Robust Degenerate Unfoldings of Cycles and Tangencies

Abstract

We construct open sets of degenerate unfoldings of heterodimensional cycles of any co-index $c>0$ and homoclinic tangencies of arbitrary codimension $c>0$. These type of sets are known to be the support of unexpected phenomena in families of diffeomorphisms, such as the Kolmogorov typical co-existence of infinitely many attractors. As a prerequisite, we also construct robust homoclinic tangencies of large codimension which cannot be inside a strong partially hyperbolic set.

Appendix A: Estimates for the Folding Manifold of Example 3.2

We consider the $(ss+c)$-dimensional $C^\infty$-manifold of Example 3.2 given by

$$\begin{aligned} \mathcal {S}:[-2,2]^{ss}\times [-\epsilon ,\epsilon ]^c \rightarrow \mathbb {R}^m, \ \ \mathcal {S}(x,t)=(x,(t_1,\dots ,t_u),h(t)) \in \mathbb {R}^{ss}\times \mathbb {R}^u\times \mathbb {R}^c \end{aligned}$$

where $t=(t_1,\dots ,t_u,\dots ,t_c) \in [-\epsilon ,\epsilon ]^c$ and $h(t)=(h_1(t),\dots ,h_c(t))$ with

$$\begin{aligned} h_i(t)= \sum _{j=0}^{u-1} t_{j+1} t_{ju+i} \quad \text {for} i=1,\dots ,u \quad \hbox {and} \quad h_i(t)=t_i \quad \text {for} i=u+1,\dots ,c. \end{aligned}$$

Let $t=t(E)$ be the $C^{r-1}$-function on $\mathcal {C}^{{G}}$ computes in Example 3.2. We have that:

Proposition A.1

For every $\varepsilon >0$ suffices small there is a neighborhood $\mathcal {C}^{{G}}$ of $E^u$ such that

$$\begin{aligned} C=\Vert Dh\Vert _\infty \cdot \max \{1,\Vert Dt\Vert _\infty \} \le 1+\varepsilon \quad \text {over}\,\mathcal {C}^{{G}}. \end{aligned}$$

Thus, $\mathcal {S}$ is a $({\alpha ,}\nu ,\delta )$-folding manifold for any $\nu >0$ such that $(1+\varepsilon )\nu <\delta$.

Proof

First of all, notice that for all $t\in [-\epsilon ,\epsilon ]^c$, it holds that

$$\begin{aligned} \Vert Dh(t)\Vert _{\infty } {\mathop {=}\limits ^{\scriptscriptstyle \mathrm def}}\max _{i=1,\dots , c} | Dh_i(t)| =\max \bigg \{ \,1, \ \max _{i=1,\dots ,u} \big |\sum _{j=0}^{u-1} t_{j+1}+t_{ju+i} \big |\,\bigg \}. \end{aligned}$$

Hence, taking $\epsilon >0$ small enough we have that $\Vert Dh\Vert _\infty =1$. On the other hand, by Cramer’s rule we get that the solution of the linear system $At= \mathbf {c}$ is given by

$$\begin{aligned} t_i = \frac{\det A_i^*}{\det A} \qquad \text {for}\,i=1,\dots ,c \end{aligned}$$

where $A^*_{i}$ is the matrix formed by replacing the ith column of A by the column vector $\mathbf {c}$. In order to compute Dt we write the variable of t by $E=\langle v_{1},\dots ,v_u\rangle$ with $v_k=(a_k,b_k,c_k)\in \mathbb {R}^{ss}\times \mathbb {R}^{u}\times \mathbb {R}^c$ and then

$$\begin{aligned} Dt=(\partial _{v_1}t,\dots ,\partial _{v_u}t) \quad \text {with} \ \ \ \partial _{v_k} t=(\partial _{a_k}t,\partial _{b_k}t,\partial _{c_k}t) \quad \text {for}\, k=1,\dots ,u. \end{aligned}$$

Moreover, since the matrix A does not depend on the variables $a_k$ we get that $\partial _{a_k}t=0$. In the sequel we will use the symbol $D_{k}$ to denote any partial derivative of the form $\partial _{b_{k\iota }}$ or $\partial _{c_{k\iota }}$. For each $i=1,\dots ,c$, using Jacobi’s formula it follows that

$$\begin{aligned} D_kt_i =\frac{D_k(\det A^*_i)-\mathrm {tr}(A^{-1} \cdot D_kA ) \cdot \det A^*_i}{\det A}. \end{aligned}$$

In particular, since $\mathbf {c}(E^u)=0$ then $\det A^*_i(E^u)=0$, and $\det A(E^u)=2$, we obtain that

$$\begin{aligned} D_kt_i(E^u)=\frac{1}{2} \, D\det A^*_i|_{E^u} = \frac{1}{2} \, \mathrm {tr}(\mathrm {Adj}(A^*_i) \cdot D_kA^*_i)|_{E^u} \end{aligned}$$

where $\mathrm {Adj}(A^*_i)$ is the adjugate matrix of $A^*_i$. Notice that

$$\begin{aligned}&\mathrm {Adj}(A^*_i(E^u))=(C_{\ell j})^T \quad \text {with} \ \ C_{\ell j} =0 \\&\quad \mathrm{if} \ j\not =i \quad \mathrm{and} \quad C_{\ell i}=(-1)^{\ell +i}\cdot \sigma _i \quad \mathrm{for}\,\ell =1,\dots ,c \end{aligned}$$

where $\sigma _i=2$ if $i\not =1$ and $\sigma _i=1$ otherwise ($i=1$). From this follows that

$$\begin{aligned} \mathrm {tr}(\mathrm {Adj}(A^*_i) \cdot D_kA^*_i)|_{\,(E^u)}= (C_{1i},\dots ,C_{ci})\cdot D_k\mathbf {c}(E^u). \end{aligned}$$

If $D_k$ is either, $\partial _{b_{k\iota }}$ for $\iota =1,\dots ,u$ or $\partial _{c_{k\iota }}$ for $\iota =u+1,\dots ,c$ then $D_k\mathbf {c}=0$ and thus $D_kt_i(E^u)=0$. Otherwise,

$$\begin{aligned} (C_{1i},\dots ,C_{ci})\cdot D_k\mathbf {c}(E^u) = C_{(k-1)u+\iota \, i} \quad \text {and hence} \quad D_kt_i(E^u)=\frac{1}{2} \, C_{(k-1)u+\iota \, i}. \end{aligned}$$

Therefore

$$\begin{aligned}&\Vert Dt(E^u)\Vert _{\infty }=\frac{1}{2} \ \max _{i=1,\dots ,c} \ \max _{k=1,\dots ,u} \\&\quad \big |\sum _{\iota =1}^{u} C_{(k-1)u+\iota \, i} \big |= \frac{1}{2} \ \max _{i=1,\dots ,c} \ \max _{k=1,\dots ,u} \ \big |\sum _{\iota =1}^u (-1)^{i+(k-1)u+\iota } \sigma _i \big | \le 1. \end{aligned}$$

Since Dt varies continuously with respect to E we have that $\Vert Dt(E)\Vert _\infty$ is close to $\Vert Dt(E^u)\Vert _\infty$ for any E close enough to $E^u$. Thus, shrinking $\mathcal {C}^{{G}}$ if necessary, this implies that $\Vert Dt\Vert _\infty \le 1+\varepsilon$ over $\mathcal {C}^{{G}}_\alpha$. Hence, $C=\Vert Dh\Vert _\infty \cdot \max \{1,\Vert Dt\Vert _\infty \} \le 1+\varepsilon$ for a fixed but arbitrarily small $\varepsilon >0$.

This $({\alpha ,}\nu ,\delta )$-folding $C^r$-manifold $\mathcal {S}$ induces a $C^{r-1}$-disc

$$\begin{aligned} \mathcal {H}^{{G}}: [-2,2]^{ss}\times \mathcal {C}^{{G}}_\alpha \longrightarrow G_u(\mathbb {R}^m), \qquad \mathcal {H}^{{G}}(x,E)=(\mathcal {S}(x,t), E) \in \overline{\mathcal {B}}\times \mathcal {C}^{{G}}= \overline{\mathcal {B}^{{G}}} \end{aligned}$$

Set $h^{{G}}=\mathscr {P}\circ \mathcal {H}^{{G}}$. Here $\mathscr {P}$ denotes the standard projection onto $\mathbb {R}^c$. We consider

$$\begin{aligned} \widehat{h}^{{G}}(J(\xi ^{{G}}))=J(h^{{G}}\circ \xi ^{{G}}) \quad \text {for}\, \ \ \xi ^{{G}}=(\xi ^{{G}}_a)_a\in C^d(\mathbb {R}^k,\mathbb {R}^{ss}\times G(u,m)) \end{aligned}$$

with

$$\begin{aligned} J(\xi ^{{G}})=((\xi _0,E_0),\partial ^1\xi ^{{G}},\dots ,\partial ^d\xi ^{{G}}) \in \big ([-2,2]^{ss}\times \mathcal {C}^{{G}}\big ) \times \overline{B}_\rho (0) \end{aligned}$$

where $\overline{B}_\rho (0)$ denotes a closed ball of radius $\rho >0$ at 0 velocity of the jets over $\mathbb {R}^{ss}\times G(u,m)$.

Proposition A.2

For every $\varepsilon >0$ suffices small there are $\rho >0$ and a neighborhood $\mathcal {C}^{{G}}$ of $E^u$ so that

$$\begin{aligned} \Vert D\widehat{h}^{{G}}\Vert _\infty \le 1+\varepsilon \quad \text {over} \ \ \big ([-2,2]^{ss}\times \mathcal {C}^{{G}}\big ) \times \overline{B}_\rho (0). \end{aligned}$$

Thus, $\widehat{\mathcal {H}}^{{G}}$ is a $({\alpha ,}\nu ,\delta )$-horizontal disc for any $\nu >0$ such that $(1+\varepsilon )\nu <\delta$.

Proof

Observe that $h^{{G}}(E)=h(t(E))$. In this way,

$$\begin{aligned} \widehat{h}^{{G}}(J(E))=J(h^{{G}}\circ E) =(h(t(E_a)),\partial ^1_a h(t(E_a)),\dots ,\partial ^d_a h(t(E_a)))_{|_{\,a=0}} \end{aligned}$$

(A.1)

for $E=(E_a)_a \in C^d(\mathbb {I}^k,G(u,m))$ with $E_0\in \mathcal {C}^{{G}}_\alpha$. Denoting $t_a=t(E_a)$ for all $a\in \mathbb {I}^k$, we can rewrite (A.1) as

$$\begin{aligned} \widehat{h}^{{G}}(J(t))=(h(t_a),\partial ^1_ah(t_a),\dots ,\partial ^d_ah(t_a))_{|_{\,a=0}} \quad \text {where}\,t=(t_a)_a \in C^d(\mathbb {I}^k,\mathbb {R}^c) \end{aligned}$$

with $t_0$ small enough in norm. Therefore,

$$\begin{aligned} D\widehat{h}^{{G}}= \frac{d\widehat{h}^{{G}}}{dJ(t)} \cdot \frac{dJ(t)}{dJ(E)}= \frac{d\widehat{h}^{{G}}}{dJ(t)} \cdot \frac{d \widehat{t}}{dJ(E)} \end{aligned}$$

(A.2)

where $\widehat{t}(J(E))=J(t\circ E)$. We want to compute $\Vert D\widehat{h}^{{G}}(J(E^u))\Vert _{\infty }$ where $E^u=(E^u_a)_a$ with $E^u_a=E^u_0=\{0^{ss}\}\times \mathbb {R}^u\times \{0^c\}$ for all $a\in \mathbb {I}^k$. Hence $J(E^u)=(E^u_0,0)$. By a straightforward calculation using Faà di Bruno’s formula, we have

$$\begin{aligned} \big \Vert \frac{d\widehat{t}}{dJ(E)}(J(E^u))\big \Vert _{\infty } = \Vert Dt(E^u_0)\Vert _\infty \le 1. \end{aligned}$$

(A.3)

By means of a similar computation we can show that

$$\begin{aligned} \big \Vert \frac{d\widehat{h}^{G}}{dJ(t)}(J(t^{u}))\big \Vert _{\infty } = \Vert Dh(0)\Vert _\infty = 1 \end{aligned}$$

(A.4)

where $t^{u}=(t^{u}_a)_a$ with $t^{u}_a=t(E^u_a)=t(E^u_0)=0$ for all $a\in \mathbb {I}^k$ and hence $J(t^{u})=0$. Finally putting together (A.2)–(A.4) we get that

$$\begin{aligned} \Vert D\widehat{h}^{{G}}(J(E^u))\Vert _\infty \le \big \Vert \frac{d\widehat{h}^{{G}}}{dJ(t)}(J(t^{{G}})) \big \Vert \cdot \big \Vert \frac{dJ(t)}{dJ(E)} (J(E^u)) \big \Vert \le 1. \end{aligned}$$

By continuity with respect to J(E), shrinking $\mathcal {C}^{{G}}$ if necessary and taking $\rho >0$ small enough, we have $\Vert D\widehat{h}^{{G}}\Vert _{\infty } \le 1+\varepsilon$ over $\big ([-2,2]^{ss}\times \mathcal {C}^{{G}}_\alpha \big ) \times \overline{B}_\rho (0)$ for a fixed but arbitrarily small $\varepsilon >0$. This completes the proof.