The Classification of Lattices of Determinant 4 and Rank 5

Abstract

In this paper, we give the classification of the positive definite lattices with determinant 4 and rank 5.

1. Introduction

A lattice is a free Abelian group of finite rank with a rational valued symmetric bilinear form. We think of it as embedded in a rational vector space with the same basis. A lattice is rectangular if it has an orthogonal basis. The dual of a lattice M is denoted by $M^{\ast }=\{x\in V|(x,y)\in \mathbb{Z}, forally\in M\}$. The determinant of a lattice is $det\left(G\right)$, where G is any Gram matrix, meaning the square matrix $\left((x_i,x_j)\right)$ of size $rank\left(M\right)$ where $\left\{x_i\right\}$ is any basis. Now assume that M is integral, i.e., $(x,y)$ are integers for all $x,y\in M$. We call M even if $(x,x)$ is an even integer, for all $x\in M$. If M is not even, it is odd. The discriminant group of M is $\mathcal{D}\left(M\right)=M^{\ast }/M$. We call M unimodular if $det\left(M\right)=1$ or $-1$. When $det\left(M\right)\ne 0$, $\left|\mathcal{D}\right(M\left)\right|=|det(M\left)\right|$. We say that M is positive definite if $(x,x)>0$ for all $x\in M,x\ne 0$.

Lattice theory, as a fundamental discipline in modern mathematics, provides essential tools for studying discrete structures with profound applications across number theory, geometry, combinatorics, representation theory, and Lie theory [1]. Integral lattices play particularly important roles in classifying finite groups [2,3], constructing error-correcting codes [4,5], and Optimizing Algorithms [6]. The classification of positive definite lattices by rank and determinant thus represents a core problem with both theoretical significance and practical applications.

As to lattices’ classification, most research works are about unimodular lattices. There are some research works on the even unimodular lattices (see [7,8,9]). Mordell gave the classification of all positive definite rank 8 unimodular lattices in [10]. His proof depends on two deep theorems on quadratic forms. R.L. Griess Jr. classified these lattices by using only elementary techniques in [11]. Then he classified positive definite lattices of rank at most 7 and determinant 2, rank at most 6 and determinant 3, and rank at most 4 and determinant 4 in his book [12]. And Borcherds, R.E. also studied the positive defined lattices in [13]. He gave the classification of the 665 25-dimensional unimodular positive definite lattices and the 121 even 25-dimensional positive definite lattices of determinant 2.

Despite these advances, classifications for moderate ranks with small determinants (>1) remain incomplete. Lattices of rank 5 and determinant 4 occupy a particularly important gap; they represent the lowest rank where unimodularity constraints no longer apply, yet avoid the combinatorial explosion of higher dimensions. Their complete classification would bridge classical low-dimensional results and uncharted territory in lattice theory. Furthermore, such lattices encode solutions to challenging Diophantine equations [14] and optimize designs in communication networks [15], highlighting their multidisciplinary relevance.

In this paper, we try to give the classification of the positive definite integral lattices with determinant 4 and rank 5 by employing the local analysis approach [16].

2. Preliminaries

In this section, we give some notations and lemmas, which are useful for the later use and could be found in [12].

Definition 1

([12] pp. 33, 34). Let $L:={\mathbb{Z}}^{n+1}=\left\{(x_0,x_1,\dots ,x_n)\right|x_i\in \mathbb{Z}\}$ and $v=(1,1,\dots ,1)\in L$. The lattice $A_n$ is defined as the annihilator of $\mathbb{Z}v$ in L, i.e., $A_n:=an{n}_L\left(\mathbb{Z}v\right)=\{x\in L|(x,v)=0$ for all $v\in \mathbb{Z}v\}=an{n}_L\left(v\right)$, where $(·,·)$ denotes the bilinear form on L.

Further:

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

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

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

([12] p. 17, Theorem 2.3.3). Let L be a lattice, and M a sublattice of L of finite index $|L:M|$. Then ${det\left(L\right)|L:M|}^2=det\left(M\right).$

Lemma 2

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

Lemma 3

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

Lemma 4

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

Lemma 5

([12] p. 57, Lemma 5.3.1). Suppose that X is an integral lattice which has rank 4 and determinant 4. Then X is isometric to one of the following lattices:

(1) $2\mathbb{Z}\perp {\mathbb{Z}}^3$;

(2) $A_1\perp A_1\perp {\mathbb{Z}}^2$;

(3) $A_3\perp \mathbb{Z}$;

(4) $D_4$.

Lemma 6

([12] p. 18, Theorem 2.4.1). Let L be a nonsingular integral lattice and M a sublattice which ia a direct summand of L. Then

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

(2) define π to be the composition of ψ and the quotient map $M^{\ast }\mapsto M^{\ast }/M,$ then we have $Ker\left(\pi \right)=M+an{n}_{L^{\ast }}\left(M\right)$ 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 7

([12] p. 49, Theorem 5.0.3). Let $H(n,d):={\left({\displaystyle \frac{4}{3}}\right)}^{{\displaystyle \frac{n-1}{2}}}·d^{{\displaystyle \frac{1}{n}}}$ 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 Result

In the following, we classify the positive definite lattice which has rank 5 and determinant 4 on the basis of professor Griess’ result (Lemma 5).

Theorem 1.

Suppose that X is an positive definite lattice which has rank 5 and determinant 4. Then X is isometric to one of the following lattices:

(1) $2\mathbb{Z}\perp {\mathbb{Z}}^4$;

(2) $A_1\perp A_1\perp {\mathbb{Z}}^3$;

(3) $A_3\perp {\mathbb{Z}}^2$;

(4) $D_4\perp \mathbb{Z}$;

(5) $D_5$.

Proof. 

If X contains unit vector $u_1$, then $X=\mathbb{Z}{u}_1\perp an{n}_X\left(u_1\right)$ since the natural map $X\to \mathcal{D}\left(\mathbb{Z}{u}_1\right)$ is onto. So X is isometric to one of $2\mathbb{Z}\perp {\mathbb{Z}}^4$, $A_1\perp A_1\perp {\mathbb{Z}}^3$, $A_3\perp {\mathbb{Z}}^2$ and $D_4\perp \mathbb{Z}$ by Lemma 5.

If there is no unit vector in X, then X has a root u since $H(5,4)=2.3458\dots$ and Lemma 7. Noting the natural map $X\to \mathcal{D}\left(\mathbb{Z}u\right)$, when $X/(\mathbb{Z}u\perp an{n}_X\left(u\right))\cong 1$, we see that $an{n}_X\left(u\right)$ has determinant 2 by Lemma 1. Thus $an{n}_X\left(u\right)$ contains a vector of norm 1 by Lemma 3, a contradiction. So the natural map $X\to \mathcal{D}\left(\mathbb{Z}u\right)$ is onto. Thus $an{n}_X\left(u\right))$ is an integral lattice of rank 4 and determinant 8 by using Lemma 1 again. Since $H(4,8)=2.5893$, $an{n}_X\left(u\right))$ contains a root v. Define $P:=\mathbb{Z}u\perp \mathbb{Z}v,Q:=an{n}_X\left(P\right)$. Considering the natural map $X\to \mathcal{D}\left(P\right)$, if $|X:(P\perp Q\left)\right|=1$, then Q is of determinant 1 and has unit vector, a contradiction. So $X/(P\perp Q)\cong 2$ or $2\times 2$.

Case 1. $X/(P\perp Q)\cong 2$.

Q is of determinant 4 and rank 3. Then by Lemma 5, we see $Q\cong A_3$. Set $Q=Span\{x,y,x\}$ with $(x,x)=(y,y)=(z,z)=2,(x,z)=0,(x,y)=(y,z)=-1.$ The nontrivial coset of $P\perp Q$ contains an element of the form ${\displaystyle \frac{1}{2}}(e+f)$, where $e\in span\{u,v\}$ and $f\in span\{x,y,z\}$. We may furthermore arrange for $e=iu+jv,f=kx+ly+mz$, where $i,j,k,l,m\in \{0,1\}$. For the norm of ${\displaystyle \frac{1}{2}}(e+f)$ to be an integer and greater than 1, we see $i^2+j^2+k^2+l^2+m^2-kl-lm\equiv 0(mod2)$ and $(i^2+j^2+k^2+l^2+m^2-kl-lm)/2>1$. Then $(i,j,k,l,m)=(1,1,1,0,1)$. So $X=span\{u,v,x,y,{\displaystyle \frac{u+v+x+z}{2}}\}$, which is type (5) lattice in the theorem.

Case 2. $X/(P\perp Q)\cong 2\times 2$.

Q is of determinant 16 and rank 3. Since $H(3,16)=3.35978\dots$, $\mu \left(Q\right)=2$ or 3.

Case 2.1. $\mu \left(Q\right)=2$.

There is a root w in Q. Define $P_1:=\mathbb{Z}u\perp \mathbb{Z}v\perp \mathbb{Z}w,Q_1:=an{n}_X\left(P_1\right)$. $P_1$ is a direct summand of X. In fact, if $P_1$ is contained in $P^{\prime }$, which is a direct summand, and $|P^{\prime }:P_1|=t$. Then $det\left(P^{\prime }\right)={\displaystyle \frac{det\left(P_1\right)}{t^2}}={\displaystyle \frac{2^2·3}{t^2}}=2$. Then $P^{\prime }$ is rectangular by Lemma 3. So $P^{\prime }$ has unit vector, a contradiction.

Considering the natural map $X\to \mathcal{D}\left(P_1\right)$, $X/(P_1\perp Q_1)\lesssim 2\times 2\times 2$. If $|X/(P_1\perp Q_1\left)\right|=1$, then $8|4$, a contradiction. If $X/(P_1\perp Q_1)\cong 2$, then $Q_1$ is of rank 2 and determinant 2. $Q_1$ has a unit vector, a contradiction. If $X/(P_1\perp Q_1)\cong 2\times 2\times 2$, then the image of X in $\mathcal{D}\left(Q_1\right)$ is $2\times 2\times 2$, which contradicts $rank\left(Q_1\right)=2$. So $X/(P_1\perp Q_1)\cong 2\times 2$. Then $Q_1$ is of determinant 8 and rank 2. And the image of X in $\mathcal{D}\left(Q_1\right)$ is $2\times 2$. So $\mathcal{D}\left(Q_1\right)\cong 2\times 4$. For any $g\in Q_1$, ${\displaystyle \frac{1}{2}}g\in Q^{\ast }$. Then $Q_1=\sqrt{2}K$, where K is an integral lattice of rank 2 and determinant 2. Therefore, $Q_1$ has a rectangular basis $x,y$, with respective norms $2,4$.

The nontrivial coset of $P_1\perp Q_1$ contains an element of the form ${\displaystyle \frac{1}{2}}(e+f)$, where $e\in span\{u,v,w\}$ and $f\in span\{x,y\}$. We may furthermore arrange for $e=iu+jv+kw$, $f=lx+my$, where $i,j,k,l,m\in \{0,1\}$. For the norm of ${\displaystyle \frac{1}{2}}(e+f)$ to be an integer and greater than 1, we see $i^2+j^2+k^2+l^2+2m^2\equiv 0(mod2)$ and $(i^2+j^2+k^2+l^2+2m^2)/2>1$. Then $(i,j,k,l,m)\in \left\{\right(0,0,1,1,1),(0,1,0,1,1),(0,1,1,0,1),(1,0,0,1,1),(1,0,1,0,1),(1,1,0,0,1)$,

$(1,1,1,1,0),(1,1,1,1,1)\}$. So X is one of $X_1=span\{u,v,w,x,y,{\displaystyle \frac{w+x+y}{2}},{\displaystyle \frac{u+v+y}{2}}\},$

$X_2=span\{u,v,w,x,y,{\displaystyle \frac{v+x+y}{2}},{\displaystyle \frac{u+w+y}{2}}\}$, or $X_3=span\{u,v,w,x,y,{\displaystyle \frac{v+w+y}{2}},{\displaystyle \frac{u+x+y}{2}}\}$. $X_1$ and $X_2$ are isometric by the isometry defined by $u\mapsto u,v\mapsto w,w\mapsto v,x\mapsto x,y\mapsto y.$ And $X_2$ and $X_3$ are also isometric by the isometry defined by $u\mapsto u,v\mapsto v,w\mapsto x,x\mapsto w,y\mapsto y.$ So in this case 2.1, X is isometric to $span\{u,v,w,{\displaystyle \frac{w+x+y}{2}},{\displaystyle \frac{u+v+y}{2}}\}$, which is type (5) lattice in the theorem.

Case 2.2. $\mu \left(Q\right)=3$.

There is a vector w of norm 3 in Q. Define $P_1:=\mathbb{Z}u\perp \mathbb{Z}v\perp \mathbb{Z}w,Q_1:=an{n}_X\left(P_1\right)$. $P_1$ is a direct summand of X. In fact, if $P_1$ is contained in $P^{\prime }$, which is a direct summand, and $|P^{\prime }:P_1|=t$. Then $det\left(P^{\prime }\right)={\displaystyle \frac{det\left(P_1\right)}{t^2}}={\displaystyle \frac{2^2·3}{t^2}}=3$. Then $P^{\prime }$ is rectangular or $A_2\perp \mathbb{Z}$ by Lemma 4. So $P^{\prime }$ has unit vector, a contradiction.

Considering the natural map $X\to \mathcal{D}\left(P_1\right)$, $X/(P_1\perp Q_1)\lesssim 2\times 2\times 3$. Since $det\left(X\right)|X/(P_1\perp Q_1){|}^2=det\left(P_1\right)det\left(Q_1\right)$ and $|X/(P_1\perp Q_1\left)\right|$ is an integer, we see $X/(P_1\perp Q_1)\cong 3,2\times 3$ or $2\times 2\times 3.$

Case 2.2.1. $X/(P_1\perp Q_1)\cong 3$.

If $X/(P_1\perp Q_1)\cong 3$, then $Q_1$ is of rank 2 and determinant 3. $Q_1$ has unit vector, a contradiction.

Case 2.2.2. $X/(P_1\perp Q_1)\cong 2\times 3$.

So $X/(P_1\perp Q_1)\cong 2\times 3$. Then $Q_1$ is of determinant 12 and rank 2. Since $H(2,12)=4$, $\mu \left(Q_1\right)=3$ or 4.

Case 2.2.2.1. $\mu \left(Q_1\right)=3$.

Set $x\in Q_1$ is of norm 3. Considering the natural map $Q_1\to \mathcal{D}\left(\mathbb{Z}x\right)$, $Q_1/(\mathbb{Z}x\perp an{n}_{Q_1}\left(x\right))\lesssim 3$. Then we claim that $Q_1$ has rectangular basis $x,y$ with respective norm $3,4$. In fact, when $Q_1/(\mathbb{Z}x\perp an{n}_{Q_1}\left(x\right))\cong 1$, it is obvious. When $Q_1/(\mathbb{Z}x\perp an{n}_{Q_1}\left(x\right))\cong 3$, $an{n}_{Q_1}\left(x\right)=\mathbb{Z}z$ is of determinant 36. A nontrivial coset of $\mathbb{Z}x\perp an{n}_{Q_1}\left(x\right))$ contains an element of the form ${\displaystyle \frac{1}{3}}(e+f)$, where $e\in span\left\{x\right\}$ and $f\in span\left\{z\right\}$. We may furthermore arrange for $e=ix,f=jz$, where $i,j\in \{0,1,2\}$. For the norm of ${\displaystyle \frac{1}{3}}(e+f)$ to be an integer and greater than 2, we see $3i^2+36j^2\equiv 0(mod9)$ and $(3i^2+36j^2)/9>2$. Then $(i,j)=(0,1)$, $(0,2)$. So $Q_1=span\{x,z,{\displaystyle \frac{z}{3}}\}=span\{x,{\displaystyle \frac{z}{3}}\}$. Set $y:={\displaystyle \frac{z}{3}}$. Then $Q_1$ has rectangular basis $x,y$ with respective norm $3,4$.

A nontrivial coset of $P_1\perp Q_1$ contains an element g of the form ${\displaystyle \frac{1}{6}}(e+f)$, where $e\in span\{u,v,w\}$ and $f\in span\{x,y\}$. We may furthermore arrange for $e=iu+jv+kw$, $f=lx+my$, where $i,j,k,l,m\in \{0,1,2,3,4,5\}$. For the norm of ${\displaystyle \frac{1}{2}}(e+f)$ to be an integer and greater than 1, we see $2i^2+2j^2+3k^2+3l^2+4m^2\equiv 0(mod36)$ and $(2i^2+2j^2+3k^2+3l^2+4m^2)/36>1$.

Assume that $X/(P_1\perp Q_1)=\langle g+(P_1\perp Q_1)\rangle$, which is an Abelian group of order 6. Then for any $t\in \{1,2,3,4,5\}$, $tg+(P_1\perp Q_1)$ is also a nontrivial coset of $P_1\perp Q_1$ 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 $2i^{\prime 2}+2j^{\prime 2}+3k^{\prime 2}+3l^{\prime 2}+4m^{\prime 2}\equiv 0(mod36)$ and $(2i^{\prime 2}+2j^{\prime 2}+3k^{\prime 2}+3l^{\prime 2}+4m^{\prime 2})/36>1$.

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

$(3,2,1,3,2),(3,4,5,3,4),$$(2,3,1,3,2),(4,3,5,3,4),$$(4,3,3,5,4),(2,3,3,1,2)\}$. So X is one of

$X_1=span\{u,v,w,x,y,{\displaystyle \frac{5v+3w+x+5y}{6}}\}$, $X_2=span\{u,v,w,x,y,$${\displaystyle \frac{5v+w+3x+5y}{6}}\}$,

$X_3=span\{u,v,w,x,y,{\displaystyle \frac{5u+w+3x+5y}{6}}\}$, $X_4=span\{u,v,w,x,y,{\displaystyle \frac{5u+3w+x+5y}{6}}\}$,

$X_5=span\{u,v,w,x,y,{\displaystyle \frac{3u+2v+3w+x+2y}{6}}\}$, $X_6=span\{u,v,w,x,y,{\displaystyle \frac{3u+2v+w+3x+2y}{6}}\}$,

$X_7=span\{u,v,w,x,y,{\displaystyle \frac{2u+3v+w+3x+2y}{6}}\}$ and $X_8=span\{u,v,w,x,y,{\displaystyle \frac{2u+3v+3w+x+2y}{6}}\}$.

Considering the isometries defined by $u\mapsto u,v\mapsto v,w\mapsto x,x\mapsto w,y\mapsto y$ and $u\mapsto v,v\mapsto u,w\mapsto w,x\mapsto x,y\mapsto y$, we see that $X_1,X_2,X_3$ and $X_4$ are isometric and $X_5,X_6,X_7$ and $X_8$ are isometric. So X is isometric to $span\{u,v,w,y,{\displaystyle \frac{5v+3w+x+5y}{6}}\}$ or $span\{u,v,w,y,{\displaystyle \frac{3u+2v+3w+x+2y}{6}}\}$. But $(v,{\displaystyle \frac{5v+3w+x+5y}{6}})={\displaystyle \frac{5}{3}}$ and $(y,{\displaystyle \frac{3u+2v+3w+x+2y}{6}})={\displaystyle \frac{4}{3}}$, which are contradictory to the fact that X is integral.

Case 2.2.2.2. $\mu \left(Q_1\right)=4$.

Set $x\in Q_1$ is of norm 4. We claim that $Q_1\cong span\{x,y|(x,x)=(y,y)=4,(x,y)=2\}$.

In fact, considering the natural map $Q_1\to \mathcal{D}\left(\mathbb{Z}x\right)$, $Q_1/(\mathbb{Z}x\perp an{n}_{Q_1}\left(x\right))\cong 1,2,$ or 4. If $Q_1/(\mathbb{Z}x\perp an{n}_{Q_1}\left(x\right))\cong 1$, then $Q_1$ has a vector of norm 3, which contradict $\mu \left(Q_1\right)=4$. When $Q_1/(\mathbb{Z}x\perp an{n}_{Q_1}\left(x\right))\cong 2$, $an{n}_{Q_1}\left(x\right)=\mathbb{Z}z$ is of determinant 12. A nontrivial coset of $\mathbb{Z}x\perp an{n}_{Q_1}\left(x\right))$ contains an element of the form ${\displaystyle \frac{1}{2}}(e+f)$, where $e\in span\left\{x\right\}$ and $f\in span\left\{z\right\}$. We may furthermore arrange for $e=ix,f=jz$, where $i,j\in \{0,1\}$. For the norm of ${\displaystyle \frac{1}{2}}(e+f)$ to be an integer and greater than 3, we see $4i^2+12j^2\equiv 0(mod4)$ and $(4i^2+12j^2)/4>3$. Then $(i,j)=(1,1)$. So $Q_1$ is isometric to $span\{x,z,{\displaystyle \frac{x+z}{2}}\}\cong span\{x,y|(x,x)=(y,y)=4,(x,y)=2\}$. When $Q_1/(\mathbb{Z}x\perp an{n}_{Q_1}\left(x\right))\cong 4$, $an{n}_{Q_1}\left(x\right)=\mathbb{Z}z$ is of determinant 48. A nontrivial coset of $\mathbb{Z}x\perp an{n}_{Q_1}\left(x\right))$ contains an element of the form ${\displaystyle \frac{1}{2}}(e+f)$, where $e\in span\left\{x\right\}$ and $f\in span\left\{z\right\}$. We may furthermore arrange for $e=ix,f=jz$, where $i,j\in \{0,1,2,3\}$. For the norm of ${\displaystyle \frac{1}{2}}(e+f)$ to be an integer and greater than 3, we see $4i^2+48j^2\equiv 0(mod16)$ and $(4i^2+48j^2)/16>3$. Then $(i,j)=(0,2)$ or $(0,3)$. This is incompatible with $Q_1/(\mathbb{Z}x\perp an{n}_{Q_1}\left(x\right))\cong 4$. So $Q_1\cong span\{x,y|(x,x)=(y,y)=4$, $(x,y)=2\}$.

A nontrivial coset of $P_1\perp Q_1$ contains an element g of the form ${\displaystyle \frac{1}{6}}(e+f)$, where $e\in span\{u,v,w\}$ and $f\in span\{x,y\}$. We may furthermore arrange for $e=iu+jv+kw,f=lx+my$, where $i,j,k,l,m\in \{0,1,2,3,4,5\}$. For the norm of ${\displaystyle \frac{1}{2}}(e+f)$ to be an integer and greater than 1, we see $2i^2+2j^2+3k^2+4l^2+4m^2+4lm\equiv 0(mod36)$ and $(2i^2+2j^2+3k^2+4l^2+4m^2+4lm)/36>1$.

Assume that $X/(P_1\perp Q_1)=\langle g+(P_1\perp Q_1)\rangle$, which is an Abelian group of order 6. Then for any $t\in \{1,2,3,4,5\}$, $tg+(P_1\perp Q_1)$ is also a nontrivial coset of $P_1\perp Q_1$ 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 $2i^{\prime 2}+2j^{\prime 2}+3k^{\prime 2}+4l^{\prime 2}+4m^{\prime 2}+4l^{\prime }{m}^{\prime }\equiv 0(mod36)$ and $(2i^{\prime 2}+2j^{\prime 2}+3k^{\prime 2}+4l^{\prime 2}+4m^{\prime 2}+4l^{\prime }{m}^{\prime })/36>1$.

But there is no $(i,j,k,l,m)$ such that the above conditions are met. So in Case 2.2.2, there is no lattice of rank 5 and determinant 4.

Case 2.2.3. $X/(P_1\perp Q_1)\cong 2\times 2\times 3$.

In this case, we see $det{Q}_1=48$ and the image of X in $\mathcal{D}\left(Q_1\right)$ is $2\times 6$. Then the Sylow 2-subgroup of $\mathcal{D}\left(Q_1\right)$ has exponent $<16$. So $Q_1:=\sqrt{2}K$, where K has rank 2 and determinant 12. Since $\mu \left(Q_1\right)\ge 3$ and $H(2,12)=4$, we see $\mu \left(K\right)=2,3$ or 4.

If $\mu \left(K\right)=3,4$, then $K\cong span\{x,y|(x,x)=(y,y)=4,(x,y)=2\}$ or $span\{x,y|(x,x)=3$, $(y,y)=4$, $(x,y)=0\}$ by the same analysis in the case 2.2.2.

If $\mu \left(K\right)=2$, set $x\in K$ is of norm 2. We claim that $K\cong span\{x,y|(x,x)=2$, $(y,y)=6$, $(x,y)=0\}$.

In fact, considering the natural map $K\to \mathcal{D}\left(\mathbb{Z}x\right)$, $K/(\mathbb{Z}x\perp an{n}_K\left(x\right))\cong 1,2$. If $K/(\mathbb{Z}x\perp an{n}_K\left(x\right))\cong 1$, then $K\cong span\{x,y|(x,x)=2,(y,y)=6,(x,y)=0\}$. When $K/(\mathbb{Z}x\perp an{n}_K\left(x\right))\cong 2$, $an{n}_K\left(x\right)=\mathbb{Z}z$ is of determinant 24. A nontrivial coset of $\mathbb{Z}x\perp an{n}_K\left(x\right))$ contains an element of the form ${\displaystyle \frac{1}{2}}(e+f)$, where $e\in span\left\{x\right\}$ and $f\in span\left\{z\right\}$. We may furthermore arrange for $e=ix,f=jz$, where $i,j\in \{0,1\}$. For the norm of ${\displaystyle \frac{1}{2}}(e+f)$ to be an integer and greater than 1, we see $2i^2+24j^2\equiv 0(mod4)$ and $(2i^2+24j^2)/4>1$. Then $(i,j)=(0,1)$. So K is isometric to $span\{x,{\displaystyle \frac{z}{2}}\}\cong span\{x,y|(x,x)=2,(y,y)=6,(x,y)=0\}$.

So $Q_1\cong span\{x,y|(x,x)=4,(y,y)=12,(x,y)=0\}$, $span\{x,y|(x,x)=6$, $(y,y)=8$, $(x,y)=0\}$ or $span\{x,y|(x,x)=8,(y,y)=8,(x,y)=4\}$.

Case 2.2.3.1. $Q_1\cong span\{x,y|(x,x)=4,(y,y)=12,(x,y)=0\}$.

A nontrivial coset of $P_1\perp Q_1$ contains an element g of the form ${\displaystyle \frac{1}{6}}(e+f)$, where $e\in span\{u,v,w\}$ and $f\in span\{x,y\}$. We may furthermore arrange for $e=iu+jv+kw$, $f=lx+my$, where $i,j,k,l,m\in \{0,1,2,3,4,5\}$. For the norm of ${\displaystyle \frac{1}{6}}(e+f)$ to be an integer and greater than 1, we see $2i^2+2j^2+3k^2+4l^2+12m^2\equiv 0(mod36)$ and $(2i^2+2j^2+3k^2+4l^2+12m^2)/36>1$.

Assume that $\langle g+(P_1\perp Q_1)\rangle$ is an Abelian subgroup of order 6 of $X/(P_1\perp Q_1)$. Then for any $t\in \{1,2,3,4,5\}$, $tg+(P_1\perp Q_1)$ is also a nontrivial coset of $P_1\perp Q_1$ 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 $2i^{\prime 2}+2j^{\prime 2}+3k^{\prime 2}+4l^{\prime 2}+12m^{\prime 2}\equiv 0(mod36)$ and $(2i^{\prime 2}+2j^{\prime 2}+3k^{\prime 2}+4l^{\prime 2}+12m^{\prime 2})/36>1$. Then $(i,j,k,l,m)\in A:=\{$ $(0,2,0,2,5),(0,2,4,2,3),(0,4,0,4,1),(0,4,2,4,3),(1,3,0,1,2),$

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

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

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

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

Since X is integral, we see $(u,g),(v,g),(w,g),(x,g)$ are integers and then $3|i,3|j,2|k$ and $3|l$. Thus there is no element in A such that this condition is met, which is a contradiction.

Case 2.2.3.2. $Q_1\cong span\{x,y|(x,x)=6,(y,y)=8,(x,y)=0\}$.

A nontrivial coset of $P_1\perp Q_1$ contains an element g of the form ${\displaystyle \frac{1}{6}}(e+f)$, where $e\in span\{u,v,w\}$ and $f\in span\{x,y\}$. We may furthermore arrange for $e=iu+jv+kw$, $f=lx+my$, where $i,j,k,l,m\in \{0,1,2,3,4,5\}$. For the norm of ${\displaystyle \frac{1}{6}}(e+f)$ to be an integer and greater than 1, we see $2i^2+2j^2+3k^2+6l^2+8m^2\equiv 0(mod36)$ and $(2i^2+2j^2+3k^2+6l^2+8m^2)/36>1$.

Assume that $\langle g+(P_1\perp Q_1)\rangle$ is an Abelian subgroup of order 6 of $X/(P_1\perp Q_1)$. Then for any $t\in \{1,2,3,4,5\}$, $tg+(P_1\perp Q_1)$ is also a nontrivial coset of $P_1\perp Q_1$ 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 $2i^{\prime 2}+2j^{\prime 2}+3k^{\prime 2}+6l^{\prime 2}+8m^{\prime 2}\equiv 0(mod36)$ and $(2i^{\prime 2}+2j^{\prime 2}+3k^{\prime 2}+6l^{\prime 2}+8m^{\prime 2})/36>1$. Then $(i,j,k,l,m)\in A:=\{$ $(0,0,2,4,3),(0,0,4,2,3),(0,3,2,1,3),(0,3,4,5,3),(1,1,0,4,1),$

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

Since X is integral, we see $(u,g),(v,g),(w,g),(x,g)$ are integers and then $3|i,3|j,2|k$ and $3|m$. So $(i,j,k,l,m)\in \{$ $(0,0,2,4,3),(0,0,4,2,3),(0,3,2,1,3),(0,3,4,5,3),$

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

Furthermore, if $\langle g+(P_1\perp Q_1)\rangle$ is an Abelian subgroup of order 2 of $X/(P_1\perp Q_1)$, then $(i,j,k,l,m)\in \{$ $(0,0,0,0,3),(0,3,0,3,3),(3,0,0,3,3),(3,3,0,0,3)\}$. Thus there is no Sylow 2-group in $X/(P_1\perp Q_1)$, a contradiction.

Case 2.2.3.3. $Q_1\cong span\{x,y|(x,x)=8,(y,y)=8,(x,y)=4\}$.

A nontrivial coset of $P_1\perp Q_1$ contains an element g of the form ${\displaystyle \frac{1}{6}}(e+f)$, where $e\in span\{u,v,w\}$ and $f\in span\{x,y\}$. We may furthermore arrange for $e=iu+jv+kw,f=lx+my$, where $i,j,k,l,m\in \{0,1,2,3,4,5\}$. For the norm of ${\displaystyle \frac{1}{6}}(e+f)$ to be an integer and greater than 1, we see $2i^2+2j^2+3k^2+8l^2+8m^2+8lm\equiv 0(mod36)$ and $(2i^2+2j^2+3k^2+8l^2+8m^2+8lm)/36>1$.

Assume that $\langle g+(P_1\perp Q_1)\rangle$ is an Abelian subgroup of order 6 of $X/(P_1\perp Q_1)$. Then for any $t\in \{1,2,3,4,5\}$, $tg+(P_1\perp Q_1)$ is also a nontrivial coset of $P_1\perp Q_1$ 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 $2i^{\prime 2}+2j^{\prime 2}+3k^{\prime 2}+8l^{\prime 2}+8m^{\prime 2}+8l^{\prime }{m}^{\prime }\equiv 0(mod36)$ and $(2i^{\prime 2}+2j^{\prime 2}+3k^{\prime 2}+8l^{\prime 2}+8m^{\prime 2}+8l^{\prime }{m}^{\prime })/36>1$. Then $(i,j,k,l,m)\in A:=\{$ $(0,0,2,1,4),(0,0,2,2,5),(0,0,2,4,1),(0,0,2,5,2),$$(0,0,2,5,5),$

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

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

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

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

Furthermore, if $\langle g+(P_1\perp Q_1)\rangle$ is an Abelian subgroup of order 2 of $X/(P_1\perp Q_1)$, then $(i,j,k,l,m)\in \{$ $(0,0,0,3,0),(0,0,0,0,3),(0,0,0,3,3),$ $(3,3,0,3,3),(3,3,0,3,0),(3,3,0,0,3)\}$.

Thus the Abelian 2-group of type $2\times 2$ is

$A_1:=\langle (0,0,0,3,0),(0,0,0,0,3),(0,0,0,3,3)\rangle ,$

$A_2:=\langle (0,0,0,3,0),(3,3,0,3,3),(3,3,0,0,3)\rangle ,$

$A_3:=\langle (0,0,0,0,3),(3,3,0,3,3),(3,3,0,3,0)\rangle$ or

$A_4:=\langle (0,0,0,3,3),(3,3,0,3,0),(3,3,0,0,3)\rangle .$

If $\langle g+(P_1\perp Q_1)\rangle$ is an Abelian subgroup of order 3 of $X/(P_1\perp Q_1)$, 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 $B_2:=\langle (0,0,4,4,4),(0,0,2,2,2)\rangle .$

Since $(0,0,2,4,4)+(0,0,0,3,3)=(0,0,2,1,1)$ and ${\displaystyle \frac{1}{6}}(2w+x+y)$ has norm 1, we see X is isometric to

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

$X_2:=span\{u,v,w,x,y,{\displaystyle \frac{1}{6}}\left(3y\right),{\displaystyle \frac{1}{6}}(3u+3v+3x),{\displaystyle \frac{1}{6}}(4w+2x+2y)\}$,

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

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

$X_5:=span\{u,v,w,x,y,{\displaystyle \frac{1}{6}}\left(3y\right),{\displaystyle \frac{1}{6}}(3u+3v+3x),{\displaystyle \frac{1}{6}}(2w+2x+2y)\}$ or

$X_6:=span\{u,v,w,x,y,{\displaystyle \frac{1}{6}}(3x+3y),{\displaystyle \frac{1}{6}}(3u+3v+3x),{\displaystyle \frac{1}{6}}(2w+2x+2y)\}$.

Considering the isometry defined by $u\mapsto u,v\mapsto v,w\mapsto w,x\mapsto y,y\mapsto x$, we see $X_1\cong X_2$ and $X_4\cong X_5$. By calculation, we find unit vectors ${\displaystyle \frac{x+y}{2}}-{\displaystyle \frac{1}{3}}(w+x+y)={\displaystyle \frac{1}{6}}(x+y-2w)\in X_3$, $X_6$, and ${\displaystyle \frac{y}{2}}-{\displaystyle \frac{w+x+y}{3}}={\displaystyle \frac{y}{2}}-({\displaystyle \frac{2w+x+y}{3}}-w)={\displaystyle \frac{1}{6}}(y-2x-2w)\in X_1$,$X_5$, which are contradictions. So in Case 2.2, there is no lattice of rank 5 and determinant 4. The proof is complete. □

4. Conclusions

In this paper, we have classified the positive definite integral lattices of rank 5 and determinant 4. We showed that there are exactly five such lattices up to isometry, as listed in Theorem 1. These lattices are $2\mathbb{Z}\perp {\mathbb{Z}}^4$, $A_1\perp A_1\perp {\mathbb{Z}}^3$, $A_3\perp {\mathbb{Z}}^2$, $D_4\perp \mathbb{Z}$, and $D_5$.

This classification avoids the combinatorial explosion of higher dimensions. And it would bridge classical low-dimensional results and uncharted territory in lattice theory. Future work might include the classification of lattices with higher rank or determinant, or studying the automorphism groups of these lattices.