A few days ago I attended a lecture given by Amie Wilkinson. She presented a proof of Furstenberg’s theorem on the Lyapunov exponents of random products of matrices in .

Let be a probability measure on , be the product measure on . Let be the shift map on , and be the projection. We consider the induced skew product on . The (largest) Lyapunov exponent of is defined to be the value such that for -a.e. .

To apply the ergodic theory, we first assume . Then is well defined. There are cases when :

(1) the generated group is compact;

(2) there exists a finite set of lines that is invariant for all .

Furstenberg proved that the above cover all cases with zero exponent:

for all other .

1. Let be a dynamical system, be a measurable and integrable function on . We define the Birkhoff sum , and the Birkhoff average whenever the limit exits.

Lemma. Suppose for -a.e. . Then for -a.e. .

That is, the only possible growth rate of a Birkhoff sum is linear. More generally, assume is ergodic and on a subset of positive measure. (1) One can apply the above lemma to the return map of on the induced map . (2) for a.e. , and hence a.e.. So Lemma 1 applies.

Proof. WLG we assume is ergodic. Then it suffices to show that . To derive a contradiction we assume , which forces .

Intuitively, we may integrate and get if one can change with . The following proof can be found in Bochi’s notes.

Let’s consider the following set , and .

Step 1. for any , there exists and such that . Otherwise, for any and for any , there exists such that . Pick the sequence inductively by , , and partition the sum to the segments , we see that , contradicting the assumption that .

Conclusion: the set is of full measure.

Step 2. For any , for some . Then we partition the segment into short segments according to the moments that . Then . Divide both sides by and let : . Integrate both sides over : .

Note that is -invariant and containing the set . So we acctually get , and hence for any . Therefore, , which contradicts with Step 1 and hence completes the proof.

2. Let be the space of lines through the origin, and each induces naturally an action on by sending to . Then we have the map , . For each probability measure on , we define the convolution be the push-forward of the measure . Then is said to be -stationary, if .

Lemma. Any -stationary measure is non-atomic.

To show , it sufficient to show that for some -stationary measure . Then a sufficient condition is

Lemma. for for all but one and -a.e. .

The line actually gives the stable bundle of at .