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Assume that $(T, A)$ is a linear cocycle such that $T:X\rightarrow X$ is a homemorphism on compact metric space $X$ and $A:X\rightarrow SL(2, \mathbb{R})$ is a continuous function.

We say that an invariant probability measure $\mu$ is aperiodic for $T$ if set of periodic points of has zero measure.

There is a famous result due to Bochi-Mane that said :

Given any invertible aperiodic ergodic system $(T,\mu)$ on a compact Hausdorff space, every continuous cocycle $A:X→SL(2, \mathbf{R})$ which is not uniformly hyperbolic can be approximated in the $C^{0}$ topology by another whose Lyapunov exponents vanish at $\mu$-almost every point.

Consider the following examples: $$ A_1=diag(2, \frac{1}{2}) , A_{2}=diag(1, 1) $$ $$ A_1=diag(2, \frac{1}{2}) , A_{2}=antidiag(1, 1)$$ Would one show me how to approximate the above linear cocycles such that it has zero Lyapunov exponent?

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I’m guessing you mean the base dynamical system to be a Bernoulli shift with the unperturbed cocycle being $A(x)=A_{x_0}$? For the second one, no perturbation is necessary. The Lyapunov exponents are already 0.

For the first system, pick an $N$. Now if $x_{-k-1}\ldots x_{N-k}=1222\ldots 21$ for some $0\le k\le N-1$, replace $I$ by a rotation by $\pi/(2N)$, so that the product over the block of 2’s is $\text{antidiag}(1,-1)$. This has zero Lyapunov exponents just as above.

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  • $\begingroup$ Thanks. It is nice example. I got it. $\endgroup$ – Adam Nov 26 '19 at 17:48

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