Proof. The $k=1$ case goes back to [Reference Bardelli and PirolaBP89]; the more general version stated above is similar.

First of all, in order to prove ${\mathcal {B}} \simeq {\mathcal {A}}^k$ over $S$ , it suffices to show that the variation of integral Hodge structure $ R^1g_\ast \mathbb Z$ on $S$ is isomorphic to $(R^1f_\ast \mathbb Z)^{\oplus k}$ . The matrix $\alpha$ has full rank, because $\psi _t$ is an isogeny and $\psi _t^\ast (H^1(B_t,\mathbb Z)) = \alpha \cdot H^1(A_t,\mathbb Z)^{\oplus k}$ by assumption. Hence, fibrewise multiplication by $\alpha$ yields an embedding of integral Hodge structures $\alpha \colon (R^1f_\ast \mathbb Z)^{\oplus k} \to (R^1f_\ast \mathbb Z)^{\oplus k}$ . Similarly, fibrewise push-forward along the family of isogenies $\psi$ yields an embedding $ \psi ^\ast \colon R^1g_\ast \mathbb Z \to (R^1f_\ast \mathbb Z)^{\oplus k}$ . To prove that $ R^1g_\ast \mathbb Z\simeq (R^1f_\ast \mathbb Z)^{\oplus k}$ , it thus suffices to prove that the images of the above embeddings coincide, i.e. that the following equality holds:

\begin{equation*} {\mathrm {im}}(\psi ^\ast \colon R^1g_\ast \mathbb Z \to (R^1f_\ast \mathbb Z)^{\oplus k})={\mathrm {im}}(\alpha \colon (R^1f_\ast \mathbb Z)^{\oplus k} \to (R^1f_\ast \mathbb Z)^{\oplus k} ). \end{equation*}

Since $S$ is connected, the above identity can be checked at the single point $t\in S$ , where it holds by assumption. This concludes the proof of the lemma.

Proof. By Lemma 2.1, it suffices to show that the induced map $\psi _t^\ast \colon H^1(B_{t},\mathbb Z) \to H^1(A_{t},\mathbb Z)^{\oplus k}$ has image $ \alpha \cdot H^1(A_{t},\mathbb Z)^{\oplus k} \subset H^1(A_{t},\mathbb Z)^{\oplus k}$ for some $\alpha \in \mathrm {M}_k(\mathbb Z)$ . This is clear: as ${\mathrm {End}}(A_{t}) = \mathbb Z$ , the composition

\begin{equation*} A_{t}^k \stackrel {\psi _t} \longrightarrow B_{t} \simeq A_{t}^k \end{equation*}

is given by a matrix $\alpha \in \mathrm {M}_k(\mathbb Z)$ , and hence the same is true on cohomology.

Proof. This is well known (see e.g. [Reference Bardelli and PirolaBP89, Remark 3.1.2]).

Proof. Let $p,q, p', q' \in C$ . Then $[p - q] = [p' - q'] \in JC$ implies that the divisors $p+q'$ and $p'+q$ on $C$ are linearly equivalent. Hence they are equal, because $C$ is not hyperelliptic.

Proof. Assume that there is a very general line $\ell \subset \mathbb P^2$ and points $p,q,p',q' \in C\cap \ell$ with $p\neq q$ and $p'\neq q'$ such that $m[p-q] = m[p'-q'] \in JC$ for some $m \geq 1$ . We need to show $p=p'$ and $q=q'$ . For a contradiction, we assume that this is not the case.

The line $\ell$ is spanned by the two points $p$ and $q$ . Since $\ell$ is very general, so is $(p,q)\in C\times C$ . In a first step, we note that $p=q'$ and $q=p'$ is impossible, as it leads to $2m[p-q]=0\in JC$ for very general $(p,q)\in C\times C$ , which is absurd. Since $(p,q)\neq (p',q')$ by assumption, we get that $\{p,q\}\neq \{p',q'\}$ .

We specialize the line $\ell$ to a line $\ell _0$ so that $p$ and $q$ collapse to a single very general point $p_0\in C$ and $\ell _0$ is the tangent line of $C$ at $p_0$ . We denote the limit points of $p'$ and $q'$ by $p'_0$ and $q'_0$ , respectively. Since $\{p,q\}\neq \{p',q'\}$ , the equality $p'_0= q'_0$ would imply that $\ell _0$ is a bitangent, which is impossible as their number is finite (there are exactly 28 such lines). Hence, $p'_0\neq q'_0$ . Moreover, the identity $m[p-q] = m[p'-q'] \in JC$ specializes to the identity $0 = m [p_0' - q_0'] \in JC$ . We can further specialize the point $p_0\in C$ at which $\ell _0$ is tangent and find that there is a one-dimensional family of points $(p_0',q_0')\in C\times C$ such that $0 = m [p_0' - q_0'] \in JC$ and $p_0' \neq q_0'$ . Since the $m$ -torsion points of $JC$ are discrete, taking the closure of the above one-dimensional family yields a curve in $C\times C$ which is different from the diagonal and which is contracted by the difference map $C\times C\to C-C$ . This contradicts Lemma 2.4 and hence concludes the proof.

Proof. By Lemma 2.3, the morphism $s\colon C\times C\to s(C\times C)$ is generically finite. Hence, the commutativity of diagram (2.3) shows that assertion (2) follows from assertion (1).

Let us prove assertion (1). By Lemma 2.4, the map $a$ is birational, so it remains to prove that the map $m \colon C-C \to m(C-C)$ is birational. Since diagram (2.3) commutes and $a$ is birational, we get that $\deg (s) = \deg (G_1) = \deg (m) \cdot \deg (G_m)$ . To prove $\deg (m)=1$ it thus suffices to show $\deg (s) \leq \deg (G_m)$ . Assume first that $g \geq 4$ . Let $p \in C$ be any point, and let $q \in C$ be a point such that $[p-q] \in JC$ is not $2m$ -torsion. Then, on the one hand, $m[p-q] \neq m[q-p] \in JC$ and, on the other hand, $G_m(m[p-q]) = G_m(m[q-p]) \in \mathrm {Gr}(2, T_0JC)$ . Therefore $2 \leq \deg (G_m)$ , and since $\deg (s) = 2$ by Lemma 2.3, we get $\deg (s) \leq \deg (G_m)$ , proving what we want.

Next, assume that $g = 3$ , and consider the canonical embedding $C \hookrightarrow \mathbb P^2$ . Let $\ell \subset \mathbb P^2$ be a very general line, so that $\ell \cap C = \left \{ p_1, p_2, p_3, p_4 \right \} \subset C$ for distinct points $p_1, p_2,p_3\mbox{ and }p_4$ on $C$ . The elements $m[p_i-p_j] \in m(C-C)$ for $i \neq j$ are all sent to the same element in $\mathrm {Gr}(2, T_0JC)$ under the rational map $G_m$ . Moreover, the twelve points $m[p_i-p_j] \in JC$ $(i\neq j \in \left \{ 1,2,3,4 \right \})$ are pairwise distinct by Lemma 2.5. We conclude that $12 \leq \deg (G_m)$ , and since $\deg (s) = 12$ by Lemma 2.3, we get $\deg (s) \leq \deg (G_m)$ and we are done.

Proof. Assume that $C$ is non-hyperelliptic; our goal is to arrive at a contradiction. Recall that $m \colon C - C \to m(C-C)$ is birational; see statement (1) of Proposition 2.6. Assume first that $f(X)$ is contained in the branch locus $B(m)$ of $m$ . The commutativity of diagram (2.3) yields a rational map $X \dashrightarrow B(s)$ defined as the composition

This rational map is non-constant as $B(m)$ must be a curve (as $f$ is non-constant) and $G_m$ is generically finite; see statement (2) in Proposition 2.6. As $B(s)$ is birational to $C$ by Lemma 2.3, one obtains a non-constant morphism $X \to C$ , proving that $C$ is hyperelliptic (see e.g. [Reference SchoenSch90, Lemma 1.1]), which yields the desired contradiction.

Therefore, the curve $f(X) \subset m(C-C)$ is not contained in $B(m)$ , and we obtain a non-constant rational map $X \dashrightarrow C - C$ defined as the composition

Note that $R(a) = \Delta _C \subset C \times C$ and $B(a) = \left \{ 0 \right \} \subset C - C$ . Consequently, composing the non-constant rational map $m^{-1} \circ f$ with the rational map $a^{-1} \colon C - C \dashrightarrow C \times C$ , one obtains a non-constant rational map $X \dashrightarrow C$ defined as the composition

\begin{equation*} a^{-1} \circ m^{-1} \circ f \colon X \dashrightarrow m(C - C) \dashrightarrow C - C \dashrightarrow C \times C. \end{equation*}

Thus, $X$ admits a non-constant morphism $X \to C$ . This is a contradiction, and we are done.

Proof. The first statement follows from the fact that ${\mathrm {Hom}}(T_1,A_2) = 0$ . For the second statement, consider the following commutative diagram with exact rows.

Applying ${\mathrm {Hom}}(-,{\mathbb G}_m)$ to this diagram and using the fact that for an abelian variety $A$ , the isomorphism $\mathrm {Ext}(A,{\mathbb G}_m) = A^\vee$ is functorial in $A$ , the commutativity of (2.6) follows.

Proof. To prove (1), first recall the following fact. Let $Y$ be a path-connected one-dimensional CW complex with basepoint $y_0$ , a $0$ -cell. Then every loop in $Y$ is homotopic to a loop consisting of a finite sequence of edges traversed monotonically; see [Reference HatcherHat02, § 1.1, Exercise 19]. Consequently, homology classes of the form (2.9) generate $H_1(\Gamma ,\mathbb Z)$ , yielding the lemma. (We do not require that $S$ be an integral basis, because not any generating set of a free $\mathbb Z$ -module contains a basis.)

To prove (2), we note that in the notation used above (2.9), we have

\begin{equation*} c^t(\gamma ) = ( P^+_{i_1} - P^-_{i_{m-1}}, P^+_{i_2} - P^-_{i_1}, P^+_{i_3} - P^-_{i_2}, \dotsc , P^+_{i_{m}} - P^-_{i_{m-1}} ) \in \prod _{j = 1}^m J {\widetilde {X}}_{i_j} \subset J {\widetilde {X}}. \end{equation*}

The lemma follows.

Proof. By the description of the map $c^t$ , we have $c^t(\gamma ) = p-q$ , where $p\mbox{ and }q$ denote the points on the normalization $\widetilde {X}$ of $X$ that are glued to form the nodal curve $X$ . As $X$ is hyperelliptic, the hyperelliptic involution on $\widetilde {X}$ descends to an involution on $X$ , which implies that $q=\iota (p)$ .

Proof. Let $T = {\mathrm {Ker}}(JX \to J{\widetilde {X}})$ and consider the extension

(2.11) \begin{align} 0 \longrightarrow T \longrightarrow JX \longrightarrow J {\widetilde {X}} \longrightarrow 0 . \end{align}

Since $X$ is very general, $J {\widetilde {X}}$ is simple. Hence any positive-dimensional abelian subvariety of $JX$ must be isogenous to $J{\widetilde {X}}$ . If such a subvariety exists, then the extension (2.11) splits up to isogeny, and so it suffices to exclude the latter. By Lemma 2.13, the isomorphism $\mathrm {Ext}(J {\widetilde {X}}, T) \simeq J {\widetilde {X}}$ (cf. § 2.3) identifies $[JX]$ with $x-\iota (x)$ for some $x \in {\widetilde {X}}$ , so (2.11) splits up to isogeny if and only if $x-\iota (x) \in J {\widetilde {X}}$ is torsion. As the one-nodal hyperelliptic curve $X$ is very general, this is not the case.

Proof. See for instance [Reference Peters and SteenbrinkPS08, Proposition C.11 and Remark C.12.ii].

Proof. See [Reference de JongJon96, Lemma 3.2], the main point being that we can find a resolution $\widetilde X$ of $X$ such that the fibres of $\widetilde X\to D$ are reduced.

Proof. This is a straightforward generalization of [Reference VoisinVoi07, Theorem 2.16].

Proof. This is a straightforward generalization of [Reference VoisinVoi07, Corollary 2.17].

Proof. This is a straightforward generalization of [Reference VoisinVoi07, Theorem 3.16].

Proof. To prove the lemma, we may assume that $\mathcal {X}$ is regular; see Lemma 2.16. Suppose that the central fibre $X_0$ has $k$ nodes, and let $\delta _1, \dotsc , \delta _k \in H^1(X_t,\mathbb Z)$ be the cohomology classes attached to the vanishing spheres $S_1, \dotsc , S_k \subset X_t$ ; see Lemma 2.17. By Lemma 2.19, we have the following generalization of the Picard–Lefschetz formula: for each $\alpha \in H^1(X_t,\mathbb Z)$ , one has

\begin{equation*} T(\alpha ) = \alpha + \sum _{i = 1}^k (\alpha \cdot \delta _i)\delta _i \in H^1(X_t,\mathbb Z). \end{equation*}

On the one hand, the monodromy operator $T'$ on $H^1(X'_{t'},\mathbb Z)=H^1(X_{t},\mathbb Z)$ satisfies $T'=T^{m}$ . On the other hand, we have

\begin{equation*} T^m(\alpha ) = \alpha + m \cdot \sum _{i = 1}^k (\alpha \cdot \delta _i)\delta _i \in H^1(X_t,\mathbb Z), \quad \alpha \in H^1(X_t,\mathbb Z). \end{equation*}

This implies that for $\alpha \in H^1(X_t,\mathbb Z)$ one has

\begin{equation*} T'(\alpha ) = \alpha \quad \iff \quad m \cdot \sum _{i = 1}^k (\alpha \cdot \delta _i)\delta _i = 0 \quad \iff \quad \sum _{i = 1}^k (\alpha \cdot \delta _i)\delta _i = 0 \quad \iff \quad T(\alpha ) = \alpha . \end{equation*}

This proves the lemma.

Proof. Let $\mathcal C_i$ ( $i\in I$ ) denote the irreducible components of $\mathcal {C}$ . By flatness of $q$ , the induced morphism $q_i \colon {\mathcal {C}}_i \to D$ is surjective. We claim that the general fibre of $q_i$ is irreducible. Clearly, it suffices to prove this after shrinking the disc $D$ . The main point is then that $\mathcal C\to D$ is a family of nodal curves, and hence the fibre above $0$ is reduced. This in turn implies that for each irreducible component $C_{0,j}$ of the special fibre of $q$ , there is up to shrinking a section of $q$ that passes through a general point of $C_{0,j}$ . This shows that the component $\mathcal C_i$ of $\mathcal C$ that contains $C_{0,j}$ admits a section and hence has irreducible general fibre. Running through all components $C_{0,j}$ of the special fibre of $q$ , we get in this way that each $q_i \colon {\mathcal {C}}_i \to D$ has irreducible general fibre.

Since the general fibre of $q\colon \mathcal C\to D$ is of compact type, the general fibre of $q_i$ is smooth and the index set $I$ forms the vertices of a tree that indicates which components of a general fibre of $q$ meet. If $i,j\in I$ are joined by an edge $e_{i,j}$ , or equivalently if the general fibres of $\mathcal C_i$ and $\mathcal C_j$ are glued at a point, then we get a section $e_{i,j}:D^*\to \mathcal C$ , which has to extend across the puncture by properness of $q$ . Since the arithmetic genus of the fibres of each $q_i\colon \mathcal C_i\to D$ is constant, and because the same holds for $q\colon \mathcal C\to D$ , we see that $\mathcal C$ is given by the quotient

\begin{equation*} \mathcal C=\biggl(\coprod _{i\in I} \mathcal C_i\biggr)/\sim \end{equation*}

where we glue for each edge $e_{i,j}$ between some indices $i,j\in I$ according to the section constructed above. Since the fibres of $q$ are nodal, we see moreover that the points that are glued via $\sim$ on the special fibre lie in the smooth locus of $\bigsqcup _{i\in I} \mathcal C_i\to D$ .

A simple Mayer–Vietoris argument now reduces us to showing that

\begin{equation*} H^1({\mathcal {C}}_i,\mathbb Z) \longrightarrow H^1(C_{it},\mathbb Z)^T \end{equation*}

is surjective for each $i\in I$ . In other words, we have reduced the result to the case where $q$ is smooth over the punctured disc $D^*$ . In this case we apply Lemma 2.16 and get a modification $\tau :\widetilde {\mathcal C}\to \mathcal C$ given by successive blow-ups of the singular points in the central fibre, such that the fibres of $\widetilde {\mathcal C}\to D$ are reduced and hence nodal curves. By [Reference KollárKol93, Theorem 7.8], $ \tau _\ast \colon \pi _1({\widetilde {{\mathcal {C}}}}) \to \pi _1({\mathcal {C}})$ is an isomorphism. Passing to the abelianization and applying ${\mathrm {Hom}}(-,\mathbb Z)$ , we find that $\tau ^\ast \colon H^1(\mathcal C,\mathbb Z)\to H^1(\widetilde {\mathcal C},\mathbb Z)$ is an isomorphism. This reduces us to the case where $\mathcal C$ is regular and $q$ is smooth over $D^*$ . By Corollary 2.18, $H_1(C_t,\mathbb Z) \to H_1({\mathcal {C}},\mathbb Z)$ is surjective. It follows that $H^1({\mathcal {C}},\mathbb Z) \to H^1(C_t,\mathbb Z)$ is injective with torsion-free cokernel. By the local invariant cycle theorem (cf. [Reference MorrisonMor84]), the map $H^1({\mathcal {C}},\mathbb Z) \to H^1(C_t,\mathbb Z)^T$ becomes surjective after tensoring with $\mathbb Q$ ; as its cokernel is torsion-free, it is surjective. This concludes the proof of the lemma.

Proof. Let $a_{n-1},\dots ,a_{2g-1}\in {\mathbb C}$ be general complex numbers, and consider the following equation:

\begin{equation*} y^2=((x-a_{n-1})^2-z_1)\cdot ((x-a_n)^2-z_2)\cdot \prod _{i=1}^{n-2}((x-z_{n+i}-a_{n+i})^2-z_{i+2}) \cdot \prod _{j=2n-1}^{2g-1}(x-z_j-a_j). \end{equation*}

(The slight asymmetry in the quadratic terms reflects automorphisms of $\mathbb P^1$ and stems from the fact that we want to have a family that depends on $2g-1$ parameters $z_1,\dots ,z_{2g-1}$ .) This defines a family of affine hyperelliptic curves over ${\mathbb A}^{2g-1}$ , branched at the $2g+2$ points

\begin{equation*} x=\infty ,\ \ x=\pm \sqrt {z_1} + a_{n-1},\ \ x=\pm \sqrt {z_2}+a_n,\ \ x=\pm \sqrt {z_{i+2}}+z_{n+i}+a_{n+i},\ \ x=z_j+a_j \end{equation*}

for $i=1,\dots ,n-2$ and $j=2n-1,\dots ,2g-1$ . This extends to a projective family of hyperelliptic curves, and we denote by $U\subset {\mathbb A}^{2g-1}$ the open subset where the corresponding hyperelliptic curve attains at most one node. The corresponding projective family of hyperelliptic curves is denoted by $p \colon {\mathcal {X}} \to U$ , and we note that $\mathcal {X}$ is regular. By construction, for $u = (z_1, \dotsc , z_{2g-1}) \in U$ , the fibre $X_u = p^{-1}(u)$ either is smooth or attains exactly one node; the latter happens if and only if $z_i = 0$ for some $i \in \left \{ 1, \dotsc , n \right \}$ . Let $\Delta _i := U\cap \{z_i=0\}$ for $i\in \left \{ 1,\dots ,n \right \}$ . The above description of the ramification points of the hyperelliptic covering $X_u\to \mathbb P^1$ for $u\in U$ shows that the moduli map $\Delta _i\to \overline {\mathcal M}_g$ is generically finite onto its image (because the hyperelliptic locus in $\mathcal M_g$ has dimension $2g-1$ ). Altogether this proves assertion (1) in the lemma.

For $i\in \left \{ 1, \dotsc , n \right \}$ , let $D_i \subset U$ be a disc that intersects $\Delta$ transversally in a general point $0_i \in \Delta _i$ . Up to shrinking the disc $D_i$ , we can assume that the total space ${\mathcal {X}}|_{D_i}$ of the restriction ${\mathcal {X}}|_{D_i} \to D_i$ is regular: the only possible singularity is at the node of the central fibre, where analytically locally an equation of $\mathcal X|_{D_i}$ is given by $t=x^2-y^2$ , which yields a regular surface. Thus, ${\mathcal {X}}|_{D_i} \to D_i$ is a Lefschetz degeneration, proving assertion (2).

The fibre $X_t$ is a double covering $X_t\to \mathbb P^1$ branched along $2g+2$ points $p_0,p_1,\dots ,p_{2g+1}$ . By construction, $\mathcal X\to U$ is given by an equation of the form $y^2=\prod _{i=1}^{n}f_i(x)\cdot \prod _{j=2n-1}^{2g-1} g_j(x)$ , where $g_j(x)=x-z_j-a_j$ and $f_i(x)$ is quadratic in $x$ . Up to reordering, we can assume that $\{p_{2i-1},p_{2i}\}$ corresponds to the roots of $f_i(x)$ . In particular, $p_{2i-1}$ and $p_{2i}$ collide along the Lefschetz degeneration over the disc $D_i$ for $i=1,\dots ,n$ .

We pick a path of shortest distance between $p_{2i-1}$ and $p_{2i}$ on $\mathbb P^1$ for $i=1,\dots ,g$ and note that the preimage of this path in the hyperelliptic curve $X_t$ gives rise to a homology class in $H_1(X_t,\mathbb Z)$ (well-defined up to sign) whose Poincaré dual $\delta _i\in H^1(X_t,\mathbb Z)$ is for $i=1,\dots ,n$ the vanishing cycle that corresponds to colliding $p_{2i-1}$ and $p_{2i}$ . The classes $\delta _1,\dots ,\delta _g$ are orthogonal to each other and can be completed to a symplectic basis, proving assertion (3).

Proof. For each $i \in \left \{ 1, \dotsc , n \right \}$ , define

\begin{align*} H_i' := \mathrm {Im}\left ( H^1(X_{0_i},\mathbb Z) \hookrightarrow H^1(X_t,\mathbb Z) \right ) \quad \text {and}\quad V_i := \mathrm {Im}\left (W_0H^1(X_{0_i},\mathbb Z) \hookrightarrow H^1(X_t,\mathbb Z) \right ). \end{align*}

Consider the monodromy operator $ T_i \colon H^1(X_t,\mathbb Z) \to H^1(X_t,\mathbb Z)$ . In view of statements (2) and (3) in Lemma 2.23, $T_i$ is given by the formula

\begin{align*} T_i(\alpha )=\alpha + (\alpha \cdot \delta _i)\delta _i, \quad \alpha \in H^1(X_t,\mathbb Z), \end{align*}

where the $\delta _i$ are the vanishing cycles $\delta _i \in H^1(X_t,\mathbb Z)$ attached to the Lefschetz degenerations ${\mathcal {X}}|_{D_i} \to D_i$ and the path $\rho _i$ . It is well known that

(2.16) \begin{align} V_i \otimes \mathbb Q = \mathrm {Im}(T_i - {\mathrm {id}}) \otimes \mathbb Q \subset {\mathrm {Ker}}(T_i - {\mathrm {id}}) \otimes \mathbb Q = H_i' \otimes \mathbb Q \subset H^1(X_t,\mathbb Q). \end{align}

In fact, (2.16) holds integrally because the respective spaces are saturated in $H^1(X_t,\mathbb Z)$ . In particular, (2.14) follows. Note that $(T_i - {\mathrm {id}})(\alpha ) = (\alpha \cdot \delta _i)\delta _i$ for each $\alpha \in H^1(X_t,\mathbb Z)$ . As we have $\gamma _i \cdot \delta _i = -1$ , this gives $ \left (T_i - {\mathrm {id}} \right )(\gamma _i) = (\gamma _i \cdot \delta _i) \delta _i = -\delta _i.$ Therefore, $ V_i = \mathrm {Im}\left (T_i - {\mathrm {id}}\right ) = \left \langle \delta _i \right \rangle = \mathbb Z \cdot \delta _i \subset H_i'$ , proving (2.15) and thereby the lemma.