MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider a finitely generated group. Assume that the first Betti number of the ball of radius n in the Cayley graph is at most polynomial in n. This property is satisfied by free groups and groups of polynomial growth. Are there any others?

share|cite|improve this question
The property seems to depend on the generating set. Consider the free group $\langle a,b\rangle$ with the generating set $\\{a,b,ab\\}$. The Cayley graph is a tree-graded space where pieces are triangles. A ball of radius $k$ has exponentially many triangles, each contributing 1 to the Betti number. So it seems that the growth is exponential if you use this generating set. Perhaps you should correct the question. – Mark Sapir May 4 '13 at 1:11
@Anton: My comment is equivalent to your answer. – Mark Sapir May 4 '13 at 1:18
@Mark: well, almost equivalent, almost simultaneous :) – Anton Petrunin May 4 '13 at 1:48

Assume the group is not free. Then its Cayley graph has a nontrivial loop. Applying shifts of the group you get a loop which starts at any element.

If the group grows faster then polynomially, then the number of disjoint cycles and therefore first Betti number grows faster than polynomial.

So, the answer is "yes" --- only free groups and groups of polynomial growth.

share|cite|improve this answer

Your Answer


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.