Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Are there any procedures which given a nonnegative nondecreasing function on the integers will construct a finitely generated group with the same growth up to the usual equivalence of growth functions? Can this at least be done for nice functions? For example Bergman gives an explicit construction of semigroups of any growth between quadratic and cubic.

Also, is there an algorithmic way to do this if your function is recursive and given as input by a Turing machine?

share|improve this question
2  
Why is this question so terrible? Constructing groups with intermediate growth of a somewhat prescribed type is very difficult. We don't even know if there is a smallest intermediate growth. –  Benjamin Steinberg Dec 8 '13 at 2:23
1  
You might like this classical paper of Bergman math.berkeley.edu/~gbergman/papers/unpub/growth.pdf –  Benjamin Steinberg Dec 8 '13 at 2:40
    
@BenjaminSteinberg Do you feel that the reason for closure is invalid? –  Todd Trimble Dec 8 '13 at 3:02
3  
@BenjaminSteinberg : In case you miss my meta comment, here it is again. I agree that one could ask a reasonable question about prescribing the growth of groups. However, this is not it. For instance, I don't know what the OP means by "manifold", and I don't know exactly what the OP means with regards to a growth function, e.g. what's the equivalence relation? For instance, as written he might have a specific function and want a group with a specific finite generating set realizing that specific function on the nose. Someone (you or the OP) should rewrite the question before it is reopened. –  Andy Putman Dec 8 '13 at 3:16
5  
Up to equivalence there is only one exponential type of growth in which case the answer is trivial. In polynomial growth the answer follows from various old theorems (notably of Wolf and Gromov): the possible growth are $n^d$ for $d$ non-negative integer and things are classified. What's remaining is subexponential growth, and the first examples with growth determined up to equivalence were obtained quite recently by Bartholdi and Erschler (Inventiones arxiv.org/abs/1011.5266, 2012, see also their more recent arxiv.org/abs/1110.3650) –  YCor Dec 8 '13 at 17:12

1 Answer 1

up vote 14 down vote accepted

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 with growth determined up to equivalence were obtained quite recently by Bartholdi and Erschler (Inventiones, arxiv 1011.5266, 2012, see also their more recent arxiv 1110.3650). Although the growth of the first Grigorchuk group $\Gamma$ is not yet known up to equivalence, $\Gamma$ is used (in a permutational wreath product construction) in the construction of the Bartholdi-Erschler groups.

share|improve this answer
3  
In particular it is not known whether there is a group with growth function equivalent to $e^{\sqrt{n}}$. –  Mustafa Gokhan Benli Dec 8 '13 at 22:22

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.