Applications of Descriptive Set Theory to Complex Analysis
aa r X i v : . [ m a t h . C V ] J a n Applications of Descriptive Set Theory toComplex Analysis
Christopher J. Caruvana ∗ School of SciencesIndiana University KokomoKokomo, Indiana 46902 Robert R. Kallman † January 20, 2021
Abstract
Descriptive set theory can be used to prove new results in classicalcomplex analysis. Let A (Ω) be the set of complex analytic functionson an open subset Ω ⊆ C endowed with the usual topology of uniformconvergence on compact subsets. A (Ω) is a Polish ring with the operationsof point-wise addition and point-wise multiplication. Inspired by L. Bers’algebraic characterization of the relation of conformality, we show thatthe topology on A (Ω) is the only Polish topology for which A (Ω) is aPolish ring for arbitrary open Ω. One thing that we deduce from this isthat, even though C has many different Polish field topologies, as long asit sits inside another Polish ring with enough complex-analytic functions,it must have its usual topology. In a different direction, we show that thebounded complex-analytic functions on the unit disk admits no Polishtopology for which it is a Polish ring. Along these lines, a corollary ofour general result is that the abstract field of meromorphic functions onan open Ω cannot be made into a Polish field. We also study the Lie ringstructure on A (Ω) which turns out to be a Polish Lie ring with the usualtopology. In this case, we restrict our attention to those domains Ω thatare connected. We extend a result of I. Amemiya to see that the Lie ringstructure is determined by the conformal structure of Ω. In a similar veinto our ring considerations, we see that, for certain domains Ω of usualinterest, the Lie ring A (Ω) has a unique Polish topology for which it is aPolish Lie ring. Again, the Lie ring A (Ω) imposes topological restrictionson C . That is, C must have its usual topology when sitting inside anyPolish Lie ring isomorphic to A (Ω). ∗ email: [email protected] † email: [email protected] Introduction
The purpose of this paper is to give numerous applications of descriptive settheory to the algebras, rings and Lie rings of mathematical objects that areubiquitous in classical complex analysis. The objects in question are Polish(complete separable metrizable) algebras, rings and Lie rings in their well knownand standard topologies. A typical theorem in this paper states that an algebraicisomorphism between one of these complex analysis objects and a similar genericPolish object is necessarily a topological isomorphism. Interesting corollariesflow from these theorems. Precedents and inspiration for this paper can befound in the work of Kakutani ([8], [9]) and Bers ([5]), Shanks and Pursell([21]), Kallman and McLinden ([11]) and I. Amemiya ([1]).This paper assumes a familiarity with descriptive set theory, especially Polishspaces, analytic sets, sets with the Baire property and Borel spaces, as may befound in Parasarathy ([18]), Kechris ([13]), Becker and Kechris ([4]) and Mackey([16]).A number of the propositions in the following discussion are given withoutproofs being presented. In every case either references are given or the proofsare easy and left to the reader.We always assume unless otherwise stated that if Ω ⊆ C is open then ∅ 6 = Ω. The following general theorems and corollaries will prove to be useful.
Theorem 1 ([4, Theorem 1.2.6]) . Let ϕ : G → H be a Baire Property measur-able homomorphism between Polish groups. Then ϕ is continuous. If moreover, ϕ [ G ] is not meager, then ϕ is also open. Theorem 2 ([13, Theorem 12.17]) . Let G be a Polish group and H a closedsubgroup of G . Then there exist a Borel subset B of G so that B meets every H -coset at exactly one point. Proposition 3 ([2, Proposition 5]) . Suppose G is a multiplicative Polish group, H is an analytic subgroup of G , and A ⊆ G is an analytic subset of G so that A meets every H -coset at exactly one point and G = AH . Then H is closed. Corollary 4.
Let G be a multiplicative Polish group, H and K be subgroups of G so that H and K are analytic sets, G = HK , and H ∩ K = { e } . Then both H and K are closed. Recall that if X is a set then a family F of subsets of X separates points iffor every pair x = y ∈ X , there exists A ∈ F so that x ∈ A and y A . Theorem 5 (Mackey [16, Theorem 3.3]) . Let X be a Polish space and suppose F is a countable family of Borel sets which separates points. Then the family F generates the Borel structure of X . That is, the smallest σ -algebra containingall members of F is precisely B ( X ) . Q ⊂ R = C as complex numbers. Corollary 6.
Let F be a countable family of Borel subsets of C , each withnonempty interior and with the property that, for every ε > , there is some B ∈ F with diam( B ) < ε . Then the countable collection { q + B | q ∈ Q , B ∈ F } generates the Borel structure of C , i.e., generates B ( C ) .Proof. This follows from Theorem 5 since it is easy to check that the countablecollection { q + B | q ∈ Q , B ∈ F } separates points.In what follows D = { ζ ∈ C | | ζ | < } will denote the open unit disk inthe complex plane and cl( D ) = { ζ ∈ C | | ζ | ≤ } is the closed unit disk in thecomplex plane. Corollary 7.
The σ -algebra generated by the sets { β + D | β ∈ Q } is B ( C ) .Similarly, the σ -algebra generated by the sets { β + cl( D ) | β ∈ Q } is B ( C ) .Proof. Let δ > δ ∈ Q be very small. Then D ∩ (2 − δ + D ) is a nonempty openset that fits inside a δ × √ δ − δ box. That is, first translate D to be centeredover 2 and then move it slightly left. Now use Corollary 6. The proof for cl( D )is virtually the same. Lemma 8.
Let X be a Polish space, ϕ a homeomorphism of X and B ⊆ X asubset with the Baire property. Then ϕ [ B ] has the Baire property.Proof. If C ⊆ X is a closed set with empty interior, then ϕ [ C ] is a closed setwith empty interior. Therefore if S ⊆ X is meager, then S is contained in theunion of a sequence closed nowhere dense sets, ϕ [ S ] is also contained in theunion of a sequence of closed nowhere dense sets and therefore ϕ [ S ] is meager.Choose an open set U so that B △ U is meager, so ϕ [ B ] △ ϕ [ U ] = ϕ [ B △ U ] isalso meager and therefore ϕ [ B ] is a set with the Baire property. Corollary 9. If R is a Polish ring, a , b ∈ R with a invertible, B ⊆ R a subsetwith the Baire property, then aB + b is also a subset with the Baire property.Proof. ϕ ( x ) = ax + b is a homeomorphism since a is invertible. Now use Lemma8. Though it is consistent with ZFC that the continuous image of a set with theBaire property fails to have the Baire property (see [17] for the existence of ∆ sets that lack the Baire property), this corollary suggests the following question.If R is a Polish ring, a ∈ R and B ⊆ R is a set with the Baire property, does aB have the Baire property? Lemma 10.
Let α, β ∈ C with α = 0 and f ( ζ ) = αζ + β . Then, for any ζ ∈ C and r > , f [ B ( ζ, r )] = B ( f ( ζ ) , | α | r ) . Lemma 11.
Let Ω ⊆ C be a Borel set with interior (Ω) = ∅ and interior (Ω c ) = ∅ . Then U = { α Ω + β | α, β ∈ Q , α = 0 } is a countable family of Borel sets that separates points. roof. U is a countable family of Borel sets since Borel sets are invariant un-der homeomorphisms. We must see that U separates points. Let ζ ∈ Q ∩ interior(Ω), ζ ∈ Q ∩ interior(Ω c ), and r > B ( ζ , r ) ⊆ interior(Ω)and B ( ζ , r ) ⊆ interior(Ω c ).Now, let w , w ∈ C be arbitrary so that w = w and pick ε > ε < | w − w | and ε | ζ − ζ | + 2 rε < | w − w | r. Pick λ j ∈ B ( w j , ε ) ∩ Q where 1 ≤ j ≤
2. Notice that B ( w , ε ) ∩ B ( w , ε ) = ∅ since 2 ε < | w − w | and therefore 0 < | λ − λ | . Also easily check that | w − w | < | λ − λ | + 2 ε by the triangle inequality.Let ϕ ( ζ ) = λ − λ ζ − ζ ( ζ − ζ ) + λ = αζ + β , where α = λ − λ ζ − ζ and β = λ ζ − λ ζ ζ − ζ are both complex rationals. Notice that ϕ ( ζ ) = λ and ϕ ( ζ ) = λ .We will prove that α Ω + β separates w and w . ϕ ( ζ j ) = λ j and therefore ϕ [ B ( ζ j , r )] = B (cid:18) λ j , | λ − λ | r | ζ − ζ | (cid:19) by Lemma 10. | w j − λ j | < ε < ( | w − w | − ε ) r | ζ − ζ | < | λ − λ | r | ζ − ζ | and therefore w j ∈ ϕ [ B ( ζ j , r )],1 ≤ j ≤ w ∈ α Ω + β and w α Ω + β , Proposition 12.
Let R be a Polish ring and ϕ : R → C be an abstract ringisomorphism. If there exists a Borel set Ω ⊆ C with interior (Ω) = ∅ andinterior (Ω c ) = ∅ so that ϕ − [Ω] has the Baire property, then ϕ is a topologicalisomorphism.Proof. First, apply Lemma 11 and then Theorem 5 to see that { α Ω + β | α, β ∈ Q , α = 0 } generates the Borel structure of C . Then, since ϕ − [Ω] has the Baireproperty and R is a Polish ring, we have that ϕ − ( α ) ϕ − [Ω] + ϕ − ( β ) is a setwith the Baire property for every α, β ∈ C , α = 0 by Lemma 9. That is, ϕ isa BP -measurable isomorphism so Theorem 1 applies to conclude that ϕ is atopological isomorphism. This section is devoted to some general facts about Polish rings and Polish C -vector spaces. Proposition 13.
Let R by any Polish ring with unity and let I R = { x ∈ R | x has a right inverse } . Then I R is an analytic set. I R is a Borel set if R is commutative.Proof. Notice that I R = { x ∈ R | there exists y ∈ R with xy = 1 R } . A = {h x, y i ∈ R | xy = 1 R } and π : R → R be the projection h x, y i 7→ x . A is closed in R since multiplication is continuous. I R is an analytic set sinceit is the range of π .Suppose R is commutative. To see that I R is Borel, it suffices to checkthat π ↾ A is injective. To see this, suppose that h x, y i , h x, y i ∈ A . Then xy = 1 R = xy = ⇒ y = y xy = xy y = y .Is I R always a Borel set? Proposition 14.
Let R be a commutative Polish ring with unity. Then themap x x − , I R → I R , is continuous if and only if I R is a G δ subset of R .Proof. First assume that x x − , I R → I R , is continuous. Let A = {h x, y i ∈ R | xy = 1 R } and notice that A is closed in R . The canonical projection π : R → R defined by h x, y i 7→ x is injective when restricted to A . Since x x − , I R → I R , is continuous, we see that x
7→ h x, x − i , I R → A , iscontinuous. So π ↾ A is a homeomorphism which implies that I R is Polishableand thus is a G δ subset of R .Conversely, suppose that G = I R is a G δ subset of R . Therefore G isPolishable. Notice that the map x x − , G → G , is a multiplicative groupisomorphism. Let A = {h x, y i ∈ G | xy = 1 R } and notice that A is closed in G since ( x, y ) xy , G → G , is continuous.The mappings A → G , h x, x − i 7→ x and h x, x − i 7→ x − are continuousbijections. Therefore the mapping x
7→ h x, x − i is a Borel mapping by theLuzin-Suslin Theorem. Hence, the mapping x
7→ h x, x − i 7→ x − is a Borelmapping. Hence, x x − , I R → I R , is continuous.From Proposition 14 we see that, for any Polish field F , the multiplicativeinversion x x − , F \ { } → F \ { } , is continuous. Proposition 15.
Let R be a Polish ring. If M ⊆ R is a principal one-sided-ideal, then M is an analytic set.Proof. The left ideal and right ideal cases are similar. For the right-ideal case, let y ∈ R be the generator for M . That is, M = { yx | x ∈ R } . Since multiplicationis continuous, the mapping x yx , R → R , is continuous and M is the imageof R under x yx . Therefore, M is an analytic set. Proposition 16.
Let X be an arbitrary complex topological vector space and = x ∈ X . The mapping λ λx , C → X , is a homeomorphism onto itsrange and its range is closed in X .Proof. This is a special case of the uniqueness theorem for finite dimensionalspaces ([14], 7.3, page 59).The ring/field of complex numbers C has 2 c ring/field automorphisms, onlytwo of which, namely the identity and complex conjugation, are continuous. Theorem 17 ([12]) . Any automorphism of C which is bounded on an F σ subsetof the plane of positive inductive dimension is necessarily continuous. The Ring of Analytic Functions and SpecialCases
Let Ω ⊆ C be open. A (Ω) denotes the collection of complex analytic functionson Ω endowed with the compact-open topology, i.e., the topology of uniformconvergence on compact sets. A (Ω) is a commutative Polish C -algebra withunity and therefore a commutative Polish ring with unity with the algebraicoperations of point-wise addition and point-wise multiplication. If λ ∈ C let c λ ∈ A (Ω) be the constant function taking the value λ . The mapping λ c λ , C → A (Ω) is continuous and is therefore a homeomorphism onto its range byProposition 16. In the future λ will be identified algebraically and topologicallywith c λ .Several of the following propositions are either well known or elementaryand their proof is left to the reader. Proposition 18.
Let Ω be open. Then Ω is connected if and only if A (Ω) isan integral domain. Proposition 19.
Let
Ω = S n<κ Ω n ⊆ C be open, where each Ω n is open andconnected, where the Ω n ’s are pairwise disjoint and where κ ℵ . The rings A (Ω) and Q n<κ A (Ω n ) are topologically ring isomorphic. Ideals in A (Ω) will play a key role in what follows, especially the followingdistinguished class of ideals. For α ∈ Ω let M α = { f ∈ A (Ω) | f ( α ) = 0 } be the associated ideal.Note that f ∈ A (Ω) is invertible if and only if f is zero-free. Let I = { f ∈A (Ω) | f is zero-free } , an abbreviation of I A (Ω) .Recall that if f ∈ A (Ω), α ∈ Ω and f ( α ) = 0, then there exists a unique g ∈ A (Ω) so that f = ( z − α ) g . More generally, for any f ∈ A (Ω) and α ∈ Ω,there exists a unique g ∈ A (Ω) so that f = ( z − α ) g + f ( α ). Lemma 20. I ⊆ A (Ω) is a proper principal maximal ideal if and only if thereexists α ∈ Ω so that I = M α .Proof. M α is a proper principal ideal since it is generated by z − α and is amaximal ideal since A (Ω) /M α is C .Conversely, let f be the generator for an ideal I in A (Ω). If f is zero-free,then f is invertible in A (Ω), so the constant 1 function is in I and I = A (Ω)contradicting the fact that I is proper. Hence, there is α ∈ Ω for which f ( α ) = 0and there is g ∈ A (Ω) such that f = ( z − α ) g . So I = f A (Ω) = ( z − α ) g A (Ω) ⊆ ( z − α ) A (Ω) = M α .I = M α since I was assumed to be maximal.6 emma 21. Let Λ be a countable dense subset of C and Ω be open and con-nected. Then, for f ∈ A (Ω) , f is constant if and only f ∈ \ λ ∈ Λ ( λ + ( I ∪ { } )) . Proof.
The implication is trivial if f is constant.Conversely, suppose f ∈ A (Ω) is non-constant. Then, by the Open MappingTheorem, there is some α ∈ Ω and some λ ∈ Λ so that f ( α ) = λ . It follows that f − λ has a zero which means f − λ is not invertible and, since f is non-constant, f − λ is not identically zero. That is, f − λ
6∈ I ∪ { } . Lemma 22.
Let Ω ⊆ C be open and connected, let R be a Polish ring and let ϕ : R → A (Ω) be an abstract isomorphism of rings. Then ϕ − [ C ] , ϕ − [ C \ Ω] and ϕ − [Ω] are Borel sets.Proof. I R = ϕ − [ I ] is a Borel set by Proposition 13. ϕ − [ C ] = \ λ ∈ Λ ( ϕ − ( λ ) + ( I R ∪ { } ))is a countable intersection of Borel sets and so is a Borel set.Next, C \ Ω = C ∩ { α ∈ C | z − α ∈ I} = C ∩ ( z + I )since −I = I . Therefore ϕ − [ C \ Ω] = ϕ − [ C ] ∩ ( ϕ − ( z ) + I R )is a Borel set.Finally, ϕ − [Ω] = ϕ [ C ] \ ϕ [ C \ Ω] is a Borel set.
Lemma 23.
Let Ω ⊆ C be open and connected, R be a Polish ring and ϕ : R →A (Ω) be an isomorphism of rings. Then(i) ϕ − [ C ] is closed in R and if M ⊆ R is a proper principal maximal ideal,then M is closed;(ii) for a proper principal maximal ideal M , h x, y i 7→ x + y, M ⊕ ϕ − [ C ] → R, is an additive group isomorphism and a homeomorphism.Proof. (i) Let M ⊆ R be a proper principal maximal ideal. It follows that thereexists α ∈ Ω so that ϕ [ M ] = M α and A (Ω) ∼ = M α ⊕ C as additive groups.Hence, R ∼ = M ⊕ ϕ − [ C ] as additive groups. As M ∩ ϕ − [ C ] = { } , M is ananalytic set by Proposition 15, and ϕ − [ C ] is a Borel set, Corollary 4 impliesthat both ϕ − [ C ] and M are closed in R .(ii) Since ϕ − [ C ] and M are closed in R , they are also additive Polish groupsso M ⊕ ϕ − [ C ] is an additive Polish group. The mapping h x, y i 7→ x + y , M ⊕ ϕ − [ C ] → R , is a homemorphism by Theorem 1 since it is continuous andan additive group isomorphism. 7 emma 24. Let Ω be open and connected and ϕ : R → A (Ω) be a ring isomor-phism. For each α ∈ Ω , the mapping x ϕ − ( ϕ ( x )( α )) , R → ϕ − [ C ] , is continuous.Proof. For α ∈ Ω, let x
7→ h x α , x ∗ α i , R → ϕ − [ M α ] ⊕ ϕ − [ C ], be the homeomor-phic isomorphism of additive groups guaranteed by Lemma 23. Since projectionis continuous, we see that the mapping x x ∗ α , R → ϕ − [ C ], is continuous.Since x = x α + x ∗ α , ϕ ( x ) = ϕ ( x α )+ ϕ ( x ∗ α ), ϕ ( x )( α ) = ϕ ( x α )( α )+ ϕ ( x ∗ α )( α ) =0 + ϕ ( x ∗ α )( α ) = ϕ ( x ∗ α )( α ) = ϕ ( x ∗ α ) since ϕ ( x α ) ∈ M α and ϕ ( x ∗ α ) ∈ C . Therefore x x ∗ α = ϕ − ( ϕ ( x ∗ α )) = ϕ − ( ϕ ( x )( α )) is continuous. Theorem 25.
Let Ω be open and connected. If R is a Polish ring and ϕ : R →A (Ω) is an abstract ring isomorphism so that ϕ ↾ ϕ − [ C ] is BP -measurable, then ϕ is a homeomorphism.Proof. ϕ − [ C ] is a closed additive subgroup of R by Lemma 23 and thereforeis itself an additive Polish group. ϕ ↾ ϕ − [ C ] is an additive group topologicalisomorphism since it is BP -measurable. Therefore for each α ∈ Ω the mapping x x ∗ α ϕ ( x ∗ α ) = ϕ ( x )( α ) is continuous. Let Λ ⊆ Ω be a countable dense setand let ψ : A (Ω) → Q λ ∈ Λ C be defined by ψ ( f ) = Q λ ∈ Λ f ( λ ). ψ is a continuousinjection, its range is a Borel set and ψ is a Borel isomorphism onto its range.Define Θ : R → A (Ω) by Θ( x ) = ψ − ( Q λ ∈ Λ ϕ ( x )( λ )). Θ is a Borel linearbijection and therefore a topological isomorphism between R and A (Ω). Corollary 26.
Let A be any Polish C -algebra and ϕ : A → A (Ω) be an ab-stract isomorphism of C -algebras, where Ω is open and connected. Then ϕ is ahomeomorphism.Proof. By Proposition 16, we have that λ λϕ − (1), C → A , is a homeomor-phism onto its range. ϕ ( λϕ − (1)) = λ since ϕ is an isomorphism of C -algebras.Hence ϕ ↾ ϕ − [ C ] is continuous so Theorem 25 applies to ensure that ϕ is ahomeomorphism.Let R be a Polish ring. R is said to be algebraically determined if, givena Polish ring S and an abstract ring isomorphism, ϕ : S → R , then ϕ is alsoa topological isomorphism. C is not an algebraically determined Polish ringsince it has many discontinuous automorphisms. On the other hand R is analgebraically determined Polish ring ([10]), a not totally trivial fact. Similardefinitions can obviously be made for algebraically determined Polish groups,Polish algebras, Polish Lie rings, or Polish Lie algebras. The previous corollarystates that A (Ω) is an algebraically determined C -algebra. Theorem 27.
Suppose Ω is open, connected and interior (Ω c ) = ∅ . Then A (Ω) is an algebraically determined Polish ring. In particular, this means that A ( D ) is algebraically determined as a Polish ring. roof. Let R be a Polish ring and let ϕ : R → A (Ω) be an algebraic iso-morphism. ϕ − [ C ] is a closed and therefore Polish subring of R by Lemma23. ϕ − [Ω] is a Borel set by Lemma 22. Proposition 12 now guarantees that ϕ ↾ ϕ − [ C ] is a Borel mapping. Conclude the proof by applying Theorem 25. Proposition 28.
Let κ ℵ , suppose R n is an algebraically determined Polishring with unity for each n < κ , and let R = Q n<κ R n . Then R is an algebraicallydetermined Polish ring.Proof. Let S be a Polish ring and let ϕ : S → R be an algebraic isomorphism.For each n let e n ∈ R be defined by e n ( ℓ ) = 0 R n if ℓ = n , e n ( n ) = 1 R n and let S n = ϕ − ( e n ) S = Sϕ − ( e n ). Each S n is an analytic set and subring of S . If f n = Q ℓ<κ R ℓ − e n , let T n = ϕ − ( f n ) S = Sϕ − ( f n ). Then T n is an analyticset and subring of S , S = S n T n = T n S n and S n ∩ T n = Q n<κ { } . Hence,Corollary 4 implies that each S n is a closed subring of S and therefore itself is aPolish ring. Let ϕ n = ϕ ↾ S n . Then ϕ n : S n → R n is an algebraic isomorphismand therefore is a topological isomorphism since R n is algebraically determined.Therefore Q n<κ ϕ n : Q n<κ S n → Q n<κ R n is a topological isomorphism.The proof will therefore be complete if we prove that S and Q n<κ S n aretopologically isomorphic. Next, note that S and S n × T n are homeomorphic sincethe natural mapping S n × T n → S is a continuous bijection of additive Polishgroups. Thus there is a natural continuous ring bijection p : S → Q n<κ S n which therefore is a topological isomorphism. Corollary 29.
Suppose Ω is open and disconnected. Then A (Ω) is an alge-braically determined Polish ring.Proof. Choose κ ℵ so that Ω = S { Ω n | n < κ } where the Ω n ’s are open,connected for n < κ and pairwise disjoint. Therefore Proposition 19 implies that A (Ω) and Q n<κ A (Ω n ) are algebraically and topologically isomorphic. Sinceinterior(Ω cn ) = ∅ for each n , Theorem 27 implies that each A (Ω n ) is algebraicallydetermined for each n < κ . Now apply Proposition 28 to conclude the proof.The results of this section, in particular Corollary 26, Theorem 27, andCorollary 29 may be considered as test cases and exercises in the techniquesdeveloped in this section and are preliminary to the general result in the nextsection. The purpose of this section is to prove that if Ω ⊆ C is open, then A (Ω) is analgebraically determined Polish ring. First recall some basic, easily proved factsabout conformal mappings. z , as usual denotes the identity mapping. Recall that a conformal mapbetween two open subsets Ω , Ω of C is a bijection γ : Ω → Ω so that9oth γ and γ − are analytic. A bijective map γ : Ω → Ω is said to beanti-conformal if γ : Ω → z [Ω ] is a conformal mapping. Here, of course,the overline denotes complex conjugation. A simple example of a conformalmapping is f ( ζ ) = αζ + β , where α , β ∈ C and α = 0. Lemma 30.
Let Ω be connected open and define Ω = z [Ω ] . Then the map ϕ : A (Ω ) → A (Ω ) defined by ϕ ( f ) = z ◦ f ◦ ( z ↾ Ω ) is a topological isomorphism of Polish rings along with the property that ϕ ( f ) ′ = ϕ ( f ′ ) for each f ∈ A (Ω ) . Proposition 31.
Suppose γ : Ω → Ω is a conformal bijection where Ω , Ω are both open. Then ϕ : A (Ω ) → A (Ω ) defined by ϕ ( f ) = f ◦ γ − is atopological isomorphism of Polish rings. More generally, A (Ω ) and A (Ω ) aretopologically isomorphic as Polish rings if Ω and Ω are open subsets of C alongwith the property that they are conformally or anti-conformally equivalent. From this we see that deciding whether or not A (Ω) is algebraically deter-mined for open and connected Ω ⊆ C reduces to deciding it for some conformalor anti-conformal representative. Definition 32.
Let R be a Polish ring and let S ⊆ R . Let ls ( S ) = { x ∈ R | x n ∈ S for some n ≥ and x m = x n for all ≤ m < n } and li ( S ) = { x ∈ R | x n ∈ S for all n ≥ and x m = x n for all ≤ m < n } . Comment: the notation is supposed to bring to mind limsup and liminf.
Lemma 33. If R is a Polish ring and S ⊆ R is a Borel set (respectively, ananalytic set), then ls ( S ) and li ( S ) are both Borel sets (respectively, analyticsets).Proof. ls ( S ) = [ n ≥ { x ∈ R | x n ∈ S } ∩ [ ≤ m Let cl( D ) ⊆ Ω ⊆ C be open, λ ∈ li (Ω) and A ∗ (Ω) = A (Ω) \ { } .Then the following are equivalent:(i) | λ | > ;(ii) { λ n | n ≥ } is relatively discrete in Ω ;(iii) there exists f ∈ A ∗ (Ω) such that f ( λ n ) = 0 for all n ≥ .Proof. (i) = ⇒ (ii). Suppose that λ ∈ li (Ω) with | λ | > 1. Then an easyinduction shows that | λ | n ≥ n ( | λ | − 1) (Bernoulli’s Inequality). Therefore {| λ n | | n ∈ N } is a sequence of strictly increasing numbers which tends towardsinfinity. This guarantees that { λ n | n ∈ N } has no accumulation points in C letalone in Ω.(ii) = ⇒ (iii). This follows immediately from Theorem 34.(iii) = ⇒ (i). Suppose λ ∈ li (Ω) and | λ | ≤ 1. Then the distinct sequence { λ n } n ≥ has a limit point in cl( D ) ⊆ Ω, which in turn implies that f is identicallyzero, a contradiction. Theorem 36. If Ω ⊆ C is open, then A (Ω) is an algebraically determinedPolish ring.Proof. Let R be a Polish ring and let ϕ : R → A (Ω) be an algebraic isomor-phism. Our goal is to prove that ϕ is a topological isomorphism. We can anddo assume that Ω is connected by Corollary 29.We can assume that cl( D ) ⊆ Ω ⊆ C . If not apply a simple conformalmapping of the form ζ αζ + β and apply Proposition 31. K = ϕ − [ C ] is a closed (and therefore Polish) subring of R by Lemma 23.To prove the theorem it suffices to show that ϕ : K → C is BP -measurableby Theorem 25. This will be accomplished by Proposition 12 if we prove that ϕ − [cl( D )] ⊆ K has the Baire property. We know already that ϕ − [ C \ Ω] ⊆ K and ϕ − [Ω] ⊆ K are Borel sets by Lemma 22. Furthermore ls ( C \ Ω) is a Borelset by Lemma 33. ( C \ Ω) ∩ cl( D ) = ∅ , so if λ n ∈ C \ Ω, then | λ | > ls ( C \ Ω) ∩ cl( D ) = ∅ . Notice that cl( D ) c = ls ( C \ Ω) ∪ { λ ∈ li (Ω) | | λ | > } . Wewill be done if we prove that ϕ − [ { λ ∈ li (Ω) | | λ | > } ] has the Baire property.Define A = {h x, a, y, b i ∈ ( R × ϕ − [Ω] × R × K ) | x = ( ϕ − ( z ) − a ) y + b } and consider the continuous map π : R × ϕ − [Ω] × R × K → R × ϕ − [Ω] definedby π ( x, a, y, b ) = h x, a i . A is relatively closed in R × ϕ − [Ω] × R × K and thereforeis a Borel subset of R × ϕ − [Ω] × R × K .We will show that π ↾ A is an injective mapping onto R × ϕ − [Ω]. Suppose h x, a, y , b i , h x, a, y , b i ∈ A. 11t follows that ( ϕ − ( z ) − a ) y + b = ( ϕ − ( z ) − a ) y + b necessitating( z − ϕ ( a )) ϕ ( y ) + ϕ ( b ) = ( z − ϕ ( a )) ϕ ( y ) + ϕ ( b ) . Since ϕ ( b ), ϕ ( b ) ∈ C , if we evaluate at ϕ ( a ), we see that ϕ ( b ) = ϕ ( b ).Consequently,( z − ϕ ( a )) ϕ ( y ) = ( z − ϕ ( a )) ϕ ( y ) = ⇒ ϕ ( y ) = ϕ ( y ) . Hence, y = y and b = b which establishes that π ↾ A is an injection.To see that π ↾ A is a surjection onto R × ϕ − [Ω], let h x, a i ∈ R × ϕ − [Ω]. Let f = ϕ ( x ) and ϕ ( a ) = α , then there exists g ∈ A (Ω) so that f = ( z − α ) g + f ( α ).If y = ϕ − ( g ) and b = ϕ − ( f ( α )), then x = ( ϕ − ( z ) − a ) y + b , h x, a, y, b i ∈ A and π ( x, a, y, b ) = h x, a i .Now, as π ↾ A is a continuous bijection onto R × ϕ − [Ω], let π ∗ : R × ϕ − [Ω] → R × ϕ − [Ω] × R × K be the Borel mapping π ∗ = ( π ↾ A ) − . Let p : R × ϕ − [Ω] × R × K → K be the projection onto the fourth coordinate, p ( x, a, y, b ) = b .For each n ≥ 1, notice that h x, a i 7→ h x, a n i , R × K → R × K , is continuous.Lemma 33 implies that li ( ϕ − [Ω]) is a Borel set since ϕ − [Ω] is a Borel set.Note that a n ∈ li ( ϕ − [Ω]) for all n ≥ a ∈ li ( ϕ − [Ω]).With all this in hand, we now define B = {h x, a i ∈ R ∗ × li ( ϕ − [Ω]) | ( p ◦ π ∗ )( x, a n ) = 0 for all n ≥ } where R ∗ = R \ { } . Notice that B is a Borel subset of R × li ( ϕ − [Ω]) as R ∗ × ϕ − [Ω] is a Borel subset of R × K and h x, a i 7→ ( p ◦ π ∗ )( x, a n ), R × ϕ − [Ω] → K , is a Borel mapping for each n ≥ 1. Hence, letting P : R × ϕ − [Ω] → ϕ − [Ω]be the projection onto the second coordinate, we see that P [ B ] is an analyticset.The next step is to see that P [ B ] = ϕ − [ { λ ∈ li (Ω) | | λ | > } ] . Toward this end, we show that ( p ◦ π ∗ )( x, a ) = ϕ − ( ϕ ( x )( ϕ ( a ))). Let h x, a, y, b i = π ∗ ( x, a ) and notice that ϕ ( x ) = ( z − ϕ ( a )) ϕ ( y )+ ϕ ( b ) = ⇒ ϕ ( x )( ϕ ( a )) = ϕ ( b ) = ⇒ b = ϕ − ( ϕ ( x )( ϕ ( a ))) . This establishes that p ◦ π ∗ ( x, a ) = ϕ − ( ϕ ( x )( ϕ ( a ))).Notice that ϕ − [ li (Ω)] = li ( ϕ − [Ω]). Now suppose a ∈ P [ B ] and let x ∈ R ∗ be so that h x, a i ∈ B . By our definition of B , a ∈ li ( ϕ − [Ω]) which, by ourinitial observation, implies that ϕ ( a ) ∈ li (Ω). By virtue of h x, a i ∈ B , we havethat 0 = ( p ◦ π ∗ )( x, a n ) = ϕ − ( ϕ ( x )( ϕ ( a n ))) = ϕ − ( ϕ ( x )( ϕ ( a ) n ))= ⇒ ϕ ( x )( ϕ ( a ) n ) = 0for each n ≥ 1. Hence, since x = 0 necessitates ϕ ( x ) = 0, we apply Lemma35 to conclude that | ϕ ( a ) | > 1. Therefore a ∈ P [ B ] implies that a ∈ ϕ − [ { λ ∈ li (Ω) | | λ | > } ] and P [ B ] ⊆ ϕ − [ { λ ∈ li (Ω) | | λ | > } ].12n the other hand suppose that a ∈ ϕ − [ { λ ∈ li (Ω) | | λ | > } ] then ϕ ( a ) ∈ li (Ω). Pick a non-zero f ∈ A (Ω) so that f ( ϕ ( a ) n ) = 0 for each n ≥ h ϕ − ( f ) , a i ∈ B , so a ∈ P [ B ], ϕ − [ { λ ∈ li (Ω) | | λ | > } ] ⊆ P [ B ], P [ B ] = ϕ − [ { λ ∈ li (Ω) | | λ | > } ] and ϕ − [ { λ ∈ li (Ω) | | λ | > } ] is an analyticsubset of K . The purpose of this section is to prove by general principles a theorem of Bers[5] as an easy application of Theorem 36. Proposition 37. Suppose γ : Ω → Ω is a conformal mapping where Ω and Ω are open and Ω is connected. Then ϕ : A (Ω ) → A (Ω ) defined by ϕ ( f ) = f ◦ γ − is a topological isomorphism of rings. The same is true if γ isanti-conformal. Lemma 38. Let ϕ : A (Ω ) → A (Ω ) be a ring isomorphism where Ω and Ω are connected. Then ϕ [ C ] = C and ϕ is either the identity or complexconjugation on C .Proof. ϕ (1) = 1, therefore ϕ ( m ) = m for m ∈ Z and furthermore ϕ ( r ) = r forall r ∈ Q . ϕ is continuous by Theorem 36 and therefore ϕ ( x ) = x for all x ∈ R . ϕ ( i ) = ϕ ( i ) = ϕ ( − 1) = − ϕ ( i ) = ± i . Hence ϕ [ C ] = C and ϕ behaveseither like the identity or complex conjugation on C . Lemma 39. Let ϕ : A (Ω ) → A (Ω ) be an abstract ring isomorphism where Ω and Ω are open and Ω is connected. Then there is a bijection γ : Ω → Ω such that ϕ ( f ( α )) = ϕ ( f )( γ ( α )) for any f ∈ A (Ω ) and α ∈ Ω .Proof. For α ∈ Ω and β ∈ Ω , let M α = { f ∈ A (Ω ) | f ( α ) = 0 } and M ∗ β = { f ∈ A (Ω ) | f ( β ) = 0 } . Now, let α ∈ Ω be arbitrary and choose γ ( α ) ∈ Ω , using Lemma 20, so that M ∗ γ ( α ) = ϕ [ M α ] .γ : Ω → Ω is a bijection since ϕ is an isomorphism of rings.Notice that f − f ( α ) ∈ M α . Then ϕ ( f ) − ϕ ( f ( α )) ∈ M ∗ γ ( α ) . Since ϕ ( f ( α )) ∈ C by Lemma 38, we see that ϕ ( f )( γ ( α )) − ϕ ( f ( α )) = 0. That is, ϕ ( f ( α )) = ϕ ( f )( γ ( α )). Theorem 40 (Bers [5]) . Let ϕ : A (Ω ) → A (Ω ) be an abstract ring iso-morphism where Ω is open and connected. Then there exists a conformal oranti-conformal mapping γ : Ω → Ω so that ϕ ( f ) = f ◦ γ − or ϕ ( f ) = f ◦ γ − . roof. ϕ is a homeomorphism by 36. Let γ : Ω → Ω be the bijection providedby Lemma 39. We proceed now by cases. Case I. Suppose ϕ ↾ C = z . Then, note that, for α ∈ Ω , α = z ( α ) = ϕ ( z ( α )) = ϕ ( z )( γ ( α )) . Hence, ϕ ( z ) ◦ γ = z ↾ Ω . Since γ is a bijection, we see that γ − = ϕ ( z ) so γ isa conformal mapping.Now, for any f ∈ A (Ω ) and α ∈ Ω , notice that f ( α ) = ϕ ( f ( α )) = ϕ ( f )( γ ( α )) = ⇒ f = ϕ ( f ) ◦ γ = ⇒ ϕ ( f ) = f ◦ γ − . Case II. Suppose ϕ ↾ C = z . Then, note that, for α ∈ Ω , α = z ( α ) = ϕ ( z ( α )) = ϕ ( z )( γ ( α )) . Hence, ϕ ( z ) ◦ γ = z ↾ Ω . It follows that γ − = ϕ ( z ) so γ is an anti-conformalmapping.Now, for any f ∈ A (Ω ) and α ∈ Ω , notice that f ( α ) = ϕ ( f ( α )) = ϕ ( f )( γ ( α )) = ⇒ z ◦ f = ϕ ( f ) ◦ γ = ⇒ z ◦ f ◦ γ − = ϕ ( f )= ⇒ ϕ ( f ) = f ◦ γ − . Recall that Liouville’s Theorem informs us that the only bounded entire func-tions are the constants. So in this trivial case the bounded entire functionsform a Polish ring. But what about the bounded analytic functions on otherdomains?Let B ( D ) be the abstract ring of bounded analytic functions on D . Let H ∞ be the ring B ( D ) endowed with the topology of uniform convergence (thetopology compatible with the sup norm metric) and identify each scalar λ ∈ C with the constant function taking value λ . In this section we assume that theabstract ring B ( D ) is given a fixed Polish ring topology and we will show thatthis leads to a contradiction.The following proposition must be well known. A detailed proof, hinted at inhttps://math.stackexchange.com/questions/1689215/h-infty-is-not-separable, isincluded for the convenience of the reader. Proposition 41. The space H ∞ is complete metrizable but not separable.Proof. The sup norm produces a complete metric on H ∞ . To see that it’s notseparable, let S = { λ ∈ C | | λ | = 1 } , and then, for each λ ∈ S , let f λ = exp (cid:18) z + λz − λ (cid:19) A = { f λ | λ ∈ S } .First, we need to see that f λ ∈ H ∞ for each λ ∈ S . Suppose λ = α + iβ where α + β = 1 and x + iy ∈ D . That is, x + y < 1. Then ℜ (cid:20) x + iy + α + iβx + iy − ( α + iβ ) (cid:21) = ℜ (cid:20) x + α + i ( y + β )( x − α ) + i ( y − β ) · ( x − α ) − i ( y − β )( x − α ) − i ( y − β ) (cid:21) = ( x + α )( x − α ) + ( y + β )( y − β )( x − α ) + ( y − β ) = x − α + y − β ( x − α ) + ( y − β ) = x + y − x − α ) + ( y − β ) < . So z + λz − λ takes D to { x + iy | x < } . Then, for x + iy so that x < e x + iy = e x (cos( y ) + i sin( y )) and, since x < e x < 1. That is, exp takes { x + iy | x < } to D . It follows that k f λ k ∞ ≤ f λ ∈ H ∞ .We will now show that A is relatively discrete in H ∞ . First, let t ∈ (0 , λ ∈ S . Notice that tλ + λtλ − λ = t + 1 t − ∈ ( −∞ , . As t → 1, we have that t + 1 t − → −∞ . Hence, exp (cid:18) t + 1 t − (cid:19) → 0. Now, take α ∈ S with α = λ . Notice that (cid:12)(cid:12)(cid:12)(cid:12) exp (cid:18) tλ + αtλ − α (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) − (cid:12)(cid:12)(cid:12)(cid:12) exp (cid:18) tλ + λtλ − λ (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) exp (cid:18) tλ + αtλ − α (cid:19) − exp (cid:18) tλ + λtλ − λ (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) = | f α ( tλ ) − f λ ( tλ ) | k f α − f λ k ∞ . Taking t → 1, we see that (cid:12)(cid:12)(cid:12)(cid:12) exp (cid:18) λ + αλ − α (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) k f α − f λ k ∞ . To finish off, let λ = a + ib and α = x + iy where a + b = x + y = 1. Then,observe that ℜ (cid:20) a + ib + x + iya + ib − ( x + iy ) (cid:21) = ℜ (cid:20) ( a + x ) + i ( b + y )( a − x ) + i ( b − y ) · ( a − x ) − i ( b − y )( a − x ) − i ( b − y ) (cid:21) = ( a + x )( a − x ) + ( b + y )( b − y )( a − x ) + ( b − y ) = a + b − ( x + y )( a − x ) + ( b − y ) = 0 . (cid:12)(cid:12)(cid:12)(cid:12) exp (cid:18) λ + αλ − α (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) = 1 k f α − f λ k ∞ . Since A is now transparently of power c and is relatively discrete in H ∞ , ametric space, we conclude that H ∞ is not separable. Lemma 42. For any f ∈ B ( D ) and α ∈ D , there exists g ∈ B ( D ) so that f = ( z − α ) g + f ( α ) . Consequently, M α = { f ∈ B ( D ) | f ( α ) = 0 } is a principalmaximal ideal.Proof. We need only check the statement for non-constant functions as thestatement obviously holds for constant functions ( g is taken to be zero when f is constant). Define g : D → C by the rule g ( ζ ) = f ( ζ ) − f ( α ) ζ − α , ζ = α ; f ′ ( α ) , ζ = α.g is analytic in D so we need only check it is bounded. Let U be a connectedopen set so that α ∈ U ⊆ cl( U ) ⊆ D and, for each ζ ∈ U , (cid:12)(cid:12)(cid:12)(cid:12) f ( ζ ) − f ( α ) ζ − α (cid:12)(cid:12)(cid:12)(cid:12) < | f ′ ( α ) | + 1 . Let D = min {| ζ − α | | ζ ∈ D \ U } and notice that D > U is an open setcontaining α . Now, for any ζ ∈ Z , | g ( ζ ) | max (cid:26) | f ′ ( α ) | + 1 , k f k ∞ D (cid:27) . and therefore g is bounded.Lastly, as the ring homomorphism f f ( α ), B ( D ) → C , has kernel M α , wesee that M α is a principal maximal ideal in B ( D ).The next lemma finds inspiration from a comment in the proof of [19, Propo-sition 3] where H. L. Royden suggests an algebraic characterization of the con-stant functions in any ring consisting of meromorphic functions. Lemma 43. Let ϕ : B ( D ) → H ∞ be the identity map. Then ϕ − [ C ] is ananalytic set.Proof. Let Λ be a countable dense subset of C . First, we will see that f isconstant if and only if, for all λ ∈ Λ, there exists g ∈ B ( D ) such that g = f − λ .If f is a constant function, then so is f − λ and f − λ has a square root.Now suppose f is non-constant and pick α ∈ D so that f ′ ( α ) = 0. Pick r > ζ ∈ cl( B ( α, r )), f ′ ( ζ ) = 0. Since f is analytic, f [ B ( α, r )] is openso pick λ ∈ f [ B ( α, r )] ∩ Λ and ζ ∈ B ( α, r ) so that f ( ζ ) = λ . Suppose we have16 ∈ B ( D ) so that g = f − λ . Notice that g ( ζ ) = f ( ζ ) − λ = 0 and g ( ζ ) = 0.Next 2 gg ′ = f ′ which gives 2 g ( ζ ) g ′ ( ζ ) = f ′ ( ζ ) = 0 , contradicting the fact that g ( ζ ) = 0. So f − λ has no square root.Now, notice that A λ = {h f, g i ∈ B ( D ) | g = f − λ } is closed so the projection π : B ( D ) → B ( D ) onto the first coordinate guar-antees that π [ A λ ] is an analytic set. Finally, analytic sets are closed undercountable intersections so A = \ { π [ A λ ] | λ ∈ Λ } is an analytic set. We are done since A = ϕ − [ C ]. Lemma 44. Let ϕ : B ( D ) → H ∞ be the identity map and let α ∈ D . Then(i) M α is closed in B ( D ) ;(ii) ϕ − [ C ] is closed in B ( D ) ;(iii) f f ( α ) , B ( D ) → ϕ − [ C ] , is continuous.Proof. (i)-(ii) Lemma 42 implies that M α is a principal ideal in B ( D ), so M α is an analytic set. Since Lemma 43 establishes that ϕ − [ C ] is an analytic addi-tive subgroup of B ( D ), both M α and ϕ − [ C ] are closed subsets of B ( D ) by anapplication of Lemma 4.(iii) The natural decomposition f 7→ h f − f ( α ) , f ( α ) i , B ( D ) → M α ⊕ ϕ − [ C ],is a homeomorphism qua a continuous group isomorphism between additivePolish groups. Lastly, f f ( α ), B ( D ) → ϕ − [ C ], is continuous. Theorem 45. The abstract ring of bounded analytic functions on the disk can-not be made into a Polish ring.Proof. Let F n = { f ∈ B ( D ) | k f k ∞ n } . Since the mapping B ( D ) → C , f f ( α ) is continuous for each α ∈ D , each F n is closed in B ( D ). Since B ( D ) = S n ≥ F n , the Baire Category Theorem implies that some F n , say F m ,contains a nonempty open subset U ⊆ F m ⊆ B ( D ). But then 0 ∈ V = U − U ⊆ F m − F m ⊆ F m , where V is now an open neighborhood of 0 ∈ B ( D ).Furthermore, if µ > 0, then 0 ∈ µ · V ⊆ µ · F m = F µ m , where µ · V is anopen neighborhood of 0 ∈ B ( D ) since multiplication by µ is a homeomorphismin B ( D ). From this it is elementary to check that every open subset of H ∞ isopen in B ( D ), i.e., the identity mapping B ( D ) → H ∞ is continuous. But thecontinuous image of a separable space is separable, contradicting Proposition41. 17 The Field of Meromorphic Functions In what follows M (Ω) denotes the abstract field of meromorphic functions onΩ. Theorem 46. The abstract field M (Ω) cannot be made into a Polish field.Proof. Suppose that M (Ω) can be given a Polish field topology. Then G = M (Ω) \ α ∈ Ω let n ( f, α ) ∈ Z be the order (positive for zeros, negative for poles, and 0 otherwise)of f at α . Each of the mappings f n ( f, α ), G → ( Z , +), is a homomorphism ofgroups which is continuous by a theorem of Dudley [7]. Since arbitrary subsetsof the discrete space Z are both open and closed, then A (Ω) \ { } = T α ∈ Ω { f ∈ G | n ( f, α ) ≥ } is closed in G . Therefore A (Ω) is a closed subset and hencePolish subring of M (Ω). Theorem 36 implies that A (Ω) ⊆ M (Ω) has its usualtopology. Then z + z/n → z in the usual topology, but 1 = n ( z + z/n, n ( z , 0) = 2, a contradiction. The purpose of this section is to give a somewhat terse discussion of an ex-trapolation of previous results to the the case of several complex variables. Weassume a familiarity with the basics as may be found in many texts, for example[15] and [20], where one can find a systematic treatment of the function theoryof several complex variables. Throughout this section we always assume that n ≥ Definition 47. Let Ω ⊆ C n be open and let π j : Ω → C for j n bedefined by π j ( α , α , . . . , α n ) = α j . For α = h α , α , . . . , α n i ∈ Ω and j n , let Ω α,j = { λ ∈ C | h α , α , . . . , α j − , λ, α j +1 , . . . , α n i ∈ Ω } and define γ α,j : Ω α,j → Ω by γ α,j ( λ ) = h α , α , . . . , α j − , λ, α j +1 , . . . , α n i . Observe that Ω α,j is an open subset of C for each α ∈ Ω and j n . Wesay that a function f : Ω → C is holomorphic provided that, for every α ∈ Ω and j n , the function f ◦ γ α,j : Ω α,j → C is holomorphic in the standardone-dimensional sense. For any domain of holomorphy Ω ⊆ C n , let H (Ω) be the collection of allholomorphic functions on Ω. By a theorem of Hartogs ([15, Theorem 1.2.5]),every holomorphic f : Ω → C is also continuous, so H (Ω) sits naturally inside of C (Ω , C ), the continuous functions from Ω to C . With the operations of point-wise addition and point-wise multiplication, H (Ω) is a commutative ring with18nity. Endow H (Ω) with the compact-open topology. This topology makes H (Ω) into a Polish ring and a Polish algebra over C since every holomorphicfunction on Ω has a local power series expansion. As before, identify the fieldof complex numbers C with the constant functions.In this section we largely work not in arbitrary open connected subsets of C n but with domains of holomorphy because if every holomorphic function onan open connected set Ω ⊂ C n can be extended to a holomorphic function on aproperly larger open connected set Ω ′ ⊆ C n and if ζ ∈ Ω ′ \ Ω, then M ζ would bea proper maximal ideal in H (Ω) = H (Ω ′ ) that does not correspond to a pointin Ω. Lemma 48. Suppose Λ ⊆ C is a countable dense set. For f ∈ H (Ω) , f is constant if and only if ( f − λ ) ∈ I ∪ { } for all λ ∈ Λ where I ⊆ H (Ω) is the collection of zero-free functions.Proof. This follows as before from the Open Mapping Theorem of several com-plex variables.From this characterization, as before, the following is immediate. Lemma 49. Let R be a Polish ring and ϕ : R → H (Ω) be a ring isomorphism.Then ϕ − [ C ] is a Borel subring of R . As in the one-dimensional case, we have the following useful result. Theorem 50. Let Ω be a domain of holomorphy, G be a Polish group, and ϕ : G → H (Ω) be an isomorphism of additive groups. If there exists a countabledense set D ⊆ Ω so that g ϕ ( g )( λ ) , G → C , is BP -measurable for each λ ∈ D , then ϕ is a homeomorphism.Proof. Let D = { λ i } i ≥ ⊆ Ω be dense with the desired property that each map g ϕ ( g )( λ i ), G → C , is BP -measurable. It follows that the map Ψ : G → Q i ≥ C defined by Ψ( g ) = Q i ≥ ϕ ( g )( λ i ) is BP -measurable. Observe thatthe map Φ : H (Ω) → Q i ≥ C defined by Φ ( f ) = Q i ≥ f ( λ i ) is a continuousinjection. Hence, by Suslin’s Theorem, its image B = Φ ( H (Ω)) is a Borelsubset of Q i ≥ C and there exists Φ : B → H (Ω) so that Φ(Φ ( f )) = f andΦ (Φ( f )) = f . Observe that Φ is a Borel bijection.Therefore Ψ( ϕ − ( f )) = Φ ( f ) so Φ ◦ Ψ = ϕ . That is, ϕ is a BP -measurablegroup isomorphism between additive Polish groups. Therefore, by Theorem 1, ϕ is a homeomorphism.If α ∈ Ω, let M α = { f ∈ H (Ω) | f ( α ) = 0 } . M α is an ideal in H (Ω).Since f = ( f − f ( α )) + f ( α ) and M α is the kernel of the ring homomorphism f f ( α ), H (Ω) → C , we see that M α is a maximal ideal and that M α ⊕ C isisomorphic to H (Ω) as additive groups.19 efinition 51. For α ∈ Ω and j n , define g α,j : Ω → C by g α,j ( β ) = π j ( β ) − π j ( α ) . We say that the ideal M α is of Gleason type if M α is the idealgenerated by { g α,j } ≤ j ≤ n . We say that a domain of holomorphy Ω is a Gleasondomain if, for every α ∈ Ω , M α is of Gleason type. Consider the collection J = X ≤ j ≤ n f j g α,j | f , f , . . . , f n ∈ H (Ω) . Clearly, J is an ideal containing { g α,j } ≤ j ≤ n and any ideal that contains { g α,j } ≤ j ≤ n contains J . That is, J is the ideal generated by { g α,j } ≤ j ≤ n . Notice that h f g α, , f g α, , . . . , f n g α,n i 7→ X ≤ j ≤ n f j g α,j , M {H (Ω) g α,j | j n } → J, is an isomorphism of additive groups. In other words, Ω is a Gleason domain ifand only if, for every α ∈ Ω, M α ∼ = M {H (Ω) g α,j | ≤ j ≤ n } as additive groups.As demonstrated in [3, Corollary 2.3], there exist domains of holomorphywhich are not Gleason domains. Lemma 52. Let Ω be a domain of holomorphy, R be a Polish ring, and ϕ : R → H (Ω) be an isomorphism of rings.(i) If, for α ∈ Ω , M α is of Gleason type, ϕ − [ M α ] is a closed ideal in R ;(ii) If there exists α ∈ Ω so that M α is of Gleason type, ϕ − [ C ] is a closedsubring of R .Proof. Let y j = ϕ − ( g α,j ) for 1 j n and notice that ϕ − [ M α ] = { x y + x y + · · · + x n y n | x , x , . . . , x n ∈ R } is an analytic set and an ideal in R . By Lemma 49, ϕ − [ C ] is a Borel subringof R . Since ϕ − [ M α ] ∩ ϕ − [ C ] = { } , Lemma 4 provides us with the fact thatboth ϕ − [ M α ] and ϕ − [ C ] are closed in R . Definition 53. We say that a domain of holomorphy Ω is almost Gleason ifthere exists a dense set D ⊆ Ω such that M α is of Gleason type for every α ∈ D . Lemma 54. Let Ω be a domain of holomorphy, R be a Polish ring, and ϕ : R → H (Ω) be a ring isomorphism. Suppose Ω is almost Gleason and let D ⊆ Ω be dense so that, for each α ∈ D , M α is of Gleason type. Then the map x ϕ − ( ϕ ( x )( α )) , R → ϕ − [ C ] , is continuous for each α ∈ D . roof. For α ∈ D , Lemma 52 informs us that ϕ − [ M α ] is a closed ideal in R , ϕ − [ C ] is a closed subring of R , and R ∼ = ϕ − [ M α ] ⊕ ϕ − [ C ] as additivePolish groups. Let the isomorphism be x 7→ h x α , x ∗ α i , R → ϕ − [ M α ] ⊕ ϕ − [ C ],and notice that it’s a homeomorphism. As projection is continuous, x x ∗ α , R → ϕ − [ C ], is continuous.The last thing to show is that ϕ ( x ∗ α ) = ϕ ( x )( α ). Notice that x α + x ∗ α = x which necessitates ϕ ( x ) = ϕ ( x α ) + ϕ ( x ∗ α ) . Since x ∗ α ∈ ϕ − [ C ], we see that ϕ ( x ∗ α ) ∈ C . Evaluating at α illustrates that ϕ ( x )( α ) = ϕ ( x ∗ α )since ϕ ( x α ) ∈ M α . Theorem 55. Let Ω be a domain of holomorphy, R be a Polish ring, and ϕ : R → H (Ω) be an abstract ring isomorphism. If Ω is almost Gleason and ϕ ↾ ϕ − [ C ] is BP -measurable, ϕ is a homeomorphism.Proof. Let D ⊆ Ω be dense so that M α is of Gleason type for every α ∈ D .Then appeal to Lemma 54 to obtain that the mapping x ϕ ( ϕ − ( ϕ ( x )( α ))) = ϕ ( x )( α ) is BP -measurable since it is the composition of a BP -measurablefunction with a continuous function. Now apply Theorem 50. Corollary 56. Let Ω be an almost Gleason domain of holomorphy. If A is aPolish C -algebra and ϕ : A → H (Ω) is an isomorphism of C -algebras, then ϕ isa homeomorphism.Proof. By Proposition 16, λ λϕ − (1), C → A , is a homeomorphism onto itsrange. In particular, ϕ − [ C ] = { λϕ − (1) | λ ∈ C } . Since ϕ is an isomorphismof C -algebras, ϕ ( λϕ − (1)) = λ . That is, ϕ ↾ ϕ − [ C ] is continuous. Now applyTheorem 55. 10 General Preliminaries on Lie Rings This section and succeeding sections are concerned with the Lie ring of analyticfunctions on a domain and its relation to Polish Lie rings, analogous to thediscussion in previous sections relating the ring of analytic functions on a domainand its relation to Polish rings. Here we will show that the continuity of a Liering isomorphism is determined by its behavior on the “abstract scalars” as wasthe case in Section 4 as witnessed by Theorem 25. It will follow from this thatthe Polish Lie rings of analytic functions on the disk and of entire functionsare algebraically determined. We will also show that Lie ring isomorphismsbetween the Lie rings of analytic functions on two different domains is inducedby a conformal or anti-conformal mapping of the underlying domains, a variationon a theorem of I. Amemiya appearing in [1] and an analogue to Theorem 40.We assume a familiarity with the basic facts about Lie rings, including thebracket operation, Lie subrings, Lie ideals. A simple example of a Lie ring21s a Lie algebra where use of all scalar operations is suppressed. The mostimportant fact to recall is the Jacobi identity: If L is a Lie ring and x, y, z ∈ L ,then [ x, [ y, z ]] + [ y, [ z, x ]] + [ z, [ x, y ]] = 0. Let L be a Lie ring. For every x ∈ L let ad ( x ) : L → L be defined by ad ( x )( y ) = [ x, y ]. For all x, y, z ∈ L we have that ad ( x ) is additive, ad ( x )([ y, z ]) = [ ad ( x )( y ) , z ] + [ y, ad ( x )( z )], and[ ad ( x ) , ad ( y )] = ad ( x ) ◦ ad ( y ) − ad ( y ) ◦ ad ( x ) = ad ([ x, y ]). In general, one canproduce a Lie ring from any ring R by defining [ · , · ] : R → R by the rule[ x, y ] = xy − yx . ( R, + , [ · , · ]) is then a Lie ring.If R is a commutative ring, then the Lie bracket is trivial. In fact, if weadd the trivial Lie bracket (which gives all brackets zero) to any Abelian group G , the result is a Lie ring. Unfortunately, there is no information to be gainedfrom trivial Lie brackets. So, in general, for an interesting Lie bracket to beobtained from a ring in this way, we would want the ring to be non-commutative.Fortunately, non-trivial brackets can be obtained from commutative rings if wehave a derivation. If R is a ring, we say that D : R → R is a derivation if D is a homomorphism of additive groups and satisfies the Leibniz rule: for each x, y ∈ R , D ( xy ) = D ( x ) y + xD ( y ). So a typical example of a derivation of a ringis the ring of analytic functions on a domain with D ( f ) = f ′ . Check that if R isa ring so that there exist x, y ∈ R with xy = 0, then any derivation D : R → R is not the identity on R . Similarly, if L is a Lie ring, we say that D : L → L is aderivation if D is a homomorphism of additive groups and satisfies the Leibnizrule: for each x, y ∈ L , D ([ x, y ]) = [ D ( x ) , y ] + [ x, D ( y )]. Every nontrivial Liering L has nontrivial derivations, for example each ad ( x ) is a derivation on L . Proposition 57. Let R be a commutative ring and D : R → R be a derivation.Define [ · , · ] : R → R by the rule [ x, y ] = xD ( y ) − D ( x ) y . Then ( R, + , [ · , · ]) is aLie ring. Moreover, D ([ x, y ]) = [ D ( x ) , y ] + [ x, D ( y )] .Proof. Again, as we already know ( R, +) is an Abelian group, we need only checkthe properties for the Lie bracket. Notice that [ x, x ] = xD ( x ) − D ( x ) x = 0, as R is commutative.Next, [ x + y, z ] = ( x + y ) D ( z ) − D ( x + y ) z = xD ( z ) + yD ( z ) − D ( x ) z − D ( y ) z = xD ( z ) − D ( x ) z + yD ( z ) − D ( y ) z = [ x, z ] + [ y, z ] . Similarly, [ x, y + z ] = xD ( y + z ) − D ( x )( y + z )= xD ( y ) + xD ( z ) − D ( x ) y − D ( x ) z = xD ( y ) − D ( x ) y + xD ( z ) − D ( x ) z = [ x, y ] + [ x, z ] . x, [ y, z ]] = [ x, yD ( z ) − D ( y ) z ]= xD ( yD ( z ) − D ( y ) z ) − D ( x )( yD ( z ) − D ( y ) z )= xD ( yD ( z )) − xD ( D ( y ) z ) − D ( x ) yD ( z ) + D ( x ) D ( y ) z = x ( D ( y ) D ( z ) + yD ( z )) − x ( D ( y ) z + D ( y ) D ( z )) − D ( x ) yD ( z ) + D ( x ) D ( y ) z = xD ( y ) D ( z ) + xyD ( z ) − xD ( y ) z − xD ( y ) D ( z ) − D ( x ) yD ( z ) + D ( x ) D ( y ) z = xyD ( z ) − xD ( y ) z − D ( x ) yD ( z ) + D ( x ) D ( y ) z ;[ y, [ z, x ]] = yzD ( x ) − yD ( z ) x − D ( y ) zD ( x ) + D ( y ) D ( z ) x = D ( x ) yz − xyD ( z ) − D ( x ) D ( y ) z + xD ( y ) D ( z );[ z, [ x, y ]] = zxD ( y ) − zD ( x ) y − D ( z ) xD ( y ) + D ( z ) D ( x ) y = xD ( y ) z − D ( x ) yz − xD ( y ) D ( z ) + D ( x ) yD ( z ) . For convenience of the reader, we rearrange the terms and align them in such away to make the cancellations more manifest:[ x, [ y, z ]] = xyD ( z ) − xD ( y ) z − D ( x ) yD ( z ) + D ( x ) D ( y ) z ;[ y, [ z, x ]] = − xyD ( z ) + D ( x ) yz + xD ( y ) D ( z ) − D ( x ) D ( y ) z ;[ z, [ x, y ]] = xD ( y ) z − D ( x ) yz + D ( x ) yD ( z ) − xD ( y ) D ( z ) . Finally, combining the terms yields [ x, [ y, z ]] + [ y, [ z, x ]] + [ z, [ x, y ]] = 0.To check that D satisfies the Leibniz rule, notice that D ([ x, y ]) = D ( xD ( y ) − D ( x ) y )= D ( xD ( y )) − D ( D ( x ) y )= D ( x ) D ( y ) + xD ( y ) − D ( x ) y − D ( x ) D ( y )= xD ( y ) − D ( x ) y and that[ D ( x ) , y ] + [ x, D ( y )] = D ( x ) D ( y ) − D ( x ) y + xD ( y ) − D ( x ) D ( y )= xD ( y ) − D ( x ) y. Now we now define the particular Lie ring with which we will be concerned.Let [ · , · ] : A (Ω) → A (Ω) be defined by [ f, g ] = f g ′ − f ′ g . Then A (Ω), alongwith point-wise addition and this bracket operation, is a Lie ring by Proposition57. Moreover, A (Ω) is a Polish Lie ring and even a Polish Lie algebra with thecompact-open topology. 23 We will prove the analogue to Theorem 40 in the context of Lie rings. The ana-logue itself is Theorem 74 and will be developed by a few intermediate results. Proposition 58. Let L be a Lie ring and M be a Lie subring. Define M ⋆ = { x ∈ M | [ x, y ] ∈ M for all y ∈ L} . Then M ⋆ is a Lie ideal in M .Proof. First, let’s see that M ⋆ is an additive subgroup of M . Let x , y ∈ M ⋆ and z ∈ L be arbitrary. It follows that[ x + y, z ] = [ x, z ] + [ y, z ] ∈ M , establishing that x + y ∈ M ⋆ .Let x ∈ M and y ∈ M ⋆ be arbitrary. To see that [ x, y ] ∈ M ⋆ , let z ∈ L bearbitrary. It follows that[ x, [ y, z ]] + [ y, [ z, x ]] + [ z, [ x, y ]] = 0 ∈ M . Notice that [ y, z ] ∈ M , which provides us with [ x, [ y, z ]] ∈ M , and that[ y, [ z, x ]] ∈ M . It follows that[[ x, y ] , z ] = − [ z, [ x, y ]] ∈ M and therefore [ x, y ] ∈ M ⋆ . Hence, M ⋆ is a Lie ideal in M . Lemma 59. For open Ω ⊆ C , Ω is connected if the following condition holds: f ′ = 0 for all f ∈ A (Ω) if and only if f is constant. Lemma 60. Let Ω ⊆ C be open and connected and let D ⊆ Ω be relativelydiscrete. Then Ω \ D is arcwise connected and therefore connected. Lemma 61. Suppose Ω is open and connected. Then for f , g ∈ A (Ω) , f = 0 , [ f, g ] = 0 if and only if there exists a unique λ ∈ C so that g = λf .Proof. Notice that [ f, λf ] = f ( λf ) ′ − f ′ λf = λ ( f f ′ − f ′ f ) = 0.Let D ⊆ Ω be the set of zeros of f . D is an at most countable relativelydiscrete subset of Ω. Now, suppose that [ f, g ] = 0. For α ∈ Ω with f ( α ) = 0 wehave (cid:18) gf (cid:19) ′ ( α ) = f ( α ) g ′ ( α ) − g ( α ) f ′ ( α ) f ( α )= 0It follows that g/f is constant on Ω \ D by Lemma 59 and Lemma 60 so thereis λ ∈ C so that g = λf on Ω \ D . It follows that g = λf on Ω.24 Recall that M α = { f ∈ A (Ω) | f ( α ) = 0 } . We will show that { M α | α ∈ Ω } isprecisely the collection Lie subalgebras of A (Ω) of co-dimension one. Lemma 62. Let Ω be open and, for a Lie subring L of A (Ω) , definealg ( L ) = X ≤ j ≤ n λ j f j | n ≥ and h λ j , f j i ∈ C × L for all ≤ j ≤ n . Then alg ( L ) is a Lie subalgebra of A (Ω) so that L ⊆ alg ( L ) and that, for anyother Lie subalgebra M of A (Ω) , if L ⊆ M , then alg ( L ) ⊆ M .Proof. It is immediate that L ⊆ alg( L ), that alg( L ) is a vector subspace of A (Ω)and that λf ∈ alg( L ) for all λ ∈ C and for all f ∈ alg( L ).To see that alg( L ) is a Lie subring, observe that X ≤ j ≤ n α j f j , X ≤ k ≤ m β k g k = X ≤ j ≤ n X ≤ k ≤ m α j β k [ f j , g k ] . As L is a Lie subring, we see that alg( L ) is a Lie subalgebra.Now, suppose M is any Lie subalgebra of A (Ω) so that L ⊆ M . Let λ f + · · · + λ n f n ∈ alg( L ) be arbitrary and notice that f j ∈ M for each 1 j n by assumption. It follows that λ j f j ∈ M for each 1 j n as M is a Liesubalgebra. So we have that λ f + · · · + λ n f n ∈ M .What we see here is that, given any Lie subring L , we can easily generate aLie subalgebra tightly related to L . Lemma 63. Let L be a Lie subalgebra of A (Ω) of co-dimension one where Ω is open. Then, for any f 6∈ L , A (Ω) ∼ = L ⊕ C f as additive groups.Proof. Let f 6∈ L and consider the space H = { g + λf | g ∈ L and λ ∈ C } .Note that H is a vector subspace of A (Ω) and that L ⊆ H . Since f 6∈ L and L is of co-dimension one, we see that H = A (Ω) so A (Ω) ∼ = L ⊕ C f as additivegroups. Lemma 64. Let Ω be open.(i) For α ∈ Ω , M α is a Lie subalgebra of A (Ω) of co-dimension one;(ii) If L is a Lie subalgebra of A (Ω) of co-dimension one and 6∈ L , thereexists α ∈ Ω so that L = M α .Proof. (i) To see that, for α ∈ Ω, M α is a Lie subring, notice that[( z − α ) f, ( z − α ) g ] = ( z − α ) ( f g ′ − f ′ g ) ∈ M α . M α is definitely closed under scalar multiplication and addition, we see that M α is a Lie subalgebra. To see that A (Ω) ∼ = M α ⊕ C as additive groups, recallthat any function f can be decomposed as follows: f = ( z − α ) g + f ( α )where g ∈ A (Ω). That is, M α is of co-dimension one.(ii) Since L is a subalgebra of co-dimension one and 1 6∈ L , Lemma 63 informsus that A (Ω) ∼ = L ⊕ C as additive groups. Thus, there exists α ∈ C so that z − α ∈ L .Let β ∈ C be so that ( z − α ) − β ∈ L . Notice that( z − α ) + β = [( z − α ) , ( z − α ) − β ] ∈ L . It follows that( z − α ) = 12 (cid:0) (( z − α ) − β ) + (( z − α ) + β ) (cid:1) ∈ L and that any scalar multiple of ( z − α ) is an element of L .Next, by way of contradiction, suppose α Ω. Then ( z − α ) − ∈ A (Ω).There exists β ∈ C so that ( z − α ) − − β ∈ L . Observe that2( z − α ) − − β = − [( z − α ) , ( z − α ) − − β ] ∈ L . It follows that( z − α ) − = (cid:0) z − α ) − − β (cid:1) − (cid:0) ( z − α ) − − β (cid:1) ∈ L . and that 1 = − 13 [( z − α ) , ( z − α ) − ] ∈ L , a contradiction. So we must have that α ∈ Ω.We will now check that L = M α . Let f ∈ A (Ω) be arbitrary and pick β ∈ C so that f − β ∈ L . Also, let λ ∈ C be so that ( z − α ) f − λ ∈ L . Check that3( z − α ) f =[( z − α ) , ( z − α ) f − λ ] − [( z − α ) , f − β ]+ 2 β ( z − α ) + (( z − α ) f − λ ) ∈ L , and therefore M α ⊆ L .Now, for f ∈ L , let g ∈ A (Ω) be so that f = ( z − α ) g + f ( α ). It follows that f ( α ) = f − ( z − α ) g ∈ L and hence f ( α ) = 0. Therefore L ⊆ M α .To finish off our analysis of the Lie subalgebras of finite co-dimension, weprovide the following lemma whose proof is adapted from the proof of [1, The-orem 3]. Notice that A (Ω) has infinitely many distinct subalgebras of finiteco-dimension, viz., if α , . . . , α n ∈ Ω are distinct, then L = T ≤ ℓ ≤ n M α ℓ is ofco-dimension n . 26 emma 65. Let Ω be open and L be a Lie subalgebra of A (Ω) of finite co-dimension. If ∈ L , then L = A (Ω) .Proof. First, note that the set P of polynomials in A (Ω) is a Lie subalgebraand is not of finite dimension. Hence, since L is a Lie subalgebra of finite co-dimension, L contains polynomials of arbitrarily large order. As 1 ∈ L and[1 , f ] = f ′ , we have that z ∈ L . Let L ⋆ = { f ∈ L | [ f, g ] ∈ L for all g ∈ A (Ω) } . L ⋆ is a Lie ideal of L by Proposition 58.For f ∈ L , we can define T f : A (Ω) / L → A (Ω) / L by T f ( g + L ) = [ f, g ] + L . Notice that T f is a well defined linear transformation. Let L ( A (Ω) / L ) be thecollection of self-linear transformations of A (Ω) / L with the structure of point-wise addition.Now define ϕ : L → L ( A (Ω) / L ) by ϕ ( f ) = T f . To see that ϕ is a lineartransformation, notice that T f + g ( h + L ) = [ f + g, h ] + L = [ f, h ] + [ g, h ] + L = T f ( h ) + T g ( h )and that, for α ∈ C \ { } , αT f ( g + L ) = α · ([ f, g ] + L ) = [ αf, g ] + L = T αf ( g ) . Notice that L ⋆ = ker( ϕ ). It follows that L ⋆ is of finite co-dimension in L and,hence, of finite co-dimension in A (Ω).Let X ∈ L \ { } and consider the set M X = { f ∈ L ⋆ | f X ∈ L ⋆ } . Since f f X , L ⋆ → A (Ω), is an additive vector space homomorphism, f f X + L ⋆ , L ⋆ → A (Ω) / L ⋆ , is an additive vector space homomorphism with kernel M X .Since L ⋆ is of finite co-dimension in A (Ω), we see that M X is of finite co-dimension in L ⋆ and, hence, in A (Ω). Also, note that for f ∈ M X and g ∈ A (Ω)we have[ f, gX ] − [ f X, g ] = f gX ′ + f g ′ X − f ′ gX − f g ′ X + f gX ′ + f ′ gX = 2 f gX ′ ∈ L . (1)Now, let M = { f ∈ A (Ω) | f g ∈ L for all g ∈ A (Ω) } . We will show that M z ⊆ M so let f ∈ M z and g ∈ L be arbitrary. By(1), we see that f g ∈ L which establishes that M z ⊆ M . Hence, M is of27nite co-dimension in A (Ω). It follows that M ∩ P = ∅ so let p ∈ M ∩ P and f ∈ A (Ω). Since p ∈ M , we see that pf ∈ L and, since 1 ∈ L , we see that[1 , pf ] = pf ′ + p ′ f ∈ L . As pf ′ ∈ L as p ∈ M , it follows that p ′ f ∈ M and, as f was arbitrary, we havethat p ′ ∈ M . Continuing in this manner we obtain that 1 ∈ M and therefore L = A (Ω). Lemma 66. Let Ω ⊆ C be open. If L is a Lie subalgebra of A (Ω) , then L is ofco-dimension one if and only if there exists α ∈ Ω so that L = M α .Proof. If α ∈ Ω, then M α is of co-dimension one by Lemma 64.On the other hand, if L is of co-dimension one, then Lemma 65 implies that1 6∈ L and Lemma 64 furnishes some α ∈ Ω so that L = M α . 13 The Relationship with Conformality We will show first that conformal equivalence induces an isomorphism of Lierings. Lemma 67. Let Ω be open and connected and define Ω = z [Ω ] = { α | α ∈ Ω } . Then the map ϕ : A (Ω ) → A (Ω ) defined by ϕ ( f ) = z ◦ f ◦ ( z ↾ Ω ) is an isomorphism of Lie rings.Proof. By Lemma 30, we need only check that ϕ preserves the Lie bracket.Lemma 30 also provided us with the fact that ϕ ( f ) ′ = ϕ ( f ′ ). Hence, ϕ ([ f, g ]) = ϕ ( f g ′ − f ′ g ) = ϕ ( f ) ϕ ( g ′ ) − ϕ ( f ′ ) ϕ ( g )= ϕ ( f ) ϕ ( g ) ′ − ϕ ( f ) ′ ϕ ( g ) = [ ϕ ( f ) , ϕ ( g )] . Proposition 68. Suppose γ : Ω → Ω is a conformal mapping where Ω isopen and connected. Then ϕ : A (Ω ) → A (Ω ) defined by ϕ ( f ) = f ◦ γ − ( γ − ) ′ , is a Lie ring isomorphism. More generally, if Ω and Ω are open and connectedsubsets of the plane along with the property that they are conformally or anti-conformally equivalent, then A (Ω ) and A (Ω ) are isomorphic as Lie rings. roof. Suppose γ : Ω → Ω is a conformal mapping. Note that, since γ − isan injective analytic function, (cid:0) γ − (cid:1) ′ is zero-free. This allows us to define ϕ asprescribed.To see that ϕ is an additive group homomorphism, observe that ϕ ( f + g ) = ( f + g ) ◦ γ − ( γ − ) ′ = f ◦ γ − ( γ − ) ′ + g ◦ γ − ( γ − ) ′ = ϕ ( f ) + ϕ ( g ) . To see that ϕ preserves the Lie bracket, witness that[ ϕ ( f ) , ϕ ( g )] = ϕ ( f ) ϕ ( g ) ′ − ϕ ( f ) ′ ϕ ( g )= f ◦ γ − ( γ − ) ′ · (cid:0) γ − (cid:1) ′ ( g ′ ◦ γ − ) (cid:0) γ − (cid:1) ′ − ( g ◦ γ − ) (cid:0) γ − (cid:1) ′′ ( γ − ) ′ ( γ − ) ′ ! − g ◦ γ − ( γ − ) ′ · (cid:0) γ − (cid:1) ′ ( f ′ ◦ γ − ) (cid:0) γ − (cid:1) ′ − ( f ◦ γ − ) (cid:0) γ − (cid:1) ′′ ( γ − ) ′ ( γ − ) ′ ! = ( f ◦ γ − )( g ′ ◦ γ − )( γ − ) ′ − (cid:0) γ − (cid:1) ′′ ( f ◦ γ − )( g ◦ γ − )( γ − ) ′ ( γ − ) ′ ( γ − ) ′ − ( f ′ ◦ γ − )( g ◦ γ − )( γ − ) ′ + (cid:0) γ − (cid:1) ′′ ( f ◦ γ − )( g ◦ γ − )( γ − ) ′ ( γ − ) ′ ( γ − ) ′ = ( f ◦ γ − )( g ′ ◦ γ − )( γ − ) ′ − ( f ′ ◦ γ − )( g ◦ γ − )( γ − ) ′ = ( f g ′ − f ′ g ) ◦ γ − ( γ − ) ′ = ϕ ([ f, g ]) . To see that ϕ is an injection, behold that ϕ ( f ) = ϕ ( g ) = ⇒ f ◦ γ − ( γ − ) ′ = g ◦ γ − ( γ − ) ′ = ⇒ f = g. Now, to see that ϕ is surjective, let g ∈ A (Ω ) be arbitrary. Then g · (cid:0) γ − (cid:1) ′ ∈A (Ω ) and g · (cid:0) γ − (cid:1) ′ ◦ γ ∈ A (Ω ). Undoubtedly, ϕ (cid:16)(cid:16) g · (cid:0) γ − (cid:1) ′ (cid:17) ◦ γ (cid:17) = g. That is, ϕ is a surjection. Conclusively, ϕ is an isomorphism of Lie rings.Now, suppose γ : Ω → Ω is an anti-conformal mapping. Then z ◦ γ : Ω → z [Ω ] is a conformal mapping so, by the above argument, A (Ω ) and A ( z [Ω ])are Lie ring isomorphic. By Lemma 67, we have that A ( z [Ω ]) and A (Ω ) areLie ring isomorphic. Therefore, A (Ω ) is Lie ring isomorphic to A (Ω ), finishingthe proof. 29hat is perhaps a bit surprising about the Proposition 68 is that the na¨ıveapproach of letting ϕ ( f ) = f ◦ γ − doesn’t produce a Lie ring isomorphism ingeneral. 14 Producing a Ring Automorphism of C Now we will show that a Lie ring isomorphism actually induces a ring automor-phism (and therefore a field automorphism) of C . Lemma 69. Let Ω and Ω be connected open sets and suppose ϕ : A (Ω ) →A (Ω ) is an isomorphism of Lie rings. There exists a field automorphism ψ : C → C so that, ϕ ( λf ) = ψ ( λ ) ϕ ( f ) for all λ ∈ C and for all f ∈ A (Ω ) Proof. Let 0 = f ∈ A (Ω ) and λ ∈ C . Notice that[ ϕ ( λf ) , ϕ ( f )] = ϕ ([ λf, f ]) = ϕ (0) = 0 . Then Lemma 61 allows us to define ψ f : C → C by the rule ψ f ( λ ) = ϕ ( λf ) ϕ ( f ) . It is immediate that ψ f (1) = 1, that ψ f is an additive group homomorphismand that its kernel is { } . Hence, ψ f is actually an injective additive grouphomomorphism of C .Observe that ψ [ f,g ] ( αβ ) ϕ [ f, g ] = ϕ ( αβ [ f, g ])= ϕ ([ αf, βg ])= [ ϕ ( αf ) , ϕ ( βg )]= [ ψ f ( α ) ϕ ( f ) , ψ g ( β ) ϕ ( g )]= ψ f ( α ) ψ g ( β ) ϕ [ f, g ] . So, ψ [ f,g ] ( αβ ) = ψ f ( α ) ψ g ( β ). Since [1 , λf ] = λf ′ = [ λ, f ], we have that ψ λf ′ ( α ) = ψ [1 ,λf ] ( α ) = ψ ( α ) ψ λf (1) = ψ ( α )and ψ λf ′ ( α ) = ψ [ λ,f ] ( α ) = ψ λ (1) ψ f ( α ) = ψ f ( α ) . Therefore, ψ ( α ) = ψ f ( α ). This allows us to define ψ : C → C by ψ ( α ) = ψ ( α )and observe that ψ has the property that ϕ ( λf ) = ψ ( λ ) ϕ ( f ).30ow, consider the following fact: ψ ( αβ ) = ψ ( αβ )= ϕ ( αβ ) ϕ (1)= ψ β ( α ) ϕ ( β ) ϕ (1)= ψ ( α ) ψ ( β )= ψ ( α ) ψ ( β ) . Ergo, ψ is an injective ring homomorphism of C .To see that ψ is surjective, let λ ∈ C be non-zero and notice that λϕ (1) ∈A (Ω ). Hence, there exists f ∈ A (Ω ) so that ϕ ( f ) = λϕ (1). Notice that0 = [ λϕ (1) , ϕ ( f )] = λ [ ϕ (1) , ϕ ( f )] = λϕ ([1 , f ]) = λϕ ( f ′ ) . It follows that f ′ = 0 which means f is constant. Hence, λϕ (1) = ϕ ( f ) = ψ ( f ) ϕ (1) which establishes that λ = ψ ( f ). Lemma 70. Suppose ϕ : A (Ω ) → A (Ω ) is a Lie ring isomorphism where Ω and Ω are open. Then the function ϕ (1) is zero-free.Proof. To facilitate readability in this proof, we will write E α ( f ) to denote thevalue of f evaluated at α . First, notice that[1 , zf ] − [ z, f ] = ( zf ′ + f ) − ( zf ′ − f ) = 2 f. Now, behold that ϕ ( f ′ ) = ϕ ([1 , f ]) = [ ϕ (1) , ϕ ( f )] = ϕ (1) ϕ ( f ) ′ − ϕ (1) ′ ϕ ( f ) (2)and, since2 ϕ ( f ) = [ ϕ (1) , ϕ ( zf )] − [ ϕ ( z ) , ϕ ( f )]= ϕ (1) ϕ ( zf ) ′ − ϕ (1) ′ ϕ ( zf ) − ϕ ( z ) ϕ ( f ) ′ + ϕ ( z ) ′ ϕ ( f ) , we have that ϕ ( f ) · (2 − ϕ ( z ) ′ ) = ϕ (1) ϕ ( zf ) ′ − ϕ (1) ′ ϕ ( zf ) − ϕ ( z ) ϕ ( f ) ′ . (3)Suppose, towards a contradiction, that ϕ (1) has a zero of order n at α . Write ϕ (1) = ( z − α ) n g where E α ( g ) = 0. Note that ϕ (1) ′ = ( z − α ) n − · (( z − α ) g ′ + ng ) . ϕ ( z ′ ) = ϕ (1) = ϕ (1) ϕ ( z ) ′ − ϕ (1) ′ ϕ ( z ) ⇐⇒ ϕ (1) ′ ϕ ( z ) = ϕ (1) · ( ϕ ( z ) ′ − . It follows that( z − α ) n − · ϕ ( z ) · (( z − α ) g ′ + ng ) = ( z − α ) n · g · ( ϕ ( z ) ′ − ϕ ( z ) · (( z − α ) g ′ + ng ) = ( z − α ) · g · ( ϕ ( z ) ′ − . Evaluating at α , we see that nE α ( g ) E α ( ϕ ( z )) = 0 which implies that E α ( ϕ ( z )) =0 as E α ( g ) = 0. Write ϕ ( z ) = ( z − α ) G where G ∈ A (Ω ). Then ϕ ( z ) ′ =( z − α ) G ′ + G . Plugging in,( z − α ) · G · (( z − α ) g ′ + ng ) = ( z − α ) · g · (( z − α ) G ′ + G − G · (( z − α ) g ′ + ng ) = g · (( z − α ) G ′ + G − . Evaluating, again, at α gives us nE α ( G ) E α ( g ) = E α ( g )( E α ( G ) − 1) = ⇒ − n ) E α ( G ) . So we see that E α ( G ) = 0 and n > 1. Recalling (3), witness that, for any f ∈ A (Ω ), ϕ ( f ) · (2 − ( z − α ) G ′ − G ) = ( z − α ) n · g · ϕ ( zf ) ′ − ( z − α ) n − · (( z − α ) g ′ + ng ) ϕ ( zf ) − ( z − α ) Gϕ ( f ) ′ = ( z − α ) H where H = ( z − α ) n − gϕ ( zf ) ′ − ( z − α ) n − (( z − α ) g ′ + ng ) ϕ ( zf ) − Gϕ ( f ) ′ . Evaluating at α , E α ( ϕ ( f )) · (2 − E α ( G )) = 0 . Since this holds for arbitrary f , E α ( G ) = 2. But 1 = (1 − n ) E α ( G ) = 2 − n implies n = 1 / , a contradiction. Therefore, ϕ (1) is zero-free. Lemma 71. Suppose ϕ : A (Ω ) → A (Ω ) is a Lie ring isomorphism where both Ω and Ω are open and connected. Then ϕ induces a bijection γ : Ω → Ω .Moreover, for any f ∈ A (Ω ) and α ∈ Ω , ϕ ( f )( γ ( α )) = ψ ( f ( α )) · ϕ (1)( γ ( α )) where ψ is the field automorphism of C provided by Lemma 69. roof. For α ∈ Ω let M α = { f ∈ A (Ω ) | f ( α ) = 0 } and, for β ∈ Ω , let M ∗ β = { f ∈ A (Ω ) | f ( β ) = 0 } . Observe that, for λ ∈ C , ϕ ( λ ) = ϕ ( λ 1) = ψ ( λ ) ϕ (1).That is, ϕ [ C ] = C · ϕ (1) . Let α ∈ Ω and notice that A (Ω ) ∼ = ϕ [ M α ] ⊕ C · ϕ (1)as additive groups establishing that ϕ [ M α ] is of co-dimension one. Hence, byLemma 66, we see that there exists β ∈ Ω so that M ∗ β = ϕ [ M α ] . Define γ ( α ) = β . As ϕ is an isomorphism of Lie rings, we see that γ : Ω → Ω is a bijection.Now, let f ∈ A (Ω ) and α ∈ Ω . Notice that f − f ( α ) ∈ M α so ϕ ( f − f ( α )) = ϕ ( f ) − ϕ ( f ( α )) = ϕ ( f ) − ψ ( f ( α )) · ϕ (1) ∈ M ∗ γ ( α ) . Evaluating at γ ( α ), we see that ϕ ( f )( γ ( α )) − ψ ( f ( α )) · ϕ (1)( γ ( α )) = 0 ⇐⇒ ϕ ( f )( γ ( α )) = ψ ( f ( α )) · ϕ (1)( γ ( α )) . Using Weierstrass’ Theorem, we obtain the following. Lemma 72. Let Ω and Ω be open and connected. Suppose ϕ : A (Ω ) → A (Ω ) is a Lie ring isomorphism and let γ : Ω → Ω be the bijection necessitated fromLemma 71. Then A ⊆ Ω is relatively discrete if and only if γ [ A ] ⊆ Ω isrelatively discrete.Proof. By symmetry, we need only demonstrate the only if direction. Let A ⊆ Ω be relatively discrete. By Weierstrass’ Theorem, there is a function f ∈A (Ω ) whose set of zeros is exactly A . Lemma 71 provides that, for α ∈ Ω , ϕ ( f )( γ ( α )) = ψ ( f ( α )) · ϕ (1)( γ ( α )). Since ϕ (1) is zero-free by Lemma 70, ϕ ( f )( γ ( α )) = 0 ⇐⇒ ψ ( f ( α )) = 0 ⇐⇒ f ( α ) = 0 . Since the set of zeros of ϕ ( f ) is exactly γ [ A ], we see that γ [ A ] is relativelydiscrete in Ω . Lemma 73. Let Ω and Ω be open and connected. Suppose ϕ : A (Ω ) → A (Ω ) is a Lie ring isomorphism. Then ψ , the field automorphism of C produced byLemma 69, is continuous.Proof. Let α ∈ Ω and ε > B ( α, ε )) ⊆ Ω . Suppose, towards acontradiction, that ψ [cl( B ( α, ε ))] is unbounded. Then, let A = { β n } n ≥ ⊆ cl( B ( α, ε ))33e so that | ψ ( β n ) | > n for each n ≥ 1. Let γ : Ω → Ω be the bijectionguaranteed by Lemma 71 and notice that ψ ( β n ) = ψ ( z ( β n )) = ϕ ( z ) ϕ (1) ( γ ( β n )) . for each n ≥ 1. Since A is not relatively discrete, Lemma 72 informs us that γ [ A ] is not relatively discrete. As ϕ ( z ) /ϕ (1) is a continuous function on Ω ,it must be the case that ψ [ γ [ A ]] has an accumulation point, contradicting ourchoice of A . Hence, ψ [cl( B ( α, ε ))] is bounded and we conclude that, by applyingTheorem 17, ψ is continuous.In [1], an analogous statement to the following is proved for complex mani-folds but requires that the isomorphism be an isomorphism of Lie algebras. Withthe additional assumption that the isomorphism be of Lie algebras, Lemma 69and most of its consequences are sidestepped. Theorem 74. Let Ω and Ω be open connected sets. Suppose that ϕ : A (Ω ) →A (Ω ) is a Lie ring isomorphism. Then there exists a conformal or anti-conformal mapping γ : Ω → Ω so that ϕ ( f ) = f ◦ γ − ( γ − ) ′ or ϕ ( f ) = f ◦ γ − (cid:16) γ − (cid:17) ′ , respectively.Proof. Let γ : Ω → Ω be the bijection guaranteed by Lemma 71. We needto see that γ is conformal or anti-conformal. So let ψ : C → C be the fieldisomorphism produced by Lemma 69 and note that ψ is continuous by Lemma73. We proceed by cases. Case I. Suppose ψ = z . Then, for α ∈ Ω , α = ψ ( α ) = ψ ( z ( α )) = ϕ ( z ) ϕ (1) ◦ γ ( α ) = ⇒ γ − = ϕ ( z ) ϕ (1) . Hence, γ is a conformal mapping as ϕ ( z ) /ϕ (1) ∈ A (Ω ).Now, observe that( γ − ) ′ = ϕ (1) ϕ ( z ) ′ − ϕ (1) ′ ϕ ( z ) ϕ (1) = [ ϕ (1) , ϕ ( z )] ϕ (1) = ϕ ([1 , z ]) ϕ (1) = 1 ϕ (1)and that, for any α ∈ Ω , f ( α ) = ψ ( f ( α )) = ϕ ( f ) ϕ (1) ◦ γ ( α ) . It follows that ϕ ( f ) = ϕ (1) · (cid:0) f ◦ γ − (cid:1) = f ◦ γ − ( γ − ) ′ . ase II. Suppose ψ = z . Then, for α ∈ Ω , α = ψ ( α ) = ψ ( z ( α )) = ϕ ( z ) ϕ (1) ◦ γ ( α ) = ⇒ γ − = ϕ ( z ) ϕ (1) . Hence, γ is an anti-conformal mapping as ϕ ( z ) /ϕ (1) ∈ A (Ω ). We now see that (cid:16) γ − (cid:17) ′ = 1 ϕ (1) . Witness that, for any α ∈ Ω , f ( α ) = ψ ( f ( α )) = ϕ ( f ) ϕ (1) ◦ γ ( α )which offers ϕ ( f ) = ϕ (1) · (cid:16) f ◦ γ − (cid:17) = f ◦ γ − (cid:16) γ − (cid:17) ′ . 15 Relationships with Polish Lie Rings Here, we turn our attention to more abstract properties about Polish Lie ringswhen they are isomorphic to the Lie ring of analytic functions on an open subsetof C . One could ask if this case can somehow be reduced to the considerationsand results from Section 4. It turns out that we can if we can show that “abstractmultiplication by the identity map z ” is continuous. Proposition 75. Suppose Ω is open, L is a Polish Lie ring, and ϕ : L → A (Ω) is an isomorphism of Lie rings. If x ϕ − ( zϕ ( x )) , L → L , is BP -measurable,then h x, y i 7→ ϕ − ( ϕ ( x ) ϕ ( y )) , L → L , is continuous.Proof. Let T z : L → L be defined by T z ( x ) = ϕ − ( zϕ ( x )) and assume that T z is BP -measurable. Notice that T z ( x + y ) = ϕ − ( zϕ ( x + y ))= ϕ − ( z ( ϕ ( x ) + ϕ ( y )))= ϕ − ( zϕ ( x ) + zϕ ( y ))= ϕ − ( zϕ ( x )) + ϕ − ( zϕ ( y ))= T z ( x ) + T z ( y ) . So we have that T z is a BP -measurable homomorphism between additive Polishgroups. Thus, by Theorem 1, we see that T z is continuous.Consider the identity: z [ f, g ] − [ zf, g ] = z ( f g ′ − f ′ g ) − zf g ′ + ( zf ′ + f ) g = f g. ∗ : L → L defined by x ∗ y = T z ([ x, y ]) − [ T z ( x ) , y ] = ϕ − ( zϕ ([ x, y ])) − [ ϕ − ( zϕ ( x )) , y ]is continuous. Observe that ϕ ( x ∗ y ) = zϕ ([ x, y ]) − [ zϕ ( x ) , ϕ ( y )]= z [ ϕ ( x ) , ϕ ( y )] − [ zϕ ( x ) , ϕ ( y )]= ϕ ( x ) ϕ ( y ) . Therefore, h x, y i 7→ ϕ − ( ϕ ( x ) ϕ ( y )), L → L , is continuous.From this, if L is a Polish Lie ring and ϕ : L → A (Ω) is a Lie ring iso-morphism so that x ϕ − ( zϕ ( x )), L → L , is continuous, we can define amultiplication ∗ : L → L by x ∗ y = ϕ − ( ϕ ( x ) ϕ ( y )). By Proposition 75,( L , + , ∗ ) is a Polish Lie ring and ϕ : L → A (Ω) is now an isomorphism of rings.In general, it’s not easy to check the hypotheses of Proposition 75 so we willhave to develop other tools.A useful result which comes to us more directly in this scenario than in themultiplicative ring case is that the “abstract scalars” are a closed subspace. Lemma 76. Let L be a Polish Lie ring and ϕ : L → A (Ω) be an isomorphismof Lie rings where Ω is open and connected. Then ϕ − [ C ] is a closed Lie subringof L . More generally, the space ϕ − [ C · f ] is a closed Lie subring of L for any f ∈ A (Ω) .Proof. First, notice that [1 , f ] = f ′ and that f is constant if and only if itsderivative is zero since Ω is connected. Then note that x ∈ ϕ − [ C ] ⇐⇒ ϕ ( x ) ∈ C ⇐⇒ [1 , ϕ ( x )] = 0 ⇐⇒ [ ϕ − (1) , x ] = 0 . It follows that ϕ − [ C ] = { x ∈ L | [ ϕ − (1) , x ] = 0 } . Hence, ϕ − [ C ] is a closed Lie subring of L .For what remains, notice that, by Lemma 61, x ∈ ϕ − [ C · f ] ⇐⇒ [ ϕ − ( f ) , x ] = 0 . We will now derive some consequences of Lemma 76. Proposition 77. Let L be a Polish Lie ring, ϕ : L → A (Ω) be a Lie ringisomorphism where Ω is open and connected, and define D : L → L by D ( x ) =[ ϕ − (1) , x ] . Then the subgroup of “abstract derivatives” D = { [ ϕ − (1) , x ] | x ∈L} is an additive Borel subgroup of L and there exists a Borel map S : D →L which chooses “abstract anti-derivatives.” That is, the function S satisfies D ◦ S ( x ) = x for each x ∈ D . roof. D is a continuous homomorphism of additive groups since D ( x + y ) = [ ϕ − (1) , x + y ] = [ ϕ − (1) , x ] + [ ϕ − (1) , y ] = D ( x ) + D ( y ) . Lemma 76 informs us that K = ϕ − [ C ] is a closed Lie subring so, by Theorem2, we can find a Borel set B ⊆ L that meets each K -coset at exactly one point.Let D = { [ ϕ − (1) , x ] | x ∈ L} and notice that D is immediately an analyticsubgroup of L as it is the image of L under D .We wish to check that D ↾ B is injective. Toward this end, let x , y ∈ B beso that D ( x ) = D ( y ). Then, notice that D ( x ) = D ( y ) = ⇒ [ ϕ − (1) , x ] = [ ϕ − (1) , y ]= ⇒ [1 , ϕ ( x )] = [1 , ϕ ( y )]= ⇒ ϕ ( x ) ′ = ϕ ( y ) ′ = ⇒ there exists λ ∈ C such that ϕ ( x ) − ϕ ( y ) = λ = ⇒ x − y ∈ K . Since B only intersects each K -coset at one point, we see that it must be thecase that x = y .Now, we will show that D ↾ B surjects onto D . Let [ ϕ − (1) , x ] ∈ D bearbitrary and pick y ∈ B ∩ ( x + K ). Let a ∈ K be so that y = x + a and observethat D ( y ) = [ ϕ − (1) , y ] = [ ϕ − (1) , x + a ] = [ ϕ − (1) , x ] . That is, D ↾ B is a surjection onto D .From the facts that D : L → L is continuous, B ⊆ L is a Borel set, D ↾ B isinjective, and D ( B ) = D , we see that D is an additive Borel subgroup of L .This allows us to define S : D → B where S = ( D ↾ B ) − . S is a Borel mapand, for x ∈ D , it chooses S ( x ) ∈ B so that [ ϕ − (1) , S ( x )] = x .Lemma 76 also allows us to reconstruct the ring structure on the abstractscalars in a continuous way. Proposition 78. Let Ω be a connected open set, L be a Polish Lie ring, and ϕ : L → A (Ω) be a Lie ring isomorphism. Then there exists a continuous map ∗ : ϕ − [ C ] → ϕ − [ C ] so that ϕ ( a ∗ b ) = ϕ ( a ) ϕ ( b ) .Proof. By Lemma 76, K = ϕ − [ C ] is a closed Lie subring, hence, an additivePolish group. Let L = ϕ − [ C · z ] and notice that L is also closed by Lemma76 and, hence, also an additive Polish group. Define D : L → K by the rule D ( x ) = [ ϕ − (1) , x ] and notice that, for any λ ∈ C , D ( ϕ − ( λz )) = [ ϕ − (1) , ϕ − ( λz )] = ϕ − ( λ ) . Also, notice that D is a continuous isomorphism of additive Polish groups.Hence, D has a continuous inverse D − : K → L .37ow, we define ∗ : K → K by the rule a ∗ b = [ a, D − ( b )] and immediatelyobserve that ∗ is continuous. To see that it has the additional desired qualities,let a = ϕ − ( α ) and b = ϕ − ( β ). Then ϕ ( a ∗ b ) = ϕ ([ a, D − ( b )])= [ α, βz ]= αβ = ϕ ( a ) ϕ ( b ) . Therefore, we see that ( K , + , ∗ ) is a Polish field as a subspace of L .For the last direct consequence of the abstract scalars being closed, we seethat the pre-image of Ω is reasonably definable. Proposition 79. Let Ω be a connected open set, L be a Polish Lie ring, and ϕ : L → A (Ω) be a Lie ring isomorphism. Then ϕ − [Ω] is an analytic subset of L .Proof. Let K = ϕ − [ C ] and recall that K is a closed Lie subring of L by Lemma76. Also, let L ∗ = L \ { } and, since L ∗ is open in L , it follows that L ∗ isPolishable. Now, consider the set A = {h a, x i ∈ K × L ∗ | [ ϕ − ( z ) − a, x ] + 2 x = 0 } . Notice that A is closed in K × L ∗ . Hence, the set A K = { a ∈ K | ∃ x ∈ L ∗ ([ ϕ − ( z ) − a, x ] + 2 x = 0) } is analytic in L .Now we just need to see that A K = ϕ − [ C \ Ω]. Suppose α ∈ C \ Ω andnotice that (cid:20) z − α, z − α (cid:21) + 2 · z − α = − ( z − α )( z − α ) − z − α + 2 z − α = 0 . This provides (cid:20) ϕ − ( z ) − ϕ − ( α ) , ϕ − (cid:18) z − α (cid:19)(cid:21) + 2 ϕ − (cid:18) z − α (cid:19) = 0 , which is that ϕ − ( α ) ∈ A K .Now, suppose α ∈ C is so that ϕ − ( α ) ∈ A K and pick x ∈ L ∗ so that[ ϕ − ( z ) − ϕ − ( α ) , x ] + 2 x = 0 . Applying ϕ , we see that0 = [ z − α, ϕ ( x )] + 2 ϕ ( x )= ( z − α ) ϕ ( x ) ′ − ϕ ( x ) + 2 ϕ ( x )= ( z − α ) ϕ ( x ) ′ + ϕ ( x ) . z − α ) ϕ ( x )) ′ = ( z − α ) ϕ ( x ) ′ + ϕ ( x ) = 0 . It follows that ( z − α ) ϕ ( x ) = λ for some λ ∈ C . Since ϕ ( x ) = 0, we see that λ = 0. It follows that ϕ ( x ) = λ ( z − α ) − , which establishes that α Ω. 16 Simply Connected Regions We first develop our results in the case of simply connected regions since thiscase provides the fundamental ideas. Here, we will always assume our simplyconnected sets are connected open sets. The difficulty when the underlyingplanar domain is not simply connected will reveal itself to be that we can’t pickglobal anti-derivatives for every function.Some of the techniques in this section owe inspiration to [11]. Definition 80. We say that a function γ : [0 , → C is a Jordan curve if γ isa simple closed curve. By the Jordan Curve Theorem, we define the interior ofa Jordan curve γ , denoted interior ( γ ) , to be the region bounded by γ . What will be useful to us here is that every analytic function on a simplyconnected domain has an anti-derivative which is also an analytic function onthat domain, see [6]. Lemma 81. Suppose Ω is a connected open set and that f ∈ A (Ω) satisfies R γ f = 0 for every piecewise differentiable Jordan curve γ in Ω . Then thereexists F ∈ A (Ω) so that F ′ = f . Lemma 82. Suppose Ω is simply connected. Then, for every f ∈ A (Ω) , thereexists F ∈ A (Ω) so that F ′ = f . This fact actually gives us a useful way to represent a function f which isanalytic on a simply connected domain. Lemma 83. Let Ω be simply connected and α ∈ Ω . For every f ∈ A (Ω) thereexists a unique g ∈ A (Ω) so that f = ( z − α ) g + f ( α ) . Moreover, there exists a G ∈ A (Ω) which is unique up to additive constant so that f = ( z − α ) G ′ + f ( α ) .Proof. Apply well known elementary facts and Lemma 82. Lemma 84. Let Ω be simply connected and α ∈ Ω . For every f ∈ A (Ω) thereexists a G ∈ A (Ω) which is unique up to additive constant so that f = G + [ z − α, G ] + f ( α ) . Proof. Let G be so that f = ( z − α ) G ′ + f ( α ), which exists by Lemma 83. Then,notice that G + [ z − α, G ] + f ( α ) = G + ( z − α ) G ′ − G + f ( α ) = ( z − α ) G ′ + f ( α ) = f. emma 85. Let Ω be simply connected, L be a Polish Lie ring, and ϕ : L →A (Ω) be a Lie ring isomorphism. For α ∈ Ω , the map x ϕ − ( ϕ ( x )( α )) , L → ϕ − [ C ] , is continuous.Proof. Recall that Lemma 76 guarantees that K = ϕ − [ C ] is a closed Lie subringof L . Now, for α ∈ Ω, let G α = {h x, y, b i ∈ L × K | x = y + [ ϕ − ( z − α ) , y ] + b } ,C α = { } × K × { } , and notice that both G α and C α are closed additive subgroups of L . To seethat C α ⊆ G α , let λ ∈ C and observe that0 = λ + [ z − α, λ ] + 0 = ⇒ ϕ − ( λ ) + [ ϕ − ( z − α ) , ϕ − ( λ )] + 0 . So h , ϕ − ( λ ) , i ∈ G α , establishing that C α ⊆ G α .Now, we appeal to Theorem 2 to produce a Borel set B ⊆ G α so that B intersects each C α -coset of G α at exactly one point. Let π : G α → L be thecanonical projection onto the first coordinate and p : G α → K be the canonicalprojection onto the third coordinate.We will now show that π ↾ B is injective. Suppose h x, y , b i , h x, y , b i ∈ B .It follows that0 = ϕ ( y ) − ϕ ( y ) + [ z − α, ϕ ( y )] − [ z − α, ϕ ( y )] + ϕ ( b ) − ϕ ( b )= ϕ ( y ) − ϕ ( y ) + ( z − α ) ϕ ( y ) ′ − ϕ ( y ) − ( z − α ) ϕ ( y ) ′ + ϕ ( y ) + ϕ ( b ) − ϕ ( b )= ( z − α )( ϕ ( y ) ′ − ϕ ( y ) ′ ) + ϕ ( b ) − ϕ ( b ) . As ϕ ( b ) , ϕ ( b ) ∈ C , evaluating at α guarantees that ϕ ( b ) = ϕ ( b ) so b = b .Now, we have that ( z − α )( ϕ ( y ) ′ − ϕ ( y ) ′ ) = 0 which guarantees that ϕ ( y ) ′ = ϕ ( y ) ′ . Then there is some λ ∈ C so that ϕ ( y ) = ϕ ( y ) + λ . It follows that h x, y , b i − h x, y , b i ∈ C α so, by definition of B , y = y . Conclusively, π ↾ B is an injection.To see that π ↾ B is also a surjection, we invoke Lemma 84. Hence, we candefine π ∗ : L → B so that π ∗ = ( π ↾ B ) − . As π ↾ B is injective, we see that π ∗ is a Borel mapping. From this, we obtain that p ◦ π ∗ is a Borel mapping.To see that p ◦ π ∗ is continuous, we will show that it is a homomorphism ofgroups. Toward this end, suppose h , y, b i ∈ G α and note that0 = ϕ ( y ) + [ z − α, ϕ ( y )] + ϕ ( b )= ϕ ( y ) + ( z − α ) ϕ ( y ) ′ − ϕ ( y ) + ϕ ( b )= ( z − α ) ϕ ( y ) ′ + ϕ ( b ) . Since ϕ ( b ) ∈ C , we see that ϕ ( b ) = 0. Hence, ( z − α ) ϕ ( y ) ′ = 0 which impliesthat ϕ ( y ) ′ = 0, providing some λ ∈ C so that ϕ ( y ) = λ . From this, we see that,provided with h , y, b i ∈ G α , we actually have that h , y, b i ∈ C α .40et x , x ∈ L and let y , y , y ∈ L and b , b , b ∈ K be so that π ∗ ( x ) = h x , y , b i , π ∗ ( x ) = h x , y , b i , and π ∗ ( x + x ) = h x + x , y , b i .Witness that π ∗ ( x ) + π ∗ ( x ) − π ∗ ( x + x ) = h , y + y − y , b + b − b i ∈ G α . Since the first coordinate is zero, we conclude that b + b − b = 0 whichestablishes that p ◦ π ∗ ( x + x ) = b = b + b = p ◦ π ∗ ( x ) + p ◦ π ∗ ( x ) . Ergo, p ◦ π ∗ is continuous qua a BP -measurable homomorphism of Polishgroups.The last thing we need to show is that p ◦ π ∗ ( x ) = ϕ − ( ϕ ( x )( α )). Let h y, b i ∈ L × K be so that π ∗ ( x ) = h x, y, b i . Then realize that, since x = y + [ ϕ − ( z − α ) , y ] + b , ϕ ( x ) = ϕ ( y ) + [ z − α, ϕ ( y )] + ϕ ( b )= ϕ ( y ) + ( z − α ) ϕ ( y ) ′ − ϕ ( y ) + ϕ ( b )= ( z − α ) ϕ ( y ) ′ + ϕ ( b ) . Hence, ϕ ( x )( α ) = ϕ ( b ) which leads us to conclude that p ◦ π ∗ ( x ) = ϕ − ( ϕ ( x )( α )). Theorem 86. Let Ω be simply connected, L be a Polish Lie ring, and ϕ : L →A (Ω) be a Lie ring isomorphism. If ϕ ↾ ϕ − [ C ] is BP -measurable, then ϕ is ahomeomorphism.Proof. The proof of this theorem is completely analogous to the proof of Theo-rem 25. Corollary 87. Suppose Ω is simply connected, L is a Polish Lie algebra, and ϕ : L → A (Ω) is an isomorphism of Lie algebras. Then ϕ is a homeomorphism.Proof. By Proposition 16 we see that λ λϕ − (1), C → L , is a homeomor-phism onto its range and that ϕ − ( C ) = { λϕ − (1) | λ ∈ C } is closed in L . As ϕ is an isomorphism of Lie algebras, ϕ ( λϕ − (1)) = λ . That is, ϕ ↾ ϕ − ( C ) is contin-uous so we appeal to Theorem 86 to conclude that ϕ is a homeomorphism. Theorem 88. Let Ω = C be simply connected. Then the Polish Lie ring A (Ω) is algebraically determined. In particular, A ( D ) is an algebraically determinedPolish Lie ring.Proof. By Theorem 74 and the Riemann Mapping Theorem, it suffices to checkthat A ( D ) is algebraically determined. Toward this end, let L be any Polish Liering and let ϕ : L → A ( D ) be a Lie ring isomorphism. Then Proposition 79yields that ϕ − [ D ] is an analytic set. Thus, Proposition 12 applies to give usthat ϕ ↾ ϕ − [ C ] is continuous since ϕ − [ C ] is a closed Lie subring by Lemma 76.Finally, appeal to Theorem 86 to see that ϕ is continuous.41he simply connected case actually affords us an application of Proposition75. Proposition 89. Suppose Ω is simply connected, L is a Polish ring, and ϕ : L → A (Ω) is a Lie ring isomorphism. Then we can define ∗ : L → L so that ( L , + , ∗ ) is a Polish ring and ϕ : L → A (Ω) is an isomorphism of rings.Proof. Since Ω is simply connected, every function has an anti-derivative. Par-ticularly, the map D : L → L defined by D ( x ) = [ ϕ − (1) , x ] is a surjection. Let S : L → L be as in Proposition 77 where we saw that S is a Borel mapping.Now, define T : L → L by the rule T ( x ) = [ ϕ − ( z ) , S ( x )] + S ( x ) and notice that T is a Borel mapping. Since S ( x ) satisifes D ◦ S ( x ) = [ ϕ − (1) , S ( x )] = x , wehave that ϕ ( x ) = [1 , ϕ ◦ S ( x )] = ( ϕ ◦ S ( x )) ′ . Now, witness that ϕ ◦ T ( x ) = [ z, ϕ ◦ S ( x )] + ϕ ◦ S ( x )= zϕ ( x ) − ϕ ◦ S ( x ) + ϕ ◦ S ( x )= zϕ ( x ) . Hence, we have that T ( x ) = ϕ − ( zϕ ( x )). To finish the proof, apply Proposition75. Theorem 90. The Lie ring of entire functions is algebraically determined.Proof. Let L be a Polish Lie ring and suppose ϕ : L → A ( C ) is a Lie ringisomorphism. Define ∗ : L → L to be x ∗ y = ϕ − ( ϕ ( x ) ϕ ( y )) so that ∗ iscontinuous by Proposition 89. As ϕ is now an isomorphism of rings, we appealto Theorem 36 to see that ϕ is a homeomorphism. Corollary 91. If Ω is simply connected, then A (Ω) is algebraically determined.Proof. Splice together Theorem 88 and Theorem 90. 17 Other Regions Definition 92. We say that an open connected set Ω is nicely complemented ifthere exist a finite collection of pairwise disjoint connected compact subsets K ,. . . , K n ⊂ C \ Ω such that Ω ⋆ = Ω ∪ [ ≤ j ≤ n K j is open, connected and simply connected, and such that for every counter clock-wise oriented piecewise differentiable Jordan curve γ in Ω , there are counterclockwise oriented piecewise differentiable Jordan curves { γ } ≤ j ≤ n in Ω suchthat each K j is in the interior of γ j and such that all of the curves γ and the γ j ’s are pairwise disjoint. We use this notation in the rest of this section. C , D , or an annulus with a finite number of pointsor closed line segments removed. It is clear in these cases how to construct the γ j ’s. Lemma 93. Suppose Ω is nicely complemented. Pick β j ∈ K j for each j n . Then, for each f ∈ A (Ω) , there exists a unique h λ , . . . , λ n i ∈ C n so that Z γ f − X ≤ j ≤ n λ j z − β j = 0 for every piecewise differentiable Jordan curve γ contained in Ω .Proof. Given γ , construct the γ j ’s provided in Definition 92. The K j ’s and the γ j ’s do not intersect γ . Note that the winding numbers n ( γ j , β ) = n ( γ j , β j ) forall β ∈ K j since K j is connected. Therefore the particular choice of β j ∈ K j isimmaterial. Next, Z γ j z − β j = 2 πi. Let λ j = 12 πi · Z γ j f and notice that Z γ j (cid:20) f − λ j z − β j (cid:21) = 2 πiλ j − πiλ j = 0 . Partition the indices 1 ≤ j ≤ n into two camps, A = { j | β j is inside γ } and B = { j | β j is outide γ } . If j ∈ A , then n ( γ, β j ) = 1 and n ( γ j , β j ) = 1. If j ∈ B , then n ( γ, β j ) = 0. If i = j , then n ( γ j , β i ) = 0. Finally, if β ∈ C \ Ω ⋆ ,then n ( γ, β ) = 0 and n ( γ j , β ) = 0. Therefore γ ∼ P j ∈ A γ j , i.e., γ and the chain P j ∈ A γ j are homologous. Therefore Z γ f − X ≤ i ≤ n λ i z − β i = Z γ ( f − X i ∈ A λ i z − β i ) = X j ∈ A Z γ j ( f − X i ∈ A λ i z − β i ) = X j ∈ A Z γ j (cid:26) f − λ j z − β j (cid:27) = 0 . Lemma 94. Suppose Ω is nicely complemented, L is a Polish Lie ring, and ϕ : L → A (Ω) is a Lie ring isomorphism. Then the additive subgroup D = { [ ϕ − (1) , x ] | x ∈ L} is closed in L . roof. Define λ : A (Ω) → C n so that Z γ f − X ≤ j ≤ n λ ( f ) j z − β j = 0for each Jordan curve γ in Ω by Lemma 93. Using Lemma 81, we see that,letting H j = C · z − β j and H = X ≤ j ≤ n g j | g j ∈ H j for all 1 j n , L ∼ = D ⊕ H. Then, as H is an analytic additive subgroup of L , D is Borel by Proposition77, and D ∩ H = { } , we see that both D and H are closed in L by Corollary4. Lemma 95. Let Ω be nicely complemented, L be a Polish Lie ring, and ϕ : L →A (Ω) be a Lie ring isomorphism. For each α ∈ Ω , the map x ϕ − ( ϕ ( x )( α )) , L → ϕ − [ C ] , is continuous.Proof. Define f j = z − αz − β j . In the notation of Lemma 94 H j = ϕ − [ C · f j ] isa closed additive subgroup of L and H = H × · · · × H n is a closed additivesubgroup of L n . Lemma 76 also guarantees that K = ϕ − [ C ] is a closed additivesubgroup of L .Define G α = {h x, y, w, b i ∈ L × H × L × K | x − ( y + · · · + y n ) = w + [ ϕ − ( z − α ) , w ] + b } and C α = { } n +1 × K × { } . Observe that G α is a closed additive subgroup of L n +3 and that C α is a closedadditive subgroup of G α . So we can appeal to Theorem 2 to find a Borel set B ⊆ G α which meets every C α -coset of G α at exactly one point.Let π : G α → L be the projection π ( x, y, w, b ) = x . To see that π ↾ B is asurjection onto L , let f ∈ A (Ω), choose g ∈ A (Ω) so that f = ( z − α ) g + f ( α ),and combine Lemmas 81 and 93 to pick G ∈ A (Ω) so that G ′ = g − X ≤ j ≤ n λ j z − β j λ , . . . , λ n ∈ C . Then observe that G + [ z − α, G ] + f ( α ) = G + ( z − α ) G ′ − G + f ( α )= ( z − α ) · g − X ≤ j ≤ n λ j z − β j + f ( α )= f − X ≤ j ≤ n λ j f j . By adding a suitable constant to G , we see that π ↾ B is a surjection onto L .Now we will see that π ↾ B is injective. Let h x, y, w , b i , h x, v, w , b i ∈ B .It follows that y + · · · + y n + w +[ ϕ − ( z − α ) , w ]+ b = v + · · · + v n + w +[ ϕ − ( z − α ) , w ]+ b which holds if and only if ϕ ( y )+ · · · + ϕ ( y n )+( z − α ) ϕ ( w ) ′ + ϕ ( b ) = ϕ ( v )+ · · · + ϕ ( v n )+( z − α ) ϕ ( w ) ′ + ϕ ( b ) . By our definition of H and f j , notice that, for each 1 ≤ j ≤ n , ϕ ( y j )( α ) = ϕ ( v j )( α ) = 0. So, since ϕ ( b ), ϕ ( b ) ∈ C , we see that ϕ ( b ) = ϕ ( b ) whichprovides b = b .Now, pick λ ,j ∈ C so that ϕ ( y j ) = λ ,j f j and λ ,j ∈ C so that ϕ ( v j ) = λ ,j f j . Witness that this provides the equality ϕ ( w ) ′ − ϕ ( w ) ′ = λ , − λ , z − β + · · · + λ ,n − λ ,n z − β n . By integrating along γ j , we see that λ ,j = λ ,j . This conclusively demonstratesthat ϕ ( w ) ′ = ϕ ( w ) ′ which provides some λ ∈ C so that ϕ ( w ) = ϕ ( w ) + λ .Hence, by our choice of B , we see that w = w . That is, that h x, y, w , b i = h x, v, w , b i , establishing the injectivity of π ↾ B .Now we can define a Borel map π ∗ : L → B , π ∗ = ( π ↾ B ) − , and, for theprojection p : G α → K , p ( x, y, w, b ) = b , we have that p ◦ π ∗ : L → K is aBorel mapping. The last thing to see is that p ◦ π ∗ ( x ) = ϕ − ( ϕ ( x )( α )) sinceit’s clear that x ϕ − ( ϕ ( x )( α )), L → K , is a Borel (and therefore continuous)homomorphism of additive Polish groups. So let h x, y, w, b i = π ∗ ( x ) and noticethat ϕ ( x ) − X ≤ j ≤ n ϕ ( y j ) = ( z − α ) ϕ ( w ) ′ + ϕ ( b ) . Again, by our definition of f j and the fact that ϕ ( b ) ∈ C , we see that ϕ ( x )( α ) = ϕ ( b ) which guarantees that p ◦ π ∗ ( x ) = ϕ − ( ϕ ( x )( α )). Theorem 96. Suppose Ω is nicely complemented, L is a Polish Lie ring, and ϕ : L → A (Ω) is a Lie ring isomorphism. If ϕ ↾ ϕ − [ C ] is BP -measurable, then ϕ is a homeomorphism.Proof. Combine Lemma 95 and the method of proof of Theorem 25.45 orollary 97. Suppose Ω is nicely complemented, L is a Polish Lie algebra, and ϕ : L → A (Ω) is an isomorphism of Lie algebras. Then ϕ is a homeomorphism.Proof. As before, recall that, by Proposition 16, λ λϕ − (1), C → L , is ahomeomorphism onto its range and that ϕ − [ C ] = { λϕ − (1) | λ ∈ C } is closedin L . As ϕ is an isomorphism of Lie algebras, ϕ ( λϕ − (1)) = λ . That is, ϕ ↾ ϕ − [ C ] is continuous so Theorem 96 implies that ϕ is a homeomorphism. Theorem 98. Suppose Ω is nicely complemented and (Ω ⋆ ) c = ∅ . Then the Liering A (Ω) is algebraically determined.Proof. Let L be a Polish Lie ring and ϕ : L → A (Ω) be a Lie ring isomorphism.As (Ω ⋆ ) c = ∅ and Ω ⋆ is simply connected, it must be the case that Ω ⋆ isconformally equivalent to D . We may suppose that Ω ⋆ = D by Proposition68, since the Lie ring isomorphism constructed there is obviously continuousand therefore a topological isomorphism. With this reduction it is clear thatinterior(Ω c ) = ∅ . ϕ − [Ω] is an analytic set by Proposition 79. Since ϕ − [ C ] is a Polish ringby Proposition 78, Proposition 12 implies that ϕ ↾ ϕ − [ C ] is continuous. Lastly,Theorem 96 guarantees that ϕ is a homeomorphism. Theorem 99. Let Ω = C \ { λ , . . . , λ n } . Then A (Ω) is an algebraically deter-mined Lie ring.Proof. Many details of the proof of this theorem are very similar to the proofof Theorem 36, but a complete proof is given for the convenience of the reader.By scaling by an appropriate real number and Proposition 68, we may as-sume that { λ , . . . , λ n } ⊆ D . Let L be a Polish Lie ring and ϕ : L → A (Ω) bea Lie ring isomorphism. Notice that ϕ − [Ω] is a Borel subset of L by Lemma76 since we’ve only taken away finitely many points from K = ϕ − [ C ]. Lemma94 informs us that the additive subgroup D = { [ ϕ − (1) , x ] | x ∈ L} is closedin L . Recall that Proposition 77 furnishes a Borel map S : D → L so that, foreach x ∈ D , S ( x ) satisfies [ ϕ − (1) , S ( x )] = x = ⇒ ϕ ( S ( x )) ′ = [1 , ϕ ( S ( x ))] = ϕ ([ ϕ − (1) , S ( x )]) = ϕ ( x ).Define T : K × D → L by the rule T ( a, x ) = [ ϕ − ( z ) − a, S ( x )] + S ( x )and notice that T is a Borel map. Also, observe that ϕ ( T ( a, x )) = ( z − ϕ ( a )) ϕ ( x ) . Now define A = {h x, a, y, b i ∈ L × ϕ − [Ω] × D × K | x = T ( a, y ) + b } and notice that A is a Borel subset of ( L × K ) . Observe that the projection π : ( L × K ) → L × K , π ( x, a, y, b ) = h x, a i , is continuous.46o see that π ↾ A is injective, let h x, a, y , b i , h x, a, y , b i ∈ A . It followsthat T ( a, y ) + b = T ( a, y ) + b = ⇒ ( z − ϕ ( a )) ϕ ( y ) + ϕ ( b ) = ( z − ϕ ( a )) ϕ ( y ) + ϕ ( b ) . Since ϕ ( a ) ∈ Ω, we see that ϕ ( b ) = ϕ ( b ). From this, we see that b = b and,moreover, that y = y . Hence, π ↾ A is an injection. So π [ A ] is a Borel subsetof L × K .With this in hand, we can define π ∗ : π [ A ] → A so that π ∗ = ( π ↾ A ) − , aBorel map. Now let p : ( L × K ) → K be the projection p ( x, a, y, b ) = b . Itfollows that p ◦ π ∗ is a Borel map.Let h x, a i ∈ π [ A ] and h x, a, y, b i = π ∗ ( x, a ). It follows that ϕ ( x ) = ( z − ϕ ( a )) ϕ ( y ) + ϕ ( b )which proves that p ◦ π ∗ ( x, a ) = b = ϕ − ( ϕ ( x )( ϕ ( a ))) . Let L ∗ = L \ { } . li ( ϕ − [Ω]) is a Borel subset of K by Lemma 33. Hence, B = {h x, a i ∈ L ∗ × li ( ϕ − [Ω)]) | h x, a k i ∈ π [ A ] and p ◦ π ∗ ( x, a k ) = 0) for all k ≥ } is a Borel subset of L × K . Let P : L × K → K be the projection P ( x, a ) = a . P [ B ] is an analytic set and therefore a set with the Baire property. We will nowprove that P [ B ] = ϕ − [ { ζ ∈ C | | ζ | > } ] . First, suppose h x, a i ∈ B . Since ϕ ( x ) is a non-zero analytic function whichvanishes at each ϕ ( a k ) ( k ≥ { ϕ ( a k ) } k ≥ must be outside ofcl( D ). In particular, | ϕ ( a ) | > a ∈ K is so that | ϕ ( a ) | > 1. Observe that a ∈ li ( K ). Moreover,since { λ , . . . , λ n } ⊆ D , we see that a k ∈ ϕ − [Ω] for each k ≥ 1. By theWeierstrass Theorem, let f be a non-zero entire function so that f ( ϕ ( a k )) = 0for each k ≥ 1. Now, consider x = ϕ − ( f ↾ Ω ). For any k ≥ f = ( z − ϕ ( a k )) g for some entire function g . Since g is entire, y = ϕ − ( g ↾ Ω ) ∈ L and we observethat h x, a k , y, i ∈ A . As k was arbitrary, h x, a i ∈ B .Conclusively, P [ B ] = ϕ − [ { ζ ∈ C | | ζ | > } ] = ϕ − [(cl( D )) c ] and, since K is aPolish ring algebraically isomorphic to C , we see that ϕ − [cl( D )] is a co-analyticset and therefore a set with the Baire property. We appeal to Proposition 12 todeduce that ϕ ↾ K is continuous. Finally, apply Theorem 96 to see that ϕ is ahomeomorphism. References [1] Ichiro Amemiya. 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