Mixing
A question on orientation in manifolds (with or without boundary)
$begingroup$
I am currently reading Prasolov and Sossinsky's Knots, Links, Braids and 3-Manifolds. In their proof of the Dehn-Lickorish theorem there are some arguments that confuse me. They begin with a statement that any self-homeomorphism of a surface with boundary that fixes a boundary component is necessarily orientation preserving, which seems logical since the homeomorphism will induce an isomorphism on the second relative homology group so it can be orientation preserving or reversing and since it fixes a part of the manifold then it must be preserving orientations. However later in the proof we see a composition of twists that reverses an inner meridian's orientation. But I was under the impression that since twists can be performed away of the boundary then it would fix it. Can someone please explain what is it I am missing?
PS: By fixing, I mean identically fixing
EDIT: I don't know if it's relevant or not but I found in Matveev and Fomenko's Algorithmic and Computational Methods a reference that an orientation preserving homeomorphism can either preserve or reverse meridians (it also says that about the fibers). Does this answer my question?
geometric-topology
asked Jan 23 at 10:37
AmontilladoAmontillado
449313
$endgroup$
add a comment |
$begingroup$
I am currently reading Prasolov and Sossinsky's Knots, Links, Braids and 3-Manifolds. In their proof of the Dehn-Lickorish theorem there are some arguments that confuse me. They begin with a statement that any self-homeomorphism of a surface with boundary that fixes a boundary component is necessarily orientation preserving, which seems logical since the homeomorphism will induce an isomorphism on the second relative homology group so it can be orientation preserving or reversing and since it fixes a part of the manifold then it must be preserving orientations. However later in the proof we see a composition of twists that reverses an inner meridian's orientation. But I was under the impression that since twists can be performed away of the boundary then it would fix it. Can someone please explain what is it I am missing?
PS: By fixing, I mean identically fixing
EDIT: I don't know if it's relevant or not but I found in Matveev and Fomenko's Algorithmic and Computational Methods a reference that an orientation preserving homeomorphism can either preserve or reverse meridians (it also says that about the fibers). Does this answer my question?
geometric-topology
asked Jan 23 at 10:37
AmontilladoAmontillado
449313
$endgroup$
$begingroup$
What's an "inner meridian"? Assuming it's not a boundary component, I don't see a contradiction.
$endgroup$
– Lee Mosher
Jan 24 at 0:25
$begingroup$
Yes, I'm sorry about the expression. I meant that it was not a boundary component. I didn't think there was contradiction, I'm just trying to understand the concept. So, an orientation preserving homeomorphism can reverse a meridian that isn't on the boundary? Also, would it be possible to look at this question as well since I did not get any answers math.stackexchange.com/questions/3068167/… ?
$endgroup$
– Amontillado
Jan 24 at 9:45
add a comment |
$begingroup$
I am currently reading Prasolov and Sossinsky's Knots, Links, Braids and 3-Manifolds. In their proof of the Dehn-Lickorish theorem there are some arguments that confuse me. They begin with a statement that any self-homeomorphism of a surface with boundary that fixes a boundary component is necessarily orientation preserving, which seems logical since the homeomorphism will induce an isomorphism on the second relative homology group so it can be orientation preserving or reversing and since it fixes a part of the manifold then it must be preserving orientations. However later in the proof we see a composition of twists that reverses an inner meridian's orientation. But I was under the impression that since twists can be performed away of the boundary then it would fix it. Can someone please explain what is it I am missing?
PS: By fixing, I mean identically fixing
EDIT: I don't know if it's relevant or not but I found in Matveev and Fomenko's Algorithmic and Computational Methods a reference that an orientation preserving homeomorphism can either preserve or reverse meridians (it also says that about the fibers). Does this answer my question?
geometric-topology
asked Jan 23 at 10:37
AmontilladoAmontillado
449313
$endgroup$
I am currently reading Prasolov and Sossinsky's Knots, Links, Braids and 3-Manifolds. In their proof of the Dehn-Lickorish theorem there are some arguments that confuse me. They begin with a statement that any self-homeomorphism of a surface with boundary that fixes a boundary component is necessarily orientation preserving, which seems logical since the homeomorphism will induce an isomorphism on the second relative homology group so it can be orientation preserving or reversing and since it fixes a part of the manifold then it must be preserving orientations. However later in the proof we see a composition of twists that reverses an inner meridian's orientation. But I was under the impression that since twists can be performed away of the boundary then it would fix it. Can someone please explain what is it I am missing?
PS: By fixing, I mean identically fixing
EDIT: I don't know if it's relevant or not but I found in Matveev and Fomenko's Algorithmic and Computational Methods a reference that an orientation preserving homeomorphism can either preserve or reverse meridians (it also says that about the fibers). Does this answer my question?
geometric-topology
geometric-topology
asked Jan 23 at 10:37
AmontilladoAmontillado
449313
asked Jan 23 at 10:37
AmontilladoAmontillado
449313
edited Jan 23 at 13:01
Amontillado
asked Jan 23 at 10:37
AmontilladoAmontillado
449313
asked Jan 23 at 10:37
AmontilladoAmontillado
449313
asked Jan 23 at 10:37
AmontilladoAmontillado
449313
449313
$begingroup$
What's an "inner meridian"? Assuming it's not a boundary component, I don't see a contradiction.
$endgroup$
– Lee Mosher
Jan 24 at 0:25
$begingroup$
Yes, I'm sorry about the expression. I meant that it was not a boundary component. I didn't think there was contradiction, I'm just trying to understand the concept. So, an orientation preserving homeomorphism can reverse a meridian that isn't on the boundary? Also, would it be possible to look at this question as well since I did not get any answers math.stackexchange.com/questions/3068167/… ?
$endgroup$
– Amontillado
Jan 24 at 9:45
add a comment |
$begingroup$
What's an "inner meridian"? Assuming it's not a boundary component, I don't see a contradiction.
$endgroup$
– Lee Mosher
Jan 24 at 0:25
$begingroup$
Yes, I'm sorry about the expression. I meant that it was not a boundary component. I didn't think there was contradiction, I'm just trying to understand the concept. So, an orientation preserving homeomorphism can reverse a meridian that isn't on the boundary? Also, would it be possible to look at this question as well since I did not get any answers math.stackexchange.com/questions/3068167/… ?
$endgroup$
– Amontillado
Jan 24 at 9:45
$begingroup$
What's an "inner meridian"? Assuming it's not a boundary component, I don't see a contradiction.
$endgroup$
– Lee Mosher
Jan 24 at 0:25
$begingroup$
What's an "inner meridian"? Assuming it's not a boundary component, I don't see a contradiction.
$endgroup$
– Lee Mosher
Jan 24 at 0:25
$begingroup$
Yes, I'm sorry about the expression. I meant that it was not a boundary component. I didn't think there was contradiction, I'm just trying to understand the concept. So, an orientation preserving homeomorphism can reverse a meridian that isn't on the boundary? Also, would it be possible to look at this question as well since I did not get any answers math.stackexchange.com/questions/3068167/… ?
$endgroup$
– Amontillado
Jan 24 at 9:45
$begingroup$
Yes, I'm sorry about the expression. I meant that it was not a boundary component. I didn't think there was contradiction, I'm just trying to understand the concept. So, an orientation preserving homeomorphism can reverse a meridian that isn't on the boundary? Also, would it be possible to look at this question as well since I did not get any answers math.stackexchange.com/questions/3068167/… ?
$endgroup$
– Amontillado
Jan 24 at 9:45
add a comment |
1 Answer
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Here's one way to think about it.
On an oriented surface $S$, consider a simple closed curve $c$ in the interior. If you assign $c$ an orientation then there is a unique transverse orientation such that the orientation of $c$ and the orientation transverse to $c$ add up to form the orientation of $S$. There can certainly be an orientation preserving homeomorphism of $S$ which takes $C$ to itself, and reverses the orientation of $c$, and reverses the orientation **transverse to* $c$. One effect of this orientation is that it will swap the two sides of $c$. For example, if you model $c$ by the unit circle $S^1$ and if you model a neighborhood of $c$ by $S^1 times (-1,+1)$ then near $c$ that homeomorphism could be modelled by the map $f : S^1 times (-1,+1) to S^1 times (-1,+1)$ given by the formula $f(z,t)=(-z,-t)$.
One way to think of this intuitively is that the orientation of $c$ is given by an arrow pointing along $c$, and the transverse orientation is given by an arrow pointing transverse to $c$. For any orientation preserving homeomorphism which preserves $c$ there are two possibilities: both the arrow along $c$ and the arrow transverse to $c$ are taken to themselves; or both of those arrows are reversed. That's exactly what happens when you rotate both arrows by $180^circ$, and that rotation preserves orientation.
answered Jan 24 at 13:17
Lee MosherLee Mosher
50.5k33787
$endgroup$
$begingroup$
Thank you very much. That was most helpful.
$endgroup$
– Amontillado
Jan 24 at 19:10
add a comment |
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$begingroup$
Here's one way to think about it.
On an oriented surface $S$, consider a simple closed curve $c$ in the interior. If you assign $c$ an orientation then there is a unique transverse orientation such that the orientation of $c$ and the orientation transverse to $c$ add up to form the orientation of $S$. There can certainly be an orientation preserving homeomorphism of $S$ which takes $C$ to itself, and reverses the orientation of $c$, and reverses the orientation **transverse to* $c$. One effect of this orientation is that it will swap the two sides of $c$. For example, if you model $c$ by the unit circle $S^1$ and if you model a neighborhood of $c$ by $S^1 times (-1,+1)$ then near $c$ that homeomorphism could be modelled by the map $f : S^1 times (-1,+1) to S^1 times (-1,+1)$ given by the formula $f(z,t)=(-z,-t)$.
One way to think of this intuitively is that the orientation of $c$ is given by an arrow pointing along $c$, and the transverse orientation is given by an arrow pointing transverse to $c$. For any orientation preserving homeomorphism which preserves $c$ there are two possibilities: both the arrow along $c$ and the arrow transverse to $c$ are taken to themselves; or both of those arrows are reversed. That's exactly what happens when you rotate both arrows by $180^circ$, and that rotation preserves orientation.
answered Jan 24 at 13:17
Lee MosherLee Mosher
50.5k33787
$endgroup$
$begingroup$
Thank you very much. That was most helpful.
$endgroup$
– Amontillado
Jan 24 at 19:10
add a comment |
$begingroup$
Here's one way to think about it.
On an oriented surface $S$, consider a simple closed curve $c$ in the interior. If you assign $c$ an orientation then there is a unique transverse orientation such that the orientation of $c$ and the orientation transverse to $c$ add up to form the orientation of $S$. There can certainly be an orientation preserving homeomorphism of $S$ which takes $C$ to itself, and reverses the orientation of $c$, and reverses the orientation **transverse to* $c$. One effect of this orientation is that it will swap the two sides of $c$. For example, if you model $c$ by the unit circle $S^1$ and if you model a neighborhood of $c$ by $S^1 times (-1,+1)$ then near $c$ that homeomorphism could be modelled by the map $f : S^1 times (-1,+1) to S^1 times (-1,+1)$ given by the formula $f(z,t)=(-z,-t)$.
One way to think of this intuitively is that the orientation of $c$ is given by an arrow pointing along $c$, and the transverse orientation is given by an arrow pointing transverse to $c$. For any orientation preserving homeomorphism which preserves $c$ there are two possibilities: both the arrow along $c$ and the arrow transverse to $c$ are taken to themselves; or both of those arrows are reversed. That's exactly what happens when you rotate both arrows by $180^circ$, and that rotation preserves orientation.
answered Jan 24 at 13:17
Lee MosherLee Mosher
50.5k33787
$endgroup$
$begingroup$
Thank you very much. That was most helpful.
$endgroup$
– Amontillado
Jan 24 at 19:10
add a comment |
$begingroup$
Here's one way to think about it.
On an oriented surface $S$, consider a simple closed curve $c$ in the interior. If you assign $c$ an orientation then there is a unique transverse orientation such that the orientation of $c$ and the orientation transverse to $c$ add up to form the orientation of $S$. There can certainly be an orientation preserving homeomorphism of $S$ which takes $C$ to itself, and reverses the orientation of $c$, and reverses the orientation **transverse to* $c$. One effect of this orientation is that it will swap the two sides of $c$. For example, if you model $c$ by the unit circle $S^1$ and if you model a neighborhood of $c$ by $S^1 times (-1,+1)$ then near $c$ that homeomorphism could be modelled by the map $f : S^1 times (-1,+1) to S^1 times (-1,+1)$ given by the formula $f(z,t)=(-z,-t)$.
One way to think of this intuitively is that the orientation of $c$ is given by an arrow pointing along $c$, and the transverse orientation is given by an arrow pointing transverse to $c$. For any orientation preserving homeomorphism which preserves $c$ there are two possibilities: both the arrow along $c$ and the arrow transverse to $c$ are taken to themselves; or both of those arrows are reversed. That's exactly what happens when you rotate both arrows by $180^circ$, and that rotation preserves orientation.
answered Jan 24 at 13:17
Lee MosherLee Mosher
50.5k33787
$endgroup$
Here's one way to think about it.
On an oriented surface $S$, consider a simple closed curve $c$ in the interior. If you assign $c$ an orientation then there is a unique transverse orientation such that the orientation of $c$ and the orientation transverse to $c$ add up to form the orientation of $S$. There can certainly be an orientation preserving homeomorphism of $S$ which takes $C$ to itself, and reverses the orientation of $c$, and reverses the orientation **transverse to* $c$. One effect of this orientation is that it will swap the two sides of $c$. For example, if you model $c$ by the unit circle $S^1$ and if you model a neighborhood of $c$ by $S^1 times (-1,+1)$ then near $c$ that homeomorphism could be modelled by the map $f : S^1 times (-1,+1) to S^1 times (-1,+1)$ given by the formula $f(z,t)=(-z,-t)$.
One way to think of this intuitively is that the orientation of $c$ is given by an arrow pointing along $c$, and the transverse orientation is given by an arrow pointing transverse to $c$. For any orientation preserving homeomorphism which preserves $c$ there are two possibilities: both the arrow along $c$ and the arrow transverse to $c$ are taken to themselves; or both of those arrows are reversed. That's exactly what happens when you rotate both arrows by $180^circ$, and that rotation preserves orientation.
answered Jan 24 at 13:17
Lee MosherLee Mosher
50.5k33787
answered Jan 24 at 13:17
Lee MosherLee Mosher
50.5k33787
answered Jan 24 at 13:17
Lee MosherLee Mosher
50.5k33787
answered Jan 24 at 13:17
Lee MosherLee Mosher
50.5k33787
50.5k33787
$begingroup$
Thank you very much. That was most helpful.
$endgroup$
– Amontillado
Jan 24 at 19:10
add a comment |
$begingroup$
Thank you very much. That was most helpful.
$endgroup$
– Amontillado
Jan 24 at 19:10
$begingroup$
Thank you very much. That was most helpful.
$endgroup$
– Amontillado
Jan 24 at 19:10
$begingroup$
Thank you very much. That was most helpful.
$endgroup$
– Amontillado
Jan 24 at 19:10
add a comment |
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$begingroup$
What's an "inner meridian"? Assuming it's not a boundary component, I don't see a contradiction.
$endgroup$
– Lee Mosher
Jan 24 at 0:25
$begingroup$
Yes, I'm sorry about the expression. I meant that it was not a boundary component. I didn't think there was contradiction, I'm just trying to understand the concept. So, an orientation preserving homeomorphism can reverse a meridian that isn't on the boundary? Also, would it be possible to look at this question as well since I did not get any answers math.stackexchange.com/questions/3068167/… ?
$endgroup$
– Amontillado
Jan 24 at 9:45