2.1 Hermitian spaces and lattices

Let $\mathbf{k}=\mathbb{Q}(\sqrt{D_{\mathbf{k}}})$ be an imaginary quadratic number field of discriminant $D_{\mathbf{k}}$ , with $D_{\mathbf{k}}$ a square-free negative integer. Let ${\mathcal{O}}_{\mathbf{k}}\subset \mathbf{k}$ be the ring of integers in $\mathbf{k}$ . Denote by $\mathfrak{d}_{\mathbf{k}}$ the different ideal and by $\mathfrak{d}_{\mathbf{k}}^{-1}$ the inverse different ideal.

We shall consider $\mathbf{k}$ as a subset of the complex numbers $\mathbb{C}$ and denote by $\unicode[STIX]{x1D6FF}_{\mathbf{k}}$ the square root of the discriminant, with the usual choice of the complex square root. Then, $\mathfrak{d}_{\mathbf{k}}$ is given by $\unicode[STIX]{x1D6FF}_{\mathbf{k}}{\mathcal{O}}_{\mathbf{k}}$ and $\mathfrak{d}_{\mathbf{k}}^{-1}$ by $\unicode[STIX]{x1D6FF}_{\mathbf{k}}^{-1}{\mathcal{O}}_{\mathbf{k}}$ .

Let $V=V_{\mathbf{k}}$ be an indefinite Hermitian space over $\mathbf{k}$ of signature $(1,n+1)$ , endowed with a nondegenerate Hermitian form denoted $\langle \cdot ,\cdot \rangle$ , linear in the left and conjugate linear in the right argument. A complex Hermitian space $V_{\mathbb{C}}=V\otimes _{\mathbf{k}}\mathbb{C}$ is obtained by extension of scalars. We denote by $V_{\mathbb{Q}}$ the $\mathbb{Q}$ -vector space underlying $V$ , which bears the structure of a quadratic space of signature $(2,2n+2)$ with the quadratic form $q(\cdot )$ defined by $q(x):=\langle x,x\rangle$ . Similarly, the real quadratic space underlying $V_{\mathbb{C}}$ is denoted $V_{\mathbb{R}}$ . We have $V_{\mathbb{R}}=V_{\mathbb{Q}}\otimes _{\mathbb{Q}}\mathbb{R}$ .

Let $L$ be a lattice in $V$ , with $L\otimes _{{\mathcal{O}}_{\mathbf{k}}}\mathbf{k}=V$ . We denote by $L^{\prime }$ the $\mathbb{Z}$ -dual of $L$ , defined as the set

$$\begin{eqnarray}\displaystyle L^{\prime } & = & \displaystyle \{x\in V;\langle x,y\rangle \in \mathfrak{d}_{\mathbf{k}}^{-1}\quad \text{for all }y\in L\}\nonumber\\ \displaystyle & = & \displaystyle \{x\in V;\operatorname{Tr}_{\mathbf{k}/\mathbb{Q}}\langle x,y\rangle \in \mathbb{Z}\quad \text{for all }y\in L\}.\nonumber\end{eqnarray}$$

Naturally, $L^{\prime }$ is a lattice in $V$ , too. If $L\subseteq L^{\prime }$ , the lattice $L$ is called integral. If further for all $x\in L$ , $\langle x,x\rangle \in \mathbb{Z}$ , then $L$ is called even. Finally, $L$ is unimodular, if $L^{\prime }=L$ . The quotient $L^{\prime }/L$ is referred to as the discriminant group of $L$ .

More generally in the context of this paper, by a Hermitian lattice we mean a discrete subgroup $M$ of $V$ , for which the ring of multipliers ${\mathcal{O}}(M)$ is an order in $k$ . (A multiplier of $M$ is a complex number $\unicode[STIX]{x1D6FC}$ with $\unicode[STIX]{x1D6FC}M\subset M$ .) Most lattices will occur here as sublattices of a fixed lattice $L$ , with $L$ as above, of full rank, Hermitian and even.

Denote by $\text{U}(V)$ the unitary group of $V$ , and by $\text{SU}(V)$ the special unitary group. The isometry group of a lattice $L$ in $\text{U}(V)$ is denoted $\text{U}(L)$ , similarly for $\text{SU}(L)$ . The discriminant kernel $\unicode[STIX]{x1D6E4}_{L}$ is the subgroup of finite index in $\text{SU}(L)$ which acts trivially on the discriminant group of $L$ . We refer to subgroups of finite index in $\unicode[STIX]{x1D6E4}_{L}$ as unitary modular groups. In the following, $\unicode[STIX]{x1D6E4}$ will always denote a unitary modular group.

2.2 A symmetric domain

Viewing $\text{U}(V)$ as an algebraic group, its set of real points, denoted $\text{U}(V)(\mathbb{R})$ , is the unitary group of $V_{\mathbb{C}}$ . A symmetric domain for the action of $\text{U}(V)(\mathbb{R})$ on $V_{\mathbb{C}}$ is given by the quotient $\mathbb{D}=\text{U}(V)(\mathbb{R})/{\mathcal{K}}$ with a maximal compact subgroup ${\mathcal{K}}$ . Denote by $\mathbb{P}V_{\mathbb{C}}$ the projective space of $V_{\mathbb{C}}$ . A projective model for $\mathbb{D}$ is given by the positive cone

$$\begin{eqnarray}{\mathcal{C}}=\{[v]\in \mathbb{P}V_{\mathbb{C}};\langle v,v\rangle >0\}.\end{eqnarray}$$

We briefly review the construction of an affine model. Denote by $\text{Iso}(V)$ the set of one-dimensional isotropic subspaces of $V_{\mathbf{k}}$ . Its elements are in one-to-one correspondence with the rational boundary components of the symmetric domain. In particular, we fix an element $I\in \text{Iso}(V)$ by choosing a primitive isotropic lattice vector $\ell \in L$ and setting $I=\mathbf{k}\ell$ . Further, we choose a primitive vector $\ell ^{\prime }\in L^{\prime }$ such that $\langle \ell ,\ell ^{\prime }\rangle \neq 0$ . We shall assume that $\ell ^{\prime }$ is isotropic, too. Note that this is a nontrivial assumption about the Hermitian lattice $L$ and its dual.

For $a\in V$ , we denote by $a^{\bot }$ the orthogonal complement with respect to $\langle \cdot ,\cdot \rangle$ . We set $D:=L\cap \ell ^{\bot }\cap {\ell ^{\prime }}^{\bot }$ . Equipped with the restriction of $\langle \cdot ,\cdot \rangle$ , $D$ is a definite Hermitian lattice of signature $(0,n)$ . Denote by $W=W_{\mathbf{k}}$ the subspace $D\otimes _{{\mathcal{O}}_{\mathbf{k}}}\mathbf{k}$ , and let $W_{\mathbb{C}}=W\otimes _{\mathbf{k}}\mathbb{C}$ .

Now, an affine model for $\mathbb{D}$ , called the Siegel domain model, is given by the following generalized upper-half-plane:

$$\begin{eqnarray}{\mathcal{H}}_{\ell ,\ell ^{\prime }}=\{(\unicode[STIX]{x1D70F},\unicode[STIX]{x1D70E})\in \mathbb{C}\times W_{\mathbb{C}};2\Im (\unicode[STIX]{x1D70F})|\unicode[STIX]{x1D6FF}_{\mathbf{k}}||\langle \ell ,\ell ^{\prime }\rangle |^{2}>-\langle \unicode[STIX]{x1D70E},\unicode[STIX]{x1D70E}\rangle \}.\end{eqnarray}$$

For $(\unicode[STIX]{x1D70F},\unicode[STIX]{x1D70E})\in {\mathcal{H}}_{\ell ,\ell ^{\prime }}$ , we set

$$\begin{eqnarray}z=z(\unicode[STIX]{x1D70F},\unicode[STIX]{x1D70E}):=\ell ^{\prime }-\unicode[STIX]{x1D6FF}_{\boldsymbol{ k}}\unicode[STIX]{x1D70F}\langle \ell ^{\prime },\ell \rangle \ell +\unicode[STIX]{x1D70E}.\end{eqnarray}$$

Clearly, under the canonical projection $\unicode[STIX]{x1D70B}_{V}:V_{\mathbb{C}}\rightarrow \mathbb{P}V_{\mathbb{C}}$ , we have $\unicode[STIX]{x1D70B}_{V}(z)\in {\mathcal{C}}$ for all $(\unicode[STIX]{x1D70F},\unicode[STIX]{x1D70E})\in {\mathcal{H}}_{\ell ,\ell ^{\prime }}$ . Conversely, every $[v]\in {\mathcal{C}}$ contains a representative of the form $z(\unicode[STIX]{x1D70F},\unicode[STIX]{x1D70E})$ for some pair $(\unicode[STIX]{x1D70F},\unicode[STIX]{x1D70E})\in {\mathcal{H}}_{\ell ,\ell ^{\prime }}$ . Usually, in the following, since $\ell$ and $\ell ^{\prime }$ are fixed, we shall simply write ${\mathcal{H}}={\mathcal{H}}_{\ell ,\ell ^{\prime }}$ .

The isotropic line $I_{\mathbb{C}}=I\otimes _{\mathbf{k}}\mathbb{C}=[\ell ]$ corresponds to the cusp at infinity of  ${\mathcal{H}}$ .

2.3 Stabilizer of the cusp

Next, we describe the stabilizer in $\unicode[STIX]{x1D6E4}$ of the cusp $[\ell ]$ . Consider the following transformations corresponding to elements of $\text{SU}(V)$ :

(2) $$\begin{eqnarray}\displaystyle & \displaystyle [h,0]:v\mapsto v-\langle v,\ell \rangle \unicode[STIX]{x1D6FF}_{\mathbf{k}}h\ell \quad \text{for }h\in \mathbb{Q}, & \displaystyle\end{eqnarray}$$

(3) $$\begin{eqnarray}\displaystyle & \displaystyle [0,t]:v\mapsto v+\langle v,\ell \rangle t-\langle v,t\rangle \ell -{\textstyle \frac{1}{2}}\langle v,\ell \rangle \langle t,t\rangle \ell \quad \text{for }t\in W. & \displaystyle\end{eqnarray}$$

Clearly, these transformations stabilize the isotropic subspace $\mathbf{k}\ell$ . Their action on ${\mathcal{H}}$ is given as follows:

$$\begin{eqnarray}\displaystyle & \displaystyle [h,0]:(\unicode[STIX]{x1D70F},\unicode[STIX]{x1D70E})\mapsto (\unicode[STIX]{x1D70F}+h,\unicode[STIX]{x1D70E}), & \displaystyle \nonumber\\ \displaystyle & \displaystyle \hspace{0.0pt}[0,t]:(\unicode[STIX]{x1D70F},\unicode[STIX]{x1D70E})\mapsto \biggl(\unicode[STIX]{x1D70F}+\frac{\langle \unicode[STIX]{x1D70E},t\rangle }{\unicode[STIX]{x1D6FF}_{\mathbf{k}}\langle \ell ^{\prime },\ell \rangle }+\frac{1}{2}\frac{\langle t,t\rangle }{\unicode[STIX]{x1D6FF}_{\mathbf{k}}},\unicode[STIX]{x1D70E}+\langle \ell ^{\prime },\ell \rangle t\biggr). & \displaystyle \nonumber\end{eqnarray}$$

The Heisenberg group attached to $\ell$ , denoted $\text{Heis}_{\ell }$ , is the set of pairs $[h,t]$ with group law given by

(4) $$\begin{eqnarray}[h,t]\circ [h^{\prime },t^{\prime }]=\biggl[h+h^{\prime }+\frac{\Im \langle t^{\prime },t\rangle }{|\unicode[STIX]{x1D6FF}_{\mathbf{k}}|},t+t^{\prime }\biggr].\end{eqnarray}$$

Here, we follow the convention that $([h,t]\circ [h^{\prime },t^{\prime }])v=[h,t]([h^{\prime },t^{\prime }]\,v)$ for $v\in V_{\mathbf{k}}$ . The center of the Heisenberg group consists of transformations of type (2).

We denote by $\unicode[STIX]{x1D6E4}_{\ell }$ the subgroup of $\unicode[STIX]{x1D6E4}$ given by the intersection $\unicode[STIX]{x1D6E4}\cap \text{Heis}_{\ell }$ , its center we denote by $\unicode[STIX]{x1D6E4}_{\ell ,T}$ . The full stabilizer of the cusp in $\unicode[STIX]{x1D6E4}$ is given by the semidirect product

$$\begin{eqnarray}\unicode[STIX]{x1D6E4}_{\ell }\ltimes (\text{U}(W)\cap \unicode[STIX]{x1D6E4})=\operatorname{Stab}_{\unicode[STIX]{x1D6E4}}(\ell ).\end{eqnarray}$$

Note that $\unicode[STIX]{x1D6E4}_{\ell }$ has finite index in the stabilizer. The elements of $\unicode[STIX]{x1D6E4}_{\ell }$ can be described as follows (this is well-known):

2.4 Boundary components

The modular variety $X_{\unicode[STIX]{x1D6E4}}$ is given by the quotient

$$\begin{eqnarray}\unicode[STIX]{x1D6E4}\backslash \mathbb{D}\simeq \unicode[STIX]{x1D6E4}\backslash \text{U}(V)(\mathbb{R})/{\mathcal{K}}\simeq \unicode[STIX]{x1D6E4}\backslash {\mathcal{H}}.\end{eqnarray}$$

Note that $X_{\unicode[STIX]{x1D6E4}}$ is noncompact. The usual Baily–Borel compactification $X_{\unicode[STIX]{x1D6E4},BB}^{\ast }$ is obtained by introducing a topology and a complex structure on the quotient

$$\begin{eqnarray}\unicode[STIX]{x1D6E4}\backslash ({\mathcal{H}}\cup \{I_{\mathbb{R}};I\in \text{Iso}(V)\}).\end{eqnarray}$$

We sketch this for the cusp at infinity, defined by $[\ell ]$ . The following sets constitute a system of neighborhoods of the cusp

(5) $$\begin{eqnarray}U_{\unicode[STIX]{x1D716}}(\ell )=\left\{[z]\in {\mathcal{C}};\frac{\langle z,z\rangle }{|\langle z,\ell \rangle |^{2}}|\langle \ell ^{\prime },\ell \rangle |^{2}>\frac{1}{\unicode[STIX]{x1D716}}\right\}\quad (\unicode[STIX]{x1D716}>0).\end{eqnarray}$$

A subset $V$ of ${\mathcal{C}}\cup \{[\ell ]\}$ is called open if $V\cap {\mathcal{C}}$ is open in the usual sense and further if $[\ell ]\in V$ implies $U_{\unicode[STIX]{x1D716}}(\ell )\subset V$ for some $\unicode[STIX]{x1D716}>0$ .

Through the quotient topology, this construction yields a topology on $\unicode[STIX]{x1D6E4}\backslash ({\mathcal{C}}\cup \{[\ell ]\})$ . The complex structure is defined though the pullback under the canonical projection ${\mathcal{C}}\cap \{I_{\mathbb{R}};I\in \text{Iso}(V)\}\rightarrow X_{\unicode[STIX]{x1D6E4},BB}^{\ast }$ , locally for each cusp, see [Reference Hofmann8] for details. This way, one gets the structure of a normal complex space on $X_{\unicode[STIX]{x1D6E4},BB}^{\ast }$ . In general, however, there are still singularities at the boundary points.

This difficulty can be avoided by using a toroidal compactification, instead. We recall the construction briefly; see [Reference Hofmann8, Chapter 1.1.5] and, in particular [Reference Bruinier, Howard and Yang5, Section 4.3] for more details. In the following, identify the sets $U_{\unicode[STIX]{x1D716}}(\ell )\subset {\mathcal{C}}$ with the corresponding sets of representatives in ${\mathcal{H}}_{\ell ,\ell ^{\prime }}$ . Clearly, the Heisenberg group $\unicode[STIX]{x1D6E4}_{\ell }$ operates on $U_{\unicode[STIX]{x1D716}}(\ell )$ . For sufficiently small $\unicode[STIX]{x1D716}$ , there is an open immersion

$$\begin{eqnarray}\unicode[STIX]{x1D6E4}_{\ell }\backslash U_{\unicode[STIX]{x1D716}}(\ell )\rightarrow X_{\unicode[STIX]{x1D6E4}}.\end{eqnarray}$$

Recall that for the center $C(\unicode[STIX]{x1D6E4}_{\ell })=\unicode[STIX]{x1D6E4}_{\ell ,T}$ , we have $\unicode[STIX]{x1D6E4}_{\ell ,T}\simeq \mathbb{Z}N_{\ell ,\unicode[STIX]{x1D6E4}}$ . We set $q_{\ell }:=\exp (2\unicode[STIX]{x1D70B}i\unicode[STIX]{x1D70F}/N_{\ell ,\unicode[STIX]{x1D6E4}})$ . The quotient $\unicode[STIX]{x1D6E4}_{\ell ,T}\backslash U_{\unicode[STIX]{x1D716}}(\ell )$ can now be viewed as bundle of punctured disks over $W_{\mathbb{C}}$ :

$$\begin{eqnarray}V_{\unicode[STIX]{x1D716}}(\ell ):=\unicode[STIX]{x1D6E4}_{\ell ,T}\backslash U_{\unicode[STIX]{x1D716}}(\ell )\simeq \left\{(q_{\ell },\unicode[STIX]{x1D70E});0<|q_{\ell }|<\exp \biggl(\frac{\unicode[STIX]{x1D70B}\langle \unicode[STIX]{x1D70E},\unicode[STIX]{x1D70E}\rangle +\unicode[STIX]{x1D716}^{-1}}{|\unicode[STIX]{x1D6FF}_{\mathbf{k}}|^{2}|\langle \ell ^{\prime },\ell \rangle |^{2}}\biggr)\right\}.\end{eqnarray}$$

Adding the center to each disk, we get the disk bundle

$$\begin{eqnarray}\widetilde{V_{\unicode[STIX]{x1D716}}}(\ell ):=\left\{(q_{\ell },\unicode[STIX]{x1D70E});|q_{\ell }|<\exp \biggl(\frac{\unicode[STIX]{x1D70B}\langle \unicode[STIX]{x1D70E},\unicode[STIX]{x1D70E}\rangle +\unicode[STIX]{x1D716}^{-1}}{|\unicode[STIX]{x1D6FF}_{\mathbf{k}}|^{2}|\langle \ell ^{\prime },\ell \rangle |^{2}}\biggr)\right\}.\end{eqnarray}$$

The action of $\unicode[STIX]{x1D6E4}_{\ell }$ is well-defined at each center, leaving the divisor $q=0$ fixed. Also, if $\unicode[STIX]{x1D6E4}$ is sufficiently small, the operation is free, hence we get an open immersion

(6) $$\begin{eqnarray}\unicode[STIX]{x1D6E4}_{\ell }\backslash U_{\unicode[STIX]{x1D716}}(\ell )\rightarrow (\unicode[STIX]{x1D6E4}_{\ell }/\unicode[STIX]{x1D6E4}_{\ell ,T})\backslash \widetilde{V_{\unicode[STIX]{x1D716}}}(\ell ),\end{eqnarray}$$

by which the right-hand side can be glued to $X_{\unicode[STIX]{x1D6E4}}$ , yielding a partial compactification. For a point $(0,\unicode[STIX]{x1D70E}_{0})\in \widetilde{V_{\unicode[STIX]{x1D716}}}(\ell )$ , we define a system of open sets

$$\begin{eqnarray}B_{\unicode[STIX]{x1D6FF}}(0,\unicode[STIX]{x1D70E}_{0})=\{(q_{\ell },\unicode[STIX]{x1D70E})\in \widetilde{V_{\unicode[STIX]{x1D716}}}(\ell );\langle \unicode[STIX]{x1D70E}-\unicode[STIX]{x1D70E}_{0},\unicode[STIX]{x1D70E}-\unicode[STIX]{x1D70E}_{0}\rangle <\unicode[STIX]{x1D6FF},|q_{\ell }|<\unicode[STIX]{x1D6FF}\}\quad (\unicode[STIX]{x1D6FF}>0).\end{eqnarray}$$

Under the immersion (6) the images of these sets form a system of open neighborhoods for the boundary point at $(0,\unicode[STIX]{x1D70E}_{0})$ .

Repeating this construction and the gluing procedure for every $I\in \unicode[STIX]{x1D6E4}\backslash \text{Iso}(V)$ yields a compactification of $X_{\unicode[STIX]{x1D6E4}}$ , which we denote $X_{\unicode[STIX]{x1D6E4},\text{tor}}^{\ast }$ .

3.1 Bilinear forms in the cohomology

In this subsection, we examine the image of certain bilinear forms in the cohomology. All calculations are carried out using the standard inhomogeneous complex of group cohomology (cf. [Reference Shimura10, Chapter 8]).

To a bilinear form in $\mathsf{BIL}$ we can associate an element of $\text{H}^{2}(\unicode[STIX]{x1D6E4}_{\ell },{\mathcal{O}}_{\unicode[STIX]{x1D716}})$ . Define the two-cocycle in $C^{2}(\unicode[STIX]{x1D6E4}_{\ell },{\mathcal{O}}_{\unicode[STIX]{x1D716}})$ by setting

(11) $$\begin{eqnarray}B([h,t],[h^{\prime },t^{\prime }]):=B(t,t^{\prime })\quad ([h,t],[h^{\prime },t^{\prime }]\in \unicode[STIX]{x1D6E4}_{\ell }).\end{eqnarray}$$

The class $[B]$ of this cocycle is the image of $B$ in the cohomology. For $B\in \mathsf{BIL}_{\mathbb{Z}}$ we also define a two-cocycle in $\text{C}^{2}(\unicode[STIX]{x1D6E4}_{\ell },\mathbb{Z})$ and the attached element in $\text{H}^{2}(\unicode[STIX]{x1D6E4}_{\ell },\mathbb{Z})$ . Thus, composing with the natural map $\text{H}^{2}(\unicode[STIX]{x1D6E4}_{\ell },\mathbb{Z})\rightarrow \text{H}^{2}(\unicode[STIX]{x1D6E4}_{\ell },{\mathcal{O}}_{\unicode[STIX]{x1D716}})$ from (7) we have a sequence

(12)

The composition of the two maps in (12) is just the restriction to $\mathsf{BIL}_{\mathbb{Z}}$ of the map $\mathsf{BIL}\rightarrow \text{H}^{2}(\unicode[STIX]{x1D6E4}_{\ell },{\mathcal{O}}_{\unicode[STIX]{x1D716}})$ defined by (11). It turns out that the sequence is exact:

Now that we know the sequence (12) to be exact, we study the first map $\mathsf{BIL}_{\mathbb{Z}}\rightarrow \text{H}^{2}(\unicode[STIX]{x1D6E4}_{\ell },\mathbb{Z})$ . It is far from being injective. The following lemma and its proof are essentially due to Freitag, a sketch is contained in (E. Freitag, personal communication, 2007).

Lemma 3.1. The kernel of the map $\mathsf{BIL}_{\mathbb{Z}}\rightarrow \text{H}^{2}(\unicode[STIX]{x1D6E4}_{\ell },\mathbb{Z})$ is the cyclic group generated by the antisymmetric bilinear form

$$\begin{eqnarray}\frac{1}{N_{\ell ,\unicode[STIX]{x1D6E4}}}\frac{\Im \langle \cdot ,\cdot \rangle }{|\unicode[STIX]{x1D6FF}_{\mathbf{k}}|}.\end{eqnarray}$$

In particular, the image of an element $B\in \mathsf{BIL}_{\mathbb{Z}}$ is a torsion element in $\text{H}^{2}(\unicode[STIX]{x1D6E4}_{\ell },\mathbb{Z})$ if and if $B$ and $|\unicode[STIX]{x1D6FF}_{\mathbf{k}}|^{-1}\Im \langle \cdot ,\cdot \rangle$ are linear dependent over $\mathbb{Z}$ .

Proof. The proof uses a transgression map, which we introduce next. First note that, since the action of $\mathbb{Z}$ is trivial, the map $\mathsf{BIL}_{\mathbb{Z}}\rightarrow \text{H}^{2}(\unicode[STIX]{x1D6E4}_{\ell },\mathbb{Z})$ factors over $\text{H}^{2}(D_{\ell ,\unicode[STIX]{x1D6E4}},\mathbb{Z})=\text{H}^{2}((\unicode[STIX]{x1D6E4}_{\ell }/N_{\ell ,\unicode[STIX]{x1D6E4}}\mathbb{Z}),\mathbb{Z})$ . With Proposition 3.1 we have:

Now, the transgression $\operatorname{tg}$ is defined as the map for which the sequence

(13)

becomes exact. Thus, the kernel of the map into $\text{H}^{2}(\unicode[STIX]{x1D6E4}_{\ell },\mathbb{Z})$ is generated by the image of the identity map $\mathbf{1}:\mathbb{Z}\rightarrow \mathbb{Z}$ under $\operatorname{tg}$ . The image $\operatorname{tg}(\mathbf{1})$ is represented by a coboundary $(t,t^{\prime })\mapsto (\unicode[STIX]{x2202}u)([\cdot ,t],[\cdot ,t^{\prime }])$ , with a one-cochain $u:\unicode[STIX]{x1D6E4}_{\ell }\rightarrow \mathbb{Z}$ , which has to satisfy two conditions:

1. (1) $u([N_{\ell ,\unicode[STIX]{x1D6E4}}\,h,0])=h$ for all $h\in \mathbb{Z}$ and

2. (2) $(\unicode[STIX]{x2202}u)([h,t],[h^{\prime },t^{\prime }])$ does not depend on $h$ or $h^{\prime }$ .

A suitable $u$ is obtained by setting $u([N_{\ell ,\unicode[STIX]{x1D6E4}}h,t]):=h$ . We get

$$\begin{eqnarray}\displaystyle (\unicode[STIX]{x2202}u)([h,t],[h^{\prime },t^{\prime }]) & = & \displaystyle [h,t]u([h^{\prime },t^{\prime }])-u([h,t][h^{\prime },t^{\prime }])+u([h,t])\nonumber\\ \displaystyle & = & \displaystyle -\frac{1}{N_{\ell ,\unicode[STIX]{x1D6E4}}}\frac{\Im \langle t^{\prime },t\rangle }{|\unicode[STIX]{x1D6FF}_{\mathbf{k}}|}.\nonumber\end{eqnarray}$$

Hence, $\operatorname{tg}(\mathbf{1})$ is represented by the cocycle

$$\begin{eqnarray}(t,t^{\prime })\longmapsto \frac{1}{N_{\ell ,\unicode[STIX]{x1D6E4}}}\frac{\Im \langle t,t^{\prime }\rangle }{|\unicode[STIX]{x1D6FF}_{\mathbf{k}}|},\end{eqnarray}$$

any integer multiple of which is then contained in the kernel. Thus, for any $B\in \mathsf{BIL}_{\mathbb{Z}}$ the image $[B]$ is a torsion element precisely if it is linear dependent to $\operatorname{tg}(\mathbf{1})$ over $\mathbb{Z}$ .◻

The linear dependence condition in the lemma can more conveniently be formulated thus: if $B\in \mathsf{BIL}_{\mathbb{Z}}$ , the image is a torsion element if and only if there is a rational number $Q$ , such that for all $t,t^{\prime }\in D_{\ell ,\unicode[STIX]{x1D6E4}}$ , the following equation holds:

(14) $$\begin{eqnarray}B(t,t^{\prime })-Q\frac{\Im \langle t,t^{\prime }\rangle }{|\unicode[STIX]{x1D6FF}_{\mathbf{k}}|}=0.\end{eqnarray}$$

Since $D_{\ell ,\unicode[STIX]{x1D6E4}}$ has full rank in $W_{\mathbf{k}}$ , by linear extension, equivalently, the equation holds for all $t,t^{\prime }\in W$ ; similarly for all $t,t^{\prime }\in W_{\mathbb{C}}$ . As an example for this, we give an application to Hermitian forms.

Remark 3.2. Let $H$ be a $\mathbf{k}$ -valued Hermitian form on $W_{\mathbf{k}}$ , and assume that ${|\unicode[STIX]{x1D6FF}_{\mathbf{k}}|}^{-1}\Im H\in \mathsf{BIL}_{\mathbb{Z}}$ . Further assume that $H$ is linear in its left argument (otherwise, invert the sign in the equation below).

Then, the map $H\mapsto {|\unicode[STIX]{x1D6FF}_{\mathbf{k}}|}^{-1}\Im H$ defines a torsion element in $\text{H}^{2}(\unicode[STIX]{x1D6E4}_{\ell },\mathbb{Z})$ if and only if the following equation holds for all $t,t^{\prime }\in D_{\ell ,\unicode[STIX]{x1D6E4}}$ :

(15) $$\begin{eqnarray}H(t,t^{\prime })+\frac{\operatorname{Tr}H}{n}\Im \langle t,t^{\prime }\rangle =0,\end{eqnarray}$$

where the trace $\operatorname{Tr}$ is taken over a normalized orthogonal basis for $\langle \cdot ,\cdot \rangle$ .

Proof. Taking the imaginary part of (15), we see that indeed, if the equation holds, $\Im H$ is a rational multiple of $\Im \langle \cdot ,\cdot \rangle$ and thus ${|\unicode[STIX]{x1D6FF}_{\mathbf{k}}|}^{-1}\Im H$ defines a torsion element by the lemma.

Conversely, assume that the image is a torsion element in $\text{H}^{2}(\unicode[STIX]{x1D6E4}_{\ell },\mathbb{Z})$ . Then, by the lemma, the form has to be linear dependent to ${|\unicode[STIX]{x1D6FF}_{\mathbf{k}}|}^{-1}\Im \langle \cdot ,\cdot \rangle$ and satisfies an equation of the form (14). Since by linear extension, the equation holds for all $t,t^{\prime }\in W_{\mathbb{C}}$ , we may replace $t$ by a purely imaginary multiple. The resulting equation, equivalent to the first, is the following:

$$\begin{eqnarray}\Re H(t,t^{\prime })-Q\cdot \Re \langle t,t^{\prime }\rangle =0,\end{eqnarray}$$

valid for all $t,t^{\prime }\in W_{\mathbb{C}}$ . Whence by linear combination of the two equations, we find $H(t,t^{\prime })=Q\langle t,t^{\prime }\rangle$ for all $t,t^{\prime }\in W_{\mathbb{C}}$ . To determine the factor of proportionality $Q$ , we take the trace. We have $\operatorname{Tr}\langle \cdot ,\cdot \rangle \mid _{W_{\mathbb{C}}}=-n$ , and get $Q=-(1/n)\operatorname{Tr}H$ .◻

4.1 Local Heegner divisors

First, we recall the usual definition of Heegner divisors on ${\mathcal{H}}$ (cf. [Reference Hofmann9, Section 6]), and introduce local Heegner divisors in the neighborhoods $U_{\unicode[STIX]{x1D716}}(\ell )$ of the cusp $[\ell ]$ .

Let $\unicode[STIX]{x1D706}\in L^{\prime }$ be a lattice vector of negative norm, that is $\langle \unicode[STIX]{x1D706},\unicode[STIX]{x1D706}\rangle <0$ . The (primitive) Heegner divisor $\mathbf{H}(\unicode[STIX]{x1D706})$ attached to $\unicode[STIX]{x1D706}$ is a divisor on ${\mathcal{H}}$ given by

$$\begin{eqnarray}\mathbf{H}(\unicode[STIX]{x1D706}):=\{(\unicode[STIX]{x1D70F},\unicode[STIX]{x1D70E})\in {\mathcal{H}};\langle \unicode[STIX]{x1D706},z(\unicode[STIX]{x1D70F},\unicode[STIX]{x1D70E})\rangle =0\},\end{eqnarray}$$

with $z(\unicode[STIX]{x1D70F},\unicode[STIX]{x1D70E})=\ell ^{\prime }-\unicode[STIX]{x1D70F}\unicode[STIX]{x1D6FF}_{\mathbf{k}}\langle \ell ,\ell ^{\prime }\rangle \ell +\unicode[STIX]{x1D70E}$ (see Section 2.2). Clearly, the divisor $\mathbf{H}(\unicode[STIX]{x1D706})$ intersects $U_{\unicode[STIX]{x1D716}}(\ell )$ for every $\unicode[STIX]{x1D716}>0$ , if and only if $\langle \unicode[STIX]{x1D706},\ell \rangle =0$ . In the following, we denote by $\ell ^{\bot }$ the (orthogonal) complement of $\ell$ with respect to $\langle \cdot ,\cdot \rangle$ .

Thus, let $\unicode[STIX]{x1D706}\in L^{\prime }\cap \ell ^{\bot }$ . Then, $\unicode[STIX]{x1D706}=\unicode[STIX]{x1D706}_{\ell }\ell +\unicode[STIX]{x1D706}_{D}$ with $\unicode[STIX]{x1D706}_{D}\in W_{\mathbf{k}}$ and $\mathbf{H}(\unicode[STIX]{x1D706})$ is given by an equation of the form

$$\begin{eqnarray}\unicode[STIX]{x1D706}_{\ell }\langle \ell ,\ell ^{\prime }\rangle +\langle \unicode[STIX]{x1D706}_{D},\unicode[STIX]{x1D70E}\rangle =0.\end{eqnarray}$$

Consider the orbit of $\unicode[STIX]{x1D706}$ under $\unicode[STIX]{x1D6E4}_{\ell }$ . Since $\unicode[STIX]{x1D6E4}$ is a modular group, the Heisenberg group $\unicode[STIX]{x1D6E4}_{\ell }$ operates trivially on the discriminant group $L^{\prime }/L$ and thus $[h,t]\unicode[STIX]{x1D706}\equiv \unicode[STIX]{x1D706}\hspace{0.6em}({\rm mod}\hspace{0.2em}L)$ for all $[h,t]\in \unicode[STIX]{x1D6E4}_{\ell }$ . Also, since $\unicode[STIX]{x1D706}\in \ell ^{\bot }$ , it remains fixed under $[h,0]$ for all $h\in N_{\ell ,\unicode[STIX]{x1D6E4}}\mathbb{Z}$ , and $\unicode[STIX]{x1D6E4}_{\ell ,T}$ acts trivially.

For an Eichler element $[0,t]$ with $t\in D_{\ell ,\unicode[STIX]{x1D6E4}}$ , we have

$$\begin{eqnarray}[0,t]\unicode[STIX]{x1D706}=\unicode[STIX]{x1D706}-\langle \unicode[STIX]{x1D706}_{D},t\rangle \ell =(\unicode[STIX]{x1D706}_{\ell }-\langle \unicode[STIX]{x1D706}_{D},t\rangle )\ell +\unicode[STIX]{x1D706}_{D}.\end{eqnarray}$$

Thus, the orbit of $\unicode[STIX]{x1D706}$ under $\unicode[STIX]{x1D6E4}_{\ell }/\unicode[STIX]{x1D6E4}_{\ell ,T}\simeq D_{\ell ,\unicode[STIX]{x1D6E4}}$ is given by $\unicode[STIX]{x1D706}-\mathfrak{T}\ell$ , where $\mathfrak{T}$ denotes the set

$$\begin{eqnarray}\mathfrak{T}=\mathfrak{T}(\unicode[STIX]{x1D706}):=\{\langle \unicode[STIX]{x1D706},t\rangle ;t\in D_{\ell ,\unicode[STIX]{x1D6E4}}\}.\end{eqnarray}$$

Note that $\mathfrak{T}\subseteq \mathfrak{d}_{\mathbf{k}}^{-1}$ (as a fractional ideal), since $D_{\ell ,\unicode[STIX]{x1D6E4}}\subseteq D$ .

Hence, the group $\unicode[STIX]{x1D6E4}_{\ell }$ operates on the set $\unicode[STIX]{x1D706}+\mathfrak{d}_{\mathbf{k}}^{-1}\ell$ with only finitely many orbits and thus, the divisor

(18) $$\begin{eqnarray}\mathbf{H}_{\infty }(\unicode[STIX]{x1D706}):=\mathop{\sum }_{\unicode[STIX]{x1D6FC}\in \mathfrak{d}_{\mathbf{k}}^{-1}}\mathbf{H}(\unicode[STIX]{x1D706}+\unicode[STIX]{x1D6FC}\ell )\end{eqnarray}$$

is invariant under $\unicode[STIX]{x1D6E4}_{\ell }$ and defines an element of $\operatorname{Div}(\unicode[STIX]{x1D6E4}_{\ell }\backslash U_{\unicode[STIX]{x1D716}}(\ell ))$ .

Heegner divisors with index

Now, let $\unicode[STIX]{x1D6FD}\in L^{\prime }/L$ be an element of the discriminant group and $m$ a negative integer. Then, the Heegner divisor of index $(\unicode[STIX]{x1D6FD},m)$ , defined as the (locally finite) sum

(19) $$\begin{eqnarray}\mathbf{H}(\unicode[STIX]{x1D6FD},m)=\mathop{\sum }_{\substack{ \unicode[STIX]{x1D706}\in L^{\prime } \\ q(\unicode[STIX]{x1D706})=m \\ \unicode[STIX]{x1D706}+L=\unicode[STIX]{x1D6FD}}}\mathbf{H}(\unicode[STIX]{x1D706}),\end{eqnarray}$$

is a $\unicode[STIX]{x1D6E4}$ -invariant divisor on ${\mathcal{H}}$ . Under the canonical projection $\mathbf{H}(\unicode[STIX]{x1D6FD},m)$ is the inverse image of a divisor on $X_{\unicode[STIX]{x1D6E4}}$ . Also note that $\mathbf{H}(\unicode[STIX]{x1D6FD},m)=\mathbf{H}(-\unicode[STIX]{x1D6FD},m)$ .

Through the open immersion $\unicode[STIX]{x1D6E4}_{\ell }\backslash U_{\unicode[STIX]{x1D716}}(\ell ){\hookrightarrow}\unicode[STIX]{x1D6E4}\backslash {\mathcal{H}}=X_{\unicode[STIX]{x1D6E4}}$ from Section 2.4, the inclusion $U_{\unicode[STIX]{x1D716}}(\ell )\subset {\mathcal{H}}$ and the projection maps, we get a commutative diagram

We denote by $\mathbf{H}_{\ell }(\unicode[STIX]{x1D6FD},m)$ the image in $\operatorname{Div}(\unicode[STIX]{x1D6E4}_{\ell }\backslash U_{\unicode[STIX]{x1D716}}(\ell ))$ of the divisor $\mathbf{H}(\unicode[STIX]{x1D6FD},m)\in \operatorname{Div}(X_{\unicode[STIX]{x1D6E4}})$ . The corresponding $\unicode[STIX]{x1D6E4}_{\ell }$ -invariant divisor in $\operatorname{Div}(U_{\unicode[STIX]{x1D716}}(\ell ))$ is also denoted by $\mathbf{H}_{\ell }(\unicode[STIX]{x1D6FD},m)$ .

For sufficiently small $\unicode[STIX]{x1D716}$ , the divisor $\mathbf{H}_{\ell }(\unicode[STIX]{x1D6FD},m)$ is given by the restriction to $U_{\unicode[STIX]{x1D716}}(\ell )$ of the sum on the right-hand side of (19). Then, only $\unicode[STIX]{x1D706}$ ’s perpendicular to $\ell$ contribute. In particular, if $\mathbf{H}_{\ell }(\unicode[STIX]{x1D6FD},m)$ is nonzero, then $\unicode[STIX]{x1D6FD}$ is contained in the subgroup

$$\begin{eqnarray}\displaystyle {\mathcal{L}} & := & \displaystyle \{\!\unicode[STIX]{x1D6FE}\in L^{\prime }/L;2\Re \langle \unicode[STIX]{x1D6FE},\ell \rangle \equiv 0\hspace{0.2em}{\rm mod}\hspace{0.2em}M_{1}\nonumber\\ \displaystyle & & \displaystyle \quad \text{and }|\unicode[STIX]{x1D6FF}_{\mathbf{k}}|\Im \langle \unicode[STIX]{x1D6FE},\ell \rangle \equiv 0\hspace{0.2em}{\rm mod}\hspace{0.2em}M_{2}\!\}\subseteq L^{\prime }/L,\nonumber\end{eqnarray}$$

where $M_{1}$ , $M_{2}$ are the unique integers given by $2\Re \langle L,\ell \rangle =M_{1}\mathbb{Z}$ and by $|\unicode[STIX]{x1D6FF}_{\mathbf{k}}|\Im \langle L,\ell \rangle =M_{2}\mathbb{Z}$ .

With $\unicode[STIX]{x1D6FD}\in {\mathcal{L}}$ the local divisor $\mathbf{H}_{\ell }(\unicode[STIX]{x1D6FD},m)$ can be written in the form

(20) $$\begin{eqnarray}\mathbf{H}_{\ell }(\unicode[STIX]{x1D6FD},m)=\mathop{\sum }_{\substack{ \unicode[STIX]{x1D705}\in D \\ q(\unicode[STIX]{x1D705}+\dot{\unicode[STIX]{x1D6FD}})=m}}\mathbf{H}_{\infty }(\unicode[STIX]{x1D705}+\dot{\unicode[STIX]{x1D6FD}}).\end{eqnarray}$$

Here, we adopt the notation of [Reference Bruinier and Freitag4, Section 4], by which $\dot{\unicode[STIX]{x1D6FD}}$ denotes a representative of $\unicode[STIX]{x1D6FD}$ with $\dot{\unicode[STIX]{x1D6FD}}\in L^{\prime }\cap \ell ^{\bot }$ , fixed once and for all for every $\unicode[STIX]{x1D6FD}\in {\mathcal{L}}$ . Note that a surjective homomorphism is given by

$$\begin{eqnarray}\unicode[STIX]{x1D70B}:{\mathcal{L}}\longrightarrow D^{\prime }/D,\qquad \unicode[STIX]{x1D6FD}\longmapsto \dot{\unicode[STIX]{x1D6FD}}_{D},\end{eqnarray}$$

where $\dot{\unicode[STIX]{x1D6FD}}_{D}$ denotes the definite part of $\dot{\unicode[STIX]{x1D6FD}}$ .

4.2 Local Borcherds products

In this section, our aim is to use local Borcherds products to describe the position of Heegner divisors in the cohomology. Given a lattice vector $\unicode[STIX]{x1D706}$ of negative norm with $\unicode[STIX]{x1D706}\in L^{\prime }\cap \ell ^{\bot }$ we can realize the local Heegner divisor attached to $\unicode[STIX]{x1D706}$ through an infinite product with factors of the form $(1-e(\langle z,[0,t]\unicode[STIX]{x1D706}\rangle ))$ with $[0,t]\in \unicode[STIX]{x1D6E4}_{\ell ,T}$ .

If for the Heegner divisor $\mathbf{H}_{\infty }(\unicode[STIX]{x1D706})$ as in (18), we set

$$\begin{eqnarray}\mathop{\prod }_{\unicode[STIX]{x1D6FD}\in \mathfrak{d}_{\mathbf{k}}^{-1}}[1-e(\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D6FD})\langle z,\unicode[STIX]{x1D706}-\unicode[STIX]{x1D6FD}\ell \rangle )],\quad \text{with }\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D6FD})\in \{\pm 1\},\end{eqnarray}$$

we get an infinite product with (zero-)divisor $\mathbf{H}(\infty )$ . For $\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D6FD})\equiv 1$ the product would be $\unicode[STIX]{x1D6E4}_{\ell }$ -invariant. However, to assure absolute convergence, we must define the sign $\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D6FD})$ depending on $\Im \unicode[STIX]{x1D6FD}$ . Then, the product is no longer fully invariant. Instead, the operation of Eichler transformations gives rise to a nontrivial automorphy factor, which we use to determine the position of $\mathbf{H}_{\infty }(\unicode[STIX]{x1D706})$ in the local Picard group.

We remark that this is not a particularly serious restriction, as under the assumptions concerning $\ell$ and $\ell ^{\prime }$ from Section 2.1, it is always possible to choose $\ell ^{\prime }$ suitably.

Now, keeping in mind that $\mathfrak{d}_{\mathbf{k}}=\unicode[STIX]{x1D6FF}_{\mathbf{k}}^{-1}{\mathcal{O}}_{\mathbf{k}}$ and $\overline{{\mathcal{O}}}_{\mathbf{k}}={\mathcal{O}}_{\mathbf{k}}=-{\mathcal{O}}_{\mathbf{k}}$ , we define the local Borcherds products as follows:

Definition 4.1. Let $\unicode[STIX]{x1D706}\in L^{\prime }$ be a negative-norm lattice vector in the orthogonal complement of $\ell$ . The local Borcherds product $\unicode[STIX]{x1D6F9}_{\unicode[STIX]{x1D706}}(z)$ attached to $\mathbf{H}_{\infty }(\unicode[STIX]{x1D706})$ is defined as

$$\begin{eqnarray}\unicode[STIX]{x1D6F9}_{\unicode[STIX]{x1D706}}(z):=\mathop{\prod }_{\unicode[STIX]{x1D6FC}\in {\mathcal{O}}_{\mathbf{k}}}\biggl[1-e\biggl(\unicode[STIX]{x1D70E}(\Im \unicode[STIX]{x1D6FC})\biggl(\langle z,\unicode[STIX]{x1D706}\rangle +\frac{\unicode[STIX]{x1D6FC}}{|D_{\mathbf{k}}|}\biggr)\biggr)\biggr],\end{eqnarray}$$

with a sign $\unicode[STIX]{x1D70E}(\Im \unicode[STIX]{x1D6FC})$ defined as follows:

$$\begin{eqnarray}\unicode[STIX]{x1D70E}(\Im \unicode[STIX]{x1D6FC})=\left\{\begin{array}{@{}ll@{}}1\quad & \text{if }\Im \unicode[STIX]{x1D6FC}\geqslant 0,\\ -1\quad & \text{otherwise.}\end{array}\right.\end{eqnarray}$$

Clearly, $\unicode[STIX]{x1D6F9}_{\unicode[STIX]{x1D706}}(z)$ is an absolutely convergent infinite product with divisor $\mathbf{H}_{\infty }(\unicode[STIX]{x1D706})$ . With $\mathfrak{d}_{\mathbf{k}}^{-1}=\unicode[STIX]{x1D6FF}_{\mathbf{k}}^{-1}(\mathbb{Z}+\unicode[STIX]{x1D701}\mathbb{Z})$ , where $\Im \unicode[STIX]{x1D701}=\frac{1}{2}\unicode[STIX]{x1D6FF}_{\mathbf{k}}$ and $2\Re \unicode[STIX]{x1D701}\equiv D_{\mathbf{k}}\hspace{0.6em}({\rm mod}\hspace{0.2em}4)$ , we can write $\unicode[STIX]{x1D6F9}_{\unicode[STIX]{x1D706}}(z)$ in the following form

$$\begin{eqnarray}\unicode[STIX]{x1D6F9}_{\unicode[STIX]{x1D706}}(z)=\mathop{\prod }_{\substack{ p\hspace{0.2em}{\rm mod}\hspace{0.2em}|D_{\boldsymbol{ k}}| \\ q\in \mathbb{Z}}}\biggl[1-e\biggl(\unicode[STIX]{x1D70E}(q)\biggl(\langle z,\unicode[STIX]{x1D706}\rangle +\frac{1}{|D_{\mathbf{k}}|}(p+\unicode[STIX]{x1D701}q)\biggr)\biggr)\biggr],\end{eqnarray}$$

with $\unicode[STIX]{x1D70E}(q)=\operatorname{sign}(q)$ if $q\neq 0$ and $\unicode[STIX]{x1D70E}(0)=+1$ .

Note that $\unicode[STIX]{x1D6F9}_{\unicode[STIX]{x1D706}}$ is invariant under translations in $\unicode[STIX]{x1D6E4}_{\ell ,T}$ , while the operation of Eichler transformations, $[0,t]$ with $t\in D_{\ell ,\unicode[STIX]{x1D6E4}}$ , gives rise to the (nontrivial) automorphy factor

(21) $$\begin{eqnarray}J_{\unicode[STIX]{x1D706}}([h,t],z)=\frac{\unicode[STIX]{x1D6F9}_{\unicode[STIX]{x1D706}}([0,t]z)}{\unicode[STIX]{x1D6F9}_{\unicode[STIX]{x1D706}}(z)}\quad ([h,t]\in \unicode[STIX]{x1D6E4}_{\ell }).\end{eqnarray}$$

Proposition 4.1. The automorphy factor $J_{\unicode[STIX]{x1D706}}$ attached to $\mathbf{H}_{\infty }(\unicode[STIX]{x1D706})$ takes the form

$$\begin{eqnarray}J_{\unicode[STIX]{x1D706}}([h,t],z)=e(-2|D_{\mathbf{k}}|\langle z,\unicode[STIX]{x1D706}\rangle \Re \langle t,\unicode[STIX]{x1D706}\rangle -2(\Re \langle t,\unicode[STIX]{x1D706}\rangle )^{2}\unicode[STIX]{x1D701}+\Re \langle t,\unicode[STIX]{x1D706}\rangle (\unicode[STIX]{x1D701}+1)),\end{eqnarray}$$

with $\unicode[STIX]{x1D701}$ such that ${\mathcal{O}}_{\mathbf{k}}=Z+\unicode[STIX]{x1D701}\mathbb{Z}$ . Note that $J_{\unicode[STIX]{x1D706}}$ is independent of the choice of $\unicode[STIX]{x1D701}$ .

Proof. Since $\langle \ell ^{\prime },\ell \rangle =-{\unicode[STIX]{x1D6FF}_{\mathbf{k}}}^{-1}$ , by (3) we have $\langle [0,t]z,\unicode[STIX]{x1D706}\rangle =\langle z,\unicode[STIX]{x1D706}\rangle -{\unicode[STIX]{x1D6FF}_{\mathbf{k}}}^{-1}\langle t,\unicode[STIX]{x1D706}\rangle$ . Since $\langle t,\unicode[STIX]{x1D706}\rangle =\langle t,\unicode[STIX]{x1D706}_{D}\rangle \in \mathfrak{d}_{\mathbf{k}}^{-1}$ we can write

$$\begin{eqnarray}\langle [0,t]z,\unicode[STIX]{x1D706}\rangle =\langle z,\unicode[STIX]{x1D706}\rangle +\frac{1}{|D_{\mathbf{k}}|}(r+\unicode[STIX]{x1D701}s),\quad \text{with }r,s\in \mathbb{Z}.\end{eqnarray}$$

We note that $s=2\Re \langle t,\unicode[STIX]{x1D706}\rangle$ . Now, after permuting representatives modulo $|D_{\mathbf{k}}|$ and a shift in the index $q$ , the automorphy factor from (21) takes the form

(22) $$\begin{eqnarray}J_{\unicode[STIX]{x1D706}}([h,t],z)=\mathop{\prod }_{p\hspace{0.2em}{\rm mod}\hspace{0.2em}|D_{\mathbf{k}}|}\mathop{\prod }_{q\in \mathbb{Z}}\frac{1-e(\unicode[STIX]{x1D70E}(q-s)(\langle z,\unicode[STIX]{x1D706}\rangle +|D_{\mathbf{k}}|^{-1}(p+q\unicode[STIX]{x1D701})))}{1-e(\unicode[STIX]{x1D70E}(q)(\langle z,\unicode[STIX]{x1D706}\rangle +|D_{\mathbf{k}}|^{-1}(p+q\unicode[STIX]{x1D701})))}.\end{eqnarray}$$

Only factors with $\unicode[STIX]{x1D70E}(q-s)\neq \unicode[STIX]{x1D70E}(q)$ contribute to the product. There are two cases: either we have $s>q\geqslant 0$ , or $s\leqslant q<0$ . We examine the first case. By applying the elementary identity

$$\begin{eqnarray}\frac{1-e(-z)}{1-e(z)}=-e(-z).\end{eqnarray}$$

we get

$$\begin{eqnarray}\displaystyle J_{\unicode[STIX]{x1D706}}([h,t],z) & = & \displaystyle \mathop{\prod }_{p\hspace{0.2em}{\rm mod}\hspace{0.2em}|D_{\mathbf{k}}|}\mathop{\prod }_{0\leqslant q<s}-e\biggl(-\langle z,\unicode[STIX]{x1D706}\rangle -\frac{1}{|D_{\mathbf{k}}|}(p+q\unicode[STIX]{x1D701})\biggr)\nonumber\\ \displaystyle & = & \displaystyle \mathop{\prod }_{p\hspace{0.2em}{\rm mod}\hspace{0.2em}|D_{\mathbf{k}}|}(-1)^{s}e\biggl(-s\langle z,\unicode[STIX]{x1D706}\rangle -\frac{s}{|D_{\mathbf{k}}|}\biggl(p+\frac{s-1}{2}\unicode[STIX]{x1D701}\biggr)\biggr)\nonumber\\ \displaystyle & = & \displaystyle e\biggl(-s|D_{\mathbf{k}}|\langle z,\unicode[STIX]{x1D706}\rangle -\frac{s(s-1)}{2}\unicode[STIX]{x1D701}-\frac{s(|D_{\mathbf{k}}|-1)}{2}+\frac{s|D_{\mathbf{k}}|}{2}\biggr)\nonumber\\ \displaystyle & = & \displaystyle e\biggl(-s|D_{\mathbf{k}}|\langle z,\unicode[STIX]{x1D706}\rangle -\frac{s^{2}}{2}\unicode[STIX]{x1D701}+\frac{s}{2}\unicode[STIX]{x1D701}+\frac{s}{2}\biggr)\nonumber\end{eqnarray}$$

Hence, recalling that $s=2\Re \langle t,\unicode[STIX]{x1D706}\rangle$ , we have

(23) $$\begin{eqnarray}\displaystyle J_{\unicode[STIX]{x1D706}}([h,t],z) & = & \displaystyle e(\!-2|D_{\mathbf{k}}|\langle z,\unicode[STIX]{x1D706}\rangle \Re \langle t,\unicode[STIX]{x1D706}\rangle \nonumber\\ \displaystyle & & \displaystyle -\,2(\Re \langle t,\unicode[STIX]{x1D706}\rangle )^{2}\unicode[STIX]{x1D701}+\Re \langle t,\unicode[STIX]{x1D706}\rangle \unicode[STIX]{x1D701}+\Re \langle t,\unicode[STIX]{x1D706}\rangle ).\end{eqnarray}$$

We remark that the last term is determined only up to sign, since $2\Re \langle t,\unicode[STIX]{x1D706}\rangle \in \mathbb{Z}$ . Finally, we note that as the second term in (23) is a quarter-integer while $2\Re \unicode[STIX]{x1D701}$ is only determined modulo $4$ , the automorphy factor is independent of the choice for $\Re \unicode[STIX]{x1D701}$ .

The second case ( $s\leqslant q<0$ ) can be treated similarly, yielding the same result for the automorphy factor $J_{\unicode[STIX]{x1D706}}([h,t],z)$ .◻

4.3 The Chern class of a Heegner divisor $\mathbf{H}_{\infty }(\unicode[STIX]{x1D706})$

From the automorphy factor $J_{\unicode[STIX]{x1D706}}$ we now determine a two-cocycle representing the Chern class of the Heegner divisor $\mathbf{H}_{\infty }(\unicode[STIX]{x1D706})$ .

Proposition 4.2. The Chern class $\unicode[STIX]{x1D6FF}(\mathbf{H}_{\infty }(\unicode[STIX]{x1D706}))$ of the local Heegner divisor $\mathbf{H}_{\infty }(\unicode[STIX]{x1D706})$ in $\text{H}^{2}(\unicode[STIX]{x1D6E4}_{\ell },\mathbb{Z})$ is determined by the cocycle

$$\begin{eqnarray}[c_{\unicode[STIX]{x1D706}}]:([h,t][h^{\prime },t^{\prime }])\longmapsto -2|\unicode[STIX]{x1D6FF}_{\boldsymbol{ k}}|\Re \langle t,\unicode[STIX]{x1D706}\rangle \Im \langle t^{\prime },\unicode[STIX]{x1D706}\rangle =\Im (-|\unicode[STIX]{x1D6FF}_{\boldsymbol{ k}}|F_{\unicode[STIX]{x1D706}}(t,t^{\prime })),\end{eqnarray}$$

where $F_{\unicode[STIX]{x1D706}}(t,t^{\prime }):=2\Re \langle t,\unicode[STIX]{x1D706}\rangle \langle t^{\prime },\unicode[STIX]{x1D706}\rangle$ .

Proof. To calculate the Chern class, we must realize the connecting homomorphism $\unicode[STIX]{x1D6FF}:\text{H}^{1}(\unicode[STIX]{x1D6E4}_{\ell },{\mathcal{O}}_{\unicode[STIX]{x1D716}}^{\ast })\rightarrow \text{H}^{2}(\unicode[STIX]{x1D6E4}_{\ell },\mathbb{Z})$ . Thus, let $A(g,z)$ be a holomorphic function satisfying $J_{\unicode[STIX]{x1D706}}(g,z)=e(A(g,z))$ and set

(24) $$\begin{eqnarray}c(g,g^{\prime })=A(gg^{\prime },z)-A(g,g^{\prime }z)-A(g^{\prime },z)\quad \text{for all }g,g^{\prime }\in \unicode[STIX]{x1D6E4}_{\ell }.\end{eqnarray}$$

Then, the two-cocycle defined by the map $(g,g)\mapsto c(g,g^{\prime })$ is a representative for the Chern class in $\text{H}^{2}(\unicode[STIX]{x1D6E4}_{\ell },\mathbb{Z})$ . Note that while $A(g,g^{\prime })$ is not uniquely determined, $c(g,g^{\prime })$ is independent of this choice; also, multiplying $J_{\unicode[STIX]{x1D706}}$ with a trivial automorphy factor changes $c(g,g^{\prime })$ only by a coboundary.

Clearly, it suffices to calculate $c(g,g^{\prime })$ for Eichler transformations $g=[0,t]$ and $g^{\prime }=[0,t^{\prime }]$ . From (24) we see that the last two terms in (23), being linear in $t$ , cancel. We calculate

$$\begin{eqnarray}\displaystyle & & \displaystyle A([0,t+t^{\prime }],z)-A([0,t],[0,t^{\prime }]z)-A([0,t^{\prime }],z)\nonumber\\ \displaystyle & & \displaystyle \quad =2|D_{\mathbf{k}}|\langle [0,t^{\prime }]z-z,\unicode[STIX]{x1D706}\rangle \Re \langle t,\unicode[STIX]{x1D706}\rangle -4\Re \langle t,\unicode[STIX]{x1D706}\rangle \Re \langle t^{\prime },\unicode[STIX]{x1D706}\rangle \unicode[STIX]{x1D701}\nonumber\\ \displaystyle & & \displaystyle \quad =2\unicode[STIX]{x1D6FF}_{\mathbf{k}}\Re \langle t,\unicode[STIX]{x1D706}\rangle \langle t^{\prime },\unicode[STIX]{x1D706}\rangle -2\Re \langle t,\unicode[STIX]{x1D706}\rangle \Re \langle t^{\prime },\unicode[STIX]{x1D706}\rangle \unicode[STIX]{x1D6FF}_{\mathbf{k}}-4\Re \langle t,\unicode[STIX]{x1D706}\rangle \Re \langle t^{\prime },\unicode[STIX]{x1D706}\rangle \Re \unicode[STIX]{x1D701}\nonumber\\ \displaystyle & & \displaystyle \quad =-2|\unicode[STIX]{x1D6FF}_{\mathbf{k}}|\Re \langle t,\unicode[STIX]{x1D706}\rangle \Im \langle t^{\prime },\unicode[STIX]{x1D706}\rangle -4\Re \langle t,\unicode[STIX]{x1D706}\rangle \Re \langle t^{\prime },\unicode[STIX]{x1D706}\rangle \Re \unicode[STIX]{x1D701},\nonumber\end{eqnarray}$$

since $\Im \unicode[STIX]{x1D701}=\frac{1}{2}|\unicode[STIX]{x1D6FF}_{\mathbf{k}}|$ . Now, consider the second term. We know $J_{\unicode[STIX]{x1D706}}$ does not depend on the choice of $\Re \unicode[STIX]{x1D701}$ , thus this term contributes at most a torsion element in the cohomology or vanishes entirely. It can hence be ignored.

Note also that the remaining first term is an integer for all $[0,t]$ , $[0,t^{\prime }]$ in  $\unicode[STIX]{x1D6E4}_{\ell ,T}$ .◻

It is worth noting that the bilinear form $F_{\unicode[STIX]{x1D706}}(\cdot ,\cdot )$ introduced in Proposition 4.2 can be written in the form

$$\begin{eqnarray}F_{\unicode[STIX]{x1D706}}(a,b)=\Re \langle a,\unicode[STIX]{x1D706}\rangle \langle b,\unicode[STIX]{x1D706}\rangle =\langle b,\unicode[STIX]{x1D706}\rangle \langle a,\unicode[STIX]{x1D706}\rangle +\langle b,\unicode[STIX]{x1D706}\rangle \langle \unicode[STIX]{x1D706},a\rangle \quad (a,b\in W_{\mathbf{k}}).\end{eqnarray}$$

Clearly, the first term is a complex bilinear form, while the second term is a Hermitian form, we denote them by $B_{\unicode[STIX]{x1D706}}(a,b)$ and $H_{\unicode[STIX]{x1D706}}(a,b)$ , respectively. Note that $H_{\unicode[STIX]{x1D706}}(a,b)$ is linear in its second argument. Further, we remark that $F_{\unicode[STIX]{x1D706}}(a,b)=F_{\unicode[STIX]{x1D706}_{D}}(a,b)$ for all $a,b\in W_{\mathbf{k}}$ .

4.4 Torsion criteria for Heegner divisors

Up to here, we have only worked on Heegner divisors attached to individual lattice vectors, that is $\mathbf{H}_{\infty }(\unicode[STIX]{x1D706})$ , for $\unicode[STIX]{x1D706}\in L^{\prime }$ with $q(\unicode[STIX]{x1D706})<0$ . Next, we consider linear combinations of Heegner divisors. We will be mainly interested in the Heegner divisors $\mathbf{H}_{\ell }(\unicode[STIX]{x1D6FD},m)$ .

For general linear combinations of Heegner divisors, we have the following lemma:

Lemma 4.1. Let $\mathbf{H}$ be a finite linear combination of Heegner divisors of the form

$$\begin{eqnarray}\mathbf{H}=\mathop{\sum }_{\substack{ \unicode[STIX]{x1D706}\in L^{\prime }\cap \ell ^{\bot } \\ q(\unicode[STIX]{x1D706})<0}}a(\unicode[STIX]{x1D706})\mathbf{H}_{\infty }(\unicode[STIX]{x1D706}),\quad (a(\unicode[STIX]{x1D706})\in \mathbb{Z}\text{ for every }\unicode[STIX]{x1D706}).\end{eqnarray}$$

Then, the Chern class $\unicode[STIX]{x1D6FF}(\mathbf{H})$ of $\mathbf{H}$ is a torsion element in $\text{H}^{2}(\unicode[STIX]{x1D6E4}_{\ell },\mathbb{Z})$ if and only if for all $t,t^{\prime }\in D_{\ell ,\unicode[STIX]{x1D6E4}}$ the following equation holds

$$\begin{eqnarray}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D706}\in L^{\prime }\cap \ell ^{\bot } \\ q(\unicode[STIX]{x1D706})<0}}a(\unicode[STIX]{x1D706})\biggl[F_{\unicode[STIX]{x1D706}}(t,t^{\prime })-\frac{\langle \unicode[STIX]{x1D706},\unicode[STIX]{x1D706}\rangle }{n}\langle t^{\prime },t\rangle \biggr]=0.\end{eqnarray}$$

From the proof of this lemma we also get the following necessary condition (where we use the same notation as in the lemma):

Corollary 4.1. If $\unicode[STIX]{x1D6FF}(\mathbf{H})$ is a torsion element, then for the bilinear form $B_{\unicode[STIX]{x1D706}}(a,b)=\langle a,\unicode[STIX]{x1D706}\rangle \langle b,\unicode[STIX]{x1D706}\rangle$ we have

(25) $$\begin{eqnarray}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D706}\in L^{\prime }\cap \ell ^{\bot } \\ q(\unicode[STIX]{x1D706})<0}}a(\unicode[STIX]{x1D706})\operatorname{Tr}B_{\unicode[STIX]{x1D706}}=0,\end{eqnarray}$$

where the trace is taken over a normal orthogonal basis with respect to $\langle \cdot ,\cdot \rangle$ .

Proof. The Chern class $\unicode[STIX]{x1D6FF}(\mathbf{H})$ is given by a linear combination of cocycles $[c_{\unicode[STIX]{x1D706}}]$ in $\text{H}^{2}(\unicode[STIX]{x1D6E4}_{\ell },\mathbb{Z})$ . By Proposition 4.2, each $[c_{\unicode[STIX]{x1D706}}]$ is represented by the two-cocycle

$$\begin{eqnarray}(t,t^{\prime })\mapsto -\Im [|\unicode[STIX]{x1D6FF}_{\boldsymbol{ k}}|F_{\unicode[STIX]{x1D706}}(t,t^{\prime })].\end{eqnarray}$$

Through (8) and the exactness of the sequence in (7), the image of $[c_{\unicode[STIX]{x1D706}}]$ in $\text{H}^{2}(\unicode[STIX]{x1D6E4}_{\ell },{\mathcal{O}}_{\unicode[STIX]{x1D716}})$ vanishes. By the results of Section 3, $\unicode[STIX]{x1D6FF}(\mathbf{H})$ is a torsion element in $\text{H}^{2}(\unicode[STIX]{x1D6E4}_{\ell },\mathbb{Z})$ if and only if there is a rational number $Q$ such that the equation

$$\begin{eqnarray}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D706}\in L^{\prime }\cap \ell ^{\bot } \\ q(\unicode[STIX]{x1D706})<0}}a(\unicode[STIX]{x1D706})|\unicode[STIX]{x1D6FF}_{\mathbf{k}}|\cdot \Im F_{\unicode[STIX]{x1D706}}(t,t^{\prime })=Q\frac{\Im \langle t^{\prime },t\rangle }{|\unicode[STIX]{x1D6FF}_{\mathbf{k}}|}\end{eqnarray}$$

holds for all $t,t^{\prime }\in D_{\ell ,\unicode[STIX]{x1D6E4}}$ . Since $D_{\ell ,\unicode[STIX]{x1D6E4}}$ has full rank in $W_{\mathbf{k}}$ , by extension of scalars, the equation holds for all pairs of vectors in $W_{\mathbf{k}}$ . Both sides of the equation are linear in $t^{\prime }$ . Thus replacing $t^{\prime }$ with a purely imaginary multiple gives a second, equivalent equation:

$$\begin{eqnarray}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D706}\in L^{\prime }\cap \ell ^{\bot } \\ q(\unicode[STIX]{x1D706})<0}}a(\unicode[STIX]{x1D706})|\unicode[STIX]{x1D6FF}_{\mathbf{k}}|\cdot \Re F_{\unicode[STIX]{x1D706}}(t,t^{\prime })=Q\frac{\Re \langle t^{\prime },t\rangle }{|\unicode[STIX]{x1D6FF}_{\mathbf{k}}|}.\end{eqnarray}$$

By linear combination of the two equations, we get

(26) $$\begin{eqnarray}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D706}\in L^{\prime }\cap \ell ^{\bot } \\ q(\unicode[STIX]{x1D706})<0}}a(\unicode[STIX]{x1D706})|\unicode[STIX]{x1D6FF}_{\mathbf{k}}|F_{\unicode[STIX]{x1D706}}(t,t^{\prime })=Q\frac{\langle t^{\prime },t\rangle }{|\unicode[STIX]{x1D6FF}_{\mathbf{k}}|}.\end{eqnarray}$$

To determine $Q$ , we take the trace of both sides of (26), using an orthogonal basis of $W_{\mathbb{C}}$ with respect to $\langle \cdot ,\cdot \rangle$ , say $\{{e_{l}\}}_{l=1,\ldots ,n}$ with $\langle e_{l},e_{m}\rangle =-\unicode[STIX]{x1D6FF}_{l,m}$ . Now, $\operatorname{Tr}\langle \cdot ,\cdot \rangle =-n$ and the trace of $H_{\unicode[STIX]{x1D706}}$ is $-\langle \unicode[STIX]{x1D706},\unicode[STIX]{x1D706}\rangle$ , hence

(27) $$\begin{eqnarray}Q(-n)=\mathop{\sum }_{\substack{ \unicode[STIX]{x1D706}\in L^{\prime }\cap \ell ^{\bot } \\ q(\unicode[STIX]{x1D706})<0}}a(\unicode[STIX]{x1D706})|D_{\mathbf{k}}|(-\langle \unicode[STIX]{x1D706},\unicode[STIX]{x1D706}\rangle +\operatorname{Tr}_{\{e_{l}\}}B_{\unicode[STIX]{x1D706}}).\end{eqnarray}$$

It turns out the trace of $B_{\unicode[STIX]{x1D706}}$ does not contribute to $Q$ . Indeed, if we take the trace of (26) over an orthogonal basis of $W_{\mathbb{C}}$ obtained from $\{e_{l}\}$ by rescaling with the complex unit $i$ , that is  $\{i{e_{l}\}}_{l=1,\ldots ,n}$ , the traces of the Hermitian forms $\langle \cdot ,\cdot \rangle$ and $H_{\unicode[STIX]{x1D706}}$ remain unchanged while that of $B_{\unicode[STIX]{x1D706}}$ switches sign. Comparing this result with (27), we obtain

$$\begin{eqnarray}Q=\mathop{\sum }_{\substack{ \unicode[STIX]{x1D706}\in L^{\prime }\cap \ell ^{\bot } \\ q(\unicode[STIX]{x1D706})<0}}a(\unicode[STIX]{x1D706})\cdot Q_{\unicode[STIX]{x1D706}}\quad \text{with }Q_{\unicode[STIX]{x1D706}}:=|D_{\mathbf{k}}|\frac{\langle \unicode[STIX]{x1D706},\unicode[STIX]{x1D706}\rangle }{n}.\end{eqnarray}$$

Together with (26) the statement follows. Further, since the contribution of $B_{\unicode[STIX]{x1D706}}$ to the trace vanishes, we get the necessary condition

$$\begin{eqnarray}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D706}\in L^{\prime }\cap \ell ^{\bot } \\ q(\unicode[STIX]{x1D706})<0}}a(\unicode[STIX]{x1D706})\operatorname{Tr}B_{\unicode[STIX]{x1D706}}=0.\end{eqnarray}$$

This proves the corollary, as well. ◻

4.5 The main result

We can now turn to the object of our main interest, Heegner divisors of the form $\mathbf{H}_{\ell }(\unicode[STIX]{x1D6FD},m)$ . We want to describe their position in the local Picard group. Recall that by (20) the divisors $\mathbf{H}_{\ell }(\unicode[STIX]{x1D6FD},m)$ can be written using divisors of the type $\mathbf{H}_{\infty }(\unicode[STIX]{x1D706})$ . Thus, any finite linear combination $\mathbf{H}$ of Heegner divisors $\mathbf{H}_{\ell }(\unicode[STIX]{x1D6FD},m)$ , can be written as a locally finite sum of Heegner divisors $\mathbf{H}_{\infty }(\unicode[STIX]{x1D706})$ . Also, note that for a divisor of this type, the Chern class $\unicode[STIX]{x1D6FF}(\mathbf{H}_{\infty }(\unicode[STIX]{x1D706}))$ depends only on the projection $\unicode[STIX]{x1D706}_{D}$ . With this notation, we formulate the following theorem.

Theorem 4.1. Consider a finite linear combination of local Heegner divisors of the form

(28) $$\begin{eqnarray}\mathbf{H}=\frac{1}{2}\mathop{\sum }_{\unicode[STIX]{x1D6FD}\in {\mathcal{L}}}\mathop{\sum }_{\substack{ m\in \mathbb{Z}+q(\unicode[STIX]{x1D6FD}) \\ m<0}}c(\unicode[STIX]{x1D6FD},m)\mathbf{H}_{\ell }(\unicode[STIX]{x1D6FD},m),\end{eqnarray}$$

with integral coefficients $c(\unicode[STIX]{x1D6FD},m)$ , satisfying $c(\unicode[STIX]{x1D6FD},m)=c(-\unicode[STIX]{x1D6FD},m)$ .

Then, $\mathbf{H}$ is torsion element in the Picard group $\text{Pic}(\unicode[STIX]{x1D6E4}_{\ell }\backslash U_{\unicode[STIX]{x1D716}}(\ell ))$ if and only if for all $t,t^{\prime }\in D_{\ell ,\unicode[STIX]{x1D6E4}}$ the following equation holds

(29) $$\begin{eqnarray}\mathop{\sum }_{\unicode[STIX]{x1D6FD}\in {\mathcal{L}}}\;\mathop{\sum }_{\substack{ m\in \mathbb{Z}+q(\unicode[STIX]{x1D6FD}) \\ m<0}}c(\unicode[STIX]{x1D6FD},m)\mathop{\sum }_{\substack{ \unicode[STIX]{x1D706}\in D^{\prime } \\ \unicode[STIX]{x1D706}+D\equiv \unicode[STIX]{x1D70B}(\unicode[STIX]{x1D6FD}) \\ q(\unicode[STIX]{x1D706})=m}}\biggl[F_{\unicode[STIX]{x1D706}}(t,t^{\prime })-\frac{\langle \unicode[STIX]{x1D706},\unicode[STIX]{x1D706}\rangle }{n}\langle t^{\prime },t\rangle \biggr]=0.\end{eqnarray}$$

Further, a necessary conditions for this to be the case is that the following identity holds, with $B_{\unicode[STIX]{x1D706}}(x,y)=\langle x,\unicode[STIX]{x1D706}\rangle \langle y,\unicode[STIX]{x1D706}\rangle$ :

(30) $$\begin{eqnarray}\mathop{\sum }_{\unicode[STIX]{x1D6FD}\in {\mathcal{L}}}\;\mathop{\sum }_{\substack{ m\in \mathbb{Z}+q(\unicode[STIX]{x1D6FD}) \\ m<0}}c(\unicode[STIX]{x1D6FD},m)\mathop{\sum }_{\substack{ \unicode[STIX]{x1D706}\in D^{\prime } \\ \unicode[STIX]{x1D706}+D\equiv \unicode[STIX]{x1D70B}(\unicode[STIX]{x1D6FD}) \\ q(\unicode[STIX]{x1D706})=m}}\operatorname{Tr}B_{\unicode[STIX]{x1D706}}=0.\end{eqnarray}$$

Here, the trace is taken over an orthogonal basis with respect to $\langle \cdot ,\cdot \rangle$ .

We note that by (17) a linear combination of Heegner divisors $\mathbf{H}$ is a torsion element in $\text{Pic}(\unicode[STIX]{x1D6E4}_{\ell }\backslash U_{\unicode[STIX]{x1D716}}(\ell ))$ if and only if it is a torsion element in the local Picard group $\text{Pic}(X_{\unicode[STIX]{x1D6E4}},\ell )$ .

Proof. If $H$ is a torsion element, the equation (29) follows from Lemma 4.1. Also, from the proof of that lemma and Corollary 4.1, it is clear that in this case, the identity (30) holds.

For the converse, assume that (29) holds for all $t,t^{\prime }\in D_{\ell ,\unicode[STIX]{x1D6E4}}$ . We show that $\mathbf{H}$ is a torsion element in the Picard group. By extension of scalars, the equation remains valid for all $t,t^{\prime }\in W_{\mathbb{C}}$ . Using (23), an automorphy factor describing $\mathbf{H}$ in $\text{Pic}(\unicode[STIX]{x1D6E4}_{\ell }\backslash U_{\unicode[STIX]{x1D716}}(\ell ))$ is given by the following (finite) product (for $g=[h,t]\in \unicode[STIX]{x1D6E4}_{\ell }$ , $z\in U_{\unicode[STIX]{x1D716}}=U_{\unicode[STIX]{x1D716}}(\ell )$ ):

(31) $$\begin{eqnarray}\displaystyle J_{\mathbf{H}}(g,z) & = & \displaystyle \mathop{\prod }_{\substack{ \unicode[STIX]{x1D6FD}\in {\mathcal{L}} \\ m\in \mathbb{Z}+q(\unicode[STIX]{x1D6FD}) \\ m<0}}\mathop{\prod }_{\substack{ \unicode[STIX]{x1D705}\in D \\ q(\unicode[STIX]{x1D705}+\dot{\unicode[STIX]{x1D6FD}})=m}}J_{\unicode[STIX]{x1D705}+\dot{\unicode[STIX]{x1D6FD}}}(g,z)^{c(\unicode[STIX]{x1D6FD},m)/2}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\prod }_{\unicode[STIX]{x1D6FD},m}\mathop{\prod }_{\unicode[STIX]{x1D705}}e(\!-2|D_{\mathbf{k}}|\langle z,\unicode[STIX]{x1D705}+\dot{\unicode[STIX]{x1D6FD}}\rangle \Re \langle t,\unicode[STIX]{x1D705}+\dot{\unicode[STIX]{x1D6FD}}_{D}\rangle \nonumber\\ \displaystyle & & \displaystyle -\,2\unicode[STIX]{x1D701}(\Re \langle t,\unicode[STIX]{x1D705}+\dot{\unicode[STIX]{x1D6FD}}_{D}\rangle )^{2}+\Re \langle t,\unicode[STIX]{x1D705}+\dot{\unicode[STIX]{x1D6FD}}_{D}\rangle (\unicode[STIX]{x1D701}+1))\!^{c(\unicode[STIX]{x1D6FD},m)/2}.\end{eqnarray}$$

Since $c(\unicode[STIX]{x1D6FD},m)=c(-\unicode[STIX]{x1D6FD},m)$ , terms which are linear in the $\unicode[STIX]{x1D705}+\dot{\unicode[STIX]{x1D6FD}}_{D}$ cancel. The remaining factors are of the form

$$\begin{eqnarray}e(-2|D_{\mathbf{k}}|\langle z,\unicode[STIX]{x1D705}+\dot{\unicode[STIX]{x1D6FD}}\rangle \Re \langle t,\unicode[STIX]{x1D705}+\dot{\unicode[STIX]{x1D6FD}}_{D}\rangle -2\unicode[STIX]{x1D701}(\Re \langle t,\unicode[STIX]{x1D705}+\dot{\unicode[STIX]{x1D6FD}}_{D}\rangle )^{2})^{c(\unicode[STIX]{x1D6FD},m)/2}.\end{eqnarray}$$

Now, we write $\langle z,\unicode[STIX]{x1D705}+\dot{\unicode[STIX]{x1D6FD}}\rangle =\langle z,\unicode[STIX]{x1D705}+\dot{\unicode[STIX]{x1D6FD}}_{D}\rangle +\langle z,\dot{\unicode[STIX]{x1D6FD}}-\dot{\unicode[STIX]{x1D6FD}}_{D}\rangle$ . Since $\langle \dot{\unicode[STIX]{x1D6FD}},\ell \rangle =0$ , the second part depends only on the constant $\ell ^{\prime }$ -component of $z$ . We get

$$\begin{eqnarray}\displaystyle & & \displaystyle \left[e\left(-2\unicode[STIX]{x1D6FF}_{\mathbf{k}}\overline{\dot{\unicode[STIX]{x1D6FD}}_{\ell }}\Re \langle t,\unicode[STIX]{x1D706}_{D}\rangle \right)\right.\nonumber\\ \displaystyle & & \displaystyle \quad \left.\cdot \,e\left(-2|D_{\mathbf{k}}|\langle \unicode[STIX]{x1D70E},\unicode[STIX]{x1D705}+\dot{\unicode[STIX]{x1D6FD}}_{D}\rangle \Re \langle t,\unicode[STIX]{x1D705}+\dot{\unicode[STIX]{x1D6FD}}_{D}\rangle -2\unicode[STIX]{x1D701}(\Re \langle t,\unicode[STIX]{x1D705}+\dot{\unicode[STIX]{x1D6FD}}_{D}\rangle )^{2}\right)\right]^{c(\unicode[STIX]{x1D6FD},m)/2}.\nonumber\end{eqnarray}$$

We ignore the first factor for the time being and examine the second factor. There, the first term in the exponential is $-2|D_{\mathbf{k}}|F_{\unicode[STIX]{x1D705}+\dot{\unicode[STIX]{x1D6FD}}_{D}}(t,\unicode[STIX]{x1D70E})$ while the second term is equal to $-2\unicode[STIX]{x1D701}\Re F_{\unicode[STIX]{x1D705}+\dot{\unicode[STIX]{x1D6FD}}_{D}}(t,t)$ . We apply (29) to both terms, and can rewrite this factor in the form

(32) $$\begin{eqnarray}\displaystyle & & \displaystyle e\biggl(2D_{\mathbf{k}}\frac{\langle \unicode[STIX]{x1D705}+\dot{\unicode[STIX]{x1D6FD}}_{D},\unicode[STIX]{x1D705}+\dot{\unicode[STIX]{x1D6FD}}_{D}\rangle }{n}\biggl[\langle \unicode[STIX]{x1D70E},t\rangle -\frac{\unicode[STIX]{x1D701}}{D_{\mathbf{k}}}\langle t,t\rangle \biggr]\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad =e\biggl(2D_{\mathbf{k}}\frac{q(\unicode[STIX]{x1D705}+\dot{\unicode[STIX]{x1D6FD}}_{D})}{n}\biggl[\langle \unicode[STIX]{x1D70E},t\rangle -\frac{1}{2\unicode[STIX]{x1D6FF}_{\mathbf{k}}}q(t)\biggr]\biggr)e\biggl(-2\Re \unicode[STIX]{x1D701}\cdot \frac{q(\unicode[STIX]{x1D706})}{n}q(t)\biggr).\end{eqnarray}$$

Clearly, the last factor in (32) has finite order and is a torsion element in $\text{Pic}(\unicode[STIX]{x1D6E4}_{\ell }\backslash U_{\unicode[STIX]{x1D716}})$ . Now, we claim that the first factor is actually a trivial automorphy factor. To see this, consider the invertible function $f(z)=e(c\unicode[STIX]{x1D70F})$ with $c\in \mathbb{Q}^{\times }$ ; under the operation of $\unicode[STIX]{x1D6E4}_{\ell ,T}$ , it gives rise to the following trivial automorphy factor

$$\begin{eqnarray}j_{1}([0,t],z)=\frac{f([h,t]z)}{f(z)}=e\biggl(c\biggl(-\langle \unicode[STIX]{x1D70E},t\rangle +\frac{1}{2\unicode[STIX]{x1D6FF}_{\mathbf{k}}}\langle t,t\rangle \biggr)\biggr).\end{eqnarray}$$

Hence the first factor in (32) is indeed trivial. Now, we return to the previously excluded factor

(33) $$\begin{eqnarray}e(-2\unicode[STIX]{x1D6FF}_{\mathbf{k}}\overline{\dot{\unicode[STIX]{x1D6FD}}_{\ell }}\Re \langle t,\unicode[STIX]{x1D705}+\dot{\unicode[STIX]{x1D6FD}}_{D}\rangle )=e(-(|\unicode[STIX]{x1D6FF}_{\mathbf{k}}|\Im \dot{\unicode[STIX]{x1D6FD}}_{\ell }+\unicode[STIX]{x1D6FF}_{\mathbf{k}}\Re \dot{\unicode[STIX]{x1D6FD}}_{\ell })2\Re \langle t,\unicode[STIX]{x1D705}+\dot{\unicode[STIX]{x1D6FD}}_{D}\rangle ).\end{eqnarray}$$

Since $|\unicode[STIX]{x1D6FF}_{\mathbf{k}}|\Im \dot{\unicode[STIX]{x1D6FD}}_{\ell }$ is rational (actually, half-integer), this term contributes only a torsion element in the Picard group. Consider the invertible function $g(z)=e(\langle \unicode[STIX]{x1D70E},\unicode[STIX]{x1D707}\rangle )$ with $\unicode[STIX]{x1D707}\in W_{\mathbf{k}}$ from which we get the trivial automorphy factor

$$\begin{eqnarray}j_{2}(z,[0,t])=e(\unicode[STIX]{x1D6FF}_{\mathbf{k}}^{-1}\langle t,\unicode[STIX]{x1D707}\rangle ).\end{eqnarray}$$

Setting $\unicode[STIX]{x1D707}=\unicode[STIX]{x1D705}+\dot{\unicode[STIX]{x1D6FD}}_{D}$ , we multiply (33) with a suitable power of $j_{2}$ to kill the term in $\Re \langle t,\unicode[STIX]{x1D705}+\dot{\unicode[STIX]{x1D6FD}}_{D}\rangle$ . Then, only torsion elements remain, as $|\unicode[STIX]{x1D6FF}_{\mathbf{k}}|\Im \langle t,\unicode[STIX]{x1D705}+\dot{\unicode[STIX]{x1D6FD}}_{D}\rangle$ and $\Re \dot{\unicode[STIX]{x1D6FD}}_{\ell }$ are rational numbers.

Thus, we find that each of the finitely may factors of $J_{\mathbf{H}}$ from (31) can be expressed through suitable powers of trivial automorphy factors of the types $j_{1}$ and $j_{2}$ and factors of finite order. Hence, it follows that $\mathbf{H}$ is a torsion element in $\text{Pic}(\unicode[STIX]{x1D6E4}_{\ell }\backslash U_{\unicode[STIX]{x1D716}})$ .◻

5.1 Relationship to global obstruction theory and the work of Bruinier and Freitag

Since the statement of Theorem 5.1 holds for all sufficiently small $\unicode[STIX]{x1D716}$ , passing to the direct limit we get the statement for the local Picard group at the cusp $\ell$ . Now Theorem 5.1 formally resembles a global obstruction statement for unitary groups from [Reference Hofmann9] in the style of Borcherds [Reference Borcherds2]. It can be stated as follows, from [Reference Hofmann9, Lemma 5, Theorem 4].

Theorem 5.2. A Heegner divisor of the form

$$\begin{eqnarray}\mathbf{H}=\frac{1}{2}\mathop{\sum }_{\unicode[STIX]{x1D6FD}\in L^{\prime }/L}\mathop{\sum }_{\substack{ m\in \mathbb{Z}+q(\unicode[STIX]{x1D6FD}) \\ m<0}}c(\unicode[STIX]{x1D6FD},m)\mathbf{H}(\unicode[STIX]{x1D6FD},m)\end{eqnarray}$$

is the divisor of a Borcherds product if and only if

$$\begin{eqnarray}\mathop{\sum }_{\unicode[STIX]{x1D6FD}\in L^{\prime }/L}\mathop{\sum }_{\substack{ m\in \mathbb{Z}+q(\unicode[STIX]{x1D6FD}) \\ m<0}}c(\unicode[STIX]{x1D6FD},m)b(\unicode[STIX]{x1D6FD},-m)=0,\end{eqnarray}$$

for every cusp form $g\in {\mathcal{S}}_{k}(\unicode[STIX]{x1D70C}_{L}^{\ast })$ with Fourier coefficients $b(\unicode[STIX]{x1D6FD},-m)$ .

Since by results of Bruinier [Reference Bruinier3], the local obstruction space

$$\begin{eqnarray}\mathbf{H}_{\ell }:=\frac{1}{2}\mathop{\sum }_{\unicode[STIX]{x1D6FD}\in {\mathcal{L}}}\;\mathop{\sum }_{\substack{ m\in \mathbb{Z}+q(\unicode[STIX]{x1D6FD}) \\ m<0}}c(\unicode[STIX]{x1D6FD},m)\mathbf{H}_{\ell }(\unicode[STIX]{x1D6FD},m)\end{eqnarray}$$

is a torsion element in the local Picard group. In fact, the same argument applies for every cusp.

The results of Bruinier and Freitag [Reference Bruinier and Freitag4, Theorem 5.1], in the setting of orthogonal groups are very similar to Theorem 5.1 above. The definition of their theta series is (essentially) the same. Indeed, if we look at the rational quadratic space $V_{\mathbb{Q}}$ underlying $V_{\mathbf{k}}$ and the lattices $L$ and $D$ as quadratic modules in $V_{\mathbb{Q}}$ , the obstruction spaces are the same.

In this case, through the embedding theory from [Reference Hofmann8, Reference Hofmann9] we can pull back Heegner divisors on the modular variety of the orthogonal group to Heegner divisors for the modular variety of the unitary group. As sketched in Remark 4.3 above, this also works locally. Thus if $H$ is the Heegner divisor of a Borcherds product for the orthogonal group, then it is trivial at generic boundary components in the sense of [Reference Bruinier and Freitag4, Definition 5.3], that is, locally torsion, and hence restricts to a torsion element in the local Picard group for every cusp $I$ of the unitary modular variety. Similarly, by pulling back the Borcherds product itself, one gets a Borcherds product for the unitary group with the pullback of $H$ as its divisor, and through Theorem 5.2, again, the corresponding local Heegner divisors are torsion elements.

We also remark that the obstruction space in Theorem 5.2 is the same as that from Borcherds’s [Reference Borcherds2] in the orthogonal situation. Hence, if $\mathbf{H}$ is the Heegner divisor of a Borcherds product for the unitary group, one can find a Heegner divisor on the orthogonal side which restricts to $\mathbf{H}$ and is the divisor of a Borcherds product.