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Are all torsion-free finitely generated linear groups over $\mathbb{C}$ left orderable? In particular, are torsion-free congruence subgroups of $SL_n(\mathbb{Z})$ left orderable?

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up vote 11 down vote accepted

The answer is no for congruence subgroups of $SL(n,\mathbb{Z})$ for $n \geq 3$. This is a theorem of Dave Witte-Morris; see

MR1198459 (95a:22014) Witte, Dave(1-MIT) Arithmetic groups of higher Q-rank cannot act on 1-manifolds. (English summary) Proc. Amer. Math. Soc. 122 (1994), no. 2, 333–340.

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@Andy: Thank you! – Mark Sapir Nov 16 '11 at 19:44
2  
There is a folk conjecture that a discrete, torsion free group with Kazhdan's property (T), cannot be left orderable. – Alain Valette Nov 16 '11 at 22:16

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