Equivalence of slice semi-regular functions via Sylvester operators
aa r X i v : . [ m a t h . C V ] D ec EQUIVALENCE OF SLICE SEMI-REGULAR FUNCTIONS VIA SYLVESTEROPERATORS
A. ALTAVILLA † AND C. DE FABRITIIS † Abstract.
The aim of this paper is to study some features of slice semi-regular functions RM (Ω) ona circular domain Ω contained in the skew-symmetric algebra of quaternions H via the analysis of afamily of linear operators built from left and right ∗ -multiplication on RM (Ω); this class of operatorsincludes the family of Sylvester-type operators S f,g . Our strategy is to give a matrix interpretationof these operators as we show that RM (Ω) can be seen as a 4-dimensional vector space on the field RM R (Ω). We then study the rank of S f,g and describe its kernel and image when it is not invertible.By using these results, we are able to characterize when the functions f and g are either equivalentunder ∗ -conjugation or intertwined by means of a zero divisor, thus proving a number of statements onthe behaviour of slice semi-regular functions. We also provide a complete classification of idempotentsand zero divisors on product domains of H . Contents
1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12. Preliminary results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.1. R as a 4-rank free module over R R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2. Semi-regular functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.3. RM as a 4-dimensional vector space over RM R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.4. Zero divisors and idempotents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83. RM R -linear endomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94. Matrix representation of L F , G -type equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115. The rank of the Sylvester operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146. The solution of the Sylvester equation in the non-singular case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177. Sylvester operators of rank 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198. Applications of the rank 2 case to function theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239. Sylvester operators of rank 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271. Introduction
The aim of this article is to investigate the behaviour of slice semi-regular functions defined on acircular domain Ω contained in the skew-symmetric algebra of quaternions H via the study of a family Mathematics Subject Classification.
Primary 30G35; secondary 15A24, 15A33, 15A54, 39B42, 47A56.
Key words and phrases.
Slice-regular functions, Sylvester equation, ∗ -product of slice-regular functions, semi-regularfunctions, idempotent functions. † GNSAGA of INdAM. The authors acknowledge with pleasure the support of FBK-Cirm, Trento, where part of thispaper was written. of Sylvester-type operators, and related equations; in particular, we single out such a family in a moregeneral class of operators which are obtained as generalizations of left and right ∗ -multiplication. Oneof our main motives for this analysis is the fact that these operators are of crucial importance in theinvestigation of the orbit of slice (semi)-regular functions under conjugation; as an example, we remarkthat this strategy could be applied in the research on the problem exp ∗ ( g ) = f , given a never vanishingregular function f . In such manner, the interplay between algebra and operator theory gives new andunexpected results under the function theoretical viewpoint.In the most common use, Sylvester equations are particular matrices equations, introduced by Sylvesterhimself [27], which are used in several subjects, including similarity, commutativity, control theory anddifferential equation (see [7]). In the quaternionic setting, such equations were studied with differentpurposes: without claiming any completeness of references, we point out the works of Bolotnikov [8, 9] andJanovsk´a–Opfer [24] regarding the quaternionic matricial equation and He–Liu–Tam [23] and referencestherein for the multitude of employments in applied sciences. For the operatorial equation in quaternionicfunction spaces we mention [1, Chapter 4] and references therein.In our paper, given f and g in the space of slice semi-regular funtions RM (Ω), we make a large useof a detailed analysis of the Sylvester operator S f,g (and in particular of its kernel in the case whenit is not of maximal rank) in order to understand when these functions belong to the same conjugacyclass under the action of an invertible element of RM (Ω). The deep interlacement between the functiontheory in RM (Ω) and the behaviour of Sylvester operators answers several open questions concerningslice semi-regular functions; in particular it gives a necessary and sufficient conditions on a function inorder it is conjugated to a one-slice preserving function (see Proposition 8.1).We now give an outline of the plan of the paper. Section 2 contains definitions and preliminarymaterial: here we recall properties of slice semi-regular functions, the definition of the ∗ -product and theinterpretation given in terms of the operators h ., . i ∗ and ∧ * defined and developed in [4, 5]. Moreover,following the approach originally due to Colombo, Gonzales-Cervantes and Sabadini, we prove that thefamily RM (Ω) of slice semi-regular functions on a symmetric domain is in fact a vector space over the set RM R (Ω) of slice semi-regular functions that preserves all the complex lines in H (see Proposition 2.10).Thanks to this result we can write any slice semi-regular function f as a sum f = f + f v , where f ∈ RM R (Ω) can be interpreted as the “real part” of f and f v as the “vector part” of f . Afterwardswe deal with idempotents for the ∗ -product: in particular we prove (see Proposition 2.13), that anysemi-regular idempotent f ∈ RM (Ω) is regular and that f is an idempotent if and only if it is a zerodivisor whose “real part” f is identically equal to . This characterization allows us to describe all zerodivisors in terms of idempotents in Propositions 2.14.In the following section we define the class of RM R (Ω)-linear operators L F , G : RM (Ω) → RM (Ω) as L F , G ( χ ) := f [1] ∗ χ ∗ g [1] + · · · f [ N ] ∗ χ ∗ g [ N ] , for any N -tuples F = ( f [1] . . . , f [ N ] ) , G = ( g [1] , . . . , g [ N ] ) ⊂ ( RM (Ω) \ { } ) N . We then study the initialcase N = 1, that is the multiplicative operators given by L f,g ( χ ) = f ∗ χ ∗ g ; in particular we provethat L f,g is an isomorphism if and only if f and g are not zero divisors (see Proposition 3.2) and inTheorem 3.3 and Proposition 3.5 we describe the image and the kernel of this operator when it is not anisomorphism.In Section 4 we present an interpretation of L F , G via coordinates, being thus able to find necessaryand sufficient conditions on F , G in order to have that L F , G is an isomorphism. We later turn to thestudy of the Sylvester operators, which corresponds to the case F = ( f,
1) and G = (1 , g ), thus giving S f,g ( χ ) = f ∗ χ + χ ∗ g . After defining the equivalence relation ≃ given by f ≃ g when there exists aninvertible h such that g = h −∗ ∗ f ∗ h , we prove that S f,g is not an isomorphism if and only if either f ≃ − g or there exist a zero divisor σ such that f ∗ σ + σ ∗ g = 0. QUIVALENCE OF SLICE SEMI-REGULAR FUNCTIONS VIA SYLVESTER OPERATORS 3
Section 5 contains a detailed analysis of the rank of the Sylvester operator according to the featuresof the functions f and g . By accurately scrutinizing the matrix associated to S f,g we prove that its rankis always strictly greater than 1 and show that S f,g is not an isomorphism if and only if( f + g ) [( f + g ) + 2( f sv + g sv )] + ( f sv − g sv ) ≡ , where f sv and g sv denote the symmetrized functions of f v and g v , respectively (see formula (2.1)). Inparticular we prove that Proposition 1.1. If Ω is a slice domain, then the following conditions are equivalent • f ≃ g , • f = g and f s = g s , • S f, − g is not an isomorphism. We then show (see Proposition 5.3 and Theorem 5.6) the following characterization of the rank of S f,g in terms of the “real parts” of the functions f and g . Proposition 1.2.
Suppose that S f,g is not an isomorphism. If f + g ≡ the operator S f,g has rank ,otherwise it has rank . The succeeding section is devoted to the study of the Sylvester operators of maximal rank. In thiscase we are able to write explicitly the solution of the equation S f,g ( χ ) = b in terms of suitable functions λ L and λ R built by means of f and g .Section 7 contains the final characterization of the equivalence relation ≃ : after describing the kernelof S f,g when f = − g and f sv = g sv , we show (see Theorem 7.1) that it contains invertible elements. Thisproves that the relation f ≃ g holds if and only if f = g and f sv = g sv , even when Ω is a product domain.We are also able to say under which conditions on the functions f and g the kernel of the operator S f,g contains zero divisors and to give a detailed picture of the image of S f,g both through an intensional andan extensional description.Thanks to the results obtained on Sylvester operators of rank 2, in Section 8 we are able to characterizewhen a slice semi-regular function is equivalent to a one-slice preserving function, namely f is conjugatedto a one slice-preserving function if and only if f sv has a square root. In particular this implies that allidempotents are equivalent. Last result allows us to give a different and more detailed description of thekernel of L f,g when both f and g are idempotents.Finally, Section 9 contains a detailed description of the couples of functions f, g such that S f,g hasrank 3. First of all we show that, if f + g ≡ S f,g is not an isomorphism, then we can find τ ∈ RM R (Ω) such that f sv = (cid:0) J (cid:0) τ + (cid:1)(cid:1) and g sv = (cid:0) J (cid:0) τ − (cid:1)(cid:1) , where J is the function given inDefinition 2.4. If τ = ± , this gives us the possibility to represent the kernel and the image of S f,g up toa conjugation which simplifies the form of f and g to one-slice preserving functions, while the case when f sv g sv ≡ f, g ) and the behaviour of the Sylvester operator S f,g , we summarize the results of Sections 5 −
10 inthe following statement:
Main Theorem.
Let f, g ∈ RM (Ω) \RM R (Ω) . Then rk ( S f,g ) is always strictly greater than . Moreoverwe have • rk ( S f,g ) = 4 ⇔ ( f + g ) [( f + g ) + 2( f sv + g sv )] + ( f sv − g sv ) ; • rk ( S f,g ) = 3 ⇔ ( f + g ) [( f + g ) + 2( f sv + g sv )] + ( f sv − g sv ) ≡ and f + g ⇔ ker( S f,g ) contains only zero divisors (this case can occur only if Ω is a product domain); • rk ( S f,g ) = 2 ⇔ f ≃ − g ⇔ f + g ≡ and f sv ≡ g sv ⇔ ker( S f,g ) contains at least an invertibleelement in RM (Ω) . A. ALTAVILLA AND C. DE FABRITIIS
In last case, ker( S f,g ) contains also zero divisors if and only if Ω is a product domain and one of thefollowing holds(1) f v = g v and f sv has a square root;(2) f v = g v and ( f v − g v ) s ≡ ;(3) ( f v − g v ) s and f sv has a square root. Preliminary results
In this section we recall some basic notion and result on slice regular and semi-regular functions andprove a couple of preliminary results. We start by recalling some relevant subset of H and the family ofdomains where we will define our functions. In the space of quaternions we denote by i, j, k the usualdefining basis, so that any quaternion q ∈ H can be written as q = q + q i + q j + q k , where q ℓ ∈ R , ℓ = 0 , , ,
3, and i, j, k satisfy i = j = k = − ij = − ji = k . If q = q + q i + q j + q k , thenits usual quaternionic conjugate will be denoted by q c = q − ( q i + q j + q k ). The square norm of q isthen given by | q | = qq c . The set of imaginary units, i.e. the set of quaternions whose square equals − S : S := { q ∈ H | q = − } . For any q = q + q i + q j + q k ∈ H , we define its vector part as q v = ( q − q c ) /
2, hence q = q + q v .Moreover, if q v = 0, we can also write q = q + | q v | q v | q v | and (cid:16) q v | q v | (cid:17) = −
1. Thus, for any q ∈ H , we have q = x + Iy , where I ∈ S , x = q , y = | q v | ∈ R . It is then clear that the space of quaternions can beunfolded as H = ∪ I ∈ S C I , where C I := Span R (1 , I ) = { x + Iy | x, y ∈ R } . Given q = x + Iy ∈ H , we set S q := { x + Jy | J ∈ S } . Definition 2.1.
We say that a domain Ω ⊂ H is circular , if, for any q = x + Iy ∈ Ω, we have that S q ⊂ Ω.If Ω ∩ R = ∅ , a circular domain Ω is called a slice domain , otherwise it is called a product domain .For any circular set Ω ⊂ H and I ∈ S , we write Ω I = Ω ∩ C I and Ω + I = Ω ∩ C + I , where C + I := { x + Iy | x ∈ R , y > } . A subset of Ω of the form Ω I (respectively Ω + I ) will be called a slice (respectivelya semi-slice ) of Ω. Notice that, if Ω is a product domain, then, for any I ∈ S , we have Ω = Ω + I × S .We have now set up all the notation we need to recall the definition of regularity (for an extensiveapproach to the subject of slice regular functions see [12, 13, 16]). Definition 2.2.
Let Ω ⊂ H be a circular domain. A function f : Ω → H is said to be slice regular if allits restrictions f I = f | Ω I are holomorphic, i.e., for any I ∈ S , it holds12 (cid:18) ∂∂x + I ∂∂y (cid:19) f I ( x + Iy ) ≡ . The family of slice regular functions over a fixed circular domain Ω will be denoted by R (Ω).Examples of slice regular functions are given by polynomials with quaternionic coefficient on the right.Among the several properties that have been proved for slice regular functions we will make use of theso-called Identity Principle [2, 10, 16] stating that if a regular function f equals 0 on a set containing twoaccumulation points living in two different semi-slices then f ≡
0. In particular, if f vanishes identicallyon an open set, then it vanishes everywhere.It is well known that pointwise product does not preserve slice regularity. This issue can be solvedby working with the ∗ -product which generalizes the usual product of polynomials on a ring. Given QUIVALENCE OF SLICE SEMI-REGULAR FUNCTIONS VIA SYLVESTER OPERATORS 5 f, g ∈ R (Ω), we define f ∗ g ∈ R (Ω) as( f ∗ g )( q ) = ( , if f ( q ) = 0 ,f ( q ) g ( f ( q ) − qf ( q )) , otherwise . In general, the ∗ -product is not commutative, however, if f and g are such that there exists I ∈ S forwhich f (Ω I ) ⊂ C I and g (Ω I ) ⊂ C I , then f ∗ g = g ∗ f . Moreover, if f is such that for any I ∈ S f (Ω I ) ⊂ C I , then f ∗ g = g ∗ f = f g , for any g ∈ R (Ω). The previous properties characterize tworemarkable sets of slice regular functions. Definition 2.3.
A function f ∈ R (Ω), such that there exists I ∈ S for which f (Ω I ) ⊂ C I is said to be one slice preserving or C I - preserving ; the set of C I -preserving regular functions is denoted by R I (Ω). Afunction f ∈ R (Ω) such that f (Ω I ) ⊂ C I , for any I ∈ S , is said to be slice preserving ; the set of slicepreserving regular functions is denoted by R R (Ω).A special regular function that will be widely used next is presented in the following definition. Definition 2.4.
We define the slice regular function J : H \ R → H as J ( q ) = q v | q v | , for all q ∈ H \ R .It is easily seen that J is slice preserving and slice constant in the sense of [2, Definition 13]. Moreover,notice that J ∗ = J = − Remark 2.5.
The function J given in Definition 2.4 can be interpreted in the sense of stem functions(see [18]) as follows: let us consider the stem function J : C \ R → H C J ( z ) := ( ı, if z ∈ C + − ı, if z ∈ C − ;then J induces the slice regular function J := I ( J ).2.1. R as a 4-rank free module over R R . Let (1 , i, j, k ) be any basis of H . We recall from [11,Proposition 3.12] and [17, Lemma 6.11], that any slice regular function f ∈ R (Ω) can be written in aunique way as a sum f = f + f i + f j + f k , where f ℓ ∈ R R (Ω), and ℓ = 0 , , ,
3. In particular R (Ω)is a 4-rank free module on R R (Ω). Given f ∈ R (Ω), by means of the previous formalism, it is possible towrite the regular conjugate f c and the symmetrized function f s (see [16, Definition 1.33]), as(2.1) f c = f + f i c + f j c + f k c , f s = f ∗ f c . Moreover, if (1 , i, j, k ) ⊂ H is an orthonormal basis, the previous formulas simplifies as explained in [4,Remark 2.2] as f c = f − ( f i + f j + f k ) , f s = f + f + f + f . A further consequence of this result is a more intuitive representation of the ∗ -product, similar to the usualquaternionic product in its “scalar-vector” form. First of all, given f ∈ R (Ω), notice that f = ( f + f c ) / f v = f − f (in particular f ≡ f ≡ − f c ). For any regular function f wewill sometimes informally call f as its “real part” and f v as its “vector part”, even if f and f v arequaternionic valued and not real or pure-imaginary valued functions. If g = g + g v is another elementof R (Ω), we have [4, Proposition 2.7](2.2) f ∗ g = f g − h f v , g v i ∗ + f g v + g f v + f v ∧ * g v , where h ., . i ∗ and ∧ * are defined as follows(2.3) h f, g i ∗ ( q ) = ( f ∗ g c ) ( q ) , ( f ∧ * g )( q ) = ( f v ∧ * g v )( q ) = ( f ∗ g )( q ) − ( g ∗ f )( q )2 . The following remark can be interpreted as a non degeneracy result of the “scalar product” h· , ·i ∗ givenin formula (2.3). A. ALTAVILLA AND C. DE FABRITIIS
Remark 2.6.
Notice that ( f δ ) ≡ δ ∈ H with | δ | = 1 if and only if f ≡
0. Indeed if we choosean orthonormal basis { , i, j, k } of H and write f = f + f i + f j + f k , we have( f · ≡ f , ( f · i ) ≡ − f , ( f · j ) ≡ − f , ( f · k ) ≡ − f , and hence f ≡ ∗ -product give in formula (2.2) makes possible to prove the following resultwhich will be useful in some of the computations to come. Lemma 2.7.
Let f and g be regular functions defined on the same domain Ω . Then we have ( f + g ) s = f s + g s + 2 h f, g i ∗ . Proof.
The following chain of equalities yields the thesis( f + g ) s = ( f + g ) ∗ ( f c + g c ) = f ∗ f c + f ∗ g c + g ∗ f c + g ∗ g c = f s + g s + f ∗ g c + ( f ∗ g c ) c = f s + g s + 2( f ∗ g c ) = f s + g s + 2 h f, g i ∗ . (cid:3) Semi-regular functions.
Another interesting property of a regular function f is the structure ofits zero set V ( f ) [15, 16, 18, 20] and of its singularities [16, 21, 22, 26]. Ghiloni, Perotti and Stoppatoproved the following statement in [22, Theorem 3.5], generalizing results due to several authors. Theorem 2.8 (Ghiloni-Perotti-Stoppato) . Assume that Ω is either a slice or a product domain and let f ∈ R (Ω) . • If f then the intersection V ( f ) ∩ C + J is closed and discrete in Ω J for all J ∈ S with at mostone exception J , for which it holds f | Ω + J ≡ . • If f s then the set V ( f ) is a union of isolated points or isolated spheres of the form S q . In the same paper, Ghiloni, Perotti and Stoppato also developed a theory of singualarities for sliceregular functions, which is a consequence of a detailed study of Laurent expansions around spheres S q and real points; the notion of meromorphic function can thus be translated in this context as that of semi-regular function . We now briefly recall the notions of removable singularity and pole at non realpoints; the case of real points is completely analogous. For more detailed statements and complete proofssee [22, Section 6].Let Ω be a circular domain and p ∈ Ω \ R . Any f ∈ R (Ω \ S p ) can be written around S p as f ( q ) = X n ∈ Z ( q − p ) ∗ n b n , f ( q ) = X ν ∈ Z ∆ νp ( q )( qu ν + v ν ) , with b n , u ν , v ν ∈ H , for any n and ν . The point p is called a removable singularity if f extends to a sliceregular function in a circular open set containing S p . If it is not a removable singularity, the point p issaid to be a pole for f if there exists an n ≥ b n = 0 for all n < n ; the minimum of such n is called the order of the pole and denoted as ord f ( p ). If p is neither a removable singularity nor a pole,then it is called an essential singularity for f and ord f ( p ) is set to be + ∞ . Finally, the spherical order of f at S p is the smallest even natural number 2 ν such that u ν = v ν = 0 for all ν < − ν . If no such ν exists, then we set ord f ( S p ) = + ∞ .Non-real singularities for slice regular functions can be classified as follows (see [22, Theorem 6.4]).Let Ω be a circular domain, p ∈ Ω \ R and set e Ω := Ω \ S p . If f ∈ R ( e Ω) then one of the following holds: • every point of S p is a removable singularity for f ; in this case ord f ( S p ) = 0 = ord f ( w ), for any w ∈ S p ; QUIVALENCE OF SLICE SEMI-REGULAR FUNCTIONS VIA SYLVESTER OPERATORS 7 • every point of S p is a non removable pole for f . There exists n ∈ N \ { } such that the function∆ np ( q ) f ( q ) extends to a slice regular function g defined on Ω that has at most one zero in S p ;in this case ord f ( S p ) = 2 k ; moreover, ord f ( w ) = k and lim Ω ∋ x → w | f ( x ) | = + ∞ for all w ∈ S p except the possible zero of g , which must have order less than k ; • every point of S p , except at most one, is an essential singularity for f ; in this case ord f ( S p ) = + ∞ and there exists at most one point w ∈ S p such that ord f ( w ) < ∞ .In the particular case of a slice preserving function f , for any point ˜ p belonging to the sphere S p , it holdsord f ( S p ) = ord f (˜ p ), i.e. all the points of S p have the same order.Notice that, the set of singularities has different structure with respect to the zero set: indeed there areno non-real isolated singular points for a slice regular function. We now give the definition of semi-regularfunction. Definition 2.9.
A function f is said to be slice semi-regular in a nonempty circular domain Ω, if thereexists a circular open subset e Ω ⊆ Ω such that f ∈ R ( e Ω) and such that each point of Ω \ e Ω is either apole or a removable singularity for f . The set of slice semi-regular functions on Ω will be denoted as RM (Ω); the sets of slice preserving and of C I -preserving (for some I ∈ S ) semi-regular functions on Ωas RM R (Ω) and RM I (Ω), respectively.2.3. RM as a 4-dimensional vector space over RM R . We now pass to analyze some algebraicproperties of RM (Ω). First of all consider the action R R (Ω) ×R (Ω) → R (Ω), given by ( f, g ) f ∗ g = f g .Thanks to the Identity Principle and the fact that the zero set of a non-constant regular function hasempty interior, the equality f g ≡ f or g is identically zero (this is a particularcase of [22, Proposition 3.8]). In particular ( R R (Ω) , + , ∗ ) is an integral domain and RM R (Ω) is a field.Moreover, recalling [22, Theorem 6.6], we have that if Ω is a slice domain then RM (Ω) is a divisionalgebra and, also when Ω is a product domain, any f ∈ RM (Ω) such that f s f −∗ = ( f s ) − f c .In the case of semi-regular functions, we can describe the structure of the algebra RM (Ω) adjustingto this situation the already mentioned results given in [11, Proposition 3.12] and [17, Lemma 6.11]. Proposition 2.10.
Let (1 , i, j, k ) be a basis of H . The map ( f , f , f , f ) ∋ ( RM R (Ω)) f + f i + f j + f k ∈ RM (Ω) is a RM R (Ω) -linear isomorphism. In particular RM (Ω) is a -dimensional vector space on RM R (Ω) .Proof. Let f ∈ RM (Ω). Let Ω ′ be a circular subdomain of Ω such that f ∈ R (Ω ′ ) and such that everypoint of Ω \ Ω ′ is either a pole or a removable singularity for f . Proposition 3.12 in [11] guarantees theexistence of a unique 4-tuple f , f , f , f ∈ R R (Ω ′ ) such that f = f + f i + f j + f k . We are left to provethat f , . . . , f ∈ RM R (Ω). If S q is a spherical pole of f then there exists m ∈ N such that ∆ mq · f extendsregularly in an open circular neighborhood U ⊂
Ω of the sphere S q . Now consider the function ∆ mq · f and apply again [11, Proposition 3.12], finding g , . . . , g ∈ R R ( U ) such that ∆ mq · f = g + g i + g j + g k .Nonetheless we also have ∆ mq · f = ∆ mq · f + ∆ mq · f i + ∆ mq · f j + ∆ mq · f k on U \ S q and the uniquenessgiven in [11, Proposition 3.12] ensures ∆ mq · f n = g n on U \ S q , for n = 0 , , ,
3. Last equality shows that f , . . . f have a pole (or a removable singularity) at S q . The case of a real pole is treated analogously,showing that f , . . . f belong to RM R (Ω). (cid:3) The uniqueness of the above statement gives as an immediate consequence that RM R (Ω) is the centerof RM (Ω) and that R R (Ω) is the center of R (Ω). Remark 2.11.
The above proof shows that if f = f + f i + f j + f k ∈ RM (Ω) has a sphere of poles S q of spherical order 2 m , then any point of S q is either a pole of spherical order at most 2 m or a removablesingularity for each of the functions f , . . . , f and that ord f ( S q ) = max { ord f ( S q ) , . . . , ord f ( S q ) } . A. ALTAVILLA AND C. DE FABRITIIS
Zero divisors and idempotents.
From [22, Theorem 6.6] we have that RM (Ω) contains zerodivisors if and only if Ω is a product domain (for a thorough study of the zero set of zero divisors see [19],while [3, Example 3] contains explicit computations for relevant examples; for an interesting applicationof idempotents to function spaces see [25] which sets questions raised in [14]). In this case f is a zerodivisor if and only if f s ≡
0. In the sequel of this paper, we will often make use of the “basic” idempotentsgiven in the following definition.
Definition 2.12.
Let Ω be any product domain and I ∈ S . We define ℓ + ,I : Ω → H and ℓ − ,I : Ω → H as ℓ + ,I ( x + Jy ) = 1 − J I , ℓ − ,I ( x + Jy ) = 1 + J I , where y > ℓ + ,I and ℓ − ,I are idempotents and that the following equalities hold (see [6,Remark2.4]): (cid:0) ℓ + ,I (cid:1) c = 1 − ℓ + ,I = ℓ − ,I , ( ℓ + ,I ) s = ( ℓ − ,I ) s = ℓ + ,I ∗ ℓ − ,I ≡ RM (Ω) showing in particular that they have only removable singu-larities (and therefore, by a slight abuse of notation, we say they are regular). Proposition 2.13.
Let f ∈ RM (Ω) \ { , } . The function f is an idempotent for the ∗ -product if andonly if f belongs to R (Ω) and it is a zero divisor such that f ≡ (and thus f sv ≡ − ).Proof. Suppose f ∈ RM (Ω) is an idempotent, that is f ∗ = f . The previous equality can be written as( f − ∗ f ≡ f is a zero divisor (since f , f = f + f v andthe fact that f v ∗ f v = − f sv the equality f ∗ = f is equivalent to the system(2.4) ( f − f sv = f f f v = f v . Last equality can be also written as (2 f − f v ≡ f v ≡ f ≡ . The firstcase cannot hold since RM R (Ω) does not contain zero divisors; thus f ≡ and the first equality ofsystem (2.4) becomes f sv ≡ − . Then we are left to prove that f is regular. Since f ≡ wheneverdefined, it can be extended regularly to the function on the domain Ω, so it only has removablesingularities. Now suppose f v has a spherical pole in S q of order k , thus there exists a function g v regular on a neighborhood U of S q which has at most one possible isolated zero in S q of order ˜ k < k ,such that(2.5) g v = ∆ kq · f v , on U \ S q (see [22, Theorem 6.4 (2)]). Thanks to [18, Theorem 22 and Remark 14] we can also write(2.6) g v = ( q − w ) ∗ · · · ∗ ( q − w ˜ k ) ∗ γ, where w , . . . , w ˜ k ∈ S q , w n +1 = w cn ( n = 1 , . . . , ˜ k −
1) and γ is never vanishing on S q . Computing thesymmetrized function g sv from equalities (2.5) and (2.6), we obtain∆ kq γ s = g sv = ∆ kq f sv = −
14 ∆ kq . Since γ s is never vanishing on S q , we then obtain ˜ k = k which is a contradiction to the above inequality.The case of a real pole is treated analogously. This shows that f v has no poles and thus f belongs to R (Ω).Straightforward computations show that if f ∈ R (Ω) is such that f ≡ and f s ≡ f sv ≡ − ),then f is an idempotent. (cid:3) QUIVALENCE OF SLICE SEMI-REGULAR FUNCTIONS VIA SYLVESTER OPERATORS 9
The above statement allows us to give an explicit characterization of zero divisors in RM (Ω). Proposition 2.14.
Let f ∈ RM (Ω) be a zero divisor. For any δ ∈ H such that | δ | = 1 and ( f δ ) ,there exits σ = σ ( δ ) ∈ R (Ω) idempotent, such that (2.7) f = 2( f δ ) σδ c . In particular, if f , we can write f = (2 f ) σ for a suitable idempotent σ .Proof. Assume first that f
0, then f −∗ = f − ∈ RM (Ω). Thus, if f = f + f v , we have that f = (2 f ) σ , where σ = 12 + (2 f ) − f v . As f s = 4 f σ s ≡
0, we also have that σ s ≡
0, proving that σ is a zero divisor. Moreover, σ ≡ andProposition 2.13 shows that σ ∈ R (Ω) is an idempotent. Now choose δ ∈ H with | δ | = 1 be such that( f δ )
0; such a δ always exists thanks to Remark 2.6. The fact that ( f δ ) s ≡ f s ≡ f δ isa zero divisor and therefore we can apply the above reasoning obtaining f δ = 2( f δ ) σ, for a suitable idempotent σ and the thesis follows by multiplying both member of the last equality onthe right by δ c . (cid:3) Remark 2.15.
We notice that the proof of the above proposition shows that formula (2.7) can be writtenas soon as ( f δ )
0. If δ and ˜ δ are unitary quaternions such that ( f δ ) f ˜ δ )
0, then we have f = 2( f δ ) σδ c = 2( f ˜ δ ) ˜ σ ˜ δ c , for σ and ˜ σ suitable idempotents. Thus we can write˜ σ = γσδ ′ = σγδ ′ , where γ = ( f ˜ δ ) − ( f δ ) ∈ RM R (Ω) and δ ′ = δ c ˜ δ is a unitary quaternion. Remark 2.16.
Given f ∈ RM (Ω) a zero divisor and η a unitary quaternion such that ( f η )
0, fromformula (2.7), we can also write(2.8) f = 2( f η ) ση c = 2( f η ) η c η ∗ σ ∗ η c = 2( f η ) η c ∗ ρ, where ρ = η ∗ σ ∗ η c is again an idempotent.The proof of Proposition 2.14 shows that if f is a zero divisor with f
0, then we can choose δ = 1and therefore formula (2.7) simplifies to f = 2 f σ .3. RM R -linear endomorphisms The aim of this section is to study a class of RM R (Ω)-linear operators in the space of slice semi-regularfunctions; they will be represented via suitable matrices in Section 4. The class of linear operators weare interested in is described as follows. Definition 3.1.
Consider two N -tuples F := ( f [1] , . . . , f [ N ] ) and G := ( g [1] , . . . , g [ N ] ) ⊂ RM (Ω) \ { } .We denote by L F , G : RM (Ω) → RM (Ω) the RM R (Ω)-linear operator given by(3.1) L F , G ( χ ) := f [1] ∗ χ ∗ g [1] + · · · f [ N ] ∗ χ ∗ g [ N ] . In particular the analysis of the image and the kernel of such operators will give complete informationon the existence and uniqueness of the solution of the equation f [1] ∗ χ ∗ g [1] + · · · f [ N ] ∗ χ ∗ g [ N ] = b , for b ∈ RM (Ω).Since RM R (Ω) is the center of RM (Ω), then ( RM R (Ω) \ { } ) N acts on the N -tuples F and G ofsemi-regular functions as follows: given α = ( α [1] , . . . , α [ N ] ) ∈ ( RM R (Ω) \ { } ) N we denote by α (cid:7) F =( α [1] f [1] , . . . , α [ N ] f [ N ] ) and α ♦ G = ( α − g [1] , . . . , α − N ] g [ N ] ). A straightforward computation shows that L F , G = L α (cid:7) F ,α ♦ G , so that, when needed, we can suppose that G contains only regular functions withoutreal and spherical zeroes.We start our investigation from the easiest case N = 1; to simplify notation we denote L { f } , { g } by L f,g . Our first result classifies the functions f and g such that L f,g is an isomorphism and gives explicitlythe solution of L f,g ( χ ) = b in the case the operator is an isomorphism. Proposition 3.2.
Let f, g ∈ RM (Ω) \ { } .(1) Provided g ∈ R (Ω) has neither real nor spherical zeroes, then L f,g maps R (Ω) to R (Ω) if andonly if f ∈ R (Ω) .(2) The operator L f,g is an isomorphism if and only if neither f nor g are zero divisors.(3) If L f,g is an isomorphism, for any b ∈ RM (Ω) the equation L f,g ( χ ) = b has the unique solution χ = f −∗ ∗ b ∗ g −∗ .(4) If L f,g is an isomorphism, then the solution of L f,g ( χ ) = b belongs to R (Ω) for any b ∈ R (Ω) ifand only if f and g are never vanishing.Proof. (1) If f ∈ R (Ω), then trivially L f,g ( R (Ω)) ⊆ R (Ω). Vice versa, if L f,g ( R (Ω)) ⊆ R (Ω), inparticular we have that L f,g (1) = f ∗ g ∈ R (Ω). Since g has neither real nor spherical zeroes, then f hasneither real nor spherical poles and therefore f ∈ R (Ω), too. (2) If f is a zero divisor, then there exists χ f f ∗ χ f ≡ L f,g ( χ f ) = 0 so that L f,g is not an isomorphism; the same holds for g . Vice versa, assume that L f,g is not an isomorphism;then there exists χ ∈ RM (Ω) \ { } such that L f,g ( χ ) = f ∗ χ ∗ g = 0. If f ∗ χ = 0, then f is a zerodivisor; otherwise the equality ( f ∗ χ ) ∗ g = 0 gives that g is a zero divisor. (3) Since L f,g is an isomorphism, then f and g are not zero divisors and f −∗ and g −∗ belong to RM (Ω). A direct computation shows that L f,g ( f −∗ ∗ b ∗ g −∗ ) = b . (4) If f, g ∈ R (Ω) are never vanishing, then (3) shows that the unique solution of L f,g ( χ ) = b belongsto R (Ω) for any b ∈ R (Ω). Vice versa, if f −∗ ∗ b ∗ g −∗ belongs to R (Ω) for any b ∈ R (Ω), by taking b = g we obtain that f −∗ ∈ R (Ω), implying that f has no zeroes; the same holds for g . (cid:3) Notice that if Ω is a slice domain, then L f,g is always an isomorphism thanks to (2) of the aboveproposition.In the case L f,g is not an isomorphism we give a necessary and sufficient condition on the function b in order it belongs to the image of L f,g . Theorem 3.3.
Let f, g ∈ RM (Ω) \ { } be such that L f,g is not an isomorphism. If f is a zero divisor,for a suitable unitary δ ∈ H , we denote by σ f the idempotent given in formula (2.7) . If g is a zero divisor,for a suitable unitary η ∈ H , we denote by ρ g the idempotent given in formula (2.8) . Then there exists χ such that L f,g ( χ ) = b if and only if b = σ f ∗ b , if f is a zero divisor, and b = b ∗ ρ g , if g is a zero divisor. Remark 3.4.
The relation b = σ f ∗ b can also be written as (1 − σ f ) ∗ b ≡ σ cf ∗ b ≡ δ appearingin formula (2.7). Indeed, if ˜ σ f is another such idempotent, we know that ˜ σ f = σ f γδ ′ for a suitable γ ∈ RM R (Ω) \ { } and δ ′ unitary quaternion, so that σ cf ∗ b ≡ σ cf ∗ b ≡ QUIVALENCE OF SLICE SEMI-REGULAR FUNCTIONS VIA SYLVESTER OPERATORS 11
Proof of Theorem 3.3. If f is a zero divisor and there exists χ such that L f,g ( χ ) = b , then f ∗ χ ∗ g = b and thus f c ∗ b = f c ∗ f ∗ χ ∗ g = f s χ ∗ g ≡
0. Now write f = 2( f δ ) σ f δ c for a suitable unitary quaternion δ and idempotent σ f . The equality f c ∗ b = 2( f δ ) δσ cf ∗ b ≡ σ cf ∗ b ≡
0. As σ cf = 1 − σ f weobtain b = σ f ∗ b . Analogous considerations hold if g is a zero divisor, showing that b = b ∗ ρ g .Vice versa if f is a zero divisor, b = σ f ∗ b and g is not a zero divisor, we have the following chain ofequalities b = σ f ∗ b = (cid:2) f δ ) σ f δ c ((2( f δ ) ) − δ (cid:3) ∗ b ∗ g −∗ ∗ g = f ∗ (cid:2) (2( f δ ) ) − δ ∗ b ∗ g −∗ (cid:3) ∗ g = L f,g ((2( f δ ) ) − δ ∗ b ∗ g −∗ ) , which shows that L f,g ( χ ) = b admits a solution. If f is not a zero divisor, g is a zero divisor and b = b ∗ ρ g ,the thesis follows by reasoning as before.If both f and g are zero divisors, b = σ f ∗ b = b ∗ ρ g , writing f = 2( f δ ) σ f δ c and g = 2( gη ) η c ∗ ρ g ,the following chain of equalities yields the thesis b = σ f ∗ b = (cid:2) f δ ) σ f δ c ((2( f δ ) ) − δ (cid:3) ∗ b = f ∗ (cid:2) (2( f δ ) ) − δ (cid:3) ∗ b = f ∗ (cid:2) (2( f δ ) ) − δ (cid:3) ∗ b ∗ ρ g = f ∗ (cid:2) (2( f δ ) ) − δ (cid:3) ∗ b ∗ (cid:2) (2( gη ) ) − η ∗ ( gη ) η c (cid:3) ∗ ρ g = f ∗ (cid:2) (2( f δ ) ) − δ (cid:3) ∗ b ∗ (cid:2) (2( gη ) ) − η (cid:3) ∗ g = L f,g (cid:0)(cid:2) (2( f δ ) ) − δ (cid:3) ∗ b ∗ (cid:2) (2( gη ) ) − η (cid:3)(cid:1) . (cid:3) We now describe the kernel of L f,g when the operator is not an isomorphism. Proposition 3.5.
Let f, g ∈ RM (Ω) \ { } be such that L f,g is not an isomorphism. If f is a zerodivisor, for a suitable unitary η ∈ H , we denote by ρ f the idempotent given in formula (2.8) . If g is azero divisor, for a suitable unitary δ ∈ H , we denote by σ g the idempotent given in formula (2.7) . Then χ ∈ ker( L f,g ) if and only if(1) ρ f ∗ χ ≡ , if f is a zero divisor and g is not a zero divisor;(2) χ ∗ σ g ≡ , if g is a zero divisor and f is not a zero divisor;(3) ρ f ∗ χ ∗ σ g ≡ if both f and g are zero divisors.Proof. (1) As g is not a zero divisor, then χ ∈ ker( L f,g ) if and only if f ∗ χ ≡
0. Choose a unitaryquaternion η such that ( f η ) f = 2( f η ) η c ∗ ρ f as given in formula (2.8). Now f ∗ χ =2( f η ) η c ∗ ρ f ∗ χ ≡ ρ f ∗ χ ≡ f η ) ∈ RM R (Ω) \ { } and η = 0. (2) This second case is obtained as in (1) by using formula (2.7). (3)
By definition χ ∈ ker( L f,g ) if and only if f ∗ χ ∗ g ≡
0. Choose two unitary quaternion δ and η such that ( f η )
0, ( gδ ) f = 2( f η ) η c ∗ ρ f , as given in formula (2.8), and g = 2( gδ ) σ g δ c as in formula (2.7). Now f ∗ χ ∗ g = 4( f η ) ( gδ ) η c ∗ ρ f ∗ χ ∗ σ g δ c ≡ ρ f ∗ χ ∗ σ g ≡ f η ) , ( gδ ) ∈ RM R (Ω) \ { } and η, δ = 0. (cid:3) Matrix representation of L F , G -type equations The techniques used in the previous section to study the case N = 1 are not powerful enough even tostudy the next step N = 2. To tackle the general case we need to represent the linear equations we aredealing with by means of square matrices in the same spirit of [24].Since we want to use coordinates for RM (Ω) over RM R (Ω), from now on we choose an orthonormalbasis B := (1 , i, j, k ) of H (which by Proposition 2.10 is a basis for RM (Ω) over RM R (Ω), too). Given f = f + f i + f j + f k , we will denote by F B : RM (Ω) → ( RM R (Ω)) the usual coordinates isomorphism F B ( f ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) f f f f (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . Definition 4.1.
For any f = f + f i + f j + f k ∈ RM (Ω) we define ı L ( f ) := (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) f − f − f − f f f − f f f f f − f f − f f f (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , ı R ( f ) := (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) f − f − f − f f f f − f f − f f f f f − f f (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . Lemma 4.2.
For any f, g, h ∈ RM (Ω) , the following equalities hold. ı R ( f ∗ g ) = ı R ( g ) ı R ( f ) ,ı L ( f ) ı R ( g ) = ı R ( g ) ı L ( f ) ,F B ( f ∗ g ) = ı L ( f ) F B ( g ) = ı R ( g ) F B ( f ) ,F B ( f ∗ g ∗ h ) = ı L ( f ) ı L ( g ) F B ( h ) = ı R ( h ) ı R ( g ) F B ( h ) ,F B ( f ∗ g ∗ h ) = ı L ( f ) ı R ( h ) F B ( g ) = ı R ( h ) ı L ( f ) F B ( g ) , (4.1) det ( ı L ( f )) = det ( ı R ( f )) = ( f s ) . (4.2) Proof.
The proof of all equalities can be performed by direct inspection. (cid:3)
Thanks to formula (4.1), for any two N -tuples F = ( f [1] , . . . , f [ N ] ) , G = ( g [1] , . . . , g [ N ] ) ⊂ RM (Ω) \ { } ,the linear operator L F , G given in formula (3.1) can be written as F B ( L F , G )( χ ) = N X n =1 ı L ( f [ n ] ) ı R ( g [ n ] ) ! F B ( χ ) , and since F B is an isomorphism, the solvability of L F , G ( χ ) = b is equivalent to the solvability of F B ( L F , G )( χ ) = F B ( b ). This interpretation allows us to characterize the cases in which the operator L F , G is an isomorphism. Proposition 4.3.
The linear operator L F , G is an isomorphism if and only if det N X n =1 ı L ( f [ n ] ) ı R ( g [ n ] ) ! . Remark 4.4.
Last proposition gives a more algebraic interpretation of Proposition 3.2 (2) . Indeed, when N = 1 we have that L f,g is an isomorphism if and only if det( ı L ( f ) ı R ( g )) = det( ı L ( f )) det( ı R ( g )) ı L ( f )) det( ı R ( g )) = ( f s ) ( g s ) , and the second term is identically zero if and only if either f s or g s vanish identically, which is thecondition that characterizes zero divisors and identically zero functions.From now on, we focus our attention on a particular class of L F , G , namely the cases when N = 2, F = ( f, G = (1 , g ). QUIVALENCE OF SLICE SEMI-REGULAR FUNCTIONS VIA SYLVESTER OPERATORS 13
Definition 4.5.
Let f, g ∈ RM (Ω). The
Sylvester operator S f,g associated to f and g is the RM R (Ω)-linear operator given by S f,g ( χ ) := L ( f, , (1 ,g ) = f ∗ χ + χ ∗ g. The associated
Sylvester equation with “constant term” b , is the RM R (Ω)-linear equation given by(4.3) S f,g ( χ ) = b . The name of “Sylvester operator” is due to the fact that, when dealing with matrices, equation (4.3)is usually called
Sylvester equation . Remark 4.6.
In the case when a , a , b , b ∈ H \ { } , it is always possible to write the expression a qb + a qb as a ( a − a q + qb b − ) b and then the solvability of a qb + a qb = p is equivalent to thesolvability of ( a − a ) q + q ( b b − ) = a − pb − , which is the Sylvester equation associated to a − a and b b − . In the case of slice (semi-)regular functions, the possible presence of zero divisors and the factthat the ∗ -inverse of a regular function is not always a regular function is an obstruction to the reductionof the general case to the Sylvester case.The following proposition shows that the Sylvester equation associated to f and g is also associatedto a wider family of functions. Proposition 4.7.
Let f, g ∈ RM (Ω) \ { } . Then for any α ∈ RM R (Ω) , we have (4.4) S f,g = S f + α,g − α . Proof.
Indeed, for any χ ∈ RM (Ω), we have S f + α,g − α ( χ ) = f ∗ χ + α ∗ χ + χ ∗ g + χ ∗ ( − α ) = f ∗ χ + χ ∗ g = S f,g ( χ ) , since RM R (Ω) is the center of RM (Ω). (cid:3) We notice that, if g v ≡
0, then S f,g = S f + g , = L f + g , ; analogously, if f v ≡
0, then S f,g = S ,f + g = L ,f + g . Since the operators of the class L f,g were thoroughly studied in Section 3, from now on, withoutloss of generality, we shall work under the following Assumption 4.8.
We consider S f,g where neither f nor g belong to RM R (Ω).We now give two definitions that will be useful to study the invertibility of S f,g . Definition 4.9.
Let f, g ∈ RM (Ω). We say that f and g are equivalent and write f ≃ g if there existsa ∗ -invertible h ∈ RM (Ω), such that f = h −∗ ∗ g ∗ h. Lemma 4.10. If f ≃ g , then f ≡ g and f s ≡ g s (that implies also f sv ≡ g sv ). In particular, if f ≃ g ,then f is a zero divisor if and only if g is.Proof. If we write g = g + g v , we then have f = h − s h c ∗ g ∗ h = h − s h c ∗ ( g + g v ) ∗ h = h − s h c g h + h − s h c ∗ g v ∗ h = g + h − s h c ∗ g v ∗ h. Then, in order to prove that f = g , it is enough to show that ( h − s h c ∗ g v ∗ h ) ≡
0. As h − s ∈ RM R (Ω),we are left to show that ( h c ∗ g v ∗ h ) ≡
0; indeed we have( h c ∗ g v ∗ h ) c = h c ∗ g cv ∗ h = − h c ∗ g v ∗ h, and the equality f = g is proven. The equality f s = g s is now straightforward.Last assertion follows immediately from the fact that f is a zero divisor if and only if f s ≡ g . (cid:3) An accurate study of the operator S f,g will show that, if f, g
6∈ RM R (Ω), then the equalities f = g and f sv = g sv imply f ≃ g (see Corollary 5.2 if the domain is slice and Corollary 7.2 in the general case).We now pass to the announced second definition. Definition 4.11.
Let f, g ∈ RM (Ω). We say that the couple ( f, g ) intertwines with (a zero divisor) σ ,if there exists a zero divisor σ such that f ∗ σ = σ ∗ g. Example 4.12.
Let Ω be a product domain and choose f and g such that f = g f v a zero divisorand g v ≡
0, then we have f sv = g sv ≡
0. We claim that f g and that the couple of functions ( f, g )intertwines with an idempotent. Indeed, if there exists h ∈ RM (Ω) invertible such that f = h −∗ ∗ g ∗ h , as g v ≡ f ≡ g , which contradicts the fact that f v is a zero divisor. Now, write f v = 2( f v η ) η c ∗ ρ for a suitable unitary η ∈ H and ρ idempotent as in equation (2.8). Since ρ ∗ ρ c ≡ g = g = f , wehave f ∗ ρ c = ( f + f v ) ∗ ρ c = f ρ c + 2( f v η ) η c ∗ ρ ∗ ρ c = f ρ c = g ρ c = ρ c ∗ g. Next proposition characterizes the non-invertibility of S f,g in terms of the previous definitions. Proposition 4.13.
Given f, g ∈ RM (Ω) , then S f,g is not an isomorphism if and only if one of the twofollowing conditions holds(1) f ≃ − g ;(2) there exist a zero divisor χ such that ( f, − g ) intertwines with χ .Proof. The operator S f,g is not an isomorphism if and only if there exists χ ∈ RM (Ω) \ { } such that f ∗ χ + χ ∗ g ≡
0. If χ is not a zero divisor, then it is invertible in RM (Ω) and − g = χ −∗ ∗ f ∗ χ exactlymeans f ≃ − g . If χ is a zero divisor, then f ∗ χ + χ ∗ g ≡ f, − g )intertwines with χ . (cid:3) Notice that the first condition says that there exists an invertible χ ∈ ker( S f,g ), while the second onemeans that a zero divisor belongs to ker( S f,g ). Remark 4.14.
Trivially, if Ω is a slice domain, for any f, g ∈ RM (Ω), the kernel of S f,g cannot containzero divisors, so (2) . can never take place and thus S f,g is not an isomorphism if and only if f ≃ − g .Together with the previous remark, the following examples show that the two cases stated in Propo-sition 4.13 are not related. Example 4.15.
Let Ω be a product domain and set f = 1 − J i, g = f j = (1 − J i ) j = j − J k. It is easily seen that χ = f c ∈ ker( S f,g ), while f and − g have different “real parts” and therefore, thanksto Lemma 4.10, they are not equivalent. Example 4.16.
Let σ ∈ R (Ω) be an idempotent and set f = σ, g = − σ . Trivially any χ ∈ RM R (Ω)belongs to ker S f,g , as well as χ = σ c .5. The rank of the Sylvester operator
We begin this section with a characterization of the invertibility of S f,g by means of the matrixrepresentation given in Section 4. To simplify notation, from now on, we set S f,g = ı L ( f ) + ı R ( g ) . QUIVALENCE OF SLICE SEMI-REGULAR FUNCTIONS VIA SYLVESTER OPERATORS 15
Proposition 5.1.
Given f = f + f v , g = g + g v ∈ RM (Ω) , the characteristic polynomial of the RM R (Ω) -linear operator S f,g is given by p ( λ ) = ( f + g − λ ) [( f + g − λ ) + 2( f sv + g sv )] + ( f sv − g sv ) . In particular S f,g is an isomorphism if and only if (5.1) ( f + g ) [( f + g ) + 2( f sv + g sv )] + ( f sv − g sv ) . Moreover, the rank of S f,g is always strictly greater than .Proof. First of all, given f = f + f i + f j + f k and g = g + g i + g j + g k , we write(5.2) S f,g := ı L ( f ) + ı R ( g ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) f + g − ( f + g ) − ( f + g ) − ( f + g ) f + g f + g − ( f − g ) f − g f + g f − g f + g − ( f − g ) f + g − ( f − g ) f − g f + g (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . A long but straightforward computation gives p ( λ ) = det( S f,g − λ ) =( f + g − λ ) + 2( f + g − λ ) ( f + f + f + g + g + g )+ ( f − g + f − g + f − g ) =( f + g − λ ) [( f + g − λ ) + 2( f sv + g sv )] + ( f sv − g sv ) (5.3) = λ − (cid:2) f + g ) (cid:3) λ + (cid:2) f sv + g sv + 3( f + g ) ) (cid:3) λ − (cid:2) f + g )(( f + g ) + f sv + g sv ) (cid:3) λ + ( f + g ) [( f + g ) + 2( f sv + g sv )] + ( f sv − g sv ) Thanks to Proposition 4.3, we have that S f,g is an isomorphism if and only if det( S f,g ) = ( f + g ) [( f + g ) + 2( f sv + g sv )] + ( f sv − g sv ) S f,g has rank less than 2. Then λ = 0 is an eigenvalue of algebraic multiplicity atleast 3, which gives(5.4) f sv + g sv + 3( f + g ) ≡ f + g )(( f + g ) + f sv + g sv ) ≡ f + g ) [( f + g ) + 2( f sv + g sv )] + ( f sv − g sv ) ≡ . The second equation is equivalent to either f + g ≡ f + g ) + f sv + g sv ≡
0. In the first case,since either f v + g v or f v − g v are not identically zero because of Assumption 4.8, we can find a 2 × S f,g with determinant different from zero, which is a contradiction. In the second case, thefirst equation of system (5.4) together with ( f + g ) + f sv + g sv ≡ ( f sv + g sv + 3( f + g ) ≡ f + g ) + f sv + g sv ≡ , which again entails f + g ≡ (cid:3) Last proposition allows us to prove that in the case of slice domains the relation f ≃ g means exactly f = g and f s ≡ g s . Indeed this holds even for product domains, but the proof of this fact will requirea much deeper investigation on the kernel of S f,g . Corollary 5.2.
Let f, g ∈ RM (Ω) and Ω be a slice domain. Then f ≃ g if and only if f ≡ g and f s ≡ g s (that is f sv ≡ g sv ). Proof.
The necessity of the condition was shown in Lemma 4.10. To prove its sufficiency, we noticethat, if f ≡ g and f s ≡ g s , then det( S f, − g ) ≡
0, hence S f, − g is not an isomorphism and thereforeker( S f, − g ) = ∅ . As Ω contains real points, there are no zero divisors in RM (Ω) and therefore ker( S f, − g )contains an invertible χ , which shows that f ≃ g . (cid:3) Next result gives a more precise characterization of the rank of S f,g when f + g = 0. Proposition 5.3.
Let f, g ∈ RM (Ω) be such that f = − g , then rk ( S f,g ) = 2 if and only if f sv = g sv .In particular if f ≃ − g , then rk ( S f,g ) = 2 .Proof. Since f = − g , Proposition 5.1 gives that rk( S f,g ) = 4 if and only if f sv g sv . So we are left tocompute the rank of S f,g when f sv = g sv . The hypotheses f = − g implies that S f,g is skew symmetric,then it is enough to compute the determinants of the first ( m, n )-minors D m,n , with 1 ≤ m < n ≤ D , = ( f − g )( f sv − g sv ) = 0 D , = ( g − f )( f sv − g sv ) = 0 D , = ( f − g )( f sv − g sv ) = 0 D , = ( f + g )( f sv − g sv ) = 0 D , = ( f + g )( f sv − g sv ) = 0 D , = ( f + g )( f sv − g sv ) = 0then the rank of S f,g is less than or equal to 2. As we proved in Proposition 5.1 that the rank of S f,g isalways strictly greater than 1, we are done. (cid:3) We now give two examples in which S f,g is not an isomorphism and f + g Example 5.4.
Let Ω be a product domain and set f = J i and g = 1 + 2 J k . Then f + g = 1, f sv ≡ − g sv ≡ −
4. A direct computation shows that the characteristic polynomial in this case is equalto λ − λ − λ + 16 λ , thus λ = 0 has algebraic multiplicity 1 and rk( S f,g ) = 3. Example 5.5.
Let Ω be a product domain and define f and g as in Example 4.15. Then f = g = 1, f = g = −J , f = f = g = g ≡ f sv = − g sv ≡
0. A direct computation shows that thecharacteristic polynomial is equal to λ − λ + 4 λ , thus λ = 0 has algebraic multiplicity 2. Nonethelessa direct computation of S f,g shows that also in this case we have rk( S f,g ) = 3.We underline that in both examples, rk( S f,g ) equals 3; nonetheless in the first case the eigenvalue 0has algebraic multiplicity equal to 1, whilst in the second one it has algebraic multiplicity equal to 2.Inspired by these instances, we prove that if S f,g is not an isomorphism and f + g
0, then the rankof S f,g is always equal to 3. Theorem 5.6.
Let f, g ∈ RM (Ω) be such that S f,g is not an isomorphism. Then f + g if andonly if S f,g has rank .Proof. If f + g ≡ S f,g is equal to 2.Now suppose that f + g S f,g . If 0 is an eigenvalueof algebraic multiplicity 1, then trivially the rank of S f,g is equal to 3.Therefore we are left to deal with the case in which 0 is an eigenvalue of algebraic multiplicity atleast 2, which by formula (5.3) and f + g ( ( f + g ) + f sv + g sv ≡ f + g ) [( f + g ) + 2( f sv + g sv )] + ( f sv − g sv ) ≡ , which is equivalent to(5.5) ( ( f + g ) + f sv + g sv ≡ f sv g sv ≡ . QUIVALENCE OF SLICE SEMI-REGULAR FUNCTIONS VIA SYLVESTER OPERATORS 17
Since RM R (Ω) is a field, then either f sv or g sv is identically zero. We perform the computation in thefirst case, being the second one completely analogous. Thus System (5.5) gives f sv ≡ f + g ) + g sv ≡ . Since rk( S f,g ) = 3 if and only if the cofactor matrix of S f,g is not identically zero, we suppose bycontradiction that cof( S f,g ) = 0 which in particular implies cof( S f,g ) + cof( S f,g ) T = 0. Up to a factor2( f + g )
0, the elements of this matrix in positions (1 , ,
3) and (1 ,
4) give the following systemof equalities g f − f g ≡ g f − f g ≡ g f − f g ≡ , which means f v ∧ * g v ≡
0. By [4, Proposition 2.10] this entails that f v and g v are linearly dependent over RM R (Ω). Nonetheless f sv ≡ g sv = − ( f + g )
0. As f v
0, this is a contradiction which showsthat rk( S f,g ) = 3. (cid:3) Remark 5.7.
Notice that, the fact that S f,g has rank 3 is symmetric in f and g . Indeed, Proposition 5.1,via Formula (5.1), guarantees that S f,g is an isomorphism if and only if S g,f is. Now it is enough tohighlight that the condition on the sum of the “real parts” given in Theorem 5.6 is symmetric.6. The solution of the Sylvester equation in the non-singular case
In this section, we study the case in which S f,g is an isomorphism, looking for the solution of theSylvester equation L f,g ( χ ) = b , given f, g, b ∈ RM (Ω). Some of the tools we introduce are inspired bythe work of Bolotnikov [8, 9].First of all, we notice that Proposition 4.7 allows us to consider the Sylvester equation only in thecases in which neither f nor g are zero divisors, as a consequence of the following Lemma 6.1.
For any f, g ∈ RM (Ω) there exists α ∈ RM R (Ω) such that neither f + α nor g − α arezero divisors.Proof. If neither f nor g are zero divisors, we can take α ≡
0. If f is a zero divisor, then f s = f + f sv ≡ f + α ) s = 2 αf + α = α (2 f + α ) ≡ α ≡ α ≡ − f . Since ( g − α ) s = α − g α + g s , it is enough to choose α any real number such that α = 0, α
6≡ − f and α − g α + g s f + α nor g − α are zero divisors. (cid:3) Notice that Lemma 6.1 and equality (4.4) only deal with “real parts” of the functions f and g , whileAssumption 4.8 only deals with their “vectorial parts”, so that they are independent. Assumption 6.2.
Without any loss of generality, in this section we shall consider only Sylvester operatorsassociated to functions f, g
6∈ RM R (Ω) none of which is a zero divisor. Definition 6.3.
Let f = f + f v , g = g + g v ∈ RM (Ω). If f is not a zero divisor, we define λ L ∈ RM (Ω),as λ L := 2 g + f + g s f −∗ . If g is not a zero divisor, we define λ R ∈ RM (Ω), as λ R := 2 f + g + f s g −∗ . Notice that, if f is not a zero divisor, then λ L ≡ λ L ∗ f ≡ f ∗ +2 g f + g s ≡
0. Analogously, if g is not a zero divisor, then λ R ≡ g ∗ + 2 f g + f s ≡ Proposition 6.4.
Let f, g ∈ RM (Ω) be such that f ≃ − g . If f (and then g ) is not a zero divisor, then λ L = λ R ≡ .Proof. Thanks to Lemma 4.10, we know that f is a zero divisor if and only if − g is; moreover, f ≡ − g and f sv ≡ g sv .If f is not a zero divisor, then λ L ≡ f ∗ + 2 g f + g s ≡
0. The following chain ofequalities yields the first part of the equality in the statement f ∗ + 2 g f + g s = f − f sv + 2 f f v + 2 g f + 2 g f v + g + g sv = ( f + g ) + 2( f + g ) f v + g sv − f sv ≡ . The second part follows by similar computations. (cid:3)
We now give a partial converse of the previous proposition.
Proposition 6.5.
Let Ω be a slice domain and f = f + f v , g = g + g v ∈ RM (Ω) \ { } . Then f ≃ − g if and only if λ L ≡ if and only if λ R ≡ .Proof. First of all notice that, being Ω a slice domain and f, g
0, both λ L and λ R are well defined.Thanks to Proposition 6.4, we are left to prove that λ L ≡ f ≃ − g . If λ L ≡
0, we have that f ∗ + 2 g f + g s ≡
0. Last quantity can also be written as f − f sv + 2 f f v + 2 g f + 2 g f v + g + g sv andhence, by splitting in “real” and “vector” parts, we obtain the following system of equations(6.1) ( f − f sv + 2 g f + g + g sv ≡ f + g ) f v ≡ . Since RM R (Ω) is a field, the second equation is satisfyied if and only if either f ≡ − g or f v ≡
0. If f ≡ − g , the first equation of system (6.1) becomes − f sv + g sv ≡
0, that is g sv ≡ f sv and corollary 5.2 entails f ≃ − g . If f + g
0, then f v ≡
0. The first equation of system (6.1) then becomes ( f + g ) + g sv ≡ f + g ) ≥ g sv ≥ f + g ) = 0 only occurs on a discrete set. (cid:3) If S f,g is an isomorphism we are now able to write explicitly the solution of S f,g ( χ ) = b . Recall that,by Assumption 6.2, neither f nor g are zero divisors. Theorem 6.6.
Let f, g ∈ RM (Ω) be such that S f,g is an isomorphism. Then for any b ∈ RM (Ω) , theunique solution of S f,g ( χ ) = b is given by χ = λ −∗ L ∗ ( b + f −∗ ∗ b ∗ g c ) = ( b + f c ∗ b ∗ g −∗ ) ∗ λ −∗ R , where λ L and λ R are given by Definition 6.3.Proof. As f and g are not zero divisors, then both λ L and λ R are well defined. We now prove that both λ sL and λ sR are not identically zero. Since f is not a zero divisor, then λ L is invertible if and only if λ sL f ∗ λ L ) s
0. Now we have3( f ∗ λ L ) s =(2 g f + f ∗ + g s ) s = 4 g f s + f s + g s + 4 g h f, f ∗ i ∗ + 4 g g s f + 2 g s ( f ∗ ) =4 g f + 4 g f sv + f + 2 f f sv + g + 2 g g sv + 4 g f + 4 g f f sv + 4 g f + 4 f g g sv + 2 g f − g f sv + 2 f g sv + ( f sv ) − f sv g sv + ( g sv ) =( f + g ) + 2[( g + f + 2 f g ) f sv + ( g + 2 g f + f ) g sv ] + ( f sv − g sv ) =( f + g ) [( f + g ) + 2( f sv + g sv )] + ( f sv − g sv ) . As S f,g is an isomorphism, by Proposition 5.1 we have that last term is not identically zero and hence λ L is invertible. An analogous computation gives that λ R is invertible. QUIVALENCE OF SLICE SEMI-REGULAR FUNCTIONS VIA SYLVESTER OPERATORS 19
Now, for any χ ∈ RM (Ω) we have the following chain of equalities f −∗ ∗ S f,g ( χ ) ∗ g c + S f,g ( χ ) = f −∗ ∗ ( f ∗ χ + χ ∗ g ) ∗ g c + f ∗ χ + χ ∗ g = χ ∗ g c + f −∗ ∗ χ ∗ g s + f ∗ χ + χ ∗ g = χ ( g + g c ) + g s f −∗ ∗ χ + f ∗ χ = 2 g χ + g s f −∗ ∗ χ + f ∗ χ = (2 g + g s f −∗ + f ) ∗ χ = λ L ∗ χ. Therefore, if χ is the solution of S f,g ( χ ) = b , we obtain f −∗ ∗ b ∗ g c + b = λ L ∗ χ , which gives χ = λ −∗ L ∗ ( f −∗ ∗ b ∗ g c + b ) . The second equality of the statement is obtained analogously. (cid:3) Sylvester operators of rank S f,g has rank 2; by Proposition 5.3 and Theo-rem 5.6 this means exactly that f = − g and f sv = g sv (we recall that, by Assumption 4.8, both f v and g v are not identically zero). Next statement describes the kernel of S f,g under the conditions f = − g and f sv = g sv . Theorem 7.1.
Let f, g ∈ RM (Ω) be such that f = − g and f sv = g sv . Then (7.1) ker( S f,g ) = { f ∗ h + h ∗ g c | h ∈ RM (Ω) } . Moreover, it is possible to find a basis of ker( S f,g ) consisting of invertible elements.Proof. Notice that, since f = − g , for any h ∈ RM (Ω) we have S f,g = S f v ,g v and f ∗ h + h ∗ g c = f v ∗ h − h ∗ g v . Then S f,g ( f v ∗ h − h ∗ g v ) = f v ∗ ( f v ∗ h − h ∗ g v ) − ( f v ∗ h − h ∗ g v ) ∗ g v = − f sv ∗ h − f v ∗ h ∗ g v + f v ∗ h ∗ g v + h ∗ g sv ≡ . The hypotheses on f and g together with Proposition 5.3 guarantee that in order to prove the equalityof the two subspaces in formula (7.1), it is enough to show that the RM R (Ω)-linear subspace { f v ∗ h − h ∗ g v | h ∈ RM (Ω) } has dimension at least 2. If h = h + h v we have f v ∗ h − h ∗ g v = h ( f v − g v ) − h f v , h v i ∗ + f v ∧ * h v + h g v , h v i ∗ − h v ∧ * g v = h g v − f v , h v i ∗ + [ h ( f v − g v ) + ( f v + g v ) ∧ * h v ] , where the first summand belongs to RM R (Ω) and the second has “real part” equal to zero. If f v = g v we take δ ∈ S such that h g v − f v , δ i ∗
0. Then f v ∗ − ∗ g v and f v ∗ δ − δ ∗ g v are linearly independentsince the first has “real part” equal to zero and it is not identically zero, while the second has “realpart” equal to h g v − f v , δ i ∗
0. If f v = g v , we have f v ∗ h − h ∗ g v = 2 f v ∧ * h v . As f v
0, we canfind two imaginary units
I, J ∈ S , such that 2 f v ∧ * I and 2 f v ∧ * J are linearly independent, showing that { f v ∗ h − h ∗ g v | h ∈ RM (Ω) } has dimension at least 2 and thus proving equality (7.1).We now prove the existence of a basis of invertible elements. We start by computing explicitly ( f v ∗ h − h ∗ g v ) s ; for h ∈ RM (Ω) we have( f v ∗ h − h ∗ g v ) s = f sv h s + g sv h s − h f v ∗ h, h ∗ g v i ∗ = 2( f sv h s − h f v ∗ h, h ∗ g v i ∗ ) . For any unitary δ ∈ H , we set h ≡ δ and find( f v ∗ δ − δ ∗ g v ) s = 2( f sv − h f v ∗ δ, δ ∗ g v i ∗ ) = 2( f sv − h f v , δ ∗ g v ∗ δ c i ∗ ) . First of all we want to show that there exists an invertible element in ker( S f,g ). Indeed, if this is not, wehave that ( f v ∗ δ − δ ∗ g v ) s ≡ δ ∈ H . In particular, choosing δ = 1 , i, j, k , we obtain f sv ≡ h f v , g v i ∗ ≡ f g + f g + f g f sv ≡ h f v , − i ∗ g v ∗ i i ∗ ≡ f g − f g − f g f sv ≡ h f v , − j ∗ g v ∗ j i ∗ ≡ − f g + f g − f g f sv ≡ h f v , − k ∗ g v ∗ k i ∗ ≡ − f g − f g + f g . Adding up all four equations we find f sv (= g sv ) ≡
0. Adding up the first equation with the second, thirdand fourth one, we find f g ≡ f g ≡ f g ≡
0. Since RM R (Ω) is a field, at least one between f v and g v has two components which are identically zero. This, together with f sv (= g sv ) ≡
0, implies thateither f v ≡ g v ≡
0, contradicting Assumption 4.8.Since we found an invertible element τ ∈ ker( S f,g ) we can complete it to a basis ( τ , τ ). If both τ and τ are invertible, we are done. Otherwise consider the following linear combination ατ + τ which islinearly independent from τ for any α ∈ RM R (Ω). We have( ατ + τ ) s = α τ s + 2 α h τ , τ i ∗ = α ( ατ s + 2 h τ , τ i ∗ ) . Therefore it is enough to chose α α τ − s h τ , τ i ∗ to obtain the required basis. (cid:3) The full strength of Theorem 7.1 discloses in the following corollary which states that two functions f, g ∈ RM (Ω) \ RM R (Ω) are equivalent if and only if f ≡ g and f sv ≡ g sv . Indeed, the existence of aninvertible element in ker( S f,g ) implies that f and g are equivalent; thus an operatorial result is appliedto function theory in order to give a necessary and sufficient condition for the equivalence of a couple ofslice semi-regular functions. Corollary 7.2.
Let f, g ∈ RM (Ω) \ RM R (Ω) be such that f ≡ g and f sv ≡ g sv . Then f ≃ g .Proof. Consider the operator S f, − g . Theorem 7.1 guarantees the existence of an invertible h ∈ ker( S f, − g ),that is S f, − g ( h ) = f ∗ h − h ∗ g ≡
0. This equality can also be written as h −∗ ∗ f ∗ h = g , i.e. f ≃ g . (cid:3) Under suitable hypotheses, it is possible to describe ker( S f,g ) in a simpler way. Corollary 7.3.
Let f, g ∈ RM (Ω) be such that f ≃ − g and ( f v − g v ) s . Then ker( S f,g ) = Span RM R (Ω) ( f v − g v , f sv + g sv + 2 f v ∗ g v ) . Proof. As f v − g v = f ∗ ∗ g c and f sv + g sv + 2 f v ∗ g v = 2 f sv + 2 f v ∗ g v = f ∗ ( − f v ) + ( − f v ) ∗ g c , wehave that Span RM R (Ω) ( f v − g v , f sv + g sv + 2 f v ∗ g v ) ⊆ ker( S f,g ) . To show the equality it is sufficient to prove that f v − g v , f sv + f v ∗ g v are linearly independent. Since f v − g v f sv + g sv + 2 f v ∗ g v = f sv + g sv − h f v , g v i ∗ + 2 f v ∧ * g v has “real part”equal to 2( f sv − h f v , g v i ∗ ) = ( f v − g v ) s
0, then we are done. (cid:3)
The above result allows us to understand under which conditions on f and g , the kernel of S f,g containsa zero divisor; obviously what follows is of interest only if Ω is a product domain. Proposition 7.4.
Let f, g ∈ RM (Ω) \ RM R (Ω) be such that f ≃ − g . Then ker( S f,g ) contains a zerodivisor if and only if one of the following conditions holds(1) f v = g v and f sv has a square root;(2) f v = g v and ( f v − g v ) s ≡ ;(3) ( f v − g v ) s and f sv has a square root. QUIVALENCE OF SLICE SEMI-REGULAR FUNCTIONS VIA SYLVESTER OPERATORS 21
Proof. If f v = g v then ker( S f,g ) = ker( S f v ,f v ) = { f v ∗ h − h ∗ f v | h ∈ RM (Ω) } = { f v ∧ * h v | h ∈ RM (Ω) } .Since f v
0, we can choose an orthonormal basis (1 , i, j, k ) ⊂ H such that f
0. Thus a basis ofker( S f,g ) is given by f v ∧ * j = − f i + f k and f v ∧ * k = f i − f j . Now suppose that ker( S f,g ) contains azero divisor. If f v ∧ * j is a zero divisor, then f + f ≡ f sv = f + f + f = f has a squareroot. If f v ∧ * j is not a zero divisor, then there exists α ∈ RM R (Ω) such that α ( f v ∧ * j ) + f v ∧ * k is a zerodivisor which can also be written as0 ≡ ( α ( f v ∧ * j ) + f v ∧ * k ) s = (( f − αf ) i − f j + αf k ) s = α ( f + f ) − αf f + f + f . By multiplying last term by f + f we equivalently obtain ( α ( f + f ) − f f ) = − f ( f + f + f ) =( J f ) f sv , showing that f sv has a square root. Vice versa, if f + f ≡
0, then f v ∧ * j is a zero divisor whichbelongs to ker( S f,g ). Otherwise, if f + f f sv has a square root ρ , a long but straightforwardcomputation of its symmetrized function shows that( f f + J f ρ ) f v ∧ * j + ( f + f ) f v ∧ * k is a zero divisor which belongs to ker( S f,g ).Now assume f v = g v . If ( f v − g v ) s ≡
0, then f v − g v is a zero divisor which belongs to ker( S f,g ).Finally, if ( f v − g v ) s
0, Corollary 7.3 states that f v − g v , f sv + f v ∗ g v is a basis of ker( S f,g ).Then, there exists α ∈ RM R (Ω) such that α ( f v − g v ) + f sv + f v ∗ g v is a zero divisor if and only if( α ( f v − g v ) + f sv + f v ∗ g v ) s ≡
0. We first compute h f v − g v , f sv + f v ∗ g v i ∗ . Since f v − g v has no “realpart”, we have h f v − g v , f sv + f v ∗ g v i ∗ = h f v − g v , f sv − h f v , g v i ∗ + f v ∧ * g v i ∗ = h f v − g v , f v ∧ * g v i ∗ ≡ . As a consequence we obtain( α ( f v − g v )+ f sv + f v ∗ g v ) s = α ( f v − g v ) s +( f sv + f v ∗ g v ) s = α ( f v − g v ) s + f sv ( f v − g v ) s = ( f v − g v ) s ( α + f v ) . Since ( f v − g v ) s
0, there exists α ∈ RM R (Ω) such that α ( f v − g v ) + f sv + f v ∗ g v is a zero divisor if andonly if α + f sv ≡ J , last equality is equivalent to saying that f sv has a squareroot. (cid:3) For a detailed study of the existence of a square root for slice preserving functions see [4, Section 3].We now describe the image of S f,g , giving necessary and sufficient conditions on b for the existence ofa solution of the equation S f,g ( χ ) = b together with an explicit description of a particular solution. Proposition 7.5.
Let f, g ∈ RM (Ω) with f = − g and f sv = g sv . Then S f,g ( χ ) = b has a solution ifand only if f c ∗ b + b ∗ g ≡ . Proof. If χ is a solution of S f,g ( χ ) = b , then b = f ∗ χ + χ ∗ g . We now have f c ∗ b + b ∗ g = f c ∗ ( f ∗ χ + χ ∗ g ) + ( f ∗ χ + χ ∗ g ) ∗ g = f s χ + f c ∗ χ ∗ g + f ∗ χ ∗ g + χ ∗ g ∗ = f s χ + 2 f χ ∗ g + χ ∗ g ∗ = χ ∗ ( f + f sv + 2 f g + 2 f g v + g − g sv + 2 g g v ) ≡ , since the fact that f ≃ − g entails f = − g and f sv = g sv .Assume now that f c ∗ b + b ∗ g ≡
0. We prove that b belongs to the image of S f,g by giving a differentdescription of this linear subspace via the matrix S f,g . Thanks to our hypothesis and to Proposition 5.3,we have that S f,g is skew symmetric and has rank 2. We now look for a square matrix M whose kernel coincides with the image of S f,g , which means rk M = 2 and M · S f,g = 0. Then b belongs to the imageof S f,g if and only if it belongs to ker M . Since f sv = g sv , a straightforward computation shows that M = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) f − g − ( f + g ) f + g f − g − ( f + g ) f + g f − g f + g − ( f + g )0 f − g f − g f − g (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , satisfies M · S f,g = 0. In particular the image of S f,g is contained in the kernel of M which therefore hasrank less or equal than 2. Since at least one between f v and g v is not identically zero, then, by directinspection we have that rk M = 2 which ensures that the image of S f,g coincides with ker M .Then writing b = b + b i + b j + b k we obtain that S f,g ( χ ) = b has a solution if and only if F B ( b ) ∈ ker M , that is(7.2) ( f − g ) b − ( f + g ) b +( f + g ) b = 0( f − g ) b − ( f + g ) b +( f + g ) b = 0( f − g ) b +( f + g ) b − ( f + g ) b = 0( f − g ) b +( f − g ) b +( f − g ) b = 0We now claim that the above system is a translation in coordinates of the equality f c ∗ b + b ∗ g ≡ f = − g , the equality f c ∗ b + b ∗ g ≡ f v ∗ b − b ∗ g v ≡ b = b + b v and splitting the “real” and “vector” parts of f v ∗ b − b ∗ g v ≡
0, we obtain theequivalent system ( h f v , b v i ∗ − h g v , b v i ∗ ≡ b f v + f v ∧ * b v − b g v − b v ∧ * g v ≡ . The properties of the scalar product h ., . i ∗ and of the ∧ * -product yield ( h f v − g v , b v i ∗ ≡ b ( f v − g v ) + ( f v + g v ) ∧ * b v ≡ . A direct check shows that, up to a rearrangements of lines, this last system coincides with system (7.2) (cid:3)
Next proposition describes a family of particular solutions of the equation S f,g ( χ ) = b . Proposition 7.6.
Let f, g ∈ RM (Ω) with f ≃ − g . If f c ∗ b + b ∗ g ≡ , then for any h = h v , k = k v ∈RM (Ω) , such that h f v , h v i ∗ + h g v , k v i ∗ , we have that χ = − (2 h f v , h v i ∗ + 2 h g v , k v i ∗ ) − ( h ∗ b + b ∗ k ) is a solution of S f,g ( χ ) = b .Proof. Being h f v , h v i ∗ + h g v , k v i ∗ ∈ RM R (Ω) \ { } , then − (2 h f v , h v i ∗ + 2 h g v , k v i ∗ ) − ( h ∗ b + b ∗ k ) is welldefined. As f = − g and f v ∗ b = b ∗ g v , the thesis is an immediate consequences of the following chainof equalities S f,g ( h ∗ b + b ∗ k ) = f ∗ ( h ∗ b + b ∗ k ) + ( h ∗ b + b ∗ k ) ∗ g = f ( h ∗ b + b ∗ k ) + g ( h ∗ b + b ∗ k ) + f v ∗ ( h ∗ b + b ∗ k ) + ( h ∗ b + b ∗ k ) ∗ g v = f v ∗ h ∗ b + f v ∗ b ∗ k + h ∗ b ∗ g v + b ∗ k ∗ g v = f v ∗ h ∗ b + b ∗ g v ∗ k + h ∗ f v ∗ b + b ∗ k ∗ g v = ( f v ∗ h v + h v ∗ f v ) ∗ b + b ∗ ( g v ∗ k v + k v ∗ g v )= − h f v , h v i ∗ ∗ b − b ∗ h g v , k v i ∗ = − h f v , h v i ∗ + h g v , k v i ∗ ) b . (cid:3) QUIVALENCE OF SLICE SEMI-REGULAR FUNCTIONS VIA SYLVESTER OPERATORS 23
Remark 7.7.
Notice that there always exist h, k ∈ RM (Ω), with h = k = 0, such that the condition h f v , h v i ∗ + h g v , k v i ∗ f v
0, it is enough to take k v ≡ h = h v ≡ δ ∈ S such that ( f v δ ) = −h f v , δ i ∗ Corollary 7.8.
Let f, g ∈ RM (Ω) be such that f ≃ − g and assume f c ∗ b + b ∗ g ≡ .(1) If f v is not a zero divisor, then χ = − (2 f sv ) − ( f v ∗ b ) is a solution of S f,g ( χ ) = b .(2) For any δ ∈ S such that ( f δ ) , then χ = − (2 f δ ) − ( δ ∗ b ) is a solution of S f,g ( χ ) = b .Proof. In case (1) take h = f v and k ≡ (2) take h ≡ δ and k ≡ (cid:3) Applications of the rank 2 case to function theory
The following result, which allows us to classify all idempotents up to ∗ -conjugation, is a first appli-cation of the characterization of the equivalence relation ≃ in terms of “real” and “vector” parts of thefunctions, namely Corollary 7.2. Proposition 8.1.
Let f ∈ RM (Ω) \ RM R (Ω) ; then f is equivalent to a one-slice preserving function g ∈ RM (Ω) \ RM R (Ω) if and only if f sv has a square root. Moreover, all idempotents in R (Ω) areequivalent.Proof. By Corollary 7.2, the function f is equivalent to g if and only iff f = g and f sv = g sv . Then itis enough to notice that for a one-slice preserving function g / ∈ RM R (Ω) we have g v = γI for a suitable I ∈ S and γ ∈ RM R (Ω) \ { } .As for the second part of the statement, given an idempotent σ and any i ∈ S , we have σ = ℓ + ,i = and σ sv = ( ℓ + ,iv ) s = − , so that σ ≃ ℓ + ,i . (cid:3) The previous proposition gives us the possibility to give a necessary and sufficient condition in orderthat the product of an idempotent with a function is identically zero. It is worth comparing this resultwith the statement of Proposition 3.5 in which the kernel of L f,g is characterized via a condition, whilenext theorem gives an extensional description. Theorem 8.2.
Given an idempotent σ ∈ R (Ω) and ρ ∈ RM (Ω) , then(1) σ ∗ ρ ≡ if and only if there exists i, j ∈ S with i ⊥ j , α, β ∈ RM R (Ω) and f ∈ RM (Ω) invertiblesuch that σ = f ∗ ℓ + ,i ∗ f −∗ and ρ = f ∗ ℓ − ,i ∗ ( α + βj ) ∗ f −∗ . In particular, ρ is an idempotentif and only if α = 1 .(2) σ ∗ ρ ∗ σ c ≡ if and only if there exists i, j ∈ S with i ⊥ j , α , α , β ∈ RM R (Ω) and f ∈ RM (Ω) invertible such that σ = f ∗ ℓ + ,i ∗ f −∗ and ρ = f ∗ ( α + α i + βℓ − ,i ∗ j ) ∗ f −∗ . In particular, ρ isan idempotent if and only if α = and α = − .(3) σ ∗ ρ ∗ σ ≡ if and only if there exists i, j ∈ S with i ⊥ j , α , β , β ∈ RM R (Ω) and f ∈ RM (Ω) invertible such that σ = f ∗ ℓ + ,i ∗ f −∗ and ρ = f ∗ ( αℓ − ,i + ( β + β i ) ∗ j ) ∗ f −∗ . In particular, ρ is an idempotent if and only if α = 1 and β + β ≡ .Proof. (1) . A direct computation shows that, if σ = f ∗ ℓ + ,i ∗ f −∗ and ρ = f ∗ ℓ − ,i ∗ ( α + βj ) ∗ f −∗ , then σ ∗ ρ = f ∗ ℓ + ,i ∗ ℓ − ,i ∗ ( α + βj ) ∗ f −∗ ≡ ℓ + ,i ∗ ℓ − ,i ≡ σ is a idempotent, there exist f ∈ RM (Ω) invertible suchthat σ = f ∗ ℓ + ,i ∗ f −∗ . As σ ∗ ρ ≡ f −∗ ∗ σ ∗ ρ ∗ f ≡
0, we can reduce ourselves to the case f = 1, that is σ = ℓ + ,i . Now set ρ = ρ + ρ i + ρ j + ρ k and compute ℓ + ,i ∗ ρ = 12 (1 − J i ) ∗ ( ρ + ρ v ) = 12 [ ρ + hJ i, ρ v i ∗ + ρ v − J ρ i − J i ∧ * ρ v ]= 12 [ ρ + J ρ + ( ρ i + ρ j + ρ k ) − J ρ i − J ( − ρ j + ρ k )]= 12 [ ρ + J ρ + ( ρ − J ρ ) i + ( ρ + J ρ ) j + ( ρ − J ρ ) k ] . Hence we obtain that ℓ + ,i ∗ ρ ≡ ρ + J ρ ≡ ,ρ − J ρ ≡ ,ρ + J ρ ≡ ρ − J ρ ≡ . This system is equivalent to ρ = J ρ and ρ = J ρ and these last two equalities give ρ = ρ + J ρ i + ρ j + ρ J k = ρ (1 + J i ) + ρ (1 + J i ) j ;by setting α = 2 ρ and β = 2 ρ we get ρ = ℓ − ,i ∗ ( α + βj ). Finally, ρ = f ∗ ℓ − ,i ∗ ( α + βj ) ∗ f −∗ is anidempotent if and only if ℓ − ,i ∗ ( α + βj ) is, and a straightforward computation shows that this holds ifand only if α = 1. (2) . Again a direct computation shows that the condition is sufficient.Vice versa, as above we can suppose that σ = ℓ + ,i ; writing ρ = ρ + ρ i + ρ j + ρ k we obtain, since ℓ + ,i is an idempotent and ℓ + ,i ∗ ℓ − ,i ≡ ℓ + ,i ∗ ρ ∗ ℓ − ,i = ℓ + ,i ∗ ( ρ + ρ i + ρ j + ρ k ) ∗ ℓ − ,i = ℓ + ,i ∗ ( ρ + ρ i ) ∗ ℓ − ,i + ℓ + ,i ∗ ( ρ j + ρ k ) ∗ ℓ − ,i = ( ρ + ρ i ) ∗ ℓ + ,i ∗ ℓ − ,i + ρ ℓ + ,i ∗ j ∗ ℓ − ,i + ρ ℓ + ,i ∗ k ∗ ℓ − ,i = ρ ℓ + ,i ∗ ℓ + ,i ∗ j + ρ ℓ + ,i ∗ ℓ + ,i ∗ k = ρ ℓ + ,i ∗ j + ρ ∗ ℓ + ,i ∗ k = ℓ + ,i ∗ ( ρ + ρ i ) ∗ j. Thus ℓ + ,i ∗ ρ ∗ ℓ − ,i ≡ ℓ + ,i ∗ ( ρ + ρ i ) ≡ (1) , gives the existence of asuitable β ∈ RM R (Ω) such that ( ρ + ρ i ) ∗ j = βℓ − ,i ∗ j and thus proves the first part of the assertion.Again ρ = f ∗ ( α + α i + βℓ − ,i ∗ j ) ∗ f −∗ is an idempotent if and only if α + α i + βℓ − ,i ∗ j is and thisis equivalent to α = and α = − . (3) . The sufficiency of the condition is proved by direct inspection, as above.We only give a short summary of the computations, since the procedure is the same as in case (2) ℓ + ,i ∗ ρ ∗ ℓ + ,i = ℓ + ,i ∗ ( ρ + ρ i + ρ j + ρ k ) ∗ ℓ + ,i = ℓ + ,i ∗ ( ρ + ρ i ) ∗ ℓ + ,i + ℓ + ,i ∗ ( ρ j + ρ k ) ∗ ℓ + ,i = ( ρ + ρ i ) ∗ ℓ + ,i ∗ ℓ + ,i + ρ ℓ + ,i ∗ j ∗ ℓ + ,i + ρ ℓ + ,i ∗ k ∗ ℓ + ,i = ( ρ + ρ i ) ∗ ℓ + ,i + ρ ℓ + ,i ∗ ℓ − ,i ∗ j + ρ ℓ + ,i ∗ ℓ − ,i ∗ k = ( ρ + ρ i ) ∗ ℓ + ,i . Thus ℓ + ,i ∗ ρ ∗ ℓ + ,i ≡ ρ + ρ i ) ∗ ℓ + ,i ≡ ρ + ρ i = αℓ − ,i for asuitable α ∈ RM R (Ω). (cid:3) Remark 8.3.
The above proposition classifies, up to conjugation, all functions σ, ρ such that σ is anidempotent and σ ∗ ρ ≡ σ = ℓ + ,i and ρ = ℓ − ,i ∗ ( α + βj ) with i, j ∈ S , i ⊥ j , α, β ∈ RM R (Ω). Notice that for these functions ρ ∗ σ can be different from 0. Indeed, ρ ∗ σ ≡ ℓ − ,i ∗ ( α + βj ) ∗ ℓ + ,i = αℓ − ,i ∗ ℓ + ,i + βℓ − ,i ∗ j ∗ ℓ + ,i ≡
0. Since ℓ − ,i ∗ ℓ + ,i ≡ ρ ∗ σ ≡ βℓ − ,i ∗ j ∗ ℓ + ,i ≡
0. As j is orthogonal to ℓ − ,i we get j ∗ ℓ + ,i = ℓ − ,i ∗ j and thus βℓ − ,i ∗ j ∗ ℓ + ,i ≡ QUIVALENCE OF SLICE SEMI-REGULAR FUNCTIONS VIA SYLVESTER OPERATORS 25 equivalent to βℓ − ,i ∗ ℓ − ,i ∗ j = βℓ − ,i ∗ j ≡
0, since ℓ − ,i is an idempotent. Thus ρ ∗ σ ≡ β ≡
0, whichis equivalent to ρ = αℓ − ,i . Again, ρ is an idempotent if and only if α = 1, that is ρ = σ c .9. Sylvester operators of rank S f,g has rank 3.Thanks to Theorem 5.6, this corresponds to the fact that f + g S f,g is not an isomorphism.We recall that by Remark 4.14 and Proposition 5.3, this can happen only if Ω is a product domain. Since f + g ∈ RM R (Ω) \ { } is invertible, with no loss of generality we can study the kernel and the imageof the operator associated to the functions ff + g and gf + g , that is we can assume f + g ≡ f and g in order that S f,g is not an isomorphism. Proposition 9.1.
Assume that f + g ≡ and S f,g is not an isomorphism. Then there exists τ ∈RM R (Ω) such that f sv = (cid:0) J (cid:0) τ − (cid:1)(cid:1) and g sv = (cid:0) J (cid:0) τ + (cid:1)(cid:1) ; in particular both f sv and g sv both have asquare root in ∈ RM R (Ω) .Proof. Under the assumption on f + g , the determinant of S f,g becomes 1 + 2( f sv + g sv ) + ( f sv − g sv ) which can also be written as ( f sv + g sv + 1) − f sv g sv .As S f,g is not an isomorphism, then we have ( f sv + g sv + 1) − f sv g sv ≡
0, which implies that f sv g sv hasa square root µ ∈ RM R (Ω).Up to a possible change of sign of µ we have that the following system holds ( f sv + g sv + 1 = 2 µ,f sv g sv = µ The first equality gives g sv = 2 µ − − f sv , and thanks the second one, we obtain f sv (2 µ − − f sv ) = µ .Last equation is equivalent to ( f sv ) − µ − ) f sv + µ − µ + ≡ − µ + which can also be written as (cid:0) f sv − µ + (cid:1) = − µ + , thus showing that − µ + has a square root τ ∈ RM R (Ω). Again, up to apossible change of sign of τ , we find f sv − µ + = τ , that is f sv = µ − + τ . As µ − = − τ − wefinally obtain that f sv = − τ −
14 + τ = − (cid:18) τ − (cid:19) = (cid:18) J (cid:18) τ − (cid:19)(cid:19) which therefore proves that f sv has a square root. Since g sv = 2 µ − − f sv , we have g sv = 2 (cid:18) − τ − (cid:19) + (cid:18) τ − (cid:19) = − τ − − τ = − (cid:18) τ + 12 (cid:19) = (cid:18) J (cid:18) τ + 12 (cid:19)(cid:19) , showing that g sv has also the required form and admits a square root. (cid:3) Last proposition gives us the possibility to study more accurately which are the functions f and g suchthat f + g ≡ S f,g is not invertible. The crucial point is that this analysis must be splitted in twoparts, corresponding to Examples 5.4 and 5.5: indeed the main difference we will find is that in the firstcase f sv g sv
0, thus ensuring that the eigenvalue 0 has algebraic multiplicity 1, while in the second one f sv g sv ≡
0, which entails that the eigenvalue 0 has algebraic multiplicity greater than 1.
Proposition 9.2.
Assume that f + g ≡ and S f,g is not an isomorphism. If f sv g sv there exist h, ˜ h ∈ RM (Ω) invertible, τ ∈ RM R (Ω) \ (cid:8) ± (cid:9) , i ∈ S such that h −∗ ∗ f v ∗ h = J (cid:0) τ − (cid:1) i and ˜ h ∗ g v ∗ ˜ h −∗ = J (cid:0) τ + (cid:1) i . Then for any j ∈ S such that i ⊥ j and k = ij we have ker( S f,g ) = n αh ∗ ( −J j + k ) ∗ ˜ h | α ∈ RM R (Ω) o and S f,g = b has a solution if and only if h h −∗ ∗ b ∗ ˜ h −∗ , J j − k i ∗ ≡ . Proof.
Thanks to Propositions 9.1 and 8.1 we can find h, ˜ h ∈ RM (Ω) invertible, τ ∈ RM R (Ω) \ (cid:8) ± (cid:9) , i ∈ S such that h −∗ ∗ f v ∗ h = J (cid:0) τ − (cid:1) i and ˜ h ∗ g v ∗ ˜ h −∗ = J (cid:0) τ + (cid:1) i . Thus, by a straightforwardcomputation, it is enough to study the Sylvester operator S f,g when f v = J (cid:0) τ + (cid:1) i and g v = J (cid:0) τ − (cid:1) i and to recover kernel and image in the general case from the kernel and the image associated to theseparticular functions. The matrix S f,g in Formula 5.2 is given by (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − J τ J τ −J J (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ;as τ = we easily obtain that ker( S f,g ) is spanned by J j + k and that the image of S f,g is spanned by1, i and j + J k . Last assertion can also be rephrased by saying that b belongs to the image of S f,g ifand only if h b , J j − k i ∗ ≡ (cid:3) We recall that, thanks to Remark 5.7, S f,g has rank 3 if and only if S g,f has rank 3. By Corollary 7.2,this condition is equivalent to the fact that ker( S f,g ) contains only zero divisors. Under this hypothesis,notice that, if there exists a zero divisor in ker( S f,g ) whose “real part” is not identically zero, thenker( S f,g ) contains exactly one idempotent. Quite surprisingly, this property is not symmetric in f and g : in particular, we can find f, g ∈ RM (Ω) such that f + g = 1 and ker( S f,g ) contains an idempotentwhile ker( S g,f ) only contains zero divisor with “real part” equal to zero.With the same notation as in the statement of Proposition 9.2, we have thatker( S g,f ) = n α ˜ h −∗ ∗ ( J j + k ) ∗ h −∗ | α ∈ RM R (Ω) o = n α ˜ h c ∗ ( J j + k ) ∗ h c | α ∈ RM R (Ω) o . Let us compute the “real part” of the elements of ker( S f,g ) and ker( S g,f ). Factoring out the slicepreserving function α we have,( h ∗ ( −J j + k ) ∗ ˜ h ) = (( h + h v ) ∗ ( −J j + k ) ∗ (˜ h + ˜ h v )) = (cid:16) ( −h h v , −J j + k i ∗ + h ( −J j + k ) + h v ∧ * ( −J j + k )) ∗ (˜ h + ˜ h v ) (cid:17) = − ˜ h h h v , −J j + k i ∗ − h h ( −J j + k ) + h v ∧ * ( −J j + k ) , ˜ h v i ∗ = − ˜ h h h v , −J j + k i ∗ − h h ( −J j + k ) , ˜ h v i ∗ − det (cid:12)(cid:12) h v ( −J j + k ) ˜ h v (cid:12)(cid:12) . (9.1)Analogously we have(9.2) (˜ h c ∗ ( J j + k ) ∗ h c ) = − ˜ h h h v , J j + k i ∗ − h h ( J j + k ) , ˜ h v i ∗ − det (cid:12)(cid:12) h v ( J j + k ) ˜ h v (cid:12)(cid:12) . Example 9.3.
Take h = ( J −
1) + i + j and ˜ h = i + k . Then h = ( J − h v = i + j , ˜ h = 0, ˜ h v = i + k , h s = ( J − + 1 + 1 = 2 − J and ˜ h s = 2. Then equation (9.1) gives( h ∗ ( −J j + k ) ∗ ˜ h ) = − ( J − · − ( −J + 1) = 0 , while equation (9.2) gives(˜ h c ∗ ( J j + k ) ∗ h c ) = − ( J − · − ( J + 1) = − J 6 = 0 . Thus for any τ ∈ RM R (Ω), given f = 1 + h ∗ ( J (cid:0) τ − (cid:1) i ) ∗ h −∗ and g = ˜ h −∗ ∗ ( J (cid:0) τ + (cid:1) i ) ∗ ˜ h , wehave that ker( S f,g ) contains only zero divisors with vanishing “real part”, while ker( S g,f ) contains anidempotent.We are now left to deal with the condition f sv g sv ≡
0. We will examine thoroughly the case g sv ≡ f sv ≡ QUIVALENCE OF SLICE SEMI-REGULAR FUNCTIONS VIA SYLVESTER OPERATORS 27
Proposition 9.4.
Assume that f + g ≡ and S f,g is not an isomorphism. If g sv ≡ , then (9.3) ker( S f,g ) = { (1 − f v ) ∗ X ∗ g v | X ∈ RM (Ω) } and S f,g = b has a solution if and only if (1 − f v ) ∗ b ∗ g v ≡ .Proof. Thanks to Propositions 9.1 and 8.1 we can find h ∈ RM (Ω) invertible and i ∈ S such that h −∗ ∗ f v ∗ h = −J i , and hence 1+ f v is a zero divisor. Moreover, since S f,g ( χ ) = ( f + g ) χ + f v ∗ χ + χ ∗ g v =1 · χ + f v ∗ χ + χ ∗ g v = (1 + f v ) ∗ χ + χ ∗ g v , a trivial computation shows that for any X ∈ RM (Ω) thefollowing chain of equality holds S f,g ((1 − f v ) ∗ X ∗ g v ) = (1 + f v ) ∗ (1 − f v ) ∗ X ∗ g v + (1 − f v ) ∗ X ∗ g v ∗ g v = ((1 + f v ) ∗ (1 − f v )) ∗ X ∗ g v + (1 − f v ) ∗ X ∗ ( g v ∗ g v )= (1 + f v ) s ∗ X ∗ g v + (1 − f v ) ∗ X ∗ ( − g sv ) = 0 , and therefore (1 − f v ) ∗ X ∗ g v ∈ ker( S f,g ) for any X ∈ RM (Ω).We now claim that there exist X ∈ RM (Ω) such that (1 − f v ) ∗ X ∗ g v is not identically zero. Indeed,since g v i ∈ S such that g v ∗ i has non-zero real part, so there exists ˜ h ∈ RM (Ω) invertiblethat ˜ h −∗ ∗ g v ∗ i ∗ ˜ h is a non-zero “real” multiple of 1 − J i . Moreover we already know that there exists h ∈ RM (Ω) invertible such that h −∗ ∗ (1 + f v ) ∗ h = 1 − J i . Thus (1 − f v ) ∗ X ∗ g v h −∗ ∗ (1 − f v ) ∗ X ∗ ( g v ∗ i ) ∗ ˜ h
0, so that last inequality is equivalent to (1 − J i ) ∗ h −∗ X ∗ ˜ h ∗ (1 − J i ) σ ∗ h −∗ X ∗ ˜ h ∗ σ σ = (1 − J i ) and taking X = h ∗ σ ∗ ˜ h −∗ gives σ ∗ σ ∗ σ = σ S f,g ) has dimension 1 and (1 − f v ) ∗ X ∗ g v is different from zero for some X ∈ RM (Ω), theequality in Formula 9.3 is established.We are now left to consider the image of the operator S f,g . First of all notice that, given χ ∈ RM (Ω)we have that (1 − f v ) ∗ S f,g ( χ ) ∗ g v = (1 − f v ) ∗ ((1 + f v ) ∗ χ + χ ∗ g v ) ∗ g v = (1 − f v ) ∗ (1 + f v ) ∗ χ ∗ g v + (1 − f v ) ∗ χ ∗ g v ∗ g v = (1 + f v ) s χ ∗ g v + (1 − f v ) ∗ χ ∗ ( g sv ) = 0 , because both 1 + f v and g v are zero divisors. Thus if S f,g ( χ ) = b has a solution then (1 − f v ) ∗ b ∗ g v ≡ S f,g is contained in the linear subspace { b ∈ RM (Ω) | (1 − f v ) ∗ b ∗ g v ≡ } .Reasoning as before, Theorem 8.2 ensures that the dimension of { b ∈ RM (Ω) | (1 − f v ) ∗ b ∗ g v ≡ } is equal to 3, and hence the image of S f,g coincides with { b ∈ RM (Ω) | (1 − f v ) ∗ b ∗ g v ≡ } , thus com-pleting the proof of the statement. (cid:3) References
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E-mail address : [email protected] Chiara de Fabritiis: Dipartimento di Ingegneria Industriale e Scienze Matematiche, Universit`a Politecnicadelle Marche, Via Brecce Bianche, 60131, Ancona, Italia
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