2.1 Dominated splitting properties

Throughout this section, we adapt some of the main properties about dominated splitting which appear in [Reference Crovisier and Potrie10] in the diffeomorphisms context to the context of endomorphisms displaying critical points.

Let f be an endomorphism displaying critical points and $E\oplus F$ be a dominated splitting over f-invariant subset $\Lambda $ of $M_f$ .

The uniqueness of the dominated splitting is guaranteed by the following.

The proof needs some preliminaries.

Proof. We can, without loss of generality, choose $\ell \geq 1$ such that $E\prec _{\ell } F$ and $G\prec _{\ell } H$ . Moreover, it follows from Remark 2.4 that $Df\mid _{H}$ is an isomorphism and $\ker (Df^m_{x_i})$ must be contained in $E(x_i)$ and $G(x_i)$ , for each $(x_i)_i \in \Lambda $ and $m\geq 1$ .

To conclude the proof of the proposition, it remains to show that if $u \in E(x_i)$ such that $Df^m(u)\neq 0$ , for every $m\geq 1$ , then $u \in G(x_i)$ . Then, for every $u \in E(x_i)$ , one can decompose $u=u_G+u_H$ in a unique way where $u_G \in G(x_i)$ and $u_H \in H(x_i)$ . Analogously, one can decompose $u_H=u^{\prime }_E+u^{\prime }_F$ , where $u^{\prime }_E \in E(x_i)$ and $u^{\prime }_F \in F(x_i)$ . Then $u^{\prime }_F$ must be zero. Otherwise, we get that

$$ \begin{align*} \|Df^{k\ell}(u)\|&\geq \|Df^{k\ell}(u_H)\|-\|Df^{k\ell}(u_G)\|\geq \bigg(1-\frac{1}{2^{k}}\bigg)\|Df^{k\ell}(u_H)\|\\ &\geq \bigg(1-\frac{1}{2^{k}}\bigg) (\|Df^{k\ell}(u^{\prime}_F)\|-\|Df^{k\ell}(u^{\prime}_E)\| )\\ &\geq \bigg(1-\frac{1}{2^{k}}\bigg)^2 \|Df^{k\ell}(u^{\prime}_F)\|, \end{align*} $$

implying that $\|Df^{k\ell }(u)\|$ and $\|Df^{k\ell }(u^{\prime }_F)\|$ have the same growth, which is impossible. Therefore, $u_H \in E(x_i)\cap H(x_i)$ and $u_G \in E(x_i)\cap G(x_i)$ . Symmetrically, we deduce that if $v \in G(x_i)$ whose $v=v_E+v_F$ , then $v_E \in G(x_i)\cap E(x_i)$ and $v_F \in G(x_i)\cap F(x_i)$ . Moreover, since $\dim E(x_i)\leq \dim G(x_i)$ , we have that either $G(x_i)=E(x_i)$ or $G(x_i)\cap F(x_i)\neq \{0\}$ .

Take non-zero vectors $u \in E(x_i)\cap H(x_i)$ and $v \in G(x_i)\cap F(x_i)$ . Then, as $G \prec H$ , we deduce that $\|Df^{\ell }(u)\|$ grows faster than $\|Df^{\ell }(v)\|$ , which contradicts the fact that ${E \prec F}$ . Thus, at least one of these intersections $E(x_i)\cap H(x_i)$ and $G(x_i)\cap F(x_i)$ is trivial. Thus, we obtain that $E(x_i)\cap H(x_i)=\{0\}$ . It implies that $E(x_i)$ is contained in $G(x_i)$ . To conclude the proof, we should observe that $G(x_i)\cap F(x_i)$ is invariant and ${E(x_i)\prec G(x_i)\cap F(x_i) \prec H(x_i)}$ .

As a direct consequence of the previous lemma, we conclude the first part of the uniqueness of the dominated splitting.

To complete the proof of Proposition 2.6, we state the following.

Proof. First, we assume that $(x_i)_i \in \Lambda $ with $x_i \notin \mathrm {Cr}(f)$ for all $i \in \mathbb {Z}$ . Then, since $Df_{x_i}:T_{x_i}M \to T_{x_{i+1}}M$ is an isomorphism, one has that for every unit vectors $u \in E(x_i)$ and $v \in F(x_i)$ ,

$$ \begin{align*} \|Df^{-\ell}(u)\|\geq \frac{1}{\|Df^{\ell}\!\mid_{E}\!\|}\geq 2\frac{1}{\operatorname{{\mathfrak{m}}}(Df^{\ell}\!\mid_{F})}\geq 2\|Df^{-\ell}(v)\|. \end{align*} $$

In particular, $F\prec _{\ell } E$ for $Df^{-1}$ along the orbit $(x_i)_i$ and, similarly, $H\prec _{\ell } E$ for $Df^{-1}$ . Therefore, applying Corollary 2.8, one concludes that $F(x_i)=H(x_i)$ since ${\dim F=\dim H}$ .

In the general case, for a non-zero vector $u \in F(x_i)$ , one deduces that there are two vectors $w \in E(x_i)$ and $v \in H(x_i)$ , with $v\neq 0$ , such that $u=w+v$ . Assume without loss of generality that $i=0$ . Then, using that $Df\mid _F$ and $Df\mid _H$ are isomorphisms, we can find the sequences $(u_j)_{j\leq 0}\subseteq F\backslash \{0\}, \, (w_i)_{j\leq 0} \subseteq E$ , and $(v_j)_{j\leq 0} \subseteq H\backslash \{0\}$ such that ${u_j=w_j+v_j}$ and $Df(u_{j-1})=u_j$ . Note that if $Df^m(w_{j-1})=0$ , for some $1\leq m\leq |j|$ , there is nothing to be proved since it directly implies that $u \in H(x_j)$ . Hence, we can assume that $Df(\ast _{i-1})=\ast _i$ , where $\ast =u,v,\text { and }w$ and $\ast _0=\ast $ .

Denote $V_i=\mathrm {span}\{w_i,v_i\}$ as the span of $w_i,v_i$ ; and define $A_i:V_{i-1} \to V_i$ as the restriction of $Df_{x_{i-1}}$ to $V_{i-1}$ for each $i\leq 0$ . Observe that defining $A_i^n:V_i \to V_{i+n}$ by $A_i^n=A_{i+n}^{-1}\cdots A_i^{- 1}$ for every $n \leq 0$ , we can repeat the previous arguments to show that $V_i$ admits two dominated splittings for $A_i^{-1}$ , which are $V_i=\mathrm {span}\{u_i\}\oplus \mathrm {span}\{w_i\}$ with $\mathrm {span}\{u_i\}\prec _{\ell } \mathrm {span}\{w_i\}$ , and $V_i=\mathrm {span}\{v_i\}\oplus \mathrm {span}\{w_i\}$ with $\mathrm {span}\{v_i\}\prec _{\ell } \mathrm {span}\{w_i\}$ since $u_i \in V_i$ and the following holds:

$$ \begin{align*}&\|A_i{\mid}_{\mathrm{span}\{w_i\}}{\kern-1.5pt}\|\leq \|Df{\mid}_{E}{\kern-1.5pt}\|, \ \|A_i{\mid}_{\mathrm{span}\{u_i\}}{\kern-1.5pt}\|\geq \operatorname{{\mathfrak{m}}}(Df{\mid}_{F}), \quad \text{and}\\ &\qquad\qquad\qquad\ \ \|A_i{\mid}_{\mathrm{span}\{v_i\}}{\kern-1.5pt}\|\geq \operatorname{{\mathfrak{m}}}(Df{\mid}_{H}).\end{align*} $$

Therefore, we repeat the argument in the proof of Lemma 2.7 to get that $v_0$ and $u_0$ are parallel. This completes the proof.

We can use the uniqueness of the dominated splitting to show the continuity and extend it to the closure.

Remark 2.11. It is well known and follows from the proof of uniqueness that if $E\oplus F$ is a dominated splitting, then the subbundle E only depends on the forward orbits. That is, for all points $(x_i)_i$ and $(y_i)_i$ in $\Lambda $ , one has that

(2.5) $$ \begin{align} E(x_i)=E(y_i)\quad \text{for all } i \geq 0, \ \text{ whenever } x_0=y_0. \end{align} $$

In particular, by Propositions 2.6 and 2.10 together with (2.5), the subbundle E of $TM_f$ induces an invariant continuous subbundle of $TM$ which will also be denoted by E.

2.2 Cone criterion

In this section, we will show that the existence of dominated splitting for endomorphisms displaying critical points can also be characterized in terms of cone fields as well as in the diffeomorphisms context.

A cone-field $\mathscr {C}$ of dimension k on M is a continuous family of convex closed non-vanishing cones $\mathscr {C}(x)$ in $T_xM$ such that the subspaces of maximal dimension inside of $\mathscr {C}(x)$ are k-dimensional. The closure of the complement of $\mathscr {C}(x)$ in $T_xM$ is a cone of dimension $d-k$ , called a dual cone at x and denoted by $\mathscr {C}^{\kern2pt\ast }(x)$ . We will call a dual cone-field of $\mathscr {C}$ the family $\mathscr {C}^{\kern2pt\ast }=\{\mathscr {C}^{\kern2pt\ast }(x):x \in M\}$ . We say that a cone-field $\mathscr {C}$ is invariant by an endomorphism f if there is an integer $\ell>0$ such that $Df^{\ell }(\mathscr {C}(x))$ is contained in $\mathrm {int}(\mathscr {C}(f^{\ell }(x)))\cup \{0\}$ , where $\mathrm {int}(\mathscr {C}(x))$ denotes the interior of $\mathscr {C}(x)$ in $T_xM$ . We also can say that $\mathscr {C}$ is $\ell $ -invariant if we want to emphasize the role of $\ell $ .

We next state an equivalent notion of dominated splitting for endomorphisms ${f:M\to M}$ exhibiting critical points.

In the proof, we will first define the cone-field, and then will prove that it is invariant and transversal to the kernel.

By Remark 2.11, E is an invariant continuous subbundle of $TM$ and, by Remark 2.4, we have that $\ker (Df^n_x)$ is contained in $E(x)$ for all $x \in M$ . Then, for every $x \in M$ , we can define a cone-field $\mathscr {C}_{E}$ on M of dimension $\kappa $ and length $\eta>0$ by

(2.6) $$ \begin{align} \mathscr{C}_{E}(x,\eta)=\{(u_1,u_2) \in E(x)\oplus E(x)^{\perp}:\|u_2\|\leq \eta \|u_1\|\}. \end{align} $$

Note that $u=(u_1,u_2) \in E(x)\oplus E^{\perp }(x)$ satisfies that

then, using that cosine is decreasing on $[0,\pi /2]$ , the cone-field can be rewritten as

From now on, since $\arctan : (0,+\infty ) \to (0,\pi /2)$ is a homeomorphism preserving the orientation, we will make an abuse of notation rewriting, for each $\eta \in (0,\pi /2)$ ,

(2.7)

Recall that the dual cone-field of $\mathscr {C}_{E}(x,\eta )$ is the closure of $T_xM\backslash \mathscr {C}_{E}(x,\eta )$ which is a $(d-\kappa )$ -dimensional cone-field and will be written as

(2.8)

It is clear that $E(x) \subseteq \mathscr {C}_E(x,\eta )$ for each $x \in M$ and $\eta \in (0,\pi /2)$ . Thus, we can conclude that $\ker (Df_x^n)\cap \mathscr {C}_E^{\kern2pt\ast }(x,\eta )=\{0\}$ for each $x \in M$ and $n\geq 1$ , since ${\ker (Df_x^n) \subseteq E(x)}$ .

Therefore, to conclude the proof of Proposition 2.12, we state the following.

Indeed, since the angle between E and F is bounded away from zero, there is a number $\alpha>0$ small enough satisfying , for each $i \in \mathbb {Z}$ in all the orbit $(x_i)_i$ . In particular, we have that the direction $F(x_0)$ associated to any orbit $(x_i)_i$ with $x_0=x$ is contained in $\mathscr {C}_E^{\kern2pt\ast }(x,\alpha )$ . Moreover, there is a constant $c>0$ (depending only on $\alpha $ ) such that any decomposition of a vector $w \in \mathscr {C}_E^{\kern2pt\ast }(x,\alpha )$ as $w=w_E+w_F$ with $w_E \in E(x)$ and $w_F$ in direction F at x satisfies $\|w_F\|\geq c\|w_E\|$ . Otherwise, w cannot belong to $\mathscr {C}_E^{\kern2pt\ast }(x,\alpha )$ . Thus, we obtain that

$$ \begin{align*} \|Df^{k\ell}(w_E)\|&\leq \|Df^{k\ell}\!\mid_{E}\!\|\|w_E\|\leq\big(\tfrac{1}{2}\big)^k\operatorname{{\mathfrak{m}}}(Df^{k\ell}\!\mid_{F})\|w_E\|\\ &\leq \big(\tfrac{1}{2}\big)^k \|Df^{k\ell}(w_F)\|\dfrac{\|w_E\|}{\|w_F\|}\leq \big(\tfrac{1}{2}\big)^k c^{-1} \|Df^{k\ell}(w_F)\| \end{align*} $$

and hence, we can take $k>0$ large enough such that $\mathscr {C}_E^{\kern2pt\ast }(x,\alpha )$ is k-invariant for every $x \in M$ . Taking $\mathscr {C}$ as the dual cone-field $\mathscr {C}_E^{\kern2pt\ast }$ , we conclude the proof.

Conversely, we will show that the existence of an invariant cone-field transversal to the kernel is sufficient to get a dominated splitting. More precisely, we state the following.

To prove Proposition 2.13, some preliminaries are needed. For a linear transformation $\Phi : V \to W$ , where V and W are d-dimensional vector spaces with an inner product, we denote by $\Phi ^{\ast }:W \to V$ its adjoint. It is well known that if the kernel of $\Phi $ is r-dimensional, then there are $\{v_1,\ldots ,v_d\}$ and $\{w_1,\ldots ,w_d\}$ orthonormal bases of V and W, respectively, such that $\Phi (v_i)=0,\ \ \Phi ^{\ast }(w_i)=0$ for each $1\leq i \leq r$ ; and $\Phi (v_i)=\sigma _i w_i, \ \ \Phi ^{\ast }(w_i)=\sigma _i v_i$ with $\sigma _i>0$ for each $r<i\leq d$ . Moreover, we have that $\sigma _i=\|\Phi (v_i)\|=\|\Phi ^{\ast }(w_i)\|$ for each $i=1,\ldots ,d$ .

We call the $\sigma _i$ terms the singular values of $\Phi $ . Note that the singular values of $\Phi $ are the eigenvalues of $\sqrt {\Phi ^{\ast }\Phi }$ associated to the eigenvectors $v_i$ , and we will order them as the following:

$$ \begin{align*}\sigma_1 = \cdots = \sigma_r=0 < \sigma_{r+1}\leq \cdots \leq \sigma_d.\end{align*} $$

When the kernel of $\Phi $ is trivial (that is, $r=0$ ), we have $0<\sigma _1\leq \sigma _2 \leq \cdots \leq \sigma _d$ .

Note that the $\sigma _i$ terms are also the singular values of $\Phi ^{\ast }$ . Moreover, the singular values are typically ordered from largest to smallest, but we are taking the opposite convention. For more information, see [Reference Horn and Johnson13, Theorem 7.3.2].

We would also like to introduce a useful ‘minimax’ characterization of the singular values, which appears in [Reference Bochi, Potrie and Sambarino6, Appendix A]:

(2.9) $$ \begin{align} \begin{split} \sigma_{j}&=\min\{\|\Phi\mid_{P}\!\!\|: \, P \in \mathrm{Grass}_j(V)\}; \quad \text{and} \\ \sigma_{j+1}&=\max\{\operatorname{{\mathfrak{m}}}(\Phi\mid_{U}): \, U \in \mathrm{Grass}_{d-j}(V)\}, \end{split} \end{align} $$

where $\mathrm {Grass}_r(V)$ denotes the r-dimensional Grassmannian of V.

From now on, we denote by $\sigma _1(x,n)\leq \cdots \leq \sigma _d(x,n)$ the singular values of $Df^n_x$ and by $\{e_1^n(x),e_2^n(x),\ldots ,e_d^n(x)\}$ an orthogonal basis of $T_xM$ so that $\|Df^n(e_i^n(x))\|=\sigma _i(x,n)$ for all $x\in M, \, n\geq 1$ and $1\leq i \leq d$ . To simplify notation, we omit the dependence on x and write $e_i$ instead of $e_i(x)$ .

To prove Proposition 2.13, we define $E_n(x)$ as the subspace of $T_xM$ spanned by the vectors $e_i^n$ on the basis such that $\|Df^n(e_i^n)\|=\sigma _i(x,n)< \sigma _{\kappa +1}(x,n)$ ; and $E^{\perp }_n(x)$ as the orthogonal complement of $E_n(x)$ . It should be noted that $E_n(x)$ is at most $\kappa $ -dimensional and $E^{\perp }_n(x)$ is the fastest direction of $Df^n$ , which means that $\operatorname {{\mathfrak {m}}}(Df^n\!\mid _{E^{\perp }_n(x)})\geq \|Df^n(v)\|$ for every unit vector $v \notin E^{\perp }_n(x)$ .

Since the cone-field $\mathscr {C}$ is $(d-\kappa )$ -dimensional and $\ker (Df_x^n)\cap \mathscr {C}(x)=\{0\}$ for each ${x \in M}$ and $n\geq 1$ , we obtain that $\dim \ker (Df^n_x) \leq \kappa $ for each $x \in M$ and $n\geq 1$ . Otherwise, the intersection between $\ker (Df^n_x)$ and the cone $\mathscr {C}(x)$ would be non-trivial. Moreover, as $\dim \ker (Df^n_x)=r$ implies that $\sigma _i(x,n)=0$ for each $1\leq i\leq r$ , we can conclude that $\ker (Df^n_x)$ is contained in $E_n(x)$ .

We will prove that for all $n\geq 1$ , the subspaces $E_n(x)$ are $\kappa $ -dimensional, implying that they are in $\mathrm {Grass}_{\kappa }(T_xM)$ . After that, we will show that the family $(E_n)_n$ is a Cauchy sequence and, recalling that $\mathrm {Grass}_{\kappa }(T_xM)$ is a compact space with the usual distance (2.4), we will get the limit $E(x)$ for each $x \in M$ . Finally, we will prove that E is $Df$ -invariant.

Note that the subspace $E(x)$ which is limit of $E_n(x)$ satisfies that $\ker (Df^n_x) \subseteq E(x)$ for each $n\geq 1$ . Indeed, since $\ker (Df^n_x) \subseteq E_n(x)$ and $\ker (Df^j_x) \subseteq \ker (Df^{j+1}_x)$ , we have that $\ker (Df^j_x) \subseteq E_n(x)$ for each $1\leq j \leq n$ and, then, we get that $\ker (Df^j_x) \subseteq E(x)$ for each $j\geq 1$ .

To define the direction F, we observe that the assumptions

$$ \begin{align*} Df_x^{\ell}(\mathscr{C}(x))\subseteq \mathrm{int}(\mathscr{C}(f^{\ell}(x))) \quad \text{and} \quad \ker(Df_x^n)\cap \mathscr{C}(x)=\{0\} \end{align*} $$

for each $x \in M$ and each $n\geq 1$ , allow us to claim the following, for which the proof can be found in [Reference Crovisier and Potrie10].

Claim 1. For each orbit $(x_i)_i$ , there is a $(d-\kappa )$ -dimensional subbundle F defined by

$$ \begin{align*}F(x_i)=\bigcap_{n\geq 0} Df^{n\ell}(\mathscr{C}(x_{i-n\ell})).\end{align*} $$

Moreover, it is contained in $\mathrm {int}(\mathscr {C}(x_i))$ and is also $Df^\ell $ -invariant.

To complete the proof of Proposition 2.13, we will prove that the subbundle E is transversal to the cone-field, implying that $E\cap F=\{0\}$ ; and, finally, we will show that $E\prec F$ .

These arguments will be divided into a series of lemmas. Before stating the first one, we need to introduce some notation.

Since the subbundle F, as in Claim 1, is in the cone-field $\mathscr {C}$ away from the boundary, we can choose $\delta _0>0$ small enough such that for every orbit $(x_i)_i$ , the family of $(d-\kappa )$ -dimensional cones $\{\mathcal {K}(x_i,\delta _0):i \in \mathbb {Z}\}$ ,

$$ \begin{align*}\mathcal{K}(x_i,\delta_0):=\{(u,v) \in F^{\perp}(x_i)\oplus F(x_i):\|u\|\leq\delta_0\|v\|\},\end{align*} $$

satisfies that $\mathcal {K}(x_i,\delta _0)$ is contained in the interior of $\mathscr {C}(x_i)$ for each $i \in \mathbb {Z}$ .

We are getting this alternative family of cones because that family is centered in the direction F along the orbit $(x_i)_i$ , which does not necessarily happen for the cone-field $\mathscr {C}$ , and this will be useful to make the computation.

Now, we state the first lemma toward the proof of Proposition 2.13. Roughly speaking, the lemma guarantees that the directions out of the cone $\mathcal {K}(x_i,\delta )$ are dominated by the direction F.

Lemma 2.14. For every $0<\delta <\delta _0$ and $\varepsilon>0$ small enough, there is an integer $N>0$ such that every orbit $(x_i)_i$ holds

(2.10) $$ \begin{align} Df^{N\ell}(\mathcal{K}(x_i,\delta))\subseteq \mathrm{int}(\mathcal{K}(x_{i+N\ell},\varepsilon))\cup\{0\}. \end{align} $$

Moreover, for every $0<\unicode{x3bb} <1$ and every $0<\delta <\delta _0$ small enough, we can choose $N>0$ such that if $w \notin \mathcal {K}(x_i,\delta )$ with $\|w\|=1$ , then at least one of the following holds:

1. (i) $Df^{N\ell }(w) \in \mathcal {K}(x_{i+N\ell },\delta )$ ; or

2. (ii) $\|Df^{N\ell }(w)\|\leq \unicode{x3bb} \operatorname {{\mathfrak {m}}}(Df^{N\ell }\!\mid _{F(x_i)})$ .

Proof. Since $\mathcal {K}(x_i,\delta ) \subseteq \mathscr {C}(x_i)$ along the orbit $(x_i)_i$ and by the definition of F, we have that for every $0<\varepsilon <\delta $ , there is a large number $N>0$ such that

$$ \begin{align*}Df^{N\ell}(\mathcal{K}(x_i,\delta))\subseteq Df^{N\ell}(\mathscr{C}(x_i)) \subseteq \mathcal{K}(x_{i+N\ell},\varepsilon)\end{align*} $$

for each $i \in \mathbb {Z}$ . This proves the first part of the lemma.

To prove the second part, let us fix any $\varepsilon $ where $0<\varepsilon <({\delta }/({1+\delta }))({\delta \unicode{x3bb} }/ ({1+\delta \unicode{x3bb} }))$ and $N>0$ satisfying (2.10). We suppose by contradiction that there is a unit vector $w \notin \mathcal {K}(x_i,\delta )$ such that

$$ \begin{align*}\|Df^{N\ell}(w)\|\geq \unicode{x3bb} \operatorname{{\mathfrak{m}}}(Df^{N\ell}\!\mid_{F(x_i)}) \quad \text{and} \quad Df^{N\ell}(w) \notin \mathcal{K}(x_{i+N\ell},\delta).\end{align*} $$

Then, we will prove that (2.10) does not happen, contradicting the first part of the lemma.

Indeed, write $Df^{N\ell }(w)=au_1+bv_1$ , where $u_1 \in F^{\perp }(x_{i+N\ell })$ and $v_1 \in F(x_{i + N\ell })$ are unit vectors such that $|a|\geq \delta |b|$ . Then, we can bound $|a|$ from below and $|b|$ from above as follows:

$$ \begin{align*}|a|\bigg(1+\dfrac{1}{\delta}\bigg)\geq|a|+|b|\geq \|Df^{ N\ell}(w)\| \quad \text{and} \quad \|Df^{ N\ell}(w)\|\geq |b|,\end{align*} $$

which gives

$$ \begin{align*}|a| \geq \bigg(\dfrac{\delta}{1+\delta}\bigg) \|Df^{ N\ell}(w)\|.\end{align*} $$

Take a unit vector $v \in F(x_i)$ satisfying $\|Df^{N\ell }(v)\|=\operatorname {{\mathfrak {m}}}(Df^{N\ell }\!\mid _{F(x_i)})$ and choose a number $c \in \mathbb {R}$ such that $w'=cw+v$ belongs to the boundary of $\mathcal {K}(x_i,\delta )$ , which is contained in $\mathscr {C}(x_i)$ . Note that the constant c satisfies $|c|\geq \delta $ , since w is a unit vector out of the cone $\mathcal {K}(x_i,\delta )$ and v is a unit vector in $F(x_i)$ .

We now decompose $Df^{N\ell }(w')=ca u_1+ cbv_1+\operatorname {{\mathfrak {m}}}(Df^{N\ell }\!\mid _{F(x_i)})v_0$ , where $v_0=Df^{N\ell }(v)/\|Df^{N\ell }(v)\| \in F(x_{i+N\ell })$ and then we can get that

$$ \begin{align*} \dfrac{\|acu_1\|}{\|bcv_1+\operatorname{{\mathfrak{m}}}(Df^{N\ell}\!\mid_{F(x_i)})v_0\|}& \geq\dfrac{|a||c|}{|b||c|+\operatorname{{\mathfrak{m}}}(Df^{N\ell}\!\mid_{F(x_i)})}\\ &\geq\bigg(\dfrac{\delta}{1+\delta}\bigg) \dfrac{\|Df^{N\ell}(w)\||c|}{\|Df^{N\ell}(w)\||c|+\operatorname{{\mathfrak{m}}}(Df^{N\ell}\!\mid_{F(x_i)})}\\ &\geq \bigg(\dfrac{\delta}{1+\delta}\bigg)\bigg(\dfrac{|c|}{|c|+\unicode{x3bb}^{-1}}\bigg)\\ &\geq \bigg(\dfrac{\delta}{1+\delta}\bigg) \bigg(\dfrac{\unicode{x3bb}|c|}{\unicode{x3bb}|c|+1}\bigg)\\ &\geq \bigg(\dfrac{\delta}{1+\delta}\bigg) \bigg(\dfrac{\unicode{x3bb}}{\unicode{x3bb}+|c|^{-1}}\bigg)\\ & \geq \bigg(\dfrac{\delta}{1+\delta}\bigg) \bigg(\dfrac{\unicode{x3bb}}{\unicode{x3bb}+\delta^{-1}}\bigg) \geq \bigg(\dfrac{\delta}{1+\delta}\bigg)\bigg(\dfrac{\delta\unicode{x3bb}}{1+\delta\unicode{x3bb}}\bigg)>\varepsilon. \end{align*} $$

Therefore, we obtain that $Df^{N\ell }(w')$ does not belong to $\mathcal {K}(x_{i+N\ell },\varepsilon )$ which contradicts (2.10) and therefore completes the proof of the lemma.

From now on, we fix $\unicode{x3bb} , \delta $ , and $\varepsilon $ as in Lemma 2.14 and, to simplify notation, we will assume that $N=1$ .

We will prove that there is an ‘exponential gap’ which means the quotient between $\sigma _{\kappa }(x,n)$ and $\sigma _{\kappa +1}(x,n)$ decreases exponentially fast to zero. It will be useful to prove that $(E_n(x))_n$ is a Cauchy sequence.

Lemma 2.15. There is $c>0$ such that

$$ \begin{align*}\sigma_{\kappa}(x,n) < c \unicode{x3bb}^n \sigma_{\kappa+1}(x,n)\end{align*} $$

for all $x \in M$ and $n\geq 1$ .

Proof. We will first prove the lemma for n multiples of $\ell $ .

We will prove for an orbit $(x_i)_i$ that

(2.11) $$ \begin{align} \sigma_{\kappa}(x_i,n\ell)\leq \unicode{x3bb}^n \sigma_{\kappa+1}(x_i,n\ell) \end{align} $$

for every $n\geq 1$ .

Consider $P_0=E_{n\ell }(x_i)$ and $P_j=Df^{j\ell }(E_{n\ell }(x_i))$ for each $1\leq j \leq n$ . We will first show that $P_j \cap \mathcal {K}(x_{i+j\ell },\delta ) =\{0\}$ for each $0\leq j \leq n$ and, by invariance of the family $\{\mathcal {K}(x_i,\delta ):i \in \mathbb {Z}\}$ , it is enough to show this for $P_n$ . Note that $P_n$ is orthogonal to $Df^{n\ell }(E_{n\ell }^\perp (x_i))$ by definition and, since $\dim E_{n\ell }^\perp (x_i)= \dim Df^{n\ell }(E_{n\ell }^\perp (x_i))$ is at least the dimension of the cone, it is enough to show that $Df^{n\ell }(E_{n\ell }^\perp (x_i)) \subset \mathcal {K}(x_{i+n\ell },\delta )$ . If this were not the case, we would find a unit vector $v \in E_{n\ell }^{\perp }(x)$ which does not belong to the cone $\mathcal {K}(x_i,\delta )$ such that $Df^{n\ell }(v)\notin \mathcal {K}(x_{i+n\ell },\delta )$ . We have that $\|Df^{n\ell }(v)\|\geq \sigma _{\kappa +1}(x_i,n\ell )\geq \operatorname {{\mathfrak {m}}}(Df^{n\ell }\mid _{F(x_i)})$ by definition of $E_{n\ell }(x_i)^{\perp }$ . Then, a computation similar to that in Lemma 2.14 gives the contradiction that $Df^{n\ell }(\mathcal {K}(x_i,\delta ))$ is not contained in $\mathcal {K}(x_{i+n\ell },\varepsilon ))$ .

To prove (2.11), we use the ‘minimax’ characterization in (2.9), which says

$$ \begin{align*} \dfrac{\sigma_{\kappa}(x_i,n\ell)} {\sigma_{\kappa+1}(x_i,n\ell)} =\dfrac{\min\{\|Df^{n\ell}\!\mid_{P}\!\|:P\in \mathrm{Grass}_{\kappa}(T_{x_{i}}M)\}} {\max\{\operatorname{{\mathfrak{m}}}(Df^{n\ell}\!\mid_{U}):U \in \mathrm{Grass}_{d-\kappa}(T_{x_{i}}M)\}}\\ \leq \dfrac{\|Df^{n\ell}\!\mid_{E_{n\ell}(x_{i})}\!\|} {\operatorname{{\mathfrak{m}}}(Df^{n\ell}\!\mid_{F(x_i)})}\leq \Pi_{j=0}^{n-1}\dfrac{\|Df^{\ell}\!\mid_{P_j}\!\|} {\operatorname{{\mathfrak{m}}}(Df^{\ell}\!\mid_{F(x_{i+j\ell})})}. \end{align*} $$

Thus, for every $w \in P_j$ , we have that $Df^{\ell }(w) \notin \mathcal {K}(x_{i+j\ell },{\delta })$ and, by Lemma 2.14, we obtain that $\|Df^{\ell }\!\mid _{P_j}\!\|\leq \unicode{x3bb} \operatorname {{\mathfrak {m}}}(Df^{\ell }\!\mid _{F(x_{i+j\ell })})$ .

Finally, by continuity of $\mathscr {C}$ and by $\ker (Df^j)\cap \mathscr {C}=\{0\}$ , we can get a uniform constant $c>0$ such that

$$ \begin{align*}\frac{\max\{\|Df^j_x\|:x \in M\}}{\min\{\operatorname{{\mathfrak{m}}}(Df^j_{x}\!\mid_{U}):U\in \mathrm{Grass}_{d-\kappa}(T_x) \text{ and } U \subseteq \mathscr{C}(x)\}}<c\end{align*} $$

for each $1\leq j<\ell $ and then conclude that $\sigma _{\kappa }(x_i,n)\leq c \unicode{x3bb} ^n \sigma _{\kappa +1}(x_i,n)$ .

The following result follows as immediate consequence of the previous lemma, since $\sigma _{\kappa }(x,n)< \sigma _{\kappa +1}(x,n)$ for every n large enough and $x \in M$ .

Let us remind the reader what we have done so far to prove Proposition 2.13. By Corollary 2.16, we can suppose without loss of generality that $E_n(x)$ is $\kappa $ -dimensional for all $n\geq 1$ and we will next show that the sequence $(E_n(x))_n$ is of Cauchy on $\mathrm {Grass}_{\kappa }(T_xM)$ endowed with the distance (2.4).

We now state the following result (see [Reference Bochi, Potrie and Sambarino6, Lemma A.4] and also see [Reference Bochi and Gourmelon4, Reference Quas, Thieullen and Zarrabi24, Reference Zhu26] for related results).

Before the proof of the lemma, we first need two useful lemmas.

Proof. The proof will follow from item (ii) in Lemma 2.14. Indeed, suppose by contradiction that $Df^n(E_n^{\perp })$ is not contained in the cone. In particular, there is a vector $v \in E_n^{\perp }(x)$ such that $Df^{j\ell }(Df^r(v))$ does not belongs to the cone with $n=j\ell + r$ and $0\leq r \leq \ell -1$ , satisfying:

$$ \begin{align*} \|Df^n(v)\|\leq \unicode{x3bb}^j\operatorname{{\mathfrak{m}}}(Df^{j\ell}\!\mid_{F(x)})\|Df^r(v)\|. \end{align*} $$

Thus, since j goes to $+\infty $ as n goes to $+\infty $ , we can get n large enough such that $\|Df^n(v)\|<\operatorname {{\mathfrak {m}}}(Df^{j\ell }\!\mid _{F(x)})$ . This contradicts the definition of $E_n^{\perp }(x)$ to be the fastest direction of $Df^n$ .

Lemma 2.19. There is a constant $K_1>0$ such that for every $x \in M$ and n large, the following inequality holds:

$$ \begin{align*} \min\{\|Df_{f^n(x)}\|\sigma_{\kappa+1}(x,n), \|Df_x||\sigma_{\kappa+1}(f(x),n)\} \geq \sigma_{\kappa+1}(x,n+1)\geq K_1\sigma_{\kappa+1}(x,n). \end{align*} $$

Proof. We take the constant $K_1=\min \{\|Df\!\mid _{\mathscr {C}(x)}\!\|: x \in M\}$ . Since the cone-field $\mathscr {C}$ depends continuously on x and is transversal to $\ker (Df^n_x)$ for all $n\geq 1$ , we have that $K_1>0$ .

We will first prove that

$$ \begin{align*} \|Df_{f^n(x)}\|\sigma_{\kappa+1}(x,n)\geq \sigma_{\kappa+1}(x,n+1) \geq K_1\sigma_{\kappa+1}(x,n). \end{align*} $$

For that, we take V as the subspace of $T_xM$ generated by $E_n(x)\cup \{e_{\kappa +1}^n\}$ . Then,

$$ \begin{align*} \|Df_{f^n(x)}\|\sigma_{\kappa+1}(x,n)&\geq \|Df_x^{n+1}\!\mid_V\!\|\\ &=\sigma_{\kappa+1}(x,n+1)\\ &\geq \operatorname{{\mathfrak{m}}}(Df^{n+1}_x\!\mid_{E^{\perp}_n(x)})\\ &\geq \operatorname{{\mathfrak{m}}}(Df_{f^n(x)}\!\mid_{Df^n(E^{\perp}_n(x))}) \operatorname{{\mathfrak{m}}}(Df^n_x\!\mid_{E^{\perp}_n(x)}). \end{align*} $$

Thus, we conclude the proof using that $Df^n_x\!\mid _{E^{\perp }_n(x)}$ is contained in $\mathscr {C}(f^n(x))$ given by Lemma 2.18.

To complete the proof, it is sufficient to show that $\|Df_x\|\sigma _{\kappa +1}(f(x),n)\geq \sigma _{\kappa +1}(x, n+1)$ , since we have already known that $\sigma _{\kappa +1}(x,n+1)\geq K_1\sigma _{\kappa +1}(x,n)$ .

To do that, we will use that

$$ \begin{align*} \sigma_j(f(x),n)=\min\{\|(Df^n)^{\ast}\!\mid_{P}\!\|: P \in \mathrm{Grass}_j(T_{f^{n+1}(x)}M)\}; \quad \text{and} \\ \sigma_{j+1}(f(x),n)=\max\{\operatorname{{\mathfrak{m}}}((Df^n)^{\ast}\!\mid_{U}): U \in \mathrm{Grass}_{d-j}(T_{f^{n+1}(x)}M)\}. \end{align*} $$

We consider two orthogonal bases $\{e_1^{n},\ldots ,e_d^{n}\}$ of $T_{f(x)}M$ and $\{w_1^{n},\ldots ,w_d^{n}\}$ of $T_{f^{n+1}(x)}M$ satisfying $Df^{n}(e_j^{n})=\sigma _j(f(x),n)w_j^{n}$ and $(Df^{n})^{\ast }(w_j^{n})=\sigma _j(f(x),n)e_j^{n}$ . After that, taking $P=\{w_1^n,\ldots ,w_{\kappa +1}^n\}$ , we can verify that $\|Df_x\|=\|(Df_x)^{\ast }\|$ and

$$ \begin{align*} \|(Df_x)^{\ast}\|\sigma_{\kappa+1}(f(x),n)&\geq \|(Df_x)^{\ast}\|\|(Df_{f(x)}^n)^{\ast}\!\mid_{P}\!\|\\ &\geq\|(Df_x^{n+1})^{\ast}\!\mid_{P}\!\|\geq\sigma_{\kappa+1}(x,n+1).\\[-3pc] \end{align*} $$

Finally, we can prove that $(E_n(x))_n$ is a Cauchy sequence.

Proof of Lemma 2.17

Suppose $n<m$ and observe that

$$ \begin{align*}\mathrm{dist}(E_m(x),E_n(x)) \leq \sum_{j=1}^{m-n} \mathrm{dist}(E_{n+j-1}(x),E_{n+j}(x)).\end{align*} $$

Then, to prove the lemma, it is sufficient to show that the distance $\mathrm {dist}(E_{n+1}(x),E_n(x))$ tends to zero exponentially fast with n and uniformly in x.

We will omit the point x and simply rewrite $E_n$ and $E_n^{\perp }$ throughout the proof. To estimate the distance between $E_n$ and $E_{n+1}$ , we take $w\in E_n$ as the farthest unit vector from $E_{n+1}$ and decompose $w=u+v$ in a unique way where $u \in E_{n+1}$ and $v \in E_{n+1}^{\perp }$ . Then, we obtain that .

On the one hand, since the images of $E_{n+1}$ and $E_{n+1}^{\perp }$ by $Df^{n+1}$ are orthogonal, we have that

$$ \begin{align*} \|Df^{n+1}(w)\|&=\|Df^{n+1}(u)\|+\|Df^{n+1}(v)\|\\ &\geq \|Df^{n+1}(v)\|\geq \sigma_{\kappa+1}(x,n+1)\|v\|. \end{align*} $$

On the other hand,

$$ \begin{align*} \|Df^{n+1}(w)\|\leq K \|Df^n(w)\|\leq K\sigma_{\kappa}(x,n), \end{align*} $$

where $K= \sup _x \|Df_x\| < \infty $ . Thus, by Lemmas 2.19 and 2.15, one concludes that

It should be noted that $(E_n(x))_n$ is a Cauchy for every $x \in M$ . Thus, we can define a continuous $\kappa $ -dimensional subbundle of $TM$ by

$$ \begin{align*}E(x)=\lim E_n(x)\quad \text{for all } x \in M.\end{align*} $$

Recall that $\ker (Df_x^j) \subseteq E(x)$ for all $x \in M$ and all $j\geq 1$ since $\ker (Df_x^j)\subseteq E_n(x)$ for each $1\leq j\leq n$ .

We next prove that E is transversal to the cone-field $\mathscr {C}$ and is a $Df$ -invariant subbundle.

Proof. We first prove that $E(x)\cap \mathscr {C}(x)=\{0\}$ . For that, recall that $Df^n(E_n(x))$ and $Df^n(E_n^{\perp }(x))$ are perpendicular and, by Lemma 2.18, $Df^n(E_n^{\perp }(x))$ is contained in $\mathscr {C}(x)$ . Then, the intersection between $Df^n(E_n(x))$ and the cone-field is trivial, implying that $E_n(x)\cap \mathscr {C}(x)=\{0\}$ . Thus, we can conclude that the intersection between $E(x)$ and $\mathscr {C}(x)$ is also trivial.

To show the invariance property, we will prove that the distance between $Df(E_{n+1}(x))$ and $E_n(f(x))$ goes to zero as n tends to infinity. Before, we fix $\rho>0$ such that

$$ \begin{align*}\|Df(v)\|\geq \rho \|v\|\end{align*} $$

for every $v \in \ker (Df_x)^{\perp }$ for every $x \in M$ .

We start the proof of invariance, taking $v \in E(x)$ such that $Df(v)=0$ . Then, clearly, $Df(v) \in E(f(x))$ .

Now, we take a unit vector $v \in \ker (Df_x)^{\perp }\cap E(x)$ and choose a sequence of unit vectors $v_n \in E_{n+1}(x)\cap \ker (Df_x)^{\perp }$ such that $v_n \to v$ . Then, we write $Df(v_n)=u_n+z_n$ , where $u_n \in E_n(f(x))$ and $z_n \in E_n^{\perp }(f(x))$ .

Consider two orthogonal bases $\{e_1^n,\ldots ,e_d^n\}$ and $\{w_1^n,\ldots ,w_d^n\}$ of $T_{f(x)}M$ and $T_{f^{n+1}(x)}M$ , respectively, so that, for each $1\leq i \leq d$ , $Df_{f(x)}^n(e_i^n)=\sigma _i(f(x),n) w_i^n$ and $(Df_{f(x)}^n)^{\ast }(w_i^n)=\sigma _i(f(x),n) e_i^n$ . Take a vector $\tilde {z}_n \in \mathrm {span}\{w_{\kappa +1}^n,\cdots , w_d^n\}$ such that $(Df_{f(x)}^n)^{\ast }(\tilde {z}_n)=z_n$ and then, $\|z_n\|\geq \sigma _{\kappa +1}(f(x),n)\|\tilde {z}_n\|$ . Thus, fixing $K=\sup \|Df_x\|$ and using Lemma 2.19, we obtain that

Then, by Lemmas 2.14 and 2.19, we obtain that

Since , we have that $Df(v_n)$ and $E_n^{\perp }(f(x))$ are getting perpendicular when n tends to infinity. Thus, $Df(v)=\lim Df(v_n)$ is perpendicular to $E(f(x))^{\perp }$ and, consequently, $Df(v)$ belongs to $E(f(x))$ .

So far, we have shown that $E\oplus F$ is non-trivial splitting on $TM_f$ . Finally, we will next show that it is a dominated splitting for f and so prove Proposition 2.13.

Proof of Proposition 2.13

We have proved so far that under the assumption of Proposition 2.13, the splitting $E\oplus F$ satisfy for each orbit $(x_i)_i$ that $Df(E(x_i)) \subseteq E(f(x_i))$ and $Df^{\ell }(F(x_i))=F(f^{\ell }(x_i))$ . Moreover, as the intersection between E and the cone-field $\mathscr {C}$ is trivial, we conclude by item (ii) in Lemma 2.14 (recall that we are assuming $N=1$ ) that

$$ \begin{align*}\dfrac{\|Df_{x_i}^{n\ell}\!\mid_{E(x_i)}\!\|}{\operatorname{{\mathfrak{m}}}(Df_{x_i}^{n\ell}\!\mid_{F(x_i)})}<\unicode{x3bb}^n.\end{align*} $$

Thus, $E\oplus F$ is a dominated splitting for $f^{\ell }$ .

To complete the proof, we use that $E\oplus Df(F)$ is also a dominated splitting for $f^{\ell }$ and get that $Df(F)=F$ by the uniqueness of dominated splitting (see Lemma 2.9).

3.1 Proof of Corollary 1.1

Let $M=S^{2n}$ be an even-dimensional sphere and a robustly transitive map $f: M \to M$ . Let $E \subseteq TM$ be a $\kappa $ -dimensional subbundle, that is, a map $E: M \to \mathrm {Grass}_{\kappa }(TM)$ so that $E(x) \in \mathrm {Grass}_{\kappa }(T_xM)$ for every $x \in M$ . Note that we can associate an Euler class to E which is an element of $H^{\kappa }(M),$ the $\kappa $ th-dimensional de Rham cohomology of M. Suppose $E\neq TM$ (or equivalently, $\kappa \neq 2n=\dim (M)$ ), then we have that its Euler class is zero since all intermediate homologies vanish on spheres. However, we have that $E^\perp $ is another subbundle of dimension $2n-\kappa $ complementary to E in $TM$ , then it follows that $E \oplus E^\perp = TM$ . Therefore, the Euler class of $TM$ (which is the Euler characteristic of M times the class of the volume form M in $H^{2n}(M)$ ) is the product of the Euler classes of E and $E^\perp $ . Hence, the Euler characteristic of M is zero, but this is a contradiction since the Euler characteristic of an even-dimensional sphere is $2$ . This argument shows that even-dimensional spheres do not admit non-trivial subbundles. This argument is classical and is taken from [Reference Avila and Bochi2], but it can be also found in [Reference Milnor and Stasheff21, Property 9.6].

Now, we can apply Theorem A which implies that if an endomorphism f is robustly transitive, then it cannot have critical points as it would imply the existence of a non-trivial subbundle. Therefore, we conclude that it is a local diffeomorphism (covering map) on M. However, since M is simply connected, the endomorphism f should be a diffeomorphism. We can use now [Reference Bonatti, Díaz and Pujals8] to conclude the corollary.

4.1 Domination property near the critical points

Our goal here is to prove Lemma 4.3. For that, we will first make some preliminaries that will be needed in the proof.

Denote by $0\leq \sigma _{1}(x)\leq \sigma _{2}(x) \leq \cdots \leq \sigma _d(x)$ the singular values of $Df$ at $x \in M$ . Using that $\mathcal {F}$ accumulates at $f_0$ (in $C^1$ -topology) together with the fact that $\ker (Df_0)$ at x is $\kappa $ -dimensional for each $x \in \mathrm {Cr}_{\kappa }(f_0)$ , we can find a neighborhood $\mathcal {U}_0$ of $f_0$ and a neighborhood $U_0$ of $\mathrm {Cr}_{\kappa }(f_0)$ so that $\mathrm {Cr}_{\kappa }(f)$ is contained in $U_0$ for every f in $\mathcal {F}\cap \mathcal {U}_0$ . In particular, the singular values of $Df$ at $x \in \mathrm {Cr}_{\kappa }(f)$ are the following:

$$ \begin{align*}0=\sigma_{1}(x)=\cdots=\sigma_{\kappa}(x)< \sigma_{\kappa+1}(x)\leq \cdots \leq \sigma_d(x).\end{align*} $$

By continuity of $\sigma _i$ terms on M since f is a $C^1$ -endomorphism, we can assume that for all $x \in U_0$ , one has that

(4.8) $$ \begin{align} \sigma_{1}(x)\leq \cdots \leq\sigma_{\kappa}(x)< \sigma_{\kappa+1}(x)\leq \cdots \leq \sigma_d(x). \end{align} $$

Denote by $V_f(x)$ and $W_f(x)$ the subspaces of $T_xM$ generated by the eigenvectors associated to $\sigma _1(x),\ldots , \sigma _{\kappa }(x)$ and $\sigma _{\kappa +1}(x), \ldots , \sigma _d(x)$ , respectively. These subspaces satisfy the following properties:

1. (1) $V_f(x)\oplus W_f(x)=T_xM$ for every $x \in U_0$ ;

2. (2) $V_{f}(x)\perp W_{f}(x)$ and $Df(V_{f}(x))\perp Df(W_{f}(x))$ for every $x \in U_0$ ;

3. (3) if $f \in \mathcal {F}$ and $(x_i)_i \in \Lambda _f$ with $x_i \in \mathrm {Cr}_{\kappa }(f)$ , then $V_f(x_i)=\ker (Df_{x_i})=E(x_i)$ and $Df_{x_i}(W_f(x_i))=\mathrm {Im}(Df_{x_i})=F(x_i)$ .

Furthermore, it follows from the continuous dependence on the singular values of $Df$ at x that the following maps are continuous:

(4.9) $$ \begin{align} \mathcal{U}_0\times U_0 \ni (f,x) \mapsto V_f(x) \quad \text{and} \quad \mathcal{U}_0\times U_0 \ni (f,x) \mapsto W_f(x). \end{align} $$

For every $x \in U_0$ and $\eta \in (0,\pi /2)$ , define the following cone-field:

and so . Furthermore, it follows from 4.1 that $\mathscr {C}_{W_f}(x,\pi /2-\eta )=\mathscr {C}^{\kern2pt\ast }_{V_f}(x,\eta )$ .

Lemma 4.5. For all $\eta , \theta \in (0,\pi /2)$ small enough, up to shrinking the neighborhoods $\mathcal {U}_0$ and $U_0$ , we have that for every $f \in \mathcal {U}_0$ and $x \in U_0$ , the following holds.

For every $u \in \mathscr {C}^{\kern2pt\ast }_{V_f}(x,\eta )$ , writing $u=(v,w) \in V_f(x)\oplus W_f(x)$ , one has that

(4.10)

The next lemma provides a relation between the splittings $E\oplus F$ and $V\oplus W$ , which guarantees a domination property around of a component of the critical set whose kernel is k-dimensional. For simplicity, the set $\mathcal {U}$ will denote a neighborhood of $f_0$ contained in $\mathcal {U}_0$ and U denotes a neighborhood of $\mathrm {Cr}_{\kappa }(f_0)$ contained $U_0$ .

Lemma 4.6. Given $\eta , \theta \in (0,\pi /2)$ small enough, there exist neighborhoods $\mathcal {U}$ and U so that for each $f \in \mathcal {F}\cap \mathcal {U}$ and each $(x_i)_i \in \Lambda _f$ , one has that if $x_i \in U$ , then for all $u_1 \in E(x_i)$ and $u_2 \in F(x_i)$ , writing $u_j=(v_j,w_j) \in V_f(x_i)\oplus W_f(x_i)$ for $j=1,2$ , the following hold:

(4.11)

Proof. Recall that $\varepsilon _0>0$ and $\alpha>0$ are fixed as in (4.5). Assume that $2\eta , 2\theta <\min \{\varepsilon _0, \alpha \}/2$ . By Lemma 4.5, there exist neighborhoods $\mathcal {U}$ and U such that for every $f \in \mathcal {U}$ and $x \in U$ , the inequality (4.10) holds and moreover, for every $f \in \mathcal {F}\cap \mathcal {U}$ , the component $\mathrm {Cr}_{\kappa }(f)$ is contained in U.

We notice that from the inequality (4.10), it follows that to prove the lemma, it is sufficient to show that for every $(x_i)_i \in \Lambda _f$ whose $x_i \in U$ , we have

(4.12) $$ \begin{align} E(x_i) \subseteq \mathscr{C}_{V_f}(x_i,\eta) \quad \mathrm{and} \quad F(x_i) \subseteq \mathscr{C}^{\kern2pt\ast}_{V_f}(x_i,\eta). \end{align} $$

Thus, our current goal is to prove (4.12). For that, observe that as , we have that the vectors of $E(x_i)$ and $F(x_i)$ are neither in $\mathscr {C}_{V_f}(x_i,\eta )$ nor in $\mathscr {C}_{V_f}^{\kern2pt\ast }(x_i,\eta )$ simultaneously. Thus, only one of the following can happen:

• • either $E(x_i) \subseteq \mathscr {C}_{V_f}(x_i,\eta ) \quad \mathrm {and} \quad F(x_i) \subseteq \mathscr {C}^{\kern2pt\ast }_{V_f}(x_i,\eta )$ ; or

• • $E(x) \subseteq \mathscr {C}^{\kern2pt\ast }_{V_f}(x_i,\eta ) \quad \mathrm {and} \quad F(x_i) \subseteq \mathscr {C}_{V_f}(x_i,\eta )$ .

If $E(x_i) \subseteq \mathscr {C}^{\kern2pt\ast }_{V_f}(x_i,\eta )$ and $F(x_i) \subseteq \mathscr {C}_{V_f}(x_i,\eta )$ , then there is a vector $u \in \mathscr {C}^{\kern2pt\ast }_{V_f}(x_i,\eta )$ , where and . Since $2\eta ,2\theta $ are smaller than $\alpha $ , we can take two rotations $R_0$ and $R_1$ of angles less than $\alpha $ and $u' \in F(x_i)$ satisfying $u \in R_0(F(x_i))$ and $R_1(Df_x(u))\in E(f(x_i))$ . Recall from the discussion in (4.5) to conclude that it is incompatible with (4.3).

Therefore, $E(x_i) \subseteq \mathscr {C}_{V_f}(x_i,\eta )$ and $F(x_i) \subseteq \mathscr {C}^{\kern2pt\ast }_{V_f}(x_i,\eta )$ , which concludes the proof.