# Which lattices are rotatable into their scaled copy?

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.

• 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 $O(d)$. Dec 17 '18 at 22:29

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 $$m^d=\det(G) \in \mathbb{Z}$$.
• 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 $G$? Dec 19 '18 at 8:37
• $K$ 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. Dec 19 '18 at 10:08
• I see! And yes, I definitely agree that $G$ is a step forward from rotations, I just hope there's something even better. Dec 19 '18 at 10:40