An application of a global lifting method for homogeneous Hörmander vector fields to the Gibbons conjecture

Abstract

In this paper we exploit a global lifting method for homogeneous Hörmander vector fields in order to extend the Gibbons conjecture to any second-order differential operator $\mathcal {L}_X = \sum _{j = 1}^mX_j^2$, where the $X_j$’s are linearly independent smooth vector fields on $\mathbb {R}^n$ satisfying Hörmander’s rank condition and fulfilling a suitable homogeneity property with respect to a family of non-isotropic dilations. The class of these operators comprehends the sub-Laplacians on Carnot groups, the smooth Grushin-type operators and the smooth $\Delta _\lambda$-Laplacians studied by Franchi, Lanconelli and Kogoj. We also establish a comparison result for the solutions of the semi-linear equation $\mathcal {L}_Xu+f(u) = 0$ under suitable assumptions on the function f.

Appendix: A brief review on Carnot groups

In this appendix we review the basic definitions and facts on homogeneous Carnot groups which have been used in the paper. We follow the exposition in [14], to which we refer for all the omitted proofs and the details.

Let $*$ be a (binary) group operation on Euclidean space $\mathbb {R}^N$. We say that $\mathbb {G}= (\mathbb {R}^N,*)$ is a Lie group (on $\mathbb {R}^N$) if the map

$$\begin{aligned} *:\mathbb {R}^N\times \mathbb {R}^N\longrightarrow \mathbb {R}^N, \qquad (z,\zeta )\mapsto z*\zeta , \end{aligned}$$

is smooth. For every fixed $z\in \mathbb {G}\equiv \mathbb {R}^N$ we indicate, respectively, by $\tau _\alpha$ and $\rho _\alpha$ the left-translation and the right-translation by z, on $\mathbb {G}$, that is,

• (a) $\tau _z:\mathbb {R}^N\rightarrow \mathbb {R}^N, \qquad \tau _z(\zeta ) := z*\zeta$;

• (b) $\rho _z:\mathbb {R}^N\rightarrow \mathbb {R}^N, \qquad \rho _z(\zeta ) := \zeta *z$.

Given a Lie group $\mathbb {G}= (\mathbb {R}^N,*)$ and a smooth vector field $X = \sum _j a_j(z)\,\partial _{z_j}$ on $\mathbb {R}^N$, we say that X is left-invariant on $\mathbb {G}$ if

$$\begin{aligned} X\big (u\circ \tau _\alpha ) = (Xu)\circ \tau _\alpha , \end{aligned}$$

(A.1)

for every $u\in C^\infty (\mathbb {R}^N,\mathbb {R})$ and every $\alpha \in \mathbb {R}^N$. We denote by $\mathrm {Lie}(\mathbb {G})$ the set of all left-invariant vector fields on $\mathbb {G}$ and we call it the Lie algebra of $\mathbb {G}$.

We then have the following very standard result.

Proposition A.1

Let $\mathbb {G}= (\mathbb {R}^N,*)$ be a Lie group with neutral element e, and let $\mathrm {Lie}(\mathbb {G})$ be the Lie algebra of $\mathbb {G}$. Then the following facts hold.

• 1. $\mathrm {Lie}(\mathbb {G})$ is a Lie algebra and $\dim _{\mathbb {R}}(\mathrm {Lie}(\mathbb {G})) = N$;

• 2. for every $i \in \{1,\ldots ,N\}$ there exists precisely one $J_i\in \mathrm {Lie}(\mathbb {G})$ such that

  $$\begin{aligned} J_iu(e) = \frac{\partial u}{\partial z_i}(e) \qquad \big (\text {for every } u\in C^\infty (\mathbb {R}^N,\mathbb {R})\big ); \end{aligned}$$

  more precisely, we have

  $$\begin{aligned} J_i = \sum _{j = 1}^N a_{ij}(z)\,\frac{\partial }{\partial z_j}, \quad \text {where } \begin{pmatrix} a_{i1}(z) \\ \vdots \\ a_{iN}(z) \end{pmatrix} = \mathcal {J}_{\tau _z}(e)\cdot e_i \end{aligned}$$

  (and $\mathcal {J}_{\tau _x}$ denotes the Jacobian matrix of $\tau _z$).

• 3. If $J_1,\ldots ,J_N$ are as above, then $\mathcal {J} = \{J_1,\ldots ,J_N\}$ is a (linear) basis of $\mathrm {Lie}(\mathbb {G})$, which is usually called the Jacobian basis of $\mathrm {Lie}(\mathbb {G})$.

Let now $\mathbb {G}= (\mathbb {R}^N,*)$ be a Lie group on $\mathbb {R}^N$, and let us assume that there exists a N-tuple $(\upsilon _1,\ldots , \upsilon _N)$ of positive real numbers such that

$$\begin{aligned} D_\lambda :\mathbb {R}^N\longrightarrow \mathbb {R}^N,\qquad D_\lambda (z) := (\lambda ^{\upsilon _1}z_1,\ldots ,\lambda ^{\upsilon _N}z_N), \end{aligned}$$

(A.2)

is an automorphism of $\mathbb {G}$ for every fixed $\lambda > 0$, that is,

$$\begin{aligned} D_\lambda (z*\zeta ) = D_\lambda (z)*D_\lambda (\zeta ), \qquad \text {for every }z,\zeta \in \mathbb {R}^N. \end{aligned}$$

Then, the triple $\mathbb {G}= (\mathbb {R}^N,*,D_\lambda )$ is called a homogeneous (Lie) group (on $\mathbb {R}^N$) and $\{D_\lambda \}_{\lambda > 0}$ is usually referred to as the family of dilations of $\mathbb {G}$.

Remark A.2

Let $\mathbb {G}= (\mathbb {R}^N,*,D_\lambda )$ be a homogeneous Lie group on $\mathbb {R}^N$, with $D_\lambda$ as in (A.2). We aim to show that, by eventually performing an ad-hoc change of coordinates in $\mathbb {R}^N$, it is not restrictive to assume

$$\begin{aligned} 1 = \upsilon _1\le \cdots \le \upsilon _N. \end{aligned}$$

(A.3)

In fact, let $\alpha$ be the unique permutation of $\{1,\ldots ,N\}$ such that

• (a) $\upsilon _{\alpha (1)}\le \cdots \le \upsilon _{\alpha (N)}$, whence $\upsilon _{\alpha (1)} = \underline{\upsilon }$;

• (b) if $\upsilon _{\alpha (i)} = \upsilon _{\alpha (j)}$ for some i, j with $i < j$, then $\alpha (i) < \alpha (j)$;

moreover, let $T:\mathbb {R}^N\rightarrow \mathbb {R}^N$ be the map defined as follows

$$\begin{aligned} T(z) := \big (z_{\alpha (1)},\ldots ,z_{\alpha (N)}\big ) \qquad (z\in \mathbb {R}^N). \end{aligned}$$

(A.4)

Obviously, T is a smooth diffeomorphism of $\mathbb {R}^N$ and

$$\begin{aligned} T^{-1}(a) = \big (a_{\beta (1)},\ldots ,a_{\beta (N)}\big ),\qquad \text { where }\beta = \alpha ^{-1}; \end{aligned}$$

furthermore, each component of T is $D_\lambda$-homogeneous of degree $\upsilon _{\alpha (i)}$, i.e.,

$$\begin{aligned} T_i\big (D_\lambda (z)\big ) = \lambda ^{\upsilon _{\alpha (i)}}\,T_i(z)\quad \forall \,z\in \mathbb {R}^N\,\,\text {and}\,\,\forall \,\,i = 1,\ldots ,N. \end{aligned}$$

(A.5)

By making use of this map T, we define a new group operation $\circ$ in $\mathbb {R}^N$ and a new family of dilations $\{D'_\lambda \}_{\lambda > 0}$ in the following (standard) way:

$$\begin{aligned}&\qquad a \circ b := T\Big (T^{-1}(a)*T^{-1}(b)\Big ) \qquad (\text {for }a,b\in \mathbb {R}^N);&\end{aligned}$$

(A.6)

$$\begin{aligned}&\qquad \qquad D'_\lambda (a) := T\Big (D_{\lambda ^{1/\underline{\upsilon }}} \big (T^{-1}(a)\big )\Big ) \qquad (\text {for }a\in \mathbb {R}^N\text { and }\lambda > 0).&\end{aligned}$$

(A.7)

Notice that, by combining (A.5) with (A.7), we have

$$\begin{aligned} D'_{\lambda }(a) = (\lambda ^{\upsilon '_1}a_1,\ldots ,\lambda ^{\upsilon '_N}a_N), \quad \text {where }\upsilon '_i = \upsilon _{\alpha (i)}/\underline{\upsilon }; \end{aligned}$$

(A.8)

as a consequence, by the choice of $\alpha$ in (a) one has

$$\begin{aligned} 1 = \upsilon '_1 \le \upsilon '_2\le \cdots \le \upsilon '_N. \end{aligned}$$

(A.9)

On account of (A.8), and taking into account the fact that $\mathbb {G}$ is a homogeneous Lie group, it is not difficult to recognize that

$$\begin{aligned} \mathbb {G}' = \mathbb {G}'_\alpha := (\mathbb {R}^N,\circ ,D'_\lambda ), \end{aligned}$$

is a homogeneous Lie group on $\mathbb {R}^N$ as well; moreover, the map T turns out to be an isomorphism of Lie groups, that is, $T\in C^\infty (\mathbb {R}^N,\mathbb {R}^N)$ and

$$\begin{aligned} T(z*\zeta ) = T(z)\circ T(\zeta ),\qquad \text {for every }z,\zeta \in \mathbb {R}^N. \end{aligned}$$

We also point out that, by the very definition of $D'_\lambda$, one has

$$\begin{aligned} T\big (D_\lambda (z)\big ) = D'_{\lambda ^{\overline{\upsilon }}}\big (T(z)\big ), \quad \text {for every }z\in \mathbb {R}^N\hbox { and every }\lambda > 0. \end{aligned}$$

We now take a look at the Lie algebras of $\mathbb {G}$ and $\mathbb {G}'$. To this end, we first introduce a notation: given any smooth vector field X on $\mathbb {R}^N$ (acting in the z variables), we denote by $T_*(X)$ the vector field on $\mathbb {R}^N$ (acting in the a variables) defined by

$$\begin{aligned} T_*(X)(u) := X(u\circ T)\circ T^{-1}, \qquad \text {for every }u\in C^\infty (\mathbb {R}^N,\mathbb {R}). \end{aligned}$$

Bearing in mind that T is an isomorphism of Lie groups, we have the following well-known facts (for a proof see, e.g., [7, Appendix C]):

• (i) $T_*(X)\in \mathrm {Lie}(\mathbb {G}')$ for every $X\in \mathrm {Lie}(\mathbb {G})$;

• (ii) the map $\mathrm {d}T:\mathrm {Lie}(\mathbb {G})\rightarrow \mathrm {Lie}(\mathbb {G}')$ defined by

  $$\begin{aligned} \mathrm {d}T(X) := T_*(X) \end{aligned}$$

  (A.10)

  is an isomorphism of Lie algebras, that is, $\mathrm {d}T$ is a (linear) isomorphism of vector spaces and, for every $X,Y\in \mathrm {Lie}(\mathbb {G})$, one has the identity

  $$\begin{aligned} \mathrm {d}T\big ([X,Y]\big ) = [\mathrm {d}T(X),\mathrm {d}T(Y)]. \end{aligned}$$

In particular, if $\mathcal {J} = \{J_1,\ldots ,J_N\}$ denotes the Jacobian basis of $\mathrm {Lie}(\mathbb {G})$, a direct computation shows that (for every $i = 1,\ldots ,N$)

$$\begin{aligned} \mathrm {d}T(J_i)(u)(0) = \frac{\partial u}{\partial a_{j}}(0),\qquad \text {with }j = \beta (i); \end{aligned}$$

as a consequence, from Proposition A.1-(2) we infer that

$$\begin{aligned} \mathrm {d}T(J_i) = J'_{\beta (i)}, \qquad \text {for every }i = 1,\ldots ,N, \end{aligned}$$

(A.11)

where $\mathcal {J}' = \{J'_1,\ldots ,J'_N\}$ denotes the Jacobian basis of $\mathrm {Lie}(\mathbb {G}')$. Summing up, the groups $\mathbb {G}$ and $\mathbb {G}'$ are isomorphic (via the map T) and their Lie algebras are isomorphic via $\mathrm {d}T$; furthermore, the matrix representing $\mathrm {d}T$ with respect to the Jacobian basis $\mathcal {J}$ and $\mathcal {J}'$ is the permutation matrix

$$\begin{aligned} P_\alpha = \begin{pmatrix} e_{\alpha (1)}\\ \vdots \\ e_{\alpha (N)} \end{pmatrix}. \end{aligned}$$

In view of all the computations just performed, it seems very natural to identify the group $\mathbb {G}$ with $\mathbb {G}'$; however, we prefer to avoid this identification since the homogeneous groups naturally associated with the homogeneous Hörmander operators considered in Sect. 2 do not satisfy (A.3).

Let $\mathbb {G}= (\mathbb {R}^N,*,D_\lambda )$ be a homogeneous group on $\mathbb {R}^N$ [with $D_\lambda$ as in (A.2)] and let $\mathcal {J} = \{J_1,\ldots ,J_N\}$ be the Jacobian basis of $\mathrm {Lie}(\mathbb {G})$. It is very easy to recognize that $J_i$ is $D_\lambda$-homogeneous of degree $\upsilon _i$ (for every fixed $i = 1,\ldots ,N$), that is,

$$\begin{aligned} J_i(u\circ D_\lambda ) = \lambda ^{\upsilon _i}\,(J_i u)\circ D_\lambda , \end{aligned}$$

(A.12)

for every $u\in C^\infty (\mathbb {R}^N,\mathbb {R})$ and every $\lambda > 0$; we then set

$$\begin{aligned} \underline{\upsilon } := \min _{1\le i\le N}\upsilon _i, \qquad \overline{\upsilon } := \max _{1\le i\le N}\upsilon _i \end{aligned}$$

(A.13)

and we define the horizontal layer $V_1(\mathbb {G})$ of $\mathbb {G}$ as follows:

$$\begin{aligned} V_1(\mathbb {G}) := \mathrm {span}\big \{J_i:\,\upsilon _i = \underline{\upsilon }\big \}\subseteq \mathrm {Lie}(\mathbb {G}). \end{aligned}$$

(A.14)

Remark A.3

Let $\mathbb {G}= (\mathbb {R}^N,*,D_\lambda )$ be a homogeneous group and let $V_1(\mathbb {G})$ be the horizontal layer of $\mathbb {G}$. Since $V_1(\mathbb {G})$ is spanned by vector fields which are $D_\lambda$-homogeneous of degree $\underline{\upsilon }$, it is straightforward to recognize that

$$\begin{aligned} X\in V_1(\mathbb {G})\,\,\Longrightarrow \,\, X\text { is }D_\lambda \text {-homogeneous of degree }\underline{\upsilon }. \end{aligned}$$

If $\mathbb {G}= (\mathbb {R}^N,*,D_\lambda )$ is a homogeneous Lie group on $\mathbb {R}^N$ and if

$$\begin{aligned} \mathrm {Lie}\big (V_1(\mathbb {G})\big ) = \mathrm {Lie}(\mathbb {G}), \end{aligned}$$

we say that $\mathbb {G}$ is a (homogeneous) Carnot group (here, by $\mathrm {Lie}(V_1(\mathbb {G}))$ we mean the smallest Lie sub-algebra of $\mathrm {Lie}(\mathbb {G})$ containing $V_1(\mathbb {G})$).

Remark A.4

Let $\mathbb {G}= (\mathbb {R}^N,*,D_\lambda )$ be a homogeneous Lie group on $\mathbb {R}^n$ and let $\mathbb {G}' = (\mathbb {R}^N,\circ ,D'_\lambda )$ be the group (isomorphic to $\mathbb {G}$) constructed in the previous Remark A.2.

Bearing in mind that the Lie algebras of $\mathbb {G}$ and of $\mathbb {G}'$ are isomorphic via the map $\mathrm {d}T$ defined in (A.10), one has

$$\begin{aligned} \mathrm {d}T\big (V_1(\mathbb {G})\big )&\,\,=\,\, \mathrm {span}\big \{\mathrm {d}T(J_i):\,\upsilon _i = \underline{\upsilon }\big \} {\mathop {=}\limits ^{(A.11)}} \mathrm {span}\big \{J'_{\beta (i)}:\,\upsilon _i = \underline{\upsilon }\big \}\\&\,\,=\,\, \mathrm {span}\big \{J'_p:\,\upsilon '_p = 1\big \} \qquad \big (\text {see (A.8) and (A.9)}\big )\\&\,\,=\,\,V_1(\mathbb {G}') \qquad \big (\text {since }\min _{j}\upsilon '_j = 1,\text { see (A.9)}\big ) \end{aligned}$$

(here, as usual, $\mathcal {J} = \{J_1,\ldots ,J_N\}$ and $\mathcal {J}' = \{J'_1,\ldots ,J'_N\}$ denote the Jacobian basis of $\mathrm {Lie}(\mathbb {G})$ and $\mathrm {Lie}(\mathbb {G}')$, respectively). As a consequence, one gets

$$\begin{aligned} \mathrm {d}T\Big (\mathrm {Lie}\big (V_1(\mathbb {G})\big )\Big ) = \mathrm {Lie}\Big (\mathrm {d}T\big (V_1(\mathbb {G})\big )\Big ) = \mathrm {Lie}\big (V_1(\mathbb {G}')\big ), \end{aligned}$$

and thus $\mathbb {G}$ is a Carnot group if and only if $\mathbb {G}'$ is.

Let $\mathbb {G}= (\mathbb {R}^N,*,D_\lambda )$ be a homogeneous Carnot group [with $D_\lambda$ as in (A.2)] and let $V_1(\mathbb {G})$ be the horizontal layer of $\mathbb {G}$. The (integer) number

$$\begin{aligned} m = m(\mathbb {G}) := \dim _{\mathbb {R}}\big (V_1(\mathbb {G})\big ) = \mathrm {card}\big \{i\in \{1,\ldots ,N\}:\,\upsilon _i = \underline{\upsilon } \big \}, \end{aligned}$$

is usually referred to as the number of generators of $\mathbb {G}$. If $\mathcal {Z} = \{Z_1,\ldots ,Z_m\}$ is any linear basis of $V_1(\mathbb {G})$, the second-order differential operator

$$\begin{aligned} \Delta _\mathcal {Z} := \sum _{j = 1}^mZ_j^2, \end{aligned}$$

is called a sub-Laplacian on $\mathbb {G}$. In the particular case when

$$\begin{aligned} \mathcal {Z} = \{J_i:\,\upsilon _i = \underline{\upsilon }\} \end{aligned}$$

(where $\{J_1,\ldots ,J_N\}$ is the Jacobian basis of $\mathrm {Lie}(\mathbb {G})$), we write $\Delta _\mathbb {G}$ in place of $\Delta _\mathcal {Z}$ and we call $\Delta _\mathbb {G}$ the canonical sub-Laplacian on $\mathbb {G}$.

We conclude this section by stating a theorem by Birindelli and Lanconelli which proves the Gibbons conjecture on any Carnot group. To this end, we first introduce another useful notation which shall be adopted throughout the sequel.

Let $\mathbb {G}= (\mathbb {R}^N,*,D_\lambda )$ be a homogeneous Carnot group, with $D_\lambda$ as in (A.2); given any $i\in \{1,\ldots ,N\}$, we say that $z_i$ is a $\mathbb {G}$-horizontal variable if

$$\begin{aligned} \upsilon _i = \underline{\upsilon } = \min _{1\le i\le N}\upsilon _i. \end{aligned}$$

Theorem A.5

([11, Theorems 1.3 and 1.4]) Let $\mathbb {G}= (\mathbb {R}^N,*,D_\lambda )$ be a homogeneous Carnot group with $m = m_\mathbb {G}\ge 1$ generators and let $\mathcal {Z} = \{Z_1,\ldots ,Z_m\}$ be a fixed linear basis of the first layer $V_1(\mathbb {G})$ of $\mathbb {G}$.

Moreover, let $f:[-1,1]\rightarrow \mathbb {R}$ be a Lipschitz-continuous function satisfying the following two properties:

• (f1) $f(-1) = f(1) = 0$;

• (f2) there exists a real $\varepsilon < 1$ such that

  $$\begin{aligned} f\text { is non-increasing on }[-1,-1+\varepsilon ] \text { and on }[1-\varepsilon ,1]. \end{aligned}$$

Finally, let $v\in C^2(\mathbb {R}^N,\mathbb {R})$ be a solution of

$$\begin{aligned} \Delta _\mathcal {Z}v + f(v) = \sum _{j = 1}^mZ_j^2v + f(v) = 0\qquad \text {on }\mathbb {G}\equiv \mathbb {R}^N, \end{aligned}$$

(A.15)

which satisfies the next two conditions:

• (a) $|v(x)|\le 1$ for every $x\in \mathbb {R}^N$;

• (b) there exists $i\in \{1,\ldots ,N\}$ such that

  $$\begin{aligned} \lim _{z_i\rightarrow \pm \infty }v(z) = \pm 1, \end{aligned}$$

  uniformly for $\hat{z} = (z_1,\ldots ,z_{i-1},z_{i+1},\ldots ,z_N)\in \mathbb {R}^{N-1}$.

Then the following facts hold true.

• (1) If $z_i$ is a $\mathbb {G}$-horizontal variable, then there exists $V\in C^2(\mathbb {R},\mathbb {R})$ such that

  $$\begin{aligned} v(z) = V(z_i)\,\,\text {for every }z\in \mathbb {R}^N\quad \text {and}\quad V' > 0\,\,\text {on }\mathbb {R}. \end{aligned}$$

  (A.16)

• (2) If $z_i$ is not a $\mathbb {G}$-horizontal variable, then one can say that

  $$\begin{aligned} \frac{\partial v}{\partial z_i}(z)\ge 0\quad \text {for every }z\in \mathbb {R}^N, \end{aligned}$$

  (A.17)

  and the inequality is strict if $\partial _{z_i}$ commutes with $Z_1,\ldots ,Z_m$.

As a matter of fact, Theorem A.5 is proved in [11] under the additional assumption that the exponents of $D_\lambda$ fulfill (A.3); for the sake of completeness, we show here how one can pass from a general Carnot group (whose family of dilations does not fulfill (A.3)) to the case considered in [11].

Proof

Let $\mathbb {G},\,f,\,v$ be as in the statement of the theorem and let

$$\begin{aligned} \mathbb {G}' = (\mathbb {R}^N,\circ ,D'_\lambda ) \end{aligned}$$

be the homogeneous group (isomorphic to $\mathbb {G}$) constructed in Remark A.2.

Since $\mathbb {G}$ is a Carnot group, we know from Remark A.4 that the same is true of $\mathbb {G}'$; moreover, since $\mathcal {Z}$ is a basis of $V_1(\mathbb {G})$ and since the Lie algebras of $\mathbb {G}$ and of $\mathbb {G}'$ are isomorphic via the map $\mathrm {d}T$ in (A.10), the set

$$\begin{aligned} \mathcal {Z}' := \{Y_1,\ldots ,Y_N\}, \qquad \text {where } Y_i := \mathrm {d}T(Z_i) \end{aligned}$$

is a basis of $V_1(\mathbb {G}') = \mathrm {d}T(V_1(\mathbb {G}))$. As a consequence, the differential operator

$$\begin{aligned} \Delta _{\mathcal {Z}'} := \sum _{j = 1}^N Y_j^2, \end{aligned}$$

is a sub-Laplacian on $\mathbb {G}'$. Now, if T is the map defined in (A.4) (which is an isomorphism between $\mathbb {G}$ and $\mathbb {G}'$), we consider the following function

$$\begin{aligned} w:\mathbb {R}^N\longrightarrow \mathbb {R}^N, \qquad w(a) := v\big (T^{-1}(a)\big ). \end{aligned}$$

Obviously, $w\in C^2(\mathbb {R}^N,\mathbb {R})$; moreover, from the very definition of $\mathrm {d}T$ (and since v solves (A.15) and satisfies assumptions (a)-(b)), we easily infer that:

• $\Delta _{\mathcal {Z}'}w+f(w) = 0$ on $\mathbb {G}'\equiv \mathbb {R}^N$;

• $|w(a)|\le 1$ for every $a\in \mathbb {R}^N$;

• setting $j := \beta (i)$ (where $\beta = \alpha ^{-1}$, see Remark A.2), we have

  $$\begin{aligned} \lim _{a_j\rightarrow \pm \infty }w(a) = \pm 1, \end{aligned}$$

  uniformly for $\hat{a} = (a_1,\ldots ,a_{j-1},a_{j+1},\ldots ,a_N)\in \mathbb {R}^{N-1}$.

Since Theorem A.5 holds in the group $\mathbb {G}'$ [see [11] and remember that the exponents $\upsilon '_1,\ldots ,\upsilon '_N$ of $D'_\lambda$ satisfy (A.9)], we have the following two cases:

• (1) $z_i$ is a $\mathbb {G}$-horizontal variable. In this case, since $\upsilon _i = \underline{\upsilon }$ we have

  $$\begin{aligned} \upsilon '_{j}\,\,=\,\,\upsilon '_{\beta (i)} {\mathop {=}\limits ^{(A.8)}} 1, \end{aligned}$$

  (A.18)

  and thus $a_j = a_{\beta (i)}$ is a $\mathbb {G}'$-horizontal variable. By Theorem A.5 (applied to the group $\mathbb {G}'$) we then infer the existence of $V\in C^2(\mathbb {R},\mathbb {R})$ such that

  • (i) $V' > 0$ throughout $\mathbb {R}$;

  • (ii) $v\big (a_{\beta (1)},\ldots ,a_{\beta (N)}\big ) = w(a) = V(a_j) = V(a_{\beta (i)})$ for every $a\in \mathbb {R}^N$;

  as a consequence, $v(z) = V(z_i)$ for every $z\in \mathbb {R}^N$, as desired.

• (2) $z_i$ is not a $\mathbb {G}$-horizontal variable. In this case, $a_j = a_{\beta (i)}$ is not a $\mathbb {G}'$-horizontal variable [see (A.18)] and thus, by applying Theorem A.5 to the group $\mathbb {G}'$, we obtain

  $$\begin{aligned} \frac{\partial w}{\partial a_j}(a) = \frac{\partial v}{\partial z_i}\big (T^{-1}(a)\big ) \ge 0, \quad \text {for every }a\in \mathbb {R}^N; \end{aligned}$$

  since T is a diffeomorphism, this proves that $\partial _{z_i}v\ge 0$ on $\mathbb {R}^N$. Finally, if $\partial _{z_i}$ commutes with $Z_1,\ldots ,Z_N$, a direct computation shows that

  $$\begin{aligned}{}[\partial _{a_j},Y_k] = [\partial _{a_j}, \mathrm {d}T(Z_k)] = 0, \quad \text {for all }k = 1,\ldots ,N, \end{aligned}$$

  and thus, again by Theorem A.5, we get $\partial _{a_j}w = \partial _{z_i}u\circ T^{-1} > 0$ on $\mathbb {R}^N$.

This ends the proof. $\square$