Are all torsionfree finitely generated linear groups over $\mathbb{C}$ left orderable? In particular, are torsionfree 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 WitteMorris; see
MR1198459 (95a:22014) Witte, Dave(1MIT) Arithmetic groups of higher Qrank cannot act on 1manifolds. (English summary) Proc. Amer. Math. Soc. 122 (1994), no. 2, 333–340.

2$\begingroup$ There is a folk conjecture that a discrete, torsion free group with Kazhdan's property (T), cannot be left orderable. $\endgroup$ – Alain Valette Nov 16 '11 at 22:16

$\begingroup$ My own conjecture would be that there exists a nontrivial leftorderable group with Property T. Some little evidence comes from this answer: leftorderable 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 leftorderable groups. $\endgroup$ – YCor Feb 16 '20 at 17:39