# Vanishing of cohomology and parameter rigidity of actions of solvable Lie
groups

Hirokazu Maruhashi<EMAIL_ADDRESS>
###### Abstract

We give a sufficient condition for parameter rigidity of actions of solvable
Lie groups, by vanishing of (uncountably many) first cohomologies of the orbit
foliations. In some cases, we can prove that vanishing of finitely many
cohomologies is sufficient. For this purpose we use a rigidity property of
quasiisometry.

As an application we prove some actions of 2-step solvable Lie groups on
mapping tori are parameter rigid. Special cases of these actions are
considered in a paper of Matsumoto and Mitsumatsu [6].

We also remark on the relation between transitive locally free actions of
solvable Lie groups and lattices in solvable Lie groups, and apply results in
rigidity theory of lattices in solvable Lie groups to construct transitive
locally free actions with some properties.

###### Contents

  1. 1 Introduction
  2. 2 General sufficient condition for parameter rigidity
    1. 2.1 Proof of Proposition 2
    2. 2.2 Proof of Theorem 3
  3. 3 Sufficient condition by the method of Matsumoto and Mitsumatsu
  4. 4 Sufficient condition given by large scale geometry of solvable Lie groups
    1. 4.1 Key Proposition to the method
    2. 4.2 Proof of Theorem 5
  5. 5 Parameter rigidity of suspension-like actions on mapping tori
    1. 5.1 Cohomology
    2. 5.2 Application of large scale geometry
    3. 5.3 Mayer–Vietoris argument
    4. 5.4 Vanishing of the cohomology
  6. 6 Parameter rigidity of transitive locally free actions and rigidity of lattices
    1. 6.1 Relations between transitive locally free actions and lattices
    2. 6.2 A counterexample to Theorem 1 for solvable Lie groups
    3. 6.3 A locally parameter rigid action of a contractible group which is not parameter rigid

##  1 Introduction

Let $\rho_{0}$ be a $C^{\infty}$ right action of a connected simply connected
solvable Lie group $S$ on a closed $C^{\infty}$ manifold $M$. We always assume
$\rho_{0}$ is locally free, that is, every isotropy subgroup is discrete in
$S$. Then we have a foliation ${\mathcal{F}}$ of $M$ by the orbits of
$\rho_{0}$, which is called the orbit foliation of $\rho_{0}$. We say that
$\rho_{0}$ is parameter rigid if any $C^{\infty}$ locally free action $\rho$
of $S$ on $M$ whose orbit foliation coincides with ${\mathcal{F}}$ is
parameter equivalent to $\rho_{0}$. Here parameter equivalence means the
following; there are an automorphism $\Phi$ of $S$ and a diffeomorphism $F$ of
$M$ such that

$F\left(\rho_{0}(x,s)\right)=\rho\left(F(x),\Phi(s)\right)$

for all $x\in M$ and $s\in S$, and $F$ preserves each leaf of ${\mathcal{F}}$
and is $C^{0}$ homotopic to the identity via $C^{\infty}$ maps preserving each
leaf.

First we review a criterion for parameter rigidity when $S$ is nilpotent.
Instead of using $S$, let $N$ denote the acting group. We have the leafwise
cohomology $H^{*}({\mathcal{F}})$ of the foliation ${\mathcal{F}}$, which is
defined in a similar way to the usual de Rham cohomology. Using the action
$\rho_{0}$ we can define a canonical injection
$H^{*}({\mathfrak{n}})\hookrightarrow H^{*}({\mathcal{F}})$. We will always
use Fraktur for the corresponding Lie algebras. The author of this article
proved in [4] and [5] the following:

######  Theorem 1.

$\rho_{0}$ is parameter rigid if and only if
$H^{1}({\mathcal{F}})=H^{1}({\mathfrak{n}})$.

This theorem reduces proving parameter rigidity — which looks a nonlinear
problem at first sight — to a linear one, that is, calculation of first
cohomology.

However this criterion is no longer true when the acting group is a general
solvable Lie group. In fact there are a solvable Lie group $S$ and a lattice
$\Gamma$ in $S$ for which the natural transitive action $\Gamma\backslash
S\curvearrowleft S$ is not parameter rigid while $H^{1}(\Gamma\backslash
S)=H^{1}({\mathfrak{s}})$. We will see it in Section 6 together with an
example of a locally parameter rigid action of a contractible group which is
not parameter rigid. These are obtained by looking at the relations between
transitive locally free actions and lattices in solvable Lie groups.

Therefore, the formulation should be changed and we will do it by using
twisted leafwise cohomologies. Let us return to the first notations. For the
action $M\stackrel{{\scriptstyle\rho_{0}}}{{\curvearrowleft}}S$ and a
representation $\pi$ of ${\mathfrak{s}}$ on a finite dimensional real vector
space $V$, we will define the leafwise cohomology
$H^{*}\left({\mathcal{F}};{\mathfrak{s}}\stackrel{{\scriptstyle\pi}}{{\curvearrowright}}V\right)$
of ${\mathcal{F}}$ with coefficient $\pi$ as follows. Let $\omega_{0}$ denote
the canonical 1–form of $\rho_{0}$, ie for any $x\in M$,
$(\omega_{0})_{x}\colon T_{x}{\mathcal{F}}\to{\mathfrak{s}}$ is the inverse of
the derivative at the identity of the map $S\to M$ which sends $g$ to
$\rho_{0}(x,g)$. This is a leafwise ${\mathfrak{s}}$–valued 1–form satisfying

$d_{\mathcal{F}}\omega_{0}+[\omega_{0},\omega_{0}]=0,$

where $d_{\mathcal{F}}$ is the leafwise exterior derivative of
${\mathcal{F}}$. By composing with $\pi$ we get an
$\mathop{\mathrm{End}}\nolimits(V)$–valued leafwise 1-form $\pi\omega_{0}$
satisfying

$d_{\mathcal{F}}\pi\omega_{0}+\left[\pi\omega_{0},\pi\omega_{0}\right]=0.$

Therefore, the trivial vector bundle $M\times V\to M$ carries the flat
leafwise connection whose connection form is $\pi\omega_{0}$. Here this
connection form is relative to any global frame of the bundle which has
constant $V$ components. The square of the exterior derivative with respect to
this connection on the space of leafwise $V$–valued forms
$\Omega^{*}({\mathcal{F}};V)$ is zero by flatness of the connection. So we
obtain the cohomology
$H^{*}\left({\mathcal{F}};{\mathfrak{s}}\stackrel{{\scriptstyle\pi}}{{\curvearrowright}}V\right)$.
On the other hand the cohomology
$H^{*}\left({\mathfrak{s}};{\mathfrak{s}}\stackrel{{\scriptstyle\pi}}{{\curvearrowright}}V\right)$
of the Lie algebra ${\mathfrak{s}}$ with coefficient $\pi$ is defined from the
complex
$\mathop{\mathrm{Hom}}\nolimits\left(\bigwedge^{*}{\mathfrak{s}},V\right)$.
Consider the map

$\mathop{\mathrm{Hom}}\nolimits\left(\bigwedge^{*}{\mathfrak{s}},V\right)\hookrightarrow\Omega^{*}({\mathcal{F}};V)$

mapping $\varphi$ to $\omega_{0}^{*}\varphi$, where $\omega_{0}^{*}$ means
pullback by $\omega_{0}$. This is a cochain map, so that it induces the map
between cohomologies.

######  Proposition 2.

The induced map
$H^{*}\left({\mathfrak{s}};{\mathfrak{s}}\stackrel{{\scriptstyle\pi}}{{\curvearrowright}}V\right)\to
H^{*}\left({\mathcal{F}};{\mathfrak{s}}\stackrel{{\scriptstyle\pi}}{{\curvearrowright}}V\right)$
is injective.

We will prove this in Section 2.1. Hereafter we shall regard
$H^{*}\left({\mathfrak{s}};{\mathfrak{s}}\stackrel{{\scriptstyle\pi}}{{\curvearrowright}}V\right)$
as a subspace of
$H^{*}\left({\mathcal{F}};{\mathfrak{s}}\stackrel{{\scriptstyle\pi}}{{\curvearrowright}}V\right)$.

Next we will specify which representations are needed for our sufficient
condition. Let ${\mathfrak{n}}$ be the nilradical of ${\mathfrak{s}}$, that
is, the largest nilpotent ideal of ${\mathfrak{s}}$. Then ${\mathfrak{n}}$
contains $[{\mathfrak{s}},{\mathfrak{s}}]$ since
$[{\mathfrak{s}},{\mathfrak{s}}]$ is a nilpotent ideal. Take any subspace
${\mathfrak{h}}$ satisfying

$[{\mathfrak{s}},{\mathfrak{s}}]\subset{\mathfrak{h}}\subset{\mathfrak{n}}.$

Since ${\mathfrak{h}}$ is a nilpotent ideal, the descending central series of
${\mathfrak{h}}$ terminates at some term:
${\mathfrak{h}}\supset{\mathfrak{h}}^{2}\supset\cdots\supset{\mathfrak{h}}^{d}\supset
0$. The adjoint representation $\mathop{\mathrm{ad}}\nolimits$ of
${\mathfrak{s}}$ has the following invariant filtration:

${\mathfrak{s}}\stackrel{{\scriptstyle\mathop{\mathrm{ad}}\nolimits}}{{\curvearrowright}}{\mathfrak{s}}\supset{\mathfrak{h}}\supset{\mathfrak{h}}^{2}\supset\cdots\supset{\mathfrak{h}}^{d}\supset
0.$

We take the associated graded quotient of this filtration:

${\mathfrak{s}}\stackrel{{\scriptstyle\mathop{\mathrm{ad}}\nolimits}}{{\curvearrowright}}\mathop{\mathrm{Gr}}\nolimits_{\mathfrak{h}}({\mathfrak{s}})={\mathfrak{s}}/{\mathfrak{h}}\oplus\bigoplus_{i=1}^{d}{\mathfrak{h}}^{i}/{\mathfrak{h}}^{i+1}.$

Note that ${\mathfrak{s}}/{\mathfrak{h}}$ is a direct sum of trivial
representations. On this associated graded quotient ${\mathfrak{h}}$ acts
trivially, so that we actually have a representation
${\mathfrak{s}}/{\mathfrak{h}}\stackrel{{\scriptstyle\mathop{\mathrm{ad}}\nolimits}}{{\curvearrowright}}\mathop{\mathrm{Gr}}\nolimits_{\mathfrak{h}}({\mathfrak{s}})$.
Let ${\mathcal{X}}$ denote the set of all surjective homomorphisms
$\varphi\colon{\mathfrak{s}}\to{\mathfrak{s}}/{\mathfrak{h}}$ of Lie algebras.
An element $\varphi\in{\mathcal{X}}$ is just a surjective linear map which
vanishes on $[{\mathfrak{s}},{\mathfrak{s}}]$ since
${\mathfrak{s}}/{\mathfrak{h}}$ is abelian. For any $\varphi\in{\mathcal{X}}$
we form a representation
${\mathfrak{s}}\stackrel{{\scriptstyle\mathop{\mathrm{ad}}\nolimits\circ\varphi}}{{\curvearrowright}}\mathop{\mathrm{Gr}}\nolimits_{\mathfrak{h}}({\mathfrak{s}})$.

######  Theorem 3 (Sufficient condition for parameter rigidity).

If

$H^{1}({\mathcal{F}};{\mathfrak{s}}\stackrel{{\scriptstyle\mathop{\mathrm{ad}}\nolimits\circ\varphi}}{{\curvearrowright}}\mathop{\mathrm{Gr}}\nolimits_{\mathfrak{h}}({\mathfrak{s}}))=H^{1}({\mathfrak{s}};{\mathfrak{s}}\stackrel{{\scriptstyle\mathop{\mathrm{ad}}\nolimits\circ\varphi}}{{\curvearrowright}}\mathop{\mathrm{Gr}}\nolimits_{\mathfrak{h}}({\mathfrak{s}}))$

for some ${\mathfrak{h}}$ and any $\varphi\in{\mathcal{X}}$, then $\rho_{0}$
is parameter rigid.

In general, $[{\mathfrak{s}},{\mathfrak{s}}]\neq{\mathfrak{n}}$ and we choose
${\mathfrak{h}}$ between them depending on the situations. For instance, we
take ${\mathfrak{h}}={\mathfrak{n}}$ in Theorem 4 while we take
${\mathfrak{h}}=[{\mathfrak{s}},{\mathfrak{s}}]$ in Theorem 5.

The assumption requires vanishing of the cohomologies for every
$\varphi\in{\mathcal{X}}$ and this sometimes causes a difficulty when applying
the theorem.111We will see such a situation in Maruhashi [5].  But it is very
likely that for most cases vanishing of the cohomologies for almost all
$\varphi\in{\mathcal{X}}$ is unnecessary, that is, vanishing for finitely many
$\varphi$ is sufficient. To prove such results we have two approaches. First
one is the method appearing in Matsumoto and Mitsumatsu [6, Section 6] which
uses a volume form of the manifold $M$. Here we generalize this method to
obtain the next theorem:

######  Theorem 4.

Assume the following four conditions.

  * •

${\mathfrak{s}}$ is nonunimodular.

  * •

$\dim{\mathfrak{s}}/{\mathfrak{n}}=1$.

  * •

$M$ is orientable.

  * •

$H^{1}\left({\mathcal{F}};{\mathfrak{s}}\stackrel{{\scriptstyle\mathop{\mathrm{ad}}\nolimits}}{{\curvearrowright}}\mathop{\mathrm{Gr}}\nolimits_{\mathfrak{n}}({\mathfrak{s}})\right)=H^{1}\left({\mathfrak{s}};{\mathfrak{s}}\stackrel{{\scriptstyle\mathop{\mathrm{ad}}\nolimits}}{{\curvearrowright}}\mathop{\mathrm{Gr}}\nolimits_{\mathfrak{n}}({\mathfrak{s}})\right)$.

Then $\rho_{0}$ is parameter rigid.

Recall that ${\mathfrak{s}}$ is unimodular if and only if
$\mathop{\mathrm{tr}}\nolimits\mathop{\mathrm{ad}}\nolimits X=0$ for every
$X\in{\mathfrak{s}}$. Note that we have used ${\mathfrak{n}}$ for
${\mathfrak{h}}$. In this theorem only vanishing for the natural projection
${\mathfrak{s}}\twoheadrightarrow{\mathfrak{s}}/{\mathfrak{n}}\in{\mathcal{X}}$
is required by assuming the first three conditions.

The next approach is a new one in which we use large scale geometry of
solvable Lie groups. This can also be applied to unimodular groups. We can
obtain several theorems using this method. Here we appeal to a theorem of
Ogasawara [8]. For $i=1,\ldots,k$, let

$A_{i}=\begin{pmatrix}\alpha^{(i)}_{1}&&\\\ &\ddots&\\\
&&\alpha^{(i)}_{n}\end{pmatrix}$

be a diagonal matrix with positive diagonal entries and put
$A(t)=A_{1}^{t_{1}}\cdots A_{k}^{t_{k}}$ for
$t=(t_{1},\ldots,t_{k})\in{\mathbb{R}}^{k}$. Here $A_{i}^{t_{i}}$ means

$\begin{pmatrix}\left(\alpha^{(i)}_{1}\right)^{t_{i}}&&\\\ &\ddots&\\\
&&\left(\alpha^{(i)}_{n}\right)^{t_{i}}\end{pmatrix}.$

Consider $S_{A}={\mathbb{R}}^{n}\rtimes_{A(t)}{\mathbb{R}}^{k}$. Let $W_{S}$
denote the image of the natural map
$\mathop{\mathrm{Aut}}\nolimits(S)\to\mathop{\mathrm{GL}}\nolimits\left(S/[S,S]\right)$.
We take ${\mathfrak{h}}=[{\mathfrak{s}},{\mathfrak{s}}]$ in the next theorem.

######  Theorem 5.

Assume $S=S_{A}$ satisfies the next condition: For any $j$ there exists $i$
such that $\alpha^{(i)}_{j}\neq 1$. If

$H^{1}\left({\mathcal{F}};{\mathfrak{s}}\stackrel{{\scriptstyle\mathop{\mathrm{ad}}\nolimits\circ\varphi}}{{\curvearrowright}}\mathop{\mathrm{Gr}}\nolimits_{[{\mathfrak{s}},{\mathfrak{s}}]}({\mathfrak{s}})\right)=H^{1}\left({\mathfrak{s}};{\mathfrak{s}}\stackrel{{\scriptstyle\mathop{\mathrm{ad}}\nolimits\circ\varphi}}{{\curvearrowright}}\mathop{\mathrm{Gr}}\nolimits_{[{\mathfrak{s}},{\mathfrak{s}}]}({\mathfrak{s}})\right)$

for all $\varphi\in W_{S}$, then $\rho_{0}$ is parameter rigid.

First $\varphi\in W_{S}$ is regarded as an element of ${\mathcal{X}}$ in the
following way. By pulling back through
$\exp\colon{\mathfrak{s}}/[{\mathfrak{s}},{\mathfrak{s}}]\simeq S/[S,S]$,
$\varphi\colon S/[S,S]\to S/[S,S]$ is regarded as
$\varphi\colon{\mathfrak{s}}/[{\mathfrak{s}},{\mathfrak{s}}]\to{\mathfrak{s}}/[{\mathfrak{s}},{\mathfrak{s}}]$.
Composing with the natural projection
${\mathfrak{s}}\twoheadrightarrow{\mathfrak{s}}/[{\mathfrak{s}},{\mathfrak{s}}]$
we get
$\varphi\colon{\mathfrak{s}}\twoheadrightarrow{\mathfrak{s}}/[{\mathfrak{s}},{\mathfrak{s}}]\in{\mathcal{X}}$.
The second remark is about the condition: For any $j$ there exists $i$ such
that $\alpha^{(i)}_{j}\neq 1$. This is equivalent to $[S,S]={\mathbb{R}}^{n}$.
In this theorem, if we put some genericity condition on $A$, the parameter set
$W_{S}$ becomes finite.

Here we have used a theorem of Ogasawara, but we can also use other theorems
treating rigidity of quasiisometry. We will give other applications in a
forthcoming paper.

Finally we give an application of this method to get parameter rigid actions.
We describe somewhat generalized version of the usual construction of
suspensions of actions. Let
$M_{0}\stackrel{{\scriptstyle\rho_{0}}}{{\curvearrowleft}}H$ be a smooth
locally free action of a connected Lie group $H$ on a closed $C^{\infty}$
manifold $M_{0}$, let $G\stackrel{{\scriptstyle\Phi}}{{\curvearrowright}}H$ be
a smooth action of a connected Lie group $G$ on $H$ by automorphisms and let
$\Gamma\stackrel{{\scriptstyle\lambda}}{{\curvearrowright}}M_{0}$ be a smooth
action of a cocompact lattice $\Gamma$ of $G$ on $M_{0}$. Assume these three
actions satisfy the following compatibility condition:

$\lambda\left(\gamma,\rho_{0}(x,h)\right)=\rho_{0}\left(\lambda(\gamma,x),\Phi_{\gamma}(h)\right)$

for any $\gamma\in\Gamma,x\in M_{0}$ and $h\in H$. Let $H\rtimes_{\Phi}G$ be
the semidirect product whose multiplication is defined by

$(h_{1},g_{1})(h_{2},g_{2})=\left(h_{1}\Phi_{g_{1}}(h_{2}),g_{1}g_{2}\right)$

for $h_{1}$, $h_{2}\in H$ and $g_{1}$, $g_{2}\in G$. We define two actions
$\Gamma\curvearrowright M_{0}\times G\curvearrowleft H\rtimes_{\Phi}G$. The
action of $\Gamma$ is defined diagonally:

$\gamma(x,g)=\left(\lambda(\gamma,x),\gamma g\right)$

for $\gamma\in\Gamma$, $x\in M_{0}$ and $g\in G$. The action of
$H\rtimes_{\Phi}G$ is defined like the multiplication rule of a semidirect
product:

$(x,g)(h,g^{\prime})=\left(\rho_{0}\left(x,\Phi_{g}(h)\right),gg^{\prime}\right)$

for $x\in M_{0}$, $g$, $g^{\prime}\in G$ and $h\in H$. Then these two actions
commute by the compatibility condition. So we get an action

$\Gamma\backslash(M_{0}\times
G)\stackrel{{\scriptstyle\rho}}{{\curvearrowleft}}H\rtimes_{\Phi}G.$

This is locally free and the fiber bundle $\Gamma\backslash(M_{0}\times
G)\to\Gamma\backslash G$ with a typical fiber $M_{0}$ is
$\left(H\rtimes_{\Phi}G\to G\right)$–equivariant. The case in which $H$ is
trivial is the usual construction of suspensions.

We deal with a special case of the above construction. Consider
$A\in\mathop{\mathrm{GL}}\nolimits(n,{\mathbb{Z}})$ and an $A$–invariant
subspace $V$ of ${\mathbb{R}}^{n}$. Take a one parameter subgroup
$\Phi\colon{\mathbb{R}}\to\mathop{\mathrm{GL}}\nolimits(V)$ satisfying
$\Phi_{1}=A|_{V}$. With respect to the above notation, we let

$M_{0}={\mathbb{T}}^{n}={\mathbb{Z}}^{n}\backslash{\mathbb{R}}^{n},\quad
H=V,\quad G={\mathbb{R}}\quad\text{and}\quad\Gamma={\mathbb{Z}},$

and given three actions are
${\mathbb{Z}}^{n}\backslash{\mathbb{R}}^{n}\curvearrowleft V$ by translations,
${\mathbb{Z}}\curvearrowright{\mathbb{T}}^{n}$ by $1\in{\mathbb{Z}}$ acting as
$A$ and ${\mathbb{R}}\stackrel{{\scriptstyle\Phi}}{{\curvearrowright}}V$,
which are compatible in the above sense. The acting group
$S=V\rtimes_{\Phi}{\mathbb{R}}$ is solvable and two actions
${\mathbb{Z}}\curvearrowright{\mathbb{T}}^{n}\times{\mathbb{R}}\curvearrowleft
S$ are defined by

$1(x,t)=(Ax,t+1)\quad\text{and}\quad(x,t)(v,s)=\left(x+\Phi_{t}(v),t+s\right).$

The resulting action is
$M={\mathbb{Z}}\backslash\left({\mathbb{T}}^{n}\times{\mathbb{R}}\right)\curvearrowleft
S$, where $M$ is the mapping torus of $A$. Note that the images
$\Phi_{\mathbb{Z}}$ and $\Phi_{\mathbb{R}}$ of ${\mathbb{Z}}$ and
${\mathbb{R}}$ by $\Phi$ lie in a real algebraic group
$\mathop{\mathrm{GL}}\nolimits(V)$.

######  Theorem 6.

We assume the following four conditions:

  * •

$V$ is Diophantine in ${\mathbb{R}}^{n}$.

  * •

$\Phi_{\mathbb{Z}}$ is Zariski dense in $\Phi_{\mathbb{R}}$, meaning
$\overline{\Phi_{\mathbb{Z}}}=\overline{\Phi_{\mathbb{R}}}$.

  * •

$1$ is not an eigenvalue of $A|_{V}$.

  * •

$A|_{V}$ has an eigenvalue whose absolute value is not $1$.

Then $M\curvearrowleft S$ is parameter rigid.

The Diophantus condition means that there are a basis $v_{1},\ldots,v_{p}$ of
$V$ and positive constants $C$ and $\alpha$ satisfying

$\mathop{\mathrm{max}}\limits_{i}\left|m\cdot
v_{i}\right|\geq\frac{C}{\|m\|^{\alpha}}$

for all $m\in{\mathbb{Z}}^{n}\setminus\\{0\\}$.

This theorem222In a future work, I plan to generalize this and prove it with a
simpler, different calculation.  is a generalization of a theorem of Matsumoto
and Mitsumatsu [6]. They deal with the case when $A$ is hyperbolic with its
characteristic polynomial irreducible over ${\mathbb{Q}}$ and without
eigenvalues on the interval $(-1,0)$, $V$ is the intersection with
${\mathbb{R}}^{n}$ of the direct sum of all eigenspaces of
$A\colon{\mathbb{C}}^{n}\to{\mathbb{C}}^{n}$ with eigenvalues of modulus less
than $1$, and $\Phi$ is the most naturally defined one. In this case $A$ is
diagonalizable since its characteristic polynomial has no multiple roots. So
they consider the suspension Anosov flow of the Anosov diffeomorphism $A$ on
${\mathbb{T}}^{n}$ and its weak stable foliation which will be the orbit
foliation of the action of $V\rtimes_{\Phi}{\mathbb{R}}$. In this setting the
group $V\rtimes_{\Phi}{\mathbb{R}}$ is always non unimodular and Theorem 4 can
be applied.

In our situation $A$ may have nontrivial Jordan blocks, may have the reducible
characteristic polynomial such as $A=\begin{pmatrix}A_{1}&\\\
&A_{2}\end{pmatrix}$ for some
$A_{1}\in\mathop{\mathrm{GL}}\nolimits(k,{\mathbb{Z}})$,
$A_{2}\in\mathop{\mathrm{GL}}\nolimits(n-k,{\mathbb{Z}})$ or $V$ can be some
smaller part of stable directions or a mixture of some stable and unstable
directions, in particular $V\rtimes_{\Phi}{\mathbb{R}}$ can be unimodular. The
method we use to prove Theorem 6 is that using large scale geometry instead of
that by Matsumoto and Mitsumatsu.

## Acknowledgements

This is a part of the Ph.D. Thesis of the author. Most part of the paper was
written when the author was a Research Fellow of the Japan Society for the
Promotion of Science. I would like to thank the advisor, Masayuki Asaoka, and
an anonymous referee, who pointed out the surjectivity in Proposition 33.

##  2 General sufficient condition for parameter rigidity

###  2.1 Proof of Proposition 2

As in Introduction, let
$M\stackrel{{\scriptstyle\rho_{0}}}{{\curvearrowleft}}S$ be an action and
${\mathfrak{s}}\stackrel{{\scriptstyle\pi}}{{\curvearrowright}}V$ be a
representation. Let $S\stackrel{{\scriptstyle\Pi}}{{\curvearrowright}}V$
denote the representation whose differentiation is $\pi$. We define an action
$M\times V\curvearrowleft S$ by

$(x,v)s=\left(\rho_{0}(x,s),\Pi\left(s^{-1}\right)v\right)$

for $(x,v)\in M\times V$ and $s\in S$ and this turns $M\times V\to M$ into an
$S$–equivariant vector bundle. We equip $V$ with a norm which comes from an
inner product. Then the space $\Gamma_{\mathrm{c}ont}(V)$ of all continuous
sections of the trivial vector bundle $M\times V\to M$ is a Banach space with
the supremum norm, and on it, we have a natural representation
$S\curvearrowright\Gamma_{\mathrm{c}ont}(V)$ defined by

$(s\xi)(x)=\Pi(s)\xi\left(\rho_{0}(x,s)\right)$

for $s\in S$, $\xi\in\Gamma_{\mathrm{c}ont}(V)$ and $x\in M$. We regard $V$ as
a subspace of $\Gamma_{\mathrm{c}ont}(V)$ consisting of constant sections.

######  Lemma 7.

There is an $S$–equivariant continuous linear map
$\mu\colon\Gamma_{\mathrm{c}ont}(V)\to V$ which is the identity on $V$.

###### Proof.

Since $S$ is amenable and $M$ is compact, there exists a $\rho_{0}$–invariant
Borel probability measure $\mu$ on $M$. We define
$\mu\colon\Gamma_{\mathrm{c}ont}(V)\to V$ by $\xi\mapsto\int_{M}\xi d\mu$.
Then it is easy to show $\left\|\int_{M}\xi d\mu\right\|\leq\dim
V\left\|\xi\right\|_{\infty}$ and $\mu(s\xi)=\Pi(s)\mu(\xi)$ for all
$\xi\in\Gamma_{\mathrm{c}ont}(V)$ and $s\in S$. ∎

Using the map $\mu$, we define a map
$r\colon\Omega^{*}({\mathcal{F}};V)\to\mathop{\mathrm{Hom}}\nolimits\left(\bigwedge^{*}{\mathfrak{s}},V\right)$
which, on the $p$-th degree, takes $\eta$ to $r(\eta)$ defined by

$r(\eta)(X_{1},\ldots,X_{p})=\mu\left(\eta(X_{1},\ldots,X_{p})\right)$

for $X_{1},\ldots,X_{p}\in{\mathfrak{s}}$. Here $X_{1},\ldots,X_{p}$ are
regarded as vector fields on $M$ using $\rho_{0}$. Namely $X_{i}$ is regarded
as a vector field $x\mapsto(\omega_{0})_{x}^{-1}(X_{i})$ tangent to the
foliation ${\mathcal{F}}$, where $\omega_{0}$ is the canonical $1$–form of
$\rho_{0}$. We will always do this identification during the paper.

Before proving that $r$ is a cochain map, let us look at our connection
closer. Denote by $\nabla$ the covariant derivative with respect to the flat
leafwise connection of the trivial bundle $M\times V\to M$, defined in Section
1. Since the connection form is $\pi\omega_{0}$, we have
$\nabla\xi={d_{\mathcal{F}}}\xi+\pi\omega_{0}\xi$ for a section
$\xi\in\Omega^{0}({\mathcal{F}};V)$. Let
$D\colon\Omega^{p}({\mathcal{F}};V)\to\Omega^{p+1}({\mathcal{F}};V)$ denote
the covariant exterior derivative arising from $\nabla$. Then it is easy to
check that $D\eta={d_{\mathcal{F}}}\eta+\pi\omega_{0}\wedge\eta$ for
$\eta\in\Omega^{p}({\mathcal{F}};V)$.

Next let us see which sections are parallel, that is, sections
$\xi\in\Omega^{0}({\mathcal{F}};V)$ satisfying $\nabla\xi=0$. Fix a point
$x_{0}\in M$ and a vector $v\in V$. Define $\xi_{0}$ locally along the leaf
passing through $x_{0}$ by
$\xi_{0}\left(\rho_{0}(x_{0},s)\right)=\Pi\left(s^{-1}\right)v$ for $s\in S$
close to the identity. Then for any $y=\rho_{0}(x_{0},s_{0})$ with small
$s_{0}\in S$ and any $Y\in{\mathfrak{s}}$, we have

$\displaystyle\nabla_{\left.\frac{d}{dt}\rho_{0}\left(y,e^{tY}\right)\right|_{t=0}}\xi_{0}$
$\displaystyle={d_{\mathcal{F}}}\xi_{0}\left(\left.\frac{d}{dt}\rho_{0}\left(y,e^{tY}\right)\right|_{t=0}\right)+\pi(Y)\xi_{0}(y)$
$\displaystyle=\left.\frac{d}{dt}\Pi\left(e^{-tY}s_{0}^{-1}\right)v\right|_{t=0}+\pi(Y)\Pi\left(s_{0}^{-1}\right)v$
$\displaystyle=0.$

Therefore $\nabla\xi_{0}=0$ and this means the directions of orbits of the
action $M\times V\curvearrowleft S$ is horizontal for the leafwise connection.
So we have

$\displaystyle\left(\nabla_{X}\xi\right)(x)$ $\displaystyle=\lim_{t\rightarrow
0}\frac{\Pi\left(e^{tX}\right)\xi\left(\rho_{0}\left(x,e^{tX}\right)\right)-\xi(x)}{t}$
$\displaystyle=\lim_{t\rightarrow
0}\frac{\left(e^{tX}\xi\right)(x)-\xi(x)}{t}$

for any $\xi\in\Omega^{0}({\mathcal{F}};V)$, $X\in{\mathfrak{s}}$ and $x\in
M$.

######  Lemma 8.

$\frac{e^{tX}\xi-\xi}{t}$ converges uniformly to $\nabla_{X}\xi$ as
$t\rightarrow 0$.

###### Proof.

Take a basis $v_{1},\ldots,v_{l}$ of $V$ and write
$\left(e^{tX}\xi\right)(x)=\sum_{i=1}^{l}f_{i}(t,x)v_{i}$ for some real valued
functions $f_{i}$. Then
$\left(\nabla_{X}\xi\right)(x)=\sum_{i=1}^{l}f_{i}^{\prime}(0,x)v_{i}$. The
function $f_{i}(t,x)$ has the Taylor expansion

$f_{i}(t,x)=f_{i}(0,x)+tf_{i}^{\prime}(0,x)+\frac{t^{2}}{2}f_{i}^{\prime\prime}\left(\theta_{i,x,t},x\right),$

where $\theta_{i,x,t}$ is a number between $0$ and $t$. Since

$\frac{\left(e^{tX}\xi\right)(x)-\xi(x)}{t}-\left(\nabla_{X}\xi\right)(x)=\frac{t}{2}\sum_{i=1}^{l}f_{i}^{\prime\prime}\left(\theta_{i,x,t},x\right)v_{i}$

and $f_{i}^{\prime\prime}(\theta,x)$ is bounded for $-1\leq\theta\leq 1$ and
$x\in M$, we get the conclusion. ∎

Recall that we have a cochain map
$\omega_{0}^{*}\colon\mathop{\mathrm{Hom}}\nolimits\left(\bigwedge^{*}{\mathfrak{s}},V\right)\hookrightarrow\Omega^{*}({\mathcal{F}};V)$.

######  Lemma 9.

The map $r$ is a cochain map and $r\circ\omega_{0}^{*}$ is the identity.
Therefore $\omega_{0}^{*}$ induces the injective map between cohomologies.

###### Proof.

Using definitions we verify easily that $r\circ\omega_{0}^{*}$ is the
identity.

By Lemma 8, we have

$\displaystyle\mu\left(\nabla_{X}\xi\right)$ $\displaystyle=\lim_{t\rightarrow
0}\mu\left(\frac{e^{tX}\xi-\xi}{t}\right)$ $\displaystyle=\lim_{t\rightarrow
0}\frac{\Pi\left(e^{tX}\right)\mu(\xi)-\mu(\xi)}{t}$
$\displaystyle=\pi(X)\mu(\xi)$

for $X\in{\mathfrak{s}}$ and $\xi\in\Omega^{0}({\mathcal{F}};V)$. By this
property of $\mu$ and a direct calculation, we see that $r$ is a cochain map.
∎

###  2.2 Proof of Theorem 3

Let $M\stackrel{{\scriptstyle\rho_{0}}}{{\curvearrowleft}}S$ be an action,
${\mathcal{F}}$ be its orbit foliation and $\omega_{0}$ be the canonical
$1$–form of $\rho_{0}$. The set of all smooth actions $M\curvearrowleft S$
with the orbit foliation ${\mathcal{F}}$ is denoted by $A({\mathcal{F}},S)$.
We will prove that any $\rho\in A({\mathcal{F}},S)$ is parameter equivalent to
$\rho_{0}$ under the assumption of Theorem 3. Let $\omega$ be the canonical
$1$–form of $\rho$. To show that $\rho$ is parameter equivalent to $\rho_{0}$,
it is sufficient to prove the existence of some $C^{\infty}$ map $P\colon M\to
S$ and some endomorphism $\Phi\colon S\to S$ satisfying

$\omega=\mathop{\mathrm{Ad}}\nolimits\left(P^{-1}\right)\Phi_{*}\omega_{0}+P^{*}\Theta,$

where $\Theta\in\Omega^{1}(S;{\mathfrak{s}})$ is the left Maurer–Cartan form
of $S$. See, for instance, Asaoka [1, Proposition 1.4.4]. In other words, we
will show that the ${\mathfrak{s}}$–valued cocycle $\omega$ is cohomologous to
a constant ${\mathfrak{s}}$–valued cocycle $\Phi_{*}\omega_{0}$. We call
$\omega$ an ${\mathfrak{s}}$–valued cocycle because it is the infinitesimal
version of a usual $S$–valued cocycle over $\rho_{0}$. In our situation, usual
$K$–valued cocycles over $\rho_{0}$ for some connected simply connected Lie
group $K$ are in one-to-one correspondence with elements
$\eta\in\Omega^{1}({\mathcal{F}};{\mathfrak{k}})$ satisfying

${d_{\mathcal{F}}}\eta+[\eta,\eta]=0.$

Two ${\mathfrak{k}}$–valued cocycles $\eta_{1}$, $\eta_{2}$ are cohomologous
if and only if

$\eta_{1}=\mathop{\mathrm{Ad}}\nolimits\left(P^{-1}\right)\eta_{2}+P^{*}\Theta_{K}$

for some smooth $P\colon M\to K$, where $\Theta_{K}$ denotes the left
Maurer–Cartan form on $K$. Also, $\eta$ is a constant cocycle if and only if
$\eta=\Phi_{*}\omega_{0}$ for some homomorphism $\Phi\colon S\to K$. See
Asaoka [1, Section 1.4.1] for more details.

Now recall that we have a filtration

${\mathfrak{s}}\supset{\mathfrak{h}}\supset{\mathfrak{h}}^{2}\supset\cdots\supset{\mathfrak{h}}^{d}\supset
0.$

Fix complementary subspaces $V_{i}$ for $i=0,\ldots,d$ so that

${\mathfrak{s}}=V_{0}\oplus{\mathfrak{h}}\quad\text{and}\quad{\mathfrak{h}}^{i}=V_{i}\oplus{\mathfrak{h}}^{i+1}.$

The adjoint representations
${\mathfrak{s}}\curvearrowright{\mathfrak{s}}/{\mathfrak{h}}$ and
${\mathfrak{s}}\curvearrowright{\mathfrak{h}}^{i}/{\mathfrak{h}}^{i+1}$ are
canonically identified with representations333Here we do not assume $V_{i}$
are invariant under $\mathop{\mathrm{ad}}\nolimits$.  of ${\mathfrak{s}}$ on
$V_{i}$, which we call $\pi_{i}$. The representation $\pi_{0}$ is a multiple
of the trivial representation. For any element $X\in{\mathfrak{s}}$, let
$X^{i}$ be the $V_{i}$–component with respect to the decomposition
${\mathfrak{s}}=\bigoplus_{i=0}^{d}V_{i}$, so that we have
$X=X^{0}+X^{1}+\cdots+X^{d}$. In this section we use this upper right symbol
$i$ as the projection operator onto $V_{i}$, or to indicate the element
belongs to $V_{i}$. Accordingly, $\omega$ is decomposed as
$\omega=\omega^{0}+\cdots+\omega^{d}$.

By looking at the $V_{0}$–component of the equation
${d_{\mathcal{F}}}\omega+[\omega,\omega]=0$, we get
${d_{\mathcal{F}}}\omega^{0}=0$. Our assumption
$H^{1}({\mathcal{F}};V_{0})=H^{1}({\mathfrak{s}};V_{0})$ implies that there
are a linear map $\varphi\colon{\mathfrak{s}}\to V_{0}$ vanishing on
$[{\mathfrak{s}},{\mathfrak{s}}]$ and a smooth map $h\colon M\to V_{0}$
satisfying

$\omega^{0}=\varphi\omega_{0}+{d_{\mathcal{F}}}h.$ (1)

We also write $\varphi_{\rho}$ for $\varphi$. This $\varphi_{\rho}$ plays an
important role (or causes a trouble) in our problem444By (1),
$\varphi_{\rho}=r\left(\omega^{0}\right)$ for $r$ defined in the previous
section. By this formula, we can define $\varphi_{\rho}$ without any
assumption on the cohomology. But in this case the definition might depend on
the choice of an $S$–invariant Borel probability measure $\mu$. .
Unexpectedly, to determine $\varphi_{\rho}$ is not so easy. If we can show
$\varphi_{\rho}$ has some restricted form, we can weaken the assumption of
vanishing of cohomologies to a smaller subset of ${\mathcal{X}}$. This is what
we will do in Section 3 and Section 4. Here let us see some properties of
$\varphi_{\rho}$.

Let $a\colon M\times S\to S$ be the unique smooth map satisfying

$\rho_{0}(x,s)=\rho\left(x,a(x,s)\right)\quad\text{and}\quad a(x,1)=1.$

This is defined since $\rho_{0}$ and $\rho$ have the same orbit foliation. The
map $a$ is a cocycle over $\rho_{0}$ and is important in our problem.

######  Lemma 10.

For any $x\in M$ and $s\in S$,

$\int_{1}^{s}\varphi_{\rho}\Theta+h\left(\rho_{0}(x,s)\right)-h(x)=\int_{1}^{a(x,s)}\Theta^{0}.$

###### Proof.

Since $\Theta$ satisfies $d\Theta+[\Theta,\Theta]=0$, we see that
$\varphi_{\rho}\Theta$ and $\Theta^{0}$ are $V_{0}$–valued closed $1$–forms on
$S$. Fix $x\in M$ and $s\in S$. Then by (1),

$\int_{x}^{\rho_{0}(x,s)}\left(\varphi_{\rho}\omega_{0}+{d_{\mathcal{F}}}h\right)=\int_{x}^{\rho\left(x,a(x,s)\right)}\omega^{0}.$

These integrals are along a curve contained in a leaf. Take a curve
$\gamma(t)$, $0\leq t\leq 1$, on $S$ connecting $1$ and $s$. Then the left
hand side of the above equation is

$\displaystyle\int_{0}^{1}\left\\{\varphi_{\rho}\omega_{0}\left(\frac{d}{dt}\rho_{0}\left(x,\gamma(t)\right)\right)+{d_{\mathcal{F}}}h\left(\frac{d}{dt}\rho_{0}\left(x,\gamma(t)\right)\right)\right\\}dt$
$\displaystyle=\int_{0}^{1}\varphi_{\rho}\Theta\left(\frac{d}{dt}\gamma(t)\right)dt+h\left(\rho_{0}(x,s)\right)-h(x)$

and to compute the right hand side, take a curve $\gamma_{1}(t)$ connecting
$1$ and $a(x,s)$ and then

$\int_{0}^{1}\omega^{0}\left(\frac{d}{dt}\rho\left(x,\gamma_{1}(t)\right)\right)dt=\int_{0}^{1}\Theta^{0}\left(\frac{d}{dt}\gamma_{1}(t)\right)dt.$

∎

######  Lemma 11.

The map $\varphi\colon{\mathfrak{s}}\to V_{0}$ is surjective.

###### Proof.

Let $q\colon{\mathfrak{s}}\to{\mathfrak{s}}/\ker\varphi_{\rho}$ be the natural
projection and let
$\bar{\varphi}_{\rho}\colon{\mathfrak{s}}/\ker\varphi_{\rho}\to V_{0}$ be the
induced map from $\varphi_{\rho}$. The map $S\to V_{0}$ mapping $s$ to
$\int_{1}^{s}\Theta^{0}$ is surjective because
$\int_{1}^{e^{X}}\Theta^{0}=X^{0}$ for every $X\in{\mathfrak{s}}$. Fix a point
$x\in M$. The map $S\to S$ defined by $s\mapsto a(x,s)$ is bijective. See
Asaoka [1, Lemma 1.4.6]. By the above lemma, we have

$\bar{\varphi}_{\rho}\left(\int_{1}^{s}q\Theta\right)=\int_{1}^{a(x,s)}\Theta^{0}-h\left(\rho_{0}(x,s)\right)+h(x).$

This and the boundedness of $h$ show $\bar{\varphi}_{\rho}$ is surjective. ∎

Therefore we can regard $\varphi$ as an element of ${\mathcal{X}}$.

######  Lemma 12.

Set $P=e^{h}\colon M\to S$. Then

$\mathop{\mathrm{Ad}}\nolimits(P)\left(\omega-P^{*}\Theta\right)=\varphi\omega_{0}+\bar{\omega}^{1}+\cdots+\bar{\omega}^{d}$

for some leafwise $1$–forms $\bar{\omega}^{i}$ with values in $V_{i}$. So we
have

$\omega=\mathop{\mathrm{Ad}}\nolimits\left(P^{-1}\right)\left(\varphi\omega_{0}+\bar{\omega}^{1}+\cdots+\bar{\omega}^{d}\right)+P^{*}\Theta.$
(2)

###### Proof.

First we show

$\displaystyle P^{*}\Theta$
$\displaystyle=\sum_{j=0}^{\infty}(-1)^{j}\frac{(\mathop{\mathrm{ad}}\nolimits
h)^{j}}{(j+1)!}{d_{\mathcal{F}}}h$
$\displaystyle={d_{\mathcal{F}}}h-\frac{1}{2}(\mathop{\mathrm{ad}}\nolimits
h){d_{\mathcal{F}}}h+\cdots.$

This is because for any point $x\in M$ and $X\in T_{x}{\mathcal{F}}$, we have

$\displaystyle\left(P^{*}\Theta\right)(X)$
$\displaystyle=\Theta\left.\frac{d}{dt}e^{h(x(t))}\right|_{t=0}=\left(L_{e^{-h(x)}}\right)_{*}\left.\frac{d}{dt}e^{h(x(t))}\right|_{t=0}$
$\displaystyle=\left.\frac{d}{dt}e^{-h(x)}e^{h(x)+h(x(t))-h(x)}\right|_{t=0}$
$\displaystyle=\sum_{j=0}^{\infty}(-1)^{j}\frac{(\mathop{\mathrm{ad}}\nolimits
h(x))^{j}}{(j+1)!}Xh,$

where $x(t)$ is a curve satisfying $\left.\frac{d}{dt}x(t)\right|_{t=0}=X$.

So
$\omega-P^{*}\Theta=\varphi\omega_{0}+\bar{\bar{\omega}}^{1}+\cdots+\bar{\bar{\omega}}^{d}$
for some $\bar{\bar{\omega}}^{i}$ taking values in $V_{i}$. Since
${\mathfrak{s}}/{\mathfrak{h}}$ is an abelian Lie algebra,
$S\stackrel{{\scriptstyle\mathop{\mathrm{Ad}}\nolimits}}{{\curvearrowright}}{\mathfrak{s}}/{\mathfrak{h}}$
is trivial. Therefore
$\mathop{\mathrm{Ad}}\nolimits(P)\left(\omega-P^{*}\Theta\right)=\varphi\omega_{0}+\bar{\omega}^{1}+\cdots+\bar{\omega}^{d}$
for some $\bar{\omega}^{i}$. ∎

By this lemma we can replace $\omega$ by a cohomologous cocycle whose
$V_{0}$–component is constant. We say an element of
$\Omega^{*}({\mathcal{F}};W)$ for some vector space $W$ is constant if it lies
in the image of
$\omega_{0}^{*}\colon\mathop{\mathrm{Hom}}\nolimits\left(\bigwedge^{*}{\mathfrak{s}},W\right)\hookrightarrow\Omega^{*}({\mathcal{F}};W)$.
Replacing by cohomologous cocycles, we will gradually make components
constant, and finally get a constant cocycle. So now we may assume
$\omega=\varphi\omega_{0}+\omega^{1}+\cdots+\omega^{d}$ and proceed to the
next step.

######  Lemma 13.

Assume that

$\omega=\varphi\omega_{0}+\varphi^{1}\omega_{0}+\cdots+\varphi^{k-1}\omega_{0}+\omega^{k}+\cdots+\omega^{d}$

for some linear maps $\varphi^{i}\colon{\mathfrak{s}}\to V_{i}$, that is,
$\omega$ is already constant up to the $V_{k-1}$–component. Then we can choose
some smooth $P\colon M\to S$ so that

$\mathop{\mathrm{Ad}}\nolimits(P)\left(\omega-P^{*}\Theta\right)=\varphi\omega_{0}+\varphi^{1}\omega_{0}+\cdots+\varphi^{k-1}\omega_{0}+\varphi^{k}\omega_{0}+\bar{\omega}^{k+1}+\cdots+\bar{\omega}^{d}$

for some linear map $\varphi^{k}\colon{\mathfrak{s}}\to V_{k}$ and
$\bar{\omega}^{i}$.

###### Proof.

Looking at the $V_{k}$–component of the equation
${d_{\mathcal{F}}}\omega+[\omega,\omega]=0$, we obtain

$\displaystyle 0$
$\displaystyle={d_{\mathcal{F}}}\omega^{k}+\left[\varphi\omega_{0}+\varphi^{1}\omega_{0}+\cdots+\varphi^{k-1}\omega_{0}+\omega^{k},\varphi\omega_{0}+\varphi^{1}\omega_{0}+\cdots+\varphi^{k-1}\omega_{0}+\omega^{k}\right]^{k}$
$\displaystyle={d_{\mathcal{F}}}\omega^{k}+\pi_{k}\varphi\omega_{0}\wedge\omega^{k}+\text{constant
form}.$

The $k$ appearing in $[\cdots,\cdots]^{k}$ in the first line of the above
denotes the projection onto $V_{k}$. Let
$D\colon\Omega^{p}({\mathcal{F}};V_{k})\to\Omega^{p+1}({\mathcal{F}};V_{k})$
be the covariant exterior derivative arising from the leafwise connection
defined by connection form $\pi_{k}\varphi\omega_{0}$. We saw
$D={d_{\mathcal{F}}}+\pi_{k}\varphi\omega_{0}\wedge$ in the previous section,
so that

$D\omega^{k}=\omega_{0}^{*}\psi$ (3)

for some
$\psi\in\mathop{\mathrm{Hom}}\nolimits\left(\bigwedge^{2}{\mathfrak{s}},V_{k}\right)$
by the above computation. Recall $r$ which is defined in the previous section.
We set $\theta=r\left(\omega^{k}\right)\colon{\mathfrak{s}}\to V_{k}$. Then

$\psi=r\left(\omega_{0}^{*}\psi\right)=r\left(D\omega^{k}\right)=Dr\left(\omega^{k}\right)=D\theta.$
(4)

Here $D$ also denotes the differential of
$\mathop{\mathrm{Hom}}\nolimits\left(\bigwedge^{*}{\mathfrak{s}},V_{k}\right)$.
By (3) and (4), we get

$D\left(\omega^{k}-\omega_{0}^{*}\theta\right)=0.$

By Lemma 11 and by our assumption, we have
$H^{1}\left({\mathcal{F}};{\mathfrak{s}}\stackrel{{\scriptstyle\pi_{k}\varphi}}{{\curvearrowright}}V_{k}\right)=H^{1}\left({\mathfrak{s}};{\mathfrak{s}}\stackrel{{\scriptstyle\pi_{k}\varphi}}{{\curvearrowright}}V_{k}\right)$.
Therefore there exist a linear map $\theta^{\prime}\colon{\mathfrak{s}}\to
V_{k}$ and a smooth map $h\colon M\to V_{k}$ satisfying

$\displaystyle\omega^{k}$
$\displaystyle=\theta\omega_{0}+\theta^{\prime}\omega_{0}+{d_{\mathcal{F}}}h+\pi_{k}\varphi\omega_{0}h$
$\displaystyle=\varphi^{k}\omega_{0}+{d_{\mathcal{F}}}h+\pi_{k}\varphi\omega_{0}h.$

Here we set $\varphi^{k}=\theta+\theta^{\prime}$. As before we let $P=e^{h}$
and then

$\omega-P^{*}\Theta=\varphi\omega_{0}+\varphi^{1}\omega_{0}+\cdots+\varphi^{k-1}\omega_{0}+\left(\varphi^{k}\omega_{0}+\pi_{k}\varphi\omega_{0}h\right)+\bar{\bar{\omega}}^{k+1}+\cdots+\bar{\bar{\omega}}^{d}$

for some $\bar{\bar{\omega}}^{i}$. Finally we compute as follows:

$\displaystyle\mathop{\mathrm{Ad}}\nolimits(P)\left(\omega-P^{*}\Theta\right)$
$\displaystyle=e^{\mathop{\mathrm{ad}}\nolimits
h}\left(\varphi\omega_{0}+\varphi^{1}\omega_{0}+\cdots+\varphi^{k}\omega_{0}+\pi_{k}\varphi\omega_{0}h\right)+\text{higher
terms}$
$\displaystyle=\varphi\omega_{0}+\varphi^{1}\omega_{0}+\cdots+\varphi^{k}\omega_{0}+\pi_{k}\varphi\omega_{0}h-\pi_{k}\varphi\omega_{0}h+\text{higher
terms}$
$\displaystyle=\varphi\omega_{0}+\varphi^{1}\omega_{0}+\cdots+\varphi^{k}\omega_{0}+\text{higher
terms}.$

∎

Applying this lemma repeatedly, we see that given $\omega$ is cohomologous to
a cocycle of the form
$\omega^{\prime}=\varphi\omega_{0}+\varphi^{1}\omega_{0}+\cdots+\varphi^{d}\omega_{0}$.
Set
$\Phi_{*}=\varphi+\varphi^{1}+\cdots+\varphi^{d}\colon{\mathfrak{s}}\to{\mathfrak{s}}$.
Then $\omega^{\prime}=\Phi_{*}\omega_{0}$ is a constant cocycle, because the
equation
${d_{\mathcal{F}}}\omega^{\prime}+\left[\omega^{\prime},\omega^{\prime}\right]=0$
implies that $\Phi_{*}$ is an endomorphism of the Lie algebra
${\mathfrak{s}}$. This completes the proof of Theorem 3.

Let ${\mathcal{X}}_{\rho_{0}}$ be the set of all $\varphi\in{\mathcal{X}}$
which can be written as $\varphi_{\rho}$ for some $\rho\in A({\mathcal{F}},S)$
(using the isomorphism ${\mathfrak{s}}/{\mathfrak{h}}\simeq V_{0}$.) What we
actually proved in this section is the following:

######  Theorem 14.

Assume $H^{1}({\mathcal{F}})=H^{1}({\mathfrak{s}})$, so that we can define
${\mathcal{X}}_{\rho_{0}}$ as a subset of ${\mathcal{X}}$. If

$H^{1}\left({\mathcal{F}};{\mathfrak{s}}\stackrel{{\scriptstyle\mathop{\mathrm{ad}}\nolimits\circ\varphi}}{{\curvearrowright}}\mathop{\mathrm{Gr}}\nolimits_{\mathfrak{h}}({\mathfrak{s}})\right)=H^{1}\left({\mathfrak{s}};{\mathfrak{s}}\stackrel{{\scriptstyle\mathop{\mathrm{ad}}\nolimits\circ\varphi}}{{\curvearrowright}}\mathop{\mathrm{Gr}}\nolimits_{\mathfrak{h}}({\mathfrak{s}})\right)$

for all $\varphi\in{\mathcal{X}}_{\rho_{0}}$, then $\rho_{0}$ is parameter
rigid.

The next task is trying to prove the set ${\mathcal{X}}_{\rho_{0}}$ is small.

######  Remark 15.

Although we deal with only ${\mathfrak{s}}$–valued cocycles arising from
actions $\rho\in A({\mathcal{F}},S)$, what we do in this section is actually
valid for any ${\mathfrak{k}}$–valued cocycles over $\rho_{0}$ for any
connected simply connected solvable Lie group $K$ and a subspace
${\mathfrak{h}}$ between $[{\mathfrak{k}},{\mathfrak{k}}]$ and the nilradical
of ${\mathfrak{k}}$. Also the acting group $S$ need not be solvable, we need
only assume that the action has an invariant Borel probability measure.
Solvability is used only for the value group $K$. This might be useful for
purposes other than parameter rigidity.

######  Remark 16.

For actions of semisimple Lie groups or groups with property (T),
${\mathbb{R}}$–valued cocycle rigidity implies $K$–valued cocycle rigidity for
any connected simply connected solvable Lie groups. This is shown by an
obvious argument and valid for actions in broader categories.

######  Remark 17.

As explained in Asaoka [1, Section 1.4.4],
$H^{1}\left({\mathcal{F}};{\mathfrak{s}}\stackrel{{\scriptstyle\mathop{\mathrm{ad}}\nolimits}}{{\curvearrowright}}{\mathfrak{s}}\right)\left/H^{1}\left({\mathfrak{s}};{\mathfrak{s}}\stackrel{{\scriptstyle\mathop{\mathrm{ad}}\nolimits}}{{\curvearrowright}}{\mathfrak{s}}\right)\right.$
can be viewed as the formal tangent space at $\rho_{0}$ in
$A({\mathcal{F}},S)/\text{(parameter equivalence)}$. We can show that
$H^{1}\left({\mathcal{F}};{\mathfrak{s}}\stackrel{{\scriptstyle\mathop{\mathrm{ad}}\nolimits}}{{\curvearrowright}}\mathop{\mathrm{Gr}}\nolimits_{\mathfrak{h}}({\mathfrak{s}})\right)=H^{1}\left({\mathfrak{s}};{\mathfrak{s}}\stackrel{{\scriptstyle\mathop{\mathrm{ad}}\nolimits}}{{\curvearrowright}}\mathop{\mathrm{Gr}}\nolimits_{\mathfrak{h}}({\mathfrak{s}})\right)$
implies

$H^{1}\left({\mathcal{F}};{\mathfrak{s}}\stackrel{{\scriptstyle\mathop{\mathrm{ad}}\nolimits}}{{\curvearrowright}}{\mathfrak{s}}\right)\left/H^{1}\left({\mathfrak{s}};{\mathfrak{s}}\stackrel{{\scriptstyle\mathop{\mathrm{ad}}\nolimits}}{{\curvearrowright}}{\mathfrak{s}}\right)\right.=0$

by an argument using spectral sequences. But the converse seems false.

##  3 Sufficient condition by the method of Matsumoto and Mitsumatsu

We prove Theorem 4 here. Let
$M\stackrel{{\scriptstyle\rho_{0}}}{{\curvearrowleft}}S$ be an action which we
consider and take any $\rho\in A({\mathcal{F}},S)$. According to Theorem 14,
what we need to show is $\varphi_{\rho}$, defined in Section 2.2, coincides
with the natural projection
${\mathfrak{s}}\twoheadrightarrow{\mathfrak{s}}/{\mathfrak{n}}$. In this
section we also use the notation $\rho_{0}^{s}(x)=\rho_{0}(x,s)$ and
$\rho^{s}(x)=\rho(x,s)$. As in Section 2.2 we have an $S$–valued cocycle
$a\colon M\times S\to S$ over $\rho_{0}$ satisfying
$\rho_{0}^{s}(x)=\rho^{a(x,s)}(x)$ for all $x\in M$ and $s\in S$. For any
$X\in{\mathfrak{s}}$, $s\in S$ and $x\in M$,

$\displaystyle\left(\rho_{0}^{s}\right)_{*}X_{x}$
$\displaystyle=\left.\frac{d}{dt}\rho_{0}\left(x,e^{tX}s\right)\right|_{t=0}$
$\displaystyle=\left.\frac{d}{dt}\rho_{0}\left(\rho_{0}^{s}(x),e^{t\mathop{\mathrm{Ad}}\nolimits\left(s^{-1}\right)X}\right)\right|_{t=0}$
$\displaystyle=\left(\mathop{\mathrm{Ad}}\nolimits\left(s^{-1}\right)X\right)_{\rho_{0}^{s}(x)}.$

Take a basis of ${\mathfrak{s}}$ and its dual basis of ${\mathfrak{s}}^{*}$.
The dual basis of ${\mathfrak{s}}^{*}$, after being pulled back by the
canonical $1$–form $\omega_{0}$ of $\rho_{0}$, is regarded as a global frame
of the bundle $T^{*}{\mathcal{F}}$. Let $\Omega_{0}\in\Omega^{\dim
S}({\mathcal{F}})$ be the wedge product of the global frame, which is a
leafwise volume form. By the above computation, we see

$\left(\rho_{0}^{s}\right)^{*}\Omega_{0}=\det\mathop{\mathrm{Ad}}\nolimits\left(s^{-1}\right)\Omega_{0}$

for all $s\in S$. We do the same thing for $\rho$, getting another leafwise
volume form $\Omega\in\Omega^{\dim S}({\mathcal{F}})$ which satisfies

$\left(\rho^{s}\right)^{*}\Omega=\det\mathop{\mathrm{Ad}}\nolimits\left(s^{-1}\right)\Omega$

for all $s\in S$. Here we must use the canonical $1$–form $\omega$ of $\rho$
rather than $\omega_{0}$. Fix a complementary subbundle $E$ to
$T{\mathcal{F}}$ in $TM$; $TM=T{\mathcal{F}}\oplus E$. Since $M$ is
orientable, we can choose a nowhere vanishing smooth section
$\Omega_{\text{tr}}$ of $\bigwedge^{\dim
E}\left(TM/T{\mathcal{F}}\right)^{*}$. We have natural projections
$T{\mathcal{F}}\leftarrow TM\to TM/T{\mathcal{F}}$ defined by $E$. Let
${\overline{\Omega_{0}}}$, ${\overline{\Omega}}\in\Omega^{\dim S}(M)$ and
${\overline{\Omega_{\text{tr}}}}\in\Omega^{\dim E}(M)$ be the pull backs of
$\Omega_{0}$, $\Omega$ and $\Omega_{\text{tr}}$ by the projections. In this
section, bars written over something stand for pulling back something by the
projections. Both
${\overline{\Omega_{0}}}\wedge{\overline{\Omega_{\text{tr}}}}$,
${\overline{\Omega}}\wedge{\overline{\Omega_{\text{tr}}}}\in\Omega^{\dim
M}(M)$ are volume forms of $M$. So there is a smooth map $c\colon M\times
S\to{\mathbb{R}}_{>0}$ satisfying

$(\rho^{s})^{*}\left({\overline{\Omega}}\wedge{\overline{\Omega_{\text{tr}}}}\right)=c(\cdotp,s)\left({\overline{\Omega}}\wedge{\overline{\Omega_{\text{tr}}}}\right)$

for all $s\in S$. It is easy to see that $c$ is a cocycle over $\rho$, that
is,

$c(x,ss^{\prime})=c(x,s)c\left(\rho(x,s),s^{\prime}\right)$

for all $x\in M$ and $s,s^{\prime}\in S$. Our assumption
$H^{1}({\mathcal{F}})=H^{1}({\mathfrak{s}})$ is equivalent to
${\mathbb{R}}$–valued cocycle rigidity of $\rho$. See Maruhashi [4, Section 2]
for instance. Thus we can find a homomorphism $\alpha\colon
S\to{\mathbb{R}}_{>0}$ and a smooth map $P\colon M\to{\mathbb{R}}_{>0}$ so
that

$c(x,s)=P(x)^{-1}\alpha(s)P\left(\rho(x,s)\right)$

holds for all $x\in M$ and $s\in S$. Then for any $s\in S$,

$\displaystyle(\rho^{s})^{*}\left(P^{-1}\left({\overline{\Omega}}\wedge{\overline{\Omega_{\text{tr}}}}\right)\right)$
$\displaystyle=P\left(\rho^{s}(\cdotp)\right)^{-1}c(\cdotp,s)\left({\overline{\Omega}}\wedge{\overline{\Omega_{\text{tr}}}}\right)$
$\displaystyle=\alpha(s)P^{-1}\left({\overline{\Omega}}\wedge{\overline{\Omega_{\text{tr}}}}\right).$

By integrating over $M$, we get $\alpha(s)=1$ for all $s\in S$. By replacing
$P^{-1}\Omega_{\text{tr}}$ by $\Omega_{\text{tr}}$ we may assume that

$(\rho^{s})^{*}\left({\overline{\Omega}}\wedge{\overline{\Omega_{\text{tr}}}}\right)={\overline{\Omega}}\wedge{\overline{\Omega_{\text{tr}}}}$
(5)

for all $s\in S$. For any $x\in M$ and $s\in S$, two maps

$\left(\rho_{0}^{s}\right)_{*},\left(\rho^{a(x,s)}\right)_{*}\colon\left(TM/T{\mathcal{F}}\right)_{x}\to\left(TM/T{\mathcal{F}}\right)_{\rho_{0}^{s}(x)=\rho^{a(x,s)}(x)}$

coincide. This is because $\rho_{0}$ and $\rho$ have the same orbit foliation
and it is easy to see this for small $s\in S$. For two small $s$,
$s^{\prime}\in S$, we have

${\left(TM/T{\mathcal{F}}\right)_{x}}$${\left(TM/T{\mathcal{F}}\right)_{\rho_{0}^{s}(x)}}$${\left(TM/T{\mathcal{F}}\right)_{\rho_{0}^{ss^{\prime}}(x)}}$${\left(TM/T{\mathcal{F}}\right)_{x}}$${\left(TM/T{\mathcal{F}}\right)_{\rho_{0}^{s}(x)}}$${\left(TM/T{\mathcal{F}}\right)_{\rho_{0}^{ss^{\prime}}(x)},}$$\scriptstyle{\left(\rho_{0}^{s}\right)_{*}}$${\circlearrowright}$$\scriptstyle{\left(\rho_{0}^{s^{\prime}}\right)_{*}}$${\circlearrowright}$$\scriptstyle{\left(\rho^{a(x,s)}\right)_{*}}$$\scriptstyle{\left(\rho^{a\left(\rho_{0}^{s}(x),s^{\prime}\right)}\right)_{*}}$

so that

$\displaystyle\left(\rho_{0}^{ss^{\prime}}\right)_{*}$
$\displaystyle=\left(\rho_{0}^{s^{\prime}}\right)_{*}\circ\left(\rho_{0}^{s}\right)_{*}=\left(\rho^{a\left(\rho_{0}^{s}(x),s^{\prime}\right)}\right)_{*}\circ\left(\rho^{a(x,s)}\right)_{*}$
$\displaystyle=\left(\rho^{a(x,s)a\left(\rho_{0}^{s}(x),s^{\prime}\right)}\right)_{*}=\left(\rho^{a\left(x,ss^{\prime}\right)}\right)_{*}.$

Using this we can prove for general $s\in S$. Therefore

$\left(\left(\rho_{0}^{s}\right)^{*}\Omega_{\text{tr}}\right)_{x}=\left(\left(\rho^{a(x,s)}\right)^{*}\Omega_{\text{tr}}\right)_{x}$
(6)

for all $x\in M$ and $s\in S$. Let $\beta\colon M\times S\to{\mathbb{R}}_{>0}$
be the smooth map satisfying
$(\rho^{s})^{*}\Omega_{\text{tr}}=\beta(\cdotp,s)\Omega_{\text{tr}}$ for all
$s\in S$.

######  Lemma 18.

We have $\beta(\cdotp,s)=\det\mathop{\mathrm{Ad}}\nolimits(s)$ for all $s\in
S$. Therefore

$(\rho^{s})^{*}\Omega_{\text{tr}}=\det\mathop{\mathrm{Ad}}\nolimits(s)\Omega_{\text{tr}}.$
(7)

###### Proof.

By (5),

${\overline{\Omega}}\wedge{\overline{\Omega_{\text{tr}}}}=(\rho^{s})^{*}\left({\overline{\Omega}}\wedge{\overline{\Omega_{\text{tr}}}}\right)=(\rho^{s})^{*}{\overline{\Omega}}\wedge(\rho^{s})^{*}{\overline{\Omega_{\text{tr}}}}.$

Since

${T_{x}M}$${T_{\rho^{s}(x)}M}$${\left(TM/T{\mathcal{F}}\right)_{x}}$${\left(TM/T{\mathcal{F}}\right)_{\rho^{s}(x)},}$$\scriptstyle{(\rho^{s})_{*}}$${\circlearrowright}$$\scriptstyle{(\rho^{s})_{*}}$

we have
$(\rho^{s})^{*}{\overline{\Omega_{\text{tr}}}}={\overline{(\rho^{s})^{*}\Omega_{\text{tr}}}}$.
On the other hand, the diagram

${T_{x}M}$${T_{\rho^{s}(x)}M}$${T_{x}{\mathcal{F}}}$${T_{\rho^{s}(x)}{\mathcal{F}}}$$\scriptstyle{(\rho^{s})_{*}}$$\scriptstyle{(\rho^{s})_{*}}$

does not commute unless $E$ is $\rho$–invariant, so usually
$(\rho^{s})^{*}{\overline{\Omega}}\neq{\overline{(\rho^{s})^{*}\Omega}}$. But

$(\rho^{s})^{*}{\overline{\Omega}}\wedge(\rho^{s})^{*}{\overline{\Omega_{\text{tr}}}}={\overline{(\rho^{s})^{*}\Omega}}\wedge{\overline{(\rho^{s})^{*}\Omega_{\text{tr}}}}$

holds, because $(\rho^{s})^{*}{\overline{\Omega}}$ and
${\overline{(\rho^{s})^{*}\Omega}}$ coincide when only vectors in
$T{\mathcal{F}}$ are substituted, and
${\overline{(\rho^{s})^{*}\Omega_{\text{tr}}}}$ vanishes if we substitute a
vector in $T{\mathcal{F}}$. Therefore

$\displaystyle{\overline{\Omega}}\wedge{\overline{\Omega_{\text{tr}}}}$
$\displaystyle={\overline{(\rho^{s})^{*}\Omega}}\wedge{\overline{(\rho^{s})^{*}\Omega_{\text{tr}}}}={\overline{\det\mathop{\mathrm{Ad}}\nolimits\left(s^{-1}\right)\Omega}}\wedge{\overline{\beta(\cdotp,s)\Omega_{\text{tr}}}}$
$\displaystyle=\det\mathop{\mathrm{Ad}}\nolimits\left(s^{-1}\right)\beta(\cdotp,s){\overline{\Omega}}\wedge{\overline{\Omega_{\text{tr}}}}$

and the claim follows. ∎

Combining (6) and (7), we get

$\left(\rho_{0}^{s}\right)^{*}\Omega_{\text{tr}}=\det\mathop{\mathrm{Ad}}\nolimits\left(a(\cdotp,s)\right)\Omega_{\text{tr}}.$

Then by the same argument in the proof of the above lemma,

$\displaystyle\left(\rho_{0}^{s}\right)^{*}\left({\overline{\Omega_{0}}}\wedge{\overline{\Omega_{\text{tr}}}}\right)$
$\displaystyle=\left(\rho_{0}^{s}\right)^{*}{\overline{\Omega_{0}}}\wedge\left(\rho_{0}^{s}\right)^{*}{\overline{\Omega_{\text{tr}}}}$
$\displaystyle={\overline{\left(\rho_{0}^{s}\right)^{*}\Omega_{0}}}\wedge{\overline{\left(\rho_{0}^{s}\right)^{*}\Omega_{\text{tr}}}}$
$\displaystyle={\overline{\det\mathop{\mathrm{Ad}}\nolimits\left(s^{-1}\right)\Omega_{0}}}\wedge{\overline{\det\mathop{\mathrm{Ad}}\nolimits\left(a(x,s)\right)\Omega_{\text{tr}}}}$
$\displaystyle=\det\mathop{\mathrm{Ad}}\nolimits\left(s^{-1}\right)\det\mathop{\mathrm{Ad}}\nolimits\left(a(x,s)\right){\overline{\Omega_{0}}}\wedge{\overline{\Omega_{\text{tr}}}}.$
(8)

This formula will be the key to prove Theorem 4.

######  Lemma 19.

For any $s\in S$,

$\det\mathop{\mathrm{Ad}}\nolimits(s)=e^{\mathop{\mathrm{tr}}\nolimits\mathop{\mathrm{ad}}\nolimits\int_{1}^{s}p\Theta},$

where $p\colon{\mathfrak{s}}\twoheadrightarrow{\mathfrak{s}}/{\mathfrak{n}}$
denotes the natural projection and $\mathop{\mathrm{ad}}\nolimits$ is also
used for
${\mathfrak{s}}/{\mathfrak{n}}\stackrel{{\scriptstyle\mathop{\mathrm{ad}}\nolimits}}{{\curvearrowright}}\mathop{\mathrm{Gr}}\nolimits_{\mathfrak{n}}({\mathfrak{s}})$.

###### Proof.

Any $s\in S$ can be written as $s=e^{X_{1}}\cdots e^{X_{k}}$ for some
$X_{1},\ldots,X_{k}\in{\mathfrak{s}}$. It is easy to see

$\int_{1}^{s}p\Theta=p(X_{1})+\cdots+p(X_{k}).$

Since $\mathop{\mathrm{tr}}\nolimits\mathop{\mathrm{ad}}\nolimits
p(X)=\mathop{\mathrm{tr}}\nolimits\mathop{\mathrm{ad}}\nolimits X$ for
$X\in{\mathfrak{s}}$, we have

$\displaystyle\det\mathop{\mathrm{Ad}}\nolimits(s)$
$\displaystyle=e^{\mathop{\mathrm{tr}}\nolimits\mathop{\mathrm{ad}}\nolimits
X_{1}}\cdots e^{\mathop{\mathrm{tr}}\nolimits\mathop{\mathrm{ad}}\nolimits
X_{k}}$
$\displaystyle=e^{\mathop{\mathrm{tr}}\nolimits\mathop{\mathrm{ad}}\nolimits\left(p(X_{1})+\cdots+p(X_{k})\right)}$
$\displaystyle=e^{\mathop{\mathrm{tr}}\nolimits\mathop{\mathrm{ad}}\nolimits\int_{1}^{s}p\Theta}.$

∎

Since we have
$\int_{1}^{s}\varphi_{\rho}\Theta+h\left(\rho_{0}(x,s)\right)-h(x)=\int_{1}^{a(x,s)}p\Theta$
as in Lemma 10, the factor appearing in the formula (8) will be

$\displaystyle\det\mathop{\mathrm{Ad}}\nolimits\left(s^{-1}\right)\det\mathop{\mathrm{Ad}}\nolimits\left(a(x,s)\right)$
$\displaystyle=e^{-\mathop{\mathrm{tr}}\nolimits\mathop{\mathrm{ad}}\nolimits\int_{1}^{s}p\Theta}e^{\mathop{\mathrm{tr}}\nolimits\mathop{\mathrm{ad}}\nolimits\int_{1}^{a(x,s)}p\Theta}$
$\displaystyle=\exp\left\\{\mathop{\mathrm{tr}}\nolimits\mathop{\mathrm{ad}}\nolimits\int_{1}^{s}\left(\varphi_{\rho}-p\right)\Theta+\mathop{\mathrm{tr}}\nolimits\mathop{\mathrm{ad}}\nolimits\left(h\left(\rho_{0}(x,s)\right)-h(x)\right)\right\\}.$

Assume $\varphi_{\rho}\neq p$ and we will get a contradiction as follows. We
have $X\in{\mathfrak{s}}$ so that $A:=(\varphi_{\rho}-p)(X)\neq 0$. Since
$\dim{\mathfrak{s}}/{\mathfrak{n}}=1$ and ${\mathfrak{s}}$ is not unimodular,
$\mathop{\mathrm{tr}}\nolimits\mathop{\mathrm{ad}}\nolimits A\neq 0$. We may
assume $\mathop{\mathrm{tr}}\nolimits\mathop{\mathrm{ad}}\nolimits A>0$ by
replacing $X$. Then for $s=e^{TX}$ with $T>0$ large, the function

$\det\mathop{\mathrm{Ad}}\nolimits\left(s^{-1}\right)\det\mathop{\mathrm{Ad}}\nolimits\left(a(x,s)\right)=e^{T\mathop{\mathrm{tr}}\nolimits\mathop{\mathrm{ad}}\nolimits
A}e^{\mathop{\mathrm{tr}}\nolimits\mathop{\mathrm{ad}}\nolimits\left(h\left(\rho_{0}(x,s)\right)-h(x)\right)}$

on $M$ must be large because $h$ is bounded. This contradicts the formula (8)
by integrating over $M$. Therefore $\varphi_{\rho}=p$ and this completes the
proof of Theorem 4.

######  Remark 20.

The assumptions $\dim{\mathfrak{s}}/{\mathfrak{n}}=1$ and nonunimodularity of
${\mathfrak{s}}$ are used only in the final part of the proof, so we have
proved the following. If $M$ is orientable,
$\mathop{\mathrm{tr}}\nolimits\mathop{\mathrm{ad}}\nolimits\left(\varphi_{\rho}-p\right)(X)=0$
for all $X\in{\mathfrak{s}}$, where $\mathop{\mathrm{ad}}\nolimits$ denotes
${\mathfrak{s}}/{\mathfrak{n}}\stackrel{{\scriptstyle\mathop{\mathrm{ad}}\nolimits}}{{\curvearrowright}}\mathop{\mathrm{Gr}}\nolimits_{\mathfrak{n}}({\mathfrak{s}})$.

##  4 Sufficient condition given by large scale geometry of solvable Lie
groups

###  4.1 Key Proposition to the method

Let $X$, $B$ be metric spaces. We say a surjective map $p\colon X\to B$ is a
distance respecting projection if for any two points $b$, $b^{\prime}\in B$,

$d(b,b^{\prime})=d\left(p^{-1}(b),p^{-1}(b^{\prime})\right)=d_{\mathcal{H}}\left(p^{-1}(b),p^{-1}(b^{\prime})\right)$

holds. Here $d_{\mathcal{H}}$ denotes the Hausdorff distance. Let $p\colon
X\to B$ and $p^{\prime}\colon X^{\prime}\to B^{\prime}$ be distance respecting
projections. For a given diagram

${X}$${X^{\prime}}$${B}$${B^{\prime}}$$\scriptstyle{f}$$\scriptstyle{p}$$\scriptstyle{p^{\prime}}$$\scriptstyle{\varphi}$

we say $f$ is fiber respecting over $\varphi$ if there is $C>0$ such that

$d_{\mathcal{H}}\left(f\left(p^{-1}(b)\right),\left(p^{\prime}\right)^{-1}\left(\varphi(b)\right)\right)<C$

for all $b\in B$.

######  Lemma 21.

Let $G$ be a Lie group and $H$ be a closed subgroup of $G$.

(1) Left invariant Riemannian metrics on $G/H$ are in one-to-one
correspondence with inner products on ${\mathfrak{g}}/{\mathfrak{h}}$
invariant under
$H\stackrel{{\scriptstyle\mathop{\mathrm{Ad}}\nolimits}}{{\curvearrowright}}{\mathfrak{g}}/{\mathfrak{h}}$
by the canonical identification ${\mathfrak{g}}/{\mathfrak{h}}\simeq
T_{H}G/H$.

(2) Assume that there exists an invariant inner product for
$H\stackrel{{\scriptstyle\mathop{\mathrm{Ad}}\nolimits}}{{\curvearrowright}}{\mathfrak{g}}/{\mathfrak{h}}$.
Take an inner product of ${\mathfrak{g}}$ for which the restriction
${\mathfrak{h}}^{\perp}\stackrel{{\scriptstyle\sim}}{{\to}}{\mathfrak{g}}/{\mathfrak{h}}$
of the projection ${\mathfrak{g}}\to{\mathfrak{g}}/{\mathfrak{h}}$ is an
isometry. Endow $G$ and $G/H$ with left invariant Riemannian metrics defined
by these inner products. Let $\pi\colon G\to G/H$ be the projection. Then, for
every $g\in G$, the restriction
$(L_{g})_{*}{\mathfrak{h}}^{\perp}\stackrel{{\scriptstyle\sim}}{{\to}}T_{gH}G/H$
of $\pi_{*}\colon T_{g}G\to T_{gH}G/H$ is an isometry, and the kernel of
$\pi_{*}\colon T_{g}G\to T_{gH}G/H$ is $(L_{g})_{*}{\mathfrak{h}}$.

(3) Assume that $G$ is connected. Under the assumption of 2), the map
$\pi\colon G\to G/H$ is a distance respecting projection.

###### Proof.

It is easy to check (1) and (2).

(3) Take arbitrary two points $b$, $b^{\prime}\in G/H$.

######  Claim.

$d\left(\pi^{-1}(b),\pi^{-1}\left(b^{\prime}\right)\right)\geq
d\left(b,b^{\prime}\right)$.

For any $g\in\pi^{-1}(b)$ and $g^{\prime}\in\pi^{-1}\left(b^{\prime}\right)$,
take a minimal geodesic $\gamma\colon[0,1]\to G$ connecting $g$ and
$g^{\prime}$. Since
$\left\|(\pi\circ\gamma)^{\prime}(t)\right\|=\left\|\pi_{*}\gamma^{\prime}(t)\right\|\leq\left\|\gamma^{\prime}(t)\right\|$
by the property of our metric, we have

$d\left(g,g^{\prime}\right)=\int_{0}^{1}\left\|\gamma^{\prime}(t)\right\|dt\geq\int_{0}^{1}\left\|(\pi\circ\gamma)^{\prime}(t)\right\|dt\geq
d\left(b,b^{\prime}\right).$

######  Claim.

$d\left(\pi^{-1}(b),\pi^{-1}\left(b^{\prime}\right)\right)\leq
d\left(b,b^{\prime}\right)$.

Take a minimal geodesic $\gamma\colon[0,1]\to G/H$ connecting $b$ and
$b^{\prime}$. Fix a point $g\in\pi^{-1}(b)$. Then there exists a curve
$\tilde{\gamma}\colon[0,1]\to G$ starting from $g$ such that
$\pi\circ\tilde{\gamma}=\gamma$ and
$\tilde{\gamma}^{\prime}(t)\in(L_{\tilde{\gamma}(t)})_{*}{\mathfrak{h}}^{\perp}$
for all $t\in[0,1]$ by a standard argument. For this curve, we have

$d\left(b,b^{\prime}\right)=\int_{0}^{1}\left\|\gamma^{\prime}(t)\right\|dt=\int_{0}^{1}\left\|\tilde{\gamma}^{\prime}(t)\right\|dt\geq
d\left(\pi^{-1}(b),\pi^{-1}\left(b^{\prime}\right)\right).$

######  Claim.

$d\left(b,b^{\prime}\right)=d_{\mathcal{H}}\left(\pi^{-1}(b),\pi^{-1}\left(b^{\prime}\right)\right)$.

We have $B\left(\pi^{-1}(b),C\right)=\pi^{-1}\left(B(b,C)\right)$ and
$B\left(\pi^{-1}(b^{\prime}),C\right)=\pi^{-1}\left(B(b^{\prime},C)\right)$ by
the above discussion, this is obvious. ∎

######  Corollary 22.

Let $G$ be a connected Lie group and $H$ be a connected normal closed subgroup
of $G$. Take an inner product of ${\mathfrak{g}}$. Endow
${\mathfrak{g}}/{\mathfrak{h}}$ with the inner product for which the
restriction
${\mathfrak{h}}^{\perp}\stackrel{{\scriptstyle\sim}}{{\to}}{\mathfrak{g}}/{\mathfrak{h}}$
of the projection ${\mathfrak{g}}\to{\mathfrak{g}}/{\mathfrak{h}}$ is an
isometry. Consider left invariant Riemannian metrics on $G$ and $G/H$
corresponding to these inner products. Then the projection $\pi\colon G\to
G/H$ is a distance respecting projection.

###### Proof.

This is because
$H\stackrel{{\scriptstyle\mathop{\mathrm{Ad}}\nolimits}}{{\curvearrowright}}{\mathfrak{g}}/{\mathfrak{h}}$
is trivial. ∎

Let us return to our previous setting:
$M\stackrel{{\scriptstyle\rho_{0}}}{{\curvearrowleft}}S$. Take an action
$\rho\in A({\mathcal{F}},S)$ and let $a\colon M\times S\to S$ be the cocycle
over $\rho_{0}$ defined by $\rho$, that is,
$\rho_{0}(x,s)=\rho\left(x,a(x,s)\right)$.

######  Lemma 23.

For any $x\in M$, the map $S\to S$ taking $s$ to $a(x,s)$ is a biLipschitz
diffeomorphism for any left invariant Riemannian metric on $S$.

###### Proof.

See Asaoka [1, Lemma 1.4.6] for a proof that the map is a diffeomorphism.

To see that the map is biLipschitz, equip $T{\mathcal{F}}$ with the metric
$\langle\cdotp,\cdotp\rangle_{\rho_{0}}$ for which $(\omega_{0})_{x}\colon
T_{x}{\mathcal{F}}\stackrel{{\scriptstyle\sim}}{{\to}}{\mathfrak{s}}$ is
isometric for all $x\in M$. We can easily verify that the metric obtained by
pulling back $\langle\cdotp,\cdotp\rangle_{\rho_{0}}$ through the map $S\to M$
taking $s$ to $\rho_{0}(x,s)$ is the original Riemannian metric on $S$.
Similarly we also have the metric $\langle\cdotp,\cdotp\rangle_{\rho}$ on
$T{\mathcal{F}}$ defined by using $\rho$. Since $M$ is compact, there is a
constant $C>1$ such that $C^{-1}\|\cdotp\|_{\rho_{0}}\leq\|\cdotp\|_{\rho}\leq
C\|\cdotp\|_{\rho_{0}}$. For any $s,s^{\prime}\in S$, take a minimal geodesic
$\gamma\colon[0,1]\to S$ connecting $s$ and $s^{\prime}$. Then

$\displaystyle d\left(s,s^{\prime}\right)$
$\displaystyle=\int_{0}^{1}\left\|\gamma^{\prime}(t)\right\|dt=\int_{0}^{1}\left\|\frac{d}{dt}\rho_{0}\left(x,\gamma(t)\right)\right\|_{\rho_{0}}dt$
$\displaystyle=\int_{0}^{1}\left\|\frac{d}{dt}\rho\left(x,a\left(x,\gamma(t)\right)\right)\right\|_{\rho_{0}}dt$
$\displaystyle\geq\frac{1}{C}\int_{0}^{1}\left\|\frac{d}{dt}\rho\left(x,a\left(x,\gamma(t)\right)\right)\right\|_{\rho}dt$
$\displaystyle=\frac{1}{C}\int_{0}^{1}\left\|\frac{d}{dt}a\left(x,\gamma(t)\right)\right\|dt$
$\displaystyle\geq\frac{1}{C}d\left(a(x,s),a\left(x,s^{\prime}\right)\right).$

The inverse is also Lipschitz by the same argument. ∎

Recall that ${\mathfrak{h}}$ is a subspace between
$[{\mathfrak{s}},{\mathfrak{s}}]$ and ${\mathfrak{n}}$. Let $K_{\rho}$ and $H$
be the Lie subgroups corresponding to $\ker\varphi_{\rho}$ and
${\mathfrak{h}}$ and $\tilde{\varphi}_{\rho}\colon S/K_{\rho}\to S/H$ be the
map induced by
$\varphi_{\rho}\colon{\mathfrak{s}}\to{\mathfrak{s}}/{\mathfrak{h}}$. Both
$S/K_{\rho}$ and $S/H$ are vector groups and $\tilde{\varphi}_{\rho}$ is a
linear isomorphism. The following proposition is the key for our method to
reduce the set ${\mathcal{X}}_{\rho_{0}}$.

######  Proposition 24.

For any $\rho\in A({\mathcal{F}},S)$ and $x\in M$, consider the next diagram

${S}$${S}$${S/K_{\rho}}$${S/H,}$$\scriptstyle{a(x,\cdotp)}$$\scriptstyle{\tilde{q}}$$\scriptstyle{\tilde{p}}$$\scriptstyle{\sim}$$\scriptstyle{\tilde{\varphi}_{\rho}}$

where the vertical maps are the natural projections. We give $S$ a left
invariant Riemannian metric and $S/K_{\rho}$, $S/H$ the left invariant
Riemannian metrics considered in Corollary 22. Then $a(x,\cdotp)$ is a fiber
respecting biLipschitz diffeomorphism over $\tilde{\varphi}_{\rho}$.

###### Proof.

Let $q\colon{\mathfrak{s}}\to{\mathfrak{s}}/\ker\varphi_{\rho}$ and
$p\colon{\mathfrak{s}}\to{\mathfrak{s}}/{\mathfrak{h}}$ be the natural
projections. Then we have

$\displaystyle e^{\int_{1}^{s}q\Theta}$ $\displaystyle=\tilde{q}(s),$
$\displaystyle e^{\int_{1}^{s}p\Theta}$ $\displaystyle=\tilde{p}(s)$

for all $s\in S$. Indeed, writing $s=e^{X_{1}}\cdots e^{X_{k}}$ for some
$X_{1},\ldots,X_{k}\in{\mathfrak{s}}$ and then

$\displaystyle e^{\int_{1}^{s}q\Theta}$
$\displaystyle=\exp\left(\int_{1}^{e^{X_{1}}}q\Theta+\int_{e^{X_{1}}}^{e^{X_{1}}e^{X_{2}}}q\Theta+\cdots+\int_{e^{X_{1}}\cdots
e^{X_{k-1}}}^{s}q\Theta\right)$ $\displaystyle=e^{q(X_{1})+\cdots+q(X_{k})}.$

Let
$\bar{\varphi}_{\rho}\colon{\mathfrak{s}}/\ker{\varphi_{\rho}}\to{\mathfrak{s}}/{\mathfrak{h}}$
be the map induced by $\varphi_{\rho}$. By taking $\exp$ to the equation
$\bar{\varphi}_{\rho}\left(\int_{1}^{s}q\Theta\right)+h\left(\rho_{0}(x,s)\right)-h(x)=\int_{1}^{a(x,s)}p\Theta$
from Lemma 10, we get

$\tilde{\varphi}_{\rho}\tilde{q}(s)e^{h\left(\rho_{0}(x,s)\right)-h(x)}=\tilde{p}\left(a(x,s)\right).$

Since $M$ is compact, there is a constant $C>0$ such that

$d\left(\tilde{\varphi}_{\rho}\tilde{q}(s),\tilde{p}\left(a(x,s)\right)\right)=d\left(1,e^{h\left(\rho_{0}(x,s)\right)-h(x)}\right)<C$
(9)

for any $s\in S$. This means that the diagram commutes up to bounded distance.
Let $f$ be the inverse map of $a(x,\cdotp)$. By Lemma 10,

$\int_{1}^{f(s)}q\Theta+\bar{\varphi}_{\rho}^{-1}\left(h\left(\rho_{0}\left(x,f(s)\right)\right)-h(x)\right)=\bar{\varphi}_{\rho}^{-1}\left(\int_{1}^{s}p\Theta\right).$

Taking $\exp$ to both sides of the equation,

$\tilde{q}f(s)e^{\bar{\varphi}_{\rho}^{-1}\left(h\left(\rho_{0}\left(x,f(s)\right)\right)-h(x)\right)}=\tilde{\varphi}_{\rho}^{-1}\tilde{p}(s).$

As before there is a constant $C^{\prime}>0$ such that

$d\left(\tilde{q}f(s),\tilde{\varphi}_{\rho}^{-1}\tilde{p}(s)\right)<C^{\prime}.$
(10)

Let $C^{\prime\prime}>0$ be a constant satisfying
$d\left(a(x,s_{1}),a(x,s_{2})\right)\leq C^{\prime\prime}d(s_{1},s_{2})$ for
all $s_{1}$, $s_{2}\in S$. Take any $b\in S/K_{\rho}$. Then we can show that

$d_{\mathcal{H}}\left(a\left(x,\tilde{q}^{-1}(b)\right),\tilde{p}^{-1}\left(\tilde{\varphi}_{\rho}(b)\right)\right)\leq\mathop{\mathrm{max}}\limits\left\\{C,C^{\prime}C^{\prime\prime}\right\\}$

in the following way:

######  Claim.

$a\left(x,\tilde{q}^{-1}(b)\right)\subset
B\left(\tilde{p}^{-1}\left(\tilde{\varphi}_{\rho}(b)\right),C\right)$.

For any $s\in\tilde{q}^{-1}(b)$,
$d\left(\tilde{\varphi}_{\rho}(b),\tilde{p}\left(a(x,s)\right)\right)<C$ by
(9). Therefore $a(x,s)\in
B\left(\tilde{p}^{-1}\left(\tilde{\varphi}_{\rho}(b)\right),C\right)$.

######  Claim.

$\tilde{p}^{-1}\left(\tilde{\varphi}_{\rho}(b)\right)\subset
B\left(a\left(x,\tilde{q}^{-1}(b)\right),C^{\prime}C^{\prime\prime}\right)$.

For any $s\in\tilde{p}^{-1}\left(\tilde{\varphi}_{\rho}(b)\right)$,
$d\left(\tilde{q}f(s),b\right)<C^{\prime}$ by (10), so that we can find
$s^{\prime}\in\tilde{q}^{-1}(b)$ satisfying $d\left(s^{\prime},f(s)\right)\leq
C^{\prime}$. Then $d\left(a\left(x,s^{\prime}\right),s\right)\leq
C^{\prime\prime}d\left(s^{\prime},f(s)\right)\leq C^{\prime\prime}C^{\prime}$.
∎

This proposition puts a strong restriction on the map $\tilde{\varphi}_{\rho}$
which we want to know, as in the next section or Section 5.2.

###  4.2 Proof of Theorem 5

As in Introduction, let

$A_{i}=\begin{pmatrix}\alpha^{(i)}_{1}&&\\\ &\ddots&\\\
&&\alpha^{(i)}_{n}\end{pmatrix}$

for $i=1,\ldots,k$ be diagonal matrices with positive diagonal entries and set
$S_{A}={\mathbb{R}}^{n}\rtimes_{A(t)}{\mathbb{R}}^{k}$, where
$A(t)=A_{1}^{t_{1}}\cdots A_{k}^{t_{k}}$ for
$t=(t_{1},\ldots,t_{k})\in{\mathbb{R}}^{k}$. Ogasawara [8] proved the
following theorem to classify up to quasiisometry groups obtained from
${\mathbb{Z}}^{n}$ by performing certain HNN extensions several times, which
is a generalization of a result of Farb and Mosher [3] on quasiisometric
rigidity of abelian-by-cyclic groups.

######  Theorem 25.

If

${S_{A}}$${S_{A^{\prime}}}$${{\mathbb{R}}^{k}}$${{\mathbb{R}}^{k}}$$\scriptstyle{f}$$\scriptstyle{\varphi}$

is a diagram in which vertical maps are natural projections and $f$ is a fiber
respecting quasiisometry over some linear map $\varphi$, then there exists a
permutation matrix $P\in\mathop{\mathrm{GL}}\nolimits(n,{\mathbb{R}})$ such
that $PA(t)=A^{\prime}\left(\varphi(t)\right)P$ for every
$t\in{\mathbb{R}}^{k}$. In particular, there exists a diagram

${S_{A}}$${S_{A^{\prime}}}$${{\mathbb{R}}^{k}}$${{\mathbb{R}}^{k},}$$\scriptstyle{\sim}$${\circlearrowright}$$\scriptstyle{\varphi}$

where the above horizontal map is an isomorphism of Lie groups taking $(v,t)$
to $\left(Pv,\varphi(t)\right)$.

Consider an action $M\stackrel{{\scriptstyle\rho_{0}}}{{\curvearrowleft}}S$
for $S=S_{A}$ and assume that for any $j$ there is some $i$ for which
$\alpha^{(i)}_{j}\neq 1$. It is easy to check that this is equivalent to
$[{\mathfrak{s}},{\mathfrak{s}}]={\mathbb{R}}^{n}$. We take
${\mathfrak{h}}=[{\mathfrak{s}},{\mathfrak{s}}]$. For any $\rho\in
A({\mathcal{F}},S)$, $\ker\varphi_{\rho}=[{\mathfrak{s}},{\mathfrak{s}}]$ and
then $K_{\rho}={\mathbb{R}}^{n}$. Therefore by Proposition 24, $a(x,\cdotp)$
is a fiber respecting biLipschitz diffeomorphism over $\tilde{\varphi}_{\rho}$
in

${S}$${S}$${{\mathbb{R}}^{k}}$${{\mathbb{R}}^{k}.}$$\scriptstyle{a(x,\cdotp)}$$\scriptstyle{\tilde{\varphi}_{\rho}}$

Hence by the above theorem we get $\tilde{\varphi}_{\rho}\in W_{S}$. Using
Theorem 14, this completes the proof of Theorem 5.

######  Remark 26.

There are solvable Lie groups for which this method does not work. Consider
the solvable Lie group $S={\mathbb{R}}^{2}\rtimes_{R(t)}{\mathbb{R}}$ for
$R(t)=\begin{pmatrix}\cos 2\pi t&-\sin 2\pi t\\\ \sin 2\pi t&\cos 2\pi
t\end{pmatrix}$. Then the standard Euclidean metric on ${\mathbb{R}}^{3}$
identified with $S$ is a left invariant Riemannian metric on $S$. Hence for
any $c\in{\mathbb{R}}\setminus\\{0\\}$ there is a biLipschitz diffeomorphism
$f\colon S\to S$ such that

${S}$${S}$${{\mathbb{R}}}$${{\mathbb{R}}.}$$\scriptstyle{f}$${\circlearrowright}$$\scriptstyle{\times
c}$

So we do not have any restriction on $c$.

##  5 Parameter rigidity of suspension-like actions on mapping tori

We prove Theorem 6 here. The proof is based on Theorem 14, Proposition 24 and
a theorem of Farb and Mosher [3]. To calculate cohomologies we use
Mayer–Vietoris exact sequences.

The action which we consider is
$M={\mathbb{Z}}\backslash\left({\mathbb{T}}^{n}\times{\mathbb{R}}\right)\curvearrowleft
S$, where $S=V\rtimes_{\Phi}{\mathbb{R}}$. Let ${\mathcal{F}}$ be its orbit
foliation.

###  5.1 Cohomology

The bracket operation of the Lie algebra
${\mathfrak{s}}=V\rtimes_{\Phi_{*}}{\mathbb{R}}$ is

$\displaystyle\left[(v_{1},t_{1}),(v_{2},t_{2})\right]$
$\displaystyle=\left([v_{1},v_{2}]+\Phi_{*}(t_{1})v_{2}-\Phi_{*}(t_{2})v_{1},[t_{1},t_{2}]\right)$
$\displaystyle=\left(\Phi_{*}(t_{1})v_{2}-\Phi_{*}(t_{2})v_{1},0\right).$

Let $T=(0,1)\in{\mathfrak{s}}$. We will use the notation
$\mathop{\mathrm{ad}}\nolimits^{0}X=\mathop{\mathrm{ad}}\nolimits X|_{V}$ for
$X\in{\mathfrak{s}}$. Then $\mathop{\mathrm{ad}}\nolimits^{0}T=\Phi_{*}(1)$,
so that $\Phi_{t}=e^{t\mathop{\mathrm{ad}}\nolimits^{0}T}$ and
$A|_{V}=e^{\mathop{\mathrm{ad}}\nolimits^{0}T}$. Since we assume $1$ is not an
eigenvalue of $A|_{V}$, $\mathop{\mathrm{ad}}\nolimits^{0}T\colon V\to V$ does
not have $0$ as an eigenvalue, hence it is invertible. So we have
$[{\mathfrak{s}},{\mathfrak{s}}]=V$.

Let us examine which cohomologies to compute. We take
$[{\mathfrak{s}},{\mathfrak{s}}]=V$ as a subspace ${\mathfrak{h}}$ appearing
in Theorem 3. We consider representations
${\mathfrak{s}}\stackrel{{\scriptstyle\mathop{\mathrm{ad}}\nolimits\circ\varphi}}{{\curvearrowright}}\mathop{\mathrm{Gr}}\nolimits_{V}({\mathfrak{s}})={\mathfrak{s}}/V\oplus
V$ for $\varphi=\varphi_{\rho}$ for some $\rho\in A({\mathcal{F}},S)$. The
first component
${\mathfrak{s}}\stackrel{{\scriptstyle\mathop{\mathrm{ad}}\nolimits\circ\varphi}}{{\curvearrowright}}{\mathfrak{s}}/V$
is just the trivial representation. Let $\bar{T}$ be the image of $T$ under
the natural projection ${\mathfrak{s}}\twoheadrightarrow{\mathfrak{s}}/V$ and
put $\varphi(T)=c\bar{T}$. Then the second component
${\mathfrak{s}}\stackrel{{\scriptstyle\mathop{\mathrm{ad}}\nolimits\circ\varphi}}{{\curvearrowright}}V$
is given by
$(v,t)v^{\prime}=ct\left(\mathop{\mathrm{ad}}\nolimits^{0}T\right)v^{\prime}=c\left(\mathop{\mathrm{ad}}\nolimits^{0}(v,t)\right)v^{\prime}$,
so that it is just ${\mathfrak{s}}\stackrel{{\scriptstyle
c\mathop{\mathrm{ad}}\nolimits^{0}}}{{\curvearrowright}}V$. Therefore we must
compute

$H^{1}({\mathcal{F}})\quad\text{and}\quad
H^{1}\left({\mathcal{F}};{\mathfrak{s}}\stackrel{{\scriptstyle
c\mathop{\mathrm{ad}}\nolimits^{0}}}{{\curvearrowright}}V\right)$

for all $c\in{\mathbb{R}}$. But we will see that vanishing of
$H^{1}({\mathcal{F}})$ and
$H^{1}\left({\mathcal{F}};{\mathfrak{s}}\stackrel{{\scriptstyle\pm\mathop{\mathrm{ad}}\nolimits^{0}}}{{\curvearrowright}}V\right)$
suffices by the method of large scale geometry and this reduction fits our
assumption of Zariski density nicely.

###  5.2 Application of large scale geometry

Here we use rigidty of quasiisometries between solvable Lie groups found by
Farb and Mosher. They prepare the following theorem as a tool to prove
quasiisometic rigidity of abelian-by-cyclic groups. In [3] a slightly
different statement (Theorem 5.2) is given, but the following is also proved.

######  Theorem 27.

Let $\Phi^{(1)}_{t}$ and $\Phi^{(2)}_{t}$ be one-parameter subgroups of
$\mathop{\mathrm{GL}}\nolimits(p,{\mathbb{R}})$ and
$S_{i}={\mathbb{R}}^{p}\rtimes_{\Phi^{(i)}}{\mathbb{R}}$. If

${S_{1}}$${S_{2}}$${{\mathbb{R}}}$${{\mathbb{R}}}$$\scriptstyle{\phi}$$\scriptstyle{\mathop{\mathrm{id}}\nolimits}$

is a diagram in which vertical maps are natural projections and $\phi$ is a
fiber respecting quasiisometry over $\mathop{\mathrm{id}}\nolimits$. Then
absolute Jordan forms of $\Phi^{(1)}_{1}$ and $\Phi^{(2)}_{1}$ coincide up to
permutation of Jordan blocks. Here an absolute Jordan form refers to a matrix
obtained by replacing the diagonal entries of a Jordan normal form over the
complex field with their absolute values.

Returning to our situation, we have

${S}$${S}$${{\mathbb{R}}}$${{\mathbb{R}},}$$\scriptstyle{a(x,\cdotp)}$$\scriptstyle{\times
c}$

where $a(x,\cdotp)$ is a fiber respecting biLipschitz diffeomorphism by
Proposition 24. Composing with

${S}$${V\rtimes_{\Phi_{ct}}{\mathbb{R}}}$${{\mathbb{R}}}$${{\mathbb{R}},}$$\scriptstyle{\sim}$${\circlearrowright}$$\scriptstyle{\times\frac{1}{c}}$

where the above horizontal map takes $(v,t)$ to $\left(v,\frac{t}{c}\right)$,
we get

${S}$${V\rtimes_{\Phi_{ct}}{\mathbb{R}}}$${{\mathbb{R}}}$${{\mathbb{R}},}$$\scriptstyle{\mathop{\mathrm{id}}\nolimits}$

in which the above horizontal map is a fiber respecting biLipschitz
diffeomorphism over $\mathop{\mathrm{id}}\nolimits$. Then by Theorem 27,
absolute Jordan forms of $\Phi_{1}$ and $\Phi_{c}$ coincide. Let
$\alpha_{1},\ldots,\alpha_{p}$ be the eigenvalues of
$\mathop{\mathrm{ad}}\nolimits^{0}T$, so that eigenvalues of
$\Phi_{t}=e^{t\mathop{\mathrm{ad}}\nolimits^{0}T}$ are
$e^{t\alpha_{1}},\ldots,e^{t\alpha_{p}}$. Therefore the sets
$\left\\{\left|e^{\alpha_{1}}\right|,\ldots,\left|e^{\alpha_{p}}\right|\right\\}$
and
$\left\\{\left|e^{\alpha_{1}}\right|^{c},\ldots,\left|e^{\alpha_{p}}\right|^{c}\right\\}$
coincide with multiplicity. Since $\Phi_{1}$ has an eigenvalue of absolute
value not equal to $1$, we must have $c=\pm 1$. So we need to verify only
vanishing of $H^{1}({\mathcal{F}})$ and
$H^{1}\left({\mathcal{F}};{\mathfrak{s}}\stackrel{{\scriptstyle\pm\mathop{\mathrm{ad}}\nolimits^{0}}}{{\curvearrowright}}V\right)$
to show parameter rigidity of the action.

###  5.3 Mayer–Vietoris argument

Let ${\mathfrak{s}}\stackrel{{\scriptstyle\pi}}{{\curvearrowright}}W$ be a
representation on a finite dimensional real vector space satisfying $\pi(v)=0$
for all $v\in V$. We will try to calculate the first cohomology
$H^{1}\left({\mathcal{F}};{\mathfrak{s}}\stackrel{{\scriptstyle\pi}}{{\curvearrowright}}W\right)$.

Let $U_{1}$ and $U_{2}$ be the projections to $M$ of
${\mathbb{T}}^{n}\times(0,1)$ and
${\mathbb{T}}^{n}\times(-\frac{1}{2},\frac{1}{2})$, so that $M=U_{1}\cup
U_{2}$. Then we have a short exact sequence of cochain complexes:

$0\to\Omega^{*}({\mathcal{F}};W)\to\Omega^{*}\left({\mathcal{F}}|_{U_{1}};W\right)\oplus\Omega^{*}\left({\mathcal{F}}|_{U_{2}};W\right)\to\Omega^{*}\left({\mathcal{F}}|_{U_{1}\cap
U_{2}};W\right)\to 0.$

The second map maps $\xi$ to $\left(\xi|_{U_{1}},\xi|_{U_{2}}\right)$ and the
third map maps $(\xi_{1},\xi_{2})$ to $\xi_{2}|_{U_{1}\cap
U_{2}}-\xi_{1}|_{U_{1}\cap U_{2}}$. Hence we obtain a exact sequence of the
cohomology

$H^{0}\left({\mathcal{F}}|_{U_{1}};\pi\right)\oplus
H^{0}\left({\mathcal{F}}|_{U_{2}};\pi\right)\stackrel{{\scriptstyle
P}}{{\to}}H^{0}\left({\mathcal{F}}|_{U_{1}\cap U_{2}};\pi\right)\\\ \to
H^{1}({\mathcal{F}};\pi)\to H^{1}\left({\mathcal{F}}|_{U_{1}};\pi\right)\oplus
H^{1}\left({\mathcal{F}}|_{U_{2}};\pi\right)\stackrel{{\scriptstyle
Q}}{{\to}}H^{1}\left({\mathcal{F}}|_{U_{1}\cap U_{2}};\pi\right),$

so that we have

$H^{1}({\mathcal{F}};\pi)\simeq\mathop{\mathrm{coker}}\nolimits P\oplus\ker
Q.$ (11)

To compute $H^{1}({\mathcal{F}};\pi)$, we first compute
$H^{1}\left({\mathcal{F}}|_{U_{1}};\pi\right)$. Let ${\mathcal{G}}$ be the
orbit foliation of the translation action ${\mathbb{T}}^{n}\curvearrowleft V$.
Define $\iota\colon{\mathbb{T}}^{n}\hookrightarrow U_{1}$ by
$\iota(x)=\left(x,\frac{1}{2}\right)$. The map
$\iota^{*}\colon\Omega^{*}\left({\mathcal{F}}|_{U_{1}};W\right)\to\Omega^{*}({\mathcal{G}};W)$
obtained by pull back is a cochain map555However,
$p^{*}\colon\Omega^{*}({\mathcal{G}};W)\to\Omega^{*}\left({\mathcal{F}}|_{U_{1}};W\right)$
is not a cochain map, where $p\colon U_{1}\to{\mathbb{T}}^{n}$ maps $(x,t)$ to
$x$.  between
$\left(\Omega^{*}\left({\mathcal{F}}|_{U_{1}};W\right),{d_{\mathcal{F}}}+\pi\omega_{0}\wedge\right)$
and $\left(\Omega^{*}({\mathcal{G}};W),d_{\mathcal{G}}\right)$, where
$\omega_{0}$ is the canonical 1–form of the action $M\curvearrowleft S$. In
fact

$\iota^{*}\left({d_{\mathcal{F}}}+\pi\omega_{0}\wedge\right)\xi=d_{\mathcal{G}}\iota^{*}\xi+\iota^{*}\left(\pi\omega_{0}\wedge\xi\right)=d_{\mathcal{G}}\iota^{*}\xi$

for $\xi\in\Omega^{k}\left({\mathcal{F}}|_{U_{1}};W\right)$ due to our
assumption on $\pi$. Thus we get the induced map on cohomologies:

$\iota^{*}\colon H^{*}\left({\mathcal{F}}|_{U_{1}};\pi\right)\to
H^{*}({\mathcal{G}};W).$

Note that $W$ in $H^{*}({\mathcal{G}};W)$ is just a direct sum of the trivial
coefficient ${\mathbb{R}}$.

######  Lemma 28.

$\iota^{*}\colon H^{1}\left({\mathcal{F}}|_{U_{1}};\pi\right)\to
H^{1}({\mathcal{G}};W)$ is an isomorphism.

###### Proof.

We first show injectivity:

Take $[\eta]\in H^{1}\left({\mathcal{F}}|_{U_{1}};\pi\right)$ satisfying
$\iota^{*}[\eta]=0$. We must show the existence of a smooth $\alpha\colon
U_{1}\to W$ such that

$\eta={d_{\mathcal{F}}}\alpha+\pi\omega_{0}\alpha.$ (12)

Such $\alpha$ must satisfy $\eta(T)=T\alpha+\pi(T)\alpha$, where $T$ is
regarded as a vector field using the action. Note that $T$ is just the vector
field $\frac{\partial}{\partial t}$, where $(x,t)\in U_{1}$. We have a smooth
map $h\colon{\mathbb{T}}^{n}\to W$ satisfying
$\iota^{*}\eta=d_{\mathcal{G}}h$. Using this we let

$\alpha(x,t)=e^{-\left(t-\frac{1}{2}\right)\pi(T)}\left(\int^{t}_{\frac{1}{2}}e^{\left(s-\frac{1}{2}\right)\pi(T)}\eta(T)(x,s)ds+h(x)\right)$

for $x\in{\mathbb{T}}^{n}$ and $t\in(0,1)$. Then this $\alpha$ satisfies
$\frac{\partial\alpha}{\partial t}(x,t)=-\pi(T)\alpha(x,t)+\eta(T)(x,t)$, ie
$\eta(T)=T\alpha+\pi(T)\alpha$ and $\alpha\left(x,\frac{1}{2}\right)=h(x)$. To
show (12) we have to prove

$\eta(X)-X\alpha=0$

for all $X\in V$. Note that

$\left(\eta(X)-X\alpha\right)\left(x,\frac{1}{2}\right)=0$ (13)

for $X\in V$. Since

$0=\left({d_{\mathcal{F}}}\eta+\pi\omega_{0}\wedge\eta\right)(T,X)=T\eta(X)-X\eta(T)-\eta\left([T,X]\right)+\pi(T)\eta(X),$

we have

$\displaystyle T\left(\eta(X)-X\alpha\right)$
$\displaystyle=X\eta(T)+\eta\left([T,X]\right)-\pi(T)\eta(X)-X\left(\eta(T)-\pi(T)\alpha\right)-[T,X]\alpha$
$\displaystyle=-\pi(T)\left(\eta(X)-X\alpha\right)+\eta\left([T,X]\right)-[T,X]\alpha.$
(14)

Choose a basis of $V$ which turns $\mathop{\mathrm{ad}}\nolimits^{0}T$ into a
real Jordan normal form

$\bigoplus\begin{pmatrix}a&1&&\\\ &\ddots&\ddots&\\\ &&\ddots&1\\\
&&&a\end{pmatrix}\oplus\bigoplus\begin{pmatrix}b&-c&1&&&&&\\\ c&b&&1&&&&\\\
&&\ddots&&\ddots&&&\\\ &&&\ddots&&\ddots&&\\\ &&&&\ddots&&1&\\\
&&&&&\ddots&&1\\\ &&&&&&b&-c\\\ &&&&&&c&b\end{pmatrix},$

where $a,b,c\in{\mathbb{R}}$, $c\neq 0$. Let $X_{1},\ldots,X_{k}$ be the basis
of the subspace corresponding to the above

$\begin{pmatrix}a&1&&\\\ &\ddots&\ddots&\\\ &&\ddots&1\\\ &&&a\end{pmatrix},$

so that relations

$\displaystyle[T,X_{1}]$ $\displaystyle=aX_{1}$ $\displaystyle[T,X_{2}]$
$\displaystyle=aX_{2}+X_{1}$ $\displaystyle\vdots$ $\displaystyle[T,X_{k}]$
$\displaystyle=aX_{k}+X_{k-1}$

hold. By (14),
$T\left(\eta(X_{1})-X_{1}\alpha\right)=\left(a-\pi(T)\right)\left(\eta(X_{1})-X_{1}\alpha\right)$.
The solution is

$\left(\eta(X_{1})-X_{1}\alpha\right)(x,t)=e^{\left(t-\frac{1}{2}\right)\left(a-\pi(T)\right)}\left(\eta(X_{1})-X_{1}\alpha\right)\left(x,\frac{1}{2}\right).$

So we get $\eta(X_{1})-X_{1}\alpha=0$ by (13). For $X_{2}$ we also have
$T\left(\eta(X_{2})-X_{2}\alpha\right)=\left(a-\pi(T)\right)\left(\eta(X_{2})-X_{2}\alpha\right)$,
so that $\eta(X_{2})-X_{2}\alpha=0$. Repeating this we get
$\eta(X_{i})-X_{i}\alpha=0$ for $i=1,\ldots,k$.

Let $X_{1},Y_{1},\ldots,X_{k},Y_{k}$ be the basis of the subspace
corresponding to

$\begin{pmatrix}b&-c&1&&&&&\\\ c&b&&1&&&&\\\ &&\ddots&&\ddots&&&\\\
&&&\ddots&&\ddots&&\\\ &&&&\ddots&&1&\\\ &&&&&\ddots&&1\\\ &&&&&&b&-c\\\
&&&&&&c&b\end{pmatrix}.$

We proceed similarly. The relations are

$\displaystyle[T,X_{1}]$ $\displaystyle=bX_{1}+cY_{1},$
$\displaystyle[T,Y_{1}]$ $\displaystyle=-cX_{1}+bY_{1},$
$\displaystyle[T,X_{2}]$ $\displaystyle=bX_{2}+cY_{2}+X_{1},$
$\displaystyle[T,Y_{2}]$ $\displaystyle=-cX_{2}+bY_{2}+Y_{1},$ $\displaystyle\
\ \vdots$ $\displaystyle\ \ \vdots$

The first equation to solve is

$T\begin{pmatrix}\eta(X_{1})-X_{1}\alpha\\\
\eta(Y_{1})-Y_{1}\alpha\end{pmatrix}=\begin{pmatrix}b-\pi(T)&c\\\
-c&b-\pi(T)\end{pmatrix}\begin{pmatrix}\eta(X_{1})-X_{1}\alpha\\\
\eta(Y_{1})-Y_{1}\alpha\end{pmatrix},$

and the solution is

$\begin{pmatrix}\eta(X_{1})-X_{1}\alpha\\\
\eta(Y_{1})-Y_{1}\alpha\end{pmatrix}(x,t)=e^{\left(t-\frac{1}{2}\right)\begin{pmatrix}b-\pi(T)&c\\\
-c&b-\pi(T)\end{pmatrix}}\begin{pmatrix}\eta(X_{1})-X_{1}\alpha\\\
\eta(Y_{1})-Y_{1}\alpha\end{pmatrix}\left(x,\frac{1}{2}\right).$

So we have $\eta(X_{1})-X_{1}\alpha=0$ and $\eta(Y_{1})-Y_{1}\alpha=0$.
Repeating, we eventually get $\eta(X_{i})-X_{i}\alpha=0$ and
$\eta(Y_{i})-Y_{i}\alpha=0$ for $i=1,\ldots,k$. This proves the injectivity.

We now prove surjectivity:

Take any $[\xi]\in H^{1}({\mathcal{G}};W)$. We must construct
$\eta\in\Omega^{1}\left({\mathcal{F}}|_{U_{1}};W\right)$ satisfying

${d_{\mathcal{F}}}\eta+\pi\omega_{0}\wedge\eta=0$ (15)

and $\iota^{*}\eta=\xi$. We will construct $\eta$ requiring the additional
property $\eta(T)=0$. In order that equation (15) is satisfied, for any $X\in
V$ we should construct $\eta(X)$ such that

$T\eta(X)-\eta\left([T,X]\right)+\pi(T)\eta(X)=0$ (16)

holds. Fix a basis $X_{1},\ldots,X_{p}$ of $V$ and let $(a_{ij})$ be the
matrix representing $\mathop{\mathrm{ad}}\nolimits^{0}T$ with respect to
$X_{1},\ldots,X_{p}$; $[T,X_{j}]=\sum^{p}_{i=1}a_{ij}X_{i}$. The $\eta(X_{j})$
should satisfy
$T\eta(X_{j})-\sum^{p}_{i=1}a_{ij}\eta(X_{i})+\pi(T)\eta(X_{j})=0$ and
$\eta(X_{j})\left(x,\frac{1}{2}\right)=\xi\left(X^{0}_{j}\right)$. Here, for
$X\in V$, $X^{0}$ denotes the section of $T{\mathcal{G}}$ satisfying
$\iota_{*}X^{0}=X|_{t=\frac{1}{2}}$. So we must solve

$T\begin{pmatrix}\eta(X_{1})\\\ \vdots\\\
\eta(X_{p})\end{pmatrix}=\left((a_{ji})-\begin{pmatrix}\pi(T)&&\\\ &\ddots&\\\
&&\pi(T)\end{pmatrix}\right)\begin{pmatrix}\eta(X_{1})\\\ \vdots\\\
\eta(X_{p})\end{pmatrix}.$

The solution is

$\begin{pmatrix}\eta(X_{1})\\\ \vdots\\\
\eta(X_{p})\end{pmatrix}(x,t)=\exp\left\\{\left(t-\frac{1}{2}\right)\left((a_{ji})-\begin{pmatrix}\pi(T)&&\\\
&\ddots&\\\
&&\pi(T)\end{pmatrix}\right)\right\\}\begin{pmatrix}\xi\left(X^{0}_{1}\right)\\\
\vdots\\\ \xi\left(X^{0}_{p}\right)\end{pmatrix}(x).$

We define $\eta(X_{i})$ by this. Then $\eta(X)$ for any $X\in V$ is defined by
linearity, so that we have defined
$\eta\in\Omega^{1}\left({\mathcal{F}}|_{U_{1}};W\right)$. Then (16) and
$\iota^{*}\eta=\xi$ are satisfied. To see that (15) holds, we only have to
show $\left({d_{\mathcal{F}}}\eta+\pi\omega_{0}\wedge\eta\right)(X,Y)=0$ for
all $X,Y\in V$. Put $\theta={d_{\mathcal{F}}}\eta+\pi\omega_{0}\wedge\eta$.
Then

$\displaystyle T\left(\theta(X,Y)\right)$ $\displaystyle=TX\eta(Y)-TY\eta(X)$
$\displaystyle=X\left(\eta\left([T,Y]\right)-\pi(T)\eta(Y)\right)+[T,X]\eta(Y)$
$\displaystyle\ \ \ \ \ \ \
-Y\left(\eta\left([T,X]\right)-\pi(T)\eta(X)\right)-[T,Y]\eta(X)$
$\displaystyle=-\pi(T)\theta(X,Y)+\theta\left(X,[T,Y]\right)+\theta\left([T,X],Y\right)$

and
$\theta(X,Y)\left(x,\frac{1}{2}\right)=X^{0}\xi\left(Y^{0}\right)-Y^{0}\xi\left(X^{0}\right)=d_{\mathcal{G}}\xi\left(X^{0},Y^{0}\right)=0$.
As in the proof of injectivity we can show $\theta(X_{i},X_{j})=0$, so that
(15) is satisfied. This completes the proof. ∎

By the assumption that $V$ is a Diophantine subspace of ${\mathbb{R}}^{n}$,
the leafwise cohomology $H^{1}({\mathcal{G}};W)$ vanishes, ie
$H^{1}({\mathcal{G}};W)=H^{1}(V;W)=V^{*}\otimes W$. Thus
$H^{1}\left({\mathcal{F}}|_{U_{1}};\pi\right)\simeq V^{*}\otimes W$. Let
$q\colon{\mathfrak{s}}\to V$ be the natural projection.

######  Lemma 29.

An element $\varphi\otimes w\in V^{*}\otimes W$ corresponds to

$\left(e^{\left(t-\frac{1}{2}\right)\left(\mathop{\mathrm{ad}}\nolimits^{0}T\right)^{*}}\varphi\right)q\omega_{0}\otimes
e^{-\left(t-\frac{1}{2}\right)\pi(T)}w\in\Omega^{1}\left({\mathcal{F}}|_{U_{1}};W\right)$

by the above isomorphism, where
$\left(\mathop{\mathrm{ad}}\nolimits^{0}T\right)^{*}\colon V^{*}\to V^{*}$ is
the adjoint of $\mathop{\mathrm{ad}}\nolimits^{0}T$.

###### Proof.

Put
$\eta=\left(e^{\left(t-\frac{1}{2}\right)\left(\mathop{\mathrm{ad}}\nolimits^{0}T\right)^{*}}\varphi\right)q\omega_{0}\otimes
e^{-\left(t-\frac{1}{2}\right)\pi(T)}w$. Then $\iota^{*}\eta=\varphi\otimes w$
and $\eta(T)=0$. For $X\in V$,
$\eta(X)=\left(e^{\left(t-\frac{1}{2}\right)\left(\mathop{\mathrm{ad}}\nolimits^{0}T\right)^{*}}\varphi\right)(X)e^{-\left(t-\frac{1}{2}\right)\pi(T)}w$
satisfies $T\eta(X)=\eta\left([T,X]\right)-\pi(T)\eta(X)$. So we see
${d_{\mathcal{F}}}\eta+\pi\omega_{0}\wedge\eta=0$ as in the proof of Lemma 28.
∎

$U_{1}\cap U_{2}$ is the disjoint union of the projections to $M$ of
${\mathbb{T}}^{n}\times\left(0,\frac{1}{2}\right)$ and
${\mathbb{T}}^{n}\times\left(\frac{1}{2},1\right)$. We define the maps

$\displaystyle\iota_{\frac{1}{2}}$
$\displaystyle\colon{\mathbb{T}}^{n}\hookrightarrow U_{1},$
$\displaystyle\iota_{\frac{1}{2}}(x)$
$\displaystyle=\left(x,\frac{1}{2}\right),$ $\displaystyle\iota_{0}$
$\displaystyle\colon{\mathbb{T}}^{n}\hookrightarrow U_{2},$
$\displaystyle\iota_{0}(x)$ $\displaystyle=(x,0),$
$\displaystyle\iota_{\frac{1}{4}}$
$\displaystyle\colon{\mathbb{T}}^{n}\hookrightarrow U_{1}\cap U_{2},$
$\displaystyle\iota_{\frac{1}{4}}(x)$
$\displaystyle=\left(x,\frac{1}{4}\right),$ $\displaystyle\iota_{\frac{3}{4}}$
$\displaystyle\colon{\mathbb{T}}^{n}\hookrightarrow U_{1}\cap U_{2},$
$\displaystyle\iota_{\frac{3}{4}}(x)$
$\displaystyle=\left(x,\frac{3}{4}\right).$

We will calculate the bottom map of the next commutative diagram in which the
vertical maps are isomorphisms:

${H^{1}\left({\mathcal{F}}|_{U_{1}};\pi\right)\oplus
H^{1}\left({\mathcal{F}}|_{U_{2}};\pi\right)}$${H^{1}\left({\mathcal{F}}|_{U_{1}\cap
U_{2}};\pi\right)}$${(V^{*}\otimes W)\oplus(V^{*}\otimes W)}$${(V^{*}\otimes
W)\oplus(V^{*}\otimes
W).}$$\scriptstyle{Q}$$\scriptstyle{\iota_{\frac{1}{2}}^{*}\oplus\iota_{0}^{*}}$${\circlearrowright}$$\scriptstyle{\iota_{\frac{1}{4}}^{*}\oplus\iota_{\frac{3}{4}}^{*}}$

######  Lemma 30.

The bottom map of the above diagram is

$\begin{pmatrix}-e^{-\frac{1}{4}(\mathop{\mathrm{ad}}\nolimits^{0}T)^{*}}\otimes
e^{\frac{1}{4}\pi(T)}&e^{\frac{1}{4}(\mathop{\mathrm{ad}}\nolimits^{0}T)^{*}}\otimes
e^{-\frac{1}{4}\pi(T)}\\\
-e^{\frac{1}{4}(\mathop{\mathrm{ad}}\nolimits^{0}T)^{*}}\otimes
e^{-\frac{1}{4}\pi(T)}&e^{-\frac{1}{4}(\mathop{\mathrm{ad}}\nolimits^{0}T)^{*}}\otimes
e^{\frac{1}{4}\pi(T)}\end{pmatrix}.$

###### Proof.

We define
$F\colon{\mathbb{T}}^{n}\times{\mathbb{R}}\to{\mathbb{T}}^{n}\times{\mathbb{R}}$
by $F(x,t)=\left(A^{-1}x,t-1\right)$. Take an element
$\left(\varphi_{1}\otimes w_{1},\varphi_{2}\otimes w_{2}\right)$ of
$(V^{*}\otimes W)\oplus(V^{*}\otimes W)$. This is mapped by the vertical map
to

$\left(\left(e^{\left(t-\frac{1}{2}\right)\left(\mathop{\mathrm{ad}}\nolimits^{0}T\right)^{*}}\varphi_{1}\right)q\omega_{0}\otimes
e^{-\left(t-\frac{1}{2}\right)\pi(T)}w_{1},\left(e^{t\left(\mathop{\mathrm{ad}}\nolimits^{0}T\right)^{*}}\varphi_{2}\right)q\omega_{0}\otimes
e^{-t\pi(T)}w_{2}\right),$

which is in turn mapped by $Q$ to

$\left(\left(e^{t\left(\mathop{\mathrm{ad}}\nolimits^{0}T\right)^{*}}\varphi_{2}\right)q\omega_{0}\otimes
e^{-t\pi(T)}w_{2}-\left(e^{\left(t-\frac{1}{2}\right)\left(\mathop{\mathrm{ad}}\nolimits^{0}T\right)^{*}}\varphi_{1}\right)q\omega_{0}\otimes
e^{-\left(t-\frac{1}{2}\right)\pi(T)}w_{1},\right.\\\
\left.F^{*}\left(\left(e^{t\left(\mathop{\mathrm{ad}}\nolimits^{0}T\right)^{*}}\varphi_{2}\right)q\omega_{0}\otimes
e^{-t\pi(T)}w_{2}\right)-\left(e^{\left(t-\frac{1}{2}\right)\left(\mathop{\mathrm{ad}}\nolimits^{0}T\right)^{*}}\varphi_{1}\right)q\omega_{0}\otimes
e^{-\left(t-\frac{1}{2}\right)\pi(T)}w_{1}\right).$

Since

$F^{*}\left(\left(e^{t\left(\mathop{\mathrm{ad}}\nolimits^{0}T\right)^{*}}\varphi_{2}\right)q\omega_{0}\otimes
e^{-t\pi(T)}w_{2}\right)=\left(e^{(t-1)\left(\mathop{\mathrm{ad}}\nolimits^{0}T\right)^{*}}\varphi_{2}\right)q\omega_{0}\otimes
e^{-(t-1)\pi(T)}w_{2},$

the above element equals to

$\left(\left(e^{\left(t-\frac{1}{4}\right)\left(\mathop{\mathrm{ad}}\nolimits^{0}T\right)^{*}}e^{\frac{1}{4}\left(\mathop{\mathrm{ad}}\nolimits^{0}T\right)^{*}}\varphi_{2}\right)q\omega_{0}\otimes
e^{-\left(t-\frac{1}{4}\right)\pi(T)}e^{-\frac{1}{4}\pi(T)}w_{2}\right.\\\
-\left(e^{\left(t-\frac{1}{4}\right)\left(\mathop{\mathrm{ad}}\nolimits^{0}T\right)^{*}}e^{-\frac{1}{4}\left(\mathop{\mathrm{ad}}\nolimits^{0}T\right)^{*}}\varphi_{1}\right)q\omega_{0}\otimes
e^{-\left(t-\frac{1}{4}\right)\pi(T)}e^{\frac{1}{4}\pi(T)}w_{1},\\\
\left(e^{\left(t-\frac{3}{4}\right)\left(\mathop{\mathrm{ad}}\nolimits^{0}T\right)^{*}}e^{-\frac{1}{4}\left(\mathop{\mathrm{ad}}\nolimits^{0}T\right)^{*}}\varphi_{2}\right)q\omega_{0}\otimes
e^{-\left(t-\frac{3}{4}\right)\pi(T)}e^{\frac{1}{4}\pi(T)}w_{2}\\\
\left.-\left(e^{\left(t-\frac{3}{4}\right)\left(\mathop{\mathrm{ad}}\nolimits^{0}T\right)^{*}}e^{\frac{1}{4}\left(\mathop{\mathrm{ad}}\nolimits^{0}T\right)^{*}}\varphi_{1}\right)q\omega_{0}\otimes
e^{-\left(t-\frac{3}{4}\right)\pi(T)}e^{-\frac{1}{4}\pi(T)}w_{1}\right)$

and finally this is mapped by the vertical arrow to

$\left(e^{\frac{1}{4}\left(\mathop{\mathrm{ad}}\nolimits^{0}T\right)^{*}}\varphi_{2}\otimes
e^{-\frac{1}{4}\pi(T)}w_{2}-e^{-\frac{1}{4}\left(\mathop{\mathrm{ad}}\nolimits^{0}T\right)^{*}}\varphi_{1}\otimes
e^{\frac{1}{4}\pi(T)}w_{1},\right.\\\
\left.e^{-\frac{1}{4}\left(\mathop{\mathrm{ad}}\nolimits^{0}T\right)^{*}}\varphi_{2}\otimes
e^{\frac{1}{4}\pi(T)}w_{2}-e^{\frac{1}{4}\left(\mathop{\mathrm{ad}}\nolimits^{0}T\right)^{*}}\varphi_{1}\otimes
e^{-\frac{1}{4}\pi(T)}w_{1}\right).$

∎

By using the canonical isomorphism $V^{*}\otimes
W\simeq\mathop{\mathrm{Hom}}\nolimits(V,W)$, the map in the above lemma
becomes a map

$\bar{Q}\colon\mathop{\mathrm{Hom}}\nolimits(V,W)\oplus\mathop{\mathrm{Hom}}\nolimits(V,W)\to\mathop{\mathrm{Hom}}\nolimits(V,W)\oplus\mathop{\mathrm{Hom}}\nolimits(V,W)$

which sends $\begin{pmatrix}\alpha\\\ \beta\end{pmatrix}$ to

$\begin{pmatrix}-e^{\frac{1}{4}\pi(T)}\circ\alpha\circ
e^{-\frac{1}{4}\mathop{\mathrm{ad}}\nolimits^{0}T}+e^{-\frac{1}{4}\pi(T)}\circ\beta\circ
e^{\frac{1}{4}\mathop{\mathrm{ad}}\nolimits^{0}T}\\\
-e^{-\frac{1}{4}\pi(T)}\circ\alpha\circ
e^{\frac{1}{4}\mathop{\mathrm{ad}}\nolimits^{0}T}+e^{\frac{1}{4}\pi(T)}\circ\beta\circ
e^{-\frac{1}{4}\mathop{\mathrm{ad}}\nolimits^{0}T}\end{pmatrix}.$

######  Lemma 31.

If $\begin{pmatrix}\alpha\\\ \beta\end{pmatrix}$ is in the kernel of
$\bar{Q}$, then $\alpha$ and $\beta$ satisfy
$e^{\pi(T)}\circ\alpha=\alpha\circ e^{\mathop{\mathrm{ad}}\nolimits^{0}T}$ and
$e^{\pi(T)}\circ\beta=\beta\circ e^{\mathop{\mathrm{ad}}\nolimits^{0}T}$.

###### Proof.

The pair $\begin{pmatrix}\alpha\\\ \beta\end{pmatrix}$ is in the kernel if and
only if $-e^{\frac{1}{2}\pi(T)}\circ\alpha+\beta\circ
e^{\frac{1}{2}\mathop{\mathrm{ad}}\nolimits^{0}T}=0$ and $-\alpha\circ
e^{\frac{1}{2}\mathop{\mathrm{ad}}\nolimits^{0}T}+e^{\frac{1}{2}\pi(T)}\circ\beta=0$.
By eliminating $\beta$ or $\alpha$ we get the conclusion. ∎

Next we will calculate the map
$H^{0}\left({\mathcal{F}}|_{U_{1}};\pi\right)\oplus
H^{0}\left({\mathcal{F}}|_{U_{2}};\pi\right)\stackrel{{\scriptstyle
P}}{{\to}}H^{0}\left({\mathcal{F}}|_{U_{1}\cap U_{2}};\pi\right)$. The group
$H^{0}\left({\mathcal{F}}|_{U_{1}};\pi\right)$ consists of all smooth
functions $f\colon U_{1}\to W$ satisfying
${d_{\mathcal{F}}}f+\pi\omega_{0}\wedge f=0$, which is equivalent to equations
$Tf+\pi(T)f=0$ and $Xf=0$ for all $X\in V$. Since the action
${\mathbb{T}}^{n}\curvearrowleft V$ has a dense orbit, such $f$ must be
constant along the directions of tori; write $f(x,t)=f(t)$. Solving the
differential equation, we get
$f(t)=e^{-\left(t-\frac{1}{2}\right)\pi(T)}f\left(\frac{1}{2}\right)$. Thus we
have an isomorphism $H^{0}\left({\mathcal{F}}|_{U_{1}};\pi\right)\simeq W$
which sends $f$ to $f\left(\frac{1}{2}\right)$. The bottom arrow of the
diagram

${H^{0}\left({\mathcal{F}}|_{U_{1}};\pi\right)\oplus
H^{0}\left({\mathcal{F}}|_{U_{2}};\pi\right)}$${H^{0}\left({\mathcal{F}}|_{U_{1}\cap
U_{2}};\pi\right)}$${W\oplus W}$${W\oplus
W}$$\scriptstyle{P}$$\scriptstyle{\iota_{\frac{1}{2}}^{*}\oplus\iota_{0}^{*}}$${\circlearrowright}$$\scriptstyle{\iota_{\frac{1}{4}}^{*}\oplus\iota_{\frac{3}{4}}^{*}}$$\scriptstyle{\bar{P}}$

sends an element $(w_{1},w_{2})$ as follows:

$\displaystyle(w_{1},w_{2})$
$\displaystyle\mapsto\left(e^{-\left(t-\frac{1}{2}\right)\pi(T)}w_{1},e^{-t\pi(T)}w_{2}\right)$
$\displaystyle\mapsto\left(e^{-t\pi(T)}w_{2}-e^{-\left(t-\frac{1}{2}\right)\pi(T)}w_{1},F^{*}\left(e^{-t\pi(T)}w_{2}\right)-e^{-\left(t-\frac{1}{2}\right)\pi(T)}w_{1}\right)$
$\displaystyle=\left(e^{-t\pi(T)}w_{2}-e^{-\left(t-\frac{1}{2}\right)\pi(T)}w_{1},e^{-(t-1)\pi(T)}w_{2}-e^{-\left(t-\frac{1}{2}\right)\pi(T)}w_{1}\right)$
$\displaystyle\mapsto\left(e^{-\frac{1}{4}\pi(T)}w_{2}-e^{\frac{1}{4}\pi(T)}w_{1},e^{\frac{1}{4}\pi(T)}w_{2}-e^{-\frac{1}{4}\pi(T)}w_{1}\right).$

An element $(w_{1},w_{2})$ is in $\ker\bar{P}$ if and only if
$w_{2}=e^{\frac{1}{2}\pi(T)}w_{1}$ and $w_{1}=e^{\frac{1}{2}\pi(T)}w_{2}$. So

$\ker\bar{P}\simeq\ker\left(\mathop{\mathrm{id}}\nolimits-e^{\pi(T)}\right)\subset
W$ (17)

by $(w_{1},w_{2})\mapsto w_{2}$.

Finally we calculate the cohomology $H^{1}({\mathfrak{s}};\pi)$ of the Lie
algebra. A 1–cocycle is a linear map $\varphi\colon{\mathfrak{s}}\to W$
satisfying $\pi(T)\varphi(X)-\varphi\left([T,X]\right)=0$ for all $X\in V$. So
the space of 1–cocycles is isomorphic to
$\mathop{\mathrm{Hom}}\nolimits_{T}(V,W)\oplus W$ by the isomorphism which
sends $\varphi$ to $\left(\varphi|_{V},\varphi(T)\right)$, where
$\mathop{\mathrm{Hom}}\nolimits_{T}(V,W)$ denotes the space of
$\left(\mathop{\mathrm{ad}}\nolimits^{0}T,\pi(T)\right)$–equivariant linear
maps from $V$ to $W$. On the other hand a 1–coboundary maps
$Z\in{\mathfrak{s}}$ to $\pi(Z)c\in W$ for some $c\in W$. So a linear map
$\varphi\colon{\mathfrak{s}}\to W$ is a 1–coboundary if and only if
$\varphi(X)=0$ for all $X\in V$ and
$\varphi(T)\in\mathop{\mathrm{im}}\nolimits\pi(T)$. The space of
1–coboundaries is isomorphic to $0\oplus\mathop{\mathrm{im}}\nolimits\pi(T)$
by the above isomorphism. Therefore we have

$H^{1}({\mathfrak{s}};\pi)\simeq\mathop{\mathrm{Hom}}\nolimits_{T}(V,W)\oplus\left(W/\mathop{\mathrm{im}}\nolimits\pi(T)\right).$
(18)

###  5.4 Vanishing of the cohomology

As we saw in Section 5.2 we must show vanishing of the following cohomologies:

  * (a)

The trivial representation ${\mathfrak{s}}\curvearrowright{\mathbb{R}}$.

  * (b)

The restriction of $\mathop{\mathrm{ad}}\nolimits$ to $V$ and its negative,
${\mathfrak{s}}\stackrel{{\scriptstyle\pm\mathop{\mathrm{ad}}\nolimits^{0}}}{{\curvearrowright}}V$.

(a) By (18) in last section,

$H^{1}({\mathfrak{s}})\simeq\left\\{\varphi\colon V\to{\mathbb{R}}\ \middle|\
\varphi\circ\mathop{\mathrm{ad}}\nolimits^{0}T=0\right\\}\oplus{\mathbb{R}}.$

By (11) and (17), $H^{1}({\mathcal{F}})\simeq{\mathbb{R}}\oplus\ker\bar{Q}$.
If $(\alpha,\beta)$ is in $\ker\bar{Q}$, by Lemma 31,
$\alpha=\alpha\circ\Phi_{k}$ for all $k\in{\mathbb{Z}}$. By the assumption of
Zariski density, $\alpha=\alpha\circ\Phi_{t}$ for all $t\in{\mathbb{R}}$. Thus
$\alpha\circ\mathop{\mathrm{ad}}\nolimits^{0}T=0$ by differentiation. The same
thing holds for $\beta$ and then we have $\alpha=\beta$. So

$\ker\bar{Q}\simeq\left\\{\varphi\colon V\to{\mathbb{R}}\ \middle|\
\varphi\circ\mathop{\mathrm{ad}}\nolimits^{0}T=0\right\\}$

and $H^{1}({\mathcal{F}})=H^{1}({\mathfrak{s}})$.

b) By (18),

$H^{1}\left({\mathfrak{s}};\pm\mathop{\mathrm{ad}}\nolimits^{0}\right)\simeq\left\\{\varphi\colon
V\to V\ \middle|\
\varphi\circ\mathop{\mathrm{ad}}\nolimits^{0}T=\pm\mathop{\mathrm{ad}}\nolimits^{0}T\circ\varphi\right\\}.$

By (11) and (17),
$H^{1}\left({\mathcal{F}};\pm\mathop{\mathrm{ad}}\nolimits^{0}\right)\simeq\ker\bar{Q}$.
If $(\alpha,\beta)$ is in $\ker\bar{Q}$, by Lemma 31, $\Phi_{\pm
k}\circ\alpha=\alpha\circ\Phi_{k}$ for all $k\in{\mathbb{Z}}$. By the
assumption of Zariski density, $\Phi_{\pm t}\circ\alpha=\alpha\circ\Phi_{t}$
for all $t\in{\mathbb{R}}$. Thus
$\pm\mathop{\mathrm{ad}}\nolimits^{0}T\circ\alpha=\alpha\circ\mathop{\mathrm{ad}}\nolimits^{0}T$
by differentiation. Note that here we use the benefit of large scale geometry.
Since the map

$\ker\bar{Q}\to\left\\{\varphi\colon V\to V\ \middle|\
\varphi\circ\mathop{\mathrm{ad}}\nolimits^{0}T=\pm\mathop{\mathrm{ad}}\nolimits^{0}T\circ\varphi\right\\}$

mapping $(\alpha,\beta)$ to $\alpha$ is injective, we have
$H^{1}\left({\mathcal{F}};\pm\mathop{\mathrm{ad}}\nolimits^{0}\right)=H^{1}\left({\mathfrak{s}};\pm\mathop{\mathrm{ad}}\nolimits^{0}\right)$.

This completes the proof of Theorem 6.

##  6 Parameter rigidity of transitive locally free actions and rigidity of
lattices

###  6.1 Relations between transitive locally free actions and lattices

Let $S$ be a connected simply connected solvable Lie group, $\Gamma_{0}$ be a
cocompact lattice in $S$ and $M=\Gamma_{0}\backslash
S\stackrel{{\scriptstyle\rho_{0}}}{{\curvearrowleft}}S$ be the action by right
multiplication.

We put

$\displaystyle{\mathcal{A}}(\Gamma_{0},S)$
$\displaystyle:=\left\\{M\stackrel{{\scriptstyle\rho}}{{\curvearrowleft}}S\
\middle|\ \text{$\rho$ is transitive locally free}\right\\}$
$\displaystyle{\mathcal{H}}(\Gamma_{0},S)$
$\displaystyle:=\left\\{\alpha\colon\Gamma_{0}\to S\ \middle|\ \text{$\alpha$
is an injective homomorphism and $\alpha(\Gamma_{0})$ is a lattice}\right\\}.$

Then

${\mathcal{H}}(\Gamma_{0},S)/\mathop{\mathrm{Aut}}\nolimits(\Gamma_{0})=\left\\{\Gamma\subset
S\ \middle|\ \text{$\Gamma$ is a cocompact lattice isomorphic to
$\Gamma_{0}$}\right\\}.$

######  Proposition 32.

There is a one-to-one correspondence between

${\mathcal{A}}(\Gamma_{0},S)/\left(\text{$C^{\infty}$--conjugacy $+$
$\mathop{\mathrm{Aut}}\nolimits(S)$}\right)$

and

$\mathop{\mathrm{Aut}}\nolimits(S)\backslash{\mathcal{H}}(\Gamma_{0},S)/\mathop{\mathrm{Aut}}\nolimits(\Gamma_{0})$
$=\mathop{\mathrm{Aut}}\nolimits(S)\backslash\left\\{\Gamma\subset S\
\middle|\ \text{$\Gamma$ is a cocompact lattice isomorphic to
$\Gamma_{0}$}\right\\}$

taking $\rho$ to the isotropy subgroup of $\rho$ at the point
$x_{0}=\Gamma_{0}$.

###### Proof.

Well-definedness and injectivity are easy. For surjectivity, take any
cocompact lattice $\Gamma$ in $S$ isomorphic to $\Gamma_{0}$. Then by a
theorem of Mostow (See for example Raghunathan [9, Theorem 3.6]),
$\Gamma\backslash S$ and $\Gamma_{0}\backslash S=M$ are diffeomorphic. So the
map is also surjective. ∎

######  Proposition 33.

There is a one-to-one correspondence between

${\mathcal{A}}(\Gamma_{0},S)/\text{parameter equivalence}$

and

$\mathop{\mathrm{Aut}}\nolimits(S)\backslash{\mathcal{H}}(\Gamma_{0},S).$

###### Proof.

The definition of the map is as follows: For a transitive locally free action
$M\stackrel{{\scriptstyle\rho}}{{\curvearrowleft}}S$, let $a_{\rho}\colon
M\times S\to S$ be the smooth map defined uniquely by
$\rho_{0}(x,s)=\rho\left(x,a_{\rho}(x,s)\right)$ and $a_{\rho}(x,1)=1$. Then
$a_{\rho}$ is a cocycle over $\rho_{0}$:

$a_{\rho}\left(x,ss^{\prime}\right)=a_{\rho}(x,s)a_{\rho}\left(\rho_{0}(x,s),s^{\prime}\right).$

Let $\Gamma_{\rho}$ be the isotropy subgroup of $\rho$ at $x_{0}$. Then we
have
$a_{\rho}(x_{0},\cdotp)\colon\Gamma_{0}\stackrel{{\scriptstyle\sim}}{{\to}}\Gamma_{\rho}\subset
S$. So we will define the map by $\rho\mapsto a_{\rho}(x_{0},\cdotp)$.

Let us see the well-definedness of this map. Take two transitive locally free
actions $M\stackrel{{\scriptstyle\rho_{i}}}{{\curvearrowleft}}S$ $(i=1,2)$
which are parameter equivalent. So there are
$\Phi\in\mathop{\mathrm{Aut}}\nolimits(S)$ and a diffeomorphism $F\colon M\to
M$ which is homotopic to the identity such that

$F\left(\rho_{1}(x,s)\right)=\rho_{2}\left(F(x),\Phi(s)\right).$ (19)

Let $b\colon M\times S\to S$ be the cocycle over $\rho_{1}$ defined by
$\rho_{1}(x,s)=\rho_{0}\left(x,b(x,s)\right)$. Note that
$s=a_{\rho_{1}}\left(x,b(x,s)\right)$. We have
$\rho_{1}(x,s)=\rho_{2}\left(x,a_{\rho_{2}}\left(x,b(x,s)\right)\right)$ and
then $a_{\rho_{2}}\left(x,b(x,s)\right)$ is a cocycle over $\rho_{1}$. Since
$F$ is homotopic to the identity we can define a smooth map $P\colon M\to S$
by $F(x)=\rho_{2}(x,P(x)^{-1})$. By (19), we see

$\displaystyle a_{\rho_{2}}\left(x,b(x,s)\right)$
$\displaystyle=P(x)^{-1}\Phi(s)P\left(\rho_{1}(x,s)\right)$
$\displaystyle=P(x)^{-1}\Phi\left(a_{\rho_{1}}\left(x,b(x,s)\right)\right)P\left(\rho_{0}\left(x,b(x,s)\right)\right),$

so that

$a_{\rho_{2}}(x,s)=P(x)^{-1}\Phi\left(a_{\rho_{1}}(x,s)\right)P\left(\rho_{0}(x,s)\right)$

for all $s\in S$ since $b(x,\cdotp)\colon S\to S$ is invertible. Taking
$x=x_{0}$ and $s=\gamma\in\Gamma_{0}$, we have

$a_{\rho_{2}}(x_{0},\gamma)=P(x_{0})^{-1}\Phi\left(a_{\rho_{1}}(x_{0},\gamma)\right)P(x_{0}).$

So the map is well-defined.

Next let us see the injectivity. Take two transitive locally free actions
$M\stackrel{{\scriptstyle\rho_{i}}}{{\curvearrowleft}}S$ for $i=1,2$ for which
there exists $\Phi\in\mathop{\mathrm{Aut}}\nolimits(S)$ such that
$a_{\rho_{2}}(x_{0},\gamma)=\Phi\left(a_{\rho_{1}}(x_{0},\gamma)\right)$ for
every $\gamma\in\Gamma_{0}$. Take $b\colon M\times S\to S$ satisfying
$\rho_{2}(x,s)=\rho_{0}\left(x,b(x,s)\right)$, so that
$s=b\left(x,a_{\rho_{2}}(x,s)\right)$. We have $S$–equivariant diffeomorphisms
$\Gamma_{\rho_{i}}\backslash S\stackrel{{\scriptstyle\sim}}{{\to}}M$ mapping
$\Gamma_{\rho_{i}}s$ to $\rho_{i}(x_{0},s)$. Using these define a
diffeomorphism $F$ by

${M}$${\Gamma_{\rho_{1}}\backslash S}$${M}$${\Gamma_{\rho_{2}}\backslash
S.}$$\scriptstyle{F}$${\circlearrowright}$$\scriptstyle{\Phi}$$\scriptstyle{\sim}$$\scriptstyle{\sim}$

Here $\Phi$ denotes the map
$\Gamma_{\rho_{1}}s\mapsto\Gamma_{\rho_{2}}\Phi(s)$. Obviously
$F\left(\rho_{1}(x,s)\right)=\rho_{2}\left(F(x),\Phi(s)\right)$. To see that
$F$ is homotopic to the identity, take any $\gamma\in\Gamma_{0}=\pi_{1}(M)$
and a curve $c\colon[0,1]\to S$ connecting $1$ and $\gamma$ and consider the
curve $\rho_{0}\left(x_{0},c(\cdotp)\right)\colon[0,1]\to M$. Since
$\rho_{0}\left(x_{0},c(t)\right)=\rho_{1}\left(x_{0},a_{\rho_{1}}\left(x_{0},c(t)\right)\right)$,
we have

$\displaystyle F\left(\rho_{0}\left(x_{0},c(t)\right)\right)$
$\displaystyle=\rho_{2}\left(x_{0},\Phi\left(a_{\rho_{1}}\left(x_{0},c(t)\right)\right)\right)$
$\displaystyle=\rho_{0}\left(x_{0},b\left(x_{0},\Phi\left(a_{\rho_{1}}\left(x_{0},c(t)\right)\right)\right)\right).$

Therefore $F_{*}\colon\pi_{1}(M)\to\pi_{1}(M)$ maps $\gamma$ to

$b\left(x_{0},\Phi\left(a_{\rho_{1}}(x_{0},\gamma)\right)\right)=b\left(x_{0},a_{\rho_{2}}(x_{0},\gamma)\right)=\gamma.$

So $F$ must be homotopic to the identity.

Finally, we see the surjectivity. Take any $\alpha\colon\Gamma_{0}\to S$. Put
$\Gamma:=\alpha(\Gamma_{0})$ and $y_{0}:=\Gamma\in\Gamma\backslash S$. We have
an isomorphism
$\alpha\colon\Gamma_{0}\stackrel{{\scriptstyle\sim}}{{\to}}\Gamma$ of the
fundamental groups of $(M,x_{0})$ and $(\Gamma\backslash S,y_{0})$. By Witte
[10, Theorem 7.4] which is a refinement of Mostow’s theorem used in the proof
of the previous proposition, we can find a diffeomorphism
$F\colon(M,x_{0})\to(\Gamma\backslash S,y_{0})$ which induces
$\alpha\colon\Gamma_{0}\stackrel{{\scriptstyle\sim}}{{\to}}\Gamma$ on the
fundamental groups. Define
$M\stackrel{{\scriptstyle\rho}}{{\curvearrowleft}}S$ by
$\rho(x,s)=F^{-1}\left(F(x)s\right)$. Then $\rho$ is a transitive locally free
action. We have $F\left(\rho_{0}(x_{0},s)\right)=y_{0}a_{\rho}(x_{0},s)$ for
all $s\in S$. Take any $\gamma_{0}\in\Gamma_{0}$. Choose a curve
$s\colon[0,1]\to S$ connecting $1$ and $\gamma_{0}$. Then the curve
$\rho_{0}\left(x_{0},s(t)\right)$ represents $\gamma_{0}$ in
$\pi_{1}(M,x_{0})$. Thus the curve $y_{0}a_{\rho}\left(x_{0},s(t)\right)$
represents $F_{*}(\gamma_{0})=\alpha(\gamma_{0})$ in $\pi_{1}(\Gamma\backslash
S,y_{0})$. Since the curve $a_{\rho}\left(x_{0},s(t)\right)$ in $S$ is the
lift of the curve $y_{0}a_{\rho}\left(x_{0},s(t)\right)$ starting from $1$, we
get $a_{\rho}(x_{0},\gamma_{0})=\alpha(\gamma_{0})$. This proves the
surjectivity. ∎

######  Proposition 34.

The map

${\mathcal{A}}(\Gamma_{0},S)\to{\mathcal{H}}(\Gamma_{0},S)$

defined in Proposition 33 is continuous. Here ${\mathcal{A}}(\Gamma_{0},S)$ is
endowed with the topology induced from the $C^{\infty}$ compact-open topology
of $C^{\infty}(M\times S,M)$ and ${\mathcal{H}}(\Gamma_{0},S)$ has the
topology of pointwise convergence.

###### Proof.

Take any $\gamma_{0}\in\Gamma_{0}$. We must show that
$a_{\rho}(x_{0},\gamma_{0})\in S$ is continuous with respect to $\rho$. The
map ${\mathcal{A}}(\Gamma_{0},S)\to C^{\infty}(S,M)$ which sends $\rho$ to
$\rho(x_{0},\cdot)$ is continuous. Take a $C^{\infty}$–curve $c\colon[0,1]\to
S$ connecting $1$ and $\gamma_{0}$. Let $c_{\rho}\colon[0,1]\to S$ be the lift
of the curve $\rho_{0}\left(x_{0},c(t)\right)$ with respect to the covering
$\rho(x_{0},\cdot)\colon S\to M$ starting at $1$. Then
$a_{\rho}(x_{0},\gamma_{0})=c_{\rho}(1)$. Then the map
${\mathcal{A}}(\Gamma_{0},S)\to C^{\infty}\left([0,1],S\right)$ taking $\rho$
to $c_{\rho}$ is continuous. In particular, $\rho\mapsto
c_{\rho}(1)=a_{\rho}(x_{0},\gamma_{0})$ is continuous. ∎

######  Proposition 35.

We have a commutative diagram

${{\mathcal{A}}(\Gamma_{0},S)/\text{parameter
equivalence}}$${\mathop{\mathrm{Aut}}\nolimits(S)\backslash{\mathcal{H}}(\Gamma_{0},S)}$${{\mathcal{A}}(\Gamma_{0},S)/\left(\text{$C^{\infty}$--conjugacy
$+$
$\mathop{\mathrm{Aut}}\nolimits(S)$}\right)}$${\mathop{\mathrm{Aut}}\nolimits(S)\backslash{\mathcal{H}}(\Gamma_{0},S)/\mathop{\mathrm{Aut}}\nolimits(\Gamma_{0}),}$${\circlearrowright}$$\scriptstyle{\sim}$$\scriptstyle{\sim}$

where the horizontal maps are defined above and the vertical surjective maps
are defined obviously.

###### Proof.

This is obvious. ∎

By combining Proposition 33 with Theorem 3, we get the following.

######  Corollary 36.

Let $S$ be a connected simply connected solvable Lie group, $\Gamma_{0}$ be a
lattice in $S$ and ${\mathfrak{h}}$ be a subspace between
$[{\mathfrak{s}},{\mathfrak{s}}]$ and ${\mathfrak{n}}$. If

$H^{1}\left(\Gamma_{0}\backslash
S;{\mathfrak{s}}\stackrel{{\scriptstyle\mathop{\mathrm{ad}}\nolimits\circ\varphi}}{{\curvearrowright}}\mathop{\mathrm{Gr}}\nolimits_{\mathfrak{h}}({\mathfrak{s}})\right)=H^{1}\left({\mathfrak{s}};{\mathfrak{s}}\stackrel{{\scriptstyle\mathop{\mathrm{ad}}\nolimits\circ\varphi}}{{\curvearrowright}}\mathop{\mathrm{Gr}}\nolimits_{\mathfrak{h}}({\mathfrak{s}})\right)$

for all surjective Lie algebra homomorphisms
$\varphi\colon{\mathfrak{s}}\to{\mathfrak{s}}/{\mathfrak{h}}$, then any
injective homomorphism $\alpha\colon\Gamma_{0}\to S$ whose image is a lattice,
is transformed into the inclusion $\Gamma_{0}\hookrightarrow S$ by an element
of $\mathop{\mathrm{Aut}}\nolimits(S)$. In particular, if $\Gamma$ is a
lattice in $S$ isomorphic to $\Gamma_{0}$, there is an isomorphism of $S$
which transforms $\Gamma$ into $\Gamma_{0}$.

###  6.2 A counterexample to Theorem 1 for solvable Lie groups

Now we see a counterexample to Theorem 1 for solvable Lie groups. The next
example is taken from Example 2.2 of Baues and Klopsch [2], which is due to
Milovanov [7]. Consider $A_{0}=\begin{pmatrix}0&-1\\\
1&3\end{pmatrix}\in\mathop{\mathrm{SL}}\nolimits(2,{\mathbb{Z}})$ which has
$\lambda=\frac{3+\sqrt{5}}{2}$ and $\lambda^{-1}=\frac{3-\sqrt{5}}{2}$ as
eigenvalues and let

$X(t)=\begin{pmatrix}\lambda^{t}\cos 2\pi t&-\lambda^{t}\sin 2\pi t&&\\\
\lambda^{t}\sin 2\pi t&\lambda^{t}\cos 2\pi t&&\\\ &&\lambda^{-t}&\\\
&&&\lambda^{-t}\end{pmatrix}$

for $t\in{\mathbb{R}}$. Since $X(1)$ and $A=\begin{pmatrix}A_{0}&\\\
&A_{0}\end{pmatrix}\in\mathop{\mathrm{SL}}\nolimits(4,{\mathbb{Z}})$ are
conjugate, the solvable Lie group
$S={\mathbb{R}}^{4}\rtimes_{X(t)}{\mathbb{R}}$ contains a lattice $\Gamma_{0}$
isomorphic to ${\mathbb{Z}}^{4}\rtimes_{A}{\mathbb{Z}}$. Then the set

${\mathcal{H}}(\Gamma_{0},S)/\mathop{\mathrm{Aut}}\nolimits(S)$

is uncountable as in Baues and Klopsch [2]. Therefore by Proposition 33 the
set

${\mathcal{A}}(\Gamma_{0},S)/\text{parameter equivalence}$

is uncountable, so that the natural action $\Gamma_{0}\backslash
S\curvearrowleft S$ is not parameter rigid. However by computing
$\Gamma_{0}/[\Gamma_{0},\Gamma_{0}]$ using
$\Gamma_{0}\simeq{\mathbb{Z}}^{4}\rtimes_{A}{\mathbb{Z}}$ we can show
$H^{1}(\Gamma_{0}\backslash S)=H^{1}({\mathfrak{s}})$.

###  6.3 A locally parameter rigid action of a contractible group which is not
parameter rigid

Let $M\stackrel{{\scriptstyle\rho_{0}}}{{\curvearrowleft}}S$ be an action with
the orbit foliation ${\mathcal{F}}$. The set of all smooth actions
$M\curvearrowleft S$ with the orbit foliation ${\mathcal{F}}$ is denoted by
$A({\mathcal{F}},S)$, which is endowed with the topology induced from the
$C^{\infty}$ compact-open topology of $C^{\infty}(M\times S,M)$. An action
$M\stackrel{{\scriptstyle\rho_{0}}}{{\curvearrowleft}}S$ is said to be locally
parameter rigid if any $\rho\in A({\mathcal{F}},S)$ which is close enough to
$\rho_{0}$ is parameter equivalent to $\rho_{0}$. In [1, Section 1.1.2],
Asaoka comments that there is no known locally parameter rigid action of a
contractible group which is not parameter rigid. Here we give an example of
such an action. In [2, Example 2.5], Baues and Klopsch give an example of a
connected simply connected solvable Lie group $S$ which has a Zariski dense
lattice $\Gamma_{0}$ such that
$\mathop{\mathrm{Aut}}\nolimits(S)\backslash{\mathcal{H}}(\Gamma_{0},S)$ is
countably infinite. Here Zariski density means that the Zariski closures of
$\mathop{\mathrm{Ad}}\nolimits(\Gamma_{0})$ and
$\mathop{\mathrm{Ad}}\nolimits(S)$ in
$\mathop{\mathrm{GL}}\nolimits({\mathfrak{s}})$ coincide. If we take such $S$
and $\Gamma_{0}$ then the action $\Gamma_{0}\backslash
S\stackrel{{\scriptstyle\rho_{0}}}{{\curvearrowleft}}S$ by right
multiplication is not parameter rigid by Proposition 33. Let us see that this
action is locally parameter rigid. Since $\Gamma_{0}$ is a Zariski dense
lattice in $S$, the inclusion $\iota\colon\Gamma_{0}\hookrightarrow S$ is
locally rigid, that is, the $\mathop{\mathrm{Aut}}\nolimits(S)$–orbit of
$\iota$ in ${\mathcal{H}}(\Gamma_{0},S)$ is a neighborhood of $\iota$. See for
example Baues and Klopsch [2, Theorem 1.9]. Take an open neighborhood $U$ of
$\iota$ in ${\mathcal{H}}(\Gamma_{0},S)$ which is contained in the
$\mathop{\mathrm{Aut}}\nolimits(S)$–orbit of $\iota$. Let $V$ be the inverse
image of $U$ by the map
${\mathcal{A}}(\Gamma_{0},S)\to{\mathcal{H}}(\Gamma_{0},S)$, then $V$ is an
open neighborhood of $\rho_{0}$ in ${\mathcal{A}}(\Gamma_{0},S)$ by
continuity. Since $U$ projects to a one-point set in
$\mathop{\mathrm{Aut}}\nolimits(S)\backslash{\mathcal{H}}(\Gamma_{0},S)$, $V$
also projects to a one-point set in
${\mathcal{A}}(\Gamma_{0},S)/\text{(parameter equivalence)}$. Therefore
$\rho_{0}$ is locally parameter rigid.

## References

  * [1] M. Asaoka. Deformation of locally free actions and leafwise cohomology. Foliations: Dynamics, Geometry and Topology, 1–40, Adv. Courses Math. CRM Barcelona, Birkhäuser/Springer, Basel, 2014.
  * [2] O. Baues and B. Klopsch. Deformations and rigidity of lattices in solvable Lie groups. J. Topol. 6(2013), 823–856.
  * [3] B. Farb and L. Mosher. On the asymptotic geometry of abelian-by-cyclic groups. Acta Math. 184(2000), 145–202.
  * [4] H. Maruhashi. Parameter rigid actions of simply connected nilpotent Lie groups. Ergod. Th. & Dynam. Sys. 33(2013), 1864–1875.
  * [5] H. Maruhashi. Vanishing of cohomology and parameter rigidity of actions of solvable Lie groups, II. Ergod. Th. & Dynam. Sys. (2020), 1–37. doi:10.1017/etds.2020.97
  * [6] S. Matsumoto and Y. Mitsumatsu. Leafwise cohomology and rigidity of certain Lie group actions. Ergod. Th. & Dynam. Sys. 23(2003), 1839–1866.
  * [7] M. V. Milovanov. The extension of automorphisms of uniform discrete subgroups of solvable Lie groups. (Russian) Dokl. Akad. Nauk BSSR. 17(1973), 892–895, 969.
  * [8] N. Ogasawara. Quasiisometric classification of groups obtained from ${\mathbb{Z}}^{n}$ by HNN extension performed several times. (Japanese) Master Thesis, Kyoto University, 2012.
  * [9] M. S. Raghunathan. Discrete subgroups of Lie groups. Springer-Verlag, 1972.
  * [10] D. Witte. Superrigidity of lattices in solvable Lie groups. Invent. math. 122(1995), 147–193.