Timeline for Elements of finite order of $\mathrm{PGL}(n,\mathbb{Q})$
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
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| Aug 31, 2014 at 19:53 | answer | added | Will Sawin | timeline score: 1 | |
| Aug 31, 2014 at 11:49 | answer | added | David E Speyer | timeline score: 4 | |
| Jul 1, 2014 at 22:22 | comment | added | i. m. soloveichik | See E. C. Dade's paper, Illinois J. Math. 1965, Maximal Finite Subgroups of 4 by 4 integral matrices. | |
| Jun 3, 2014 at 7:20 | comment | added | Jérémy Blanc | Yes, good question. I do not know... | |
| Jun 2, 2014 at 13:31 | comment | added | YCor | This is not my question: my question is: is it true, given $n$ and $d$ that if $PGL_n(Q)$ has an element of order $d$, then $GL_n(Q)$ has an element of finite order whose image in $PGL_n(Q)$ has order $d$. For instance this is true for $n=2$. | |
| Jun 2, 2014 at 4:35 | comment | added | Jérémy Blanc | @Ycor Maybe I did not investigate enough, but $(1)$ does not seem so easy at first sight for elements not coming from elements of finite order of $\mathrm{GL}_n$. Note: the order of elements coming from elements of finite order of $\mathrm{GL}_n$ can be different: if $n=2$ the orders in $\mathrm{GL}_n$ are $1,2,3,4,6$ but in $\mathrm{PGL}_n$ are $1,2,3$. I do not know if the orders in $\mathrm{PGL}_n$ are always a subset of those of $\mathrm{GL}_n$. | |
| Jun 1, 2014 at 18:24 | comment | added | YCor | @JérémyBlanc: to motivate my previous question: your problem splits into two distinct problems: (1) determine possible orders of elements, and (2) classify conjugacy classes of cyclic subgroups of a given possible order. Do you consider (1) as an issue or is it easy? | |
| Jun 1, 2014 at 16:38 | comment | added | YCor | Is it expected that the set of orders of finite order elements in $PGL_n(Q)$ is reduced to the (a priori smaller) set of orders of elements in $PGL_n(Q)$ that are images of elements of finite order in $GL_n(Q)$? | |
| Jun 1, 2014 at 16:32 | comment | added | Geoff Robinson | @Jeremy: OK, I see what you mean. | |
| Jun 1, 2014 at 16:19 | comment | added | Jérémy Blanc | @GeoffRobinson I do not see what you mean here. If $A$ is an element of $\mathrm{GL}(n,\mathbb{Q})$ corresponding to an element of order $d$ in $\mathrm{PGL}$, then $A^d=\lambda I$ and $\det(A)^d=\lambda^n$. So if $d,n$ are coprime, then $\lambda$ is some $d$-th power and you can multiply $A$ by a scalar to assume that it is of finite order in $\mathrm{GL}(n,\mathbb{Q})$. | |
| Jun 1, 2014 at 16:13 | comment | added | Jérémy Blanc | Thanks for all your comments. Thanks Yves for your suggestion, it is basically what I wanted to do. I am still wondering if this was not done somewhere already. Thanks for the reference on Beauville's paper, that I knewed but which does the case of dimension $2$ (but is rather trivial over $\mathbb{Q}$). | |
| Jun 1, 2014 at 16:10 | history | edited | Jérémy Blanc | CC BY-SA 3.0 |
added 407 characters in body
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| Jun 1, 2014 at 13:01 | answer | added | Geoff Robinson | timeline score: 0 | |
| Jun 1, 2014 at 13:01 | answer | added | Alireza Abdollahi | timeline score: -1 | |
| Jun 1, 2014 at 12:14 | comment | added | Matthias Wendt | There is a paper of Beauville on the case $PGL_2(K)$, arXiv:0909.3942. The paper deals with the classification of conjugacy classes of finite subgroups in general, but the methods are (as you mention) mostly Galois theory. There is only one subtlety related to classifying elements of order $n$ in $PGL_n(K)$. Conjugacy classes correspond to $n$-power residues in $K$, the only case that is not "combinatorial" using cyclotomic polynomials. | |
| Jun 1, 2014 at 11:37 | comment | added | YCor | @Jérémy: my guess is that a natural approach would be to start up to conjugation in $PGL_n(C)$, and then to understand when two elements in $PGL_n(Q)$ are conjugate over $C$ are conjugate over $Q$: since this problem is trivial in $GL_n$, one can expect it to we well encoded (in Galois cohomology?) in $PGL_n$. | |
| Jun 1, 2014 at 11:30 | comment | added | Geoff Robinson | @Yves Cornulier: I just meant that that lcm is not so easy to explicitly evaluate, although theoretically, as you say, it is a formality. | |
| Jun 1, 2014 at 11:26 | comment | added | YCor | @GeoffRobinson: what's so subtle in $GL_n(Q)$? you need to list $L_n$, the set of all cyclotomic polynomials of degree $\le n$, then from $L_n$ you list $L'_n$, the set of all subsets of $L_n$ whose sum of degrees is $\le n$. For each $\{\Phi_{n_1},\dots,\Phi_{n_k}\}$ in $L'_n$, you get an element of order lcm$(n_1,\dots,n_k\}$ in $GL_n(Q)$. Of course I don't claim it's algorithmically efficient when $n$ is very large. | |
| Jun 1, 2014 at 11:01 | comment | added | Geoff Robinson | Isn't this just a question about rational canonical form? The question is certainly easier for elements of finite order in ${\rm GL}(n,\mathbb{Q}),$ but even there, calculating the maximum possible order of an element of finite order in ${\rm GL}(n,\mathbb{Q})$ is quite subtle. | |
| Jun 1, 2014 at 10:32 | comment | added | Dima Pasechnik | there seems to be a lot known about finite subgroups of $GL_n(\mathbb{Q})$, cf. e.g. references in ams.org/journals/proc/1997-125-12/S0002-9939-97-04283-4/… E.g. each of them is conjugate to one in $GL_n(\mathbb{Z})$. | |
| Jun 1, 2014 at 10:22 | history | asked | Jérémy Blanc | CC BY-SA 3.0 |