Timeline for A question on linear groups
Current License: CC BY-SA 3.0
20 events
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Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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Jul 27, 2016 at 12:16 | vote | accept | erz | ||
Jul 19, 2016 at 11:59 | answer | added | Thomas | timeline score: 2 | |
Jun 30, 2016 at 23:53 | answer | added | YCor | timeline score: 6 | |
Jun 30, 2016 at 14:15 | answer | added | Venkataramana | timeline score: 2 | |
Jun 30, 2016 at 13:42 | comment | added | Wilberd van der Kallen | Maybe it helps to observe that the commutator subgroup of $G$ is the same as the commutator subgroup of $\mathbb{R}^{*}G$. And the latter commutator subgroup is $SL_n(\mathbf{R})$. | |
Jun 30, 2016 at 9:44 | comment | added | YCor | @erz I said that I refer to Q2. In Q1 I know you don't assume you homotheties, but still my argument shows that the subgroup of $GL_n$ generated by $O(n)$ and $T$ will contain $SL_n(\mathbf{R})$ as soon as $T$ does not belong to the group of similarities. Precisely it is equal to the group of elements whose determinant is of the form $\pm\det(T)^n$ for some $n\in\mathbf{Z}$. | |
Jun 30, 2016 at 9:31 | comment | added | erz | @YCor: No, I understand what is $\mathbb{R}^{*}O_n(\mathbb{R})$, but my group $G$ does not contain $\mathbb{R}^{*}O_n(\mathbb{R})$, only $O_n(\mathbb{R})$. Thank you for your explanation, but is there a more elementary proof? Or could you point a reference, where I could learn enough of Lie algebra theory to understand yours? Please forgive me my ignorance, as I've said algebra is not my area. | |
Jun 30, 2016 at 9:25 | comment | added | YCor | How to prove all this stuff? If $G$ is a subgroup of $SL_n(\mathbf{R})$ strictly containing $SO(n)$, then $G$ is a subgroup of $SL_n(\mathbf{R})$ which contains $SO_n$ as well as a distinct conjugate of it. The set of $x$ in the Lie algebra such that $\exp(tx)$ belongs to $G$ for $t$ small enough is a Lie subalgebra of $\mathfrak{sl}_n$ strictly containing $\mathfrak{so}_n$ and Lie algebra theory implies that it's the whole $\mathfrak{sl}_n$. Thus $G$ is open in $SL_n$ and hence by openness, $G$ is the whole group. | |
Jun 30, 2016 at 9:21 | comment | added | YCor | @erz: did you misunderstand my comment? when I say and you rewrite $\mathbf{R}^*O_n(\mathbf{R})$, the left-hand $\mathbf{R}^*$ denotes the group of homotheties. Thus $\mathbb{R}^*O_n(\mathbf{R})$ denotes the whole group of similarities. | |
Jun 30, 2016 at 8:27 | comment | added | erz | @YCor: the group doesn't contain all homotheties... | |
Jun 30, 2016 at 8:23 | comment | added | erz | @Wilberd van der Kallen: Yes, sorry, I forgot about the condition $n>2$. Thank you for catching that. But nevertheless, how to prove all of this stuff? | |
Jun 30, 2016 at 8:22 | history | edited | erz | CC BY-SA 3.0 |
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Jun 30, 2016 at 8:12 | comment | added | YCor | Indeed. Still for all $n$ it's true that a subgroup containing the orthogonal group and not contained in the group of similarities, will contain $SL_n(R)$. Thus Q1 is true for $n\ge 3$. | |
Jun 30, 2016 at 8:00 | history | edited | YCor |
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Jun 30, 2016 at 7:54 | comment | added | Wilberd van der Kallen | I should have said that Q1 is wrong for $n=2$. | |
Jun 30, 2016 at 7:50 | comment | added | YCor | @WilberdvanderKallen I first misread your comment and I refer to Q2 | |
Jun 30, 2016 at 7:50 | comment | added | YCor | Q2 holds true: indeed here determinant does not matter since the subgroup already contains all homotheties and contains elements of all determinants. After modding out by homotheties, the question boils down to whether $PO(n)(R)$ is maximal in $PGLn(R)$, and this is true and follows from the fact that $SO(n)(R)$ is maximal in $SL_n(R)$. Here $R$ are the reals. | |
Jun 30, 2016 at 7:25 | comment | added | Wilberd van der Kallen | For $n=2$ it is wrong. Take $T$ of determinant one. | |
Jun 30, 2016 at 7:01 | history | asked | erz | CC BY-SA 3.0 |