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[NCG] So I'm trying to pin down a fairly well-known bit of noncommutative-geometric folklore that says that for $\Theta \in M_N(\mathbb{Q})$ skew-symmetric, the corresponding noncommutative $N$-torus $C(\mathbb{T}^N_\Theta)$ is isomorphic to $C(\mathbb{T}^N,\mathcal{M}_\Theta)$, where $\mathcal{M}_\Theta \to \mathbb{T}^N$ is a bundle of full matrix algebras. Due to the way $C(\mathbb{T}^N_\Theta)$ is constructed, such a result presumably arises from solving the following Diophantine(-ish?) problem: [/NCG]

So, let $\Theta \in M_N(\mathbb{Q})$ be skew-symmetric. For which $r \in \mathbb{Z}^N$ is $\Theta r \in \mathbb{Z}^N$?

The case when $N = 2$ is well known: we can write $$ \Theta = \begin{pmatrix} 0 & -p/q \\ p/q & 0 \end{pmatrix} $$ where $q > 0$ and $p$ are coprime integers, whence $\Theta r \in \mathbb{Z}^2$ if and only if $r \in (q\mathbb{Z})^2$. The moment that $N > 2$, I'm afraid I don't even know the appropriate jargon to look up this type of (I guess) Diophantine problem myself.

Of course, if this does turn out to have been already answered here or at MathOverflow, I'd be immensely grateful just to know it is a duplicate, and of what.


EDIT: In the context I care about, $\Theta$ is invertible WLOG. Hence (thank you, Will Jagy!) what I'm really after is the lattice $$ \Lambda_\Theta := \mathbb{Z}^N \cap \Theta^{-1}\mathbb{Z}^N. $$ By a basic result in lattice theory, since $(\mathbb{Z}^N)^\ast = \mathbb{Z}^N$ and $(\Theta^{-1}\mathbb{Z}^N)^\ast = \Theta \mathbb{Z}^N$ (since $\Theta$ is skew-symmetric and invertible), it follows that $$ \Lambda_\Theta = ((\mathbb{Z}^N)^\ast \cup (\Theta^{-1}\mathbb{Z}^N)^\ast)^\ast = (\mathbb{Z}^N \cup \Theta \mathbb{Z}^N)^\ast; $$ since the columns of $I_N$ are a basis for $\mathbb{Z}^N$ and the columns of $\Theta$ are a basis for $\Theta \mathbb{Z}^N$, it apparently follows that a basis for $\Lambda_\Theta^\ast$ is given by the columns of the Hermite normal form of the augmented matrix $$ [I_N \mid \Theta]. $$ Now, are there any overall simplifications that happen with this Hermite normal form, given that $\Theta$ is skew-symmetric and invertible, or is it really just whatever it is?

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    $\begingroup$ It's a lattice in any case. How does 3 by 3 work out? I know some stuff by G. L. Watson but I'm not sure it adapts to skew symmetric matrices. In case of interest, it's in Watson_Transformations_1.pdf at zakuski.math.utsa.edu/~kap/forms.html $\endgroup$
    – Will Jagy
    Sep 25, 2013 at 2:55
  • $\begingroup$ Right. I'm this context, I can actually assume WLOG that $\Theta$ is invertible, so that I'm just computing the intersection of the lattices $\mathbb{Z}^N$ and $\Theta^{-1}\mathbb{Z}^N$, which is apparently a fairly standard procedure. Thank you! $\endgroup$ Sep 25, 2013 at 3:11

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