Simple conjecture about rational orthogonal matrices and lattices

Proof

Let $R$ be any matrix. We have the obvious exact sequence

$$ 0 \longrightarrow\mathbb{R}^N \xrightarrow[\left(\begin{matrix} I \\ R \end{matrix}\right)]{} \mathbb{R}^N \oplus \mathbb{R}^N \xrightarrow[\left(\begin{matrix} I & -R^{-1} \end{matrix}\right)]{} \mathbb{R}^N \longrightarrow 0 $$

This contains as a subsequence

$$ 0 \longrightarrow\mathbb{Z}^N \cap R^{-1} \mathbb{Z}^N \longrightarrow \mathbb{Z}^N \oplus \mathbb{Z}^N \longrightarrow \mathbb{Z}^N + R^{-1} \mathbb{Z}^N \longrightarrow 0 $$

By definition, $\mathbb{Z}^N \cap R^{-1} \mathbb{Z}^N = \Lambda$. Taking the dual of this equation, $\mathbb{Z}^N + R^T \mathbb{Z}^N = \Lambda^\star$. At this point, we need to invoke the fact that $R$ is orthogonal, so that $R^T = R^{-1}$. Then taking the quotient of the two sequences yields

$$ 0 \longrightarrow \frac{\mathbb{R}^N}{\Lambda} \longrightarrow \frac{\mathbb{R}^N}{\mathbb{Z}^N} \oplus \frac{\mathbb{R}^N}{\mathbb{Z}^N} \longrightarrow \frac{\mathbb{R}^N}{\Lambda^\star} \longrightarrow 0 $$

Suppose that every vector of $\Lambda$ has even length-squared. Then $(1, \dots, 1)$ has even inner product with every vector in $\Lambda$, so $\tfrac{1}{2} (1, \dots, 1)^T \in \Lambda^\star$. This tells us that $\tfrac{1}{2}(1, \dots, 1) \oplus 0$, viewed as an element of the middle group above, maps to zero. By exactness, it must therefore be the image of some $v \in \mathbb{R}^N$. So we have

$$ \begin{align} v &= \tfrac{1}{2}(1, \dots, 1) \mod \mathbb{Z}^N \\ Rv &= 0 \mod \mathbb{Z}^N \end{align} $$

Comparing length-squareds,

$$ \underbrace{v^2 \vphantom{)^2}}_{\frac{N}{4} \text{ mod } 2} = \; \underbrace{(Rv)^2}_{\vphantom{\frac{N}{4}} 0 \text{ mod } 1} $$

we immediately read off that $N$ is a multiple of $4$.

A connection

A closely related problem is considered here. For an $N \times N$ rational orthogonal matrix $R$, and a sublattice $L \subseteq \mathbb{Z}^N$, define the coincidence index

$$ \Sigma_L(R) := [ L : L \cap R L ] $$

It can be shown that

$$ \frac{\Sigma_{\mathbb{Z}^N}(R)}{\Sigma_{D_N}(R)} \in \{1, 2\} $$

where $D_N$ is understood to mean the sublattice of vectors with even component sum. The original conjecture is equivalent to saying that

$$ 4 \nmid N \; \implies \; \Sigma_{\mathbb{Z}^N}(R) = \Sigma_{D_N}(R) $$

The special case $N = 3$ is a known result, stated in the paper as Fact 3.