2.1 Representation spaces

Let $L^2(M)$ be the Hilbert space of complex-valued square-integrable functions with respect to the $\mathsf H$ -invariant volume form for M. By the Stone–von Neumann theorem, the space $L^2(M)$ decomposes into an orthogonal sum of irreducible, unitary representations that are unitarily equivalent to certain one-dimensional or infinite-dimensional models that we describe at the top of §§2.2 and 2.3. Moreover, by irreducibility, Sobolev spaces $W^s(M)$ are also decomposable in the above sense, because vector fields in $\operatorname {\mathfrak h}$ split into irreducible, unitary representation spaces, and the infinitesimal representations of $\operatorname {\mathfrak h}$ extend to representations of the enveloping algebra. For this reason, we may prove our Sobolev estimates concerning coboundary operators (Theorems 1.2 and 1.3) in simpler, orthogonal components of $W^s(M)$ , and then we glue the estimates together at the end (see (50)).

2.2 Finite-dimensional representations

The one-dimensional representations are unitarily equivalent to characters $\rho _{\mathbf {m}}$ of ${ \mathbb R}^{2n}$ in $L^2({ \mathbb T}^{2n})$ , for $\mathbf {m} \in { \mathbb Z}^{2n}$ , and are given by

(11) $$ \begin{align} \rho_{\mathbf{m}}(\mathbf{x}, \boldsymbol \xi, t) f = e^{2\pi i \mathbf{m} \cdot (\mathbf{x}, \boldsymbol \xi)} f. \end{align} $$

For each integer $1 \leq j \leq n$ , the derived representations of $\rho _{\mathbf {m}}$ are

$$ \begin{align*} X_j = 2\pi i m_j, \quad \Lambda_j = 2\pi i m_{n + j}, \quad Z = 0. \end{align*} $$

Write

$$ \begin{align*} \rho := \bigoplus_{\mathbf{m} \in { \mathbb Z}^4} \rho_{\mathbf{m}}. \end{align*} $$

So, given $f \in L^2({ \mathbb T}^{2n})$ , we have the orthogonal decomposition

$$ \begin{align*} f(\mathbf{x}, \boldsymbol \xi) = \sum_{\mathbf{m} \in { \mathbb Z}^4} f_{\mathbf{m}} e^{2\pi i \mathbf{m} \cdot (\mathbf{x}, \boldsymbol \xi)}, \end{align*} $$

where $\triangle $ acts on irreducible, unitary representations of $L^2({ \mathbb T}^{2n})$ by

$$ \begin{align*} \rho_{\mathbf{m}}(\triangle) = 4\pi^2 \mathbf{m} \cdot \mathbf{m}. \end{align*} $$

For $s> 0$ , the subspace of s-differentiable functions is $W^s({ \mathbb T}^{2n}) \subset L^2({ \mathbb T}^{2n})$ , defined to be the maximal domain of the operator $(I + \rho (\triangle ))^{s/2}$ on $L^2({ \mathbb T}^{2n})$ with inner product and norm given by (2). In particular,

(12) $$ \begin{align} \Vert f \Vert_s^2 = \sum_{\mathbf{m} \in { \mathbb Z}^{2n}} (1 + 4\pi^2 \mathbf{m} \cdot \mathbf{m})^s \vert f_{\mathbf{n}} \vert^2. \end{align} $$

We denote the space of smooth functions in $L^2({ \mathbb T}^{2n})$ by

$$ \begin{align*} W^{\infty}({ \mathbb T}^{2n}) := \bigcap_{s \geq 0} W^s({ \mathbb T}^{2n}). \end{align*} $$

Furthermore, for every s, we have $W^s({ \mathbb T}^{2n}) = \mathbb C \langle 1 \rangle \oplus W_0^s({ \mathbb T}^{2n})$ , where $W_0^s({ \mathbb T}^{2n})$ is the Sobolev space of s-differentiable, zero-average functions on ${ \mathbb T}^{2n}$ . So, it follows that

$$ \begin{align*} W^{\infty}({ \mathbb T}^{2n}) = \mathbb C \langle 1\rangle \oplus W_0^{\infty}({ \mathbb T}^{2n}), \end{align*} $$

where $W_0^{\infty }({ \mathbb T}^{2n}) = \bigcap _{s \geq 0} W_0^s({ \mathbb T}^{2n})$ .

The two propositions below establish Theorems 1.2 and 1.3 in the case of finite-dimensional representations. The proofs are straightforward and deferred to the appendix.

Proposition 2.1. There is a constant $C_{\boldsymbol {\tau }, \boldsymbol {\eta }}> 0$ such that for any zero-average $f, g \in W_0^{\infty }({ \mathbb T}^{2n})$ that satisfy $L_{\boldsymbol {\tau }} g = L_{\boldsymbol {\eta }} f$ , there is a solution $P \in W^{\infty }({ \mathbb T}^{2n})$ such that

$$ \begin{align*} L_{\boldsymbol{\tau}} P = f \quad\text{and}\quad L_{\boldsymbol{\eta}} P = g \end{align*} $$

and, for any $s \geq 0$ ,

$$ \begin{align*} \| P \|_{s} \leq C_{\boldsymbol{\tau}, \boldsymbol{\eta}} (\| f \|_{s+2\gamma} + \Vert g \Vert_{s + 2\gamma}). \end{align*} $$

Proposition 2.2. There is a constant $C_{\boldsymbol \tau , \boldsymbol \eta }> 0$ such that for any $\phi \in W^{\infty }({ \mathbb T}^{2n})$ and any non-constant zero-average functions $f, g\in W_0^{\infty }({ \mathbb T}^{2n})$ that satisfy $L_{\boldsymbol {\eta }} f - L_{\boldsymbol {\tau }} g = \phi $ , there is a non-constant function $P \in W^{\infty }({ \mathbb T}^{2n})$ such that for any $s \geq 0$ ,

$$ \begin{align*} \begin{array}{l} \|g - L_{\boldsymbol{\eta}} P\|_s \leq C_{\boldsymbol \tau, \boldsymbol \eta} \|\phi \|_{s+2\gamma} , \\ \|f - L_{\boldsymbol{\tau}} P\|_s \leq C_{\boldsymbol \tau, \boldsymbol \eta} \|\phi \|_{s+2\gamma} , \\ \| P \|_s \leq C_{\boldsymbol \tau, \boldsymbol \eta} (\Vert f \Vert_{s + 2\gamma} + \| g \|_{s+2\gamma}). \end{array} \end{align*} $$

2.3 Schrödinger representations

Next we consider the irreducible, infinite-dimensional representations. By the Stone–von Neumann theorem, any infinite-dimensional representation is unitarily equivalent to a Schrödinger representation of $\mathsf H$ on $L^2({ \mathbb R}^n)$ with a parameter $h \neq 0$ . When acting on the right, this is

(13) $$ \begin{align} (\mu_{h}(\mathbf{x},\boldsymbol \xi, t) \phi)(y) = e^{-i h t + i \epsilon \vert h \vert^{1/2} \boldsymbol \xi\cdot y - ({1}/{2}) i h \boldsymbol \xi\cdot \mathbf{x}} \phi(y-\vert h \vert^{1/2} \mathbf{x}), \end{align} $$

where $\epsilon = \text {sign}(h) = \pm 1.$ For integers $1 \leq j \leq n$ , we have

$$ \begin{align*} \mu_h(X_j) = -\vert h \vert^{1/2} \frac{\partial}{\partial y_j}, \ \ \ \mu_h(\Lambda_j) = i \epsilon \vert h \vert^{1/2} y_j, \ \ \ \mu_h(Z) = -i h. \end{align*} $$

The derived representation extends to the enveloping algebra of the Lie algebra of $\mathsf H$ .

In §1.1, we noted that we will work with the standard lattice and, in this case, the Schr $\ddot {\textrm {o}}$ dinger representations are parameterized by $h \in 2\pi { \mathbb Z}\setminus \{0\}$ (see [Reference Cosentino and Flaminio1, §3.2]). Then observe that

$$ \begin{align*} \vert Z \vert = \vert h \vert \end{align*} $$

and define the operator $\Box $ in the model $\mu _h$ to be

$$ \begin{align*} \mu_h(\Box) & := \frac{1}{2\pi}\bigg( \vert \mu_h(Z) \vert -\sum_{i = 1}^n \mu_h(X_i^2) + \mu_h(\Lambda_i^2) \bigg)\\ & = \frac{\vert h \vert}{2\pi} \bigg(1 + \sum_{i = 1}^n y_i^2 -\frac{\partial^2}{\partial y_i^2}\bigg), \end{align*} $$

which is homogeneous in $\vert h \vert $ . The operator $\mu _h(\Box )$ is related to $\mu _h(\triangle )$ by

(14) $$ \begin{align} \mu_h(\Box) = \frac{1}{2\pi}(\mu_h(\triangle) + \vert \mu_h(Z) \vert + \mu_h(Z)^2 ). \end{align} $$

Define $W^s(\mu _h, { \mathbb R}^n) \subset L^2({ \mathbb R}^n)$ to be Hilbert Sobolev space of s-differentiable functions, that is, the maximal domain of the operator $\mu _h(\Box )^s$ on $L^2({ \mathbb R}^n)$ with inner product

$$ \begin{align*} \langle \mu_h(\Box)^s f, g \rangle_{L^2({ \mathbb R}^n)} = \bigg(\frac{\vert h \vert}{2\pi}\bigg)^s \bigg\langle \bigg(I + \sum_{i = 1}^n y_i^2 -\frac{\partial^2}{\partial y_i^2}\bigg)^s f, g\bigg\rangle_{L^2({ \mathbb R}^n)}. \end{align*} $$

Denote the Sobolev norm of this operator by

(15) $$ \begin{align} |\mkern-1.5mu|\mkern-1.5mu | f |\mkern-1.5mu |\mkern-1.5mu |_s := \langle \mu_h(\Box)^s f, f\rangle_{L^2({ \mathbb R}^n)}. \end{align} $$

Clearly, the space of smooth functions in $L^2({ \mathbb R}^n)$ with respect to $\mu _h(\Box )$ is the Schwartz space

$$ \begin{align*} \mathscr{S}({ \mathbb R}^n) = \bigcap_{s \geq 0} W^s(\mu_h, { \mathbb R}^n). \end{align*} $$

Following [Reference Cosentino and Flaminio1], estimates of linear operators with respect to the Laplacian (1) are a consequence of such estimates in the homogeneous norm. This estimate differs from Lemma 3.15 of [Reference Cosentino and Flaminio1] in that our homogeneous operator $\mu _h(\Box )$ includes the term $\vert \mu _h(Z)\vert $ , whereas that operator in [Reference Cosentino and Flaminio1] does not. The argument below is the same as in [Reference Cosentino and Flaminio1, Lemma 3.15].

Lemma 2.3. Let $T: \mathscr {S}({ \mathbb R}^n) \to \mathscr {S}({ \mathbb R}^n)$ be a linear map for the representation $\mu _h$ such that for every $s \geq 0$ , there are a constant $C_s> 0$ and some $t \geq 0$ satisfying

$$ \begin{align*} |\mkern-1.5mu|\mkern-1.5mu| T f |\mkern-1.5mu|\mkern-1.5mu |_s \leq C_s |\mkern-1.5mu|\mkern-1.5mu| f |\mkern-1.5mu|\mkern-1.5mu|_{s + t}. \end{align*} $$

Then, for every $s \geq 0$ , there is a constant $C_{s, t} = 2^{(s + t)/2} \max _{0 \leq k \leq s} \{C_k\}> 0$ such that

$$ \begin{align*} \Vert T f \Vert_s \leq C_{s, t} \Vert f \Vert_{s + t}. \end{align*} $$

Proof First let $s \geq 0$ be an integer. Then

(16) $$ \begin{align} \Vert T f \Vert_s^2 & = \bigg\langle \bigg(I -\mu_h(Z)^2 -\sum_{i = 1}^n \mu_h(X_i)^2 + \mu_h(\Lambda_i)^2\bigg)^s T f, T f\bigg\rangle \notag \\ & \leq \sum_{k = 0}^s {s \choose k} (1 + h^2)^{s - k} \langle \mu_h(\Box)^k T f, T f\rangle. \end{align} $$

By the definition of $|\mkern -1.5mu|\mkern -1.5mu|T f |\mkern -1.5mu|\mkern -1.5mu|_k$ , by (14) and because all terms are positive, we have

(17) $$ \begin{align} (16) & \leq \sum_{k = 0}^s {s \choose k} (1 + h^2)^{s - k} C_k^2 \langle \mu_h(\Box)^{k + t} f, f\rangle \notag \\ & = \max_{0 \leq k \leq s} \{C_k^2\} \langle (I - \mu_h(Z)^2 + \mu_h(\Box))^{s} \mu_h(\Box)^{t/2} f , \mu_h(\Box)^{t/2} f\rangle. \end{align} $$

Because $\mu _h(Z)^2$ is a constant, $(I - \mu _h(Z)^2 + \mu _h(\Box ))$ and $\mu _h(\Box )$ commute. Then, using that $- \mu _h(Z)^2$ is positive again, we have

$$ \begin{align*} (17) & \leq \max_{0 \leq k \leq s} \{C_k^2\} \langle (I - \mu_h(Z)^2 + 2\pi \mu_h(\Box))^{s + t} f , f\rangle \\ & = \max_{0 \leq k \leq s} \{C_k^2\} \langle (I + \mu_h(\triangle) + 2\pi \vert\mu_h(Z)\vert )^{s + t} f , f\rangle \\ & \leq 2^{s + t} \max_{0 \leq k \leq s} \{C_k^2\} \Vert f \Vert_{s + t}^2. \end{align*} $$

The estimate for $s \geq 0$ follows by interpolation.

We will use the above lemma to reduce our estimates to the case $h = 1$ . Because the norm (15) is homogeneous in h, by rescaling by the factor $\vert h \vert ^{s/2}$ from $|\mkern -1.5mu|\mkern -1.5mu|f|\mkern -1.5mu|\mkern -1.5mu|_{s}$ , we can restrict ourselves to the case $ \vert h \vert = 2\pi $ , as in [Reference Cosentino and Flaminio1]. In what follows, we set $h = 2\pi $ , as the argument for $h = -2\pi $ is analogous.

Then, to simplify notation, we write

$$ \begin{align*} X_j = -\frac{\partial}{\partial y_j}, \quad \Lambda_j = i y_j , \quad Z = -i \end{align*} $$

and we refer to the Schrödinger representation on $L^2({ \mathbb R}^n)$ as

$$ \begin{align*} \mu := \mu_{2\pi}. \end{align*} $$

For $s> 0$ , we denote $W^s({ \mathbb R}^n) := W^s(\mu , { \mathbb R}^n)$ .

It will be convenient to define the Sobolev space $W^s({ \mathbb R}^{n-1})$ that is the maximal domain of the operator $I + \sum _{i = 2}^{n} y_i^2 - ({\partial ^2}/{\partial y_i^2})$ on $L^2({ \mathbb R}^{n-1})$ . We use the same notation for the inner product, where in this setting

$$ \begin{align*} \langle f, g \rangle_s := \bigg\langle \bigg(I + \sum_{i = 2}^{n} y_i^2 - \frac{\partial^2}{\partial y_i^2}\bigg)^{s} f, g\bigg\rangle_{L^2({ \mathbb R}^{n-1})}. \end{align*} $$

2.3.1 Change of variable

Define

$$ \begin{align*} \tau = \sqrt{\sum_{j = 1}^{n} \tau_j^2}. \end{align*} $$

Let $A = [\mathbf a_1, \mathbf a_2, \ldots , \mathbf a_n] \in O(n)$ be an $n\times n$ matrix with orthonormal rows $\mathbf a_i$ such that

$$ \begin{align*} \mathbf a_1 = \frac{1}{\tau} (\tau_1, \tau_2, \ldots, \tau_n). \end{align*} $$

Observe that $(\tau _j)$ and $(\eta _j)$ span a two-dimensional subspace of ${ \mathbb R}^n$ , so we can choose $\mathbf a_2$ to be such that

$$ \begin{align*} (\eta_1, \eta_2, \ldots, \eta_n) \in \textrm{span}( \{\mathbf a_1, \mathbf a_2\}). \end{align*} $$

Further, choose the signs of the vectors $\mathbf a_j$ , for $2 \leq j \leq n$ , so that $A \in \operatorname {SO}(n)$ . Then A is the determinant one rotation of ${ \mathbb R}^n$ such that

(18) $$ \begin{align} \begin{array}{l} A(\tau_1, \tau_2, \ldots, \tau_n) = (\tau, 0, \ldots, 0), \\ A(\eta_1, \eta_2, \ldots, \eta_n) = (\nu_1, \nu_2, 0, \ldots, 0) \end{array} \end{align} $$

for some $(\nu _1, \nu _2) \in { \mathbb R}^2$ .

For $y = (y_1, y_2, \ldots , y_n)$ , define $z = (z_1, z_2, \ldots , z_n)$ via matrix–vector multiplication by

$$ \begin{align*} z = A y. \end{align*} $$

Therefore,

(19) $$ \begin{align} f(y) := f \circ A^{-1} (z). \end{align} $$

Clearly, because A is an orthogonal matrix, the operator $U_A: L^2({ \mathbb R}^n, dy) \to L^2({ \mathbb R}^n, dz)$ given by

$$ \begin{align*} U_A f = f \circ A^{-1} \end{align*} $$

is unitary. Let $\tilde {\mu }$ be the representation on $\mathsf H$ such that for any $g \in \mathsf H$ , $\tilde {\mu }(g): L^2({ \mathbb R}^n, dz) \to L^2({ \mathbb R}^n, dz)$ is given by

$$ \begin{align*} \tilde{\mu}(g) := U_A \mu(g) U_A^{-1}. \end{align*} $$

So, $\tilde {\mu }$ is unitarily equivalent to $\mu $ .

Now we compute a basis for $\operatorname {\mathfrak h}$ in terms of the derived representations of $\tilde {\mu }$ . For each $j \in \{1, 2, \ldots , n\}$ , let $(x_{j, t}, \lambda _{j, t}, z_{j, t})_{t\in [-1,1]}$ be smooth curves in $\mathsf H$ such that

$$ \begin{align*} X_j = \frac{d}{dt}\mu(x_{j, t}) |_{t = 0}, \quad \Lambda_j = \frac{d}{dt}\mu(\lambda_{j, t})|_{t = 0}, \quad Z_j = \frac{d}{dt}\mu(z_{j, t})|_{t = 0}. \end{align*} $$

Then set

$$ \begin{align*} \tilde{X}_j = \frac{d}{dt}\tilde{\mu}(x_{j, t}) |_{t = 0}, \quad \tilde{\Lambda}_j = \frac{d}{dt}\tilde{\mu}(\lambda_{j, t})|_{t = 0}, \quad \tilde{Z}_j = \frac{d}{dt}\tilde{\mu}(z_{j, t})|_{t = 0}. \end{align*} $$

Let $A^{-1}$ be the matrix

$$ \begin{align*} A^{-1} = (b_{i j}) \end{align*} $$

for some coefficients $b_{i j}$ . A calculation shows that for $j \in \{1, 2, \ldots , n\}$ ,

$$ \begin{align*} \tilde{X}_j = -\sum_{k = 1}^n b_{j k} \frac{\partial}{\partial z_k} , \quad \tilde{\Lambda}_j = i \sum_{j = k}^{n} b_{j k} z_k , \quad \tilde{Z} = -i. \end{align*} $$

One can check that these operators satisfy the commutation relations

$$ \begin{align*} [\tilde{X}_i, \tilde{X}_j] = 0, \ \ \ \ [\tilde{\Lambda}_i, \tilde{\Lambda}_j] = 0, \ \ \ \ [\tilde{X}_i, \tilde{\Lambda}_j] = \delta_{i j} \tilde{Z} \end{align*} $$

for $i, j \in \{1, 2, \ldots , n\}$ .

Lemma 2.5. We have

$$ \begin{align*} \tilde{\mu}(\Box) = I + \sum_{i = 1}^n z_i^2 - \frac{\partial^2}{\partial z_i^2}. \end{align*} $$

Proof By definition,

(20) $$ \begin{align} \tilde{\mu}(\Box) = I + \sum_{i = 1}^n -\tilde{X}_i^2 - \tilde{\Lambda}_i^2. \end{align} $$

Notice that

$$ \begin{align*} \tilde{X}_i^2 & = \bigg(-\sum_{j = 1}^n b_{i j} \frac{\partial}{\partial z_j}\bigg)^2 \\[4pt] & = \sum_{j, m = 1}^n b_{i j} b_{i m} \frac{\partial^2}{\partial z_j \partial z_m}. \end{align*} $$

Because the columns of $A^{-1}$ are orthonormal, we get

$$ \begin{align*} \sum_{i = 1}^n \tilde{X}_i^2 & = \sum_{i = 1}^n \sum_{j, m = 1}^n b_{i j} b_{i m} \frac{\partial^2}{\partial z_j \partial z_m} \\ & = \sum_{j, m = 1}^n \sum_{i = 1}^n b_{i j} b_{i m} \frac{\partial^2}{\partial z_j \partial z_m} \\ & = \sum_{1 \leq j \neq m \leq n} \frac{\partial}{\partial z_j \partial z_m} \sum_{i = 1}^n b_{i j} b_{i m} + \sum_{j = 1}^n \frac{\partial^2}{\partial z_j^2} \sum_{i = 1}^n b_{i j}^2 \\ & = \sum_{j = 1}^n \frac{\partial^2}{\partial z_j^2}. \end{align*} $$

Similarly,

$$ \begin{align*} \sum_{i = 1}^n \tilde{\Lambda}_i^2 & = - \sum_{j, m = 1}^n \sum_{i = 1}^n b_{i j} b_{i m} z_j z_m \\ & = -\sum_{1 \leq j \neq m \leq n} z_j z_m \sum_{i = 1}^n b_{i j} b_{i m} - \sum_{j = 1}^n z_j^2 \sum_{i = 1}^n b_{i j}^2 \\ & = - \sum_{j = 1}^n z_j^2. \end{align*} $$

Hence,

$$ \begin{align*} (20) = I + \sum_{i = 1}^n z_i^2 - \frac{\partial^2}{\partial z_i^2}.\\ \end{align*} $$

Finally, we compute the operator $\tilde \mu (\textrm{ exp}(Y_{\boldsymbol {\kappa }}))$ for $\boldsymbol \kappa \in \{\boldsymbol {\tau }, \boldsymbol \eta \}$ .

Lemma 2.6. For any $f \in L^2({ \mathbb R}^n)$ and $z \in { \mathbb R}^n$ , we have

$$ \begin{align*} &\tilde\mu(\textrm{ exp}(Y_{\boldsymbol{\tau}})) f(z) = f(z - (\tau, 0, \ldots, 0)), \\ &\tilde\mu(\textrm{ exp}(Y_{\boldsymbol{\eta}})) f(z) = \textrm{ exp}(i\nu_2 z_2) f(z) \end{align*} $$

for some $\nu _2 \in { \mathbb R}^*$ .

Proof To help keep track of which coordinate system we are working in, note that $U_A f = f\circ A^{-1}$ , where $z = A y$ . So,

$$ \begin{align*} U_A: L^2({ \mathbb R}^n, dy) \to L^2({ \mathbb R}^n, dz), \quad U_A^{-1}: L^2({ \mathbb R}^n, dz) \to L^2({ \mathbb R}^n, dy) \end{align*} $$

and of course the Schrödinger representation $\mu $ satisfies

$$ \begin{align*} \mu(g): L^2({ \mathbb R}^n, dy) \to L^2({ \mathbb R}^n, dy) \end{align*} $$

for any $g \in \mathsf H$ . Then

$$ \begin{align*} \tilde{\mu}(\textrm{ exp}(Y_{\boldsymbol \tau})) f(z) &:= U_A \mu(\textrm{ exp}(Y_{\boldsymbol \tau})) U_A^{-1} f(z) \notag \\[3pt] & = \mu(\textrm{ exp}(Y_{\boldsymbol \tau}))U_A^{-1} f(A^{-1} z) \notag \\[3pt] & = \mu(\textrm{ exp}(Y_{\boldsymbol \tau})) U_A^{-1} f(y) \notag \\[3pt] & = U_A^{-1} f(y_1 - \tau_1, \ldots , y_n - \tau_n) \notag \\[3pt] & = f(A(y_1 - \tau_1, \ldots , y_n - \tau_n)) \notag \\[3pt] & = f(Ay - A(\tau_1, \ldots , \tau_n)) \notag \\[3pt] & = f(z - (\tau, 0, \ldots, 0)). \notag \end{align*} $$

Next, recall that $\boldsymbol \eta = (\mathbf 0, \eta _1, \eta _2, \ldots , \eta _n) \in { \mathbb R}^{2n}$ and define $\underline {\boldsymbol \eta } := (\eta _1, \eta _2, \ldots , \eta _n)$ . Then $\mu (\textrm{ exp}(Y_{\boldsymbol {\eta }}))$ is the multiplication operator

$$ \begin{align*} \mu(\textrm{ exp}(Y_{\boldsymbol{\eta}})) f (y)= e^{i \underline{\boldsymbol{\eta}} \cdot y} \cdot f(y). \end{align*} $$

So,

(21) $$ \begin{align} \tilde{\mu}(Y_{\boldsymbol{\eta}}) f (z) &:= U_A \mu(\textrm{ exp}(Y_{\boldsymbol{\eta}})) U_A^{-1} f(z) \notag \\[3pt] & = \mu(\textrm{ exp}(Y_{\boldsymbol{\eta}})) U_A^{-1} f (A^{-1} z) \notag \\[3pt] & = \mu(\textrm{ exp}(Y_{\boldsymbol{\eta}})) U_A^{-1} f (y) \notag \\[3pt] & = e^{i \underline{\boldsymbol{\eta}} \cdot y} U_A^{-1} f(y) \notag \\[3pt] & = e^{i \underline{\boldsymbol{\eta}} \cdot y} f(Ay) \notag \\[3pt] & = e^{i \underline{\boldsymbol{\eta}} \cdot A^{-1} z} f(z) \notag \\[3pt] & = e^{i A \underline{\boldsymbol{\eta}} \cdot z} f(z). \end{align} $$

Now recall from (18) that $A\underline {\boldsymbol {\eta }} = (\nu _1, \nu _2, 0, \ldots , 0)$ for some $(\nu _1, \nu _2) \in { \mathbb R}^2$ , so

(22) $$ \begin{align} (21) = \textrm{ exp}(i (\nu_1 z_1 + \nu_2 z_2)) f(z). \end{align} $$

Furthermore, observe that the assumption $[Y_{\boldsymbol {\tau }}, Y_{\boldsymbol {\eta }}] = 0$ from (6) is equivalent to the condition

$$ \begin{align*} \sum_{j = 1}^n \tau_j \eta_j = 0. \end{align*} $$

We also have $A^{-1} (\tau , 0, \ldots , 0) = (\tau _j)$ , where $A^{-1} = (b_{i j})$ . Then, for all $1 \leq j \leq n$ ,

$$ \begin{align*} b_{j 1} = \frac{\tau_j}{\tau}. \end{align*} $$

Because $A \in \operatorname {SO}(n)$ , we have

$$ \begin{align*} a_{1 j} = b_{j 1} = \frac{\tau_j}{\tau}. \end{align*} $$

Hence,

(23) $$ \begin{align} \nu_1 & = (A \boldsymbol \eta)_1 = \sum_{j = 1}^n a_{1 j} \eta_j = \frac{1}{\tau}\sum_{j = 1}^n \tau_j \eta_j = 0. \end{align} $$

Because A is a rotation and $\nu _1 = 0$ , we get that $\vert \nu _2\vert = \vert \boldsymbol \eta \vert> 0$ . Finally, because A is a real matrix and $\boldsymbol \eta \in { \mathbb R}^n$ , it follows that $\nu _2 \in { \mathbb R}^*$ . The lemma now follows from (22) and (23).

For $\boldsymbol \kappa \in \{\boldsymbol {\tau }, \boldsymbol \eta \}$ , the operator $L_{\boldsymbol {\kappa }}$ is defined on functions of the $\mathbf z$ -variable by

$$ \begin{align*} L_{\boldsymbol{\kappa}} := \tilde{\mu}(\textrm{ exp}(Y_{\boldsymbol\kappa})) - I, \end{align*} $$

so, by the above lemma,

(24) $$ \begin{align} L_{\boldsymbol{\kappa}} f(z) = \begin{cases} f(z - (\tau, 0, \ldots, 0)) - f(z) & \text{ if } \boldsymbol\kappa = \boldsymbol{\tau}, \\ [\textrm{ exp}(i \nu_2 z_2 ) - 1] f(z) & \text{ if } \boldsymbol\kappa = \boldsymbol\eta. \ \end{cases} \end{align} $$

The coordinates $(z_3, z_4, \ldots , z_n)$ will not play a central role, so for any $f \in L^2({ \mathbb R}^n)$ and for any $z \in { \mathbb R}^n$ , define

$$ \begin{align*} &\mathbf z_3 := (z_3, z_4, \ldots, z_n) \in { \mathbb R}^{n-2}, \\ &f_{\mathbf z_3}(z_1, z_2) := f(z). \end{align*} $$

For $j = 1, 2$ , let $\mathcal F_j$ be the Fourier transform in the $z_j$ -variable, so

$$ \begin{align*} &\mathcal F_1 f_{\mathbf z_3}(\omega_1, z_2) := \int_{ \mathbb R} f_{\mathbf z_3}(z_1, z_2) e^{-2\pi i \omega_1 z_1}\,dz_1, \\ &\mathcal F_2 f_{\mathbf z_3}(z_1, \omega_2) := \int_{ \mathbb R} f_{\mathbf z_3}(z_1, z_2) e^{-2\pi i \omega_2 z_2}\,dz_2. \end{align*} $$

We begin with a short lemma.

Lemma 2.7. For any $s \geq 0$ and for any $\epsilon \in (0, 1)$ , there is a constant $C_{\epsilon }> 0$ such that for any $\mathbf z \in { \mathbb R}^n$ and for any $f \in W^{s+n/2+\epsilon }({ \mathbb R}^n)$ , the functions f, $\mathcal F_1 f$ and $\mathcal F_2 f$ are continuous on ${ \mathbb R}^n$ , and

(25) $$ \begin{align} & \vert f_{\mathbf z_3}(z_1, z_2) \vert \leq\dfrac {C_{\epsilon}}{\big(1 + \displaystyle\sum\nolimits_{i=1}^n z_i^2\big)^{s/2}} |\mkern-1.5mu|\mkern-1.5mu| f |\mkern-1.5mu|\mkern-1.5mu|_{s + n/2 + \epsilon}, \end{align} $$

(26) $$ \begin{align} & \vert \mathcal F_2 f_{\mathbf z_3}(z_1, \omega_2) \vert \leq \dfrac{C_{\epsilon}}{\big(1 + \omega_2^2 + \displaystyle\sum\nolimits_{\substack{1 \leq i \leq n \\ i \neq 2}} z_i^2\big)^{s/2}} |\mkern-1.5mu|\mkern-1.5mu| f |\mkern-1.5mu|\mkern-1.5mu|_{s + n/2 + \epsilon}. \end{align} $$

Similarly, for any $(\omega , z_2) \in { \mathbb R}^2$ , for any $r \geq 0$ and for any $f \in W^{s+r+ n/2+\epsilon }({ \mathbb R}^n)$ ,

(27) $$ \begin{align} & \vert f_{\mathbf z_3}(z_1, z_2) \vert \leq\frac {C_\epsilon}{(1 + z_1^2)^{r/2}\big(1 + \displaystyle\sum\nolimits_{i = 2}^n z_i^2\big)^{s/2} } |\mkern-1.5mu|\mkern-1.5mu| f |\mkern-1.5mu|\mkern-1.5mu|_{s +r+ n/2 + \epsilon}, \end{align} $$

(28) $$ \begin{align} & \vert \mathcal F_1 f_{\mathbf z_3} (\omega_1, z_2) \vert \leq \frac{C_\epsilon}{(1 + \omega_1^2)^{r/2}\big(1 + \displaystyle\sum\nolimits_{i = 2}^n z_i^2\big)^{s/2} } |\mkern-1.5mu|\mkern-1.5mu| f |\mkern-1.5mu|\mkern-1.5mu|_{r+s + n/2+\epsilon}. \end{align} $$

Proof The Sobolev embedding theorem implies that there is a constant $C_\epsilon> 0$ such that

This implies the first inequality and that f is continuous. The inequality (26) follows in the same way by applying the inverse Fourier transform $\mathcal F_2^{-1}$ . Then $\mathcal F_2 f$ is also continuous.

For (28), the Sobolev embedding theorem again gives a constant $C_\epsilon> 0$ such that

$$ \begin{align*} &\bigg\Vert (I + \omega_1^2)^{r/2} \bigg(I + \sum_{i = 2}^n z_i^2\bigg)^{s/2} \mathcal F_1 f\bigg\Vert_{ C^{0, \epsilon} ({ \mathbb R}^n)} \\ &\quad \leq C_\epsilon \bigg\Vert \bigg(I - \frac{\partial^2}{\partial \omega_1^2} - \sum_{i = 2}^n\frac{\partial^2}{\partial z_i^2}\bigg)^{(n/2 + \epsilon)/2} (I + \omega_1^2)^{r/2} \bigg(I+\sum_{i = 2}^n z_i^2\bigg)^{s/2} \mathcal F_1 f \bigg\Vert_{L^2({ \mathbb R}^n)}\\ &\quad \leq C_\epsilon \bigg\Vert \bigg(I + z_1^2 - \sum_{i = 2}^n\frac{\partial^2}{\partial z_i^2}\bigg)^{(n/2 + \epsilon)/2} \bigg(I - \frac{\partial^2}{\partial z_1^2}\bigg)^{r/2}\bigg(I + \sum_{i = 2}^n z_i^2\bigg)^{s/2} f \bigg\Vert_{L^2({ \mathbb R}^n)} \\ &\quad \leq C_\epsilon |\mkern-1.5mu|\mkern-1.5mu| f |\mkern-1.5mu|\mkern-1.5mu|_{r + s + n/2 + \epsilon}. \end{align*} $$

The estimate of (27) follows as above.

2.3.2 Invariant operators and cohomological equations

For any $m \in { \mathbb Z}$ , let $\pi _{m, \boldsymbol {\tau }}$ be the formal operator

(29) $$ \begin{align} \pi_{m, \boldsymbol{\tau}} f(z) := \mathcal F_1 f\bigg(\frac{m}{\tau}, z_2, \ldots, z_n\bigg). \end{align} $$

Now we record a decay estimate of $\vert \pi _{m, \boldsymbol {\tau }}(f) \vert _s$ with respect to m, which will be used later in the splitting result, Theorem 1.3. Recall Definition 2.4 for the meaning of $\vert f \vert _s$ .

Corollary 2.8. For any $\epsilon> 0$ , there is a constant $C_\epsilon> 0$ such that for any ${s, r \geq 0}$ and for any $m \in { \mathbb Z}$ , the operator $\pi _{m, \boldsymbol {\tau }}$ satisfies the following estimate. For any $f \in W^{r+s+3n/2-1+\epsilon }({ \mathbb R}^n)$ , we have

$$ \begin{align*} \vert \pi_{m, \boldsymbol{\tau}}(f) \vert_s \leq C_\epsilon \bigg(1 + \bigg\vert \frac{m}{\tau}\bigg\vert \bigg)^{-r} |\mkern-1.5mu|\mkern-1.5mu| f |\mkern-1.5mu|\mkern-1.5mu|_{r+s+3n/2-1+\epsilon}. \end{align*} $$

Proof First let $f \in \mathscr {S}({ \mathbb R}^n)$ . Because $\mathcal F_1$ commutes with $(I - \sum _{i = 2}^n ({\partial ^2}/{\partial z_i^2}) + z_i^2)^{s/2}$ , for any $m \in { \mathbb Z}$ , we have

(30) $$ \begin{align} \vert \pi_{m, \boldsymbol{\tau}}(f) \vert_s & = \bigg\Vert \bigg(I - \sum_{i = 2}^n \frac{\partial^2}{\partial z_i^2} + z_i^2\bigg)^{s/2} \mathcal F_1 f_{\mathbf z_3}\bigg(\frac{m}{\tau}, z_2\bigg)\bigg\Vert_{L^2({ \mathbb R}^{n-1})} \notag \\ & = \bigg\Vert \mathcal F_1\bigg(\bigg(I - \sum_{i = 2}^n \frac{\partial^2}{\partial z_i^2} + z_i^2\bigg)^{s/2} f_{\mathbf z_3}\bigg)\bigg(\frac{m}{\tau}, z_2\bigg) \bigg\Vert_{L^2({ \mathbb R}^{n-1})}. \end{align} $$

Then (28) gives

so

$$ \begin{align*} (30) & \leq C_\epsilon \bigg(1 + \bigg\vert \frac{m}{\tau}\bigg\vert \bigg)^{-r} |\mkern-1.5mu|\mkern-1.5mu| f |\mkern-1.5mu|\mkern-1.5mu|_{r+s+3n/2-1+2\epsilon} \notag.\\ \end{align*} $$

The next lemma shows that for any $m \in { \mathbb Z}$ , $\pi _{m, \boldsymbol {\tau }}$ are invariant operators for $\tilde {\mu }(\textrm{ exp}(Y_{\boldsymbol {\tau }}))$ on sufficiently regular functions.

Lemma 2.9. For any $m \in { \mathbb Z}$ and for any $\epsilon> 0$ ,

$$ \begin{align*} \pi_{m, \boldsymbol{\tau}} L_{\boldsymbol{\tau}} = 0 \end{align*} $$

holds on $W^{n/2 + \epsilon }({ \mathbb R}^n)$ .

Proof Lemma 2.7 shows that for any $m \in { \mathbb Z}$ and any $f \in W^{n/2+\epsilon }({ \mathbb R}^n)$ , $\pi _{m, \boldsymbol {\tau }} f$ is continuous on ${ \mathbb R}^{n-1}$ . Moreover,

$$ \begin{align*} \pi_{m, \boldsymbol{\tau}}\tilde{\mu}(\textrm{ exp}(Y_{\boldsymbol{\tau}}))f(z)& = \pi_{m, \boldsymbol{\tau}} f(z - (\tau, 0, \ldots, 0))\\ & = \pi_{m, \boldsymbol{\tau}}f(z).\\ \end{align*} $$

For any $s> 2$ , define

$$ \begin{align*} \textrm{ Ann}_{\boldsymbol{\tau}} := \{f \in \mathscr{S}({ \mathbb R}^n) : \pi_{m, \boldsymbol{\tau}}(f) \equiv 0 \text{ for all }m \in { \mathbb Z} \}. \end{align*} $$

Proposition 2.10. For any $f \in \textrm{ Ann}_{\boldsymbol {\tau }}$ , the cohomological equation

(31) $$ \begin{align} L_{\boldsymbol{\tau}} P = f \end{align} $$

has a unique solution P in $L^2({ \mathbb R}^n)$ and, moreover, for any $\epsilon> 0$ , there is a constant $C_{\epsilon }> 0$ such that for any $s \geq 0$ ,

$$ \begin{align*} \Vert P \Vert_s \leq \frac{C_{\epsilon}}{\tau} \Vert f \Vert_{ s + 3n/2 + 1 + \epsilon}. \end{align*} $$

Proof Let $s \geq 0$ , $f \in \mathscr {S}({ \mathbb R}^n)$ and $T: \mathscr {S}({ \mathbb R}^n) \to \mathscr {S}({ \mathbb R}^n)$ be the linear map

(32) $$ \begin{align} T f(z) = \sum_{m = 1}^{\infty} f(z_1 + m \tau, z_2, \ldots, z_n), \end{align} $$

which converges absolutely and uniformly on compact sets.

By (24), the cohomological equation (31) is

$$ \begin{align*} P(z - (\tau, 0, \ldots, 0)) - P(z) = f(z). \end{align*} $$

Clearly, there is at most one $L^2({ \mathbb R}^n)$ solution P to the above equation. Because $T f$ is a solution,

$$ \begin{align*} L_{\boldsymbol{\tau}} Tf = f \end{align*} $$

on ${ \mathbb R}^n$ .

Because $f \in \textrm{ Ann}_{\boldsymbol {\tau }}$ , the Poisson summation formula gives that for any $z \in { \mathbb R}^n$ ,

$$ \begin{align*} \sum_{m \in { \mathbb Z}} f(z_1 + m \tau, z_2, \ldots, z_n) = 0. \end{align*} $$

By combining the above equality with (32), we get that

(33) $$ \begin{align} T f(z) = \sum_{m = 0}^{\infty} f(z_1 - m\tau, z_2, \ldots, z_n), \end{align} $$

which is again convergent.

Now we estimate the homogeneous norm $ |\mkern -1.5mu|\mkern -1.5mu| T f |\mkern -1.5mu|\mkern -1.5mu|_s$ . By (25) and formula (32), we get that for all $z\in ({ \mathbb R}^+\cup \{0\}) \times { \mathbb R}^{n-1}$ ,

(34) $$ \begin{align} &\kern-15pt\bigg\vert \bigg(I + \sum_{i= 1}^n z_i^2-\frac{\partial}{\partial z_i^2}\bigg)^{s/2} T f (z) \bigg\vert\qquad\qquad\qquad\qquad\qquad\qquad\qquad \end{align} $$

(35) $$ \begin{align} & \leq \sum_{m = 0}^{\infty} \bigg\vert \bigg(I + \sum_{i= 1}^n z_i^2-\frac{\partial}{\partial z_i^2}\bigg)^{s/2} f(z_1+m\tau, z_2, \ldots, z_n) \bigg\vert \notag \\ & \leq \sum_{m = 0}^{\infty} \bigg|\bigg(I - \frac{\partial^2}{\partial (z_1+m\tau)^2} +(z_1+m\tau)^2+ \sum_{i=2}^n z_i^2 - \frac{\partial^2}{\partial z_i^2}\bigg)^{s/2} \notag \\ & \qquad\qquad\times f(z_1+m\tau, z_2, \ldots, z_n)\bigg\vert. \end{align} $$

Let $r = n + 1 + 2\epsilon $ , so equation (25) gives

Using (33), we get by a completely analogous argument that for all $z \in { \mathbb R}^- \times { \mathbb R}^{n-1}$ ,

$$ \begin{align*} \bigg\vert \bigg(I + \sum_{i= 1}^n z_i^2 - \frac{\partial}{\partial z_i^2}\bigg)^{s/2} T f (z) \bigg\vert\leq \frac{C_{\epsilon}}{\tau} \bigg(1 + \sum_{i = 1}^n z_i^2\bigg)^{-(n + \epsilon)/2} |\mkern-1.5mu|\mkern-1.5mu| f |\mkern-1.5mu|\mkern-1.5mu|_{s + 3n/2 +1+ 3\epsilon}. \end{align*} $$

It follows that

$$ \begin{align*} |\mkern-1.5mu|\mkern-1.5mu| T f |\mkern-1.5mu|\mkern-1.5mu|_{s} \leq \frac{C_{\epsilon}}{\tau} |\mkern-1.5mu|\mkern-1.5mu| f |\mkern-1.5mu|\mkern-1.5mu|_{s + 3n/2 +1+ 3\epsilon}. \end{align*} $$

Because $s \geq 0$ was arbitrary, we have shown that the above estimate in terms of the norm that is homogeneous in h holds for any $s \geq 0$ . Then we apply Lemma 2.3 to get the estimate with respect to the Laplacian. So,

$$ \begin{align*} \Vert T f \Vert_{s} \leq \frac{C_{\epsilon}}{\tau} \Vert f \Vert_{s + 3n/2 +1+ 3\epsilon}. \end{align*} $$

The lemma is now proved with $P = T f$ .

Now we find a solution with Sobolev estimates to the equation $L_{\boldsymbol \eta } P = f. $ For any $m \in { \mathbb Z}$ , define $\pi _{m, \boldsymbol \eta }$ to be the formal operator

$$ \begin{align*} \pi_{m, \boldsymbol\eta} f(z_1, z_3, \ldots, z_n) := f\bigg(z_1, \frac{2\pi m}{\nu_2}, z_3, \ldots, z_n\bigg). \end{align*} $$

We get as in Corollary 2.8 that for any $s \geq 0$ and $\epsilon> 0$ , $\pi _{m, \boldsymbol \eta } : W^{s+3n/2-1+\epsilon }({ \mathbb R}^n) \to W^{s}({ \mathbb R}^{n-1})$ and, by Lemma 2.7 that for any $f \in W^{n/2 + \epsilon }({ \mathbb R}^n)$ , $\pi _{m, \boldsymbol \eta } f$ is continuous on ${ \mathbb R}^{n-1}$ .

As in Lemma 2.9, it can be immediately verified that for any $m \in { \mathbb Z}$ , $\pi _{m, \boldsymbol \eta }$ is invariant for the operator $\tilde {\mu }(\textrm{ exp}(Y_{\boldsymbol \eta }))$ . Define

$$ \begin{align*} \textrm{ Ann}_{\boldsymbol\eta} := \{f \in \mathscr{S}({ \mathbb R}^n) : \pi_{m, \boldsymbol{\eta}}(f) \equiv 0 \text{ for all }m \in { \mathbb Z}\}. \end{align*} $$

We have a corresponding estimate for the cohomological equation $L_{\boldsymbol {\eta }} P = f$ .

Corollary 2.11. For any $f \in \textrm{ Ann}_{\boldsymbol {\eta }}$ , the equation

$$ \begin{align*} L_{\boldsymbol{\eta}} P = f \end{align*} $$

has a unique solution P in $L^2({ \mathbb R}^n)$ and, moreover, for any $\epsilon> 0$ , there is a constant $C_{\epsilon }> 0$ such that for any $s \geq 0$ ,

$$ \begin{align*} \Vert P \Vert_s \leq \frac{C_{\epsilon}}{\nu_2} \Vert f \Vert_{s + 3n/2 + 1 + \epsilon}. \end{align*} $$

Proof Writing (24) as a Fourier transform, we get

(36) $$ \begin{align} \mathcal F_2 L_{\boldsymbol{\eta}} f_{\mathbf z_3}(z_1, \omega_2) = \mathcal F_2 f_{\mathbf z_3}\bigg(z_1, \omega_2 - \frac{\nu_2}{2\pi}\bigg) - \mathcal F_2 f_{\mathbf z_3}(z_1, \omega_2). \end{align} $$

Then, setting $\tau = ({\nu _2}/{2\pi })$ , the corollary follows in the same way as Proposition 2.10, using the decay estimate (26) in place of (25).

Next, we prove Theorem 1.2 for Schrödinger representations.

Theorem 2.12. For any $f, g \in \mathscr {S}({ \mathbb R}^n)$ that satisfy $L_{\boldsymbol {\tau }} g = L_{\boldsymbol {\eta }} f$ , there is a solution $P \in \mathscr {S}({ \mathbb R}^n)$ such that

$$ \begin{align*} L_{\boldsymbol{\tau}} P = f \quad\text{and}\quad L_{\boldsymbol{\eta}} P = g. \end{align*} $$

Moreover, for any $\epsilon> 0$ , there is a constant $C_{\epsilon }> 0$ such that for any $s \geq 0$ ,

$$ \begin{align*} \Vert P \Vert_{s} \leq \frac{C_{\epsilon}}{\tau} \Vert f \Vert_{s + 3n/2 + 1 + \epsilon}. \end{align*} $$

Proof Let $m \in { \mathbb Z}$ . Because $\pi _{m, \boldsymbol {\tau }}$ is invariant for $\tilde {\mu }(\textrm{ exp}(Y_{\boldsymbol {\tau }}))$ , we have that

$$ \begin{align*} 0 \equiv \pi_{m, \boldsymbol{\tau}} L_{\boldsymbol{\tau}} g = \pi_{m, \boldsymbol{\tau}}L_{\boldsymbol{\eta}} f. \end{align*} $$

From the formulas for $\pi _{m, \boldsymbol {\tau }}$ and $L_{\boldsymbol \eta }$ , see (29) and (24), respectively, we get

$$ \begin{align*} [L_{\boldsymbol\eta}, \pi_{m, \boldsymbol{\tau}}] = 0. \end{align*} $$

Moreover, for any $(z_2, \ldots , z_n) \in { \mathbb R}^{n-1}$ ,

$$ \begin{align*} 0 = L_{\boldsymbol{\eta}} \pi_{m, \boldsymbol{\tau}}f (z_2, \ldots, z_n) = [\textrm{ exp}( i \nu_2 z_2) - 1] \mathcal F_1f\bigg(\frac{m}{\tau}, z_2, \ldots, z_n\bigg). \end{align*} $$

So, we get that off a countable set of $z_2 \in { \mathbb R}$ ,

$$ \begin{align*} \mathcal F_1f\bigg(\frac{m}{\tau}, z_2, \ldots, z_n\bigg) = 0. \end{align*} $$

Lemma 2.7 shows that $\mathcal F_1f$ is continuous, which implies that $\pi _{m, \boldsymbol {\tau }} f \equiv 0$ . Because $m \in { \mathbb Z}$ was arbitrary, we conclude that $f \in \textrm{ Ann}_{\boldsymbol {\tau }}$ .

Proposition 2.10 now implies that there is a unique function P in $L^2({ \mathbb R}^n)$ that is a solution to

$$ \begin{align*} L_{\boldsymbol{\tau}} P = f \end{align*} $$

and, for any $\epsilon> 0$ , there is a constant $C_\epsilon> 0$ such that

$$ \begin{align*} \Vert P \Vert_{s} \leq \frac{C_{\epsilon}}{\tau} \Vert f \Vert_{s + 3n/2 + 1 + \epsilon}. \end{align*} $$

Finally, because $[Y_{\boldsymbol \eta }, Y_{\boldsymbol {\tau }}] = 0$ , we use $L_{\boldsymbol {\tau }} g = L_{\boldsymbol {\eta }} f$ and get

$$ \begin{align*} L_{\boldsymbol{\tau}} g = L_{\boldsymbol{\eta}} L_{\boldsymbol{\tau}} P = L_{\boldsymbol{\tau}} L_{\boldsymbol{\eta}} P. \end{align*} $$

So, $L_{\boldsymbol {\tau }}(g- L_{\boldsymbol {\eta }} P) = 0$ and, because $g - L_{\boldsymbol {\eta }}P \in L^2({ \mathbb R}^n)$ , it follows that

$$ \begin{align*} g = L_{\boldsymbol{\eta}} P \end{align*} $$

in $L^2({ \mathbb R}^n)$ .

Now we will prove Theorem 1.3 in the case of Schrödinger representations. Recall from Lemma 2.6 that $\nu _2 \neq 0$ .

Theorem 2.13. For any $f, g, \phi \in \mathscr {S}({ \mathbb R}^n)$ that satisfy $L_{\boldsymbol {\eta }} f - L_{\boldsymbol {\tau }} g = \phi $ , there exists a non-constant function $P \in \mathscr {S}({ \mathbb R}^n)$ such that the following holds. For any $s \geq 0$ and for any $\epsilon> 0$ , there is a constant $C_{s, \epsilon }> 0$ such that

$$\begin{gather*} \|g - L_{\boldsymbol{\eta}} P\|_s \leq C_{s, \epsilon} (\tau^{-1} + \tau^{\epsilon}) \| \phi\|_{s + 3n + 1 + 2\epsilon}, \\ \|f - L_{\boldsymbol{\tau}} P\|_s \leq \frac{C_{s, \epsilon}}{\nu_2} (1 + \tau^{1+\epsilon}) \| \phi\|_{s + 3n + 1 + 2\epsilon} , \\ \| P \|_s \leq C_{s, \epsilon} (\tau^{-1} + \tau^{\epsilon}) (\| f\|_{s + 3n + 1 + \epsilon} + \| g\|_{s + 3n + 1 + 2\epsilon}). \end{gather*}$$

Proof Notice that if $f = g = 0$ , then $\phi = 0$ and the above statement holds trivially. Without loss of generality, we assume that $f \neq 0$ .

Let $\psi \in \mathscr {S}({ \mathbb R})$ , whose Fourier transform is compactly supported on $[-({1}/{2\tau }), ({1}/{2 \tau })]$ and satisfies $\mathcal F\psi (0) = 1$ . For each $m \in { \mathbb Z}$ , define the functional $\Pi _{m, \boldsymbol {\tau }}$ on $L^2({ \mathbb R}^{n-1})$ by

(37) $$ \begin{align} \Pi_{m, \boldsymbol{\tau}} F(z_2, \ldots, z_n) = e^{2\pi i z_1 m/\tau} \psi(z_1) F(z_2, \ldots, z_n). \end{align} $$

Below, $\Pi _{m, \boldsymbol {\tau }}$ will be a component of an operator that maps sufficiently regular functions to coboundaries. See (39) and Lemma 2.16 for details. For now, we have the following estimate.

Lemma 2.14. For any $s \in 2 { \mathbb N}$ , for any $m \in { \mathbb Z}$ and for any $F \in W^s({ \mathbb R}^{n-1})$ , there is a constant $C_s> 0$ such that

$$ \begin{align*} \kern-25pt|\mkern-1.5mu|\mkern-1.5mu| \Pi_{m, \boldsymbol{\tau}} F |\mkern-1.5mu|\mkern-1.5mu|_s \leq C_s \Vert \bigg(I + z_1^2- \frac{\partial^2}{\partial z_1^2}\bigg)^{s/2} \psi \Vert_{L^{\infty}({ \mathbb R})} \sum_{k = 0}^{s/2} \bigg(1+\bigg\vert\frac{m}{\tau}\bigg\vert\bigg)^{2k} \vert F \vert_{s-2k}.\\ \end{align*} $$

Proof Because $\psi $ is supported on $[-1/2, 1/2]$ , we have

(38) $$ \begin{align} |\mkern-1.5mu|\mkern-1.5mu| \Pi_{m, \tau} F |\mkern-1.5mu|\mkern-1.5mu|_s & = \bigg\Vert \bigg(\bigg(z_1^2 - \frac{\partial^2}{\partial z_1^2}\bigg)+\bigg(I +\sum_{i = 2}^n z_i^2-\frac{\partial^2}{\partial z_i^2}\bigg)\bigg)^{s/2} (e^{- 2\pi i z_1 m/\tau} \psi F)\bigg\Vert_{L^2({ \mathbb R}^n)}. \end{align} $$

Then, because $z_1^2 - ({\partial ^2}/{\partial z_1^2})$ and $(I +\sum _{i = 2}^n z_i^2-({\partial ^2}/{\partial z_i^2}))$ commute, the triangle inequality gives

$$ \begin{align*} (38) &\leq C_s \sum_{k = 0}^{s/2} \bigg\Vert \bigg(z_1^2 - \frac{\partial^2}{\partial z_1^2}\bigg)^k \bigg(I +\sum_{i = 2}^n z_i^2-\frac{\partial^2}{\partial z_i^2}\bigg)^{s/2-k} (e^{- 2\pi i z_1 m/\tau} \psi F) \bigg\Vert_{L^2({ \mathbb R}^n)} \notag \\ & \leq C_s \sum_{k = 0}^{s/2} \bigg\Vert \bigg(z_1^2 - \frac{\partial^2}{\partial z_1^2}\bigg)^k e^{- 2\pi i z_1m/\tau} \psi \bigg\Vert_{L^{\infty}({ \mathbb R})} \bigg\Vert \bigg(I +\sum_{i = 2}^n z_i^2-\frac{\partial^2}{\partial z_i^2}\bigg)^{s/2-k} F \bigg\Vert_{L^2({ \mathbb R}^{n-1})}. \notag\\ \end{align*} $$

Note that $\Pi _{m, \boldsymbol {\tau }}$ depends on $\psi $ . Then we formally define the operator $R_\psi $ on $L^2({ \mathbb R}^n)$ by

(39) $$ \begin{align} R_\psi := I - \sum_{m \in { \mathbb Z}} \Pi_{m, \boldsymbol{\tau}} \pi_{m, \boldsymbol{\tau}}. \end{align} $$

Over the next two lemmas, we describe properties of $R_\psi $ .

Lemma 2.15. For any $s \geq 0$ and for any $\epsilon> 0$ , there is a constant $C_{s, \epsilon }> 0$ such that for any non-zero $f \in W^{s+n/2+\epsilon }({ \mathbb R}^n)$ , we can choose $\psi $ such that $R_\psi f \neq 0$ and

$$ \begin{align*} \Vert R_\psi f\Vert_s \leq C_{s, \epsilon} (1 + \tau^{s + 1 + \epsilon}) \Vert f \Vert_{s + 3n/2 + \epsilon}. \end{align*} $$

Proof We first claim that we can choose $\psi $ such that $R_{\psi } f \neq 0$ and, for some universal constant $C_s^{(0)}> 0$ ,

$$ \begin{align*} \bigg\Vert \bigg(I + z_1^2- \frac{\partial^2}{\partial z_1^2}\bigg)^{s/2} \psi \bigg\Vert_{L^{\infty}({ \mathbb R})} \leq C_s^{(0)} (1 + \tau^s). \end{align*} $$

Fix $\psi $ . So, for some $C_s^{(0)}> 0$ , the above estimate holds. If $R_\psi f \neq 0$ , then the claim is holds, so suppose that $R_\psi f = 0$ . Hence,

$$ \begin{align*} f(z) = \psi(z_1) \sum_{m \in { \mathbb Z}}\textrm{ exp}(2\pi i z_1 m/\tau) \mathcal F_1 f\bigg(\frac{m}{\tau}, z_2, \ldots, z_n\bigg). \end{align*} $$

So, we can perturb $\psi $ to a smooth function compactly supported in $[-({1}/{2\tau }), ({1}/{2\tau })]$ satisfying $\mathcal F\psi _0(0) = 1, \psi _0 \neq \psi $ and $R_\psi f \neq 0$ , where also

(40) $$ \begin{align} \bigg\Vert \bigg(I + z_1^2 - \frac{\partial^2}{\partial z_1^2}\bigg)^{s/2} \psi_0 \bigg\Vert_{L^{\infty}({ \mathbb R})} \leq (C_s^{(0)} + 1) (1 + \tau^s). \end{align} $$

This proves the claim.

Now we say that $s \in { \mathbb N}$ is even and let $R_\psi f \neq 0$ , where $\psi $ satisfies (40). By the triangle inequality and Lemma 2.14, we get a constant $C_s^{(1)}> C_s^{(0)} + 1$ such that

(41) $$ \begin{align} |\mkern-1.5mu|\mkern-1.5mu| R_\psi f |\mkern-1.5mu|\mkern-1.5mu|_s & \leq |\mkern-1.5mu|\mkern-1.5mu| f |\mkern-1.5mu|\mkern-1.5mu|_s + \sum_{m \in { \mathbb Z}} |\mkern-1.5mu|\mkern-1.5mu| \Pi_{m, \tau} \pi_{m, \tau} f |\mkern-1.5mu|\mkern-1.5mu|_s \notag \\ & \leq |\mkern-1.5mu|\mkern-1.5mu| f |\mkern-1.5mu|\mkern-1.5mu|_s + C_s^{(1)} (1 + \tau^s) \sum_{m \in { \mathbb Z}} \sum_{k = 0}^{s/2} \bigg(1+\bigg\vert\frac{m}{\tau}\bigg\vert\bigg)^{2k} \vert \pi_{m, \tau} f \vert_{s-2k}. \end{align} $$

By Corollary 2.8, there is a constant $C_\epsilon> 0$ such that for any $m \in { \mathbb Z}$ ,

$$ \begin{align*} \bigg(1+\bigg\vert\frac{m}{\tau}\bigg\vert\bigg)^{2k} \vert \pi_{m, \tau} f \vert_{s-2k} & \leq C_{\epsilon} \bigg(1+\bigg\vert\frac{m}{\tau}\bigg\vert\bigg)^{2k} \bigg(1+\bigg\vert\frac{m}{\tau}\bigg\vert\bigg)^{-(2k + 1 + \epsilon)} |\mkern-1.5mu|\mkern-1.5mu| f |\mkern-1.5mu|\mkern-1.5mu|_{s + 3n/2 + \epsilon} \notag \\ & \leq C_\epsilon (1 + \tau)^{s+ 1 + \epsilon} (1 + \vert m\vert)^{-(1 + \epsilon)} |\mkern-1.5mu|\mkern-1.5mu| f |\mkern-1.5mu|\mkern-1.5mu|_{s + 3n/2 + \epsilon}. \notag \end{align*} $$

Hence, there is a constant $C_{s, \epsilon }> 0$ such that

$$ \begin{align*} \begin{array}{l@{\,}l} (41) & \leq |\mkern-1.5mu|\mkern-1.5mu| f |\mkern-1.5mu|\mkern-1.5mu|_s + C_{s, \epsilon}(1 + \tau)^{s+ 1 + \epsilon} \displaystyle\sum\limits_{m \in { \mathbb Z}} (1 + m^2)^{-(1+\epsilon)/2} |\mkern-1.5mu|\mkern-1.5mu| f |\mkern-1.5mu|\mkern-1.5mu|_{s+3n/2+\epsilon} \\ & \leq C_{s, \epsilon} (1 + \tau)^{s + 1 + \epsilon} |\mkern-1.5mu|\mkern-1.5mu| f |\mkern-1.5mu|\mkern-1.5mu|_{s+3n/2+\epsilon}. \end{array} \end{align*} $$

By interpolation, the above estimate holds for all $s \geq 0$ . Because $R_\psi $ is a linear operator and the estimate holds for all s, Lemma 2.3 gives the estimate for $s \geq 0$ .

Next we show that the operator $R_\psi $ is a projection into $\textrm{ Ann}_{\boldsymbol {\tau }}$ and it commutes with $L_{\boldsymbol {\eta }}$ .

Lemma 2.16. Let $\psi $ be as in the previous lemma. Then

$$ \begin{align*} R_\psi: \mathscr{S}({ \mathbb R}^n) \to \textrm{ Ann}_{\boldsymbol{\tau}} \end{align*} $$

and

$$ \begin{align*} R_\psi L_{\boldsymbol{\tau}} = L_{\boldsymbol{\tau}} , \quad [R_\psi, L_{\boldsymbol{\eta}}] = 0 \end{align*} $$

on $\mathscr {S}({ \mathbb R}^n)$ .

Proof Let $f \in \mathscr {S}({ \mathbb R}^n)$ . By the previous lemma, $R_{\psi } f \in \mathscr {S}({ \mathbb R}^n)$ , so we need to show that $R_{\psi } f$ is in the kernel of every $\pi _{m, \boldsymbol {\tau }}$ . Using the property that $\mathcal F\psi $ is supported on the interval $[-({1}/{2\tau }), ({1}/{2\tau })]$ and $\mathcal F\psi (0) = 1$ , we get that for any $m \in { \mathbb Z}$ ,

(42) $$ \begin{align} \pi_{m, \tau} R_\psi f & = \pi_{m, \tau}f - \sum_{k \in { \mathbb Z}} \pi_{m, \tau} \Pi_{k, \tau} \pi_{k, \tau}f \notag \\ & = \pi_{m, \tau}f - \pi_{m, \tau} \Pi_{m, \tau} \pi_{m, \tau}f \notag \\ & = \pi_{m, \tau}f - \pi_{m, \tau}f \notag \\ & = 0. \end{align} $$

This implies that $R_{\psi } :\mathscr {S}({ \mathbb R}^n) \to \textrm{ Ann}_{\boldsymbol {\tau }}$ .

By Lemma 2.9, for any $m \in { \mathbb Z}$ , $\pi _{m, \boldsymbol {\tau }} L_{\boldsymbol {\tau }} = 0$ . We have

$$ \begin{align*} R_\psi L_{\boldsymbol{\tau}} & = \bigg(I - \sum_{m \in { \mathbb Z}} \Pi_{m, \tau} \pi_{m, \tau}\bigg) L_{\boldsymbol{\tau}} \\ & = L_{\boldsymbol{\tau}}. \end{align*} $$

Finally, we prove that $[L_{\boldsymbol {\eta }}, R_\psi ] = 0$ . We have

$$ \begin{align*} \Pi_{m, \boldsymbol{\tau}} \pi_{m, \boldsymbol{\tau}} L_{\boldsymbol{\eta}} f(z) & = \textrm{ exp}(2\pi i z_1 m/\tau)\, \psi(z_1) [\pi_{m, \boldsymbol{\tau}} L_{\boldsymbol{\eta}} f](z_2, \ldots, z_n) \\ & = \textrm{ exp}(2\pi i z_1 m/\tau)\, \psi(z_1) \mathcal F_1[L_{\boldsymbol{\eta}} f]\bigg(\frac{m}{\tau}, z_2, \ldots, z_n\bigg) \\ & = (\textrm{ exp}(i \nu_2 z_2) - 1) \,\textrm{ exp}(2\pi i z_1 m/\tau)\, \psi(z_1) \mathcal F_1f\bigg(\frac{m}{\tau}, z_2, \ldots, z_n\bigg) \\ & = L_{\boldsymbol{\eta}} \,\textrm{ exp}(2\pi i z_1 m/\tau)\, \psi(z_1) \mathcal F_1 f\bigg(\frac{m}{\tau}, z_2, \ldots, z_n\bigg) \\ & = L_{\boldsymbol{\eta}} \Pi_{m, \boldsymbol{\tau}} \mathcal F_1 f\bigg(\frac{m}{\tau}, z_2, \ldots, z_n\bigg) \\ & = L_{\boldsymbol{\eta}} \Pi_{m, \boldsymbol{\tau}} \pi_{m, \boldsymbol{\tau}} f. \end{align*} $$

This proves that $[L_{\boldsymbol {\eta }}, R_\psi ] = 0$ and finishes the proof of the lemma.

Now we prove Theorem 2.13. Let

$$ \begin{align*} \phi = L_{\boldsymbol{\eta}} f - L_{\boldsymbol{\tau}} g \end{align*} $$

be as in the theorem and recall from the beginning of its proof that we take $f \neq 0$ . By Lemmas 2.15 and 2.16, we can choose $\psi $ such that there is a non-constant function P that is a solution to $R_\psi f = L_{\boldsymbol {\tau }} P$ and, for a fixed constant $C_s^{(1)}> 0$ ,

$$ \begin{align*} \bigg\Vert \bigg(I - \frac{\partial^2}{\partial z_1^2}\bigg)^{s/2} \psi \bigg\Vert_{L^{\infty}({ \mathbb R})} \leq C_s^{(1)}. \end{align*} $$

In particular, Lemma 2.16 implies that

(43) $$ \begin{align} R_\psi \phi & = R_\psi L_{\boldsymbol{\eta}} f - R_\psi L_{\boldsymbol{\tau}} g \nonumber\\ & = L_{\boldsymbol{\eta}} R_\psi f - L_{\boldsymbol{\tau}} g \nonumber\\ & = L_{\boldsymbol{\eta}} L_{\boldsymbol{\tau}} P - L_{\boldsymbol{\tau}} g\nonumber\\ & = L_{\boldsymbol{\tau}} (L_{\boldsymbol{\eta}} P - g). \end{align} $$

Then, by Proposition 2.10 and by Lemmas 2.15 and 2.5, we get that for any $s \geq 0$ and for any $\epsilon> 0$ , there is a constant $C_{s, \epsilon }> 0$ such that

(44) $$ \begin{align} \Vert L_{\boldsymbol{\eta}} P - g \Vert_s & = \Vert L_{\boldsymbol{\eta}} P - g \Vert_s \end{align} $$

(45) $$ \begin{align} & \kern129pt\leq \frac{C_{s, \epsilon}}{\tau} \Vert R_\psi \phi \Vert_{s + 3n/2 + 1 + \epsilon} \notag \\ & \kern129pt\leq C_{s, \epsilon} (\tau^{-1} + \tau^{s+\epsilon}) \Vert \phi \Vert_{s + 3n + 1 + 2\epsilon}. \end{align} $$

To estimate $\Vert L_{\boldsymbol {\tau }} P - f \Vert _s$ , because $R_\psi f = L_{\boldsymbol {\tau }} P$ and by Lemma 2.5,

(46) $$ \begin{align} \Vert L_{\boldsymbol{\tau}} P - f \Vert_s & = \Vert L_{\boldsymbol{\tau}} P - f \Vert_s \notag \\ & = \Vert L_{\boldsymbol{\tau}} P - R_\psi f + R_\psi f - f\Vert_s \notag \\ & = \Vert (R_\psi -I) f\Vert_s. \end{align} $$

Notice that by Lemma 2.16,

$$ \begin{align*} (R_\psi - I) L_{\boldsymbol{\tau}} g = 0. \end{align*} $$

Then, using $L_{\boldsymbol {\eta }} f - L_{\boldsymbol {\tau }} g = \phi $ , we get

(47) $$ \begin{align} (R_\psi - I) L_{\boldsymbol{\eta}} f & = (R_\psi - I) \phi + (R_\psi - I) L_{\boldsymbol{\tau}} g \nonumber\\ & = (R_\psi - I) \phi. \end{align} $$

Then, by Lemma 2.16 again, we get

$$ \begin{align*} L_{\boldsymbol{\eta}} (R_\psi - I) f = (R_\psi - I) \phi. \end{align*} $$

We conclude by Corollary 2.11 and Lemmas 2.15 and 2.5 that

(48) $$ \begin{align} (46) & \leq \frac{C_{s, \epsilon}}{\nu_2} \Vert (R_\psi - I) \phi \Vert_{s + 3n/2 + 1 + \epsilon} \notag \\ & \leq \frac{C_{s, \epsilon}}{\nu_2} (1 + \tau^{s + 1+\epsilon}) \Vert \phi \Vert_{s + 3n + 1 + 2\epsilon}. \end{align} $$

Finally, because $L_{\boldsymbol {\tau }} P = R_\psi f$ , Proposition 2.10 and Lemma 2.15 give

$$ \begin{align*} \Vert P \Vert_s & \leq \frac{C_{s, \epsilon}}{\tau} \Vert R_\psi f \Vert_{s + 3n/2 + 1 + \epsilon} \notag \\ & \leq C_{s, \epsilon} (\tau^{-1} + \tau^{s + \epsilon}) \Vert f \Vert_{s + 3n + 1 + 2\epsilon} \notag. \notag \end{align*} $$

At the start of the proof of Theorem 2.13, we assumed that $f \neq 0$ . If we instead choose $g \neq 0$ , then by first applying the Fourier transform, the same argument proves the above estimates in terms of $\Vert \phi \Vert _{ s + 3n + 1 + \epsilon }$ and

$$ \begin{align*} \Vert P \Vert_s \leq C_{s, \epsilon} (\tau^{-1} + \tau^{s + \epsilon}) \Vert g \Vert_{s + 3n + 1 + 2\epsilon}. \end{align*} $$

This completes the proof of Theorem 2.13.

Proof Proof of Theorem 1.2

The regular representation of $\mathsf H$ on $L^2(M)$ decomposes as

$$ \begin{align*} L^2(M) = \mathbb C \langle 1 \rangle \oplus \bigoplus_{\mathbf{m} \in { \mathbb Z}^{2n} \setminus\{0\}} \mathcal P_{\mathbf{m}} \oplus \bigoplus_{h \in { \mathbb Z}} \mathcal P_h, \end{align*} $$

where each $\mathcal P_{\mathbf {m}}$ is an abelian representation of ${ \mathbb R}^{2n}$ equivalent to a character given by (11), and each $\mathcal P_h$ is equivalent to a countable collection of Schrödinger representations $\mu _h$ of $\mathsf H$ on $L^2({ \mathbb R}^n)$ given by (13). The subspace of zero-average functions in $L^2(M)$ is denoted $L_0^2(M)$ , which therefore decomposes as

$$ \begin{align*} L_0^2(M) & = \bigoplus_{\mathbf{m} \in { \mathbb Z}^{2n} \setminus\{0\}} \mathcal P_{\mathbf{m}} \oplus \bigoplus_{h \in { \mathbb Z}} \mathcal P_h \\ & = L_0^2({ \mathbb T}^{2n}) \oplus \bigoplus_{h \in { \mathbb Z}} \mathcal P_h. \end{align*} $$

As indicated in §2.1, vector fields in $\operatorname {\mathfrak h}$ split into the unitary components in the above Hilbert space. Then the decomposition of the Sobolev space of s-differentiable, zero-average functions is

(49) $$ \begin{align} W_0^s(M) = W_0^s({ \mathbb T}^{2n}) \oplus \bigoplus_{h \in { \mathbb Z}} W^s(\mathcal P_h), \end{align} $$

where $W_0^s({ \mathbb T}^{2n})$ and $W^s(\mathcal P_h)$ are s-order Sobolev spaces on the torus ${ \mathbb T}^{2n}$ and of the representation $\mathcal P_h$ , respectively.

Now, in Theorem 1.2, we are given zero-average functions $f, g \in C^{\infty }(M)$ that satisfy $L_{\boldsymbol \tau } g = L_{\boldsymbol \eta } f$ and we aim to find a solution $P \in C^{\infty }(M)$ such that

$$ \begin{align*} L_{\boldsymbol \tau} P = f , \quad L_{\boldsymbol \eta} P = g. \end{align*} $$

Write

$$ \begin{align*} f = f_t \oplus \bigoplus_{h \in { \mathbb Z}} f_h , \quad g = g_t \oplus \bigoplus_{h \in { \mathbb Z}} g_h, \end{align*} $$

where $f_t, g_t \in W_0^{\infty }({ \mathbb T}^{2n})$ and, for each $h \in { \mathbb Z}$ , $f_h, g_h \in \mathcal P_h$ . Proposition 2.1 and Theorem 2.12 give smooth solutions $P_t \in W_0^{\infty }({ \mathbb T}^{2n})$ and $\{P_h\}_{h \in { \mathbb Z}} \subset W^{\infty }(\mathcal P_h)$ satisfying the estimate of Theorem 1.2 in the finite- and infinite-dimensional representations, respectively. Define $P \in C^{\infty }(M)$ by

$$ \begin{align*} P := P_t \oplus \bigoplus_{h \in { \mathbb Z}} P_h. \end{align*} $$

So, there exists a constant $C_{s, \epsilon }:= C_{s, \epsilon , \boldsymbol {\tau }, \boldsymbol {\eta }}> 0$ such that

(50) $$ \begin{align} \Vert P \Vert_{W^s(M)}^2 & = \Vert P_t \Vert_{s}^2 + \sum_{h \in { \mathbb Z}} \Vert P_h \Vert_{s}^2 \nonumber\\ & \leq C_{s, \epsilon} (\Vert f_t \Vert_{s + 2\gamma} + \Vert g_t \Vert_{s + 2 \gamma})^2 + C_{s, \epsilon} \sum_{h \in { \mathbb Z}} \Vert f_h \Vert_{s + 3n/2 + 1 + \epsilon}^2 \nonumber\\ & \leq C_{s, \epsilon} (\Vert f \Vert_{s + \max\{2\gamma, \ 3n/2 + 1 + \epsilon\}} + \Vert g \Vert_{s + 2 \gamma})^2.\nonumber\\ \end{align} $$

Proof Proof of Theorem 1.3

This follows from Proposition 2.2 and Theorem 2.13 as in the proof of Theorem 1.2.

3.1 Constant cohomology for the discrete-time action

We have

$$ \begin{align*} D e^{Y_{\boldsymbol \tau} }(X_1) = X_1, \ \ D e^{Y_{\boldsymbol \tau}}(X_2) = X_2, \,\,\, D e^{Y_{\boldsymbol \tau}}(Z) = Z. \end{align*} $$

Furthermore,

$$ \begin{align*} e^{Y_{\boldsymbol \tau}} e^{t \Lambda_1} & = (e^{Y_{\boldsymbol \tau}} e^{t \Lambda_1} e^{-Y_{\boldsymbol \tau}}) e^{Y_{\boldsymbol \tau}} \\ & = \textrm{ exp}(e^{\text{ad}_{Y_{\boldsymbol \tau}}} t \Lambda_1) e^{Y_{\boldsymbol \tau}} = \textrm{ exp}(t \Lambda_1 + t [Y_{\boldsymbol \tau}, \Lambda_1]) e^{Y_{\boldsymbol \tau}}\\ & = \textrm{ exp}(t (\Lambda_1 + \tau_1Z)) e^{Y_{\boldsymbol \tau}}. \end{align*} $$

Therefore, for a constant vector field

$$ \begin{align*} H= h_1X_1+ h_2X_2+ h_3{\Lambda}_1+ h_4{\Lambda}_2+ h_5Z, \end{align*} $$

where $h_i$ are constants, we have

$$ \begin{align*}D e^{Y_\tau}(H)= h_1X_1+ h_2X_2+ h_3{\Lambda}_1+ h_4{\Lambda}_2+ (h_5-h_3\tau_1-h_4\tau_2)Z.\end{align*} $$

Another way to write this is

$$ \begin{align*}D e^{Y_{\boldsymbol \tau}}(H)= H+ [Y_{\boldsymbol \tau}, H].\end{align*} $$

A similar computation can be done for $D e^{Y_{\boldsymbol \eta }}$ . In the matrix form, we have

$$ \begin{align*} D e^{Y_{\boldsymbol \tau}} = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & \tau_1 & \tau_2 & 1 \\ \end{pmatrix}, \quad D e^{Y_{\boldsymbol \eta}} = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ -\eta_1 & -\eta_2 & 0 &0 & 1 \\ \end{pmatrix}. \end{align*} $$

A pair of constant vector fields $(F, G)\in \frak h\times \frak h$ is a cocycle over the ${ \mathbb Z}^2$ action generated by $Y_{\boldsymbol \tau }$ and $Y_{\boldsymbol \eta }$ if

$$ \begin{align*} (D e^{Y_{\boldsymbol \tau}}-\text{Id}) G= (D e^{Y_{\boldsymbol \eta}}-\text{Id}) F, \end{align*} $$

which implies that

(51) $$ \begin{align} [Y_{\boldsymbol \tau}, G]= [Y_{\boldsymbol \eta}, F]. \end{align} $$

Because H is two-step nilpotent, this condition is only on the off-center coordinates of F and G. More precisely, if $F= f_1X_1+ f_2X_2+ f_3{\Lambda }_1+ f_4{\Lambda }_2+ f_5Z$ and $G=g_1X_1+ g_2X_2+ g_3{\Lambda }_1+ g_4{\Lambda }_2+ g_5Z$ , then (51) implies that $g_i$ and $f_i$ for $i=1,2,3,4$ satisfy the relation

(52) $$ \begin{align} \tau_1g_3+\tau_2g_4+\eta_1f_1+\eta_2f_2=0. \end{align} $$

Since the constants $f_5$ and $g_5$ are arbitrary, the space of constant cocycles has dimension nine.

A pair of constant vector fields $(F, G)$ is a coboundary if

$$ \begin{align*} (D e^{Y_{\boldsymbol \tau} }-\text{Id})H= F,\quad (D e^{Y_{\boldsymbol \eta}}-\text{Id}) H= G, \end{align*} $$

that is, if

(53) $$ \begin{align} [Y_{\boldsymbol \tau}, H]= F,\quad [Y_{\boldsymbol \eta}, H]= G. \end{align} $$

This implies that off-center coordinates of both F and G must be zero, and the center ones must satisfy certain relations. More precisely, if $H= h_1X_1+ h_2X_2+ h_3{\Lambda }_1+ h_4{\Lambda }_2+ h_5Z$ , the equations (53) imply that

(54) $$ \begin{align} f_5= \tau_1h_3+\tau_2h_4,\quad g_5= -\eta_1h_1-\eta_2h_2 \end{align} $$

and these equations always have solutions for coefficients of H.

This implies that the first cohomology is seven dimensional, and each cohomology class is represented by cocycles $(F, G)$ of the following form: $F= f_1X_1+ f_2X_2+ f_3{\Lambda }_1+ f_4{\Lambda }_2$ and $G=g_1X_1+ g_2X_2+ g_3{\Lambda }_1+ g_4{\Lambda }_2$ , where the coefficients $g_i$ and $f_i$ for $i=1,2,3,4$ satisfy the relation (52).

It is clear from the above computation that the dimension of the constant cohomology over the action generated by $Y_{\boldsymbol \tau }$ and $Y_{\boldsymbol \eta }$ in the case where the manifold is a $(2n+1)$ -dimensional Heisenberg nilmanifold is parallel to what we wrote above in the case $n=2$ and that the resulting cohomology has dimension $4n-1$ .

3.2 The finite-dimensional family of $\mathbb Z^2$ algebraic actions

For easier notation, in the rest of the paper we let $Y_1=Y_{\boldsymbol \tau }$ and $Y_2=Y_{\boldsymbol \eta }$ . In what follows we will use the fact that in $\frak h$ we have $\textrm{ exp}(X+Y+ {\tfrac {1}{2}}[X, Y])= \textrm{ exp}(X)\textrm{ exp}(Y)$ . In the remainder of the paper the brackets $[\cdot , \cdot ]$ denote the bracket in the Lie algebra $\frak h$ , so each bracket which appears as a result has a vector field in the Z-direction only.

We define now a nine-dimensional family of $\mathbb Z^2$ actions $\rho ^{\unicode{x3bb} }$ on $M=\Gamma \setminus H$ generated by the following maps:

(55) $$ \begin{align} y_i^{\unicode{x3bb}}(x)=x\cdot \textrm{ exp}(Y_i+F_i^{\unicode{x3bb}}) \end{align} $$

for $i=1,2$ , subject to the commutativity relation $y_1^{\unicode{x3bb} }\circ y_2^{\unicode{x3bb} }=y_2^{\unicode{x3bb} }\circ y_1^{\unicode{x3bb} }$ , where $F_i^{\unicode{x3bb} }\in \frak h$ . In particular, at the parameter ${\unicode{x3bb} }$ equal to 0, $F_i^0=0$ and $y_i^0=y_i$ , where $y_1=\textrm{ exp}\, Y_1$ and $y_2=\textrm{ exp}\, Y_2$ are generators of our original action $\rho $ .

The following lemma is a simple computation.

If $F_i^{\unicode{x3bb} }= f_i^1X_1+ f_i^2X_2+ f_i^3{\Lambda }_1+ f_i^4{\Lambda }_2+f_i^5Z$ , the coefficients $f_i^k\in \mathbb R$ for $i=1,2$ and $k= 1,2,3,4, 5$ , commutativity implies that the coefficients are subject to the relation

(56) $$ \begin{align} \tau_1f_2^3+\tau_2f_2^4+\eta_1f_1^1+\eta_2f_1^2=f_2^1f_1^3+f_2^2f_1^4 -f_2^3f_1^1-f_2^4f_1^2. \end{align} $$

Therefore, the parameter ${\unicode{x3bb} }$ is understood here as a vector in the nine-dimensional space:

$$ \begin{align*} \{f_i^k \in \mathbb R, i=1,2,\, k= 1,2,3,4, 5, \text{ subject to relation}\,\, (56) \}.\end{align*} $$

Within this nine-dimensional family of actions we impose identifications via conjugacies obtained by constant vector fields. More precisely, if

$$ \begin{align*}y_i^{\unicode{x3bb}}(x)= x\cdot \textrm{ exp}(Y_i+ F_i^{\unicode{x3bb}})\end{align*} $$

and, for some $H\in \frak h$ and $h(x):=x\cdot \textrm{ exp}\, H$ , we have

$$ \begin{align*}h\circ y_i^{\unicode{x3bb}}(x) = y_i\circ h,\end{align*} $$

then

$$ \begin{align*}x\cdot \textrm{ exp}(Y_i+F_i^{\unicode{x3bb}})\,\textrm{ exp}\, H= x\cdot \textrm{ exp}\, H\,\textrm{ exp}\, Y_i.\end{align*} $$

This implies that

$$ \begin{align*}Y_i+F_i^{\unicode{x3bb}} +H +{\tfrac{1}{2}} [Y_i+F_i^{\unicode{x3bb}}, H]= Y_i+H+{\tfrac{1}{2}}[H, Y_i].\end{align*} $$

Thus,

$$ \begin{align*}F_i^{\unicode{x3bb}} +{\tfrac{1}{2}} [F_i^{\unicode{x3bb}}, H]+ [Y_i, H]=0.\end{align*} $$

This implies that in the off-center direction the components of $F_i^{\unicode{x3bb} }$ are trivial and, for the center direction, we have

$$ \begin{align*}(F_i^{\unicode{x3bb}})_c= -[Y_i, H].\end{align*} $$

In particular, this defines a coordinate change H which produces conjugate algebraic actions in the family, and each conjugacy class is two dimensional, determined only by the values $(F_i^{\unicode{x3bb} })_c$ ( $i=1,2$ ).

So, the nine-dimensional family of algebraic actions modulo the algebraic conjugacy classes gives a seven-dimensional family of non-conjugate algebraic actions. This is the family $\rho ^{\unicode{x3bb} }$ in Theorem 1.7.

In the case where the manifold is a $(2n+1)$ -dimensional Heisenberg nilmanifold, the dimension of the family of non-conjugate algebraic actions in Theorem 1.7 is $4n-1$ and the family is described as in (55).

3.3 The commutator operator

Now we analyze the commutator operator for non-algebraic perturbations of translations which generate $\rho $ . Recall that $\rho $ is the $\mathbb Z^2$ action generated by the translation maps $y_i$ , $i=1,2$ , where $y_i(x)=x\cdot \textrm{ exp}(Y_i)$ and $Y_i$ are the two commuting elements in $\frak H$ .

Lemma 3.2. Let $F, G\in \textrm{ Vect}^{\infty }{M}$ be two sufficiently small vector fields so that the maps $f(x)= x\cdot \textrm{ exp}(Y_1+F(x))$ and $g(x)= x\cdot \textrm{ exp}(Y_2+G(x))$ are in $\textrm{ Diff}^{\infty }{M}$ . If f and g commute, then the vector fields F and G satisfy the following nonlinear equation:

(57) $$ \begin{align} F\circ y_2-F + \tfrac{1}{2}[Y_2, F\circ y_2+F]-(G\circ y_1-G + \tfrac{1}{2}[Y_1, G\circ y_1+G])+E(F, G)=0, \end{align} $$

where

(58) $$ \begin{align} E(F, G)&= (F\circ g-F\circ y_2)- (G\circ f-G\circ y_1)\nonumber \\ &\quad+ {\tfrac{1}{2}} [Y_2, F\circ g-F\circ y_2]- {\tfrac{1}{2}}[Y_1, G\circ f-G\circ y_1]\nonumber\\ &\quad+ {\tfrac{1}{2}} [G, F\circ g]-{\tfrac{1}{2}} [F, G\circ f]. \end{align} $$

Proof The commutation $f\circ g= g\circ f$ implies that

$$ \begin{align*} x \cdot \textrm{ exp}(Y_2+G)\,\textrm{ exp}(Y_1+ F\circ g)=x \cdot\textrm{ exp}(Y_1+F)\,\textrm{ exp}(Y_2+ G\circ f). \end{align*} $$

Hence,

$$ \begin{align*} x \cdot &\textrm{ exp}(Y_1+ Y_2+ G+ F\circ g+ {\tfrac{1}{2}} [Y_2, F\circ g]- {\tfrac{1}{2}} [Y_1, G]+ {\tfrac{1}{2}} [G, F\circ g])\\ &= x\cdot\textrm{ exp}(Y_1+ Y_2+ F+ G\circ f+ {\tfrac{1}{2}} [Y_1, G\circ f]- {\tfrac{1}{2}} [Y_2, F]+ {\tfrac{1}{2}} [F, G\circ f]). \end{align*} $$

The above implies the nonlinear equation directly due to the following very simple fact: $x \cdot \textrm{ exp}(Y+ F)= x \cdot \textrm{ exp}(Y+G)$ with $Y\in \frak h$ and $F, G\in \textrm{ Vect}^{\infty }{M}$ if and only if $F=G$ .

The following immediate consequence of the lemma above will be used later.

3.4 The conjugation operator

Here we analyze the conjugation operator for conjugacies close to the identity, we derive the linear part of the conjugacy operator and estimate the error.

Lemma 3.4. Let $f(x)=x\cdot \textrm{ exp}(Y +F(x))$ be a diffeomorphism of M, where $Y\in {\frak h}$ and $F\in \textrm{ Vect}^{\infty }{M}$ is a smooth vector field. Let $h\in \textrm{ Diff}^{\infty }{M}$ be a diffeomorphism close to the identity given by a small vector field $H\in \textrm{ Vect}^{\infty }{M}$ via $h(x)=x\cdot \textrm{ exp}\, H(x)$ . Then $g:=h^{-1}\circ f\circ h$ is a diffeomorphism close to f given by $G\in \textrm{ Vect}^{\infty }{M}$ via $g(x)=x\cdot \textrm{ exp}(Y +G(x))$ and

$$ \begin{align*} G\kern1.5pt{=}\kern1.5pt H\kern1.5pt{-}\kern1.5pt H\kern1pt{\circ}\kern1pt g\kern1.5pt{+}\kern1.5pt {\tfrac{1}{2}}[H\kern1.5pt{+}\kern1.5pt H\kern1pt{\circ}\kern1pt g, Y] \kern1.5pt{+}\kern1.5ptF\kern1pt{\circ}\kern1pt h\kern1.5pt{+}\kern1.5pt [H, F\kern1pt{\circ}\kern1pt h]\kern1.5pt{+}\kern1.5pt{\tfrac{1}{2}} [H, H\kern1pt{\circ}\kern1pt g]\kern1.5pt{-}\kern1.5pt{\tfrac{1}{2}} [F\kern1pt{\circ}\kern1pt h, H\kern1pt{\circ}\kern1pt g]. \end{align*} $$

Proof We have

$$ \begin{align*} h^{-1}\circ & f\circ h(x) =x\cdot \textrm{ exp}\, H(x)\,\textrm{ exp}(Y+F\circ h(x))\,\textrm{ exp}(-H\circ h^{-1}\circ f\circ h(x)) \\ & =x\cdot \textrm{ exp}(H+Y+F\circ h+{\tfrac{1}{2}}[H, Y]+[H, F\circ h])(x)\,\textrm{ exp}(-H\circ g)(x) \\ & =x\cdot \textrm{ exp}(H+Y-H\circ g+F\circ h+ {\tfrac{1}{2}}[H, Y]+ [H, F\circ h] \\ & \quad-{\tfrac{1}{2}} [H, -H\circ g]-{\tfrac{1}{2}}[Y, H\circ g]-{\tfrac{1}{2}} [F\circ h, H\circ g])(x). \end{align*} $$

This implies the equality claimed for G.

3.5 Linearizations of the conjugacy and the commutator operators: first and second coboundary operators on vector fields, splitting

The linear part of the nonlinear equation (57) defines the second coboundary operator on vector fields over the action $\rho $ generated by $y_1$ and $y_2$ .

Definition 3.5. Let ${\mathbf {d}}_2: \textrm{ Vect}^{\infty }{M}\times \textrm{ Vect}^{\infty }{M} \to \textrm{ Vect}^{\infty }{M}$ be the linear operator defined by

$$ \begin{align*} {\mathbf{d}}_2(F, G)= F\circ y_2-F + \tfrac{1}{2}[Y_2, F\circ y_2+F]-(G\circ y_1-G + \tfrac{1}{2}[Y_1, G\circ y_1+G]). \end{align*} $$

We say that a pair of smooth vector fields $(F, G)$ generates a cocycle over the action $\rho $ if $(F, G)\in \text {Ker}\, {\mathbf {d}}_2$ .

The first coboundary operator on vector fields over the action $\rho $ is given by the following definition.

Definition 3.6. Let $H\in \textrm{ Vect}^{\infty }{M}.$ Then we define ${\mathbf {d}}_1:\textrm{ Vect}^{\infty }{M}\to \text { Vect}^{\infty }{M} \times \textrm{ Vect}^{\infty }{M}$ by

$$ \begin{align*} {\mathbf{d}}_1(H)=(H\circ y_1-H + \tfrac{1}{2}[Y_1, H\circ y_1+H], H\circ y_2-H + \tfrac{1}{2}[Y_2, H\circ y_2+H]). \end{align*} $$

It is an easy exercise to check that $\rm {Im}\, {\mathbf {d}}_1\subset \text {Ker}\, {\mathbf {d}}_2$ . The first cohomology over $\rho $ with coefficients in vector fields is the quotient space $H^1_{\rho }(\textrm{ Vect}^{\infty }{M}):= \rm Ker\, {\mathbf {d}}_2/\rm Im\, {\mathbf {d}}_1$ . Notice that for constant vector fields $H\in \frak h$ cocycles and coboundaries defined here coincide with those defined in §3.1. The subsequent proposition has as a corollary that for our fixed action $\rho $ the cohomology $H^1_{\rho }(\textrm{ Vect}^{\infty }{M})$ is the same as the cohomology $H^1_{\rho }(\frak h)$ with coefficients in the constant vector fields $\frak h$ which was computed in §3.1.

Proposition 3.7. If the two vector fields $F, G\in \textrm{ Vect}^{\infty }{M}$ satisfy ${\mathbf {d}}_2(F, G)=\Phi $ and $(\textrm{ Ave}\, F, \textrm{ Ave}\, G) $ is in the trivial cohomology class in $H_{\rho }^1(\frak n)$ , then there exist $H, \tilde {F}, \tilde {G}\in \textrm{ Vect}^{\infty }{M}$ such that $(F, G)= {\mathbf {d}}_1H+ (\tilde {F}, \tilde {G})$ and the following estimates hold:

(59) $$ \begin{align} \|\tilde{F}\|_s &\leq C_{s} \| \Phi\|_{s+\sigma}, \notag\\ \|\tilde{G}\|_s &\leq C_{s}\| \Phi\|_{s+\sigma}, \notag\\ \| H \|_s &\leq C_{s, r, S, T, \Gamma}\| F\|_{s+\sigma}. \end{align} $$

Proof Recall that we disintegrate an arbitrary vector field $H\in \textrm{ Vect}^{\infty }{M}$ into $H= H_c+ H_T$ , where $ H_c$ is the component of H in the direction of Z and $H_T$ is the component of H in all the directions other than Z. So, one can view $H_T$ as $H_T=\sum h_t^ iX_i+\bar h_t^i\Lambda _i$ , where $h_t^ i, \bar h_t^i$ are smooth functions.

The equation ${\mathbf {d}}_2(F_T, G_T)=\Phi $ (since ${\mathbf {d}}_2$ is a linear operator) splits then in the off-center directions into finitely many functional equations each of which has a form

$$ \begin{align*}f\circ y_2-f-(g\circ y_1-g)=\phi.\end{align*} $$

Since by assumption $(\textrm{ Ave}\, F, \textrm{ Ave}\, G) $ is assumed to be in the trivial constant cohomology class, it implies in particular that all the off-center components are 0 (see §3.1).

Now we may apply Theorem 1.3, which for each of these finitely many equations gives as an output smooth functions $\tilde {f}, \tilde {g}, h$ such that

$$ \begin{align*}f=h\circ y_1-f+ \tilde{f}, \quad g= g\circ y_2-g+ \tilde{g}\end{align*} $$

such that the corresponding estimates hold. Putting these coordinate functions all together gives functions $\tilde {F}_T, \tilde {G}_T, H_t$ such that $({\mathbf {d}}_2(\tilde {F}_T, \tilde {G}_T))_T=\Phi _T$ , $(F_T, G_T)=({\mathbf {d}}_1H_T)_T + (\tilde {F}_T, \tilde {G}_T)$ and $\tilde {F}_T, \tilde {G}_T, H_T$ satisfy the estimates (59).

Now let $ F_c, G_c$ be the components of F and G in the center direction. Then, because the Z components within the brackets do not contribute, we have

$$ \begin{align*} {\mathbf{d}}_2(F_c, G_c)& = F_c\circ y_2-F_c + \tfrac{1}{2}[Y_2, F_T\circ y_2+F_T] \\ & \quad-\big(G_c\circ y_1-G_c + \tfrac{1}{2}[Y_1, G_T\circ y_1+G_T]\big)= \Phi_c. \end{align*} $$

Since we already have $(F_T, G_T)=({\mathbf {d}}_1H_T)_T + (\tilde {F}_T, \tilde {G}_T)$ , we can substitute this in the above expression to obtain

(60) $$ \begin{align} &\big(F_c-{\tfrac{1}{2}}[Y_1, H_T\circ y_1+H_T]\big)\circ y_2 -\big(F_c-{\tfrac{1}{2}}[Y_1, H_T\circ y_1+H_T]\big) \nonumber\\ &\quad-\big(\big(G_c-{\tfrac{1}{2}}[Y_2, H_T\circ y_2+H_T]\big)\circ y_1 -\big(G_c-{\tfrac{1}{2}}[Y_2, H_T\circ y_2+H_T]\big)\big)=\Phi^{\prime}_c, \end{align} $$

where

$$ \begin{align*} \Phi^{\prime}_c= \Phi_c-{\tfrac{1}{2}} [Y_2, \tilde{F}_T\circ y_2+ \tilde{F}_T ]+ {\tfrac{1}{2}} [Y_1, \tilde{G}_T\circ y_1+ \tilde{G}_T ]. \end{align*} $$

Clearly, since for any s we have $\|\tilde {F}_T\|_s,\|\tilde {G}_T\|_s\le \|\tilde {\Phi }_T\|_{s+\sigma }$ , it follows that $\| \Phi ^{\prime }_c\|_s\le \|\tilde {\Phi }\|_{s+\sigma }$ , where $\sigma $ is re-defined to be $\sigma + 1$ .

The vector field $H_T$ is determined only up to a constant vector field, so we may choose $H_T$ so that $\textrm{ Ave}\, F_c= \textrm{ Ave}[Y_1, H_T]$ . This forces $F_c-{\tfrac {1}{2}}[Y_1, H_T\circ y_1+H_T]$ to have the average 0. Moreover, because of the assumption that $(\textrm{ Ave}\, F, \textrm{ Ave}\, G) $ is in the trivial cohomology class, we also have that $\textrm{ Ave}\, F_c= \textrm{ Ave}\, G_c$ .

So, the equation (60) is again the same type of equation as in Theorem 1.3. By applying the theorem, we get $\tilde {F}_c, \tilde {G}_c, H_c$ such that

$$ \begin{align*} & F_c-{\tfrac{1}{2}}[Y_1, H_T\circ y_1+H_T]= H_c\circ y_1-H_c + \tilde{F}_c, \\ & G_c-{\tfrac{1}{2}}[Y_2, H_T\circ y_2+H_T]= H_c\circ y_2-H_c + \tilde{G}_c. \end{align*} $$

This clearly implies that

$$ \begin{align*} (F_c, G_c)= ({\mathbf{d}}_1H)_c + (\tilde{F}_c, \tilde{G}_c). \end{align*} $$

Putting the c- and T-components together gives the solution. Estimates (59) are direct consequences of coordinate-wise estimates which are obtained already in Theorem 1.3.

3.6 Set-up of the perturbative problem and the iterative scheme

We will frequently refer here to [Reference Damjanović and Katok7], so we recommend that the reader has that paper at hand.

We consider here a family of perturbations $\tilde {\rho }^{\unicode{x3bb} }$ of $\rho ^{\unicode{x3bb} }$ , which are generated by commuting maps $\tilde {y}_1^{\unicode{x3bb} }$ and $\tilde {y}_2^{\unicode{x3bb} }$ , where, for $i=1,2$ ,

(61) $$ \begin{align} \tilde{y}_i^{\unicode{x3bb}}(x)=x\cdot \textrm{ exp}(Y_i+ \tilde{F}_i^{\unicode{x3bb}}). \,\,\, \end{align} $$

Here $\tilde {F}_i^{\unicode{x3bb} }$ are small vector fields such that $\tilde {y}_1^{\unicode{x3bb} }$ and $\tilde {y}_2^{\unicode{x3bb} }$ commute.

Now let h be a diffeomorphism of the manifold, close to the identity, defined via the smooth vector field H as follows:

$$ \begin{align*} h(x)= { \Gamma\theta}\cdot \textrm{ exp}\, H(x). \end{align*} $$

The iterative step consists of the following: given the perturbation $\tilde {\rho }^{\unicode{x3bb} }$ of $\rho ^{\unicode{x3bb} }$ , define a new perturbation $\bar \rho ^{\unicode{x3bb} }$ which is a conjugation of $\tilde {\rho }^{\unicode{x3bb} }$ via h, so $\bar \rho ^{\unicode{x3bb} }$ is generated by two diffeomorphisms $\bar y_i^{\unicode{x3bb} }$ , $i\in \{1,2\}$ , defined by

$$ \begin{align*} \bar y_i^{\unicode{x3bb}}= h^{-1}\circ \tilde{y}_i^{\unicode{x3bb}}\circ h. \end{align*} $$

In each iterative step this is done for the $\lambda $ parameter in some ball, and it is shown that in that ball there is a parameter for which the new family of perturbations is much (quadratically) closer to $\rho $ for parameters in some smaller ball. The next proposition shows that that this process is controlled in the sequence of $C^r$ norms.

We will need to control derivatives of each perturbed family $\tilde {\rho }^{\unicode{x3bb} }$ in the direction of the parameter $\lambda $ as well, so we will use the following norms for a family of vector fields $\tilde {F}_i^{\unicode{x3bb} }$ : $\|\tilde {F}_i^{\unicode{x3bb} }\|_{0, k}$ stands for the supremum of the $C^k$ norms of $\tilde {F}_i^{\unicode{x3bb} }$ in the ${\unicode{x3bb} }$ variable. $\|\tilde {F}_i^{\unicode{x3bb} }\|_{r, k}$ is the same only taken over all the derivatives of $\tilde {F}_i^{\unicode{x3bb} }$ in the manifold direction. As before, we reserve the notation $\|\tilde {F}_i^{\unicode{x3bb} }\|_{r}$ for the usual $C^r$ norm on M of the vector field $ \tilde {F}_i^{\unicode{x3bb} }\in \textrm{ Vect}^{\infty }{M}$ for a fixed parameter ${\unicode{x3bb} }$ .

The following is an immediate corollary of the classical implicit function theorem and we will use it for the maps which compute averages of vector fields for actions in the perturbed family.

Now we state the main iterative step proposition where we show that one can obtain indeed estimates which are needed for the convergence of the process to a smooth conjugation map.

Proposition 3.9. There exist constants $\bar C$ and $r_0$ such that the following holds.

Given the family $\tilde {\rho }_n^{\unicode{x3bb} }$ of perturbations of $\rho ^0$ generated by $\tilde {y}_{i, n}^{\unicode{x3bb} }$ ( $i=1,2$ ), assume that for all ${\unicode{x3bb} }$ in a ball B centered at 0, for $r\in \mathbb N$ and $t>0$ :

1. (1) $\| \tilde {F}_{i, n}^{\unicode{x3bb} }\|_0\le {\varepsilon }_n<1$ ;

2. (2) the map ${\unicode{x3bb} }\mapsto \tilde {F}_{i, n}^{\unicode{x3bb} }$ is $C^2$ in ${\unicode{x3bb} }$ , $\|\tilde {F}_{i, n}^{\unicode{x3bb} }\|_{0, 1}\le {\varepsilon }_n$ and $\|\tilde {F}_{i, n}^{\unicode{x3bb} }\|_{0, 2}\le K_n$ ;

3. (3) the map $\Phi ^n : {\unicode{x3bb} }\mapsto \textrm{ Ave}(\tilde {F}_{i, n}^{\unicode{x3bb} })$ is in $\mathcal O$ and has a zero at ${\unicode{x3bb} }_n$ ;

4. (4) $\| \tilde {F}_{i, n}^{\unicode{x3bb} }\|_{r_0+r}\le \delta _{r, n}$ ;

5. (5) $t^{r_0}{\varepsilon } _n^{1-({1}/({r+r_0}))}\delta _{r}^{({1}/({r+r_0}))}<\bar C$ .

There exists a $H_n\in \textrm{ Vect}^{\infty }{M}$ such that $h_n$ defined by $ h(x)= x\cdot \textrm{ exp}\, H_n(x) $ is such that the newly formed family of perturbations $\tilde {\rho }_{n+1}^{\unicode{x3bb} }$ of $\rho $ , generated by $\tilde {y}_{i, n+1}^{\unicode{x3bb} }= h^{-1}\circ \tilde {y}_{i, n}^{\unicode{x3bb} }\circ h$ , with $\tilde {y}_{i, n+1}^{\unicode{x3bb} }(x)= x\,\textrm{ exp}(Y_i+ \tilde {F}_{i, n+1}^{\unicode{x3bb} }(x))$ , satisfies the following:

1. (a) $\|H_n\|_r\le C_rt^{2r_0}\|\tilde {F}_i^{{\unicode{x3bb} }_n}\|_r$ ;

2. (b) $ \|\tilde {F}_{i, n+1}^{{\unicode{x3bb} }}\|_0\le K_n\|{\unicode{x3bb} }-{\unicode{x3bb} }_n\|+ {\rm Err}(t, r)$ , where

   $$ \begin{align*}\textrm{ Err}_{n+1}(t,r)&:=C {\varepsilon }_n^2 +C\delta_{r, n}^{{(r_0+1)}/({r_0+r})}{\varepsilon }_n^{2-(({r_0+1})/({r_0+r}))}+C_{r}t^{-r}\delta_{r,n}\\&\quad+C_{r_0}t^{2r_0}{\varepsilon }_n^{2-({1}/({r_0+r}))}\delta_{r,n}^{{1}/({r_0+r})}+C_{r_0}t^{2r_0}{\varepsilon }_n^{3-({1}/({r_0+r}))}\delta_{r,n}^{{2}/({r_0+r})}; \end{align*} $$

3. (c) $\| \tilde {F}_{i, n+1}^{\unicode{x3bb} }\|_{r_0+r}\le C_rt^{2r_0}\delta _{r, n}:=\delta _{r, n+1} $ ;

4. (d) the map $\Phi ^{n+1}:={\unicode{x3bb} }\mapsto \textrm{ Ave}(\tilde {F}_{i, n+1}^{\unicode{x3bb} })$ satisfies

$$ \begin{align*} \|\Phi^{n+1}-\Phi^n\|_{(0)}&\le \textrm{ Err}_{n+1}(t,r),\\ \|\Phi^{n+1}-\Phi^n\|_{(1)}&\le K_nt^{r_0}{\varepsilon }_n+\textrm{ Err}_{n+1}(t,r). \end{align*} $$

If $\Phi ^{n+1}$ is in $\mathcal O$ , then it has a zero at ${\unicode{x3bb} }_{n+1}\in B$ which satisfies

$$ \begin{align*}\|{\unicode{x3bb}}_{n+1}-{\unicode{x3bb}}_{n}\|\le C\textrm{ Err}_{n+1}(t,r)+CK_n( K_nt^{r_0}{\varepsilon }_n+\textrm{ Err}_{n+1}(t,r))^2;\end{align*} $$

1. (e) $\tilde {F}_{i, n+1}^{\unicode{x3bb} }$ is $C^2$ in ${\unicode{x3bb} }$ and

   $$ \begin{align*}\| \tilde{F}_{i, n+1}^{\unicode{x3bb}}\|_{0, 2}\le (1+Ct^{r_0}{\varepsilon }_n^{1-({1}/({r+r_0}))}\delta_{r,n}^{{1}/({r+r_0})})K_n=:K_{n+1}(t,r).\end{align*} $$

Proof As was mentioned in [Reference Damjanović and Katok7, Remark 6.3], the proof of the iterative step is universal given tame splitting for vector fields (Proposition 3.7). We repeat the main points here for the sake of completeness with fewer details than in the proof of the corresponding proposition in [Reference Damjanović and Katok7, Proposition 6.2].

In this proof, as is customary whenever there is a loss of regularity for solutions of linearized equations, we will use the smoothing operators. For the construction of smoothing operators on $C^{\infty }(M)$ , see the following in [Reference Hamilton10]: Example 1.1.2(2), Definition 1.3.2, Theorem 1.3.6 and Corollary 1.4.2. There exists a collection of smoothing operators $S_t:C^{\infty }(M)\to C^{\infty }(M)$ , $t>0$ , such that the following holds:

(62) $$ \begin{align} \|S_tF\|_{s+s'}&\le C_{s,s'}t^{s'}\|F\|_s,\nonumber\\ \|(I-S_t)F\|_{s-s'}&\le C_{s,s'}t^{-s'}\|F\|_s. \end{align} $$

Smoothing operators on $C^{\infty }(M)$ clearly induce smoothing operators on $\textrm{ Vect}^{\infty }{M}$ via smoothing operators applied to coordinate maps.

It is easy to see that averages of F with respect to the Haar measure on M, in various directions in the tangent space, do not affect the properties of smoothing operators listed above, so without loss of generality we may assume that $S_t$ are such that averages of $S_tF$ are the same as those of F.

Given ${ \tilde {F}_{i, n}^{\unicode{x3bb} }}$ , we first apply the smoothing operators to it and write ${ \tilde {F}_{i, n}^{\unicode{x3bb} }}= S_t{ \tilde {F}_{i, n}^{\unicode{x3bb} }}+ (I-S_t) { \tilde {F}_{i, n}^{\unicode{x3bb} }}$ . Now $\textrm{ Ave}({ \tilde {F}_{i, n}^{\unicode{x3bb} }})= \textrm{ Ave}(S_t{ \tilde {F}_{i, n}^{\unicode{x3bb} }})$ . From the commutativity of ${ \tilde {F}_{i, n}^{\unicode{x3bb} }}$ for $i=1$ and $i=2$ (see Corollary 3.3), it follows that $|\textrm{ Ave}[Y_2, \tilde {F}_{1, n}]-\textrm{ Ave}[Y_1, \tilde {F}_{2, n}]|\le C\|\tilde {F}_{1, n}\|_1\|\tilde {F}_{2, n}\|_1\le C{\varepsilon }_n^2$ and clearly the same holds after application of the corresponding smoothing operators. Now we can apply Proposition 3.7 to $S_t{ \tilde {F}_{i, n}^{\unicode{x3bb} }}- \textrm{ Ave}({ \tilde {F}_{i, n}^{\unicode{x3bb} }})_T$ , $i=1, 2$ (recall that $\textrm{ Ave}({ \tilde {F}_{i, n}^{\unicode{x3bb} }})_T$ are averages in the off-center directions). Proposition 3.7 gives existence of $H_n$ such that

$$ \begin{align*} \|(S_t\tilde{F}_{1, n}- \textrm{ Ave}(\tilde{F}_{1, n})_T, S_t\tilde{F}_{2, n}- \textrm{ Ave}(\tilde{F}_{2, n})_T ) - \delta_1 H_n\| _r\le C\|\Phi\|_{r+\sigma}, \end{align*} $$

where (see (58))

$$ \begin{align*} \Phi:= E(\tilde{F}_{1, n}, \tilde{F}_{2, n}). \end{align*} $$

From the expression for E in (58), we have the following estimate for $\Phi $ :

$$ \begin{align*} \|\Phi\|_r\le C \|{ \tilde{F}_{i, n}^{\unicode{x3bb}}}\|_r\|{ \tilde{F}_{i, n}^{\unicode{x3bb}}}\|_{r+1}, \end{align*} $$

where we use short notation $ \|{ \tilde {F}_{i, n}^{\unicode{x3bb} }}\|_r$ for the maximum of the norms for $i=1$ and $i=2$ . Also, from Proposition 3.7, we have

$$ \begin{align*} \| H \|_r \leq C\| S_t\tilde{F}_{i, n}^{\unicode{x3bb}}-\textrm{ Ave}(\tilde{F}_{1, n})_T\|_{r+\sigma}\le Ct^{\sigma} \|\tilde{F}_{i, n}^{\unicode{x3bb}}\|_r. \end{align*} $$

From Lemma 3.4, it follows that if we define $h_n$ by $h_n(x) = x\cdot \textrm{ exp}\, H_n(x)$ , and $\tilde {y}_{i, n+1}^{\unicode{x3bb} }= h^{-1}\circ \tilde {y}_{i, n}^{\unicode{x3bb} }\circ h$ , with $\tilde {y}_{i, n+1}^{\unicode{x3bb} }(x)= x\cdot \textrm{ exp}(Y_i+ \tilde {F}_{i, n+1}^{\unicode{x3bb} }(x))$ , then $ \tilde {F}_{i, n+1}^{\unicode{x3bb} }$ satisfy the following, after applying the interpolation estimates and the smoothing estimates and assumptions (2) and (3) (compare to [Reference Damjanović and Katok7, (6.7)]):

(63) $$ \begin{align} \|\tilde{F}_{i, n+1}^{\unicode{x3bb}}\|_0&\le K_n\|{\unicode{x3bb}}-{\unicode{x3bb}}_n\| + C {\varepsilon }_n^2+C\delta_{r, n}^{({r_0+1})/({r_0+r})}{\varepsilon }_n^{2-(({r_0+1})/({r_0+r}))}\nonumber\\ &+C_{r}t^{-r}\delta_{r,n}+C_{r_0}t^{2r_0}{\varepsilon }_n^{2-({1}/({r_0+r}))}\delta_{r,n}^{{1}/({r_0+r})}+C_{r_0}t^{2r_0}{\varepsilon }_n^{3-({1}/({r_0+r}))}\delta_{r,n}^{{2}/({r_0+r})}. \end{align} $$

For the $C^{r_0+r}$ norm of the new error $\tilde {F}_{i, n+1}^{\unicode{x3bb} }$ , as usual in this type of proofs, we only need a ‘linear’ bound with respect to the corresponding norm of the old error. This follows easily from the conjugacy relation and we obtain for any $s\ge 0$ :

$$ \begin{align*} \|\tilde{F}_{i, n+1}^{\unicode{x3bb}}\|_s \le C_st^{2r_0}\|\tilde{F}_{i, n}^{\unicode{x3bb}}\|_s, \end{align*} $$

which, as in [Reference Damjanović and Katok7], implies that

$$ \begin{align*} \|\tilde{F}_{i, n+1}^{\unicode{x3bb}}\|_s \le C_st^{2r_0}\delta_{r, n}. \end{align*} $$

The remaining two statements (e) and (d) follow exactly in the same way as in the proof of [Reference Damjanović and Katok7, Proposition 6.2].

Given Proposition 3.9 (compare to [Reference Damjanović and Katok7, Proposition 6.2]), we can now apply the convergence of the successive iterative scheme proved in [Reference Damjanović and Katok7, §7]. Consequently, we obtain the following theorem, which is a more precise statement of our main transversal local rigidity result in Theorem 1.7.

Theorem 3.10. There exist $l>0$ , $\epsilon>0$ , $R>0$ such that if a family $\tilde {\rho }^{\unicode{x3bb} }$ of perturbations of $\rho $ generated by $\tilde {y}_i^{\unicode{x3bb} }$ is $\epsilon $ close to $\rho $ in the $C^l$ norm for parameters $\lambda $ in an R-ball around 0, and in the $C^1$ norm in the parameter $\lambda $ direction, then there exists a small parameter $\bar \lambda $ such that the action $\tilde {\rho }^{\bar {\unicode{x3bb} }}$ is conjugate to $\rho $ via h, that is, for $i=1, 2$ , we have

$$ \begin{align*} h\circ y_i= \tilde{y}_i^{\bar {\unicode{x3bb}}}\circ h, \end{align*} $$

where h is a smooth diffeomorphism order of $\epsilon $ close to the identity in the $C^1$ norm.