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The basic question is in the title, but I am interested in both necessary and sufficient conditions.

I know the Tits' alternative and Malcev's result that finitely generated linear groups are residually finite, but I don't know any purely group-theoretical sufficient conditions.

Though it's not purely group-theoretical, what if the group acts simplicially on a (finite) simplicial complex? Does this imply linearity? Does it imply linearity over $\mathbb{C}$?

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Deciding whether or not a particular group is linear can be extremely hard, as seen in the 2001/2002 papers by Bigelow and Krammer proving linearity of braid groups. In particular, necessary or sufficient conditions to be linear will probably depend on first characterizing the type of groups you have in mind as narrowly as possible. – Jim Humphreys Jun 11 '13 at 15:20
What do you mean by "what if the group acts simplicially on a (finite) simplicial complex"? – David Cohen Jun 11 '13 at 15:24
Since this question doesn't seem to admit a unique answer, I think it should be community-wiki. – HJRW Jun 11 '13 at 15:33
Also, you might be interested in Yves Cornulier's excellent answer to this question:… . – HJRW Jun 11 '13 at 15:47
See also the reference to Lubotzky's criterion of linearity in… – Misha Jun 11 '13 at 16:15

Since this is not purely group theoretical and not a complete answer, this maybe should be more of a comment, but since you mentioned simplicial complexes perhaps you should check out the following paper to get you started:

Haglund, Frédéric, and Daniel T. Wise. "Special cube complexes." Geometric and Functional Analysis 17.5 (2008): 1551-1620.

Which is concerned with the fundamental groups of certain square complexes (VH complexes whose 1-cells are divided into two classes, "horizontal" and "vertical", and the attaching maps of squares alternate v-h-v-h). For instance, they prove that any fundamental group of a compact virtually clean (clean means attaching maps are embeddings, and this implies that the group splits as a clean graph of groups, as studied in [1]) VH-complex is linear.

Although this isn't purely group theoretic, it is at least mostly presentation theoretic, and the result itself isn't too hard to apply if you have for instance a finitely presented group. In this case, there is often an easy algorithm to check whether such a group has a VH-subdivision. (For instance, there is an example of such a procedure outlined in my paper with Wise [2]). After you still have to check the virtually clean condition, which may or may not be so easy depending on what you are doing.

[1] Wise, Daniel T. "The residual finiteness of negatively curved polygons of finite groups." Inventiones mathematicae 149.3 (2002): 579-617.

[2] Polák and Wise, "A Note on VH Subdivisions", To appear.

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I think this is a great answer! (I was going to post something similar, but you beat me to it.) I don't know what a 'complete answer' would be, other than perhaps the observation that linearity is an algorithmically undecidable property. – HJRW Jun 11 '13 at 17:57
Linearity of fundamental groups of special cube complexes (with finitely many walls) follows from embedding in right-angled reflection groups. In turn, linearity is proved for reflection groups by showing that they act discretely on a certain convex cone in projective space, with fundamental domain a finite-sided polyhedron. – Ian Agol Jun 14 '13 at 3:44

Here are some more purely group theoretical conditions. This is also not a complete answer, since it gives just some necessary conditions for certain groups to be linear.

  1. Schur: Suppose that $G$ is a finitely generated linear group, such that all elements have finite order. Then $G$ is finite.

  2. Jordan: Suppose that $G$ is a finite linear group of degree $n$ over a field of characteristic zero. Then there exists an integer-valued function $\beta(n)$ such that G contains an abelian normal subgroup of finite index at most $\beta(n)$.

  3. Malcev: Suppose that $G$ is a finitely-generated linear group. Then $G$ is residually finite. If $G$ is simple, then $G$ is finite.

  4. Platonov: Suppose that $G$ is a linear group of degree $n$ of finite Pruefer rang $r$ over a field of characteristic $p > 0$. Then $G$ contains an abelian normal subgroup of finite index bounded in terms of $r, n$, and $p$.

  5. Malcev: Suppose that $G$ is a solvable linear group of degree $n$ over an algebraically closed field. Then $G$ contains a triangularizable normal subgroup of finite index bounded by a function of $n$.

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1. is false in characteristic $p$, you need further hypotheses (e.g. that $G$ is finitely generated). 2. is dramatically false in characteristic $p$ (there is a more complicated substitute by Weisfeiler and others). In 3. "furthermore" is rather an immediate corollary. 4.: how do you define "periodic"? one definition I know is that all elements have finite order, but I don't expect this is what you mean. 5. you don't define triangularizable; the statement seems to be true only if the field is algebraically closed. – YCor Jun 13 '13 at 9:45
Yes, I apologize. I have corrected the statements (hopefully). – Dietrich Burde Jun 13 '13 at 11:26
You need to fix 4. once more: a solvable group of finite Prufer rank in characteristic zero need not be abelian-by-finite (counterexamples are the solvable Baumslag-Solitar groups, and polycyclic groups such as the integral Heisenberg group). This is only true in positive characteristic. (Also in 2 and 4 you don't say what $n$ is; I guess it's the dimension of a faithful linear rep... in 5 you call $n$ the degree, is that standard? it can be confused with the solvability length.) – YCor Jun 13 '13 at 11:58
@Yves, thank you. Next time I am more careful. – Dietrich Burde Jun 13 '13 at 14:06
here's one more condition in the same spirit: in a group, call centralizer a subgroup which is centralizer of some subset. Then in a linear group any chain of centralizers is finite, and actually of length bounded by $n^2$, where $n$ is the dimension. This is just because the centralizer is the intersection of a subalgebra of matrices with the group. – YCor Jun 13 '13 at 15:05

One more necessary condition. Let $T$ be a matrix from $SL_n(\mathbb{C})$. Then the set of all matrices $B$ such that $\lim_{n\to \infty} T^{n} BT^{-n} = 1$ is a nilpotent subgroup (an exercise, first noticed by Margulis, I think, a proof can be found here.). This implies, for example, that the group $\langle a,b,t \mid tat^{-1}=a^2, tbt^{-1}=b^2\rangle$ is not linear (this group is residually finite).

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This is not a purely group-theoretic property (it depends on a given embedding and uses the topology of $\mathbf{C}$). It there is a group-theoretic criterion, it is hidden in your "This implies". – YCor Jun 16 '13 at 14:20
here's a group theoretic statement (P): for every $g\in G$, the set $G[g]$ of $x\in G$ of infinite order such that $gx^ag^{-1}=x^b$ for some $0<a<b$, generates a torsion-free nilpotent subgroup. Then any group linear over $\mathbf{C}$ satisfies (P). Actually, if $G$ is linear in characteristic $p>0$, then a stronger statement holds: for every $g$ the set $G(g)$ is empty. – YCor Jun 16 '13 at 18:14
Yes, this can be found in the text I gave a link to in the question. – Mark Sapir Jun 17 '13 at 11:54

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