Timeline for Subgroup of $SL(n,\mathbb{R})$ with positive entries
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Sep 2, 2013 at 17:32 | vote | accept | Li Yutong | ||
| Sep 2, 2013 at 16:55 | answer | added | Benjamin Steinberg | timeline score: 7 | |
| Sep 2, 2013 at 16:09 | comment | added | Pietro Majer | @Li Yutong : just a geometrical argument. The matrix of a linear map $L$ in a given basis has non-negative entries iff the convex cone $C$ spanned by the basis is $L$-invariant: $L(C)\subset C$. This implies that for elements of $G$ $L(\mathbb{R}_{\le0}^n) = \mathbb{R}_{\le0}^n$, whence my conclusion above. | |
| Sep 2, 2013 at 16:03 | comment | added | Li Yutong | @dke Thank you! I think this answered my question! | |
| Sep 2, 2013 at 15:59 | comment | added | dke | math.stackexchange.com/a/214574/19786 | |
| Sep 2, 2013 at 15:58 | history | edited | Li Yutong | CC BY-SA 3.0 |
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| Sep 2, 2013 at 15:58 | comment | added | Li Yutong | @PietroMajer So basically, this means it almost $S_n$ up to the scaling of coefficients? What is your reason? | |
| Sep 2, 2013 at 15:57 | comment | added | Johannes Hahn | The definition should read "$M$, $M^{-1}$ both have non-negative entries" instead of "positive entries". | |
| Sep 2, 2013 at 15:55 | comment | added | Li Yutong | @Asaf yeah, but I don't want those "trivial" examples, but if I restrict to $SL(n,\mathbb{Z})$, then I also want the coefficient to be real. | |
| Sep 2, 2013 at 15:55 | comment | added | Pietro Majer | I think elements of G are exactly diagonal matrices up to permutations. | |
| Sep 2, 2013 at 15:53 | comment | added | Asaf | Still, take the split torus $A=diag(e^{t},e^{-t})$. | |
| Sep 2, 2013 at 15:53 | history | edited | Li Yutong | CC BY-SA 3.0 |
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| Sep 2, 2013 at 15:51 | comment | added | Li Yutong | You are right, I should restrict to $SL(n,\mathbb{R})$. | |
| Sep 2, 2013 at 15:45 | comment | added | user6976 | The group is uncountable. It has a homomorphism $\det$ whose image includes the multiplicative group $\mathbb{R}_{> 0}$. | |
| Sep 2, 2013 at 15:42 | history | asked | Li Yutong | CC BY-SA 3.0 |