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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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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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    $\begingroup$ There is a folk conjecture that a discrete, torsion free group with Kazhdan's property (T), cannot be left orderable. $\endgroup$
    – Alain Valette
    Commented Nov 16, 2011 at 22:16
  • $\begingroup$ My own conjecture would be that there exists a nontrivial left-orderable group with Property T. Some little evidence comes from this answer: left-orderable is compatible with some weak form of Property T. Also the evidence in the opposite direction is not convincing to me: arithmetic lattices were initially seen as prototypical examples of Property T groups, but they're only some very special (though important) examples. Same for local indicability, and subgroups of diffeomorphism groups, which are only very special instances of left-orderable groups. $\endgroup$
    – YCor
    Commented Feb 16, 2020 at 17:39

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