Examples of Compact Simply Connected Holomorphic Symplectic Manifolds Which Are Not Formal

Abstract

In this paper, we prove that the complex four dimensional compact holomorphic symplectic manifold we found earlier is not formal. This gives another strong consequence that it is not a topological Kähler manifold. We also conjecture that this is true for the higher dimensional ones.

1. Introduction

Every Kähler structure on a smooth manifold also provides a symplectic structure on that manifold. On the other hand, examples have been known for some time of symplectic manifolds that do not admit any Kähler structure [1,2]. Much work has been conducted probing this difference between Kähler versus symplectic manifolds. A major step was taken by D. McDuff [3], who gave the first examples of simply connected symplectic manifolds that do not admit any Kähler structure. Her example was of a real dimension ten. In both Thurston’s and McDuff’s examples, the criterion used to determine non-admittance of a Kähler structure was cohomological.

At a deeper level than cohomology, the rational homotopy structure of Kähler manifolds was elucidated in [4], where it was shown that compact, simply connected Kähler manifolds are formal. In [4], the authors apply the work of the fourth author on relating the real homotopy type of a compact manifold to its algebra of differential forms [5,6,7] to Kähler manifolds. Using the full strength of Hodge theory, the authors show in particular that the real homotopy type of a simply connected compact Kähler manifold M is entirely determined by its cohomology ring [8].

Formality for a simply connected manifold M means that its rational homotopy type is determined by its cohomology algebra. Let us recall its definition (see [4,9] for more details). Let X be a simply connected smooth manifold and consider its algebra of differential forms $({\mathrm{\Omega }}^{*}\left(X\right),d)$. Let $\varphi :(\wedge V,d)\to ({\mathrm{\Omega }}^{*}\left(X\right),d)$ be a minimal model for this algebra [4]. Then X is formal if there is a quasi-isomorphism $\varphi :(\wedge V,d)\to (H^{*}\left(X\right),d=0)$, i.e., a morphism of differential algebras, inducing the identity on cohomology.

In [10,11] (see also [12]), we have found irreducible compact holomorphic symplectic manifolds of dimension $2n$ for $n\ge 2$ which do not admit any Kähler structure. Those manifolds are simply connected. Actually, on those manifolds, there is a quadratic form on the second cohomology such that the n-th power of it is proportional to the 2n-th product of the element (in [10]). This quadratic form has a kernel generated by an element $\mathbf{b}$ ($\beta$ in [11] or the element $\gamma$ in [10]). Therefore, $\mathbf{b}$ is in the kernel of the 2-Lefschetz map with any element (or symplectic structure). That is, the Lefschetz condition always fails for any given real symplectic structure. In particular, when $n=2$, our example actually gave the first example of a compact simply connected real symplectic manifold of dimension eight which is not Kähler. Many four- and six-dimensional examples were eventually found later on by Gompf [13].

Motivated by this construction from [10,11], M. Fernandez and V. Muñoz [14], as mentioned in their paper, found an eight-dimensional compact simply connected real symplectic manifold which is not formal after [15] solved the same problem for manifolds with dimensions greater or equal to ten.

In this paper, we find that

Theorem 1. 

When $n=2$, our original eight-dimensional manifold is also not formal.

For any n, there are two other elements, a and c, such that $a{b}^{n-1},b^n,c{b}^{n-1}$ are exact. Therefore, we can define the $b^{n-1}$ relative Massey product $d=(b^{n-1}:a,b,c)$ as in [16]. We proved that for $n=2$, d is nonzero. Therefore, we proved that when $n=2$ our original simply connected real eight-dimensional holomorphic symplectic manifold is not formal.

Let M be a complex manifold of dimension $2n$. A holomorphic symplectic structure or form is a closed holomorphic 2-form $\omega$ on M with maximal rank (see [17], p. 47).

People might ask under what condition a compact holomorphic symplectic manifold is Kählerian, i.e., it admits a positive closed (1,1) form. For example, by [18,19], we know that every K-3 surface is Kählerian. In [20], Todorov asked if every irreducible compact holomorphic symplectic manifold of a dimension more than four is Kählerian. Some counterexamples have been found in [10,11]. Those are the manifolds we deal with in this paper.

2. Preliminary

2.1. On Compact Kähler Holomorphic Symplectic Manifolds

Here, we collect some results on compact Kähler holomorphic symplectic manifolds, i.e., compact complex manifolds with both holomorphic symplectic structures and Kähler structures. From [21,22,23], we have the following:

Proposition 1. 

Let X be a compact Kähler holomorphic symplectic manifold. Then, M admits a hyperkähler structure, i.e., there is a Kähler structure which is Ricci flat and the holomorphic symplectic structure is parallel with regard to this Kähler structure. In addition, there exists a finite unramified Galois covering $\tilde{X}⟶X$ such that $\tilde{X}$ is isomorphic to a product $T\times Y_1\times \cdots \times Y_r$, where T is a complex torus and $Y_i,1\le i\le r$, are simply connected Kähler symplectic manifolds with $h^{2k,0}\left(Y_i\right)=1$ and $h^{2k-1,0}\left(Y_i\right)=0$ for any $k\ge 1$. Here, the direct factors are uniquely determined by X up to permutation. Moreover, if for each i we let $g_i\left(x\right)$ be the homogeneous form of degree $2n_i(n_i={dim}_{\mathbf{C}}{Y}_i)$ on $H^2(Y_i,\mathbf{Q})$ defined by $g_i\left(x\right)=x^{2n_i}\left[Y_i\right],x\in H^2(Y_i,\mathbf{Q})$ where $\left[Y_i\right]$ denotes the evaluation on $Y_i$, then there exists a constant $c\in {\mathbf{Q}}^{+}$ and a nondegenerate quadratic form $f_i$ of signature $(3,b_i-3)$ on $H^2(Y_i,\mathbf{Q})$ such that $g_i=c{f}_i^{n_i}$, where $b_i=b_2\left(Y_i\right)$. If we let ${\phi }_i$ be a holomorphic 2-form on $Y_i$ such that ${\left({\phi }_i{\overline{\phi }}_i\right)}^{n_i}\left[Y_i\right]=1$, then $f_i$ can be written as

$${\displaystyle \frac{n_i}{2}}{\left({\phi }_i{\overline{\phi }}_i\right)}^{n_i-1}{x}^2\left[Y_i\right]+(1-n_i){\phi }_i^{n_i-1}{\overline{\phi }}_i^{n_i}x\left[Y_i\right]·{\phi }_i^{n_i}{\overline{\phi }}_i^{n_i-1}x\left[Y_i\right].$$

up to a multiple of a constant.

This, for example, comes from [22], Theorem 5 in [21] and Theorem 4.7 in [23] (see also [11,17]). We gave a simpler proof and a generalization for the last part (or Fujiki Theorem [23]) of this proposition in Section 5 in [10].

2.2. Compact Holomorphic Symplectic Surfaces

It is well known that every simply connected holomorphic symplectic surface is a K-3 surface and its second Betti number is 22. There are three different classes of holomorphic symplectic surfaces, i.e., K-3 surface K, complex torus A and Kodaira–Thurston surface S (see [24], p. 188). The holomorphic symplectic 2-forms come from any trivial canonical sections. We are more interested in the surface S here. We know that $b_1\left(S\right)=3,b_2\left(S\right)=4$. Let $K^{\left[r\right]},A^{\left[r\right]}$ and $S^{\left[r\right]}$ are the Hilbert scheme which parameterizes the finite subsets Z with $\left|Z\right|=r$ (see [21,25,26,27]). Then,

Proposition 2 

(Cf. [21,28]). $K^{\left[r\right]},A^{\left[r\right]}$ and $S^{\left[r\right]}$ are compact holomorphic symplectic manifolds and $b_1\left(K^{\left[r\right]}\right)=0,b_1\left(A^{\left[r\right]}\right)=4,b_1\left(S^{\left[r\right]}\right)=3$. Moreover, $K^{\left[r\right]},A^{\left[r\right]}$ are Kähler and $S^{\left[r\right]}$ is not Kähler.

$S^{\left[r\right]}$ is not Kähler since $b_1\left(S^{\left[r\right]}\right)$ is not even.

2.3. Compact Parallelizable Manifolds

A parallelizable manifold is the quotient of a real Lie group by a discrete subgroup. It is a solvmanifold or nilmanifold according to whether the Lie group is either solvable or nilpotent. We have the following Nomizu’s Theorem [29]:

Proposition 3. 

Let $M=G/H$ be a compact parallelizable nilmanifold, then the deRham cohomology can be calculated by the complex of G-right invariant forms, which is isomorphic to the complex of the Lie algebra.

Notice that if G has a G-right invariant complex structure, which induces a complex structure on M, there is a question on the Dolbeault cohomology part. However, in [10,11], similarly in this paper, we only deal with the case in which we apply a similar version to this proposition to S, which is a complex torus over a complex torus. That is, our complex structures on the nilmanifolds are rational in [30], similarly to the nilmanifolds in this paper. Actually, we do not need the Dolbeault cohomology in this paper. Therefore, there is an advantage of the method in this paper to [10,11].

A Kodaira–Thurston surface is a nilmanifold with a complex structure that comes from a right G-invariant complex structure on G. For example, we consider $S_r=G/H_r$ with

$$G=\{(x,y)=\left(\begin{array}{ccc}1& x& y\\ 0& 1& \overline{x}\\ 0& 0& 1\end{array}\right)|x,y\in \mathbf{C}\}$$

and

$$H_r=\{(x,y)\in G|x\in r(\mathbf{Z}+\mathbf{Z}i),y\in r(\mathbf{Z}+\mathbf{Z}i)\}$$

where $r\in \mathbf{Z}$. They admit a holomorphic symplectic structure $\omega =dx\wedge dy$. There is a covering $M=G/T$ of $S_r$ with $T=\{(0,y)\in H_r\}$. On M, the function x defines the Hamiltonian vector field $X_x=\partial /\partial y$ by $dx=\omega (,X_x)$. This structure induces a holomorphic symplectic reduction in $S_r^{\left[r\right]}$ (see [11]).

Proposition 4 

(Cf. [11,21]). Let R be either $A^{\left[r\right]},A=T\times T$ with T a torus, or $S_r^{\left[r\right]}$; let L be either the torus which is the left torus in the definition of A, or the center of S (which is generated by $y)$. Then, the L action on R which is induced by the diagonal action of L on the product of L is a subgroup of the holomorphic symplectic automorphism group. If $R_r$ is the symplectic quotient of R under L, then $R_r$ is a holomorphic symplectic orbifold and there is a $r^2$ to 1 covering from a compact simply connected holomorphic symplectic manifold $R_{\left[r\right]}$ to $R_r$. Furthermore, $R_{\left[2\right]}$ is a K-3 surface. If $r>2$, we have that in the case of A, $b_2\left(R_{\left[r\right]}\right)=7$ and $R_{\left[r\right]}$ is Kähler; in the case of $S_r$, $b_2\left(R_{\left[r\right]}\right)=6$ and $R_{\left[r\right]}$ is not Kähler.

3. General Construction from Parallelizible Manifolds

Now, following [10], we start to construct some examples of simply connected compact holomorphic symplectic manifolds from parallelizible manifolds. We call a parallelizible manifold a parallelizible holomorphic symplectic manifold if it admits a holomorphic symplectic structure which is induced by a right invariant 2-form on G. As one can see from the last part of the proof of the Theorem A in [31], these manifolds must be solvmanifolds:

Proposition 5. 

Every compact parallelizible symplectic manifold is a solvmanifold.

The method which we use here (from [10]) generalizes the method in [11]. Starting from a holomorphic symplectic nilmanifold (or solvmanifold) M (see [10] for more examples of them), we try to find a faithful representation of a finite group $S$ as a subgroup (we denote this group also by $S$ if there is no confusion) of the automorphism of the Lie group which preserves the isotropic group and the complex structure as well as the holomorphic symplectic structure. One might find this easily by the condition that the complex structure equation is preserved by $S$. Suppose that the subset $D=\{m\in M|s\left(m\right)=m\mathrm{for}\mathrm{some}s\in S\}$ has only codimension 2 irreducible components and ${\left({\pi }_1\left(M\right)\right)}^S=1$, then a desingulation of the quotient $M/S$ is probably a simply connected holomorphic symplectic manifold which is not Kählerian. For example, we can construct the examples $R_{\left[r\right]}$ in [11] directly from the nilmanifold $Q^{\left[r\right]}$ there.

Instead of $Q^{\left[r\right]}$, we consider $M_{r,t}$ with structure equation

$$\begin{array}{c}d{x}_i=0\hfill \\ d{y}_i=-(r-1)t{x}_i\wedge {\overline{x}}_i+2t\sum _j{x}_j\wedge {\overline{x}}_j+t\sum _{j\ne k}{x}_j\wedge {\overline{x}}_k\hfill \end{array}$$

where $(x_1,x_2,\cdots ,x_r,y_1,y_2,\cdots ,y_r)$ are the coordinates of the Lie algebra of $M_{r,t}$, $r\in \mathbf{N}t\in \mathbf{Z}$. It is not difficult to check that $Q^{\left[r\right]}=M_{r-1,1}$.

Now we define the symmetric group $S_{r+1}$ action on $M_{r,t}$. $S_{r+1}$ is generated by ${\tau }_{(1,r+1)}=(1,r+1)$ and $S_r$. The group $S_r$ acts on $M_{r,t}$ just as $S_r$ and

$$\begin{array}{c}\hfill {\displaystyle {\tau }_{(1,r+1)}(x_1,x_2,\cdots ,x_r,y_1,y_2,\cdots ,y_r)=}\hfill \\ \hfill (-x_1-x_2-\cdots -x_r.x_2,\cdots ,x_r,\hfill \\ \hfill -y_1-y_2-\cdots -y_r,y_2,\cdots ,y_r).\hfill \end{array}$$

It is not difficult to check that $S_{r+1}$ preserves the structure equations and the set $D_{r,t}$ has only codimension 2 irreducible components. Hence, we can desingular $M_{r,t}/S_{r+1}$ as in [11], which is achieved through the resolution provided by the Hilbert schemes for the singularities of the symmetry product, and we denote the desingulation by $Q_{r,t}$; then, we have the following:

Proposition 6 

(Cf [11], Theorem 1). The desingulation $Q_{r,t}$ of $M_{r,t}/S_{r+1}$ is a compact simply connected holomorphic symplectic manifold which is a K-3 surface if $r=1$ and is not Kählerian with $b_2=6$ if $r>1$.

The simply connectedness basically comes from

$${\pi }_1\left(Q_{r,t}\right)={\left({\pi }_1\left(M_{r,t}\right)\right)}^{S_{r+1}}=0.$$

See also [12].

An alternative proof of the nonkählerness comes from the fact that the pair of $H^2\left(M_{r,t}\right)$ and $H^2\left(M_{r,t}\right)\wedge {\omega }^{r-1}\wedge {\overline{\omega }}^{r-1}$ (or $H^2\wedge {\omega }^{2r-2}$ for any given real symplectic structure $\omega$) is not a perfect pair.

The possible closed (1, 1) $S_{r+1}$-invariant classes are linear combinations of $\alpha =a{\sum }_j$$y_j\wedge {\overline{y}}_j+b{\sum }_{j\ne k}{y}_j\wedge {\overline{y}}_k$ and $\beta =c{\sum }_j{x}_j\wedge {\overline{y}}_j+d{\sum }_{j\ne k}{x}_j\wedge {\overline{y}}_k$, $\delta =g{\sum }_j{y}_j\wedge {\overline{x}}_j+h{\sum }_{j\ne k}{y}_j\wedge {\overline{x}}_k$, $\gamma =e{\sum }_j{x}_j\wedge {\overline{x}}_j+f{\sum }_{j\ne k}{x}_j\wedge {\overline{x}}_k$. Since they are invariant under the action of ${\tau }_{(1,r+1)}$, we have $a=2b,c=2d,e=2f,g=2h$. Furthermore, from

$$\overline{\partial }\alpha =\overline{\partial }\beta =\overline{\partial }\gamma =\overline{\partial }\delta =0$$

we obtain $a=0$, that is, $\alpha =0$. There is no $S_{r+1}$ invariant $(1,0)$ form, and we find that ${\left(H^{1,1}\left(M_{r,t}\right)\right)}^{S_{r+1}}$ is generated by $\beta ,\delta ,\gamma$ with $c=e=g=2$. We see that $H^{1,1}\left(Q_{r,t}\right)$ has dimension $3+1=4$. In the same way, we find that $H^{2,0}\left(Q^{r,t}\right)$ is generated by $\omega =2{\sum }_j{x}_j\wedge y_j+{\sum }_{j\ne k}{x}_j\wedge y_k$ and $H^{0,2}\left(Q_{r,t}\right)$ is generated by $\overline{\omega }$. We see that $H^2(Q_{r,t},\mathbf{C})=H^{2,0}+H^{1,1}+H^{0,2}$.

We let $2b=i\gamma$, then b is the $\beta$ in [11]. By [10], even in the nonkähler case with the property that $H^2=H^{2,0}+H^{1,1}+H^{0,2}$ we still have the topological quadratic and b is in the kernel.

In the rest of this paper, we assume that $t=1$.

We let $4a=\beta +\omega -\delta +\overline{\omega }$. Then a is the same as the ${\sum }_j{z}_j\wedge u_j$ in [11].

We also let $4ic=\omega -\beta -\delta -\overline{\omega }$. Then c is the same as the ${\sum }_j{z}_j\wedge v_j$ in [11].

Similarly, when $r=2$, one could calculate that the orbifold contribution of the de Rham cohomology in $H^3$ only comes from

$$\sum _{i\ne j}{x}_i\wedge ((y_j+{\overline{y}}_j)\wedge {\overline{x}}_i+(y_1+y_2+{\overline{y}}_1+{\overline{y}}_2)\wedge {\overline{x}}_j).$$

4. Several Massey Products

4.1. Our Main Result

Theorem 2. 

Let $(M,\omega )$ be the compact simply connected holomorphic symplectic manifold constructed above complex dimension $2n$, for the technic reason we assume that $t=1$, then 1. $b^{n-1}a,b^n,b^{n-1}c$ are exact. 2. When $n=2$, the quadruple Massey product $(a,b,b,c)$ is nonzero. 3. When $n=2$, the relative Massey product $(b:a,b,c)$ is nonzero. 4. $M^2$ is nonformal.

We notice that our notation of both the Massey quadruple product and the relative b Massey product is a little bit different from those in [14,16]. The reason is that our definitions (see later in this section) are more of a form than a class although they do represent a cohomology class.

4.2. Some Calculations

For the definitions of Massey quadruple products and the relative Massey products, one could check with [14]. We shall also provide some simple definitions later on in the process of our proof of this Theorem.

Since we work on $H^2$ instead of the Hodge cohomology, we shall retain the notation from [11] instead of that from [10] in this section.

To start with, we have

$$b=2\sum _j{z}_j\wedge w_j+\sum _{j\ne k}{z}_j\wedge w_k$$

$$a=\sum _j{z}_j\wedge u_j,c=\sum _j{w}_j\wedge u_j.$$

$$d{z}_i=d{w}_i=d{u}_i=0,d{v}_i=-2(\sum _{j\ne i}{z}_j\wedge w_j+\sum _{j\ne k}{z}_j\wedge w_k).$$

The symmetry group $S_{n+1}$ is generated by $S_n$ and $(1,n+1)$ with $(1,n+1)$:

$$(z_1,w_1)\to (-z_1-z_2-\cdots -z_n,w_1-w_2-\cdots -w_n)$$

$$(z_i,w_i)\to (z_i,w_i)i>1$$

$$(u_1,v_1)\to (-u_1,-v_1)$$

$$(u_i,v_i)\to (u_i-u_1,v_i-v_1)i>1.$$

We first check that

$$b^n=(n+1)!{\wedge }_i(z_i\wedge w_i)$$

$$b^{n-1}=(n-1)!(n\sum _j\left({\wedge }_{i\ne j}(z_i\wedge w_i)\right)-\sum _{j\ne k}\left({\wedge }_{i\notin (j,k)}(z_i\wedge w_i)\right)\wedge z_j\wedge w_k)$$

$$b^{n-1}a=(n-1)!\sum _j\left({\wedge }_{i\ne j}(z_i\wedge w_i)\right)\wedge z_j\wedge (n{u}_j-\sum _{k\ne j}{u}_k)$$

Let $\delta ={\sum }_j\left({\wedge }_{i\ne j}(z_i\wedge w_i)\right)\wedge {\sum }_{k\ne j}{v}_k$, then

$$d\delta =-2n(n-1){\wedge }_j(z_j\wedge w_j)=A$$

let $\gamma ={\sum }_{k\ne j}\left({\wedge }_{i\notin (j,k)}(z_i\wedge w_i)\right)\wedge z_j\wedge u_j\wedge ((n-2)v_k-n{v}_j)$, then

$$d\gamma =2(n-1)\sum _j\left({\wedge }_{i\ne j}(z_i\wedge w_i)\right)\wedge z_j\wedge (n{u}_j-\sum _{k\ne j}{u}_k)=B$$

Combining these calculations we have

$$d\overline{\delta }=A,d\overline{\gamma }=B$$

with $\overline{\delta }$ (or $\overline{\gamma }$) the average of $\delta$ (or $\gamma$). This concludes the proof of the exact property of $b^n$ and $b^{n-1}a$. Similarly for $b^{n-1}c$, we exchange z and w. Then we have 1. in our Theorem.

In practice, to calculate the average, we notice that $S_{n+1}={\cup }_{i=1}^{n+1}S\left(i\right)$ with $S\left(i\right)=(n+1,i)S_n$.

4.3. Massey Quadruple Products

We define the Massey products by the following: If $d{a}_{i,j}={\sum }_{k=0}^{j-i-1}{\overline{a}}_{i,i+k}\wedge a_{i+k+1,j}$ for all $1\le i<j\le 4$ with $(i,j)\ne (1,4)$ and $\overline{s}={(-1)}^{deg s+1}s$, then

$$(a_{1,1},a_{2,2},a_{3,3},a_{4,4})={\overline{a}}_{1,1}\wedge a_{2,4}+{\overline{a}}_{1,2}\wedge a_{3,4}+{\overline{a}}_{1,3}\wedge a_{4,4}$$

is a cohomology class. Notice that our definition here is a form although it represents a cohomology class.

Now, we prove 2.: Let $a_{1,1}=2a,a_{2,2}=a_{3,3}=2b,a_{4,4}=2c$, and

$$\delta =\sum _{i=1}^2{z}_i\wedge (u_i\wedge v_i-u_i\wedge v_{i^{*}}-u_{i^{*}}\wedge v_i)$$

$$\omega =\sum _{i=1}^2{z}_i\wedge (w_i\wedge (v_i-2v_{i^{*}})-w_{i^{*}}\wedge (v_i+v_{i^{*}}))$$

$$\gamma =\sum _{i=1}^2{w}_i\wedge (u_i\wedge v_i-u_i\wedge v_{i^{*}}-u_{i^{*}}\wedge v_i)$$

where $\{i,i^{*}\}=\{1,2\}$. Then $a_{1,2}=-2\delta ,a_{2,3}=-2\omega ,a_{3,4}=-2\gamma$. $a_{1,3}=a_{2,4}=0$.

$$16(a,b,b,c)={\overline{a}}_{1,2}\wedge a_{3,4}=4\gamma \delta =V_{uv}b\ne 0.$$

where $V_{uv}=d{u}_1\wedge d{u}_2\wedge d{v}_1\wedge d{v}_2$. We notice that the nonzero property of our class does imply the nonformal property of the manifold. The reason is that 1. $a_{1,3}=a_{2,4}=0$, i.e., only the middle term in our formula is nonzero (as a form). 2. The property of the only nonzero invariant $H^3$ cohomology class from the orbifold in the last sentence of last section is in a different format from $\gamma$ and $\delta$—it only involves $z,w$ and u but not v, i.e., this is similar to the case in [14] in which $H^3=0$.

4.4. Relative Massey Product

We define the relative Massey product (following [16], page 580): If $d{b}_i=b\wedge a_i$ with $1\le i\le 3$, then

$$(b:a_1,a_2,a_3)=b_1\wedge b_2\wedge a_3+b_2\wedge b_3\wedge a_1+b_3\wedge b_1\wedge a_2.$$

Now we prove 3. We notice that from [16] page 581 Lemma 2.6, if $(a_1,b,a_2)=(a_2,b,a_3)$$=(a_1,b,a_3)=0$ as cohomology classes, then the given relative Massey product as a cohomology class does not depend on the choice of $b_i$.

Let $a_1=2a,a_2=2b,a_3=2c$, then $b_1=-\delta ,b_2=-\omega ,b_3=-\gamma$. Other than what we have above, we have $(a_1,b,a_3)=d\left(V_{uv}\right)$.

$$(b:2a,2b,2c)=-12V_{zw}{V}_{uv}=-12V\ne 0,$$

where $V_{zw}=d{z}_1\wedge d{z}_2\wedge d{w}_1\wedge d{w}_2$.

From 2. or 3. we obtain 4.