Is there any way to find all matrices $G \in SL(n,\mathbb Z)$ such that there exists a matrix $A \in GL(n,\mathbb R)$ satisfying $$ AGA^{-1} \in SO(n,\mathbb R)? $$
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5$\begingroup$ Yes, these are matrices with finite order. It's an exercise, not really research level. $\endgroup$– YCorCommented Feb 6, 2019 at 3:12
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These are exactly the finite-order elements:
If $G \in \text{SL}(n,\mathbb{Z})$ has finite order, then there exists an inner product preserved by $G$ (take an arbitrary inner product and add up its images under all powers of $G$). Changing bases to an orthonormal basis for this invariant inner product has the effect of conjugating $G$ into the orthogonal group.
Conversely, if $G \in \text{SL}(n,\mathbb{Z})$ is such that there exists some $A \in \text{GL}(n,\mathbb{R})$ with $A G A^{-1} \in \text{SO}(n,\mathbb{R})$, then since $A \cdot \text{SL}(n,\mathbb{Z}) \cdot A^{-1}$ is a discrete subgroup of $\text{GL}(n,\mathbb{R})$, its intersection with the compact group $\text{SO}(n,\mathbb{R})$ is a finite group, and thus $G$ has finite order.
answered Feb 6, 2019 at 2:54
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$\begingroup$ Are there any partial results about the classification of all finite order elements of $GL(n,\mathbb Z)$, up to conjugacy? $\endgroup$– TotoroCommented Feb 6, 2019 at 20:13
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1$\begingroup$ @Totoro: It's a vast field. A good place to start is MR1903154 (2003c:20057) Reviewed Kuzmanovich, James(1-WKFR); Pavlichenkov, Andrey Finite groups of matrices whose entries are integers. Amer. Math. Monthly 109 (2002), no. 2, 173–186. $\endgroup$ Commented Feb 8, 2019 at 4:54