Random products of $SL(2,R)$ matrices and Furstenberg's theorem

Question

I was recently learning Furstenberg's theorem on random products of $SL(2,R)$ matrices, and came across with a simple example that confused me:

Considering random products of two matrices $A=\begin{pmatrix} 2 &0\\ 0 &1/2 \end{pmatrix}$ and $B=\begin{pmatrix} 0 &1\\ -1 &0 \end{pmatrix}$ with probability $1/2$ and $1/2$, it is mentioned in literature that the Lyapunov exponent for this random product is $0$. See, e.g., page 29 in Damanik's survey article: https://arxiv.org/abs/1410.2445

I understand that if we focus on a specific matrix element of the product, it has no well-defined limit. However, if we study its norm, say the Frobenius norm $||A||=\sqrt{\sum_{i,j}|a_{ij}|^2}$. I found that by taking average, this norm indeed grows exponentially, and the Lyapunov exponent is positive.

So my question is, why it is said "the lyapunov exponent for this example is $0$" in literature? Thanks!

Answer 1

Because its Lyapunov exponent is zero, but that is not what you are computing.

Instead of looking at random walks on $SL(2, \mathbb{R})$, let me focus on random walks on $\mathbb{R}_+^*$, as there is the same issue. Let $(X_n)$ be i.i.d. in $\mathbb{R}_+^*$, and to make things simple, assume that there are only finitely many values. Let $P_n := X_n \ldots X_1$.

The Lyapunov exponent of this random walk is the real $\Lambda$ such that

$$\lim_{n \to + \infty} \frac{\ln (P_n)}{n} = \Lambda.$$

By the law of large numbers, $\Lambda = \mathbb{E} (\ln(X_1))$. For instance, if $X_1$ takes values $2$ and $1/2$ each with probability $1/2$, the Lyapunov exponent is $0$: the Markov chain $(P_n)$ will oscillate between very large and very low values.

However, if you compute the expectation of the norm, a short computation gets you $\mathbb{E} (P_n) = (5/4)^n$, which grows exponentially fast. But that does not mean that the Lyapunov exponent is $\ln (5/4)$. The issue is merely that the exponential does not commute with the expectation:

$$1 = e^{\mathbb{E}(\ln(P_n))} \leq \mathbb{E} (e^{\ln(P_n)}) = (5/4)^n.$$

To get back to a general random walk, and very roughyl, we have $\ln(P_n) \simeq \mathcal{N} (n\mu, n\sigma^2)$. The Lyapunov exponent is the constant $\mu$. However,

$$\mathbb{E} (P_n) \simeq \mathbb{E} (e^{\mathcal{N} (n\mu, n\sigma^2)}) = \mathbb{E} (e^{n(\mu+\frac{\sigma^2}{2})}),$$

so taking the norm as you get gives you an error coming from the diffusion of the random walk (well, in practice, the exact value of $\sigma^2/2$ for this error is wrong, but I don't think the heuristics is too bad at this level).