-   or ${\mathit{LE}}(A,T)=0$.

We prove that if $X$ is a $C^r$ manifold ($r\ge 1$) then the *$C^r$-generic map $A$ satisfies the continuous dichotomy*.

Next we give the precise statements.

The measurable case
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To define richness we need first to introduce some notation. If $\nu$ is a measure in ${\mathbb{G}}$ and $v \in {\mathbb{P}}^1$, then the push-forwards of $\nu$ by the maps $$M \in {\mathbb{G}}\mapsto M^{-1} \in {\mathbb{G}}\quad \text{and} \quad
M \in {\mathbb{G}}\mapsto M \cdot v \in {\mathbb{P}}^1$$ are indicated by $\nu^{-1}$ and $\nu * v$, respectively. If $n \in {\mathbb{N}}$, the push-forward of $\nu^n$ by the map $$(M_1, \ldots, M_n) \in {\mathbb{G}}^n \; \mapsto \; M_n \cdots M_1 \in {\mathbb{G}}$$ (i.e., the $n$-th convolution power) is indicated by $\nu^{*n}$.

\[d.rich\] Let $\nu$ be a finite measure on ${\mathbb{G}}$ of bounded support. Then $\nu$ is called *$N$-rich*, for $N \in {\mathbb{N}}$, if there exists $\kappa>0$ such that for every $v \in {\mathbb{P}}^1$ we have $$\nu^{*N} * v \ge \kappa m
\quad \text{and} \quad
(\nu^{*N})^{-1} * v \ge \kappa m,$$ where $m$ denotes Lebesgue measure in ${\mathbb{P}}^1$. The measure $\nu$ is called *rich* if it is $N$-rich for some $N\in{\mathbb{N}}$.

The richness property is studied in § \[ss.richness\], where the following 