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Let $L=\{\sum_i n_iv_i\mid n_i\in\mathbb Z\}$ be some lattice generated by $d$ independent vectors $(v_i)_1^d$ from $\mathbb R^d$. Call $L$ rotatable if for some $M$, a scalar multiple of some rotation (orthogonal transformation that is other than the identity), we have $M(L)\subset L$. For example, $\mathbb Z^d$ is rotatable, and so is $\{n_1(1,0)+n_2(0,\sqrt 2)\}$, but $\{n_1(1,0)+n_2(0,\pi)\}$ is not. I'm sure this is some well-known notion with several equivalent definitions, but I couldn't find anything, so I would be grateful for any pointers/results about them.

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    $\begingroup$ You're asking about lattices $L\subset \mathbb R^d$ such that the group of $\mathbb Q$-linear automorphisms of $L\otimes_{\mathbb Z}\mathbb Q$ has a non-trivial intersection with the orthogonal group $O(d)$. One could ask something stronger: namely, that $\mathrm{Aut}(L\otimes_{\mathbb Z}\mathbb Q)\cap O(d)$ be the set of $\mathbb Q$-rational points of a form of $O(d)$. $\endgroup$ Dec 17 '18 at 22:29
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If one let F be a $d \times d$ matrix whose columns form an integral basis of $L$, $K \in O(d)$ be the rotation and $m>0$ be the scaling factor, then in order for $mKF$ to generate a sub-lattice, we must have $mKF=FG$ where $G$ is a $d \times d$ integral matrix and $|\det(G)|$ is the index. Letting $A=F^TF$ be the Gram matrix of $L$,we have the necessary condition $m^2A=G^TAG$, which is also sufficient because given $A$, one can always recover $F$, for example by completing squares for the quadratic form $x^TAx$. So $m$ must be a real algebraic integer satisfying $m^d=\det(G) \in \mathbb{Z}$.

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  • $\begingroup$ How did you make $K$ disappear from the necessary condition? Also, is there possibly some nicer characterization? I mean this requires a witness $G$. Can we decide given $F$ whether there's such a $G$? $\endgroup$
    – domotorp
    Dec 19 '18 at 8:37
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    $\begingroup$ $K$ is an othogonal matrix, so $K^TK=I_d$. $K$ and $G$ are related by $K=(FGF^{-1})/m$. This is the usual characterization in terms of quadratic form. Perhaps it is easier to work with the discrete $G$ than the continuous rotation. $\endgroup$
    – Chua KS
    Dec 19 '18 at 10:08
  • $\begingroup$ I see! And yes, I definitely agree that $G$ is a step forward from rotations, I just hope there's something even better. $\endgroup$
    – domotorp
    Dec 19 '18 at 10:40

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