Application of Birkhoff Ergodic Theorem; Asymptotic direction of shortest geodesics at a fixed point in Riemannian manifold

Question

Assume that $M=\mathbb{R}^n$ is Riemannian manifold with $G=\mathbb{Z}^n$-isometric action s.t. $G\cdot o$ is the set of integer points where $o$ is origin. Here $\exp_x\ tv$ is a unit speed geodesic starting at $x$ Assume that there is no conjugate point on $M$(I do not know that this condition is indispensable. At any rate we assume this). Hence $M$ has no cut point.

On unit tangent bundle $UT\mathbb{R}^n$, define $$ D(x,v) = \lim_{t\rightarrow \infty} \ \frac{\exp_x\ tv - o}{t}$$

problem : Prove that for a.e. $(x,v)\in UT\mathbb{R}^n$, $D$ is defined.

Proof : In the paper, it is stated that this can be proved by Birkhoff Ergodic theorem :

When $T : X\rightarrow X$ is a map on probability space $X$ i.e. $\mu(X)=1$ and it is ergodic i.e.

1. measure preserving and

2. $T^{-1}(E)=E$ implies $\mu (E)=0$ or $1$,

then for any $ f : \rightarrow \mathbb{R}$, $$ \frac{1}{n} \sum_{j=1}^n \ f\circ T^j (x)\rightarrow \int_X\ fd\mu $$

From John B's answer, we know that $$T (x,v)=(\exp_x\ v,\frac{d}{dt}\bigg|_{t=1}\exp_x\ tv)$$ from $UT\mathbb{R}^n$ to itself, and $f :UT\mathbb{R}^n\rightarrow \mathbb{R}^n,\ f(x,v)=\exp_x\ v-x$ is a function.

Since $T$ is a measure-preserving (cf Geodesic flow), so $UT\mathbb{R}^n$ has a canonical volume form but it has infinite volume.

Hence how can we apply this theorem to our problem ?

Answer 1

Consider the quotient of $M$ by $G$ -- the torus $\mathbb{T}^n$ with induced metric, and its unit tangent bundle. Then $T([x, v])=([\exp_x(v), \frac{d}{dt}\bigg|_{t=1}\exp_x\ tv])$ is independent of the choices of $x, v$ in the equivalence class, and so is well-defined; moreover it coincides with the geodesic flow on $UT\mathbb{T}^n$ and hence is a measure preserving transformation of $UT\mathbb{T}^n$. We consider the (vector valued) function $f:UT\mathbb{T}^n \to \mathbb{R}^n$ given by $f([x, v])=\exp_x v -x$. This is independent of the choice of $x, v$ in the equivalence class, hence well-defined. By (first, non-ergodic part) of Birkhoff theorem the limit

$\lim_{m\to \infty}\frac{1}{m} \sum_{i=0}^{m-1} f(T^i([x, v])=\lim_{m\to \infty} \frac{1}{m}(\exp_x{mv}-x)$ exists for almost all $[x, v]\in UT\mathbb{T}^n$. Hence the limit $\lim_{m\to \infty} \frac{1}{m}(\exp_x{mv}-x)$ exists for almost all $(x,v)$ in $TUM$. Since $|\exp_y u|$ is bounded on $UTM$, it follows that $\lim_{t\to \infty} \frac{1}{t}(\exp_x{tv}-x)$ for continuous $t$ also exists for the same $x, v$ (alternatively, one could use a continuous version of Birkhoff's theorem). Finally, of course, since $x-o$ is constant and $t$ grows, $\lim_{t\to \infty} \frac{1}{t}(\exp_x{tv}-o)$ exists for the same $(x, v)$.

Answer 2

The measure on the unit tangent bundle need not be ergodic, but you don't need that for Birkhoff's ergodic theorem (actually only the measure on the base is important in this example). Just take $$f(o,v)=\exp_ov-o\quad\text{and}\quad T(o,v)=(\exp_ov,v). $$ Then $$ \frac{1}{m} \sum_{j=0}^{m-1}f(T^j (x))=\frac{\exp_o(mv) - o}{m}, $$ which implies the result in your problem for almost all $(o,v)$.