[columns=3]

# On the convergence of certain indefinite theta series

Xiaoyu Zhang  Universität Duisburg-Essen, Fakultät für Mathematik,
Mathematikcarrée Thea-Leymann-Straße 9, 45127 Essen, Germany
<EMAIL_ADDRESS>
###### Abstract.

We give an elementary proof of the convergence of indefinite theta series
associated to an inner space of signature $(n,2)$ conjectured in the work of
Alexandrov,Banerjee,Manschot and Pioline ([ABMP18b, C]) and show that the
incidence conditions in loc.cit are also necessary for the convergence.

###### 2010 Mathematics Subject Classification:

11F03,11F27

###### Contents

  1. 1 Introduction
  2. 2 Convergence of indefinite theta series

##  1\. Introduction

Theta series are one of the most fundamental examples in the theory of modular
forms. The theta series associated to a lattice $(L,Q)$ with integer valued
positive definite quadratic form $Q$ is classically known to converge
absolutely termwise and define a holomorphic modular form:

$\theta(\tau)=\sum_{x\in L}q^{Q(x)},\quad q=e^{-2i\pi\tau}\text{ and
}\tau=u+vi\in\mathbb{C}\text{ with }v>0.$

Write $(-,-)$ for the symmetric bilinear form associated to $Q$ such that
$Q(x)=\frac{1}{2}(x,x)$. When the quadratic space is indefinite, the theta
series is no longer absolutely convergent (because the isometry group of
$(L,Q)$ is infinite) and there are several different approaches to take care
of the convergence problem. One approach is to sum over a subset of $L$. In
this case the theta series is holomorphic but generally no longer a modular
form. However, in some situations the series can be completed to a non-
holomorphic modular form. This direction of research began with the work of
Göttsche and Zagier [GZ98] and Zwegers ([Zwe02]) for $V$ of signature $(n,1)$:
fix two negative vectors $C_{1},C_{2}$ in $V$ (that is,
$(C_{1},C_{1}),(C_{2},C_{2})$ are negative) such that $(C_{1},C_{2})<0$ and
take $\mu\in L^{\vee}$ (the dual of $L$), then put

$\theta_{\mu}(\tau;\\{C_{1},C_{2}\\})=\sum_{x\in\mu+L}\Phi(x;\\{C_{1},C_{2}\\})q^{(x,x)/2}$

where
$\Phi(x;\\{C_{1},C_{2}\\})=\mathrm{sgn}((x,C_{1}))-\mathrm{sgn}((x,C_{2}))$.
Then Zwegers shows that this theta series is termwise absolutely convergent
and that $\theta_{\mu}$ be completed to a modular form using certain
complementary error function (we refer the reader to[Zag10] for an overview).
In [ABMP18a], Alexandrov, Banerjee, Manschot and Pioline established an
analogue of $\Phi(x;\\{C_{1},C_{2}\\})$ for $V$ of signature $(n,2)$ and later
on, in [ABMP18b], based on motivations from physics, they consider a variant
of $\Phi(x;\\{C_{1},C_{2}\\})$ for $V$ of signature $(n,2)$ and conjecture the
absolute convergence of the corresponding theta series. This is what we will
consider in this note.

More precisely, take $N$ non-zero vectors
$\mathcal{C}=\\{C_{1},\cdots,C_{N}\\}$ in $V$ and consider the following
incidence conditions on these vectors:

(I.1) $\displaystyle(C_{j},C_{j})\leq 0,$ (I.2)
$\displaystyle(C_{j},C_{j})(C_{j+1},C_{j+1})-(C_{j},C_{j+1})^{2}\begin{cases*}>0,&if
$(C_{j},C_{j})(C_{j+1},C_{j+1})>0$,\\\ =0,&if
$(C_{j},C_{j})(C_{j+1},C_{j+1})=0$,\end{cases*}$ (I.3)
$\displaystyle\begin{cases*}(C_{j},C_{j})(C_{j-1},C_{j+1})-(C_{j},C_{j-1})(C_{j},C_{j+1})<0,&if
$(C_{j},C_{j})<0$,\\\ (C_{j-1},C_{j+1})>0,&if $(C_{j},C_{j})=0$.\end{cases*}$

To make formulas in this note more readable, we set in the following
$\mathrm{sgn}(v,w)=\mathrm{sgn}((v,w))$ for $v,w\in V$. Then we put (following
the notation in [FK22], up to a sign change)

$\mathbf{w}(x;\mathcal{C})=\sum_{j=1}^{N}\mathrm{sgn}(x,C_{j})\mathrm{sgn}(x,C_{j+1}),\quad
x\in V.$

Here by convention, $C_{N+1}=C_{1}$. Take a negative vector $v\in V$ and set

$\Phi(x;\mathcal{C})=\mathbf{w}(x;\mathcal{C})-\mathbf{w}(v;\mathcal{C}).$

We then define

$\theta_{\mu}(\tau)=\theta_{\mu}(\tau;\mathcal{C})=\sum_{x\in\mu+L}\Phi(x;\mathcal{C})q^{(x,x)/2}.$

The main result of this note is

######  Theorem 1.1.

  1. (1)

Suppose $\mathcal{C}$ satisfies (I.1)-(I.3), then $\mathbf{w}(x;\mathcal{C})$
is constant on $V_{<0}$ (the set of negative vectors in $V$) and the
indefinite theta series $\theta_{\mu}(\tau)$ is termwise absolutely
convergent.

  2. (2)

Suppose $\mathcal{C}$ satisfies (I.1)-(1.2) and the indefinite theta series
$\theta_{\mu}(\tau)$ is termwise absolutely convergent. If moreover no three
consecutive vectors in $\mathcal{C}$ are all null-vectors, then $\mathcal{C}$
also satisfies (I.3).

The value $\mathbf{w}(v;\mathcal{C})$ has the following obvious
interpretation: write $v^{\perp}$ for the orthogonal complement of $v$ (a
hyperplane in $V$). Then

  1. (1)

$(v,C_{j})(v,C_{j+1})\geq 0$ if and only if the straight segment $\gamma_{j}$
in $V$ connecting $C_{j}$ to $C_{j+1}$ passes through the hyperplane
$v^{\perp}$;

  2. (2)

$(v,C_{j})(v,C_{j+1})<0$ if and only if this segment lies outside this
hyperplane.

Write $\gamma(\mathcal{C})$ for the sum
$\gamma_{1}+\gamma_{2}+\cdots+\gamma_{N}$, a piecewise smooth loop in $V$.
Then the value $\mathbf{w}(v;\mathcal{C})$ measures how many times this loop
$\gamma(\mathcal{C})$ passes through the hyperplane $v^{\perp}$, assuming that
$(v,C_{j})\neq 0$ for any $j=1,\cdots,N$. If $\gamma(\mathcal{C})$ passes
through $r$ times through $v^{\perp}$, then one has
$\mathbf{w}(v;\mathcal{C})=N-4r$.

The first part of the theorem is also proved in [FK22] under the further
condition that $(C_{j},C_{j})<0$ for all $j=1,\cdots,N$. As explained in
loc.cit, (I.3) means that the $2$-dimensional oriented plane $[C_{j},C_{j+1}]$
spanned by $C_{j}$ and $C_{j+1}$ all lie in the same connected component of
$D(V)$, the Grassmannian of oriented negative $2$-planes in $V$. This part
corrects the conjecture in [ABMP18b, C], which does not include the term
$\mathbf{w}(v;\mathcal{C})$ in the definition of $\Phi(x;\mathcal{C})$: as
explained in [FK22], for $N=3$, $\mathbf{w}(v;\mathcal{C})$ is non-zero (we
can also use the preceding interpretation of $\mathbf{w}(v;\mathcal{C})$) and
therefore $\theta_{\mu}(\tau)$ can not be termwise absolutely convergent
without this term.

The theta series $\theta_{\mu}$ defines a mock modular form of weight
$(n+2)/2$, whose modular completion is given by

$\widehat{\theta}_{\mu}(\tau)=\sum_{x\in\mu+L}\left(-\mathbf{w}(v;\mathcal{C})+\sum_{j=1}^{N}E_{2}(C_{j},C_{j+1};\sqrt{2}x)\right)q^{(x,x)/2},$

where

$E_{2}(C,C^{\prime};x)=\int_{z}e^{\pi(y-\mathrm{pr}_{z}(x),y-\mathrm{pr}_{z}(x))}\mathrm{sgn}(y,C)\mathrm{sgn}(y,C^{\prime})dy$

with $z$ the subspace of $V$ generated by $C$ and $C^{\prime}$ and
$\mathrm{pr}_{z}(x)$ the projection of $x$ to $z$. We refer the reader to
[ABMP18b, C] for more details.

We say a vector $v\in V$ is regular at $\mathcal{C}$ if $(v,C_{j})\neq 0$ for
any $j=1,\cdots,N$. Write $\mathrm{Reg}(\mathcal{C})$ for the subset of $V$
consisting of $v$ regular at $\mathcal{C}$, which is an open dense subset of
$V$ and also a finite disjoint union of polyhedral cones.

The strategy of proof of 1.1(1) is based on the following simple observation:
the function $\mathbf{w}(x;\mathcal{C})$ is locally constant on
$\mathrm{Reg}(\mathcal{C})$ and it is constant on $V_{<0}$. So we only need to
take care of the connected components of $\mathrm{Reg}(\mathcal{C})$ whose
closure intersect with the closure $\overline{V}_{<0}$ only at $0$. On such
components, $(x,x)$ takes positive values and in fact is bounded below by some
positive definite inner product $(x,x)_{\mathcal{C}}$.

##  2\. Convergence of indefinite theta series

As in the introduction, let $(V,(-,-))$ be an indefinite inner product space
over $\mathbb{R}$ of signature $(n,2)$. Take $N$ non-zero vectors
$\mathcal{C}=\\{C_{1},\cdots,C_{N}\\}$ in $V$ satisfying conditions
(I.1)-(I.3).

######  Lemma 2.1.

The restriction of the map $\mathbf{w}(-;\mathcal{C})$ to
$\mathrm{Reg}(\mathcal{C})$ is locally constant.

###### Proof.

This is because each $\mathrm{sgn}(x,C_{j})$ is locally constant on
$\mathrm{Reg}(\mathcal{C})$. ∎

This is also true without $\mathcal{C}$ satisfying (I.1)-(I.3).

$C_{j}$$C_{j-1}$$C_{j+1}$$v^{\perp}\cap
V^{\prime}$$\mathbb{R}C_{j}$$v$$V^{\prime}$ Figure 1.

The following three lemmas are the technical heart of this note. The main idea
can be explained as follows (see Figure.1 for an illustration): suppose
$C_{j-1},C_{j},C_{j+1}$ lie in the same negative plane $V^{\prime}$ (this is
possible under the conditions (I.1) and (I.2)), then (I.3) implies that
$C_{j-1}$ and $C_{j+1}$ lie in different connected components (half planes) of
$V^{\prime}\backslash\mathbb{R}C_{j}$. Therefore, for a negative vector $v$,
when $(v,C_{j})$ is very small (compared to the norms of $v,C_{j\pm
1},C_{j},(C_{j},C_{j\pm 1})$), the intersection $V^{\prime}\cap v^{\perp}$
separates the two vectors $C_{j-1},C_{j+1}$. As a result, the value
$\mathrm{sgn}(v,C_{j-1})\mathrm{sgn}(v,C_{j+1})$ is negative. This means that
when we perturb $v$ slightly such that $(v,C_{j})$ changes signs (assuming the
signs $\mathrm{sgn}(v,C_{i})$ remain unchanged for all $i\neq j$), the value
$\mathbf{w}(v;\mathcal{C})$ is unchanged.

We fix a basis $(e_{1},e_{2},\cdots,e_{n+2})$ of $V$ such that the bilinear
form $(-,-)$ is represented by the diagonal matrix
$\mathrm{diag}(-1_{2},1_{n})$.

######  Lemma 2.2.

Fix $j=1,\cdots,N$. For any vector $v\in V_{<0}$ such that $(v,C_{j})=0$, we
have

$\mathrm{sgn}(v,C_{j-1})\mathrm{sgn}(v,C_{j+1})<0.$

###### Proof.

First we assume that $C_{j}$ is not a null vector. We can suppose that the
vectors $C_{j}$ and $v$ are of the form: $C_{j}=(1,0,\cdots)$ and
$v=(0,1,0,\cdots)$. Then we can write $C_{j-1}=(a,b,c,d,\cdots)$ and
$C_{j+1}=(a^{\prime},b^{\prime},c^{\prime},d^{\prime},\cdots)$. (I.2) and
(I.3) give

$\displaystyle b^{2}-c^{2}-d^{2}-\cdots$ $\displaystyle\geq 0,$
$\displaystyle(b^{\prime})^{2}-(c^{\prime})^{2}-(d^{\prime})^{2}-\cdots$
$\displaystyle\geq 0,$ $\displaystyle
bb^{\prime}-cc^{\prime}+dd^{\prime}+\cdots$ $\displaystyle\leq 0.$

From this, we get

$(v,C_{j-1})(v,C_{j+1})=bb^{\prime}<0.$

Next assume $(C_{j},C_{j})=0$, then (I.2) implies

$(C_{j},C_{j-1})=0=(C_{j},C_{j+1}).$

As in the above, we can assume $C_{j}=(1,0,1,0,\cdots)$ and
$v=(0,1,0,\cdots)$. Then we can write $C_{j-1}=(a,b,a,c,d,\cdots)$ and
$C_{j+1}=(a^{\prime},b^{\prime},a^{\prime},c^{\prime},d^{\prime},\cdots)$.
(I.1) and (I.3) imply

$\displaystyle-b^{2}+(c^{2}+d^{2}+...)$ $\displaystyle\leq 0$
$\displaystyle-(b^{\prime})^{2}+((c^{\prime})^{2}+(d^{\prime})^{2}+...)$
$\displaystyle\leq 0$ $\displaystyle-
bb^{\prime}+(cc^{\prime}+dd^{\prime}+...)$ $\displaystyle>0.$

We deduce that $bb^{\prime}<0$ and thus
$(v,C_{j-1})(v,C_{j+1})=bb^{\prime}<0$. ∎

For two vectors $v,w\in V$, we write

$\gamma_{v,w}(s)=(1-s)v+sw,\quad s\in[0,1]$

for the path/segment in $V$ connecting $v$ to $w$. For each $j=1,\cdots,N$, we
write

$\gamma_{j}(s)=\gamma_{C_{j},C_{j+1}}(s).$

Then (I.1) and (I.2) imply that $\gamma_{j}(s)$ is a non-positive vector for
any $s\in[0,1]$.

######  Lemma 2.3.

Let $\gamma\colon[0,1]\rightarrow V_{<0}$ be a continuous map such that there
are only finitely many $t\in[0,1]$ with $\gamma(t)\in
V_{<0}\backslash\mathrm{Reg}(\mathcal{C})$. Then
$\mathbf{w}(\gamma(t);\mathcal{C})$ is a constant map on $t\in[0,1]$.

###### Proof.

If $\gamma([0,1])\subset\mathrm{Reg}(\mathcal{C})$, then the conclusion holds
thanks to Lemma 2.1. We can then assume there is only one $t_{0}\in[0,1]$ such
that $\gamma(t_{0})\notin\mathrm{Reg}(\mathcal{C})$. We can put $t_{0}=1$ and
$(\gamma(1),C_{j_{1}})=\cdots=(\gamma(1),C_{j_{p}})=0$ for indices
$j_{1}<j_{2}<\cdots<j_{p}$ (we have $j_{r}+1\neq j_{r+1}$ for any
$r=1,\cdots,p-1$ by Lemma 2.2) and $(v,C_{j})\neq 0$ for other indices $j$. We
can then assume that $\mathrm{sgn}(\gamma(t),C_{j_{r}})<0$ for any
$t\in[0,1[$. By Lemma 2.2, we have
$\mathrm{sgn}(\gamma(t),C_{j_{r}-1})\mathrm{sgn}(\gamma(t),C_{j_{r}+1})<0$ for
any $t\in[0,1]$. As a result,

$\mathrm{sgn}(\gamma(t),C_{j_{r}-1})\mathrm{sgn}(\gamma(t),C_{j_{r}})+\mathrm{sgn}(\gamma(t),C_{j_{r}})\mathrm{sgn}(\gamma(t),C_{j_{r}+1})=0.$

For $j\neq j_{r}-1,j_{r}$, the cardinal of the intersection
$\gamma(t)^{\perp}\cap\gamma_{j}(]0,1[)$ is constant on $t\in[0,1]$ and
therefore $\mathrm{sgn}(\gamma(t),C_{j})\mathrm{sgn}(\gamma(t),C_{j+1})$ is
constant on $t\in[0,1]$. So we conclude that
$\mathbf{w}(\gamma(t);\mathcal{C})$ is indeed constant on $t\in[0,1]$. ∎

######  Lemma 2.4.

For any negative vectors $v,w\in V_{<0}$, we have
$\mathbf{w}(v;\mathcal{C})=\mathbf{w}(w;\mathcal{C})$.

###### Proof.

It suffices to construct a path $\gamma\colon[0,1]\rightarrow V_{<0}$ as in
Lemma 2.3 connecting $v$ and $w$. Since $\mathrm{Reg}(\mathcal{C})\cap V_{<0}$
is open dense in $V_{<0}$, we can find $v_{1},w_{1}\in
V_{<0}\cap\mathrm{Reg}(\mathcal{C})$ in sufficiently small open neighborhoods
of $v$ and $w$ such that the paths $\gamma_{v,v_{1}}$ and $\gamma_{w_{1},w}$
satisfy Lemma 2.3. Moreover we can always find a negative vector $u\in V_{<0}$
such that $\gamma_{v_{1},u}$ and $\gamma_{u,w_{1}}$ both lie in $V_{<0}$ and
satisfy Lemma 2.3: indeed, under a suitable basis of $V$ such that the
quadratic form $(-,-)$ is represented by the matrix
$\mathrm{diag}(-1_{2},1_{n})$, we can write $v_{1}$ and $w_{1}$ in coordinates
as $v_{1}=(1,0,\cdots,0)$ and $w_{1}=(a_{1},a_{2},\cdots,a_{n+2})$ with
$(w_{1},w_{1})=-1$. Then we can always find a vector
$u=(x_{1},x_{2},0,\cdots,0)\in V$ such that $x_{1}^{2}+x_{2}^{2}=-(u,u)=1$,
$x_{1}^{2}=(v_{1},u)^{2}<1$ ($v_{1}$ and $u$ span a negative subspace of
dimension $2$) and $(a_{1}x_{1}+a_{2}x_{2})^{2}=(u,w_{1})^{2}<1$ ($u$ and
$w_{1}$ span a negative subspace of dimension $2$). Now it is easy to see that
the path $\gamma_{v,v_{1}}+\gamma_{v_{1},u}+\gamma_{u,w_{1}}+\gamma_{w_{1},w}$
in $V_{<0}$ satisfies Lemma 2.3. ∎

We write $\mathbf{w}_{\mathcal{C}}=\mathbf{w}(v;\mathcal{C})$ for some $v\in
V_{<0}$.

###### Proof of Theorem1.1(1).

The above proposition gives the first part of Theorem 1.1(1). For the
convergence, we proceed by a case-by-case study of the vectors in $V$, the
main idea is that, as long as a vector $x$ lies in (the boundary of) a
connected component of $\mathrm{Reg}(\mathcal{C})$ which has non-empty
intersection with $V_{<0}$, we can approximate $x$ with a vector $x^{\prime}$
in this component and these two values $\mathbf{w}(x;\mathcal{C})$ and
$\mathbf{w}(x^{\prime};\mathcal{C})$ should be equal.

It is easy to see that any non-zero vector $x$ in $V$ belongs to one and only
one of the following cases:

  1. (1)

$x\in U$ for some connected component $U$ (a polyhedral cone) of
$\mathrm{Reg}(\mathcal{C})$ such that $U\cap V_{<0}=\emptyset$. Since
$\mathbf{w}(x;\mathcal{C})$ is constant on $U$ and also on $V_{<0}$, so
$\mathbf{w}(x;\mathcal{C})=\mathbf{w}_{\mathcal{C}}$.

  2. (2)

$x$ does not satisfy the preceding case but lies in
$\overline{U}_{1}\cap\overline{U}_{2}$ for two distinct connected components
$U_{1},U_{2}$ of $\mathrm{Reg}(\mathcal{C})$ such that $U_{1}\cap
V_{<0},U_{2}\cap V_{<0}\neq\emptyset$. We can assume that
$\overline{U}_{1}\cap\overline{U}_{2}$ is contained in $C_{j_{i}}^{\perp}$ for
$j_{1}<j_{2}<\cdots<j_{k}$ and not in any other $C_{i}^{\perp}$. Then these
vectors $C_{j_{1}},\cdots,C_{j_{k}}$ are all parallel and it is easy to see
$j_{i+1}-j_{i}>1$ for any $i$. Note that
$\overline{U}_{1}\cap\overline{U}_{2}$ is a polyhedral cone in a subspace of
$V$ of co-dimension $1$. First we assume that $x$ lies in the interior of the
polyhedral cone $\overline{U}_{1}\cap\overline{U}_{2}$. So we can choose two
vectors $x_{1}\in U_{1}$ and $x_{2}\in U_{2}$ in a sufficiently small open
neighborhood of $x$ such that

$\displaystyle\mathrm{sgn}(x_{1},C_{j_{i}})=-\mathrm{sgn}(x_{2},C_{j_{i}})$
$\displaystyle=1,\,\quad\forall i=1,\cdots,k$
$\displaystyle\mathrm{sgn}(x_{1},C_{j})=\mathrm{sgn}(x_{2},C_{j})$
$\displaystyle=\mathrm{sgn}(x,C_{j}),\,\quad\forall
j\notin\\{j_{1},\cdots,j_{k}\\}.$

By definition of $\mathbf{w}(x_{1};\mathcal{C})$ and
$\mathbf{w}(x_{2};\mathcal{C})$, one gets

$\displaystyle 0$
$\displaystyle=\mathbf{w}(x_{1};\mathcal{C})-\mathbf{w}(x_{2};\mathcal{C})$
$\displaystyle=\sum_{i=1}^{k}\mathrm{sgn}(x_{1},C_{j_{i}-1})\mathrm{sgn}(x_{1},C_{j_{i}})+\mathrm{sgn}(x_{1},C_{j_{i}+1})\mathrm{sgn}(x_{1},C_{j_{i}})$
$\displaystyle\quad-\mathrm{sgn}(x_{2},C_{j_{i}-1})\mathrm{sgn}(x_{2},C_{j_{i}})-\mathrm{sgn}(x_{2},C_{j_{i}+1})\mathrm{sgn}(x_{2},C_{j_{i}})$
$\displaystyle=2\sum_{i=1}^{k}\mathrm{sgn}(x_{1},C_{j_{i}-1})\mathrm{sgn}(x_{1},C_{j_{i}})+\mathrm{sgn}(x_{1},C_{j_{i}+1})\mathrm{sgn}(x_{1},C_{j_{i}})$

On the other hand,

$\displaystyle\mathbf{w}(x_{1};\mathcal{C})-\mathbf{w}(x;\mathcal{C})$
$\displaystyle=$
$\displaystyle\sum_{i=1}^{k}\mathrm{sgn}(x_{1},C_{j_{i}-1})\mathrm{sgn}(x_{1},C_{j_{i}})+\mathrm{sgn}(x_{1},C_{j_{i}+1})\mathrm{sgn}(x_{1},C_{j_{i}})$

Thus $\mathbf{w}(x;\mathcal{C})=\mathbf{w}_{\mathcal{C}}$.

If $x$ lies in the boundary of the polyhedral cone
$\overline{U}_{1}\cap\overline{U}_{2}$, we can repeat the above argument by
choosing points $x_{1}^{\prime},x_{2}^{\prime}$ from
$\overline{U}_{1}\backslash U_{1},\overline{U}_{2}\backslash U_{2}$ (we have
already shown that
$\mathbf{w}(x_{1}^{\prime};\mathcal{C})=\mathbf{w}(x_{2}^{\prime};\mathcal{C})=\mathbf{w}_{\mathcal{C}}$).
This process will continue if $x$ lies in the boundary of the boundary of
$\overline{U}_{1}\cap\overline{U}_{2}$, etc. It will terminate in finitely
many steps as there are only finitely many connected components of
$\mathrm{Reg}(\mathcal{C})$.

  3. (3)

$x$ does not satisfy the preceding cases but lies in a connected component $U$
of $\mathrm{Reg}(\mathcal{C})$ such that $U\cap V_{<0}=\emptyset$ and
$\overline{U}\cap\overline{V}_{<0}\neq\emptyset$. Then there is another
connected component $U^{\prime}$ of $\mathrm{Reg}(\mathcal{C})$ such that
$U^{\prime}\cap V_{<0}\neq\emptyset$ and
$\overline{U}\cap\overline{U^{\prime}}\cap\overline{V}_{<0}\neq\emptyset$.
Take $x_{0}\in\overline{U}\cap\overline{U^{\prime}}\cap\overline{V}_{<0}$ and
$x^{\prime}\in U^{\prime}$ and assume that $x,x^{\prime}$ are in a
sufficiently small open neighborhood of $x_{0}$ (note that
$\mathbf{w}(x;\mathcal{C})$ is constant on $U$). Suppose that
$\overline{U}\cap\overline{U^{\prime}}$ is contained in $C_{j}^{\perp}$ for
some $j$. There are two cases to consider: $(C_{j},C_{j})$ is null or not.

If $(C_{j},C_{j})<0$, we can write

$\displaystyle C_{j}$ $\displaystyle=(0,1,0,\cdots),$ $\displaystyle
x_{0}=(1,0,1,0,\cdots),$ $\displaystyle C_{j-1}$
$\displaystyle=(a,b,c,d,\cdots),$ $\displaystyle
C_{j+1}=(a^{\prime},b^{\prime},c^{\prime},d^{\prime},\cdots).$

Conditions (I.2) and (I.3) give

$\displaystyle a^{2}-c^{2}-d^{2}-\cdots$ $\displaystyle\geq 0,$
$\displaystyle(a^{\prime})^{2}-(c^{\prime})^{2}-(d^{\prime})^{2}-\cdots$
$\displaystyle\geq 0,$ $\displaystyle
aa^{\prime}-cc^{\prime}-dd^{\prime}-\cdots$ $\displaystyle<0.$

In particular, we have $aa^{\prime}<0$ and thus
$(x_{0},C_{j-1})(x_{0},C_{j+1})=(-a+c)(-a^{\prime}+c^{\prime})<0$. So
$\mathrm{sgn}(x_{0},C_{j-1})+\mathrm{sgn}(x,C_{j+1})=0$.

If $(C_{j},C_{j})=0$ and $x_{0}$ is not parallel to $C_{j}$, then we proceed
as above and get $\mathrm{sgn}(x_{0},C_{j-1})+\mathrm{sgn}(x_{0},C_{j+1})=0$.

If $(C_{j},C_{j})=0$ and $x_{0}$ is parallel to $C_{j}$, then (I.2) gives

$\mathrm{sgn}(x,C_{j-1})+\mathrm{sgn}(x,C_{j+1})=0+0=0.$

Since $x,x^{\prime}$ and $x_{0}$ are all sufficiently near to each other, we
can use the argument in (2) to conclude that
$\mathbf{w}(x;\mathcal{C})=\mathbf{w}(x_{0};\mathcal{C})=\mathbf{w}(x^{\prime};\mathcal{C})=\mathbf{w}_{\mathcal{C}}$.

  4. (4)

$x$ does not satisfy the preceding cases but lies in
$\overline{U}_{1}\cap\overline{U}_{2}$ for two distinct connected components
$U_{1},U_{2}$ of $\mathrm{Reg}(\mathcal{C})$ such that $U_{1}\cap
V_{<0}\neq\emptyset$ and $U_{2}\cap V_{<0}=\emptyset$. Note that we have
$\overline{U}_{2}\cap\overline{V}_{<0}\neq\emptyset$. From (3), we know that
on $U_{1}$ and $U_{2}$, $\mathbf{w}(-;\mathcal{C})$ is equal to
$\mathbf{w}_{\mathcal{C}}$. Now we use the argument in (2) to conclude that
$\mathbf{w}(x;\mathcal{C})=\mathbf{w}_{\mathcal{C}}$.

  5. (5)

$x$ does not satisfy the preceding cases but lies in $U$ for some connected
component $U$ of $\mathrm{Reg}(\mathcal{C})$ such that
$\overline{U}\cap\overline{V}_{<0}=\\{0\\}$. In this case we fix a positive
definite inner product $(-,-)_{\mathcal{C}}$ on $V$ and consider the quotient
$(x,x)/(x,x)_{\mathcal{C}}$, which is a continuous function on
$\overline{U}\backslash\\{0\\}$ and moreover it is homogeneous of degree $0$.
Since it is everywhere positive on $\overline{U}\backslash\\{0\\}$, it admits
a positive infimum $r_{U,\mathrm{inf}}$, that is, for any
$x\in\overline{U}\backslash\\{0\\}$,

$(x,x)\geq r_{U,\mathrm{inf}}(x,x)_{\mathcal{C}}.$

We write $\mathcal{U}$ for the set of $U$ as in (5) above, fix the inner
product $(-,-)_{\mathcal{C}}$ for all $U\in\mathcal{U}$ and set
$r_{\mathrm{inf}}=\min_{U\in\mathcal{U}}r_{U,\mathrm{inf}}>0$. We deduce
immediately that if $x$ is not in $U$ for any $U\in\mathcal{U}$, then
$\mathbf{w}(x;\mathcal{C})=\mathbf{w}_{\mathcal{C}}$; if $x$ is in $U$ for
some $U\in\mathcal{U}$, then $(x,x)\geq r_{\mathrm{inf}}(x,x)_{\mathcal{C}}$.
From this one get the absolute convergence of $\theta_{\mu}$:

$\displaystyle|\sum_{x\in\mu+L}\Phi(x;\mathcal{C})q^{(x,x)/2}|$
$\displaystyle\leq 2N\sum_{x\in U\text{for some }U\in\mathcal{C}}e^{-\pi
vr_{\mathrm{inf}}(x,x)_{\mathcal{C}}/2}$ $\displaystyle\leq
2N\sum_{x\in\mu+L}e^{-\pi vr_{\mathrm{inf}}(x,x)_{\mathcal{C}}/2}.$

∎

We have shown that for any set $\mathcal{C}$ of $N$ non-zero vectors
satisfying (I.1)-(I.3), the map $\mathbf{w}(v;\mathcal{C})$ is constant on
$v\in V_{<0}$. In fact, we have the following partial converse

######  Proposition 2.5.

For any set of $N$ non-zero vectors $\mathcal{C}=\\{C_{1},\cdots,C_{N}\\}$ in
$V$ satisfying (I.1) and (I.2), if $\mathbf{w}(v;\mathcal{C})$ is constant on
$v\in V_{<0}$, then $\mathcal{C}$ also satisfies (I.3).

###### Proof.

Take $v\in V_{<0}\cap\mathrm{Reg}(\mathcal{C})$. We can choose $N$ non-zero
vectors $\mathcal{C}^{\prime}=\\{C_{1}^{\prime},\cdots,C_{N}^{\prime}\\}$ with
$C_{i}^{\prime}$ in a sufficiently small neighborhood of $C_{i}$ for all
$i=1,\cdots,N$ such that the following conditions are satisfied

  1. (1)

$\mathbf{w}(v;\mathcal{C})=\mathbf{w}(v;\mathcal{C}^{\prime})$,

  2. (2)

the vectors $C_{j}^{\prime}$ are not parallel to each other,

  3. (3)

$\mathcal{C}$ satisfies (I.1)-(I.3) if and only if $\mathcal{C}^{\prime}$
satisfies (I.1)-(I.3).

Therefore we can assume from the onset that the vectors $C_{j}$ are not
parallel to each other.

Fix an index $j=1,\cdots,N$. Take a negative vector $v\in V$ such that
$(v,C_{j})=0$ and $(v,C_{i})\neq 0$ for any $i\neq j$ (this is always possible
since the $C_{j}$ are not parallel to each other). Then we can find another
negative vector $w\in V$ in a sufficiently small open neighborhood of $v$ such
that for any $s\in[0,1]$, the following conditions hold

  1. (1)

the vector $\gamma_{w,v}(s)$ is a negative vector for any $s\in[0,1]$;

  2. (2)

$(\gamma_{w,v}(s),C_{i})\neq 0$ unless $i=j$ and $s=1$;

  3. (3)

for any $i\neq j-1,j$, $v^{\perp}\cap\gamma_{i}(]0,1[)=\emptyset$ if and only
if $\gamma_{w,v}(s)^{\perp}\cap\gamma_{i}(]0,1[)=\emptyset$ for any
$s\in[0,1]$;

  4. (4)

$\mathrm{sgn}(v,C_{i})\mathrm{sgn}(\gamma_{w,v}(s),C_{i})=1$ for $i=j-1,j+1$
and for any $s\in[0,1]$.

Thus
$\mathrm{sgn}(\gamma_{w,v}(s),C_{i})\mathrm{sgn}(\gamma_{w,v}(s),C_{i+1})$ are
all constant on $s\in[0,1]$ (for any $i\neq j-1,j$), we deduce that

$\mathrm{sgn}(\gamma_{w,v}(s),C_{j-1})\mathrm{sgn}(\gamma_{w,v}(s),C_{j})+\mathrm{sgn}(\gamma_{w,v}(s),C_{j})\mathrm{sgn}(\gamma_{w,v}(s),C_{j+1})$

is constant on $s\in[0,1]$. Setting $s=1$, we know that this map is equal to
$0$. Now take $s\neq 1$ and we must have
$\mathrm{sgn}(\gamma_{w,v}(s),C_{j-1})\mathrm{sgn}(\gamma_{w,v}(s),C_{j+1})=-1$
and therefore $(v,C_{j-1})(v,C_{j+1})<0$.

If $(C_{j},C_{j})<0$, we can write

$\displaystyle v=(1,0,\cdots),$ $\displaystyle\quad C_{j}=(0,1,0,\cdots),$
$\displaystyle C_{j-1}=(a,b,c,d,\cdots),$ $\displaystyle\quad
C_{j+1}=(a^{\prime},b^{\prime},c^{\prime},d^{\prime},\cdots).$

(I.2) gives

$a^{2}-c^{2}-d^{2}-\cdots>0,\quad(a^{\prime})^{2}-(c^{\prime})^{2}-(d^{\prime})^{2}-\cdots>0.$

Since $(v,C_{j-1})(v,C_{j+1})=aa^{\prime}<0$, we have
$aa^{\prime}-cc^{\prime}-dd^{\prime}-\cdots<0$ and thus (I.3) holds for $j$.

If $(C_{j},C_{j})=0$, we can write

$\displaystyle v=(1,0,\cdots),$ $\displaystyle\quad C_{j}=(0,1,1,0,\cdots),$
$\displaystyle C_{j-1}=(a,b,b,0,\cdots)+D,$ $\displaystyle\quad
C_{j+1}=(a^{\prime},b^{\prime},b^{\prime},0,\cdots)+D^{\prime}$

where $D,D^{\prime}\in\mathbb{R}(e_{3},\cdots,e_{n+2})$. (I.1) gives

$a^{2}-(D,D)\geq 0,\quad(a^{\prime})^{2}-(D^{\prime},D^{\prime})\geq 0.$

Combined with $(v,C_{j-1})(v,C_{j+1})=aa^{\prime}<0$, we get that
$(C_{j-1},C_{j+1})=-aa^{\prime}+(D,D^{\prime})\geq 0$, and the equality holds
if and only if $a^{2}=(D,D)$, $(a^{\prime})^{2}=(D^{\prime},D^{\prime})$ and
$D$ is parallel to $D^{\prime}$. In particular, if the equality holds, we have
$(C_{j-1},C_{j-1})=-a^{2}+(D,D)=0$ and similarly $(C_{j+1},C_{j+1})=0$,
contradicting the assumption that no three consecutive vectors in
$\mathcal{C}$ are all null vectors. Therefore we must have
$(C_{j-1},C_{j+1})>0$. ∎

###### Proof of Theorem 1.1(2).

The absolute convergence of $\theta_{\mu}(\tau)$ implies that the discrete-
valued map $\Phi(x;\mathcal{C})$ is identically zero if $(x,x)<0$. Then we
apply the above proposition. ∎

######  Remark 2.6.

The strategy of proof of 1.1 can be easily adapted to the case of indefinite
theta series $\theta_{\mu}(\tau;\\{C_{1},C_{2}\\})$ for an integral lattice
$(L,(-,-))$ of signature $(n,1)$ mentioned in the introduction. In particular,
for two negative vectors $C_{1},C_{2}$ in $V$, if
$\theta_{\mu}(\tau;\\{C_{1},C_{2}\\})$ are termwise absolutely convergent,
then $(C_{1},C_{2})<0$.

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