Ergodicity of the geodesic flow on surfaces of nonpositive curvature

Introduction

This paper provides a full proof of the following fact. Here $C^{3+\mathrm{Lip}}$ means that the third derivatives of the metric are locally Lipschitz.

Write $SM$ for the unit tangent bundle, $\pi: SM \to M$ for its projection, and $g^s$ for the geodesic flow. Normalize area to $\bar a(M)=1$; Liouville probability measure is $dm=d\bar a\,d\theta/(2\pi)$. Following Pesin’s ergodicity theorem on the regular set [7], it suffices to prove that the singular set

has zero Liouville measure. Indeed, $g^s$ restricted to $SM\setminus S$ is ergodic for the restricted Liouville measure.

We have also proved that the geodesic flow of every closed, connected, smooth rank-one manifold of nonpositive sectional curvature and dimension at least two is ergodic with respect to Liouville measure. Here rank one means that some geodesic admits no nonzero parallel normal Jacobi field; see [1]. The proof of this higher-dimensional result will be presented separately.

Throughout, $\lesssim$ denotes an inequality up to a constant depending only on the fixed metric. In estimates involving a Fourier degree $k$ or a heat time $t$, this constant is independent of both unless explicitly stated. All $L^2$ norms are taken with respect to the normalized measures above.

Fourier modes and the connection Laplacian

Assume, towards a contradiction, that $m(S)>0$, and set $f=\mathbf{1}_S$. The set $S$ is closed, invariant under $g^s$, and invariant under $v\mapsto-v$. If $X$ is the geodesic vector field, then

Let $V$ generate rotation in each oriented unit tangent circle. For $k\in\mathbb{Z}$, let $\Omega_k$ be the closed subspace of $L^2(SM)$ on which $V=ik$. In a local oriented orthonormal frame,

These local expressions define global Fourier components. Reversal invariance and Parseval’s identity give

The zeroth component

is nonnegative, and $|f_k(x,v)|\le f_0(x)$. Consequently, with

every $f_k$, viewed as a section over $M$, is supported in $F$. Since $f_0=0$ on $\{K<0\}$, continuity of $K$ gives $K|_F=0$. The Gauss–Bonnet theorem and the genus assumption imply $\{K<0\}\ne\varnothing$. Thus $f_0$ vanishes on a nonempty open set, whereas $\int_M f_0\,d\overline{a}=m(S)>0$; in particular, $f_0$ is not constant.

The metric regularity implies $K\in C^{1+\operatorname{Lip}}$. Every point of $F$ is a maximum of $K$, so $\nabla K|_F=0$. Taylor’s theorem with Lipschitz first derivative therefore yields

These estimates hold globally by compactness.

Put $H=[V,X]$. Thus $H$ moves the base point in the direction $Jv$, parallel transporting $v$, where $J$ denotes rotation by $\pi/2$. With the conventions of Guillemin–Kazhdan [2],

The three vector fields preserve Liouville measure. Define the Guillemin–Kazhdan operators

Then $X=\eta_++\eta_-$, $\eta_\pm:\Omega_k\to\Omega_{k\pm1}$, and $\eta_+^*=-\eta_-$. The degree-$(2j+1)$ component of (1) is

Such Fourier recurrences are used by Guillemin–Kazhdan [3], Paternain–Salo–Uhlmann [6 Sections 2 and 5] and, for invariant functions in nonpositive curvature, by Lefeuvre [5, Section 4.2.

From (5), $[\eta_+,\eta_-]=\frac{i}{2}KV$. Integration by parts gives the Guillemin–Kazhdan energy identity [2]:

In particular,

The $H^1$ statement follows from the identity on differentiable sections by density.

Consider

The bracket identities give

so $L$ preserves every $\Omega_k$. Let $L_k$ be the restriction of $L$ to $\Omega_k$. For an oriented orthonormal frame $(e_1,e_2)$, write $\nabla e_1=\omega\otimes e_2$. Horizontal differentiation of $u=a(x)e^{ik\theta}$ is represented by

Since $v,Jv$ form an orthonormal basis of $T_xM$,

Thus $L_k=\nabla_k^*\nabla_k$ is elliptic. We use its nonnegative self-adjoint realization defined by the closed quadratic form $\|\nabla_k u\|_2^2$, and set

These are contractions on $\Omega_k$, preserve the Fourier degree, and converge strongly to the identity as $t\downarrow0$. Their construction uses the spectral theorem for nonnegative self-adjoint operators.

In an orthonormal eigenbasis $(\phi_\ell)_\ell$ the expression of $P_t^k f_k$ is

where $f_k=\sum_\ell\langle f_k,\phi_\ell\rangle\phi_\ell$ and the eigenvalues $\lambda_\ell$ are nonnegative. For later use, the same spectral theorem and $\sup_{\lambda\ge0}\lambda e^{-2t\lambda}\lesssim t^{-1}$ give

Both bounds are uniform in $k$. In particular $P_t^k f_k$ belongs to $H^1(SM)\cap\Omega_k$ for $t>0$, so (8) applies to it.

Weighted heat estimates

We prove the two estimates that will be used with (4). We will use the wave equation for the same connection Laplacian:

By spectral calculus,

At the zero eigenvalue the first multiplier is interpreted as $r$. Energy conservation, or directly the identity $\cos^2(r\sqrt{\lambda})+\sin^2(r\sqrt{\lambda})=1$, gives

Proof. The distance function $d_F$ is Lipschitz, with $|\nabla d_F|\le1$ almost everywhere. Set

First take smooth initial data, without imposing a support condition. The spectral formula gives $w_{rr},\nabla_kw_r\in L^2$, so $E$ is absolutely continuous. Differentiating and using $w_{rr}=-\nabla_k^*\nabla_kw$, integration by parts and the product rule give, for almost every $r$,

Here the Lipschitz weight times $w_r$ is an admissible $H^1$ test section, and the inequality uses $|\nabla d_F|\le1$. Thus $E(r)\le E(0)$ for every $r\ge 0.

Now choose smooth $f_{k,N} \in\Omega_k$ converging to $f_k$ in $L^2$, and let $E_N$ be the corresponding weighted energy, with the same set $F$. By (14) applied to differences of solutions and Cauchy–Schwarz,

Passing to the limit in $E_N(r) \le E_N(0)$ gives $E(r) \le E(0)$. No support condition on the approximants is needed.

Since the original datum is supported in $F$,

Hence $E(r)=0$, so $z(r)=w_r(r)$ and $\nabla_k w(r)$ vanish almost everywhere on $\{d_F>r\}$. Finally, $w(r)=\int_0^r z(s)\,ds$ in $L^2$, and $F_s\subset F_r$ for $0\le s\le r$, so $w(r)$ also vanishes outside $F_r$. □

Proof. The heat–wave formula of Kannai [4] is

which follows directly from the scalar Gaussian cosine transform

and the spectral theorem. On the support of $z(r)$, $d_F\le r$; thus (14) and (17) imply

For the derivative estimate, integrate (17) by parts in wave time:

The boundary terms vanish, since $w(0)=0$ and $\|w(r)\|_2\le r\|f_k\|_2$. By (14) and (15),

The integral in (18) converges in the form domain; taking its weak covariant derivative therefore gives

All constants came from scalar Gaussian integrals and the wave energy, so they are independent of $k$. □

The heat commutator

For an operator mapping degree $k$ to degree $k\pm1$, use the notation

Proof. The Guillemin–Kazhdan relations (2.5)–(2.10) give, on degree $k$,

Indeed, for $u\in\Omega_k$, we have $Vu=iku$. Since $\eta_{\pm}u\in\Omega_{k\pm1}$,

Using $L=-2(\eta_{+}\eta_{-}+\eta_{-}\eta_{+})$ and the Guillemin–Kazhdan relation $[\eta_{+},\eta_{-}]=\frac{i}{2}KV$, the product rule for commutators gives

Likewise,

In each calculation, the last equality uses the product rule for $\eta_{\pm}(Ku)$. This computation makes explicit that the commutator is first order and involves only $K$ and its first derivatives. In particular,

Although its defining expression has higher derivatives, the right-hand side of (20) or (21) defines the commutator on the form domain.

Apply Proposition 3 and the Taylor estimates (4). They give separately

Consequently,

Duhamel’s formula gives

The integral belongs to $L^2$ by (23) and the $L^2$ contraction of the heat operators.

Finally, the $L^2$ contraction of $P_{t-\tau}^{k\pm1}$ and (23) imply

Completion of the proof

Put $u_k(t)=P_t^k f_k$. First, there are $c>0$ and $t_0>0$ such that

Otherwise some $t_n\downarrow0$ would satisfy $\eta_{+}P_{t_n}^{0}f_0\to0$ in $L^2$. Since $P_{t_n}^{0}f_0\to f_0$ in $L^2$, this would imply $\eta_{+}f_0=0$ distributionally. The function $f_0$ is real, so (6) gives $Xf_0=Hf_0=0$. These two horizontal derivatives determine the base differential of a degree-zero function; hence $df_0=0$. On the connected surface $M$, a function with vanishing distributional differential is constant almost everywhere, contrary to the construction of $f_0$. This proves (25).

Heat smoothing need not preserve the flow invariance $Xf=0$. The resulting error is $XP_tf=[X,P_t]f$, where $P_tf=\sum_k u_k(t)$. We denote its degree-$(2\ell+1)$ Fourier component by

Thus $r_\ell(t)$ measures the failure of the recurrence (7) after smoothing. That recurrence gives

so Proposition 4 bounds precisely this error.

Since every Fourier component $f_k$ is supported in $F$, Proposition 4 gives

The Guillemin–Kazhdan inequality (9) and the triangle inequality imply

Fix $t>0$. Applying this inequality successively for $\ell=0,\ldots,j-1$, we obtain

By the Cauchy–Schwarz inequality and Parseval’s identity (2), (26) gives

Choose one $a>0$, with $a\le t_0$, sufficiently small that the last bound at $t=a/j$ is at most $c/2$, for every $j\ge1$. Combining (25), (27), (28), and (12) yields

Consequently $\|f_{2j}\|_2^2\gtrsim j^{-1}$, with a positive constant independent of $j$. A final application of Parseval’s identity gives the contradiction

Hence $m(S)=0$. Pesin’s theorem on the regular set [7] proves Theorem 1.

AI use acknowledgement

We used ChatGPT (OpenAI) to assist with mathematical arguments, exposition, and reference searches. The authors take full responsibility for the content of this paper.