# Example of zero Lyapunov exponentes

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?

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.