Erratum to: The Frölicher spectral sequence can be arbitrarily non-degenerate

1 Erratum to: Math. Ann. (2008) 341:623–628 DOI 10.1007/s00208-007-0206-z

2 Introduction

In the original publication the second author constructed for $n\ge 2$ a series of compact complex manifolds $X_n$ and an element $[\beta _1]$ in the $E_n$-term of the Frölicher spectral sequence claiming that $d_n([\beta _1])\ne 0$ (Lemma 2 in loc.cit.). This claim is incorrect: we explain in Remark 2 that on the contrary $\beta _1$ induces a class in $E_\infty$.

However, the main result of the original publication remains true (up to a change in the dimension of the examples).

Theorem 1

For every $n\ge 2$ there exist a complex $4n-2$-dimensional compact complex manifold $X_n$ such that the Frölicher spectral sequence does not degenerate at the $E_n$ term, i.e., $d_n\ne 0$.

The method of construction has remained the same, but we needed to introduce some extra counting variables.

We believe that in every dimension there are examples of nilmanifolds with left-invariant complex structure where the maximal possible non-degeneracy occurs, but the structure equations might be quite complicated.

3 Construction of the example

Consider the space $G_n:={\mathbb C}^{4n-2}$ with coordinates

$$\begin{aligned} x_1, \dots , x_{n-1}, y_1, \dots , y_n, z_1,\dots , z_{n-1}, w_1, \dots , w_n. \end{aligned}$$

Endow $G_n$ with the structure of a real nilpotent Lie-group by identifying it with the subgroup of $\mathrm {Gl}(2n+2, {\mathbb C})$ consisting of upper triangular matrices of the form

$$\begin{aligned} A= \left( \!\!\!\! \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} &{} 1 &{} 0 &{} &{}&{}&{}&{}\dots &{}&{}0&{}\bar{y}_1&{}w_1\\ &{} &{}1 &{} 0 &{}\dots &{}0&{}\bar{z}_1 &{} - x_1 &{}0 &{} \dots &{} 0 &{}w_2\\ &{}&{} &{} \ddots &{} &{} &{} &{} \ddots &{} &{} &{} \vdots &{}\vdots \\ &{}&{}&{}&{}1 &{} 0 &{}\dots &{}0&{} \bar{z}_{n-1} &{}- x_{n-1} &{} 0 &{} w_n\\ &{}&{}&{}&{}&{}1 &{} 0 &{}&{}\dots &{} &{} 0 &{} y_1\\ &{}&{}&{}&{}&{}&{}\ddots &{} &{}&{} &{} \vdots &{} \vdots \\ &{}&{}&{}&{}&{}&{}&{}&{}&{}&{}&{}&{}\\ &{}&{}&{}&{}&{}&{}&{}&{}&{}&{}&{}&{}\\ &{}&{}&{}&{}&{}&{}&{}&{}\ddots &{} &{} \vdots &{} \vdots \\ &{}&{}&{}&{}&{}&{}&{}&{}&{} 1&{}0&{}y_n\\ &{}&{}&{}&{}&{}&{}&{}&{}&{} &{}1&{}z_1\\ &{}&{}&{}&{}&{}&{}&{}&{}&{} &{}&{}1 \end{array} \!\!\!\!\right) \end{aligned}$$

Let $\varGamma = G_n \cap \mathrm {Gl}(2n+2, {\mathbb Z}[i])$, which is a lattice in the real Lie-group $G$. Note that if $g\in G_n$ is a fixed element then the action on the left, $g' \mapsto gg'$, is holomorphic with respect to the complex structure on ${\mathbb C}^{4n-2}$. The quotient

$$\begin{aligned} X_n = \varGamma \slash G_n \end{aligned}$$

is a compact complex manifold; more precisely, it is a compact nilmanifold with left-invariant complex structure.

Remark 1

The manifold $X_n$ admits a simple geometric description in terms of principal holomorphic torus bundles: the centre of $G_n$ is given by the matrices for which all $x_i$, $y_i$ and $z_i$ vanish and hence isomorphic (as a Lie group) to ${\mathbb C}^n$. This yields an exact sequence of real Lie-groups

$$\begin{aligned} 0\rightarrow {\mathbb C}^n\rightarrow G_n\rightarrow {\mathbb C}^{3n-2}\rightarrow 0 \end{aligned}$$

which is compatible with the action of $\varGamma$. Denoting by $T_k$ the quotient ${\mathbb C}^k/{\mathbb Z}[i]^k$ the exact sequence induces a $T_n$ principal bundle structure on $X_n\rightarrow T_{3n-2}$.

The space of left-invariant 1-forms $U$ is spanned by the components of ${A}^{-1} dA$ and their complex conjugates, so a basis for the forms of type $(1,0)$ is given by

$$\begin{aligned} dx_1, \dots , dx_{n-1},dy_1, \dots , dy_n, dz_1, \dots , dz_{n-1}, \omega _1,\dots , \omega _n \end{aligned}$$

where

$$\begin{aligned} \omega _1&= dw_1 -\bar{y}_1 dz_1,\\ \omega _k&= dw_k -\bar{z}_{k-1}dy_{k-1} + x_{k-1} d y _k \quad (k = 2, \dots , n). \end{aligned}$$

For later reference we calculate the differentials of the above basis vectors:

$$\begin{aligned} d(dx_i)&= d(dz_i) = 0 \qquad&(i=1\cdots n-1)\\ d(dy_i)&= 0 \qquad&(i=1\cdots n)\\ d\omega _1&= - d\bar{y}_1 \wedge dz_1\\ d\omega _i&= dx_{i-1}\wedge dy_i+ dy_{i-1}\wedge d\bar{z}_{i-1} \end{aligned}$$

The following lemma shows that the Frölicher spectral sequence of $X_n$ has non-vanishing differential $d_n$ thus proving our Theorem.

Lemma 1

The differential form $\beta _1=\bar{\omega }_1\wedge d\bar{z}_2\wedge \cdots \bar{d}z_{n-1}$ defines a class $[\beta _1]_n\in E_n^{0,n-1}$ and

$$\begin{aligned} d_n([\beta _1]_n)=(-1)^{n-2}[dx_1\wedge \dots \wedge dx_{n-1}\wedge dy_n]_n\ne 0 \quad \text {in } E_n^{n,0}. \end{aligned}$$

Proof

By Remark 1 the projection to the $(x,y,z)$-coordinates endows $X_n$ with the structure of holomorphic principal torus bundle over a complex torus. By the results of [2] the inclusion of left-invariant forms into the double complex $({\mathcal A}^{p,q}(X_n), \partial , {\bar{\partial }})$ induces an isomorphism on the $E_1$-terms of the respective spectral sequences. Thus for our purpose we may work with left-invariant forms only, that is, start with the $E_0$-term

$$\begin{aligned} E_0^{p,q} = \Lambda ^p U\otimes \Lambda ^q\bar{U}. \end{aligned}$$

A $(p,q)$-form $\alpha$ lives to $E_r$ if it represents a class in $E_r^{p,q}$, which is a subquotient of $E_0^{p,q}$; the resulting class will be denoted by $[\alpha ]_r$.

$\beta _1$ defines a class in $E_n$. As explained in [1, §14, p.161ff] this is equivalent to the existence of a zig-zag of length $n$, that is, a collection of elements $\beta _2, \dots , \beta _{n}$ such that

$$\begin{aligned} \beta _i\in E_0^{p+i, q-i},\quad {\bar{\partial }} \beta _1=0, \quad \partial \beta _{i-1}+{\bar{\partial }} \beta _i=0 \quad (i=2, \dots n). \end{aligned}$$

Consider the following differential forms $\beta _k$ of bidegree $(k-1, n-k)$:

$$\begin{aligned} \beta _2&= \omega _2 \wedge d\bar{z}_2 \wedge \dots \wedge \bar{z}_{n-1}\\ \beta _k&= dx_1 \wedge \dots \wedge dx_{k-2} \wedge \omega _k\wedge d\bar{z}_k \wedge \dots \wedge \bar{z}_{n-1} \qquad (3\le k\le n-1)\\ \beta _n&= dx_1 \wedge \dots \wedge dx_{n-2} \wedge \omega _n \end{aligned}$$

A simple calculation shows that

$$\begin{aligned}&\displaystyle {\bar{\partial }} \beta _1=0,\nonumber \\&\displaystyle \partial \beta _1=-dy_1\wedge d\bar{z}_1\wedge \dots \wedge d\bar{z}_{n-1}=-{\bar{\partial }} \beta _2, \end{aligned}$$

and for $2\le k\le n-1$

$$\begin{aligned} \partial \beta _{k}= (-1)^{k-2} dx_1\wedge \dots \wedge dx_{k-1}\wedge dy_k\wedge d\bar{z}_k\wedge \dots \wedge d\bar{z}_{n-1}=-{\bar{\partial }} \beta _{k+1}. \end{aligned}$$

Therefore these elements define a zig-zag and $\beta _1$ defines a class in $E_n^{0,n-1}$.

It remains to prove that

$$\begin{aligned} d_n[\beta _1]_n=[\partial \beta _n]_n = (-1)^{n-2}[dx_1\wedge \dots \wedge dx_{n-1}\wedge dy_n]_n \end{aligned}$$

defines a non-zero class in $E_n^{n,0}$, or equivalently, that $\beta _1$ does not live to $E_{n+1}$. In other words, we have to prove that does not exist a zig-zag of length $n+1$ for $\beta _1$. Since we are in a first quadrant double complex we have $E_0^{n, -1}=0$ and there exists a zig-zag of length $n+1$ if and only if there exists a zig-zag $(\beta _1, \beta _2', \dots , \beta _n')$ of length $n$ such that $\partial \beta _n'= 0$.

To see that this cannot happen we put

$$\begin{aligned} U_1 \!=\! \langle dx_1, \dots , dx_{n-1},dy_1, \dots , dy_n, dz_1, \dots , dz_{n-1}\rangle _{\mathbb C}\quad \text {and}\quad U_2 \!=\! \langle \omega _1,\dots , \omega _n\rangle _{\mathbb C}, \end{aligned}$$

such that $U = U_1\oplus U_2$. The above basis of $U$ and its complex conjugate induce a basis on each exterior power and a decomposition

$$\begin{aligned} \Lambda ^{n} (U\oplus \bar{U}) = \Lambda ^n(U_1\oplus \bar{U}_1) \oplus S^n, \end{aligned}$$

where $S^n$ is spanned by wedge products of basis elements, at least one of which is in $U_2\oplus \bar{U}_2$.

The elements $\beta _k$ and $(-1)^{k}\partial \beta _k$ are basis vectors and we decompose

$$\begin{aligned} E^{k-1, n-k}_0 = \beta _k{\mathbb C}\oplus V_k\quad \text {and} \quad E^{k, n-k}_0 = \partial \beta _k{\mathbb C}\oplus W_k, \end{aligned}$$

where $V_k$ (resp. $W_k$) is spanned by all other basis elements of type $(k-1, n-k)$ (resp. $(k, n-k)$). Let $\xi _k$ be the element of the dual basis such that $\xi _k\lrcorner \partial \beta _k = 1$ and the contraction with any other basis element is zero.

The differentials $\partial$ and ${\bar{\partial }}$ respect this decomposition, in the sense that

$$\begin{aligned} \partial (V_k)\subset W_k\quad \text {and} \quad {\bar{\partial }}(V_k)\subset W_{k-1}. \end{aligned}$$

(1)

More precisely, let $\alpha$ be one of the forms in our chosen basis for $\Lambda ^{n-1}(U\oplus \bar{U})$. Recall that $\alpha$ is a decomposable form. Then

$$\begin{aligned} \partial \alpha \notin W_k&\iff \xi _k \lrcorner \partial \alpha \ne 0 \iff \alpha =\beta _k,\\ {\bar{\partial }} \alpha \notin W_k&\iff \xi _k \lrcorner {\bar{\partial }} \alpha \ne 0 \iff \alpha =\beta _{k+1}, \end{aligned}$$

which implies (1). If $\alpha \in \Lambda ^{n-1}(U_1\oplus \bar{U}_1)$ then $d \alpha = 0$ and the claim is trivial. If the form $\alpha$ contains at least two basis elements of $U_2\oplus \bar{U}_2$ then each summand of $d\alpha$ with respect to the basis contains at least one element of $U_2\oplus \bar{U}_2$, in other words, $d\alpha \in S^n$. Since $\xi _k\lrcorner S^n = 0$ the claim is true also for those elements.

The remaining elements of the basis are of the form $\pm \alpha '\wedge \omega _i$ or $\pm \alpha '\wedge \bar{\omega }_i$ for some $\alpha '$ in our chosen basis for $\Lambda ^{n-2} (U_1\oplus \bar{U}_1)$. Paying special attention to the counting variable $dy_k$ this case is easily checked by looking for solutions of the equation

$$\begin{aligned} \partial (\alpha '\wedge \omega _i)&= (-1)^{n-1} \alpha '\wedge \partial \omega _i\\&= (-1)^{k-2} dx_1\wedge \dots \wedge dx_{k-1}\wedge dy_k\wedge d\bar{z}_k\wedge \dots \wedge d\bar{z}_{n-1}= \partial \beta _k , \end{aligned}$$

and similarly in the other cases involving either ${\bar{\partial }}$ or $\bar{\omega }_i$.

Thus if $(\beta _1, \beta _2', \dots , \beta _n')$ is any zig-zag of length $n$ for $\beta _1$ then $\beta _k' \equiv \beta _k \mod V_k$ by (1) and, in particular,

$$\begin{aligned} \partial \beta _n' \equiv \partial \beta _n \not \equiv 0 \mod W_n. \end{aligned}$$

Thus $\beta _1$ does not live to $E_{n+1}$ and $d_n[\beta _1]_n$ is non-zero as claimed. $\square$

Remark 2

In the original publication we constructed a compact complex manifold in a very similar way and an element $[\beta _1]_n \in E_n^{0, n-1}$. However, our claim that $d_n([\beta _1]_n)\ne 0$ was wrong: while the constructed zig-zag could not be extended, the sequence of elements (in the notation of [original publication, Lem. 2])

$$\begin{aligned} (\beta _1, dx_1\wedge \bar{\omega }_2\wedge d\bar{x}_3 \wedge \dots \wedge d \bar{x}_{n-1}, 0,0,\dots ) \end{aligned}$$

gives an infinite zig-zag for $\beta _1$. In other words, the element considered gives an element of $E_\infty$ and thus a de Rham cohomology class.