Mixing
Unstable and stable manifold of Anosov on torus intersect
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I have the following question : given some Anosov $T : mathbb{T}^2 to mathbb{T}^2$ preserving the Lebesgue measure on the torus, is it true that for any arbitrary $x, y in mathbb{T}^2$, we have that $W^s(x) cap W^u(y) neq emptyset$ ? I know that locally, this is true, so I suspect that (since $mathbb{T}^2$ is connected) it remains true globally...
Help grealty appreciated !
dynamical-systems
asked Jan 20 at 12:45
Noam EluarNoam Eluar
469
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add a comment |
$begingroup$
I have the following question : given some Anosov $T : mathbb{T}^2 to mathbb{T}^2$ preserving the Lebesgue measure on the torus, is it true that for any arbitrary $x, y in mathbb{T}^2$, we have that $W^s(x) cap W^u(y) neq emptyset$ ? I know that locally, this is true, so I suspect that (since $mathbb{T}^2$ is connected) it remains true globally...
Help grealty appreciated !
dynamical-systems
asked Jan 20 at 12:45
Noam EluarNoam Eluar
469
$endgroup$
add a comment |
$begingroup$
I have the following question : given some Anosov $T : mathbb{T}^2 to mathbb{T}^2$ preserving the Lebesgue measure on the torus, is it true that for any arbitrary $x, y in mathbb{T}^2$, we have that $W^s(x) cap W^u(y) neq emptyset$ ? I know that locally, this is true, so I suspect that (since $mathbb{T}^2$ is connected) it remains true globally...
Help grealty appreciated !
dynamical-systems
asked Jan 20 at 12:45
Noam EluarNoam Eluar
469
$endgroup$
I have the following question : given some Anosov $T : mathbb{T}^2 to mathbb{T}^2$ preserving the Lebesgue measure on the torus, is it true that for any arbitrary $x, y in mathbb{T}^2$, we have that $W^s(x) cap W^u(y) neq emptyset$ ? I know that locally, this is true, so I suspect that (since $mathbb{T}^2$ is connected) it remains true globally...
Help grealty appreciated !
dynamical-systems
dynamical-systems
asked Jan 20 at 12:45
Noam EluarNoam Eluar
469
asked Jan 20 at 12:45
Noam EluarNoam Eluar
469
asked Jan 20 at 12:45
Noam EluarNoam Eluar
469
asked Jan 20 at 12:45
Noam EluarNoam Eluar
469
asked Jan 20 at 12:45
Noam EluarNoam Eluar
469
469
add a comment |
add a comment |
1 Answer
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Much more is true, namely $W^s(x)$ and $W^u(y)$ are each dense on $mathbb T^2$, and their intersection $W^s(x) cap W^u(y)$ is also dense.
To prove this, one can use the theorem (I think due to Roy Adler) which says that $T$ is topologically conjugate to a linear Anosov automorphism, for which the properties I wrote are more or less obvious: the stable foliation has a constant irrational slope, and the unstable foliation has a different constant irrational slope.
answered Jan 28 at 5:57
Lee MosherLee Mosher
49.9k33686
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Much more is true, namely $W^s(x)$ and $W^u(y)$ are each dense on $mathbb T^2$, and their intersection $W^s(x) cap W^u(y)$ is also dense.
To prove this, one can use the theorem (I think due to Roy Adler) which says that $T$ is topologically conjugate to a linear Anosov automorphism, for which the properties I wrote are more or less obvious: the stable foliation has a constant irrational slope, and the unstable foliation has a different constant irrational slope.
answered Jan 28 at 5:57
Lee MosherLee Mosher
49.9k33686
$endgroup$
add a comment |
$begingroup$
Much more is true, namely $W^s(x)$ and $W^u(y)$ are each dense on $mathbb T^2$, and their intersection $W^s(x) cap W^u(y)$ is also dense.
To prove this, one can use the theorem (I think due to Roy Adler) which says that $T$ is topologically conjugate to a linear Anosov automorphism, for which the properties I wrote are more or less obvious: the stable foliation has a constant irrational slope, and the unstable foliation has a different constant irrational slope.
answered Jan 28 at 5:57
Lee MosherLee Mosher
49.9k33686
$endgroup$
add a comment |
$begingroup$
Much more is true, namely $W^s(x)$ and $W^u(y)$ are each dense on $mathbb T^2$, and their intersection $W^s(x) cap W^u(y)$ is also dense.
To prove this, one can use the theorem (I think due to Roy Adler) which says that $T$ is topologically conjugate to a linear Anosov automorphism, for which the properties I wrote are more or less obvious: the stable foliation has a constant irrational slope, and the unstable foliation has a different constant irrational slope.
answered Jan 28 at 5:57
Lee MosherLee Mosher
49.9k33686
$endgroup$
Much more is true, namely $W^s(x)$ and $W^u(y)$ are each dense on $mathbb T^2$, and their intersection $W^s(x) cap W^u(y)$ is also dense.
To prove this, one can use the theorem (I think due to Roy Adler) which says that $T$ is topologically conjugate to a linear Anosov automorphism, for which the properties I wrote are more or less obvious: the stable foliation has a constant irrational slope, and the unstable foliation has a different constant irrational slope.
answered Jan 28 at 5:57
Lee MosherLee Mosher
49.9k33686
answered Jan 28 at 5:57
Lee MosherLee Mosher
49.9k33686
answered Jan 28 at 5:57
Lee MosherLee Mosher
49.9k33686
answered Jan 28 at 5:57
Lee MosherLee Mosher
49.9k33686
49.9k33686
add a comment |
add a comment |
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