## Question

Up to equivalence, there are two positive-definite even unimodular lattices in $16$ dimensions: $D_{8}^{+}\oplus D_{8}^{+}$ and $D_{16}^{+}$. As observed by [Witt](witt) in 1941, the theory of modular forms implies the "strange and interesting" fact that these inequivalent lattices have identical theta functions. He comments that while modular forms make this fact obvious, geometrically it remains opaque, seemingly a pure coincidence.

Is there any known geometric explanation for this fact? For example, is it possible to construct a (geometrically meaningful) explicit length-preserving bijection between $D_{8}^{+}\oplus D_{8}^{+}$ and $D_{16}^{+}$?

**Reference**: Witt, Ernst. "Eine Identität zwischen Modulformen zweiten Grades." Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg. Vol. 14. No. 1. Springer-Verlag, 1941.

## Background

### Definitions

By *lattice* I mean a free abelian group $L$ of finite rank $n$ equipped with a positive-definite inner product. Two lattices are *equivalent* if they are isomorphic as groups by an isomorphism which preserves the inner product.

The *theta function* of a lattice $L$ is the generating function $\theta_{L}(q)$ counting the number of lattice points of a given squared-length in $L$: 
$$\theta_{L}(q):=\sum_{x\in L}q^{x\cdot x}=\sum_{\alpha\in\left\{ x\cdot x\mid x\in L\right\} }N_{\alpha}q^{\alpha},\quad N_{\alpha}:=\#\left\{ x\in L\bigm|x\cdot x=\alpha\right\} .$$

The $D_{n}$ root lattice is the sublattice of $\mathbb{Z}^{n}$ consisting of all vectors whose components sum to an even number. Let $c_{n}\in\mathbb{Z}^{n}$ denote the characteristic vector $(1,1,\ldots,1)$. Then $D_{n}	 =\left\{ x\in\mathbb{Z}^{n}\bigm|c_{n}\cdot x\equiv0\pmod{2}\right\}$, and we define

$$
D_{n}^{+}	 :=D_{n}\cup\left(D_{n}+\tfrac{1}{2}c_{n}\right).
$$ 

Note that $D_{8}^{+}$ is the usual $E_{8}$ lattice.

## Collected results

#### The lattices are even.

$D_{n}^{+}$ is a lattice when $n\equiv0\pmod{2}$. If additionally $n\equiv0\pmod{4}$, then $D_{n}^{+}$ is integral: $\left\{ x\cdot x\mid x\in L\right\} \subset\mathbb{Z}$. If moreover $n\equiv0\pmod{8}$, then $D_{n}^{+}$ is even: $\left\{ x\cdot x\mid x\in L\right\} \subset2\mathbb{Z}$.

#### The lattices are unimodular.

Both $\mathbb{Z}^{n}$ and $D_{n}^{+}$ have index $2$ in $D_{n}$. Thus $\det\mathbb{Z}^{n}=1\implies\det D_{n}=2^{2}\implies\det D_{n}^{+}=1$, so $D_{n}^{+}$ is unimodular.

#### The lattices have the same theta functions.

As a consequence of Poisson summation, if $L$ is even and unimodular with dimension $n$, then $\theta_{L}$ is a modular form of level $1$ and weight $n/2$. 

The unital ring of modular forms of level $1$ is graded by weight and is freely generated by the Eisenstein series

$$
\begin{align}
E_{4}(q)	&=1+240q^{2}+2160q^{4}+\cdots, \\
E_{6}(q)	&=1-504q^{2}-16632q^{4}+\cdots.
\end{align}
$$

Thus the modular forms of weight $8$ are spanned by $E_{4}(q)^{2}=1+480q^{2}+61920q^{4}+\cdots$, and any positive-definite even unimodular lattice of dimension $16$ must have $E_{4}(q)^{2}$ as its theta series.

#### The lattices are inequivalent.

The 480 vectors of length $\sqrt{2}$ in $D_{16}^{+}$ generate a proper sublattice, while those in $D_{8}^{+}\oplus D_{8}^{+}$ generate the full lattice. Concretely, the 480 vectors of length $\sqrt{2}$ in $D_{16}^{+}$ are the $2^{2}\cdot\binom{16}{2}$ vectors of the form $\left\{ \pm e_{i}\pm e_{j}\mid1\leq i<j\leq16\right\}$. The vector $\tfrac{1}{2}c_{16}$ is excluded since it has length $\sqrt{4}$. The included vectors span $D_{16}\subset D_{16}^{+}$. On the other hand, the 240 vectors with length $\sqrt{2}$ in $D_{8}^{+}$ are $2^{2}\cdot\binom{8}{2}+2^{7}$, where $2^{7}$ represents all vectors of the form $\left(\pm\tfrac{1}{2},\cdots\pm\tfrac{1}{2}\right)$ with an even number of minus signs. Since $\tfrac{1}{2}c_8$ is included, the span is all of $D_8^+$.

## Motivation

In [Four-Dimensional Lattices With the Same Theta Series](http://neilsloane.com/doc/nipp.pdf), Conway and Sloane construct an explicit length-preserving bijection between isospectral lattices (changing the sign of the first coordinate which is divisible by 3). I'm curious if a similar construction is possible in the above case.

## EDIT: Attempted solution

Perhaps there is some clever way to slice up, reflect/rotate, and recombine the two lattices into each other.

One possible way to slice them up is into cosets. For this we need a common sublattice. An obvious candidate is the index 4 sublattice $D_8\oplus D_8$ given by

$$
\begin{align*}
0\rightarrow D_{8}\oplus D_{8} & \rightarrow D_{8}^{+}\oplus D_{8}^{+}\rightarrow\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}\rightarrow0,\\
0\rightarrow D_{8}\oplus D_{8} & \rightarrow D_{16}^{+}\rightarrow\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}\rightarrow0.
\end{align*}
$$

The rightmost map on the top is obvious, while the rightmost map on the bottom measures
1. the parity of the sum of the first eight coordinates
2. whether $\tfrac{1}{2}$ appears in some/every coordinate.

Then we can start computing the theta functions of the cosets. Recall the standard formulas $\theta_{D_n}=\tfrac{1}{2}(\theta_3^n+\theta_4^n)$ and $\theta_{D_n^+}=\tfrac{1}{2}(\theta_2^n+\theta_3^n+\theta_4^n)$ (see SPLAG 2.3.37).

Denoting the theta functions of the $\mathbb{Z}_2\oplus\mathbb{Z}_2$ cosets $(a,b)$ with superscripts $\theta^{a,b}$,

\begin{align*}
\theta_{D_{8}^{+}\oplus D_{8}^{+}}^{0,0}=\theta_{D_{16}^{+}}^{0,0}=\theta_{D_{8}}^{2} & =1+224q^{2}+14816q^{4}+260736q^{6}+\cdots,\\
\theta_{D_{8}^{+}\oplus D_{8}^{+}}^{0,1}=\theta_{D_{8}^{+}\oplus D_{8}^{+}}^{1,0}=\theta_{D_{8}}(\theta_{D_{8}^{+}}-\theta_{D_{8}}) & =128q^{2}+15360q^{4}+263680q^{6}+\cdots,\\
\theta_{D_{8}^{+}\oplus D_{8}^{+}}^{1,1}=(\theta_{D_{8}^{+}}-\theta_{D_{8}})^{2} & =16384q^{4}+262144q^{6}+\cdots,\\
\theta_{D_{16}^{+}}^{0,1}\stackrel{?}{=}\theta_{D_{16}^{+}}^{1,1}\stackrel{?}{=}(\theta_{D_{8}^{+}}-\theta_{D_{8}})^{2} & =16384q^{4}+262144q^{6}+\cdots,\\
\theta_{D_{16}^{+}}^{1,0}\stackrel{?}{=}(\theta_{D_{8}^{+}}-\theta_{D_{8}})(3\theta_{D_{8}}-\theta_{D_{8}^{+}}) & =256q^{2}+14336q^{4}+265216q^{6}+\cdots.
\end{align*}

The last two lines were done with a computer up to $q^8$, hence marked with "$\stackrel{?}{=}$".

The theta functions of the cosets don't match up, but they are related by very simple linear dependencies, so it seems plausible that this strategy may eventually succeed. (For example, note that $3\theta_{D_{8}}-\theta_{D_{8}^{+}}=\theta_{D_{4}^2}$.)