Proving conditions for a periodic orbit to be a hyperbolic set
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Let $M$ be a Riemannian manifold, $U subset M$ an open subset and $f : U to M$ a $C^1$ diffeomorphism. I'm trying to show that a periodic orbit with period $n$ is a hyperbolic set iff $Df^n$ at any point of the orbit has no eigenvalue with absolute value $1$.
Let $Lambda = p, f(p), ..., f^n-1(p)$ be a periodic orbit with period $n$. Suppose that $Lambda$ is a hyperbolic set. Then there are constants $lambda in (0, 1), c > 0$ and for every $x in Lambda$, $T_x M = E^s_x oplus E^u_x$ such that
$$ Df_x(E^s_x) = E^s_f(x), Df_x(E^u_x) = E^u_f(x) \
| Df^k_x(v) | leq c lambda^k | v |, textfor every ; v in E^s_x, k geq 0 \
| Df^-k_x(v) | leq c lambda^k | v |, textfor every ; v in E^u_x, k geq 0.$$
Let $x = f^i(p) in Lambda,; 0 le i le n-1$. Suppose that $Df^n_x$ has an eigenvalue of absolute value $1$, i.e there are $mu$ with $|mu| = 1$ and $v neq 0, v = v^s + v^u$, $v^s in E^s_x, v^u in E^u_x$, such that $Df^n_x(v) = mu v$.
I can't obtain a contradiction. Can someone give me a hint?
linear-algebra dynamical-systems
asked Aug 15 at 9:43
g.pomegranate
1,058415
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Let $M$ be a Riemannian manifold, $U subset M$ an open subset and $f : U to M$ a $C^1$ diffeomorphism. I'm trying to show that a periodic orbit with period $n$ is a hyperbolic set iff $Df^n$ at any point of the orbit has no eigenvalue with absolute value $1$.
Let $Lambda = p, f(p), ..., f^n-1(p)$ be a periodic orbit with period $n$. Suppose that $Lambda$ is a hyperbolic set. Then there are constants $lambda in (0, 1), c > 0$ and for every $x in Lambda$, $T_x M = E^s_x oplus E^u_x$ such that
$$ Df_x(E^s_x) = E^s_f(x), Df_x(E^u_x) = E^u_f(x) \
| Df^k_x(v) | leq c lambda^k | v |, textfor every ; v in E^s_x, k geq 0 \
| Df^-k_x(v) | leq c lambda^k | v |, textfor every ; v in E^u_x, k geq 0.$$
Let $x = f^i(p) in Lambda,; 0 le i le n-1$. Suppose that $Df^n_x$ has an eigenvalue of absolute value $1$, i.e there are $mu$ with $|mu| = 1$ and $v neq 0, v = v^s + v^u$, $v^s in E^s_x, v^u in E^u_x$, such that $Df^n_x(v) = mu v$.
I can't obtain a contradiction. Can someone give me a hint?
linear-algebra dynamical-systems
asked Aug 15 at 9:43
g.pomegranate
1,058415
Either $v^sne0$ or $v^une0$. But note that $mu$ might not be real. You need to do something about it.
– John B
Aug 15 at 11:15
I will fix it, but, firstly, I want to suppose that $mu$ is real. If $mu$ is real, how can I obtain a contradiction?
– g.pomegranate
Aug 15 at 11:44
Why do you want to prove by contradiction? If the periodic orbit of $x$ is hyperbolic then $Df^n(x)$ restricted to $E^s_x$ has spectrum inside the unit circle and $Df^n(x)$ restricted to $E^u_x$ has spectrum outside the unit circle. As $E^s_x$ and $E^u_x$ are complementary, the spectrum of $Df^n(x)$ is disjoint with the unit circle, QED.
– user539887
Aug 15 at 20:37
add a comment |Â
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Let $M$ be a Riemannian manifold, $U subset M$ an open subset and $f : U to M$ a $C^1$ diffeomorphism. I'm trying to show that a periodic orbit with period $n$ is a hyperbolic set iff $Df^n$ at any point of the orbit has no eigenvalue with absolute value $1$.
Let $Lambda = p, f(p), ..., f^n-1(p)$ be a periodic orbit with period $n$. Suppose that $Lambda$ is a hyperbolic set. Then there are constants $lambda in (0, 1), c > 0$ and for every $x in Lambda$, $T_x M = E^s_x oplus E^u_x$ such that
$$ Df_x(E^s_x) = E^s_f(x), Df_x(E^u_x) = E^u_f(x) \
| Df^k_x(v) | leq c lambda^k | v |, textfor every ; v in E^s_x, k geq 0 \
| Df^-k_x(v) | leq c lambda^k | v |, textfor every ; v in E^u_x, k geq 0.$$
Let $x = f^i(p) in Lambda,; 0 le i le n-1$. Suppose that $Df^n_x$ has an eigenvalue of absolute value $1$, i.e there are $mu$ with $|mu| = 1$ and $v neq 0, v = v^s + v^u$, $v^s in E^s_x, v^u in E^u_x$, such that $Df^n_x(v) = mu v$.
I can't obtain a contradiction. Can someone give me a hint?
linear-algebra dynamical-systems
asked Aug 15 at 9:43
g.pomegranate
1,058415
Let $M$ be a Riemannian manifold, $U subset M$ an open subset and $f : U to M$ a $C^1$ diffeomorphism. I'm trying to show that a periodic orbit with period $n$ is a hyperbolic set iff $Df^n$ at any point of the orbit has no eigenvalue with absolute value $1$.
Let $Lambda = p, f(p), ..., f^n-1(p)$ be a periodic orbit with period $n$. Suppose that $Lambda$ is a hyperbolic set. Then there are constants $lambda in (0, 1), c > 0$ and for every $x in Lambda$, $T_x M = E^s_x oplus E^u_x$ such that
$$ Df_x(E^s_x) = E^s_f(x), Df_x(E^u_x) = E^u_f(x) \
| Df^k_x(v) | leq c lambda^k | v |, textfor every ; v in E^s_x, k geq 0 \
| Df^-k_x(v) | leq c lambda^k | v |, textfor every ; v in E^u_x, k geq 0.$$
Let $x = f^i(p) in Lambda,; 0 le i le n-1$. Suppose that $Df^n_x$ has an eigenvalue of absolute value $1$, i.e there are $mu$ with $|mu| = 1$ and $v neq 0, v = v^s + v^u$, $v^s in E^s_x, v^u in E^u_x$, such that $Df^n_x(v) = mu v$.
I can't obtain a contradiction. Can someone give me a hint?
linear-algebra dynamical-systems
asked Aug 15 at 9:43
g.pomegranate
1,058415
asked Aug 15 at 9:43
g.pomegranate
1,058415
asked Aug 15 at 9:43
g.pomegranate
1,058415
asked Aug 15 at 9:43
g.pomegranate
1,058415
1,058415
Either $v^sne0$ or $v^une0$. But note that $mu$ might not be real. You need to do something about it.
– John B
Aug 15 at 11:15
I will fix it, but, firstly, I want to suppose that $mu$ is real. If $mu$ is real, how can I obtain a contradiction?
– g.pomegranate
Aug 15 at 11:44
Why do you want to prove by contradiction? If the periodic orbit of $x$ is hyperbolic then $Df^n(x)$ restricted to $E^s_x$ has spectrum inside the unit circle and $Df^n(x)$ restricted to $E^u_x$ has spectrum outside the unit circle. As $E^s_x$ and $E^u_x$ are complementary, the spectrum of $Df^n(x)$ is disjoint with the unit circle, QED.
– user539887
Aug 15 at 20:37
add a comment |Â
Either $v^sne0$ or $v^une0$. But note that $mu$ might not be real. You need to do something about it.
– John B
Aug 15 at 11:15
I will fix it, but, firstly, I want to suppose that $mu$ is real. If $mu$ is real, how can I obtain a contradiction?
– g.pomegranate
Aug 15 at 11:44
Why do you want to prove by contradiction? If the periodic orbit of $x$ is hyperbolic then $Df^n(x)$ restricted to $E^s_x$ has spectrum inside the unit circle and $Df^n(x)$ restricted to $E^u_x$ has spectrum outside the unit circle. As $E^s_x$ and $E^u_x$ are complementary, the spectrum of $Df^n(x)$ is disjoint with the unit circle, QED.
– user539887
Aug 15 at 20:37
Either $v^sne0$ or $v^une0$. But note that $mu$ might not be real. You need to do something about it.
– John B
Aug 15 at 11:15
Either $v^sne0$ or $v^une0$. But note that $mu$ might not be real. You need to do something about it.
– John B
Aug 15 at 11:15
I will fix it, but, firstly, I want to suppose that $mu$ is real. If $mu$ is real, how can I obtain a contradiction?
– g.pomegranate
Aug 15 at 11:44
I will fix it, but, firstly, I want to suppose that $mu$ is real. If $mu$ is real, how can I obtain a contradiction?
– g.pomegranate
Aug 15 at 11:44
Why do you want to prove by contradiction? If the periodic orbit of $x$ is hyperbolic then $Df^n(x)$ restricted to $E^s_x$ has spectrum inside the unit circle and $Df^n(x)$ restricted to $E^u_x$ has spectrum outside the unit circle. As $E^s_x$ and $E^u_x$ are complementary, the spectrum of $Df^n(x)$ is disjoint with the unit circle, QED.
– user539887
Aug 15 at 20:37
Why do you want to prove by contradiction? If the periodic orbit of $x$ is hyperbolic then $Df^n(x)$ restricted to $E^s_x$ has spectrum inside the unit circle and $Df^n(x)$ restricted to $E^u_x$ has spectrum outside the unit circle. As $E^s_x$ and $E^u_x$ are complementary, the spectrum of $Df^n(x)$ is disjoint with the unit circle, QED.
– user539887
Aug 15 at 20:37
add a comment |Â
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Either $v^sne0$ or $v^une0$. But note that $mu$ might not be real. You need to do something about it.
– John B
Aug 15 at 11:15
I will fix it, but, firstly, I want to suppose that $mu$ is real. If $mu$ is real, how can I obtain a contradiction?
– g.pomegranate
Aug 15 at 11:44
Why do you want to prove by contradiction? If the periodic orbit of $x$ is hyperbolic then $Df^n(x)$ restricted to $E^s_x$ has spectrum inside the unit circle and $Df^n(x)$ restricted to $E^u_x$ has spectrum outside the unit circle. As $E^s_x$ and $E^u_x$ are complementary, the spectrum of $Df^n(x)$ is disjoint with the unit circle, QED.
– user539887
Aug 15 at 20:37