January 26, 2026
Our main result is the determination of the respective groups \(\mathop{\mathrm{Aut}}_\mathbb{Z}(S)\) of cohomologically trivial automorphisms and \(\mathop{\mathrm{Aut}}_\mathbb{Q}(S)\) of numerically trivial automorphisms for the reducible fake quadrics, that is, the surfaces \(S\) isogenous to a product with \(q=p_g=0\).
In this way we produce new record winning examples: a surface \(S\) with \(|\mathop{\mathrm{Aut}}_\mathbb{Q}(S)| =192\), and a surface whose cohomology has torsion with nontrivial \(\mathop{\mathrm{Aut}}_\mathbb{Z}(S) \cong \mathbb{Z}/2.\)
As in [1] let \(\mathop{\mathrm{Aut}}_R(S): = \{ g \in \mathop{\mathrm{Aut}}(S)| g^* = 1 \;on\;H^*(S,R)\}\) denote the group of those automorphism of \(S\) acting trivially on the cohomology with coefficients in the ring \(R\).
In this paper we continue (cf. [1], [2], [3]) the study of cohomologically trivial automorphisms \(\mathop{\mathrm{Aut}}_\mathbb{Z}(S)\) and numerically trivial automorphisms \(\mathop{\mathrm{Aut}}_\mathbb{Q}(S)\) of surfaces. The former group \(\mathop{\mathrm{Aut}}_\mathbb{Z}(S)\) is a subgroup of the latter, and for minimal non-ruled surfaces there are no examples where the group \(\mathop{\mathrm{Aut}}_\mathbb{Z}(S)\) is finite with cardinality \(\geq 4\); the first authors who dealt with the subtleties of the subgroup \(\mathop{\mathrm{Aut}}_\mathbb{Z}(S)\) were Mukai and Namikawa [4].
The basic questions we consider in this article are the followings.
Question 1. (I) What is \(max | \mathop{\mathrm{Aut}}_\mathbb{Z}(S)|\) for a minimal surface \(S\) of general type?
(II) What is \(max | \mathop{\mathrm{Aut}}_\mathbb{Q}(S)|\) for a minimal surface \(S\) of general type?
Since \(\mathop{\mathrm{Aut}}_\mathbb{Z}(S) \subset \mathop{\mathrm{Aut}}_\mathbb{Q}(S)\), that such maxima exist follows from a result of Xin Jing Cai [5] (beware that Cai uses the term cohomologically trivial for the weaker notion, introduced later, of numerically trivial).
Theorem 2 (Cai). For surfaces \(S\) of general type, there is an absolute constant \(C\) such that \[| \mathop{\mathrm{Aut}}_\mathbb{Q}(S)| \leq C .\]
There is a well known example for question (II): for the classical Beauville surface \(| \mathop{\mathrm{Aut}}_\mathbb{Q}(S)| = 75\) and \(| \mathop{\mathrm{Aut}}_\mathbb{Z}(S)| = 1\), as we shall show later on2.
Recall that the above is a surface isogenous to a product, \(S = (C\times C) / G\), \(G \cong (\mathbb{Z}/5)^2\), where \(C\) is the Fermat quintic curve, and the action is free. \(\mathop{\mathrm{Aut}}_\mathbb{Q}(S)\) is induced by \((G \rtimes \mathbb{Z}/3) \times \mathop{\mathrm{Id}}\), and we prove that in this case \(| \mathop{\mathrm{Aut}}_\mathbb{Z}(S)| = 1 .\)
We produce in this paper an example of a surface isogenous to a product with \(| \mathop{\mathrm{Aut}}_\mathbb{Q}(S)|=192\), hence we ask the following.
Question 3. Is \(\max | \mathop{\mathrm{Aut}}_\mathbb{Q}(S)| = 192\) for a minimal surface \(S\) of general type?
The second aim that we accomplish here is to provide an example of a surface of general type \(S\) with \(| \mathop{\mathrm{Aut}}_\mathbb{Z}(S)| = 2 ,\) and \(H^*(S,\mathbb{Z})\) having nontrivial torsion (an example where there was no torsion had been previously obtained by Cai [7]).
More precisely, this is our main result, summarizing Theorems 15 and 16.
Theorem 4. Let \(S=(C_1\times C_2)/G\) be a surface isogenous to a product of unmixed type, with \(q(S)= p_g(S)=0\).
Then \(|\mathop{\mathrm{Aut}}_\mathbb{Q}(S)| \leq 192\), and equality is attained for \(G = (\mathbb{Z}/2)^3\).
If \(G\) is abelian, \(|\mathop{\mathrm{Aut}}_\mathbb{Q}(S)|\) reaches \(160\) for \(G = (\mathbb{Z}/2)^4\), \(72\) for \(G = (\mathbb{Z}/3)^2\), and \(|\mathop{\mathrm{Aut}}_\mathbb{Q}(S)| = 75\) for the classical Beauville surface; while \(|\mathop{\mathrm{Aut}}_\mathbb{Q}(S)| \leq 32\) for \(G\) not abelian.
Moreover, \(\mathop{\mathrm{Aut}}_\mathbb{Z}(S)\) is trivial, except for the case of \(G=D_4\times \mathbb{Z}/2\), where \(\mathop{\mathrm{Aut}}_\mathbb{Z}(S) \cong \mathbb{Z}/2\) and the intersection form is even.
The reader may ask for the reasons to consider surfaces isogenous to a product.
There are several, all pointing out to the phenomenon that, when \(|\mathop{\mathrm{Aut}}_\mathbb{Q}(S)|\) achieves its maximum, then \(S\) is isogenous to a product.
We review in fact in Section 3 the status of our two main questions for irregular surfaces, that is, with \(q>0\), i.e., \(H_1(S, \mathbb{Z})\) infinite.
In [8] and [9] it is proved that irregular surfaces with \(q(S) \geq 2\) and nontrivial \(\mathop{\mathrm{Aut}}_\mathbb{Q}(S)\) are necessarily isogenous to a product of unmixed type, with \(q(S)=2\); and then \(\mathop{\mathrm{Aut}}_\mathbb{Q}(S) = \mathbb{Z}/2\).
Later [10] proved that irregular surfaces with \(q(S)=1\) have \(|\mathop{\mathrm{Aut}}_\mathbb{Q}(S)| \leq 4\) and if equality holds, \(S\) is necessarily isogenous to a product of unmixed type.
Finally, in [5] Cai showed that \(|\mathop{\mathrm{Aut}}_\mathbb{Q}(S)| \leq 4\) if \(\chi(\mathcal{O}_S)> 188\), and recently in [11] the case \(|\mathop{\mathrm{Aut}}_\mathbb{Q}(S)| = 4\) has been investigated for regular surfaces. In particular, they showed that if \(K_S\) is ample and equality is attained then \(\mathop{\mathrm{Aut}}_\mathbb{Q}(S) = (\mathbb{Z}/2)^2\) and \(S\) is isogenous to a product of curves of unmixed type. Moreover, they provide an infinite series of examples having \(K^2\) arbitrary high.
These are good reasons for us to investigate in this article the groups \(\mathop{\mathrm{Aut}}_\mathbb{Z}(S), \mathop{\mathrm{Aut}}_\mathbb{Q}(S)\) for surfaces isogenous to a product.
We recall some definitions and facts essentially introduced in [12], and we refer to it for further details.
Definition 5. A smooth surface is said to be isogenous to a (higher) product if there is a finite group \(G\) and curves \(C_1, C_2\) of respective genera \(g_1, g_2 \geq 2\) such that \(G\) acts freely on \(C_1 \times C_2\), and such that \(S = (C_1 \times C_2)/G\).
The word “higher” emphasises that the respective genera of \(C_1, C_2\) are \(\geq 2\), in particular this implies that \(S\) is of general type (and with ample canonical divisor). However, for commodity, from now on we shall drop the word “higher”.
There are two types of surfaces isogenous to a product: the mixed type is when there are elements of \(G\) swapping the factors; and the unmixed type is when \(G\) acts on each curve and diagonally on the product.
Remark 6. (I) If \(S\) is isogenous to a product, it is of unmixed type unless possibly if \(C_1 \cong C_2 \cong C\).
(II) In the mixed case there is an exact sequence \(1 \rightarrowG^0 \rightarrowG \rightarrow\mathbb{Z}/2\) such that \[S^0 : = (C \times C)/G^0\] is of unmixed type, and there is an element in \(G \setminus G^0\) which is an automorphism of the form \(\tau' (x,y) = (y , \tau(x))\), with \(\tau \in G^0\).
(III) In all cases we can take a minimal realization \(S = (C_1 \times C_2)/G\), this means that no element of \(G\) acts trivially on one of the two curves.
(IV) We have an exact sequence of fundamental groups associated to the minimal realization: \[1 \rightarrow\pi_{g_1} \times \pi_{g_2} \rightarrow\pi_1(S) \rightarrowG \rightarrow1,\] which is left invariant by every automorphism of \(S\).
By the lifting condition (a continuous map lifts to an unramified covering space if and only if it leaves invariant the associated subgroup) every automorphism of \(S\) lifts to the minimal realization \(C_1 \times C_2\).
The conclusion of Remark 6-(IV) is that \(\mathop{\mathrm{Aut}}(S)\) lifts to \(C_1 \times C_2\), and we see that \[\mathop{\mathrm{Aut}}(S) = N(G) / G , \qquad N(G) < \mathop{\mathrm{Aut}}(C_1 \times C_2) \text{ being the normalizer of } G.\]
From now on we restrict to surfaces isogenous to a product of unmixed type.
Definition 7. (i) Denote for simplicity \(A_1 : = \mathop{\mathrm{Aut}}(C_1), A_1 : = \mathop{\mathrm{Aut}}(C_2)\).
(ii) say moreover that we are in the strongly unmixed case if \(C_1\) is not isomorphic to \(C_2\).
Then, in the strongly unmixed case \(\mathop{\mathrm{Aut}}(C_1 \times C_2) = A_1 \times A_2\), whereas if \(\sigma(x,y) = (y,x)\), then \[\mathop{\mathrm{Aut}}(C \times C) = A^2 \rtimes \langle \sigma\rangle \cong A^2 \rtimes \mathbb{Z}/2.\]
The description of surfaces isogenous to a product is accomplished through the Riemann existence theorem, which allows to construct Galois coverings between projective curves (we refer to [13] for more details on this part).
Definition 8. Let \(H\) be a finite group, and let \(g\geq 0 and m_1, \ldots, m_r > 1\) be integers. A Hurwitz generating vector for \(H\) of type \((g;m_1,\ldots ,m_r)\) is a \((2g+r)\)-tuple of elements of \(H\): \[V:=(a_1,b_1,\ldots, a_{g},b_{g};c_1, \ldots, c_r)\] such that
the elements of \(V\) generate \(H\),
\(\prod_{i=1}^{g}[a_i,b_i]\cdot c_1\cdot c_2\cdots c_r=1\) and
\(\mathrm{ord}(c_i)=m_i\) for \(i=1,\ldots, r\).
By Riemann’s existence theorem, any curve \(C\) together with an action of a finite group \(H\) on it, such that \(C/H\) is a curve \(C'\) of genus \(g\), is essentially determined by a generating vector \(V\) for \(H\) of signature \((g;m_1,\ldots ,m_r)\), where the mapping class group of the punctured base curve acts, in particular the orders \(m_1, \dots, m_r\) are defined only up to a permutation \(\sigma \in \mathfrak{S}_r\).
We call the elements \(a_1,b_1,\ldots, a_{g},b_{g}\) the global monodromies (they correspond to the choice of a basis of \(\pi_1(C')\)) while \(c_1, \ldots, c_r\) are called the local monodromies and correspond to geometric loops around the branch points of \(C\to C/H\). Moreover, they determine the set of elements of \(H\) which fix points on \(C\): \[\Sigma(V):= \bigcup_{h\in G} \bigcup_{i=1}^r \bigcup_{j=1}^{\mathrm{ord}(c_i)} \{ h\cdot c_i^j\cdot h^{-1}\}\,,\] is called the set of stabilizers for the action of \(H\) on \(C\).
A surface \(S=(C_1\times C_2)/G\) isogenous to a product of unmixed type is then described via a pair \((V_1,\, V_2)\) of generating vectors of \(G\), which are disjoint, in the sense that \(\Sigma(V_1)\cap \Sigma(V_2)=\{ 1\}\).
To avoid confusion, since \(G\) acts diagonally on \(C_1 \times C_2\), we view \(G < A_j\), for \(j=1,2\), and \(G\) acting on \(C_1 \times C_2\) as the diagonal subgroup \(\Delta_G < A_1 \times A_2\).
Then look at the normalizer \(N^0(G): = N(G) \cap (A_1 \times A_2)\) of \(\Delta_G < G \times G < A_1 \times A_2\) inside \(A_1 \times A_2\).
We have, setting \(\mathop{\mathrm{Ad}}(\gamma) g : = \gamma g \gamma^{-1}\), and letting \(N_ {A_j}(G)\) be the normalizer of \(G\) inside \(A_j\), \[N^0(G) = \{ (\gamma_1, \gamma_2) \in N_ {A_1}(G)\times N_ {A_2}(G) \mid \mathop{\mathrm{Ad}}(\gamma_1) = \mathop{\mathrm{Ad}}(\gamma_2)\} < N_ {A_1}(G) \times N_ {A_2}(G).\] Clearly \(\Delta_G < N^0(G)\) and we have an exact sequence of groups \[1 \rightarrow\Delta_G \rightarrowN^0(G) \cap (G \times G) \rightarrowZ(G) \rightarrow1,\] where \(Z(G)\) is the centre of \(G\) and the last map \(G \times G \rightarrowG, \;(g_1, g_2) \mapsto g_1^{-1} g_2\) is a map of sets (if \(G\) is not abelian), but a homomorphism if \(g_1^{-1} g_2 \in Z(G)\).
In other words, setting \(N_j : = N_ {A_j}(G)\), \(N^0(G)\) is the inverse image of the diagonal \(\Delta_{\mathop{\mathrm{Aut}}(G)}\) under the exact sequence \[1 \rightarrow{\mathcal{C}}_1 \times {\mathcal{C}}_2 \rightarrowN_1 \times N_2 \rightarrowH_1 \times H_2 < \mathop{\mathrm{Aut}}(G)^2 ,\] where \({\mathcal{C}}_j\) is the centralizer of \(G\) in \(A_j\), \(H_j : = N_j /{\mathcal{C}}_j\).
Summing up, if we set \(H := H_1 \cap H_2\), we have the exact sequence \[1 \rightarrow{\mathcal{C}}_1 \times {\mathcal{C}}_2 \rightarrowN^0(G) \rightarrowH \rightarrow1.\] Dividing by \(\Delta_G\) and defining \(\mathop{\mathrm{Aut}}^*(S) : = N^0(G) / \Delta_G\) we get \[\label{eq95autS} 1 \rightarrow({\mathcal{C}}_1 \times {\mathcal{C}}_2) / \Delta_{Z(G)} \rightarrow\mathop{\mathrm{Aut}}^*(S) \rightarrowH / \Delta_{\mathop{\mathrm{Inn}}(G)} \rightarrow1,\tag{1}\] since \(({\mathcal{C}}_1 \times {\mathcal{C}}_2) \cap \Delta_G = \Delta_{Z(G)} \cong Z(G)\) and \(G / Z(G) = \mathop{\mathrm{Inn}}(G)\).
Observe that in the strongly unmixed case, \(\mathop{\mathrm{Aut}}^*(S) = \mathop{\mathrm{Aut}}(S).\)
In the unmixed case with \(C_1 \cong C_2\), we have that \(N^0(G) < N(G)\) has index at most 2, and equal to \(2\) if and only if there is an automorphism \[\gamma(x,y) = (\gamma_1(y), \gamma_2(x))\] normalizing the subgroup \(\Delta_G\), which acts on \(C_1 \times C_2\) via \(g(x,y) = (g(x), g(y))\) (but where we do not identify \(C_1\) with \(C_2\) under a given isomorphism, hence we do not get the diagonal \(\Delta_G\) of \(G \times G\)).
This means that \(\gamma_1 : C_2 \rightarrowC_1, \gamma_2 : C_1 \rightarrowC_2,\) satisfy that for each \(g\in G\) there is \(g'\) such that \(\gamma_2^{-1} g \gamma_2(x) = g'(x)\), \(\gamma_1^{-1} g \gamma_1(y) = g'(y)\).
To make things more concrete, identify now \(C_1\) and \(C_2\) with the same curve \(C\) in view of a chosen isomorphism.
Then we get an automorphism \(\psi : A : = \mathop{\mathrm{Aut}}(C) \rightarrowA\) such that \[\Delta_G = \{ (g , \psi (g))| g \in G\}.\] Now \(\gamma(x,y) = (\gamma_1 (y), \gamma_2 (x))\) normalizes \(\Delta_G\) if and only if \[\gamma_1^{-1} g \gamma_1 = \psi (\gamma_2^{-1} \psi(g ) \gamma_2).\]
We do not carry out further these calculations, since in the unmixed case if we consider numerically trivial automorphisms, which cannot exchange the two fibrations of \(S\), hence they lie in \(\mathop{\mathrm{Aut}}^*(S)\).
Recall that \(N_j < A_j : = \mathop{\mathrm{Aut}}(C_j)\) is the normalizer of \(G\), hence \(G < N_j\) and \(N'_j : = N_j /G\) descends to the subgroup of automorphisms of \(C'_j = C_j /G\) which lift to \(C_j\).
Let \({\mathcal{B}}\) be the branch locus of the quotient map \(C_j \rightarrowC'_j\), so that the covering is given by an epimorphism \(\pi_1 (C'_j \setminus {\mathcal{B}}, y_0) \rightarrowG\).
Then \[N'_j : = \{ \phi \in \mathop{\mathrm{Aut}}(C'_j) \mid \phi ({\mathcal{B}}) = {\mathcal{B}}, \;\exists \Phi \in \mathop{\mathrm{Aut}}(G), \Phi \circ \mu_0 = \mu_1 \circ \phi_* \},\] Here \[\phi_* : \pi_1 (C'_j \setminus {\mathcal{B}}, y_0) \rightarrow\pi_1 (C'_j \setminus {\mathcal{B}}, \phi (y_0)),\] and \(\mu_0, \mu_1\) are the respective monodromies with base points \(y_0\), respectively \(y_1 : = \phi (y_0)\).
Since changing the path the isomorphism of the two fundamental groups changes up to inner automorphism, \(\Phi\) is only well defined up to \(\mathop{\mathrm{Inn}}(G)\).
The set of local monodromies of the covering is the so called Nielsen assignment of the conjugacy class \(Conj (\mu(b))\) (in \(G\)) to a point \(b \in {\mathcal{B}}\). And \(\phi\) induces a permutation of \({\mathcal{B}}\) preserving the Nielsen assignment.
\(\mu_0\) determines the orbifold fundamental group sequence \[1 \rightarrow\pi_1(C_j , x_0) \rightarrow\pi_1^{orb} (C'_j, y_0) \rightarrowG \rightarrow1,\] which in turn determines, via \(\phi_*, \Phi,\) the action of \(N'_j\) on \(H_1(C_j, \mathbb{Z})\).
The elements of the centralizer of \(G\) inside \(A_j\) induce an automorphism of \(C_j\) commuting with the action of \(G\), hence descending to \(C'_j\), and leaving fixed the monodromy.
In particular, if the local monodromies are pairwise distinct, then these elements induce the identity on the branch set. Hence, if the base curve \(C'_j\) has genus \(\leq 1\), then these elements induce the identity on \(C'_j\), hence \({\mathcal{C}}_j = Z(G)\).
To simplify things, assume that we are concentrating on \(\mathop{\mathrm{Aut}}_{\mathbb{Z}}(S)\). Assume that we have an automorphism \(\gamma\in \mathop{\mathrm{Aut}}_{\mathbb{Z}}(S)\). Then it induces an automorphism \(\phi\) of \(C'_j\), which is in \(N'_j\). \(\phi\) is then acting trivially on the cohomology of \(C'_j\), and we are going to show that \(\phi\) must be the identity, with three possible exceptions, where \(N''_j\), the image of \(\mathop{\mathrm{Aut}}_\mathbb{Z}(S)\) in \(N'_j\), has order equal to 2.
By Lefschetz’ theorem, this is clear if \(C'_j\) has genus at least \(2\).
If instead the genus of \(C'_j\) is 1, then \(\phi\) could be a translation, but then it should permute the branch points in \({\mathcal{B}}\). By [1], there can be only two branch points, with multiplicity \(m=2\), and the translation should be of order \(2\).
Example 9 (Exception I). The group \(G\) is generated by elements \(a,b,c : = c_1\), which are the monodromy images \(a: = \mu(\alpha), b: = \mu(\beta), c_1 := \mu (\gamma_1)\) \[\mu : \langle \alpha, \beta, \gamma_1, \gamma_2| \gamma_1^2 = \gamma_2^2= [\alpha, \beta]\gamma_1 \gamma_2 =1\rangle \rightarrowG.\] And there exists an automorphism \(\Phi : G \rightarrowG\) such that \[\Phi (a)=a, \Phi (b)=b, \Phi (c_1)=c_2, \Phi (c_2) = c_2^{-1} c_1 c_2.\]
If instead the genus of \(C'_j\) is 0, then \(\phi\) can exchange, again by the cited principle, at most two points of \({\mathcal{B}}\), which have multiplicities \(2\), while the other have odd multiplicity and are fixed. Hence we have at most 4 branch points.
Example 10 (Exception II). The group \(G\) is generated by elements \(c_1, c_2,c_3\), which are the monodromy images \(c_1: = \mu(\gamma_1), c_2 := \mu (\gamma_2), c_3 := \mu (\gamma_3)\) \[\mu : \langle \gamma_1, \gamma_2, \gamma_3 | \gamma_1^2 = \gamma_2^2= \gamma_1 \gamma_2 \gamma_3 =\gamma_3^{2n+1} = 1\rangle \rightarrowG.\] And there exists an automorphism \(\Phi : G \rightarrowG\) such that \[\Phi (c_3)=c_3 , \Phi (c_1)=c_2, \Phi (c_2) = c_2^{-1} c_1 c_2 = c_3 c_2.\] Observe that the Abelianization of \(G\) is \(\mathbb{Z}/2\), with cyclic kernel of order exactly \(2n+1\).
This case exists, since the group \(\mathbb{Z}/2 * \mathbb{Z}/2\) is generated by elements \(c'_1, c'_2\) of order two, and since \(c_3= c_2 c_1\), it suffices to take \((\mathbb{Z}/2 * \mathbb{Z}/2 )/ {\mathcal{K}}\), with \({\mathcal{K}}\) normally generated by \((c'_2 c'_1)^{2n+1}\).
Example 11 (Exception III). The group \(G\) is generated by elements \(c_1, c_3, c_4\), which are the monodromy images \(c_1: = \mu(\gamma_1), c_3 := \mu (\gamma_3), c_4 := \mu (\gamma_4)\) \[\mu : \langle \gamma_1, \gamma_2, \gamma_3, \gamma_4 | \gamma_1^2 = \gamma_2^2= \gamma_1 \gamma_2 \gamma_3 \gamma_4 =\gamma_3^{2n+1} = \gamma_4^{2h+1} =1\rangle \rightarrowG.\] And there exists an automorphism \(\Phi : G \rightarrowG\) such that \[\Phi (c_3)=c_3 , \Phi (c_4)=c_4 , , \Phi (c_1)=c_2, \Phi (c_2) = c_2^{-1} c_1 c_2 .\]
Observe that in the Abelianization \(G^{ab}\) of \(G\) (multiply by \((2n+1)(2h+1)\)) \([c_1]= [c_2]\), hence \([c_3]= -[c_4]\), therefore \(G^{ab}\) is a quotient of \(\mathbb{Z}/2 \oplus \mathbb{Z}/ (2r+1)\), where \(2r + 1 = \mathrm{GCD} (2n+1, 2h+1)\) and \(G^{ab}\) can be equal to \(\mathbb{Z}/2 \oplus \mathbb{Z}/ (2r+1)\).
Remark 12. We have \(q(S) = g'_1 + g'_2\), where \(g'_j\) is the genus of \(C'_j = C_j /G\).
In the case where \(q(S)=0\), then we can apply [2] and infer that all the branch points are left fixed, hence \(N''_1, N''_2\) are trivial.
More generally, if for instance \(g'_2=0\), then \(N''_1\) is trivial since the relative irregularity of the fibration \(S\to C'_1\) is zero (and similarly, if \(g'_1=0\), then \(N''_2\) is trivial).
Surfaces isogenous to a product with \(q= p_g=0\) carry two fibrations overs curves (see [14] for the mixed case). Therefore, in the unmixed case, \(\mathop{\mathrm{Aut}}_\mathbb{Q}(S)\) equals the group \(\mathop{\mathrm{Aut}}^*(S)\) which does not interchange these two fibrations, namely the natural isotrivial fibrations \(S\to C'_j:=C_j/G\), \(j=1,2\). In fact, \(H^j(S, \mathbb{Q})=0\) for \(j=1,3\), while \(H^2(S, \mathbb{Q})\) has rank 2 and is generated by the classes of the respective fibres of the two fibrations.
We want to study \(\mathop{\mathrm{Aut}}_\mathbb{Q}(S)\) and \(\mathop{\mathrm{Aut}}_\mathbb{Z}(S)\) for surfaces isogenous to a product of unmixed type with \(q= p_g=0\). These surfaces have been classified in [15]:
Theorem 13. Let \(S = (C_1\times C_2)/G\) be a surface isogenous to a product of unmixed type, with \(p_g(S) = 0\), then \(G\) is one of the groups in the Table 1 and the types are listed in the table. For each case in the list we have an irreducible component of the moduli space, whose dimension is denoted by \(D\).
| \(G\) | \(Id(G)\) | \(T_1\) | \(T_2\) | \(D\) | \(H_1(S,\mathbb Z)\) |
|---|---|---|---|---|---|
| \(\mathfrak{A}_5\) | \(\langle 60,5\rangle\) | \([2,5,5]\) | \([3,3,3,3]\) | 1 | \((\mathbb Z/3)^2\times (\mathbb Z/{15})\) |
| \(\mathfrak{A}_5\) | \(\langle 60,5\rangle\) | \([5,5,5]\) | \([2,2,2,3]\) | 1 | \((\mathbb Z/{10})^2\) |
| \(\mathfrak{A}_5\) | \(\langle 60,5\rangle\) | \([3,3,5]\) | \([2,2,2,2,2]\) | 2 | \((\mathbb Z/2)^3\times \mathbb Z/6\) |
| \(\frak S_ 4 \times \mathbb Z/2\) | \(\langle 48,48 \rangle\) | \([2,4,6]\) | \([2,2,2,2,2,2]\) | 3 | \((\mathbb Z/2)^4\times \mathbb Z/4\) |
| G(32) | \(\langle 32,27 \rangle\) | \([2,2,4,4]\) | \([2,2,2,4]\) | 2 | \((\mathbb Z/2)^2\times \mathbb Z/4\times \mathbb Z/8\) |
| \((\mathbb Z/5)^2\) | \(\langle 25,2\rangle\) | \([5,5,5]\) | \([5,5,5]\) | 0 | \((\mathbb Z/5)^3\) |
| \(\mathfrak{S}_4\) | \(\langle 24,12\rangle\) | \([3,4,4]\) | \([2,2,2,2,2,2]\) | 3 | \((\mathbb Z/2)^4\times \mathbb Z/8\) |
| G(16) | \(\langle 16,3\rangle\) | \([2,2,4,4]\) | \([2,2,4,4]\) | 2 | \((\mathbb Z/2)^2\times \mathbb Z/4\times \mathbb Z/8\) |
| \(D_4\times \mathbb Z/2\) | \(\langle 16,11\rangle\) | \([2,2,2,4]\) | \([2,2,2,2,2,2]\) | 4 | \((\mathbb Z/2)^3\times (\mathbb Z/4)^2\) |
| \((\mathbb Z/2)^4\) | \(\langle 16,14\rangle\) | \([2,2,2,2,2]\) | \([2,2,2,2,2]\) | 4 | \((\mathbb Z/4)^4\) |
| \((\mathbb Z/3)^2\) | \(\langle 9,2\rangle\) | \([3,3,3,3]\) | \([3,3,3,3]\) | 2 | \((\mathbb Z/3)^5\) |
| \((\mathbb Z/2)^3\) | \(\langle 8,5\rangle\) | \([2,2,2,2,2]\) | \([2,2,2,2,2,2]\) | 5 | \((\mathbb Z/2)^4\times (\mathbb Z/4)^2\) |
Remark 14. We read the Hurwitz generating vectors associated to the surfaces in Table 1 from [16] and [15].
\(G=\frak A_5\):
\(V_1=[(24)(35), (13452),(12345)]\), \(V_2=[(123),(345),(243),(152)]\);
\(V_1=[(12534),(12453),(12345)]\), \(V_2=[(12)(34),(24)(35),(14)(35),(243)]\);
\(V_1=[(123),(345),(15432)]\), \(V_2=[(12)(34),(13)(24),(14)(23),(14)(25),(14)(25)]\).
\(G=\frak S_ 4 \times \mathbb{Z}/2\):
\(V_1=[((12),0), ((1234), 1), ((243),1) ]\),
\(V_2=[((12)(34),1),((12),1),((34),1),((14)(23),1),((23),1),((14),1)]\).
\(G=G(32)=\langle x_1,x_2,x_3,x_4,x_5\mid x_j^2, [x_1,x_2]x_4,[x_1,x_3]x_5,[x_j,x_k] \text{ for } (j,k)\neq(1,2),(1,3) \rangle\):
\(V_1=[ x_2 x_3 x_4, x_2, x_1 x_2 x_3 x_5, x_1 x_2 ]\),
\(V_2= [ x_1 x_4 x_5, x_2 x_3 x_4 x_5, x_2 x_4 x_5, x_1 x_3 x_4]\).
\(G=(\mathbb{Z}/5)^2\): \(V_1=[e_1,e_2, -(e_1+e_2) ]\), \(V_2=[ e_1-e_2,e_1+2e_2,-2e_1-e_2]\).
\(G=\mathfrak {S}_4\): \(V_1=[(123),(1234),(1243) ]\), \(V_2=[(12),(12),(23),(23),(34),(34)]\).
\(G=G(16)=(\mathbb{Z}/4\times \mathbb{Z}/2)\rtimes \mathbb{Z}/2= \langle x,y,z\mid x^4,y^2,z^2, [x,y],[y,z], [z,x]y \rangle\):
\(V_1=[z,z,x,x^{-1}]\), \(V_2=[zx^2y,zx^2y,xyz,(xyz)^{-1}]\).
\(G=D_4\times \mathbb{Z}/2= \langle r,s \mid r^4,s^2, (sr)^2\rangle \times \mathbb{Z}/2\):
\(V_1=[(1,1),(s,1),(rs,0),(r,0) ]\), \(V_2=[(s,0),(sr,1),(sr^2,0),(sr,1),(r^2,1),(r^2,1) ]\).
\(G=(\mathbb{Z}/2)^4\): \(V_1=[e_1, e_2, e_3, e_4, e:=e_1+e_2+e_3+e_4 ]\), \(V_2= [ e + e_1, e + e_2, e_1 + e_3, e_2 + e_4 , e_3 + e_4]\).
\(G=(\mathbb{Z}/3)^2\): \(V_1=[e_1, e_2, -e_1, - e_2]\), \(V_2=[ e_1+e_2 , e_1 - e_2, -e_1 - e_2 , -e_1 + e_2]\).
\(G=(\mathbb{Z}/2)^3\): \(V_1=[e_1+ e_2, e_1 + e_3, e_2+ e_3, e:= e_1 +e_2 + e_3, e]\), \(V_2= [e_1, e_2, e_3, e_1, e_2, e_3]\).
Theorem 15. Let \(S\) be a surface isogenous to a product of unmixed type, with \(q(S)= p_g(S)=0\).
Then \(|\mathop{\mathrm{Aut}}_\mathbb{Q}(S)| \leq 192\), and equality is attained for \(G = (\mathbb{Z}/2)^3\).
If \(G\) is abelian, \(|\mathop{\mathrm{Aut}}_\mathbb{Q}(S)|\) reaches \(160\) for \(G = (\mathbb{Z}/2)^4\), \(72\) for \(G = (\mathbb{Z}/3)^2\), while \(|\mathop{\mathrm{Aut}}_\mathbb{Q}(S)| = 75\) for the classical Beauville surface (the sixth case in the table).
Moreover, if \(G\) is not abelian, then \(|\mathop{\mathrm{Aut}}_\mathbb{Q}(S)| \leq 32\).
Proof. We have already observed that \(\mathop{\mathrm{Aut}}_\mathbb{Q}(S)\) is the subgroup \(\mathop{\mathrm{Aut}}^*(S)\) of \(\mathop{\mathrm{Aut}}(S)\) not exchanging the two fibrations \(S \rightarrowC'_j= C_j /G\).
We have also shown that we have an exact sequence \[1 \rightarrow( {\mathcal{C}}_1 \times{\mathcal{C}}_2)/ Z(G) \rightarrow\mathop{\mathrm{Aut}}^*(S) \rightarrowH/\mathop{\mathrm{Inn}}(G) \rightarrow1.\]
If, for each \(C_j \rightarrowC'_j\) there are three points with local monodromies different from all the other local monodromies, then our previous argument shows that \({\mathcal{C}}_j = Z(G)\), hence \[( {\mathcal{C}}_1 \times{\mathcal{C}}_2)/ Z(G) \cong Z(G).\]
The case of the classical Beauville surface. Here \(G = (\mathbb{Z}/5)^2\), and the local monodromies are3 the two triples \([e_1, e_2 , -e] : = [(1,0), (0,1), (-1,-1)]\), respectively \[\Psi [e_1, e_2 , -e] : = [(1,-1), (1,2), (-2,-1)] = : [e'_1, e'_2, -e'].\] In both cases we have three elements which generate \(G\) with sum equal to zero.
Hence \(N'_j \cong \frak S_3\), and the map \(N'_j \rightarrowH_j < \mathop{\mathrm{Aut}}(G)\) is obtained if we consider the induced linear map \(\Phi\) associated to a permutation of the three vectors.
The five nontrivial maps in \(H_1\) are the maps sending \((a,b)\in G\) respectively to \[(b,a), \; (-a,-a+b), \;(a-b, -b) , \;(-b, a-b) , \;(b-a, -a).\]
The last two maps correspond to the 3-cycles of \(\frak S_3\) and cyclically permute the vectors \(e'_1, e'_2, -e'\). On the other hand, looking at the image of \(e_1'=(1,-1)\) under the first three maps, we obtain the vectors \((-1,1), (-1,-2),(2,1)\), none of which is in the set \(\{e'_1, e'_2, -e'\}\).
Therefore, \(H\) has order 3 and \(|\mathop{\mathrm{Aut}}^*(S)|= 75\).
The case with \(G= (\mathbb{Z}/2)^4\). The local monodromies are \([e_1, e_2, e_3, e_4, e:=e_1+e_2+e_3+e_4]\) and \[\Psi [e_1, e_2, e_3, e_4, e] = : [e'_1, e'_2, e'_3, e'_4, e'] : = [ e + e_1, e + e_2, e_1 + e_3, e_2 + e_4 , e_3 + e_4].\]
We see that the permutation \(\sigma_1\) exchanging \(e_1\) with \(e_2\) and \(e_3\) with \(e_4\) yields the permutation exchanging \(e'_1\) with \(e'_2\) and \(e'_3\) with \(e'_4\).
In fact, the first quintuple consists of 4 vectors of weight \(1\) and one of weight \(4\), the second quintuple consists of 3 vectors of weight \(2\) and 2 of weight \(3\). Hence, every non-trivial permutation of \(\frak S_4\) permuting the elements \(e_1, e_2, e_3, e_4\) leaves the weight invariant, therefore, if it leaves invariant the set \({\mathcal{E}}' : = \{ e + e_1, e + e_2, e_1 + e_3, e_2 + e_4 , e_3 + e_4\}\) it must swap \(e'_1\) with \(e'_2\), hence \(e_1\) with \(e_2\), and then \(e_3\) with \(e_4\).
If instead we have a permutation in \(\frak S_5 \setminus \frak S_4\), then \(e\) is exchanged with some \(e_j\).
If we exchange \(e\) with \(e_1\), then we must exchange \(e_2\) with \(e_3\), this leaves invariant the set \({\mathcal{E}}'\).
If we exchange \(e\) with \(e_3\), then we must permute \(\{e_1, e_2\}\) with \(\{e_1, e_4\}\), and the only possibility of having \({\mathcal{E}}'\) invariant is that \(e_2\) is permuted with \(e_4\), \(e_1\) is fixed.
Now, using the two above permutations of order two, and their products with \(\sigma_1\), we see that the orbit of \(e\) equals the whole set \({\mathcal{E}}: = \{e_1, e_2, e_3, e_4, e\}\).
Hence the group of permutations of the set \({\mathcal{E}}: = \{e_1, e_2, e_3, e_4, e\}\) inducing a linear map permuting the set \({\mathcal{E}}'\) has order 10, hence \(H \cong D_{5}\), and \(\mathop{\mathrm{Aut}}^*(S)\) has order equal to \(16 \cdot 10 = 160\).
The case with \(G= (\mathbb{Z}/3)^2\).
Here, the four-tuples of monodromies are \([e_1, e_2, -e_1, - e_2]\), \([e_1+e_2 , e_1 - e_2, -e_1 - e_2 , -e_1 + e_2]\).
Now, the admissible permutations of the first 4-tuple which induce an automorphism of \(G\) are leaving the partition \(\{e_1, - e_1\} \cup \{e_2, - e_2\}\) invariant.
These permutations permute clearly the complementary set \({\mathcal{E}}^c\) of \({\mathcal{E}}=\{e_1, e_2, -e_1, - e_2\}\) (inside \(G \setminus \{0\}\)), hence here \(\mathop{\mathrm{Aut}}^*(S)\) has order equal to \(9 \cdot 8 = 72\).
The next case that we consider is the record winning case.
The case with \(G= (\mathbb{Z}/2)^3\).
Here, the 6-tuple of monodromies is \([e_1, e_2, e_3, e_1, e_2, e_3]\), while the 5-tuple of monodromies is \([e_1+ e_2, e_1 + e_3, e_2+ e_3, e:= e_1 +e_2 + e_3, e]\).
We assume without loss of generality that the first 3 branch points are \(\infty, 0, 1\).
We claim that, for a proper choice of the branch points, the group \(N'_1\) surjects onto the group \(\frak S_3\) of automorphisms of \(G\), permuting the three vectors \(e_1, e_2, e_3\).
To achieve this, we need to show that, for each permutation of \(\{1,2,3\}\), there is a projectivity permuting accordingly the pairs \(\{\infty, P_4\}\), \(\{0 , P_5\}\), \(\{1, P_6\}\).
A way to show this is to observe that \({\mathcal{B}}_1: = \{\infty, 0, 1\}\) is an orbit for the natural subgroup, isomorphic to \(\frak S_3\), preserving \({\mathcal{B}}_1\). Then \({\mathcal{B}}_2 : = \{P_4, P_5, P_6\}\) must be another orbit, in particular, given the tranposition \(z \mapsto 1/z\) exchanging \(\infty\) and \(0\), we may take \(P_6\) as the other fixpoint \(-1\). Arguing like this, we find that \(P_4, P_5, P_6\) must be \(1/2, 2, -1\).
The kernel of this surjection consists of projectivities \(\tau\) fixing each of the sets \(\{\infty, P_4\}\), \(\{0 , P_5\}\), \(\{1, P_6\}\).
Clearly since a projectivity with four fixpoints is the identity, the kernel is a subgroup of \((\mathbb{Z}/2)^3\) such that \(\tau\) different from the identity has at most one coordinate different from \(1\). Hence the subgroup is either \((\mathbb{Z}/2)(1,1,1)\) or is contained in \((1,1,1)^{\perp}\).
In both cases, if there is such a nontrivial \(\tau\), there is a transformation permuting the two elements of a couple, without loss of generality \(1, -1\). The only projectivities \(\tau\) with such a property are of the form \(z \mapsto \frac{a z + b }{-b z -a}\), which have order 2. If \(a=0\), then we obtain that \(0, \infty\) are exchanged, which contradicts that \(\tau\) is in the kernel. Hence we may assume that \(a=1\), and then \(\tau (0) =0 \Leftrightarrow b=0\) and so \(\tau(2) = -2\), hence this case is excluded and we must have that \(\tau\) transposes \(0\) with \(2\), that is, \(a=1, b= -2\); we see now that \(\tau (z) = \frac{ z -2 }{2 z -1}\) transposes \(0\) with \(2\), \(\infty\) with \(1/2\), and \(1\) with \(-1\).
In this way we have found a group \(N'_1\) of cardinality \(6\cdot 2=12\).
For each other choice of branch points \(P_1, \dots, P_6\) we have a homorphism \(N'_1 \rightarrow\frak S_3\) whose kernel has either 4 or 2 elements.
The cardinality 12 is maximal unless \(N'_1\) surjects onto \(\frak S_3\) and the kernel has order \(4\).
In this case \(N'_1\) would be a non abelian subgroup of \(\mathbb{P}GL(2, \mathbb{C})\) of order \(24\), hence it would be isomorphic to \(\frak S_4\). There is indeed a surjection of \(\frak S_4\) onto \(\frak S_3\) with kernel the Klein group \({\mathcal{K}}\) of double transpositions.
There is according to Klein only one action of \(\frak S_4\) on \(\mathbb{P}^1\), corresponding to a Galois covering of \(\mathbb{P}^1\) branched in three points, with local monodromies of respective orders \(2,3,4\).
Then all the orbits have \(24\) elements, except the special orbits, with respectively \(12,8, 6\) elements.
Hence \({\mathcal{B}}\), which is \(N'_1\)-invariant, is a union of orbits and should be the orbit with 6 elements, each having as stabilizer a cyclical permutation of order \(4\).
The partition of the orbit as the union of three pairs corresponds to the choice to the three Sylow subgroups of order \(8\), which are dihedral groups \(D_4\), inverse images of the three transpositions of \(\frak S_3\). To each pair is then associated an element of the set \(\{e_1, e_2, e_3\}\) and in this way the monodromy homomorphism to \(G\) is determined.
Concerning the second monodromy, we see that we have a surjection of \(N'_2\) onto \(\frak S_3\) which permutes the three different monodromies, and leaves \(e\) fixed. And the kernel must be trivial.
In both cases the previous criterion does not apply, because there are repeated monodromies, but we see that \(N'_2\) consists of \(6\) elements, and for each one of them there is a permutation in \(N'_1\), consisting of \(12\) or \(24\) elements, corresponding to the given action on \(G\); hence \(\mathop{\mathrm{Aut}}^*(S)\) has order equal to \(8 \cdot 12\) or \(8 \cdot 24\), that is, \(96\) or \(192\) elements.
The non-abelian cases.
We shall give more crude estimates for the case where the group \(G\) is not abelian.
Recalling that \(H = H_1 \cap H_2 \supset \mathop{\mathrm{Inn}}(G)\), we have \(\frac{ H}{\mathop{\mathrm{Inn}}(G)} \subseteq \frac{ H_j }{\mathop{\mathrm{Inn}}(G)}\) and the exact sequence \[1\rightarrow{\mathcal{C}}_j/Z(G) \rightarrowN_j'\rightarrowH_j/\mathop{\mathrm{Inn}}(G)\to 1,\] which together with 1 provide the rough estimation:
\[|\mathop{\mathrm{Aut}}^*(S)|\leq \min\{|N_1'|\cdot |{\mathcal{C}}_2| , |N_2'|\cdot |{\mathcal{C}}_1| \}.\]
The group \(N'_j\), being a finite subgroup of \(\mathbb{P}GL(2, \mathbb{C})\) is either:
cyclic \(\mathbb{Z}/n\), with two orbits of cardinality 1, the others of cardinality \(n\);
dihedral \(D_n\), with three special orbits of cardinalities \((n,n,2)\) and the others of cardinality \(2n\);
\(\frak A_4\), with three special orbits of cardinalities \((6,4,4)\) and the others of cardinality \(12\);
\(\frak S_4\), with three special orbits of cardinalities \((12, 8,6)\) and the others of cardinality \(24\);
\(\frak A_5\), with three special orbits of cardinalities \((30,20,12)\) and the others of cardinality \(60\).
Since \({\mathcal{B}}_j\) is a union of orbits for \(N_j'\), we observe that if there is an orbit of length 1, then \(N'_j\) must be cyclic.
Looking at Table 1, we see that the case \(\frak A_5\) never happens, and if \(N_j'=\frak S_4\), then \({\mathcal{B}}_j\) is a single orbit with \(6\) elements, and the six monodromies have the same order.
Similarly, if \(N_j'=\frak A_4\), then \({\mathcal{B}}_j\) is a single orbit with either \(4\) or \(6\) elements, and in both cases all branching order are equal.
In the case where \(N'_j\) is cyclic, then its order is bounded by \(|{\mathcal{B}}_j|\) if all the branching orders are equal, otherwise by \(|{\mathcal{B}}_j|-1\).
Similarly, if \(N'_j\) is dihedral, its order is bounded by \(2| {\mathcal{B}}_j|\) if all the branching orders are equal, otherwise by \(2|{\mathcal{B}}_j|-4\).
Using these considerations we obtain the following estimations:
if \(G=\frak A_5\), \(T_1=[2,5,5]\) and \(T_2=[3,3,3,3]\), then \(|N_1'|\leq 2\) and \(|N_2'|\leq 12\).
If \(G=\frak A_5\), \(T_1=[5,5,5]\) and \(T_2=[2,2,2,3]\), then \(|N_1'|\leq 6\) and \(|N_2'|\leq 3\).
If \(G=\frak A_5\), \(T_1=[3,3,5]\) and \(T_2=[2,2,2,2,2]\), then \(|N_1'|\leq 2\) and \(|N_2'|\leq 10\).
If \(\mathfrak S_ 4 \times \mathbb{Z}/2\), \(T_1=[2,4,6]\) and \(T_2=[2,2,2,2,2,2]\), then \(|N_1'|=1\) and \(|N_2'|\leq 24\).
If \(G=G(32)\), \(T_1=[2,2,4,4]\) and \(T_2=[2,2,2,4]\), then \(|N_1'|\leq 4\) and \(|N_2'|\leq 3\).
If \(G=\frak S_4\), \(T_1=[3,4,4]\) and \(T_2=[2,2,2,2,2,2]\), then \(|N_1'|\leq 2\) and \(|N_2'|\leq 24\).
If \(G=G(16)\), \(T_1=[2,2,4,4]\) and \(T_2=[2,2,4,4]\), then \(|N_1'|\leq 4\) and \(|N_2'|\leq 4\).
If \(G=D_4\times \mathbb{Z}/2\), \(T_1=[2,2,2,4]\) and \(T_2=[2,2,2,2,2,2]\), then \(|N_1'|\leq 3\) and \(|N_2'|\leq 24\).
We look now at the cardinalities \(|{\mathcal{C}}_1| , |{\mathcal{C}}_2 |\), and recall that \(Z(G)\) is trivial for \(\frak A_5\) and \(\frak S_4\), has order \(2\) for \(\frak S_4 \times \mathbb{Z}/2\), and equals \((\mathbb{Z}/2)^2\) for \(G(32), G(16)\) and \(D_4 \times \mathbb{Z}/2\).
As remarked above, if for \(C_j \rightarrowC'_j\) there are three points with local monodromies different from all the other local monodromies, then \({\mathcal{C}}_j = Z(G)\).
By inspecting the local monodromies reported in Remark 14, we see that \(|{\mathcal{C}}_j| = |Z(G)|\) (\(j=1,2\)), except possibly if \(G=\frak S_4\) or \(G=G(16)\) or \(G=D_4\times \mathbb{Z}/2\).
Thus, if \(G\) is not one of these 3 groups, we achieve the bound \(|\mathop{\mathrm{Aut}}^*(S)|\leq 12\).
For \(G=\frak S_4\), it holds \({\mathcal{C}}_1=Z(G)\), and \(|\mathop{\mathrm{Aut}}^*(S)| \leq |N_2'|\leq 24\).
Let us consider \(G=G(16)\), whose centre is generated by \(x^2,y\) (see Remark 14). The local monodromies are \([z,z,x,x^{-1}]\) and \([zx^2y,zx^2y,xyz,(xyz)^{-1}]\), so that \(|{\mathcal{C}}_j/Z(G)|\leq 2\), whence \(|\mathop{\mathrm{Aut}}^*(S)|\leq 32\).
Finally, let us consider \(G=D_4\times \mathbb{Z}/2\), whose centre has order 4 and the local monodromies are: \[V_1=[(1,1),(s,1),(rs,0),(r,0) ], \;V_2=[(s,0),(sr,1),(sr^2,0),(sr,1),(r^2,1),(r^2,1) ].\]
so that \({\mathcal{C}}_1=Z(G)\) and \(|{\mathcal{C}}_2/Z(G)|\leq 2\), whence \(|\mathop{\mathrm{Aut}}^*(S)|\leq 24\). ◻
We consider now \(\mathop{\mathrm{Aut}}_\mathbb{Z}(S)\).
By Remark 12, it follows that \(\mathop{\mathrm{Aut}}_{\mathbb{Z}}(S) < Z(G) = N^0(G) \cap (G \times G) / \Delta_G\), because the image of \(\mathop{\mathrm{Aut}}_\mathbb{Z}(S)\) in \(N'_j=N_j/G\) is trivial for \(j=1,2\). This means, that \(\mathop{\mathrm{Aut}}_{\mathbb{Z}}(S)\) is a group of automorphisms induced by a subgroup \(H < Z(G) \times\{1_G\} < G \times G\).
Theorem 16. Let \(S\) be a surface isogenous to a product of unmixed type, with \(q(S)= p_g(S)=0\).
Then \(\mathop{\mathrm{Aut}}_\mathbb{Z}(S)\) is trivial, except for the case of \(G=D_4\times \mathbb{Z}/2\), where \(\mathop{\mathrm{Aut}}_\mathbb{Z}(S) \cong \mathbb{Z}/2\) and the intersection form is even.
Proof. The statement is clear for the cases with \(G=\frak A_5\) and \(G=\frak S_4\).
In the remaining case, we use a MAGMA [17] script to determine which automorphisms in \(Z(G)\) act trivially on \(H_1(S,\mathbb{Z})\) (see Appendix 4).
The computations below show that there are no non-trivial elements acting trivially on \(H_1(S,\mathbb{Z})\) in all cases except for \(G=D_4\times \mathbb{Z}/2\).
For \(G=D_4\times \mathbb{Z}/2\) the element \((r^2,0)\) acts trivially on \(H_1(S,\mathbb{Z}) \cong (\mathbb{Z}/2)^3\times (\mathbb{Z}/4)^2\).
There remains to show that \((r^2,0)\) acts trivially on \(H^2(S,\mathbb{Z})\).
In view of the exact sequence \[1 \rightarrow\mathop{\mathrm{Tors}}(S) \cong (\mathbb{Z}/2)^3\times (\mathbb{Z}/4)^2 \rightarrowH^2(S,\mathbb{Z}) \rightarrow\mathop{\mathrm{Num}}(S) = H^2(S,\mathbb{Z})/ \mathop{\mathrm{Tors}}(S) \cong \mathbb{Z}^2\rightarrow0,\] and since \((r^2,0)\) acts trivially on \(\mathop{\mathrm{Tors}}(S)\), and on the quotient group \(\mathop{\mathrm{Num}}(S)\), it suffices to find a splitting \(\mathop{\mathrm{Num}}(S) \rightarrowH^2(S,\mathbb{Z})\) which is \((r^2,0)\)-invariant.
As in [18] we consider the two fibrations \(f_i : S \rightarrowC_i /G\), and denote by \(F_i\) the respective fibres, and argue as follows.
Because of the multiplicities of the fibres, we find \(Z(G)\) invariant divisors \(\Phi_1, \Phi_2\) such that the following linear equivalences hold true: \(F_1 \equiv 4 \Phi_1, F_2 \equiv 2 \Phi_2\).
Observe that \(\Phi_1 \cdot \Phi_2 = |G|/ 8 = 2.\)
Moreover \(K_S\) is numerically equivalent to \(\Phi_1 + 2 \Phi_2\), hence the intersection form on \(S\) is even if the numerical class of \(\Phi_1\) is divisible by 2, and then if \(2 \Phi'_1\) is numerically equivalent to \(\Phi_1\), then \(\Phi'_1 , \Phi_2\) are a basis of \(\mathop{\mathrm{Num}}(S)\), with \(\Phi'_1 \cdot \Phi_2 =1, (\Phi'_1)^2 = \Phi_2^2 =0\).
To find \(\Phi'_1\), consider the surjection \(G = D_4 \times \mathbb{Z}/2 \twoheadrightarrow (\mathbb{Z}/2)^3\), and its kernel \(H = \langle r^2 \rangle \cong \mathbb{Z}/2\).
Then, setting \[B = C_1 / (\mathbb{Z}/2) , E : = C_1 / H,\] \(B\) is a \(D_4\)-covering of \(\mathbb{P}^1\) with branching multiplicities \((2,2,4)\), hence \(B\) is a projective line, while \(E\) is an elliptic curve, a \((\mathbb{Z}/2)^3\) covering of \(\mathbb{P}^1\) branched in 4 points, and \(C_1 \rightarrowE\) is branched in 4 points.
Hence \(C: = C_1\) is a hyperelliptic curve of genus \(3\), and \(K_C = 2 {\mathcal{H}}\), where \({\mathcal{H}}\) is the hyperelliptic divisor (inverse image of a point in \(B\)), whose linear equivalence class is invariant for every automorphism of \(C\).
We consider then a divisor \(\hat{\Phi}_1\) on \(C_1 \times C_2\) corresponding to the divisor \({\mathcal{H}}\) on \(C_1\). More precisely, consider the \(G\)-orbit \({\mathcal{O}}\) in \(C\) corresponding to the last branch point of \(C \rightarrow\mathbb{P}^1\), which is indeed the \(G / \langle r\rangle \cong \mathbb{Z}/2 \times \mathbb{Z}/2\)-orbit.
\({\mathcal{O}}= \{ P, (0,1)P, (s,0) P, (s,1)P\}\) consists of the ramification points for \(C \rightarrowE\), hence \[K_C \equiv (P+ (0,1)P) + ( (s,0) P + (s,1)P) \equiv {\mathcal{H}}+ {\mathcal{H}},\] since \({\mathcal{H}}\) is the pull back of a point in \(B \cong \mathbb{P}^1\).
In particular, \[(P+ (0,1)P) \equiv ( (s,0) P + (s,1)P).\]
On \(C_1 \times C_2\) consider the divisor \(\hat{\Phi}_1: = P \times C_2 + (0,1)P\times C_2.\)
For each automorphism in \(G \times G\) the divisor \(\hat{\Phi}_1\) is either left fixed, or sent to \((s,0) \hat{\Phi}_1\): then the previous linear equivalence shows that the linear equivalence class of \(\hat{\Phi}_1\) is left invariant.
By our choice, the effective divisor \(\hat{\Phi}_1\) on \(C_1 \times C_2\) is invariant for the action of the group \(G'': = \langle r \rangle \times \mathbb{Z}/2\), hence the corresponding line bundle has a \(G''\)-linearization, therefore \(\hat{\Phi}_1\) descends to a divisor \(\Phi''_1\) on \(S'' : = (C_1 \times C_2) / G''\) of which it is the inverse image.
Likewise \(\hat{\Phi}_1 + (s,0) \hat{\Phi}_1\) is \(G\)-invariant, and is the full inverse image of \(\Phi_1\).
We have an unramified double cover \(p: S'' \rightarrowS = S'' / s\), and both \(\Phi''_1, s \Phi''_1\) satisfy \[p_* (\Phi''_1 ) = p_* (s \Phi''_1) = \Phi_1.\]
Now, consider the line bundle \({\mathcal{L}}: = \mathcal{O}_{S''}( \Phi''_1)\). Then, since also \((s,0) \hat{\Phi}_1\) is \(G''\)-invariant and linearly equivalent to \(\hat{\Phi}_1\), \(s^*({\mathcal{L}}) \cong {\mathcal{L}}\), and we can consider the associated Theta group \[1 \rightarrow\mathbb{C}^* \rightarrow\Theta({\mathcal{L}}) \rightarrow\mathbb{Z}/2 = \langle s \rangle \rightarrow1,\] which is classified as a central extension by \(\epsilon\in H^2(\mathbb{Z}/2 , \mathbb{C}^*) = 1.\)
Hence the Theta group splits and there is a a lifting of \(s\) to an automorphism of \({\mathcal{L}}\), which yields a quotient line bundle on \(S\), hence a divisor class \(\Phi'_1\) on \(S\) such that \(p^*(\Phi'_1) \equiv \Phi''_1.\)
But then \(2 \Phi'_1 = p_*(p^*(\Phi'_1)) \equiv p_*( \Phi''_1) = \Phi_1.\) ◻
Remark 17. That the intersection form is even for \(D_4 \times \mathbb{Z}/2\), this was proven also by Kyoung-Seog-Lee and his collaborators (personal communication).
In the paper [8] it was shown that, for a surface \(S\) with \(q(S) \geq 2\), \(\mathop{\mathrm{Aut}}_{\mathbb{Q}}(S)\) is trivial, unless \(q(S)=2\) and the surface \(S\) is isogenous to a product of unmixed type, with \(C'_j = C_j /G\) having genus \(1\) for \(j=1,2\), the group \(G\) is abelian with \[G = (\mathbb{Z}/2m) \oplus (\mathbb{Z}/2mn), \quad {\rm or } \quad \;G = (\mathbb{Z}/2) \oplus (\mathbb{Z}/2m) \oplus (\mathbb{Z}/2mn),\] and all the branch multiplicities are equal to \(2\).
Using the notation of [8] we let the number of branch points be \(2k\) for the first covering, and \(2l\) for the second covering. Moreover, the local monodromies of the first covering are all equal to \(\gamma\), while the local monodromies of the second covering all are equal to \(\gamma'\), with \(\gamma\neq \gamma'\).
If \(k \geq 2\), then \(N'_1\) is trivial, similarly \(N'_2\) is trivial if \(l \geq 2\).
In loc. cit. it was proven that \(\mathop{\mathrm{Aut}}_{\mathbb{Q}}(S)\) is the involution \(\iota\) induced by \(\gamma\times \gamma' \in G \times G\).
We shall now see whether \(\iota\) belongs to \(\mathop{\mathrm{Aut}}_{\mathbb{Z}}(S)\).
Question 18. If \(S\) is an irregular surface with \(q(S)=2\), then can \(\mathop{\mathrm{Aut}}_{\mathbb{Z}}(S)\) be nontrivial? Equivalently, is it possible that \(\mathop{\mathrm{Aut}}_{\mathbb{Z}}(S) = \langle \iota \rangle \cong \mathbb{Z}/2\)?
The first author, in joint work with Gromadtzki in summer 2015 ([19], or handwritten notes by the first author), has studied the action of \(\iota\) on \(H_1(S,\mathbb{Z})\), showing that for certain values of \(m,n\) the action is trivial, and for others it is nontrivial (for the first case, triviality amounts to \(m\) being even).
The difficulty is however to determine the triviality of the action on \(H^2(S, \mathbb{Z})\), because the above surfaces have large \(p_g\).
The paper [10] showed that for a surface \(S\) with \(q(S) =1\), if \(\mathop{\mathrm{Aut}}_{\mathbb{Q}}(S)\) is non-trivial, then it has order at most 4, and that if equality holds, then \(S\) is isogenous to a product of unmixed type; the authors also give examples where \(\mathop{\mathrm{Aut}}_{\mathbb{Q}}(S)\) has order 4.
If \(q=1\), then any element \(h \in \mathop{\mathrm{Aut}}_\mathbb{Z}(S)\) acts as the identity on the elliptic curve \(A = \mathop{\mathrm{Alb}}(S)\), by [2].
Assume now that \(S\) is isogenous to a product of unmixed type: is the case \(|\mathop{\mathrm{Aut}}_\mathbb{Z}(S)|=4\), \(q(S)=1\) possible?
The surface \(S\) admits a second fibration \(f : S \rightarrowC_2/G = \mathbb{P}^1\), which has multiple fibres of multiplicities \(m_1 \leq m_2\leq \dots \leq m_r\) with \(r \geq 3\), and \(h\) acts as the identity on \(C_2/G\), by [1] unless \(m_1=m_2=2\), \(m_j\) is odd for \(j=3,4\), \(r = 4\) and the action on \(C_2/G\) has order 2.
Hence, if \(\mathop{\mathrm{Aut}}_{\mathbb{Z}}(S)\) had order 4, there would be an element \(h=(h_1,h_2)\) acting as the identity on \(C_2/G\), and acting as the identity on \(C_1\), so that necessarily \(h_2 \in Z(G)\).
But, again, how to deal with the transcendental cycles in \(H^2(S, \mathbb{Z})\)?
The script can be run at http://magma.maths.usyd.edu.au/calc/.
/* Input: i) the group: G; ii) the monodromy images of pi_orb(C_i)-> G: mon1 and mon 2; iii) the genus of the quotient curves C_i/G: h1 and h2;
Output: the elements of Z(G), acting trivially on H_1(S,Z)*/
// Given the monodromy images, the next function constructs // the group pi^orb and the monodromy map pi^orb–>>G
Orbi:=function(gr,mon, h) F:=FreeGroup(#mon); Rel:=; G:=Id(F); for i in 1..h do G:=G*(F.(2*i-1)^-1,F.(2*i)^-1); end for; for i in 2*h+1..#mon do G:=G*F.(i); Include( Rel,F.(i)^(Order(mon[i]))); end for; Include( Rel,G); P:=quo<F|Rel>; return P, hom<P->gr|mon>; end function;
// MapProd computes given two maps f,g:A->B the map product induced by the product on B
MapProd:=function(map1,map2) seq:=[]; A:=Domain(map1); B:=Codomain(map1); if Category(A) eq GrpPC then n:=NPCgens(A); else n:=NumberOfGenerators(A); end if; for i in [1..n] do Append( seq, map1(A.i)*map2(A.i)); end for; return hom<A->B|seq>; end function;
// This is the main routine of the script
TrivialActionH1:=function(G,mon1, h1, mon2, h2) // First of all we construct the monodromy maps pi_j^orb–>>G
T1,f1:=Orbi(G,mon1, h1); T2,f2:=Orbi(G,mon2, h2);
T1xT2,inT,proT:=DirectProduct([T1,T2]); GxG,inG:=DirectProduct(G,G); Diag:=MapProd(inG[1],inG[2])(G); f:=MapProd(proT[1]*f1*inG[1],proT[2]*f2*inG[2]); Pi1S:=Rewrite(T1xT2,Diag@f?); // This is the fundamental group of S=(C1xC2)/G H1S,q:=AbelianQuotient(Pi1S); //H_1(S,Z)
triv:=G!1;
for z in Center(G) do // we consider the automorphisms in Z(G) and their action on H_1(S,Z) act:=; // in act we collect the differences z*h-h, where h is a generator of H_1(S,Z) z1:=(inG[1](z)@f?); // lifts of z=(z,1) to the productT1xT2 for h in Generators(H1S) do h1:=h@q?; Include( act,q(z1*h1*z1^-1)-h); // con end for; // check if g in Z(G)=Aut_Q(S), acts trivially on H_1(S,Z) if act eq H1S!0 then Include( triv,z); end if; end for;
return triv; end function;
We use the previous script, to check, which elements of \(Z(G)\) act non-trivially on \(H_1(S,\mathbb{Z})\), for the algebraic data listed in Remark 14.
G:=SmallGroup(48,48); // S4 x Z/2 mon1:=[ G.1 * G.4 * G.5, G.1 * G.2 * G.3^2 * G.5, G.2 * G.3 ]; mon2:=[ G.2 * G.5, G.2 * G.4 * G.5, G.1 * G.2 * G.3^2 * G.4, G.1 * G.2 * G.3 * G.5, G.1 * G.2 * G.3 * G.5, G.1 * G.2 * G.3^2 ]; TrivialActionH1(G,mon1,0,mon2,0); > Id(G)
G:=SmallGroup(32,27); // G(32) mon1:=[ G.2 * G.3 * G.4, G.2, G.1 * G.2 * G.3 * G.5, G.1 * G.2 ]; mon2:=[ G.1 * G.4 * G.5, G.2 * G.3 * G.4 * G.5, G.2 * G.4 * G.5, G.1 * G.3 * G.4]; TrivialActionH1(G,mon1,0,mon2,0); > Id(G)
G:=SmallGroup(25,2); // (Z/5)^2 mon1:=[ G.1, G.2, G.1^4 * G.2^4 ]; mon2:=[ G.1*G.2^4, G.1 * G.2^2, G.1^3 * G.2^4 ]; TrivialActionH1(G,mon1,0,mon2,0); > Id(G)
G:=SmallGroup(16,3); // G(16) mon1:=[ G.2, G.2, G.1, G.1^3 ]; mon2:=[ G.2 * G.1^2* G.3 , G.2 * G.1^2* G.3, G.1*G.3*G.2, (G.1*G.3*G.2)^-1 ]; TrivialActionH1(G,mon1,0,mon2,0); > Id(G)
G:=SmallGroup(16,11); // D4 x Z/2 mon1:=[ G.2 * G.3, G.3 * G.4, G.1 * G.3, G.1 * G.2 * G.3 * G.4 ]; mon2:=[ G.1 * G.4, G.1 * G.4, G.3, G.3, G.2 * G.4, G.2 * G.4 ]; TrivialActionH1(G,mon1,0,mon2,0); > Id(G), G.4 // G.4=(r^2,0) acts trivially
G:=SmallGroup(16,14); // (Z/2)^4 mon1:=[ G.1, G.2, G.3, G.4, G.1 * G.2 * G.3 * G.4 ]; mon2:=[ G.2 * G.3 * G.4, G.1 * G.3 * G.4, G.1 * G.3, G.2 * G.4, G.3 * G.4 ]; TrivialActionH1(G,mon1,0,mon2,0); > Id(G)
G:=SmallGroup(9,2); // (Z/3)^2 mon1:=[ G.1, G.2, G.1^2, G.2^2 ]; mon2:=[ G.1 * G.2, G.1 * G.2^2, G.1^2 * G.2^2, G.1^2 * G.2 ]; TrivialActionH1(G,mon1,0,mon2,0); > Id(G)
G:=SmallGroup(8,5); // (Z/2)^3 mon1:=[ G.1 * G.2, G.1 * G.3, G.2 * G.3, G.1 * G.2 * G.3, G.1 * G.2 * G.3 ]; mon2:=[ G.1, G.2, G.3, G.1, G.2, G.3 ]; TrivialActionH1(G,mon1,0,mon2,0); > Id(G)