On the Positive Definite Lattice with Determinant 5

Abstract

This paper is devoted to providing a classification of positive definite lattices with determinant 5 and a rank less than or equal to 6.

1. Introduction

Lattice theory, as a foundational branch of modern mathematics, serves as a crucial tool for the study of discrete structures, with far-reaching applications in many branches of mathematics. Among its subfields, integral lattices hold special significance as they play key roles in finite groups [1,2], optimizing algorithms [3], and constructing error-correcting codes [4,5]. In addition, such lattices serve to encode solutions for intricate Diophantine equations [6] and contribute to the optimization of communication network designs [7], thereby emphasizing their cross-disciplinary importance. Consequently, the classification of positive definite lattices based on rank and determinant has developed into a central research problem, which carries both theoretical worth and practical relevance.

In the realm of lattice classification, a large portion of existing research has centered on unimodular lattices. Specifically, multiple studies have delved into even unimodular lattices (see [8,9,10]). In [11], Mordell completed the classification of all positive definite unimodular lattices of rank 8, where his proof drew on "two deep theorems" in quadratic forms. Subsequently, R.L. Griess Jr. adopted solely elementary methods to reclassify these lattices in [12]. Later, in his monograph [13], he further extended this work: he classified positive definite lattices with a determinant of 2 and a rank at most of 7, those with a determinant of 3 and a rank at most of 6, and those with a determinant of 4 and a rank at most of 4. Beyond this, Wang Ruiqing investigated additively indecomposable positive definite integral lattices in [14]. Additionally, R.E. Borcherds conducted research on positive definite lattices in [15]: he classified 665 positive definite unimodular lattices in 25 dimensions and identified 121 even lattices with a determinant of 2. More recently, Libo Zhao et al. presented the classification of positive definite lattices of determinant 4 and rank 5 in [16]. Nevertheless, despite these progressions, the classification of lattices with moderate ranks and small determinants (>1) remains incomplete.

Determinant 5, as a small prime, imposes strict arithmetic constraints (e.g., on the discriminant group, vector norms, and duality relations), making it ideal for exploring the interplay between a lattice’s arithmetic and geometric properties. Existing studies on lattices with small prime determinants (e.g., 2, 3) lay a foundation, and extending this to determinant 5 helps build a comprehensive understanding of prime-determinant lattices. Rank 6 is a critical dimension in low-dimensional positive definite lattice research. It balances “simplicity” (facilitating explicit classification) and “non-triviality” (exhibiting structural properties that connect to broader fields like arithmetic geometry and group theory). So lattices of determinant 5 and rank 6 occupy a particularly significant gap.

In this paper, we aim to classify positive definite lattices of determinant 5 and a rank ≤6.

The terminology and the notation in this paper are standard, which can be referred to in [13].

2. Preliminaries

In this section, we give some notations and lemmas, which are useful for the later use and could be found in [13].

A lattice is termed rectangular when it possesses an orthogonal basis. The dual of a lattice M is denoted as $M^{*}=\{x\in V\mid (x,y)\in \mathbb{Z}\mathrm{for}\mathrm{every}y\in M\}$.

The determinant of a lattice is defined to be $det\left(G\right)$, where G denotes an arbitrary Gram matrix. In particular, a Gram matrix is a square matrix of order $\mathrm{rank}\left(M\right)$ (the rank of lattice M), whose entries are given by $\left((x_i,x_j)\right)$ here, and $\left\{x_i\right\}$ denotes any basis of the lattice.

Now, suppose that M is an integral lattice, meaning that $(x,y)$ is an integer for all $x,y\in M$. We say that M is even if $(x,x)$ is an even integer for each $x\in M$; if M fails to meet this condition, it is called an odd lattice.

The discriminant group of M, denoted as $\mathcal{D}\left(M\right)$, is defined as $\mathcal{D}\left(M\right)=M^{*}/M$.

A lattice M is called unimodular when $det\left(M\right)=1$ or $det\left(M\right)=-1$. For $det\left(M\right)\ne 0$, the order of the discriminant group $\mathcal{D}\left(M\right)$ (denoted as $\left|\mathcal{D}\right(M\left)\right|$) equals the absolute value of the lattice’s determinant, that is, $\left|\mathcal{D}\right(M\left)\right|=|det(M\left)\right|$.

A positive definite lattice is a free $\mathbb{Z}$-module $L\subseteq {\mathbb{R}}^n$ with a positive definite inner product $\langle ·,·\rangle :L\times L\to \mathbb{Z}$.

And we use $\mathbb{Z}v$ to denote the integer lattice generated by vector v. For $v\in {\mathbb{R}}^n$ (where n is the dimension of the ambient Euclidean space), $\mathbb{Z}v=\{kv\mid k\in \mathbb{Z}\}$ consists of all integer scalar multiples of v, forming a one-dimensional sublattice.

Definition 1. 

([13] pp. 33, 34) Let $L:={\mathbb{Z}}^{n+1}=\left\{(x_0,x_1,...,x_n)\right|x_i\in \mathbb{Z}\}$ and $v=(1,1,...,1)\in L$. $A_n:=an{n}_L\left(\mathbb{Z}v\right)=\{x\in L|(x,v)=0$ for all $v\in \mathbb{Z}v\}=an{n}_L\left(v\right)$, where $(·,·)$ denotes the bilinear form on L.

Further,

$D_n:=\{(x_1,...,x_n)\in {\mathbb{Z}}^n|\sum x_i\in 2\mathbb{Z}\}$;

$E_8:=D_8+\mathbb{Z}{\displaystyle \frac{1}{2}}(1,1,...,1)$;

The $E_7$ lattice is the annihilator in $E_8$ of any $A_1$ sublattice;

The $E_6$ lattice is the annihilator in $E_8$ of any $A_2$ sublattice.

Lemma 1. 

([13] p. 17, Theorem 2.3.3) Suppose that L is a lattice, and M is a sublattice of L of index $|L:M|$. Then, ${det\left(L\right)|L:M|}^2=det\left(M\right).$

Lemma 2. 

([13] p. 55, Theorem 5.2.1) Suppose that L is a positive definite lattice of determinant 1 and a rank no more than 8. Then, $L\cong {\mathbb{Z}}^n$ or $L\cong E_8$.

Lemma 3. 

([13] p. 55, Theorem 5.2.2) Suppose that L is a positive definite integral lattice of determinant 2 and rank 7. Then, L is rectangular or $L\cong E_7$.

Lemma 4. 

([13] p. 56, Theorem 5.2.3) Suppose that L is a positive definite integral lattice of determinant 3 and rank 6. Then, L is rectangular, or $L\cong A_2\perp {\mathbb{Z}}^4$, or $L\cong E_6$.

Lemma 5. 

([13] p. 18, Theorem 2.4.1) Suppose that L is a nonsingular integral lattice and M is a sublattice which is a direct summand of L. Then,

(1) The natural map $\psi :L^{*}\mapsto M^{*}$ is onto.

(2) Define π to be the composition of the quotient map $M^{*}\mapsto M^{*}/M$ and ψ, then we have $Ker\left(\pi \right)=M+an{n}_{L^{*}}\left(M\right)$ and $Ker\left(\pi {|}_L\right)=M+an{n}_L\left(M\right)$.

(3) ${\pi |}_L$ is surjective if $\left|\mathcal{D}\right(L\left)\right|$ and $\left|\mathcal{D}\right(M\left)\right|$ are relatively prime.

Lemma 6. 

([13] pp.49, Theorem 5.0.3) Let $H(n,d):={\left({\displaystyle \frac{4}{3}}\right)}^{\frac{n-1}{2}}·d^{\frac{1}{n}}$ and $\mu \left(L\right):=min\left\{\right|(x,x)\left|\right|x\in L,x\ne 0\}$. If L is a rational lattice of rank n, then $\mu \left(L\right)\le H(n,|det\left(L\right)\left|\right)$.

3. Main Results

This section focuses on providing the classification of lattices that have a determinant of 5 and a rank of 6. Since rank 1 lattices are straightforward, we initiate our analysis with the rank 2 scenario.

Theorem 1. 

Let X be a positive definite lattice with determinant 5 and rank 2. Then X is rectangular or isometric to $L_2:=span\{u,v\}$, where $(u,u)=2,(v,v)=3,(u,v)=1$.

Proof. 

If there exists an unit vector $u_1$ in X, then, by using Lemma 5, the map $X\to \mathcal{D}\left(\mathbb{Z}{u}_1\right)$ is onto. And so $X=\mathbb{Z}{u}_1\perp an{n}_X\left(u_1\right)$. Thus X is rectangular.

When there exists no unit vector in X, then, using Lemma 6, we see that X has a root vector u since $H(2,5)=2.58198\dots$. The natural map $X\to \mathcal{D}\left(\mathbb{Z}u\right)$ is onto since $(detX,det\mathbb{Z}u)=1$. Thus, $X/(\mathbb{Z}u\perp an{n}_X\left(u\right))\cong \mathcal{D}\left(\mathbb{Z}u\right)$ and then $an{n}_X\left(u\right)=\mathbb{Z}v$ has a determinant of 10 by Lemma 1. A nontrivial coset of $\mathbb{Z}u\perp \mathbb{Z}v$ contains ${\displaystyle \frac{1}{2}}(iu+jv)$, where $i,j\in \{0,1\}$. For the norm of ${\displaystyle \frac{1}{2}}(iu+jv)$ to be an integer and $\ge 1$, we see $2i^2+10j^2\equiv 0(mod4)$ and $(2i^2+10j^2)/4>1$. Then $(i,j)=(1,1)$. So $X=span\{u,v,{\displaystyle \frac{u+v}{2}}\}=span\{u,{\displaystyle \frac{u+v}{2}}\}$, which is lattice $L_2$ in the theorem. □

Theorem 2. 

Let X be a positive definite lattice with determinant 5 and rank 3. Then X is rectangular or isometric to $L_2\perp \mathbb{Z}$.

Proof. 

If there exists an unit vector $u_1$ in X, then $X=\mathbb{Z}{u}_1\perp an{n}_X\left(u_1\right)$ because the map $X\to \mathcal{D}\left(\mathbb{Z}{u}_1\right)$ is onto. So X is rectangular or $L_2\perp \mathbb{Z}$ by Theorem 1.

When there exists no unit vector in X, since $H(3,5)=2.2799679\dots$, X contains a root, say u. By considering $X/(\mathbb{Z}u\perp an{n}_X\left(u\right))\cong 3$, we see that $det\left(an{n}_X\left(u\right)\right)=10$. Since $H(2,10)=3.65148\dots$, $an{n}_X\left(u\right)$ has a vector v of norm 2 or 3. Define $P:=\mathbb{Z}u\perp \mathbb{Z}v$ and $R:=an{n}_X\left(P\right)=\mathbb{Z}w$.

P is a direct summand of X. If not, P is contained in some direct summand $P^{\prime }$ and $|P^{\prime }:P|=t>1$. Then $det\left(P^{\prime }\right)={\displaystyle \frac{det\left(P\right)}{t^2}}=1$. This implies that $P^{\prime }$ contains a unit vector, which is a contradiction.

Suppose that $(v,v)=2$. Also $X/(P\perp R)\cong 2\times 2$. Then $detR=20$. A nontrivial coset of $P\perp R$ contains an element with the form ${\displaystyle \frac{1}{2}}(iu+jv+kw)$, where $i,j\in \{0,1\}$. Since the norm of ${\displaystyle \frac{1}{2}}(iu+jv+kw)$ must be an integer and $\ge 1$, we see $2i^2+2j^2+20k^2\equiv 0(mod4)$ and $(2i^2+2j^2+20k^2)/4>1$. Then $(i,j,k)=\left\{\right(0,0,1),(1,1,1\left)\right\}$. This is incompatible with $X/(P\perp R)\cong 2\times 2$. Thus, $(v,v)=3$.

We see that $X/(P\perp R)\cong 2\times 3$. Then $detR=30$. A nontrivial coset of $P\perp R$ contains an element g with form ${\displaystyle \frac{1}{6}}(iu+jv+kw)$, where $i,j\in \{0,1,2,3,4,5\}$. For the norm of ${\displaystyle \frac{1}{6}}(iu+jv+kw)$ to be an integer and greater than 1, we see $2i^2+3j^2+30k^2\equiv 0(mod36)$ and $(2i^2+3j^2+30k^2)/36>1$. Then $(i,j,k)=(3,0,3)$. This is incompatible with $X/(P\perp R)\cong 2\times 3$. □

Theorem 3. 

Let X be a positive definite lattice with determinant 5 and rank 4. Then X is rectangular or isometric to $L_2\perp {\mathbb{Z}}^2$ or $A_4$.

Proof. 

If there exists an unit vector $u_1$ in X, then $X=\mathbb{Z}{u}_1\perp an{n}_X\left(u_1\right)$ because the map $X\to \mathcal{D}\left(\mathbb{Z}{u}_1\right)$ is onto. Then X is rectangular or $L_2\perp {\mathbb{Z}}^2$ by Theorem 2.

When there exists no unit vector in X, since $H(4,5)=2.30224\dots$ and $H(3,10)=2.872579\dots$, X contains two orthogonal roots $u,v$. Let $P:=\mathbb{Z}u\perp \mathbb{Z}v$. The natural map $X\mapsto \mathcal{D}P$ is onto by $(detX,detP)=1$. Then, $R=an{n}_X\left(P\right)$ is of determinant 20 and the image of X in $\mathcal{D}R$ is $2\times 2$. So $R^{*}/R\cong 2\times 10$. Then for any vector $g\in R,{\displaystyle \frac{1}{2}}g\in R^{*}$ and $(g,{\displaystyle \frac{1}{2}}g)\in \mathbb{Z}$. Thus $(g,g)\in 2\mathbb{Z}$. Let $K:={\displaystyle \frac{1}{\sqrt{2}}}R$. $detK={\displaystyle \frac{detR}{2^2}}=5$. So K is rectangular or isometric to $L_2$ by Theorem 1.

Suppose that K is rectangular. Then $R:=\mathbb{Z}w\perp \mathbb{Z}x$ with $(w,w)=2,(x,x)=10$.

A nontrivial coset of $P\perp R$ contains an element g of the form ${\displaystyle \frac{1}{2}}(iu+jv+kw+lx)$, where $i,j,k,l\in \{0,1\}$. For the norm of ${\displaystyle \frac{1}{2}}(iu+jv+kw+lx)$ to be an integer and greater than 1, we see $2i^2+2j^2+2k^2+10l^2\equiv 0(mod4)$ and $(2i^2+2j^2+2k^2+10l^2)/4>1$. Then $(i,j,k,l)\in \left\{\right(0,0,1,1),(0,1,0,1),(1,0,0,1),(1,1,1,1\left)\right\}$. Let $A:=\left\{\right(0,0,1,1),(0,1,0,1),(1,0,0,1),(1,1,1,1\left)\right\}$.

$X/(P\perp R)$ is an abelian group of order 4 and exponent 2. Assume that $X/(P\perp R)=\langle g+(P\perp R),q+(P\perp R)\rangle$, $q=i_{1}u+j_{1}v+k_{1}w+l_{1}x$, where $i_1,j_1,k_1,l_1\in \{0,1\}$. Then $g+q+(P\perp R)$ is nontrivial in $P\perp R$ and contains the element $g^{\prime }={\displaystyle \frac{1}{2}}(i^{\prime }u+j^{\prime }v+k^{\prime }w+l^{\prime }x)$, where $i^{\prime },j^{\prime },k^{\prime },l^{\prime }\in \{0,1\}$ and $i^{\prime }\equiv i+i_1(mod2),j^{\prime }\equiv j+j_1(mod2),k^{\prime }\equiv k+k_1(mod2),l^{\prime }\equiv l+l_1(mod2).$ So $2i^{\prime 2}+2j^{\prime 2}+3k^{\prime 2}+3l^{\prime 2}\equiv 0(mod4)$ and $(2i^{\prime 2}+2j^{\prime 2}+3k^{\prime 2}+3l^{\prime 2})/4>1$. Then $(i^{\prime },j^{\prime },k^{\prime },l^{\prime })$ is also in A. But the sum of any two elements is not in A, which is a contradiction.

So K is isometric to $L_2$ and $R:=span\{w,x\}$ with $(w,w)=4,(x,x)=6,(w,x)=2.$ A nontrivial coset of $P\perp R$ has an element g with the form ${\displaystyle \frac{1}{2}}(iu+jv+kw+lx)$, where $i,j,k,l\in \{0,1\}$. For the norm of ${\displaystyle \frac{1}{2}}(iu+jv+kw+lx)$ to be an integer and greater than 1, we see $2i^2+2j^2+4k^2+6l^2+4kl\equiv 0(mod4)$ and $(2i^2+2j^2+4k^2+6l^2+4kl)/4>1$. Then $(i,j,k,l)\in \left\{\right(0,1,0,1\left)\right(0,1,1,1),(1,0,0,1),(1,0,1,1),(1,1,1,0\left)\right\}$. So X is isometric to $X_1=span\{u,v,w,x,{\displaystyle \frac{u+x}{2}},{\displaystyle \frac{u+v+w}{2}}\}$ or $X_2=span\{u,v,w,x,{\displaystyle \frac{v+x}{2}},{\displaystyle \frac{u+v+w}{2}}\}$. Then $X_1$ is isometric to $X_2$ by the isometry $u\mapsto v,v\mapsto u,w\mapsto w,x\mapsto x$. So $X\cong span\{u,v,w,x,{\displaystyle \frac{u+x}{2}},{\displaystyle \frac{u+v+w}{2}}\}=span\{u,v,{\displaystyle \frac{u+x}{2}},{\displaystyle \frac{u+v+w}{2}}\}\cong span\{e_1-e_2,e_4-e_3,e_1-e_5,e_1-e_3\}\cong A_4$. The proof is complete. □

Theorem 4. 

Let X be a positive definite lattice with determinant 5 and rank 5. Then X is rectangular or isometric to $L_2\perp {\mathbb{Z}}^3$ or $A_4\perp \mathbb{Z}$.

Proof. 

If there exists a unit vector $u_1$ in X, then $X=\mathbb{Z}{u}_1\perp an{n}_X\left(u_1\right)$ bacause the natural map $X\to \mathcal{D}\left(\mathbb{Z}{u}_1\right)$ is onto. Then X is rectangular or isometric to $L_2\perp {\mathbb{Z}}^3$ or $A_4\perp \mathbb{Z}$ by Theorem 3.

When there exists no unit vector in X, since $H(5,5)=2.4528527\dots$, $H(4,10)=2.737840\dots$, and $H(3,20)=3.6192234\dots$, X contains three orthogonal vectors, $u,v,w$, with norms $2,2,2$ or $2,2,3$, respectively. Define $P:=\mathbb{Z}u\perp \mathbb{Z}v\perp \mathbb{Z}w$, $Q:=an{n}_X\left(P\right)$. We claim that P is a direct summand of X. In fact, if $P\subset P^{\prime }$ (where $P^{\prime }$ is a direct summand) then $|P^{\prime }:P|=m$. Then $det{P}^{\prime }={\displaystyle \frac{detP}{m^2}}=2$ or 3. Then $P^{\prime }$ is rectangular or $A_2\perp \mathbb{Z}$ and so $P^{\prime }$ has a unit vector by Lemma 4 and Lemma 3, which is a contradiction.

The map $X\mapsto \mathcal{D}\left(P\right)$ is onto by $(detX,detP)=1$. Thus, the image of X in $\mathcal{D}\left(Q\right)$ is $2\times 2\times 2$ or $2\times 2\times 3$. And Q has a rank of 2, so we see $\mu \left(Q\right)\ge 3$, $(w,w)=3$ and $X/(P\perp Q)\cong 2\times 6$.

Then Q has a determinant of 60. And the Sylow 2-subgroup of $\mathcal{D}\left(Q\right)$ has an exponent of 2. So $Q:=\sqrt{2}K$, where K has a rank of 2 and a determinant of 15. Since $\mu \left(Q\right)\ge 3$ and $H(2,12)=4.472135\dots$, we see $\mu \left(K\right)=2,3$ or 4.

If $\mu \left(K\right)=2$, let $x\in K$ with norm 2. Then we claim that $K\cong span\{x,y|(x,x)=2,(y,y)=8,(x,y)=1\}$. In fact, the map $K\to \mathcal{D}\left(\mathbb{Z}x\right)$ is onto. Then $K/(\mathbb{Z}x\perp an{n}_K\left(x\right))\cong 2$, $an{n}_K\left(x\right)=\mathbb{Z}z$ is with determinant 30. The nontrivial coset $\mathbb{Z}x\perp an{n}_K\left(x\right))$ has an element with the form ${\displaystyle \frac{1}{2}}(e+f)$, where $e\in span\left\{x\right\}$ and $f\in span\left\{z\right\}$. Furthermore, we may arrange for $e=ix,f=jz$, where $i,j\in \{0,1\}$. Since the norm of ${\displaystyle \frac{1}{2}}(e+f)$ must be an integer and $>1$, we see $2i^2+30j^2\equiv 0(mod4)$ and $(2i^2+30j^2)/4>1$. Then $(i,j)=(1,1)$. So K is isometric to $span\{x,{\displaystyle \frac{x+z}{2}}\}\cong span\{x,y|(x,x)=2,(y,y)=8,(x,y)=1\}$.

If $\mu \left(K\right)=3$, let $x\in K$ be of norm 3. Then we claim that $K\cong span\{x,y|(x,x)=3,(y,y)=5,(x,y)=0\}$. In fact, by noting the map $K\to \mathcal{D}\left(\mathbb{Z}x\right)$, we see $K/(\mathbb{Z}x\perp an{n}_K\left(x\right))\cong 1$ or 3. If $K/(\mathbb{Z}x\perp an{n}_K\left(x\right))\cong 1$, then $K\cong \{x,y|(x,x)=3,(y,y)=5,(x,y)=0\}$. If $K/(\mathbb{Z}x\perp an{n}_K\left(x\right))\cong 3$, then $an{n}_K\left(x\right)=\mathbb{Z}z$ is with determinant 45. A nontrivial coset of $\mathbb{Z}x\perp an{n}_K\left(x\right))$ has an element with form ${\displaystyle \frac{1}{2}}(e+f)$, where $e\in span\left\{x\right\}$, and $f\in span\left\{z\right\}$. Furthermore we may arrange for $e=ix,f=jz$, where $i,j\in \{0,1,2\}$. Since the norm of ${\displaystyle \frac{1}{3}}(e+f)$ must be an integer and $>1$, we see $3i^2+45j^2\equiv 0(mod9)$ and $(3i^2+45j^2)/9>1$. Then $(i,j)=(0,1),(0,2)$. So K is isometric to $span\{x,{\displaystyle \frac{z}{3}}\}\cong span\{x,y|(x,x)=3,(y,y)=5,(x,y)=0\}$.

If $\mu \left(K\right)=4$, let $x\in K$ with norm 4. Then we claim that $K\cong span\{x,y|(x,x)=4,(y,y)=4,(x,y)=1\}$. In fact, because the natural map $K\to \mathcal{D}\left(\mathbb{Z}x\right)$ is onto, we see $K/(\mathbb{Z}x\perp an{n}_K\left(x\right))\cong 4$. Then $an{n}_K\left(x\right)=\mathbb{Z}z$ is with determinant 60. A nontrivial coset of $\mathbb{Z}x\perp an{n}_K\left(x\right))$ has an element with form ${\displaystyle \frac{1}{4}}(e+f)$, where $e\in span\left\{x\right\}$ and $f\in span\left\{z\right\}$. Furthermore we may arrange for $e=ix,f=jz$, where $i,j\in \{0,1,2,3\}$. Since the norm of ${\displaystyle \frac{1}{4}}(e+f)$ must be an integer and $>1$, we see $4i^2+60j^2\equiv 0(mod16)$ and $(4i^2+60j^2)/16>1$. Then $(i,j)=(0,2),(1,1),(1,3),(2,2)$ or $(3,3)$. So K is isometric to $span\{x,{\displaystyle \frac{x+z}{4}}\}\cong span\{x,y|(x,x)=4,(y,y)=4,(x,y)=1\}$.

So $Q\cong span\{x,y|(x,x)=4,(y,y)=16,(x,y)=2\}$, $span\{x,y|(x,x)=6,(y,y)=10,(x,y)=0\}$ or $span\{x,y|(x,x)=8,(y,y)=8,(x,y)=2\}$.

Case 1. $Q\cong span\{x,y|(x,x)=4,(y,y)=16,(x,y)=2\}$

A nontrivial coset of $P\perp Q$ has an element g with form ${\displaystyle \frac{1}{6}}(e+f)$, where $e\in span\{u,v,w\}$ and $f\in span\{x,y\}$. Furthermore we may arrange for $e=iu+jv+kw,f=lx+my$, where $i,j,k,l,m\in \{0,1,2,3,4,5\}$. Since the norm of ${\displaystyle \frac{1}{6}}(e+f)$ must be an integer and $>1$, $2i^2+2j^2+3k^2+4l^2+16m^2+4lm\equiv 0(mod36)$ and $(2i^2+2j^2+3k^2+4l^2+16m^2+4lm)/36>1$.

Assume that $\langle g+(P\perp Q)\rangle$ is an abelian subgroup of order 6 of $X/(P\perp Q)$. Then for any $t\in \{1,2,3,4,5\}$, $tg+(P\perp Q)$ is also a nontrivial coset of $P\perp Q$ and contains an element $g^{\prime }={\displaystyle \frac{1}{6}}(i^{\prime }u+j^{\prime }v+k^{\prime }w+l^{\prime }x+m^{\prime }y)$, where $i^{\prime },j^{\prime },k^{\prime },l^{\prime },m^{\prime }\in \{0,1,2,3,4,5\}$ and $i^{\prime }\equiv ti(mod6),j^{\prime }\equiv tj(mod6),k^{\prime }\equiv tk(mod6),l^{\prime }\equiv tl(mod6),m^{\prime }\equiv tm(mod6).$ So $2i^{\prime 2}+2j^{\prime 2}+3k^{\prime 2}+4l^{\prime 2}+16m^{\prime 2}+4l^{\prime }{m}^{\prime }\equiv 0(mod36)$ and $(2i^{\prime 2}+2j^{\prime 2}+3k^{\prime 2}+4l^{\prime 2}+16m^{\prime 2}+4l^{\prime }{m}^{\prime })/36>1$.

Then $(i,j,k,l,m)\in A:= \{$

$(0,0,2,2,5),(0,0,2,4,1),(0,0,2,5,5),(0,0,4,1,1),(0,0,4,2,5)$

$(0,0,4,4,1),(3,3,2,1,1),(3,3,2,1,4),(3,3,2,2,5),(3,3,2,4,1),$

$(3,3,2,5,2),(3,3,2,5,5),(3,3,4,1,1),(3,3,4,1,4),(3,3,4,2,5),$

$(3,3,4,4,1),$$(3,3,4,5,2),(3,3,4,5,5)\}$.

Furthermore, if $\langle g+(P\perp Q)\rangle$ is of order 2, then $(i,j,k,l,m)\in \{$ $(0,0,0,0,3),(0,0,0,3,3),$ $(3,3,0,3,3),(3,3,0,3,0),(3,3,0,0,3)\}$.

Therefore the abelian 2-group with type $2\times 2$ is $A_1:=\langle (0,0,0,0,3),(3,3,0,3,3),(3,3,0,3,0)\rangle$ or $A_2:=\langle (0,0,0,3,3),(3,3,0,3,0),(3,3,0,0,3)\rangle .$

If $\langle g+(P\perp Q)\rangle$ is of order 3 in $X/(P_1\perp Q_1)$, then $(i,j,k,l,m)\in \left\{\right(0,0,4,2,2),(0,0,4,4,4),(0,0,2,2,2),(0,0,2,4,4)$ }. Thus the group of order 3 is

$B_1:=\langle (0,0,4,2,2),(0,0,2,4,4)\rangle$ or $B_2:=\langle (0,0,4,4,4),(0,0,2,2,2)\rangle .$

So $X_1:=span\{u,v,w,x,y,{\displaystyle \frac{1}{6}}\left(3y\right),{\displaystyle \frac{1}{6}}(3u+3v+3x),{\displaystyle \frac{1}{6}}(4w+2x+2y)\}$,

$X_2:=span\{u,v,w,x,y,{\displaystyle \frac{1}{6}}\left(3y\right),{\displaystyle \frac{1}{6}}(3u+3v+3x),{\displaystyle \frac{1}{6}}(2w+2x+2y)\}$,

$X_3:=span\{u,v,w,x,y,{\displaystyle \frac{1}{6}}(3x+3y),{\displaystyle \frac{1}{6}}(3u+3v+3x),{\displaystyle \frac{1}{6}}(4w+2x+2y)\}$

$X_4:=span\{u,v,w,x,y,{\displaystyle \frac{1}{6}}(3x+3y),{\displaystyle \frac{1}{6}}(3u+3v+3x),{\displaystyle \frac{1}{6}}(2w+2x+2y)\}$.

By calculation, we find unit vectors ${\displaystyle \frac{y}{2}}-{\displaystyle \frac{1}{3}}(2w+x+y)+w={\displaystyle \frac{1}{6}}(y-2x+2w)\in X_1$,${\displaystyle \frac{y}{2}}-{\displaystyle \frac{1}{3}}(w+x+y)={\displaystyle \frac{1}{6}}(y-2x-2w)\in X_1$ ${\displaystyle \frac{x+y}{2}}-{\displaystyle \frac{1}{3}}(w+x+y)={\displaystyle \frac{1}{6}}(x+y-2w)\in X_4$, and ${\displaystyle \frac{x+y}{2}}-({\displaystyle \frac{2w+x+y}{3}}-w)={\displaystyle \frac{1}{6}}(x+y+w)\in X_3$, which are contradictions.

Case 2. $Q\cong span\{x,y|(x,x)=6,(y,y)=10,(x,y)=0\}$

A nontrivial coset of $P\perp Q$ has an element g with form ${\displaystyle \frac{1}{6}}(e+f)$, where $e\in span\{u,v,w\}$ and $f\in span\{x,y\}$. Furthermore we may arrange for $e=iu+jv+kw,f=lx+my$, where $i,j,k,l,m\in \{0,1,2,3,4,5\}$. Since the norm of ${\displaystyle \frac{1}{6}}(e+f)$ must be an integer and $>1$, we see that $2i^2+2j^2+3k^2+6l^2+10m^2\equiv 0(mod36)$ and $(2i^2+2j^2+3k^2+6l^2+10m^2)/36>1$.

Assume that $\langle g+(P\perp Q)\rangle$ is an abelian subgroup of order 6 of $X/(P\perp Q)$. Then for each $t\in \{1,2,3,4,5\}$, $tg+(P\perp Q)$ is a nontrivial coset of $P\perp Q$ and has an element $g^{\prime }={\displaystyle \frac{1}{6}}(i^{\prime }u+j^{\prime }v+k^{\prime }w+l^{\prime }x+m^{\prime }y)$, where $i^{\prime },j^{\prime },k^{\prime },l^{\prime },m^{\prime }\in \{0,1,2,3,4,5\}$ and $i^{\prime }\equiv ti(mod6),j^{\prime }\equiv tj(mod6),k^{\prime }\equiv tk(mod6),l^{\prime }\equiv tl(mod6),m^{\prime }\equiv tm(mod6).$ So $2i^{\prime 2}+2j^{\prime 2}+3k^{\prime 2}+6l^{\prime 2}+10m^{\prime 2}\equiv 0(mod36)$ and $(2i^{\prime 2}+2j^{\prime 2}+3k^{\prime 2}+6l^{\prime 2}+10m^{\prime 2})/36>1$.

Then $(i,j,k,l,m)\in \{$

$(0,0,2,1,3),(0,0,4,5,3),(0,1,0,4,1),(0,3,2,4,3),(0,3,4,2,3)$

$(0,5,0,2,5),(1,0,0,4,1),(1,3,0,5,1),(2,3,0,1,2),(2,3,0,5,2),$

$(3,0,2,4,3),(3,0,4,2,3),(3,1,0,5,1),(3,2,0,1,2),(3,2,0,5,2),$

$(3,3,2,1,3),(3,3,4,5,3),(3,4,0,1,4),(3,4,0,5,4),(3,5,0,1,5),$

$(4,3,0,1,4),(4,3,0,5,4),(5,0,0,2,5),(5,3,0,1,5)$}.

Since $(u,g),(v,g),(w,g),(y,g)\in \mathbb{Z}$, we see $3|i,3|j,3|m,2|k$. Then

$(i,j,k,l,m)\in \{$$(0,0,2,1,3),(0,0,4,5,3),(0,3,2,4,3),(0,3,4,2,3)$$(3,0,2,4,3),$

$(3,0,4,2,3),(3,3,2,1,3),(3,3,4,5,3),$}.

Furthermore, if $\langle g+(P\perp Q)\rangle$ is of order 2 in $X/(P\perp Q)$, then $(i,j,k,l,m)\in \{$ $(0,0,0,3,3),(0,3,0,0,3),(3,0,0,0,3),(3,3,0,3,3)\}$. Thus there exists no abelian 2-group with type $2\times 2$, which is a contradiction.

Case 3. $Q\cong span\{x,y|(x,x)=8,(y,y)=8,(x,y)=2\}$

A nontrivial coset of $P\perp Q$ has an element g with form ${\displaystyle \frac{1}{6}}(e+f)$, where $e\in span\{u,v,w\}$ and $f\in span\{x,y\}$. Thus we may arrange for $e=iu+jv+kw,f=lx+my$, where $i,j,k,l,m\in \{0,1,2,3,4,5\}$. Since the norm of ${\displaystyle \frac{1}{6}}(e+f)$ must be an integer and >1, $2i^2+2j^2+3k^2+8l^2+8m^2+4lm\equiv 0(mod36)$ and $(2i^2+2j^2+3k^2+8l^2+8m^2+4lm)/36>1$.

Assume that $\langle g+(P\perp Q)\rangle$ is an abelian subgroup of order 6. For each $t\in \{1,2,3,4,5\}$, $tg+(P\perp Q)$ is a nontrivial coset in $P\perp Q$ and contains $g^{\prime }={\displaystyle \frac{1}{6}}(i^{\prime }u+j^{\prime }v+k^{\prime }w+l^{\prime }x+m^{\prime }y)$, where $i^{\prime },j^{\prime },k^{\prime },l^{\prime },m^{\prime }\in \{0,1,2,3,4,5\}$ and $i^{\prime }\equiv ti(mod6),j^{\prime }\equiv tj(mod6),k^{\prime }\equiv tk(mod6),l^{\prime }\equiv tl(mod6),m^{\prime }\equiv tm(mod6).$ So $2i^{\prime 2}+2j^{\prime 2}+3k^{\prime 2}+8l^{\prime 2}+8m^{\prime 2}+4l^{\prime }{m}^{\prime }\equiv 0(mod36)$ and $(2i^{\prime 2}+2j^{\prime 2}+3k^{\prime 2}+8l^{\prime 2}+8m^{\prime 2}+4l^{\prime }{m}^{\prime })/36>1$. Thus $(i,j,k,l,m)=(1,1,4,1,1)$ or $(5,5,2,5,5)$, which contradicts to $X/(P\perp Q)\cong 2\times 6$. The proof is complete. □

Theorem 5. 

Let X be a positive definite lattice with determinant 5 and rank 6. Then X is rectangular or isometric to $L_2\perp {\mathbb{Z}}^4$ or $A_4\perp {\mathbb{Z}}^2$.

Proof. 

If there exists a unit vector $u_1$ in X, then $X=\mathbb{Z}{u}_1\perp an{n}_X\left(u_1\right)$ because the natural map $X\to \mathcal{D}\left(\mathbb{Z}{u}_1\right)$ is onto. Then X is rectangular, $L_2\perp {\mathbb{Z}}^4$ or $A_4\perp {\mathbb{Z}}^2$ by induction and Theorem 4.

In the following, we try to prove that X contains a unit vector.

Let $u\in X^{*}$ so that $u+X$ generates $\mathcal{D}X$. Then $(u,u)={\displaystyle \frac{k}{5}}\notin \mathbb{Z}$ (or else the $X^{*}$ is an integer while $det\left(X^{*}\right)={\displaystyle \frac{1}{5}}$), where k is an integer.

When $k\equiv 1(mod5)$, we define a new lattice $M=span\left\{m\right\},(m,m)=5$. Then ${\displaystyle \frac{2m}{5}}\in M^{*}$ has norm ${\displaystyle \frac{4}{5}}$. Set $P:=X\perp M+\mathbb{Z}(u+{\displaystyle \frac{2m}{5}})$. We see that $|P:(X\perp M\left)\right|=5$ and so P is a unimodular integral lattice by Lemma 1. Thus $P\cong {\mathbb{Z}}^7$ by using Lemma 2, so we identify P in ${\mathbb{Z}}^7$. Then $X=an{n}_P\left(y\right)$ for some vector y of norm 5. Then the only possibilities for $y\in P$ are $(2,1,0^5),(1^5,0^2)$, up to monomial transformations. Therefore, X contains a unit vector.

When $k\equiv 2(mod5)$, we define a new lattice M whose basis $e,f$ gives the Gram matrix $\left(\begin{array}{cc}2& 1\\ 1& 3\end{array}\right)$. Then $m={\displaystyle \frac{1}{5}}(3e-f)\in M^{*}$ whose norm is ${\displaystyle \frac{3}{5}}$. Thus $P:=X\perp M+\mathbb{Z}(u+m)$ is a unimodular integral lattice. Thus $P\cong {\mathbb{Z}}^8$ or $E_8$. Since $E_8$ is an even lattice and there is a vector with norm 3 in P, we see $P\ncong E_8$. So we identify P with ${\mathbb{Z}}^8$. Then $X=an{n}_P(e,f)$ where $(e,e)=2,(f,f)=3,(e,f)=1$. The only possibility for such $e\in P$ is $(1^2,0^6)$ up to monomial transformations. And $f\in P$ is $(1^3,0^5)$ up to monomial transformations. Since $\left|supp\right(e)\cup supp(f\left)\right|=4<8$, X contains a unit vector.

The remaining cases $k\equiv 4(mod5)$ and $k\equiv 3(mod5)$ are similar to cases $k\equiv 1(mod5)$ and $k\equiv 2(mod5)$, respectively. □

4. Conclusions

We have classified positive definite integral lattices characterized by determinant 5 and rank $\le 6$ in this paper. The main result (Theorem 5) states that such a lattice X must be isometric to one of the following three types:

• Rectangular lattices: Lattices with an orthogonal basis, i.e., their Gram matrices are diagonal.
• ${\mathit{L}}_{\mathbf{2}}\perp {\mathbb{Z}}^{\mathbf{4}}$: Here $L_2$ is the rank-2 lattice defined by the Gram matrix $\left(\begin{array}{cc}2& 1\\ 1& 3\end{array}\right)$ (as in Theorem 3.1).
• ${\mathit{A}}_{\mathbf{4}}\perp {\mathbb{Z}}^{\mathbf{2}}$: This is where $A_4$ is the root lattice of rank 4 (Definition 1).

• The proof relies on an inductive analysis of lower-rank cases (Theorems 1–4) and a key reduction showing that X must contain a unit vector (otherwise, embedding X into a unimodular lattice leads to a contradiction). This classification bridges a gap in the literature and provides a complete solution for lattices of moderate rank with a small determinant.

5. Future Work and Open Problems

• Extend to Higher Ranks: Classify positive definite lattices with determinant 5 and rank $>6$. Do structural patterns (e.g., the number of isomorphism classes, automorphism group properties) from rank 6 generalize?
• Extend to Higher Determinants: Classify positive definite lattices of rank 6 with larger determinants (e.g., $det=6,7$). Applications in coding theory and Diophantine equations may be explored further.
• Applications: Applications in coding theory and Diophantine equations may be explored further.