Cycles

In Example , we showed a robust saddle connection between equilibria of the same isotropy type. Another possibility is that we may have robust saddle connections between equilibria of different isotropy type. This in turn can lead to cycles of connections.

Given $x \in V$, and let $\omega(x)$ and $\alpha(x)$ respectively denote the omega- and alpha-limit sets of the maximal integral curve through x. Thus, if $\phi: (a,b) \mbox{$\rightarrow$ }V$ is the maximal integral curve through x, then $y \in \omega(x)$ if there exists a strictly increasing sequence $(t_i)\subset \mbox{$\mathbb{R} $ }$ such that $t_i \mbox{$\rightarrow$ }b$, $\phi(t_i) \mbox{$\rightarrow$ }y$ as $i \mbox{$\rightarrow$ }\infty$. It follows from (,), that if $\Gamma_x = H$, then $\omega(x), \alpha(x) \subset V^H$. In particular, any point in $\omega(x),\alpha(x)$ must have isotropy group as least as big as H. Thus, even though isotropy is constant along integral curves, it is possible in principle for symmetry to increase at limit points of a trajectory. As we shall now explain, this observation underlies the formation of robust heteroclinic cycles in equivariant dynamics.

  

Figure: A connection in V^H

Suppose that $\alpha(x) = \{p\}$, $\omega(x) = \{q\}$. Necessarily, p, q must be equilibria of X. We assume that that p,q are hyperbolic. Obviously, $W^u(p)\cap W^s(q) \cap V^H$contains the trajectory through x (see Figure ). If $p,q \in V^H \setminus V^{(H)}$, then $\Gamma_p, \Gamma_q \supsetneq H = \Gamma_x$. In the simplest case, we might suppose that $\Gamma_p \ne \Gamma_q$, the fixed point spaces of $\Gamma_p, \Gamma_q$are one-dimensional ( $\Gamma_p, \Gamma_q$ are maximal isotropy groups) and V^H is two-dimensional (as in Figure ). Note that the connection between p,q is robust to $\Gamma$-perturbations of X even though the intersection of W^u(p) and W^s(q) can never be transverse.

Suppose that p,q are symmetry related. That is, there exists $\gamma \in \Gamma$such that $\gamma p = q$ ( $\Gamma_p, \Gamma_q$ are conjugate subgroups of $\Gamma$). It follows by $\Gamma$-equivariance that $DX(q) = \gamma^{-1} DX(p) \gamma$. In particular, DX(q) has a positive eigenvalue and the corresponding eigenspace must lie in the two-dimensional fixed point space $\gamma V = V^{\gamma H \gamma^{-1}}$. By equivariance, we see that there must be a connection between q and $\gamma q$ in $V^{\gamma H \gamma^{-1}}$ (note that $\gamma q \ne p$, else $V^H = \gamma V^H$ and the indices of DX(p)|V^H, DX(q)|V^H would be equal). Iterating this construction, we obtain connections between $\gamma^n p$ and $\gamma^n q$, for all $n \in \mbox{$\mathbb{Z} $ }$. Since we assume $\Gamma$ is finite, there exists a smallest N > 0 such that $\gamma^N q = p$. The resulting cycle of connections between equilibria lying on the same $\Gamma$-orbit is called a homoclinic cycle. If we had constructed a cycle between equilibria lying on different $\Gamma$-orbits, we call the cycle a heteroclinic cycle (we refer to [24] where this type of distinction was first made and to [15] for generalities on homoclinic and heteroclinic cycles).

Generalizing the previous argument, it is easy to prove the following result.

Proposition 3.1   Suppose that $\Gamma$ is finite and that p,q are hyperbolic equilibria of the smooth equivariant vector field X on V. Suppose that there exists $x \in V$, $\gamma \in \Gamma$ such that

(a)

$\alpha(x) = \{p\}$, $\omega(x) = \{q\}$.

(b)

$\Gamma_p, \Gamma_q \supsetneq \Gamma_x$.

(c)

W^u(p) intersects W^s(q) transversally within $V^{\Gamma_x}$.

(d)

$\gamma p = q$.

If N denotes the order of $\gamma$, then $N \geq 3$ and there exists a robust $\mbox{$\mathbb{Z} $ }_N$-invariant homoclinic cycle for X connecting adjacent equilibria on the $\mbox{$\mathbb{Z} $ }_N$-orbit of p.

One can perhaps think of the flow-invariant and fixed point subspaces of V as `pathways' that allow robust communication between parts of the phase space; pathways that would normally be non-generic in a non-equivariant system. Viewed this way, one might guess that an abundance of fixed point subspaces would lead to many homoclinic and heteroclinic cycles. In fact the opposite is true! The more fixed point subspaces there are, the less likely are there to be cycles or periodic phenomena. In order to see many robust cycles it better to have few rather than many fixed point spaces. The difficulty is that if there are many fixed point spaces, then equilibria forced by symmetry will typically lie in many different fixed point spaces. As a result there may be strong restrictions on the linearizations at these equilibria. In addition, fixed point spaces can also act `barriers' to connections between equilibria. Both phenomena occur in the next example.

  

Figure: Invariant subspaces for symmetry group of cube

Example 3.2   Let $\Delta_3 = \mbox{$\mathbb{Z} $ }_2^3$ denote the group of diagonal matrices with entries $\pm 1$ and S₃ denote the symmetric group on three symbols. Let both groups act in the natural way on $\mbox{$\mathbb{R} $ }^3$. It is well known [15] that the full symmetry group of the three dimensional cube is the semi-direct product $\Gamma = \Delta_3 \rtimes S_3$. Although $\Gamma$ contains many subgroups isomorphic to $\mbox{$\mathbb{Z} $ }_3, \mbox{$\mathbb{Z} $ }_4$(required by Proposition  if there are to be cycles), it is not difficult to show that $\Gamma$-equivariant vector fields on $\mbox{$\mathbb{R} $ }^3$ do not possess any robust homoclinic cycles. Referring to Figure , we show the fixed point subspaces of $\Gamma$ that pass through the z-axis. Suppose that X is a $\Gamma$-equivariant vector field and p is a nonzero hyperbolic equilibrium point on the z-axis. Observe that in the direction transverse to the z-axis, there are four invariant lines. Hence any eigenspace for the linearization of X at p which does not contain the z-axis (radial direction) must have geometric multiplicity two. Hence it is not possible to connect p to any point on the $\Gamma$-orbit of p by a trajectory lying within a 2-dimensional fixed point plane. Similar remarks hold for non-zero equilibria lying on the fixed point space of S₃ (the diagonal line $\{(t,t,t)\mbox{$\;\vert\;$ }t \in \mbox{$\mathbb{R} $ }\}$). For somewhat different reasons, a nonzero equilibrium $p \in L = \{(t,t,0) \mbox{$\;\vert\;$ }t \in \mbox{$\mathbb{R} $ }\}$ cannot be connected to an equilibrium in the group orbit of p by a connection lying within a fixed point plane. Indeed, two fixed point planes intersect along L: the coordinate plane z = 0 and the plane x = y. In Figure  (a), we show the one-dimensional fixed point spaces contained in the plane z = 0. The invariant lines x = 0 and y = 0form a barrier to connections between p and -p or equilibria on x = -y. In Figure  (b), we show fixed point spaces contained in the plane x = y. In this case, the only line of symmetry in the $\Gamma$-orbit of L that meets the plane is L. Note that the fixed point lines $z=\pm x$ preclude any connections between p and -p. Since we cannot construct robust cycles connecting points with trivial isotropy by Theorem  and Remarks  (2), it follows that X cannot have any robust homoclinic cycles.

  

Figure: Invariant planes through the line x = y, z = 0.

$\heartsuit$