2. Shapes of number fields

This section introduces notation and reviews background material on the shapes of number fields. Fix an integer $r \geq 2$, and let $M_r(\mathbb{R})$ denote the space of real $r \times r$ matrices. The group $\operatorname{GL}_r(\mathbb{R})$ consists of all invertible elements of $M_r(\mathbb{R})$, and denote by $\operatorname{I}_r$ the identity matrix. For any matrix $M \in M_r(\mathbb{R})$, write $M^T$ for its transpose. Let $V \simeq \mathbb{R}^r$ be an inner product space with standard orthonormal basis $\{e_1, \dots, e_r\}$, so that

\begin{equation*}(e_i, e_j) = \delta_{i,j}=\begin{cases} 1 & \text{if }i=j,\\ 0 & \text{if }i\neq j. \end{cases}\end{equation*}

Let $\operatorname{GO}_r(\mathbb{R})$ be the subgroup of $\operatorname{GL}_r(\mathbb{R})$ consisting of matrices $M$ satisfying $M^T M = \lambda \operatorname{I}_r$ for some scalar $\lambda \in \mathbb{R}_{ \gt 0}$; these are precisely the invertible linear transformations preserving the inner product up to a positive scaling.

Definition 2.1. Let $\Lambda \subset V$ be a lattice. The shape of $\Lambda$, denoted $\operatorname{sh}_\Lambda$, is its equivalence class under the action of rotations, reflections, and positive scalar dilations. The space of all such shapes is the double coset space

\begin{equation*} \mathcal{S}_r := \operatorname{GL}_r(\mathbb{Z}) \backslash \operatorname{GL}_r(\mathbb{R}) / \operatorname{GO}_r(\mathbb{R}). \end{equation*}

Given a basis $\{b_1, \dots, b_r\}$ for $\Lambda$, we express each $b_i$ in terms of the standard basis as $b_i = \sum_{j=1}^r a_{i,j} e_j$. The matrix $(a_{i,j}) \in \operatorname{GL}_r(\mathbb{R})$ then determines a point in $\mathcal{S}_r$, namely the shape of $\Lambda$. This space is equipped with a natural measure $\mu_{\mathcal{S}_r}$ induced from the Haar measure on $\operatorname{GL}_r(\mathbb{R})$.

Given any measurable subset $Y \subset \mathcal{S}_r$, we write $\mu(Y) := \mu_{\mathcal{S}_r}(Y)$ for its measure. It is often convenient to describe $\mathcal{S}_r$ as the quotient of a discrete group acting on a homogeneous space $\mathcal{H}_r$. Let $\mathcal{P}_r$ denote the space of positive definite $r \times r$ symmetric real matrices. The group $\operatorname{GL}_r(\mathbb{Z})$ acts on $\mathcal{P}_r$ by $G \mapsto M^T G M$, while $\lambda\in \mathbb{R}^\times$ acts by scaling, $G \mapsto \lambda^2 G$. Set $\mathcal{H}_r:=\mathcal{P}_r/\mathbb{R}^\times$ and note that $\mathcal{H}_r$ inherits a natural $\operatorname{GL}_r(\mathbb{Z})$-action. We identify $\mathcal{H}_r$ with a subset of affine space $\mathbb{A}^{r^2}=\operatorname{M}_r(\mathbb{R})$ as follows:

\begin{equation*}\mathcal{H}_r=\{M\in \operatorname{M}_r(\mathbb{R})\mid M\text{is positive definite and symmetric with }\det M=1\}.\end{equation*}

There is a natural bijection $\operatorname{GL}_r(\mathbb{R}) / \operatorname{GO}_r(\mathbb{R}) \xrightarrow{\sim} \mathcal{H}_r$ given by $M \mapsto |\det M|^{-2/r} M^T M$, inducing an identification $\mathcal{S}_r \simeq \operatorname{GL}_r(\mathbb{Z}) \backslash \mathcal{H}_r$. This identification can be made explicit as follows. If $\Lambda$ is a lattice with basis $\{b_1, \dots, b_r\}$, its Gram matrix $\operatorname{Gr}(\Lambda)$ is defined by $(\langle b_i, b_j \rangle)_{i,j}$, and the shape of $\Lambda$ corresponds to the class of $\operatorname{Gr}(\Lambda)$ in $\mathcal{S}_r$. This class is invariant under change of basis by $M \in \operatorname{GL}_r(\mathbb{Z})$ and rescaling by $\lambda \in \mathbb{R}^\times$, as is seen from the following relation

\begin{equation*} \operatorname{Gr}(\lambda M \cdot \Lambda) = \lambda^2 M^T \operatorname{Gr}(\Lambda) M, \end{equation*}

demonstrating that $\operatorname{sh}_\Lambda$ is indeed well defined.

The description of $\mathcal{H}_2$ is quite simple, since it consists of matrices $\left({\matrix{a&b \cr b&c \cr}}\right)\in {M_2}\left(\mathbb{R}\right)$ with $a \gt 0$ and $ac-b^2=1$. Let $\mathbb{H}$ denote the upper half-plane in $\mathbb{C}$. Let $\operatorname{GL}_2^+(\mathbb{R})$ be the subgroup of $\operatorname{GL}_2(\mathbb{R})$ consisting of invertible matrices with positive determinant. There is a natural action of $\operatorname{GL}_2(\mathbb{R})$ on $\mathbb{H}$ by fractional linear transformations:

\begin{equation*} \left( {\begin{array}{cc} \alpha & \beta \\ \gamma & \delta \\ \end{array} } \right)(z):=\left(\frac{\alpha z+\beta}{\gamma z+\delta}\right).\end{equation*}

This action is transitive and the stabilizer of $i$ consists of matrices in $\operatorname{GO}_2^+(\mathbb{R})$. This gives an identification

\begin{equation*}\operatorname{GL}_2(\mathbb{R})/\operatorname{GO}_2(\mathbb{R})\xrightarrow{\sim}\mathbb{H}\end{equation*}

induced by the map $\left({\matrix{ \alpha&\beta\cr \gamma&\delta\cr }}\right)\mapsto\left({{{\alpha z + \beta }\over{\gamma z + \delta }}}\right)$. Consider the matrix

\begin{equation*}M= \left( {\begin{array}{cc} 1 & x \\ 0 & 1 \\ \end{array} } \right) \left( {\begin{array}{cc} y^{1/2} & 0 \\ 0 & y^{-1/2} \\ \end{array} } \right),\end{equation*}

where $y \gt 0$ and let

\begin{equation*} \left( {\begin{array}{cc} a & b \\ b & c \\ \end{array} } \right)=MM^T= \left( {\begin{array}{cc} (y+y^{-1}x^2) & y^{-1}x \\ y^{-1}x & y^{-1} \\ \end{array} } \right).\end{equation*}

There is a $\operatorname{GL}_2(\mathbb{R})$-equivariant identification $\mathcal{H}_2\xrightarrow{\sim} \mathbb{H}$ sending $ \left({\matrix{ a&b \cr b&c \cr }}\right)$ to $z=x+iy\in \mathbb{H}$, where $x=\frac{b}{c}$ and $y=\sqrt{\frac{a}{c}-x^2}$. A fundamental domain for the action of $\operatorname{GL}_2(\mathbb{Z})$ on $\mathbb{H}$ is then given by

\begin{equation*}\mathcal{F}_2=\{x+iy\mid x\in [0, 1/2), \quad y \gt 0, \quad x^2+y^2\geq 1\}.\end{equation*}

For $r \gt 2$, describing a fundamental domain for the action of $\operatorname{GL}_r(\mathbb{Z})$ on $\mathcal{H}_r$ becomes significantly more intricate. We focus here on the case $r = 3$, and give an explicit description of a fundamental domain for the action of $\operatorname{GL}_3(\mathbb{Z})$ on $\mathcal{H}_3$. Since $\operatorname{SL}_3(\mathbb{Z})$ is a subgroup of index two in $\operatorname{GL}_3(\mathbb{Z})$, with the two cosets distinguished by determinant $\pm 1$, it suffices to describe a fundamental domain for the action of $\operatorname{SL}_3(\mathbb{Z})$ on $\mathcal{H}_3$.

The resulting fundamental domain $\mathcal{F}_3$ can be viewed as a higher-rank generalization of the classical fundamental domain for the $\operatorname{SL}_2(\mathbb{Z})$-action on the upper half-plane. To realize $\mathcal{H}_3$ as an analogue of an upper half-space, we adopt Iwasawa coordinates, expressing each point $Y \in \mathcal{H}_3$ as

\begin{equation*} Y = \begin{pmatrix} 1 & x_1 & x_2 \\ 0 & 1 & x_3 \\ 0 & 0 & 1 \end{pmatrix}^T\begin{pmatrix} y_1 & 0 & 0 \\ 0 & y_2 & 0 \\ 0 & 0 & y_3 \end{pmatrix} \begin{pmatrix} 1 & x_1 & x_2 \\ 0 & 1 & x_3 \\ 0 & 0 & 1 \end{pmatrix}, \end{equation*}

where $y_j \gt 0$, $x_j \in \mathbb{R}$, and $\prod_{j=1}^3 y_j = 1$.

A fundamental domain for the action of $\operatorname{SL}_3(\mathbb{Z})$ on $\mathcal{H}_3$, due to Minkowski, is given by the region

\begin{equation*} \mathcal{M}_3 = \left\{Y = (y_{ij}) \in \mathcal{H}_3 \,\left|\, \begin{aligned} & y_{11} \le y_{22} \le y_{33}, \quad 0 \le y_{12} \le \tfrac{1}{2} y_{11}, \\ & 0 \le y_{23} \le \tfrac{1}{2} y_{22}, \quad |y_{13}| \le \tfrac{1}{2} y_{11}, \\ & e^t Y e \ge y_{33} \quad \text{for all } e = (\pm 1, \pm 1, \pm 1) \end{aligned}\right.\right\}. \end{equation*}

It follows that

(1)\begin{equation} 0 \leqslant x_1 \leqslant \frac{1}{2}, \quad\left|x_2\right| \leqslant \frac{1}{2}, \quad 0 \leqslant y_1 x_1 x_2+y_2 x_3 \leqslant \frac{1}{2}\left(y_1 x_1^2+y_2\right), \end{equation}

and

(2)\begin{equation} -\frac{1}{4} y_1 / y_2 \leqslant x_3 \leqslant \frac{1}{2}+(3 / 8) y_1 / y_2. \end{equation}

Hence, we get

\begin{align*} \mathcal{F}_3 = \{v =(x_1, x_2, x_3, y_1, y_2) \in \mathbb{R}^5 \ \ \mid \ \ v \ \text{satisfies (2.1) and (2.2)}. \end{align*}

Now we discuss the shape of a number field $K$ with $[K:\mathbb{Q}]=n$ and ring of integers $\mathcal{O}_K$. Let $ \{1, \alpha_1, \dots, \alpha_{n-1}\} $ be an integral basis. Let $ \{\sigma_i\}_{1 \leq i \leq n} $ denote the set of embeddings of $ K $ into $ \mathbb{C} $. The Minkowski embedding

\begin{equation*} \begin{aligned} j: K &\to \mathbb{C}^{n} \\ \alpha &\mapsto \big( \sigma_1(\alpha), \dots, \sigma_n(\alpha) \big) \end{aligned} \end{equation*}

identifies $ K $ with an $ n $-dimensional subspace of $ \mathbb{C}^n $. The real span of the image, denoted $ K_{\mathbb{R}} $, inherits a natural inner-product structure, and the restriction of $ j $ to $ \mathcal{O}_{K} $ yields a rank- $ n $ lattice in $ K_{\mathbb{R}} $. Although one might be inclined to define the shape of $ K $ as the shape of this lattice, such a definition presents difficulties when studying the distribution of shapes across a family of number fields. This is because each lattice necessarily contains the vector $ j(1) $. This introduces a nontrivial constraint, disrupting randomness in the distribution of shapes. To circumvent this issue, we instead define the shape of $ K $ in terms of a sublattice obtained by projecting $ j(\mathcal{O}_K) $ onto the orthogonal complement of $ j(1) $. This is accomplished via the map

\begin{equation*} \alpha^{\perp} := n\alpha - \operatorname{tr}(\alpha), \end{equation*}

which sends elements of $ \mathcal{O}_K $ to those of trace zero. The set $ \mathcal{O}_K^{\perp} $, consisting of the images of $ \mathcal{O}_K $ under this map, forms a rank $r:=(n-1)$ submodule of $ \mathcal{O}_K $.

Given a basis $ \{1, \alpha_1, \dots, \alpha_{n-1}\} $ of $ \mathcal{O}_K $, the corresponding basis for $ \mathcal{O}_K^{\perp} $ is given by $\{\alpha_1^{\perp}, \dots, \alpha_{n-1}^{\perp} \} $. As we discussed above, the Gram matrix is given as follows:

(3)\begin{equation} \operatorname{Gr}\left(j\left(\mathcal{O}_{K}^{\perp}\right)\right) = \big( \langle j(\alpha_i^{\perp}), j(\alpha_j^{\perp}) \rangle \big). \end{equation}

3. Integral basis and shapes of pure quartic number fields

Let $m \in \mathbb{Z}$ be fourth-power-free, and let $\beta_m$ be a root of $X^4 - m \in \mathbb{Q}[X]$, with $\arg(\beta_m) = 0$ if $m \gt 0$, and $\arg(\beta_m) = \pi/4$ if $m \lt 0$. The field $K_m:=\mathbb{Q}(\beta_m)$ is called a pure quartic field. Up to replacing $m$ by $m^3$, we assume in this section that $ m = abc^3, $ where:

1. (i) $a \ne 1$, and $a, b, c$ are pairwise coprime, squarefree integers;

2. (ii) $b, c \gt 0$;

3. (iii) $|a| \ge c$ if $a$ is odd;

4. (iv) $c$ is odd;

5. (v) $m \ne -4$.

Condition (v) excludes the reducible case $X^4 + 4$. The assumptions (iii) and (iv) are in place in order to compute an integral basis and thus describe the shape of $K_m$. Since isomorphic number fields have the same shape, we shall in the next section replace (i)–(v) above with the following:

1. (i.’) $a \ne 1$, and $a, b, c$ are pairwise coprime, squarefree integers;

2. (ii.’) $b, c \gt 0$;

3. (iii.’) $|a| \ge c$;

4. (iv.’) $m \ne -4$.

We henceforth assume $m$ is in this normal form, and define

\begin{equation*} \alpha = \frac{\beta_m^2}{bc}, \quad \beta = \beta_m, \quad \gamma = \frac{\beta_m^3}{bc^2}. \end{equation*}

An explicit integral basis for $K_m = \mathbb{Q}(\beta_m)$ was computed by Funakura [Reference Funakura8], as recalled below.

Theorem 3.1 (Funakura)

Given an integer $m$ as above, $\mathcal{B}=\{1,\lambda,\mu,\nu\}$ given below forms an integral basis of $K_m$:

\begin{equation*} \begin{aligned} &\begin{array}{|c|c|c|c|c|} \hline m & 1 & \lambda & \mu & \nu \\ \hline 1(\bmod 8) & 1 & \frac{1+\alpha}{2} & \beta & \frac{a b+\alpha+b \beta+\gamma}{4} \\ \hline \begin{array}{c} 2(\bmod 4) \\ 3(\bmod 4) \end{array} & 1 & \alpha & \beta & \gamma \\ \hline \begin{array}{c} 4(\bmod 16) \\ 5(\bmod 8) \end{array} & 1 & \frac{1+\alpha}{2} & \beta & \frac{\beta+\gamma}{2} \\ \hline 12(\bmod 32) & 1 & \alpha & \frac{1+\alpha+\beta}{2} & \frac{\beta+\gamma}{2} \\ \hline 28(\bmod 32) & 1 & \alpha & \frac{1+\alpha+\beta}{2} & \frac{4 \alpha+b \beta+2 \gamma}{8} .\\ \hline \end{array} \end{aligned} \end{equation*}

We shall denote the integral basis by $\mathcal{B}_\ast$ when $K_m$ is of Type $\ast$. As a corollary to the above result, Funakura computes the discriminant $\Delta_m$ of $K_m$.

Corollary 3.2. (Funakura)

With respect to the notation above, we have that

\begin{equation*}\Delta_m=\begin{cases} -2^2a^3 b^2 c^3 &\text{if }m\equiv 1\mod{8}, \quad 28\mod{32},\\ -2^4a^3 b^2 c^3 &\text{if }m\equiv 4\mod{16},\quad 5\mod{8},\quad 12\mod{32},\\ -2^8a^3 b^2 c^3 &\text{if }m\equiv 2,3\mod{4}. \end{cases}\end{equation*}

Theorem 3.1 provides an explicit integral basis in all cases under consideration, since $m$ is not divisible by 8. For each such case, we now determine the shape of the corresponding lattice $K_m$ by computing the Gram matrix. The table above suggests that the cases $m \equiv 2,\,3 \pmod{4}$ admit particularly simple bases. We begin by computing the Gram matrix in this setting and subsequently use change-of-basis matrices to treat the remaining cases.

If $m \gt 0$, $K_m$ has two real embeddings. Let $\sigma_1$ (resp. $\sigma_2$) be the embedding $K\hookrightarrow \mathbb{C}$ sending $\beta$ to $\beta$ (resp. $-\beta$). Let $\tau$ be the complex embedding mapping $\beta\mapsto i \beta$. On the other hand, when $m \lt 0$, let $\tau_1$ (resp. $\tau_2$) be the embedding defined by $\beta\mapsto i\beta$ (resp. $\beta\mapsto -\beta$). Let $j_1$ and $j_2$ denote the embeddings of $K$ into $\mathbb{C}^4$ corresponding to the cases $m \gt 0$ and $m \lt 0$, respectively. These are defined by

\begin{equation*} j_1(a) = \big(\sigma_1(a), \sigma_2(a), \tau(a), \overline{\tau(a)}\big), \quad \text{and} \quad j_2(a) = \big(\tau_1(a), \tau_2(a), \overline{\tau_2(a)}, \overline{\tau_1(a)}\big). \end{equation*}

Let $\mathcal{P}_{\ast, \pm}$ denote the Gram matrix when $m$ is of Type $\ast$ and sign $\pm$.

Proposition 3.3. The Gram matrix of the basis $ \mathcal{B}_{\rm{II}} = \{1,\alpha,\beta,\gamma\}$ is given by:

\begin{equation*} \mathcal{P}_{\rm{II}, +}= \begin{pmatrix} 4 & 0 & 0 & 0 \\ 0 & 4\alpha^2 & 0 & 0 \\ 0 & 0 & 4\beta^2 & 0 \\ 0 & 0 & 0 & 4\gamma^2 \end{pmatrix},\ \ \mathcal{P}_{\rm{II}, -}= \begin{pmatrix} 4 & 0 & 0 & 0 \\ 0 & 4|\alpha|^2 & 0 & 0 \\ 0 & 0 & 4|\beta|^2 & 0 \\ 0 & 0 & 0 & 4|\gamma|^2 \end{pmatrix}. \end{equation*}

Proof. Let $m \gt 0$. Then we have:

1. (1) $j_1(1) = (1,1,1,1)$,

2. (2) $j_1(\alpha) = (\alpha, \alpha, -\alpha, -\alpha)$,

3. (3) $j_1(\beta) = (\beta, -\beta, i\beta, -i\beta)$,

4. (4) $j_1(\gamma) = (\gamma, -\gamma, -i\gamma, i\gamma)$.

Taking the standard Hermitian inner product on $\mathbb{C}^4$, we compute:

\begin{equation*} \langle j_1(\lambda), j_1(\lambda) \rangle = 4\lambda^2 \end{equation*}

for each $\lambda \in \{1, \alpha, \beta, \gamma\}$, and

\begin{equation*} \langle j_1(\lambda_1), j_1(\lambda_2) \rangle = 0 \end{equation*}

for all distinct $\lambda_1, \lambda_2 \in \{1, \alpha, \beta, \gamma\}$. That is, the vectors $j_1(1), j_1(\alpha), j_1(\beta), j_1(\gamma)$ form an orthogonal set, and the corresponding Gram matrix is diagonal with entries $4\lambda^2$ on the diagonal, as claimed.

Now suppose $m \lt 0$. Then:

1. (1) $j_2(1) = (1, 1, 1, 1)$,

2. (2) $j_2(\beta) = (\beta, i\beta, -i\bar{\beta}, \bar{\beta})$,

3. (3) $j_2(\alpha) = \left( \frac{\beta^2}{bc}, -\frac{\beta^2}{bc}, -\frac{\bar{\beta}^2}{bc}, \frac{\bar{\beta}^2}{bc} \right)$,

4. (4) $j_2(\gamma) = \left( \frac{\beta^3}{bc^2}, i\frac{\beta^3}{bc^2}, -i\frac{\bar{\beta}^3}{bc^2}, \frac{\bar{\beta}^3}{bc^2} \right)$.

It is easy to see that the vectors $j_2(\alpha), j_2(\beta), j_2(\gamma)$ are mutually orthogonal with respect to the standard Hermitian inner product. The squared norm of $j_2(\alpha)$ is:

\begin{equation*} \begin{aligned} \langle j_2(\alpha), j_2(\alpha) \rangle &= \left\langle \left( \frac{\beta^2}{bc}, -\frac{\beta^2}{bc}, -\frac{\bar{\beta}^2}{bc}, \frac{\bar{\beta}^2}{bc} \right), \left( \frac{\beta^2}{bc}, -\frac{\beta^2}{bc}, -\frac{\bar{\beta}^2}{bc}, \frac{\bar{\beta}^2}{bc} \right) \right\rangle \\ &= \frac{1}{(bc)^2} \left( |\beta^2|^2 + |\beta^2|^2 + |\bar{\beta}^2|^2 + |\bar{\beta}^2|^2 \right) = 4 \left( \frac{|\beta|^2}{bc} \right)^2 = 4|\alpha|^2. \end{aligned} \end{equation*}

A similar computation gives:

\begin{equation*} \langle j_2(\beta), j_2(\beta) \rangle = 4|\beta|^2, \quad \langle j_2(\gamma), j_2(\gamma) \rangle = 4|\gamma|^2. \end{equation*}

Hence, the Gram matrix of the set $\{j_2(1), j_2(\alpha), j_2(\beta), j_2(\gamma)\}$ is diagonal with entries $4|\lambda|^2$, and the vectors are pairwise orthogonal as desired.

Proposition 3.4. The Gram matrix of the basis $ \mathcal{B}_{\rm{I}} = \{1, \frac{1 +\alpha}{2}, \beta, \frac{ab + \alpha + b \beta +\gamma}{4} \}$ is given by:

\begin{align*} \mathcal{P}_{\rm{I}, +} & = \begin{pmatrix} 4 & 2 & 0 & ab \\ 2 & 1 + \alpha^2 & 0 & \frac{ab + \alpha^2}{2} \\ 0 & 0 & 4\beta^2 & b\beta^2 \\ ab & \frac{ab + \alpha^2}{2} & b\beta^2 & \frac{a b^2 + b \alpha^2 + b^2 \beta^2 + \gamma^2}{4} \end{pmatrix},\\ \mathcal{P}_{\rm{I}, -} & = \begin{pmatrix} 4 & 2 & 0 & ab \\ 2 & 1 + |\alpha|^2 & 0 & \frac{ab + |\alpha|^2}{2} \\ 0 & 0 & 4|\beta|^2 & b|\beta|^2 \\ ab & \frac{ab + |\alpha|^2}{2} & b|\beta|^2 & \frac{a b^2 + b |\alpha|^2 + b^2 |\beta|^2 + |\gamma|^2}{4} \end{pmatrix}. \end{align*}

Proof. The change of basis matrix from $\mathcal{B}_{\rm{II}}$ to $\mathcal{B}_{\rm{I}}$ is given as

\begin{equation*} C_{\rm{II}}:= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 1/2 & 1/2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ ab/4 & 1/4 & b/4 & 1/4 \end{pmatrix}. \end{equation*}

Therefore,

\begin{align*} \mathcal{P}_{\rm{I}, +} & = C_{\rm{II}} \, \mathcal{P}_{\rm{II}, +} \, C_{\rm{II}}^T\\ & = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 1/2 & 1/2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ ab/4 & 1/4 & b/4 & 1/4 \end{pmatrix} \begin{pmatrix} 4 & 0 & 0 & 0 \\ 0 & 4\alpha^2 & 0 & 0 \\ 0 & 0 & 4\beta^2 & 0 \\ 0 & 0 & 0 & 4\gamma^2 \end{pmatrix} \begin{pmatrix} 1 & 1/2 & 0 & ab/4 \\ 0 & 1/2 & 0 & 1/4 \\ 0 & 0 & 1 & b/4 \\ 0 & 0 &0 & 1/4 \end{pmatrix} \\ & = \begin{pmatrix} 4 & 2 & 0 & ab \\ 2 & 1 + \alpha^2 & 0 & \frac{ab + \alpha^2}{2} \\ 0 & 0 & 4\beta^2 & b\beta^2 \\ ab & \frac{ab + \alpha^2}{2} & b\beta^2 & \frac{a b^2 + b \alpha^2 + b^2 \beta^2 + \gamma^2}{4} \end{pmatrix}. \end{align*}

Similarly,

\begin{align*} \mathcal{P}_{\rm{I}, -} = C_{\rm{II}} \, \mathcal{P}_{\rm{II}, -} \, C_{\rm{II}}^T &= \begin{pmatrix} 4 & 2 & 0 & ab \\ 2 & 1 + |\alpha|^2 & 0 & \frac{ab + |\alpha|^2}{2} \\ 0 & 0 & 4|\beta|^2 & b|\beta|^2 \\ ab & \frac{ab + |\alpha|^2}{2} & b|\beta|^2 & \frac{a b^2 + b |\alpha|^2 + b^2 |\beta|^2 + |\gamma|^2}{4} \end{pmatrix}. \end{align*}

Proposition 3.5. The Gram matrix of the basis $\mathcal{B}_{\rm{III}} = \{1,\frac{1 + \alpha}{2},\beta,\frac{\beta + \gamma}{2}\}$ is given by:

\begin{equation*} \mathcal{P}_{\rm{III}, +}= \begin{pmatrix} 4 & 2 & 0 & 0 \\ 2 & 1 + \alpha^2 & 0 & 0 \\ 0 & 0 & 4\beta^2 & 2\beta^2 \\ 0 & 0 & 2\beta^2 & \beta^2 + \gamma^2 \end{pmatrix}, \ \ \mathcal{P}_{\rm{III}, -} = \begin{pmatrix} 4 & 2 & 0 & 0 \\ 2 & 1 + |\alpha|^2 & 0 & 0 \\ 0 & 0 & 4|\beta|^2 & 2|\beta|^2 \\ 0 & 0 & 2|\beta|^2 & |\beta|^2 + |\gamma|^2 \end{pmatrix} \end{equation*}

Proof. The change of basis matrix from $\mathcal{B}_{\rm{II}}$ to $\mathcal{B}_{\rm{III}}$ is given as

\begin{equation*} C_{\rm{III}}:= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 1/2 & 1/2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 1/2 & 1/2 \end{pmatrix}. \end{equation*}

Therefore,

\begin{align*} \mathcal{P}_{\rm{III},+} & = C_{\rm{III}} \, \mathcal{P}_{\rm{II},+} \, C_{\rm{III}}^T\\ & = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 1/2 & 1/2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 1/2 & 1/2 \end{pmatrix} \begin{pmatrix} 4 & 0 & 0 & 0 \\ 0 & 4\alpha^2 & 0 & 0 \\ 0 & 0 & 4\beta^2 & 0 \\ 0 & 0 & 0 & 4\gamma^2 \end{pmatrix} \begin{pmatrix} 1 & 1/2 & 0 & 0 \\ 0 & 1/2 & 0 & 0 \\ 0 & 0 & 1 & 1/2 \\ 0 & 0 &0 & 1/2 \end{pmatrix} \\ & = \begin{pmatrix} 4 & 2 & 0 & 0 \\ 2 & 1 + \alpha^2 & 0 & 0 \\ 0 & 0 & 4\beta^2 & 2\beta^2 \\ 0 & 0 & 2\beta^2 & \beta^2 + \gamma^2 \end{pmatrix}. \end{align*}

A similar relation holds when $m$ is negative.

Proposition 3.6. The Gram matrix of the basis $\mathcal{B}_{\rm{IV}} = \{1,\alpha,\frac{1 + \alpha +\beta}{2}, \frac{\beta + \gamma}{2} \}$ is given by:

\begin{equation*} \mathcal{P}_{\rm{IV}, +}= \begin{pmatrix} 4 & 0 & 2 & 0 \\ 0 & 4\alpha^2 & 2\alpha^2 & 0 \\ 2 & 2\alpha^2 & 1 + \alpha^2 + \beta^2 & \beta^2 \\ 0 & 0 & \beta^2 & \gamma^2 \end{pmatrix}, \ \ \mathcal{P}_{\rm{IV}, -}= \begin{pmatrix} 4 & 0 & 2 & 0 \\ 0 & 4|\alpha|^2 & 2|\alpha|^2 & 0 \\ 2 & 2|\alpha|^2 & 1 + |\alpha|^2 + |\beta|^2 & |\beta|^2 \\ 0 & 0 & |\beta|^2 & |\gamma|^2 \end{pmatrix}. \end{equation*}

Proof. The change of basis matrix from $\mathcal{B}_{\rm{II}}$ to $\mathcal{B}_{\rm{IV}}$ is given as

\begin{equation*} C_{\rm{IV}}:= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 1/2 & 1/2 & 1/2 & 0 \\ 0 & 0 & 1/2 & 1/2 \end{pmatrix}. \end{equation*}

Therefore,

\begin{align*} \mathcal{P}_{\rm{IV}, +} & = C_{\rm{IV}} \, \mathcal{P}_{\rm{II}, +} \, C_{\rm{IV}}^T\\ & = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 1/2 & 1/2 & 1/2 & 0 \\ 0 & 0 & 1/2 & 1/2 \end{pmatrix} \begin{pmatrix} 4 & 0 & 0 & 0 \\ 0 & 4\alpha^2 & 0 & 0 \\ 0 & 0 & 4\beta^2 & 0 \\ 0 & 0 & 0 & 4\gamma^2 \end{pmatrix} \begin{pmatrix} 1 & 0 & 1/2 & 0 \\ 0 & 1 & 1/2 & 0 \\ 0 & 0 & 1/2 & 1/2 \\ 0 & 0 & 0 & 1/2 \end{pmatrix} \\ & = \begin{pmatrix} 4 & 0 & 2 & 0 \\ 0 & 4\alpha^2 & 2\alpha^2 & 0 \\ 2 & 2\alpha^2 & 1 + \alpha^2 + \beta^2 & \beta^2 \\ 0 & 0 & \beta^2 & \gamma^2 \end{pmatrix}. \end{align*}

A similar relation holds when $m$ is negative.

Proposition 3.7. The Gram matrix of the basis $\mathcal{B}_{\rm{V}} = \{1,\alpha,\frac{1 + \alpha + \beta}{2},\frac{4\alpha + b \beta + 2 \gamma}{8}\}$ is given by:

\begin{equation*} \mathcal{P}_{\rm{V}, +}= \begin{pmatrix} 4 & 0 & 2 & 0 \\ 0 & 4\alpha^2 & 2\alpha^2 & 2\alpha^2 \\ 2 & 2\alpha^2 & 1 + \alpha^2 + \beta^2 & \alpha^2 + \frac{b}{4}\beta^2 \\ 0 & 2\alpha^2 & \alpha^2 + \frac{b}{4}\beta^2 & \alpha^2 + \frac{b^2}{64}\beta^2 + \frac{1}{4}\gamma^2 \end{pmatrix} \end{equation*}

and

\begin{equation*} \mathcal{P}_{\rm{V}, -}= \begin{pmatrix} 4 & 0 & 2 & 0 \\ 0 & 4|\alpha|^2 & 2|\alpha|^2 & 2|\alpha|^2 \\ 2 & 2|\alpha|^2 & 1 + |\alpha|^2 + |\beta|^2 & |\alpha|^2 + \frac{b}{4}|\beta|^2 \\ 0 & 2|\alpha|^2 & |\alpha|^2 + \frac{b}{4}|\beta|^2 & |\alpha|^2 + \frac{b^2}{64}|\beta|^2 + \frac{1}{4}|\gamma|^2 \end{pmatrix}. \end{equation*}

Proof. The change of basis matrix from $\mathcal{B}_{\rm{II}}$ to $\mathcal{B}_{\rm{V}}$ is given as

\begin{equation*} C_{\rm{V}}:= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 1/2 & 1/2 & 1/2 & 0 \\ 0 & 1/2 & b/8 & 1/4 \end{pmatrix}. \end{equation*}

Therefore,

\begin{align*} \mathcal{P}_{\rm{V}, +} & = C_{\rm{V}} \, \mathcal{P}_{\rm{II}, +} \, C_{\rm{V}}^T \\ & = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 1/2 & 1/2 & 1/2 & 0 \\ 0 & 1/2 & b/8 & 1/4 \end{pmatrix} \begin{pmatrix} 4 & 0 & 0 & 0 \\ 0 & 4\alpha^2 & 0 & 0 \\ 0 & 0 & 4\beta^2 & 0 \\ 0 & 0 & 0 & 4\gamma^2 \end{pmatrix} \begin{pmatrix} 1 & 0 & 1/2 & 0 \\ 0 & 1 & 1/2 & 1/2 \\ 0 & 0 & 1/2 & b/8 \\ 0 & 0 & 0 & 1/4 \end{pmatrix} \\ & = \begin{pmatrix} 4 & 0 & 2 & 0 \\ 0 & 4\alpha^2 & 2\alpha^2 & 2\alpha^2 \\ 2 & 2\alpha^2 & 1 + \alpha^2 + \beta^2 & \alpha^2 + \frac{b}{4}\beta^2 \\ 0 & 2\alpha^2 & \alpha^2 + \frac{b}{4}\beta^2 & \alpha^2 + \frac{b^2}{64}\beta^2 + \frac{1}{4}\gamma^2 \end{pmatrix}. \end{align*}

A similar relation holds when $m$ is negative.

We summarize the results above.

Proposition 3.8. When $K_m$ is of Type $\ast$ and $m$ has sign $\pm$, there is an invertible matrix $C_\ast$ such that

\begin{equation*}\mathcal{P}_{\ast, \pm}=C_\ast\mathcal{P}_{\rm{II}, \pm}C_\ast^T.\end{equation*}

Here, the matrix $C_\ast$ is given by:

\begin{equation*} \begin{split} & C_{\rm{I}}= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 1/2 & 1/2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ ab/4 & 1/4 & b/4 & 1/4 \end{pmatrix},\\ & C_{\rm{II}}= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix},\\ & C_{\rm{III}}=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 1/2 & 1/2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 1/2 & 1/2 \end{pmatrix}, \\ & C_{\rm{IV}}=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 1/2 & 1/2 & 1/2 & 0 \\ 0 & 0 & 1/2 & 1/2 \end{pmatrix},\\ & C_{\rm{V}}= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 1/2 & 1/2 & 1/2 & 0 \\ 0 & 1/2 & b/8 & 1/4 \end{pmatrix}. \end{split} \end{equation*}

Proof of Theorem 1.3

By Proposition 3.8, the shape of $K_m$ lies in one of five distinct loci, determined by the Type of $m$. These sets are disjoint since distinct torus orbits in the space of shapes are disjoint. Thus if

\begin{equation*} \operatorname{sh}(K_{m_1})=\operatorname{sh}(K_{m_2}), \end{equation*}

then $m_1$ and $m_2$ must have the same type. Write

\begin{equation*} m_i = a_i b_i^2 c_i^3, \qquad i=1,2, \end{equation*}

where $a_i,b_i,c_i$ are pairwise coprime squarefree integers with $b_i \gt 0$ and $c_i \gt 0$. Equality of shapes implies

\begin{equation*} \lambda_1(m_1)=\lambda_1(m_2), \qquad \lambda_2(m_1)=\lambda_2(m_2), \end{equation*}

and therefore

\begin{equation*} \frac{c_1}{|a_1|}=\frac{c_2}{|a_2|}, \qquad b_1=b_2. \end{equation*}

Thus $a_1 c_2 = \pm a_2 c_1$. Since $(a_i,c_i)=1$ for $i=1,2$, it follows that

\begin{equation*} a_1=\pm a_2,\qquad b_1=b_2,\qquad c_1=c_2. \end{equation*}

If $a_1=a_2$, then $m_1=m_2$ and we are done. Suppose instead that $a_1=-a_2$, so that $m_1=-m_2$. Because $m_1$ and $m_2=-m_1$ have the same Type (as their shapes coincide), the sign change does not move $m_1$ out of its Type. By inspection, this occurs only when $m_1 \equiv 2 \pmod{4}$. This proves the theorem.

4. Distribution results

For the sake of discussion, assume that $m \gt 0$ and $m\equiv 2, 3\pmod{4}$. The shape $\operatorname{sh}_{m}$ of $j(\mathcal{O}_{K_m})^\perp\in \mathcal{S}_3$, is given by

\begin{equation*}\operatorname{sh}_m= \left( {\begin{array}{ccc} \alpha^2 & & \\ & \beta^2 & \\ & & \gamma^2 \\ \end{array} } \right)= \left( {\begin{array}{ccc} \beta^4/(bc)^2 & & \\ & \beta^2 & \\ & & \beta^6/(bc^2)^2 \\ \end{array} } \right).\end{equation*}

After permuting the variables, we have that:

\begin{equation*}\operatorname{sh}_m= \left( {\begin{array}{ccc} \gamma_1^2 & & \\ & \gamma_2^2 & \\ & & \gamma_3^2 \\ \end{array} } \right)\end{equation*}

where $\gamma_1:=\beta$, $\gamma_2:=\beta^2/bc$ and $\gamma_3:=\beta^3/bc^2$. We scale this matrix by dividing by $(abc)$. Let $\lambda_i=\gamma_i^2/(abc)$, we find that $\operatorname{sh}_m=\operatorname{diag}\left(\lambda_1, \lambda_2, \lambda_3\right)$,

\begin{equation*} \begin{split} & \lambda_1^2= \frac{\beta^4}{(abc)^2}=\frac{ab^2c^3}{a^2b^2c^2}=\frac{c}{a},\\ & \lambda_2= \frac{\beta^4}{(bc)^2(abc)}=\frac{ab^2c^3}{ab^3c^3}=\frac{1}{b},\\ & \lambda_3^2=\frac{\beta^{12}}{(bc^2)^4(abc)^2}=\frac{a^3b^6c^9}{a^2b^{6}c^{10}}=\frac{a}{c}. \end{split} \end{equation*}

Since $\beta$ is real and $a,b,c \gt 0$, we find that $\lambda_i \gt 0$ for $i=1, 2, 3$, and therefore,

\begin{equation*}\lambda_1=\sqrt{\frac{c}{a}},\quad \lambda_2=\frac{1}{b}\quad \text{and}\quad \lambda_3=\sqrt{\frac{a}{c}}.\end{equation*}

We find that

\begin{equation*}\operatorname{sh}_m= \left( {\begin{array}{ccc} \sqrt{\frac{c}{a}} & & \\ & \frac{1}{b} & \\ & & \sqrt{\frac{a}{c}} \\ \end{array} } \right),\end{equation*}

we refer to the pair

\begin{equation*}\lambda(m)=(\lambda_1(m), \lambda_2(m))=\left(\sqrt{\frac{c}{a}}, \frac{1}{b}\right)\end{equation*}

as the shape parameter. In the more general case, suppose that $m$ is of Type $\ast$. Assume that the conditions (i’)–(iv’) are satisfied and set

\begin{equation*}\lambda(m)=(\lambda_1(m), \lambda_2(m))=\left(\sqrt{\frac{c}{|a|}}, \frac{1}{b}\right).\end{equation*}

The shape of $K_m$ is then given by the projection of the matrix

\begin{equation*} C_\ast \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & \sqrt{c/|a|} & 0 & 0 \\ 0 & 0 & 1/b & 0 \\ 0 & 0 & 0 & \sqrt{|a|/c} \end{pmatrix} C_\ast^T \end{equation*}

to the orthogonal complement of the vector $(1, 1, 1, 1)$. Let $\mathcal{R} \subset \mathbb{R}^2$ be a closed and bounded region whose boundary is rectifiable and has total length $L$. The result below is often attributed to Lipschitz, with an early reference appearing in the work of Bachmann [Reference Bachmann1], who credits Lipschitz as the source (see also [Reference Davenport6]).

Lemma 4.1. (Lipschitz Principle)

Let $\mathcal{R} \subset \mathbb{R}^2$ be as above. Then the number of integer lattice points contained in $\mathcal{R}$ satisfies

\begin{equation*} \left| \#(\mathcal{R} \cap \mathbb{Z}^2) - \operatorname{Area}(\mathcal{R}) \right| \leq 4(L + 1). \end{equation*}

Consider the torus orbit $\mathcal{T} \subset \mathcal{H}_3$ consisting of diagonal matrices of the form

\begin{equation*} D(x_1, x_2) = x_2^{1/3} \begin{pmatrix} x_1^{-1} & & \\ & x_2^{-1} & \\ & & x_1 \end{pmatrix}. \end{equation*}

Define

\begin{equation*} \psi(x) := \sum_{n = 1}^{\lfloor x \rfloor} \alpha(n), \quad \text{where} \quad \alpha(n) := \begin{cases} n^{-2/3} \displaystyle\prod_{\ell \mid n} \left( \dfrac{\ell - 1}{\ell + 1} \right) & \text{if } n \text{ is squarefree}, \\ 0 & \text{otherwise}, \end{cases} \end{equation*}

and the product is over prime divisors $\ell$ of $n$. Let $\hat{\mu}$ denote the measure $\frac{\alpha(\lfloor x_2 \rfloor)}{x_1}\, dx_1\, dx_2$ on $\mathcal{T}$.

Given a pair $(R_1, R_2)$, define the region

\begin{equation*} W(R_1, R_2) := \left\{D(x_1, x_2) \in \mathcal{T} \;\left|\; x_1^2 \in [1, R_1], \; x_2 \in [1, R_2] \right.\right\}. \end{equation*}

Then we find that

\begin{equation*} \hat{\mu}\left(W(R_1, R_2)\right) = \int_1^{R_2} \int_1^{\sqrt{R_1}} \frac{\alpha(\lfloor x_2 \rfloor)}{x_1} \, dx_1 \, dx_2 = \frac{1}{2} \log R_1 \cdot \psi(R_2). \end{equation*}

Let $\mathcal{R}(N, R_1, R_2)$ be defined as follows:

\begin{equation*}\mathcal{R}(N, R_1, R_2):=\{(a,b,c)\in \mathbb{R}^3_{\geq 1}\mid ab^{2/3}c \lt N, \quad \frac{a}{c}\in [1,R_1]\quad \text{and }\quad b\in [1, R_2]\}.\end{equation*}

Let $\mathcal{R}_{\mathbb{Z}^3}(N, R_1, R_2) := \mathcal{R}(N, R_1, R_2) \cap \mathbb{Z}^3$. For $M, R \geq 1$, define

\begin{equation*} \mathcal{S}(M, R) := \left\{(a, c) \in \mathbb{R}_{\geq 1}^2 \,\left|\, ac \lt M,\; \frac{a}{c} \in [1, R] \right.\right\}, \end{equation*}

and let $\mathcal{S}_{\mathbb{Z}^2}(M, R) := \mathcal{S}(M, R) \cap \mathbb{Z}^2$. Then we have the identity

\begin{equation*} \mathcal{R}_{\mathbb{Z}^3}(N, R_1, R_2) = \bigsqcup_{b=1}^{\lfloor R_2\rfloor} \mathcal{S}_{\mathbb{Z}^2}\left(\frac{N}{b^{2/3}}, R_1\right), \end{equation*}

where the union is disjoint.

Lemma 4.2. The area of $\mathcal{S}(M, R)$ is given by:

\begin{equation*} \operatorname{Area}\left(\mathcal{S}(M, R)\right)=\begin{cases} \frac{M}{2}\log R -\frac{(R-1)}{2} &\text{if }M\geq R;\\ \frac{M}{2}\log M-\frac{(M-1)}{2} & \text{if }M \lt R.\\ \end{cases}\end{equation*}

Proof. Note that for any fixed $c \in [1, \sqrt{M}]$, the condition $ac \lt M$ and $\frac{a}{c} \in [1, R]$ together imply that

\begin{equation*}a \in \left[c, \min(Rc, M/c)\right].\end{equation*}

We begin by analyzing the case $M \geq R$. Accordingly, we split the region of integration into two parts:

\begin{equation*} \begin{aligned} \operatorname{Area}(\mathcal{S}(M, R)) &= \int_{1}^{\sqrt{M/R}} \int_{c}^{Rc} \, da\, dc + \int_{\sqrt{M/R}}^{\sqrt{M}} \int_{c}^{M/c} \, da\, dc \\ &= \int_{1}^{\sqrt{M/R}} (R - 1)c \, dc + \int_{\sqrt{M/R}}^{\sqrt{M}} \left( \frac{M}{c} - c \right) dc \\ &= \frac{R - 1}{2} \left(\frac{M}{R}\right) - \frac{1}{2}\left(M - \frac{M}{R}\right) + \frac{M}{2} \log R \\ &= \frac{M}{2} \log R - \frac{R - 1}{2}. \end{aligned} \end{equation*}

Next, suppose $M \lt R$. In this case, for all $c \in [1, \sqrt{M}]$, we have $M/c \leq Rc$, so the upper bound of the inner integral is simply $M/c$. Thus,

\begin{equation*} \begin{aligned} \operatorname{Area}(\mathcal{S}(M, R)) &= \int_{1}^{\sqrt{M}} \int_{c}^{M/c} \, da\, dc = \int_{1}^{\sqrt{M}} \left( \frac{M}{c} - c \right) dc \\ &= M \log \sqrt{M} - \frac{1}{2}(M - 1) = \frac{M}{2} \log M - \frac{M - 1}{2}. \end{aligned} \end{equation*}

Lemma 4.3. Let $L=L(M, R)$ be the length of the boundary of $\mathcal{S}(M, R)$. Then, we have that

\begin{equation*}L\leq \left(\sqrt{2}+\sqrt{1+R^2}\right)(\sqrt{M}-1)+(R-1).\end{equation*}

Proof. Assume that $M \geq R$. Then the total boundary length is given by $ L = L_1 + L_2 + L_3 + L_4$, where the individual contributions are as follows:

• • $L_1=\operatorname{length}\{(c, c) \mid c \in [1, \sqrt{M}]\} = \sqrt{2}(\sqrt{M} - 1)$,

• • $ L_2 = \text{length}\{(c, Rc) \mid c \in [1, \sqrt{M/R}]\} = \sqrt{1 + R^2}(\sqrt{M/R} - 1) $,

• • $ L_3 = \operatorname{length} \{(c, M/c) \mid c \in [\sqrt{M/R}, \sqrt{M}]\} = \int_{\sqrt{M/R}}^{\sqrt{M}} \sqrt{1 + \left( \frac{M}{c^2} \right)^2} \, dc\leq \sqrt{1+R^2}\left(\sqrt{M}-\sqrt{M/R}\right) $,

• • $L_4=R-1$ is the length of the vertical segment from $(1,1)$ to $(1,R)$.

Thus, the total length is given by

\begin{equation*}L\leq \left(\sqrt{2}+\sqrt{1+R^2}\right)(\sqrt{M}-1)+(R-1).\end{equation*}

Similar analysis applies in the case when $M \lt R$ to give the result.

Proposition 4.4. We have that

\begin{equation*}\# \mathcal{R}_{\mathbb{Z}^3}(N, R_1, R_2)= \frac{N}{2}\log R_1\left(\sum_{n=1}^{\lfloor R_2\rfloor}n^{-2/3}\right)+O_{R_1, R_2}(\sqrt{N}). \end{equation*}

Proof. From Lemma 4.1 and Lemmas 4.2 and 4.3, it follows that

\begin{equation*}\left|\# \left(\mathcal{S}(M, R)\cap \mathbb{Z}^2\right)-\frac{M}{2}\log R\right|\leq 4(L+1)\leq C R \sqrt{M},\end{equation*}

where $C \gt 0$ is an absolute constant. Noting that

\begin{equation*}\# \mathcal{R}_{\mathbb{Z}^3}(N, R_1, R_2)=\sum_{b=1}^{\lfloor R_2\rfloor} \# \mathcal{S}_{\mathbb{Z}^2}(N/b^{2/3}, R_1),\end{equation*}

we deduce that

\begin{equation*}\left|\# \mathcal{R}_{\mathbb{Z}^3}(N, R_1, R_2)-\frac{N}{2}\left(\sum_{b=1}^{\lfloor R_2\rfloor}b^{-2/3}\right) \log R_1 \right|\leq C R_1\sqrt{R_2} \sqrt{N}.\end{equation*}

Definition 4.6. Let $\ell$ be a prime number. A triple $(a,b,c)\in \mathbb{Z}_{ \gt 0}^3$ is said to be $\ell$-carefree if

\begin{equation*} \ell^2 \nmid ab,\quad \ell^2 \nmid bc,\quad \ell^2 \nmid ca. \end{equation*}

The triple $(a,b,c)$ is said to be carefree if each of the products $ab$, $bc$, and $ca$ is squarefree. Equivalently, $(a,b,c)$ is carefree if and only if it is $\ell$-carefree for all primes $\ell$.

Given a positive integer $n$, we say that a triple $(a,b,c) \in \mathbb{Z}_{ \gt 0}^3$ is $n$-carefree if it is $\ell$-carefree for all primes $\ell$ dividing $n$. Fix a residue class $\tau$ modulo $32$ such that $8\nmid \tau$. Let $\mathcal{C}^\tau(N, R_1, R_2)$ (respectively, $\mathcal{C}_n^\tau(N, R_1, R_2)$) denote the set of triples $(a,b,c) \in \mathcal{R}_{\mathbb{Z}^3}(N, R_1, R_2)$ that are carefree (respectively, $n$-carefree) and $ab^2c^3\equiv \tau\pmod{32}$. Then

\begin{equation*} \mathcal{C}^\tau(N, R_1, R_2) = \bigcap_{\ell} \mathcal{C}_\ell^\tau(N, R_1, R_2), \end{equation*}

where the intersection is taken over all prime numbers $\ell$.

Let $n \gt 0$ be squarefree and odd. Define $\Omega_n$ to be the set of residue classes $(\bar{a}, \bar{b}, \bar{c}) \in (\mathbb{Z}/32n^2\mathbb{Z})^3$ such that for each prime $\ell$ dividing $n$, one has that $\ell^2 \nmid \bar{a} \bar{b},\quad \ell^2 \nmid \bar{b} \bar{c},\quad \ell^2 \nmid \bar{c} \bar{a} $, and moreover, $\bar{a}\bar{b}^2\bar{c}^3\equiv \tau\pmod{32}$. Given an odd prime $\ell$, let $\mathcal{A}_\ell$ denote $(\bar{a}, \bar{b}, \bar{c}) \in (\mathbb{Z}/\ell^2\mathbb{Z})^3$ such that $\ell^2 \nmid \bar{a} \bar{b},\quad \ell^2 \nmid \bar{b} \bar{c},\quad \ell^2 \nmid \bar{c} \bar{a} $. Denote by $\mathcal{A}_2^\tau$ the set of $(\bar{a}, \bar{b}, \bar{c}) \in (\mathbb{Z}/32\mathbb{Z})^3$ such that $4\nmid \bar{a}$ and $\bar{a}\bar{b}^2\bar{c}^3\equiv \tau\pmod{32}$. Clearly, by the Chinese Remainder Theorem,

\begin{equation*} \Omega_n = \mathcal{A}_2^\tau\times \prod_{\ell \mid n} \mathcal{A}_\ell, \end{equation*}

where $\ell$ ranges over prime divisors of $n$. Define

\begin{equation*} \mathfrak{d}_\ell := \frac{\# \mathcal{A}_\ell}{\# (\mathbb{Z}/\ell^2\mathbb{Z})^3} = \frac{\# \mathcal{A}_\ell}{\ell^6}, \end{equation*}

if $\ell$ is an odd prime. On the other hand, set

\begin{equation*}\mathfrak{d}_2^\tau := \frac{\# \mathcal{A}_2^\tau}{\# (\mathbb{Z}/32\mathbb{Z})^3} = \frac{\# \mathcal{A}_2^\tau}{32^3}.\end{equation*}

For each squarefree natural number $b$, define $n_\tau(b)$ to be the number of pairs $(\bar{a}, \bar{c})\in (\mathbb{Z}/32\mathbb{Z})^2$ such that $\tau=\bar{a}\bar{b}^2\bar{c}^3$.

Lemma 4.7. For any odd prime $\ell$, we have

\begin{equation*} \# \mathcal{A}_\ell = (\ell^2-\ell)^2\bigl(\ell^2+2\ell-3\bigr), \qquad \text{and hence} \qquad \mathfrak{d}_\ell = \left(1-\frac{1}{\ell}\right)^2 \left(1+\frac{2}{\ell}-\frac{3}{\ell^2}\right). \end{equation*}

Proof. The number of tuples $(\bar{a},\bar{b},\bar{c})\in(\mathbb{Z}/\ell^2\mathbb{Z})^3$ such that $\ell\nmid \bar{a}\bar{b}\bar{c}$ is $(\ell^2-\ell)^3$, since each coordinate must be a unit modulo  $\ell$. The number of triples for which exactly one coordinate is divisible by  $\ell$ is

\begin{equation*} 3\cdot \ell(\ell-1)\,(\ell^2-\ell)^2, \end{equation*}

where the factor $\ell(\ell-1)$ counts elements divisible by $\ell$ but not by $\ell^2$ in $\mathbb{Z}/\ell^2\mathbb{Z}$, and the factor $3$ accounts for the choice of the coordinate.

In total, the number of triples $(\bar{a},\bar{b},\bar{c})$ for which at most one coordinate is divisible by  $\ell$ is

\begin{equation*} (\ell^2-\ell)^3 \;+\; 3\ell(\ell-1)(\ell^2-\ell)^2 = (\ell^2-\ell)^2\bigl(\ell^2+2\ell-3\bigr). \end{equation*}

Dividing by $\ell^6$, we obtain

\begin{equation*} \mathfrak{d}_\ell=\frac{(\ell^2-\ell)^2(\ell^2+2\ell-3)}{\ell^6} = \left(1-\frac{1}{\ell}\right)^2 \left(1+\frac{2}{\ell}-\frac{3}{\ell^2}\right). \end{equation*}

Proposition 4.8. Let $n$ be an odd squarefree natural number which is divisible by all primes $\ell\leq R_2$ and $\tau$ be a residue class mod $32$ such that $8\nmid \tau$. Then for $N \gt n^4 R_1 R_2$,

\begin{align*} \# \mathcal{C}_n^\tau (N, R_1, R_2)& =\frac{N}{2\times 32^2}\prod_{\ell|n} \left(1-\frac{3}{\ell^2}+\frac{2}{\ell^3}\right)\log R_1\\ & \quad\times\sum_{\substack{b=1,\\ b\,\square\text{-free}}}^{\lfloor R_2\rfloor} b^{-2/3}n_\tau(b)\prod_{\ell|b}\left(\frac{\ell}{\ell+2}\right)+O_{n, R_1, R_2}(\sqrt{N}),\end{align*}

where $O_{n,R_1, R_2}(\cdot)$ indicates that the constant in the error bound depends on $R_1$, $R_2$ and $n$.

Proof. Let $\Lambda_n$ be the lattice $(32 n^2 \mathbb{Z})^3\subset \mathbb{Z}^3$. Note that $\Omega_n\subset \mathbb{Z}^3/\Lambda_n$ choose a set of representatives $\widetilde{\Omega}_n\subset ([0,32 n^2)\cap \mathbb{Z})^3$. We find that

\begin{equation*} \begin{split}\mathcal{C}_n^\tau(N, R_1, R_2)=&\bigsqcup_{z\in \widetilde{\Omega}_n} (z+\Lambda_n)\cap \mathcal{R}(N, R_1, R_2),\\ =&\bigsqcup_{z\in \widetilde{\Omega}_n} \Lambda_n \cap \left(\mathcal{R}(N, R_1, R_2)-z\right). \end{split} \end{equation*}

Observe that

\begin{equation*}\# \left(\Lambda_n \cap \left(\mathcal{R}(N, R_1, R_2)-z\right)\right)=\# \left(\mathbb{Z}^3 \cap \left(\frac{1}{32 n^2}\cdot \mathcal{R}(N, R_1, R_2)-\frac{z}{32 n^2}\right)\right)\end{equation*}

and we estimate

\begin{align*} & \# \left(\mathbb{Z}^3 \cap \left(\frac{1}{32 n^2}\cdot \mathcal{R}(N, R_1, R_2)-\frac{z}{32 n^2}\right)\right)\\ & \quad =\# \left\{\vphantom{\left[\frac{-1}{32 n^2}, \frac{R_2}{32 n^2}\right]}(a-z_1/32 n^2,b-z_2/32 n^2,c-z_3/32 n^2)\in \mathbb{Z}_{\geq 1}^3\mid ab^{2/3}c\right.\\ & \quad\left. \lt \frac{N}{(32n^2)^{8/3}},\, a/c\in [1, R_1]\,\text{and } b\in \left[\frac{-1}{32 n^2}, \frac{R_2}{32 n^2}\right]\right\}\\ & \quad =\sum_{b'=0 }^{\lfloor (R_2-z_2)/32n^2\rfloor} \# \left\{\vphantom{ \lt \frac{N}{32^2n^4(32n^2b'+z_2)^{2/3}}}(a-z_1/32n^2, c-z_3/32n^2)\in \mathbb{Z}_{\geq 1}^2\mid ac\right.\\ & \quad \left. \lt \frac{N}{32^2n^4(32n^2b'+z_2)^{2/3}},\, a/c\in [1,R_1]\right\}\\ & \quad = \sum_{b'=0 }^{\lfloor (R_2-z_2)/32^2n^4\rfloor} \#\left(\mathcal{S}\left( \frac{N}{32^2n^4(32n^2b'+z_2)^{2/3}},R_1\right)-\left(\frac{z_1}{32n^2}, \frac{z_3}{32n^2}\right)\right)\cap \mathbb{Z}^2 \\ & \quad = \sum_{b'=0 }^{\lfloor (R_2-z_2)/32n^2\rfloor} \operatorname{Area}\left(\mathcal{S}\left( \frac{N}{32^2n^4(32n^2b'+z_2)^{2/3}},R_1\right)\right)\\ &\qquad +O\left(\sum_{b'=0 }^{\lfloor (R_2+z_2)/32n^2\rfloor} L\left( \frac{N}{32^2n^4(32n^2b'+z_2)^{2/3}},R_1\right)\right), \end{align*}

where in the above chain of equalities, we set $b':=b-z_2/32n^2$. Note that in the sum, $b$ does not take integer values, but $b'$ does. From Lemma 4.2, we find that the area of $\mathcal{S}(M, R)$ is given by:

\begin{equation*} \operatorname{Area}\left(\mathcal{S}(M, R)\right)=\begin{cases} \frac{M}{2}\log R -\frac{(R-1)}{2} &\text{if }M\geq R;\\ \frac{M}{2}\log M-\frac{(M-1)}{2} & \text{if }M \lt R.\\ \end{cases}\end{equation*}

Since $N \gt n^4R_1R_2$, we have that

\begin{equation*} \begin{split} &\# \left(\mathbb{Z}^3 \cap \left(\frac{1}{32n^2}\cdot \mathcal{R}(N, R_1, R_2)-\frac{z}{32n^2}\right)\right) \\ & \quad = \frac{N\log R_1}{2}\frac{1}{32^2n^4}\left\{\sum_{\substack{b=1,\\ b\equiv z_2\mod{32 n^2}}}^{\lfloor R_2\rfloor} b^{-2/3}\right\}+O_{n, R_1, R_2}(\sqrt{N}) \\ \end{split} \end{equation*}

Thus we find that

\begin{align*} & \#\mathcal{C}_n^\tau(N, R_1, R_2)\\ & \quad = \frac{N\log R_1}{2}\frac{1}{32^2n^4}\sum_{\substack{b=1,\\ b\,\square\text{-free}}}^{\lfloor R_2\rfloor} \frac{\#\{z\in \widetilde{\Omega}_n\mid z_2=b\pmod{32n^2}\}}{b^{2/3}}+O_{n, R_1, R_2}(\sqrt{N})\\ &\quad = \frac{N\log R_1 }{2 \times 32^2}\frac{1}{n^4}\sum_{\substack{b=1,\\ b\,\square\text{-free}}}^{\lfloor R_2\rfloor} \frac{n_\tau(b)\times\prod_{\ell|b}(\ell^2-\ell)^2\times \prod_{\ell|n, \ell\nmid b} \left(\ell^4-3\ell^2+2\ell\right)}{b^{2/3}}+O_{n, R_1, R_2}(\sqrt{N})\\ & \quad= \frac{N }{2\times 32^2}\prod_{\ell|n} \left(1-\frac{3}{\ell^2}+\frac{2}{\ell^3}\right)\log R_1\sum_{\substack{b=1,\\ b\,\square\text{-free}}}^{\lfloor R_2\rfloor} n_\tau(b)b^{-2/3}\prod_{\ell|b}\left(\frac{\ell}{\ell+2}\right)+O_{n, R_1, R_2}(\sqrt{N}).\\ \end{align*}

Let $\mathcal{C}_\ell'(N, R_1, R_2)$ be the complement of $\mathcal{C}_\ell(N, R_1, R_2)$ in $\mathcal{R}_{\mathbb{Z}^3}(N, R_1, R_2)$.

Lemma 4.9. We have that

\begin{equation*}\frac{\# \mathcal{C}_\ell'(N, R_1, R_2)}{N}=O\left(\frac{1}{\ell^{4/3}}+\frac{1}{\sqrt{N}}\right),\end{equation*}

where the implied constant depends only on $R_1$ and $R_2$, and not on $N$ or $\ell$.

Proof. Suppose $(a,b,c)\in \mathcal{C}_\ell'(N, R_1, R_2)$, then there are the following cases to consider:

1. (1) $\ell^2|a$,

2. (2) $\ell^2|b$,

3. (3) $\ell^2|c$,

4. (4) $\ell|a$ and $\ell|b$,

5. (5) $\ell|a$ and $\ell|c$,

6. (6) $\ell|b$ and $\ell|c$.

For $i=1, \dots, 6$, let $\mathcal{F}_i(N, R_1, R_2)$ be the subset of $\mathcal{C}_\ell'(N, R_1, R_2)$ consisting of $(a,b,c)$ satisfying $(i)$ above. First we bound $\#\mathcal{F}_1(N, R_1, R_2)$ we write $a=\ell^2 a'$. Note that $(a',b,c)$ lies in the set

\begin{equation*}\{(a',b,c)\in \mathbb{Z}_{\geq 1}^3\mid a'b^{2/3}c\leq N/\ell^2,\quad a'/c\in [1/\ell^2, R_1/\ell^2],\quad b\in [1, R_2]\}.\end{equation*}

We find that

\begin{equation*} \begin{split}\# \mathcal{F}_1(N, R_1, R_2)=& \sum_{b=1}^{\lfloor R_2\rfloor} \# \{(a',c)\in \mathbb{Z}_{\geq 1}^2\mid a'c\leq \frac{N}{\ell^2b^{2/3}},\quad a'/c\in [1/\ell^2, R_1/\ell^2]\}\\ =& \sum_{b=1}^{\lfloor R_2\rfloor} O\left( \operatorname{Area}\left(\{(a',c)\in \mathbb{Z}_{\geq 1}^2\mid a'c\leq \frac{N}{\ell^2b^{2/3}},\quad a'/c\in [1/\ell^2, R_1/\ell^2]\}\right)\right)\\ + & \sum_{b=1}^{\lfloor R_2\rfloor} O\left( L\left(\{(a',c)\in \mathbb{Z}_{\geq 1}^2\mid a'c\leq \frac{N}{\ell^2b^{2/3}},\quad a'/c\in [1/\ell^2, R_1/\ell^2]\}\right)\right)\\ =& O_{R_1, R_2}(N/\ell^2), \end{split} \end{equation*}

where the area and length calculation is similar to that in Lemmas 4.2 and 4.3. We find that

\begin{align*} \# \mathcal{F}_2(N, R_1, R_2)=& \sum_{b'=1}^{\lfloor R_2/\ell^2\rfloor} \# \{(a,c)\in \mathbb{Z}_{\geq 1}^2\mid ac\leq \frac{N}{\ell^{4/3}b'^{2/3}},\quad a/c\in [1, R_1]\}\\ =& \sum_{b'=1}^{\lfloor R_2/\ell^2\rfloor} O\left( \operatorname{Area}\left(\{(a,c)\in \mathbb{Z}_{\geq 1}^2\mid ac\leq \frac{N}{\ell^{4/3}b'^{2/3}},\quad a/c\in [1, R_1]\}\right)\right)\\ + & \sum_{b'=1}^{\lfloor R_2/\ell^2\rfloor} O\left( L\left(\{(a,c)\in \mathbb{Z}_{\geq 1}^2\mid ac\leq \frac{N}{\ell^{4/3}b'^{2/3}},\quad a/c\in [1, R_1]\}\right)\right)\\ =& O_{R_1, R_2}\left(\frac{N}{\ell^{4/3}}\right) + O_{R_1, R_2}(\sqrt{N}), \\ \# \mathcal{F}_3(N, R_1, R_2)=& \sum_{b=1}^{\lfloor R_2\rfloor} \# \{(a,c')\in \mathbb{Z}_{\geq 1}^2\mid ac'\leq N/\ell^2b,\quad a/c'\in [1, \ell^2 R_1]\}\\ =& O_{R_1, R_2}(N \log \ell/\ell^2)+ O_{R_1, R_2}(\sqrt{N}),\\ \# \mathcal{F}_4(N, R_1, R_2)=& \sum_{b'=1}^{\lfloor R_2/\ell \rfloor} \# \{(a',c)\in \mathbb{Z}_{\geq 1}^2\mid a'c\leq N/\ell^{5/3}b',\quad a'/c\in [1, R_1/\ell]\}\\ =& O_{R_1, R_2}(N /\ell^{5/3})+ O_{R_1, R_2}(\sqrt{N}),\\ \# \mathcal{F}_5(N, R_1, R_2)=& \sum_{b=1}^{\lfloor R_2 \rfloor} \# \{(a',c')\in \mathbb{Z}_{\geq 1}^2\mid a'c'\leq N/\ell^{2}b,\quad a'/c'\in [1, R_1]\}\\ =& O_{R_1, R_2}(N /\ell^2)+ O_{R_1, R_2}(\sqrt{N}),\\ \# \mathcal{F}_6(N, R_1, R_2)=& \sum_{b'=1}^{\lfloor R_2/\ell \rfloor} \# \{(a,c')\in \mathbb{Z}_{\geq 1}^2\mid ac'\leq N/\ell^{5/3} b',\quad a/c'\in [1, \ell R_1]\}\\ =& O_{R_1, R_2}\left(N \log \ell /\ell^{5/3}\right)+ O_{R_1, R_2}(\sqrt{N}).\\ \end{align*}

Thus, putting it all together, one finds that

\begin{equation*}\frac{\# \mathcal{C}_\ell'(N, R_1, R_2)}{N}=\sum_{i=1}^6 \frac{\mathcal{F}_i(N, R_1, R_2)}{N}=O\left(\ell^{-4/3}+\frac{1}{\sqrt{N}}\right).\end{equation*}

Lemma 4.10. Let $Y \gt 0$ be a positive constant and let $\mathcal{D}_Y(N, R_1, R_2):=\bigcup_{\ell \gt Y} \mathcal{C}_\ell'(N, R_1, R_2)$. Then we find that

\begin{equation*}\frac{\# \mathcal{D}_Y(N, R_1, R_2)}{N}=O\left(\sum_{\ell \gt Y} \ell^{-4/3}+ \frac{1}{\log N}\right).\end{equation*}

Proof. Clearly $\mathcal{C}_\ell'(N, R_1, R_2)=\emptyset$ unless $\ell\ll \sqrt{N}$. Thus from Lemma 4.9,

\begin{equation*}\frac{\# \mathcal{D}_Y(N, R_1, R_2)}{N}=O\left(\sum_{\ell \gt Y} \ell^{-4/3}+ \frac{\pi(\sqrt{N})}{\sqrt{N}}\right).\end{equation*}

The result follows from the prime number theorem.

Proposition 4.11. We have that

\begin{equation*} \begin{split}\lim_{N\rightarrow \infty}\frac{\# \mathcal{C}^\tau (N, R_1, R_2)}{N}=& \frac{1}{2048}\prod_\ell\left(1-\frac{3}{\ell^2}+\frac{2}{\ell^3}\right)\log R_1\sum_{\substack{b=1,\\ b\,\square\text{-free}}}^{\lfloor R_2\rfloor} n_\tau(b)b^{-2/3}\prod_{\ell|b}\left(\frac{\ell}{\ell+2}\right),\\ =& \frac{1}{2048}\prod_{\ell}\left(1-\frac{3}{\ell^2}+\frac{2}{\ell^3}\right)\log R_1\psi_\tau(R_2),\end{split} \end{equation*}

where

\begin{equation*}\psi_\tau(x):=\sum_{n=1}^{\lfloor x\rfloor} n_\tau(n)\alpha(n).\end{equation*}

Proof. Let $n$ be a squarefree natural number which is divisible by all primes $\ell\leq R_2$. Then for $N \gt n^4 R_1 R_2$, Proposition 4.8 asserts that

\begin{align*} \# \mathcal{C}_n^\tau(N, R_1, R_2)& =\frac{N}{2048}\prod_{\ell|n} \left(1-\frac{3}{\ell^2}+\frac{2}{\ell^3}\right)\log R_1\\ & \quad \times\sum_{\substack{b=1,\\ b\,\square\text{-free}}}^{\lfloor R_2\rfloor} n_\tau(b)b^{-2/3}\prod_{\ell|b}\left(\frac{\ell}{\ell+2}\right)+O_{n, R_1, R_2}(\sqrt{N}).\end{align*}

For $Y \gt 0$, set $n_Y:=\prod_{\ell\leq Y} \ell$. We find that

\begin{align*} \limsup_{N\rightarrow \infty} \frac{\# \mathcal{C}^\tau(N, R_1, R_2)}{N}&\leq \lim_{N\rightarrow \infty} \frac{\# \mathcal{C}_{n_Y}^\tau(N, R_1, R_2)}{N}\\ & =\frac{1}{2048}\prod_{\ell\leq Y} \left(1-\frac{3}{\ell^2}+\frac{2}{\ell^3}\right)\log R_1\sum_{\substack{b=1,\\ b\,\square\text{-free}}}^{\lfloor R_2\rfloor} n_\tau(b)b^{-2/3}\prod_{\ell|b}\left(\frac{\ell}{\ell+2}\right)\!.\end{align*}

Taking the limit of the right-hand side as $Y\rightarrow \infty$, we find that

\begin{equation*}\limsup_{N\rightarrow \infty} \frac{\# \mathcal{C}^\tau(N, R_1, R_2)}{N}\leq \frac{1}{2048}\prod_\ell\left(1-\frac{3}{\ell^2}+\frac{2}{\ell^3}\right)\log R_1\sum_{\substack{b=1,\\ b\,\square\text{-free}}}^{\lfloor R_2\rfloor} n_\tau(b)b^{-2/3}\prod_{\ell|b}\left(\frac{\ell}{\ell+2}\right).\end{equation*}

On the other hand, since

\begin{equation*}\mathcal{C}_{n_Y}^\tau(N, R_1, R_2)\subseteq \mathcal{C}^\tau(N, R_1, R_2)\cup \mathcal{D}_Y(N, R_1, R_2), \end{equation*}

we find that

(4)\begin{align} &\liminf_{N\rightarrow \infty} \frac{\# \mathcal{C}^\tau(N, R_1, R_2)}{N}\nonumber\\ & \quad \geq \lim_{N\rightarrow \infty} \frac{\# \mathcal{C}_{n_Y}(N, R_1, R_2)}{N}-\limsup_{N\rightarrow \infty} \frac{\# \mathcal{D}_Y(N, R_1, R_2)}{N}\nonumber\\ & \quad \geq \frac{1}{2048}\prod_{\ell\leq Y} \left(1-\frac{3}{\ell^2}+\frac{2}{\ell^3}\right)\log R_1\sum_{\substack{b=1,\\ b\,\square\text{-free}}}^{\lfloor R_2\rfloor} n_\tau(b)b^{-2/3}\prod_{\ell|b}\left(\frac{\ell}{\ell+2}\right)- C\sum_{\ell \gt Y} \ell^{-4/3}\end{align}

for some positive constant $C \gt 0$. Taking the limit of the right-hand side with $Y \rightarrow \infty$, we find that

\begin{equation*}\liminf_{N\rightarrow \infty} \frac{\# \mathcal{C}^\tau(N, R_1, R_2)}{N}\geq \frac{1}{2048}\prod_\ell\left(1-\frac{3}{\ell^2}+\frac{2}{\ell^3}\right)\log R_1\sum_{\substack{b=1,\\ b\,\square\text{-free}}}^{\lfloor R_2\rfloor} n_\tau(b)b^{-2/3}\prod_{\ell|b}\left(\frac{\ell}{\ell+2}\right).\end{equation*}

Therefore, we have shown that

\begin{equation*}\lim_{N\rightarrow \infty} \frac{\# \mathcal{C}^\tau(N, R_1, R_2)}{N}= \frac{1}{2048}\prod_\ell\left(1-\frac{3}{\ell^2}+\frac{2}{\ell^3}\right)\log R_1\sum_{\substack{b=1,\\ b\,\square\text{-free}}}^{\lfloor R_2\rfloor} n_\tau(b)b^{-2/3}\prod_{\ell|b}\left(\frac{\ell}{\ell+2}\right).\end{equation*}

We now give a proof of the main result.

Proof of Theorem 1.4

Assume that $m \gt 0$ and is of Type $\mathrm{II}$. The argument in the remaining cases proceeds similarly. Write $m = ab^2c^3$, where $(a,b,c)$ is a carefree triple with $a, b, c \gt 0$, and suppose further that $c/a \gt 1$. This condition ensures that the isomorphism class of the number field $K_m$ is uniquely determined. Recall from Corollary 3.2 that the discriminant of $K_m$ is given by $\Delta_m = -2^8 a^3 b^2 c^3$. Setting $X = 2^8 N^3$, the condition $|\Delta_m| \leq X$ becomes equivalent to

\begin{equation*} ab^{2/3}c \leq N. \end{equation*}

Consider the four rectangles:

\begin{equation*} \begin{split} \mathcal{R}_1 &:= [1,R_1] \times [1,R_2], \\ \mathcal{R}_2 &:= [1,R_1] \times [1,R_2'], \\ \mathcal{R}_3 &:= [1,R_1'] \times [1,R_2], \\ \mathcal{R}_4 &:= [1,R_1'] \times [1,R_2'], \end{split} \end{equation*}

so that by inclusion–exclusion,

\begin{equation*} \int_{\mathcal{R}} = \int_{\mathcal{R}_1} - \int_{\mathcal{R}_2} - \int_{\mathcal{R}_3} + \int_{\mathcal{R}_4}. \end{equation*}

It therefore suffices to prove the theorem under the simplifying assumption $R_1' = R_2' = 1$, in which case the claim follows directly from Proposition 4.11.