Quaternionic slice regularity beyond slice domains

After Gentili and Struppa introduced in 2006 the theory of quaternionic slice regular function, the theory has focused on functions on the so-called slice domains. The present work defines the class of speared domains, which is a rather broad extension of the class of slice domains, and proves that the theory is extremely interesting on speared domains. A Semi-global Extension Theorem and a Semi-global Representation Formula are proven for slice regular functions on speared domains: they generalize and strengthen some known local properties of slice regular functions on slice domains. A proper subclass of speared domains, called hinged domains, is defined and studied in detail. For slice regular functions on a hinged domain, a Global Extension Theorem and a Global Representation Formula are proven. The new results are based on a novel approach: one can associate to each slice regular function $f:\Omega\to\mathbb{H}$ a family of holomorphic stem functions and a family of induced slice regular functions. As we tighten the hypotheses on $\Omega$ (from an arbitrary quaternionic domain to a speared domain, to a hinged domain), these families represent $f$ better and better and allow to prove increasingly stronger results.


Introduction
This work aims at significantly extending the theory of quaternionic slice regular functions introduced in [13,14] (see Definition 2.2).The extension performed is manifold.
While the theory has focused thus far on the so-called slice domains (see Definition 4.2), an important achievement here is the definition of a wider class of quaternionic domains, called speared domains (see Definition 4.1).Slice regular functions on speared domains turn out to inherit many of the nice properties proven over slice domains in [10,11].We point out that carefully choosing the domains of definition of slice regular functions is crucial, as pathological examples can be constructed on quaternionic domains that do not intersect the real axis.Slice regular functions on domains included in H \ R need not be continuous, see [12,Example 1.11].
The new approach adopted to study slice regularity on a general speared domain Ω is so powerful that it allows to strengthen the Local Extension Theorem and the Local Representation Formula of [11] to what we call the Semi-global Extension Theorem 4.11 and the Semi-global Representation Formula (Corollary 4.14).These new results are stronger than the aforementioned local results even when Ω happens to be a slice domain.
Another significant novelty concerns the Extension Theorem, which was proven in [11] for the so-called simple slice domains (see Definition 6.4) and is known to be false for general slice domains, as a counterexample appeared in [4, .To see explicitly that that specific example is not a simple slice domain, see [10,Example 4.4].Another related article is [5].In the present work, we prove the Global Extension Theorem 5.12 on the class of hinged domains (see Definition 5.10), which is considerably wider than the class of simple slice domains.The result is not obtained directly under the global hypothesis that the domain is hinged.Instead, several preliminary steps are taken on a general speared domain Ω.These include constructing an equivalence relation ∼ on Ω and proving, in Theorem 5.9, that if f is a slice regular function on Ω and x, x ′ are equivalent points of Ω, then the local extensions of f near x and x ′ are consistent.Only later we define hinged domains by means of the equivalence relation ∼ and prove that a slice regular function on a hinged domain Ω automatically extends to a unique slice regular function on the so-called symmetric completion of Ω (see Subsection 2.1).
The paper is organized as follows.Section 2 is devoted to preliminary material: Subsection 2.1 sets up the complex and quaternionic algebras; Subsection 2.2 recalls the definition and the first properties of slice regular functions; Subsection 2.3 recalls the alternate approach to regularity proposed in [16] by introducing the function class SR.
Section 3 provides a novel interpretation of the General Extension Formula of [3].This makes it possible to associate to a slice regular function f on an arbitrary domain Ω ⊆ H a double-index family of elements of SR strictly related to f .Section 4 defines the concept of speared domain.For a slice regular function f on a speared domain Ω ⊆ H, the double-index family associated to f reduces to a single-index family.This allows us to prove the aforementioned Semi-global Extension Theorem and Representation Formula, which open the path for a detailed analysis of slice regularity on speared domains.
Section 5 is divided into two parts.Subsection 5.1 constructs, on an arbitrary speared domain Ω ⊆ H, the aforementioned equivalence relation ∼.For two equivalent points x, x ′ , the corresponding elements of the single-index family are strictly related.This allows to prove that the extension of f near x is consistent with the extension of f near x ′ , as we already mentioned.Subsection 5.2 defines hinged domains as those for which the equivalence relation ∼ becomes trivial.The Global Extension Theorem for hinged domains is therefore an immediate consequence of the work done in the previous subsection.A Global Representation Formula follows.
Section 6 studies speared domains and hinged domains in full detail.Subsection 6.1 proves that the class of speared domains strictly includes the class of slice domains and provides a large family of examples.It also proves that every speared domain can be locally shrunk to a slice domain.The final Subsection 6.2 presents several subclasses of the class of hinged domains: spear-simple domains, including simple slice domains; S-connected speared domains; and speared domains having a main sail.A wealth of examples is provided, to show that all these subclasses are distinct and that there exist hinged domains belonging to none of these subclasses.

Real quaternions and their complexification
Let C, H denote the real * -algebras of complex numbers and quaternions, respectively.As explained in [6,23], they can be built from the real field R by means of the so-called Cayley-Dickson construction: On the one hand, this construction endows the two real vector spaces with a bilinear multiplicative operation, which makes each of them a (real) algebra.By construction, each of them is unitary, that is, it has a multiplicative neutral element 1; and R is identified with the subalgebra generated by 1.For the purpose of the present work, every time we will speak of an algebra or of a subalgebra, we will automatically imply that it is unitary.It is well-known that H is associative but not commutative.The center {r ∈ H | rx = xr ∀ x ∈ H} of the algebra H is R.
On the other hand, the Cayley-Dickson construction endows each of C, H with a * -involution, i.e., a (real) linear transformation x → x c with the following properties: (x c ) c = x and (xy) c = y c x c for every x, y; x c = x for every x ∈ R. Thus, C and H are * -algebras.The trace and norm functions, defined by the formulas t(x) := x + x c and n(x) are real-valued.Moreover, in C ≃ R 2 and H ≃ R 4 , the expression 1 2 t(xy c ) coincides with the standard scalar product ⟨x, y⟩ and n(x) coincides with the squared Euclidean norm ∥x∥ 2 .In particular, these algebras are nonsingular, i.e., n(x) = xx c = 0 implies x = 0.The trace function t vanishes on every commutator [x, y] := xy − yx (see [19,Lemma 5.6]).It holds n(xy) = n(x)n(y), or equivalently ∥xy∥ = ∥x∥ ∥y∥.We point out that (r + v) c = r − v for all r ∈ R and all v in the Euclidean orthogonal complement of R.
Every nonzero element x of C or H has a multiplicative inverse, namely x −1 = n(x) −1 x c = x c n(x) −1 .For all elements x, y ̸ = 0, the equality (xy) −1 = y −1 x −1 holds.As a consequence, each of the algebras C and H is a division algebra and has no zero divisors.A famous result due to Frobenius states that R, C, H are the only (finite-dimensional) associative division algebras.
Let us consider the 2-sphere of quaternionic imaginary units The * -subalgebra generated by any I ∈ S, i.e., C I = span(1, I), is * -isomorphic to the complex field C (endowed with the standard multiplication and conjugation) through the * -isomorphism The union The algebra H has the following useful property.
• [Splitting property] For each imaginary unit I ∈ S, there exists J ∈ H such that {1, I, J, IJ} is a real vector basis of H, called a splitting basis of H associated to I.Moreover, J can be chosen to be an imaginary unit and to make the basis orthonormal.
We consider on C and H the natural Euclidean topology and differential structure as a finitedimensional real vector space.Similarly, we will adopt on C and H the natural structure of real analytic manifold.The relative topology on each C J with J ∈ S clearly agrees with the topology determined by the natural identification ϕ J between C and C J .A subset T of H is called (axially) symmetric (or circular ) if it is symmetric with respect to the real axis R; i.e., if for all x = α + βI ∈ T (with α, β ∈ R and I ∈ S), the whole set is included in T .We observe that S x = {x} if x ∈ R. On the other hand, for x ∈ H \ R, the set S x is obtained by real translation and dilation from the 2-sphere S. For each T ⊆ H, the symmetric completion of T is the set For each E ⊆ C, the circularization Ω E of E is defined as the following subset of H: For instance, given x = α + βI ∈ H, we have that S x is both the symmetric completion of the singleton {x} ⊂ H and the circularization of the singleton {α + iβ} ⊂ C.
In this work, we will also use the complexified algebra x, y ∈ H}, endowed with the following product: In addition to the complex conjugation x + ıy = x − ıy, H C is endowed with a * -involution x + ıy → (x + ıy) c := x c + ıy c , which makes it a * -algebra.This * -algebra is still associative and it has center R C = R + ıR.If we identify R C with C and if I ∈ S, then the previously defined * -isomorphism ϕ I : C → C I extends to Throughout the paper, we will keep referring to For instance: ı commutes with every element of H C , while i does not.We point out that the extended ϕ I is still a surjective * -algebra morphism, but is no longer injective.For future reference, we make the following remark.We recall that: a constant complex structure on R n0 is an R-linear endomorphism J 0 : R n0 → R n0 such that Remark 2.1.Left multiplication by ı is a constant complex structure on H C , which preserves R C .
For each I ∈ S, left multiplication by I is a constant complex structure on H, which preserves C I .The map ϕ I is a holomorphic map from (H C , ı) to (H, I) and its restriction to R C is a biholomorphic map from (R C , ı) to (C I , I).

Slice regular functions
As usual, a nonempty open connected subset of C is called a (complex) domain.Similarly, a nonempty connected open subset of H is called a (quaternionic) domain.The following definition was given in [3] and is a slight modification of [13,14].It has been adopted as standard in the monograph [12].
Definition 2.2.Let Ω ⊆ H be a domain and let f : Ω → H be a function.For each I ∈ S, let us set vanishes identically in Ω I .The function f is termed slice regular if, for each I ∈ S, f I is holomorphic.
More precisely, the definition of ∂ I should be stated as follows: where ϕ I denotes the restriction of ϕ I to ϕ −1 I (Ω I ) (see [15]).In other words, f I is called holomorphic if it is a holomorphic map from (Ω I , I) to (H, I).The so-called "splitting lemma" follows: after fixing a splitting basis {1, I, J, IJ} of H and letting F, G be the functions Ω I → C I such that f I = F + GJ, we have that f I is holomorphic if, and only if F, G are holomorphic functions from (Ω I , I) to (C I , I).
For the purposes of this paper, we will also need the next definition.x m a m = a 0 +xa 1 +. . .x n a n with coefficients a 0 , . . ., a n ∈ H is a slice regular function on H.
Every power series of the form n∈N x n a n converges absolutely and uniformly on compact sets in B(0, R) for some R ∈ [0, +∞], determined by the equality 1/R = lim sup n→+∞ |a n | 1/n .If R > 0, then the sum of the series is a slice regular function on B(0, R).
Conversely, if R ∈ (0, +∞] and if f : B(0, R) → H is a slice regular function, then there exists a sequence {a n } n∈N ⊂ H such that A slice regular function f is called slice preserving if, for each I ∈ S, it maps the "slice" Ω I into C I .A polynomial or power series n∈N x n a n is slice preserving if, and only if, {a n } n∈N ⊂ R. The following result derives from [8,9].Lemma 2.5.Let R ∈ (0, +∞] and let f : B(0, R) → H be a slice regular function.For each α, β ∈ R with α 2 + β 2 < R 2 , there exist quaternions b, c such that for all J, K ∈ S with J ̸ = K.The same article [3] proved a result called General Extension Formula, upon which we will elaborate in Section 3.

Slice functions
Let D be a nonempty subset of R C that is invariant under the complex conjugation z = α + iβ → z = α − iβ: then for each J ∈ S the map ϕ J (α + iβ) = α + βJ naturally embeds D into a "slice" ϕ J (D) of Ω D = J∈S ϕ J (D) ⊆ H.The following definitions were given in [16].
Definition 2.6.Assume D ⊆ R C to be nonempty and preserved by complex conjugation.A function If we consider a decomposition F = F 1 + ıF 2 into H-valued components F 1 and F 2 , F is a stem function if, and only if, F 1 (z) = F 1 (z) and F 2 (z) = −F 2 (z) for every z ∈ D. Definition 2.7.Assume D ⊆ R C to be nonempty and preserved by complex conjugation.A function f : Ω D → H is called a (left) slice function if there exists a stem function commutes for each J ∈ S. In this situation, we say that f is induced by F and we write f = I(F ).
If F is R C -valued, then we say that the slice function f is slice preserving.The real vector space of slice functions on Ω D is denoted as S(Ω D ).
It is easy to check that each slice function f is induced by a unique stem function F .For each slice function f : Ω D → H, it is useful to define the slice function f • s : Ω D → H, called spherical value of f , and the slice function The works [3,16] showed that these functions are constant on each sphere , where the function Im is a slice preserving element of S(H).The function f is slice preserving if, and only if, f • s and f ′ s are real-valued.Some special subclasses of the algebra S(Ω D ) of slice functions have been singled out in [16]: the nested subspaces S 0 (Ω D ), S 1 (Ω D ), S ω (Ω D ), SR(Ω D ) of slice functions induced by continuous, continuously differentiable, real analytic and holomorphic stem functions F : D → H C , respectively.By holomorphic, here, we mean that F is a holomorphic map from (D, ı) to (H C , ı).For all of these definitions except the first one, Ω D is assumed to be an open subset of It has been proven in [16] that, when Ω = Ω D is a symmetric slice domain, a function f : Ω D → H belongs to SR(Ω D ) if, and only if, it is slice regular.We will provide a new proof of this fact in the forthcoming Section 3. On the other hand, when Ω D is a product domain, the class SR(Ω D ) (studied, for instance, in [1]) is strictly included in the class slice regular functions Ω D → H.To see this, it suffices to restrict to Ω D the slice regular function appearing in the next example.
Example 2.9.Fix I 0 ∈ S and let f : H \ R → H be defined as follows: The function f is slice regular because each restriction f I is constant, whence holomorphic.It belongs to neither SR(H \ R) nor S(H \ R).
More work on the relation between SR(Ω D ) and the class of slice regular functions on Ω D includes [15].The recent paper [22] generalizes the class SR, dropping the symmetry hypothesis on Ω D and imposing the existence of local zonal decompositions.This provides an extension of the theory of slice regular functions beyond the classical setup of slice domains, distinct from the one undertaken in the present work.

General extension formula 3.1 General extension formula, revisited
This subsection elaborates the ideas behind the General Extension Formula, namely [3, Theorem 4.2].This will produce new powerful results in the forthcoming Subsection 3.2.
For all I ∈ S and all T ⊆ H, we will use the notations The following properties hold: • p J = r and p K = s.
• For each I ∈ S, it holds As a consequence, if r is holomorphic in Proof.By direct inspection in the definition of p I , we find that Analogous computations prove that p K coincides with s in Ω ⩾ K .Another direct inspection in the definition of p I yields that Moreover, The definition The next result will be extremely useful in the sequel.It follows from the classical Schwarz Reflection Principle if we identify R C with C and H C with C 4 .We include a direct proof for the sake of completeness.We recall that x + ıy = x − ıy for all x, y ∈ H and we set, for all D ⊆ R C , the notations Overall, we have proven that, for each We are now ready to associate to each couple of holomorphic functions r : Ω ⩾ J → H and s : Ω ⩾ K → H a holomorphic stem function.Theorem 3.3 (General extension formula).Let Ω be a symmetric open subset of H and take distinct J, K ∈ S. Consider holomorphic functions r : Ω ⩾ J → H and s : Ω ⩾ K → H and assume the equality r(α) = s(α) to hold for all α ∈ Ω ∩ R (if any).Let us set as well as F (z) := F (z). Then F is a well-defined holomorphic stem function.It induces an element of SR(Ω) (whence a slice regular function), namely f Moreover, for all α ∈ Ω ∩ R, it holds Now, r(α) = r(α) because the map x + ıy = x − ıy preserves all x ∈ H. Thus, F (α) = F (α) for all α ∈ D ∩ R. By applying Lemma 3.2, we get that the function F : D → H C is a well-defined holomorphic stem function, as desired.
If we set f := I(F ), then for all , as desired.

Double-index slice regular family
The work done in the previous subsection allows us to associate to each slice regular function a family of holomorphic stem functions depending on two indices J, K ∈ S. We recall that, for every E ⊆ R C , the symbol E denotes the image of E through conjugation z → z.
Let Ω ⊆ H be a domain and let f : Ω → H be a slice regular function.For each choice of distinct J, K ∈ S, we set If E (whence D J,K and Ω J,K ) is not empty, we can apply Theorem 3.3 to r := f | ϕ J (E) and to as well as F J,K (z) := F J,K (z).The family {F J,K } J,K∈S,J̸ =K,D J,K ̸ =∅ is called the double-index holomorphic stem family associated to f .The double-index slice regular family associated to f is the family {f J,K } J,K∈S,J̸ =K,Ω J,K ̸ =∅ of the elements f J,K = I(F J,K ) of SR(Ω J,K ) induced by the F J,K 's.
We will soon study the double-index families associated to f .Before doing so, we need the next definition and result.
Remark 3.6.If we adopt the notations of Definition 3.4, then for all distinct J, K ∈ S: if D J,K is not empty, then it includes the set spine R C (Ω).It follows that spine(Ω) is contained in the circularization Ω J,K of D J,K , which is the domain of f J,K .
We are now ready for the announced study.Proposition 3.7.Let Ω ⊆ H be a domain and let f : Ω → H be a slice regular function.Let The function f J,K coincides with f in spine(Ω).Moreover, f and f J,K coincide with

4.
For each connected open subset C of D J ′ ,K ′ ∩ D J,K that includes a real point (if any), the function ).Up to precomposing the function f with the real translation x → x + α 0 , we may suppose α 0 = 0. Lemma 2.5 implies that for all α + βI ∈ B(0, R).The thesis now follows from the definition of spine(Ω).
3. By property 2., f J,K = I(F J,K ) and As a byproduct, we obtain a new proof of two well-known results: the General Representation Formula for symmetric slice domains, proven in [3], and the fact, proven in [16], that the class of slice regular functions on a symmetric slice domain Ω D coincides with SR(Ω D ) (whence is included in the class of real analytic functions).We merge all of these results into a single statement.
Theorem 3.8 (Global representation formula for symmetric slice domains).Let Ω D be a symmetric slice domain in H and let f : Ω D → H be a slice regular function.Then f ∈ SR(Ω D ) and, in particular, f is real analytic.Moreover, for all distinct J, K ∈ S, it holds f = f J,K and for all I ∈ S and all α, β ∈ R with β ≥ 0 such that α + βI ∈ Ω D .
Proof.We assume, without loss of generality, D to be preserved by conjugation z → z.Let {F J,K } J,K∈S,J̸ =K denote the double-index holomorphic stem family associated to f : Ω D → H. Since Ω D is a symmetric slice domain, all domains D J,K (whence all intersections D J,K ∩ D J ′ ,K ′ ) coincide with D. By hypothesis, the latter set is connected and intersects the real axis.By property 4. in Proposition 3.7, all f J,K : Ω D → H coincide and form a single element g of SR(Ω D ).Moreover, property 1. in the same theorem guarantees that g coincides with f in (Ω D ) ⩾ J for all J ∈ S, whence g = f .As a consequence, the formula holds for all α + βI ∈ Ω D with α, β ∈ R, β ≥ 0 and I ∈ S.
4 Speared domains and local representation formula

Speared domains and their cores
Let us introduce some new terminology.
We remark that, if Ω ⊆ H is a speared domain, then Ω ∩ R ̸ = ∅.In particular, the sets D J,K and Ω J,K of Definition 3.4 are nonempty for each choice of distinct J, K ∈ S.
As we will see in the forthcoming Section 6, the class of speared open sets is a rather broad extension of the class of slice domains, whose definition we recall from [12, Definition 1.12].Definition 4.2.Let Ω be a domain in H. Then Ω is called a slice domain if it intersects the real axis R and if, for all J ∈ S, the "slice" Ω J is connected.
However, if a speared open subset U of H happens to be symmetric, we are only slightly generalizing the notion of symmetric slice domain: indeed, like every other symmetric open subset of H, U is a disjoint union of symmetric slice domains or product domains.Now, every symmetric slice domain is speared and every product domain is not.It follows that our U is a disjoint union of symmetric slice domains.By applying Theorem 3.8 in each connected component of U , we can make the next remark.
If U is a general open subset of H, we will still be able to apply Remark 4.3 to an appropriately chosen symmetric speared open subset of U .This choice will be possible after some preliminary steps.We begin with the next remark, which uses the fact that every open subset of R C or R ⩾ C is locally path-connected.• each 2-sphere (or singleton) for all I ∈ S and all α, β ∈ R with β ≥ 0 such that α + βI ∈ core(U ).As a consequence, for all Besides its independent interest, we will use Lemma 4.6 to prove the forthcoming Theorem 4.8.

Slice regular families on speared domains
The aim of this subsection is proving that the double-index families associated to a slice regular function f : Ω → H reduce to single-index families when Ω is a speared domain.
We observe that C ε is the union of open disks of radii ε centered at arbitrary points z 0 ∈ C, whence C ε is an open neighborhood of C. Since C is path-connected and since any point z ∈ C ε can be joined to a point z 0 ∈ C through a line segment included in C ε , it follows that C ε is path-connected.
We finally show that ϕ As a consequence, ϕ K (z) ∈ Y , as desired.
We are now ready to prove that the double index reduces to a single index if we are dealing with a speared domain.We recall that, for every E ⊆ R C , the symbol E denotes the image of E through conjugation z → z.
Theorem 4.8.Let Ω ⊆ H be a speared domain, let f : Ω → H be a slice regular function and let {F J,K : D J,K → H C } J,K∈S,J̸ =K denote the double-index holomorphic stem family associated to f .Let us fix J ∈ S and consider the speared open set Then there exists a holomorphic stem function F J : D J → H C with the following property: for each z ∈ D J , there exist a real number ε z > 0 and a speared open neighborhood In case z ∈ core R C (Ω), we can assume U z to be core R C (Ω) and ε z = 3 (whence Cap(J, ε z ) = S).
We observe, for future reference, that γ(t 0 ) belongs to the boundary of core R C (Ω) and that Now, C is compact and path-connected; moreover, ϕ J (C) ⊂ Ω > J ⊂ Ω \ R. If we apply Lemma 4.7 to C and to Y := Ω \ R, we conclude that there exists a real number ε > 0 such that the path-connected open neighborhood for all K ∈ Cap(J, ε).We define U z := core R C (Ω) ∪ C ε and observe that z ∈ U z ⊆ D J ′ ,K ′ for all distinct J ′ , K ′ ∈ Cap(J, ε).We also remark that, since C intersects the boundary of core R C (Ω) at γ(t 0 ), its open neighborhood C ε intersects core R C (Ω) at some point p 0 .We can prove that the open set U z is speared by applying Remark 4.4.Indeed, we can exhibit for every p ∈ U z a path γ p : [0, 1] → U z with γ p (0) ∈ R and γ p (1) = p: By applying property 4. of Proposition 3.7 to each connected component of U z , we obtain that We conclude by setting ε z := ε.
We are now ready to define F J (z) as the constant value of the map Cap * (J, ε z ) → H C , K → F J,K (z).In particular, F J (z) = lim Cap * (J,εz)∋K→J F J,K (z) .
We notice that the value F J (z) does not depend on the choice of ε z nor on the choice of U z .Furthermore, F J is holomorphic near z because, for K ∈ Cap * (J, ε z ), it coincides in U z with the holomorphic stem function F J,K .
We have therefore defined a holomorphic We conclude the proof by setting U z := U z and ε z := ε z for all z ∈ (D J ) ⩾ .We can do so because for all distinct J ′ , K ′ ∈ Cap(J, ε z ), as a consequence of the fact that F J, ′ K ′ and F J are stem functions.Definition 4.9.In the situation described in Theorem 4.8, {F J : D J → H C } J∈S is called the holomorphic stem family associated to f .For each J ∈ S, let us consider the symmetric speared open set Ω J := Ω D J : the slice regular family associated to f is the family {f J : Ω J → H} J∈S of the elements f J = I(F J ) of SR(Ω J ) induced by the F J 's.
For future use, we prove the next lemma.
Lemma 4.10.Let Ω ⊆ H be a speared domain, let f : Ω → H be a slice regular function and let {F J : D J → H C } J∈S be the holomorphic stem family associated to f .Fix J 0 , J 1 ∈ S. The holomorphic stem functions Proof.The first statement concerning core R C (Ω) immediately follows from Theorem 4.8 and Lemma 4.6.Now let us prove the second statement about x 0 and x 1 belonging to the same connected component C of (α + βS) ∩ Ω.It suffices to pick any J : [0, 1] → S such that α + βJ is a path within C and to show that there exists an open neighborhood U of α + βı in D J (0),J (1)  such that F J (0) , F J (1) coincide in U .
For every J ∈ S such that α + βJ ∈ Ω, Theorem 4.8 guarantees that there exist an an open neighborhood U J of the point α + βı in D J and an ε J > 0 with the following property: for any choice of distinct Consider the open cover {J −1 (Cap(J, ε J ))} J∈S of the compact interval [0, 1].Take N ∈ N large enough for 1/N to be a Lebesgue number for this cover (see [21,Lemma 7.2] for details).It follows that, for each s ∈ {0, . . ., N − 1}, there exists J s ∈ S such that J ([s/N, (s + 1)/N ]) ⊂ Cap(J s , ε J s ).As a consequence, U J s ⊆ D J (s/N ),J ((s+1)/N ) and F (1) ) and we have , whence the thesis follows.Let Ω ⊆ H be a speared domain, let f : Ω → H be a slice regular function and let {f J } J∈S be the slice regular family associated to f .Fix J 0 ∈ S.There exists a speared open set Λ with

Semi-global extension and representation for speared domains
such that f coincides with f J0 in Λ.In particular, for each x 0 ∈ Ω ⩾ J0 , f coincides with f J0 in a speared domain including x 0 : namely, the connected component of Λ including x 0 .
Proof.Let {F J : D J → H C } J∈S and {F J,K : D J,K → H C } J,K∈S,J̸ =K denote, respectively, the holomorphic stem family and the double-index holomorphic stem family associated to f .In particular, f J0 = I(F J0 ).

Corollary 4.13 (Local extension for speared domains).
If f is a slice regular function on a speared domain Ω and if x 0 ∈ Ω, then there exists a slice extension triplet ( f , N x0 , Λ x0 ) for f with Λ x0 ∋ x 0 .As a consequence, f is real analytic.
Proof.We apply Theorem 4.11 to find a speared extension triplet (f J0 , Ω J0 , Λ) for f with Λ ⊇ core(Ω)∪Ω ⩾ J0 .We define N x0 to be the connected component of Ω J0 that includes x 0 : since Ω J0 is a symmetric speared open set, it follows immediately that N x0 is a symmetric slice domain.The connected component Λ 0 of Λ that includes x 0 is a speared domain contained in N x0 .Moreover, Λ 0 is contained in Ω because Λ was.By the forthcoming Proposition 6.2, there exists a slice domain  Let Ω ⊆ H be a speared domain, let f : Ω → H be a slice regular function and let J 0 ∈ S.There exists a speared open subset Λ ⊆ Ω, including both core(Ω) and Ω ⩾ J0 , such that the formula holds for all I, J, K ∈ S and all α, β ∈ R with J ̸ = K, β ≥ 0 and α + βI, α + βJ, α + βK ∈ Λ.
The last corollary allows us to extend [16,Definition 6] and [10, Definition 3.1] to all speared domains.Definition 4.15.In the situation described in Corollary 4.14, for all α, β ∈ R with β ≥ 0 such that x 0 := α + βJ 0 ∈ Ω, we define the spherical value of f at x 0 as the quaternion which does not depend on the specific choice of J ̸ = K such that α + βJ, α + βK ∈ Λ. If, moreover, β > 0, we define the spherical derivative of f at x 0 as the quaternion which does not depend on the specific choice of J ̸ = K such that α + βJ, α + βK ∈ Λ.
Remark 4.16.Let Ω ⊆ H be a speared domain, let f : Ω → H be a slice regular function and let {F J : D J → H C } J∈S denote the holomorphic stem family associated to f .Fix J 0 ∈ S and decompose the H C -valued function F J0 as F J0 1 + ıF J0 2 , where F J0 1 , F J0 2 are H-valued.Then, for all α, β ∈ R with β ≥ 0 such that α + βJ 0 ∈ Ω, we have f As a consequence, the functions f • s : Ω → H and f ′ s : Ω \ R → H are real analytic.
Corollary 4.14 implies the following properties.We recall that the slice (or complex) derivative of any slice regular function f : Ω → H is the slice regular function f ′ c : Ω → H defined as at any α + βI ∈ Ω (see [12,Definition 1.8]).If f happens to be C 1 , let us denote the real differential of f at a point x 0 ∈ Ω as df x0 .
Let Ω ⊆ H be a speared domain and let f : Ω → H be a slice regular function.
In view of property 3. of Proposition 4.17, we point out that our Definition 4.15 is consistent with [12,Definition 11.6].

Hinged points and global extension theorem for hinged domains 5.1 Hinged points
Assumption 5.1.Throughout this subsection, we assume Ω ⊆ H to be a speared domain and we adopt the notations K for all distinct J, K ∈ S.Moreover, we assume f : Ω → H to be a slice regular function and denote by {F J,K : D J,K → H C } J,K∈S,J̸ =K the double-index holomorphic stem family associated to f and by {F J : D J → H C } J∈S the holomorphic stem family associated to f .As customary, in the previous assumption the symbol E denotes the image of a set E ⊆ R C through conjugation z → z.We now define and study several relations on Ω, which will be useful to prove our Global Extension Theorem. Figure 2: An example where (x 1 , y 1 ) shadows (x 0 , y 0 ) in Ω and p 1 is strongly hinged to p 0 in Ω.
We say that x 1 is strongly hinged to x 0 in Ω if there exists y ∈ Ω ∩ R such that (x 1 , y) shadows (x 0 , y).
The expression "strongly hinged", used here for the first time, subsumes ideas already used in the articles [4,5,10,11].Figure 2 portrays an example of shadowing and an example of point that is strongly hinged to another point.Being strongly hinged is a reflexive and symmetric relation on Ω.By construction, if x 1 is strongly hinged to x 0 , then S x1 = S x0 .
For future use, we establish the following lemma.
2. If x 1 is strongly hinged to x 0 , then there exists speared open subset of D J0,J1 including α+βı where F J0 , F J1 coincide.In particular, the connected component of D J0,J1 including α + βı intersects the real axis. Proof.
In preparation for the main result in this section, we will need one more relation.
The sequence {x s } t s=0 is called a chain of length t connecting x 0 and x t if, after erasing from T the index t and all indices s 0 where a simple step happens, there remain in T an even number 2ℓ of indices, which can be listed in a form s 1 , s 1 , . . ., s ℓ , s ℓ such that, for all m, n ∈ {1, . . ., ℓ}: • a double step happens at (s m , s m ) (whence s m < s m ); • s 1 < . . .< s ℓ ; in other words, the double steps are listed in the same order they start; and • if m < n, then either s m < s n or s m > s n ; in other words, if the nth double step starts before the mth double step is concluded, then the nth double step must also end before the mth double step does; equivalently, the double step at (s n , s n ) may be nested within the double step at (s m , s m ), but not intertwined with it.
For all points x, x ′ ∈ Ω, we say that x ′ is hinged to x in Ω, and we write x ∼ x ′ , if there exists a chain connecting x and x ′ .
An example of chain of length 4 is portrayed in the forthcoming Figure 12.We will soon prove that ∼ is an equivalence relation.Before doing so, we establish a few useful technical results.
Remark 5.6.Let us consider a finite sequence {x s } t s=0 ⊂ Ω.If there is a simple step at s, then S xs = S xs+1 .If there is a double step at (s, s), then S xs = S xs+1 and S xs+1 = S xs .This is a consequence of the definitions of simple and double step, as well as of Remark 5.3.Lemma 5.7.If {x s } t s=0 is a chain of length t ≥ 2 connecting x 0 and x t , then one of the following facts is true.
1.There exists u with 0 < u < t such that {x s } u s=0 is a chain of length u connecting x 0 and x u (whence x 0 ∼ x u ) and {x s } t s=u is a chain of length t − u connecting x u and x t (whence 2. There is a double step at (0, t − 1) and {x s } t−1 s=1 is a chain of length t − 2 connecting x 1 and x t−1 (whence x 1 ∼ x t−1 ).
The second case is excluded when t = 2.
Proof.If there is a simple step at 0, then the sequence {x s } 1 s=0 is a chain of length 1 connecting x 0 and x 1 and there exists no double step of the form (0, s).Thus, {x s } t s=1 is automatically a chain of length t − 1 connecting x 1 and x t .Overall, we are in case 1.
Suppose, instead, that there is not a simple step at 0. Then, if we list all double steps as in Definition 5.5, we find (0, s 1 ), . . ., (s ℓ , s ℓ ) for some natural number ℓ.We separate two cases.
• If s 1 < t − 1 and we set u := s 1 + 1, then we are in case 1. because {x s } u s=0 is a chain of length u connecting x 0 and x u and {x s } t s=u is a chain of length t − u connecting x u and x t .Indeed, every simple step in the original chain will still be a simple step in one of the two subsequences.Every double step (s m , s m ) with m > 1 will be a double step in the first subsequence if it is nested within (0, s 1 ) (i.e., if 0 < s m < s m < s 1 ) or a double step in the second subsequence if it starts after the double step at (0, s 1 ) is concluded (i.e., if then there is a double step at (0, t − 1).We are in case 2. because {x s } t−1 s=1 is a chain of length t − 2 connecting x 1 and x t−1 .Indeed, every simple step in the original chain will still be a simple step in the subsequence.Every double step (s m , s m ) with m > 1 will be a double step in the subsequence because it is nested within the double step at (0, t − 1) (i.e., 0 < s m < s m < t − 1).
Proof.The relation ∼ is reflexive: for each x ∈ Ω, it holds x ∼ x because we can set t = 1, x 0 = x, x 1 = x and observe that there is a simple step at 0 because x 0 , x 1 fulfill condition 1. of Definition 5.5.
The relation ∼ is symmetric: if {x s } t s=0 is a chain connecting x and x ′ , then {x t−s } t s=0 is a chain connecting x ′ and x.Indeed, if there is a simple step at s 0 in the original sequence, then there is a simple step at t − s 0 − 1 in the new sequence because conditions 1. or 2. are unchanged when we swap s and s ′ .If there is a double step at (s, s) in the original sequence, then there is a double step at (t − s − 1, t − s − 1) in the new sequence because condition 3. stays equivalent if we swap s and s ′ , while also swapping s ′′ and s ′′′ .
The relation ∼ is transitive.Assume x ∼ x ′ and x ′ ∼ x ′′ .Let {x s } t s=0 be a chain connecting x and x ′ and let {y u } t ′ u=0 be a chain connecting x ′ and x ′′ .We form a longer sequence x 0 = x, x 1 , . . ., x t = x ′ , x t+1 , . . ., x t+t ′ = x ′′ with x t+u := y u for all u ∈ {1, . . ., t ′ } and argue that {x s } t+t ′ s=0 is a chain of length t + t ′ connecting x and x ′′ , thus proving that x ∼ x ′′ .Indeed, any simple step happening at s 0 in the first short sequence will still be a simple step at s 0 in the long sequence, while any simple step happening at u 0 in the second short sequence corresponds to a simple step at t + u 0 in the long sequence.If, according to Definition 5.5, the double steps in the first short sequence can be listed as (s 1 , s 1 ), . . ., (s ℓ , s ℓ ) and the double steps in the second short sequence can be listed as (u 1 , u 1 ), . . ., (u ℓ ′ , u ℓ ′ ), then the double steps in the long sequence can be listed as (s 1 , s 1 ), . . ., (s ℓ , s ℓ ), (t + u 1 , t + u 1 ), . . ., (t + u ℓ ′ , t + u ℓ ′ ).Here, s 1 < . . .< s ℓ < t + u 1 < . . .< t + u ℓ ′ and no double steps are intertwined because they were not in either short list and because max{s 1 , . . ., s ℓ } < t ≤ t + u 1 = min{t + u 1 , . . ., t + u ℓ ′ }.
We have therefore proven the first statement.
The second statement can be proven by induction on the length t of the chain {x s } t s=0 connecting x and x ′ .If t = 1, then there is a simple step at 0: Remark 5.6 guarantees that S x = S x0 = S x1 = S x ′ .Now suppose the thesis true for all t < t 0 and let us prove it for t = t 0 .If there exists u such that {x s } t0 s=0 breaks at s = u into two chains of lengths less than t 0 , then the inductive hypothesis yields that S x = S xu = S x ′ .If there exists no such u, then Lemma 5.7 guarantees that there is a double step at (0, t 0 − 1).In such a case, Remark 5.6 guarantees that Hinged points have an extremely relevant property: if x ∼ x ′ , then the local representation of f near x ′ is consistent with the local representation of f near x.Equivalently, the corresponding holomorphic stem functions locally coincide.Theorem 5.9.Let Ω ⊆ H be a speared domain, let f : Ω → H be a slice regular function and let {F J : D J → H C } J∈S be the holomorphic stem family associated to f .Let α 0 , β 0 ∈ R (with β 0 ≥ 0) and I, I ′ ∈ S be such that Proof.Since x ∼ x ′ , there exist a natural number t ≥ 1 and a chain {x s } t s=0 connecting x and x ′ .We prove the thesis by induction on the length t of the chain.
If t = 1, then there is a simple step at 0. If x ′ = x 1 belongs to the connected component of S x ∩ Ω that includes x = x 0 , then the thesis follows immediately from Lemma 4.10.If, instead, x ′ = x 1 is strongly hinged to x = x 0 , then case 2. in Lemma 5.4 yields the thesis.Now choose t 0 ≥ 2, suppose the thesis true for all t < t 0 and let us prove it for t = t 0 .
• Let us first assume that there exists u with 0 < u < t 0 such that {x s } t0 s=0 breaks at s = u into two chains of lengths u and t 0 − u.In particular, x ∼ x u and x u ∼ x ′ , whence x u = α 0 + β 0 I ′′ for some I ′′ ∈ S. Since u and t 0 − u are both strictly less than t 0 , the inductive hypothesis yields that there exist open neighborhoods U, U ′ of α 0 + β 0 ı in D I,I ′′ , D I ′′ ,I ′ (respectively) such that F I , F I ′′ coincide in U and • Let us now assume, instead, that there exists no u with 0 < u < t 0 such that {x s } t0 s=0 breaks at s = u into two chains.Lemma 5.7 guarantees that t 0 ≥ 3 and that there is a double step at (0, t 0 − 1) and that {x s } t0−1 s=1 is a chain of length t 0 − 2 from x 1 to x t0−1 .In particular, by the definition of double step and by Remark 5.6, there exist α 1 , β 1 ∈ R (with β 1 ≥ 0) such that x 1 = α 1 + β 1 I, x t0−1 = α 1 + β 1 I ′ .Since t 0 − 2 < t 0 , the inductive hypothesis guarantees that there exists an open neighborhood of α 1 + β 1 ı in D I,I ′ where F I , F I ′ coincide.Since (x t0 , x t0−1 ) shadows (x 0 , x 1 ), point 1. in Lemma 5.4 guarantees that there exists a path-connected open subset of D I,I ′ including both α 0 +β 0 ı and α 1 +β 1 ı where F I , F I ′ coincide.
To prove the last statement, it suffices to perform the following computation, where we apply Remark 4.16: for α + βı ∈ U .

Hinged domains and extension theorem
This subsection is devoted to the class of domains described in the next definition and proves for them the Global Extension Theorem and the Representation Formula.
Definition 5.10.A speared domain Ω ⊆ H is termed hinged if, for each x ∈ Ω, it holds x ∼ x ′ for all x ′ ∈ S x ∩Ω; or, equivalently, if S x ∩Ω is an equivalence class under the equivalence relation ∼.
Over a hinged domain, we can draw the following consequence from Theorem 5.9.
Corollary 5.11.Let Ω ⊆ H be a hinged domain, let f : Ω → H be a slice regular function and let {F J : D J → H C } J∈S be the holomorphic stem family associated to f .If we set D := J∈S D J , then there exists a holomorphic stem function F : D → H C such that, for each J ∈ S, the function Proof.For each J ∈ S, we define F to equal F J in D J .The function F is well defined: by construction, for each α, β ∈ R (with β ≥ 0) such that α + βı ∈ D J ∩ D J ′ = D J,J ′ , the equivalence α + βJ ∼ α + βJ ′ holds, whence F J and F J ′ agree near α + βı by Theorem 5.9.
Since each F J is a holomorphic stem function, it follows that F is a holomorphic stem function, as desired.
We are now in a position to prove the announced Global Extension Theorem.It had been proven in [11,Theorem 4.5] only for the so-called simple slice domains.In the forthcoming Subsection 6.2, we will recall the definition of simple slice domain, prove that all simple slice domains are hinged and provide many examples of hinged domains that are not simple.Theorem 5.12 (Global extension for hinged domains).Let Ω ⊆ H be a hinged domain and let f : Ω → H be a slice regular function.Then there exists a unique slice regular function f : Ω → H that extends f to the symmetric completion Ω of its domain Ω.
Proof.We recall that {F J : D J → H C } J∈S denotes the holomorphic stem family associated to f and {f J : Ω J → H} J∈S denotes the slice regular family associated to f .Corollary 5.11 guarantees that there there exists a holomorphic stem function F from the union D := J∈S D J to H C such that, for each J ∈ S, the function F J is the restriction F | D J .We remark that Ω D = Ω and define f to be the element I(F ) of SR( Ω) induced by F .For each J ∈ S, the definition f J := I(F J ) yields that f J is the restriction of f to Ω J := Ω D J ⊆ Ω.We can now apply Theorem 4.11 to conclude that f coincides with f in J∈S Ω ⩾ J = Ω.
Theorem 5.12 immediately implies the Global Representation Formula for hinged domains.
Corollary 5.13 (Global representation formula for hinged domains).Let Ω ⊆ H be a hinged domain and let f : Ω → H be a slice regular function.For all α, β ∈ R with β ≥ 0 and all I, J, K ∈ S with J ̸ = K, such that α + βI, α + βJ, α + βK ∈ Ω, the equality holds.As a consequence, for each x ∈ Ω, the spherical value function f An equivalent statement is: if {f J : Ω J → H} J∈S is the slice regular family associated to a slice regular function f on a hinged domain Ω, then f coincides with f J in Ω ∩ Ω J .

Study of speared domains and hinged domains
This section is devoted to the study of speared domains and hinged domains.

Speared domains
We begin by proving that the class of speared domains (see Definition 4.1) includes the class of slice domains (see Definition 4.2).Proposition 6.1.Every quaternionic slice domain is a speared domain.
Proof.Let Ω ⊆ H be a domain.If Ω is not speared, then it is not a slice domain.Indeed: assume there exists J ∈ S such that Ω ⩾ J has a connected component C with C ∩ R = ∅.Necessarily, C is also a connected component of Ω J .In this situation, either Ω J is not connected or Ω J = C does not intersect R, whence Ω ∩ R = ∅.In either case, Ω is not a slice domain.
We now prove that every speared open subset can be locally shrunk to a slice domain.Proposition 6.2.Let U be a speared open subset of H.For every x 0 ∈ U , there exists a slice domain Λ x0 ⊆ U including x 0 .
Proof.If x 0 is a real point α 0 ∈ R, we can take Λ x0 to be any open ball B(α 0 , δ) contained in U .We therefore assume x 0 ∈ U \ R. Our proof borrows some techniques already used in the proofs of Theorems 4.8 and 4.11.
Let I ∈ S be such that x 0 ∈ U > I .By Remark 4.4, there exists a path γ : [0, 1] → ϕ −1 I (U ⩾ I ) such that γ(0) is a real number α and ϕ I (γ(1)) = x 0 .We can pick δ > 0 such that B = B(α, δ) ⊆ U ; let us denote by ∆ the disk of radius δ centered at α in R C .We set If we apply Lemma 4.7 to C and to Y := U \ R, we conclude that there exists a real number ε > 0 with the following property: for the path-connected open neighborhood for all K ∈ Cap(I, ε).We remark that, since C intersects the boundary of ∆ at γ(t 0 ), its open neighborhood C ε intersects ∆ at some point α 0 + β 0 ı.If we define then, by construction, Λ x0 is an open neighborhood of x 0 in U that intersects the real axis in the interval (α − δ, α + δ).Moreover, for each J ∈ S, its slice Λ x0 J is connected because it is the union of the following open connected subsets of C J : the disk B J = B −J of radius δ centered at α in C J ; if J ∈ Cap(I, ε), the path-connected set ϕ J (C ε ) (which intersects the disk at α 0 + β 0 J); and, if −J ∈ Cap(I, ε), the path-connected set ϕ −J (C ε ) (which intersects the disk at α 0 − β 0 J).
The same argument yields that Λ x0 is connected, whence a domain.
• We first prove that Ω is an open subset of H by picking α 0 +β 0 I ∈ Ω (with α 0 , β 0 ∈ R, β 0 ≥ 0 and I = y 1 i + y 2 j + y 3 k ∈ S) and finding an open neighborhood ), then we can pick α 1 such that w 1 (y 3 ) > α 1 > α 0 and set 4) and of the open subset of S defined by the inequality • Ω is connected, whence a domain, because, for J 0 := 1 − r 2 0 j +r 0 k, the C-shaped half-slice is connected and because, for each J ∈ S, every connected component of Ω ⩾ J intersects Ω ⩾ J0 in the real interval (−1, 0) or in the real interval (2, 3).
• The previous argument also proves that Ω is a speared domain.
We remark that spine The planar open sets spine R C (Ω) and core R C (Ω) are portrayed in Figure 4.It is an open question whether an interesting theory of slice regular functions on such a large class of domains is also available in several quaternionic variables.Indeed, the theory developed in [18] is restricted to (the multidimensional analogs of) symmetric slice domains and product domains.A positive answer to this open question may widen the class of quaternionic manifolds to which the direct approach of [7], based on the use of slice regular functions in several quaternionic variables, can be applied.

Hinged domains
We now prove that the class of hinged domains strictly includes the class of simple slice domains defined in [11].We rephrase the definition of simple slice domain according to the present setup, as follows.
Let us prove that Ω is S-connected by fixing α + βı ∈ C and arguing that the intersection between the sphere S := α + βS and Ω is connected.This is obviously true if α + βı ∈ R, since in such a case S is entirely included in Ω.We therefore assume α + βı ∈ [0, 2] × (3,4).In such a case, we see that For instance, the width functions portrayed in Figure 6, are continuous and concave; they both vanish at ±1 and take their maximum value 8/5 at 0; they have sum w 1 + w 2 greater than 2 exactly in (−2/3, 2/3) (corresponding to the dashed part of the figure).Let us denote the resulting Ω as Ω 2 .By our previous discussion, Ω 2 is S-connected.However, Ω 2 is not spear-simple.For instance, if we set J := - has a connected component that does not meet the real axis.This fact is visible in Figure 7 and can be proven by direct computation, as follows.We compute 4).We remark that B 0 is a connected component of the intersection A that does not intersect the real axis, as claimed; the other two connected components being the Example 6.13.Let us choose the width functions w 1 (r) := 2 − 2|r| =: w 2 (r) (for r ∈ [−1, 1]), portrayed in Figure 8, and denote the resulting speared domain Ω as Ω 0 .We remark that w 1 , w 2 are coinciding, continuous and concave functions on [−1, 1], vanishing at ±1 and taking their maximum value 2 at 0. Their sum w 1 +w 2 is greater than 2 exactly in (−1/2, 1/2) (corresponding to the dashed part of the figure).Based on the discussions in Examples 6.10, 6.11 and 6.12, the domain Ω 0 is spear-simple and S-connected; moreover, it has a main sail.portrayed in Figure 9.By direct inspection, w 1 , w 2 are lower semicontinuous, zero at ±1 and fulfill the inequality w 1 + w 2 > 2 exactly in the interval (−1/4, 1/4) (corresponding to the dashed part of the figure).We claim that Ω 3 is not spear-simple nor S-connected.
• Suppose y ∈ Ω.Since Ω D ′ \D does not intersect Ω, it follows that α + βı ∈ D. Thus, the point x λ belongs to Ω. Since Ω is a hinged domain, it follows that x λ ∼ y in Ω, whence in Ω ′ .Since we already established that x ∼ x λ in Ω ′ , it follows that x ∼ y in Ω ′ , as desired.
We can now provide examples of hinged domains that do not have a main sail.We do so by adding identical sails to previously constructed examples.• By Proposition 6.18, the speared domain Ω ′ is a hinged domain.
• There exists no J = x 1 i + x 2 j + x 3 k ∈ S such that (Ω ′ ) ⩾ J is a main sail for Ω ′ .If such a main sail existed, it would have to include both a copy ϕ J (D ′ ) of the added sail and the C-shaped figure ϕ J (C).The former inclusion would imply x 3 > ρ; the latter inclusion would imply w 1 (x 3 ) + w 2 (x 3 ) > 2; this would contradict our construction.
In particular, if we consider the spear-simple domain Ω 0 constructed in Example 6.13 (which was S-connected) and we set (Ω 0 ) ′ := Ω 0 ∪ U 1/2 , then (Ω 0 ) ′ is an example of spear-simple domain that is S-connected but has no main sail.If we consider the spear-simple domain Ω 1 constructed in Example 6.11 (which was not S-connected) and we set (Ω 1 ) ′ := Ω 1 ∪ U 3/4 , then (Ω 1 ) ′ is an example of spear-simple domain that is not S-connected nor has a main sail.If we consider the S-connected speared domain Ω 2 constructed in Example 6.12 (which was not spear-simple) and we set (Ω 2 ) ′ := Ω 2 ∪ U 2/3 , then (Ω 2 ) ′ is an example of S-connected speared domain that is not spear-simple nor has a main sail.
In our last example of hinged domain, double steps are necessary.In particular, the hinged domain we construct has no main sail and is not spear-simple nor S-connected.  is a simple step at 2 because there is a path from 1 + 7/2ı to 5/2 in E j,J ⊂ R C ; and there is a double step at (0, 3) because the line segment from 1 + 5ı to 1 + 7/2ı is entirely contained in E −J,J ⊂ R C .We remark that {x t } 3 t=1 is a chain of length 2 connecting x 1 to x 3 in Ω 3 ⊂ (Ω 3 ) ′ .Moreover, the points x 1 = 1 − 7/2J and x 3 = 1 + 7/2J cannot be connected by a chain of length 1 in (Ω 3 ) ′ ⊃ Ω 3 because: they belong to distinct connected components of (1 + 7/2S) ∩ (Ω 3 ) ′ ; they are not strongly hinged because the point 1 + 7/2ı belongs to the connected component D ′ of E −J,J ⊂ R C , which has D ′ ∩ R = ∅.
We conclude with Table 1, which recaps the properties of the speared domains Ω s and (Ω s ) ′ for s = 0, 1, 2, 3. spear-simple S-connected with a main sail hinged domain Table 1: Properties of the examples of speared domains constructed in the present subsection.
coincides with the entire algebra H.Moreover, C I ∩ C J = R for every I, J ∈ S with I ̸ = ±J.As a consequence, every element x of H \ R can be expressed as follows: x = α + βI, where α ∈ R is uniquely determined by x, while β ∈ R and I ∈ S are uniquely determined by x, but only up to sign.If x ∈ R, then α = x, β = 0 and I can be chosen arbitrarily in S. Therefore, it makes sense to define the real part Re(x) and the imaginary part Im(x) by setting Re(x) := t(x)/2 = (x + x c )/2 = α and Im(x) := x − Re(x) = (x − x c )/2 = βI.It also makes sense to call the Euclidean norm ∥x∥ = n(x) = a 2 + β 2 the modulus of x and to denote it as |x|.
a connected open subset of R C and it intersects the real line R, then Ω D is called a symmetric slice domain.If an open D does not intersect R and has two connected components switched by complex conjugation, then Ω D is called a product domain.A general open Ω D is a disjoint union of symmetric slice domains and product domains.

Remark 2 . 8 .
The elements of SR(Ω D ) are automatically slice regular functions according to Definition 2.3.Indeed, if f = I(F ) with F : (D, ı) → (H C , ı) holomorphic, then, for each J ∈ S, the composition ϕ as well as T ⩾ I := T ∩ C ⩾ I and T > I := T ∩ C > I .In the sequel, we will speak of a C 1 function r : T ⩾ I → H according to the usual convention: there exists a C 1 function r from an open neighborhood of T ⩾ I in C I to H, whose restriction to T ⩾ I equals r.Similarly, a C 1 function r : T ⩾ I → H will be called holomorphic if there exist an open neighborhood U of T ⩾ I in C I and a holomorphic map r : (U, I) → (H, I) whose restriction to T ⩾ I equals r.We point out that, if T ⩾ I is an open subset of C ⩾ I and if r : T ⩾ I → H is a C 1 function, then the real partial derivative of r in an arbitrary direction within C I is well-defined and continuous at each point of T ⩾ I , including the real points.Proposition 3.1.Let Ω = Ω D be a symmetric open subset of H and take distinct J, K ∈ S. Consider C 1 functions r : Ω ⩾ J → H and s : Ω ⩾ K → H.For each I ∈ S, we define p I : Ω ⩾ I → H by setting, for each α + βI ∈ Ω with α, β ∈ R and β ≥ 0,

Definition 3 . 5 .
Let Ω be an open subset of H.The spine of Ω is the open subset spine(Ω) of H defined as the union of all quaternionic open balls B

Remark 4 . 3 .
Let Ω D be a symmetric speared open subset of H and let f : Ω D → H be a slice regular function.Then f ∈ SR(Ω D ).Moreover, for all distinct J, K ∈ S, it holds f = f J,K and

Remark 4 . 4 .
An open subset U of H is speared if, and only if, for all I ∈ S and for all x ∈ U ⩾ I , there exists a path γ :[0, 1] → ϕ −1 I (U ⩾ I ) such that γ(0) ∈ R and ϕ I (γ(1)) = x.The latter property is equivalent to assuming the relative 0-homology group H 0 (U ⩾ I , U ∩ R; Z) to be trivial.An open subset T of R C is speared if, and only if, for all z ∈ T , there exists a path γ : [0, 1] → T such that γ(0) ∈ R and γ(1) = z.Up to restricting γ, we can assume the support of γ to be entirely contained in R ⩾ C or in R ⩽ C .For every open subset U of H, by Definition 3.5, the symmetric set spine(U ) is a speared open subset of H (nonempty if U ∩ R ̸ = ∅).Remark 4.4 implies that any union of speared open sets of H is speared.This justifies the next definition.Definition 4.5.Let U be an open subset of H.The core of U , denoted as core(U ), is the largest symmetric speared open subset of U .We define core R C (U ) to be the open subset D ⊆ R C , preserved by conjugation z → z, such that core(U ) = Ω D .Clearly, core(∅) = ∅.The core of a nonempty open subset U ⊆ H can be obtained as follows.Let us first set D := J∈S ϕ −1 J (U ⩾ J ): then the circularization Ω D is the largest symmetric open subset of U .This is a consequence of the following facts: it suffices to set ε := 3. Let us assume henceforth H \ Y ̸ = ∅.Since ϕ J (C) is a compact subset of the open set Y , the distance between H \ Y and ϕ J (C) is strictly positive.Set ℓ := max z∈C | Im(z)| and

Theorem 4 .
8 allows us to extend and strengthen[11, Theorem 3.2]  to our first main theorem.

4 . 6
allows us to conclude that f J0 coincides with f in V x0 = core(Ω) ∪ V ′ x0 .The thesis now follows setting Λ := x0∈Ω ⩾ J 0 V x0 .Let us add the next definition, which extends [10, Definition 2.14].We have recalled the notion of slice domain in Definition 4.2.Definition 4.12.Let Ω be a domain in H and let f : Ω → H be a slice regular function.Whenever N is a symmetric speared open set, f : N → H is a slice regular function and Λ is a speared open set contained in Ω ∩ N such that f |Λ = f |Λ , then( f , N, Λ) is called a speared extension triplet for f .A speared extension triplet ( f , N, Λ) for f where N is a symmetric slice domain and Λ is a slice domain is called a slice extension triplet for f .Using the last definition, Theorem 4.11 can be reformulated as follows.

Figure 10 :
Figure 10: The planar domain D r ∪ D ′ of Example 6.19.

K
coincide.Only when x 3 , y 3 > ρ: one of the connected components of A, namely the Γshaped component including R 2 and intersecting the real axis in the interval (2, 3), changes into a larger T -shaped component of B including R 2 ∪ D ′ , still intersecting the real axis in (2, 3); each further connected component of A equals one of the remaining connected components of B.

Figure 11 and
Figure 11 and contains the square D := (1/2, 3/2) × (3, 4) (dotted).Set [14,nition 2.3.Let U be an open subset of H: then every connected component of U is a domain.We call a function f : U → H slice regular if it is slice regular when restricted to each connected component of U .Polynomials and power series are examples of slice regular functions, see [14, Theorem 2.1], while every slice regular function on a Euclidean ball centered at 0 expands into power series, see[14, Theorem 2.7].We can summarize these facts as follows.For any x 0 ∈ H and R ∈ [0, +∞], we adopt the notation B(x 0 , R) := {x ∈ H | |x − x 0 | < R} .Theorem 2.4.Every polynomial of the form n m=0 and only if, it is contained in U ; • the open set U contains x + yS (which is compact) if, and only if, it contains a symmetric neighborhood of x + yS.Now, Ω D is a disjoint union of symmetric slice domains or product domains.The set core(U ) is the union of all connected components of Ω D that are symmetric slice domains.Clearly, spine(U ) and core(U ) are nested symmetric speared open subsets of U .They are nonempty if, and only if, U ∩ R ̸ = ∅ (which is true if U is a nonempty speared open subset of H).
Lemma 4.6.Let U be an open subset of H with U ∩ R ̸ = ∅ and let f : U → H be a slice regular function.Then f | core(U ) ∈ SR(core(U )).Moreover, for all distinct J, K ∈ S, it holds [11,proof is based on Lemma 4.6 and on a second lemma, which is a variant of[11, Lemma 3.1].Let us use the notation Cap(J, ε) := {K ∈ S : |K − J| < ε} for the spherical cap of radius ε centered at J in S (or the whole sphere S if ε > 2).Let Cap * (J, ε) := Cap(J, ε) \ {J} denote the same spherical cap, punctured at J. Lemma 4.7.Let Y be a nonempty open subset of H. Let C be a nonempty, compact and path-connected subset of R C ≃ C. For each positive real number ε, the set [11, slice extension triplet for f .Our final statement concerning real analyticity now follows from Theorem 3.8.The original result[11, Theorem 3.2]was exactly Corollary 4.13, under the additional hypothesis that Ω be a slice domain.Based on the same result, the article[10]developed a rather complete theory of regular functions on general slice domains.Later on,[12,  §11]defined the class of domains having the Local Extension Property and proved several properties of slice regular functions on such domains: for instance, the existence of a * -algebra structure [12, Corollary 11.16], the Strong Identity Principle [12, Corollary 11.19], the existence of the regular reciprocal f − * of a slice regular function f [12, Proposition 11.28], the Maximum Modulus Principle [12, Theorem 11.53], the Minimum Modulus Principle [12, Theorem 11.54], the Open Mapping Theorem [12, Theorem 11.57], the Local Cauchy Formula [12, Proposition 11.58] and its volumetric analog [12, Proposition 11.59], as well as the existence of local spherical series expansions [12, Theorem 11.63].Our Corollary 4.13 implies that the properties just listed are true on general speared domains.By applying Remark 4.3 to f J0 , we obtain another corollary of Theorem 4.11.The corollary strengthens[11, Theorem 3.4] and extends it to speared domains.