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Schmidt (https://projecteuclid.org/euclid.dmj/1077377618) showed that the number of $m$-dimensional subgroups of $\mathbb{Z}^{n}$ of covolume $\leq X$ is $$c_{1}\left(m,n\right)X^{n}+O\left(X^{n-\delta_{1}\left(m,n\right)}\right)$$ and that the number of primitive $m$-dimensional subgroups of $\mathbb{Z}^{n}$ of covolume $\leq X$ is $$c_{2}\left(m,n\right)X^{n}+O\left(X^{n-\delta_{2}\left(m,n\right)}\right)$$ for some positive $c_{1}\left(m,n\right)$, $c_{2}\left(m,n\right)$, $\delta_{1}\left(m,n\right)$ and $\delta_{2}\left(m,n\right)$ whose specific values are not relevant to my question.

I refer to an orthogonal transformation with determinant 1 that is neither plus or minus the identity, as a non-trivial rotation.

My question is: what can be said about the number of $m$-dimensional subgroups of $\mathbb{Z}^{n}$ (primitive or not) of covolume $\leq X$ that satisfy that they are invariant under a non-trivial rotation of their spanned subspace? I am looking for an upper bound, and am only interested in the exponent (and not in the constant).

Two remarks:

  1. When considering the similarity classes of $m$-dimensional lattices up to rotation and rescaling, these classes are identified with the points of the space $$SL_m(\mathbb{Z})/SL_m(\mathbb{R})\setminus SO_m(\mathbb{R}).$$

    The classes of the lattices which are invariant under non-trivial rotations fall in the boundary of this space (when identified with a suitable fundamental domain inside $SL_m(\mathbb{R})\setminus SO_m(\mathbb{R})$). This should hint at the fact that the number of these lattices is $o\left(X^{n}\right)$; indeed, we were able to show that it is $O\left(X^{n-\frac{1}{n-1}}\right)$, but we expect it is actually $O\left(X^{n-1}\right)$, or even below that. For example, for $n=3$ and $m=2$, this number is finite regardless of $X$.

  2. I am more than happy to know the answer for just $m=n-1$.

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    $\begingroup$ Note that for subgroups of $\mathbf{Z}^n$, "covolume" is the same as "index". $\endgroup$ – YCor Jan 20 '18 at 16:40
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    $\begingroup$ I can see two interpretations of your question. (a) you count the number of subgroups such that there is a nontrivial rotation (...) (b) you fix a nontrivial rotation and you count the number of subgroups invariant by this given rotation. Also, I'm not sure what you mean by "rotation" and "non-trivial rotation". In even dimension, does $x\mapsto -x$ account as a non-trivial rotation? as a rotation? $\endgroup$ – YCor Jan 20 '18 at 19:53
  • $\begingroup$ I mean option (a). And yes, $x\mapsto -x$ does count as a rotation in even dimension. A "non-trivial" rotation is a rotation that is not the identity, or minus the identity. If I was not clear enough, please feel free to edit it! $\endgroup$ – Tal H Jan 21 '18 at 17:49
  • $\begingroup$ Oops, thanks (I realize you already clearly gave the answer to my second question in your original post). And I realize my first comment (which I'll likely soon erase) relies on a misunderstanding. I guess that the covolume of a subgroup of $\mathbf{Z}^n$ means its covolume in the Euclidean subspace it spans, is this correct? (Hence for a finite index subgroup, this is indeed the same as the index.) $\endgroup$ – YCor Jan 21 '18 at 19:08

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