1 Introduction
Let p be a prime. Up to isomorphism, there is a unique quaternion algebra
$B_p$
over
$\mathbb Q$
ramified at exactly p and
$\infty $
. The quaternion algebra
$B_p$
comes equipped with a canonical involution
$x\mapsto \overline {x}$
, a norm 
, and a trace 
. For background on quaternion algebras and their orders, the reader may consult [Reference Brzeziński4, Reference Voight21].
There will typically be many non-isomorphic maximal orders in
$B_p$
: the number of isomorphism classes of maximal orders in
$B_p$
(the type number of a maximal order) is between
$\frac {p-1}{24}$
and
$\frac {p+13}{12}$
inclusive [Reference Voight21, Exercise 30.6, Proposition 30.9.2]. This article presents two theorems, each of which allows one to recover information about a maximal order
$\mathcal {O}$
in
$B_p$
, given only information about elements in
$\mathcal {O}$
with prescribed norms.
Theorem 1.1 Let
$\mathcal {O}$
be a maximal order in
$B_p$
, and let S be an infinite set of positive integers coprime to p. There is a generating set for
$\mathcal {O}$
as a
${\mathbb {Z}}$
-module consisting of elements with norm in S.
Remark 1.2 See Remark 2.11 for a discussion of the coprime to p condition. The theorem still holds if we take
$\mathcal {O}$
to be an Eichler order in
$B_p$
with index coprime to
$6$
, but is false for every other Eichler order in
$B_p$
(see Section 2.3.4).
To prove Theorem 1.1, we first establish a local–global principle for lattices having the property of being generated as a
${\mathbb {Z}}$
-module by elements of norm in a given set S (Theorem 2.2). In Section 2.3, we check that under the conditions of Theorem 1.1, all the local conditions of Theorem 2.2 are satisfied.
As a special case, we can take
$S=\{\ell ^k:k\geq 0\}$
for any prime
$\ell \neq p$
, and conclude that maximal orders are generated by elements of norm equal to a power of
$\ell $
. This has implications for the study of isogeny graphs of supersingular elliptic curves. If E is a supersingular elliptic curve over
$\overline {\mathbb F_p}$
, then
$\operatorname {\mathrm {End}}(E)$
is a maximal order in
$B_p$
, and endomorphisms of E with norm a power of
$\ell $
can be generated by finding cycles from E in the
$\ell $
-isogeny graph of supersingular curves over
$\overline {\mathbb F_p}$
(see, for instance, [Reference Bank, Camacho-Navarro, Eisenträger, Morrison, Park, Balakrishnan, Folsom, Lalín and Manes1, Reference Eisenträger, Hallgren, Lauter, Morrison, Petit, Nielsen and Rijmen9, Reference Kohel14]). Theorem 1.1 implies that such endomorphisms generate the entire endomorphism ring as an abelian group. This is used as a heuristic assumption in [Reference Eisenträger, Hallgren, Lauter, Morrison, Petit, Nielsen and Rijmen9, Section 3.3]. This question is considered by [Reference Bank, Camacho-Navarro, Eisenträger, Morrison, Park, Balakrishnan, Folsom, Lalín and Manes1], who determine when two cycles generate a full rank sublattice of
$\operatorname {\mathrm {End}}(E)$
(as an order), as well as a necessary condition for these cycles to generate
$\operatorname {\mathrm {End}}(E)$
, but show by example that these necessary conditions are not sufficient.
The second result is that the isomorphism type of
$\mathcal {O}$
is determined by the number of elements of each norm. Given a lattice
$\Lambda $
with integral quadratic form Q, we define the theta function of
$\Lambda $
,
so that the coefficient of
$q^n$
is the number of elements of norm n. This function encodes the spectrum of the Laplace operator of the Riemannian manifold
$\Lambda \otimes \mathbb R/\Lambda $
(see [Reference Nilsson, Rowlett and Rydell16] for more on this analytic interpretation). When
$\mathcal {O}$
is a maximal order in
$B_p$
,
$\theta _{\mathcal {O}}(q)$
is a modular form of weight
$2$
and level
$\Gamma _0(p)$
.
We say two lattices are isospectral if their theta functions are equal. Lattices of rank
$n\leq 3$
are uniquely determined up to isometry by their theta function [Reference Schiemann18], but in rank
$4$
and above, there exist pairs of non-isometric isospectral lattices. Even if we restrict to the set of lattices in the genus of a fixed maximal order of
$B_p$
, there may exist pairs of non-isometric integral lattices in
$B_p$
whose left and right orders are maximal and yet have the same theta function (see Section 3.1 for examples). However, we prove that this does not occur if we restrict to maximal orders in
$B_p$
.
Theorem 1.3 If
$\mathcal {O},\mathcal {O}'$
are maximal orders in
$B_p$
with
$\theta _{\mathcal {O}}=\theta _{\mathcal {O}'}$
, then
$\mathcal {O}\simeq \mathcal {O}'$
.
As an immediate consequence, a supersingular elliptic curve over
$\overline {\mathbb F_p}$
can be identified uniquely up to Frobenius twist by the number of endomorphisms of each degree.
The proof of Theorem 1.3 is divided into two steps: the first may be of independent interest so we state it here as a separate theorem. Given an order
$\mathcal {O}$
in
$B_p$
, we define its Gross lattice
This is a strict subset of the set of trace
$0$
elements in
$\mathcal {O}$
; see Section 3.3 for further details and discussion of the Gross lattice. For
$i=1,2,3$
, the ith successive minimum of
$\mathcal {O}^T$
is the minimum value
$D_i$
such that the span of all elements
$\alpha \in \mathcal {O}^T$
with
$N(\alpha )\leq D_i$
has dimension at least i (Definition 3.3).
In [Reference Chevyrev and Galbraith6], Chevyrev and Galbraith determine conditions under which the successive minima of the Gross lattice of
$\mathcal {O}$
determine the isomorphism class of
$\mathcal {O}$
. The following result is a strengthening of [Reference Chevyrev and Galbraith6, Theorem 1], and uses many of the same methods.
Theorem 1.4 Let p be an odd prime. Suppose
$\mathcal {O}_1$
,
$\mathcal {O}_2$
are orders of
$B_p$
, each of index r in some (not necessarily the same) maximal order. Suppose
$\mathcal {O}_1^T$
and
$\mathcal {O}_2^T$
have the same successive minima
$D_1\leq D_2\leq D_3$
, and that
$D_1\geq 8r^2$
. Then
$\mathcal {O}_1\simeq \mathcal {O}_2$
.
In particular, for all primes p, a maximal order in
$B_p$
is determined up to isomorphism by the successive minima of its Gross lattice: this is vacuously true if
$p=2$
or
$D_1<8$
because this information determines a unique maximal order in
$B_p$
(see Lemma 3.10), and it holds for
$D_1\geq 8$
by Theorem 1.4.
After setting up some preliminary results on the geometry of quaternion orders in Section 3, we prove Theorem 1.4 in Section 4.1. The remainder of Section 4 is used to show that the theta function of
$\mathcal {O}$
determines the successive minima of
$\mathcal {O}^T$
, allowing us to conclude Theorem 1.3.
In future work, we will explore similar questions for quaternion algebras over totally real fields.
1.1 Lattice definitions and conventions
Let
$R={\mathbb {Z}}$
or
$R={\mathbb {Z}}_\ell $
for some prime
$\ell $
, and K the fraction field of R. A lattice
$\Lambda $
is a free finite-rank R-module (so that
$\Lambda \simeq R^n$
for some positive integer n) equipped with a nondegenerate quadratic form
$Q\colon \Lambda \to K$
. A quadratic form Q is integral if it takes values in R. For
$\mathbf{x}\in \Lambda $
, we will refer to the value
$Q(\mathbf{x})$
as the norm of
$\mathbf{x}$
. Any quadratic form defines a bilinear form 
; if Q is integral than the bilinear form is valued in
$\frac 12 R$
. Given a basis
$\mathbf{v}_1,\ldots ,\mathbf{v}_n$
for
$\Lambda $
, the Gram matrix for
$\Lambda $
(we will also say “the Gram matrix of Q”) is the symmetric matrix
$\mathbf{A}_{\Lambda }\in \frac 12 M_n(R)$
defined by
If we write
$\mathbf{x}\in \Lambda $
as a vector in terms of the basis
$\mathbf{v}_1,\ldots ,\mathbf{v}_n$
, then the Gram matrix satisfies the relation
$$\begin{align*}Q(\mathbf{x})=\mathbf{x}^T\mathbf{A}_{\Lambda}\mathbf{x}.\end{align*}$$
If there is no room for confusion, the subscript of
$\mathbf{A}_\Lambda $
may be dropped. The determinant of
$\Lambda $
,
$\det \Lambda $
, is defined to be the determinant of a Gram matrix for
$\Lambda $
. When
$R={\mathbb {Z}}$
and Q is positive definite, we have
$\det \Lambda>0$
and the value does not depend on the choice of basis.
Given
$a_1,\ldots ,a_k\in \Lambda $
, we use the notation
$\langle a_1,\ldots ,a_k\rangle $
to denote the sublattice of
$\Lambda $
generated by
$a_1,\ldots ,a_k$
as an R-module. We say that a subset
$C\subseteq \Lambda $
is a generating set for
$\Lambda $
if C generates
$\Lambda $
as an R-module.
2 Generating sets for maximal orders
2.1 A local–global principle for being generated by elements of prescribed norms
Let
$Q\colon {\mathbb {Z}}^n\to {\mathbb {Z}}$
be a quadratic form with Gram matrix
$\mathbf{A}\in \frac 12 M_n({\mathbb {Z}})$
. For a prime
$\ell $
, set
$$\begin{align*}\tau_\ell=\left\{\!\begin{array}{ll} 1,&\ell>2,\\ 3,&\ell=2. \end{array}\right.\end{align*}$$
Aside from the last line, the following definition appears in Browning and Dietmann [Reference Browning and Dietmann3].
Definition 2.1 For
$s\in {\mathbb {Z}}_{>0}$
and
$\mathbf{A}\in M_n({\mathbb {Z}})$
, the pair
$(s,Q)$
satisfies the strong local solubility condition (“strong LSC”) if for every prime
$\ell ,$
there exists
$\mathbf{x}\in ({\mathbb {Z}}/\ell ^{\tau _\ell }{\mathbb {Z}})^n$
with
$Q(\mathbf{x})\equiv s\ \ \pmod {\ell ^{\tau _\ell }}$
and
$\ell \nmid \mathbf{A}\mathbf{x}$
.
If
$\mathbf{A}\in \frac 12M_n({\mathbb {Z}})\setminus M_n({\mathbb {Z}})$
, then we say
$(s,Q)$
satisfies strong LSC if
$(2s,2Q)$
does.
Note that the strong LSC condition does not depend on the basis for
${\mathbb {Z}}^n$
used to define the Gram matrix
$\mathbf{A}$
.
If for every prime
$\ell ,$
there exists
$\mathbf{x}\in {\mathbb {Z}}_\ell ^n$
with
$Q(\mathbf{x})=s$
, we say
$(s,Q)$
satisfies the weak local solubility condition (“weak LSC”). Strong LSC implies weak LSC by Hensel-lifting, but the converse does not hold (Example 2.7).
The following theorem is a local–global principle for lattices with the property of being generated by elements with norm in S.
Theorem 2.2 Let
$n\geq 4$
, Q a nondegenerate integral quadratic form on
${\mathbb {Z}}^n$
, and
$S\subseteq {\mathbb {Z}}_{>0}$
. Suppose that for all
$M\geq 0$
and all primes
$\ell ,$
there exists a generating set
$C_\ell $
for
${\mathbb {Z}}_\ell ^n$
such that for all
$\mathbf{x}\in C_\ell $
, the norm 
satisfies:
-
(a)
$s\in S$
, -
(b)
$s\geq M$
, -
(c)
$(s,Q)$
satisfies strong LSC.
Then
${\mathbb {Z}}^n$
has a generating set consisting of elements with norm in S.
We prove this in Section 2.2. Before that we draw a corollary, and then discuss the necessity of the conditions in the theorem.
A quadratic form
$Q\colon {\mathbb {Z}}^n\to {\mathbb {Z}}$
is primitive if
$\gcd (\{Q(\mathbf{x}):\mathbf{x}\in {\mathbb {Z}}^n\})=1$
.
Corollary 2.3 Let
$n\geq 4$
, Q a nondegenerate primitive integral quadratic form on
${\mathbb {Z}}^n$
, and
$S\subseteq {\mathbb {Z}}_{>0}$
an infinite set. Suppose that for all
$s\in S$
and that for all primes
$\ell $
, there exists a basis for
${\mathbb {Z}}_\ell ^n$
consisting of elements of norm s. Then
${\mathbb {Z}}^n$
has a generating set consisting of elements with norm in S.
Proof Since s can be arbitrarily large, we just need to check that
$(s,Q)$
satisfies strong LSC. We have a basis consisting of elements
$\mathbf{x}\in {\mathbb {Z}}_\ell ^n$
with
$Q(\mathbf{x})=s$
, so it suffices to show that one such basis vector has
$\ell \nmid \mathbf{A}\mathbf{x}$
(or
$\ell \nmid (2\mathbf{A})\mathbf{x}$
when
$\mathbf{A}\notin M_n({\mathbb {Z}})$
).
Let
$\ell $
be an odd prime, and for the sake of contradiction, suppose
$\ell \mid \mathbf{A}\mathbf{x}$
for all
$\mathbf{x}$
in a basis for
${\mathbb {Z}}_\ell ^n$
. Then
$\ell \mid \mathbf{A}\mathbf{x}$
for all
$\mathbf{x}\in {\mathbb {Z}}_\ell ^n$
, so
$$\begin{align*}\ell\mid \mathbf{x}^t \mathbf{A}\mathbf{x}=Q(\mathbf{x}),\end{align*}$$
contradicting the assumption that Q is primitive. Thus
$(s,Q)$
must satisfy strong LSC at
$\ell $
.
Now suppose
$\ell =2$
. If
$\mathbf{A}\in M_n({\mathbb {Z}})$
, the same argument as above applies. Now suppose
$\mathbf{A}\in \frac 12 M_n({\mathbb {Z}})\setminus M_n({\mathbb {Z}})$
, and for the sake of contradiction suppose
$2\mid (2\mathbf{A})\mathbf{x}$
for all
$\mathbf{x}$
in a basis for
${\mathbb {Z}}_2^n$
. Letting
$\mathbf{B}\in \operatorname {\mathrm {GL}}_n({\mathbb {Z}}_2)$
denote the matrix with columns corresponding to this basis, we have
$(2\mathbf{A})\mathbf{B}\in 2M_n({\mathbb {Z}}_2)$
. This implies
$\mathbf{A}\mathbf{B}\in M_n({\mathbb {Z}}_2)$
, so multiplying on the right by
$\mathbf{B}^{-1}\in \operatorname {\mathrm {GL}}_n({\mathbb {Z}}_2)$
, we have
$\mathbf{A}\in M_n({\mathbb {Z}}_2)$
. This contradicts our initial assumption on
$\mathbf{A}$
, so
$(2s,2Q)$
– and therefore also
$(s,Q)$
– satisfies strong LSC at
$2$
.
We now discuss the technical conditions in the statement of Theorem 2.2, and demonstrate through example that they cannot be removed or substantially weakened.
Remark 2.4 The conditions
$n\geq 4$
, (b), and (c) of Theorem 2.2 will be familiar to experts in the study of representability of integers by quadratic forms. When
${Q\colon {\mathbb {Z}}^n\to {\mathbb {Z}}}$
is a quadratic form with
$n\geq 4$
, Browning and Dietmann find an explicit lower bound M such that whenever
$k\geq M$
and
$(k,Q)$
satisfies strong LSC, k is representable by Q [Reference Browning and Dietmann3, Theorem 5]. In [Reference Browning and Dietmann3, Section 1.2], they discuss examples due to Watson [Reference Watson22, Section 7.7] demonstrating that a local–global principle can fail if
$k<M$
or if
$(k,Q)$
does not satisfy strong LSC.
In each of the counterexamples below, on the other hand, every value in S is globally represented by Q. Even in this setting, we show that if we drop any of the conditions
$n\geq 4$
, (b), or (c), the existence of generating sets for
${\mathbb {Z}}_\ell ^n$
with norms in S for all primes
$\ell $
is not sufficient to conclude the existence of a generating set for
${\mathbb {Z}}^n$
with norms in S.
Example 2.5 If we remove the condition
$n\geq 4$
from Theorem 2.2, a counterexample is given by the quadratic form
$$\begin{align*}Q(x,y)=x^2+21y^2\end{align*}$$
on
${\mathbb {Z}}^2$
and
$S=\{19^{2k}:k\geq 0\}$
. The elements in
${\mathbb {Z}}^2$
with norm in S generate an index
$4$
sublattice of
${\mathbb {Z}}^2$
: using the observation that
$5^2+21\cdot 4^2=19^2$
, we can factor each side of the equation
$x^2+21y^2=19^{2k}$
into prime ideals of
${\mathbb {Z}}[\sqrt {-21}]$
to show that we must have
$4\mid y$
. But for any
$k\geq 0$
and any prime
$\ell $
, there is a basis for
${\mathbb {Z}}_\ell ^2$
consisting of elements of norm
$19^{2k}$
: we have:
$$ \begin{align*}\begin{array}{lclll} Q(19^k,0) &\!\!\!\!\equiv\!\!\!\! & Q(19^k,1)&\!\!\!\!\equiv\ 19^{2k}\quad\pmod \ell & \text{for } \ell=3,7,\\ Q(1,0) &\!\!\!\!\equiv\!\!\!\! & Q(2,1)&\!\!\!\!\equiv\ 19^{2k}\quad\pmod 8, &\text{and}\\ Q(6,1)&\!\!\!\!\equiv\!\!\!\! & Q(6,-1)&\!\!\!\!\equiv\ 19^{2k}\quad\pmod {19}, \end{array} \end{align*} $$
and for remaining
$\ell $
, we can use the fact that
$x^2+21y^2-19^kt^2=0$
defines a smooth projective conic over
$\mathbb F_\ell $
to find two independent points
$(x,y)\in \mathbb F_\ell ^2$
with
$Q(x,y)\equiv 19^k\ \pmod \ell $
. In each case these Hensel-lift to a basis for
${\mathbb {Z}}_\ell ^2$
of elements with norm
$19^k$
.
We do not currently know whether or not it is sufficient to assume
$n\geq 3$
in Theorem 2.2.
Example 2.6 Condition (b) of Theorem 2.2 (that the local generators have norm at least M) can be thought of as a constraint coming from the infinite place of
$\mathbb Q$
. If we remove it, a counterexample is given by the quadratic form
$$\begin{align*}Q(x,y,z,w)=x^2+9y^2+9z^2+9w^2\end{align*}$$
on
${\mathbb {Z}}^4$
with
$S=\{37\}$
. The only
$x\in {\mathbb {Z}}$
satisfying
$x^2\equiv 37\ \ \pmod 9$
and
$x^2\leq 37$
is
$x=\pm 1$
, so the set of vectors of norm
$37$
generate an index
$16$
sublattice of
${\mathbb {Z}}^4$
with basis
$$\begin{align*}(1,2,0,0),(-1,2,0,0),(1,0,2,0),(1,0,0,2).\end{align*}$$
However, for all primes
$\ell $
, there is a basis for
${\mathbb {Z}}_\ell ^4$
consisting of elements of norm
$37$
: for odd
$\ell $
we can use the above basis because
$16$
is a unit in
${\mathbb {Z}}_\ell ^\times $
, and for
$\ell =2$
we can let u be a square root of
$\frac {11}{3}$
in
${\mathbb {Z}}_2^\times $
and use
$$\begin{align*}(1,2,0,0),(2,u,0,0),(2,0,u,0),(2,0,0,u).\end{align*}$$
Example 2.7 If we weaken condition (c) to merely requiring that
$(s,Q)$
satisfies weak LSC, a counterexample is given by the quadratic form
$$\begin{align*}Q(x,y,z,w)=3x^2+5y^2+11\cdot 15^2z^2+11\cdot 15^3w^2\end{align*}$$
on
${\mathbb {Z}}^4$
and
$S=\{3^k:k\geq 1\text { odd}\}\cup \{5^k:k\geq 1\text { odd}\}$
. The elements
$(3^{(k-1)/2},0,0,0)$
and
$(0,5^{(k-1)/2},0,0)$
show that every element of S is globally represented (and hence everywhere locally represented). Further, for any odd
$k\geq 1$
and
$\ell \neq 5$
,
${\mathbb {Z}}_\ell ^4$
has a basis of elements of norm
$5^k$
, using the observations that
$$ \begin{align*} Q(1,1,0,1) \equiv Q(0,1,1,1) \equiv Q(0,1,0,0) \equiv Q(0,0,0,1) &\equiv 5^k\quad\pmod{8},\\ Q(0,1,0,0) \equiv Q(1,1,0,0) \equiv Q(0,1,1,0) \equiv Q(0,1,0,1) &\equiv 5^k\quad\pmod{3},\\ Q(3r,0,0,0) \equiv Q(0,r,0,0) \equiv Q(0,r,1,0) \equiv Q(0,r,0,1) &\equiv 5^k\quad\pmod{11} \end{align*} $$
with
$r=5^{(k-1)/2}$
. In a similar way, we can show that for any odd
$k\geq 1$
and
$\ell \neq 3$
,
${\mathbb {Z}}_\ell ^4$
has a basis of elements of norm
$3^k$
.
However, for
$k\geq 4$
, the only elements of norm
$5^k$
in
${\mathbb {Z}}_5^4$
are in
$5{\mathbb {Z}}_5^4$
:
$$ \begin{align*} 0\equiv Q(x,y,z,w)&\equiv 3x^2\quad\pmod 5 & &\Rightarrow 5\mid x;\\ 0\equiv Q(5x_1,y,z,w)&\equiv 5y^2\quad\pmod{25} & &\Rightarrow 5\mid y;\\ 0\equiv Q(5x_1,5y_1,z,w)&\equiv 25(3x_1^2+4z^2)\quad\pmod{125} & &\Rightarrow 5\mid x_1,z;\\ 0\equiv Q(25x_2,5y_1,5z_1,w)&\equiv 125(y_1^2+2w^2)\quad\pmod{625} & &\Rightarrow 5\mid y_1,w. \end{align*} $$
Thus
$(5^k,Q)$
does not satisfy strong LSC at
$5$
. In a similar way, for
$k\geq 4,$
the only elements of norm
$3^k$
in
${\mathbb {Z}}_3^4$
are in
$3{\mathbb {Z}}_3^4$
, so
$(3^k,Q)$
does not satisfy strong LSC at
$3$
. In particular, every element of
${\mathbb {Z}}^4$
with norm in S satisfies
$15\mid z,w$
(using the argument above for
$k\geq 4$
and checking explicitly for small k), so such elements do not generate
${\mathbb {Z}}^4$
.
2.2 Proof of Theorem 2.2
The proof is a direct application of a strong approximation theorem of Sardari. We quote a special case of this theorem here.
Given an integer s, a prime
$\ell $
, an integer
$t_\ell \geq 0$
, and
$\mathbf{a}_\ell \in {\mathbb {Z}}_\ell ^n$
, define the local density
where
Given a choice of
$t_\ell $
and
$\mathbf{a}_\ell $
for all
$\ell $
with
$t_\ell =0$
for all but finitely many
$\ell $
, set 
and
$V=\prod _{\ell } \ell ^{-t_\ell }$
.
Theorem 2.8 [Reference Sardari17, Theorem 1.6]Footnote 1
Let
$n\geq 4$
,
$Q\colon {\mathbb {Z}}^n\to {\mathbb {Z}}$
a nondegenerate quadratic form, and
$\epsilon>0$
. For all primes
$\ell $
and all integers s, the number of
$\mathbf{x}\in {\mathbb {Z}}^n$
satisfying
$Q(\mathbf{x})=s$
and
$\mathbf{x}\equiv \mathbf{a}_\ell \ \ \pmod {\ell ^{t_\ell }}$
for all primes
$\ell $
is
$$\begin{align*}\gg \mathfrak{S}(s)V^{n-1}s^{\frac{n-2}{2}}\left(1+O(V^{-3(n-3)/2}s^{\epsilon-\frac{n-3}{4}})\right),\end{align*}$$
where the implied constants in
$\gg $
and O depends only on
$\epsilon $
and Q.
Lemma 2.9 Let
$\ell $
be a prime, s an integer, and
$\mathbf{a}_\ell \in {\mathbb {Z}}_\ell ^n$
satisfying
$Q(\mathbf{a}_\ell )=s$
. Let
$t_\ell =1$
and
$t_{\ell '}=0$
for
$\ell '\neq \ell $
. Suppose
$(s,Q)$
satisfies strong LSC. Then for all
$\delta>0,$
we have
$\mathfrak{S}(s)\gg |s|^{-\delta }$
, where the implicit constant depends only on Q,
$\ell $
, and
$\delta $
(not on s).
Proof Consider the modified singular series
Note that for
$\ell '\neq \ell $
, the term at
$\ell '$
is equal to the term at
$\ell '$
of
$\mathfrak{S}(s)$
. The terms at
$\ell $
are each bounded above and below by nonzero constants in s that depend on
$\ell $
(the lower bound follows by an application of Hensel’s lemma), so
$\mathfrak{S}'(s)$
and
$\mathfrak{S}(s)$
have the same rate of decay in s.
Browning and Dietmann prove [Reference Browning and Dietmann3, Proposition 2] that for any
$\delta>0$
, if
$\mathbf{A}\in M_n({\mathbb {Z}})$
and
$(s,Q)$
satisfies strong LSC, then
$$\begin{align*}\mathfrak{S}'(s,Q)\gg |s\Delta_Q|^{-\delta},\end{align*}$$
with
$\Delta _Q$
the discriminant of Q, and the implicit constant depending only on
$\delta $
.
If
$\mathbf{A}\notin M_n({\mathbb {Z}})$
, then
$(s,Q)$
satisfying strong LSC implies
$$\begin{align*}\mathfrak{S}'(2s,2Q)\gg |2s\Delta_{2Q}|^{-\delta}.\end{align*}$$
Now
$\mathfrak{S}'(2s,2Q)$
and
$\mathfrak{S}'(s,Q)$
are the same at every prime except
$2$
, where they differ by at most a constant factor. So we reach the same conclusion for
$\mathfrak{S}'(s,Q)$
.
Proof of Theorem 2.2
Fix any prime
$\ell $
, and let M be large (we will specify how large later). Let
$C_\ell $
be a generating set for
${\mathbb {Z}}_\ell ^n$
satisfying conditions (a) through (c), let
${\mathbf{a}_\ell \in C_\ell }$
, and let 
. Set
$t_\ell =1$
and
$t_{\ell '}=0$
for all primes
$\ell '\neq \ell $
. By Lemma 2.9, the corresponding singular series is asymptotically larger than
$s^{-\frac {n-2}{2}}$
. So by Theorem 2.8, if M is sufficiently large, there exists
$\mathbf{y}\in {\mathbb {Z}}^n$
satisfying
$Q(\mathbf{y})=s$
and
$\mathbf{y}\equiv \mathbf{a}_\ell \ \ \pmod {\ell }$
. Here, “sufficiently large” may depend on
$\ell $
, Q, and a choice of
$0<\epsilon <\frac {n-3}{4}$
, but these choices can all be made at the outset.
Thus, there is a set
$\widehat {C_\ell }\subseteq {\mathbb {Z}}^n$
such that for each
$\mathbf{a}_\ell \in C_\ell $
, there is a corresponding
${\mathbf{y}\in \widehat {C_\ell }}$
with
$Q(\mathbf{y})=Q(\mathbf{a}_\ell )\in S$
and
$\mathbf{y}\equiv \mathbf{a}_\ell \ \ \pmod \ell $
. Since the inclusion
${{\mathbb {Z}}^n\to {\mathbb {Z}}_\ell ^n}$
induces an isomorphism
${\mathbb {Z}}^n/\ell {\mathbb {Z}}^n\to {\mathbb {Z}}_\ell ^n/\ell {\mathbb {Z}}_\ell ^n$
, and
$C_\ell $
generates
${\mathbb {Z}}_\ell ^n$
,
$\widehat {C_\ell }$
generates
${\mathbb {Z}}^n/\ell {\mathbb {Z}}^n$
.
Since elements in
${\mathbb {Z}}^n$
with norm in S generate
${\mathbb {Z}}^n/\ell {\mathbb {Z}}^n$
for all primes
$\ell $
, we can conclude that such elements generate
${\mathbb {Z}}^n$
.
Remark 2.10 If every element of S is coprime to
$2p$
, then a much shorter proof of Theorem 2.2 can be given using Theorem 1.2 of [Reference Sardari17]. The authors were informed by Naser Sardari that a correction needs to be made to this result: as written it only requires N to be odd, but in fact N must also be relatively prime to the discriminant of Q.
2.3 Local generating sets for maximal orders
Let
$\mathcal {O}$
be a maximal order in
$B_p$
, and let s be any positive integer relatively prime to p. We will show that for all primes
$\ell $
,
$\mathcal {O}\otimes {\mathbb {Z}}_\ell $
has a basis consisting of elements of norm s, so that Theorem 1.1 follows from Corollary 2.3.
2.3.1
$\ell \neq p$
In this case
$\mathcal {O}\otimes {\mathbb {Z}}_\ell \simeq M_2({\mathbb {Z}}_\ell )$
, with the norm on
$\mathcal {O}$
inducing the determinant on
$M_2({\mathbb {Z}}_\ell )$
. The elements
$$\begin{align*}\begin{pmatrix} s&0\\0&1 \end{pmatrix},\quad \begin{pmatrix} s&1\\0&1 \end{pmatrix},\quad \begin{pmatrix} s&0\\1&1 \end{pmatrix},\quad \begin{pmatrix} 1&-1\\ s&0 \end{pmatrix}\end{align*}$$
each have norm s, and these evidently span
$M_2({\mathbb {Z}}_\ell )$
.
2.3.2
$\ell =p\neq 2$
Let
$K/\mathbb Q_p$
be the unique unramified quadratic extension, with Galois group generated by
$\sigma $
. Then
$$\begin{align*}B_p\otimes {\mathbb{Z}}_p\simeq \left\{\begin{pmatrix} u&p v\\ \sigma(v)&\sigma(u) \end{pmatrix}:u,v\in K\right\}\subseteq M_2(K)\end{align*}$$
[Reference Voight21, Corollary 13.3.14], and
$\mathcal {O}\otimes {\mathbb {Z}}_p$
is the corresponding valuation ring [Reference Voight21, Proposition 13.3.4], obtained by restricting u and v to be in the valuation ring of K. The norm on
$\mathcal {O}\otimes {\mathbb {Z}}_p$
is
$u\sigma (u)-p v\sigma (v)$
.
Remark 2.11 If u is not a multiple of p, then
$u\sigma (u)-p v\sigma (v)$
is not a multiple of p. This shows that any basis of a maximal order
$\mathcal {O}\subseteq B_p$
must contain at least two elements with norm relatively prime to p.
If
$p\neq 2,$
we have
$K\simeq \mathbb Q_p(\sqrt {a})$
for some
$a\in \mathbb Q_p$
such that a is not a square modulo p. Therefore
$\mathcal {O}\otimes {\mathbb {Z}}_p\simeq {\mathbb {Z}}_p^4$
with quadratic form
Since
$p\nmid s$
, the equation
$x^2-ay^2-st^2=0$
defines a smooth projective conic over
$\mathbb F_p$
. This curve has
$p+1\geq 3\ \mathbb F_p$
-points, none of which has
$t=0$
, and no three of which lie on a common line. Thus, there exist two linearly independent points
$(c_1,d_1),(c_2,d_2)\in \mathbb F_p^2$
with
$c_1^2-ad_1^2=c_2^2-ad_2^2=s$
in
$\mathbb F_p$
, so we have
$$\begin{align*}Q(c_1,d_1,0,0)\equiv Q(c_1,d_1,1,0)\equiv Q(c_1,d_1,0,1)\equiv Q(c_2,d_2,0,0)\equiv s\quad\pmod p.\end{align*}$$
By Hensel lifting, we obtain a basis for
${\mathbb {Z}}_p^4$
consisting of elements of norm s.
2.3.3
$\ell =p=2$
As above, the norm on
$\mathcal {O}\otimes {\mathbb {Z}}_p$
is
$u\sigma (u)-p v\sigma (v)$
, but this time, we have
$K\simeq \mathbb Q_2(\zeta _3)$
, where
$\zeta _3\in K$
satisfies
$\zeta _3^2+\zeta _3+1=0$
. Therefore
$\mathcal {O}\otimes {\mathbb {Z}}_2\simeq {\mathbb {Z}}_2^4$
with quadratic form
We have
$$ \begin{align*} Q(1,0,0,0)&\equiv Q(0,1,0,0)\equiv Q(1,1,1,0)\equiv Q(1,1,0,1)\equiv 1 &\pmod 2,\\ Q(1,1,0,0)&\equiv Q(1,0,1,1)\equiv Q(0,1,1,2)\equiv Q(0,1,2,1)\equiv 3 &\pmod 8,\\ Q(1,1,1,1)&\equiv Q(1,1,1,2)\equiv Q(1,1,2,1)\equiv Q(2,1,1,0)\equiv 5 & \pmod 8,\\ Q(1,0,1,0)&\equiv Q(0,1,1,0)\equiv Q(1,0,0,1)\equiv Q(0,1,1,3)\equiv 7 &\pmod 8, \end{align*} $$
and for each row, the four vectors define a matrix with odd determinant. So regardless of the value of s, we can Hensel lift to obtain a basis for
${\mathbb {Z}}_2^4$
consisting of elements of norm s.
2.3.4 Eichler orders
An Eichler order is an intersection of two maximal orders. If
$\mathcal {O}$
is an Eichler order in
$B_p$
, then
$\mathcal {O}\otimes {\mathbb {Z}}_p$
is maximal, and for
$\ell \neq p$
,
$\mathcal {O}\otimes {\mathbb {Z}}_\ell $
is conjugate to
$(\begin {smallmatrix} {\mathbb {Z}}_\ell & {\mathbb {Z}}_\ell \\ \ell ^{r_\ell }{\mathbb {Z}}_\ell & {\mathbb {Z}}_\ell \end {smallmatrix})$
for some
$r_\ell \geq 0$
[Reference Voight21, Section 23.4.19]. The exponent
$r_\ell $
is nonzero only for finitely many primes
$\ell $
, and the product
$\prod _\ell \ell ^{r_\ell }$
is the index of
$\mathcal {O}$
(equal to the index of
$\mathcal {O}$
in any maximal order containing it).
Let
$\ell $
be a prime dividing the index of
$\mathcal {O}$
(we necessarily have
$\ell \neq p$
). If
$\ell \mid s$
, then the elements
$$\begin{align*}\begin{pmatrix} s&0\\0&1 \end{pmatrix},\quad \begin{pmatrix} s&1\\0&1 \end{pmatrix},\quad \begin{pmatrix} s&0\\\ell^{r_\ell}&1 \end{pmatrix},\quad \begin{pmatrix} 1&0\\ 0&s \end{pmatrix}\end{align*}$$
each have norm s and form a basis for
$\mathcal {O}\otimes {\mathbb {Z}}_\ell $
. If
$\ell \nmid 6s,$
then the elements
$$\begin{align*}\begin{pmatrix} s&0\\0&1 \end{pmatrix},\quad \begin{pmatrix} s&1\\0&1 \end{pmatrix},\quad \begin{pmatrix} s&0\\\ell^{r_\ell}&1 \end{pmatrix},\quad \begin{pmatrix} 2s&0\\ 0&\frac12 \end{pmatrix}\end{align*}$$
each have norm s and contain
$(\begin {smallmatrix} 3s&0\\0&0 \end {smallmatrix})$
and
$(\begin {smallmatrix} 0&0\\0&3 \end {smallmatrix})$
in their span, so they form a basis for
$\mathcal {O}\otimes {\mathbb {Z}}_\ell $
. So if the index of
$\mathcal {O}$
is not divisible by
$2$
or
$3$
, then for all primes
$\ell ,$
there is a basis for
$\mathcal {O}\otimes {\mathbb {Z}}_\ell $
consisting of elements of norm s; by Corollary 2.3, we can conclude that
$\mathcal {O}$
has a generating set with norms in S.
On the other hand, if the index of
$\mathcal {O}$
is even, then
$\mathcal {O}$
is not generated by elements of odd norm. This is because for
$r\geq 1$
,
$\det (\begin {smallmatrix} x&y\\ 2^rz&w \end {smallmatrix})\equiv 1\ \pmod 2$
implies
$x\equiv w\equiv 1\ \ \pmod 2$
, and the span of elements of this form lie in a proper sublattice of
$\mathcal {O}\otimes {\mathbb {Z}}_2$
. For a similar reason, if the index of
$\mathcal {O}$
is a multiple of
$3$
and
$i=1$
or
$2$
, then
$\mathcal {O}$
is not generated by elements of norm congruent to
$i\ \ \pmod 3$
. So additional constraints on S are necessary if the index is not relatively prime to
$6$
.
3 Theta function determines maximal order: background and setup
We now turn to the second topic of this article: determining maximal orders of
$B_p$
by their theta functions. In this section, we discuss some examples of isospectral lattices and then set up some results about the lattice structure of quaternion orders; the main results of Theorems 1.3 and 1.4 are proven in Section 4.
3.1 Isospectral lattices
Two integral lattices
$\Lambda ,\Lambda '$
are isospectral if their theta functions are equal. Schiemann proved that there do not exist any pairs of non-isometric isospectral lattices of rank at most
$3$
[Reference Schiemann18], but there are many pairs of non-isometric but isospectral lattices of rank
$4$
, including a four-parameter family due to Conway and Sloane [Reference Conway and Sloane7].
Now, suppose we restrict to integral lattices in a quaternion algebra
$B_p$
whose left and right orders are maximal (every such lattice is locally similar to a maximal order of
$B_p$
, and every quadratic form locally similar to a maximal order can be obtained in this way up to oriented similarity [Reference Voight21, Section 19.6.7]). Equivalently, one can ask if a pair
$E,E'$
of supersingular elliptic curves over
$\overline {\mathbb F_p}$
can be identified (up to Frobenius) by counting the number of isogenies of given degree from E to
$E'$
. Even in this constrained setting, it is common to find multiple lattices with the same theta function; the first instance of this occurring is for
$p=67$
. At
$p=151,$
we even find two non-isometric right ideals of the same maximal order that have equal theta functions. These observations were explored in depth by Shiota [Reference Shiota19].
Example 3.1 We include a brief description of the isospectral right ideals in the case
$p=151$
. Take
$B_p=\mathbb Q\langle 1,i,j,k\rangle $
with
$i^2=-1$
,
$j^2=-151$
, and 
. Consider the maximal order
This order has right ideals
each of norm
$512$
, and we can check that
$I_1$
and
$I_2$
have non-isomorphic left orders. For each of
$I_1$
and
$I_2$
, we take the basis
$x_1,x_2,x_3,x_4$
given above and compute the corresponding (normalized) Gram matrix
$\frac {1}{1024}(\operatorname {\mathrm {Tr}}(x_i\overline {x_j}))_{1\leq i,j\leq 4}$
, yielding
$$\begin{align*}\mathbf{A}_1=\frac12\begin{pmatrix} 6&2&-1&1\\ 2&12&5&4\\ -1&5&16&6\\ 1&4&6&28 \end{pmatrix},\qquad \mathbf{A}_2=\frac12\begin{pmatrix} 6&0&2&3\\ 0&12&3&4\\ 2&3&14&2\\ 3&4&2&28 \end{pmatrix}.\end{align*}$$
Both matrices have determinant
$\frac {151^2}{16}$
, as expected (by [Reference Voight21, Proposition 16.4.3] and Eq. (3.1)). The theta functions of both lattices begin with
$$\begin{align*}1 + 2q^3 + 2q^6 + 2q^7 + 2q^8 + 4q^9 + 2q^{10} + 2q^{11} + 4q^{12} + 4q^{14} + 4q^{15}+\dots\end{align*}$$
and are weight
$2$
cusp forms for
$\Gamma _0(151)$
[Reference Voight21, Section 40.4.5]. One can find a basis of
$M_2(\Gamma _0(151))$
consisting of
$13$
elements, and check that under projection to the first
$13$
Fourier coefficients they remain independent. (The Sturm bound predicts that the first
$25$
coefficients are sufficient, but we can get by with fewer in this case.) Thus, if two elements of
$M_2(\Gamma _0(151))$
agree on these coefficients, they must be equal. Hence, these lattices are isospectral. However, the lattices are not isometric because the vectors of norm
$3$
are orthogonal to the vectors of norm
$6$
in the second lattice but not in the first.
Remark 3.2 Even if two non-isometric lattices have equal theta functions, the lattices may be distinguished using an enriched theta function that carries more information about the structure of the lattice than just the number of elements of each norm. For
${k\geq 1}$
, let
$\mathbb {H}_k$
be the Siegel upper half-space, consisting of symmetric matrices
${Z\in M_k(\mathbb C)}$
with positive-definite imaginary part. For an integral lattice
$\Lambda ,$
one may define the degree k theta function of
$\Lambda $
by
where
$Z\in \mathbb {H}_k$
and
$T(v_1,\ldots ,v_k)\in \frac 12 M_k({\mathbb {Z}})$
has
$v_i\cdot v_j$
in the
$i,j$
component. We recover the classical theta function by taking
$k=1$
and setting
$q=e^{2\pi iz}$
for z in the complex upper half-plane.
Kitaoka showed that for any collection of rank n lattices that are pairwise non-isometric, the corresponding degree
$n-1$
theta functions are linearly independent [Reference Kitaoka13]. Suppose we restrict our attention to the set of rank
$4$
integral lattices in a definite quaternion algebra over
$\mathbb Q$
whose left and right orders are maximal. Böcherer and Schulze-Pillot classified all linear relations between the degree
$2$
theta functions of integral lattices of this form, and showed that two such lattices are isometric if and only if their degree
$2$
theta functions are equal [Reference Böcherer and Schulze-Pillot2, Corollary 9.2]). In this light, our Theorem 1.3 shows that for maximal orders, at least in
$B_p$
, already their degree
$1$
theta function distinguishes between them.
3.2 Lattice geometry of quaternion orders
As before, let
$B_p$
denote the quaternion algebra over
$\mathbb Q$
ramified at p and
$\infty $
. There exists an isometry
$B_p\otimes \mathbb R\simeq \mathbb R^4$
, with the norm on
$B_p$
corresponding to the square of the standard Euclidean distance on
$\mathbb R^4$
, and
$\frac 12\operatorname {\mathrm {Tr}}(x\bar y)$
giving the standard inner product
$x\cdot y$
.
Let
$\Lambda $
be a lattice in
$B_p$
of rank
$1\leq k\leq 4$
; we say
$\Lambda $
is an integral lattice if for all
$x\in \Lambda ,$
we have
$N(x),\operatorname {\mathrm {Tr}}(x)\in {\mathbb {Z}}$
. Given a basis
$\{v_1,\ldots ,v_k\}$
of an integral lattice
$\Lambda ,$
we can define a Gram matrix
and determinant 
as in Section 1.1. The quaternion norm N on
$B_p$
defines an integral quadratic form on
$\Lambda $
, and
$\mathbf{A}_\Lambda \in \frac 12M_k({\mathbb {Z}})$
.
Now suppose
$\Lambda $
is an order in
$B_p$
, so it is contained in a maximal order
$\mathcal {O}$
with finite index. We define the discriminant of
$\Lambda $
to be 
. Note that the discriminant of
$\Lambda $
is a positive integer, while the determinant may not be an integer. In particular, if
$\mathcal {O}$
is a maximal order containing
$\Lambda $
then
$$ \begin{align} \operatorname{\mathrm{disc}}\Lambda=16\det\Lambda =[\mathcal{O}:\Lambda]^2p^2 \end{align} $$
[Reference Voight21, Lemma 15.2.15, Theorem 15.5.5]. If
$\Lambda $
has basis
$v_1,\ldots ,v_4$
, then we have
$$ \begin{align} \operatorname{\mathrm{disc}} \Lambda=\det(\operatorname{\mathrm{Tr}}(v_i\bar{v_j}))_{1\leq i,j\leq 4}=|\det(\operatorname{\mathrm{Tr}}(v_iv_j))_{1\leq i,j\leq 4}| \end{align} $$
[Reference Voight21, Exercise 15.13], where we call
$(\operatorname {\mathrm {Tr}}(v_iv_j))_{1\leq i,j\leq 4}$
the trace matrix of the basis
$\{v_1,\ldots ,v_4\}$
.
Definition 3.3 Let
$\Lambda $
be a lattice of rank k with a positive definite quadratic form Q. For
$1\leq i\leq k$
, the ith successive minimum of
$\Lambda $
is the minimum value
$D_i$
such that the rank of the
${\mathbb {Z}}$
-module generated by
$\{v\in \Lambda :Q(v)\leq D_i\}$
is greater than or equal to i. An ordered list
$v_1,\ldots ,v_k\in \Lambda $
attains the successive minima of
$\Lambda $
if it is linearly independent and
$Q(v_i)=D_i$
for each
$i=1,\ldots ,k$
.
Remark 3.4 This is a nonstandard definition, following the notation of [Reference Chevyrev and Galbraith6]; if
$(\Lambda ,Q)$
is embedded isometrically in a Euclidean space
$\mathbb R^n$
, then the successive minima under Definition 3.3 are the squares of the corresponding successive minima under the standard definition. There always exists a list of elements attaining the successive minima, for instance, by [Reference Cassels5, Section VIII.1.2, Lemma 1].
Lemma 3.5 Let
$\Lambda $
be a lattice of rank
$k\leq 3$
. If a list of k vectors attains the successive minima of
$\Lambda $
, then these vectors form a basis for
$\Lambda $
. The same holds true for
$k=4$
if we additionally assume that
$\Lambda $
is an order in
$B_p$
for p odd.
Note that a counterexample for
$p=2$
is given by the Hurwitz quaternions, 
where
$i^2=j^2=-1$
and
$ij=k$
. The elements
$1,i,j,k$
attain the successive minima, but
$\frac 12(1+i+j+k)$
is not contained in their span.
Proof Among all lattices of rank at most
$4$
, the rank
$4$
cubic centered lattice
$\mathbf{D}_4$
(which is isometric to
$\mathcal {Z}$
after rescaling) is the only lattice up to similarity for which an arbitrary list of vectors attaining the successive minima is not always a basis [Reference Martinet15, Corollary 6.2.3]. If
$\Lambda $
is an order in
$B_p$
for p odd then the first successive minimum is equal to
$1$
, so if
$\Lambda $
were a cubic centered lattice then its Gram matrix would have determinant
$\frac 14$
. This contradicts the fact that the determinant of an order in
$B_p$
must be an integer multiple of
$\frac {p^2}{16}$
by Eq. (3.1).
3.3 The Gross lattice
We define an additive map
$\tau \colon B_p\to B_p$
by
$$\begin{align*}\tau(x)=2x-\operatorname{\mathrm{Tr}}(x).\end{align*}$$
If we restrict
$\tau $
to an order
$\mathcal {O}\subseteq B_p$
, then the kernel of this map is
${\mathbb {Z}}$
, and the image is the Gross lattice of
$\mathcal {O}$
,
(cf. [Reference Gross, Kisilevsky and Labute11, Section 12]). The Gross lattice is a strict subset of
$\mathcal {O}^0$
, the subset of
$\mathcal {O}$
consisting of trace zero elements; more precisely, we have
$\mathcal {O}^T=\mathcal {O}^0\cap ({\mathbb {Z}}+2\mathcal {O})$
. Some relations between sublattices of
$\mathcal {O}$
are given below: inclusion arrows below are labeled by the index of one sublattice in the other, and 
denotes orthogonal direct sum.
The index of 
in
$\mathcal {O}$
can be computed by noting that 
consists of all elements of
$\mathcal {O}$
with even trace, and that
$\mathcal {O}$
must contain an element of odd trace (since
$\det \mathcal {O}=\frac {p^2}{16}$
implies the Gram matrix of
$\mathcal {O}$
cannot be an integer matrix).
Remark 3.6 Given a lattice
$\Lambda \subseteq B_p$
, define the dual lattice
$\Lambda ^\sharp $
by
While this will not be used in the rest of the article, we note that
$\mathcal {O}^T$
can be related to
$(\mathcal {O}^\sharp )^0$
, the trace zero part of the dual lattice of
$\mathcal {O}$
. (This lattice arises in the correspondence between quaternion orders and ternary quadratic forms via the Clifford algebra construction; see [Reference Voight21, Chapter 22].) Specifically, we have the equality
$$\begin{align*}(\tfrac12\mathcal{O}^T)^\sharp=(\mathcal{O}^\sharp)^0,\end{align*}$$
because when
$\operatorname {\mathrm {Tr}}(y)=0$
and
$x\in \mathcal {O}$
we have
$\operatorname {\mathrm {Tr}}(\tau (x)\bar y)=2\operatorname {\mathrm {Tr}}(x\bar y)$
. Since
$\frac 12\mathcal {O}^T$
is the orthogonal projection of
$\mathcal {O}$
onto
$B_p^0$
, we can summarize this by saying that the dual of the projection equals the restriction of the dual.
The primary motivation for studying the Gross lattice is that elements of
$\mathcal {O}^T$
correspond to embeddings of imaginary quadratic orders in
$\mathcal {O}$
. Observe that for any
$x\in B_p,$
we have the equality
$$ \begin{align} N(\tau(x))=4N(x)-\operatorname{\mathrm{Tr}}(x)^2. \end{align} $$
So for any
$\beta \in \mathcal {O}^T\setminus \{0\}$
,
$-N(\beta )$
is equal to the discriminant of the quadratic order generated by a preimage of
$\beta $
under
$\tau $
.
Given an imaginary quadratic order R of discriminant
$-D$
for
$D>0$
, an orientation of R is a choice of
$x\in R$
satisfying
$x^2+D=0$
(which we will usually denote
$\sqrt {-D}$
). Given two oriented orders
$(R,x)$
and
$(R,x')$
, an oriented isomorphism is an isomorphism
$R\to R'$
sending
$x\mapsto x'$
. Note that for every imaginary quadratic discriminant
$-D,$
there are exactly two oriented quadratic orders R up to oriented isomorphism, sent to each other by the nontrivial Galois action on R.
Given an imaginary quadratic order R and an embedding
$\phi :R\hookrightarrow \mathcal {O}$
, we say the embedding is optimal if
$\mathbb Q\phi (R)\cap \mathcal {O}=\phi (R)$
. Given a nonzero element v of a lattice L, we say v is primitive if there does not exist
$w\in L$
and
$n\geq 2$
with
$v=nw$
. The following proposition is implicit in [Reference Gross, Kisilevsky and Labute11, Proposition 12.9]; we include the proof for completeness.
Proposition 3.7 There is a one-to-one correspondence between nonzero elements
${\beta \in \mathcal {O}^T}$
and embeddings
$R\hookrightarrow \mathcal {O}$
of oriented imaginary quadratic orders R up to oriented isomorphism. Under this correspondence, we have
$\operatorname {\mathrm {disc}} R=-N(\beta )$
, and the embedding
$R\hookrightarrow \mathcal {O}$
is optimal if and only if the corresponding
$\beta $
is primitive.
Proof The imaginary quadratic order
$R={\mathbb {Z}}[\frac 12(D+\sqrt {-D})]$
has discriminant
$-D$
and orientation determined by the element
$\sqrt {-D}$
. To an embedding
$\phi \colon R\to \mathcal {O}$
, we associate the element
Conversely, given an element
$\beta =2x-\operatorname {\mathrm {Tr}}(x)$
of
$\mathcal {O}^T$
, set
$D=N(\beta )$
and associate to
$\beta $
the embedding of
${\mathbb {Z}}[\frac 12 (D+\sqrt {-D})]$
into
$\mathcal {O}$
determined by
$$ \begin{align*} \frac{D+\sqrt{-D}}{2}&\mapsto x+\frac{D-\operatorname{\mathrm{Tr}}(x)}{2}. \end{align*} $$
We have
$x+\frac {D-\operatorname {\mathrm {Tr}}(x)}{2}\in \mathcal {O}$
because
$D\equiv \operatorname {\mathrm {Tr}}(x)\ \ \pmod 2$
by Eq. (3.4). Further, both sides have trace D, and applying Eq. (3.4) to the fact that
$\tau \left (x+\frac {D-\operatorname {\mathrm {Tr}}(x)}{2}\right )=\beta $
we find that both sides have norm
$\frac 14 (D^2+D)$
. Hence, this map is well-defined. It is straightforward to verify that these two associations are inverses and that the remaining claims are satisfied.
Note that if two embeddings
$R\to \mathcal {O}$
are related by Galois conjugation on R, then the corresponding elements of
$\mathcal {O}^T$
are negatives of each other.
For any
$\beta \in \mathcal {O}^T$
, there exists a unique
$\alpha \in \mathcal {O}$
satisfying
$\tau (\alpha )=\beta $
and
$\operatorname {\mathrm {Tr}}(\alpha )\in \{0,1\}$
(explicitly, we can let
$\delta \in \{0,1\}$
satisfy
$\delta \equiv N(\beta )\ \pmod 2$
and set
$\alpha =\frac 12(\delta +\beta )$
). By Eq. (3.4), this element
$\alpha $
attains the minimal norm among all elements of
$\mathcal {O}$
mapping under
$\tau $
to
$\beta $
.
Lemma 3.8 Let
$\beta _1,\beta _2,\beta _3\in \mathcal {O}^T$
be linearly independent, and let
$\alpha _1,\alpha _2,\alpha _3\in \mathcal {O}$
satisfy
$\tau (\alpha _i)=\beta _i$
and
$\operatorname {\mathrm {Tr}}(\alpha _i)\in \{0,1\}$
for each i. The following are equivalent:
-
(a) The successive minima for
$\mathcal {O}^T$
are attained by
$\beta _1,\beta _2,\beta _3$
. -
(b) The successive minima for
$\mathcal {O}$
are attained by
$1,\alpha _1,\alpha _2,\alpha _3$
, and if
$\operatorname {\mathrm {Tr}}(\alpha _i)=0$
for some
$i=1,2,3$
, then for all
$\gamma \in \mathcal {O}$
with
$N(\gamma )=N(\alpha _i)$
that are linearly independent from
$1,\alpha _1,\ldots ,\alpha _{i-1}$
, we have
$\operatorname {\mathrm {Tr}}(\gamma )=0$
.
Proof Observe that
$1,\alpha _1,\alpha _2,\alpha _3$
are linearly independent, because applying
$\tau $
to a linear dependence would induce a dependence among
$\beta _1,\beta _2,\beta _3$
. Now assume (a). For
$i=1,2,3$
, let
$S_i$
denote the set of
$\gamma \in \mathcal {O}$
satisfying
$N(\gamma )<N(\alpha _i)$
. For any
$\gamma \in S_i,$
we have
$$\begin{align*}N(\tau(\gamma))=4N(\gamma)-\operatorname{\mathrm{Tr}}(\gamma)^2\leq 4(N(\alpha_i)-1)<4N(\alpha_i)-\operatorname{\mathrm{Tr}}(\alpha_i)^2=N(\beta_i),\end{align*}$$
and since
$N(\beta _i)$
is the ith successive minimum for
$\mathcal {O}^T$
, the span of
$\tau (\gamma )$
for all
$\gamma \in S_i$
has dimension at most
$i-1$
. Thus the span of
$S_i$
has dimension at most i, proving by induction on i that
$1,\alpha _1,\alpha _2,\alpha _3$
attain the successive minima for
$\mathcal {O}$
. Now for the sake of contradiction suppose that
$\operatorname {\mathrm {Tr}}(\alpha _i)=0$
, and there exists
$\gamma \in \mathcal {O}$
linearly independent from
$1,\alpha _1,\ldots ,\alpha _{i-1}$
with
$N(\gamma )=N(\alpha _i)$
and
$\operatorname {\mathrm {Tr}}(\gamma )\neq 0$
. Then
$\beta _1,\ldots ,\beta _{i-1},\tau (\gamma )$
are i independent elements of
$\mathcal {O}^T$
, but
$$\begin{align*}N(\tau(\gamma))=4N(\gamma)- \operatorname{\mathrm{Tr}}(\gamma)^2 <4N(\alpha_i)=N(\beta_i),\end{align*}$$
contradicting the assumption that
$\beta _i$
attains the ith successive minimum for
$\mathcal {O}^T$
. Thus (a) implies (b).
Now assume (b). For
$i=1,2,3$
, let
$S_i$
denote the set of
$x\in \mathcal {O}^T$
satisfying
${N(x)<N(\beta _i)}$
. For any
$x\in S_i,$
there exists
$\gamma \in \mathcal {O}$
with
$\tau (\gamma )=x$
and
$\operatorname {\mathrm {Tr}}(\gamma )\in \{0,1\}$
. If
${N(x)<N(\beta _i)-1}$
, then in fact
$N(x)\leq N(\beta _i)-3$
because the norm of every element of
$\mathcal {O}^T$
is either
$0$
or
$3$
mod
$4$
by Eq. (3.4), and so
$$\begin{align*}N(\gamma)=\frac14(N(\tau(\gamma))+\operatorname{\mathrm{Tr}}(\gamma)^2)\leq\frac14(N(\beta_i)-2)< N(\alpha_i).\end{align*}$$
Since
$N(\alpha _i)$
is the
$(i+1)$
st successive minimum for
$\mathcal {O}$
, this implies
$1,\alpha _1,\ldots ,\alpha _{i-1},\gamma $
must be linearly dependent. On the other hand, suppose
$N(x)=N(\beta _i)-1$
. Then
$N(\gamma )=N(\alpha _i)$
,
$\operatorname {\mathrm {Tr}}(\gamma )=1$
, and
$\operatorname {\mathrm {Tr}}(\alpha _i)=0$
, so once again
$1,\alpha _1,\ldots ,\alpha _{i-1},\gamma $
are linearly dependent. Either way we can conclude that
$x=\tau (\gamma )$
is in the
$\mathbb Q$
-span of
$\beta _1,\ldots ,\beta _{i-1}$
. This shows that the span of
$S_i$
has dimension less than i, so
$N(\beta _i)$
is indeed the ith successive minimum for
$\mathcal {O}^T$
.
Remark 3.9 It is possible for
$1,\alpha _1,\alpha _2,\alpha _3$
to attain the successive minima of
$\mathcal {O}$
, but
$\tau (\alpha _1),\tau (\alpha _2),\tau (\alpha _3)$
not attain the successive minima of
$\mathcal {O}^T$
. A simple example is given by the Hurwitz quaternions
$\mathcal {Z}= \langle 1,i,j,\frac 12(1+i+j+k)\rangle $
with
$i^2=j^2=-1$
and
$ij=k$
. The successive minima of
$\mathcal {Z}$
are all
$1$
, and the successive minima of
$\mathcal {Z}^T$
are all
$3$
. The elements
$1,i,j,k$
attain the successive minima of
$\mathcal {O}$
, but
$N(\tau (i))=N(\tau (j))=N(\tau (k))=4$
, and so
$\tau (i),\tau (j),\tau (k)$
do not realize the successive minima of
$\mathcal {Z}^T$
. And indeed, condition (b) of Lemma 3.8 does not hold; for any of
$i=1,2,3$
, we can take
$\gamma =\frac 12(1+i+j+k)$
.
Lemma 3.10 Let
$D_1<15$
. Up to isomorphism, there is at most one maximal order
${\mathcal {O}\subseteq B_p}$
such that the first successive minimum of
$\mathcal {O}^T$
is equal to
$D_1$
.
Proof If
$D_1$
is not
$0$
or
$3$
mod
$4,$
then there is no maximal order with first successive minimum
$D_1$
by Proposition 3.7. For all remaining
$D_1$
, the quadratic order of discriminant
$-D_1$
has class number
$1$
; in this case, there is a unique maximal order (up to isomorphism) in which this quadratic order embeds [Reference Voight21, Corollary 30.4.23], and therefore a unique
$\mathcal {O}$
(up to isomorphism) such that
$\mathcal {O}^T$
has an element of norm
$D_1$
.
Remark 3.11 If
$-D$
is a fundamental discriminant, an explicit maximal order admitting an optimal embedding of the ring of integers of discriminant
$-D$
can be written down explicitly using [Reference Dorman, De Koninck and Levesque8, Eq. (5)] (see also [Reference Dorman, De Koninck and Levesque8, Theorem 1]).
3.4 Constraints on short Gross lattice vectors
A key idea we will apply is that there are very strict constraints on arrangements of short elements in the Gross lattice. This idea can be made precise using a construction due to Kaneko [Reference Kaneko12], which we present as Proposition 3.12 with minor modifications for our convenience. Kaneko used this construction to prove a bound on the discriminants of quadratic orders embedding into a quaternion order, a special case of which is given by Corollary 3.14. In addition to this bound (a constraint on the norms of independent elements in
$\mathcal {O}^T$
), we also establish a constraint on the angle between two elements of
$\mathcal {O}^T$
, Corollary 3.15.Footnote
2
Together, these constraints will be sufficient to show that a quaternion order is uniquely determined up to isomorphism by the numbers of short vectors in
$\mathcal {O}^T$
.
Proposition 3.12 Let
$\mathcal {O}$
be an order in
$B_p$
, and let
$\beta _1,\beta _2\in \mathcal {O}^T$
be linearly independent. Then
$$\begin{align*}N(\beta_1)N(\beta_2)-\frac14\operatorname{\mathrm{Tr}}(\beta_1\bar{\beta_2})^2 \end{align*}$$
is a positive integer multiple of
$4p$
.
The number
$N(\beta _1)N(\beta _2)-\frac 14\operatorname {\mathrm {Tr}}(\beta _1\bar {\beta _2})^2$
is the determinant of the Gram matrix of the lattice
$\langle \beta _1,\beta _2\rangle $
, so this can be interpreted as saying that every parallelogram in
$\mathcal {O}^T$
has area
$2\sqrt {kp}$
for some positive integer k. A version of this is proven by Kaneko in [Reference Kaneko12, Section 3]; see also Equation (3.2) of [Reference Chevyrev and Galbraith6]. A similar idea also appears in [Reference Goren and Lauter10].
Proof The value is positive because it is the determinant of the Gram matrix of the lattice
$\langle \beta _1,\beta _2\rangle $
, so it suffices to show divisibility by
$4p$
. For
$i=1,2$
, set
$D_i=N(\beta _i)$
, and let
$\alpha _i\in \mathcal {O}$
be a minimal norm preimage of
$\beta _i$
under
$\tau $
; that is, take
$\delta _i\in \{0,1\}$
with
$\delta _i\equiv D_i\ \ \pmod 2$
and set
$\alpha _i=\frac 12(\beta _i+\delta _i)$
. Define an order
$$\begin{align*}\Lambda=\langle 1,\alpha_1,\alpha_2,\alpha_1\alpha_2\rangle\subseteq \mathcal{O}.\end{align*}$$
Letting
$s=\operatorname {\mathrm {Tr}}(\alpha _1\alpha _2)$
, one can compute the trace matrix of
$\Lambda $
(Eq. (3.2)),
$$\begin{align*}\mathbf{T}=\begin{pmatrix} 2 & \delta_1 & \delta_2 & s\\[2.5pt] \delta_1 & \delta_1^2-2N(\alpha_1) & s & -\delta_2N(\alpha_1)+\delta_1s\\[2.5pt] \delta_2 & s & \delta_2^2-2N(\alpha_2) & -\delta_1N(\alpha_2)+\delta_2s\\[2.5pt] s & -\delta_2N(\alpha_1)+\delta_1s & -\delta_1N(\alpha_2)+\delta_2s & s^2 - 2N(\alpha_1)N(\alpha_2) \end{pmatrix}.\end{align*}$$
Since
$N(\alpha _i)=\frac 14(\delta _i+D_i)$
, we compute
$$\begin{align*}\operatorname{\mathrm{disc}}\Lambda=|\det\mathbf{T}|=\frac{(D_1D_2-(2s-\delta_1\delta_2)^2)^2}{16}.\end{align*}$$
Noting that
$$ \begin{align} \operatorname{\mathrm{Tr}}(\beta_1\beta_2)=\operatorname{\mathrm{Tr}}((2\alpha_1-\delta_1)(2\alpha_2-\delta_2))=4\operatorname{\mathrm{Tr}}(\alpha_1\alpha_2)-2\delta_1\delta_2, \end{align} $$
we can replace
$2s-\delta _1\delta _2$
with
$\frac 12\operatorname {\mathrm {Tr}}(\beta _1\beta _2)=-\frac 12\operatorname {\mathrm {Tr}}(\beta _1\bar {\beta _2})$
. Since
$\Lambda $
is an order in
$B_p$
, we can use Eq. (3.1) to conclude that
$\frac 14(D_1D_2-\frac 14\operatorname {\mathrm {Tr}}(\beta _1\bar {\beta _2})^2)$
is an integer multiple of p.
Some immediate consequences of this calculation are as follows.
Corollary 3.13 Let
$\mathcal {O}$
be an order in
$B_p$
of discriminant
$\Delta \in {\mathbb {Z}}$
. Let
$D_1,D_2,D_3$
be the successive minima of
$\mathcal {O}^T$
. Then
$$ \begin{align*} D_1&\leq 2\Delta^{1/3},\\ \frac{4p}{D_1}\leq D_2&\leq\left(\frac{8\Delta}{D_1}\right)^{1/2}. \end{align*} $$
Proof Since
$\det \mathcal {O}=\frac \Delta {16}$
, Eq. (3.3) implies that
(Recall that
$\det \Lambda $
refers to the determinant of the Gram matrix of
$\Lambda $
, so
$\Lambda '\subseteq \Lambda $
implies
$\det \Lambda '=[\Lambda :\Lambda ']^2\det \Lambda $
.) We also have
$D_1D_2D_3\leq 2\det \mathcal {O}^T$
(a bound specific to rank
$3$
lattices) by [Reference Siegel20, Lecture XI(25)]. The desired upper bounds follow from
$D_1^3\leq D_1D_2D_3$
and
$D_1D_2^2\leq D_1D_2D_3$
. The lower bound
$D_1D_2\geq 4p$
follows from Proposition 3.12.
Corollary 3.14 If a quadratic order R has two embeddings in
$\mathcal {O}$
with distinct images, then
$\operatorname {\mathrm {disc}} R> p$
.
Proof This is a special case of [Reference Kaneko12, Theorem 2’]. Let 
, and
$\beta _1,\beta _2\in \mathcal {O}^T$
be the elements corresponding to the two embeddings of R under Proposition 3.7. Then from Proposition 3.12, we obtain
$$\begin{align*}p\mid \left(\frac{D+t}{2}\right)\left(\frac{D-t}{2}\right),\end{align*}$$
where
$t=\frac 12\operatorname {\mathrm {Tr}}(\beta _1\bar \beta _2)\in {\mathbb {Z}}$
by Eq. (3.5). Each factor is an integer:
$D+t$
and
$D-t$
have the same parity, and since their product is a multiple of
$4$
, both must be even. Thus p divides one of the factors, so
$p\leq \frac 12(D+t)$
. Since
$D^2-t^2>0,$
we have
$t<D$
, so
$p<D$
.
Recall that
$(v_1,v_2)\mapsto \frac 12\operatorname {\mathrm {Tr}}(v_1\overline {v_2})$
defines an inner product on
$B_p$
. The following result says that if two elements of
$\mathcal {O}^T$
are sufficiently small, then their norms uniquely determine the angle between them (up to negating either element). This is one of the most important conceptual ingredients of the proofs of Theorems 1.3 and 1.4: while the theta function records lengths of elements but loses all information about angles between them, this result allows us to recover information about angles from information about lengths.
Corollary 3.15 Let
$\beta _1,\beta _2$
be independent elements of
$\mathcal {O}^T$
. Suppose
$N(\beta _1)\leq p$
, and that
$\beta _2$
has minimal norm in
$\beta _2+{\mathbb {Z}}\beta _1$
. Then
$|\frac 12\operatorname {\mathrm {Tr}}(\beta _1\bar {\beta _2})|$
equals the smallest positive square root of
$N(\beta _1)N(\beta _2)$
modulo p.
Proof By Proposition 3.12, we have
$\frac 14\operatorname {\mathrm {Tr}}(\beta _1\bar {\beta _2})^2\equiv N(\beta _1)N(\beta _2)\ \ \pmod {4p}$
, so that
$|\frac 12\operatorname {\mathrm {Tr}}(\beta _1\bar {\beta _2})|$
, an integer by Eq. (3.5), is a square root of
$N(\beta _1)N(\beta _2)$
modulo p. Expanding
