# Entropy of a North South Transformation.

Let $f:\mathbb{S}^2\to\mathbb{S}^2$ be a continuous north south Transformation, in other words, the point $(0,0,1)$ is a global attractor for $f$ and $(0,0,-1)$ is a global attractor for $f^{-1}$.

How to calculate the entropy of $f$?

I do not want to use the theorem:

$$h_{top}(f)=h_{top}(f|NW).$$

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One easy way to do this is to use the variational principle to say that the topological entropy of $f$ is the supremum over all $f$-invariant probability measures $\mu$ of the metric entropy $h_\mu(f)$, that is $$h_{top}(f) = \sup_\mu h_\mu(f).$$ The problem then becomes a matter of understanding $f$-invariant probability measures. Suppose $\mu$ is an $f$-invariant probability measure. Then the Poincaré recurrence theorem implies that almost every point in the support of $\mu$ is recurrent for $f$. Since the only points in $\mathbb{S}^2$ which are recurrent for $f$ are the north and south pole, any such $\mu$ must be supported on these two points, i.e., $\mu = a\delta_{N} + b\delta_S$, where $a,b\geq 0$, $a + b = 1$, and $\delta_N$ and $\delta_S$ denote the Dirac probability measures at the north and south poles, respectively. It is easy to check that $h_\mu(f) = 0$ for any such $\mu$. Thus $h_{top}(f) = 0$.

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It's an interesting solution, but at the moment variational principle is a result that I do not like to use. –  user27456 Dec 6 '12 at 1:54

A simple way for this proof is see that any 2 power of the nort-south maps are conjugated, so fo every $n\in\mathbb{N}$ we have,

$$h_{top}(T)=h_{top}(T^n)=n\cdot h_{top}(T)$$

then we have $h_{top}(T)=0.$

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Is it known that they are really conjugate? –  Le Curious Feb 24 at 9:37