Invariant distributions, Beurling transforms and tensor tomography in higher dimensions

Abstract

In the recent articles Paternain et al. (J. Differ Geom, 98:147–181, 2014, Invent Math 193:229–247, 2013), a number of tensor tomography results were proved on two-dimensional manifolds. The purpose of this paper is to extend some of these methods to manifolds of any dimension. A central concept is the surjectivity of the adjoint of the geodesic ray transform, or equivalently the existence of certain distributions that are invariant under the geodesic flow. We prove that on any Anosov manifold, one can find invariant distributions with controlled first Fourier coefficients. The proof is based on subelliptic type estimates and a Pestov identity. We present an alternative construction valid on manifolds with nonpositive curvature, based on the fact that a natural Beurling transform on such manifolds turns out to be essentially a contraction. Finally, we obtain uniqueness results in tensor tomography both on simple and Anosov manifolds that improve earlier results by assuming a condition on the terminator value for a modified Jacobi equation.

Appendices

Appendix A: Proofs of the commutator identities

In this appendix we give the proofs of the commutator identities in Sect. 2. This is done via local coordinate computations, and we also give coordinate expressions for the relevant operators which have been defined invariantly in Sect. 2. The arguments are not new and they arise in the calculus of semibasic tensor fields as in [47]. The main points here are that the basic setting is the unit sphere bundle $SM$ instead of $TM$, and that all computations can be done on the level of vector fields instead of (higher order) semibasic tensor fields.

Vector fields on $SM$. If $x$ is a system of local coordinates in $M$, let $(x,y)$ be associated coordinates in $TM$ where tangent vectors are written as $y^j \partial _{x_j}$. One has corresponding coordinates $(x,y,X,Y)$ in $T(TM)$ where vectors of $T(TM)$ are written as $X^j \partial _{x_j} + Y^j \partial _{y_j}$. It is convenient to introduce the vector fields

$$\begin{aligned} \delta _{x_j} = \partial _{x_j} - \Gamma _{jk}^l y^k \partial _{y_l} \end{aligned}$$

where $\Gamma _{jk}^l$ are the Christoffel symbols of $(M,g)$. The Sasaki metric on $TM$ is expressed in local coordinates as

$$\begin{aligned} \langle X^j \delta _{x_j} + Y^k \partial _{y_k}, \tilde{X}^j \delta _{x_j} + \tilde{Y}^k \partial _{y_k} \rangle = g_{jk} X^j \tilde{X}^k + g_{jk} Y^j \tilde{Y}^k. \end{aligned}$$

The horizontal and vertical subbundles are spanned by $\{ \delta _{x_j} \}_{j=1}^n$ and $\{ \partial _{y_k} \}_{k=1}^n$, respectively. It will be very convenient to identify horizontal and vertical vector fields on $TM$ with vector fields on $M$ via the maps $X^j \delta _{x_j} \mapsto X^j \partial _{x_j}$ and $Y^k \partial _{y_k} \mapsto Y^k \partial _{x_k}$ (see for example [36] for more details), and we will use this identification freely below. We will also raise and lower indices with respect to the metric $g_{jk}$.

The hypersurface $SM$ in $TM$ is given by $SM = f^{-1}(1)$ where $f: TM \rightarrow {\mathbb R}$ is the function $f(x,y) = g_{jk}(x) y^j y^k$. A computation gives

$$\begin{aligned} df(X^j \delta _{x_j} + Y^k \partial _{y_k}) = 2 y_k Y^k. \end{aligned}$$

Then $T(SM)$ is the subset of $T(TM)$ given by

$$\begin{aligned} T(SM) = \{ X^j \delta _{x_j} + Y^k \partial _{y_k} \in T(TM) \,;\, (x,y) \in SM, \ y_k Y^k = 0 \}. \end{aligned}$$

To be precise, we identify vector fields $V$ on $SM$ with the corresponding fields $i_* V$ on $TM$, where $i: SM \rightarrow TM$ is the natural inclusion. Equip $SM$ with the restriction of the Sasaki metric from $TM$. The identity $d_{SM} i^* U = i^* d_{TM} U$ for functions $U$ on $TM$ implies that the gradient on $SM$ is given by

$$\begin{aligned} \langle \nabla _{SM} u, V \rangle = (\delta _{x_j} \tilde{u}) X^j + (\partial _{y_k} \tilde{u}) Y^k, \quad u \in C^{\infty }(SM), \end{aligned}$$

where $\tilde{u} \in C^{\infty }(TM)$ is any function with $\tilde{u}|_{SM} = u$ and where the vector field $V$ on $SM$ is expressed as above in the form $V = X^j \delta _{x_j} + Y^k \partial _{y_k}$.

We define vector fields on $SM$ that act on $u \in C^{\infty }(SM)$ by

$$\begin{aligned} \delta _j u&= \delta _{x_j} (u \circ p)|_{SM}, \\ \partial _k u&= \partial _{y_k} (u \circ p)|_{SM} \end{aligned}$$

where $p: TM {\setminus } \{0\} \rightarrow SM$ is the projection $p(x,y) = (x,y/|y |_{g(x)})$. We see that the decomposition $\nabla _{SM} u = (Xu)X + \overset{\mathtt{h}}{\nabla }u + \overset{\mathtt{v}}{\nabla }u$ has the following form in local coordinates:

$$\begin{aligned} Xu&= v^j \delta _j u, \\ \overset{\mathtt{h}}{\nabla }u&= (\delta ^j u - (v^k \delta _k u) v^j) \partial _{x_j}, \\ \overset{\mathtt{v}}{\nabla }u&= (\partial ^k u) \partial _{x_k}. \end{aligned}$$

Commutator formulas. Direct computations in local coordinates give the following formulas for vector fields on $TM$:

$$\begin{aligned} \,[\delta _{x_j}, \delta _{x_k}] = -R_{jkl}^{m} y^l \partial _{y_m}, \quad [\delta _{x_j}, \partial _{y_k}] = \Gamma _{jk}^l \partial _{y_l}, \quad [\partial _{y_j}, \partial _{y_k}] = 0. \end{aligned}$$

We wish to consider corresponding formulas for the vector fields $\delta _j$ and $\partial _j$ on $SM$. If $u \in C^{\infty }(SM)$, write $\tilde{u}(x,y) = (u \circ p)(x,y) = u(x,y/|y |_g)$. Homogeneity implies that

$$\begin{aligned} (\delta _j u)\,\tilde{\,}(x,y) = \delta _{x_j} \tilde{u}(x,y), \quad (\partial _k u)\,\tilde{\,}(x,y) = |y |_g \partial _{y_k} \tilde{u}(x,y). \end{aligned}$$

Since also $\delta _{x_j}(|y |_g) = 0$ and $\partial _{y_j}(|y |_g) = y_j/|y |_g$, we obtain

$$\begin{aligned} \,[\delta _j, \delta _k] = - R_{jklm} v^l \partial ^m, \quad [\delta _j, \partial _k] = \Gamma _{jk}^l \partial _l , \quad [\partial _j, \partial _k] = v_j \partial _k - v_k \partial _j, \\ \,[\partial _j, v^k] = \delta _j^{k} - v_j v^k, \quad [\delta _j, v^l] = -\Gamma _{jk}^l v^k. \end{aligned}$$

We also note that

$$\begin{aligned} v^j \partial _j = 0. \end{aligned}$$

Using the identity $\partial _{x_j} g^{ab} + g^{am} \Gamma _{jm}^b + g^{bm} \Gamma _{jm}^a = 0$ we also obtain

$$\begin{aligned} \,[\delta _j, \delta ^l] = -g^{lk} R_{jkpq} v^p \partial ^q + (\partial _{x_j} g^{lk}) \delta _k, \quad [\delta _j, \partial ^l] = -\Gamma _{jk}^l \partial ^k. \end{aligned}$$

We can now compute $[X, \overset{\mathtt{v}}{\nabla }]$. Note that

$$\begin{aligned} \overset{\mathtt{v}}{\nabla }Xu = \partial ^l(v^j \delta _j u) \partial _{x_l} = \overset{\mathtt{h}}{\nabla }u + v^j \partial ^l \delta _j u \partial _{x_l} \end{aligned}$$

and (by the formula (11.2) below)

$$\begin{aligned} X \overset{\mathtt{v}}{\nabla }u = (v^j \delta _j \partial ^l u + \Gamma _{jk}^l v^j \partial ^k u) \partial _{x_l}. \end{aligned}$$

Thus

$$\begin{aligned} \,[X, \overset{\mathtt{v}}{\nabla }]u = - \overset{\mathtt{h}}{\nabla }u + v^j ( [\delta _j, \partial ^l] u + \Gamma _{jk}^l \partial ^k u) = -\overset{\mathtt{h}}{\nabla }u. \end{aligned}$$

Moving on to $[X, \overset{\mathtt{h}}{\nabla }]$, we observe that

$$\begin{aligned} \overset{\mathtt{h}}{\nabla }X u = (\delta ^l(Xu) - (X^2 u) v^l) \partial _{x_l} \end{aligned}$$

and (again by (11.2))

$$\begin{aligned} X \overset{\mathtt{h}}{\nabla }u = (X (\delta ^l u - (Xu) v^l) + \Gamma _{jk}^l v^j (\delta ^k u - (Xu) v^k)) \partial _{x_l}. \end{aligned}$$

In the second term we have $-(Xu)(Xv^l) - \Gamma _{jk}^l v^j (Xu) v^k = 0$, and when taking the commutator the terms containing $X^2 u$ cancel. It follows that

$$\begin{aligned} \,[X, \overset{\mathtt{h}}{\nabla }] u&= (v^j \delta _j \delta ^l u - \delta ^l(v^j \delta _j u) + \Gamma _{jk}^l v^j \delta ^k u) \partial _{x_l} \\&= (v^j [\delta _j, \delta ^l] u - g^{lr} [\delta _r, v^j] \delta _j u + \Gamma _{jk}^l v^j \delta ^k u) \partial _{x_l} \\&= (- g^{lk} R_{jkpq} v^j v^p \partial ^q u + [ (\partial _j g^{lk}) v^j \delta _k u + g^{lr} \Gamma _{rm}^j v^m \delta _j u\\&\quad + g^{km} \Gamma _{jk}^l v^j \delta _m u ] ) \partial _{x_l}. \end{aligned}$$

The part in brackets is zero, which yields $[X, \overset{\mathtt{h}}{\nabla }] u = R_{abc}^{l} (\partial ^a u) v^b v^c \partial _{x_l} = R \overset{\mathtt{v}}{\nabla }u$.

Adjoints. To prove the last basic commutator formula, it is useful to have local coordinate expressions for the adjoints of $X, \overset{\mathtt{h}}{\nabla }, \overset{\mathtt{v}}{\nabla }$ on the space $\mathcal Z$. The first step is to compute the adjoints of the local vector fields $\delta _j$ and $\partial _j$: if $u, w \in C^{\infty }(SM)$ and $w$ vanishes when $x$ is outside a coordinate patch (and additionally $w$ vanishes on $\partial (SM)$ if $M$ has a boundary), we claim that in the $L^2(SM)$ inner product

$$\begin{aligned} (\delta _j u, w) = -(u, (\delta _j + \Gamma _j) w), \quad (\partial _j u, w) = -(u, (\partial _j -(n-1)v_j) w). \end{aligned}$$

(11.1)

Here $\Gamma _j = \Gamma _{jk}^k$.

Assuming these identities, one can check that the adjoint of $X$ on $C^{\infty }(SM)$ is $-X$. Moreover, if $Z \in \mathcal Z$ is written as $Z(x,v) = Z^j(x,v) \partial _{x_j}$, the vector field $XZ$ is the covariant derivative (with respect to the Levi-Civita connection in $(M,g)$)

$$\begin{aligned} XZ(x,v) = D_t(Z(\phi _t(x,v)))|_{t=0} = (XZ^j) \partial _{x_j} + \Gamma _{jk}^l v^j Z^k \partial _{x_l}. \end{aligned}$$

(11.2)

Then the adjoint of $X$ on $\mathcal Z$ is also $-X$. The adjoints of $\overset{\mathtt{h}}{\nabla }$ and $\overset{\mathtt{v}}{\nabla }$ are given in local coordinates by $-\overset{\mathtt{h}}{\text{ div }}$ and $-\overset{\mathtt{v}}{\text{ div }}$, where

$$\begin{aligned} \overset{\mathtt{h}}{\text{ div }}Z = (\delta _j + \Gamma _j) Z^j, \quad \overset{\mathtt{v}}{\text{ div }}Z = \partial _j Z^j. \end{aligned}$$

Given these expressions, we get the final commutator formula:

$$\begin{aligned} \overset{\mathtt{h}}{\text{ div }}\overset{\mathtt{v}}{\nabla }u - \overset{\mathtt{v}}{\text{ div }}\overset{\mathtt{h}}{\nabla }u&= (\delta _k + \Gamma _k)(\partial ^k u) - \partial _k(\delta ^k u - (Xu) v^k) \\&= [\delta _k, \partial ^k] u + \Gamma _k \partial ^k u + (Xu)(\partial _k v^k) \\&= (n-1) Xu. \end{aligned}$$

It remains to check (11.1). To do this it is enough to prove that

$$\begin{aligned} \partial _{x_j} \left( \int _{S_x M} u \,dS_x \right) = \int _{S_x M} \delta _j u \,dS_x, \quad \int _{S_x M} \partial _j u \,dS_x = (n-1) \int _{S_x M} u v_j \,dS_x. \end{aligned}$$

(The proof of (11.1) also uses the identity $|g |^{-1/2} \partial _{x_j} (|g |^{1/2}) = \Gamma _j$.) To show the first formula, let $u \in C^{\infty }(SM)$ vanish when $x$ is outside a coordinate patch, and define

$$\begin{aligned} f(x) = \int _{S_x M} u(x,v) \,dS_x(v). \end{aligned}$$

Write $\tilde{u}(x,y) = u(x,y/|y |_g)$, and choose $\varphi \in C^{\infty }_c((0,\infty ))$ so that $\int _0^{\infty } \varphi (r) r^{n-1} \,dr = 1$. Write $g(x)$ for the matrix of $g$ in the $x$ coordinates. We write $\omega$ for points in $S^{n-1}$, and note that $\omega \mapsto g(x)^{-1/2} \omega$ is an isometry from $S^{n-1}$ (with the metric induced by the Euclidean metric $e$ in ${\mathbb R}^n$) onto $S_x M$ (with the metric induced by Sasaki metric on $T_x M$, having volume form $dT_x = |g(x) |^{1/2} \,dx$). Therefore

$$\begin{aligned} f(x)&= \int _0^{\infty } \varphi (r) r^{n-1} \int _{S_x M} u(x,v) \,dS_x(v) \,dr \\&= \int _0^{\infty } \int _{S^{n-1}} \varphi (r) r^{n-1} \tilde{u}(x,g(x)^{-1/2} r\omega ) \,d\omega \,dr \\&= \int _{{\mathbb R}^n} \varphi (|\eta |_e) \tilde{u}(x,g(x)^{-1/2} \eta ) \,d\eta \\&= \int _{{\mathbb R}^n} \varphi (|y |_g) \tilde{u}(x,y) |g(x) |^{1/2} \,dy. \end{aligned}$$

Since $|g |^{-1/2} \partial _{x_j} (|g |^{1/2}) = \Gamma _j = \partial _{y_l}(\Gamma _{jk}^l y^k)$, we have

$$\begin{aligned} \partial _{x_j} f(x)&= \int _{{\mathbb R}^n} (\partial _{x_j} + \partial _{y_l}(\Gamma _{jk}^l y^k)) \left[ \varphi (|y |_g) \tilde{u}(x,y) \right] |g(x) |^{1/2} \,dy \\&= \int _{{\mathbb R}^n} \delta _{x_j} \left[ \varphi (|y |_g) \tilde{u}(x,y) \right] |g(x) |^{1/2} \,dy. \end{aligned}$$

Now $\delta _{x_j}(|y |_g) = 0$, and it follows by undoing the changes of variables above that

$$\begin{aligned} \partial _{x_j} f(x) = \int _{S_x} \delta _j u(x,v) \,dS_x(v) \end{aligned}$$

as required. The second formula follows from a similar computation as above: now we define

$$\begin{aligned} f(x) = \int _{S_x M} \partial _j u(x,v) \,dS_x(v) \end{aligned}$$

and compute

$$\begin{aligned} f(x)&= \int _0^{\infty } \varphi (r) r^{n-1} \int _{S_x M} \partial _j u(x,v) \,dS_x(v) \,dr \\&= \int _0^{\infty } \int _{S^{n-1}} r \varphi (r) r^{n-1} \partial _{y_j} \tilde{u}(x,g(x)^{-1/2} r\omega ) \,d\omega \,dr \\&= \int _{{\mathbb R}^n} |y |_g \varphi (|y |_g) \partial _{y_j} \tilde{u}(x,y) |g(x) |^{1/2} \,dy. \end{aligned}$$

Write $h(y) = |y |_g \varphi (|y |_g)$. Then

$$\begin{aligned} f(x)&= -\int _{{\mathbb R}^n} \partial _{y_j} h(y) \tilde{u}(x,y) |g(x) |^{1/2} \,dy \\&= -\int _{S_x M} \left[ \int _0^{\infty } \partial _{y_j} h(rv) r^{n-1} \,dr \right] u(x,v) \,dS_x(v). \end{aligned}$$

The expression in brackets is $v_j \int _0^{\infty } (\varphi (r) + r \varphi '(r))r^{n-1} \,dr = -(n-1) v_j$, and the result follows.

Appendix B: The two-dimensional case

In this section we reconsider the arguments in this paper in the special case of two-dimensional manifolds. This discussion allows to connect the present treatment with earlier work in two dimensions, in particular [38, 41].

Vector fields. Let $(M,g)$ be a compact oriented Riemann surface with no boundary (the boundary case is analogous, if we additionally assume that test functions vanish on $\partial (SM)$ if appropriate). We have $n = \dim (M) = 2$. For any vector $v \in S_x M$, there is a unique vector $iv \in S_x M$ such that $\{ v, iv \}$ is a positive orthonormal basis of $T_x M$. Let $X$ be the geodesic vector field on $SM$ as before. We can define vector fields $V$ and $X_{\perp }$ on $SM$, acting on $u \in C^{\infty }(SM)$ by

$$\begin{aligned} X_{\perp } u(x,v)&:= -\langle \overset{\mathtt{h}}{\nabla }u(x,v), iv \rangle , \\ Vu(x,v)&:= \langle \overset{\mathtt{v}}{\nabla }u(x,v), iv \rangle . \end{aligned}$$

Note that since $\langle \overset{\mathtt{v}}{\nabla }u, v \rangle = \langle \overset{\mathtt{h}}{\nabla }u, v \rangle = 0$, we have

$$\begin{aligned} \overset{\mathtt{h}}{\nabla }u&= -(X_{\perp } u) iv, \\ \overset{\mathtt{v}}{\nabla }u&= (Vu) iv. \end{aligned}$$

Note also that any $Z \in \mathcal Z$ is of the form $Z(x,v) = z(x,v) iv$ for some $z \in C^{\infty }(SM)$. If $\gamma (t)$ is a unit speed geodesic we have

$$\begin{aligned} \langle D_t [i \dot{\gamma }(t)], \dot{\gamma }(t) \rangle&= \partial _t \langle i \dot{\gamma }(t), \dot{\gamma }(t) \rangle = 0, \\ \langle D_t [i \dot{\gamma }(t)], i \dot{\gamma }(t) \rangle&= \frac{1}{2} \partial _t \langle i \dot{\gamma }(t), i \dot{\gamma }(t) \rangle = 0. \end{aligned}$$

Thus $iv$ is parallel along geodesics, and for $Z = z(x,v) iv$ we have $XZ = (Xz) iv$.

The Guillemin-Kazhdan operators [21] are defined as the vector fields

$$\begin{aligned} \eta _{\pm } := \frac{1}{2}(X \pm i X_{\perp }). \end{aligned}$$

All these vector fields have simple expressions in isothermal coordinates. Since $(M,g)$ is two-dimensional, near any point there are positively oriented isothermal coordinates $(x_{1},x_{2})$ so that the metric can be written as $ds^2=e^{2\lambda }(dx_{1}^2+dx_{2}^2)$ where $\lambda$ is a smooth real-valued function of $x=(x_{1},x_{2})$. This gives coordinates $(x_{1},x_{2},\theta )$ on $SM$ where $\theta$ is the angle between a unit vector $v$ and $\partial /\partial x_{1}$. In these coordinates we have

$$\begin{aligned} V = \frac{\partial }{\partial \theta }, \quad X = e^{-\lambda } \left[ \cos \theta \frac{\partial }{\partial x_{1}}+ \sin \theta \frac{\partial }{\partial x_{2}}+ \left( -\frac{\partial \lambda }{\partial x_{1}}\sin \theta +\frac{\partial \lambda }{\partial x_{2}}\cos \theta \right) \frac{\partial }{\partial \theta } \right] , \\ X_{\perp } =-e^{-\lambda }\left[ -\sin \theta \frac{\partial }{\partial x_{1}}+ \cos \theta \frac{\partial }{\partial x_{2}}- \left( \frac{\partial \lambda }{\partial x_{1}}\cos \theta +\frac{\partial \lambda }{\partial x_{2}}\sin \theta \right) \frac{\partial }{\partial \theta }\right] , \\ \eta _+ = e^{-\lambda } e^{i\theta } \left[ \frac{\partial }{\partial z} + i \frac{\partial \lambda }{\partial z} \frac{\partial }{\partial \theta } \right] , \quad \eta _- = e^{-\lambda } e^{-i\theta } \left[ \frac{\partial }{\partial \bar{z}} - i \frac{\partial \lambda }{\partial \bar{z}} \frac{\partial }{\partial \theta } \right] \end{aligned}$$

where $\partial /\partial z = \frac{1}{2}(\partial _{x_1} - i \partial _{x_2})$ and $\partial /\partial \bar{z} = \frac{1}{2}(\partial _{x_1} + i \partial _{x_2})$.

Commutator formulas. The above discussion shows that the commutator formula $[X, \overset{\mathtt{v}}{\nabla }] = -\overset{\mathtt{h}}{\nabla }$ reduces to

$$\begin{aligned} \,[X,V] = X_{\perp }. \end{aligned}$$

Since $(R \overset{\mathtt{v}}{\nabla }u)(x,v) = K(x)(Vu) iv$ where $K$ is the Gaussian curvature, the commutator formula $[X, \overset{\mathtt{h}}{\nabla }] = R\overset{\mathtt{v}}{\nabla }$ becomes

$$\begin{aligned} \,[X, X_{\perp }] = -KV. \end{aligned}$$

For the last commutator formula we need to compute $\overset{\mathtt{h}}{\text{ div }}$ and $\overset{\mathtt{v}}{\text{ div }}$. First observe that a local coordinate computation gives for $w \in C^{\infty }(SM)$ that

$$\begin{aligned} \int _{SM} Vw = \int _M \int _{S_x M} (iv)^j \partial _j w = 0. \end{aligned}$$

Thus the adjoint of $V$ is $-V$, and the first commutator formula implies that the adjoint of $X_{\perp }$ is $-X_{\perp }$. Consequently

$$\begin{aligned} \overset{\mathtt{h}}{\text{ div }}(z(x,v) iv)&= -X_{\perp } z, \\ \overset{\mathtt{v}}{\text{ div }}(z(x,v) iv)&= Vz. \end{aligned}$$

The commutator formula $\overset{\mathtt{h}}{\text{ div }}\,\overset{\mathtt{v}}{\nabla }-\overset{\mathtt{v}}{\text{ div }}\,\overset{\mathtt{h}}{\nabla }=(n-1)X$ thus reduces to

$$\begin{aligned} \,[V,X_{\perp }] = X. \end{aligned}$$

Spherical harmonics expansions. It is easy to express the operators $X_{\pm }$ in terms of $\eta _{\pm }$. The vertical Laplacian on $SM$ is given by

$$\begin{aligned} -\overset{\mathtt{v}}{\text{ div }}\overset{\mathtt{v}}{\nabla }u = -\overset{\mathtt{v}}{\text{ div }}((Vu) iv) = -V^2 u. \end{aligned}$$

The operator $-iV$ on $L^2(SM)$ has eigenvalues $k \in {\mathbb Z}$ with corresponding eigenspaces $E_k$. We write

$$\begin{aligned} L^2(SM) = \bigoplus _{k=-\infty }^{\infty } E_k, \quad u = \sum _{k=-\infty }^{\infty } u_k. \end{aligned}$$

Locally in the $(x,\theta )$ coordinates, elements of $E_k$ are of the form $\tilde{w}(x) e^{ik\theta }$. Writing $\Lambda _k = C^{\infty }(SM) \cap E_k$, the spherical harmonics of degree $m$ are given by

$$\begin{aligned} \Omega _m = \Lambda _m \oplus \Lambda _{-m}, \quad m \ge 0. \end{aligned}$$

If $m \ge 1$, the action of $X_{\pm }$ on $\Omega _m$ is given by

$$\begin{aligned} X_{\pm }(e_m+e_{-m}) = \eta _{\pm } e_m + \eta _{\mp } e_{-m}, \quad e_j \in \Lambda _j, \end{aligned}$$

and for $m=0$ we have $X_+|_{\Omega _0} = \eta _+ + \eta _-$, $X_-|_{\Omega _0} =0$. In the two-dimensional case it will be convenient to work with the $\Lambda _k$ spaces (the corresponding results in terms of the $\Omega _m$ spaces will follow easily).

Beurling transform and invariant distributions. Recall that $w \in \mathcal {D}'(SM)$ is called invariant if $Xw = 0$. If the geodesic flow is ergodic, these are genuinely distributions since any $w \in L^1(SM)$ that satisfies $Xw = 0$ must be constant. For Riemann surfaces, one can look at distributions with one-sided Fourier series; let us consider the case where $w_k = 0$ for $k < k_0$, for some integer $k_0 \ge 0$. For such a distribution, the equation $Xw = 0$ reduces to countably many equations for the Fourier coefficients (by parity it is enough to look at $w_{k_0+2j}$ for $j \ge 0$):

$$\begin{aligned} \eta _- w_{k_0}&= 0, \\ \eta _- w_{k_0+2}&= -\eta _+ w_{k_0}, \\ \eta _- w_{k_0+4}&= -\eta _+ w_{k_0+2}, \\&\ \,\vdots \end{aligned}$$

On a Riemann surface with genus $\ge 2$, the operator $\eta _+: \Lambda _{k-1} \rightarrow \Lambda _k$ is injective and its adjoint $\eta _-: \Lambda _k \rightarrow \Lambda _{k-1}$ is surjective for $k \ge 2$ by conformal invariance (there is a constant negative curvature metric in the conformal class, and these have no conformal Killing tensors). Also for $k \ge 2$ we have the $L^2$-orthogonal splitting

$$\begin{aligned} \Lambda _k = \mathrm {Ker}(\eta _-|_{\Lambda _k}) \oplus \eta _+ \Lambda _{k-1}. \end{aligned}$$

If $k \ge 0$, we define the Beurling transform

$$\begin{aligned} B_+: \Lambda _k \rightarrow \Lambda _{k+2}, \ \ f_k \mapsto f_{k+2} \end{aligned}$$

where $f_{k+2}$ is the unique function in $\Lambda _{k+2}$ orthogonal to $\mathrm {Ker}(\eta _-|_{\Lambda _{k+2}})$ (equivalently, the $L^2$-minimal solution) satisfying $\eta _- f_{k+2} = -\eta _+ f_k$. Note that in ${\mathbb R}^2$, one thinks of $\eta _-$ as $\overline{\partial }$ and of $\eta _+$ as $\partial$, so $B_+$ is formally the operator $-\overline{\partial }^{-1} \partial$ which is the usual Beurling transform up to minus sign. Note also that $B_+$ is the first ladder operator from [22].

If $k \ge 0$ one has the analogous operator

$$\begin{aligned} B_-: \Lambda _{-k} \rightarrow \Lambda _{-k-2}, \ \ f_{-k} \mapsto f_{-k-2} \end{aligned}$$

where $f_{-k-2}$ is the $L^2$-minimal solution of $\eta _+ f_{-k-2} = -\eta _- f_{-k}$. The relation to the Beurling transform in Sect. 5 is

$$\begin{aligned} B(f_k + f_{-k}) = B_+ f_k + B_- f_{-k}, \quad f_j \in \Lambda _j. \end{aligned}$$

On a closed surface of genus $\ge 2$, one can always formally solve the countably many equations for $w_k$. If we take the minimal energy solution for each equation, we arrive at the formal invariant distributions. We restrict our attention to $B_+$ (the case of $B_-$ is analogous).

Definition 11.10

Let $(M,g)$ be a closed oriented surface with genus $\ge 2$, let $k_0 \ge 0$, and let $f \in \Lambda _{k_0}$ satisfy $\eta _- f = 0$. The formal invariant distribution starting at $f$ is the formal sum

$$\begin{aligned} w = \sum _{j=0}^{\infty } (B_+)^j f. \end{aligned}$$

As before, it is not clear if the sum converges in any reasonable sense. However, if the surface has nonpositive curvature it does converge nicely. This follows from the fact that the Beurling transform $B_+$ is a contraction on such surfaces, and to prove this we use the Guillemin-Kazhdan energy identity [21]:

Lemma 11.11

Let $(M,g)$ be a closed Riemann surface. Then

$$\begin{aligned} ||\eta _- u ||^2 = ||\eta _+ u ||^2 - \frac{i}{2} (K V u, u), \quad u \in C^{\infty }(SM). \end{aligned}$$

Proof

In [21] one has the commutator formula

$$\begin{aligned} \,[\eta _+, \eta _-] = \frac{i}{2} K V. \end{aligned}$$

This implies that, for $u \in C^{\infty }(SM)$,

$$\begin{aligned} ||\eta _- u ||^2 = ||\eta _+ u ||^2 + ([\eta _-,\eta _+]u,u) = ||\eta _+ u ||^2 - \frac{i}{2} (K V u, u). \end{aligned}$$

$\square$

Lemma 11.12

Let $(M,g)$ be a closed surface, and assume that $K \le 0$. Then for any $k \ge 0$ we have

$$\begin{aligned} ||\eta _- u ||_{L^2} \le ||\eta _+ u ||_{L^2}, \quad u \in \Lambda _k, \end{aligned}$$

and

$$\begin{aligned} ||B_+ f ||_{L^2} \le ||f ||_{L^2}, \quad f \in \Lambda _k. \end{aligned}$$

If $k_0 \ge 0$ and if $f \in \Lambda _{k_0}$ satisfies $\eta _- f = 0$, then the formal invariant distribution $w$ starting at $f$ is an element of $L^2_x H^{-1/2-\varepsilon }_{\theta }$ for any $\varepsilon > 0$. Moreover, the Fourier coefficients of $w$ satisfy

$$\begin{aligned} ||w_k ||_{L^2} \le ||f ||_{L^2}, \quad k \ge 0. \end{aligned}$$

Proof

Lemma 11.11 implies that for $u \in \Lambda _k$ with $k \ge 0$,

$$\begin{aligned} ||\eta _- u ||^2 = ||\eta _+ u ||^2 - \frac{i}{2} (K V u, u) = ||\eta _+ u ||^2 + \frac{k}{2} (Ku,u). \end{aligned}$$

Using that $K \le 0$ and $k \ge 0$ we get $||\eta _- u || \le ||\eta _+ u ||$. The rest of the claims follow as in Sect. 5. $\square$

The next lemma considers $B$ instead of $B_+$ and shows that the constants in the first inequality are sharp for flat Riemann surfaces.

Lemma 11.13

Let $(M,g)$ be a closed surface, and assume that $K \le 0$. Then

$$\begin{aligned} ||X_- u ||_{L^2} \le \left\{ \begin{array}{ll} ||X_+ u ||_{L^2}, &{}\quad u \in \Omega _m \text { with }\, m \ge 2, \\ \sqrt{2} ||X_+ u ||_{L^2}, &{}\quad u \in \Omega _1. \end{array} \right. \end{aligned}$$

If $K=0$ the constants are sharp. If additionally the genus is $\ge 2$ (so there are no conformal Killing tensors), then

$$\begin{aligned} ||Bf ||_{L^2} \le \left\{ \begin{array}{ll} ||f ||_{L^2}, &{}\quad f \in \Omega _m \text { with }\, m \ge 1, \\ \sqrt{2} ||f ||_{L^2}, &{}\quad f \in \Omega _0. \end{array} \right. \end{aligned}$$

Proof

If $u \in \Omega _m$ with $m \ge 2$, then $u = f_m + f_{-m}$ with $f_j \in \Lambda _j$. Lemma 11.11 yields

$$\begin{aligned} ||X_- u ||^2&= ||\eta _- f_m ||^2 + ||\eta _+ f_{-m} ||^2 \\&= ||\eta _+ f_m ||^2 + ||\eta _- f_{-m} ||^2 + \frac{m}{2} ((K f_m, f_m) + (K f_{-m}, f_{-m})). \end{aligned}$$

Using that $K \le 0$, this gives $||X_- u ||^2 \le ||X_+ u ||^2$ and equality holds for all $u \in \Omega _m$ if $K = 0$.

If instead $u \in \Omega _1$ we have $u = f_1 + f_{-1}$ with $f_j \in \Lambda _j$, and Lemma 11.11 again gives

$$\begin{aligned} ||X_- u ||^2&= ||\eta _- f_1 + \eta _+ f_{-1} ||^2 \le 2(||\eta _- f_1 ||^2 + ||\eta _+ f_{-1} ||^2) \\&= 2 \left[ ||\eta _+ f_1 ||^2 + ||\eta _- f_{-1} ||^2 + \frac{1}{2} ((K f_1, f_1) + (K f_{-1}, f_{-1})) \right] . \end{aligned}$$

Since $K \le 0$ we get $||X_- u ||^2 \le 2 ||X_+ u ||^2$. If $K=0$ we have equality if and only if $\eta _- f_1 = \eta _+ f_{-1}$. Identifying $\Omega _1$ with $1$-forms on $M$, this means that the $1$-form $f_1 - f_{-1}$ is divergence free and thus $f_1 - f_{-1} = *d a_0 + h$ for some $a_0 \in C^{\infty }(M)$ and some harmonic $1$-form $h$. Consequently, if $K=0$ then equality holds exactly when $u = d a_0 + h$ for some $a_0 \in C^{\infty }(M)$ and some harmonic $1$-form $h$. (Note that $h = h_1 + h_{-1}$ is a harmonic $1$-form if and only if $\eta _- h_1 = \eta _+ h_{-1} = 0$.)

The inequalities for $B$ follow as in Sect. 5. $\square$

Remark 11.14

We note that the inequality $||X_{-}u ||\le \sqrt{2}||X_{+}u ||$ for $u\in \Omega _1$ differs by the factor $\sqrt{2}$ with inequality (5.2) in [23, p. 173] for $n=2$ and $p=1$ which gives $||X_{-}u ||\le ||X_{+}u ||$. Since our inequality in Lemma 11.13 is shown to be sharp in the flat case this indicates an algebraic mistake in the calculation of the constants in [23].

Pestov and Guillemin-Kazhdan energy identities. We conclude this section by discussing the relation between two basic energy identities. The Pestov identity from Proposition 2.2 takes the following form in two dimensions:

$$\begin{aligned} ||VXu ||^2 = ||XVu ||^2 - (KVu,Vu) + ||Xu ||^2, \quad u \in C^{\infty }(SM). \end{aligned}$$

The Guillemin-Kazhdan energy identity in Lemma 11.11 looks as follows:

$$\begin{aligned} ||\eta _- u ||^2 = ||\eta _+ u ||^2 - \frac{i}{2} (K V u, u), \quad u \in C^{\infty }(SM). \end{aligned}$$

As discussed in [38], the Pestov identity is essentially the commutator formula $[XV,VX] = -X^2 + VKV$, whereas the Guillemin-Kazhdan identity follows from the commutator formula $[\eta _+, \eta _-] = \frac{i}{2} K V$.

We now show that the Pestov identity applied to $u \in \Lambda _k$ is just the Guillemin-Kazhdan identity for $u \in \Lambda _k$. Indeed, we compute

$$\begin{aligned} ||VXu ||^2 = ||V\eta _+ u ||^2 + ||V \eta _- u ||^2 = (k+1)^2 ||\eta _+ u ||^2 + (k-1)^2 ||\eta _- u ||^2 \end{aligned}$$

and

$$\begin{aligned} ||XVu ||^2&- (KVu,Vu) + ||Xu ||^2 = k^2(||\eta _+ u ||^2 + ||\eta _- u ||^2) + ik(KVu,u)\\&+ ||\eta _+ u ||^2 + ||\eta _- u ||^2. \end{aligned}$$

The Pestov identity and simple algebra show that

$$\begin{aligned} 2k(||\eta _+ u ||^2 - ||\eta _- u ||^2) = ik(KVu,u) \end{aligned}$$

This is the Guillemin-Kazhdan identity if $k \ne 0$.

In the converse direction, assume that we know the Guillemin-Kazhdan identity for each $\Lambda _k$,

$$\begin{aligned} ||\eta _+ u_k ||^2 - ||\eta _- u_k ||^2 = \frac{i}{2}(KVu_k,u_k), \quad u \in \Lambda _k. \end{aligned}$$

Multiplying by $2k$ and summing gives

$$\begin{aligned} \sum 2k(||\eta _+ u_k ||^2 - ||\eta _- u_k ||^2) = \sum ik (KVu_k,u_k). \end{aligned}$$

On the other hand, the Pestov identity for $u = \sum _{k=-\infty }^{\infty } u_k$ reads

$$\begin{aligned}&\sum k^2 ||\eta _+ u_{k-1} + \eta _- u_{k+1} ||^2 \\&\quad = \sum (||\eta _+(Vu_{k-1}) + \eta _-(Vu_{k+1}) ||^2 + ik (KVu_k, u_k) + ||\eta _+ u_{k-1} + \eta _- u_{k+1} ||^2). \end{aligned}$$

Notice that

$$\begin{aligned} k^2 ||\eta _+ u_{k-1} \!+\! \eta _- u_{k+1} ||^2 \!=\! k^2 (||\eta _+ u_{k-1} ||^2 \!+\! ||\eta _- u_{k+1} ||^2) + 2k^2 \mathrm {Re} (\eta _+ u_{k-1}, \eta _- u_{k+1}) \end{aligned}$$

and

$$\begin{aligned}&||\eta _+(Vu_{k-1}) + \eta _-(Vu_{k+1}) ||^2 + ||\eta _+ u_{k-1} + \eta _- u_{k+1} ||^2 \\&\quad \!=\! (k^2-2k\!+\!2) ||\eta _+ u_{k-\!1} ||^2 \!+\! (k^2+2k+2) ||\eta _- u_{k+1} ||^2 \!+\! 2 k^2 \mathrm {Re} (\eta _+ u_{k-1},\! \eta _- u_{k+1}). \end{aligned}$$

Thus the Pestov identity is equivalent with

$$\begin{aligned} \sum \left[ (2k-2) ||\eta _+ u_{k-1} ||^2 - (2k+2) ||\eta _- u_{k+1} ||^2 \right] = \sum ik (KVu_k, u_k). \end{aligned}$$

This becomes the summed Guillemin-Kazhdan identity after relabeling indices.