proposition@alttheorem lemma@alttheorem corollary@alttheorem
counterexample@alttheorem definition@alttheorem question@alttheorem
conjecture@alttheorem remark@alttheorem example@alttheorem

# Polynomial and horizontally polynomial functions on Lie groups

Gioacchino Antonelli and Enrico Le Donne  Gioacchino Antonelli: Scuola Normale
Superiore, Piazza dei Cavalieri, 7, 56126 Pisa, Italy
<EMAIL_ADDRESS>Enrico Le Donne: Dipartimento di Matematica,
Università di Pisa, Largo B. Pontecorvo 5, 56127 Pisa, Italy
University of Jyväskylä, Department of Mathematics and Statistics, P.O. Box
(MaD), FI-40014, Finland
& Department of Mathematics, University of Fribourg, Chemin du Musée 23, 1700
Fribourg, Switzerland<EMAIL_ADDRESS>
###### Abstract.

We generalize both the notion of polynomial functions on Lie groups and the
notion of horizontally affine maps on Carnot groups. We fix a subset $S$ of
the algebra $\mathfrak{g}$ of left-invariant vector fields on a Lie group
$\mathbb{G}$ and we assume that $S$ Lie generates $\mathfrak{g}$. We say that
a function $f:\mathbb{G}\to\mathbb{R}$ (or more generally a distribution on
$\mathbb{G}$) is $S$-polynomial if for all $X\in S$ there exists
$k\in\mathbb{N}$ such that the iterated derivative $X^{k}f$ is zero in the
sense of distributions.

First, we show that all $S$-polynomial functions (as well as distributions)
are represented by analytic functions and, if the exponent $k$ in the previous
definition is independent on $X\in S$, they form a finite-dimensional vector
space.

Second, if $\mathbb{G}$ is connected and nilpotent we show that $S$-polynomial
functions are polynomial functions in the sense of Leibman. The same result
may not be true for non-nilpotent groups.

Finally, we show that in connected nilpotent Lie groups, being polynomial in
the sense of Leibman, being a polynomial in exponential chart, and the
vanishing of mixed derivatives of some fixed degree along directions of
$\mathfrak{g}$ are equivalent notions.

###### Key words and phrases:

Nilpotent Lie groups, polynomial maps, Leibman polynomial, polynomial on
groups, horizontally affine functions, precisely monotone sets

######  Mathematics Subject Classification:

53C17, 22E25, 22E30, 43A80, 08A40.

G.A. was partially supported by the European Research Council (ERC Starting
Grant 713998 GeoMeG ‘ _Geometry of Metric Groups_ ’). E.L.D. was partially
supported by the Academy of Finland (grant 288501 ‘ _Geometry of subRiemannian
groups_ ’ and by grant 322898 ‘ _Sub-Riemannian Geometry via Metric-geometry
and Lie-group Theory_ ’) and by the European Research Council (ERC Starting
Grant 713998 GeoMeG ‘ _Geometry of Metric Groups_ ’). The authors wish to
thank Mattia Calzi and Fulvio Ricci for fruitful discussions and constructive
feedback around the topic of the paper.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
    1. 2.1 Adjoint and convolutions on Lie groups
    2. 2.2 Distributions on Lie groups
    3. 2.3 Nilpotent Lie groups and stratified groups
  3. 3 Polynomial and $S$-polynomial distributions on Lie groups
    1. 3.1 Relations with pointwise convergence
    2. 3.2 Propagation of being $S$-polynomial
    3. 3.3 Proof of Theorem 1.1
  4. 4 The case of nilpotent Lie groups
    1. 4.1 Proof of Theorem 1.3
  5. 5 Relations between various notions of polynomial
    1. 5.1 Proof of Theorem 1.2
    2. 5.2 Proof of Section 1
  6. A Examples

##  1\. Introduction

Following the terminology of Gromov, a polarized manifold is a (connected)
manifold equipped with a choice of a subbundle of its tangent bundle, which
most of the times is assumed Lie bracket generating and called space of
horizontal directions. In subRiemannian geometry, in analysis, but also in
group theory, there are several phenomena showing that the bracket generation
property upgrades a “horizontal” property to one in every direction: this is
the case, for example, of the Chow-Rashevskii Theorem, see [Mon02], the Pansu
differentiability Theorem [Pan89] and from a more analytic point of view of
the celebrated Hörmander Theorem [Hor67]. In this paper we consider functions
on a polarized Lie group that have the property that are polynomials along
horizontal directions.

We stress that polynomial maps between groups have been studied also from an
algebraic point of view, see, e.g., [Pas68, Buc70, Sze85, Lei02, GT06, KP20]
and references therein. Even if our interest is mainly analytic and geometric,
we will also highlight the connections of our results with the algebraic point
of view.

Let $\mathbb{G}$ be a Lie group with Lie algebra $\mathfrak{g}$, seen as left-
invariant vector fields on $\mathbb{G}$. We fix a left-Haar measure $\mu$ on
$\mathbb{G}$. Let $S\subseteq\mathfrak{g}$ be a subset that is Lie bracket
generating, i.e., the only subalgebra of $\mathfrak{g}$ that contains $S$ is
$\mathfrak{g}$.

We say that a distribution $f$ on $\mathbb{G}$ is $S$-polynomial if for all
$X\in S$ there exists $k\in\mathbb{N}$ such that the iterated derivative
$X^{k}f$ is zero in the sense of distributions on $\mathbb{G}$. We say that a
distribution $f$ is $S$-polynomial with degree at most $k$ if for every $X\in
S$ we have that $X^{k}f$ is zero in the sense of distributions on
$\mathbb{G}$. For basic definitions and properties of distributions on Lie
groups we refer the reader to Section 2.2. The first main outcome of this
paper is a regularity result for $S$-polynomial distributions on arbitrary Lie
groups, see Section 3.3 and Section 3.3 for the proof of the following
statement.

######  Theorem 1.1.

Let $\mathbb{G}$ be a Lie group, let $f$ be a distribution on $\mathbb{G}$,
and let $S\subseteq\mathfrak{g}$ be a subset of the Lie algebra $\mathfrak{g}$
that Lie generates $\mathfrak{g}$. If $f$ is $S$-polynomial, then it is
represented by an analytic function. Moreover, the vector space of
$S$-polynomial distributions with degree at most $k\in\mathbb{N}$ on each
connected component of $\mathbb{G}$ is finite-dimensional.

It is natural to ask if an $S$-polynomial distribution on $\mathbb{G}$ is
actually a polynomial in some sense. Various definitions of polynomial maps
between groups have been proposed and studied in the literature, see [Pas68]
for arbitrary groups, and [Buc70, Sze85] and references therein for the case
of Abelian groups. A notion of polynomial map between arbitrary groups that
showed to be versatile has been studied, with a special attention toward the
nilpotent case, in [Lei02], see Section 3.1. In the case we deal with, i.e.,
the case of maps $f:\mathbb{G}\to\mathbb{R}$, Leibman’s definition can be
generalized for distributions. Let us define the operator $D_{g}$ acting on
distributions $f$ on $\mathbb{G}$ as follows

$D_{g}f:=f\circ R_{g}-f,$

where $R_{g}$ stands for the right translation by $g\in\mathbb{G}$ and $f\circ
R_{g}$ should be properly defined, see (2.22). We say that a distribution $f$
on $\mathbb{G}$ is polynomial à la Leibman with degree at most
$d\in\mathbb{N}$ if

$g_{1},\dots,g_{d+1}\in\mathbb{G}\Rightarrow D_{g_{1}}\cdots
D_{g_{d+1}}f\equiv 0,\quad\text{in the sense of distributions on
$\mathbb{G}$}.$ (1.1)

We stress that our main result Theorem 1.1 helps in proving that the latter
notion of being polynomial, which is “discrete” in spirit, in our setting is
equivalent to a “differential” one. We say that a distribution $f$ on
$\mathbb{G}$ is polynomial (in the differential sense) with degree at most
$d\in\mathbb{N}$ if

$X_{1},\dots,X_{d+1}\in\mathfrak{g}\Rightarrow X_{1}\cdots X_{d+1}f\equiv
0,\quad\text{in the sense of distributions on $\mathbb{G}$}.$ (1.2)

For some benefit towards the understanding of the next result we recall that,
given a Lie algebra $\mathfrak{g}$, the nilpotent residual
$\mathfrak{g}_{\infty}$ of $\mathfrak{g}$ is the intersection
$\cap_{k\in\mathbb{N}}\mathfrak{g}_{k}$, where
$\mathfrak{g}:=\mathfrak{g}_{0}\supseteq\mathfrak{g}_{1}\supseteq\dots$ is the
lower central series associated to $\mathfrak{g}$. Notice that
$\mathfrak{g}\mathrel{/}\mathfrak{g}_{\infty}$ is the biggest nilpotent
quotient of $\mathfrak{g}$. Let us denote $\mathbb{G}_{\infty}$ the closure of
the unique connected subgroup of $\mathbb{G}$ with Lie algebra
$\mathfrak{g}_{\infty}$. We call
$N_{\mathbb{G}}:=\mathbb{G}\mathrel{/}\mathbb{G}_{\infty}$ the maximal
nilpotent Lie quotient. The following statement is a corollary of Theorem 1.1
and Section 5.1, see Section 5.1.

######  Theorem 1.2.

Let $\mathbb{G}$ be a connected Lie group.

  1. (1)

A distribution on $\mathbb{G}$ is polynomial à la Leibman with degree at most
$d\in\mathbb{N}$, see (1.1), if and only if it is polynomial (in the
differential sense) with degree at most $d$, see (1.2).

  2. (2)

Every polynomial distribution is represented by an analytic function.

  3. (3)

The vector space of distributions that are polynomial with degree at most
$d\in\mathbb{N}$ is finite-dimensional.

  4. (4)

For every distribution $f$ that is polynomial of degree at most $d$, and for
every $X$ in the $d$-th element of the lower central series, we have $Xf\equiv
0$. Consequently, for every analytic function $f$ that is polynomial, there
exists a function $\widetilde{f}:N_{\mathbb{G}}\to\mathbb{R}$ that is
polynomial such that $\widetilde{f}\circ\pi=f$, where $\pi$ is the projection
onto the maximal nilpotent Lie quotient $N_{\mathbb{G}}$.

Thus, with the previous result jointly with Theorem 1.1, it becomes clear that
a continuous polynomial map $f:\mathbb{G}\to\mathbb{R}$ à la Leibman is
analytic and in particular it is $S$-polynomial, no matter what is the choice
of a Lie generating $S$. Instead, the converse may not be true in arbitrary
Lie groups. The counterexample is already found in the two-dimensional group
of the orientation-preserving affine functions of $\mathbb{R}$, called
$\mathrm{Aff}^{+}(\mathbb{R})$. With the classical choice of coordinates on
$\mathrm{Aff}^{+}(\mathbb{R})$ one can find a Lie generating $S$ such that an
$S$-polynomial function that is not polynomial is $f(x,y)=(x+1)\log y$, see
(A.1).

The last part of the statement of Theorem 1.2 tells us that polynomial maps
always factor via a nilpotent group. Hence, we shall only study polynomial
maps on nilpotent Lie groups. When $\mathbb{G}$ is a connected nilpotent Lie
group $\exp:\mathfrak{g}\to\mathbb{G}$ is an analytic and surjective map, and
thus one could also give another definition of “polynomial”, namely a map
$f:\mathbb{G}\to\mathbb{R}$ is polynomial in exponential chart if and only if
$f\circ\exp:\mathfrak{g}\to\mathbb{R}$ is a polynomial. In case $\mathbb{G}$
is a connected and nilpotent Lie group, we show that the property of being
$S$-polynomial propagates to the entire Lie algebra. Namely, we prove that a
$S$-polynomial distribution on $\mathbb{G}$ is represented by a polynomial in
exponential chart, and thus in particular it is $\mathfrak{g}$-polynomial of
some degree $k\in\mathbb{N}$, see Section 4.1. Our second main result now can
be stated as follows, see Section 4.1 for the proof of the following
statement.

######  Theorem 1.3.

Let $\mathbb{G}$ be a connected nilpotent Lie group, let $f$ be a distribution
on $\mathbb{G}$, and let $S$ be a Lie generating subset of $\mathfrak{g}$. If
$f$ is $S$-polynomial, then it is represented by a function that is polynomial
in exponential chart.

We stress that if $\mathbb{G}$ is not nilpotent, being $S$-polynomial for a
Lie generating $S$ may not imply being $\mathfrak{g}$-polynomial, or even
being polynomial in exponential chart, see (A.1) for such a counterexample in
$\mathrm{Aff}(\mathbb{R})^{+}$. For non-nilpotent groups we do not know either
if being $\mathfrak{g}$-polynomial implies being polynomial, or even if being
$\mathfrak{g}$-polynomial passes to the maximal nilpotent Lie quotient.

With the previous main result we can prove that on a connected nilpotent Lie
group $\mathbb{G}$ the different notions of being polynomial that we discussed
above are equivalent and in particular they are equivalent to being
$S$-polynomial for any Lie generating $S$. The following statement is a
corollary of Theorem 1.3, Theorem 1.2, and Section 5.2, see Section 5.2.

######  Corollary .

Let $\mathbb{G}$ be a connected nilpotent Lie group, and let $f$ be a
distribution on $\mathbb{G}$. Then the following are equivalent

  1. (1)

$f$ is an $S$-polynomial distribution for some Lie generating
$S\subseteq\mathfrak{g}$,

  2. (2)

$f$ is an $S$-polynomial distribution for all $S\subseteq\mathfrak{g}$,

  3. (3)

$f$ is represented by a function that is polynomial in exponential chart,

  4. (4)

$f$ is a polynomial distribution (in the differential sense), see (1.2).

  5. (5)

$f$ is a polynomial distribution à la Leibman, see (1.1).

A characterization of polynomial functions on stratified groups has been
provided also in [BLU07], and we stress that our result is stronger than this
characterization in [BLU07, Corollary 20.1.10], see Section 5.2. We also
notice that a characterization of harmonic polynomials in stratified groups
through Almgren’s frequency function has been provided in [GR15, Theorem 9.1].

Let us discuss the strategy of the proofs of Theorem 1.1 and Theorem 1.3. The
starting idea is to prove the results first when $f$ is smooth, and then we
recover the general statements for distributions by using convolutions with
smoothing kernels. The latter strategy can be performed since smooth functions
that are $S$-polynomials with degree at most $k\in\mathbb{N}$ form a finite-
dimensional vector space, see Section 3.3.

The idea for proving the statements when $f$ is smooth is to somehow propagate
the information about being $S$-polynomial to the directions not in $S$. Let
us give here, in a particular case, a hint of how we do this. Let $f$ be
smooth and polynomial with degree at most $2$ with respect to
$X,Y\in\mathfrak{g}$, and let us denote with $e$ the identity element of
$\mathbb{G}$. We have the following equalities for $t,s\in\mathbb{R}$,

$\begin{split}f(\exp(tX)\exp(sY))&=f(\exp(tX))+s(Yf)(\exp(tX))\\\
&=f(e)+t(Xf)(e)+s\frac{\operatorname{d}}{\operatorname{d}\varepsilon}_{|_{\varepsilon=0}}f(\underbrace{\exp(tX)\exp(\varepsilon
Y)\exp(-tX)}_{p_{\varepsilon}}\exp(tX))\\\
&=f(e)+t(Xf)(e)+s\frac{\operatorname{d}}{\operatorname{d}\varepsilon}_{|_{\varepsilon=0}}\left(f(p_{\varepsilon})+t(Xf)(p_{\varepsilon})\right)\\\
&=f(e)+t(Xf)(e)+s(\mathrm{Ad}_{\exp(tX)}Y)f(e)+st(\mathrm{Ad}_{\exp(tX)}YX)f(e),\end{split}$
(1.3)

where in the first, second and third equalities we are using that $f$ is
2-polynomial with respect to $Y$ and $X$, respectively, and in the last
equality we are using the definition of the adjoint, see Section 2.1. In the
general case, the proper generalization of the latter representation formula,
see (3.8), allows to conclude the following statement: given
$\ell\in\mathbb{N}$, the function $f$ along the concatenation of $\ell$
horizontal curves $\exp(t_{1}Y_{1})\cdots\exp(t_{\ell}Y_{\ell})$, with
$Y_{1},\dots,Y_{\ell}\in S$, is analytic in $(t_{1},\dots,t_{\ell})$ and
depends only on the value of the jet of order $\delta_{1}$ of $f$ at the
identity, where $\delta_{1}$ depends only on $\ell$ and the order of
polynomiality of $f$ along $Y_{1},\dots,Y_{\ell}$. The latter observation is
enough to conclude both the fact that $f$ is analytic and the finite-
dimensional result of Theorem 1.1. To this aim, it is important that in every
polarized Lie group there exists a chart, around every point $p$, that can be
written as the concatenation of a fixed number (twice the dimension of the
group) of flow lines, starting from $p$, of horizontal vector fields, see
Section 3.3.

In the particular case in which $\mathbb{G}$ is nilpotent, the representation
formula (1.3) gives an additional piece of information because
$\mathrm{Ad}_{\exp(tX)}$ is a “polynomial in $t$” sum of operators, see (2.1).
Thus, in case $\mathbb{G}$ is nilpotent, one concludes, from the proper
generalization of (1.3), see Section 4, the following statement: given
$\ell\in\mathbb{N}$, the function $f$ along the concatenation of $\ell$
horizontal curves $\exp(t_{1}Y_{1})\cdots\exp(t_{\ell}Y_{\ell})$, with
$Y_{1},\dots,Y_{\ell}\in S$, is a polynomial in $(t_{1},\dots,t_{\ell})$ with
degree at most $\delta_{2}$, where $\delta_{2}$ depends only on $\ell$ and the
order of polynomiality of $f$ along $Y_{1},\dots,Y_{\ell}$, and where the
coefficients of the polynomial only depend on some mixed derivatives of $f$ of
bounded order at the identity.

In order to prove Theorem 1.3, one first reduces to the case when $\mathbb{G}$
is simply connected by passing to the universal cover. Then, the first idea is
to lift the problem to free-nilpotent groups, see the proof of Theorem 4.1.
Free-nilpotent groups are stratified, and stratified Lie groups are also
called Carnot groups. We then notice that it is sufficient to prove Theorem
1.3 in the case $\mathbb{G}$ is a Carnot group. Now, if $\mathbb{G}$ is a
Carnot group, we exploit that $f$ is analytic, which we have previously
obtained in the setting of arbitrary Lie groups, and a blow-up argument, which
we can perform since $\mathbb{G}$ has a homogeneous structure, to obtain that
each term in the homogeneous Taylor expansion of $f$ at the identity is
$S^{\prime}$-polynomial of a fixed order $k\in\mathbb{N}$, where
$S^{\prime}\subseteq S$ is Lie generating, see the proof of Section 4.1. Now
the observation above according to which every $S^{\prime}$-polynomial of
order $k\in\mathbb{N}$ is a polynomial of bounded degree along
$\exp(t_{1}Y_{1})\cdots\exp(t_{\ell}Y_{\ell})$ allows us to conclude that
every $S^{\prime}$-polynomial of order $k\in\mathbb{N}$ has a polynomial
growth order of bounded degree at infinity: this is done by using Section 4.1
according to which we can bound below, up to a constant, the distance of a
point $p=\exp(t_{1}Y_{1})\cdots\exp(t_{\ell}Y_{\ell})$ from the identity with
$|t_{1}|+\dots+|t_{\ell}|$. Thus the homogeneous Taylor expansion of $f$ at
the identity cannot have terms of arbitrarily large order, and then $f$ is a
polynomial, concluding the proof.

Let us finally discuss our initial motivation for studying such a problem. If
$\mathbb{G}$ is a Carnot group of step $s$ with stratification
$\mathfrak{g}=V_{1}\oplus\dots\oplus V_{s}$, we take $S=V_{1}$ and we take a
function $f$ such that for every $X\in V_{1}$ we have $X^{2}f\equiv 0$, we
recover the notion of horizontally affine maps on a Carnot group, which has
recently been introduced and studied in [LDMR20]. The main result of [LDMR20],
i.e., [LDMR20, Theorem 1.1], is a complete characterization of horizontally
affine maps on step-2 Carnot groups. Our main result Theorem 1.3 can be seen
as a broad generalization of the part of the statement in [LDMR20, Theorem
1.1] according to which every horizontally affine map on a step-2 Carnot group
is ultimately polynomial in exponential chart. Indeed, our Theorem 1.3 holds
for arbitrary connected nilpotent Lie groups, an arbitrary degree of
polynomiality, and even if we ask the polynomial property with respect to a
finite set $S$ that Lie generates, which is not necessarily a vector subspace
of $\mathfrak{g}$.

The interest toward horizontally affine maps on Carnot groups is motivated by
the fact that they are linked to precisely monotone sets, which have been
first introduced and studied in [CK10], and [CKN11], in order to show that the
first Heisenberg group $\mathbb{H}^{1}$ with the subRiemannian distance does
not biLipschitz embed into $L^{1}(\mathbb{R},\mathscr{L}^{1})$. A precisely
monotone set $E$ in a Carnot group $\mathbb{G}$ is a subset of $\mathbb{G}$
such that $E$ and the complement $E^{c}$ are $h$-convex, see [Ric06,
Definition 3.1]. A set $E$ is $h$-convex in $\mathbb{G}$ if whenever $a,b\in
E$ and there exists $\gamma$ an integral curve of a horizontal left-invariant
vector field with extrema $a,b$, then $\gamma$ is contained in $E$. It is
simple to observe, from the very definition, that the sublevel sets of a
horizontally affine map on a Carnot group are precisely monotone sets.

The problem of classifying precisely monotone sets is rather difficult already
in the easiest Carnot groups. The precisely monotone sets have been completely
classified in the Heisenberg groups $\mathbb{H}^{n}$, see [CK10, Theorem 4.3],
and [NY18, Proposition 65], and in $\mathbb{H}^{1}\times\mathbb{R}$, see
[Mor18, Theorem 1.2]. In all the latter three cases one can prove that if
$E\notin\\{\emptyset,\mathbb{G}\\}$ is precisely monotone, then $\partial E$
is a hyperplane in exponential chart and $\mathrm{Int}(E)$ and $\overline{E}$
are one of the open, respectively closed, half-spaces bounded by $\partial E$.
On the contrary, from the results in [LDMR20], it follows that, already in the
step-2 case, there are sublevel sets of horizontally affine maps, and thus
precisely monotone sets, that are not half-spaces in exponential chart. The
results in [LDMR20, Theorem 1.1] show that in arbitrary Carnot groups there
are plenty of horizontally affine maps that are not affine, and we could
construct plenty of precisely monotone sets that are not half-spaces in
exponential chart. On the other hand, since we proved in particular that in
arbitrary Carnot groups horizontally affine maps are polynomial in exponential
chart, the precisely monotone sets constructed as sublevel sets of
horizontally affine maps result in being semialgebraic sets in exponential
chart. Also we stress that in some Carnot groups, e.g., the free Carnot group
of step 3 and rank 2, there are precisely monotone sets whose boundary is not
an algebraic variety in exponential chart, see [BLD19, Theorem 6.2]. As a
consequence, in general it is even not true that the boundary of a precisely
monotone set in a Carnot group is a zero-level set of a horizontally affine
function.

The structure of the paper is as follows.

In Section 2 we recall some basic facts about Lie groups. In particular, in
Section 2.1 we discuss the convolution in arbitrary Lie groups. In Section 2.2
we recall basic facts about distributions on arbitrary Lie groups. In Section
2.3 we fix the notation on nilpotent and stratified, also called Carnot,
groups. In Section 3 we prove Theorem 1.1. In particular, in Section 3.1 we
introduce the definition of $S$-polynomial distributions and polynomial
distributions in arbitrary Lie groups and we study how $S$-polynomiality
behaves under pointwise convergence. In Section 3.2 we prove Section 3.2 that
will allow to prove the representation formula in Section 3.2. In Section 3.3
we conclude the proof of Theorem 1.1. In Section 4 we prove Theorem 1.3. In
particular, in Section 4.1 we reduce to the case of Carnot groups and then we
conclude by exploiting Section 4, a consequence of Section 3.2. In Section 5
we give the proof of Theorem 1.2 and of Section 1. Finally, in Appendix A, we
discuss some examples.

##  2\. Preliminaries

###  2.1. Adjoint and convolutions on Lie groups

We recall here the definition of the adjoint map on an arbitrary Lie group,
mostly to fix notation and have formulas ready for later.

######  Definition  (Conjugate and adjoint).

Let $\mathbb{G}$ be a Lie group with identity $e$. Let us fix $g\in\mathbb{G}$
and define the conjugate function $C_{g}:\mathbb{G}\to\mathbb{G}$ as
$C_{g}(h):=ghg^{-1}$, and the adjoint operator
$\mathrm{Ad}_{g}:\mathfrak{g}\to\mathfrak{g}$ as
$\mathrm{Ad}_{g}(X):=(\operatorname{d}(C_{g})_{e})(X).$

######  Remark  (Formula for the adjoint).

In an arbitrary Lie group $\mathbb{G}$ the following formula holds

$\mathrm{Ad}_{\exp(X)}(Y)=\sum_{j=0}^{+\infty}\frac{1}{j!}(\mathrm{ad}_{X})^{j}Y,$
(2.1)

where the operator $\mathrm{ad}_{X}:\mathfrak{g}\to\mathfrak{g}$ is defined as
$\mathrm{ad}_{X}(Y):=[X,Y]$, see [CG90, Equation (3) page 12]. Notice that if
$\mathbb{G}$ is nilpotent then (2.1) is a finite sum.

We recall the notion of modular function on a Lie group $\mathbb{G}$. We
follow closely the presentation in [Fol95, Chapter 2]. We fix a left-Haar
measure $\mu$ on $\mathbb{G}$. Recall that every two left-invariant Haar
measures are one a constant multiple of the other, see [Fol95, Theorem 2.20].
For every $x\in\mathbb{G}$ the measure defined by the equality
$\mu_{x}:=(R_{x^{-1}})_{*}\mu$, where $R_{g}$ stands for the right translation
by $g$, is a left-invariant Haar measure. Thus there exists $\Delta(x)>0$ such
that $\mu_{x}=\Delta(x)\mu$. The function $x\mapsto\Delta(x)$ is called
modular function and it enjoys the following properties

  * •

$\Delta:\mathbb{G}\to\mathbb{R}$ is an analytic function and

$\Delta(x)=\det\mathrm{Ad}_{x},$ (2.2)

for every $x\in\mathbb{G}$, see [Fol95, Proposition 2.24 and Proposition
2.30];

  * •

we have

$(\mathrm{inv})_{*}\mu=(\Delta\circ\mathrm{inv})\mu,$ (2.3)

where $\mathrm{inv}$ stands for the inverse function on $\mathbb{G}$, see
[Fol95, Equation (2.32)].

A Lie group is unimodular if $\Delta\equiv 1$, i.e., if $\mu$ is also a right-
invariant Haar measure. Notice that every connected nilpotent Lie group is
unimodular, due to (2.2). We introduce now the convolution between two
functions.

######  Definition  (Convolution).

Let $\mu$ be a left-invariant Haar measure on a Lie group $\mathbb{G}$. Let
$f,g\in L^{1}(\mathbb{G},\mu)$. The convolution of
$f,g:\mathbb{G}\to\mathbb{R}$ is defined as

$f\ast
g(x):=\int_{\mathbb{G}}f(y)g(y^{-1}x)\operatorname{d}\mu(y),\qquad\forall
x\in\mathbb{G}.$ (2.4)

The definition is well-posed since an application of Fubini theorem implies
that the integral is absolutely convergent for every $x\in\mathbb{G}$, see
[Fol95, page 50].

It is readily seen by the very Section 2.1 that when $g$ is continuous with
compact support, the convolution $f\ast g$ is continuous. Moreover, if $g\in
C^{\infty}_{c}(\mathbb{G})$ we get that $X(f\ast g)=f\ast Xg$ for every left-
invariant vector field $X$. Thus if $g\in C^{\infty}_{c}(\mathbb{G})$ we
conclude that $f\ast g\in C^{\infty}(\mathbb{G})$. In addition one has that
$\mathrm{supp}(f\ast
g)\subseteq\overline{\mathrm{supp}(f)\cdot\mathrm{supp}(g)}$, and then if $f$
and $g$ have compact support, then also $f\ast g$ has compact support.

The following change of variable formula holds, taking into account [Fol95,
Equation (2.36)],

$f\ast
g(x):=\int_{\mathbb{G}}f(y)g(y^{-1}x)\operatorname{d}\mu(y)=\int_{\mathbb{G}}f(xy^{-1})g(y)\Delta(y^{-1})\operatorname{d}\mu(y),$
(2.5)

whenever $f,g\in L^{1}(\mathbb{G},\mu)$ and $x\in\mathbb{G}$, since (2.3)
holds. Let us introduce a variant of the convolution $\ast$ introduced before.
Given $f,g\in L^{1}(\mathbb{G},\mu)$, we define

$f\,\widetilde{\ast}\,g(x):=\int_{\mathbb{G}}f(y)g(yx)\operatorname{d}\mu(y)=\int_{\mathbb{G}}f(yx^{-1})g(y)\Delta(x^{-1})\operatorname{d}\mu(y),$
(2.6)

where the first integral is absolutely convergent since $f,g\in
L^{1}(\mathbb{G},\mu)$ and the second formula holds taking
$(R_{x})_{*}\mu=\Delta(x^{-1})\mu$ into account. It is readily seen that, when
$f$ and $g$ have compact support, then $f\,\widetilde{\ast}\,g$ has compact
support and if $g\in C^{\infty}_{c}(\mathbb{G})$ we can write
$X(f\,\widetilde{\ast}\,g)=f\,\widetilde{\ast}\,Xg$ whenever $X$ is a left-
invariant vector field. As a consequence if $g\in C^{\infty}_{c}(\mathbb{G})$
then $f\,\widetilde{\ast}\,g\in C^{\infty}(\mathbb{G})$. Let us notice the
following fact, which comes from (2.3), $(R_{x^{-1}})_{*}\mu=\Delta(x)\mu$,
(2.5) and (2.6)

$f\,\widetilde{\ast}\,g=\check{(f\Delta)}\ast g$, where, for a function
$h:\mathbb{G}\to\mathbb{R}$, we define $\check{h}:=h\circ\mathrm{inv}$. (2.7)

We recall that, when the following integral makes sense for functions
$f,g:\mathbb{G}\to\mathbb{R}$, we denote

$\langle f,g\rangle:=\int_{\mathbb{G}}f(x)g(x)\operatorname{d}\mu(x).$ (2.8)

We claim that the following formula holds

$\langle f\ast h,g\rangle=\langle h,f\,\widetilde{\ast}\,g\rangle=\langle
h,\check{(f\Delta)}\ast g\rangle,\quad\text{for every $f,g,h$ continuous with
compact support.}$ (2.9)

Indeed, taking (2.7) into account, by Fubini theorem and the fact that $\mu$
is left-invariant we have

$\begin{split}\langle f\ast
h,g\rangle&=\int_{\mathbb{G}}\left(\int_{\mathbb{G}}f(y)h(y^{-1}x)\operatorname{d}\mu(y)\right)g(x)\operatorname{d}\mu(x)\\\
&=\int_{\mathbb{G}}\left(\int_{\mathbb{G}}g(x)h(y^{-1}x)\operatorname{d}\mu(x)\right)f(y)\operatorname{d}\mu(y)\\\
&=\int_{\mathbb{G}}\left(\int_{\mathbb{G}}g(yx)h(x)\operatorname{d}\mu(x)\right)f(y)\operatorname{d}\mu(y)\\\
&=\int_{\mathbb{G}}\left(\int_{\mathbb{G}}f(y)g(yx)\operatorname{d}\mu(y)\right)h(x)\operatorname{d}\mu(x)=\langle
h,f\,\widetilde{\ast}\,g\rangle=\langle h,\check{(f\Delta)}\ast
g\rangle.\end{split}$ (2.10)

Notice that if we are on a unimodular Lie group, (2.9) implies the well-known
formula $\langle f\ast h,g\rangle=\langle h,\check{f}\ast g\rangle$ when the
integral makes sense.

Let us end this section by computing, for every left-invariant vector field
$X$ on $\mathbb{G}$, the action of the adjoint $X^{\top}$ of $X$ with respect
to $\langle\cdot,\cdot\rangle$, on the smooth functions with compact support.
We recall that if $\varphi\in C^{\infty}_{c}(\mathbb{G})$, the operator
$X^{\top}$ acting on $\varphi$ is the unique smooth function $X^{\top}\varphi$
such that

$\langle
X^{\top}\varphi,\psi\rangle=\langle\varphi,X\psi\rangle,\qquad\text{for all
$\psi\in C^{\infty}_{c}(\mathbb{G})$}.$

For every $f,g\in C^{\infty}_{c}(\mathbb{G})$, every $X\in\mathfrak{g}$ and
$t>0$ the following equality holds

$\int_{\mathbb{G}}f(y\exp(tX))g(y)\operatorname{d}\mu(y)=\int_{\mathbb{G}}f(y)g(y\exp(-tX))\Delta(\exp(-tX))\operatorname{d}\mu(y),$

due to the fact that $(R_{x^{-1}})_{*}\mu=\Delta(x)\mu$. We deduce that
$X^{\top}=-\Delta(e)X-(X\Delta)(e)\mathrm{id}=-X-(X\Delta)(e)\mathrm{id}$,
since $\Delta(e)=1$. Namely

$\langle Xf,g\rangle=:\langle f,X^{\top}g\rangle=\langle
f,-Xg-(X\Delta)(e)g\rangle,\quad\text{for every $f,g\in
C^{\infty}_{c}(\mathbb{G})$}.$ (2.11)

###  2.2. Distributions on Lie groups

In this section we recall some basic facts about the theory of distributions
on a Lie group $\mathbb{G}$, see also [Ehr56]. We stress that we will use a
variant of the definition of convolution between functions and distributions
with respect to [Ehr56]. We fix a left-invariant Haar measure $\mu$ on
$\mathbb{G}$. We remark that every vector space considered will be a vector
space over $\mathbb{R}$.

We denote with $\mathcal{D}(\mathbb{G})\equiv C^{\infty}_{c}(\mathbb{G})$ the
topological vector space of real-valued $C^{\infty}$ functions with compact
support on $\mathbb{G}$ equipped with the final locally convex topology with
respect to the immersions
$C^{\infty}_{K}(\mathbb{G})\hookrightarrow\mathcal{D}(\mathbb{G})$, where $K$
is a compact subset of $\mathbb{G}$ and $C^{\infty}_{K}(\mathbb{G})$ is the
space of real-valued $C^{\infty}$ functions with compact support contained in
$K$. Let us recall that, if we fix $\\{X_{1},\dots,X_{n}\\}$ a basis of the
Lie algebra $\mathfrak{g}$, the (countably many) seminorms that induce the
locally convex topology on $C^{\infty}_{K}(\mathbb{G})$ are

$\|f\|_{j_{1},\dots,j_{r},K}:=\sup_{x\in K}|X_{j_{1}}\dots X_{j_{r}}f(x)|,$

where $r\in\mathbb{N}$ and $j_{1},\dots,j_{r}\in\\{1,\dots,n\\}$. Let us
denote with $\mathcal{D}^{\prime}(\mathbb{G})$ the dual of
$\mathcal{D}(\mathbb{G})$, i.e., the set of continuous linear functionals from
$\mathcal{D}(\mathbb{G})$ to $\mathbb{R}$, equipped with the locally convex
weak* topology. If $\Phi\in\mathcal{D}^{\prime}(\mathbb{G})$, we recall that
by $\langle\Phi,\varphi\rangle$ we mean the evaluation of $\Phi$ at
$\varphi\in\mathcal{D}(\mathbb{G})$. Let us remark that if $f\in
L^{1}_{\mathrm{loc}}(\mathbb{G},\mu)$, there is a canonical way of seeing $f$
as a distribution, by means of

$\langle
f,\varphi\rangle:=\int_{\mathbb{G}}f(x)\varphi(x)\operatorname{d}\mu(x),\qquad\text{for
all $\varphi\in\mathcal{D}(\mathbb{G})$},$

so that the notation of the evaluation $\langle\cdot,\cdot\rangle$ for a
distribution is consistent with the one introduced in (2.8).

We recall that if $f_{j}\in\mathcal{D}(\mathbb{G})$, and
$f\in\mathcal{D}(\mathbb{G})$, then $f_{j}\to f$ in $\mathcal{D}(\mathbb{G})$
if and only of there exists a compact subset $K\subseteq\mathbb{G}$ such that
all the supports of $f_{j}$’s are contained in $K$ and for every
$r\in\mathbb{N}$ and every $j_{1},\dots,j_{r}\in\\{1,\dots,n\\}$ we have
$X_{j_{1}}\dots X_{j_{r}}f_{j}\to X_{j_{1}}\dots X_{j_{r}}f$ uniformly on $K$.
On the other hand if $\Phi_{j},\Phi\in\mathcal{D}^{\prime}(\mathbb{G})$, we
have that $\Phi_{j}\to\Phi$ in the weak* topology of
$\mathcal{D}^{\prime}(\mathbb{G})$ if and only if
$\langle\Phi_{j},f\rangle\to\langle\Phi,f\rangle$ for every
$f\in\mathcal{D}(\mathbb{G})$.

######  Definition  (Derivative of a distribution and convolution with a
function).

If $X$ is a left-invariant vector field on $\mathbb{G}$ and
$\Phi\in\mathcal{D}^{\prime}(\mathbb{G})$ we define

$\langle X\Phi,g\rangle:=\langle\Phi,X^{\top}g\rangle,\qquad\forall
g\in\mathcal{D}(\mathbb{G}),$

where the action of the adjoint $X^{\top}$ of $X$ on $\mathcal{D}(\mathbb{G})$
has been explicitly computed in (2.11).

Moreover, if $f\in\mathcal{D}(\mathbb{G})$ and
$\Phi\in\mathcal{D}^{\prime}(\mathbb{G})$ we define

$\langle
f\ast\Phi,g\rangle:=\langle\Phi,f\,\widetilde{\ast}\,g\rangle=\langle\Phi,\check{(f\Delta)}\ast
g\rangle,\quad\forall g\in\mathcal{D}(\mathbb{G}),$ (2.12)

where $f\,\widetilde{\ast}\,g$ is defined in (2.6), see also (2.7).

######  Remark  (Derivative and convolution of a distribution).

We notice that the definitions given in Section 2.2 are consistent with the
case in which $\Phi$ is a function, see (2.11) and (2.9). Moreover, if
$\\{g_{n}\\}_{n\in\mathbb{N}}\subseteq\mathcal{D}(\mathbb{G})$ and $g_{n}\to
g\in\mathcal{D}(\mathbb{G})$ in the topology of $\mathcal{D}(\mathbb{G})$,
then $X^{\top}g_{n}\to X^{\top}g$ and $f\,\widetilde{\ast}\,g_{n}\to
f\,\widetilde{\ast}\,g$ in the topology of $\mathcal{D}(\mathbb{G})$, due to
the explicit expressions in (2.11), and (2.6), respectively. Thus $X\Phi$ and
$f\ast\Phi$ are well defined distributions whenever
$\Phi\in\mathcal{D}^{\prime}(\mathbb{G})$, $f\in\mathcal{D}(\mathbb{G})$, and
$X$ is a left-invariant vector field.

Actually, we claim that for every $f\in\mathcal{D}(\mathbb{G})$ and every
$\Phi\in\mathcal{D}^{\prime}(\mathbb{G})$, the distribution $f\ast\Phi$
coincides with the $C^{\infty}(\mathbb{G})$ function defined as follows

$f\ast\Phi(x):=\langle\Phi,y\mapsto f(xy^{-1})\Delta(y^{-1})\rangle.$

Indeed, it can be verified111Compare with [Ehr56, Proposition 6]. Let us call
$\tau^{f}_{x}(y):=f(xy^{-1})\Delta(y^{-1})$. It is sufficient to prove that,
if $f\in\mathcal{D}(\mathbb{G})$ is fixed, the map $x\mapsto\tau_{x}^{f}$ is
continuous from $\mathbb{G}$ to $\mathcal{D}(\mathbb{G})$ and that
$t^{-1}(\tau_{\exp(tX)x}^{f}-\tau_{x}^{f})\to\tau^{X^{R}f}_{x}$ in
$\mathcal{D}(\mathbb{G})$ as $t$ goes to zero. that $f\ast\Phi(x)$ is
$C^{\infty}(\mathbb{G})$ and for every $h\in\mathcal{D}(\mathbb{G})$ we have

$\int_{\mathbb{G}}f\ast\Phi(x)h(x)\operatorname{d}\mu(x)=\langle\Phi,y\mapsto\int_{\mathbb{G}}f(xy^{-1})h(x)\Delta(y^{-1})\operatorname{d}\mu(x)\rangle=\langle\Phi,f\,\widetilde{\ast}\,h\rangle=\langle\Phi,\check{(f\Delta)}\ast
g\rangle,$

where the last two equalities follow from (2.6) and (2.7), respectively.

######  Remark  (Derivative of a convolution).

Let us prove that if $f\in\mathcal{D}(\mathbb{G})$,
$\Phi\in\mathcal{D}^{\prime}(\mathbb{G})$ and $X\in\mathfrak{g}$ we have

$X(f\ast\Phi)=f\ast(X\Phi).$ (2.13)

Indeed, for every $g\in\mathcal{D}(\mathbb{G})$,

$\begin{split}\langle X(f\ast\Phi),g\rangle&=\langle
f\ast\Phi,X^{\top}g\rangle=\langle\Phi,f\,\widetilde{\ast}\,(X^{\top}g)\rangle\\\
&=\langle\Phi,X^{\top}(f\,\widetilde{\ast}\,g)\rangle=\langle
X\Phi,f\,\widetilde{\ast}\,g\rangle=\langle f\ast X\Phi,g\rangle,\end{split}$
(2.14)

where we only exploited Section 2.2 and in the third equality we used the fact
that $X^{\top}=-X-(X\Delta)(e)\mathrm{id}$, see (2.11), and
$X(f\,\widetilde{\ast}\,g)=f\,\widetilde{\ast}\,(Xg)$.

######  Definition  (Approximate identity).

We say that a sequence
$\\{\varphi_{n}\\}_{n\in\mathbb{N}}\subseteq\mathcal{D}(\mathbb{G})$ is an
approximate identity if the following holds

$\varphi_{n}\ast\Phi\to\Phi,\quad\mbox{in the topology of
$\mathcal{D}^{\prime}(\mathbb{G})$ for all
$\Phi\in\mathcal{D}^{\prime}(\mathbb{G})$.}$ (2.15)

We now construct an approximate identity on a Lie group $\mathbb{G}$. On
$\mathbb{G}$ we fix an arbitrary right-invariant distance $d$ that induces the
manifold topology, which for example can be taken to be Riemannian. We fix
$\mu$ a left-invariant Haar measure on $\mathbb{G}$ as well. We know that
$\exp:\mathfrak{g}\to\mathbb{G}$ is a local analytic diffeomorphism around
$0\in\mathfrak{g}$, see [CG90, page 11]. By a classical choice of smooth
functions that are compactly supported in a sufficiently small neighbourhood
of $0\in\mathfrak{g}$, and by reading these functions on $\mathbb{G}$ through
the exponential map, we can readily build a family of positive functions
$\\{\varphi_{n}\\}_{n\in\mathbb{N}}\subseteq\mathcal{D}(\mathbb{G})$ such that
the following conditions hold

  1. (1)

$\int_{\mathbb{G}}\varphi_{n}(y)\operatorname{d}\mu(y)=1$, for all
$n\in\mathbb{N}$;

  2. (2)

The family $\\{\mathrm{supp}(\varphi_{n})\\}_{n\in\mathbb{N}}$ is a
fundamental system of compact neighbourhoods of the identity element
$e\in\mathbb{G}$ contained in a common compact neighbourhood of $e$.

######  Proposition .

With the notation and the setting above, whenever
$f\in\mathcal{D}(\mathbb{G})$, it holds $\varphi_{n}\,\widetilde{\ast}\,f\to
f$ in the topology of $\mathcal{D}(\mathbb{G})$, where $\,\widetilde{\ast}\,$
is defined in (2.6). Thus, whenever $\Phi\in\mathcal{D}^{\prime}(\mathbb{G})$,
we conclude $\varphi_{n}\ast\Phi\to\Phi$ in the topology of
$\mathcal{D}^{\prime}(\mathbb{G})$ and then
$\\{\varphi_{n}\\}_{n\in\mathbb{N}}$ is an approximate identity.

###### Proof.

By the definitions of convolution with a distribution, see (2.12), and of
convergence in $\mathcal{D}^{\prime}(\mathbb{G})$, the second part of the
statement readily follows from the first part.

Hence, let us fix $f\in\mathcal{D}(\mathbb{G})$ and let $K$ be the support of
$f$. Since $\\{\mathrm{supp}(\varphi_{n})\\}_{n\in\mathbb{N}}$ is a
fundamental system of compact neighbourhoods of the identity element
$e\in\mathbb{G}$ there exists an infinitesimal decreasing sequence
$\varepsilon_{n}\to 0$ such that

$\text{if $x\in\mathbb{G}$ is such that $d(x,e)>\varepsilon_{n}$ then
$\varphi_{n}(x)=0$}.$ (2.16)

Moreover, since
$\operatorname{supp}(\varphi_{n}\,\widetilde{\ast}\,f)\subseteq\overline{\operatorname{supp}(\varphi_{n})^{-1}\cdot\operatorname{supp}f}$,
there exists a compact set $\widetilde{K}\supseteq K$ such that

$\mathrm{supp}(f),\mathrm{supp}(\varphi_{n}\,\widetilde{\ast}\,f)\subseteq\widetilde{K},\qquad\forall
n\in\mathbb{N}.$ (2.17)

Moreover there exists a compact set $K^{\prime}\supseteq\widetilde{K}$ such
that

$\text{$\operatorname{supp}(\varphi_{n})\cdot\widetilde{K}\subseteq
K^{\prime}$},\qquad\forall n\in\mathbb{N}.$ (2.18)

Let us fix $x\in\widetilde{K}$. Then from the fact that for all
$n\in\mathbb{N}$ it holds
$\int_{\mathbb{G}}\varphi_{n}(y)\operatorname{d}\mu(y)=1$, and by the very
definition of the convolution $\,\widetilde{\ast}\,$ we get, for all
$n\in\mathbb{N}$, that the following inequality holds

$\displaystyle|\varphi_{n}\,\widetilde{\ast}\,f(x)-f(x)|$
$\displaystyle=\left|\int_{\mathbb{G}}\varphi_{n}(y)(f(yx)-f(x))\operatorname{d}\mu(y)\right|$
(2.19)
$\displaystyle\leq\int_{\\{y\in\mathbb{G}:y\in\operatorname{supp}(\varphi_{n})\\}}\varphi_{n}(y)|f(yx)-f(x)|\operatorname{d}\mu(y).$

Now, from (2.18), we get that whenever $x\in\widetilde{K}$ and
$y\in\operatorname{supp}(\varphi_{n})$ we have $\\{x,yx\\}\subseteq
K^{\prime}$. Since $f$ is uniformly continuous on the compact set
$K^{\prime}$, taking into account also (2.16), we get that for every
$\varepsilon>0$ there exists $n_{0}=n_{0}(\varepsilon)$ such that for every
$n\geq n_{0}$ the following holds

if $x\in\widetilde{K}$ and $y\in\operatorname{supp}(\varphi_{n})$, then
$d(yx,x)=d(y,e)\leq\varepsilon_{n}$ and thus $|f(yx)-f(x)|\leq\varepsilon$,
(2.20)

where we used that $d$ is right-invariant. Thus if we fix $\varepsilon>0$ we
can use (2.20) and continue (2.19) in order to obtain that for all
$x\in\widetilde{K}$ and for every $n\geq n_{0}(\varepsilon)$ the following
inequality holds

$|\varphi_{n}\,\widetilde{\ast}\,f(x)-f(x)|\leq\int_{\\{y\in\mathbb{G}:y\in\operatorname{supp}(\varphi_{n})\\}}\varphi_{n}(y)\varepsilon\operatorname{d}\mu(y)\leq\int_{\mathbb{G}}\varphi_{n}(y)\varepsilon\operatorname{d}\mu(y)=\varepsilon.$

The previous equation implies that

$\lim_{n\to+\infty}\sup_{x\in\widetilde{K}}|\varphi_{n}\,\widetilde{\ast}\,f(x)-f(x)|=0.$

Let us fix $\\{X_{1},\dots,X_{n}\\}$ a basis of $\mathfrak{g}$. By exploiting
the fact that $X(f\,\widetilde{\ast}\,g)=f\,\widetilde{\ast}\,(Xg)$ for every
left-invariant vector field $X$ and every $f,g\in\mathcal{D}(\mathbb{G})$, we
get arguing exactly as before, that for every multi-index
$\alpha:=(j_{1},\dots,j_{r})$ with $r\in\mathbb{N}$ and
$j_{1},\dots,j_{r}\in\\{1,\dots,n\\}$,

$\lim_{n\to+\infty}\sup_{x\in\widetilde{K}}|X^{\alpha}(\varphi_{n}\,\widetilde{\ast}\,f)(x)-X^{\alpha}f(x)|=\lim_{n\to+\infty}\sup_{x\in\widetilde{K}}|(\varphi_{n}\,\widetilde{\ast}\,X^{\alpha}f)(x)-X^{\alpha}f(x)|=0,$
(2.21)

where $X^{\alpha}:=X_{j_{1}}\dots X_{j_{r}}$. Hence (2.17) and (2.21)
precisely means that $\varphi_{n}\,\widetilde{\ast}\,f\to f$ in the topology
of $\mathcal{D}(\mathbb{G})$ that was what we wanted to prove. ∎

We define now the precomposition of a distribution with a right translation.

######  Definition  (Right translation of a distribution).

Let $f$ be a distribution on a Lie group $\mathbb{G}$ and let $R_{g}$ denote
the right translation by $g\in\mathbb{G}$. Then we define

$\langle f\circ R_{g},\varphi\rangle:=\langle f,\Delta(g^{-1})\varphi\circ
R_{g^{-1}}\rangle,\qquad\text{for all $\varphi\in\mathcal{D}(\mathbb{G})$}.$
(2.22)

By using the fact that $(R_{g})_{*}\mu=\Delta(g^{-1})\mu$ for all
$g\in\mathbb{G}$, we see that the above definition is consistent with (2.8)
when $f$ is a smooth function. Moreover, let us define, for every
$g\in\mathbb{G}$, the operator $D_{g}$ acting on distributions $f$ on
$\mathbb{G}$ as follows

$D_{g}f:=f\circ R_{g}-f.$ (2.23)

######  Remark  (Commutation of convolution and right translation).

We claim that if $\varphi\in\mathcal{D}(\mathbb{G})$,
$f\in\mathcal{D}^{\prime}(\mathbb{G})$ and $g\in\mathbb{G}$ then the following
equality holds

$\langle(\varphi\ast f)\circ R_{g},\psi\rangle=\langle\varphi\ast(f\circ
R_{g}),\psi\rangle,\qquad\text{for all $\psi\in\mathcal{D}(\mathbb{G})$}.$
(2.24)

Indeed, we have

$\begin{split}\langle(\varphi\ast f)\circ
R_{g},\psi\rangle&=\langle\varphi\ast f,\Delta(g^{-1})\psi\circ
R_{g^{-1}}\rangle=\langle
f,\Delta(g^{-1})\varphi\,\widetilde{\ast}\,(\psi\circ R_{g^{-1}})\rangle\\\
&=\langle f,\Delta(g^{-1})(\varphi\,\widetilde{\ast}\,\psi)\circ
R_{g^{-1}}\rangle=\langle f\circ
R_{g},\varphi\,\widetilde{\ast}\,\psi\rangle\\\ &=\langle\varphi\ast(f\circ
R_{g}),\psi\rangle,\end{split}$

where in the first and the fourth equalities we used the definition in (2.22),
in the second and the fifth equalities we used the definition in (2.12), and
in the third equality we used $\varphi\,\widetilde{\ast}\,(\psi\circ
R_{g^{-1}})=(\varphi\,\widetilde{\ast}\,\psi)\circ R_{g^{-1}}$, which we now
prove. Indeed, for every $x\in\mathbb{G}$,

$\varphi\,\widetilde{\ast}\,(\psi\circ
R_{g^{-1}})(x)=\int\varphi(y)\psi(yxg^{-1})\operatorname{d}\mu(y)=(\varphi\,\widetilde{\ast}\,\psi)\circ
R_{g^{-1}}(x),$

where we used the definition in (2.6). As a consequence of (2.24) we obtain
the following equality, for every $\varphi\in\mathcal{D}(\mathbb{G})$,
$f\in\mathcal{D}^{\prime}(\mathbb{G})$ and $g\in\mathbb{G}$

$D_{g}(\varphi\ast f)=\varphi\ast D_{g}f.$ (2.25)

###  2.3. Nilpotent Lie groups and stratified groups

We now focus the attention on special classes of Lie groups. First we discuss
nilpotent groups, and then we discuss stratified groups, also called Carnot
groups. For the notions in this section we refer the reader to classical books
and recent surveys or lecture notes, e.g., [FS82, Var84, CG90, BLU07, LD17].
We stress that every vector space (or Lie algebra) considered will be a vector
space (or Lie algebra) over $\mathbb{R}$. We recall that a connected Lie group
is nilpotent if and only if its algebra is nilpotent. It is well known that
the exponential map $\exp\colon\mathfrak{g}\to\mathbb{G}$ is a global analytic
diffeomorphism whenever $\mathbb{G}$ is a simply connected nilpotent Lie
group, while in general, if $\mathbb{G}$ is connected and nilpotent but not
necessarily simply connected, it is an analytic and surjective map, see
[Var84, Theorem 3.6.1].

We say that a Lie algebra $\mathfrak{g}$ is stratifiable if there exist a
stratification of it, namely there exist subspaces $V_{1},\dots,V_{s}$ of the
Lie algebra $\mathfrak{g}$ such that

$\mathfrak{g}=V_{1}\oplus\dots\oplus
V_{s},\qquad[V_{1},V_{j}]=V_{j+1}\quad\forall
j=1,\dots,s-1,\qquad[V_{1},V_{s}]=\\{0\\}.$

When one of such stratifications is fixed we say that $\mathfrak{g}$ is a
stratified Lie algebra. Recall that two stratifications of a stratifiable Lie
algebra differ by an automorphism, see [LD17, Proposition 2.17]. Notice that
every stratified Lie algebra is nilpotent. A stratified group (also called a
Carnot group) $\mathbb{G}$ is a simply connected Lie group whose Lie algebra
$\mathfrak{g}$ is stratifiable and one such stratification is fixed. If the
stratifications of $\mathbb{G}$ are made with vector subspaces
$V_{1},\dots,V_{s}$, we call $s$ the step of $\mathbb{G}$, while
$m\coloneqq\dim(V_{1})$ is called _rank_ of $\mathbb{G}$. For every
$i=1,\dots,s$, we call $V_{i}$ the $i$-th layer of the stratification.

Every Carnot group has a one-parameter family of dilations that we denote by
$\\{\delta_{\lambda}:\lambda>0\\}$. These dilations act on $\mathfrak{g}$ as

$(\delta_{\lambda})_{|_{V_{i}}}:=\lambda^{i}(\mathrm{id})_{|_{V_{i}}},\qquad\forall\lambda>0,\quad\forall
1\leq i\leq s,$

and are extended linearly. We will indicate with $\delta_{\lambda}$ both the
dilations on $\mathfrak{g}$ and the group automorphisms corresponding to them
via the exponential map. For some features of the general theory of
homogeneous Lie groups we refer the reader to [FS82, Chapter 1, Section A].
For recent developments we refer the reader to [LDNG19].

We recall that $\|\cdot\|$ is a homogeneous norm on a Carnot group
$\mathbb{G}$ if it is continuous from $\mathbb{G}$ to $[0,+\infty)$ and

$\begin{split}\|g\|&=0,\qquad\mbox{if and only if $g=0$},\\\
\|\delta_{\lambda}g\|&=\lambda\|g\|,\qquad\forall\lambda>0,\quad\forall
g\in\mathbb{G},\\\ \|g\|&=\|g^{-1}\|,\qquad\forall g\in\mathbb{G}.\end{split}$
(2.26)

On a Carnot group a homogeneous norm always exists and moreover two arbitrary
homogeneous norms are bi-Lipschitz equivalent. Moreover if $\|\cdot\|$ is a
homogeneous norm on a Carnot group $\mathbb{G}$ there exists $C>0$ such that
$\|xy\|\leq C(\|x\|+\|y\|)$ for every $x,y\in\mathbb{G}$, see [FS82,
Proposition 1.6]. We can always construct, on an arbitrary Carnot group, a
homogeneous norm such that the previous $C$ is $1$: it suffices to take any
Carnot-Carathéodory distance from the identity element of $\mathbb{G}$, see
[LD17, Section 3.3].

On a Carnot group $\mathbb{G}$ with a stratification
$\mathfrak{g}=V_{1}\oplus\dots\oplus V_{s}$, let us set $m_{0}\coloneqq 0$ and
$m_{j}\coloneqq\dim{V_{j}}$ for any $j=1,\dots,s$. We stress that $m=m_{1}$.
Let us define $n_{0}\coloneqq 0$, and
$n_{j}\coloneqq\sum_{\ell=1}^{j}m_{\ell}$ for any $j=1,\dots,s$. The ordered
set $(X_{1},\dots,X_{n})$ is an adapted basis for $\mathfrak{g}$ if the
following facts hold.

  * (i)

The vector fields $X_{n_{j}+1},\dots,X_{n_{j+1}}$ are chosen among the
iterated commutators of order $j$ of the vector fields $X_{1},\dots,X_{m}$,
for every $j=1,\dots,s-1$.

  * (ii)

The set $\\{X_{n_{j}+1},\dots,X_{n_{j+1}}\\}$ is a basis for $V_{j+1}$ for
every $j=0,\dots,s-1$.

If we fix an adapted basis $(X_{1},\dots,X_{n})$, and
$\ell\in\\{1,\dots,n\\}$, we define the holonomic degree of $\ell$ to be the
unique $j^{*}\in\\{1,\dots,s\\}$ such that $n_{j^{*}-1}+1\leq\ell\leq
n_{j^{*}}$. We denote $\deg\ell\coloneqq j^{*}$ and we also say that $j^{*}$
is the holonomic degree of $X_{\ell}$, i.e., $\deg(X_{\ell})\coloneqq j^{*}$.
If an adapted basis $(X_{1},\dots,X_{n})$ of the Lie algebra $\mathfrak{g}$ of
a Carnot group $\mathbb{G}$ is fixed, we identify $x\in\mathbb{G}$ with a
point of $\mathbb{R}^{n}$ through exponential coordinates of the first kind as
follows

$x\equiv(x_{1},\dots,x_{n})\leftrightarrow\exp(x_{1}X_{1}+\dots+x_{n}X_{n}).$

We recall that the homogeneous degree of the monomial
$x_{1}^{a_{1}}\cdot\dots\cdot x_{n}^{a_{n}}$ in exponential coordinates of the
first kind associated to the adapted basis $(X_{1},\dots,X_{n})$ is
$\sum_{\ell=1}^{n}a_{\ell}\cdot\deg\ell$.

We recall here the definition of free-nilpotent Lie algebras see [BLU07,
Definition 14.1.1].

######  Definition  (Free-nilpotent Lie algebras of step $s$ with $m$
generators).

Let $m\geq 2$ be an integer number. We say that $\mathfrak{f}_{m,s}$ is the
free-nilpotent Lie algebra of step $s$ with $m$ generators
$X_{1}^{\prime},\dots,X_{m}^{\prime}$ if the following facts hold.

  * (i)

$\mathfrak{f}_{m,s}$ is a Lie algebra generated by the elements
$X_{1}^{\prime},\dots,X_{m}^{\prime}$, i.e., $\mathfrak{f}_{m,s}$ is the
smallest subalgebra of $\mathfrak{f}_{m,s}$ containing
$\\{X_{1}^{\prime},\dots,X_{m}^{\prime}\\}$;

  * (ii)

$\mathfrak{f}_{m,s}$ is nilpotent of step $s$, i.e., nested Lie brackets of
length $s+1$ are always $0$;

  * (iii)

for every nilpotent Lie algebra $\mathfrak{g}$ of step $s$ and for every map
$\Psi\colon\\{X_{1}^{\prime},\dots,X_{m}^{\prime}\\}\to\mathfrak{g}$, there
exists a unique homomorphism of Lie algebras
$\overline{\Psi}:\mathfrak{f}_{m,s}\to\mathfrak{g}$ that extends $\Psi$.

We stress that every free-nilpotent Lie algebra is stratifiable, see [LD17,
Example 2.5], with $\mathrm{span}\\{X_{1}^{\prime},\dots,X_{m}^{\prime}\\}$
being the first layer of a stratification. Thus there exists a unique Carnot
group $\mathbb{F}_{m,s}$ such that its Lie algebra is the free-nilpotent Lie
algebra of step $s$ and with $m$ generators.

##  3\. Polynomial and $S$-polynomial distributions on Lie groups

In Section 3.1 we introduce the notion of polynomial distribution with respect
to a subset $S$ of the Lie algebra $\mathfrak{g}$ of an arbitrary Lie group
$\mathbb{G}$. Roughly speaking we say that a distribution $f$ is polynomial
with respect to $S$, or briefly $S$-polynomial, when for every $X\in S$ there
exists $k$ such that $X^{k}f\equiv 0$ in the sense of distributions on
$\mathbb{G}$, see Section 3.1. When $k$ is independent on $X\in S$ we say that
$f$ is $k$-polynomial with respect to $S$, or, that is the same,
$S$-polynomial with degree at most $k$. We introduce also the definition of
polynomial distribution on arbitrary Lie groups $\mathbb{G}$, see Section 3.1:
namely, a distribution on $\mathbb{G}$ is polynomial if there exists
$k\in\mathbb{N}$ such that for all $X_{1},\dots,X_{k}\in\mathfrak{g}$ we have
$X_{1}\dots X_{k}f\equiv 0$ in the sense of distributions on $\mathbb{G}$.
This latter definition happens to be consistent with the definition of
polynomial map between groups introduced by Leibman in [Lei02], see Section
3.1. We conclude the subsection by proving a lemma about the pointwise limit
of smooth $k$-polynomial functions with respect to $X$, where
$X\in\mathfrak{g}$, see Section 3.1, and Section 3.1. These latter results
about convergence will come into play in the proof of Section 4.1.

In Section 3.2 we prove formula (3.6), see Section 3.2. Namely, given a smooth
function $f:\mathbb{G}\to\mathbb{R}$ for which $X^{k}f\equiv 0$ for some
$X\in\mathfrak{g}$ and $k\in\mathbb{N}$, we show that knowing $f$ on an open
set $U\subseteq\mathbb{G}$ completely determines $f$ on the set
$U\cdot\exp(\mathbb{R}X)$. Then we use Section 3.2 to prove the fundamental
representation formula (3.8) in Section 3.2 according to which if $f$ is
$S$-polynomial with degree at most $k$, with a Lie generating $S$, then it is
completely determined by the jet of some sufficiently big order of $f$ at the
identity.

In Section 3.3 we prove Theorem 1.1. The proof of Theorem 1.1, see Section 3.3
and Section 3.3, is reached by means of Section 3.3, according to which in a
connected Lie group $\mathbb{G}$ around every point there exists a chart that
is a concatenation of horizontal curves, and by means of the representation
formula in Section 3.2.

Before starting the discussion, let us recall some basic facts about the
analytic structure of a Lie group. It is a classical result of Gleason,
Montgomery and Zippin that a topological group that has the structure of a
$C^{k}$-manifold for some $0\leq k\leq+\infty$ admits exactly one analytic
structure that is compatible with the $C^{k}$-structure, see the discussion in
[Var84, page 42]. Thus, on a Lie group, we can give a meaning to a function
$f:\mathbb{G}\to\mathbb{R}$ being analytic by using an analytic atlas.

###  3.1. Relations with pointwise convergence

In what follows we give the definitions of $S$-polynomial and polynomial
distributions on Lie groups.

######  Definition  ($S$-polynomial distributions on Lie groups).

Let $\mathbb{G}$ be a Lie group with Lie algebra $\mathfrak{g}$, and let us
fix a subset $S\subseteq\mathfrak{g}$. We say that a distribution
$f\in\mathcal{D}^{\prime}(\mathbb{G})$ is polynomial with respect to $S$, or
horizontally polynomial if $S$ is understood, if

$\forall X\in S\,\,\exists k\in\mathbb{N}\,\,\text{such that}\,\,X^{k}f\equiv
0\,\,\text{holds on $\mathbb{G}$ in the sense of distributions.}$

If the previous condition holds, we also say that $f$ is $S$-polynomial. If
the choice of $k$ is uniform on $X\in S$, we say that $f$ is $k$-polynomial
with respect to $S$, or horizontally $k$-polynomial if $S$ is understood. If
the previous condition holds, we also say that $f$ is $S$-polynomial with
degree at most $k$. When $k=2$ we say that $f$ is affine with respect to $S$,
or horizontally affine if $S$ is understood.

######  Remark  (Taylor expansion for $S$-polynomial smooth functions).

It is easy to notice that if $f:\mathbb{G}\to\mathbb{R}$ is a smooth function
that is $k$-polynomial with respect to $X$, then

$f(p\exp(tX))=f(p)+t(Xf)(p)+\dots+\frac{t^{k-1}}{(k-1)!}(X^{k-1}f)(p),\quad\forall
p\in\mathbb{G},\forall t\in\mathbb{R}.$ (3.1)

The previous observation comes from the fact that for an arbitrary smooth
function $f:\mathbb{G}\to\mathbb{R}$ the following equality holds

$X^{j}f(p)=\frac{\operatorname{d}^{j}}{\operatorname{d}\varepsilon^{j}}_{|_{\varepsilon=0}}f(p\exp(\varepsilon
X)),\quad\forall p\in\mathbb{G},\forall j\in\mathbb{N}.$

Moreover, let us notice that if $f:\mathbb{G}\to\mathbb{R}$ is a function such
that it is a polynomial with degree at most $k$ along the flow lines of
$X\in\mathfrak{g}$, then $X^{k}f\equiv 0$ holds on $\mathbb{G}$ in the
classical sense and we can write the expansion in (3.1). To be more precise,
if we fix $p\in\mathbb{G}$ and there exist $k$ real numbers
$a_{0},\dots,a_{k-1}$ such that

$f(p\exp(tX))=a_{0}+ta_{1}+\dots+t^{k-1}a_{k-1},\quad\forall t\in\mathbb{R},$
(3.2)

then $f$ is differentiable $k$ times along $X$ at $p$, $X^{j}f(p)=j!a_{j}$ for
every $j=0,\dots,k-1$, and $X^{k}f(p)=0$.

######  Definition .

(Polynomial distributions on Lie groups) A distribution
$f\in\mathcal{D}^{\prime}(\mathbb{G})$ is a polynomial distribution if there
exists a $k_{0}\in\mathbb{N}$ such that for every
$Y_{1},\dots,Y_{k_{0}}\in\mathfrak{g}$ we have $Y_{1}\dots Y_{k_{0}}f\equiv 0$
on $\mathbb{G}$.

######  Remark .

Due to Theorem 1.1, one can equivalently ask $f$ to be an analytic function in
Section 3.1.

######  Remark  (Comparison between Section 3.1 and Leibman’s definition in
[Lei02]).

In [Lei02] the author gives and studies the notion of polynomial map
$f:\mathbb{G}\to\mathbb{H}$ between two groups $\mathbb{G}$ and $\mathbb{H}$.
Given $g\in\mathbb{G}$ we define the operator $D_{g}$ that acts on functions
$f:\mathbb{G}\to\mathbb{H}$ as follows

$(D_{g}f)(g^{\prime}):=f(g^{\prime})^{-1}f(g^{\prime}g).$ (3.3)

According to [Lei02, Section 0.2], a map $f:\mathbb{G}\to\mathbb{H}$ between
two groups $\mathbb{G}$ and $\mathbb{H}$ is a polynomial map with degree at
most $d$, being $d\in\mathbb{N}$, if for every
$g_{1},\dots,g_{d+1}\in\mathbb{G}$ we have

$D_{g_{1}}\dots D_{g_{d+1}}f\equiv e_{\mathbb{H}},$ (3.4)

where $e_{\mathbb{H}}$ is the (function that is constantly equal to the)
identity of $\mathbb{H}$. In our case, i.e., when $\mathbb{H}=(\mathbb{R},+)$,
we can give a definition that mimics the previous one but for distributions
$f$ on $\mathbb{G}$. Let us define the operator $D_{g}$ acting on
distributions $f$ as in (2.23), thus generalizing (3.3) for distributions. We
say that a distribution $f$ on $\mathbb{G}$ is polynomial à la Leibman with
degree at most $d\in\mathbb{N}$ if for every
$g_{1},\dots,g_{d+1}\in\mathbb{G}$ we have

$D_{g_{1}}\dots D_{g_{d+1}}f\equiv 0,\qquad\text{in the sense of distributions
on $\mathbb{G}$}.$ (3.5)

Let us notice that if $f$ is continuous the two definitions in (3.4) and (3.5)
agree. However there are non-continuous functions, already from $\mathbb{R}$
to $\mathbb{R}$, that satisfy (3.3) but they cannot be seen as distributions,
see Section 5.2.

We stress that the result in Theorem 1.2 tells us that Section 3.1 and
Leibman’s definition adapted for distributions give raise to the same class of
distributions.

We prove the following lemma about the pointwise limit of functions that are
polynomials along one line in the direction of $X\in\mathfrak{g}$ emanating
from a fixed point $p\in\mathbb{G}$. We are going to prove that the pointwise
limit of such functions, whenever it exists, is still polynomial along the
same line.

######  Lemma .

Let $\mathbb{G}$ be a Lie group. Let $p\in\Omega\subseteq\mathbb{G}$, where
$\Omega$ is open, let $X$ be a left-invariant vector field on $\mathbb{G}$,
and let $k\in\mathbb{N}$. Let $0\in I\subseteq\mathbb{R}$ be an interval such
that $p\exp(tX)\in\Omega$ for all $t\in I$. Let $\\{f_{n}\\}_{n\in\mathbb{N}}$
be a sequence of functions $f_{n}:\Omega\to\mathbb{R}$ such that, for every
$n\in\mathbb{N}$, there exists a sequence of $k$-uples of real numbers
$\\{(a_{n,0},\dots,a_{n,k-1})\\}_{n\in\mathbb{N}}$ such that

$f_{n}(p\exp(tX))=a_{n,0}+ta_{n,1}+\dots+t^{k-1}a_{n,k-1},\qquad\forall t\in
I.$

Let us assume that there exists a function $f:\Omega\to\mathbb{R}$ such that
$f_{n}\to f$ pointwise on the set $\\{p\exp(tX):t\in I\\}$, as $n\to+\infty$.
Then there exists a $k$-uple of real numbers $(a_{0},\dots,a_{k-1})$ such that
$a_{n,i}\to a_{i}$ for all $i=0,\dots,k-1$ and for $n\to+\infty$, and

$f(p\exp(tX))=a_{0}+ta_{1}+\dots+t^{k-1}a_{k-1},\qquad\forall t\in I.$

###### Proof.

Let us fix $k$ pairwise distinct nonzero real numbers in $I$ and let us call
them $r_{0},\dots,r_{k-1}$. By hypothesis we get that, for every
$n\in\mathbb{N}$, the following holds

$\sum_{i=0}^{k-1}r_{j}^{i}a_{n,i}\xrightarrow{n\to+\infty}f(p\exp(r_{j}X)),\qquad\forall
j=0,\dots,k-1.$

For every $n\in\mathbb{N}$ we denote with $\overline{a}_{n}$ the column
$k$-vector $(a_{n,0},\dots,a_{n,k-1})$, and with $\overline{f}_{p,X}$ the
column $k$-vector $(f(p\exp(r_{j}X)))_{j=0,\dots,k-1}$. Thus we can write the
previous convergence as follows

$V\cdot\overline{a}_{n}\xrightarrow{n\to+\infty}\overline{f}_{p,X},$

where $V$ is the $k\times k$ Vandermonde’s matrix
$V_{ji}:=(r_{j}^{i})_{j,i=0,\dots,k-1}$. Since $V$ is invertible, we conclude
that

$\overline{a}_{n}\xrightarrow{n\to+\infty}V^{-1}\cdot\overline{f}_{p,X},$

and thus, if we denote $(a_{0},\dots,a_{k-1})$ the components of the vector
$V^{-1}\cdot\overline{f}_{p,X}$, we conclude that $a_{n,i}\to a_{i}$ for all
$i=0,\dots,k-1$ and for $n\to+\infty$. The last part of the statement easily
follows from the latter convergence and the fact that $f_{n}\to f$ pointwise
on $\\{p\exp(tX):t\in I\\}$, as $n\to+\infty$. ∎

######  Corollary .

Let $\mathbb{G}$ be a Lie group, $\Omega\subseteq\mathbb{G}$ be open, $X$ be a
left-invariant vector field on $\mathbb{G}$, and $k\in\mathbb{N}$. Let
$\\{f_{n}\\}_{n\in\mathbb{N}}$ be a sequence of smooth functions
$f_{n}:\Omega\to\mathbb{R}$ such that $X^{k}f_{n}\equiv 0$ on $\Omega$ for
every $n\in\mathbb{N}$. If there exists a function $f:\Omega\to\mathbb{R}$
such that $f_{n}\to f$ pointwise on $\Omega$, then $X^{k}f\equiv 0$ on
$\Omega$ in the classical sense.

###### Proof.

If we fix $p\in\Omega$, since $\Omega$ is open there exists an interval $0\in
I\subseteq\mathbb{R}$ such that $\\{p\exp(tX):t\in I\\}\subseteq\Omega$. From
Section 3.1, see in particular the localized version of (3.1), we get that the
sequence $\\{f_{n}\\}_{n\in\mathbb{N}}$ satisfies the hypotheses of Section
3.1. Thus, applying Section 3.1, the function $f$ coincides with a polynomial
with degree at most $k$ in $t$ along the piece of line $\\{p\exp(tX):t\in
I\\}$, and arguing as at the end of Section 3.1, see in particular the
localized reasoning above and below (3.2), we get that $X^{k}f(p)=0$ and then
we get the conclusion since $p\in\Omega$ is arbitrary. ∎

###  3.2. Propagation of being $S$-polynomial

In the following lemma we are going to prove a formula that will be of crucial
importance in the proof of Theorem 1.1 and Theorem 1.3, since it is the main
tool to prove the representation formulas in Section 3.2, and Section 4.

######  Lemma .

Let $\mathbb{G}$ be a Lie group. Let us fix $k\in\mathbb{N}$,
$X\in\mathfrak{g}$, and let $f:\mathbb{G}\to\mathbb{R}$ be a smooth function
such that $X^{k}f\equiv 0$ on $\mathbb{G}$. Then if we fix
$r\in\\{0,1,2,\dots\\}$ and $X_{1},\dots,X_{r}\in\mathfrak{g}$, the following
formula holds

$(X_{1}\ldots
X_{r}f)(q\exp(tX))=\sum_{i=0}^{k-1}\frac{t^{i}}{i!}(\mathrm{Ad}_{\exp(tX)}(X_{1})\ldots\mathrm{Ad}_{\exp(tX)}(X_{r})X^{i}f)(q),$
(3.6)

for every $q\in\mathbb{G}$ and every $t\in\mathbb{R}$.

###### Proof.

Let us prove the statement by induction on $r$. If $r=0$ the formula holds
true by (3.1). Let us now suppose that the statement holds true for some
$r\in\\{0,1,2,\dots\\}$ and let us prove that it holds true for $r+1$. Indeed,
let us fix $X_{1},\dots,X_{r+1}\in\mathfrak{g}$, $t\in\mathbb{R}$, and
$q\in\mathbb{G}$. Then we compute the derivative of $X_{2}\dots X_{r+1}f$
along $X_{1}$ as follows

$\begin{split}X_{1}(X_{2}\ldots
X_{r+1}f)&(q\exp(tX))=\frac{\operatorname{d}}{\operatorname{d}\varepsilon}_{|_{\varepsilon=0}}(X_{2}\ldots
X_{r+1}f)(q\exp(tX)\exp(\varepsilon X_{1}))\\\
&=\frac{\operatorname{d}}{\operatorname{d}\varepsilon}_{|_{\varepsilon=0}}(X_{2}\ldots
X_{r+1}f)(\underbrace{q\exp(tX)\exp(\varepsilon
X_{1})\exp(-tX)}_{q_{\varepsilon}}\exp(tX))\\\
&=\frac{\operatorname{d}}{\operatorname{d}\varepsilon}_{|_{\varepsilon=0}}\sum_{i=0}^{k-1}\frac{t^{i}}{i!}\mathrm{Ad}_{\exp(tX)}(X_{2})\ldots\mathrm{Ad}_{\exp(tX)}(X_{r+1})X^{i}f(q_{\varepsilon})\\\
&=\sum_{i=0}^{k-1}\frac{t^{i}}{i!}\frac{\operatorname{d}}{\operatorname{d}\varepsilon}_{|_{\varepsilon=0}}\mathrm{Ad}_{\exp(tX)}(X_{2})\ldots\mathrm{Ad}_{\exp(tX)}(X_{r+1})X^{i}f(q_{\varepsilon})\\\
&=\sum_{i=0}^{k-1}\frac{t^{i}}{i!}\mathrm{Ad}_{\exp(tX)}(X_{1})\ldots\mathrm{Ad}_{\exp(tX)}(X_{r+1})X^{i}f(q),\end{split}$
(3.7)

where in the third equality we used the inductive hypothesis, and in the last
equality we used that the curve $\varepsilon\to q_{\varepsilon}$ has
$\mathrm{Ad}_{\exp(tX)}(X_{1})_{|_{q}}$ as tangent vector at $\varepsilon=0$.
∎

######  Lemma .

Let $\mathbb{G}$ be a Lie group, and let $f:\mathbb{G}\to\mathbb{R}$ be smooth
and $k$-polynomial with respect to
$\\{Y_{1},\dots,Y_{\ell}\\}\subseteq\mathfrak{g}$ for some
$k,\ell\in\mathbb{N}$. Given $t_{1},\dots,t_{\ell}\in\mathbb{R}$, let us
define, for every $j=2,\dots,\ell$, the element of the group
$g_{j}:=\exp(t_{1}Y_{1})\dots\exp(t_{j-1}Y_{j-1})$, and let us denote
$g_{1}:=e$ the identity of the group. Then the following equality holds

$f(\exp(t_{1}Y_{1})\dots\exp(t_{\ell}Y_{\ell}))=\sum_{i_{1},\dots,i_{\ell}=0}^{k-1}\frac{t_{1}^{i_{1}}\dots
t_{\ell}^{i_{\ell}}}{i_{1}!\dots
i_{\ell}!}\left((\mathrm{Ad}_{g_{\ell}}Y_{\ell})^{i_{\ell}}(\mathrm{Ad}_{g_{\ell-1}}Y_{\ell-1})^{i_{\ell-1}}\dots(\mathrm{Ad}_{g_{1}}Y_{1})^{i_{1}}f\right)(e).$
(3.8)

###### Proof.

Let us show, for simplicity, the computations only in the nontrivial case
$\ell=3$, while the general case follows along the same lines and we omit it.
For every $(t_{1},t_{2},t_{3})\in\mathbb{R}^{3}$ the following equality holds

$\begin{split}f(\exp(t_{1}Y_{1})&\exp(t_{2}Y_{2})\exp(t_{3}Y_{3}))=\sum_{i_{3}=0}^{k-1}\frac{t_{3}^{i_{3}}}{i_{3}!}(Y_{3}^{i_{3}}f)(\exp(t_{1}Y_{1})\exp(t_{2}Y_{2}))\\\
&=\sum_{i_{3}=0}^{k-1}\frac{t_{3}^{i_{3}}}{i_{3}!}\sum_{i_{2}=0}^{k-1}\frac{t_{2}^{i_{2}}}{i_{2}!}\left((\mathrm{Ad}_{\exp(t_{2}Y_{2})}Y_{3})^{i_{3}}Y_{2}^{i_{2}}f\right)(\exp(t_{1}Y_{1}))\\\
&=\sum_{i_{3}=0}^{k-1}\frac{t_{3}^{i_{3}}}{i_{3}!}\sum_{i_{2}=0}^{k-1}\frac{t_{2}^{i_{2}}}{i_{2}!}\sum_{i_{1}=0}^{k-1}\frac{t_{1}^{i_{1}}}{i_{1}!}\left((\mathrm{Ad}_{\exp(t_{1}Y_{1})}\mathrm{Ad}_{\exp(t_{2}Y_{2})}Y_{3})^{i_{3}}(\mathrm{Ad}_{\exp(t_{1}Y_{1})}Y_{2})^{i_{2}}Y_{1}^{i_{1}}f\right)(e)\\\
&=\sum_{i_{1},i_{2},i_{3}=0}^{k-1}\frac{t_{1}^{i_{1}}t_{2}^{i_{2}}t_{3}^{i_{3}}}{i_{1}!i_{2}!i_{3}!}\left((\mathrm{Ad}_{\exp(t_{1}Y_{1})\exp(t_{2}Y_{2})}Y_{3})^{i_{3}}(\mathrm{Ad}_{\exp(t_{1}Y_{1})}Y_{2})^{i_{2}}Y_{1}^{i_{1}}f\right)(e),\end{split}$

where in the first equality we used that $f$ is $k$-polynomial with respect to
$Y_{3}$; in the second equality we used (3.6) with $r=i_{3}$,
$X_{1}=\dots=X_{r}=Y_{3}$, $X=Y_{2}$, and $q=\exp(t_{1}Y_{1})$; in the third
equality we used again (3.6) with $r=i_{3}+i_{2}$,
$X_{1}=\dots=X_{i_{3}}=\mathrm{Ad}_{\exp(t_{2}Y_{2})}Y_{3}$,
$X_{i_{3}+1}=\dots=X_{i_{3}+i_{2}}=Y_{2}$, $X=Y_{1}$, and $q=e$; and in the
fourth equality we used that $\mathrm{Ad}_{g}\mathrm{Ad}_{h}=\mathrm{Ad}_{gh}$
for every $g,h\in\mathbb{G}$. ∎

###  3.3. Proof of Theorem 1.1

Before starting the proof of Theorem 1.1 we recall here a lemma that tells us
that on a connected Lie group we can find local charts by concatenating a
fixed amount of flow lines of horizontal vector fields. This is a standard
result in control theory.

######  Lemma  ([ABB19, Lemma 3.33]).

Let $\mathbb{G}$ be a connected Lie group of topological dimension $n$ and let
$S$ be a subset of the Lie algebra $\mathfrak{g}$ that Lie generates
$\mathfrak{g}$. Then there exists an open neighbourhood $U$ of the identity
$e$ and $2n$ elements $X_{1},\dots,X_{2n}\in S$ such that for all
$p\in\mathbb{G}$ and all $q\in p\cdot U$ there exist
$s_{1},\dots,s_{2n}\in\mathbb{R}$ such that

$q=p\exp(s_{1}X_{1})\dots\exp(s_{2n}X_{2n});$

more precisely there exist an open bounded set $\hat{V}\subseteq(0,1)^{n}$ and
$(\hat{s}_{1},\dots,\hat{s}_{n})\in(0,1)^{n}$ such that

$\begin{split}\hat{\psi}_{p}\colon\hat{V}&\to p\cdot U,\\\
\hat{\psi}_{p}(s_{1},\dots,s_{n})\coloneqq
p\exp(s_{1}X_{1})\dots&\exp(s_{n}X_{n})\exp(-\hat{s}_{n}X_{n+1})\dots\exp(-\hat{s}_{1}X_{2n}),\end{split}$
(3.9)

is a diffeomorphism for every $p\in\mathbb{G}$.

As a first step toward the proof of Theorem 1.1, we prove the finite-
dimensional result in the second part of Theorem 1.1 for analytic functions on
connected Lie groups. The proof of the forthcoming Section 3.3 is reached by
joining the previous representation formula proved in Section 3.2 with Section
3.3.

######  Proposition .

Let $\mathbb{G}$ be a connected Lie group of topological dimension $n$, and
let $S$ be a Lie generating subset of $\mathfrak{g}$. Then for every
$k\in\mathbb{N}$ there exists $\delta$, which depends on $k$ and $n$, such
that the vector space

$\mathscr{F}:=\\{f:\mathbb{G}\to\mathbb{R}\,\text{is analytic and
$k$-polynomial with respect to $S$}\\},$

is finite-dimensional and its dimension is bounded above by $\delta$.

###### Proof.

Let $\hat{\psi}_{e}:\hat{V}\subseteq\mathbb{R}^{n}\to U\subseteq\mathbb{G}$ be
the local chart around $e$ constructed as in (3.9). This chart induces a local
frame $(\partial_{x_{1}},\dots,\partial_{x_{n}})$ of the tangent bundle $TU$.
We denote with $\alpha:=(\alpha_{1},\dots,\alpha_{n})$ an arbitrary $n$-uple
of natural numbers, we set $\lvert\alpha\rvert:=\alpha_{1}+\dots+\alpha_{n}$
and we denote

$\partial_{x^{\alpha}}:=\partial_{\underbrace{x_{1}\dots
x_{1}}_{\alpha_{1}}\cdots\underbrace{x_{n}\dots x_{n}}_{\alpha_{n}}}.$

Recall that $\hat{\psi}_{e}$ has the following explicit expression

$\hat{\psi}_{e}(s_{1},\dots,s_{n})=\exp(s_{1}X_{1})\dots\exp(s_{n}X_{n})\exp(-\hat{s}_{n}X_{n+1})\dots\exp(-\hat{s}_{1}X_{2n}),$

for some $X_{1},\dots,X_{2n}\in S$, and
$(\hat{s}_{1},\dots,\hat{s}_{n})\subseteq(0,1)^{n}$. Moreover, since
$\mathrm{Ad}$, the exponential map, and the product on the group $\mathbb{G}$
are analytic functions, from (3.8) read in coordinates and the previous
explicit expression of $\hat{\psi}_{e}$ we get that there exists $D:=D(k,n)$
such that

$f(\hat{\psi}_{e}(s_{1},\dots,s_{n}))=\sum_{\lvert\alpha\rvert\leq
D}h_{\alpha}(s_{1},\dots,s_{n})(\partial_{x^{\alpha}}f)(e),$ (3.10)

for every $(s_{1},\dots,s_{n})\in\hat{V}$, and where, for every multi-index
$\alpha$, the function $h_{\alpha}(s_{1},\dots,s_{n}):\hat{V}\to\mathbb{R}$ is
an analytic function depending on the chart $\hat{\psi}_{e}$. Now, there
exists $\delta:=\delta(D)$ such that the number of the operators
$(\partial_{x^{\alpha}})_{|_{e}}$ with $|\alpha|\leq D$ is $\delta$. Let us
define the vector space

$\mathscr{F}^{\prime}:=\\{f_{|_{U}},\,\text{where $f:\mathbb{G}\to\mathbb{R}$
is analytic and $k$-polynomial with respect to $S$}\\},$

and let $\Psi:\mathscr{F}^{\prime}\to\mathbb{R}^{\delta}$ be the linear map
defined by $\Psi(f_{|_{U}}):=((\partial_{x^{\alpha}}f)(e))_{|\alpha|\leq D}$
for every $f_{|_{U}}\in\mathscr{F}^{\prime}$. From (3.10) we get that $\Psi$
is an injective map, and thus the dimension of $\mathscr{F}^{\prime}$ is
finite and bounded above by $\delta$.

Let $\Psi^{\prime}:\mathscr{F}\to\mathscr{F}^{\prime}$ be the linear map
defined by $\Psi^{\prime}(f)=f_{|_{U}}$ for every analytic function $f$ on
$\mathbb{G}$. By analytic continuation we deduce that whenever
$f_{|_{U}}\equiv g_{|_{U}}$ for two analytic functions, then $f\equiv g$ on
$\mathbb{G}$. Thus $\Psi^{\prime}$ is injective as well and we deduce that
$\mathscr{F}$ is finite-dimensional and its dimension is bounded above by
$\delta$. ∎

###### Proof of Theorem 1.1.

We first notice that we can reduce to work with the case $\mathbb{G}$ is
connected. Indeed, the connected component of the identity of $\mathbb{G}$ is
itself a Lie group, and if we know the result for such a connected component
we obtain the result for all the connected components by composing the
distribution $f$ to the right with some left translation, which preserves the
condition of being $S$-polynomial. Thus, from now on in the proof, we assume
$\mathbb{G}$ is connected.

Up to taking a subset of $S$ that is finite and still Lie generates
$\mathfrak{g}$, we may assume that $S$ is finite. Since now $S$ is finite,
there exists $k\in\mathbb{N}$ such that $f$ is $k$-polynomial with respect to
$S$. Let $\varphi_{n}$ be an approximate identity as in Section 2.2. Thus
$\varphi_{n}\ast f$ is a smooth function on $\mathbb{G}$, see Section 2.2, and
$X(\varphi_{n}\ast f)=\varphi_{n}\ast Xf$ for every $X\in\mathfrak{g}$, see
(2.13). Thus, iterating (2.13), $\varphi_{n}\ast f$ is smooth and
$k$-polynomial with respect to $S$. We claim that for every $n\in\mathbb{N}$
the function $\varphi_{n}\ast f$ is analytic on $\mathbb{G}$.

Indeed, for the sake of notation, let us fix $n\in\mathbb{N}$ and let us
rename $h:=\varphi_{n}\ast f$. For every $g\in\mathbb{G}$, the function
$h\circ L_{g}$, where $L_{g}$ is the left translation by $g\in\mathbb{G}$, is
still smooth and $k$-polynomial with respect to $S$. Thus in order to prove
that $h$ is analytic it is sufficient to prove that $h\circ L_{g}$ is analytic
in a neighbourhood of $e$ for every $g\in\mathbb{G}$. In conclusion, in order
to prove that $h$ is analytic on $\mathbb{G}$, we only need to prove that
every $\widetilde{h}\in C^{\infty}(\mathbb{G})$ which is $k$-polynomial with
respect to $S$ is analytic in a neighbourhood of the identity. Let
$\hat{\psi}_{e}:\hat{V}\subseteq\mathbb{R}^{n}\to U\subseteq\mathbb{G}$ be the
local chart around $e$ constructed as in (3.9) from $S$. From the fact that
for every $(s_{1},\dots,s_{n})\in\hat{V}$ we have

$\hat{\psi}_{e}(s_{1},\dots,s_{n})=\exp(s_{1}X_{1})\dots\exp(s_{n}X_{n})\exp(-\hat{s}_{n}X_{n+1})\dots\exp(-\hat{s}_{1}X_{2n}),$

for some $X_{1},\dots,X_{2n}\in S$ and
$(\hat{s}_{1},\dots,\hat{s}_{n})\subseteq\mathbb{R}^{n}$, and the fact that
the representation formula (3.10) holds with some analytic functions
$h_{\alpha}$ that only depend on the chart $\hat{\psi}_{e}$, we conclude that,
for every $\widetilde{h}\in C^{\infty}(\mathbb{G})$ that is $k$-polynomial
with respect to $S$, the function $\widetilde{h}\circ\hat{\psi}_{e}$ is an
analytic real-valued function defined on $\hat{V}$. Moreover $\hat{\psi}_{e}$
is an analytic map, since it is a composition of the exponential map with the
product of the group. Since $\hat{\psi}_{e}$ is invertible being a chart, by
the inverse function theorem we conclude that $(\hat{\psi}_{e})^{-1}$ is
analytic as well222This assertion follows from the fact that if a
diffeomorphism $\xi$ between open subsets of $\mathbb{R}^{n}$ is analytic,
then $\xi^{-1}$ is analytic as well. Indeed, there exists a complex
holomorphic extension of $\xi$, say $\widetilde{\xi}$, between open subsets of
$\mathbb{C}^{n}$. At every $x\in\mathrm{dom}(\xi)$, the Jacobian determinant
$\det((J\xi)_{x})$ is nonzero since $\xi$ is a diffeomorphism. Then, when
seeing $x$ as an element of $\mathbb{C}^{n}$, the complex Jacobian determinant
$\det((J\widetilde{\xi})_{x})$, see [FG02, page 30], is nonzero as well due to
[FG02, Chapter I, Theorem 7.2]. Thus from the complex Inverse Function
Theorem, see [FG02, Chapter I, Theorem 7.5], $\widetilde{\xi}$ is
biholomorphic in a neighbourhood of $x$ seen in $\mathbb{C}^{n}$, and then, by
restriction, $\xi^{-1}$ is analytic in a neighbourhood of $\xi(x)$. as a map
from $U$ to $\hat{V}$. Thus, for every $\widetilde{h}\in
C^{\infty}(\mathbb{G})$ that is $k$-polynomial with respect to $S$, we have
that
$\widetilde{h}=(\widetilde{h}\circ\hat{\psi}_{e})\circ(\hat{\psi}_{e})^{-1}$
is analytic as a map from $U$ to $\mathbb{R}$, since it is the composition of
analytic functions, and then the proof of the claim is concluded.

Hence $\varphi_{n}\ast f\in\mathscr{F}$, see Section 3.3, and $\mathscr{F}$ is
closed in the weak*-topology of $\mathcal{D}^{\prime}(\mathbb{G})$ since it is
finite-dimensional, see [Rud91, Theorem 1.21]. Since $\varphi_{n}\ast f\to f$
in the topology of $\mathcal{D}^{\prime}(\mathbb{G})$, because $\varphi_{n}$
is an approximate identity, we conclude that $f$ has a representative in
$\mathscr{F}$ as well, and thus $f$ is represented by an analytic function.

In order to prove the second part of Theorem 1.1, let us fix a distribution
$f$ that is $S$-polynomial with degree at most $k\in\mathbb{N}$. Hence $f$ is
represented by an analytic function from what we proved above, and the
application of Section 3.3 concludes the proof. ∎

##  4\. The case of nilpotent Lie groups

In this section we focus our attention on the case when $\mathbb{G}$ is a
connected nilpotent Lie group. We recall that the exponential map
$\exp\colon\mathfrak{g}\to\mathbb{G}$ is a global analytic diffeomorphism
whenever $\mathbb{G}$ is a simply connected nilpotent Lie group, while, when
$\mathbb{G}$ is connected and nilpotent but not necessarily simply connected,
it is an analytic and surjective map, see [Var84, Theorem 3.6.1]. Hence, we
can use the exponential map to give a natural definition of polynomial in
exponential chart. We say that a function $f\colon\mathbb{G}\to\mathbb{R}$ is
polynomial in exponential chart if $f\circ\exp$ is a polynomial, see Section
4. Such a notion, in the nilpotent setting, is consistent, see Section 4, with
both the definition of polynomial distribution on a Lie group, see Section
3.1, and with Leibman’s definition in Section 3.1.

In Section 4 we use the formula in Section 3.2 to prove that whenever a smooth
function $f$ is $k$-polynomial with respect to $S$ in a nilpotent Lie group of
nilpotency step $s$, then $f$ is polynomial along the concatenation of flows
of elements of $S$ emanating from a fixed $p\in\mathbb{G}$. More precisely we
prove that, if $p\in\mathbb{G}$ is fixed, $\ell\in\mathbb{N}$, and
$Y_{1},\dots,Y_{\ell}\in S$, the map $(t_{1},\dots,t_{\ell})\mapsto
f(p\exp(t_{1}Y_{1})\dots\exp(t_{\ell}Y_{\ell}))$ is a polynomial in the
variables $t_{1},\dots,t_{\ell}$ and we explicitly bound the degree of the
polynomial with a constant $\nu:=\nu(k,s,\ell)$.

Thus we use the latter result to give the proof of Theorem 1.3 in Section 4.1.
In order to do so we first prove that on a simply connected nilpotent Lie
group $\mathbb{G}$ a distribution that is polynomial with respect to a set $S$
that Lie generates $\mathfrak{g}$ is actually polynomial in exponential chart,
and then we reduce to the simply connected case by passing to the universal
cover. In order to prove the result for $\mathbb{G}$ nilpotent and simply
connected, we reduce ourselves to the case of Carnot groups, by lifting the
problem to a free-nilpotent Lie algebra, see the proof of Theorem 4.1. Theorem
1.3 in the case of Carnot groups is proved in Section 4.1 and the proof goes
as follows. First, we use the main result in Theorem 1.1 to obtain that a
distribution that is $k$-polynomial with respect to a set $S$ that Lie
generates the Lie algebra is represented by an analytic function. Second, we
prove that each homogeneous term in the Taylor series of $f$ around the
identity, see (4.7), is $k$-polynomial with respect to $S$ as well. Third, we
conclude because, thanks to the fundamental Section 4, and thanks to an
improvement of Section 3.3 in the setting of Carnot groups, namely Section
4.1, a smooth $k$-polynomial function with respect to $S$ has polynomial
growth of bounded order at infinity. Thus the Taylor expansion of $f$ at the
identity is finite, and from the fact that $f$ is analytic we conclude that
$f$ coincides with this finite Taylor expansion, namely $f$ is a polynomial.

We give the following notion of polynomial in exponential chart. Let us stress
that the following definition agrees with the one given in [BLU07, Definition
20.1.1] and therein studied in the more restrictive setting of stratified
groups. We also stress that the forthcoming definition does not depend on the
choice of a basis of the Lie algebra $\mathfrak{g}$.

######  Definition  (Polynomial in exponential chart on connected nilpotent
Lie groups).

Given a connected nilpotent Lie group $\mathbb{G}$ of dimension $n$ and a
basis $\\{X_{1},\dots,X_{n}\\}$ of the Lie algebra $\mathfrak{g}$, we say that
$f:\mathbb{G}\to\mathbb{R}$ is polynomial in exponential chart if

$\mathbb{R}^{n}\ni(t_{1},\dots,t_{n})\mapsto
f(\exp(t_{1}X_{1}+\dots+t_{n}X_{n})),$

is a polynomial function of the variables $t_{1},\dots,t_{n}$.

######  Remark  (Section 4, Section 3.1, and Leibman’s definition are
consistent).

When $\mathbb{G}$ is a connected and nilpotent Lie group, the result in
Section 1 tells us that the definition of polynomial in exponential chart, see
Section 4, the definition of polynomial distribution à la Leibman, see Section
3.1, and the definition of polynomial distribution, see Section 3.1, are
equivalent.

In the forthcoming Section 4 we prove that on an arbitrary nilpotent group
$\mathbb{G}$ a smooth $k$-polynomial function with respect to $S$ is
polynomial along the concatenation of lines in the directions of $S$. In order
to prove this we exploit the formula in Section 3.2 and the following remark.

######  Remark  (Formula (3.6) on nilpotent groups).

Let us notice that if $\mathbb{G}$ is a nilpotent Lie group of nilpotency step
$s$ we have that $(\mathrm{ad}_{X})^{s}(Y)\equiv 0$ for all
$X,Y\in\mathfrak{g}$. Then, from (2.1) we get

$\mathrm{Ad}_{\exp(tX)}(Y)=\sum_{j=0}^{s-1}\frac{1}{j!}(\mathrm{ad}_{tX})^{j}(Y),$

for very $X,Y\in\mathfrak{g}$ and every $t\in\mathbb{R}$. Thus, if we fix
$t\in\mathbb{R}$, $r\in\\{0,1,2,\dots\\}$, and
$X,X_{1},\dots,X_{r}\in\mathfrak{g}$, we conclude that
$\mathrm{Ad}_{\exp(tX)}(X_{1})\ldots\mathrm{Ad}_{\exp(tX)}(X_{r})$ is a sum of
left-invariant operators of the form $X_{i_{1}}\dots X_{i_{k}}$, where $k$ is
at most $rs$ and $X_{i_{1}},\dots,X_{i_{k}}\in\\{X,X_{1},\dots,X_{r}\\}$, each
one multiplied by a polynomial in $t$. This means that, if in the setting of
Section 3.2 the group $\mathbb{G}$ is nilpotent of nilpotency step $s$, the
right hand side in (3.6) can be written as the sum of at most $rs+k-1$ mixed
derivatives, in some of the directions $X,X_{1},\dots,X_{r}$, of $f$, each one
multiplied by a polynomial in $t$. Notice also that the degree of $t$ in the
right hand side of (3.6) can be at most $r(s-1)+k-1$.

######  Lemma .

For each positive integers $s,k,\ell$ there exist positive integers $D,\nu$
with the following property. Let $\mathbb{G}$ be a nilpotent Lie group of
nilpotency step $s$ and $p\in\mathbb{G}$. Let us assume that
$f:\mathbb{G}\to\mathbb{R}$ is smooth and $k$-polynomial with respect to
$Y_{1},\dots,Y_{\ell}\in\mathfrak{g}$. Then, there exists a polynomial
$P:\mathbb{R}^{\ell}\to\mathbb{R}$ with degree at most $\nu$, whose
coefficients depend on mixed derivatives along some directions of
$\\{Y_{1},\dots,Y_{\ell}\\}$ of order at most $D$ of $f$ at $p$, such that

$f(p\exp(t_{1}Y_{1})\dots\exp(t_{\ell}Y_{\ell}))=P(t_{1},\dots,t_{\ell}),$
(4.1)

for every $t_{1},\dots,t_{\ell}\in\mathbb{R}$.

###### Proof.

If $\ell=1$ the proof is straightforward from (3.1). Let us assume $\ell\geq
2$. Let us inductively construct the string of natural numbers
$\\{a_{0},\dots,a_{\ell-1}\\}$ as follows

$\displaystyle a_{\ell-1}:=k-1,\qquad a_{j-1}:=sa_{j}+k-1,\quad\forall 1\leq
j\leq\ell-1.$ (4.2)

Let us inductively define the string of natural numbers
$\\{\nu_{1},\dots,\nu_{\ell-1}\\}$ as follows

$\displaystyle\nu_{1}:=k-1+a_{1}(s-1),\qquad\nu_{j+1}:=\nu_{j}+k-1+a_{j+1}(s-1),\quad\forall
j=1,\dots,\ell-2,$ (4.3)

where if $\ell=2$ the second part of the previous equation does not come into
play. We claim that

$\nu=\nu(k,s,\ell):=\nu_{\ell-1}+k-1,\qquad D=D(k,s,\ell):=a_{0}.$ (4.4)

Let us prove an intermediate result. Let us fix
$\\{Y_{1},\dots,Y_{\ell}\\}\subseteq S$. We want to prove by induction on
$j=1,\dots,\ell-1$ the following statement. For every $0\leq m\leq a_{j}$, and
for every $X_{i_{1}},\dots,X_{i_{m}}\in\\{Y_{1},\dots,Y_{\ell}\\}$ there
exists a polynomial $P_{j}:\mathbb{R}^{j}\to\mathbb{R}$ with degree at most
$\nu_{j}$ whose coefficients depend on the mixed derivatives of $f$ at $p$ of
order at most $a_{0}$ along some directions of $\\{Y_{1},\dots,Y_{\ell}\\}$,
such that for every $t_{1},\dots,t_{j}\in\mathbb{R}$, the following equality
holds

$(X_{i_{1}}\dots
X_{i_{m}}f)(p\exp(t_{1}Y_{1})\dots\exp(t_{j}Y_{j}))=P_{j}(t_{1},\dots,t_{j}).$
(4.5)

Let us go through the base step. Let us fix $0\leq m\leq a_{1}$ and
$X_{i_{1}},\dots,X_{i_{m}}\in\\{Y_{1},\dots,Y_{\ell}\\}$, and we want to write
$(X_{i_{1}}\dots X_{i_{m}}f)(p\exp(t_{1}Y_{1}))$. In order to do this we use
(3.6) in the case $\mathbb{G}$ is a nilpotent group, see Section 4. Indeed,
from the right hand side of (3.6) and the reasoning in Section 4, we get that
$(X_{i_{1}}\dots X_{i_{m}}f)(p\exp(t_{1}Y_{1}))$ is a sum of at most
$sm+k-1\leq sa_{1}+k-1=a_{0}$ mixed derivatives of $f$ evaluated at $p$ in
some directions of the set $\\{Y_{1},\dots,Y_{\ell}\\}$ each one multiplied by
a polynomial in $t_{1}$. Moreover the degree of $t_{1}$ is at most
$k-1+m(s-1)\leq k-1+a_{1}(s-1)=\nu_{1}$, see again the right hand side in
(3.6) and the reasoning in Section 4. Thus we have proved (4.5) in the base
step $j=1$. Let us now proceed with the induction and let us assume the
statement is true for some $j=1,\dots,\ell-2$. We want to prove it true for
$j+1$. Thus, let us fix $0\leq m\leq a_{j+1}$ and
$X_{i_{1}},\dots,X_{i_{m}}\in\\{Y_{1},\dots,Y_{\ell}\\}$, and we want to write

$(X_{i_{1}}\dots
X_{i_{m}}f)(\underbrace{p\exp(t_{1}Y_{1})\dots\exp(t_{j}Y_{j})}_{p^{\prime}}\exp(t_{j+1}Y_{j+1})).$

We can thus apply (3.6) with $p^{\prime}$ instead of $q$. From the right hand
side of (3.6) and the reasoning in Section 4 we get that $(X_{i_{1}}\dots
X_{i_{m}}f)(p^{\prime}\exp(t_{j+1}Y_{j+1}))$ is a sum of at most $sm+k-1\leq
sa_{j+1}+k-1=a_{j}$, see (4.2), mixed derivatives of $f$ in some directions of
$\\{Y_{1},\dots,Y_{\ell}\\}$ evaluated at
$p^{\prime}=p\exp(t_{1}Y_{1})\dots\exp(t_{j}Y_{j})$, each one multiplied by a
polynomial in $t_{j+1}$. Moreover the degree of $t_{j+1}$ is at most
$k-1+m(s-1)\leq k-1+a_{j+1}(s-1)$, see again the right hand side of (3.6) and
the reasoning in Section 4. By the inductive hypothesis, every mixed
derivative of order at most $a_{j}$ of $f$ in some directions of
$\\{Y_{1},\dots,Y_{\ell}\\}$ evaluated at
$p^{\prime}=p\exp(t_{1}Y_{1})\dots\exp(t_{j}Y_{j})$ is a polynomial in
$(t_{1},\dots,t_{j})$ with degree at most $\nu_{j}$ whose coefficients depend
on mixed derivatives of $f$ at $p$ of order at most $a_{0}$ along some
directions of $\\{Y_{1},\dots,Y_{\ell}\\}$. Then, for all
$t_{1},\dots,t_{j+1}\in\mathbb{R}$ we can write

$(X_{i_{1}}\dots
X_{i_{m}}f)(\underbrace{p\exp(t_{1}Y_{1})\dots\exp(t_{j}Y_{j})}_{p^{\prime}}\exp(t_{j+1}Y_{j+1}))=P_{j+1}(t_{1},\dots,t_{j+1}),$

where $P_{j+1}$ is a polynomial with degree at most
$\nu_{j}+k-1+a_{j+1}(s-1)=\nu_{j+1}$, see (4.3), and whose coefficients depend
on mixed derivatives of $f$ at $p$ of order at most $a_{0}$ along some
directions of $\\{Y_{1},\dots,Y_{\ell}\\}$. Thus this concludes the proof of
(4.5) by induction.

Let us now complete the proof. By (3.1) we can write, for every
$t_{1},\dots,t_{\ell}\in\mathbb{R}$, that

$f(\underbrace{p\exp(t_{1}Y_{1})\dots\exp(t_{\ell-1}Y_{\ell-1})}_{p^{\prime\prime}}\exp(t_{\ell}Y_{\ell}))=\sum_{i=0}^{k-1}\frac{t_{\ell}^{i}}{i!}(Y_{\ell}^{i}f)(p^{\prime\prime}).$

By the induction before, see (4.5), and since $a_{\ell-1}=k-1$, see (4.2), we
conclude that, for every $0\leq i\leq k-1$,
$(Y^{i}_{\ell}f)(p^{\prime\prime})=(Y^{i}_{\ell}f)(p\exp(t_{1}Y_{1})\dots\exp(t_{\ell-1}Y_{\ell-1}))$
is a polynomial in $(t_{1},\dots,t_{\ell-1})$ with degree at most
$\nu_{\ell-1}$ whose coefficients depend on mixed derivatives of $f$ at $p$ of
order at most $a_{0}$ along some directions of $\\{Y_{1},\dots,Y_{\ell}\\}$.
Thus we conclude (4.1) where $P$ is a polynomial with degree at most
$\nu:=\nu_{\ell-1}+k-1$ whose coefficients depend on mixed derivatives of $f$
at $p$ of order at most $D=a_{0}$ along some directions of
$\\{Y_{1},\dots,Y_{\ell}\\}$. Thus we obtained the result with the constants
chosen in (4.4). ∎

###  4.1. Proof of Theorem 1.3

Before starting the proof of Theorem 1.3 we give a refinement of Section 3.3
when $\mathbb{G}$ is a Carnot group.

######  Lemma .

Let $\mathbb{G}$ be a Carnot group of topological dimension $n$, let $S$ be a
basis of its first layer, and let $d_{\mathbb{G}}$ be the subRiemannian
distance associated to $S$. Then there exist $\widetilde{C}$, and
$X_{1},\dots,X_{2n}\in S$ such that given two arbitrary points
$p,q\in\mathbb{G}$ there exist $t_{1},\dots,t_{2n}\in\mathbb{R}$ such that

$q=p\exp(t_{1}X_{1})\dots\exp(t_{2n}X_{2n}),$

and

$d_{\mathbb{G}}(p,q)\geq\widetilde{C}(\left\lvert
t_{1}\right\rvert+\dots+\left\lvert t_{2n}\right\rvert).$ (4.6)

###### Proof.

Without loss of generality we can prove the statement for $p=e$. From Section
3.3, by using the same notation therein, up to eventually reduce $\hat{V}$
there exists $\varepsilon>0$ such that
$(\hat{\psi}_{e})_{|_{\hat{V}}}:\hat{V}\to B_{\varepsilon}(e)$ is a
diffeomorphism, where $\hat{V}\subseteq(0,1)^{n}$, and $B_{\varepsilon}(e)$ is
the open ball, in the metric $d_{\mathbb{G}}$, of radius $\varepsilon$ and
centre the identity $e$. Thus we have shown that there exist $\varepsilon>0$
and $X_{1},\dots,X_{2n}\in S$ such that for all $q\in B_{\varepsilon}(e)$
there exists at least one $(t_{1},\dots,t_{n})\in\hat{V}\subseteq(0,1)^{n}$
such that

$\hat{\psi}_{e}(t_{1},\dots,t_{n})=q.$

We now claim that the Lemma holds true with $\widetilde{C}:=\varepsilon/(4n)$,
and with the $2n$ vector fields $X_{1},\dots,X_{2n}\in S$ found above. Let us
define the norm $\|g\|_{\mathbb{G}}:=d_{\mathbb{G}}(g,e)$. Let us take an
arbitrary $q\in\mathbb{G}$ and consider
$q^{\prime}:=\delta_{\varepsilon/(2\|q\|_{\mathbb{G}})}q$, where $\varepsilon$
is defined above and $\delta_{\lambda}$ is the dilation of factor $\lambda$ on
$\mathbb{G}$. Since $q^{\prime}\in B_{\varepsilon}(e)$ we have that there
exists $(t_{1},\dots,t_{n})\in(0,1)^{n}$ such that

$\exp(t_{1}X_{1})\dots\exp(t_{n}X_{n})\exp(-\hat{s}_{n}X_{n+1})\dots\exp(-\hat{s}_{1}X_{2n})=:\hat{\psi}_{e}(t_{1},\dots,t_{n})=q^{\prime},$

where $\hat{s}\in(0,1)^{n}$ is as in Section 3.3. Thus if we dilate by a
factor $k:=2\|q\|_{\mathbb{G}}/\varepsilon$ the previous equality we get,
since $X_{1},\dots,X_{n}$ are in the first layer

$\exp(kt_{1}X_{1})\dots\exp(kt_{n}X_{n})\exp(-k\hat{s}_{n}X_{n+1})\dots\exp(-k\hat{s}_{1}X_{2n})=q.$

Since $(t_{1},\dots,t_{n})\in(0,1)^{n}$ and
$(\hat{s}_{1},\dots,\hat{s}_{n})\in(0,1)^{n}$ we get that

$\left\lvert kt_{1}\right\rvert+\dots+\left\lvert
kt_{n}\right\rvert+\left\lvert k\hat{s}_{1}\right\rvert+\dots+\left\lvert
k\hat{s}_{n}\right\rvert\leq\frac{4n}{\varepsilon}\|q\|_{\mathbb{G}},$

and thus we conclude that every point $q\in\mathbb{G}$ can be connected to the
identity $e\in\mathbb{G}$ by means of a concatenation of at most $2n$
horizontal lines in the directions $X_{1},\dots,X_{2n}$, and moreover (4.6)
holds with $\widetilde{C}=\varepsilon/(4n)$. This concludes the proof. ∎

We now show that on an arbitrary Carnot group a $k$-polynomial distribution
with respect to a basis of the first layer of the Lie algebra is polynomial in
exponential chart. The forthcoming proposition is the first step to prove
Theorem 1.3. Indeed the forthcoming Section 4.1 is precisely Theorem 1.3 in
the setting of Carnot groups. Then we will prove Theorem 4.1, which is Theorem
1.3 for simply connected nilpotent groups, and eventually we conclude this
section by giving the proof of Theorem 1.3.

######  Proposition .

Let $k,n,s$ be positive integers. Then there exists a constant $\nu$ such that
the following holds. Let $\mathbb{G}$ be an arbitrary Carnot group of step $s$
and topological dimension $n$, and assume that
$f\in\mathcal{D}^{\prime}(\mathbb{G})$ is a distribution that is
$k$-polynomial with respect to a basis $S$ of the first layer of
$\mathfrak{g}$. Then $f$ is a polynomial in exponential chart, see Section 4,
with degree at most $\nu$.

###### Proof.

We stress a little abuse of notation in this proof. For a function
$g:\mathbb{G}\to\mathbb{R}$ we will write without making a distinction between
$g$ and $g\circ\exp$, since when $\mathbb{G}$ is a Carnot group $\exp$ is a
global analytic diffeomorphism. In other words we will identify
$\mathbb{G}\equiv\mathfrak{g}\equiv\mathbb{R}^{n}$ by means of the exponential
map $\exp$ and a choice for an adapted basis of $\mathfrak{g}$ that extends
$S$.

From Theorem 1.1 we get that $f$ is represented by an analytic function. Let
us fix $\Omega$ an open neighbourhood of the identity of $\mathbb{G}$ on which
$f$ coincides with his Taylor expansion. Thus in particular we have the
following equality in the pointwise sense

$f(t_{1},\dots,t_{n})=\sum_{d=0}^{\infty}p_{d}(t_{1},\dots,t_{n}),\qquad\text{for
all $(t_{1},\dots,t_{n})\in\Omega$},$ (4.7)

where $p_{d}(t_{1},\dots,t_{n})$ is the polynomial of the Taylor expansion
that is $\delta_{\lambda}$-homogeneous with degree $d$. We claim that for
every $d\geq 0$ the polynomial $p_{d}$ is $k$-polynomial with respect to $S$.
Indeed, let us prove this statement by induction on $d$. Clearly if $d=0$ the
conclusion is proved since constant functions are always $k$-polynomial with
respect to $S$. Let us now assume that $p_{i}$ is $k$-polynomial with respect
to $S$ for $i=0,\dots,d$. We want to prove that $p_{d+1}$ is $k$-polynomial
with respect to $S$. First of all, by linearity of the condition of being
$S$-polynomial with degree at most $k$, we get that

$g_{d}:=f-\sum_{i=0}^{d}p_{i}\quad\text{is $k$-polynomial with respect to
$S$.}$

Since $g_{d}$ is $k$-polynomial with respect to $S$ we also get that
$g_{d}\circ\delta_{\lambda}$ is $k$-polynomial with respect to $S$ for every
$\lambda>0$. This last assertion comes from the iteration of the equality

$X(g_{d}\circ\delta_{\lambda})(p)=\frac{\operatorname{d}}{\operatorname{d}\varepsilon}_{|_{\varepsilon=0}}(g_{d}\circ\delta_{\lambda})(p\exp(\varepsilon
X))=\lambda(Xg_{d})(\delta_{\lambda}(p)),\quad\text{for all $S$, $\lambda>0$,
$p\in\mathbb{G}$}.$

From (4.7) we get that

$\frac{g_{d}\circ\delta_{\lambda}}{\lambda^{d+1}}=p_{d+1}+\lambda\left(\sum_{i=d+2}^{+\infty}\lambda^{i-d-2}p_{i}\right)=:p_{d+1}+\lambda
R_{d+1}^{\lambda},\qquad\text{on $\delta_{\lambda^{-1}}\Omega$, for all
$\lambda>0$}.$

Since $g_{d}\circ\delta_{\lambda}$ is a $k$-polynomial function with respect
to $S$ the same is true for $(g_{d}\circ\delta_{\lambda})/\lambda^{d+1}$, and
thus for $p_{d+1}+\lambda R_{d+1}^{\lambda}$, on
$\delta_{\lambda^{-1}}\Omega$, for every $\lambda>0$, since the previous
equality holds. Let us fix $\Omega^{\prime}$ an arbitrary open bounded set of
$\mathbb{G}$. Since $\Omega^{\prime}$ is bounded there exists $\lambda_{0}$
such that $\Omega^{\prime}\subseteq\delta_{\lambda^{-1}}\Omega$ for all
$\lambda\leq\lambda_{0}$. Thus, for every $\lambda\leq\lambda_{0}$, the
function $p_{d+1}+\lambda R_{d+1}^{\lambda}$ is a $k$-polynomial function with
respect to $S$ on $\Omega^{\prime}$. Since $p_{d+1}+\lambda R_{d+1}^{\lambda}$
converges pointwise to $p_{d+1}$ on $\Omega^{\prime}$ as $\lambda\to 0$, we
can apply Section 3.1 to infer that $p_{d+1}$ is $k$-polynomial with respect
to $S$ on $\Omega^{\prime}$. Since $\Omega^{\prime}$ is arbitrary we get that
$p_{d+1}$ is $k$-polynomial with respect to $S$ on the entire $\mathbb{G}$ and
thus the induction is complete.

Let us now prove an independent result that will lead to the conclusion of the
proof. Let us prove that there exists a constant $\nu$, which depends on
$k,s,n$, such that if $f$ is an arbitrary smooth function that is
$k$-polynomial with respect to $S$ on $\mathbb{G}$, then
$f(p)=O(\|p\|_{\mathbb{G}}^{\nu})$ as $\|p\|_{\mathbb{G}}\to+\infty$, where
$\|\cdot\|_{\mathbb{G}}$ is the homogeneous norm associated to the
subRiemannian distance $d_{\mathbb{G}}$ induced by $S$, i.e.,
$\|p\|_{\mathbb{G}}:=d_{\mathbb{G}}(p,e)$, where $e$ is the identity of the
group.

Indeed, as a direct application of Section 4.1, we first get that there exist
$\widetilde{C}$, and some $X_{1},\dots,X_{2n}\in S$ such that for every point
$p\in\mathbb{G}$ we can write the following equality

$p=\exp(t_{1}X_{1})\dots\exp(t_{2n}X_{2n}),$

for some $t_{1},\dots,t_{2n}\in\mathbb{R}$, and moreover the following
inequality holds

$\|p\|_{\mathbb{G}}\geq\widetilde{C}(\left\lvert
t_{1}\right\rvert+\dots+\left\lvert t_{2n}\right\rvert).$

Moreover, as a direct application of Section 4, there exist a constant $\nu$,
which depends on $s,k,n$, and a constant $C$, which may depend also on $f$,
such that

$\left\lvert f(p)\right\rvert=\left\lvert
f(\exp(t_{1}X_{1})\dots\exp(t_{2n}X_{2n}))\right\rvert\leq C(1+\left\lvert
t_{1}\right\rvert+\dots+\left\lvert t_{2n}\right\rvert)^{\nu}\leq
C(1+\|p\|_{\mathbb{G}}/\widetilde{C})^{\nu}.$

Thus $f(p)=O(\|p\|_{\mathbb{G}}^{\nu})$, as $\|p\|_{\mathbb{G}}\to+\infty$. In
order to conclude the proof let us prove that in the sum (4.7), $p_{d}\equiv
0$ for all $d\geq\nu+1$. Let us fix $d\geq\nu+1$ and let us suppose by
contradiction that
$\max_{\\{\|x\|_{\mathbb{G}}=1\\}}p_{d}(x)=p_{d}(\overline{x})=m>0$, for some
$\overline{x}\in\mathbb{G}$ with $\|\overline{x}\|_{\mathbb{G}}=1$. Hence for
an arbitrary $\lambda>0$, since $p_{d}$ is a $\delta_{\lambda}$-homogeneous
polynomial of degree $d$, we have
$p_{d}(\delta_{\lambda}\overline{x})=\lambda^{d}p_{d}(\overline{x})=m\lambda^{d}$.
Hence, being $d\geq\nu+1$ and $m>0$, the previous inequality is a
contradiction with the fact that $p_{d}(x)=O(\|x\|_{\mathbb{G}}^{\nu})$ as
$\|x\|_{\mathbb{G}}\to+\infty$, which holds true since we proved that $p_{d}$
is $k$-polynomial with respect to $S$, and since we obtained the polynomial-
growth bound above. Thus we conclude that $p_{d}\equiv 0$ for all
$d\geq\nu+1$, and then by analytic continuation $\eqref{eqn:SERIES}$ holds
everywhere on $\mathbb{G}$, where now the sum is taken up to $\nu$. Then $f$
is a polynomial in exponential chart. Moreover, its homogeneous degree, and
thus its degree, is bounded above by $\nu$. ∎

######  Theorem 4.1.

Let $\mathbb{G}$ be a simply connected nilpotent group of nilpotency step $s$,
and let $S\subseteq\mathfrak{g}$ be a subset that Lie generates
$\mathfrak{g}$. If $f\in\mathcal{D}^{\prime}(\mathbb{G})$ is a distribution
that is $S$-polynomial then $f$ is a polynomial in exponential chart, see
Section 4.

###### Proof.

We stress a little abuse of notation in this proof. For a function
$g:\mathbb{G}\to\mathbb{R}$ we will write without making a distinction between
$g$ and $g\circ\exp$, since when $\mathbb{G}$ is a simply connected nilpotent
group, $\exp$ is a global analytic diffeomorphism. In other words we identify
$\mathbb{G}\equiv\mathfrak{g}\equiv\mathbb{R}^{n}$ by means of $\exp$ and a
choice for a basis of $\mathfrak{g}$. Up to taking a subset of $S$ that is
finite and still Lie generates $\mathfrak{g}$, we may assume that $S$ is
finite, namely $S=\\{X_{1},\dots,X_{m}\\}$ for some
$X_{1},\dots,X_{m}\in\mathfrak{g}$. Since now $S$ is finite, there exists
$k\in\mathbb{N}$ such that $f$ is $k$-polynomial with respect to $S$.

From Theorem 1.1 we get that $f$ is represented by an analytic function. Let
us consider the free Lie algebra $\mathfrak{f}_{m,s}$ of step $s$ and with $m$
generators $\\{X_{1}^{\prime},\dots,X_{m}^{\prime}\\}$ introduced in Section
2.3. By item (iii) of Section 2.3 there exists a Lie algebra homomorphism
$\varphi:\mathfrak{f}_{m,s}\to\mathfrak{g}$ such that
$\varphi(X^{\prime}_{i})=X_{i}$ for every $1\leq i\leq m$. We claim that
$f\circ\varphi$ is smooth and $k$-polynomial with respect to
$\\{X_{1}^{\prime},\dots,X_{m}^{\prime}\\}$. This latter assertion is true
since, first of all $f\circ\varphi$ is smooth since $f$ is analytic and
$\varphi$ is linear, and second because, from the fact that
$\varphi(X^{\prime}_{i})=X_{i}$ for every $1\leq i\leq m$, we conclude that

$X^{\prime}_{i}(f\circ\varphi)=X_{i}f\circ\varphi,\qquad\forall 1\leq i\leq
m,$

and thus iterating

$(X^{\prime}_{i})^{k}(f\circ\varphi)=X_{i}^{k}f\circ\varphi\equiv
0,\qquad\forall 1\leq i\leq m.$

Since $\mathrm{span}\\{X_{1}^{\prime},\dots,X_{m}^{\prime}\\}$ is the first
layer of a stratification of $\mathfrak{f}_{m,s}$, we can apply Section 4.1
and conclude that $f\circ\varphi$ is a polynomial in exponential chart with
degree at most $\delta^{\prime}$, where $\delta^{\prime}$ depends on $k,m,s$
since the topological dimension of $\mathfrak{f}_{m,s}$ is bounded above by a
function of $m$ and $s$. Since $\\{X_{1},\dots,X_{m}\\}$ Lie generates
$\mathfrak{g}$, since $\varphi(X^{\prime}_{i})=X_{i}$ for every $1\leq i\leq
m$, and since $\varphi$ is a Lie algebra homomorphism, we get that $\varphi$
is surjective. Hence there exists a linear map
$\varphi^{-1}:\mathfrak{g}\to\mathfrak{f}_{m,s}$ such that
$\varphi\circ\varphi^{-1}=\mathrm{id}_{|_{\mathfrak{g}}}$. Thus
$f=(f\circ\varphi)\circ\varphi^{-1}$ is a polynomial in exponential chart
since it is the composition of a polynomial in exponential chart with a linear
map. Notice that the degree of $f$ in exponential chart is at most
$\delta^{\prime}$ since $\varphi$ is linear. ∎

###### Proof of Theorem 1.3.

From [Var84, Theorem 3.6.1] we deduce that there exists a unique simply
connected nilpotent Lie group $\mathbb{G}^{\prime}$ with Lie algebra
$\mathfrak{g}$, and $\mathbb{G}$ is the quotient of $\mathbb{G}^{\prime}$ with
one central discrete subgroup $\Gamma$ of $\mathbb{G}^{\prime}$. Let
$\pi:\mathbb{G}^{\prime}\to\mathbb{G}^{\prime}\mathrel{/}\Gamma\simeq\mathbb{G}$
be the projection map, which is open. Then
$\pi_{*}:\mathfrak{g}\to\mathfrak{g}$ is surjective, and hence a bijection. We
claim that the map $f\circ\pi$ is $(\pi_{*})^{-1}(S)$-polynomial in
$\mathbb{G}^{\prime}$. Indeed, for every $X\in\mathfrak{(}\pi_{*})^{-1}(S)$ we
have

$X(f\circ\pi)=(\pi_{*}X)f\circ\pi,$ (4.8)

and thus iterating and using that $f$ is $S$-polynomial we get the sought
claim. Hence $f\circ\pi\circ\exp_{\mathbb{G}^{\prime}}$ is a polynomial,
according to Theorem 4.1, since $(\pi_{*})^{-1}(S)$ Lie generates
$\mathfrak{g}$ as well. But since
$f\circ\pi\circ\exp_{\mathbb{G}^{\prime}}=f\circ\exp_{\mathbb{G}}\circ\pi_{*}$,
and since $\pi_{*}$ is a bijection, we get that $f\circ\exp_{\mathbb{G}}$ is a
polynomial as well, and then we are done. ∎

######  Remark  (The constant $\nu$ in Section 4.1).

We stress that from the proof of Section 4.1 we infer that the homogeneous
degree of $f$, and thus also the degree of $f$, in the exponential chart is at
most $\nu(k,s,2n)$, where $\nu$ is explicitly provided in the proof of Section
4, see (4.4). Thus the constant $\nu$ in Section 4.1 can be taken to be
$\nu(k,s,2n)$. This in particular gives, in case $\mathbb{G}$ is connected and
nilpotent, an explicit bound on the degree of the polynomial in exponential
chart that represents a distribution $f$ that is $k$-polynomial with respect
to a Lie generating $S$, see the proofs of Theorem 4.1 and Theorem 1.3.

######  Remark  (Relaxation of the hypotheses in Theorem 1.3).

The hypothesis of $f\in\mathcal{D}^{\prime}(\mathbb{G})$ being polynomial with
respect to $S$ in Theorem 1.3 can be relaxed to the following one: for every
$X\in S$ there exists $k\in\mathbb{N}$ and a polynomial in exponential chart
$g$ such that $X^{k}f=g$ in the distributional sense on $\mathbb{G}$. Indeed,
if this is the case, there exists a finite subset
$\\{X_{1},\dots,X_{m}\\}\subseteq S$ that Lie generates $\mathfrak{g}$ and
such that $X_{i}^{k_{i}}f=g_{i}$ for $1\leq i\leq m$, where
$k_{i}\in\mathbb{N}$ and $g_{i}$ are polynomials in exponential chart. Thus,
taking Section 5.2 into account, there exists a sufficiently large
$k\in\mathbb{N}$ such that $f$ is $k$-polynomial with respect to
$\\{X_{1},\dots,X_{m}\\}$, and then we can use Theorem 1.3.

######  Remark  ($S$-polynomial implies $\mathfrak{g}$-polynomial).

Let us further notice the following non-obvious fact, which is a consequence
of Theorem 1.3 and Section 5.2. If $\mathbb{G}$ is a connected nilpotent Lie
group and $S$ Lie generates $\mathfrak{g}$, then if
$f\in\mathcal{D}^{\prime}(\mathbb{G})$ is $S$-polynomial, $f$ is
$\mathfrak{g}$-polynomial, with a degree of polynomiality $k$ uniform with
respect to $X\in\mathfrak{g}$, but that may eventually depend on $f$. This
latter assertion is true since if $f$ is $S$-polynomial, then Theorem 1.3
tells us that $f$ is a polynomial in exponential chart and thus we can apply
Section 5.2.

######  Remark .

The example in (A.1) shows that our Theorem 1.3 is sharp in the class of
connected nilpotent Lie groups.

######  Remark .

In case $\mathbb{G}$ is simply connected and nilpotent, $\exp$ is a global
analytic diffeomorphism and so the class of polynomial distributions à la
Leibman on $\mathbb{G}$ coincides, up to identifying
$\mathbb{G}\equiv\mathfrak{g}$, with the class of polynomials on
$\mathbb{R}^{n}$, where $n$ is the topological dimension of $\mathbb{G}$. If
$\mathbb{G}$ is nilpotent but not necessarily simply connected it might happen
that the class of polynomial trivializes: for example it is readily seen that
the polynomials on a torus $\mathbb{S}^{1}\times\mathbb{S}^{1}$ are just the
constant functions.

##  5\. Relations between various notions of polynomial

In this section we shall prove Theorem 1.2 and Section 1.

###  5.1. Proof of Theorem 1.2

Let us now prove Theorem 1.2, that is, let us prove that our definition of
polynomial distribution à la Leibman, see Section 3.1, is consistent with our
definition of polynomial distribution, see Section 3.1, on every connected Lie
group. We first prove this equivalence on smooth functions, and then conclude
by using convolutions with smooth kernels.

######  Proposition .

Let $\mathbb{G}$ be a connected Lie group and let
$\varphi:\mathbb{G}\to\mathbb{R}$ be a smooth function. Then $\varphi$ is a
polynomial with degree at most $d$ à la Leibman, see (3.5), if and only if for
every $X_{1},\dots,X_{d+1}\in\mathfrak{g}$ we have

$X_{1}\dots X_{d+1}\varphi\equiv 0,\qquad\text{on $\mathbb{G}$}.$ (5.1)

###### Proof.

First, let us prove that (3.5) implies that whenever
$X_{1},\dots,X_{d+1}\in\mathfrak{g}$ then
$X_{1}\dots X_{d+1}\varphi\equiv 0$ on $\mathbb{G}$. In order to show this,
let us notice that, defined $g_{d+1}(t):=\exp(tX_{d+1})$, then
$t^{-1}D_{g_{d+1}(t)}\varphi$ converges pointwise on $\mathbb{G}$ to
$X_{d+1}\varphi$ as $t\to 0$. Moreover, for every $g\in\mathbb{G}$, the
operator $D_{g}$ is continuous with respect to the pointwise convergence of
functions, i.e., if $\varphi_{n}\to\varphi$ pointwise on $\mathbb{G}$ as
$n\to+\infty$, then $D_{g}\varphi_{n}\to D_{g}\varphi$ pointwise on
$\mathbb{G}$ as $n\to+\infty$, for every $g\in\mathbb{G}$. By also using that
$D_{g}(\lambda\varphi)=\lambda D_{g}(\varphi)$ for every $g\in\mathbb{G}$ and
$\lambda\in\mathbb{R}$, and putting $g_{d+1}(t)$ in (3.5) we get

$D_{g_{1}}\dots D_{g_{d}}(t^{-1}D_{g_{d+1}(t)}\varphi)\equiv 0,\quad\text{for
all $t>0$ on $\mathbb{G}$}\Rightarrow D_{g_{1}}\dots
D_{g_{d}}X_{d+1}\varphi\equiv 0,$

for every $g_{1},\dots,g_{d}\in\mathbb{G}$, where in the previous conclusion
we are taking $t\to 0$ and we are exploiting the continuity of the operators
$D_{g}$ with respect to the pointwise convergence. Now if we iterate the
argument with $X_{d+1}\varphi$ instead of $\varphi$, we obtain

$X_{1}\dots X_{d+1}\varphi\equiv 0,\qquad\text{on $\mathbb{G}$},$

which is what we wanted.

Regarding the opposite direction, let us now denote
$\mathscr{P}^{d}_{\mathcal{L}}$ the vector space of smooth functions on
$\mathbb{G}$ that are polynomials à la Leibman with degree at most $d$, see
(3.5). Let us denote $\mathscr{P}^{d}_{\mathcal{D}}$ the vector space of
smooth functions $\varphi$ such that for every
$X_{1},\dots,X_{d+1}\in\mathfrak{g}$ we have $X_{1}\dots X_{d+1}\varphi\equiv
0$ on $\mathbb{G}$. Let us now prove the following statement

$\begin{split}&\text{for every $d\in\\{0,1,\dots\\}$,
$\mathscr{P}^{d}_{\mathcal{L}}=\mathscr{P}^{d}_{\mathcal{D}}$, and for every
$\varphi\in\mathscr{P}^{d}_{\mathcal{D}}$ and every $Y\in\mathfrak{g}$}\\\
&\text{the map $x\xmapsto{\varphi\circ R_{\exp(Y)}}\varphi(x\exp(Y))$ defined
on $\mathbb{G}$ is in $\mathscr{P}^{d}_{\mathcal{D}}$,}\end{split}$ (5.2)

where we recall that $R_{g}$ stands for the right translation by
$g\in\mathbb{G}$. It is readily seen that the first part of the previous
statement proves the proposition. Let us prove (5.2) by induction on $d$.

If $d=0$, one readily sees that
$\mathscr{P}^{0}_{\mathcal{L}}=\mathscr{P}^{0}_{\mathcal{D}}$ and they agree
with the set of constant functions on $\mathbb{G}$. Thus the statement (5.2)
is verified for $d=0$. Let us now assume that (5.2) is true for $d-1$, with
$d\geq 1$, and let us prove it true for $d$. Let us start from the second part
of the statement (5.2). Let us fix $Y\in\mathfrak{g}$ and
$\varphi\in\mathscr{P}^{d}_{\mathcal{D}}$. We have, for $x\in\mathbb{G}$, and
$X\in\mathfrak{g}$,

$\begin{split}X(\varphi\circ
R_{\exp(Y)})(x)&=\frac{\operatorname{d}}{\operatorname{d}\varepsilon}_{|_{\varepsilon=0}}\varphi\circ
R_{\exp(Y)}(x\exp(\varepsilon X))=\\\
&=\frac{\operatorname{d}}{\operatorname{d}\varepsilon}_{|_{\varepsilon=0}}\varphi(x\exp(Y)\exp(-Y)\exp(\varepsilon
X)\exp(Y))=\\\
&=\left(\mathrm{Ad}_{\exp(-Y)}(X)\varphi\right)(x\exp(Y))=\mathrm{Ad}_{\exp(-Y)}(X)\varphi\circ
R_{\exp(Y)}(x).\end{split}$ (5.3)

Since $\varphi\in\mathscr{P}^{d}_{\mathcal{D}}$ we get that
$\mathrm{Ad}_{\exp(-Y)}(X)\varphi$ is in $\mathscr{P}^{d-1}_{\mathcal{D}}$ by
the definition of $\mathscr{P}^{d}_{\mathcal{D}}$. Thus by the inductive
hypothesis in the second part of the statement (5.2) we get that
$\mathrm{Ad}_{\exp(-Y)}(X)\varphi\circ
R_{\exp(Y)}\in\mathscr{P}^{d-1}_{\mathcal{D}}$. Thus by (5.3) we get that
$X(\varphi\circ R_{\exp(Y)})\in\mathscr{P}^{d-1}_{\mathcal{D}}$ and then, by
arbitrariness of $X$, we conclude $\varphi\circ
R_{\exp(Y)}\in\mathscr{P}^{d}_{\mathcal{D}}$. This conclude the induction for
the second part of (5.2). Let us now complete the induction by proving the
first part of (5.2) with $d\geq 1$, assuming it is true for $d-1$.

First of all, the first argument in the proof of this proposition shows that
$\mathscr{P}_{\mathcal{L}}^{d}\subseteq\mathscr{P}_{\mathcal{D}}^{d}$. Let us
prove $\mathscr{P}_{\mathcal{L}}^{d}\supseteq\mathscr{P}_{\mathcal{D}}^{d}$.
Take $\varphi\in\mathscr{P}_{\mathcal{D}}^{d}$. From Theorem 1.1 we conclude
that $\varphi$ is analytic. In order to prove that
$\varphi\in\mathscr{P}^{d}_{\mathcal{L}}$ we claim that it suffices to prove
that

$D_{g_{1}}\cdots D_{g_{d}}D_{\exp(Y)}\varphi\equiv 0,\qquad\text{on
$\mathbb{G}$},$ (5.4)

for every $g_{1},\dots,g_{d}\in\mathbb{G}$ and every $Y\in\mathfrak{g}$.
Indeed, the map $(g_{1},\dots,g_{d},g_{d+1},g)\mapsto(D_{g_{1}}\dots
D_{g_{d}}D_{g_{d+1}}\varphi)(g)$ is analytic from $\mathbb{G}^{d+2}$ to
$\mathbb{R}$, since $\varphi$ and the group operation $\cdot$ are analytic,
recall (3.3). Thus, since the set
$\\{(g_{1},\dots,g_{d},g,\exp(Y)):g_{1},\dots,g_{d},g\in\mathbb{G},Y\in\mathfrak{g}\\}$
contains an open neighbourhood of the identity in $\mathbb{G}^{d+2}$, if we
show (5.4) we are done by analytic continuation. Let us prove (5.4).

Since $\varphi\in\mathscr{P}^{d}_{\mathcal{D}}$, we get that
$Y\varphi\in\mathscr{P}^{d-1}_{\mathcal{D}}$. We also have

$D_{\exp(Y)}\varphi(x)=\varphi(x\exp(Y))-\varphi(x)=\int_{0}^{1}(Y\varphi)(x\exp(tY))\operatorname{d}t,\qquad\text{for
all $x\in\mathbb{G}$},$ (5.5)

and $(Y\varphi)\circ R_{\exp(tY)}\in\mathscr{P}^{d-1}_{\mathcal{D}}$ for every
$t\in[0,1]$, since $Y\varphi\in\mathscr{P}^{d-1}_{\mathcal{D}}$ and since the
second part of (5.2), which we already proved, holds. Thus, by the inductive
hypothesis, $(Y\varphi)\circ R_{\exp(tY)}\in\mathscr{P}^{d-1}_{\mathcal{L}}$
for every $t\in[0,1]$, and since $\mathscr{P}^{d-1}_{\mathcal{L}}$ is a vector
space closed under pointwise convergence, we finally get that
$D_{\exp(Y)}\varphi\in\mathscr{P}^{d-1}_{\mathcal{L}}$ by exploiting (5.5). By
the definition of $\mathscr{P}^{d-1}_{\mathcal{L}}$ the latter conclusion
implies (5.4), and thus the induction, and the proof, are concluded. ∎

###### Proof of Theorem 1.2.

Let $f$ be a distribution such that there exists $d\in\mathbb{N}$ for which
$X_{1}\dots X_{d+1}f\equiv 0$ in the sense of distributions on $\mathbb{G}$
whenever $X_{1},\dots,X_{d+1}\in\mathfrak{g}$. In particular, the distribution
$f$ is $\mathfrak{g}$-polynomial with degree at most $d+1$ and thus, from
Theorem 1.1, we deduce that it is represented by an analytic function. Since
in particular $f$ is represented by a smooth function we can thus use Section
5.1 to obtain that $f$ is a polynomial distribution à la Leibman with degree
at most $d$.

Viceversa, let $f$ be a distribution that is polynomial à la Leibman with
degree at most $d$. Let $\\{\varphi_{n}\\}_{n\in\mathbb{N}}$ be an approximate
identity. Then, iteratively applying (2.25), the convolution $\varphi_{n}\ast
f$ is a smooth function that is polynomial à la Leibman with degree at most
$d$. Thus, from Section 5.1, we get that for every $n\in\mathbb{N}$ and every
$X_{1},\dots,X_{d+1}\in\mathfrak{g}$ we have

$0\equiv X_{1}\dots X_{d+1}(\varphi_{n}\ast f)=\varphi_{n}\ast X_{1}\dots
X_{d+1}f,\qquad\text{on $\mathbb{G}$},$

where in the second equality we are iteratively applying (2.13). Thus, letting
$n\to+\infty$ in the previous equality, since
$\\{\varphi_{n}\\}_{n\in\mathbb{N}}$ is an approximate identity we get that
$X_{1}\dots X_{d+1}f\equiv 0$ in the sense of distributions on $\mathbb{G}$
whenever $X_{1},\dots,X_{d+1}\in\mathfrak{g}$, concluding the proof of the
equivalence.

The fact that $f$ is represented by an analytic function and the conclusion
about the finite dimension of the vector space of the polynomial distributions
with degree at most $d\in\mathbb{N}$ à la Leibman is now a direct consequence
of Theorem 1.1.

Let us prove the final part of the statement of Theorem 1.2. Let us recall
that the lower central series is defined inductively as follows:
$\mathfrak{g}_{0}:=\mathfrak{g}$, and
$\mathfrak{g}_{k}:=[\mathfrak{g},\mathfrak{g}_{k-1}]$ for every $k\geq 1$. We
now prove that the distributions $f$ that are polynomial à la Leibman with
degree at most $d\in\mathbb{N}$ are invariant along the directions of
$\mathfrak{g}_{d}$, namely for every $X\in\mathfrak{g}_{d}$ we have $Xf=0$ on
$\mathbb{G}$.

Indeed, for every $d\geq 0$, every element of $\mathfrak{g}_{d}$ can be
written as a linear combination of left-invariant operators that are the
composition of at least $d+1$ left-invariant vector fields. To see the latter
property we proceed by induction on $d$. The case $d=0$ is true by definition,
so let us suppose the assertion true for $d\geq 0$ and prove it for $d+1$. If
$X\in\mathfrak{g}_{d+1}=[\mathfrak{g},\mathfrak{g}_{d}]$, then for some
$n\in\mathbb{N}$, $X=\sum_{i=1}^{n}c_{i}[X_{i},Y_{i}]$, where
$X_{i}\in\mathfrak{g}$ and $Y_{i}\in\mathfrak{g}_{d}$. By the inductive step,
for every $1\leq i\leq n$ there exists $j_{i}\in\mathbb{N}$ such that for
every $1\leq k\leq j_{i}$ there exists integers $m_{i,k}\geq d+1$ and real
numbers $\beta_{i,k}$ such that
$Y_{i}=\sum_{k=1}^{j_{i}}\beta_{i,k}X_{i,k,1}\dots X_{i,k,m_{i,k}}$, with
$X_{i,k,1},\dots,X_{i,k,m_{i,k}}\in\mathfrak{g}$. Thus expanding the
commutator in the equality $X=\sum_{i=1}^{n}c_{i}[X_{i},Y_{i}]$ and by using
the previous equalities on $Y_{i}$ we conclude. As a consequence, since every
polynomial distribution à La Leibman with degree at most $d\in\mathbb{N}$ is
such that $X_{1}\dots X_{d+1}f=0$ on $\mathbb{G}$ for every
$X_{1},\dots,X_{d+1}\in\mathfrak{g}$, see Theorem 1.2, and since every
$X\in\mathfrak{g}_{d}$ can be written as a linear combination of left-
invariant operators that are composition of at least $d+1$ elements of
$\mathfrak{g}$, we conclude that $Xf=0$ for every $X\in\mathfrak{g}_{d}$.

Thus, since the nilpotent residual $\mathfrak{g}_{\infty}$ equals the
intersection $\cap_{k\in\mathbb{N}}\mathfrak{g}_{k}$, the previous reasoning
shows that every distribution $f$ that is polynomial à la Leibman of an
arbitrary degree on $\mathbb{G}$ is $\mathfrak{g}_{\infty}$-invariant, namely
$Xf=0$ for every $X\in\mathfrak{g}_{\infty}$. As a consequence we conclude
that every polynomial à la Leibman on a connected Lie group passes to the
quotient to a polynomial à la Leibman on the nilpotent group
$\mathbb{G}\mathrel{/}\mathbb{G}_{\infty}$, where $\mathbb{G}_{\infty}$ is the
closure of the unique connected (and normal, since $\mathfrak{g}_{\infty}$ is
an ideal) Lie subgroup of $\mathbb{G}$ with Lie algebra
$\mathfrak{g}_{\infty}$. ∎

######  Remark  (A variant of the first part of Theorem 1.2).

We notice that the proofs provided for the first part of Theorem 1.2 and for
Section 5.1 can be exploited, with very little modifications, to prove the
following variant of the first part of Theorem 1.2. Let $\varphi$ be a
distribution on $\mathbb{G}$, a connected Lie group with Lie algebra
$\mathfrak{g}$, and let $S\subseteq\mathfrak{g}$ be an $\mathrm{Ad}$-closed
cone, i.e., for every $X\in S$ then $tX\in S$ for every $t\in\mathbb{R}$, and
for every $X,Y\in S$ we have $\mathrm{Ad}_{\exp(X)}Y\in S$. Then (5.1) holds
for every $X_{1},\dots,X_{d+1}\in S$ if and only if (3.5) holds for every
$g_{1},\dots,g_{d+1}\in\exp S$.

###  5.2. Proof of Section 1

The following proposition shows that when $\mathbb{G}$ is a connected
nilpotent Lie group, a polynomial in exponential chart, see Section 4, is
$k$-polynomial with respect to $\mathfrak{g}$ for some $k\in\mathbb{N}$. This
is the last main step in order to obtain Section 1.

######  Proposition .

Let $\mathbb{G}$ be a connected nilpotent Lie group and let
$f\colon\mathbb{G}\to\mathbb{R}$ be a polynomial in exponential chart on
$\mathbb{G}$, see Section 4. Then there exists $k_{0}\in\mathbb{N}$ such that
for every $Y_{1},\dots,Y_{k_{0}}\in\mathfrak{g}$ we have $Y_{1}\dots
Y_{k_{0}}f\equiv 0$ on $\mathbb{G}$.

###### Proof.

Let us first prove the statement when $\mathbb{G}$ is simply connected. In the
latter case $\exp$ is a global analytic diffeomorphism, hence we will abuse a
little the notation in the proof of this case: for a function
$g:\mathbb{G}\to\mathbb{R}$ we will write without making a distinction between
$g$ and $g\circ\exp$. Let $\\{X_{1},\dots,X_{n}\\}$ be a basis of
$\mathfrak{g}$, and let $s$ be the step of nilpotency of $\mathbb{G}$. From
the definition of free-nilpotent Lie algebras we get that there exists a Lie
algebra homomorphism $\varphi:\mathfrak{f}_{n,s}\to\mathfrak{g}$ such that
$\varphi(X_{i}^{\prime})=X_{i}$ for every $1\leq i\leq n$, and we recall that
$V_{1}^{\prime}:=\mathrm{span}\\{X_{1}^{\prime},\dots,X_{n}^{\prime}\\}$ is
the first layer of a stratification of $\mathfrak{f}_{n,s}$, where
$X_{1}^{\prime},\dots,X^{\prime}_{n}$ are the generators of
$\mathfrak{f}_{n,s}$. We stress a little abuse of notation: we will denote
with $\varphi$ also the map
$\exp\circ\varphi\circ\exp^{-1}:\mathbb{F}_{n,s}\to\mathbb{G}$.

We claim that there exists $k_{0}\in\mathbb{N}$ such that for every
$Y_{1}^{\prime},\dots,Y_{k_{0}}^{\prime}\in\\{X_{1}^{\prime},\dots,X_{n}^{\prime}\\}$,
we have $Y_{1}^{\prime}\dots Y_{k_{0}}^{\prime}(f\circ\varphi)\equiv 0$ on
$\mathbb{F}_{n,s}$. Indeed, since $\mathfrak{f}_{n,s}$ is a stratified Lie
algebra with $V_{1}^{\prime}$ as a first layer, we get that, in exponential
coordinates, every $X_{i}^{\prime}$, with $1\leq i\leq n$, is an operator of
homogeneous degree $-1$; that is to say if $p$ is a polynomial in exponential
chart on $\mathbb{F}_{n,s}$ of homogeneous degree $d$, see the last part of
Section 2.3 for the definition of homogeneous degree, then

$\text{the homogeneous degree of $X_{i}^{\prime}\,p$ is less or equal than
$d-1$ for every $1\leq i\leq n$}.$ (5.6)

The latter assertion is a simple consequence of the explicit expression of
$X^{\prime}_{i}$, for every $1\leq i\leq n$, in exponential coordinates, see
[FS82, Proposition 1.26]. In conclusion, since $f$ is a polynomial in
exponential chart and $\varphi$ is a linear map, we get that $f\circ\varphi$
is a polynomial in exponential chart as well. Thus the claim is true taking
into account (5.6), and setting $k_{0}$ to be strictly greater than the
maximum of the homogeneous degrees of the monomials of $f\circ\varphi$.

Now we claim that for every $Y_{1},\dots,Y_{k_{0}}\in\\{X_{1},\dots,X_{n}\\}$,
we have $Y_{1}\dots Y_{k_{0}}f\equiv 0$ on $\mathbb{G}$. Indeed, since
$\varphi(X_{i}^{\prime})=X_{i}$ for every $1\leq i\leq n$, we conclude that
$X_{i}f\circ\varphi=X_{i}^{\prime}(f\circ\varphi)$ on $\mathbb{F}_{n,s}$ for
every $1\leq i\leq n$. Thus, iterating, we obtain that for every
$Y_{1},\dots,Y_{k_{0}}\in\\{X_{1},\dots,X_{n}\\}$ we have

$Y_{1}\dots Y_{k_{0}}f\circ\varphi=Y_{1}^{\prime}\dots
Y_{k_{0}}^{\prime}(f\circ\varphi),\qquad\text{on $\mathbb{F}_{n,s}$}.$

Thus, since $\varphi$ is surjective, the previous equality and the first claim
proven above imply the latter claim. The proof of the proposition in the case
$\mathbb{G}$ is simply connected thus follows from the latter claim taking
into account that every $Y\in\mathfrak{g}$ can be written as a linear
combination of $X_{1},\dots,X_{n}$.

Let us now deal with the general case in which $\mathbb{G}$ is connected. As
at the beginning of the proof of Theorem 1.3 we have a unique simply connected
nilpotent $\mathbb{G}^{\prime}$, with Lie algebra $\mathfrak{g}$, such that
$\mathbb{G}$ is the quotient of $\mathbb{G}^{\prime}$ with one central
discrete subgroup $\Gamma$ of $\mathbb{G}^{\prime}$. Let
$\pi:\mathbb{G}^{\prime}\to\mathbb{G}^{\prime}\mathrel{/}\Gamma\simeq\mathbb{G}$
be the projection map, and then one has that
$\pi_{*}:\mathfrak{g}\to\mathfrak{g}$ is a bijection. If
$f:\mathbb{G}\to\mathbb{R}$ is such that $f\circ\exp_{\mathbb{G}}$ is a
polynomial, then also
$f\circ\exp_{\mathbb{G}}\circ\pi_{*}=f\circ\pi\circ\exp_{\mathbb{G}^{\prime}}$
is a polynomial, since $\pi_{*}$ is bijective. Hence $f\circ\pi$ is polynomial
in exponential chart in $\mathbb{G}^{\prime}$ and we can apply the first part
of this proof to obtain that there exists $k_{0}$ such that for every
$Y_{1}^{\prime},\dots,Y_{k_{0}}^{\prime}\in\mathrm{Lie}(\mathbb{G}^{\prime})\simeq\mathfrak{g}$
we have $Y_{1}^{\prime}\dots Y_{k_{0}}^{\prime}(f\circ\pi)\equiv 0$ on
$\mathbb{G}^{\prime}$. Thus, iteratively applying (4.8), and by using that
$\pi_{*}$ is a bijection and $\pi$ is surjective we conclude that for every
$Y_{1},\dots,Y_{k_{0}}\in\mathrm{Lie}(\mathbb{G})\simeq\mathfrak{g}$ we have
$Y_{1}\dots Y_{k_{0}}f\equiv 0$ on $\mathbb{G}$, that is the sought
conclusion. ∎

We now provide the proof of Section 1, and we conclude with a remark.

###### Proof of Section 1.

(1)$\Rightarrow$(3) is a direct consequence of Theorem 1.3.
(3)$\Rightarrow$(4) is a direct consequence of Section 5.2.
(4)$\Rightarrow$(2) and (2)$\Rightarrow$(1) are trivial by definitions.
(4)$\Leftrightarrow$(5) is Theorem 1.2. ∎

######  Remark  (Comparison with the results in [KP20] and [BLU07]).

We stress that a slightly weaker statement of the equivalence of
(3)$\Leftrightarrow$(5) of Section 1 has recently appeared in [KP20]. Indeed,
in [KP20, Theorem C] the authors prove the equivalence between being a
continuous polynomial map à la Leibman and being a polynomial in exponential
chart, in the setting of simply connected nilpotent Lie groups. The proof
given there is algebraic and is completely different from ours. Let us
moreover notice that we do not ask for the continuity of $f$ in (5) of Section
1, but we work with distributions.

Let us finally stress that, without some regularity assumption on $f$, it is
not true that every polynomial map $f:\mathbb{G}\to\mathbb{R}$ à la Leibman,
see Section 3.1, is smooth. Indeed, there exist non-continuous homomorphisms,
and thus polynomial maps with degree at most $2$ à la Leibman, from
$(\mathbb{R},+)$ to $(\mathbb{R},+)$.

Let us also stress that the equivalent notions of being polynomial on
connected nilpotent Lie groups in Section 1 agree with the one given in the
Carnot setting in [BLU07, Definition 20.1.1]. We also notice that Theorem 1.3
is a sharpening of a result contained in [BLU07]. Let $\mathbb{G}$ be an
arbitrary Carnot group of step $s$ with stratification
$\mathfrak{g}=V_{1}\oplus\dots\oplus V_{s}$ and let
$S:=\\{X_{1},\dots,X_{m}\\}$ be a basis of $V_{1}$. In [BLU07, Corollary
20.1.10] it is proved that if there exists a smooth function
$f:\mathbb{G}\to\mathbb{R}$ and a natural number $d$ such that for every
$X_{1},\dots,X_{d}\in S$ we have $X_{1}\dots X_{d}f\equiv 0$, then $f$ is a
polynomial in exponential chart. Our result Theorem 1.3 improves this
criterion in the nilpotent case by only asking that a priori $f$ could be a
distribution, and without asking anything on the mixed derivatives; i.e., it
suffices that for every $1\leq i\leq m$ there exists $d_{i}$ such that
$X_{i}^{d_{i}}f\equiv 0$ in the sense of distributions on $\mathbb{G}$.

##  Appendix A Examples

We list here some explicit examples of $S$-polynomial functions in some Lie
groups. If an adapted basis $(X_{1},\dots,X_{n})$ of the Lie algebra
$\mathfrak{g}$ of a Carnot group $\mathbb{G}$ is fixed, when we say that we
work in exponential coordinates of the second kind we mean that we are
identifying a point $x\in\mathbb{G}$ with a point of $\mathbb{R}^{n}$ as
follows

$x\equiv(x_{1},\dots,x_{n})\leftrightarrow\exp(x_{n}X_{n})\cdot\dots\cdot\exp(x_{1}X_{1}).$

On the contrary, when we say that we work in exponential coordinates of the
first kind we mean that we are identifying a point $a\in\mathbb{G}$ with a
point in $\mathbb{R}^{n}$ as follows

$a\equiv(a_{1},\dots,a_{n})\leftrightarrow\exp(a_{1}X_{1}+\dots+a_{n}X_{n}).$

Heisenberg group. Let $\mathbb{H}^{1}$ be the first Heisenberg group with Lie
algebra

$\mathfrak{h}^{1}=\mathrm{span}\\{X_{1},X_{2}\\}\oplus\mathrm{span}\\{X_{3}\\}=V_{1}\oplus
V_{2},,$

with the only nontrivial bracket relation $[X_{1},X_{2}]=X_{3}$. If we work in
exponential coordinates of the second kind $(x_{1},x_{2},x_{3})$ with respect
to the adapted basis $(X_{1},X_{2},X_{3})$ we can write

$X_{1}=\partial_{1},\qquad X_{2}=\partial_{2}+x_{1}\partial_{3},\qquad
X_{3}=\partial_{3},$

see [LDT20, page 11]. Every distribution $f$ such that
$X_{1}^{2}f=X_{2}^{2}f=0$ on $\mathbb{G}$ is represented by a polynomial, see
Theorem 1.3, and moreover one can check with straightforward computations by
using the expressions of the vector fields above that
$f\in\mathrm{span}\\{1,x_{1},x_{2},x_{3},x_{1}x_{2},x_{1}x_{3}\\}$. Notice
that in this case, for every such $f$, $X_{3}^{2}f=0$. Nevertheless it is not
true that every $\\{X,Y\\}$-affine function is affine along every direction of
the algebra: indeed, $(X_{1}+X_{2})^{k}(x_{1}x_{3})\not\equiv 0$ for all
$k\leq 3$.

If in addition to $X_{1}^{2}f=X_{2}^{2}f=0$ we ask that
$(X_{1}X_{2}+X_{2}X_{1})f=0$, the two conditions together being equivalent to
asking that $X^{2}f=0$ for every $X\in V_{1}$, we conclude that
$f\in\mathrm{span}\\{1,x_{1},x_{2},x_{3}-(1/2)x_{1}x_{2}\\}$. Notice that,
when read in exponential coordinates of the first kind, the functions
$x_{1},x_{2},x_{3}-(1/2)x_{1}x_{2}$ are precisely the coordinate functions
$a_{1},a_{2},a_{3}$, respectively. In this way we recover the already known
property that every horizontally affine function in $\mathbb{H}^{1}$ is
actually affine in exponential coordinates of the first kind. For the complete
characterization of horizontally affine maps in Carnot groups of step $2$ one
can see [LDMR20].

Engel group. Let $\mathbb{E}^{1}$ be the Engel group, i.e., the Carnot group
of topological dimension $4$ with stratified algebra

$\mathfrak{e}^{1}=\mathrm{span}\\{X_{1},X_{2}\\}\oplus\mathrm{span}\\{X_{3}\\}\oplus\mathrm{span}\\{X_{4}\\},$

the only nontrivial bracket relations being $[X_{1},X_{2}]=X_{3}$, and
$[X_{1},X_{3}]=X_{4}$. Working in exponential coordinates of the second kind
with respect to the adapted basis $(X_{1},X_{2},X_{3},X_{4})$ we can write

$X_{1}=\partial_{1},\qquad
X_{2}=\partial_{2}+x_{1}\partial_{3}+(x_{1}^{2}/2)\partial_{4},\qquad
X_{3}=\partial_{3}+x_{1}\partial_{4},\qquad X_{4}=\partial_{4},$

see [LDT20, page 13]. We notice that a distribution $f$ is a horizontally
affine function on $\mathbb{E}^{1}$, i.e., such that $X^{2}f=0$ for every
$X\in\mathrm{span}\\{X_{1},X_{2}\\}$, if and only if it satisfies the three
equalities $X_{1}^{2}f=(X_{1}X_{2}+X_{2}X_{1})f=X_{2}^{2}f=0$ on
$\mathbb{E}^{1}$. Let us write explicitly the horizontally affine maps in