MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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 the Fibonacci semi-group $<L,R|LRR=RLL>$ with a Bernoulli walk $P(R)=p, P(L)=1-p$. Is the entropy $H(p)$ an unimodal function with maximum at p=0.5? Is this true for all finitely generated semi-groups (with some symmetries)? For the free semi-group $<L,R>$ (the Shift space) it is well known and easy to prove that $H(p)=-(plog(p)+(1-p)log(1-p))$.

share|cite|improve this question
If a semigroup is finite, then the entropy is constant ($= 0$)? – Mark Sapir May 8 '12 at 17:08
Yes, in this case the entropy is zero and also for the (infinite) Pascal semi-group <R,L|RL=LR> it is zero as well. One can prove that the entropy of all random walks are bounded by exponential growth rate of the semi group. Hence for the Fibonacci semi-group the entropy is bounded by the logarithm of the golden number. In fact Bernoulli walks do not attain this entropy but there exists walks that do. – Jörg Neunhäuserer May 8 '12 at 17:55
But then the $H(p)$ is not unimodular, it is a constant? – Mark Sapir May 9 '12 at 6:03
Yes, if You want to put it in this way. It depends if You like to have a Your unimodal function to be strictly mononton or just monoton. But that is not the point. I am interested in the Fibonacci semi-group and other infinite semi-groups with positive exponetial growth rate, where H(p) is not expected to be constant. In fact I not even have a proof of this, but numerical calculations suggest it. – Jörg Neunhäuserer May 9 '12 at 11:54

Your Answer


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

Browse other questions tagged or ask your own question.