Proof. On complex valued forms, $d= {\bar \mu } + {\bar \partial } +\partial + \mu $ with components of bidegrees $(-1,2)$ , $(0,1)$ , $(1,0)$ , and $(2,-1)$ , respectively. In bidgree $(p,q)$ , $ \mathbb {I}^{-1} = (-1)^{p-q} \mathbb {I}$ , so that conjugating an operator of bidegree $(r,s)$ by $ \mathbb {I}$ acts by multiplication by $(-i)^{r-s}$ . As operators on forms of bidegree $(p,q)$ ,

$$\begin{align*}[d, \mathbb{J}] = 3i {\bar \mu} + i {\bar \partial} - i \partial - 3i \mu , \end{align*}$$

whereas

$$\begin{align*}\mathbb{I}^{-1} \, d \, \mathbb{I} = -i {\bar \mu} + i{\bar \partial} - i \partial + i \mu, \end{align*}$$

so that

$$\begin{align*}[d, \mathbb{J}] - \mathbb{I}^{-1} \, d \, \mathbb{I} = 4 i( {\bar \mu} - \mu). \end{align*}$$

This vanishes if and only if J is integrable.

Proof. Proposition 3.2 implies that d passes to $H_{d^c}$ with $d = 0 $ , and that the two right maps respect the differentials. The last statement is immediate to check.

Proof. Consider

The $d d^c$ -condition implies the first map, being the sum of two surjective maps, is surjective. Alternatively, the $d d^c$ -condition implies the last map, being the sum of two injective maps, is injective. By either argument, $\delta _k=0$ .

Proof. The sequence in Condition $(1)$ has a vanishing connecting homomorphism if and only if the long exact sequence splits into short exact sequences

which holds if and only if the diagram in Condition $(2)$ is a pushout and pullback for all $k \geq 0$ . To see the equivalence of the third condition, note that from the definitions we have

$$\begin{align*}H( {\mathcal A} / \mathrm{Im} \,d^c) = \frac{d^{-1}(\mathrm{Im} \,d^c)}{\mathrm{Im} \, d + \mathrm{Im} \,d^c} \quad \textrm{and} \quad H(\mathrm{Ker}\, \, d^c) = \frac{\mathrm{Ker}\, \, d \cap \mathrm{Ker}\, \, d^c}{ d( \mathrm{Ker}\, \,d^c)}. \end{align*}$$

Condition $(3)$ expresses that the connecting homomorphism $\delta $ has image zero.

The equivalence of Conditions $(4)$ to $(1)$ can be seen as follows. For any compact complex manifold, the long exact sequence of Theorem 3.6 implies:

$$\begin{align*}\dim \left( H^k(\mathrm{Ker}\, d^c) / \mathrm{Im} \, \delta_{k-1} \right) + \dim \left( \mathrm{Ker}\, \delta_k \right)= 2 b_k. \end{align*}$$

Then the equality in Condition $(4)$ holds if and only if Condition $(1)$ holds.

Proving that Condition $(5)$ is equivalent to Condition $(1)$ will require several steps. First, the diagram

$$\begin{align*}\delta_k: H^k ( {\mathcal A} / \mathrm{Im} \, d^c ) \to H^{k+1} ( \mathrm{Ker}\, d^c ) \end{align*}$$

can be understood as a functor on the category of bicomplexes over $\mathbb {C}$ , with values in the (linear) category of diagrams of two vector spaces and a linear map between them. Namely, for any bicomplex $(\mathcal {B}, \partial , \bar \partial )$ , we let $d = \partial + \bar \partial $ , and $d^c = i ( \bar \partial - \partial )$ , and consider the diagram above. The map $\delta _k$ is readily seen to be induced by d. This functor is linear, and it takes direct sums of bicomplexes to direct sums of vector spaces and maps between them, since $\mathrm{Ker}\, d^c$ and $\mathcal {B}/\mathrm {Im} \, d^c$ are compatible with direct sums.

From [Reference Khovanov and QiKQ20], [Reference StelzigSte21b], every (bounded) bicomplex $(\mathcal {B}, \partial , \bar \partial )$ decomposes as a direct sum of dots, squares, and zigzags, the definition of which we recall below.

To complete the proof of the claim, it suffices to check that the map

$$\begin{align*}\delta_j: H^j ( {\mathcal A} / \mathrm{Im} \, d^c ) \to H^{j+1} ( \mathrm{Ker}\, d^c ) \end{align*}$$

is zero, for all j, on all bicomplexes which contain only dots, squares, and length $3$ zigzags, and that $\delta _j$ is nonzero for some j on any bicomplex that contains an even-length zigzag, or an odd zigzag of length $5$ or higher. We summarize these groups and the map $\delta _k$ in Figure 1 below, which can be checked on a case-by-case basis, and explains the case of each row in the diagram.

Figure 1

Zigzag contributions for the connecting map ${\delta _k: H^k({\mathcal A} / \mathrm {Im} \, d^c ,d) \to H^{k+1}(\mathrm{Ker}\, d^c ,d).}$

(Dot) The case of a dot is a single vector space $\mathbb {C}$ , in some bidegree $(p,q)$ with $k=p+q$ , and zeroes elsewhere with vanishing differentials $\partial $ and ${\bar \partial }$ . We compute that ${\mathcal A} / \mathrm {Im} \, d^c = \mathrm{Ker}\, d^c = \mathbb {C}$ in degree k, with zero in all other degrees, so that $\delta _j : H^j ( {\mathcal A} / \mathrm {Im} \, d^c ) \to H^{j+1}( \mathrm{Ker}\, d^c )$ is the zero map for all j.

(Square) A square is a bicomplex whose only nonzero entries are as follows, with maps that are isomorphisms:

Let $k = p+q$ . Computing the total complexes $ ({\mathcal A} / \mathrm {Im} \, d^c ,d)$ in total degrees $k,k+1,k+2$ , we have $\mathbb {C} {\stackrel {\sim }{\smash {\longrightarrow }\rule {0pt}{0.4ex}}} \mathbb {C} \to 0$ . Similarly, the complex $(\mathrm{Ker}\, d^c, d) $ in degrees $k,k+1,k+2$ is $0 \to \mathbb {C} {\stackrel {\sim }{\smash {\longrightarrow }\rule {0pt}{0.4ex}}} \mathbb {C}$ . This shows the functors $H ( {\mathcal A} / \mathrm {Im} \, d^c )$ and $H( \mathrm{Ker}\, d^c )$ vanish on squares, and clearly, $\delta _j=0$ for all j.

(Odd length $3$ zigzag) Consider an ‘L’, that is a length $3$ zigzag with outgoing differentials that are isomorphisms:

In this case, the total complex $ ({\mathcal A} / \mathrm {Im} \, d^c ,d)$ in degrees $k,k+1$ is $\mathbb {C} {\stackrel {\sim }{\smash {\longrightarrow }\rule {0pt}{0.4ex}}} \mathbb {C}$ , so the cohomology $ H ({\mathcal A} / \mathrm {Im} \, d^c ,d)$ is zero in all degrees. Similarly, the complex $(\mathrm{Ker}\, d^c, d) $ in total degrees $(k,k+1)$ is $0 \to \mathbb {C}^2$ and the cohomology $H(\mathrm{Ker}\, d^c, d) $ is $(0,\mathbb {C}^2)$ in degrees k and $k+1$ , respectively. Thus, $\delta _j=0$ for all j.

Next, we consider a ‘reverse L’, that is a length $3$ zigzag with incoming differentials that are isomorphisms:

In this case, the total complex $ ({\mathcal A} / \mathrm {Im} \, d^c ,d)$ in total degrees $(k+1,k+2)$ is $\mathbb {C}^2 \to 0$ , so the cohomology $ H ({\mathcal A} / \mathrm {Im} \, d^c ,d)$ is $(\mathbb {C}^2,0)$ in total degrees $k+1$ and $k+2$ , respectively. Similarly, the complex $(\mathrm{Ker}\, d^c, d) $ in total degrees $(k+1,k+2)$ is $\mathbb {C}{\stackrel {\sim }{\smash {\longrightarrow }\rule {0pt}{0.4ex}}} \mathbb {C}$ and the cohomology $H(\mathrm{Ker}\, d^c, d) $ is $(0,0)$ in degrees $k+1$ and $k+2$ , respectively. Thus, $\delta _j=0$ for all j.

(General odd-length zigzag) For general odd-length zigzags, we have two cases, outgoing and incoming:

Consider the first case, which is in total degrees k and $k+1$ , with vector spaces $\mathbb {C}^m$ and $\mathbb {C}^{m+1}$ , respectively. The total complex $ ({\mathcal A} / \mathrm {Im} \, d^c ,d)$ is $\mathbb {C}^m$ in degree k, and $\mathbb {C}$ in degree $k+1$ , since $d^c$ is injective on each copy of $\mathbb {C}$ in degree k but not onto. The differential $d: \mathbb {C}^m \to \mathbb {C}$ is onto, so the total cohomology of $H({\mathcal A} / \mathrm {Im} \, d^c ,d)$ is $\mathbb {C}^{m-1}$ , and $0$ , in total degrees k and $k+1$ , respectively.

Continuing with this odd-length outgoing case, the total complex $ (\mathrm{Ker}\, d^c ,d)$ is $0$ in degree k, and $\mathbb {C}^{m+1}$ in degree $k+1$ . So the total cohomology of $H(\mathrm{Ker}\, d^c ,d)$ is $0$ and $\mathbb {C}^{m+1}$ , in total degrees k and $k+1$ , respectively. Finally, the differential $\delta _k:H^k({\mathcal A} / \mathrm {Im} \, d^c ,d)\to H^{k+1}(\mathrm{Ker}\, d^c ,d)$ is the injection $\mathbb {C}^{m-1} \to \mathbb {C}^{m+1}$ .

The case of odd-length incoming is computed similarly. The results are in Figure 1 and yield that $\delta _k :H^k({\mathcal A} / \mathrm {Im} \, d^c ,d)\to H^{k+1}(\mathrm{Ker}\, d^c ,d)$ is the surjection $\mathbb {C}^{m+1} \twoheadrightarrow \mathbb {C}^{m-1}$ , which is nonzero for $m> 1$ .

(Even-length zigzags) For general odd-length zigzags, we again have two cases, where the top-leftmost space has an outgoing or incoming map:

Suppose each complex has $\mathbb {C}^m$ in total degrees k and $k+1$ . Here, the two cases yield the same complexes in total degree. Namely, in either case, the complex $({\mathcal A} / \mathrm {Im} \, d^c ,d)$ is $\mathbb {C}^m \to 0$ with the same cohomology, and the complex $(\mathrm{Ker}\, d^c ,d)$ is $0 \to \mathbb {C}^m$ , with the same cohomology. The differential $\delta _k: H^k({\mathcal A} / \mathrm {Im} \, d^c ,d) \to H^{k+1}(\mathrm{Ker}\, d^c ,d)$ is the isomorphism $\mathbb {C}^m {\stackrel {\sim }{\smash {\longrightarrow }\rule {0pt}{0.4ex}}} \mathbb {C}^m$ , which is a nonzero for $m>0$ , that is length at least $2$ .

Proof. By Proposition 3.18, $\mathrm {pdef}(M)=n$ implies that there is a zigzag of length $2n+1$ , which, for space reasons, would have to have a nonzero component in bidegree $(n,0)$ . However, by an application of Stokes’ theorem, one may see that the only indecomposable complexes with nonzero components in degree $(n,0)$ are dots and squares (c.f. [Reference StelzigSte21b, Chapter 4]).

Proof. Degeneration at $E_1$ occurs if and only if there are no even zigzags, and purity defect at most $1$ occurs if and only if there are no odd zigzags of length greater than $3$ . The two together are equivalent to the condition that a bicomplex $({\mathcal A},\partial ,{\bar \partial })$ decomposes into a direct sum of dots, squares, and length $3$ zigzags.

Proof. This follows from Theorem 3.1 and Example 3.17 (1) (or Proposition 3.18).

Proof. Let $Z_M$ be an odd zigzag of maximal length $2m+1$ in ${\mathcal A}(M)$ and $Z_N$ an odd zigzag of maximal length $2n+1$ in ${\mathcal A}(N)$ . By Serre duality, we may assume they both have incoming outermost arrows. Now, there is an $E_1$ -isomorphism ${\mathcal A}(M\times N)\simeq _1 {\mathcal A}(N)\otimes {\mathcal A}(M)$ , and so $Z_M\otimes Z_N$ is a direct summand in ${\mathcal A}(M\times N)$ . But $Z_M\otimes Z_N\simeq _1 Z_{M\times N}$ , where $Z_{M\times N}$ is an odd zigzag of length $2(n+m)+1$ . Thus, $\mathrm {pdef}(M)+\mathrm {pdef}(N)\leq \mathrm {pdef}(M\times N)$ . The argument also works the other way, since a tensor product of an even-length zigzag with any other bicomplex does not contain odd zigzags ([Reference StelzigSte21b, Chapter 3]).

Proof. First note that the Hodge filtrations on $\mathrm{Ker}\, d^c$ (induced by row and column filtration on ${\mathcal A}(M)$ ) are compatible with the wedge product, since they are on ${\mathcal A}(M)$ . In particular, the cup product on cohomology respects these filtrations, that is $F^pH^r(\mathrm{Ker}\, d^c)\cup F^qH^s(\mathrm{Ker}\, d^c)\subseteq F^{p+q}H^{r+s}(\mathrm{Ker}\, d^c)$ and similarly for $\bar F$ . Now we argue via indecomposable bicomplexes: For I any square, or reverse L, $\mathrm{Ker}\, d^c (I)$ is contractible. For I any even-length zigzag, dot, or L, the bicomplex structure on I induces a bicomplex structure on $\mathrm{Ker}\, d^c(I)$ , which is then a direct sum of dots. Thus, in all cases that have $\mathrm {pdef}(I)\leq 1$ , the Hodge filtrations induce a pure Hodge structure on $H^k(\mathrm{Ker}\, d^c)$ .

Proof. This follows since

$$\begin{align*}H^*(\mathrm{Ker}\, d^c) = \frac{\mathrm{Ker}\, d \cap \mathrm{Ker}\, d^c}{ d( \mathrm{Ker}\, \,d^c)} \quad \quad H^*({\mathcal A}/ \mathrm{Im} \,d^c) = \frac{d^{-1} \mathrm{Im} \, d^c}{\mathrm{Im} \,d + \mathrm{Im} \, d^c}.\\[-42pt] \end{align*}$$

Proof. This follows from a general duality statement for cohomological functors. We do the first case in detail. The linear functor sending a bicomplex $(A,\partial ,{\bar \partial })$ to $H^k(\mathrm{Ker}\, d^c(A))$ sends squares to zero and commutes with arbitrary direct sums, and so defines a cohomological functor. Denote by $D{\mathcal A}$ the dual bicomplex as a bigraded vector space given by $(D{\mathcal A})^{p,q}=(A^{n-p,n-q})^{\vee }$ with differential $d_{D{\mathcal A}}=(\varphi \mapsto (-1)^{|\varphi |-1}\varphi \circ d)$ . Since M is compact and oriented, the duality map ${\mathcal A}\to D{\mathcal A}$ given by $\omega \mapsto \int _M \omega \wedge -$ induces an isomorphism on all cohomological functors [Reference StelzigSte21b, Corollary 20], so, in particular

$$\begin{align*}H^k(\mathrm{Ker}\, d^c({\mathcal A}))\cong H^k(\mathrm{Ker}\, d^c(D{\mathcal A})). \end{align*}$$

Now it is a linear algebra calculation to show that

$$\begin{align*}H^k(\mathrm{Ker}\, d^c(D{\mathcal A}))\cong (H^{2n-k}({\mathcal A}/\mathrm{Im} \, d^c))^{\vee}. \end{align*}$$

In fact, since over fields cohomology commutes with duals, this follows from the identification

$$\begin{align*}[\mathrm{Ker}\, d^c(D{\mathcal A})]^k=\{\varphi\in [{\mathcal A}^{2n-k}]^{\vee} \mid \varphi\circ d^c=0\}\cong ([{\mathcal A}/\mathrm{Im} \, d^c]^{2n-k})^{\vee}.\\[-37pt] \end{align*}$$

Proof. It remains only to show Condition $(3)$ is equivalent to Condition $(2)$ . Let us focus on the groups $H_{\urcorner }$ first, where a similar calculation can be done for $H_{\llcorner }$ , or, if ${\mathcal A}={\mathcal A}(M)$ for a compact manifold, one may appeal to Serre duality as in Proposition 3.27.

Figure 2

Zigzag contributions for the short exact sequence ${0\to H_{\urcorner } ({\mathcal A}) \to H_{BC}({\mathcal A}) \to H(\mathrm{Ker}\, \, d^c)\to 0.}$

Recall that $H_\mathrm{{dR}}({\mathcal A}) $ has a pure Hodge structure if and only if there are only squares, even zigzags, and dots, with no odd zigzags of length greater than $1$ [Reference StelzigSte21b]. The proof then proceeds by computing either $H_{\urcorner }({\mathcal A}) $ or $\phi $ on every type of indecomposable complex. The results are summarized in Figure 2.

Proof. Claim $(1)$ is in Theorem 3.28, while Claim $(3)$ is immediate. Claim $(4)$ follows from Claims $(1)$ and $(2)$ , while Claim $(5)$ follows from Claims $(2)$ , $(3)$ , and Theorem 3.9. It remains to show Claim $(2)$ . This follows from an inspection of Figure 1 and the observation that the expression $h_{{\bar \partial }}+h_{\partial }$ vanishes on squares and is equal to two on every zigzag, regardless of its length.

Proof. By the degeneration assumption, there are no even zigzags in any decomposition of ${\mathcal A}(Z(M))$ into indecomposables. It remains to rule out the possibility of odd zigzags of length greater than $3$ , which, for dimension reasons, must have length $5$ or $7$ . Length $7$ zigzags do not occur by Corollary 3.19. There are two possibilities for a length $5$ zigzag on a three-fold:

Any complex having one of these as a direct summand would have $E_2^{0,2}\neq 0$ , and $E_2^{2,0}\neq 0$ , respectively, both of which are prohibited here.

Proof. Let M be a complex manifold. Denote by $\tilde {M}$ the blow-up of M in some smooth centre Z of codimension $d\geq 2$ . Let ${\mathcal V}$ be a holomorphic vector bundle of rank r over M. Let $f:M\to N$ be a holomorphic surjection. In [Reference StelzigSte21a], it was shown that there are (chains of) $E_1$ -isomorphisms

$$ \begin{align*} {\mathcal A}(\tilde{M})&\simeq_1 {\mathcal A}(M)\oplus_{i=1}^{d-1} {\mathcal A}(Z)[i]\\ {\mathcal A}(\mathbb{P}({\mathcal V}))&\simeq_1 \sum_{i=0}^{r-1} {\mathcal A}(M)[i]\\ {\mathcal A}(M)&\simeq_1 {\mathcal A}(N)\oplus {\mathcal A}(M)/p^*{\mathcal A}(N). \end{align*} $$

Here, ${\mathcal A}(Z)[i]$ denotes the complex shifted by degree $(i,i)$ , that is ${\mathcal A}(Z)[i]^{p,q}={\mathcal A}(Z)^{p-i,q-i}$ . Since two bounded bicomplexes are $E_1$ -isomorphic if and only if all zigzags have the same multiplicity, Conditions 1, 3, and 4 follow. For Condition 2, note that by the Künneth formula, there is an $E_1$ -isomorphism ${\mathcal A}(M\times N)\simeq {\mathcal A}(M)\otimes {\mathcal A}(N)$ . If M satisfies the $d d^c$ -condition, we have ${\mathcal A}(M)\simeq _1 D$ , where D has a trivial differential, that is it is a direct sum of dots $D=\bigoplus D^{p,q}$ . Then ${\mathcal A}(M\times N)\simeq _1 \bigoplus _{p,q} D^{p,q}\otimes {\mathcal A}(N)$ , and the result follows because the tensor product of any bicomplex with a dot is isomorphic to a shift of the original bicomplex. Conversely, the purity defect is additive under products by Proposition 3.23, so $\mathrm {pdef}(M\times N)\leq 1$ implies that $\mathrm {pdef}(M)\leq 1$ and $\mathrm {pdef}(N)=0$ , or vice versa. Furthermore, the Frölicher spectral sequence of both M and N is a direct summand in that of $M\times N$ , so, if it degenerates on the product, it does on both factors.

Proof. First, we recall the well known argument (see, e.g. [Reference VoisinVoi07]) that for t sufficiently close to $0$ , the Frölicher spectral sequence of $M_t$ degenerates and the Hodge numbers are the same as those of $M_0$ : Choosing a smooth family of Hermitian metrics on the fibres $M_t$ , the Hodge numbers may be computed via the ${\bar \partial }$ -Laplacian which is an elliptic operator that varies smoothly in t. Therefore, the eigenvalues vary continuously and, in particular, the dimension of the kernel (i.e. the Hodge numbers) can only drop when passing from $M_0$ to a nearby fibre:

$$\begin{align*}h^{p,q}_{\bar \partial}(M_0)\geq h^{p,q}(M_t). \end{align*}$$

On the other hand,

$$\begin{align*}b_k(M_0)=\sum_{p+q=k}h^{p,q}(M_0)\geq \sum_{p+q=k}h^{p,q}(M_t)\geq b_k(M_t)=b_k(M_0) \end{align*}$$

so equality holds everywhere. Now, for any t, consider the inclusion of complexes $F^p{\mathcal A}(M_t)\subseteq {\mathcal A}(M_t)$ defined by $F^p{\mathcal A}(M_t)=\bigoplus _{r\geq p} {\mathcal A}^{r,s}(M_t)$ . The induced map on cohomology has image $F^pH_\mathrm{{dR}}^k(M_t)$ and kernel

$$\begin{align*}\frac{\operatorname{\mathrm{Im}\,} d\cap F^p{\mathcal A}^k_{M_t}}{d(F^p{\mathcal A}^k_{M_t})}. \end{align*}$$

Thus, the induced map on cohomology is injective if and only if the differential d is strict, which in turn is known to be equivalent to degeneration of the Frölicher spectral sequence [Reference DeligneDel71, Proposition 1.3.2]. Therefore, for small t, we have an identification $F^pH_\mathrm{{dR}}^k(M_t)=H^k(F^p{\mathcal A}(M_t))$ .

Now, because $F^p{\mathcal A}(M_t)$ is an elliptic complex (see, e.g. [Reference StelzigSte22b]), or because their collection over all t forms a resolution of the truncated complex of coherent sheaves of relative holomorphic forms

$$\begin{align*}0\to\Omega^p_{{\mathcal M}/B}\to\Omega_{{\mathcal M}/B}^{p+1}\to\cdots,\end{align*}$$

the cohomology groups $H^k(F^p{\mathcal A}(M_t))$ form a vector bundle on the base as soon as their dimensions are constant. On the other hand, because the dimension of a filtred vector space is the same as that of its associated graded, we find that

$$\begin{align*}\dim F^pH^k_{\mathrm{dR}}({\mathcal A}(M_t))=\sum_{r+s=k, r\geq p} h^{r,s}(M_t) \end{align*}$$

is constant indeed.

In summary, for sufficiently small t, we have proved that $\{F^pH_\mathrm{{dR}}^k(M_t)\}$ form a complex vector subbundle of the vector bundle $\{H_\mathrm{{dR}}^k(M_t)\}$ . The same holds for the conjugate filtration $\bar {F}^qH_\mathrm{{dR}}^k(M_t)$ . But the dimension of an intersection of two vector subbundles behaves upper semicontinuously.

Proof. It suffices to show the multiplicities of all zigzags are constant for t close to $0$ . There are no even zigzags on $M_0$ or nearby fibres by degeneration of the Frölicher spectral sequence, as in the first part of the proof of Theorem 4.8. The odd zigzags are counted by the refined Betti numbers $b_k^{p,q}(M_t)$ . Thus, we have to show that the numbers $b_k^{p,q}(M_t)$ are constant for t close to $0$ .

First, we note that condition ( $\ast $ ) has to hold for nearby fibres as well. In fact, when $F_{tot}^rH^k(M_0)=0$ for some r, then also $F_{tot}^rH^k(M_t)=0$ for all nearby fibres by Theorem 4.8. By duality, the same implication holds for the condition $F_{tot}^rH^k=H^k$ .

When $p+q=r(k)$ , we have

$$\begin{align*}b_k^{p,q}=\dim F^pH^k\cap \bar F^q H^k \end{align*}$$

by assumption, and we have seen that this number varies upper semicontinuously in Theorem 4.8. On the other hand, $b_k^{p,q}(M)=b_{2n-k}^{n-p,n-q}(M)$ and so also the numbers for $p+q-k=r(k)-1$ vary upper semicontinuously. Finally,

$$\begin{align*}b_k(M_t)=\sum_{p+q\in\{r(k),r(k)+1\}}b_k^{p,q}(M_t)\leq \sum_{p+q\in\{r(k),r(k)+1\}}b_k^{p,q}.(M_0)=b_k(M_0)=b_k(M_t). \end{align*}$$

Proof. Only the statement about the second factor of $S_{p,q}$ still requires proof. For the first claim, with $k=n$ , the bicomplex is

$$\begin{align*}\Lambda\langle\theta^{1,0},\theta^{0,1},\omega_0\rangle/(\omega_0)= \Lambda\langle\theta^{1,0},\theta^{0,1}\rangle, \end{align*}$$

and the relation $\partial \theta ^{0,1}=-{\bar \partial }\theta ^{1,0}=\frac i 2 \omega _0$ implies all differentials are zero.

For the general case, $k<n$ , we have $\partial \theta ^{0,1}=-{\bar \partial }\theta ^{1,0}=\frac i 2 \omega _0$ and ${\bar \partial }\theta ^{0,1}=\partial \theta ^{1,0}=0$ , giving the length three zigzag

in total degrees $1$ and $2$ . The same relations also give the squares in the statement, with all other differentials on these spaces zero, again, since ${\bar \partial }\theta ^{0,1}=\partial \theta ^{1,0}=0$ . Finally, the length $3$ zigzag

follows from the same relations and the fact that $\omega _0^{n-k+1}=0$ , which also implies that $\partial \theta ^{0,1}\theta ^{1,0}\omega _0^{n-k} = {\bar \partial } \theta ^{0,1}\theta ^{1,0}\omega _0^{n-k}=0$ .

Proof. Using the notation introduced before Theorem 5.3, we have:

$$\begin{align*}{\mathcal A}(V) \simeq_1 \bigoplus_{r,s\in\mathbb{Z}}S_{r,s} \end{align*}$$

by Proposition 5.2. Now, [Reference StelzigSte21b, Corollary 13] states that an $E_1$ -isomorphism induces an isomorphism on $H_{BC}$ , so by the Definition of $S_{r,s}$ and the fact that $P_{r,s}$ is a complex with trivial differentials, we have:

$$\begin{align*}H^{p,q}_{BC}(V)=\bigoplus_{r,s} P_{r,s}\otimes H^{p,q}_{BC}\left(\Lambda\langle \theta^{0,1},\theta^{1,0},\omega_0\rangle/(\omega_0^{n-(r+s)+1})\right). \end{align*}$$

The result now follows by Theorem 5.3, as the groups $H_{BC}$ on any zigzag are known to be computed by the dots (here, P) and the ‘tips’, that is the spaces X, U, and T in diagrams such as those below:

The proof for the Aeppli cohomology is the same, except $H_{A}$ is computed by the dots (again, P) and the spaces W, Y, and Z in the diagrams above.

Proof. According to [Reference Otiman and TomaOT21], the Frölicher spectral sequence degenerates on all Oeljeklaus-Toma manifolds and the purity defect of an OT manifold of type $(s,t)$ is equal to s (this follows from [Reference Angella, Dubickas, Otiman and StelzigADOS22, Theorem 9]).

Proof. As seen before, Vaisman manifolds V have a model $B\otimes \Lambda (J\theta )$ , with $B=H\otimes \Lambda (\theta )$ , where $H=H_B(V)$ denotes the basic cohomology. This satisfies the required conditions by the explicit formulae for the cohomology of Vaisman manifolds: In fact, denoting by $P_B^k\subseteq H^k $ the primitive part of the basic cohomology, one has

$$\begin{align*}H_{\mathrm{dR}}^k(V)=\begin{cases} P_B^k\oplus P_B^{k-1}\theta&\text{if }k\leq n\\ P_B^n\theta\oplus P_B^nJ\theta&\text{if }k=n+1\\ (P_B^{n-l}\omega_0^{l}\oplus P_B^{n-(l-1)}\omega_0^{l-1}\theta)J\theta&\text{if }k=n+1+l\geq n+2 \end{cases} \end{align*}$$

(see, for instance [Reference KashiwadaKas80], [Reference VaismanVai82], [Reference Ornea and VerbitskyOV22], or Theorem 5.3).

Proof. Using the notation of Theorem 5.13, it suffices to show that $p_k = 0$ for all $k \geq 4$ .

As complexes, we have $A= B\oplus Bx$ , and $Bx$ is an ideal in A satisfying $(Bx)^2=0$ . Note that, $\mathrm{Im}\, d\subseteq B$ and $\mathrm{Ker}\, d=B\oplus \mathrm{Ker}\, d\cap Bx$ . This allows to choose a splitting $A=\mathrm{Im}\, d \oplus L\oplus H$ , where $L\subseteq Bx$ , $d:L\to \mathrm{Im}\, d$ is an isomorphism, and $d|_H=0$ . Choose $i: H \to A$ to be the inclusion, that represents the cohomology of A, with projection $\pi : A \to H$ , and define a contracting homotopy $h: A \to A$ to be a projection onto $\mathrm{Im}\,(d)$ , followed by $d^{-1}:\mathrm{Im}\, d\to L$ . By construction, $\mathrm{Im}\, h$ is contained in the ideal $Bx$ , and satisfies $h(Bx) = (\mathrm{Im}\, h)^2 = 0$ .

Therefore, for $k=3$ , we have

$$\begin{align*}p_3 = m \left (i \otimes h p_2 \right) + m \left ( h p_2 \otimes i \right) \subseteq Bx, \end{align*}$$

so that $p_4=0$ , since

$$\begin{align*}m(i \otimes h p_3) = m(h p_2 \otimes h p_2 ) = m( h p_3 \otimes i )=0. \end{align*}$$

Similarly, for all $k \geq 4$ and every $1 \leq j <k$ , $m \left ( h p_j \otimes h p_{k-j} \right ) =0$ , so $p_k = 0$ .

Proof. A real model of the complex $({\mathcal A}(V), d)$ is given by

$$\begin{align*}H \otimes \Lambda\langle \theta^{0,1},\theta^{1,0}\rangle = \left( H \otimes \Lambda \theta \right) \otimes \Lambda ( J\theta), \end{align*}$$

where H is the basic cohomology, $d \theta =0$ , and $d (J \theta )= J\omega _0 \in H \otimes \Lambda \theta $ [Reference Ishida and KasuyaIK19]. Now apply Proposition 5.14 with $B = H \otimes \Lambda \theta $ and $x=J \theta $ .

Proof. First, note that the statement is clear if the diagram strictly commutes. Thus, in the above setting, we obtain a morphism of $H(B)$ -modules $H(cone(\varphi ))\to H(cone(f_A \circ \varphi ))$ and also a morphism $H(cone (\varphi '\circ f_B))\to H(cone(\varphi '))$ which intertwines the $H(B)$ , respectively, $H(B')$ module structures. Next, pick a homotopy $K:B\to A'\langle t, dt\rangle $ , such that $\epsilon _0\circ K=\varphi '\circ f_B$ and $\epsilon _1\circ K=f_A\circ \varphi $ , where $\epsilon _i$ sends $t\mapsto i, dt\mapsto 0$ , we obtain isomorphisms of $H(B)$ -modules $H(cone(f_A\circ \varphi ))\leftarrow H(cone(K))\to H(cone(\varphi '\circ f_B))$ .

Proof. Since the two long exact sequences have the same underlying spaces, they are isomorphic only if the maps $\varphi _A$ and $\varphi _H$ have the same rank on cohomology. So, the symmetry condition follows from a case-by-case inspection of every indecomposable bicomplex, for which we refer to Figure 3.

Figure 3

Ranks of maps in standard $d^c$ -diagram.

The connectivity property follows from the long exact sequence associated to the short exact sequence

$$\begin{align*}0 \to \mathrm{Ker}\, \, d^c \to {\mathcal A} \to {\mathcal A}/\mathrm{Ker}\, \, d^c \to 0 \end{align*}$$

since $H^0({\mathcal A}/\mathrm{Ker}\, \, d^c)=0$ for any complex manifold.

For the duality property, since $\varphi :=(i, \mathbb {I} \circ \pi ):\mathrm{Ker}\, d^c\to {\mathcal A}(M) \oplus H(M)$ is injective, the natural projection $\pi : cone(\varphi )\to \operatorname {Coker}(\varphi )$ , given by

$$\begin{align*}\pi(b,a) = \varphi(b) + \left( {\mathcal A}(M) \oplus H(M) \right) / \mathrm{Im}( \varphi), \end{align*}$$

induces an isomorphism $H(\pi ) : H(cone(\varphi )) \to H(\operatorname {Coker}(\varphi ))$ . By Lemma 6.4, the projection $\pi $ is compatible with the $H(\mathrm{Ker}\, d^c)$ -module-structure on cohomologies, and $\operatorname {Coker}(\varphi ) = {\mathcal A}/ \mathrm {Im} \, d^c$ , by Theorem 3.6, so the duality property follows from Proposition 3.27.

Proof. First, $H^k(B)=0$ for $k>2n$ by duality, since the cone is nonnegatively graded. The second claim follows from the long exact sequence

again, by duality $H^k(cone(\varphi )) \cong H^{2n-k}(B)$ , using exactness at $H^k(A) \oplus H^k(A)$ . The third claim follows similarly from exactness and duality, since the Euler characteristic is additive along long exact sequences.

For the fourth claim, note that the choice of a representative for a fundamental class induces a commutative diagram

where $D_{2n}$ denotes the dualization functor, defined for any complex C as $(D_{2n}C)^k=(C^{2n-k})^{\vee }$ , with differential given (up to sign) by pullback. Thus, we obtain an isomorphism of the associated long exact sequences involving the cones of $\varphi _H$ and $D_{2n}\psi _H$ .

Finally, for the last claim, we note that from the definitions there is a short exact sequence

$$\begin{align*}0\longrightarrow H\overset{\psi_H}{\longrightarrow} cone(\varphi)\longrightarrow cone(\varphi_A)\longrightarrow 0. \end{align*}$$

Thus, comparing this long exact sequence with that induced by

$$\begin{align*}0\longrightarrow cone(\varphi)\longrightarrow cone(\psi_H)\longrightarrow H[-1]\longrightarrow 0, \end{align*}$$

we see that the natural map $cone(\varphi _A)\to cone(\psi _H)$ has to be a quasi-isomorphism. The case of $\varphi _H$ and $\psi _A$ is analogous.