We give two sufficient criteria for schlichtness of envelopes of holomorphy
in terms of topology. These are weakened converses of results of Kerner and
Royden. Our first criterion generalizes a result of Hammond in dimension 2.
Along the way we also prove a generalization of Royden's theorem.Comment: v2: 3 pages; added new results, including generalization of a theorem
  of Royde

Hatcher

Munkres

Zach Teitler

English

arXiv

AbstractWe give two sufficient criteria for schlichtness of envelopes of holomorphy in terms of topology. These are weakened converses of results of Kerner and Royden. Our first criterion generalizes a result of Hammond in dimension 2. Along the way, we also prove a generalization of Royden's theorem.</jats:p

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Proceedings of the Edinburgh Mathematical Society

Topological criteria for schlichtness

We give two sufficient criteria for schlichtness of envelopes of holomorphy in terms of topology. These are weakened converses of results of Kerner and Royden. Our first criterion generalizes a result of Hammond in dimension 2. Along the way, we also prove a generalization of Royden\u27s theorem

Teitler, Zach

Boise State University - ScholarWorks

Boise State UniversityScholarWorksMathematics Faculty Publications and Presentations Department of Mathematics6-1-2013Topological Criteria for SchlichtnessZach TeitlerBoise State UniversityThis document was originally published by Cambridge University Press in Proceedings of the Edinburgh Mathematical Society. Copyright restrictions mayapply. DOI: 10.1017/S001309151300031X.Proceedings of the Edinburgh Mathematical Society (2013) 56, 637–640DOI:10.1017/S001309151300031XTOPOLOGICAL CRITERIA FOR SCHLICHTNESSZACH TEITLERDepartment of Mathematics, 1910 University Drive, Boise State University,Boise, ID 83725-1555, USA (zteitler@boisestate.edu)(Received 11 February 2011)Abstract We give two suﬃcient criteria for schlichtness of envelopes of holomorphy in terms of topology.These are weakened converses of results of Kerner and Royden. Our ﬁrst criterion generalizes a result ofHammond in dimension 2. Along the way, we also prove a generalization of Royden’s theorem.Keywords: envelope of holomorphy; schlichtness; covering space2010 Mathematics subject classiﬁcation: Primary 32D10Let Ω ⊆ Cn be a domain. The envelope of holomorphy of Ω is a pair (Ω˜, π) consisting of aconnected Stein manifold Ω˜ and a locally biholomorphic map π : Ω˜ → Cn, together witha holomorphic inclusion α : Ω → Ω˜, characterized by the following properties: π ◦ α isthe identity, and each holomorphic function f on Ω has a unique holomorphic extensionFf on Ω˜, with f = Ff ◦ α. Let Ω′ = π(Ω˜) and let i = π ◦ α : Ω → Ω′. The envelopeof holomorphy (Ω˜, π) is schlicht if π : Ω˜ → Ω′ is biholomorphic. One would like to giveconditions on Ω to have a schlicht envelope of holomorphy.Two results of Kerner and Royden lead to necessary conditions. Kerner [5] has shownthat α∗ : π1(Ω) → π1(Ω˜) is surjective. Royden [8] has shown that α∗ : H1(Ω˜;Z) →H1(Ω;Z) is injective. It follows trivially that if (Ω˜, π) is schlicht, so Ω˜ = Ω′, theni∗ : π1(Ω) → π1(Ω′) is surjective and i∗ : H1(Ω′;Z) → H1(Ω;Z) is injective.Neither of these conditions is suﬃcient, by a result of Fornæss and Zame [1] (see [2,§ 3]). Following an idea of Hammond [2], one may seek suﬃcient conditions by adjoiningto the surjectivity of i∗ (or injectivity of i∗) the assumption that π : Ω˜ → Ω′ is a coveringspace. This strong assumption is still reasonable, as covering maps certainly occur amongenvelopes of holomorphy; indeed, Fornæss and Zame show in [1] that for any coveringmap π : Ω˜ → Ω′ there is a domain Ω ⊆ Ω′ with envelope of holomorphy (Ω˜, π).Speciﬁcally, Hammond has shown that, in dimension n = 2, if i∗ : π1(Ω) → π1(Ω′) issurjective and π : Ω˜ → Ω′ is a covering map, then (Ω˜, π) is schlicht. We give an elementaryproof of Hammond’s theorem in all dimensions n  2. In addition, we give a suﬃcientcondition for schlichtness in terms of the injectivity of i∗ on cohomology, again assumingπ is a covering map. Along the way, we give an alternative proof of Royden’s theorem,which also extends it to coeﬃcient groups other than Z.c© 2013 The Edinburgh Mathematical Society 637638 Z. TeitlerTheorem 1. If π is a covering map and i∗ : π1(Ω) → π1(Ω′) is surjective, then (Ω˜, π)is schlicht.This extends the theorem of Hammond for dimension n = 2. Hammond’s proof relieson a result of Jupiter [4], which is special to dimension 2.Proof. The number of sheets of the covering map π is equal to the index of π∗(π1(Ω˜))in π1(Ω′) (see, for example, [3, Proposition 1.32]). The surjectivity of i∗ = π∗ ◦ α∗implies that π∗ is surjective. Hence, the index of the image subgroup is 1, so π : Ω˜ → Ω′is 1-sheeted, i.e. a homeomorphism. Since π is a holomorphic homeomorphism, it isbiholomorphic and so Ω˜ is schlicht. Compare with the more technical proof in [2].The cohomology in Royden’s result is Cˇech cohomology with coeﬃcients in the sheafof locally constant Z-valued functions. Since our spaces are manifolds, Cˇech cohom-ology coincides with singular cohomology (with coeﬃcients in Z); see, for example, [6,Theorem 73.2]. Recall also that by the universal coeﬃcient theorem, H1(X;G) =Hom(π1(X), G) for a path-connected space X and abelian coeﬃcient group G [3, p. 198].Before we go on, observe that this proves Royden’s theorem as a consequence ofKerner’s theorem and extends it to other coeﬃcient groups.Theorem 2 (Royden). For any abelian group G, α∗ : H1(Ω;G) → H1(Ω˜;G) is injec-tive.Proof. Since α∗ : π1(Ω) → π1(Ω˜) is surjective,α∗ : Hom(π1(Ω), G) → Hom(π1(Ω˜), G)is injective and these Hom groups coincide with H1(Ω;G), H1(Ω˜;G). Royden proves this for G = Z using Cˇech cohomology, in particular the exponentialshort exact sequence (hence the restriction to G = Z). No such result holds for highercohomology groups [1, Theorem 4].Now, we aim to give a suﬃcient criterion for schlichtness by assuming i∗ : H1(Ω′;G) →H1(Ω;G) is injective for every abelian group G, and that π is a covering map. Ourstrategy is to deduce that π∗ : π1(Ω˜) → π1(Ω′) is surjective, as in the proof of Theorem 1.This would follow if we could deduce that i∗ : π1(Ω) → π1(Ω′) is surjective, but, ingeneral, injectivity of Hom(A,G) → Hom(B,G) does not imply surjectivity of B → A.The problem is that if the image of B is a proper subgroup which is not contained in anyproper normal subgroup, then there is no non-zero f : A → G vanishing on the image ofB. For example, let S3 be the symmetric group on three letters and let B = Z/2Z bethe subgroup generated by a transposition. If f : S3 → G is any group homomorphismsuch that the restriction f | B is zero, then f itself is zero.We must solve this problem by adjoining a hypothesis to ensure that every propersubgroup of π1(Ω′) is contained in a proper normal subgroup. However, this alone is notenough. For, suppose that B ⊂ π1(Ω′) is a proper subgroup, contained in a proper normalTopological criteria for schlichtness 639subgroup N . We get a non-zero homomorphism f : π1(Ω′) → G = π1(Ω′)/N , namely thequotient map, whose restriction to B ⊆ N is zero, so Hom(π1(Ω′), G) → Hom(B,G) isnot injective. This will prove the theorem we want, but only if G is abelian, so we canidentify these Hom groups with singular cohomology.So we need to know that every proper subgroup of π1(Ω′) is not only contained in aproper normal subgroup, but in one such subgroup N whose quotient G = π1(Ω′)/N isabelian.Fortunately, this condition is more natural than it sounds. It holds if π1(Ω′) is nilpo-tent, as in that case every maximal proper subgroup is normal and has prime index(see [7, Theorem 5.40]).We get the following.Theorem 3. If π is a covering map, π1(Ω′) is nilpotent and i∗ : H1(Ω′;G) → H1(Ω;G)is injective for every abelian group G, then (Ω˜, π) is schlicht.Proof. Since i∗ = α∗ ◦ π∗ is injective, π∗ is injective as well. Via π∗, we regard π1(Ω˜)as a subgroup of π1(Ω′). Recall that if H is any nilpotent group, then every maximalproper subgroup N of H is normal and has prime index, and, in particular, H/N isabelian. If π1(Ω˜)  π1(Ω′), there exists a maximal subgroup π1(Ω˜) ⊆ N  π1(Ω′) andhence a surjection π1(Ω′) → G = π1(Ω′)/N to an abelian group with π1(Ω˜) mapping tozero. This surjection is non-zero and lies in the kernel ofπ∗ : H1(Ω′;G) = Hom(π1(Ω′), G) → Hom(π1(Ω˜), G) = H1(Ω˜;G)for the abelian group G = π1(Ω′)/N , contradicting the injectivity of π∗.It follows that π1(Ω˜) = π1(Ω′). As before, this implies that π is a degree 1 coveringmap, and hence a biholomorphism. Solvability would not be enough, as shown by the example of Z/2Z ⊂ S3. This wouldnot only obstruct the proof given above, but would actually lead to a counter-exampleto the version of the statement, with solvable in place of nilpotent.Example 4. Recall that Artin’s braid group on three strands, denoted B3, is thefundamental group of the complement of the braid arrangement A2 in C3, the union of thethree hyperplanes deﬁned by (y−x)(z−x)(z−y) = 0. Quotienting by the small diagonal,the line x = y = z = 0, B3 is the fundamental group of Ω′ ⊂ C2, the complement of theunion of three lines through the origin in C2. Let B2 ⊂ B3 be a subgroup correspondingto two of the strands, so B2 ∼= Z has index 3 in B3 and is not normal. There exists acovering space Ω˜ → Ω′ such that π1(X) = B2 ⊂ B3. Since U is a Stein manifold, so isΩ˜ [9]. By [1, Theorem 5], there exists a domain Ω ⊂ Ω′ with envelope of holomorphyΩ˜. This is not schlicht, but for every abelian group G, Hom(B3, G) → Hom(B2, G) isinjective.More generally, let H be any ﬁnitely presented group. H is the fundamental group ofa 2-complex, which may be embedded in R5, or, for that matter, C3; then, a tubularneighbourhood Ω′ of this complex (in C3) still has π1(Ω′) = H. Any subgroup K ⊂ H640 Z. Teitleroccurs as the fundamental group of a covering space Ω˜ → Ω′. Again, Ω˜ is Stein since Ω′is, and there exists a domain Ω ⊂ Ω′ with envelope of holomorphy Ω˜.It is not necessary to assume that i∗ is injective when coeﬃcients are taken in anyabelian group G. It would be enough to assume that i∗ is injective when coeﬃcients aretaken in any ﬁnite cyclic group, in any abelian quotient G of π1(Ω′) or even just in asingle abelian quotient G = π1(Ω′)/N for some proper normal subgroup N containingπ1(Ω˜).If, in addition, π : Ω˜ → Ω′ is a normal covering space, then π1(Ω˜) ⊆ π1(Ω′) is a normalsubgroup and we can take G to be an abelian quotient of π1(Ω′)/π1(Ω˜), which is thegroup of deck transformations.Corollary 5. Suppose π is a normal covering map with deck transformation group H.If there exists a non-zero abelian quotient G of H such that i∗ : H1(Ω′;G) → H1(Ω;G)is injective, then (Ω˜, π) is schlicht.Acknowledgements. The author thanks Chris Hammond for explaining his theo-rem, and Emil Straube, Craig Westerland and Jens Harlander for helpful and patientconversations.References1. J. E. Fornæss and W. R. Zame, Riemann domains and envelopes of holomorphy, DukeMath. J. 50 (1983), 273–283.2. C. Hammond, Schlicht envelopes of holomorphy and topology, Math. Z. 266 (2010),285–288.3. A. Hatcher, Algebraic topology (Cambridge University Press, 2002).4. D. Jupiter, A schlichtness theorem for envelopes of holomorphy, Math. Z. 253 (2006),623–633.5. H. Kerner, U¨berlagerungen und Holomorphiehu¨llen, Math. Annalen 144 (1961), 126–134.6. J. R. Munkres, Elements of algebraic topology (Addison-Wesley, 1984).7. J. J. Rotman, An introduction to the theory of groups, 4th edn, Graduate Texts inMathematics, Volume 148 (Springer, 1995).8. H. L. Royden, One-dimensional cohomology in domains of holomorphy, Annals Math.78 (1963), 197–200.9. K. Stein, U¨berlagerungen holomorph-vollsta¨ndiger komplexer Ra¨ume, Arch. Math. 7(1956), 354–361.

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