Determinants for integral forms in lattice type vertex operator algebras

TL;DR

This paper establishes a determinant formula for the standard integral form of lattice vertex operator algebras, advancing the understanding of their algebraic structure.

Contribution

It provides a new determinant formula specifically for the integral form of lattice vertex operator algebras, which was not previously known.

Findings

Derived a determinant formula for the integral form

Enhanced understanding of lattice vertex operator algebra structure

Potential applications in algebraic and conformal field theories

Abstract

We prove a determinant formula for the standard integral form of a lattice vertex operator algebra.

Equations

$$H[t^{-1}]t^{-1}={\oplus }_{r\ge 1}H\otimes \mathbb{C}{t}^{-r}.$$

$$M(1)={\oplus }_{n\ge 0}M(1{)}_n$$

$$E^{-}(-\alpha ,z):=\exp \left(\sum _{n>0}\frac{\alpha (-n)}{n}{z}^n\right)=\sum _{n\ge 0}{s}_{\alpha ,n}{z}^n.$$

$$B_i=\{s_{x_i,n_1}\cdots s_{x_i,n_q}|n_1\ge \cdots \ge n_q\ge 0\}$$

$$E^{-}(-n\alpha ,z)=E^{-}(-\alpha ,z{)}^n=(\sum _{m\ge 0}{s}_{\alpha ,m}{z}^m{)}^n.$$

$$spa{n}_{\mathbb{Z}}(x_1,x_2,x_3,\dots x_k)>spa{n}_{\mathbb{Z}}(p{x}_1,x_2,x_3,\dots x_k)>$$

$$spa{n}_{\mathbb{Z}}(p{x}_1,p{x}_2,x_3,\dots x_k)>\cdots >spa{n}_{\mathbb{Z}}(p{x}_1,p{x}_2,p{x}_3,\dots p{x}_k).$$

$$B(x_1,x_2,x_3,\dots x_k;a)>B(p{x}_1,x_2,x_3,\dots x_k;a)>$$

$$B(p{x}_1,p{x}_2,x_3,\dots x_k;a)>\cdots >B(p{x}_1,p{x}_2,p{x}_3,\dots p{x}_k;a)$$

$$|B(x_1,x_2,\dots ,x_k;a):B(p{x}_1,x_2,\dots ,x_k;a)|$$

$$\hat{H}=H\otimes \mathbb{C}[t,t^{-1}]\oplus \mathbb{C}K$$

$$S(k,n):=\frac{1}{k}\sum _{a\in \mathcal{A}(n)}N(k,a)=\frac{1}{k}\sum _{a=(a_j)\in \mathcal{A}(n)}\prod _{j=1}^n{a}_j\left(\genfrac{}{}{0ex}{}{a_j+k-1}{k-1}\right).$$

$$\det (B(p{x}_1,x_2,\dots ,x_k;n))=\det (B(x_1,x_2,\dots ,x_k;n))p^{2S(k,n)}$$

$$⟨e^{\alpha }\mid e^{\beta }⟩={\delta }_{\alpha +\beta ,0}$$

$$⟨\alpha (n)u\mid v⟩=-⟨u\mid \alpha (-n)v⟩$$

$$b_n:=\prod _{a=(a_1,a_2,\cdots )\in \mathcal{A}(n)}\prod _{i\ge 1}{i}^{a_i}\cdot a_i!.$$

$$\prod _{n_1,...,n_k\ge 0,\sum n_i=n}{b}_{n_1}\cdots b_{n_k}$$

$$⟨x_i(-p{)}^s\mid x_i(-p{)}^s⟩=(-1{)}^s⟨1\mid x_i(p{)}^s{x}_i(-p)1⟩$$

$$=(-1{)}^{s}s!p^s⟨1\mid 1⟩$$

$$=(-1{)}^{s}s!p^s$$

$$|\det (C_i)|=|\det (A_i)|[A_i:C_i{]}^2,\det (C_2)=\det (C_1)[C_1:C_2{]}^2$$

$$1=|\det (A_{{\mathbb{Z}}^k}(n))|=[B_K(n):B_{{\mathbb{Z}}^k}(n){]}^2|\det (A_K(n))|=T^{2kS(k,n)}|\det (A_K(n))|.$$

$$|\det (A_L(n))|=(p_1\cdots p_k{)}^{2S(k,n)}{T}^{-2kS(k,n)}=\det (L{)}^{2S(k,n)}.$$

$$(V_L{)}_{\mathbb{Z},n}={\oplus }_{m=0}^n{\oplus }_{\alpha \in Y_{2m}}{W}_n^{\alpha }$$

$$\det ((V_L{)}_{\mathbb{Z},n})=\prod _{m=0}^n\prod _{\alpha \in Y_{2m}}\det (W_m^{\alpha }).$$

$$|\det (W_n^{\alpha })|=\det (A_L(n-m){)}^2.$$

$$|\det ((V_L{)}_{\mathbb{Z},n})|=\prod _{m=0}^n\det (L{)}^{|L_{2m}|2S(k,n-m)}.$$

Full text

1 Introduction

We have studied group-invariant integral forms in vertex operator algebras [2, 3]. In this article, we study standard integral forms in lattice vertex operator algebras and give the determinant of each homogeneous piece as a particular integral power of determinant of the input positive definite even integral lattice. When the lattice is unimodular, all these homogeneous pieces have determinant 1, already proved in [2]. For lattices of other determinants, there did not seem to be an obvious answer. Borcherds stated without proof in [1] that the determinant of a homogeneous piece was some (unspecified) integral power of the input lattice.

The standard integral form $V_{L.{\mathbb{Z}}}$ for a lattice vertex operator algebra $V_{L}$ is reviewed in Section 4, Definition (4.1). Lemma (3.3) shows that our main theorem (5.1) is reduced to a study of determinants for integral forms within a certain symmetric algebra. The latter determinant is therefore our main object of study in this article.

2 Background

Lemma 2.1**.**

If $x_{1},\dots,x_{k}$ are variables, then the number of monomials $x_{1}^{a_{1}}\cdots x_{k}^{a_{k}}$, $a_{i}\in{\mathbb{Z}}_{\geq 0}$, of total degree $n$ is ${n+k-1}\choose{k-1}$.

**Proof.  **This is essentially a counting result, called Balls in Urns. Monomials correspond to the set of $k-1$ marker balls to be chosen among a set of $n+k-1$ balls arranged in a straight line. One adds marker ball 0 at the very beginning and marker ball $k$ at the very end. The sequence $a_{1},\dots,a_{k}$ gives the lengths of the gaps between successive marker balls. $\square$

Lemma 2.2**.**

If $J\leq K$ are finite rank lattices and the index $|J:K|$ is finite, then $det(K)=|J:K|^{2}det(J)$.

3 Symmetric algebas

Notation 3.1**.**

Let $H$ be a $k$-dimensional vector space over ${\mathbb{C}}$ and let $t$ be a variable. For $r\geq 1$, let $H_{r}=H\otimes{\mathbb{C}}t^{-r}$ be a copy of $H$, defined to have degree $r$, and set $h(-r)=h\otimes t^{-r}$ for $h\in H.$ We shall work in the symmetric algebra $M(1)=\mathbb{S}[H[t^{-1}]t^{-1}]={\mathbb{C}}[h(-r)|h\in H]$ where

[TABLE]

Then

[TABLE]

is graded such that $M(1)_{n}$ is spanned by $h_{1}(-r)\cdots h_{p}(-r_{p})$ for $h_{1},...,h_{p}\in H$ and $r_{1},...,r_{p}\in{\mathbb{N}}$ with $\sum_{i}r_{i}=n.$ For a sequence $a=(a_{1},\dots,a_{n},\dots)$ of nonnegative integers which is almost all zero, define $wt(a):=\sum_{j\geq 1}\ ja_{j}$. For $n\geq 0$, define $\mathcal{A}(n)$ to be the set of such $a$ of weight $n$. Note that $\mathcal{A}(0)=\emptyset$.

Suppose that $x_{1},\dots,x_{k}$ is a basis of $H$. Then $L={\mathbb{Z}}x_{1}+\cdots+{\mathbb{Z}}x_{k}$ is a free abelian group of rank $r$ in $H$ and $M(1)={\mathbb{C}}[x_{1}(-r),\cdots x_{k}(-r)|r\geq 1].$ Define $B(a):=B(x_{1},\dots x_{k};a)$ to be the ${\mathbb{Z}}$-span of all words $w_{1}\cdots w_{n}$ in $M(1)$ where $w_{i}$ is a product of length $a_{i}$ in the variables $x_{j}(-i)$, for $j\in\{1,2,\dots,k\}$. Finally, for an integer $n\geq 0$, define $B_{L}(n):=B(x_{1},\dots x_{k};n):=\oplus_{a\in\mathcal{A}(n)}B(a)$ and $B_{L}:=\oplus_{n\geq 0}B_{L}(n)={\mathbb{Z}}[x_{1}(-r),\cdots x_{k}(-r)|r\geq 1]$. Then $B_{L}$ is a subring of $R.$ Note that these objects are unchanged if $x_{1},\dots,x_{k}$ is replaced by any basis of the ${\mathbb{Z}}$ span of $x_{1},\dots,x_{k}$.

Notation 3.2**.**

Let $A_{L}$ be the ${\mathbb{Z}}$-submodule of $M(1)$ generated by $s_{\alpha,n}$ for $\alpha_{i}\in\{x_{1},\dots x_{k}\}$ and $n\geq 0$ where

[TABLE]

Although we do not use the vertex operator algebra structure on $M(1)$, we use the notations $E^{-}(-\alpha,z)$ and $s_{\alpha,n}$ from [4] and [2] here. Then $A_{L}$ has a ${\mathbb{Z}}$-base $B_{1}B_{2}\cdots B_{k}$ where

[TABLE]

and $B_{1}\cdots B_{k}=\{u_{1}u_{2}\cdots u_{k}|u_{i}\in B_{i}\}.$

We also set $A_{L}(n)=A_{L}\cap M(1)_{n}$ for all $n.$ The following result will be useful in computing the determinants for the lattice vertex operator algebras.

Lemma 3.3**.**

$B_{L}$* is a subring of $A_{L}$ and the index $[A_{L}(n):B_{L}(n)]$ is independent of the base $\{x_{1},...,x_{k}\}$ for any $n\geq 0.$*

**Proof.  **We first prove that $B_{L}$ is a subring of $A_{L}.$ It is good enough to show that $\alpha(-n)\in B_{L}$ for $\alpha\in\{x_{1},...,x_{k}\}$ and $n\geq 0.$ Note that

[TABLE]

So the coefficient $c_{n}$ of $z^{n}$ in $E^{-}(-n\alpha,z)$ lies in $A_{L}.$ Clearly, $c_{n}$ is also the coefficient of $z^{n}$ in $(\sum_{m=0}^{n}s_{\alpha,m}z^{m})^{n}.$ A straightforward computation shows that $a_{n}=\alpha(-n)+u$ where $u$ is a ${\mathbb{Z}}$-linear combination of elements of the form $s_{\alpha,m_{1}}s_{\alpha,m_{2}}\cdots$ with$m_{i}<n$ and $m_{1}+m_{2}+\cdots=n.$ As a result, $\alpha(-n)\in A_{L}.$

To show that the index $[A_{L}(n):B_{L}(n)]$ is independent of the base $\{x_{1},...,x_{k}\}$ for any $n\geq 0$, we let $\{y_{1},...,y_{k}\}$ be another basis of $H$ and $K={\mathbb{Z}}y_{1}+\cdots+{\mathbb{Z}}y_{k}.$ Then a group isomorphism $f$ from $L$ to $K$ by sending $x_{i}$ to $y_{i}$ induces a ring isomorphism $\hat{f}$ from $R$ to itself such that $\hat{f}(A_{L})=A_{K}$ and $\hat{f}(B_{L})=B_{K}.$ It is evident that $\hat{f}$ is a degree preserving map. As a result, $\hat{f}(A_{L}(n))=A_{K}(n)$ and $\hat{f}(B_{L})=B_{K}.$ Thus, $[A_{L}(n):B_{L}(n)]=[A_{K}(n):B_{K}(n)].$ $\square$

Note that both $A_{L}$ and $B_{L}$ are ${\mathbb{Z}}$-forms of $M(1).$

Lemma 3.4**.**

(i) $rank(S^{m}(H))={{m+k-1}\choose{k-1}}$;

(ii) $rank(B(a))=\prod_{j=1}^{n}{{a_{j}+k-1}\choose{k-1}}$;

(iii) $wt(B(a))=\sum_{j\geq 1}\ ja_{j}$;

(iv) $rank(B(n))=\sum_{a:wt(a)=n}rank(B(a))$.

**Proof.  **Straightforward, with Lemma (2.1). $\square$

So far, there is no bilinear form in this discussion. We shall introduce forms later, after Corollary 3.8.

We now compare what happens to the $B(a)$ when $x_{1},\dots,x_{n}$ is replaced by another basis. We already noted in Notation 3.1 that $B(x_{1},\dots x_{k};a)=B(y_{1},\dots y_{k};a)$ if $span_{{\mathbb{Z}}}(x_{1},\dots x_{k})=span_{{\mathbb{Z}}}(y_{1},\dots y_{k})$.

Using the proof of Lemma 3.3 we can easily have:

Lemma 3.5**.**

If $x_{1},\dots x_{k}$ and $y_{1},\dots y_{k}$ are bases and if $span_{{\mathbb{Z}}}(x_{1},\dots x_{k})$ contains $span_{{\mathbb{Z}}}(y_{1},\dots y_{k})$, then for any invertible linear transformation $T$ on $H$, $B(x_{1},\dots x_{k};a)/B(y_{1},\dots y_{k};a)\cong B(Tx_{1},\dots Tx_{k};a)/B(Ty_{1},\dots Ty_{k};a)$. In particular, we have equality of indices $|B(x_{1},\dots x_{k};a):B(y_{1},\dots y_{k};a)|=|B(Tx_{1},\dots Tx_{k};a):B(Ty_{1},\dots Ty_{k};a)|$.

Lemma 3.6**.**

Suppose that $p>0$ is an integer. Then $B(x_{1},x_{2},\dots,x_{k};a)$ contains $B(px_{1},px_{2},\dots,px_{k};a)$ and the index is $p^{N(k,a)}$ where $N(k,a):=\prod_{j=1}^{n}a_{j}{{a_{j}+k-1}\choose{k-1}}$.

**Proof.  **The free abelian group $B(x_{1},x_{2},\dots,x_{k};a)$ has basis consisting of monomials in the $x_{t}(-j)$. Such a monomial has a unique expression $w_{1}\cdots w_{wt(a)}$, where $w_{j}$ is a monomial in the $x_{t}(-j)$. There are ${{a_{j}+k-1}\choose{k-1}}$ such $w_{j}$. The formula for $N(k,a)$ is now clear. $\square$

Lemma 3.7**.**

Suppose that $p>0$ is an integer. Then $B(x_{1},x_{2},\dots,x_{k};a)$ contains $B(px_{1},x_{2},\dots,x_{k};a)$ and the index is $p^{{\frac{1}{k}}N(k,a)}$ where $N(k,a):=\prod_{j=1}^{n}a_{j}{{a_{j}+k-1}\choose{k-1}}$.

**Proof.  **Observe that we have a chain

[TABLE]

[TABLE]

By Lemma (3.5), the indices for each containment

[TABLE]

[TABLE]

are equal. We then deduce the result from Lemma (3.6). $\square$

Corollary 3.8**.**

In the notation of Lemma (3.7), the index

[TABLE]

is $p^{{\frac{1}{k}}\sum_{a\in\mathcal{A}(n)}N(k,a)}$ = $p^{{\frac{1}{k}}\sum_{(a_{j})=a\in\mathcal{A}(n)}\prod_{j=1}^{n}a_{j}{{a_{j}+k-1}\choose{k-1}}}$.

Now assume that $H$ has a nondegenerate symmetric bilinear form $\langle\cdot|\cdot\rangle.$ Then we can make $M(1)$ an irreducible module for the affine algebra

[TABLE]

such that $H\otimes{\mathbb{C}}[t]$ annihilates $\bf 1$ and the central element $K$ acts as $1.$ We abbrevuate $h\otimes t^{m}$ by writing $h(m)$ for $h\in H$ and $m\in{\mathbb{Z}}.$

Notation 3.9**.**

There is a unique nondegenerate symmetric bilinear form $\langle\cdot|\cdot\rangle$ on $M(1)$ such that $\langle\hbox{\bf 1}|\hbox{\bf 1}\rangle=1$ and $\langle h(m)u|v\rangle=-\langle u|h(-m)v\rangle$ for $u,v\in M(1)$ and $h\in H$ (see [5], [2]).

Furthermore, $B_{L}(n)=B(x_{1},...,x_{k};n)$ is a ${\mathbb{Z}}$-form of $M(1)_{n}.$ Note that if $L={\mathbb{Z}}x_{1}+\cdots+{\mathbb{Z}}x_{k}$ is rational lattice of $H$ in the sense that for any $\alpha,\beta\in L$ $\langle\alpha|\beta\rangle\in{\mathbb{Q}}$, then $B(x_{1},...,x_{k};n)$ is also a rational lattice, due to the form. In the notation of Corollary (3.8), we have

Corollary 3.10**.**

Assume existence of the form as in Notation 3.9. For $k\geq 1$ and $n\geq 0$, define

[TABLE]

Then $S(k,0)=0$ and

[TABLE]

for all $k\geq 1$ and $n\geq 0$.

Remark 3.11**.**

This presentation helps us understand the “homogeneous part” of the standard integral form in the symmetric algebra spanned over ${\mathbb{Z}}$ by all monomials made from a basis. The integral form involves expressions like Schur functions which have fractional coefficients so are not in the homogeneous part. We shall study the quotient of that integral form by its homogeneous part.

4 Integral forms of $M(1)$

Let $L$ be a positive definite integral lattice with basis $x_{1},\dots,x_{k}$ and we denote the form on $L$ by $\langle\cdot\mid\cdot\rangle.$ We recall the standard integral form for the lattice vertex operator algebra based on $L$.

Note from [4] that $M(1):={\mathbb{C}}[x_{i}(-n)|i=1,...,k;n>0]$ is the Heisenberg vertex operator algebra and $V_{L}=M(1)\otimes{\mathbb{C}}^{\epsilon}[L]$ is the corresponding lattice vertex operator algebra where $\epsilon$ is a bimultiplicative map from $L\times L\to\langle\pm 1\rangle$ such that $\epsilon(\alpha,\beta)\epsilon(\beta,\alpha)=(-1)^{\langle\alpha\mid\beta\rangle}$ and $\epsilon(\alpha,\alpha)=(-1)^{\langle\alpha\mid\alpha\rangle/2}$ and where ${\mathbb{C}}^{\epsilon}[L]=\oplus_{\alpha\in L}{\mathbb{C}}e^{\alpha}$ is the twisted group algebra. There is a unique nondegenerate symmetric invariant bilinear form $\langle\cdot\mid\cdot\rangle$ on $V_{L}$ such that

[TABLE]

and

[TABLE]

for all $u,v\in V_{L}$ $\alpha\in L$ and $n\in{{\mathbb{Z}}}$ (see [1], [2]).

Recall the subring $A_{L}$ from Section 3. Then $A_{L}$ is a ${\mathbb{Z}}$-form of $M(1)$ in the sense that $A_{L}$ is a vertex algebra over ${\mathbb{Z}},$ $\langle u\mid v\rangle\in{\mathbb{Z}}$ for $u,v\in A_{L}$ and $M(1)={\mathbb{C}}\otimes_{{\mathbb{Z}}}A_{L}$ [2].

We also set $(V_{L})_{\mathbb{Z}}=\oplus_{\alpha\in L}A_{L}\otimes e^{\alpha}.$ Then $(V_{L})_{\mathbb{Z}}$ is a vertex operator algebra over ${\mathbb{Z}}$ generated by $e^{\pm x_{i}}$ for $i=1,...,k$ and is a free ${\mathbb{Z}}$-module such that $V_{L}={\mathbb{C}}\otimes_{{\mathbb{Z}}}(V_{L})_{{\mathbb{Z}}}.$

Definition 4.1**.**

$(V_{L})_{\mathbb{Z}}=\oplus_{\alpha\in L}A_{L}\otimes e^{\alpha}.$* is the standard integral form in the lattice vertex operator algebra $V_{L}$.*

Let $(V_{L})_{{\mathbb{Z}},n}$ consists of vectors of weight $n$ in $(V_{L})_{{\mathbb{Z}}}.$ To get $det((V_{L})_{{\mathbb{Z}},n})$, we first understand $\det(A_{L}(n))$ in terms of $\det(L).$ Recall $B_{L}={\mathbb{Z}}[x_{i}(-n)\mid i=1,...,k;\,n>0]$ and $B_{L}(n)=B_{L}\cap M(1)_{n}$ for $n\geq 0.$ By Lemma 3.3, $[A_{L}(n):B_{L}(n)]$ only depends on the rank of $L$ and integer $n.$

Using Lemma 3.3 we can give an explicit expression of $[A_{L}(n):B_{L}(n)]$. We define numbers $b_{0}:=1$ and for $n>0$,

[TABLE]

Lemma 4.2**.**

The index $[A_{L}(n):B_{L}(n)]$ is the square root of

[TABLE]

for $n\geq 0.$

**Proof.  **By Lemma 3.3, $[A_{L}(n):B_{L}(n)]$ is independent of lattice $L.$ So we can choose $L={\mathbb{Z}}x_{1}+\cdots+{\mathbb{Z}}x_{k}$ such that $\{x_{1},...,x_{k}\}$ is an orthonormal basis of $H$ for convenience of computation. Then $A_{L}(n)$ is a unimodular lattice by Proposition 3.6 of [2]. It is easy to show that

[TABLE]

for any $i,s$. This shows that $|\det(B_{L}(n))|$ equals $\prod_{n_{1},...,n_{k}\geq 0,\,\sum n_{i}=n}b_{n_{1}}\cdots b_{n_{k}}.$ Since $|\det(B_{L}(n))|=[A_{L}(n):B_{L,n}]^{2},$ the result follows immediately. $\square$

Lemma 4.3**.**

Let $A_{1},A_{2},C_{1},C_{2}$ be lattices with the same rank such that $C_{i}\subset A_{i}$ for $i=1,2,$ $C_{2}\subset C_{1}.$ Then $|\det(A_{2})|=\frac{[C_{1}:C_{2}]^{2}[A_{1}:C_{1}]^{2}|\det(A_{1})}{[A_{2}:C_{2}]^{2}}.$ In particular, if $\det(A_{1})=1$ and $[A_{1}:C_{1}]=[A_{2}:C_{2}]$ then $|\det(A_{2})|=[C_{1}:C_{2}]^{2}.$

**Proof.  **The result follows from the following relations

[TABLE]

for $i=1,2.$ $\square$

Theorem 4.4**.**

Let $L$ be an positive definite integral lattice with a base $\{x_{1},...,x_{k}\}$ as before. Then for each $n\geq 0,$ $|\det(A_{L}(n))|$ is an integer power of $\det(L).$ In fact, $|\det(A_{L}(n))|=\det(L)^{2S(k,n)}$, where $S(k,n)$ is given by Lemma (3.10).

**Proof.  **We prove the theorem in several steps. If $K$ contains the sublattice $J$ with finite index, then one may deduce the results for $K$ from those for $J$, and conversely, from the results of Section 3.

The result $|\det(A_{L}(n))|=1$ when $L$ is unimodular was proved in [2]. Let $p$ be a positive integer.

Case (a): Let $0\neq p\in{\mathbb{Z}}$ and $L={\mathbb{Z}}pe_{1}\oplus{\mathbb{Z}}e_{2}\oplus\cdots\oplus{\mathbb{Z}}e_{k}$ be a sublattice of ${\mathbb{Z}}^{k}={\mathbb{Z}}e_{1}\oplus\cdots\oplus{\mathbb{Z}}e_{k}$ where $\{e_{1},...,e_{k}\}$ is the standard orthonormal basis of ${\mathbb{R}}^{k}.$ Using Lemmas 3.3, 4.2 with $A_{1}=A_{{\mathbb{Z}}^{k}}(n),$ $A_{2}=A_{L}(n),$ $C_{1}=B_{{\mathbb{Z}}^{k}}(n)$ and $C_{2}=B_{L}(n)$ gives $|\det(A_{L}(n))|=[B_{{\mathbb{Z}}^{k}}(n):B_{L}(n)]^{2}.$ Note that $\det(L)=p^{2}.$ By Corollary 3.10, $|\det(A_{L}(n))|=p^{2S(k,n)}=\det(L)^{2S(k,n)}.$

Case (b): Let $L={\mathbb{Z}}p_{1}e_{1}\oplus{\mathbb{Z}}p_{2}e_{2}\oplus\cdots\oplus{\mathbb{Z}}p_{k}e_{k}$ for any positive integers $p_{1},...,p_{k}.$ Then $\det(L)=p_{1}^{2}\cdots p_{k}^{2}$ and $|\det(A_{L}(n))|=\det(L)^{2S(k,n)}$ by Case (a).

Case (c): Let $T$ be a positive integer such that $L$ is a rank $k$ sublattice of $K=\frac{1}{T}{\mathbb{Z}}^{k}$, i.e., $L\subset{\mathbb{Q}}^{k}$. Then $\det(L)=[K:L]^{2}T^{-2k}.$ There exist a base $\{u_{1},...,u_{k}\}$ of $K$ and positive integers $p_{1},...,p_{k}$ such that $\{p_{1}u_{1},...,p_{k}u_{k}\}$ is a base of $L.$ This implies that $[K:L]=p_{1}\cdots p_{k}.$ From Lemma 4.3 and discussion in Case (b), we see that $|\det(A_{L}(n))|=[B_{K}(n):B_{L}(n)]^{2}|\det(A_{K}(n))|=(p_{1}\cdots p_{k})^{2S(k,n)}|\det(A_{K}(n))|.$ On the other hand,

[TABLE]

Thus

[TABLE]

Case (d): Let $L$ be an arbitrary integral lattice in Euclidean space ${\mathbb{R}}^{k}.$ The problem with applying (c) is that $L$ is not necessarily a sublattice of ${\mathbb{Q}}^{k}.$ However, we can use a sequence of rational lattices $\{L_{i}|i\in{\mathbb{N}}\}$ such that “$\lim_{i\to\infty}L_{i}=L$”. We fix a base $\{v_{1},...,v_{k}\}$ of $L$ and take linearly independent vectors $v^{i}_{1},...,v^{i}_{k}\in{\mathbb{Q}}^{k}$ such that $|v^{i}_{j}-v_{j}|<\frac{1}{i}$ for all $i,j.$ It is clear that $\lim_{i\to\infty}\det(L_{i})=\det(L)$ and $\lim_{i\to\infty}|\det(A_{L_{i}}(n))|=|\det(A_{L}(n))|$ for any $n\geq 0.$ It follows from Case (c) that $|\det(A_{L}(n))|=\det(L)^{2S(k,n)},$ as desired. $\square$

5 Integral forms of $V_{L}$

We now assume that $L$ is a positive definite even lattice. Recall that $(V_{L})_{\mathbb{Z}}=\oplus_{\alpha\in L}A_{L}\otimes e^{\alpha}.$ Also recall $(V_{L})_{{\mathbb{Z}},n}$ from Section 4. We determine $\det((V_{L})_{{\mathbb{Z}},n})$ in this section.

For $m\geq 0$ we set $L_{2m}=\{\alpha\in L|\langle\alpha\mid\alpha\rangle=2m\}.$ Define $Y_{0}:=L_{o}=\{0\}$. For $m\geq 1$, let $Y_{2m}$ be a subset of $L_{2m}$ such that $2|Y_{2m}|=|L_{2m}|$ and $L_{2m}=Y_{2m}\cup(-Y_{2m}).$ For $\alpha\in L$ we set $W^{\alpha}=M(1)_{L}\otimes e^{\alpha}+M(1)_{L}\otimes e^{-\alpha}\subset(V_{L})_{{\mathbb{Z}}}.$ Let $W^{\alpha}_{n}=W^{\alpha}\cap(V)_{L,n}.$ Then $W^{\alpha}_{n}\neq 0$ if and only if $\alpha\in L_{2m}$ and $m\leq n.$ In this case, $W^{\alpha}_{n}=A_{L}(n-m)\otimes e^{\alpha}+A_{L}(n-m)\otimes e^{-\alpha}.$ Observe that

[TABLE]

and $\langle W^{\alpha}\mid W^{\beta}\rangle=0$ if $\alpha\neq\beta.$ So

[TABLE]

From the definition of the bilinear form, we know that for $\alpha\in Y_{2m}$ with $m\neq 0$

[TABLE]

Also, $det(W_{0}^{0})=det({\mathbb{Z}}\hbox{\bf 1})=1$.

Here is our main theorem, an immediate consequence of Theorem 4.4.

Theorem 5.1**.**

For all $n\geq 0$, we have

[TABLE]

6 Acknowledgements

C. Dong was supported by China NSF grant 11871351. R. Griess was supported by funds from his Collegiate Professorship and Distinguished University Professorship at the University of Michigan.