EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 10, No. 3, 2017, 440-454 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global Narrowing Cohomology for Complex S6 Andrew McHugh Department of Mathematics and Statistics, Zayed University, Abu Dhabi, UAE Abstract. We compute Bott Chern, and Aeppli cohomology for a complex structure on the six sphere, S6. We also give a table for the hodge numbers for the Bott-Chern (and thus also Aeppli) cohomology where hodge numbers are given in terms of whole number parameters a = h2,0 ∂̄ − h1,0 ∂̄ , c = h0,2 ∂̄ ,d = h1,2 ∂̄ , h2,0 ∂̄ , h1,1 BC , and h2,2 BC . As an example, we work out the Bott-Chern hodge numbers completely in the hypothetical case that the Dolbeault cohomology has h2,0 = a = c = d = 0. 2010 Mathematics Subject Classifications: 53C56,55N99,32Q99 Key Words and Phrases: Six sphere, complex structure, Hodge numbers, Aeppli cohomology, Bott-Chern cohomology 1. Introduction The existence of a complex structure on S6 has been a persistent question for many years. In 1954, Hirzebruch[6] showed that if a complex structure on S6 does exist, then by blowing up a point, one obtains an exotic complex structure on CP3. In fact, these complex structures on S6 and CP3 are non-Kahler. In 1987 Lebrun[8] showed that a complex structure on S6 cannot be compatible with the standard metric on S6. In 1998, Campana, Demailly, and Pertenell[3] showed that a complex S6 has no global non-constant meromorphic functions. In 2000, Huckleberry, Kebekus, and Peternell showed it is not almost homogeneous. Recently in 2015, Etesi[4] has published an article which constructs a complex structure on S6. In this paper, we search for the Dolbeault, Bott-Chern and Aeppli cohomology hodge numbers for a complex S6. In 1997, Gray[5] showed that for the Dolbeault hodge numbers, we have h3,0 = h0,3 = 0 and h0,1 ≥ 1. In 2000, Ugarte essentially gave the following for the Dolbeault cohomology on S6 which we shall summarise shortly in a table. Let a = h2,0 2 where h2,0 2 = dimCE 2,0 2 from the Frohlicher spectral sequence. Ugarte shows that h2,0 2 = h2,0 − h1,0. Now, let c = h0,2, and d = h2,1. We have Ugarte’s results in the following: Email addresses: andrew.mchugh@zu.ac.ae , andrewmchugh@snet.net (A. McHugh) http://www.ejpam.com 440 c© 2017 EJPAM All rights reserved. A. McHugh / Eur. J. Pure Appl. Math, 10 (3) (2017), 440-454 441 Table 1: Ugarte: hp,q for a Complex Structure on S6 0 h1,0 + a h1,0 1 c d d− a+ 1 c+ 1 c+ 1 d− a+ 1 d c 1 h1,0 h1,0 + a 0 where 0 ≤ a ≤ c+ 1, and c ≤ d. 2. Some results on the Dolbeault cohomology of compact complex manifolds and of complex S6. We begin with the result of Gray[5], (see also Ugarte[11] and Brown[2]): Theorem 1. (Gray) Let X be a compact complex manifold of complex dimension n such that bn(X) = 0. Any complex structure on X has the property hn,0 = h0,n = 0. We will be supposing that X is a complex manifold with H1(X,Z) = H2(X,Z) = Hn(X,C) = 0. (For example S6 with a complex structure). By above we have of course, hn,0 = h0,n = 0. Note that this implies that the associated canonical bundle to the complex structure, K, is not holomorphically trivial. We also note that because H2(X,Z) = 0, we have that the first Chern class of K ( and for that matter any complex line bundle on X) is 0. It is straight forward to show for such complex n-fold X, a result of Gray on complex S6 (See also Brown[2].) Theorem 2. h0,1 ≥ 1 (Gray) Proof. This can be seen by considering the short exact sequence of sheaves: 0→ Z→ O → O∗ → 0 and (the portion of ) the resulting long exact sequence . . .→ H1(X,Z)→ H1(X,O)→ H1(X,O∗)→ H1(X,Z)→ . . . where O denotes the sheaf of holomorphic functions on X and O∗ denotes the sheaf of nowhere zero holomorphic functions on X . Also, the map Z→ O is the map, k 7→ ik and the map O → O∗ is the exponential map, f 7→ exp(f). Since H1(X,Z) = H2(X,Z) = 0 we have H1(X,O) = H1(X,O∗). Note that 1 6= K ∈ H1(X,O∗) and thus h0,1 6= 0. We now specialize to X being a three dimensional complex manifold with H1(X,Z) = H2(X,Z) = H3(X,C) = 0, i.e. topologically equivalent to S6. Lemma 1. h1,0 ≤ h2,0 A. McHugh / Eur. J. Pure Appl. Math, 10 (3) (2017), 440-454 442 Proof. We consider the portion of the Frohlicher spectral sequence: ∂ : H1,0(X)→ H2,0(X) We shall show that this is an injective map of vector spaces. Indeed, let φ be a ∂̄ closed 1, 0 form, such that ∂[φ] = 0. Thus, [∂φ] = 0 and by type we have ∂φ = 0. Therefore, dφ = (∂+ ∂̄)φ = 0. Since, b1 = 0 , we have φ = df for some complex valued, C∞ function, f . Considering type, we have ∂̄f = 0 and thus f as a global holomorphic function is a constant. Thus φ = 0. This shows ∂ induces an injective map from H1,0(X)→ H2,0(X). This can also be directly deduced from the result of Ugarte[11] that E1,0 2 = 0 and E0,0 2 = 1. More specifically, Brown[2] gives the following table derived by Ugarte for Ep,q 2 of the Frohlicher spectral sequence for a complex structure on S6: Table 2: Ep,q 2 for a Complex Structure on S6 0 a 0 1 b b 0 a a 0 b b 1 0 a 0 The bottom row of the table corresponds to the portion of the Frohlicher sequence H0,0 → H1,0 → H2,0 → H3,0 . Since H0,0 = C, E0,0 = 1, and E1,0 2 = 0 the sequence reduces to 0→ H1,0 → H2,0 → 0 . It is exact at H1,0 and thus ∂ : H1,0(X)→ H2,0(X) is injective. Note that H1,0 = H2,0 if and only if a = 0 and if H1,0 = 0 then H2,0 = a . Huckleberry, Kebekus and Peternell [7] gave a proof pointed out to them by M. Toma that h1,0 ≤ 1. We give a somewhat different but related proof here. The present au- thor is indebted to Daniel Angella for pointing out the correct statement of Huckleberry, Kebekus, Peternell and Toma’s result. Lemma 2. (Huckleberry, Kebekus, Peternell, Toma) h1,0 ≤ 1, i.e. h1,0 = 0 or 1 Proof. Indeed, if h1,0 = 2, then h2,0 ≥ 2. Let φ1 and φ2 be two linearly independent ∂̄-closed global 1, 0-forms. Let Φ1 = φ1 ∧ φ2 This is a global ∂̄-closed 2, 0-form on X that is not identically zero. Since h2,0 ≥ 2 we can select Φ2 another ∂̄-closed global 2, 0-form that is linearly independent of Φ1. A. McHugh / Eur. J. Pure Appl. Math, 10 (3) (2017), 440-454 443 We may choose a point, p ∈ X such that Φ1 and Φ2 are non-zero and linearly indepen- dent at p. Note that φ1 and φ2 are also non-zero and linearly independent of each other at p since Φ1(p) = φ1(p) ∧ φ2(p) is not zero. Let ηp ∈ T 1,0 p be linearly independent of φ1(p) and φ2(2), completing a basis for T 1,0 p . Thus for Φ(p) and some complex numbers, a, b1, b2, we have, Φ2(p) = aφ1 ∧ φ2 + b1η ∧ φ1 + b2η ∧ φ2 . Now b1 and b2 are not both zero. Without loss of generality, assume b1 6= 0. Hence Φ2(p) ∧ φ2 6= 0 and Φ2 ∧ φ2 is a non-zero holomorphic 3, 0-form on X. This contradicts h3,0 = 0. We must then have h1,0 ≤ 1. We summarize with two tables of the plausible hodge numbers (with h0,0 in the bottom lefthand corner) for Dolbeault cohomology for a complex structure on S6: Table 3: (h1,0 = 1): hp,q for a Complex Structure on S6 0 a+ 1 1 1 c d d− a+ 1 c+ 1 c+ 1 d− a+ 1 d c 1 1 a+ 1 0 where 0 ≤ a ≤ c+ 1, and c ≤ d. Table 4: (h1,0 = 0): hp,q for a Complex Structure on S6 0 a 0 1 c d d− a+ 1 c+ 1 c+ 1 d− a+ 1 d c 1 0 a 0 where 0 ≤ a ≤ c+ 1, and c ≤ d. Note that we have in both cases, a = h2,0 − h1,0. 3. Aeppli and Bott-Chern Cohomology on complex S6. The Aeppli cohomology of a complex manifold is defined by the vector spaces (see Popovici [9]) : Hp,q A = ker(∂∂̄ : C∞ p,q → C∞ p+1,q+1) im(∂ : C∞ p−1,q → C∞ p,q) + im(∂̄ : C∞ p,q−1 → C∞ p,q) A. McHugh / Eur. J. Pure Appl. Math, 10 (3) (2017), 440-454 444 The Bott-Chern cohomology of a complex manifold is defined by the vector spaces (again see Popovici [9]) : Hp,q BC = ker(∂ : C∞ p,q → C∞ p+1,q) ∩ ker(∂̄ : C∞ p,q → C∞ p,q+1) im(∂∂̄ : C∞ p−1,q−1 → C∞ p,q) On compact complex manifolds, there is a harmonic theory for each of these cohomologies which ensures that they are finite dimensional complex vector spaces. Let hp,qA = dim(Hp,q A ) and hp,qBC = dim(Hp,q BC) .We note (see Popovici[9]) that hp,qA = hq,pA , hp,qBC = hq,pBC and hp,qA = hn−p,n−qBC . The Serre duality of Bott-Chern and Aeppli cohomology is due to Schweitzer[10] We try to narrow down as much as possible the Aeppli and Bott-Chern Cohomology on complex S6. 3.1. Some long exact sequences of cohomology Consider the following sequence of maps of cohomology on a complex manifold X: 0→ Hp,0 BC /im(∂̄)−→ Hp,0 ∂̄ /(im(∂̄)+im(∂))−→ Hp,0 A ∂̄→ Hp,1 BC /im(∂̄)−→ · · · · · · /(im(∂̄)+im(∂))−→ Hp,n−1 A ∂̄→ Hp,n BC /im(∂̄)−→ Hp,n ∂̄ /(im(∂̄)+im(∂))−→ Hp,n A → 0 . We prove some claims and lemmas below about this sequence of maps. The sophisticated readers may just read the claims and lemmas skipping over their proofs if they appear to be straight forward or obvious. Lemma 3. The sequence of maps above is exact at Hp,q A . Namely, im(/(im(∂̄) + im(∂)) : Hp,q ∂̄ → Hp,q A ) = ker(∂̄ : Hp,q A → Hp,q+1 BC ) . Proof. Let [φ]A ∈ ker(∂̄ : Hp,q A → Hp,q+1 BC ) where φ is some smooth p, q-form representative of [φ]A. We have then that ∂̄([φ]A) = [∂̄(φ)]BC = 0 in Hp,q+1 BC , i.e. ∂̄(φ) = ∂̄∂θ for some p− 1, q-form θ. Thus ∂̄(φ− ∂θ) = 0 and φ− ∂θ is a ∂̄-closed p, q-form. We conclude then that [φ]A = [φ− ∂θ]A ∈ im(/(im(∂̄) + im(∂)) : Hp,q ∂̄ → Hp,q A ) A. McHugh / Eur. J. Pure Appl. Math, 10 (3) (2017), 440-454 445 and ker(∂̄ : Hp,q A → Hp,q+1 BC ) ⊆ im(/(im(∂̄) + im(∂)) : Hp,q ∂̄ → Hp,q A ) . Now let [φ]A ∈ im(/(im(∂̄) + im(∂)) : Hp,q ∂̄ → Hp,q A ) where φ is some smooth p, q-form representative of [φ]A. We may assume that φ is ∂̄-closed since [φ]A ∈ im(/(im(∂̄) + im(∂)) : Hp,q ∂̄ → Hp,q A ) . Clearly, ∂̄([φ]A) = [∂̄φ]BC = 0 in Hp,q+1 BC . Thus [φ]A ∈ ker(∂̄ : Hp,q A → Hp,q+1 BC ) and im(/(im(∂̄) + im(∂)) : Hp,q ∂̄ → Hp,q A ) ⊆ ker(∂̄ : Hp,q A → Hp,q+1 BC ) . The two inclusions give im(/(im(∂̄) + im(∂)) : Hp,q ∂̄ → Hp,q A ) = ker(∂̄ : Hp,q A → Hp,q+1 BC ) . Lemma 4. The sequence of maps above is exact at Hp,q BC . Namely, ker(/im(∂̄) : Hp,q BC → Hp,q ∂̄ ) = im(∂̄ : Hp,q−1 A → Hp,q+1 BC ) . Proof. Let [γ]BC ∈ im(∂̄ : Hp,q−1 A → Hp,q BC) where γ is some smooth p, q-form representative of [γ]BC . Since [γ]BC ∈ im(∂̄ : Hp,q−1 A → Hp,q BC) we may assume γ = ∂̄µ for some p, q − 1-form µ such that ∂∂̄µ = 0. Clearly, γ/im(∂̄) = ∂̄µ/im(∂̄) = 0/im(∂̄) and [γ] ∈ ker(/im(∂̄) : Hp,q BC → Hp,q ∂̄ ) . Thus im(∂̄ : Hp,q−1 A → Hp,q BC) ⊆ ker(/im(∂̄) : Hp,q BC → Hp,q ∂̄ ) . Now let [γ]BC ∈ ker(/im(∂̄) : Hp,q BC → Hp,q ∂̄ ) where γ is some smooth p, q-form representative of [γ]BC . Since [γ]BC ∈ ker(/im(∂̄) : Hp,q BC → Hp,q ∂̄ ) A. McHugh / Eur. J. Pure Appl. Math, 10 (3) (2017), 440-454 446 we have γ/im(∂̄) = 0/im(∂̄). Thus γ = ∂̄µ for some p, q − 1-form µ. Now, γ is ∂̄-closed and also ∂-closed. Thus, ∂γ = ∂∂̄µ = 0 . This shows [γ]BC = ∂̄([µ]A) and [γ]BC ∈ im(∂̄ : Hp,q−1 A → Hp,q BC) . Hence, ker(/im(∂̄) : Hp,q BC → Hp,q ∂̄ ) ⊆ im(∂̄ : Hp,q−1 A → Hp,q BC) and the two inclusions give the equality, ker(/im(∂̄) : Hp,q BC → Hp,q ∂̄ ) ⊆ im(∂̄ : Hp,q−1 A → Hp,q BC) . Lemma 5. If the Betti number, bp+q = 0 on our complex manifold, X, then the sequence above is exact at Hp,q ∂̄ . More specifically, im(/im(∂̄) : Hp,q BC → Hp,q ∂̄ ) = ker(/(im(∂̄) + im(∂)) : Hp,q ∂̄ → Hp,q A ) . Proof. Let [φ]∂̄ ∈ im(/im(∂̄) : Hp,q BC → Hp,q ∂̄ ) . We have dφ = ∂φ+ ∂̄φ = 0. Since bp+q = 0, we have that φ = dλ for some p+q−1-form,λ on X. Thus φ = ∂λp−1,q + ∂̄λp,q−1 where ∂λp−1,q and ∂̄λp,q−1 are the projections of λ to its p− 1, q and p, q − 1 parts respectively. Thus [φ]A = 0 in Hp,q A , and [φ]∂̄ ∈ ker(/(im(∂̄) + im(∂)) : Hp,q ∂̄ → Hp,q A ) and we have im(/im(∂̄) : Hp,q BC → Hp,q ∂̄ ) ⊆ ker(/(im(∂̄) + im(∂)) : Hp,q ∂̄ → Hp,q A ) . Now let [φ]∂̄ ∈ ker(/(im(∂̄) + im(∂)) : Hp,q ∂̄ → Hp,q A ) . Note that φ is ∂̄-closed. Since [φ]A = 0, we have φ = ∂µ+ ∂̄ν for some p− 1, q-form µ and some p, q − 1-form, ν. Since ∂̄φ = 0 we have ∂̄(∂µ) = 0 and [φ]∂̄ = [∂µ] in Hp,q ∂̄ . We also have obviously that ∂(∂µ) = 0 and thus [φ]∂̄ = (/im(∂̄))([∂µ]BC) ∈ im(/im(∂̄) : Hp,q BC → Hp,q ∂̄ ) . and hence ker(/(im(∂̄) + im(∂)) : Hp,q ∂̄ → Hp,q A ) ⊆ im(/im(∂̄) : Hp,q BC → Hp,q ∂̄ ) . A. McHugh / Eur. J. Pure Appl. Math, 10 (3) (2017), 440-454 447 The two inclusions allow us to conclude that ker(/(im(∂̄) + im(∂)) : Hp,q ∂̄ → Hp,q A ) = im(/im(∂̄) : Hp,q BC → Hp,q ∂̄ ) . Notice above that we nowhere used that bp+q = 0 in proving the inclusion, [φ]∂̄ ∈ ker(/(im(∂̄) + im(∂)) : Hp,q ∂̄ → Hp,q A ) ⊆ im(/im(∂̄) : Hp,q BC → Hp,q ∂̄ ) . This suggests that we define what we shall call BCA-cohomology: Ep,q BCA = ker(∂) ∩ ker(∂̄) ker(∂̄) ∩ im(∂) + ker(∂) ∩ im(∂̄) . This definition is along the lines of Varouchas[12] who defines similar in spirit vector spaces to create long exact sequences involving Bott-Chern, Dolbeault, and Aeppli cohomology. We refer the reader also to Angella[1] for more details. It is easy to show that if bp+q = 0 then Ep,q BCA = {0}. We have the following claim: Lemma 6. For a compact complex manifold, X, the sequence above is exact at Hp,q ∂̄ if and only if Ep,q BCA = {0}. Proof. Let us first assume that Ep,q BCA = {0}. We need to only to show im(/im(∂̄) : Hp,q BC → Hp,q ∂̄ ) ⊆ ker(/(im(∂̄) + im(∂)) : Hp,q ∂̄ → Hp,q A ) since the reverse inclusion has already been shown true in general. Let [φ]∂̄ ∈ im(/im(∂̄) : Hp,q BC → Hp,q ∂̄ ) . We may assume φ ∈ ker(∂) ∩ ker(∂̄). Since Ep,q BCA = {0}, we have φ ∈ (ker(∂̄) ∩ im(∂) + ker(∂) ∩ im(∂̄)) ⊆ (im(∂) + im(∂̄)) . Hence [φ]∂̄ ∈ ker(/(im(∂̄) + im(∂)) : Hp,q ∂̄ → Hp,q A ) and im(/im(∂̄) : Hp,q BC → Hp,q ∂̄ ) ⊆ ker(/(im(∂̄) + im(∂)) : Hp,q ∂̄ → Hp,q A ) . Thus im(/im(∂̄) : Hp,q BC → Hp,q ∂̄ ) = ker(/(im(∂̄) + im(∂)) : Hp,q ∂̄ → Hp,q A ) . and the sequence above is exact at Hp,q ∂̄ . In the other direction we assume the sequence above is exact at Hp,q ∂̄ and im(/im(∂̄) : Hp,q BC → Hp,q ∂̄ ) = ker(/(im(∂̄) + im(∂)) : Hp,q ∂̄ → Hp,q A ) . Let φ ∈ ker(∂) ∩ ker(∂̄). Then [φ]∂̄ ∈ im(/im(∂̄) : Hp,q BC → Hp,q ∂̄ ) A. McHugh / Eur. J. Pure Appl. Math, 10 (3) (2017), 440-454 448 and thus [φ]∂̄ ∈ ker(/(im(∂̄) + im(∂)) : Hp,q ∂̄ → Hp,q A ) . Specifically, [φ]A = 0. Hence φ = ∂µ+ ∂̄ν for some p − 1, q-form, µ and p, q − 1-form ν. Since ∂̄φ = 0 we have ∂̄(∂µ) = 0 and ∂µ ∈ ker(∂̄)∩im(∂). Similiarly, Since ∂φ = 0 we have ∂(∂̄ν) = 0 and ∂̄ν ∈ ker(∂)∩im(∂̄). Thus φ ∈ (ker(∂̄) ∩ im(∂) + ker(∂) ∩ im(∂̄)) and Ep,q BCA = {0}. Lemma 7. Let X be a compact complex manifold of complex dimension n. If we have the first Betti number is zero (b1 = 0), then h1,0 BC = h0,1 BC = hn,n−1 A = hn−1,n A = 0 . Proof. Consider a 1, 0-form, µ such that µ ∈ ker(∂) ∩ ker(∂̄) . We have then dµ = 0. Since b1 = 0, we have µ = df form some function f . Thus µ = ∂f + ∂̄f . We must have ∂̄f = 0 since µ is a 1, 0-form. Hence f is a holomorphic function on a compact complex manifold and thus must be a constant function. Finally, we have µ = ∂f = 0 since f is constant. Thus ker(∂) ∩ ker(∂̄) = {0} for 1, 0-forms and h1,0 BC = 0. The other equalities follow from ”complex conjugation” and Serre duality mentioned above. Lemma 8. The map of vector spaces, Hp,0 BC /im(∂̄)−→ Hp,0 ∂̄ , is injective. Proof. A p, 0-form cannot be in the image of ∂̄∂ or ∂̄. Thus, ( /im(∂̄) )([µ]BC) = [0]∂̄ if and only if µ = 0. We also note on the end of the sequence we have Hp,n A ∂̄→ Hp,n+1 BC = {0} . Now we focus again on our compact complex manifold, X, being topologically equiv- alent to S6. Noting, that b0 = b6 = 1,and bj = 0 for 1 ≤ j ≤ 5, we have, using the above lemmas, the following: Theorem 3. For a complex structure on a manifold X topologically equivalent to S6, we have the following long exact sequences of vector spaces: 0. (p = 0) 0 → H0,1 ∂̄ /(im(∂̄)+im(∂))−→ H0,1 A ∂̄→ H0,2 BC /im(∂̄)−→ H0,2 ∂̄ /(im(∂̄)+im(∂))−→ H0,2 A ∂̄→ H0,3 BC /im(∂̄)−→ H0,3 ∂̄ /(im(∂̄)+im(∂))−→ H0,3 A ∂̄→ 0 A. McHugh / Eur. J. Pure Appl. Math, 10 (3) (2017), 440-454 449 1. (p = 1) 0 → H1,0 ∂̄ /(im(∂̄)+im(∂))−→ H1,0 A ∂̄→ H1,1 BC /im(∂̄)−→ H1,1 ∂̄ /(im(∂̄)+im(∂))−→ H1,1 A ∂̄→ H1,2 BC /im(∂̄)−→ H1,2 ∂̄ /(im(∂̄)+im(∂))−→ H1,2 A ∂̄→ H1,3 BC /im(∂̄)−→ H1,3 ∂̄ /(im(∂̄)+im(∂))−→ H1,3 A ∂̄→ 0 2. (p = 2) 0 → H2,0 BC /im(∂̄)−→ H2,0 ∂̄ /(im(∂̄)+im(∂))−→ H2,0 A ∂̄→ H2,1 BC /im(∂̄)−→ H2,1 ∂̄ /(im(∂̄)+im(∂))−→ H2,1 A ∂̄→ H2,2 BC /im(∂̄)−→ H2,2 ∂̄ /(im(∂̄)+im(∂))−→ H2,2 A ∂̄→ H2,3 BC /im(∂̄)−→ H2,3 ∂̄ /(im(∂̄)+im(∂))−→ H2,3 A ∂̄→ 0 3. (p = 3) 0 → H3,0 BC /im(∂̄)−→ H3,0 ∂̄ /(im(∂̄)+im(∂))−→ H3,0 A ∂̄→ H3,1 BC /im(∂̄)−→ H3,1 ∂̄ /(im(∂̄)+im(∂))−→ H3,1 A ∂̄→ H3,2 BC /im(∂̄)−→ H3,2 ∂̄ /(im(∂̄)+im(∂))−→ 0 One of our goals is to complete as much as possible a table of hodge numbers for Bott- Chern cohomology on a complex S6. The table of hodge numbers for Aeppli cohomology is, of course, given by the Serre duality with Bott-Chern cohomology. Since hp,qBC = hq,pBC , we may concern ourselves with just the bottom triangle of the table. Using h3,0 ∂̄ = 0, we can see straight away from the long exact sequence for p = 3 that h3,0 BC = 0. We also see that h2,3 BC = h3,2 BC ≥ h 3,2 ∂̄ = h0,1 ∂̄ ≥ 1 . We also show the following: Lemma 9. h2,0 BC = h2,0 ∂̄ and h3,1 BC = h3,1 ∂̄ = c . Proof. We know that if φ is a ∂̄-closed 2,0-form, then ∂φ = 0. Thus dφ = ∂φ+ ∂̄φ = 0 . Since b1 = 0, we have φ = dη = ∂η + ∂̄η for some 1-form, η. Thus the image of the map, H2,0 ∂̄ /(im(∂̄)+im(∂))−→ H2,0 A A. McHugh / Eur. J. Pure Appl. Math, 10 (3) (2017), 440-454 450 is {0}. Hence, since the p = 2 sequence is exact, the map, H2,0 BC /im(∂̄)−→ H2,0 ∂̄ is an isomorphism and we have h2,0 BC = h2,0 ∂̄ . Notice in the above argument we could also have concluded more specifically for our ∂̄-closed 2,0-form, φ, that φ = ∂η for some 1,0-form, η. In a similar manner, we take µ to be a 0,2-form representative of an element in H0,2 BC . Since dµ = 0, we may conclude that µ = ∂̄χ for some 0,1-form, χ. Thus the image of the map, H0,2 BC /im(∂̄)−→ H0,2 ∂̄ is {0} and hence, using the fact h0,3 BC = h3,0 BC = 0, the map, H0,2 ∂̄ /(im(∂̄)+im(∂))−→ H0,2 A is an isomorphism. We have then h3,1 BC = h0,2 A = c . The fact that the image of the map, H0,2 BC /im(∂̄)−→ H0,2 ∂̄ is {0} also shows from the p=2 sequence that we have the short exact sequence 0→ H0,1 ∂̄ /(im(∂̄)+im(∂))−→ H0,1 A ∂̄→ H0,2 BC /im(∂̄)−→ 0 and thus that h3,2 BC = h0,1 A = h0,1 ∂̄ + h2,0 ∂̄ = c+ 1 + h2,0 ∂̄ . Please recall that a = h2,0 ∂̄ − h1,0 ∂̄ . In trying to be as complete as possible, we also show: Theorem 4. If X is a compact complex manifold then h0,0 BC = 1. Furthermore, if b1 = 0 then h3,3 BC = 1. A. McHugh / Eur. J. Pure Appl. Math, 10 (3) (2017), 440-454 451 Proof. A 0,0-form or function, f , on a compact complex manifold, X, such that df = 0 is a constant. Thus, H0,0 BC = C and h0,0 BC = 1. We consider now H0,0 A . We know that H0,0 ∂̄ consists of the constant functions. The sequence, 0→ H0,0 BC /im(∂̄)−→ H0,0 ∂̄ /(im(∂̄)+im(∂))−→ H0,0 A ∂̄→ H0,1 BC is not exact at H0,0 ∂̄ but it is exact at H0,0 A . Thus, since h0,1 BC = 0, we know that H0,0 A = im(/(im(∂̄) + im(∂)) : H0,0 ∂̄ → H0,0 A ) . Now if f is a constant function, then f /∈ (im(∂̄) + im(∂)) so im(/(im(∂̄) + im(∂)) : H0,0 ∂̄ → H0,0 A ) = C and H0,0 A = C. By the Serre duality, we have h3,3 BC = 1. Recall that hp,qBC = hq,pBC . We thus have so far, for our essential lower triangle for Bott-Chern cohomology 1 h2,2 BC c+ 1 + h2,0 ∂̄ h1,1 BC h2,1 BC c 1 0 h2,0 ∂̄ 0 We still have not computed h1,1 BC , h 2,2 BC and h2,1 BC . We can determine h2,1 BC in terms of the others by plugging our results so far into the following well known result for long exact sequences of vector spaces: Theorem 5. If 0→ A1 → A2 → . . .→ An → 0 is a long exact sequence of vector spaces with aj = dim(Aj) then n∑ j=1 (−1)j+1aj = 0 . We apply this to the long exact sequence for p = 1: h1,0 − (c+ 1 + h2,0) + h1,1 BC − (d− a+ 1) + h2,2 BC − h 2,1 BC + d− h2,1 BC + c− h2,0 + h2,0 = 0 . A. McHugh / Eur. J. Pure Appl. Math, 10 (3) (2017), 440-454 452 This reduces to h1,1 BC + h2,2 BC = 2h2,1 BC + 2 . Thus our essential lower triangle for Bott-Chern cohomology is 1 h2,2 BC c+ 1 + h2,0 ∂̄ h1,1 BC h1,1 BC+h2,2 BC 2 − 1 c 1 0 h2,0 ∂̄ 0 3.2. The Bott-Chern and Aeppli cohomology for a hypothetical possibil- ity of the Dolbeault cohomology on complex S6 We consider a specific possible scenario of the Dolbeault cohomology on complex S6. Namely, h2,0 = a = c = d = 0. In terms of hodge numbers, this is h1,0 = h2,0 = h0,2 = h1,2 = 0 and h0,1 = h1,1 = 1 . This is one of the Dolbeault cohomology scenarios suggested at the end of Etesi[4]. In fact, the other cohomology scenario, with h1,1 = h2,1 = 1 and h1,0 = h2,0 = a = 0 is not possible on complex S6 according to our table for Dolbeault cohomology above since h1,1 = h1,2 + 1− a . Etesi does actually in fact also show the incompatibility of this other cohomology scenario. We look at the following portion of the p = 2-long exact sequence: H2,0 A → H2,1 BC → H2,1 ∂̄ → H2,1 A → H2,2 BC → H2,2 ∂̄ → H2,2 A → H2,3 BC → H2,3 ∂̄ REFERENCES 453 Using a total abuse of notation where we write just the dimensions of the vector spaces, this is: 0→ h2,1 BC → 0→ h2,1 A → h2,2 BC → 1→ h2,2 A → 1→ 0 Since the sequence is exact, we have right away, h2,1 BC = 0 and thus h2,1 A = 0 by the Serre duality. By the Frohlicher sequence, we know that ∂ : H2,2 ∂̄ → H3,2 ∂̄ is an isomorphism. In particular, we cannot have a non zero ∂̄-harmonic 2,2-form being d-closed. Thus we cannot have H2,2 BC being isomorphic to H2,2 ∂̄ . Thus h2,2 BC = 0 and h2,2 A = h1,1 BC = 2. This completes the Bott-Chern and Aeppli cohomology for this possible Dolbeault cohomology. 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