Proof. Recall that if $\mathcal{V}$ is a vector bundle of rank $r$ on a stack $Z$ then the Bogomolov expression $\textrm {Bog}(\mathcal{V})$ is given by

\begin{align*}\textrm {Bog}(\mathcal{V}) = \frac {1}{2r}c_{1}^{2}\mathcal{V} - \textrm {ch}_{2}\mathcal{V}.\end{align*}

If $\mathcal{V}$ is a vector bundle on a family of curves, the pushforward of the Bogomolov expression to the base of the family gives the divisor class of the locus where the vector bundle restricts to an unstable bundle, assuming the general fiber is stable (or regular polystable in the case of genus $1$ curves.)

And so the proposition follows from this and Proposition 2.8.

Proof. The proof is immediate from Proposition 2.11.

Proof. The proof is immediate from Proposition 2.11.

Proof. Assume Conjecture 2.14, and let $d$ be any of the infinitely many integers satisfying Corollary 2.4. Let $\beta : B \to \mathsf{Hur}_{d,g}^{\diamond }(X)$ be a complete curve realizing the truth of Conjecture 2.14. Then, since $B$ passes through a general point and since $\varphi ^{-1}(D) \neq \underline {\mathsf{Hur}}_{d,g}(X)$ , it follows that

\begin{align*}\deg \beta ^{*}(D) = a \deg \beta ^{*} \lambda - b \deg \beta ^{*} \delta \geq 0.\end{align*}

As $\beta ^{*} \deg \lambda \geq 0$ (in fact strictly so) for a non-constant proper family of curves with smooth general member, the theorem now follows from Corollary 2.13 by letting $d \to \infty$ .

Proof. The divisors $\Delta _{0}(d), \Delta _{1}(d)$ precisely correspond to the sets $A_{d}(x)$ and $B_{d}(x)$ , respectively.

Proof. The parameter $|C_{d}(x)|$ can be identified with the number of lattices $\Lambda \subset \mathbb{Z}^{2}$ which have co-volume $d$ . We may canonically choose a basis of $\Lambda$ for which the inclusion $\Lambda \subset \mathbb{Z}^{2}$ is represented by a matrix

\begin{align*}\left(\begin{matrix} a\;\;\; & c \\ 0\;\;\; & b \end{matrix}\right),\end{align*}

where $a, b \gt 0$ , $ab=d$ and $0 \leq c \lt a$ . For each factorization $ab=d$ there are $a$ possibilities for $c$ , and hence the number of such sublattices is $\sum _{a \mid d}a = \sigma _{1}(d)$ .

Proof. Let $x \in X$ denote a marked point. The proof proceeds by counting the elements of a set $D$ in two different ways. The set $D$ is the collection of (isomorphism classes of) all degree $d$ branched covers (not necessarily primitive) $\alpha : X_{1} \cup X_{2} \to X$ where:

1. (1) $X_{1},X_{2}$ are smooth genus $1$ curves meeting at a single node $y$ ;

2. (2) $\alpha (y) = x$ .

On the one hand, if the degree of the isogeny $\alpha |_{X_{1}}:(X_{1},y) \to (X,x)$ is $h$ , then there are $\sigma _{1}(h)$ choices for this isogeny by Lemma 3.6. Similarly, there are $\sigma _{1}(d-h)$ choices for the isogeny $\alpha |_{X_{2}}:(X_{2},y) \to (X,x)$ . If $h \neq d-h$ then, accounting for labeling, it follows that there are $\sigma _{1}(h)\sigma _{1}(d-h)$ choices for a combined branched cover $\alpha :X' \to X$ where $X'$ has two components, one of which maps with degree $h$ .

If $h = d-h$ then, after accounting for labeling, there are $\binom{\sigma _{1}(h)}{2}$ choices of branched covers $\alpha :X_{1} \cup X_{2} \to X$ where the isogenies $\alpha |_{X_{1}}$ and $\alpha |_{X_{2}}$ are distinct. If we then add in the $\sigma _{1}(h)$ -many branched covers where $\alpha |_{X_{1}} = \alpha |_{X_{2}}$ , we obtain a contribution of $\sigma _{1}(h) + \binom{\sigma _{1}(h)}{2} = \frac {1}{2}\sigma _{1}(h) + \frac {1}{2}\sigma _{1}(h)^{2}$ . Adding all contributions per each $h \in \{1, \ldots , d-1\}$ yields the right-hand side of (9).

On the other hand, every element $\alpha : X_{1} \cup X_{2} \to X$ in $D$ factors canonically through an isogeny $(X_{1} \cup X_{2},y) \to (Z,z) \to (X,x)$ , $Z$ a smooth genus $1$ curve, in such a way that $X_{1} \cup X_{2} \to Z$ is primitive. By counting with respect to the degree $d'$ of the factor $X_{1}\cup X_{2} \to Z$ we obtain the left-hand side of (9). This concludes the proof.

Proof. The case $d=1$ is immediate, as both sets $A_{1}(x)$ and $M_{1}$ are clearly empty. Also, the fact that $|M_{2}| = 3$ is obvious because there are three surjections $\mathbb{Z}^{2} \to \mathbb{Z}/2\mathbb{Z}$ . So we assume $d \geq 2$ . We will produce a function $\Phi : M_{d} \to A_{d}(x)$ which will explain the remaining content in the lemma. First, we identify $\pi _{1}(X,x)$ with $\mathbb{Z}^{2}$ . Next, pick any element $(\Lambda ,\eta ) \in M_{d}$ .

By the fundamental Galois correspondence in the theory of covering spaces, the subgroup $\Lambda \subset \mathbb{Z}^{2}$ defines a degree $d$ connected, Galois covering space $\epsilon : Z \to X$ with group of deck transformations canonically isomorphic to $\mathbb{Z}^{2}/\Lambda$ . Here, $Z$ inherits from $X$ a complex structure, and therefore $Z$ is a genus 1 curve with $\epsilon$ a degree $d$ finite étale map. Pick a point $z \in \epsilon ^{-1}(x)$ . The chosen element $\eta \in \mathbb{Z}^{2}/\Lambda$ defines a second point $z' = \eta (z)$ on $Z$ . Let $Y_{(\Lambda ,\eta )}$ denote the nodal curve obtained by identifying the points $z$ and $z'$ on the curve $Z$ , then $Y_{(\Lambda ,\eta )}$ is independent of the choice of $z$ because the cover $Z \to X$ is Galois. The map $\epsilon$ descends to a finite, flat degree $d$ map

\begin{align*}\alpha :Y_{(\Lambda ,\eta )} \to X,\end{align*}

where the node of $Y_{(\Lambda ,\eta )}$ maps to $x$ . Since $\eta$ generates $\mathbb{Z}^{2}/\Lambda$ , it follows that $\alpha : Y_{(\Lambda ,\eta )} \to X$ does not factor through a non-trivial étale cover $X' \to X$ , again by the Galois correspondence.

Thus, $\alpha : Y_{(\Lambda ,\eta )} \to X$ is an element of $A_{d}(x)$ . The function $\Phi : M_{d} \to A_{d}(x)$ is defined by the rule

\begin{align*}(\Lambda ,\eta ) \mapsto [\alpha :Y_{(\Lambda ,\eta )} \to X].\end{align*}

It suffices to show that $\Phi$ is surjective and $2:1$ when $d \geq 3$ , and $1:1$ when $d=2$ . To that end, let $\alpha : Y\to X$ be an arbitrary element of $A_{d}(x)$ , let $y \in Y$ denote the node of $Y$ and let $Z \to Y$ denote the normalization of $Y$ . Here, $Z$ is a genus $1$ curve, and there are two points $z, z' \in Z$ lying above the node $y \in Y$ . Again, since $\alpha : Y \to X$ is primitive, it follows that the induced (Galois) cover $\epsilon : Z \to X$ has a cyclic Galois group $G$ . Hence, there is unique element $\eta _{1} \in G$ which satisfies $\eta (z) = z'$ . Setting $\Lambda _{1}$ to be the image of the pushforward map $\epsilon _{*}:\pi _{1}(Z,z) \to \pi _{1}(X,x)$ , we see that $\alpha = \Phi (\Lambda _{1}, \eta _{1})$ .

Now suppose $\alpha = \Phi (\Lambda _{2}, \eta _{2})$ . Then $\Lambda _{2} = \Lambda _{1}$ because both are simply the image of $\pi _{1}(Z)$ under pushforward $\epsilon _{*}$ (the normalization is intrinsic to $Y$ ). Now suppose $w,w' \in Z$ are two points in $\epsilon ^{-1}(x)$ such that $\eta _{2}(w)=w'$ . Since $\alpha = \Phi (\Lambda _{2},\eta _{2})$ , it follows by construction of $\Phi$ that $\alpha :Y \to X$ can be obtained by identifying the points $w$ and $w'$ , creating the node $y$ . By uniqueness of normalization, it follows that there is an automorphism $g: Z \to Z$ (over $X$ ) which sends the set $\{z,z'\}$ to the set $\{w,w'\}$ . Since the cover $\epsilon : Z \to X$ is Galois, it follows that $\eta _{2}$ either sends $z$ to $z'$ or $z'$ to $z$ . In the former case $\eta _{2}=\eta _{1}$ and in the latter $\eta _{2} = \eta _{1}^{-1}$ . The proof is finished as we have shown that $\Phi$ is surjective and $2:1$ when $d \geq 3$ and $1:1$ when $d=2$ .

Proof. The proposition will fall out of carefully counting the elements of $M_{d}$ and then using Lemma 3.10; the Kronecker deltas will correspond to the two edge cases in Lemma 3.10.

First, the set of lattices $\Lambda \subset \mathbb{Z}^{2}$ of index $d$ is in natural bijection with the set of integer matrices

\begin{align*}\left(\begin{matrix} a\;\;\; & c \\ 0\;\;\; & b \end{matrix}\right),\end{align*}

where $a,b \gt 0$ , $0 \leq c \lt a$ and $ab=d$ . In terms of this bijection, the condition that $\mathbb{Z}^{2}/\Lambda$ is cyclic is expressed by the condition that $\gcd (a,b,c)=1$ .

For each choice of factorization $ab=d$ , the condition $\gcd (a,b,c)=1$ can be replaced by the condition: any prime $p$ dividing both $a$ and $b$ does not divide $c$ . Thus, for a fixed $a$ dividing $d$ , the number of possible choices for $c$ is

\begin{align*} a \cdot \prod _{p \mid a \,\, {\rm and} \,\, p \mid b}\left (1 - \frac {1}{p} \right ).\end{align*}

Summing over all factorizations we get

\begin{align*} \sum _{ab=d}a\left [ \prod _{p \mid a \,\, {\rm and} \,\, p \mid b}\left (1 - \frac {1}{p} \right ) \right ] &= \prod _{p^{n}\mid \mid d}a\left [1 + \left ( \sum _{i=1}^{n-1}p^{i}\left (1-\frac {1}{p}\right ) \right ) + p^{n} \right ]\\ &= \prod _{p^{n} \mid \mid d}(p^{n}+p^{n-1}), \end{align*}

where “ $p^{n} \mid \mid d$ ” means “ $p^{n}$ is the largest power of $p$ dividing $d$ .”

Now, accounting for the $\eta$ part of the data of an element $(\Lambda ,\eta ) \in M_{d}$ , we conclude that, for $d \geq 2$ ,

\begin{align*}|M_{d}| = \phi (d)\prod _{p^{n} \mid \mid d}(p^{n}+p^{n-1}) = \prod _{p^{n} \mid \mid d}p^{2n-2}(p^{2}-1),\end{align*}

where $\phi$ is the Euler phi function, and hence by Lemma 3.10 we conclude that, when $d \geq 3$ ,

(10) \begin{align} 2|A_{d}(x)| = \prod _{p^{n} \mid \mid d}p^{2n-2}(p^{2}-1). \end{align}

When $d=1$ (10) is almost right: the left-hand side is $0$ while the right-hand side is $1$ because it is a product over an empty set. When $d=2$ (10) is also almost right: the left-hand side should be replaced by $|A_{2}(x)|$ because of the $d=2$ exception in Lemma 3.10. When we adjust $2 | A_{d}(x)|$ by the Kronecker deltas in the statement of the proposition, we find that the function

\begin{align*}F(d) := 2|A_{d}(x)|+\delta _{d,1}-3\delta _{d,2}\end{align*}

has output $\prod _{p^{n} \mid \mid d}p^{2n-2}(p^{2}-1)$ , which is plainly multiplicative in $d$ . The proposition follows.

Proof of Theorem 3.2. Recall that it suffices to prove (6), and hence by Lemma 3.5 it suffices to prove

(11) \begin{align} (5d-6)|A_{d}(x)| = 12|B_{d}(x)|. \end{align}

In order to prove (11) we will show that the numbers $\frac {5d-6}{12}|A_{d}(x)|$ satisfy the recursion (9) in Proposition 3.7 after replacing $|B_{d'}(x)|$ with $\frac {5d'-6}{12}|A_{d'}(x)|$ on the left-hand side. Denote by $\iota : \mathbb{Z}_{\gt 0} \to \mathbb{Z}$ the inclusion map, which is clearly multiplicative.

On one hand, in light of Proposition 3.11, we get

(12) \begin{align}\sum _{d' \mid d}\frac {(5d'-6)|A_{d'}(x)|}{12}\sigma _{1}(d/d')\end{align}

(13) \begin{align}=\frac {1}{24}\sum _{d' \mid d} (5d'-6)F(d')\sigma _{1}(d/d') - \frac {1}{24}(5 \times 1 - 6)\sigma _{1}(d) + \frac {1}{24}\times 3 \times (5 \times 2 - 6)\sigma _{1}(d/2)\end{align}

(14) \begin{align}= \frac {5}{24}\left [ (\iota \cdot F) * \sigma _{1} \right ](d) - \frac {1}{4}\left [ F * \sigma _{1} \right ](d) + \frac {1}{24}\sigma _{1}(d) + \frac {1}{2}\sigma _{1}(d/2).\end{align}

Here, and in what follows, $f \cdot g$ is the ordinary multiplication of two numerical functions and $f*g$ denotes the Dirichlet convolution of two numerical functions, given by $(f*g)(d) = \sum _{d' \mid d}f(d')g(d/d')$ . If $f$ and $g$ are both multiplicative then so are $f\cdot g$ and $f*g$ .

On the other hand, we can simplify the right-hand side of (9) using the Ramanujan identity

\begin{align*}\sum _{h=1}^{d-1}\sigma _{1}(h)\sigma _{1}(d-h) = \left (\frac {1}{12}-\frac {d}{2}\right )\sigma _{1}(d)+ \frac {5}{12}\sigma _{3}(d).\end{align*}

Thus it remains to prove the equality

\begin{align*} &\frac {1}{2}\left (\sigma _{1}\left(\frac{d}{2}\right) + \left (\frac {1}{12}-\frac {d}{2}\right )\sigma _{1}(d)+ \frac {5}{12}\sigma _{3}(d)\right ) \\&\quad = \frac {5}{24}\left [(\iota \cdot F) * \sigma _{1} \right ](d) - \frac {1}{4}\left [ F * \sigma _{1} \right ](d) + \frac {1}{24}\sigma _{1}(d) + \frac {1}{2}\sigma _{1}(d/2). \end{align*}

This further simplifies, and it then suffices to show the equality

(15) \begin{align}\frac {5}{12}\sigma _{3}(d) - \frac {d}{2}\sigma _{1}(d)\end{align}

(16) \begin{align}= \frac {5}{12}\left [(\iota \cdot F) * \sigma _{1} \right ](d) - \frac {1}{2}\left [ F * \sigma _{1} \right ](d).\end{align}

Equation (15) then follows from the following two simple identities, both proven using the fact that the Dirichlet convolution of multiplicative functions is multiplicative, and then by evaluating each expression at prime powers:

1. (1) $\left [(\iota \cdot F) * \sigma _{1} \right ](d) = \sigma _{3}(d)$ ;

2. (2) $\left [ F * \sigma _{1} \right ](d) = d \cdot \sigma _{1}(d)$ .

This finishes the verification that the numbers $\frac {5d-6}{12}|A_{d}(x)|$ satisfy the recursion (9) for $|B_{d}(x)|$ , and therefore it follows that (6) is satisfied, finishing the proof of Theorem 3.2.

Proof.

1. (1) The pencil $B$ can be interpreted as a linear map

   \begin{align*}\lambda : {\mathbb{P}}^{1} \to {\mathbb{P}} (H^{0}(X, \textrm {Sym}^{3} \mathcal{E} \otimes (\det \mathcal{E})^{\vee })).\end{align*}
   We may interpret $\lambda$ as a strict injection
   \begin{align*}{\mathcal{O}}_{{\mathbb{P}}^{1}}(-1) \hookrightarrow \textrm {Sym}^{3} \mathcal{E} \otimes (\det \mathcal{E})^{\vee }.\end{align*}
   But this, in turn, can be interpreted as a single section
   \begin{align*}t \in H^{0}\left ( X \times {\mathbb{P}}^{1}, \textrm {Sym}^{3} (\mathcal{E'}) \otimes (\det \mathcal{E'})^{\vee } \right ),\end{align*}
   where $\mathcal{E}'$ is short for $\mathcal{E} \boxtimes {\mathcal{O}}_{{\mathbb{P}}^{1}}(1)$ . Conversely, a general section $t$ can be interpreted as coming from a pencil $B$ , and so we obtain a general section $t$ in this way.

   The large $g$ assumption ensures that $\textrm {Sym}^{3} (\mathcal{E'}) \otimes (\det \mathcal{E'})^{\vee }$ is globally generated, in fact even very ample. It follows from Casnati and Ekedahl’s Bertini-type theorems in [Reference Casnati and EkedahlCE96, Reference CasnatiCas96] that $t$ defines a smooth surface $S \subset {\mathbb{P}} \mathcal{E}'$ which is a triple cover of $X \times B$ with Tschirnhausen bundle $\mathcal{E}'$ . The numerics of the initial bundle $\mathcal{E}$ ensures that the natural morphism $S \to {\mathbb{P}}^{1}$ is a family of arithmetic genus $g$ curves, with smooth general fiber.

   Furthermore, again as a consequence of the large $g$ condition, the set of elements of ${\mathbb{P}} (H^{0}(X, \mathcal{E}\otimes (\det \mathcal{E})^{\vee }))$ which define a worse-than-nodal curve in ${\mathbb{P}} \mathcal{E}$ has codimension at least $2$ . Therefore, a general pencil avoids this locus and so we obtain in this way a map $B \to \mathsf{Hur}_{3,g}^{\diamond }(X)$ which is clearly moving.

2. (2) We proceed in much the same way as in the previous case. A general pencil $B \subset {\mathbb{P}} (H^{0}(X, \textrm {Sym}^{2}\mathcal{E} \otimes \mathcal{F}^{\vee }))$ can instead be interpreted as a single general global section $t$ of $\left ( \textrm {Sym}^{2} \mathcal{E} \otimes \mathcal{F}^{\vee } \right ) \boxtimes {\mathcal{O}}_{{\mathbb{P}}^{1}}(1)$ on $X \times {\mathbb{P}}^{1}$ . Let $\mathcal{E}'$ and $\mathcal{F}'$ denote the bundles $\mathcal{E} \boxtimes {\mathcal{O}}_{{\mathbb{P}}^{1}}(2)$ and $\mathcal{F} \boxtimes {\mathcal{O}}_{{\mathbb{P}}^{1}}(3)$ , respectively, on $X \times {\mathbb{P}}^{1}$ . Then $t$ is a general global section of $\textrm {Sym}^{2} \mathcal{E}' \otimes (\mathcal{F}')^{\vee }$ , and $\det \mathcal{E}' = \det \mathcal{F}'$ .

   The large $g$ condition ensures $\textrm {Sym}^{2} \mathcal{E}' \otimes (\mathcal{F}')^{\vee }$ is globally generated. By Casnati and Ekedahl’s Bertini-type theorems [Reference Casnati and EkedahlCE96, Reference CasnatiCas96], $t$ defines a smooth surface $S \subset {\mathbb{P}} \mathcal{E}'$ , which is a degree $4$ branched cover of $X \times B$ with Tschirnhausen bundle $\mathcal{E}'$ . The morphism $S \to B$ gives a family of arithmetic genus $g$ curves with smooth general fiber.

   The positivity of $\textrm {Sym}^{2}\mathcal{E} \otimes \mathcal{F}^{\vee }$ implies that the subvariety of its global sections defining worse-than-nodal curves has codimension at least $2$ . So $B$ can be made to avoid this locus, and therefore $B$ defines a moving curve in $\underline {\mathsf{Hur}}_{4,g}^{\diamond }(X)$ .

3. (3) Proceed in parallel to the previous two cases, with the relevant bundles on $X \times {\mathbb{P}}^{1}$ being $\mathcal{E}' = \mathcal{E} \boxtimes {\mathcal{O}}_{{\mathbb{P}}^{1}}(5)$ and $\mathcal{F}' = \mathcal{F} \boxtimes {\mathcal{O}}_{{\mathbb{P}}^{1}}(8)$ . We omit the repetitive details.