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Open book on any closed oriented 3-manifold
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I am reading "Lectures on Open Book Decompositions and Contact Structures" by Etnyre. In that paper Etnyre proves existence of open books on closed oriented 3-manifolds by using two lemma's of Alexander.
First lemma of Alexander states that for any closed oriented 3-manifold M, there exists a branched cover $P:Mlongrightarrow S^3$ with a branched $L_M$ where $L_M$ is a link. The other one states that any link can be braided about unknot i.e. if U denotes the unknot, taking $S^3/Ucong S^1timesmathbbD^2$, $L_M$ can be isotoped such that $L_Msubset S^1timesmathbbD^2$ and $L_M$ is transverse to $ptimesmathbbD^2$ for each $pin S^1$.
Now comes the open book I constructed. Considering the notation given above set $B=P^-1(U)$. Since $P$ is a covering away from $L_M$ and U is a loop in $S^3$, by homotopy lifting, $B$ is a loop in M. Therefore one can construct an open book around B as a sequence of maps:
$alpha_1:(BtimesmathbbD^2)/Bcong Btimes(mathbbD^2/0)longrightarrow (mathbbD^2/0)$ which is the projection and
$alpha_2:mathbbD^2/0longrightarrow S^1 $ with $alpha_2(z)=fracz$.
So we get B as the binding. But I do not understand what happens around $L_M$ living in $S^3/U$. Why do the fibers around $L_M$ have as B as the binding. Thank you for any help or suggestion.
differential-topology low-dimensional-topology fibration contact-topology
asked Jul 18 at 13:22
selflearner
10715
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up vote
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I am reading "Lectures on Open Book Decompositions and Contact Structures" by Etnyre. In that paper Etnyre proves existence of open books on closed oriented 3-manifolds by using two lemma's of Alexander.
First lemma of Alexander states that for any closed oriented 3-manifold M, there exists a branched cover $P:Mlongrightarrow S^3$ with a branched $L_M$ where $L_M$ is a link. The other one states that any link can be braided about unknot i.e. if U denotes the unknot, taking $S^3/Ucong S^1timesmathbbD^2$, $L_M$ can be isotoped such that $L_Msubset S^1timesmathbbD^2$ and $L_M$ is transverse to $ptimesmathbbD^2$ for each $pin S^1$.
Now comes the open book I constructed. Considering the notation given above set $B=P^-1(U)$. Since $P$ is a covering away from $L_M$ and U is a loop in $S^3$, by homotopy lifting, $B$ is a loop in M. Therefore one can construct an open book around B as a sequence of maps:
$alpha_1:(BtimesmathbbD^2)/Bcong Btimes(mathbbD^2/0)longrightarrow (mathbbD^2/0)$ which is the projection and
$alpha_2:mathbbD^2/0longrightarrow S^1 $ with $alpha_2(z)=fracz$.
So we get B as the binding. But I do not understand what happens around $L_M$ living in $S^3/U$. Why do the fibers around $L_M$ have as B as the binding. Thank you for any help or suggestion.
differential-topology low-dimensional-topology fibration contact-topology
asked Jul 18 at 13:22
selflearner
10715
Those are singular points in the image of the fiber. Since the fiber surface of $S^3U$ intersect transversally to $L_M$. BTW $B$ may not be a loop but it can be a link as well (M is compact).
– Anubhav Mukherjee
Jul 18 at 13:59
topologically the fiber is still a surface with boundary $B$. But around those points the chart looks lile $zmapsto z^n$.
– Anubhav Mukherjee
Jul 18 at 14:01
@Anubhav Mukherjee thank you for your help. But why the transversality of $L_M$ imply singularity on fibers? Is this a theorem by Milnor or something like that?
– selflearner
Jul 18 at 16:21
1
This is not very difficult to see. First try to observe what is happening with the fiber for $S^3-U to S^1$. In this case fibers are transversal to $L_M$. So the intersection is a discrete set of points. Now Take the pull back of one fiber in $M$. Then restriction of $P$ onto this pull back is a branched cover which branching over a finite set of point. So around those point the chart looks like $zmapsto z^n$.
– Anubhav Mukherjee
Jul 18 at 17:19
@AnubhavMukherjee I got it, thanks!
– selflearner
Jul 18 at 17:28
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am reading "Lectures on Open Book Decompositions and Contact Structures" by Etnyre. In that paper Etnyre proves existence of open books on closed oriented 3-manifolds by using two lemma's of Alexander.
First lemma of Alexander states that for any closed oriented 3-manifold M, there exists a branched cover $P:Mlongrightarrow S^3$ with a branched $L_M$ where $L_M$ is a link. The other one states that any link can be braided about unknot i.e. if U denotes the unknot, taking $S^3/Ucong S^1timesmathbbD^2$, $L_M$ can be isotoped such that $L_Msubset S^1timesmathbbD^2$ and $L_M$ is transverse to $ptimesmathbbD^2$ for each $pin S^1$.
Now comes the open book I constructed. Considering the notation given above set $B=P^-1(U)$. Since $P$ is a covering away from $L_M$ and U is a loop in $S^3$, by homotopy lifting, $B$ is a loop in M. Therefore one can construct an open book around B as a sequence of maps:
$alpha_1:(BtimesmathbbD^2)/Bcong Btimes(mathbbD^2/0)longrightarrow (mathbbD^2/0)$ which is the projection and
$alpha_2:mathbbD^2/0longrightarrow S^1 $ with $alpha_2(z)=fracz$.
So we get B as the binding. But I do not understand what happens around $L_M$ living in $S^3/U$. Why do the fibers around $L_M$ have as B as the binding. Thank you for any help or suggestion.
differential-topology low-dimensional-topology fibration contact-topology
asked Jul 18 at 13:22
selflearner
10715
I am reading "Lectures on Open Book Decompositions and Contact Structures" by Etnyre. In that paper Etnyre proves existence of open books on closed oriented 3-manifolds by using two lemma's of Alexander.
First lemma of Alexander states that for any closed oriented 3-manifold M, there exists a branched cover $P:Mlongrightarrow S^3$ with a branched $L_M$ where $L_M$ is a link. The other one states that any link can be braided about unknot i.e. if U denotes the unknot, taking $S^3/Ucong S^1timesmathbbD^2$, $L_M$ can be isotoped such that $L_Msubset S^1timesmathbbD^2$ and $L_M$ is transverse to $ptimesmathbbD^2$ for each $pin S^1$.
Now comes the open book I constructed. Considering the notation given above set $B=P^-1(U)$. Since $P$ is a covering away from $L_M$ and U is a loop in $S^3$, by homotopy lifting, $B$ is a loop in M. Therefore one can construct an open book around B as a sequence of maps:
$alpha_1:(BtimesmathbbD^2)/Bcong Btimes(mathbbD^2/0)longrightarrow (mathbbD^2/0)$ which is the projection and
$alpha_2:mathbbD^2/0longrightarrow S^1 $ with $alpha_2(z)=fracz$.
So we get B as the binding. But I do not understand what happens around $L_M$ living in $S^3/U$. Why do the fibers around $L_M$ have as B as the binding. Thank you for any help or suggestion.
differential-topology low-dimensional-topology fibration contact-topology
asked Jul 18 at 13:22
selflearner
10715
asked Jul 18 at 13:22
selflearner
10715
asked Jul 18 at 13:22
selflearner
10715
asked Jul 18 at 13:22
selflearner
10715
10715
Those are singular points in the image of the fiber. Since the fiber surface of $S^3U$ intersect transversally to $L_M$. BTW $B$ may not be a loop but it can be a link as well (M is compact).
– Anubhav Mukherjee
Jul 18 at 13:59
topologically the fiber is still a surface with boundary $B$. But around those points the chart looks lile $zmapsto z^n$.
– Anubhav Mukherjee
Jul 18 at 14:01
@Anubhav Mukherjee thank you for your help. But why the transversality of $L_M$ imply singularity on fibers? Is this a theorem by Milnor or something like that?
– selflearner
Jul 18 at 16:21
1
This is not very difficult to see. First try to observe what is happening with the fiber for $S^3-U to S^1$. In this case fibers are transversal to $L_M$. So the intersection is a discrete set of points. Now Take the pull back of one fiber in $M$. Then restriction of $P$ onto this pull back is a branched cover which branching over a finite set of point. So around those point the chart looks like $zmapsto z^n$.
– Anubhav Mukherjee
Jul 18 at 17:19
@AnubhavMukherjee I got it, thanks!
– selflearner
Jul 18 at 17:28
add a comment |Â
Those are singular points in the image of the fiber. Since the fiber surface of $S^3U$ intersect transversally to $L_M$. BTW $B$ may not be a loop but it can be a link as well (M is compact).
– Anubhav Mukherjee
Jul 18 at 13:59
topologically the fiber is still a surface with boundary $B$. But around those points the chart looks lile $zmapsto z^n$.
– Anubhav Mukherjee
Jul 18 at 14:01
@Anubhav Mukherjee thank you for your help. But why the transversality of $L_M$ imply singularity on fibers? Is this a theorem by Milnor or something like that?
– selflearner
Jul 18 at 16:21
1
This is not very difficult to see. First try to observe what is happening with the fiber for $S^3-U to S^1$. In this case fibers are transversal to $L_M$. So the intersection is a discrete set of points. Now Take the pull back of one fiber in $M$. Then restriction of $P$ onto this pull back is a branched cover which branching over a finite set of point. So around those point the chart looks like $zmapsto z^n$.
– Anubhav Mukherjee
Jul 18 at 17:19
@AnubhavMukherjee I got it, thanks!
– selflearner
Jul 18 at 17:28
Those are singular points in the image of the fiber. Since the fiber surface of $S^3U$ intersect transversally to $L_M$. BTW $B$ may not be a loop but it can be a link as well (M is compact).
– Anubhav Mukherjee
Jul 18 at 13:59
Those are singular points in the image of the fiber. Since the fiber surface of $S^3U$ intersect transversally to $L_M$. BTW $B$ may not be a loop but it can be a link as well (M is compact).
– Anubhav Mukherjee
Jul 18 at 13:59
topologically the fiber is still a surface with boundary $B$. But around those points the chart looks lile $zmapsto z^n$.
– Anubhav Mukherjee
Jul 18 at 14:01
topologically the fiber is still a surface with boundary $B$. But around those points the chart looks lile $zmapsto z^n$.
– Anubhav Mukherjee
Jul 18 at 14:01
@Anubhav Mukherjee thank you for your help. But why the transversality of $L_M$ imply singularity on fibers? Is this a theorem by Milnor or something like that?
– selflearner
Jul 18 at 16:21
@Anubhav Mukherjee thank you for your help. But why the transversality of $L_M$ imply singularity on fibers? Is this a theorem by Milnor or something like that?
– selflearner
Jul 18 at 16:21
1
1
This is not very difficult to see. First try to observe what is happening with the fiber for $S^3-U to S^1$. In this case fibers are transversal to $L_M$. So the intersection is a discrete set of points. Now Take the pull back of one fiber in $M$. Then restriction of $P$ onto this pull back is a branched cover which branching over a finite set of point. So around those point the chart looks like $zmapsto z^n$.
– Anubhav Mukherjee
Jul 18 at 17:19
This is not very difficult to see. First try to observe what is happening with the fiber for $S^3-U to S^1$. In this case fibers are transversal to $L_M$. So the intersection is a discrete set of points. Now Take the pull back of one fiber in $M$. Then restriction of $P$ onto this pull back is a branched cover which branching over a finite set of point. So around those point the chart looks like $zmapsto z^n$.
– Anubhav Mukherjee
Jul 18 at 17:19
@AnubhavMukherjee I got it, thanks!
– selflearner
Jul 18 at 17:28
@AnubhavMukherjee I got it, thanks!
– selflearner
Jul 18 at 17:28
add a comment |Â
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Those are singular points in the image of the fiber. Since the fiber surface of $S^3U$ intersect transversally to $L_M$. BTW $B$ may not be a loop but it can be a link as well (M is compact).
– Anubhav Mukherjee
Jul 18 at 13:59
topologically the fiber is still a surface with boundary $B$. But around those points the chart looks lile $zmapsto z^n$.
– Anubhav Mukherjee
Jul 18 at 14:01
@Anubhav Mukherjee thank you for your help. But why the transversality of $L_M$ imply singularity on fibers? Is this a theorem by Milnor or something like that?
– selflearner
Jul 18 at 16:21
1
This is not very difficult to see. First try to observe what is happening with the fiber for $S^3-U to S^1$. In this case fibers are transversal to $L_M$. So the intersection is a discrete set of points. Now Take the pull back of one fiber in $M$. Then restriction of $P$ onto this pull back is a branched cover which branching over a finite set of point. So around those point the chart looks like $zmapsto z^n$.
– Anubhav Mukherjee
Jul 18 at 17:19
@AnubhavMukherjee I got it, thanks!
– selflearner
Jul 18 at 17:28