\le \infty$ and let $X$ be a compact $C^r$-manifold, with a smooth volume measure $\mu$. Then there is a residual set ${\mathcal{R}}\subset C^r(X,{\mathbb{G}})$ such that if $A \in {\mathcal{R}}$ then $A$ satisfies the continuous dichotomy.

Our proof of the corollary \[c.measdich\] gives an effective way to decide whether a given map $A \colon X \to {\mathbb{G}}$ satisfies the measurable dichotomy. That is not so for our proof of the continuous dichotomy theorem \[t.contdich\]. Two related questions are:

Is there any analogue of theorem \[t.measclassif\] for the continuous case?

In theorem \[t.contdich\], can we take an open and dense set, instead of a residual one?

The smooth case
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If both $T$ and $A$ are assumed to have higher regularity, then dichotomy between uniform hyperbolicity and zero exponents is no longer true – either fixing $A$ and varying $T$ or fixing $T$ and varying $A$. This is shown by the following two examples:

1.  Let $T \colon X \to X$ be a volume-preserving $C^2$ Anosov diffeomorphism. Then the exponent is positive on an open and dense subset of $C^1(X,{\mathbb{G}})$, by results of Bonatti and Viana [@BnV].

2.  Consider Schrödinger cocycles on the $d$-torus: $$\label{e.schrodinger}
    S_{\lambda,V}(\theta) =
    \begin{pmatrix} \lambda V(\theta) & - 