In this post I prove the existence of generic vectors.

Proof of Existence of Generic Vectors

We follow Wilkinson’s notes on Otal’s theorem. In particular, we solve Exercise 4.6.

Let \(\phi^t : T^1M \rightarrow T^1M\) be an ergodic flow (with respect to a measure \(\lambda\)) on the unit tangent bundle of a Riemannian manifold. We say that a vector \(v \in T^1M\) is generic with respect to the flow if for every continuous function \(k : T^1M \rightarrow \mathbb{R}\) we have

\[\lim_{T \rightarrow \infty} \frac{1}{T} \int_0^T k(\phi^t(v)) dt = \frac{1}{\lambda(T^1M)} \int_{T^1M} k(v) d\lambda(v).\]

The goal of the article is to prove the following claim.

Claim: Almost every \(v \in T^1M\) is generic.

The two main ingredients are the following.

1) Theorem: (Birkhoff Ergodic Theorem) If \(f^t : X \rightarrow X\) is an ergodic flow with respect to the measure \(m\), then for all \(k \in L^1(m)\) the following equality holds for \(m\) almost every \(x \in X\):

\[\lim_{T \rightarrow \infty}\frac{1}{T} \int_0^T k(f^t(x)) dt = \int_X f dm.\]

2) There is a countable dense subset of \(C(T^1M)\). Label this \((k_n)_{n = 1}^\infty \subseteq C(T^1M)\).

With these, we can prove the claim.

Proof: Without loss of generality, let’s just assume \(\lambda\) is a probability measure. Let

\[\Gamma(m) := \left\{v \in T^1M \ | \ \lim_{T \rightarrow \infty} \frac{1}{T} \int_0^T k_m(\phi^t(v)) dt = \int_{T^1M} k_m(v) d\lambda(v) \right\}.\]

The Birkhoff ergodic theorem tells us that this has measure \(1\). The countable intersection of sets with measure \(1\) still has measure \(1\), so let

\[\Gamma := \bigcap_{m = 1}^\infty \Gamma(m).\]

Then \(\lambda(\Gamma) = 1\). Now let \(k \in C(T^1M)\). Since \(\{k_m\}\) is dense, we can assume that we have a sequence so that \(k_m \rightarrow k\). Thus we can write

\[\int_{T^1M} k(v) d\lambda(v) \xleftarrow[]{m \rightarrow \infty} \int_{T^1M} k_m(v) d\lambda(v) \xleftarrow[]{T \rightarrow \infty} \frac{1}{T} \int_0^T k_m(\phi^t(v)) dt \xrightarrow[]{m \rightarrow \infty} \frac{1}{T} \int_0^T k(\phi^t(v)) dt.\]

We’d like to somehow relate all of these without worrying about the limits (as something may go wrong blindly taking limits). Fix \(\epsilon > 0\). We can find \(N\) so that for all \(m \geq N\) the following holds:

\[\|k_m - k\|_\infty < \epsilon.\]

Now we see that for \(v \in \Gamma\) we have

\[\left\| \int_{T^1M} k d\lambda - \frac{1}{T} \int_0^T k(\phi^t(v))dt \right\| \leq \left\| \int_{T^1M} k d\lambda - \frac{1}{T} \int_0^T k_m(\phi^t(v)) dt \right\| + \left\|\frac{1}{T} \int_0^T k_m(\phi^t(v))dt + \frac{1}{T} \int_0^T k(\phi^t(v))dt \right|.\]

Notice that the term on the right satisfies

\[\left\|\frac{1}{T} \int_0^T k_m(\phi^t(v))dt + \frac{1}{T} \int_0^T k(\phi^t(v))dt \right| \leq \frac{1}{T} \int_0^T \|k_m - k\|_\infty dt = \epsilon.\]

We can break up the term on the left again to get

\[\begin{split}\left\| \int_{T^1M} k d\lambda - \frac{1}{T} \int_0^T k_m(\phi^t(v)) dt \right\| \leq \left\| \int_{T^1M} [k - k_m] d\lambda \right\| + \left\| \int_{T^1M} k_m d\lambda - \frac{1}{T} \int_0^T k(\phi^t(v)) dt \right\| \\ \leq \epsilon + \left\| \int_{T^1M} k_m d\lambda - \frac{1}{T} \int_0^T k(\phi^t(v)) dt \right\|.\end{split}\]

We already know that as \(T \rightarrow \infty\) the term on the right tends to zero. Hence

\[\left\| \int_{T^1M} k d\lambda - \lim_{T \rightarrow \infty}\frac{1}{T} \int_0^T k(\phi^t(v))dt \right\| \leq 2 \epsilon.\]

However our choice of \(\epsilon > 0\) was arbitrary, so we let \(\epsilon \rightarrow 0\) and the result follows. Hence every \(v \in \Gamma\) is generic, so \(\lambda\)-almost every \(v \in T^1M\) is generic. \(\blacksquare\)