2005
DOI: 10.1016/j.top.2004.03.005
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A Poincaré–Hopf-type theorem for holomorphic one-forms

Abstract: We prove that if a holomorphic one-form in a neighborhood of a closed euclidian ball B 2n ⊂ C n , in the n-dimensional complex a ne space, deÿnes a distribution transverse to the boundary sphere S 2n−1 = @B 2n , then n is even and admits a sole singularity q ∈ B 2n . Moreover, this singularity is simple. ?

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Cited by 8 publications
(12 citation statements)
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“…Assume by contradiction that F is transverse to the boundary sphere S 2n−1 (1). We note that F is defined by an integrable holomorphic one-form Ω in U ⊃ D 2n and according to [9] there is a unique singular point p of Ω inside the disc D 2n and this is a non-degenerate singular point: the determinant of the matrix D(Ω)(p) of the coefficients of the linear part of Ω is different from 0. By a Möbius transformation we can assume that the origin is this singular point.…”
Section: Lemmamentioning
confidence: 99%
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“…Assume by contradiction that F is transverse to the boundary sphere S 2n−1 (1). We note that F is defined by an integrable holomorphic one-form Ω in U ⊃ D 2n and according to [9] there is a unique singular point p of Ω inside the disc D 2n and this is a non-degenerate singular point: the determinant of the matrix D(Ω)(p) of the coefficients of the linear part of Ω is different from 0. By a Möbius transformation we can assume that the origin is this singular point.…”
Section: Lemmamentioning
confidence: 99%
“…As it is known such foliation is defined by a holomorphic one-form Ω, with singular set of codimension ≥ 2, and satisfying the integrability condition Ω ∧ dΩ = 0. In previous papers [7], [8] and [9] we proved the non-existence of the foliation under some additional conditions: for instance, F is defined by an integrable homogeneous polynomial one-form, or F has a global separatrix transverse to each sphere S 2n−1 (r), 0 < r ≤ 1, or n is odd. In the case of a non-integrable holomorphic one-form, if n = 2m + 1 is odd there is no holomorphic one-form Ω such that the distribution Ker(Ω) = {Ω = 0} is transverse to the sphere S 4m+1 (1) ⊂ C 2m+1 ( [7]).…”
Section: Introductionmentioning
confidence: 99%
“…This implies that the real matrix = λ 2n + · · · + (−1) n | det A| 2 . It was shown in [4,5] that any holomorphic distribution in an odd dimensional space C 2n+1 is not transverse to the sphere S 4n+1 (1). The above proof gives a new proof for the linear distribution case, since the polynomial (2) has a positive root λ when n is odd.…”
Section: Proposition 8 Let ωmentioning
confidence: 99%
“…Secondly, if ω A is not integrable, we will give a necessary and sufficient condition of transversality between K(ω A ) and the unit sphere S 2n−1 (1) ⊂ C n , n ≥ 3. It is shown in [4,5] that a holomorphic distribution in the odd dimensional space C 2n+1 is not transverse to the sphere S 4n+1 (1) by using the fact that S 4n+1 (1) has no 2-field. If we restrict ourselves to linear distributions, it is shown that the above theorem also claims the tangency because any matrixĀA with det A = 0 has a positive eigenvalue if n is odd.…”
Section: Introductionmentioning
confidence: 99%
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