# Quartic double solids with icosahedral symmetry

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## Abstract

We study quartic double solids admitting icosahedral symmetry.

### Keywords

Fano threefold Rationality Quartic surface Icosahedron### Mathematics Subject Classification

14J45 14E05 14E07 14E08## 1 Introduction

*G*into the Cremona groupone should first describe all

*n*-dimensional

*G*-Mori fiber spaces (see [8, Definition 1.1.5]), and then to decide which of these

*G*-Mori fiber spaces are rational and which are not. To describe all such embeddings up to conjugation, one should also describe

*G*-birational maps between the resulting rational

*G*-Mori fiber spaces. A priori, all three problems (classification, rationality and conjugation) can be solved for any given group

*G*. In fact, most of the results here are in dimensions \(n=2\) (see [11] for a nearly complete classification), and \(n=3\) (where only some very special groups have been treated).

In [15], Prokhorov managed to find all isomorphism classes of finite *simple* non-abelian subgroups of \({\mathrm {Cr}}_3({\mathbb {C}})\). He proved that the six groups: \({\mathfrak {A}}_5\) (alternating group of degree 5), \({\mathrm {PSL}}_2({\mathbb {F}}_7)\), \({\mathfrak {A}}_6\), \({\mathrm {SL}}_2({\mathbb {F}}_8)\), \({\mathfrak {A}}_7\), and \({\mathrm {PSp}}_4({\mathbb {F}}_3)\) are the only non-abelian finite simple subgroups of \({\mathrm {Cr}}_3({\mathbb {C}})\). The former three of these six groups actually admit embeddings to \({\mathrm {Cr}}_2({\mathbb {C}})\), and the icosahedral group \({\mathfrak {A}}_5\) is also realized as a subgroup of \({\mathrm {Cr}}_1({\mathbb {C}})={\mathrm {PGL}}_2({\mathbb {C}})\), while the latter three groups are new three-dimensional artefacts.

*G*-Mori fiber spaces withare

*G*-Fano threefolds, i.e., Fano threefolds with terminal singularities such that all

*G*-invariant Weil divisors on them are \({\mathbb {Q}}\)-linearly equivalent to multiples of the anticanonical ones. In [15], Prokhorov classified all

*G*-Fano threefolds, where

*G*is one of the latter three groups. His results together with [2] and [7, Corollary 1.22] imply that \({\mathrm {Cr}}_3({\mathbb {C}})\) contains two (unique, respectively) subgroups isomorphic to \({\mathrm {PSp}}_4({\mathbb {F}}_3)\) (to \({\mathrm {SL}}_2({\mathbb {F}}_8)\) or to \({\mathfrak {A}}_7\), respectively) up to conjugation.

The papers [6, 7] describe several non-conjugate embeddings of the groups \({\mathrm {PSL}}_2({\mathbb {F}}_7)\) and \({\mathfrak {A}}_6\) into \({\mathrm {Cr}}_3({\mathbb {C}})\) using a similar approach, although a complete answer is not known in these two cases. The book [8] is devoted to the icosahedral subgroups in \({\mathrm {Cr}}_3({\mathbb {C}})\). In particular, it describes three non-conjugate embeddings of the group \({\mathfrak {A}}_5\) into \({\mathrm {Cr}}_3({\mathbb {C}})\).

## 2 Quartic double solids with an action of \({\mathfrak {A}}_5\)

Let \(\tau :X\rightarrow {\mathbb {P}}^3\) be double cover branched over a (possibly reducible) reduced quartic surface *S*.

*Remark 2.1*

Recall from [21, Theorem 5] and [23, Corollary 4.7 (b)] that *X* is non-rational provided that the surface *S* is smooth. Therefore, we will be mostly interested in the cases when *S* is singular.

Suppose that *X* admits a faithful action of the icosahedral group \({\mathfrak {A}}_5\).

*Remark 2.2*

*S*. Vice versa, every \({\mathfrak {A}}_5\)-action on \({\mathbb {P}}^3\) that leaves the quartic surface

*S*invariant can be lifted to the corresponding double cover

*X*. Indeed,

*X*has a natural embedding to the projectivisation of the vector bundleThe group \({{\mathfrak {A}}}_5\) acts on the projective space \({{\mathbb {P}}}^{3}\), so that either \({{\mathfrak {A}}}_5\) or its central extension Open image in new window act on the line bundle \({\mathscr {O}}_{{{\mathbb {P}}}^2}(1)\). In any case, the action of \({{\mathfrak {A}}}_5\) lifts to the line bundle \({\mathscr {O}}_{{{\mathbb {P}}}^2}(2)\) and thus to \({\mathscr {E}}\). It remains to notice that the group \({{\mathfrak {A}}}_5\) does not have non-trivial characters, so that one can write down an \({{\mathfrak {A}}}_5\)-invariant equation of

*X*in the projectivisation of \({\mathscr {E}}\).

*I*the trivial representation of \({\mathfrak {A}}_5\). Let \(W_3\) and \(W_2'\) be the two irreducible three-dimensional representations of \({\mathfrak {A}}_5\), and let \(W_4\) be the irreducible four-dimensional representation of \({\mathfrak {A}}_5\). Let \(U_2\) and \(U_2'\) be two non-isomorphic two-dimensional representations of the central extension Open image in new window of the group \({\mathfrak {A}}_5\) by Open image in new window, and let \(U_4\cong {\mathrm {Sym}}^3(U_2)\) be the faithful four-dimensional irreducible representation of Open image in new window. Then \({\mathbb {P}}^3\) equipped with a faithful action of the group \({\mathfrak {A}}_5\) can be identified with one of the following projective spaces: Open image in new window, or Open image in new window.

*Remark 2.3*

Computing the symmetric powers Open image in new window and Open image in new window, we see that the only \({\mathfrak {A}}_5\)-invariant quartic surface in Open image in new window is not reduced (it is the unique \({\mathfrak {A}}_5\)-invariant quadric taken with multiplicity two). Similarly, we see that there are no \({\mathfrak {A}}_5\)-invariant quartic surfaces in Open image in new window at all.

Since the quartic surface *S* is reduced, our \({\mathbb {P}}^3\) can be identified with either Open image in new window, or Open image in new window, or Open image in new window. Let us start with the case Open image in new window.

**Theorem 2.4**

Suppose that Open image in new window. Then \({\mathfrak {A}}_5\)-invariant quartic surfaces in \({\mathbb {P}}^3\) form a pencil \({\mathscr {P}}\). This pencil contains exactly two (isomorphic) surfaces with non-isolated singularities, and exactly two nodal surfaces.

*X*, the morphism \(\widetilde{\tau }\) is a double cover branched over the proper transform of the surface \({\mathscr {S}}\) on Open image in new window, the rational map \(\psi \) is given by the linear system of quadrics in \({\mathbb {P}}^3\) passing through \({\mathscr {C}}\), the surface

*Q*is a smooth quadric, the morphism \(\tau _{Q}\) is a double cover branched over the unique \({\mathfrak {A}}_5\)-invariant conic in \({\mathbb {P}}^2\), and \(\phi \) and \(\varphi \) are \({\mathbb {P}}^1\)-bundles. In particular, in this case

*X*is rational.

*Proof*

Restrict \(U_4\) to the subgroups of Open image in new window isomorphic to Open image in new window, Open image in new window, and Open image in new window, where \({\mathrm {D}}_{2n}\) denotes a dihedral group order 2*n*. We see that none of these representations has one-dimensional subrepresentations. Thus, the projective space \({\mathbb {P}}^3\) does not contain \({\mathfrak {A}}_5\)-orbits of lengths 5, 6, and 15. On the other hand, restricting \(U_4\) to a subgroup of Open image in new window isomorphic to Open image in new window, we see that \({\mathbb {P}}^3\) contains exactly two \({{\mathfrak {A}}}_5\)-orbits of length 10.

The lines in \({\mathbb {P}}^3\) that are tangent to the curves \({\mathscr {C}}\) and \({\mathscr {C}}'\) sweep out quartic surfaces \({\mathscr {S}}\) and \({\mathscr {S}}'\), respectively. These surfaces are \({\mathfrak {A}}_5\)-invariant. The singular loci of \({\mathscr {S}}\) and \({\mathscr {S}}'\) are the curves \({\mathscr {C}}\) and \({\mathscr {C}}'\), respectively. In particular, the surfaces \({\mathscr {S}}\) and \({\mathscr {S}}'\) are different. Their singularities along these curves are locally isomorphic to a product of Open image in new window and a cusp.

The fibers of \(\phi \) are proper transforms of the secant or tangent lines to \({\mathscr {C}}\). Moreover, the proper transforms of the tangent lines to \({\mathscr {C}}\) are mapped by \(\phi \) to the points of the unique \({\mathfrak {A}}_5\)-invariant conic *C* in \({\mathbb {P}}^2\). Let \(\tau _Q:Q\rightarrow {{\mathbb {P}}}^2\) be a double cover branched over *C*. Then *Q* is a smooth quadric surface. A preimage on *X* of a secant line to \({\mathscr {C}}\) splits as a union of two smooth rational curves, while a preimage of a tangent line to \({\mathscr {C}}\) is contained in the ramification locus of \(\tau \). This shows the existence of the commutative diagram (1) and the rationality of *X*. The same construction applies to the case \(S={\mathscr {S}}'\).

Now we suppose that \(S\ne {\mathscr {S}}\) and \(S\ne {\mathscr {S}}'\). Let *S* be singular. Note that *S* is irreducible, because \({{\mathbb {P}}}^3\) contains no \({\mathfrak {A}}_5\)-invariant surfaces of degree less than four.

We claim that *S* has isolated singularities. Indeed, suppose that *S* is singular along some \({\mathfrak {A}}_5\)-invariant curve *Z*. A general plane section of *S* is an irreducible plane quartic curve that has Open image in new window singular points, which implies that the degree of *Z* is at most three. Thus, *Z* is a twisted cubic curve, so that either \(Z={\mathscr {C}}\) or \(Z={\mathscr {C}}'\). Since neither of these curves is contained in the base locus of \({\mathscr {P}}\), this would imply that either \(S={\mathscr {S}}\) or \(S={\mathscr {S}}'\). The latter is not the case by assumption.

We see that the singularities of *S* are isolated. Hence, *S* contains at most two non-Du Val singular points (cf. [10]). This follows from [22, Theorem 1] or from Shokurov’s famous [19, Theorem 6.9] applied to the minimal resolution of singularities of the surface *S*. Since the set of all non-Du Val singular points of the surface *S* must be \({\mathfrak {A}}_5\)-invariant, we see that *S* has none of them, because \(U_4\) is an irreducible representation of the group Open image in new window. Thus, all singularities of *S* are Du Val.

*S*is nodal, the set Open image in new window consists of one \({\mathfrak {A}}_5\)-orbit, andSince \({\mathbb {P}}^3\) does not contain \({\mathfrak {A}}_5\)-orbits of lengths 5, 6, and 15, we see that Open image in new window is either an \({{\mathfrak {A}}}_5\)-orbit of length 10 or an \({{\mathfrak {A}}}_5\)-orbit of length 12.

*S*is an \({{\mathfrak {A}}}_5\)-orbit \({\Sigma }_{12}\) of length 12. Then

*S*does not contain other \({\mathfrak {A}}_5\)-orbits of length 12 by [8, Lemma 6.7.3 (iv)]. Since \({\mathscr {C}}\) is not contained in the base locus of \({\mathscr {P}}\), and \({\mathscr {C}}\) is contained in \({\mathscr {S}}\), we see that \({\mathscr {C}}\not \subset S\). Sinceand \({\Sigma }_{12}\) is the only \({\mathfrak {A}}_5\)-orbit of length at most 12 in \({\mathscr {C}}\cong {\mathbb {P}}^1\), we have \(S\cap {\mathscr {C}}={\Sigma }_{12}\). Thus,which is absurd.

Therefore, we see that the singular locus of *S* is an \({{\mathfrak {A}}}_5\)-orbit \({\Sigma }_{12}\) of length 10. Vice versa, if an \({{\mathfrak {A}}}_5\)-invariant quartic surface passes through an \({{\mathfrak {A}}}_5\)-orbit of length 10, then it is singular by [8, Lemma 6.7.1 (ii)]. We know that there are exactly two \({{\mathfrak {A}}}_5\)-orbits of length 10 in \({{\mathbb {P}}}^3\), so that there are two surfaces \({\mathscr {R}}\) and \({\mathscr {R}}'\) that are singular exactly at the points of these two \({{\mathfrak {A}}}_5\)-orbits, respectively. The above argument shows that every surface in \({\mathscr {P}}\) except \({\mathscr {S}}, {\mathscr {S}}', {\mathscr {R}}\) and \({\mathscr {R}}'\) is smooth.\(\square \)

It would be interesting to find out whether the double covers of \({\mathbb {P}}^3\) branched over nodal surfaces of the pencil \({\mathscr {P}}\) are rational or not.

*Remark 2.5*

*B*the base locus of the pencil \({\mathscr {P}}\). We claim that

*B*is an irreducible curve with 24 cusps, and its normalization has genus nine. Indeed, let \(\rho :\widehat{{\mathscr {S}}}\rightarrow {\mathscr {S}}\) be the normalization of the surface \({\mathscr {S}}\in {\mathscr {P}}\) having non-isolated singularities, let \(\widehat{{\mathscr {C}}}\) be the preimage of the curve \({\mathscr {C}}\) via \(\rho \), and let \(\widehat{B}\) be the preimage of the curve

*B*via \(\rho \). Then the action of \({\mathfrak {A}}_5\) lifts to \(\widehat{{\mathscr {S}}}\), one has Open image in new window, and Open image in new window is a divisor of bi-degree (1, 2). This shows that \(\widehat{{\mathscr {C}}}\) is a divisor of bi-degree (1, 1), and \(\widehat{B}\) is a divisor of bi-degree (4, 8). In particular, the action of \({\mathfrak {A}}_5\) on \(\widehat{{\mathscr {S}}}\) is diagonal by [8, Lemma 6.4.3 (i)], so that \(\widehat{{\mathscr {C}}}\) is irreducible by [8, Lemma 6.4.4 (i)]. Note that the curve \(\widehat{{\mathscr {C}}}\) is singular. Indeed, recall that \({\mathscr {S}}'\) denotes the second surface in \({\mathscr {P}}\) with non-isolated singularities, and its singular locus is the twisted cubic curve \({\mathscr {C}}'\). Since \({\mathscr {C}}'\) is not contained in \({\mathscr {S}}\), we see that the intersection \({\mathscr {S}}\cap {\mathscr {C}}'\) is an \({\mathfrak {A}}_5\)-orbit \({\Sigma }_{12}'\) of length 12. Similarly, we see that the intersection \({\mathscr {S}}'\cap {\mathscr {C}}\) is an \({\mathfrak {A}}_5\)-orbit \({\Sigma }_{12}\) of length 12. These \({\mathfrak {A}}_5\)-orbits are different, since two different twisted cubic curves cannot have twelve points in common. Since

*B*is the scheme theoretic intersection of the surfaces \({\mathscr {S}}\) and \({\mathscr {S}}'\), it must be singular at every point of \({\Sigma }_{12}\cup {\Sigma }_{12}'\). Denote by \(\widehat{{\Sigma }}_{12}\) and \(\widehat{{\Sigma }}_{12}'\) the preimages via \(\rho \) of the \({\mathfrak {A}}_5\)-orbits \({\Sigma }_{12}\) and \({\Sigma }_{12}'\), respectively. Thenand \(\widehat{B}\) is singular in every point of \(\widehat{{\Sigma }}_{12}'\). Moreover, \(\widehat{B}\) is smooth away of \(\widehat{{\Sigma }}_{12}'\), because its arithmetic genus is 21, and the surface \(\widehat{{\mathscr {S}}}\) does not contain \({\mathfrak {A}}_5\)-orbits of length less than 12. In particular, the genus of the normalization of \(\widehat{B}\) (and thus also of

*B*) equals nine. On the other hand, we have

*B*is an irreducible curve whose only singularities are the points of \({\Sigma }_{12}\cup {\Sigma }_{12}'\), and each such point is a cusp of the curve

*B*.

Now let us deal with the case when \({\mathbb {P}}^3\) is identified with Open image in new window. The proof of the following statement is straightforward and we leave it to the reader.

**Theorem 2.6**

*C*in \({\mathbb {P}}^3\). All \({\mathfrak {A}}_5\)-invariant quadric surfaces in \({\mathbb {P}}^3\) form a pencil \({\mathscr {P}}\). Two general quadrics from this pencil are tangent to each other along

*C*. Moreover, any reduced \({\mathfrak {A}}_5\)-invariant quartic surface in \({\mathbb {P}}^3\) is a union of two different quadrics from \({\mathscr {P}}\). Furthermore, if

*S*is such a quartic surface, then there exists a commutative diagramHere the morphism \(\sigma \) is a blow up of the conic

*C*, the morphism \(\rho \) is a blow up of the preimage of

*C*on

*X*(which is the singular locus of

*X*), the morphism \(\widetilde{\tau }\) is a double cover branched over the proper transform of the surface

*S*on Open image in new window, the morphism \(\widetilde{\sigma }\) is a blow up of the intersection curve \(\widetilde{C}\) of the proper transforms on Open image in new window of the irreducible components of

*S*, the morphism \(\widetilde{\rho }\) is a blow up of the preimage of \(\widetilde{C}\) on \(\widetilde{X}\) (which is the singular locus of \(\widetilde{X})\), the morphism \(\widehat{\tau }\) is a double cover branched over the proper transform of the surface

*S*on Open image in new window, the rational map \(\psi \) is given by the pencil \({\mathscr {P}}\), the morphism \(\tau _{{\mathbb {P}}^1}\) is a double cover, and general fibers of \(\phi \) and \(\varphi \) are smooth quadric surfaces. In particular, the threefold

*X*is rational.

We will deal with the remaining case Open image in new window in the next section.

## 3 Hashimoto’s pencil of quartic surfaces

*I*is the usual five-dimensional permutation representation of \({\mathfrak {A}}_5\). Let \(x_0,\ldots ,x_4\) be the coordinates in Open image in new window permuted by \({\mathfrak {A}}_5\). Then \(W_4\) is given in Open image in new window by the equation

*i*, putThen \(F_2\) is the unique \({\mathfrak {A}}_5\)-invariant polynomial of degree two in \(x_1,x_2,x_3,x_4\) modulo scaling. Similarly, modulo scaling, \(F_3\) is the unique \({\mathfrak {A}}_5\)-invariant polynomial of degree three in \(x_1,x_2,x_3,x_4\). Finally, modulo \(F_2\) and scaling, \(F_4\) is the unique \({\mathfrak {A}}_5\)-invariant polynomial of degree four in \(x_1,x_2,x_3,x_4\). Thus, any \({\mathfrak {A}}_5\)-invariant quartic surface in \({\mathbb {P}}^3\) is given by

*Remark 3.1*

The quartic surfaces given by (2) have been studied by Hashimoto in [13]. In particular, he described all singular surfaces in this pencil.

*Remark 3.2*

The points Open image in new window, and Open image in new window are collinear. Thus, there are ten lines in \({\mathbb {P}}^3\) such that each of them contains three points of \({\Sigma }_{10}\), and each point of \({\Sigma }_{10}\) lies on exactly three of these lines. Similarly, the points Open image in new window, Open image in new window, Open image in new window, and Open image in new window are coplanar. Hence, there are ten planes in \({\mathbb {P}}^3\) such that each of them contains four points of \({\Sigma }_{10}'\), and each point of \({\Sigma }_{10}'\) lies on exactly four of these planes. In particular, for each \({\mathfrak {A}}_5\)-orbit \({\Sigma }\in \{{\Sigma }_{10}, {\Sigma }_{10}'\}\), there exists a plane in \({\mathbb {P}}^3\) that contains at least four points of \({\Sigma }\) such that no three of them are collinear.

The \({\mathfrak {A}}_5\)-orbits \({\Sigma }_{5}, {\Sigma }_{10}, {\Sigma }_{10}'\), and \({\Sigma }_{15}\) are the only \({\mathfrak {A}}_5\)-orbits of lengths 5, 10, and 15 in \({\mathbb {P}}^3\). Moreover, there are exactly two \({\mathfrak {A}}_5\)-orbits \({\Sigma }_{12}\) and \({\Sigma }_{12}'\) in \({\mathbb {P}}^3\) of length 12, see, for example, [8, Corollary 5.2.3, Lemma 5.3.3].

By [8, Lemma 5.3.3 (xi)], the curve in \({\mathbb {P}}^3\) that is given by \(F_2=F_4=0\) is a smooth curve of genus nine. In particular, this curve does not contain the \({\mathfrak {A}}_5\)-orbits \({\Sigma }_{5}, {\Sigma }_{10}, {\Sigma }_{10}'\), and \({\Sigma }_{15}\), see, for example, [8, Lemma 5.1.4]. Therefore, there exists a unique \({\mathfrak {A}}_5\)-invariant quartic surface \(S_5\) (respectively, \(S_{10}, S_{10}', S_{15}\)) in \({\mathbb {P}}^3\) that contains the \({\mathfrak {A}}_5\)-orbit \({\Sigma }_{5}\) (respectively, \({\Sigma }_{10}, {\Sigma }_{10}', {\Sigma }_{15}\)).

**Proposition 3.3**

([13, Proposition 3.1]) The surface \(S_5\) (respectively, \(S_{10}, S_{10}', S_{15})\) is given by equation (2) with \(\lambda =13/20\) (respectively, \(\lambda =1/2\), \(\lambda =7/30\), \(\lambda =1/4)\). The surface \(S_5\) (respectively, \(S_{10}, S_{10}', S_{15})\) has nodes at the points of the \({\mathfrak {A}}_5\)-orbit \({\Sigma }_{5}\) (respectively, \({\Sigma }_{10}, {\Sigma }_{10}', {\Sigma }_{15})\), and is smooth outside of it. Moreover, \(S_5, S_{10}, S_{10}'\), and \(S_{15}\) are the only singular surfaces given by (2).

Denote by \(X_5\) (respectively, \(X_{10}, X_{10}', X_{15}\)) the double cover of \({\mathbb {P}}^3\) branched over the surface \(S_5\) (respectively, \(S_{10}, S_{10}', S_{15}\)). As we already mentioned in Remark 2.2, the action of the group \({\mathfrak {A}}_5\) lifts to each of these threefolds. In the next two theorems, we prove two guesses that were formulated in [8, Example 1.3.9].

**Theorem 3.4**

*Proof*

The threefold \(X_5\) is \({\mathbb {Q}}\)-factorial by [5, Theorem 1.8], and it is non-rational by [5, Theorem 1.2]. The preimage on \(X_5\) of a general twisted cubic that passes through \({\Sigma }_5\) is an irreducible (singular) rational curve. Thus, the existence of the commutative diagram follows from [14, Proposition 4.7].\(\square \)

In the proof of [5, Theorem 4.2], the authors constructed a birational transformation of \(X_5\) into a standard conic bundle over the smooth del Pezzo surface \(Y_5\) of degree 5. Unlike the birational map Open image in new window from Theorem 3.4, this transformation is not \({\mathfrak {A}}_5\)-equivariant (it is only \({\mathfrak {A}}_4\)-equivariant).

**Theorem 3.5**

The threefolds \(X_{10}\) and \(X_{10}'\) are \({\mathbb {Q}}\)-factorial, and \(X_{10}\) is not stably rational.

*Proof*

The \({\mathbb {Q}}\)-factoriality of \(X_{10}\) (respectively, \(X_{10}'\)) is equivalent to the fact that \({\Sigma }_{10}\) (respectively, \({\Sigma }_{10}'\)) is not contained in a quadric surface in \({\mathbb {P}}^3\), see [9, Section 3] or [12]. The latter condition is very easy to check by solving the system of linear equations. However, it can also be easily proved without any computations. Indeed, suppose that there are quadrics in \({\mathbb {P}}^3\) passing through \({\Sigma }_{10}\). Note that \({\Sigma }_{10}\) does not lie on the quadric given by \(F_2=0\) by [8, Lemma 5.3.3 (vi)]. The latter equation defines the unique \({\mathfrak {A}}_5\)-invariant quadric in \({\mathbb {P}}^3\). Thus, the intersection of all quadrics passing through \({\Sigma }_{10}\) is either an \({\mathfrak {A}}_5\)-invariant set of at most eight points, or an \({\mathfrak {A}}_5\)-invariant curve *Z* of degree at most four. The former case is clearly impossible because the set \({\Sigma }_{10}\) contains more than eight points. In the latter case one has Open image in new window, because there are no \({\mathfrak {A}}_5\)-invariant curves of degree at most three in \({\mathbb {P}}^3\) by [8, Lemma 5.3.3 (ix)]. Thus, *Z* is a complete intersection of two quadrics. On the other hand, a direct computation shows that \({\mathrm {Sym}}^2(W_4)\) does not contain two-dimensional subrepresentations of \({\mathfrak {A}}_5\). The obtained contradiction shows that \({\Sigma }_{10}\) is not contained in any quadric surface in \({\mathbb {P}}^3\). Similarly, we see that \({\Sigma }_{10}'\) is not contained in any quadric surface in \({\mathbb {P}}^3\). Hence, both threefolds \(X_{10}\) and \(X_{10}'\) are \({\mathbb {Q}}\)-factorial.

*O*, the morphism

*f*is the blow up of the point in \(X_{10}\) that is mapped to the point

*O*by the double cover \(\tau \), and \(\pi \) is a conic bundle.

*C*of the conic bundle \(\pi \). We may assume thatPlugging Open image in new window into (2) with \(\lambda =1/2\), and considering the affine equation of \(S_{10}\) in the chart \(x_4\ne 0\) with the new coordinateswe see that \(O=(0,0,0)\) in these coordinates, and the chart of the surface \(S_{10}\) in \({\mathbb {A}}^3\) is given by the equation Then every line

*L*in \({\mathbb {A}}^3\) passing through

*O*is given by

*t*is a parameter. Plugging this parametric equation into (3), dividing the resulting equation by \(t^2\), and taking the discriminant of the resulting quadratic equation, we see that the equation of

*C*in \({\mathbb {P}}^2\) iswhere we consider \(z_1, z_2\), and \(z_3\) as homogeneous coordinates on \({\mathbb {P}}^2\). Thus

*C*is a union of a line \(\ell \) that is given by \(z_3=0\), a smooth conic \(\gamma \) that is given by

*V*is a smooth projective threefold,

*U*is a smooth surface, the relative Picard group of

*V*over

*U*has rank 1, and \(\varrho \) is a birational morphism that factors as

Let \({\Delta }\) be the degeneration curve of the conic bundle \(\nu \). Then \({\Delta }\) is the proper transform of the curve *C* by [5, Theorem 4.2 (iii)]. Thus, the curve \({\Delta }\) is not connected, so that \(H^3(V,{\mathbb {Z}})\) has non-trivial 2-torsion by [24, Theorem 2], see also [1, Proposition 3]. So the threefold \(X_{10}\) is not stably rational by [1, Proposition 1].\(\square \)

We do not know whether \(X_{10}'\) is rational or not.

*Remark 3.6*

*f*is the blow up of the point in \(X_{10}'\) that is mapped to

*O*, and \(\pi \) is a conic bundle. Denote by

*C*the degeneration curve of the conic bundle \(\pi \). Making computations similar to those in the proof of Theorem 3.5, we see that

*C*can be given bywhere \(z_1, z_2\), and \(z_3\) are homogeneous coordinates on \({\mathbb {P}}^2\). One can check that

*C*is an irreducible nodal curve with exactly nine nodes. It follows from [5, Theorem 4.2 (i), (ii)] that there exists a commutative diagramwhere

*V*is a smooth projective threefold,

*U*is a smooth surface, the relative Picard group of

*V*over

*U*has rank 1, and \(\varrho \) is a blow up of nine nodes of

*C*. Moreover, the degeneration curve of the conic bundle \(\nu \) is the proper transform of the curve

*C*by [5, Theorem 4.2 (iii)]. Thus, \({\Delta }\) is an irreducible smooth elliptic curve in Open image in new window, which implies that the group \(H^3(V,{\mathbb {Z}})\) is trivial (see, for example, [24, Theorem 2]). In particular, the intermediate Jacobian of

*V*is trivial, and the approach of [1] does not work in this case either. Note that [18, Conjecture 10.3] predicts that \(X_{10}'\) is non-rational.

*Remark 3.7*

In [12, p. 354], it is claimed that the resolution of singularities of any \({\mathbb {Q}}\)-factorial nodal quartic double solid with ten nodes has non-trivial torsion in the third integral cohomology group. The computations in Remark 3.6 show that \(X_{10}'\) is a counter-example to this claim.

*Remark 3.8*

In the proof of Theorem 3.5 and in Remark 3.6, we refer to [24, Theorem 2]. Note that the notation of this theorem is a bit non-standard. Namely, the second summand on the right hand side in [24, (13)] is torsion free (see [24, Lemma 5] and [24, Lemma 7] for a detailed computation).

Recall that a normal variety *X* with an action of the finite group *G* is said to be \(G{\mathbb {Q}}\)-factorial if any *G*-invariant Weil divisor on *X* is a \({\mathbb {Q}}\)-Cartier divisor.

**Theorem 3.9**

The threefold \(X_{15}\) is not \({\mathbb {Q}}\)-factorial, it is \({\mathfrak {A}}_5{\mathbb {Q}}\)-factorial, and it is rational.

*Proof*

The threefold \(X_{15}\) is not \({\mathbb {Q}}\)-factorial by [5, Corollary 1.7]. Its rationality follows from [16, Theorem 8.1] or [5, Theorem 1.3].

*O*is trivial.

*O*is Open image in new window in the new coordinates. One hasWritewhere \(R_i\) is a form of degree

*i*. ThenThe threefold \(X_{15}\) is given in the weighted projective space Open image in new window with weighted homogeneous coordinates \(y_1, y_2, y_3, y_4\), and

*w*by equation \(w^2=R\), and \(\sigma \) acts trivially on

*w*. Identify the point \(O\in {\mathbb {P}}^3\) with the unique point of \(X_{15}\) that is mapped to

*O*by the double cover morphism. Then the tangent cone to \(X_{15}\) at

*O*is a cone over a quadric surface

*B*given by equationin a three-dimensional projective space with coordinates \(y_1, y_2, y_3\), and

*w*. The two lines \(y_1=w-2y_3=0\) and \(y_1=w+2y_3=0\) are contained in two different pencils of lines on

*B*. They are interchanged by the involution \(\sigma \), which implies that the \({\Gamma }\)-invariant local class group of the point

*O*is trivial.\(\square \)

## 4 \({\mathfrak {A}}_5\)-birational superrigidity

Let us use notation of Sect. 3. Let *S* be a quartic surface in \({{\mathbb {P}}}^3\) that is given by (2), and let \(\tau :X\rightarrow {{\mathbb {P}}}^3\) be a double cover branched over *S*. By Remark 2.2, the threefold *X* is faithfully acted on by the group \({\mathfrak {A}}_5\). Recall that *Q* is the surface in \({{\mathbb {P}}}^3\) that is given by \(F_2=0\). The surface *Q* is smooth, so that Open image in new window. Denote by *H* the class of the pull-back of the plane in \({\mathbb {P}}^3\) via \(\tau \).

*Remark 4.1*

Every \({\mathfrak {A}}_5\)-invariant Weil divisor on *X* is rationally equivalent to a multiple of *H*. This follows from Theorems 3.4, 3.5, 3.9, and the Lefschetz theorem.

Thus, *X* is an \({\mathfrak {A}}_5\)-Fano threefold. The goal of this section is to prove the following result.

**Theorem 4.2**

The threefold *X* is \({\mathfrak {A}}_5\)-birationally superrigid if and only if \(X\ne X_5\).

**Corollary 4.3**

The group \({\mathrm {Cr}}_3({\mathbb {C}})\) contains at least four non-conjugate subgroups isomorphic to \({\mathfrak {A}}_5\).

*Proof*

Since \(X_{15}\) is rational by Theorem 3.9, the required assertion follows from Theorem 4.2, [8, Remark 1.2.1, Example 1.3.9, Theorem 1.4.1].\(\square \)

**Corollary 4.4**

The group \({\mathrm {Cr}}_3({\mathbb {C}})\) contains at least three non-conjugate subgroups isomorphic to the permutation group \({\mathfrak {S}}_5\).

*Proof*

*Q*. Therefore, there is no \({\mathfrak {S}}_5\)-birational map Open image in new window by [17, Proposition A.4], see also [8, Theorem 1.1.1].

\(\square \)

We already described \({\mathfrak {A}}_5\)-orbits in \({{\mathbb {P}}}^3\) of small length in Sect. 3. Now let us describe \({\mathfrak {A}}_5\)-invariant curves in \({{\mathbb {P}}}^3\) of degree less than eight. As we will see a bit later, they all lie in the surface given by \(F_3=0\), and one of them also lies in the surface *Q*. Thus, we need to take a closer look at these surfaces.

The surface *Q* does not contain the \({\mathfrak {A}}_5\)-orbits \({\Sigma }_5, {\Sigma }_{10}, {\Sigma }_{10}'\), and \({\Sigma }_{15}\), and it does contain the \({\mathfrak {A}}_5\)-orbits \({\Sigma }_{12}\) and \({\Sigma }_{12}'\), see [8, Lemma 5.3.3 (vi)]. Moreover, since *Q* contains only two \({\mathfrak {A}}_5\)-orbits of length 12, and \(W_4\) is an irreducible representation of \({\mathfrak {A}}_5\), it follows from [8, Lemma 6.4.3 (i)] that the \({\mathfrak {A}}_5\)-action on *Q* is twisted diagonal, i.e. the quadric *Q* can be identified with Open image in new window.

Denote by \({\mathrm {S}}_3\) the surface in \({{\mathbb {P}}}^3\) that is given by \(F_3=0\), and denote by \({\mathscr {B}}_6\) the curve in \({{\mathbb {P}}}^3\) that is given by \(F_2=F_3=0\). Then \({\mathrm {S}}_3\) is a smooth surface known as the *Clebsch diagonal cubic surface*, and \({\mathscr {B}}_6\) is a smooth irreducible curve of genus four known as the *Bring curve*.

**Lemma 4.5**

The curve \({\mathscr {B}}_6\) is the only \({\mathfrak {A}}_5\)-invariant curve in *Q* of degree less than eight. It contains the \({\mathfrak {A}}_5\)-orbits \({\Sigma }_{12}\) and \({\Sigma }_{12}'\). Moreover, the set \({\Sigma }_{12}\cup {\Sigma }_{12}'\) is cut out on \({\mathscr {B}}_6\) by the equation \(F_4=0\).

*Proof*

*Q*of degree \(d<8\). Then \(d\geqslant 4\) by [8, Lemma 5.3.3 (ix)]. Let us show that \({\Gamma }={\mathscr {B}}_6\). If \({\Gamma }\) is contained in \({\mathrm {S}}_3\), then \({\Gamma }={\mathscr {B}}_6\) by construction. Thus, we may assume that this is not the case. Therefore, \(d\ne 6\) and \(d\ne 7\), becauseand the lengths of \({\mathfrak {A}}_5\)-orbits in \({\mathbb {P}}^3\) are 5, 10, 12, 15, 20, 30, and 60. Hence, either \(d=4\) or \(d=5\).

*a*,

*b*) on Open image in new window, where

*a*and

*b*are non-negative integers. Without loss of generality, we may assume that \(a\leqslant b\). Since

*a*,

*b*) must be one of the following: (0, 4), (1, 3), (2, 2), (0, 5), (1, 4), or (2, 3). The cases (0, 4) and (0, 5) are impossible by [8, Lemma 6.4.1]. Moreover,

*Q*contains no \({\mathfrak {A}}_5\)-invariant effective divisors of bi-degree (1, 3) and (1, 4) by [8, Lemma 6.4.11 (o)], because the \({\mathfrak {A}}_5\)-action on

*Q*is twisted diagonal. Thus, either \((a,b)=(2,2)\) or \((a,b)=(2,3)\). By [8, Lemma 6.4.3 (ii)], the quadric

*Q*contains a smooth rational curve \(C_{1,7}\) that is a divisor of bi-degree (1, 7). One haswhich is impossible, because the lengths of \({\mathfrak {A}}_5\)-orbits in \(C_{1,7}\cong {\mathbb {P}}^1\) are 12, 20, 30, and 60.\(\square \)

**Lemma 4.6**

The \({\mathfrak {A}}_5\)-orbits of general points of the curves \({\mathscr {B}}_6, {\mathscr {C}}, {\mathscr {C}}', {\mathscr {L}}_6\), and \({\mathscr {L}}_6'\) are of length 60. These curves are the only \({\mathfrak {A}}_5\)-invariant curves in \({\mathbb {P}}^3\) of degree less than eight.

*Proof*

The stabilizers in \({\mathfrak {A}}_5\) of general points of the curves \({\mathscr {B}}_6, {\mathscr {C}}\) and \({\mathscr {C}}'\) are trivial, because these curves are irreducible and none of them is contained in a plane in \({\mathbb {P}}^3\). Thus, the \({\mathfrak {A}}_5\)-orbits of general points of the curves \({\mathscr {B}}_6, {\mathscr {C}}\), and \({\mathscr {C}}'\) are of length 60. The stabilizer of each irreducible component of the curve \({\mathscr {L}}_6\) acts faithfully on it by [8, Corollary 5.2.3 (v)]. This implies that the \({\mathfrak {A}}_5\)-orbit of a general point of (an irreducible component of) the curve \({\mathscr {L}}_6\) is of length 60. Similarly, the \({\mathfrak {A}}_5\)-orbit of a general point of the curve \({\mathscr {L}}_6'\) is also of length 60.

Let us show that \({\mathscr {B}}_6, {\mathscr {C}}, {\mathscr {C}}', {\mathscr {L}}_6\), and \({\mathscr {L}}_6'\) are the only \({\mathfrak {A}}_5\)-invariant curves in \({\mathbb {P}}^3\) of degree less than eight. Let \({\Gamma }\) be an \({\mathfrak {A}}_5\)-invariant curve in \({\mathbb {P}}^3\) of degree *d*. If \({\Gamma }\subset {\mathrm {S}}_3\), then the required assertion follows from [8, Theorem 6.3.18]. Thus, we may assume that \({\Gamma }\not \subset {\mathrm {S}}_3\). One has \({\Gamma }\not \subset Q\) by Lemma 4.5.

*Q*are at least 12.\(\square \)

**Corollary 4.7**

*Proof*

Existence of the surfaces \(S_{{\mathscr {C}}}\) and \(S_{{\mathscr {L}}_6}\) follows from Lemma 4.6, because none of the curves \({\mathscr {C}}\) and \({\mathscr {L}}_6\) is contained in *Q* by Lemma 4.5. Moreover, the curve \({\mathscr {C}}\) (respectively, \({\mathscr {L}}_6\)) is contained in the smooth locus of the surface \(S_{{\mathscr {C}}}\) (respectively, \(S_{{\mathscr {L}}_6}\)) by Proposition 3.3. Thus, Open image in new window on the surface \(S_{{\mathscr {C}}}\), and Open image in new window on the surface \(S_{{\mathscr {L}}_6}\). Now it follows from [8, Lemma 6.7.3 (i), (ii)] that both surfaces \(S_{{\mathscr {C}}}\) and \(S_{{\mathscr {L}}_6}\) are smooth.

To prove Theorem 4.2, we also need the following technical result.

**Lemma 4.8**

*m*that pass through the set \({\Sigma }\). Then \({\mathscr {M}}\) does not have base curves and fixed components.

*Proof*

If \({\Sigma }={\Sigma }_5\), then \({\mathscr {M}}\) does not have fixed components by [8, Lemma 5.3.3 (vi)], and \({\mathscr {M}}\) does not have base curves, because \({\mathbb {P}}^3\) does not have \({\mathfrak {A}}_5\)-invariant curves of degree less than six by Lemma 4.6. Similarly, if \({\Sigma }={\Sigma }_{10}\) (respectively, \({\Sigma }={\Sigma }_{10}'\), \({\Sigma }={\Sigma }_{15}\)), then \({\Sigma }\) is a singular locus of the nodal quartic surface \(S_{10}\) (respectively, \(S_{10}', S_{15}\)). Since the singular locus of a quartic surface is cut out by cubics, we see that the base locus of \({\mathscr {M}}\) is \({\Sigma }\) in the latter case. If \({\Sigma }={\Sigma }_{12}\) or \({\Sigma }={\Sigma }_{12}'\), then \({\mathscr {M}}\) does not have base curves and fixed components by Lemma 4.5. Thus, we may assume that \(|{\Sigma }|\geqslant 20\), so that \(m=|{\Sigma }|/2\).

*Q*. Let Open image in new window be a point in \({\Sigma }\). Then \(F_2(a_0,a_1,a_2,a_3,a_4)\ne 0\). PutThen the system of equationshas finitely many solutions in \({\mathbb {P}}^3\), because the system of equations \(F_2=F_3=F_4=F_5\) has no solution in \({\mathbb {P}}^3\). Using the forms Open image in new window, Open image in new window, and Open image in new window, we can produce three surfaces in \({\mathscr {M}}\) that have only finitely many common points. This shows that the base locus of \({\mathscr {M}}\) is zero-dimensional as requested.

*R*is any surface in \({\mathbb {P}}^3\) of degree \(m-2\). Thus, the base locus of \({\mathscr {M}}\) is contained in

*Q*. Denote by

*T*the hyperplane section of

*Q*. Let \({\mathscr {M}}_Q\) be the linear system consisting of all curves in |

*mT*| that pass through \({\Sigma }\). Then \({\mathscr {M}}_Q\) is not empty, because

*Q*by a surface in \({\mathscr {M}}\), because we have a surjection

*Q*, and

*Q*is not a fixed surface of \({\mathscr {M}}\). In particular, the base loci of \({\mathscr {M}}\) and \({\mathscr {M}}_Q\) are the same.

*Z*. Then

*Z*is a divisor of bi-degree (

*a*,

*b*) on Open image in new window, where \(a\leqslant m\) and \(b\leqslant m\). One haswhich implies that

*Z*is \(a+b\leqslant 4\), which contradicts Lemma 4.6.\(\square \)

Now we are ready to prove Theorem 4.2.

*Proof of Theorem 4.2*

By Theorem 3.4, the threefold \(X_5\) is not \({\mathfrak {A}}_5\)-birationally superrigid. Thus, we may assume that \(X\ne X_5\). Suppose that *X* is not \({\mathfrak {A}}_5\)-birationally superrigid. Then it follows from [8, Corollary 3.3.3] that there exists an \({\mathfrak {A}}_5\)-invariant mobile linear system \({\mathscr {D}}\) on *X* such that the singularities of the log pair Open image in new window are not canonical, where *n* is a positive integer that is defined by \({\mathscr {D}}\sim nH\).

*Z*be an irreducible subvariety of

*X*that is a center of canonical singularities of the log pair Open image in new window, see [8, Definition 2.4.1]. Then either

*Z*is a point, or

*Z*is an irreducible curve. Let \(\{Z_1,\ldots ,Z_r\}\) be the \({\mathfrak {A}}_5\)-orbit of \(Z=Z_1\). If

*Z*is a curve, then

*D*in \({\mathscr {D}}\), see, for example, [8, Lemma 2.4.4]. If

*Z*is a smooth point, thenfor each \(Z_i\) and two general surfaces \(D_1\) and \(D_2\) in \({\mathscr {D}}\), see, for example, [8, Theorem 2.5.2]. We will use (4) to show that

*Z*is not a curve, then we will use (5) to show that

*Z*is not a smooth point of

*X*. Finally we will use [4, Theorem 1.7.20] to exclude the case when

*Z*is a singular point of

*X*.

*Z*is a curve. Put \({\Sigma }=Z_1+\cdots +Z_r\). For general surfaces \(D_1\) and \(D_2\) in \({\mathscr {D}}\), inequality (4) giveswhich implies that Open image in new window. Applying Lemma 4.6, we see thatand \(\tau ({\Sigma })\) is one of the following curves: \({\mathscr {B}}_6, {\mathscr {C}}, {\mathscr {C}}', {\mathscr {L}}_6\), or \({\mathscr {L}}_6'\). Since Open image in new window and \(\tau ({\Sigma })\) is a curve of degree six, the double cover \(\tau \) induces an isomorphism \({\Sigma }\xrightarrow { \sim \,}\tau ({\Sigma })\).

*S*, and

*S*transversally in 24 points. Hence, the preimage of the curve \({\mathscr {B}}_6\) via \(\tau \) is a smooth irreducible curve that is a double cover of \({\mathscr {B}}_6\) branched over \({\Sigma }_{12}\cup {\Sigma }_{12}'\). In particular, the preimage of the curve \({\mathscr {B}}_6\) via \(\tau \) is not isomorphic to \({\mathscr {B}}_6\), so that \(\tau ({\Sigma })\ne {\mathscr {B}}_6\).

*X*is a quartic hypersurface in the weighted projective space Open image in new window with weighted homogeneous coordinates \(x_1, x_2, x_3, x_4\) and

*w*defined by equation

*X*bywhere Open image in new window.

*Q*. Recall that this curve is smooth. In particular, Open image in new window is either a smooth

*K*3 surface or a nodal

*K*3 surface by Proposition 3.3.

*P*be a general point in \({\Sigma }\). Then its \({\mathfrak {A}}_5\)-orbit is of length 60, because the \({\mathfrak {A}}_5\)-orbit of the point \(\tau (P)\) is of length 60 by Lemma 4.6. Let

*Y*be a surface in \({\mathscr {P}}\) such that \(P\in Y\). Then \({\Sigma }\subset Y\), since otherwise we would haveHence, we have \(\tau ({\Sigma })\subset \tau (Y)\), which implies that \(\tau (Y)\ne Q\), because none of the curves \({\mathscr {C}}, {\mathscr {C}}', {\mathscr {L}}_6\), and \({\mathscr {L}}_6'\) is contained in

*Q*by Lemma 4.5. Thus, \(\tau (Y)\) is a (possibly nodal) \({\mathfrak {A}}_5\)-invariant quartic surface, and \(\tau \) induces an isomorphism \(Y\xrightarrow {\sim \,}\tau (Y)\).

By Corollary 4.7, the surface \(\tau (Y)\) is the surface \(S_{{\mathscr {C}}}\) (respectively, \(S_{{\mathscr {L}}_6}\)) in the case when \(\tau ({\Sigma })\) is one of the curves \({\mathscr {C}}\) or \({\mathscr {C}}'\) (respectively, \({\mathscr {L}}_6\) or \({\mathscr {L}}_6'\)). In particular, the surface \(\tau (Y)\) is smooth.

*Y*contains an \({\mathfrak {A}}_5\)-invariant curve \({\Sigma }'\) such that \({\Sigma }'\ne {\Sigma }\), the curves \({\Sigma }'\) and \({\Sigma }\) are isomorphic, and

*Y*of the curve \({\mathscr {C}}'\) (respectively, \({\mathscr {C}}, {\mathscr {L}}_6', {\mathscr {L}}_6\)).

*l*and \(l'\) are non-negative integers, and \({\Omega }\) is an effective divisor on the surface

*Y*whose support does not contain irreducible components of the curves \({\Sigma }\) and \({\Sigma }'\). By (6), we haveMoreover, one has Open image in new window, because \(l>n/2\) by (4). Hence, we obtain Open image in new window, so thatwhich is absurd. The obtained contradiction shows that

*Z*is not a curve.

*Z*is a point. Denote by \({\Xi }\) its \({\mathfrak {A}}_5\)-orbit. Define an integer

*m*as in Lemma 4.8. Let \({\mathscr {M}}\) be the linear system consisting of all surfaces in |

*mH*| passing through \({\Xi }\). Then \({\mathscr {M}}\) does not have base curves and fixed components by Lemma 4.8. Thus, if

*Z*is a smooth point of

*X*, then (5) givesfor a general surface

*M*in \({\mathscr {M}}\), and two general surfaces \(D_1\) and \(D_2\) in \({\mathscr {D}}\). Therefore, we see that

*Z*is a singular point of

*X*. By Proposition 3.3, either \(X=X_{10}\), or \(X=X_{10}'\), or \(X=X_{15}\). Put \(r=|{\Xi }|\), so that either \(r=10\) or \(r=15\).

*f*-exceptional surfaces. Denote by \(\widetilde{D}_1\) and \(\widetilde{D}_2\) the proper transforms via

*f*of the surfaces \(D_1\) and \(D_2\), respectively. Then

*s*. Denote by \(\widetilde{M}\) the proper transform of the surface

*M*via

*f*. Then

*t*. Actually, one can show that \(t=1\), but we will not use this. Since \({\mathscr {M}}\) does not have base curves and fixed components, the divisor \(\widetilde{M}\) is nef. Thus, we havewhich implies that Open image in new window. On the other hand, \(s>n/2\) by [4, Theorem 1.7.20]. This gives \(r=10\), so that either \(\tau ({\Xi })={\Sigma }_{10}\) or \(\tau ({\Xi })={\Sigma }_{10}'\).

*C*be a general conic in \({\Pi }\) that passes through these four points. Then its preimage on

*X*via \(\tau \) splits as a union of two smooth rational curves. Denote by \(\widetilde{C}\) the proper transform of one of these curves on

*W*via

*f*. Then \(\widetilde{C}\) is not contained in the support of \(\widetilde{D}_1\). On the other hand, we havewhich implies that \(s\leqslant n/2\). This is again impossible, because we already know that \(s>n/2\).\(\square \)

## Notes

### Acknowledgments

This work was started when Ivan Cheltsov and Constantin Shramov were visiting the Mathematisches Forschungsinstitut Oberwolfach under Research in Pairs program, and was completed when they were visiting Centro Internazionale per la Ricerca Matematica in Trento under a similar program. We want to thank these institutions for excellent working conditions, and personally Marco Andreatta and Augusto Micheletti for hospitality.

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