Erratum to the paper: Integral cohomology of the Generalized Kummer fourfold

TL;DR

This paper corrects a previous proof regarding the integral cohomology of generalized Kummer fourfolds, establishing that it is torsion free in dimension 4.

Contribution

It provides a corrected proof confirming the torsion-free nature of the integral cohomology for generalized Kummer fourfolds.

Findings

Integral cohomology of generalized Kummer fourfolds is torsion free.

Corrected proof of the previous theorem in dimension 4.

Clarification of the cohomological properties of these fourfolds.

Abstract

We provide a correct proof of arXiv:1607.03431, Theorem 5.2 in dimension 4. More precisely, we show that the integral cohomology of the generalized Kummer fourfold is torsion free.

Equations

$$\mathrm{tors}{H}^4=\mathrm{tors}{H}_4$$

$$W_{\tau }\simeq \mathbb{P}(1,1,3).$$

$$U$$

$$H^3(K_2(A),Z)=H^3(U^{\prime },Z)\text{and}{H}^5(K_2(A),Z)=H^5(U^{\prime },Z).$$

$$H^3(U^{\prime },U,Z)\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$\mathrm{tors}{H}^3(U^{\prime },Z)=\mathrm{tors}{H}^3(U,Z)\text{and}\mathrm{tors}{H}^5(U^{\prime },Z)\hookrightarrow \mathrm{tors}{H}^5(U,Z).$$

$$V:=(A\times A)\setminus A[3],$$

$$\Delta$$

$$U\simeq \tilde{V}/{\mathfrak{S}}_3.$$

$$S^{\prime }:=\left\{\xi \in A^{[2]}|\mathrm{Supp}\xi =\left\{x,-2x\right\},x\in A\right\}.$$

$${\Sigma }_{\tau }:=r^{-1}\left(\left\{\xi \in A^{[2]}|\mathrm{Supp}\xi =\left\{\tau \right\}\right\}\right).$$

$$\tilde{\mathcal{W}}:=\widetilde{A^{[2]}}\setminus \bigcup _{\tau \in A[3]}{\Sigma }_{\tau }.$$

$$\tilde{\mathcal{W}}\simeq \tilde{V}/{\mathfrak{S}}_2.$$

$$\tilde{V}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$\pi :\tilde{\mathcal{W}}\to U.$$

$$\mathrm{tors}{H}^{\ast }\left(\widetilde{A^{[2]}},Z\right)=0.$$

$$H^3(\widetilde{A^{[2]}},\tilde{\mathcal{W}},Z)\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$H^k(\widetilde{A^{[2]}},\tilde{\mathcal{W}},Z)\simeq \underset{\tau \in A[3]}{⨁}{H}^{k-4}({\Sigma }_{\tau },Z).$$

$$\tilde{V}/{\mathfrak{A}}_3\to U.$$

$$\left(\begin{array}{c}1\end{array}\right),\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)\text{and}\left(\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 0& 0& 1\end{array}\right).$$

$${\ell }_q^k(X)={\ell }_q(H^k(X,Z)).$$

$$H^k(A\times A,Z)\simeq {\wedge }^k{H}^1(A\times A,Z),$$

$$\{\begin{array}{l}(1)=(N_3\oplus N_1)\otimes (N_3\oplus N_1)=\stackrel{N_3^3}{\overbrace{N_3\otimes N_3}}\oplus N_3^2\oplus N_1=N_3^5\oplus N_1,\\ (2)=N_1,\\ (3)=N_1\otimes (N_1\oplus N_3)=N_1\oplus N_3.\end{array}$$

$$H^{\ast }(\tilde{V},Z)\simeq H^{\ast }(V,Z)\oplus t\cdot H^{\ast }(\overline{\Delta },Z),$$

$$\mathrm{tors}{H}^3(\tilde{V}/{\mathfrak{A}}_3,Z)=\mathrm{tors}{H}^3(V/{\mathfrak{A}}_3,Z)\oplus \mathrm{tors}{H}^1(\overline{\Delta },Z)=0,$$

$$\mathrm{tors}{H}^5(\tilde{V}/{\mathfrak{A}}_3,Z)=\mathrm{tors}{H}^5(V/{\mathfrak{A}}_3,Z)\oplus \mathrm{tors}{H}^3(\overline{\Delta },Z)=\mathrm{tors}{H}^3(\overline{\Delta },Z).$$

$$0\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

Full text

[TABLE]

Thus, it suffices to prove that $H^{3}(K_{2}(A),\operatorname{\mathbb{Z}})$ and $H^{5}(K_{2}(A),\operatorname{\mathbb{Z}})$ are torsion free. Moreover, since Theorem 1 is only a topological result, without loss of generality, we can assume that $A$ is an abelian surface.

Let $W_{\tau}\subset K_{2}(A)$ be the locus of subschemes supported at $\tau\in A[3]$. As it is explained in [6, Section 4], we have:

[TABLE]

Let $pt_{\tau}\in W_{\tau}$ be the singular point and $W_{\tau}^{*}:=W_{\tau}\smallsetminus pt_{\tau}$. Put

[TABLE]

[TABLE]

Moreover:

[TABLE]

By Thom’s isomorphism, $H^{3}(U^{\prime},U,\operatorname{\mathbb{Z}})=0$ and $H^{4}(U^{\prime},U,\operatorname{\mathbb{Z}})=\bigoplus_{\tau\in A[3]}H^{0}(W_{\tau}^{*},\operatorname{\mathbb{Z}})$ is torsion free. Furthermore, $H^{5}(U^{\prime},U,\operatorname{\mathbb{Z}})=\bigoplus_{\tau\in A[3]}H^{1}(W_{\tau}^{*},\operatorname{\mathbb{Z}})=0$ since, by (1), $W_{\tau}^{*}$ is the quotient of $\mathbb{P}^{2}\smallsetminus\left\{pt\right\}$ by an automorphism of order 3. It follows from (3):

[TABLE]

Then (2) concludes the proof. ∎

Hence, it remains to prove that $H^{3}(U,\operatorname{\mathbb{Z}})$ and $H^{5}(U,\operatorname{\mathbb{Z}})$ are torsion free. To do so, we consider

[TABLE]

where $A[3]$ is embedded in $A\times A$ diagonally: $A[3]\hookrightarrow A\times A:x\rightarrow(x,x)$.

Let $r:\widetilde{V}\rightarrow V$ be the blow-up of $V$ in

[TABLE]

As explained in [2, Section 7], we have:

[TABLE]

[TABLE]

The surface $S^{\prime}$ is isomorphic to the blow-up of $A$ in $A[3]$. We consider $r:\widetilde{A^{[2]}}\rightarrow A^{[2]}$ the blow-up of $A^{[2]}$ in $S^{\prime}$. For $\tau\in A[3]$, we denote

[TABLE]

The surfaces $\Sigma_{\tau}$ are Hirzebruch surfaces. We also consider

[TABLE]

We have:

[TABLE]

Indeed, if we denote by $\widetilde{A\times A}$ the blow-up of $A\times A$ in the diagonal, it is well known that $\widetilde{A\times A}/\mathfrak{S}_{2}\simeq A^{[2]}$. For $\tau\in A[3]$, we denote $\ell_{\tau}:=\left\{\left.\xi\in A^{[2]}\right|\ \operatorname{Supp}\xi=\left\{\tau\right\}\right\}$. Then, if we consider $V_{1}$ the blow-up of $V$ in $\Delta$, we have $V_{1}/\mathfrak{S}_{2}\simeq A^{[2]}\smallsetminus\bigcup_{\tau\in A[3]}\ell_{\tau}$. Therefore by [5, Corollary II 7.15], we obtain a commutative diagram:

[TABLE]

This provides (5). Then, it follows from (4) a triple cover:

[TABLE]

Hence, if we prove that $H^{3}(\widetilde{\mathcal{W}},\operatorname{\mathbb{Z}})$ and $H^{5}(\widetilde{\mathcal{W}},\operatorname{\mathbb{Z}})$ are torsion free, Lemma 2 will be proven by [1, Theorem 5.4 ].

We know that $H^{*}(A^{[2]},\operatorname{\mathbb{Z}})$ is torsion free from [9, Theorem 2.2]. Then, we deduce from [10, Theorem 7.31] (or from [4, Theorem 4.1] which is more general) that:

[TABLE]

Consider the exact sequence:

[TABLE]

By Thom’s isomorphism:

[TABLE]

Since $\Sigma_{\tau}$ is an Hirzebruch surface, $H^{3}(\widetilde{A^{[2]}},\widetilde{\mathcal{W}},\operatorname{\mathbb{Z}})=H^{5}(\widetilde{A^{[2]}},\widetilde{\mathcal{W}},\operatorname{\mathbb{Z}})=0$ and $H^{4}(\widetilde{A^{[2]}},\widetilde{\mathcal{W}},\operatorname{\mathbb{Z}})$, $H^{6}(\widetilde{A^{[2]}},\widetilde{\mathcal{W}},\operatorname{\mathbb{Z}})$ are torsion free. It follows from (6) and (7) that $H^{3}(\widetilde{\mathcal{W}},\operatorname{\mathbb{Z}})$ and $H^{5}(\widetilde{\mathcal{W}},\operatorname{\mathbb{Z}})$ are torsion free. ∎

We can also consider the double cover:

[TABLE]

Applying [1, Theorem 5.4 ] and Lemma 2, it is suffices to prove that $H^{3}(\widetilde{V}/\mathfrak{A}_{3},\operatorname{\mathbb{Z}})$ and $H^{5}(\widetilde{V}/\mathfrak{A}_{3},\operatorname{\mathbb{Z}})$ are torsion free to conclude the proof of Theorem 1.

First, we show that $\operatorname{tors}H^{3}(V/\mathfrak{A}_{3},\operatorname{\mathbb{Z}})=\operatorname{tors}H^{5}(V/\mathfrak{A}_{3},\operatorname{\mathbb{Z}})=0$. Since the action of $\mathfrak{A}_{3}$ on $V$ is free, it can be realized using the equivariant cohomology as explained in [8, Section 4]. The computation of the equivariant cohomology can be done using the Boissière–Sarti–Nieper-Wisskirchen invariants defined in [3, Section 2]. We recall their definition in our specific case. Let $T$ be a $3$-torsion-free $\operatorname{\mathbb{Z}}$-module of finite rank equipped with a linear action of $\mathfrak{A}_{3}=\left\langle\sigma_{1,2,3}\right\rangle$. We consider the action of $\mathfrak{A}_{3}$ on $T\otimes\mathbb{F}_{3}$. Then the matrix of the endomorphism $\sigma_{1,2,3}$ on $T\otimes\mathbb{F}_{3}$ admits a Jordan normal form. We can decompose $T\otimes\mathbb{F}_{3}$ as a direct sum of some $\mathbb{F}_{3}[\mathfrak{A}_{3}]$-modules $N_{q}$ of dimension $q$ with $1\leq q\leq 3$, where $\sigma_{1,2,3}$ acts on the $N_{q}$ in a suitable basis, respectively by matrices of the following form:

[TABLE]

[TABLE]

• (i)

$\ell_{1}^{1}(A\times A)=\ell_{3}^{1}(A\times A)=0$* and $\ell_{2}^{1}(A\times A)=4$;*

• (ii)

$\ell_{1}^{2}(A\times A)=10$, $\ell_{2}^{2}(A\times A)=0$ and $\ell_{3}^{2}(A\times A)=6$;

• (iii)

$\ell_{1}^{3}(A\times A)=0$, $\ell_{2}^{3}(A\times A)=16$ and $\ell_{3}^{3}(A\times A)=8$;

• (iv)

$\ell_{1}^{4}(A\times A)=19$, $\ell_{2}^{4}(A\times A)=0$ and $\ell_{3}^{4}(A\times A)=17$.

[TABLE]

for all $k$.

• (i)

For $a\in H^{1}(A,\operatorname{\mathbb{Z}})$ and a generator $A$ of $H^{0}(A,\operatorname{\mathbb{Z}})$, $\sigma_{1,2,3}^{*}(A\otimes a)=-a\otimes A,\ \text{and}\ \sigma_{1,2,3}^{*}(a\otimes A)=-A\otimes a+a\otimes A.$ It follows (i).

• (ii)

We have $\wedge^{2}(N_{2}^{4})=(\wedge^{2}N_{2})^{4}\oplus(N_{2}\otimes N_{2})^{6}=N_{1}^{4}\oplus(N_{3}\oplus N_{1})^{6}.$

• (iii)

We have $\wedge^{3}(N_{2}^{4})=(\underbrace{N_{2}\otimes N_{2}\otimes N_{2}}_{(1)})^{4}\oplus(\underbrace{\wedge^{2}N_{2}\otimes N_{2}}_{(2)})^{12}.$

Moreover, $\begin{cases}(1)=(N_{3}\oplus N_{1})\otimes N_{2}=N_{3}^{2}\oplus N_{2},\\ (2)=N_{1}\otimes N_{2}=N_{2}.\end{cases}$

• (iv)

We have $\wedge^{4}(N_{2}^{4})=\underbrace{N_{2}\otimes N_{2}\otimes N_{2}\otimes N_{2}}_{(1)}\oplus(\underbrace{\wedge^{2}N_{2}\otimes\wedge^{2}N_{2}}_{(2)})^{6}\oplus(\underbrace{\wedge^{2}N_{2}\otimes N_{2}\otimes N_{2}}_{(3)})^{12}.$

Moreover:

[TABLE]

∎

Since $\mathfrak{A}_{3}$ acts freely on $V$, the spectral sequence of equivariant cohomology provides that $H^{3}(V/\mathfrak{A}_{3},\operatorname{\mathbb{Z}})$ and $H^{5}(V/\mathfrak{A}_{3},\operatorname{\mathbb{Z}})$ are torsion free (see [8, Section 4] for a reminder about this spectral sequence). Moreover, $\widetilde{V}/\mathfrak{A}_{3}$ is the blow-up of $V/\mathfrak{A}_{3}$ in the image $\overline{\Delta}$ of $\Delta$ in $V/\mathfrak{A}_{3}$. Since $V/\mathfrak{A}_{3}$ and $\overline{\Delta}$ are smooth, by [4, Theorem 4.1] we have an isomorphism of graded $\operatorname{\mathbb{Z}}$-modules:

[TABLE]

where $t$ is a class of degree 2. Hence, in degree 3 and 5, we obtain:

[TABLE]

and

[TABLE]

It remains to show that $H^{3}(\overline{\Delta},\operatorname{\mathbb{Z}})$ is torsion free. Since $\overline{\Delta}\simeq A\smallsetminus A[3]$, it can be done considering the following exact sequence:

[TABLE]

where $H^{3}(A,\operatorname{\mathbb{Z}})$ is torsion free and $H^{4}(A,A\smallsetminus A[3],\operatorname{\mathbb{Z}})\simeq\operatorname{\mathbb{Z}}^{81}$ by Thom’s isomorphism. ∎

Acknowledgements. This work was supported by the ERC-ALKAGE, grant No. 670846.