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Apr 22, 2022 at 13:54 answer added user481018 timeline score: 0
Dec 14, 2020 at 14:29 answer added Aurel timeline score: 2
Dec 13, 2020 at 16:46 history edited Mikhail Borovoi
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Dec 13, 2020 at 16:22 answer added Geoff Robinson timeline score: 4
Dec 13, 2020 at 8:21 comment added YCor @LSpice no, I should have said $\mathbf{R}$-anisotropic. I think a connected linear algebraic $\mathbf{Q}$-group $G\subset\mathrm{GL}_n$ has $G(\mathbf{Z})$ finite iff it's ($\mathbf{R}$-anisotropic)-by-($\mathbf{Q}$-split torus).
Dec 13, 2020 at 1:44 comment added LSpice @YCor, compact = $\mathbf Q$-anisotropic?
Dec 13, 2020 at 1:23 comment added YCor 1: this is the set of $\mathbf{Z}$-points in some explicit algebraic group. This is finite iff the latter is, I believe, compact-by-($\mathbf{Q}$-split torus). I guess this can algorithmically be checked from its Lie algebra (which can easily be described by a linear system of equations with rational coefficients).
Dec 13, 2020 at 1:19 history edited YCor
edited tags
Dec 13, 2020 at 0:44 history edited LSpice CC BY-SA 4.0
Oops, messed up the title
Dec 13, 2020 at 0:39 comment added LSpice I did some proofreading, I believe without any change in meaning. \\ $\operatorname C'(K)$ is called the twisted centraliser of $K$ (twisted by the inverse-transpose automorphism $\sigma$ of $\operatorname{GL}(n, \mathbb Z)$). It is the intersection with $\operatorname{GL}(n, \mathbb Z)$ of the centraliser of the element $K \rtimes \sigma \in \operatorname{GL}(n, \mathbb Z) \rtimes \langle\sigma\rangle$. It has certainly been studied before, but I don't know what, if anything, is known about your specific questions.
Dec 13, 2020 at 0:38 history edited LSpice CC BY-SA 4.0
Proofreading
Dec 13, 2020 at 0:25 history edited YCor CC BY-SA 4.0
formatting
Dec 13, 2020 at 0:14 history asked en kuo CC BY-SA 4.0