Timeline for Finding invariant closed subspace which are also subgroups for the action of $\text{SL}(2, \Bbb Z)$ on $\Bbb R^n\times \Bbb R^n$
Current License: CC BY-SA 4.0
17 events
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| Apr 21, 2020 at 19:01 | comment | added | Moishe Kohan | Once $n\ge 3$ (maybe even $2$), this is hopeless since the action has nonempty domain of proper discontinuity. | |
| Apr 19, 2020 at 9:54 | history | edited | InsideOut | CC BY-SA 4.0 |
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| Apr 19, 2020 at 9:47 | comment | added | InsideOut | I did. In that case we have the action on $\text{SL}(2,\Bbb Z)$ on $\Bbb R\times\Bbb R\cong\Bbb R^2$. So there are lattices of course. If you take a vector $v\in\Bbb R^2$ such that $v$ is not a multiple of an integral vector (all entries are integers) then the orbit is dense in $\Bbb R^2$. If the entries are rational multiples of integers, instead, the orbit is discrete. | |
| Apr 19, 2020 at 7:49 | review | Close votes | |||
| Apr 26, 2020 at 16:44 | |||||
| Apr 19, 2020 at 7:34 | comment | added | YCor | Have you looked at the case $n=1$? It's already non-trivial. In general, the question is equivalent to describe closed subsets of the (usually non-Hausdorff) quotient space. | |
| Apr 19, 2020 at 7:33 | history | edited | YCor | CC BY-SA 4.0 |
changed tag, changed confusing "subspace"
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| Apr 19, 2020 at 7:33 | history | edited | InsideOut | CC BY-SA 4.0 |
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| Apr 19, 2020 at 7:32 | comment | added | InsideOut | Yes, I am interested in closed subspaces in particular. Right, I forgot to write it. | |
| Apr 19, 2020 at 7:30 | comment | added | YCor | Topological subspace just means any subset (with the endowed topology, but this is irrelevant to the question). You really want no further assumption such as "closed"? | |
| Apr 19, 2020 at 7:29 | comment | added | InsideOut | Yes, maybe the term "subspace" has been misleading. I actually mean topological subspaces, where $\Bbb R^n\times \Bbb R^n$ is endowed with the product of the standard euclidean topology. The point Is that I have examples with linear subspaces and lattices. | |
| Apr 19, 2020 at 7:26 | comment | added | Vladimir Dotsenko | When you say "subspace", what exactly do you mean? Presumably not vector subspace (because you mention lattices) - that was probably the context of the answer of @YCor . So what is it? Subset ? Algebraic subvariety? | |
| Apr 19, 2020 at 7:13 | history | edited | InsideOut | CC BY-SA 4.0 |
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| Apr 19, 2020 at 7:11 | comment | added | InsideOut | Hi @YCor, thanks for the answer first. I have a couple of doubts. Why do you say that the invariant subspaces for the $\text{SL}(2\Bbb Z)$ action are the same as for the $G$-action? The $n$-times product of lattice is not $G$-invariant. Am I missing something? The second doubt is about the splitting, the action is diagonal on $\Bbb R^2\times\cdots\times \Bbb R^2$. So the splitting may contain factors greater than $W_1$. | |
| Apr 19, 2020 at 7:04 | history | edited | YCor | CC BY-SA 4.0 |
formatting, changed tags
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| Apr 19, 2020 at 7:02 | comment | added | YCor | It's not an unusual action. First since this is the restriction of an algebraic action of $G=\mathrm{SL}_2(\mathbf{R})$ in which it's Zariski dense, its invariant subspaces are the same as for the $G$-action and since $G$ is semisimple this behaves well. Also this action, call it $W_n$, obviously splits coordinate-wise as $W_1^{\otimes n}$, and $W_1$ is absolutely irreducible. Describing invariant subspaces is then an exercise. | |
| Apr 19, 2020 at 6:19 | history | edited | InsideOut | CC BY-SA 4.0 |
edited title
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| Apr 19, 2020 at 5:38 | history | asked | InsideOut | CC BY-SA 4.0 |