0
$\begingroup$

Hi,

I know that in the diffeomorphism case the measure entropy of the T:M^{2}-->M^{2} (M smooth Rimannian surface) will be the same as the measure entropy of T^{-1}. But i need to know about the C^{2} endomorphism case. it seems that it can not be true, i want to know, what can be the relation between the entropy of T and its inverse? can we add some conditions which help us that the equality occurs again, maybe?

Thanks, Pooh

Improve this question
$\endgroup$
  • 1
    $\begingroup$ I don't think the question makes sense. $T^{-1}$ isn't a map and so its entropy isn't defined? $\endgroup$ – Anthony Quas Sep 27 '12 at 22:33
  • $\begingroup$ Agreed with Anthony. If $T$ is only an endomorphism and not a diffeo, then $T^{-1}$ is not defined as a map of $M$. So "entropy of the inverse map" doesn't make sense. $\endgroup$ – Vaughn Climenhaga Sep 28 '12 at 0:52
1
$\begingroup$

I don't see any problems with defining entropy of $T^{-1}$ as it can always be done in a "natural" way by passing to the natural extension (not sure the author meant this, though). However, more important is that the author is apparently not aware of the fact that coincidence of entropies of the inverse and original maps has nothing to do with diffeomorphisms, surfaces etc. and is an absolutely general property.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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