Correction to: Annals of Global Analysis and Geometry (2020) 57:519–535 https://doi.org/10.1007/s10455-020-09711-y

In the proof of [1, Theorem 1.2] (see [1, p. 531]), we said that

“Fix two integers s and t. Consider the double complexes

$$\begin{aligned} K^{\bullet ,\bullet }= & {} \bigoplus \limits _{\begin{array}{c} u+w=s\\ v+w=t \end{array}}ss(S^{\bullet ,\bullet }(X,{\mathcal {L}},u,v)\otimes _{{\mathbb {C}}} S^{\bullet ,\bullet }(Y,{\mathcal {H}},w,w)),\\ L^{\bullet ,\bullet }= & {} S^{\bullet ,\bullet }(X\times Y,{\mathcal {L}}\boxtimes {\mathcal {H}},s,t) \end{aligned}$$

and a morphism \(f=pr_1^*(\bullet )\wedge pr_2^*(\bullet ):K^{\bullet ,\bullet }\rightarrow L^{\bullet ,\bullet }\)."

In general cases, f is not a morphism.

FormalPara Example 0.1

Assume that \(X=\{pt\}\) is a single-point set and Y is an n-dimensional complex manifold. Let \(pr_1\), \(pr_2\) be projections from \(\{pt\}\times Y\) onto \(\{pt\}\), Y, respectively. Consider the double complexes

$$\begin{aligned} K^{\bullet ,\bullet }= & {} \bigoplus \limits _{\begin{array}{c} u+w=0\\ v+w=n \end{array}}ss(S^{\bullet ,\bullet }(\{pt\},{\mathbb {C}},u,v)\otimes _{{\mathbb {C}}} S^{\bullet ,\bullet }(Y,{\mathbb {C}},w,w)),\\ L^{\bullet ,\bullet }= & {} S^{\bullet ,\bullet }(\{pt\}\times Y,{\mathbb {C}},0,n) \end{aligned}$$

and a map \(f=pr_1^*(\bullet )\wedge pr_2^*(\bullet ):K^{\bullet ,\bullet }\rightarrow L^{\bullet ,\bullet }\).

In such case, we have

$$\begin{aligned} L^{a,b}= & {} A^{a,b}(\{pt\}\times Y),\\ _LD_1^{a,b}= & {} \partial ,\quad _LD_2^{a,b}={\bar{\partial }}. \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} K^{a,b}&=\bigoplus \limits _{\begin{array}{c} u+w=0\\ v+w=n \end{array}}\bigoplus \limits _{\begin{array}{c} p+q=a\\ r+s=b \end{array}}S^{p,r}(\{pt\},{\mathbb {C}},u,v)\otimes _{{\mathbb {C}}} S^{q,s}(Y,{\mathbb {C}},w,w)\\&=\bigoplus \limits _{\begin{array}{c} u+w=0\\ v+w=n \end{array}}S^{0,0}(\{pt\},{\mathbb {C}},u,v)\otimes _{{\mathbb {C}}} S^{a,b}(Y,{\mathbb {C}},w,w)\\&=S^{0,0}(\{pt\},{\mathbb {C}},-a,n-a)\otimes _{{\mathbb {C}}} S^{a,b}(Y,{\mathbb {C}},a,a)\\&={\mathcal {A}}^{0,0}(\{pt\})\otimes _{{\mathbb {C}}} {\mathcal {A}}^{a,b}(Y), \\ _KD_1^{a,b}&=\sum \limits _{\begin{array}{c} u+w=0\\ v+w=n \end{array}}\sum \limits _{\begin{array}{c} p+q=a\\ r+s=b \end{array}}d^{p,r}_{\{pt\}1} \otimes 1_{S^{q,s}(Y,{\mathbb {C}},w,w)}\\&\quad +\sum \limits _{\begin{array}{c} u+w=0\\ v+w=n \end{array}}\sum \limits _{\begin{array}{c} p+q=a\\ r+s=b \end{array}}(-1)^{p+r}1_{S^{p,r}(\{pt\},{\mathbb {C}},u,v)}\otimes d_{S^{\bullet ,\bullet }(Y,{\mathbb {C}},w,w)1}^{q,s}\\&=\sum \limits _{\begin{array}{c} u+w=0\\ v+w=n \end{array}}1_{S^{0,0}(\{pt\},{\mathbb {C}},u,v)}\otimes d_{S^{\bullet ,\bullet }(Y,{\mathbb {C}},w,w)1}^{a,b}\\&=0\quad \left( \text{ since }\,d_{S^{\bullet ,\bullet }(Y,{\mathbb {C}},w,w)1}^{a,b}=0\right) , \\ _KD_2^{a,b}&=\sum \limits _{\begin{array}{c} u+w=0\\ v+w=n \end{array}}\sum \limits _{\begin{array}{c} p+q=a\\ r+s=b \end{array}}d^{p,r}_{\{pt\}2}\otimes 1_{S^{q,s}(Y,{\mathbb {C}},w,w)}\\&\quad +\sum \limits _{\begin{array}{c} u+w=0\\ v+w=n \end{array}}\sum \limits _{\begin{array}{c} p+q=a\\ r+s=b \end{array}}(-1)^{p+r}1_{S^{p,r}(\{pt\},{\mathbb {C}},u,v)}\otimes d_{S^{\bullet ,\bullet }(Y,{\mathbb {C}},w,w)2}^{q,s}\\&=\sum \limits _{\begin{array}{c} u+w=0\\ v+w=n \end{array}}1_{S^{0,0}(\{pt\},{\mathbb {C}},u,v)}\otimes d_{S^{\bullet ,\bullet }(Y,{\mathbb {C}},w,w)2}^{a,b}\\&=\sum \limits _{w=0}^n1_{S^{0,0}(\{pt\},{\mathbb {C}},-w,n-w)}\otimes d_{S^{\bullet ,\bullet }(Y,{\mathbb {C}},w,w)2}^{a,b}\\&=1_{{\mathcal {A}}^{0,0}(\{pt\})}\otimes \left( \sum \limits _{w=0}^nd_{S^{\bullet ,\bullet }(Y,{\mathbb {C}},w,w)2}^{a,b}\right) \\&= 1\otimes {\bar{\partial }} \quad \left( \text{ since }\,\sum \limits _{w=0}^nd_{S^{\bullet ,\bullet }(Y,{\mathbb {C}},w,w)2}^{a,b}=d_{\bigoplus \limits _{w=0}^nS^{\bullet ,\bullet }(Y,{\mathbb {C}},w,w)2}^{a,b} =d_{{\mathcal {A}}^{\bullet ,\bullet }(Y)2}^{a,b}={\bar{\partial }}\right) . \end{aligned} \end{aligned}$$

We easily check that \(f\circ _KD_1=0\ne _LD_1\circ f\) and \(f\circ _KD_2=_LD_2\circ f\). So f is not a morphism of double complexes.

Example 0.1 also gives a counterexample to [1, Theorem 1.2]. Let \(X=\{pt\}\) be a single-point set, Y an n-dimensional complex manifold, \({\mathcal {L}}={\mathcal {O}}_X\), \({\mathcal {H}}={\mathcal {O}}_Y\) and \(s=0\), \(t=n\). Then

$$\begin{aligned} \begin{aligned}&\bigoplus \limits _{\begin{array}{c} a+b=c\\ u+w=0\\ v+w=n \end{array}}{\mathbb {H}}^{a}\left( X,\Omega _X^{[u,v]}({\mathcal {L}})\right) \otimes _{{\mathbb {C}}} {\mathbb {H}}^{b}\left( Y,\Omega _Y^{[w,w]}({\mathcal {H}})\right) \\&\quad =\bigoplus \limits _{\begin{array}{c} a+b=c\\ 0\le w\le n \end{array}}{\mathbb {H}}^{a}\left( \{pt\},\Omega _{\{pt\}}^{[-w,n-w]}\right) \otimes _{{\mathbb {C}}} H^{b-w}\left( Y,\Omega _Y^w\right) \\&\quad =\bigoplus \limits _{0\le w\le n}{\mathbb {H}}^{0}\left( \{pt\},\Omega _{\{pt\}}^{[0,0]}\right) \otimes _{{\mathbb {C}}} H^{c-w}\left( Y,\Omega _Y^w\right) \\&\quad =\bigoplus \limits _{0\le w\le n}H^{0}\left( \{pt\},{\mathcal {O}}_{\{pt\}}\right) \otimes _{{\mathbb {C}}} H^{c-w}\left( Y,\Omega _Y^w\right) \\&\quad =\bigoplus \limits _{p+q=c}H^{q}\left( Y,\Omega _Y^p\right) \end{aligned} \end{aligned}$$

and

$$\begin{aligned} {\mathbb {H}}^{c}\left( X\times Y,\Omega _{X\times Y}^{[0,n]}({\mathcal {L}}\boxtimes {\mathcal {H}})\right) ={\mathbb {H}}^{c}\left( \{pt\}\times Y,\Omega _{\{pt\}\times Y}^{[0,n]}\right) ={\mathbb {H}}^{c}\left( Y,\Omega _{Y}^{[0,n]}\right) =H_{dR}^{c}(Y,{\mathbb {C}}) \end{aligned}$$

However,\(\bigoplus \limits _{p+q=c}H^{q}(Y,\Omega _Y^p)\ne H_{dR}^{c}(Y,{\mathbb {C}})\) for a general complex manifold Y. So [1, Theorem 1.2] doesn’t hold.