Corollary 3.6]. A NIQUENESS UNDER SPECTRAL VARIATION IN THE SOCLE 11 similar statement is valid for multiplication on the left. We shall use this to showthat q j = p j for each j ∈ { , . . . , k } . For j ∈ { , . . . , k } we have (cf. the remarkfollowing (1.3))(2.2) q j = 12 πi Z Γ j ( λ − a ′ ) − dλ and(2.3) p j = 12 πi Z Γ j ( λ − b ′ ) − dλ, where Γ j is a small circle surrounding α j and separating it from 0 and the remainingspectrum of b ′ . From (2.1) it follows that q j a ′ = a ′ q j = 0 so, by the precedingparagraph, we have q j b ′ = q j a ′ and b ′ q j = a ′ q j and hence that(2.4) q j p j = 12 πi Z Γ j q j ( λ − b ′ ) − dλ = 12 πi Z Γ j q j ( λ − a ′ ) − dλ = q j = q j , and similarly p j q j = q j . Now, if p j a ′ = 0 then p j q j = 12 πi Z Γ j p j ( λ − a ′ ) − dλ = 12 πi Z Γ j p j λ dλ = 0which contradicts the first calculation that p j q j = q j = 0. Thus, p j a ′ = 0 fromwhich we have p j a ′ = p j b ′ . From a similar argument we have a ′ p j = b ′ p j . As in(2.4), but now using (2.2), we have p j q j = p j = q j p j . Hence, ax = α p + · · · + α k p k .Since x is invertible we can solve for a and b in (2.1) and our result follows with a j = α j p j x − . Now, since E ( a ) ∩ E ( b ) is dense and open in A we can find a sequence( x n ) ⊆ E ( a ) ∩ E ( b ) such that x n → as n → ∞ . But for each x n the first part ofthe proof shows that ax n ( bx n − ax n ) = ( bx n − ax n ) ax n = 0 . So in the limit we obtain a ( b − a ) = ( b − a ) a = 0 and hence also ab = ba . (cid:3) It is immediate from the above result that if we add to the assumptions the re-quirement that rank ( a ) = rank ( b ), then a = b . With the hypothesis of Theorem2.13 a inherits analytic properties from b : Corollary 2.14.
Suppose a and b satisfy the hypothesis of Theorem and that a = 0 . If f ( λ ) is holomorphic on a domain D containing σ ( b ) and f ( b ) = 0 , thenalso f ( a ) = 0 and f ( b − a ) = 0 . In particular, if b is a projection then so is a .Proof. If b is invertible, then by Lemma 2.12 and [5, Theorem 2.1] we have a = αb for some α ∈ C . So rank ( a ) = rank ( b ), and the comment following Theorem2.13 readily yields a = b . We may therefore assume that b / ∈ G ( A ) and thatrank ( a ) < rank ( b ) = n = 0. Moreover, we may also assume that f is not identically0. By hypothesis and the Spectral Mapping Theorem, f ( σ ( b )) = σ ( f ( b )) = { } .Hence, f has zeroes at the spectral points of b . By [8, Corollary 4.3.9] there existsa polynomial h ( λ ) without a constant term and a holomorphic function g ( λ ) on D such that f ( λ ) = h ( λ ) g ( λ ) and g ( α ) = 0 for all α ∈ σ ( b ). In particular then, g ( b )is invertible by the Spectral Mapping Theorem. Hence, since 0 = f ( b ) = h ( b ) g ( b )(by the Holomorphic Functional Calculus), we have h ( b ) = 0. Since a and b − a are orthogonal, it follows that h ( a ) = − h ( b − a ). For the sake of a contradictionsuppose that h ( a ) and h ( b − a ) are not zero. Then since a = a + · · · + a k +1 and b − a = a k +1 + · · · + a n by Theorem 2.13, it follows that there is a largest integer k + 1 ≤ m ≤ n and x , . . . , x m ∈ A such that0 = a m x m = a x + · · · + a m − x m − . Since a m has rank one the minimal right ideal a m A = a m x m A which shows that a m ∈ a A + · · · + a m − A . However, by the subadditivity of the rank we then obtainthat rank ( b ) < n which is absurd. Thus, h ( a ) = h ( b − a ) = 0 and the resultfollows from the Holomorphic Functional Calculus. The last part of the statementis obvious. (cid:3) The next result is similar in spirit to Theorem 2.13:
Theorem 2.15.
Let p be a projection of A , let q ∈ A , and suppose there exist aneighbourhood N p of p and a neighbourhood N − p of − p such that σ ( qx ) = σ ( px ) < ∞ for all x ∈ N p ∪ N − p . Then q is a scalar multiple of p or q is a scalar multiple of the identity.Proof. If p = , then by [5, Theorem 2.1] q is a scalar multiple of the identity. If p = 0 and q / ∈ G ( A ), then q ∈ Rad A = { } . If p = 0 and q ∈ G ( A ), then forall x in a neighbourhood N of 0 we have that σ ( qx ) = 1 which by the ScarcityTheorem implies that every element of A has one point spectrum. Thus, since A is semisimple, A ∼ = C and hence q is a scalar multiple of the identity. So assumethat p is neither 0 nor , and that q is not a scalar multiple of the identity. Thehypothesis implies that, for all x in some neighbourhood of , say N , we have σ A ( qpxp ) = σ A ( pxp ). Moreover, since yqpxp ∈ G ( A ) or pxp ∈ G ( A ) implies p = which contradicts our hypothesis on p , it follows that 0 belongs to both σ A ( yqpxp ) and σ A ( pxp ) for all x, y ∈ A . Hence, by Jacobson’s Lemma we mayinfer that σ A (( pqp )( pxp )) = σ A ( pxp ) for all x ∈ N . So it follows that σ ′ pAp (( pqp )( pxp )) = σ ′ pAp ( p ( pxp )) when x ∈ N . Applying the Open Mapping Theorem to the continuous linear operator x pxp from A onto pAp we have that σ ′ pAp (( pqp )( pxp )) = σ ′ pAp ( p ( pxp ))for all pxp in some neighbourhood of p in pAp . Hence, by the density of E pAp ( pqp )and E pAp ( p ) in pAp we may conclude that rank pAp ( pqp ) = rank pAp ( p ). Whence,since pAp is finite-dimensional, it follows that pqp ∈ G ( pAp ). Thus, σ pAp (( pqp )( pxp )) = σ pAp ( p ( pxp ))for all pxp in some neighbourhood of p in pAp . Hence, by [5, Theorem 2.1] it followsthat pqp = αp for some α ∈ C . On the other hand, using the hypothesis with theneighbourhood N − p and the fact that q is not a scalar multiple of the identity, itfollows, for all x in some neighbourhood of , that σ ( q ( − p ) x ) = σ (( − p ) qx ) = { } . Hence, by the Scarcity Theorem and the semisimplicity of A we get q = pq = qp .Therefore, q = αp , which completes the proof. (cid:3) Theorem 2.16.
Soc A is a minimal two-sided ideal if and only if the following areequivalent for any = a ∈ A and = b ∈ Soc A : (i) σ ′ ( ax ) ⊆ σ ′ ( bx ) for all x in some nonempty open set N . NIQUENESS UNDER SPECTRAL VARIATION IN THE SOCLE 13 (ii) σ ′ ( ax ) ⊆ σ ′ ( bx ) for all x ∈ A . (iii) a = b .Proof. Suppose first that Soc A is a minimal two-sided ideal and let 0 = a ∈ A and b ∈ Soc A . Obviously (iii) ⇒ (i). Moreover, by Lemma 2.12, (i) ⇒ (ii). So assumethat condition (ii) holds. Since (ii) implies that ρ ( ax ) ≤ ρ ( bx ) for all x ∈ A , itreadily follows from Theorem 2.5 and hypothesis that a = λb for some λ ∈ C − { } .Hence, rank ( a ) = rank ( b ), and so, by Theorem 2.13 and the remark following it, a = b . This proves the forward implication. For the other direction, we arguecontrapositively. Suppose that Soc A is not a minimal two-sided ideal. Then by[12, Lemma 3.5] we may infer the existence of two rank one projections p and q such that J p J q = J q J p = { } . However, as in the proof of Theorem 2.5 this impliesthat p = p + q and σ ′ ( px ) ⊆ σ ′ (( p + q ) x ) for all x ∈ A . Hence, (ii) (iii), whichestablishes the result. (cid:3) Moreover, we obtain a similar characterization of prime Banach algebras as wasdone in Theorem 2.9:
Theorem 2.17.
Suppose that
Soc A = { } . Then A is prime if and only if for any a, b ∈ A − { } we have that the following are equivalent: (i) σ ′ ( ax ) ⊆ σ ′ ( bx ) for all x ∈ A . (ii) a = b .Proof. If A is not prime then we may proceed as in the proof of Lemma 2.6 andexpose two elements a and b such that a = a + b and σ ′ ( ax ) ⊆ σ ′ (( a + b ) x ) forall x ∈ A . This proves the reverse implication. Conversely, if A is prime, thensince Soc A = { } it follows that Soc A is a minimal two-sided ideal and that l (Soc A ) = { } . Let a, b ∈ A − { } be arbitrary. Obviously (ii) ⇒ (i). So assumethat condition (i) holds and let y ∈ Soc A be arbitrary but fixed. Then by Theorem2.16 we may infer that ay = by . Since y ∈ Soc A was arbitrary, we conclude thatTr (( a − b ) y ) = 0 for all y ∈ Soc A . Hence, by [3, Corollary 3.6] it follows that a − b ∈ l (Soc A ) = { } . Therefore, (i) ⇒ (ii), so the theorem is true. (cid:3) To conclude we will show that if Soc A is a minimal two-sided ideal, then conditions(i) and (ii) in Theorem 2.17 are equivalent whenever b belongs to some inessentialideal ; that is, a two-sided ideal in which the spectrum of all elements contain at most0 as an accumulation point. Before that, however, we will need a little preparation: Lemma 2.18.
Let s ∈ A and for each x ∈ A suppose that σ ( sx ) contains atmost as an accumulation point for all x ∈ A . Then the Riesz projections of s corresponding to nonzero spectral values have finite rank.Proof. Let σ ′ ( s ) = { λ , λ , . . . } and set, for i ∈ N , p := p ( λ i , s ). Recall that pAp isa semisimple Banach algebra with identity p . There exists an open neighborhood V of in A such that pxp is invertible in pAp for each x ∈ V . Now suppose x ∈ V and σ A ( px ) = ∞ . Then, by Jacobson’s Lemma, σ A ( pxp ) = ∞ = σ ′ A ( pxp ),and, since p ∈ sA , it follows from our hypothesis on s that σ ′ A ( pxp ) is a sequenceconverging to 0. But this means σ pAp ( pxp ) contains a sequence converging to zero,from which it follows (since the spectrum is closed) that pxp cannot be invertiblein pAp giving a contradiction. So σ A ( px ) < ∞ for all x ∈ V and a standardapplication of the Scarcity Theorem then says σ A ( px ) < ∞ for all x ∈ A . Thusrank ( p ) < ∞ . (cid:3) Theorem 2.19.
Suppose that
Soc A is a minimal two-sided ideal. Let = a ∈ A and let = b ∈ A such that σ ( bx ) has at most as an accumulation point for all x ∈ A . Then the following are equivalent: (i) σ ′ ( ax ) ⊆ σ ′ ( bx ) for all x ∈ A . (ii) a = b .Proof. Let 0 = a ∈ A and b ∈ A . Surely, (ii) ⇒ (i). So assume that condition (i)holds. We claim that σ ( ax ) = σ ( bx ) for all x ∈ A : Let x ∈ A be arbitrary. It willsuffice to show that σ ′ ( ax ) = σ ′ ( bx ) and 0 ∈ σ ( ax ) ⇔ ∈ σ ( bx ). If σ ( bx ) = { } ,then σ ′ ( ax ) = σ ′ ( bx ) = ∅ . So assume that σ ( bx ) = { } and let λ ∈ σ ′ ( bx ). Since σ ′ ( bx ) is either finite or a sequence converging to zero, we may consider the Rieszprojection of bx associated with λ , say p := p ( λ, bx ). Now, by Lemma 2.18 it followsthat p ∈ Soc A . Consequently, by hypothesis and Theorem 2.16, we may infer that axp = bxp . Moreover, since bxp = pbx , by condition (i), Jacobson’s Lemma andTheorem 2.16 it follows that pax = pbx = axp . Hence,( ax ( − p )) ( axp ) = ( axp ) ( ax ( − p )) = 0 . Thus, since ax = ax ( − p ) + axp , it follows from [1, Chapter 3, Exercise 9] that σ ′ ( ax ) = σ ′ ( ax ( − p )) ∪ σ ′ ( axp ) = σ ′ ( ax ( − p )) ∪ σ ′ ( bxp ) . But by the Holomorphic Functional Calculus it follows that σ ′ ( bxp ) = { λ } . Hence, λ ∈ σ ′ ( ax ). This shows that σ ′ ( ax ) = σ ′ ( bx ). Suppose now that 0 / ∈ σ ( bx ). Then,by hypothesis on b it must be the case that σ ( bx ) is finite, say σ ( bx ) = { α , . . . , α r } .For each i ∈ { , . . . , r } , let p i denote the Riesz projection of bx associated with α i .By condition (i) and Theorem 2.16 it follows that axp i = bxp i for all i ∈ { , . . . , r } .But by the Holomorphic Functional Calculus p + · · · + p r = . Hence, ax = ax ( p + · · · + p r ) = axp + · · · + axp r = bxp + · · · + bxp r = bx ( p + · · · + p r ) = bx. So, 0 / ∈ σ ( ax ). Similarly, 0 / ∈ σ ( ax ) yields bx = ax and consequently 0 / ∈ σ ( bx ).Hence, 0 ∈ σ ( ax ) ⇔ ∈ σ ( bx ). This proves our claim. By Theorem 1.1 we maytherefore conclude that a = b , which completes the proof. (cid:3) References
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Department of Pure and Applied Mathematics, University of Johannesburg, SouthAfrica
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