Let $SL(n, \mathbb{R})$ be the group of $n \times n$ invertible matrices of determinant $1$ in real numbers. Let $G:=SL(n, \mathbb{R}_{\geq 0})$ be its subgroup $\{M \in SL(n,\mathbb{R}) \mid M, M^{-1} \text{ both have non-negative entries}\}$. Is there any known results on this group? I am particular interest in how big (roughly) this group could be. I know $S_n \subset SL(n, \mathbb{R}_{\geq 0})$ (considering the permutation of a basis of a $n$- dimensional vector space), but I wish it could be much bigger than that.
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3$\begingroup$ The group is uncountable. It has a homomorphism $\det$ whose image includes the multiplicative group $\mathbb{R}_{> 0}$. $\endgroup$– user6976Commented Sep 2, 2013 at 15:45
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3$\begingroup$ Still, take the split torus $A=diag(e^{t},e^{-t})$. $\endgroup$– AsafCommented Sep 2, 2013 at 15:53
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2$\begingroup$ The definition should read "$M$, $M^{-1}$ both have non-negative entries" instead of "positive entries". $\endgroup$– Johannes HahnCommented Sep 2, 2013 at 15:57
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4$\begingroup$ math.stackexchange.com/a/214574/19786 $\endgroup$– dkeCommented Sep 2, 2013 at 15:59
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1$\begingroup$ @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. $\endgroup$– Pietro MajerCommented Sep 2, 2013 at 16:09
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1 Answer
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This is answered as Pietro says in http://archive.numdam.org/ARCHIVE/CM/CM_1969__21_4/CM_1969__21_4_376_0/CM_1969__21_4_376_0.pdf
In general semigroup theorists have heavily studied maximal subgroups of semigroups of nonnegative matrices.
Again all maximal subgroups are isomorphic to the group of monomial matrices with nonnegative entries of appropriate rank.
answered Sep 2, 2013 at 16:55