Proof. For connected G, $\mathrm {Aut}(G)$ is a closed subgroup of $\mathrm {Aut}(\mathfrak {g})$ with Lie algebra $\mathrm {Der}(\mathfrak {g})$ and these coincide when G is simply connected (cf. [Reference VaradarajanVar84, Theorem 2.75]). Picking a basis of the Lie algebra, a Lie algebra automorphism is completely determined by its associated matrix with respect to this basis. The bracket relations on basis elements give us quadratic polynomial relations in these matrix coefficients. Moreover, any matrix satisfying these relations induces an automorphism. Hence, these polynomial relations provide $\mathrm {Aut}(\mathfrak {g})$ with the structure of a real affine variety in $\mathrm {GL}(\mathfrak {g})\subset \mathbb {P}_{\mathbb {R}}^{(\dim \mathfrak {g})^2}$ .

Proof. It is clear that automorphisms preserving the grading are a subgroup of automorphisms lying in $\bigoplus _{i=0}^r \mathrm {GL}(\mathfrak {g}^i)<\mathrm {GL}(\mathfrak {g})$ . It is easy to check that the subgroup $\mathrm {Aut}_g(\mathfrak {g})<\mathrm {Aut}(\mathfrak {g})$ corresponds to the subgroup $\mathrm {Aut}_g(G)<\mathrm {Aut}(G)$ under the natural isomorphism given by the previous proposition. It remains to show that $\mathrm {Aut}_g(\mathfrak {g})$ is a subvariety of $\mathrm {Aut}(\mathfrak {g})$ . However, as the graded subspaces are linear, the additional defining equations for the corresponding matrix elements are linear relations. Hence $\mathrm {Aut}_g(\mathfrak {g})$ is also a real algebraic subvariety. Lastly, groups which are real algebraic subvarieties of an algebraic group are themselves algebraic groups.

Proof. If $\mathrm {Aut}_g(\mathfrak {g})=L\mathcal {D}$ consists of homotheties of a Carnot–Carathéodory metric, then $L<\mathrm {SL}_{\pm }(\mathfrak {g}^0)$ and each element of L is uniquely determined by its action on $\mathfrak {g}^0$ . Moreover, each element of L must be an isometry with respect to the given inner product $\langle \cdot ,\cdot \rangle _0$ on $\mathfrak {g}^0$ . Hence L belongs to the compact group $\mathrm {O}(d_0,\langle \cdot ,\cdot \rangle _0)$ .

Conversely, if $\mathrm {Aut}_g(\mathfrak {g})=L\mathcal {D}$ where $L<K$ for some compact subgroup $K<\mathrm {SL}_{\pm }(d_0,\mathbb {R})$ , then K is conjugate into $\mathrm {O}(d_0,\mathbb {R})$ . Equivalently, K belongs to $\mathrm {O}(d_0,\langle \cdot ,\cdot \rangle _0)$ for the inner product on $\mathfrak {g}^0$ conjugate to the standard one. This inner product then induces a Carnot–Carathéodory metric for which $\mathrm {Aut}_g(\mathfrak {g})$ are homotheties.

Proof. By continuity of the defining polynomial relations satisfied by the matrices in $\mathrm {Aut}_g(\mathfrak {g}_k)$ , any element of the limit will satisfy the polynomial relations required to be in $\mathrm {Aut}_g(\mathfrak {g})$ . Let $H\subset \mathrm {Aut}_g(\mathfrak {g})$ be the collection of limit points. By definition this is a closed subset of $\mathrm {Aut}_g(\mathfrak {g})$ . Considered as equivalence classes of convergent sequences, it is elementary to verify that the product structure given by $[\{ a_n \}]*[\{ b_n \}]=[\{ a_n b_n \}]$ is well defined and gives H a group structure compatible with that of $\mathrm {Aut}_g(\mathfrak {g})$ . Hence H is a Lie subgroup of $\mathrm {Aut}_g(\mathfrak {g})$ .

Proof. Let $(\mathfrak {g}^k)$ be a sequence of Carnot Lie algebras converging to $\mathfrak {g}$ in $\mathcal {NL}_n$ . Write $\mathrm {Aut}_g(\mathfrak {g})=L\mathcal {D}$ where $L=\mathrm {Aut}_g(\mathfrak {g})\cap \mathrm {SL}_{\pm }(d_0,\mathbb {R})$ lies in some compact group K. Without loss of generality, we may assume after a conjugation that $L<K=\mathrm {O}(d_0,\mathbb {R})$ . Similarly, let $\mathrm {Aut}_g(\mathfrak {g}^k)=L_k\mathcal {D}$ for $L_k=\mathrm {Aut}_g(\mathfrak {g}^k)\cap \mathrm {SL}_{\pm }(d_0,\mathbb {R})$ .

By Proposition 2.9, the $L_k$ converge to a subgroup of L. We wish to show that $L_k$ belongs to a compact subgroup. We will first consider $L_k^0$ , the connected component of the identity of $L_k$ . We fix a norm on $\mathbb {R}^n$ . If the $L_k^0$ are not contained in a compact group, then the operator norm, $\lVert \cdot \rVert _{\mathrm {op}}$ is unbounded on $L_k^0$ . Then $L_k^0$ admits a path whose elements have norms taking on all values in $[1,\infty )$ , and in particular there is an element $a_k$ with $\lVert a_k \rVert _{\mathrm {op}}=2$ . The sequence $\{ a_k \}$ admits a convergent subsequence to an element of determinant $\pm 1$ , but which does not have norm one, leading to a contradiction.

By Proposition 2.5, $\mathrm {Aut}_g(\mathfrak {g}^k)$ and thus $L_k$ are algebraic groups. An algebraic group has only finitely many connected components by [Reference WhitneyWhi57, Theorem 3]. Therefore $L_k$ must lie in a compact subgroup.

Proof. Since X is horizontal it has degree at most one, and we can write it in the basis as $X(y)=\sum _i a_i(y) X_i$ for functions $a_i$ on the neighborhood U of p. Express this as $X(y)= \sum _i a_i(p)X_i+ \sum _i (a_i(y)-a_i(p)) X_i$ . Since the functions $a_i(y)-a_i(p)$ vanish at p, they have degree at most $-1$ at p and the $X_i$ have degree $1$ . Hence $\sum _i (a_i(y)-a_i(p)) X_i$ has degree at most $0$ and $\widehat {X}_p=\sum _i a_i(p)\widehat {X}_{i;p}$ .

Proof. We need only show that the isomorphism class of the tangent cones vary continuously locally. By Theorem 3.2, the nilpotent structures depend only on the horizontal distribution, not the choice of basis above used to construct the structure. Hence we may choose the initial basis freely. We now refer back to the notation at the beginning of §3.1.

For a neighborhood U of a point $p\in N$ , by the genericity assumption, we may choose the basis of fields $X_i$ for $i=1,\ldots ,d_0$ so that the commutators of $X_i$ in $E^k$ do not degenerate to $E^{k-1}$ at any point of U if they do not do so at the point p. In particular, the same indices can be used to choose the same fields $Y_i$ , for $i=1,\ldots ,n$ at each point $x\in U$ . Therefore, the construction of all the $\widehat {X}_{k,j;x}$ varies smoothly in a neighborhood of p. This implies that the structure constants of the Lie algebras of the tangent cones vary smoothly on U and hence this implies that the nilpotent structure varies continuously on the space of nilpotent structures described in §2.2. This continuity thus passes to the orbit quotient space of nilpotent isomorphism classes as well.

Lastly, since the metric on $TC_xN$ is a Carnot–Carathéodory metric for the distributions induced by the inclusion of $E(x)$ in $\mathfrak {n}_x$ , the metric also varies continuously on U, and thus on all of N.

Proof. Given a unit-speed sub-Riemannian geodesic segment $\gamma $ starting at p, let X be its tangent field extended to a smooth horizontal field on the neighborhood U of p. By [Reference Margulis and MostowMM00, Proposition 5.6], the Carnot–Carathéodory metric induced on $TC_pN$ does not depend on the choice of basis vector fields (satisfying certain assumptions). Using X as a coordinate field, we obtain

$$ \begin{align*} d_C(p,\delta_s\exp(X)(p))=d_C(p,\exp(s X)(p))=d_C(p,\gamma(s))=s. \end{align*} $$

This implies in $TC_pN$ that $d([p],[\delta _s \exp (X)(p)])=1$ by (3.2). Since the identity element in $TC_pN$ is the class $[p]$ of the constant curve, and $\widehat {X}=X$ along its integral curve, $[\delta _s \exp (t X)(p)]=e^{t\overline {X}}$ and therefore $d(1,e^{t \overline {X}})=t$ . The induced norm on the first level $\mathfrak {n}^0_p$ of the Lie algebra then gives $\|\overline {X}\|_{\mathfrak {n}^0_p}=1$ .

Thus the identification preserves the norm of vectors. Since inner products are uniquely determined by their norms, the inner products are preserved as well.

Proof. Since the statement is local, it is enough to consider any sufficiently small precompact neighborhood U of an arbitrary point $p\in N$ where f is a diffeomorphism and we use instead of f. We assume U is chosen such that there exist smooth horizontal vector fields $X_1,\ldots ,X_{d_0}$ on U which form a basis for $E(x)$ at each $x\in U$ . We let $X_i'=Df(X_i)$ for each $i=1,\ldots ,d_0$ . By hypothesis, $\{ X_1',\ldots ,X_{d_0}' \}$ forms a basis of $E'(y)$ at each $y\in U'$ . As above, these vector fields give us normal coordinates and we use them to define our respective dilatations $\delta _s$ on U and $\delta _s'$ on $U'=f(U)$ .

We note that $f:U\to U'$ is a quasi-conformal diffeomorphism since f has uniformly bounded derivatives on the compact set $\overline {U}$ . By [Reference Margulis and MostowMM95, Theorem 10.5] and its proof, f induces an isomorphism of tangent cones at each point x belonging to a certain full measure set $U^0\cap f^{-1}(\Omega _{X_1'}\cap \cdots \cap \Omega _{X_{d_0}'})$ to be defined shortly. (Note that the $f^{-1}$ was inadvertently left off of the $\Omega _{X_i'}$ in the proof of [Reference Margulis and MostowMM95].) We will show that both $U^0$ and $f^{-1}(\Omega _{X_i'})$ for any $i=1,\ldots ,d_0$ are all of U in our case.

The definition of $U^0$ given in [Reference Margulis and MostowMM95, §10.1] simplifies in our situation. It consists of the density points for the Hausdorff measure of $d_C$ (cf. [Reference Margulis and MostowMM95, Definition 2.1.4]) for the set A where the Lipschitz constant of f is uniformly bounded. However, since f is $C^1$ with uniformly bounded derivatives, $A=U$ . Moreover, every point $x\in U$ is a density point since U is open.

We now recall the definition of the $\Omega _{X_i'}$ from [Reference Margulis and MostowMM95]. For $i=1,\ldots , d_0$ , the right invariant vector fields $\overline {X}_i'\in \mathfrak {n}_y$ on $TC_yU'$ correspond to the homogeneous degree-one fields $\widehat {X}_{i;y}'$ , namely $\overline {X}_i'=\mathcal {M}_y(\widehat {X}_{i;y}')$ . Recall that the $X_i'$ were used to define the normal coordinates and the dilatation $\delta _s'$ . Hence, along the integral curve of $X_i'$ through y, we have $1/{t} (\delta _t')_* X_i'=X_i'$ , and hence $\widehat {X}_{i;y}'=\lim \nolimits _{t\to \infty } 1/{t} (\delta _t')_* X_i'=X_i'$ along the same curve. Let $e^{s \overline {X}_i'}$ denote the one-parameter subgroup in $TC_y U'$ of the right invariant field $\overline {X}_i'$ , viewed as an element of the Lie algebra $\mathfrak {n}_y$ . Furthermore, following [Reference Margulis and MostowMM95, §8.6], let $\epsilon _y$ denote the homeomorphism of a neighborhood of $1\in TC_y U'$ to $U'$ which takes the equivalence class of the path ${\{ \delta _s' z \}}_{s\in [0,1]}$ in $TC_y U'$ to the point $z\in U'$ . (The equivalence classes of the germs as $s\to 0$ of the paths $\delta _s' z$ correspond to points in the tangent cone by [Reference MitchellMit85, Lemmas 3.1 and 3.2].)

As in [Reference Margulis and MostowMM95], for $i=1,\ldots ,d_0$ let $A_{X_i'}$ be the set consisting of all points $y\in U'$ such that $s^{-1} d_C'((\exp s X_i')(y),\epsilon _y(e^{s \overline {X}_i'}))$ tends to $0$ uniformly as $s\to 0$ . Finally, the set $\Omega _{X_i'}$ consists of all the density points in $A_{X_i'}$ for the Hausdorff measure of $d_C'$ .

Finally, observe that since $\widehat {X}_{i;y}'=X_i'$ along its integral curve through y, we have

$$ \begin{align*} \delta'_t \exp(s X_i')(y)=\delta'_t \exp(s \widehat{X}_{i;y}')(y). \end{align*} $$

The latter curve in s converges in the pointwise Gromov–Hausdorff sense through the metric spaces $(U',y,({d'}/{t}))$ to the curve $e^{s\overline {X_i'}}\in TC_yN'$ as $t\to 0$ . In other words, the curve $\epsilon _y(e^{s \overline {X_i'}})$ coincides with the curve $\exp (s X_i')(y)$ for every $y\in U'$ . In particular, $A_{X_i'}=U'$ and therefore $\Omega _{X_i'}=U'$ since every point of $U'$ is a density point for the Hausdorff measure.

Proof. By Proposition 3.8 the limit of the maps $f_t$ given in Definition 3.7 exists and is $f_*:TC_pN\to TC_{f(p)}N'$ .

The push-forward under f of a horizontal vector field X is also horizontal and therefore converges under metric rescaling to a right invariant horizontal field $\overline {Df(X)}$ by [Reference Margulis and MostowMM95, Proposition 8.10]. As shown in the proof of [Reference Margulis and MostowMM95, Theorem 10.5], the image under $f_*$ of the one-parameter subgroup $e^{s\overline {X}}$ is the one-parameter subgroup of this limit push-forward field. In other words, for all $g\in TC_pN$ and $s\in \mathbb {R}$ , we have

$$ \begin{align*} f_*(e^{s \overline{X}}g)=e^{s \overline{Df(X)}}f_*(g). \end{align*} $$

Taking Lie derivatives, we have $df_*(\overline {X})=\overline {Df(X)}$ and therefore under the Métivier correspondence we have $\widehat {f}_*\widehat {X}=\widehat {Df(X)}$ as claimed.

The second statement follows directly from the Métivier correspondence and (3.1) and the fact that ${df}_*[\overline {X}_{i,k},\overline {X}_{j,l}]=[{df}_*\overline {X}_{i,k},{df}_*\overline {X}_{j,l}]$ .

Proof. The lower part of the diagram commutes as a standard consequence of Lie theory.

By Lemma 3.10, we have $\widehat {f}_*\widehat {X}=\widehat {Df(X)}$ . Since $X(p)=\widehat {X}_p(p)\in E(x)$ for a horizontal field X, we have

$$ \begin{align*} df_*(\imath(X(p))&=df_*(\overline{X})=df_*\mathcal{M}_p(\widehat{X})=\mathcal{M}_{f(p)}\widehat{f}_*\widehat{X}=\mathcal{M}_{f(p)}\widehat{Df(X)}\\&=\overline{Df(X)}=\imath'(D_{p}f(X(p))), \end{align*} $$

where the third equality is by the definition of $\widehat {f}_*$ . This establishes commutativity of the upper diagram.

The final statement follows from the commutativity of the diagram and the fact that the inclusions are isometric.

Proof. Again let $U\subset N$ be a small neighborhood around the point p and $X_1,\ldots , X_n$ be a graded basis where $X_1,\ldots , X_{d_0}$ are horizontal vector fields spanning E on U. As above, normal coordinates define a dilatation $\delta _t:U\to N$ . Then if X and Y are horizontal, by Métivier [ Reference MétivierMét76 ], X and Y are vector fields of degree at most $1$ at p and $\widehat {X}=\lim \nolimits _{t\to \infty }{1}/{t}(\delta _t)_* X$ is the homogeneous degree-one part of X. Note that $[X, Y]$ is a vector field of degree at most $2$ and we denote the homogeneous degree-two part by $[X,Y]^{(2)}=\lim \nolimits _{t\to \infty }{1}/{t^2}(\delta _t)_*[ X, Y]$ .

Without loss of generality we may assume that fields $X_1,\ldots ,X_{d_0}$ in the chosen basis have corresponding right invariant fields $\overline {X}_i$ which diagonalize $f_*$ on the first level of the Lie algebra $\mathfrak {n}$ . Indeed, by assumption, there are $\lambda _i\in \mathbb {R}$ such that $df_*\overline {X_i}=\lambda _i \overline {X_i}$ for $i=1,\ldots ,d_0$ .

By Corollary 3.11, for $i=1,\ldots ,d_0$ we have $\imath D_pf(X_i(p))=df_*(\imath (X_i(p)))=df_*(\overline {X}_i)=\lambda _i \overline {X}_i$ . Taking $\imath ^{-1}$ , we have $D_pf (X_i(p)) = \lambda _i X_i(p)$ .

By Lemma 3.10, if $[X_i,X_j]$ is a graded basis element, then $\widehat {f}_*[\widehat {X}_i,\widehat {X}_j]=[\widehat {f}_*\widehat {X}_i,\widehat {f}_*\widehat {X}_j]$ . Therefore we have

$$ \begin{align*} df_*\mathcal{M}_p[\widehat{X}_i,\widehat{X}_j]&=\mathcal{M}_p[\widehat{f}_*\widehat{X}_i,\widehat{f}_*\widehat{X}_j], \\ df_*[\mathcal{M}_p\widehat{X}_i,\mathcal{M}_p\widehat{X}_j]&=\mathcal{M}_p[\widehat{Df {X}_i},\widehat{Df {X}_j}], \\ [df_*\overline{X}_i,df_*\overline{X}_j]&=\mathcal{M}_p[{Df {X}_i},{Df {X}_j}]^{(2)},\\ \lambda_i\lambda_j[\overline{X}_i,\overline{X}_j]&=\mathcal{M}_p[{Df {X}_i},{Df {X}_j}]^{(2)}. \end{align*} $$

It then follows that

(3.3) $$ \begin{align} [{Df {X}_i},{Df {X}_j}]^{(2)} &=\lambda_i\lambda_j\mathcal{M}_p^{-1}[\overline{X}_i,\overline{X}_j]=\lambda_i\lambda_j[\widehat{X}_i,\widehat{X}_j]=\lambda_i\lambda_j[X_i,X_j]^{(2)}. \end{align} $$

By hypothesis $\{ X_1,\ldots ,X_{d_0} \}\cup \{ [X_i,X_j]:1\le i,j \le d_0 \}$ spans $E^1$ . Since the homogeneity degree of $Df[X_i,X_j]=[Df(X_i),Df(X_j)]$ and $[X_i,X_j]$ at p is at most $2$ , ${Df[X_i,X_j]-\lambda _i\lambda _j[X_i,X_j]}$ has homogeneity degree at most $1$ . Evaluating (3.3) at p, we have $[Df(X_i),Df(X_j)](p)-\lambda _i\lambda _j[X_i,X_j](p)\in E(p)=E^0(p)$ .

Continuing by induction, if $X_{i}$ and $X_{j}$ are graded basis fields of degree at most q and s respectively such that $[X_{i},X_{j}]$ is a basis field, then a nearly identical computation shows

$$ \begin{align*} [{Df {X}_i},{Df {X}_j}]^{(q+s)}=\sigma_i\sigma_j[X_i,X_j]^{(q+s)}, \end{align*} $$

where by the inductive hypothesis $\sigma _i=\lambda _{i_1}\cdots \lambda _{i_\kappa }$ and $\sigma _j=\lambda _{j_1}\cdots \lambda _{j_\tau }$ for some indices $i_1,\ldots ,i_\kappa $ and $j_1,\ldots ,j_\tau $ .

This shows that in the $\{ X_i(p) \}$ basis, $Df$ is upper triangular. If, restricted to the horizontal distribution, the diagonal values are $\lambda _1,\ldots ,\lambda _{d_0}$ , then other diagonal elements are products of the $\lambda _i$ as claimed.

Proof. Let $({X}_{0,1},\ldots ,{X}_{r,d_r})$ be a graded basis of N in a neighborhood of p. As before, we let $\displaystyle {\mathcal {M}_p: \operatorname {span}\{ \widehat {X}_{0,1},\ldots ,\widehat {X}_{r,d_r} \}\to \mathfrak {n}_p}$ be the Métivier correspondence. We will show that with respect to the basis $({X}_{0,1}(p),\ldots ,{X}_{r,d_r}(p))$ , the matrix for $D_pf$ is a block upper triangular matrix. Suppose that $f_*=\delta _{\lambda } h$ , where $\delta _\lambda $ is the dilation of scale $\lambda>0$ and h is an isometry of the Carnot–Carathéodory metric of $TC_pN$ which necessarily lies in a compact subgroup of the graded automorphism group.

By Lemma 3.10, $\widehat {f}_*\widehat {X}_{0,j}=\widehat {Df(X_{0,j})}$ . Thus,

$$ \begin{align*}\widehat{Df(X_{0,j})}=\mathcal{M}_p df_* \mathcal{M}_p^{-1}\widehat{X}_{0,j}=\mathcal{M}_p df_* \overline{X}_{0,j}=\lambda\mathcal{M}_p dh \overline{X}_{0,j}=\lambda\mathcal{M}_p dh \mathcal{M}_p^{-1}\widehat{X}_{0,j}.\end{align*} $$

Letting $\widehat {dh}=\mathcal {M}_p dh \mathcal {M}_p^{-1}$ , we can rewrite the above equation as $\widehat {Df(X_{0,j})}=\lambda \widehat {dh}\widehat {X}_{0,j}$ . Since $dh$ is a graded Lie algebra homomorphism, $\widehat {dh}$ is also a graded Lie algebra homomorphism of $ \operatorname {span}\{ \widehat {X}_{0,1},\ldots ,\widehat {X}_{r,d_r} \}$ . In particular, $\widehat {dh}$ preserves $ \operatorname {span}\{ \widehat {X}_{i,1},\ldots ,\widehat {X}_{i,d_i} \}$ for every $i=0,\ldots , r$ . It follows that $\widehat {dh}\widehat {X}_{0,j}\in \operatorname {span}\{ \widehat {X}_{0,1},\ldots ,\widehat {X}_{0,d_0} \}$ for every ${j=1, \ldots , d_0}$ . Therefore, $D_pf(\widehat {X}_{0,j}(p))=\widehat {Df(X_{0,j})}(p)=(\widehat {dh}\widehat {X}_{0,j})(p)\in \operatorname {span}\{ \widehat {X}_{0,1}(p),\ldots , \widehat {X}_{0,d_0}(p) \}=E(p)$ .

Consider $X_{1,l}$ . Suppose that $X_{1,l}=[X_{0,i},X_{0,j}]$ . By Lemma 3.10, $\widehat {f}_*\widehat {X_{1,l}}=\widehat {f}_*[\widehat {X}_{0,i},\widehat {X}_{0,j}] =[\widehat {f}_*\widehat {X}_{0,i},\widehat {f}_*\widehat {X}_{0,j}]=[\widehat {DfX}_{0,i},\widehat {DfX}_{0,j}]$ . And thus,

(3.4) $$ \begin{align} (Df[{X}_{0,i},{X}_{0,j}])^{(2)}=[{DfX}_{0,i},{DfX}_{0,j}]^{(2)}=[\widehat{DfX}_{0,i},\widehat{DfX}_{0,j}]=\lambda^2\widehat{dh}\widehat{X_{1,l}}. \end{align} $$

Set $V^{(i)}(p)=\operatorname {span}\{ {X}_{i,1}(p),\ldots ,{X}_{i,d_i}(p) \}$ so that $E^{i}(p)=V^{(0)}(p)\oplus \cdots \oplus V^{(i)}(p)$ . We note that $\widehat {dh}\widehat {X}_{0,l}\in \operatorname {span}\{ \widehat {X}_{0,1},\ldots ,\widehat {X}_{0,d_0} \}$ since $\widehat {dh}$ is a graded Lie algebra homomorphism. We observe that for every $l=1\ldots ,d_1$ , we have $\widehat {X}_{1,l}(p) \in \operatorname {span}\{ X_{1,l}(p), X_{0,1}(p), \ldots , X_{0,d_0}(p) \}$ . It follows that $(Df[{X}_{0,i},{X}_{0,j}])^{(2)}(p)\in \operatorname {span}\{ \widehat {X}_{0,1}(p),\ldots ,\widehat {X}_{0,d_0}(p), \widehat {X}_{1,1}(p),\ldots ,\widehat {X}_{1,d_1}(p) \}=E^{1}(p)$ . On the other hand,

$$ \begin{align*}(Df[{X}_{0,i},{X}_{0,j}])(p)=(Df[{X}_{0,i},{X}_{0,j}])^{(2)}(p)+(Df[{X}_{0,i},{X}_{0,j}])^{(\le 1)}(p),\end{align*} $$

where $(Df[{X}_{0,i},{X}_{0,j}])^{(\le 1)}$ is the homogeneous part of $(Df[{X}_{0,i},{X}_{0,j}])$ of degree at most 1. Evaluating a vector field of degree at most 1 at p, we get a vector in $E(p)= \operatorname {span}\{ \widehat {X}_{0,1}(p),\ldots ,\widehat {X}_{0,d_0}(p) \}$ . Therefore, $(Df[{X}_{0,i},{X}_{0,j}])(p)\in \operatorname {span}\{ \widehat {X}_{0,1}(p),\ldots , \widehat {X}_{0,d_0}(p),\widehat {X}_{1,1}(p),\ldots ,\widehat {X}_{1,d_1}(p) \}=E^{1}(p)$ .

Inductively, for $X_{m+1,l}=[X_{m,i},X_{0,j}]$ , we obtain

(3.5) $$ \begin{align} (Df[{X}_{m,i},{X}_{0,j}])^{(m+2)}&=[{DfX}_{m,i},{DfX}_{0,j}]^{(m+2)}\nonumber\\&=[({DfX}_{m,i})^{(m+1)},\widehat{DfX}_{0,j}]=\lambda^{m+2}\widehat{dh}\widehat{X}_{m+1,l}. \end{align} $$

And it follows that

(3.6) $$ \begin{align} \begin{split} (D_pf)(({X}_{m+1,l})(p))&=\lambda^{m+2}(\widehat{dh}\widehat{X}_{m+1,l})(p)+(Df{X}_{m+1,l})^{(\le m+1)}(p) \in E^{m+1}(p), \end{split} \end{align} $$

where the notation $Y^{(\le m+1)}$ indicates the sum of the homogeneous parts of any vector field Y of degree at most $m+1$ .

Therefore, with respect to the basis $({X}_{0,1}(p),\ldots ,{X}_{r,d_r}(p))$ , $D_pf$ is a block upper triangular matrix. We let $A_i$ be the blocks on the diagonal of $D_pf$ corresponding to the basis elements $\{ X_{i,1}(p),\ldots ,X_{i,d_i}(p) \}$ , for every $i=0,\ldots , r$ . We note that the size of $A_i$ is $\dim (\mathfrak {n}^{i}_p)\times \dim (\mathfrak {n}^{i}_p)$ . Since $(Df{X}_{m+1,l})^{(\le m+1)}(p)\in E^{m}(p)$ , equation (3.6) implies that

$$ \begin{align*}\operatorname{\mathrm{proj}}_{V^{(m+1)}(p)}(D_pf)(({X}_{m+1,l})(p))=\operatorname{\mathrm{proj}}_{V^{(m+1)}(p)}(\lambda^{m+2}(\widehat{dh}\widehat{X}_{m+1,l})(p)),\end{align*} $$

where $\operatorname {\mathrm {proj}}_{V^{(m+1)}(p)}$ is the projection of $T_pN$ onto $V^{(m+1)}(p)$ along the subspace $\check {V}^{(m+1)}(p):=V^{(0)}(p)\oplus \cdots \oplus V^{(m)}(p)\oplus V^{(m+2)}(p)\oplus \cdots \oplus V^{(r)}(p)$ . Thus the $(j,l)$ entry of the block matrix $A_{m+1}$ is the $X_{m+1,j}(p)$ component of $\lambda ^{m+2}(\widehat {dh}\widehat {X}_{m+1,l})(p)$ , for $j,l\in \{ 1,\ldots ,d_{m+1} \}$ .

Next, we estimate Lyapunov exponents of block matrices $A_0, \ldots , A_r$ . Applying equation (3.5) for $f^n$ instead of f, we obtain

(3.7) $$ \begin{align} (Df^n[{X}_{m,i},{X}_{0,j}])^{(m+2)}&=[{Df^nX}_{m,i},{Df^nX}_{0,j}]^{(m+2)}\nonumber\\ &=[({Df^nX}_{m,i})^{(m+1)},\widehat{Df^nX}_{0,j}]=\lambda^{n(m+2)}\widehat{dh}^n\widehat{X}_{m+1,l}. \end{align} $$

Since

$$ \begin{align*} (Df^n[{X}_{m,i},{X}_{0,j}])(p)=(Df^n[{X}_{m,i},{X}_{0,j}])^{(m+2)}(p)+(Df^n[{X}_{m,i},{X}_{0,j}])^{(\le m+1)}(p) \end{align*} $$

and

$$ \begin{align*} (Df^n[{X}_{m,i},{X}_{0,j}])^{(\le m+1)}(p)\in E^{m}(p), \end{align*} $$

the projection of $(Df^n[{X}_{m,i},{X}_{0,j}])(p)$ along $\check {V}^{(m+1)}(p)$ onto $V^{(m+1)}(p)$ is the same as the projection of $\lambda ^{n(m+2)}\widehat {dh}^n\widehat {X}_{m+1,l}(p)$ to $V^{(m+1)}(p)$ along $\check {V}^{(m+1)}(p)$ .

We claim that

(3.8) $$ \begin{align} \operatorname{span}\{ \widehat{X}_{m+1,1}(p),\ldots,\widehat{X}_{m+1,k_{m+1}}(p) \}\cap \check{V}^{(m+1)}(p)=\{0\}. \end{align} $$

Indeed, we have $\operatorname {span}\{ \widehat {X}_{0,1}(p),\ldots ,\widehat {X}_{i,d_{i}}(p) \}= \operatorname {span}\{ {X}_{0,1}(p),\ldots ,{X}_{i,d_{i}}(p) \}$ for all $i=0,\ldots ,m+1$ . Moreover, these sets of vectors form bases of their corresponding spanned vector spaces. This implies the claim.

This claim implies that for every $w\neq 0$ in $\operatorname {span}\{ \widehat {X}_{m+1,1}(p),\ldots ,\widehat {X}_{m+1,d_{m+1}}(p) \}$ , its projection to $V^{(m+1)}(p)$ along $\check {V}^{(m+1)}(p)$ is non-zero. For every $v\neq 0$ in $V^{(m+1)}(p)$ , there are $c_1,\ldots , c_{d_{m+1}}\in \mathbb {R}$ not all being zero such that $v=c_1 {X}_{m+1,1}(p) +\cdots + c_{d_{m+1}}{X}_{m+1,d_{m+1}}(p)$ . We then have

$$ \begin{align*} D_pf^nv&=c_1 (Df^n{X}_{m+1,1})(p) +\cdots + c_{k_{m+1}}(D_pf^n{X}_{m+1,k_{m+1}}(p))\\ &=c_1 (Df^n{X}_{m+1,1})^{(m+2)}(p) +\cdots + c_{k_{m+1}}(D_pf^n{X}_{m+1,k_{m+1}}^{(m+2)}(p))\\ &\quad+ c_1 (Df^n{X}_{m+1,1}^{(\le m+1)})(p) +\cdots + c_{k_{m+1}}(D_pf^n{X}_{m+1,k_{m+1}}^{(\le m+1)})(p)\\ &=\lambda^{n(m+2)}\widehat{dh}^n (c_1 \widehat{X}_{m+1,1}(p) +\cdots + c_{k_{m+1}}\widehat{X}_{m+1,k_{m+1}}(p))\\ &\quad+ c_1 (Df^n{X}_{m+1,1}^{(\le m+1)})(p) +\cdots + c_{k_{m+1}}(D_pf^n{X}_{m+1,k_{m+1}}^{(\le m+1)})(p). \end{align*} $$

By same argument as above, $D_pf^nv$ and $\lambda ^{n(m+2)}\widehat {dh}^n (c_1 \widehat {X}_{m+1,1}(p) +\cdots + c_{d_{m+1}}\widehat {X}_{m+1,d_{m+1}} (p))$ have the same projection onto $\operatorname {span}\{ {X}_{m+1,1}(p),\ldots ,{X}_{m+1,d_{m+1}}(p) \}$ along $\operatorname {span}\{ {X}_{1,1}(p),\ldots ,{X}_{m,d_{m}}(p) \}$ . Since $c_1, \dots, c_{d_{m+1}}$ are not all zero, the vector $c_1 \widehat {X}_{m+1,1}(p) +\cdots + c_{d_{m+1}}\widehat {X}_{m+1,d_{m+1}}(p)$ is non-zero, and thus has a non-zero projection on $\operatorname {span}\{ {X}_{m+1,1}(p),\ldots ,{X}_{m+1,d_{m+1}}(p) \}$ by (3.8).

Since h is in a compact subgroup of $\mathrm {GL}(n,\mathbb {R})$ , there is a subsequence $(n_k)$ such that $\widehat {dh}^{n_k}$ converges to the identity. Choose an arbitrary norm $\|\cdot \|$ on $V^{(m+1)}(p)$ . Therefore, we may find two constants $C_1,C_2>0$ independent of k such that

$$ \begin{align*}C_1\lVert v \rVert<\|\operatorname{\mathrm{proj}}_{V^{(m+1)}(p)}(\widehat{dh}^{n_k}(v))\|<C_2\lVert v \rVert.\end{align*} $$

Hence,

$$ \begin{align*}C_1\lambda^{n_k(m+2)}\|v\|<\|\operatorname{\mathrm{proj}}_{V^{(m+1)}(p)}(D_pf^{n_k}v)\|<C_2\lambda^{n_k(m+2)}\|v\|,\end{align*} $$

for every $v\neq 0$ in $V^{(m+1)}(p)$ . It follows that $A_{m+1}$ has all Lyapunov exponents equal to $(m+2)\log \lambda $ .

Finally, as is well known, the Lyapunov exponents of a block upper triangular matrix are the same as those of the corresponding block diagonal matrix.

Proof. By switching f with $f^{-1}$ we observe that it is sufficient to prove just the lower bound. Let $\{ {X}_{0,1},\ldots ,{X}_{r,d_r} \}$ be a graded basis of N in a neighborhood of p. We let $\displaystyle {\mathcal {M}_p: \operatorname {span}\{ \widehat {X}_{0,1},\ldots ,\widehat {X}_{r,d_r} \}\to \mathfrak {n}_p}$ be the Métivier correspondence. Note that $n_i=\sum _{j=0}^i d_i$ for $i=0,\ldots ,r$ .

Our assumptions imply that the induced map $f_*$ exists as a graded automorphism. Its derivative $df_*$ and the corresponding $\widehat {f}_*=\mathcal {M}_{f(p)}^{-1}\circ df_* \circ \mathcal {M}_p$ are graded Lie algebra automorphisms. With respect to the basis $(\widehat {X}_{0,1},\ldots ,\widehat {X}_{r,d_r})$ , $\widehat {f}_*$ is a block diagonal matrix. We further assume that we have chosen the initial horizontal basis $X_{0,1},\ldots ,X_{0,d_0}$ so that $\widehat {f}_*$ is in real Jordan canonical form with respect to $\widehat {X}_{0,1},\ldots ,\widehat {X}_{0,d_0}$ .

We note that for every $0\le j\le r$ , $\operatorname {span}\{ \widehat {X}_{0,1}(p) \ldots ,\widehat {X}_{j,d_j}(p) \}=\operatorname {span}\{ {X}_{0,1}(p),\ldots , {X}_{j,d_j}(p) \}\subset T_pN$ . Since $\widehat {f}_*$ is a block diagonal matrix, $(\widehat {f}_*\widehat {X}_{j,l})(p)\in \operatorname {span}\{ \widehat {X}_{j,1}(p), \ldots ,\widehat {X}_{j,d_j}(p) \}$ for every $1\le l\le d_j$ . On the other hand, $(Df{X}_{j,l} - \widehat {f}_*\widehat {X}_{j,l})(p)\in \operatorname {span}\{ \widehat {X}_{0,1}(p), \ldots ,\widehat {X}_{j-1,d_{j-1}}(p) \}$ by Lemma 3.10. Here we used the convention that $\operatorname {span}\{ \widehat {X}_{0,1}(p), \ldots ,\widehat {X}_{j-1,d_{j-1}}(p) \}=\{0\}$ when $j=0$ . By replacing f by $f^n$ , we have that $(Df^n{X}_{j,l} - \widehat {f}_*^n\widehat {X}_{j,l})(p)\in \operatorname {span}\{ \widehat {X}_{0,1}(p), \ldots ,\widehat {X}_{j-1,d_{j-1}}(p) \}$ for every $n\in \mathbb {N}$ . In particular, in the $\widehat {X}_{j,l}$ basis, $D_pf$ is block upper triangular and with diagonal blocks coinciding with the matrix of $\widehat {f}_*$ in this basis. Because $\widehat {f}_*^n\widehat {X}_{j,l}(p)\in \operatorname {span}\{ \widehat {X}_{j,1}(p),\ldots ,\widehat {X}_{j,d_j}(p) \}$ and $\operatorname {span}\{ \widehat {X}_{j,1}(p),\ldots ,\widehat {X}_{j,d_j}(p) \}\cap \operatorname {span}\{ \widehat {X}_{0,1}(p),\ldots ,\widehat {X}_{j-1,d_{j-1}}(p) \}=\{0\}$ , the growth rate of $Df^nX_{j,l}(p)=D_pf(X_{j,l}(p))$ is at least as large as the growth rate of $\widehat {f}_*^n\widehat {X}_{j,l}(p)$ for all $1\leq l \leq d_j$ . Hence for any $v=\sum _{i=1}^{d_j} a_i {X}_{j,1}(p)$ the growth rate of $Df^n v$ is at least as large as the growth rate of $\widehat {f}_*^n(w)$ where $w=\sum _{i=1}^{d_j} a_i \widehat {X}_{j,1}(p)$ . By the block upper triangular structure of $D_pf$ , it follows that the growth rate of $D_pf^n(v)$ is at least as large as the minimum of the growth rates of $\widehat {f}_*^n$ restricted to $\operatorname {span}\{ \widehat {X}_{j,1}(p), \ldots , \widehat {X}_{j,d_j}(p) \}$ .

The conclusion now follows from the fact that the growth rate of $\widehat {f}_*^n\widehat {X}_{j,i}(p)$ is at least $j+1$ times the minimum of growth rates of $\widehat {f}_*^n\widehat {X}_{0,1}(p), \ldots , \widehat {f}_*^n\widehat {X}_{0,d_0}(p)$ due to the bracket relations. (Recall from the choice of initial basis that $\widehat {f}_*$ achieves its minimal growth rate on one of the vectors $\widehat {X}_{0,1},\ldots ,\widehat {X}_{0,d_0}$ .) Since ${X}_{0,1},\ldots ,{X}_{0,d_0}$ are horizontal, $\widehat {f}_*^n\widehat {X}_{0,i}(p)=D_pf^n({X}_{0,i}(p))$ . Thus, the minimum of growth rates of $\widehat {f}_*^n\widehat {X}_{0,1}(p), \ldots , \widehat {f}_*^n\widehat {X}_{0,d_0}(p)$ is at least $\log \lambda _1$ .

Proof. The Heisenberg group $H^{2n+1}$ has a distinguished basis of right invariant vector fields, $\overline {X}_1,\ldots , \overline {X}_n, \overline {Y}_1, \ldots , \overline {Y}_n, \overline {Z}$ such that $[\overline {X}_i,\overline {Y}_j]=\delta _{ij}\overline {Z}$ and $[\overline {X}_i,\overline {X}_j]=[\overline {Y}_i,\overline {Y}_j]=[\overline {X}_i,\overline {Z}]=[\overline {Z},\overline {Y}_j]=0$ . By assumption, $TC_pN$ is isomorphic to $H^{2n+1}$ . Hence in a neighborhood of p there is a graded basis of fields $X_1,\ldots , X_n, Y_1, \ldots , Y_n,Z=[X_1,Y_1]$ such that $\mathcal {M}_p(\widehat {X}_i)=\overline {X}_i$ , $\mathcal {M}_p(\widehat {Y}_i)=\overline {Y}_i$ and $\mathcal {M}_p([\widehat {X}_i,\widehat {Y}_i])=\overline {Z}$ for $i\in \{ 1,\ldots ,n \}$ . In other words, setting $\widehat {Z}=[\widehat {X}_1,\widehat {Y}_1]$ , we have $[\widehat {X}_i,\widehat {Y}_j]=\delta _{ij}\widehat {Z}$ and $[\widehat {X}_i,\widehat {X}_j]=[\widehat {Y}_i,\widehat {Y}_j]=[\widehat {X}_i,\widehat {Z}]=[\widehat {Z},\widehat {Y}_j]=0$ for all $i,j\in \{ 1,\ldots ,n \}$ . Note that by the identity (3.1), we have $Z^{(2)}=\widehat {Z}$ .

With respect to the chosen basis we have an identification $\mathfrak {n}_p=\mathbb {R}^{2n}\oplus \mathbb {R}$ . The graded automorphism group of $H^{2n+1}$ is $\mathbb {Z}_2\ltimes (\operatorname {Sp}(2n,\mathbb {R}) \times \operatorname {Dil})$ where $\operatorname {Sp}(2n,\mathbb {R})$ is the symplectic group, $\operatorname {Dil}\cong \mathbb {R}_+$ are the dilations and the $\mathbb {Z}_2$ generator switches $\overline {X}_i$ and $\overline {Y}_i$ for $i=1,\ldots n$ [Reference TilgnerTil70]. Since $df_*:\mathfrak {n}_p\to \mathfrak {n}_p$ is a graded automorphism, there is an $A\in \operatorname {Sp}(2n,\mathbb {R})$ , a block matrix $U=\big[\begin {smallmatrix} 0 & I_n\\ I_n &0\end {smallmatrix}\big]$ or $U=I_{2n}$ , and $\sigma>0$ such that $df_*$ has the form $df_*(x,z)=(\sigma U A x, \sigma ^2z)$ . (Here the U matrix represents the corresponding element of $\mathbb {Z}_2$ .) Since Lyapunov exponents of $f^2$ are twice those of f and $(UA)^2\in \operatorname {Sp}(2n,\mathbb {R})$ , without loss of generality we may replace f with $f^2$ and assume $df_*(x,z)=(\sigma A x, \sigma ^2z)$ . Next we show that with respect to the basis $\{ X_1(p),\ldots , X_n(p),Y_1(p),\ldots ,Y_n(p),Z(p) \}$ of $T_pN$ , $D_pf=\big[\begin {smallmatrix}\sigma A&B\\0&\sigma ^2\end {smallmatrix}\big]$ for some $2n\times 1$ matrix B.

It follows that with respect to the basis $\{\widehat {X}_1,\ldots , \widehat {X}_n, \widehat {Y}_1, \ldots , \widehat {Y}_n\}$ , $\widehat {f}_*(\widehat {X}_i)=\sigma A\widehat {X}_i$ and $\widehat {f}_*(\widehat {Y}_j)=\sigma A\widehat {Y}_j$ . We note that $\widehat {X}_i(p)=X_i(p)$ and $\widehat {Y}_j(p)=Y_j(p)$ . By Lemma 3.10, $\widehat {f}_*(\widehat {X}_i)=\widehat {DfX_i}$ and $\widehat {f}_*(\widehat {Y}_j)=\widehat {DfY_i}$ . Thus with respect to the basis $\{X_1(p),\ldots , X_n(p), Y_1(p), \ldots , Y_n(p)\}$ , by restricting of $D_pf$ on $\operatorname {span}\{ X_1(p),\ldots , X_n(p), Y_1(p), \ldots , Y_n(p) \}$ , we have $D_pf(X_i(p))=AX_i(p)$ and $D_pf(Y_j(p))=AY_j(p)$ .

Moreover, we have

$$ \begin{align*} \sigma^2Z^{(2)}&=\widehat{f}_*(Z^{(2)})=\widehat{f}_*([\widehat{X}_1,\widehat{Y}_1])=[\widehat{f}_*\widehat{X}_i,\widehat{f}_*\widehat{Y}_i]\\&=[\widehat{DfX}_1,\widehat{DfY}_1]=(Df[X_1,Y_1])^{(2)}=(DfZ)^{(2)}. \end{align*} $$

Thus,

$$ \begin{align*} D_pf(Z(p))\kern1.5pt{=}\kern1.5pt DfZ(p)\kern1.5pt{=}\kern1.5pt(DfZ)^{(2)}(p)+(DfZ)^{(\le 1)}(p)\kern1.5pt{=}\kern1.5pt\sigma^2Z^{(2)}(p)+(DfZ) {}^{(\le 1)}(p). \end{align*} $$

We note that $(DfZ)^{(\le 1)}(p)\in \operatorname {span}\{ X_1(p),\ldots , X_n(p), Y_1(p), \ldots , Y_n(p) \}$ . Also, $Z=Z^{(2)}+Z^{(\leq 1)}$ with $Z^{(\leq 1)}(p)\in \operatorname {span}\{ X_1(p),\ldots , X_n(p), Y_1(p), \ldots , Y_n(p) \}$ . Hence, with respect to the basis $\{{X}_1(p),\ldots , {X}_n(p), {Y}_1(p), \ldots , {Y}_n(p), Z(p)\}$ of $T_pN$ , there is a $2n\times 1$ matrix B such that $D_pf=\big[\begin {smallmatrix}\sigma A&B\\0&\sigma ^2\end {smallmatrix}\big]$ .

Now it follows that the Lyapunov exponents of f are $2 \log \sigma $ and the logarithms of the moduli of the eigenvalues of $\sigma A$ . Since $\det (\sigma A)=\sigma ^{2n}$ , we conclude that $n\log \lambda _{2n+1}=2n\log \sigma $ is equal to the sum of all other exponents.