42

A typical example arising in algebraic geometry is the following.
Take a scheme $X$ of finite type over $\mathbb{C}$. Then there are (at least) two concepts of fundamental group for $X$: the topological fundamental group $\pi_1^{top}(X)$, i.e. the fundamental group of the underlying topological space $X(\mathbb{C})$ with the analytic topology, and the étale ...

39

Consider the class of finitely generated linear groups. Such groups $G$ satisfy certain well-known restrictions, for instance:
Every such $G$ is residually finite (Malcev, 1940). Thus, most Baumslag-Solitar groups, e.g.
$$
\langle a, b| a b^2 a^{-1} =b^3\rangle
$$
are not linear. This is the simplest example of a nonlinear f.g. group I know.
$G$ is ...

39

Yes, the fundamental group of the Hawaiian earring $\pi_1(\mathbb{H},b_0)$ is an important group which is sometimes called the free sigma product $\#_{\mathbb{N}}\mathbb{Z}$. Its is often defined in purely algebraic terms as a group of "transfinite words" in countably many letters. In many ways this group behaves like the non-abelian version of the Specker ...

25

Below are some sources of hyperbolic groups. Of course, the list is far from being exhaustive.
Groups defined by generators and relations:
Finitely generated free groups, as their Cayley graphs are simplicial trees.
If $\varphi$ is an atoroidal automorphism of a free group $\mathbb{F}_n$, then the extension $\mathbb{F}_n \rtimes_\varphi \mathbb{Z}$ is ...

24

The conjecture is indeed strictly weaker than $\mathrm{P = NP}$, in the sense that it follows from $\mathrm E=\Sigma^E_2$, which is not known to imply $\mathrm{P = NP}$. Of course, we cannot prove this unconditionally with current technology, as it would establish $\mathrm{P\ne NP}$.
Here, $\mathrm E$ denotes $\mathrm{DTIME}(2^{O(n)})$, $\mathrm{NE}$${}=\...

answered Aug 6 '18 at 12:29

Emil Jeřábek supports Monica

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24

An earlier reference for groups with this property is
J. Avenhaus and K. Madlener. Subrekursive Komplexität der Gruppen. I. Gruppen mit vorgeschriebenen Komplexität. Acta Infomat., 9 (1): 87-104, 1977/78.
There is a hierarchy of the recursive functions known as the (difficult to pronounce) Grzegorczyk Hierarchy $E_0 \subset E_1 \subset E_2 \subset \cdots$, ...

gr.group-theory computational-complexity geometric-group-theory computational-group-theory word-problem

21

The answer to Question 1 is "no". The group $\langle x, y \mid x^{12}y = yx^{18}\rangle$
is Hopﬁan but contains a non-Hopﬁan subgroup of ﬁnite index (see Baumslag, Gilbert; Solitar, Donald Some two-generator one-relator non-Hopfian groups. Bull. Amer. Math. Soc. 68 1962 199–201.).
Edit. Since the paper of Baumslag and Solitar contains almost no proofs, ...

21

(Probably the question would be more suitable for MathSE)
The answer is no. Well, the trivial group is a counterexample. Also finite groups are counterexamples. So first and for all you should have specified that you assume the group to be infinite. But then the answer is still no.
In a finitely generated group, an element $g$ is called distorted if $\lim |...

21

Bridson and Vogtmann proved a much stronger result. From the abstract: 'If $m$ is less than $n$ then [the image of] a homomorphism $\mathrm{Aut}(F_n)\to\mathrm{Aut}(F_m)$ can have cardinality at most 2.'

19

It is a famous open problem. Akhmedov in MR2424177 claimed he could prove that the answer is "no". No proof exists, so I guess he discovered a gap in his argument.

19

A usually hard problem in Algebraic Geometry is computing the fundamental group of the complement $U=\mathbb{P}^n - V$, where $V$ is a reduced hypersurface.
At least in principle, the computation of $\pi_1(U)$ can be carried out by using Seifert-Van Kampen theorem; the tricky part is that the relations in the presentation depend not only on the ...

answered Jan 26 '17 at 16:47

Francesco Polizzi

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19

As Andreas says (in his answer and his comment to it), there are groups whose word problem is undecidable and one could similarly set up a group that encodes the halting problem of a class of Turing machines where this is decidable but difficult. However, one must be careful in the encoding. In
Isoperimetric and Isodiametric Functions of Groups,
Mark V. ...

gr.group-theory computational-complexity geometric-group-theory computational-group-theory word-problem

answered Jan 23 '19 at 18:09

Benjamin Steinberg

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18

A better question is:
Given a group $\Pi$, is there a compact manifold $M$ with non-negative Ricci curvature such that $\pi_1(M)=\Pi$?
The answer is given in "On fundamental groups of manifolds of nonnegative curvature" by Wilking.
Here is the main result:
(source: psu.edu)

18

Yes. Let $H$ be a finitely generated subgroup of $F$. Let $[0,a]$ be the largest interval where all elements of $H$ are equal to the identity. Then consider the map that sends $h\in H$ to the $\log_2$ of the (right) slope of $h$ at $a$. This map is a non-trivial homomorphism of $H$ into $\mathbb{Z}$.
A more fancy (but essentialy the same) proof is this: by ...

18

Thompson's group F is an example. It's finitely presented and, according to this paper of Ken Brown, the integral homology is free abelian of rank 2 in every positive dimension.

18

Assuming the assertions your claim are true, it follows that $\mathrm{GL}_n(\mathbf{Z})$ is generated by 2 elements for all $n$; for $n$ odd (your missing case), the matrices $-s_1,s_3$ indeed generate $\mathrm{GL}_n(\mathbf{Z})$.
This follows from the following three facts:
1) Let $p$ be prime and $C_p=\langle c\rangle$ the cyclic group of order $p$. If $...

18

Yes. The following answer is inspired by Andy Putman's comment. Let $N_\infty(G)$ be the subgroup generated by elements of infinite order, in a group $G$.
Every non-elementary hyperbolic group $G$ with trivial finite radical has a useful property, which bears the ridiculous name "Pnaive", namely that for every finite subset $F$ there exists a $x$ such that ...

18

I think this is a great question, as there is still a need for an authoritative reference about (word-)hyperbolic groups. Since the textbook doesn't exist, I'd like to take the question in a slightly different direction by listing some of the material I think it should cover. (This is inevitably a personal and biased account.) I'll try to include the best ...

17

There is quite a lot that can be said about the Charney-Davis conjecture, so let me say a few things and perhaps add more later.
1) The conjecture is a discrete analogue of a well known conjecture by Hopf on the Euler characteristic of nonpositively curved manifolds in odd dimension. You consider cubical complexes and the condition of nonpositive curvature ...

17

Dehn gave at least three solutions of the conjugacy problem for surface
groups, which can be found in my translation in the book Papers on Group
Theory and Topology (Springer 1986), Papers 2, 4, and 5.
The first is based on an idea of Poincaré: lifting a curve to the universal
cover, which is the disk model of the hyperbolic plane, replacing it by the
...

answered Jul 17 '13 at 6:09

John Stillwell

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16

Presumably you've consulted the Wikipedia page on the Baumslag-Solitar group, which states that $BS(m,n)$ is not residually finite (and therefore not linear) if $|m|\neq |n|$ and $|m|>1, |n|>1$. This leaves the case that $|m|=|n|$. In this case, one can show that the group is the fundamental group of a compact Seifert-fibered 3-manifold, which is known ...

16

(1) Bridson showed that if a mapping class group of a surface (of genus at least 3) acts on a CAT(0) space, then Dehn twists act as elliptic or parabolic elements. This implies that the mapping class groups of genus $\geq 3$ are not CAT(0) (Edit: as pointed out by Misha in the comments, this was originally proved by Kapovich and Leeb, based on an observation ...

gt.geometric-topology mg.metric-geometry geometric-group-theory mapping-class-groups alexandrov-geometry

16

For some reason, the question seems to be asking for an algebraic (number- or ring-theoretical) justification for residual finiteness (and implicitly LERF, though in fact the correct statement is that every hyperbolic group is residually finite if and only if they are all QCERF, as proved by Agol--Groves--Manning). Recall that residual finiteness means that ...

16

It's undecidable.
Lemma: a group $G$ is nontrivial if and only if the free product $H=G\ast\mathbf{Z}$ has an infinitely generated derived subgroup.
Proof: assume $G$ finitely generated and nontrivial. The kernel $N$ of the canonical epimorphism $H\to\mathbf{Z}$ is isomorphic to $G^{\ast\mathbf{Z}}$ and hence is infinitely generated. Since $H/[H,H]$ is ...

15

There is never a (finite) generating set with that property.
Consider a generating set $S=\{x_1,\ldots,x_{\ell}\}$ of cardinality $\ell$. Let $B_k := B_k(S)$, $S_k := B_k \setminus B_{k-1}$, and $g := gr(S)$. Let $b_k := |B_k|$ and $s_k: = |S_k|$. Assume for simplicity that $L := \lim_{k \to \infty} \frac{b_k}{g^k}$ exists (although it shouldn't be ...

15

The following is an elaboration of the last paragraph of Misha's answer.
For me, the thing that makes non-linear (discrete) groups interesting is that we are not very good at constructing them! Linearity remains unknown for many natural examples of groups, such as mapping class groups, and if true would dramatically simplify some hard theorems. (For ...

15

First of all, it is not true that there are only two types of nondegenerate $3$-forms in $\Lambda^3(V^\ast)$ when $\dim_\mathbb{R}V=6$. There are actually $3$ such nondegenerate orbits. Here is the complete list of $\mathrm{GL}(V)$-orbit types in this space: Let $e^1,\ldots, e^6$ be a basis of $V$. Then the following $3$-forms are inequivalent under $\...

geometric-group-theory sg.symplectic-geometry dg.differential-geometry ag.algebraic-geometry complex-geometry

answered Feb 19 '13 at 15:53

Robert Bryant

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15

My own interest in automatic groups has been principally algorithmic, and I believe that this was Thurston's original motivation for studying them - they provided a method for carrying out practical computations in a variety of interesting groups with negative curvature. Once a (geodesic) automatic structure has been computed, you can compute the growth ...

15

Up to equivalence there is only one exponential type of growth in which case the answer is trivial. In polynomially bounded growth the answer follows from various old theorems (notably of Wolf and Gromov): the possible growths are $n^d$ for $d$ non-negative integer and things are classified. What's remaining is intermediate growth, and the first examples ...

15

An algorithm for computing the hyperbolic "thinness" constant $\delta$ is described in the paper:
Epstein, David B. A.; Holt, Derek F
Computation in word-hyperbolic groups.
Internat. J. Algebra Comput. 11 (2001), no. 4, 467–487.
and I did implement it. It works OK on reasonably straightforward examples like surface groups, hyperbolic triangle groups, etc, ...

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