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edited Jul 25 2011 at 22:50
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Hi, Again:
I am trying to understand an argument I wrote down a long time ago, to show that a given
element of $M_3$ (mapping-class group of a) genus-3-surface, defined on a trivially-embedded copy of $S_3$ in $S^4$ extends (into the ambient $S^4$); by trivially embedded we mean that this $S_3$ bounds a topological $H_3$, i.e., a 3-handlebody; a 3-ball with 3 handles attached.
EDIT: My previous layout may have been unclear:
The SETUP is that of a genus-3 orientable surface $S_3$ trivially-embedded ( in the sense that it bounds a 3-handlebody) in the 4-sphere $S^4$, and we have a composition $C_4C_3C_4^{-1}$ of Dehn twists defined in the embedded $S_3$ about respective curves $c_3,c_4$ in $S_3$ ; where these two (basic) curves are defined in the paragraph below. We then want to show that the composition of twists on the embedded $S_3$ extends to the whole ambient $S^4$, i.e., we want to show that there is a self-diffeomorphism Phi:$S^4$-->$S^4$ , which extends the composition of the twists $C_4C_3C_4^{-1}$ defined on $S_3$, i.e., $Phi|_{S_3}$ (the restriction of Phi to $S_3$), is the map $C_4C_3C_4^{-1}$,
So consider the orientable, genus-3 surface $S_3$, trivially-embedded in the 4-sphere $S^4$ and two basis curves $c_3$ and $c_4$, which meet at exactly one point (i.e., their algebraic intersection)intersection number is 1), e.g., a meridian and a parallel in the same sub-torus of $S_3$.
Then the map h defined on (the embedded copy of) $S_3$ which we want to be embedded have extended into $S^4$ is defined by :
h:=$C_4C_3C_4^{-1}$, where $C_i$ is the Dehn twist about $c_i$.
The argument showing that h extends had to see with a sort of surgery on $S^4$ , which where we get decompose $S^4$ by joining (identifying) identifying/gluing) a $B^3 \times S^1$ (a 3-ball and a 1-sphere)and an $S^2\times D^2$ (2-sphere and a 2-disk), along their common boundary $S^2\times S^1$ (no problem so far).
NOw######, this is how the original argument went:
We somehow consider a solid torus $D^2\times S^1$ trivially-embedded in the 3-ball $B^3$ --which -where "trivially" means that its complement in $S^3$ is also a solid torus (we can show this last using, e.g., the Hopf Fibration)Fibration, so that a small disk $d^2$in an evenly-covered neighborhood lifts to a $d^2\times S^1$; same for the complement of the $d^2$), and we consider a tubular neighborhood U of the solid torus $D^2\times S^1$, given by:
U=$D^2\times S^1 \times S^1$ of the solid torus.
The complement $S^4 -U$:= $E^4$ in $S^4$ of this tubular neighborhood U is a topological $D^2\times S^1\times S^1$ , and the boundary $\partial E^4 $ of this $E^4$ is itself a 3-torus, i.e. $\partial E^4 = S^1\times S^1\times S^1$.
We then use a result by Montesinos, whereby maps defined on the bounding $\partial E^4$, which induce certain maps in homology, extend to the whole complement $E^4$, and then we construct ,
I think, a map from $S^4$ to itself, whose restriction to $S_3$ \partial E^4$ induce a map of the Montesinos type.
But I cannot make sense of why/how this argument shows that the given map extends.
Any Ideas?
Thanks in Advance.
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edited Jul 25 2011 at 22:21
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Hi, Again:
I am trying to understand an argument I wrote down a long time ago, to show that a given
element of $M_3$ (mapping-class group of a) genus-3-surface, defined on a trivially-embedded copy of $S_3$ in $S^4$ extends (into the ambient $S^4$); by trivially embedded we mean that this $S_3$ bounds a topological $H_3$, i.e., a 3-handlebody; a 3-ball with 3 handles attached.
EDIT: My previous layout may have been unclear:
The SETUP is that of a genus-3 orientable surface $S_3$ trivially-embedded ( in the sense that it bounds a 3-handlebody) in the 4-sphere $S^4$, and we have a composition $C_4C_3C_4^{-1}$ of Dehn twists defined in the embedded $S_3$ about respective curves $c_3,c_4$ in $S_3$ ; where these two (basic) curves are defined in the paragraph below. We then want to show that the composition of twists on the embedded $S_3$ extends to the whole ambient $S^4$, i.e., we want to show that there is a self-diffeomorphism Phi:$S^4$-->$S^4$ , which extends the composition of the twists $C_4C_3C_4^{-1}$ defined on $S_3$, i.e., $Phi|_{S_3}$ (the restriction of Phi to $S_3$), is the map $C_4C_3C_4^{-1}$,
So consider the orientable, genus-3 surface $S_3$, trivially-embedded in the 4-sphere $S^4$ and two basis curves $c_3$ and $c_4$, which meet at exactly one point (algebraic intersection), e.g., a meridian and a parallel in the same sub-torus of $S_3$.
Then the map h defined on (the embedded copy of) $S_3$ to be embedded is :
h:=$C_4C_3C_4^{-1}$, where $C_i$ is the Dehn twist about $c_i$.
The argument showing that h extends had to see with a sort of surgery on $S^4$ , which we get by joining (identifying) $B^3 \times S^1$ (a 3-ball and a 1-sphere)and an $S^2\times D^2$ (2-sphere and a 2-disk), along their common boundary $S^2\times S^1$ (no problem so far).
NOw######, this is how the original argument went:
We then somehow consider a solid torus $D^2\times S^1$ trivially-embedded in the 3-ball $B^3$ --which means that its complement in $S^3$ is also a solid torus (we can show this last using, e.g., the Hopf Fibration), and we consider the a tubular neighborhood U of the solid torus $D^2\times S^1$, given by:
U=$D^2\times S^1 \times S^1$ of the solid torus.
The complement $S^4 -U$:= $E^4$ in $S^4$ of this tubular neighborhood U is a topological $D^2\times S^1\times S^1$ , and the boundary $\partial E^4 $ of this $E^4$ is itself a 3-torus, i.e. $\partial E^4 = S^1\times S^1\times S^1$.
We then use a result by Montesinos, whereby maps defined on the bounding $\partial E^4$, which induce certain maps in homology, extend to the whole complement $E^4$, and then we construct ,
I think, a map whose restriction to $S_3$ induce a map of the Montesinos type.
But I cannot make sense of why/how this argument shows that the given map extends.
Any Ideas?
Thanks in Advance.
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edited Jul 25 2011 at 22:07
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Hi, Again:
I am trying to understand an argument I wrote down a long time ago, to show that a given
element of $M_3$ (mapping-class group of a) genus-3-surface, defined on a trivially-embedded copy of $S_3$ in $S^4$ extends (into the ambient $S^4$); by trivially embedded we mean that this $S_3$ bounds a topological $H_3$, i.e., a 3-handlebody; a 3-ball with 3 handles attached.
EDIT: My previous layout may have been unclear:
The SETUP is that of an a genus-3 orientable surface $S_3$ trivially-embedded ( in the sense that it bounds a 3-handlebody) in the 4-sphere $S^4$, and we have a composition $C_4C_3C_4^{-1}$ of Dehn twists defined in the embedded $S_3$ about respective curves $c_3,c_4$ in $S_3$ ; where these two (basic) curves are defined in the paragraph below. We then want to show that the composition of twists on the embedded $S_3$ extends to the whole ambient $S^4$, i.e., we want to show that there is an automorphism a self-diffeomorphism Phi:$S^4$-->$S^4$ , which extends the composition of the twists , $C_4C_3C_4^{-1}$ defined on $S_3$, i.e., $Phi|_{S_3}$ (the restriction of Phi to $S_3$), is$C_4C_3C_4^{-1}$,is the map $C_4C_3C_4^{-1}$,
So consider the orientable, genus-3 surface $S_3$, trivially-embedded in the 4-sphere $S^4$ and two basis curves $c_3$ and $c_4$, which meet at exactly one point (algebraic intersection), e.g., a meridian and a parallel in the same sub-torus of $S_3$.
Then the map h defined on (the embedded copy of) $S_3$ to be embedded is :
h:=$C_4C_3C_4^{-1}$, where $C_i$ is the Dehn twist about $c_i$.
The argument showing that h extends had to see with a sort of surgery on $S^4$ , which we get by joining (identifying) $B^3 \times S^1$ (a 3-ball and a 1-sphere)and an $S^2\times D^2$ (2-sphere and a 2-disk), along their common boundary $S^2\times S^1$ (no problem so far).
NOw######:
We then somehow consider a solid torus $D^2\times S^1$ trivially-embedded in $B^3$ --which means that its complement in $S^3$ is also a solid torus (we can show this last using, e.g., the Hopf Fibration), and we consider the tubular neighborhood U of the solid torus $D^2\times S^1$, given by:
U=$D^2\times S^1 \times S^1$ of the solid torus.
The complement $S^4 -U$:= $E^4$ in $S^4$ of this tubular neighborhood U is a topological $D^2\times S^1\times S^1$ , and the boundary $\partial E^4 $ of this $E^4$ is itself a 3-torus, i.e. $\partial E^4 = S^1\times S^1\times S^1$.
We then use a result by Montesinos, whereby maps defined on the bounding $\partial E^4$, which induce certain maps in homology, extend to the whole complement $E^4$, and then we construct ,
I think, a map whose restriction to $S_3$ induce a map of the Montesinos type.
But I cannot make sense of why/how this argument shows that the given map extends.
Any Ideas?
Thanks in Advance.
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edited Jul 25 2011 at 22:01
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Hi, Again:
I am trying to understand an argument I wrote down a long time ago, to show that a given
element of $M_3$ (mapping-class group of a) genus-3-surface, defined on a trivially-embedded copy of $S_3$ in $S^4$ extends (into the ambient $S^4$); by trivially embedded we mean that this $S_3$ bounds a topological $H_3$, i.e., a 3-handlebody; a 3-ball with 3 handles attached.
EDIT: My previous layout may have been unclear:
The SETUP is that of an $S_3$ trivially-embedded ( in the sense that it bounds a 3-handlebody) in $S^4$, and we have a composition $C_4C_3C_4^{-1}$ of Dehn twists in the embedded $S_3$ about respective curves $c_3,c_4$ defined below. We then want to show that there is an automorphism Phi:$S^4$-->$S^4$ , which extends the composition of the twists, i.e., $Phi|_{S_3}$ (the restriction of Phi to $S_3$), is$C_4C_3C_4^{-1}$,
So consider the orientable, genus-3 surface $S_3$, trivially-embedded in the 4-sphere $S^4$ and two basis curves $c_3$ and $c_4$, which meet at exactly one point (algebraic intersection), e.g., a meridian and a parallel in the same sub-torus of $S_3$, and let S_3$.
Then the map be given by
h defined on (the embedded copy of) $C_4C_3C_4^{-1}$, S_3$ to be embedded is :
h:=$C_4C_3C_4^{-1}$, where $C_i$ is the Dehn twist about $c_i$.
The argument showing that h extends had to see with a sort of surgery on $S^4$ , which we get by joining (identifying) a $B^3 \times S^1$ (a 3-ball and a 1-sphere)and an $S^2\times D^2$ (2-sphere and a 2-disk), along their common boundary $S^2\times S^1$ (no problem so far).
NOw######:
We then somehow consider a solid torus $D^2\times S^1$ trivially-embedded in $B^3$ --which means that its complement in $S^3$ is also a solid torus (we can show this last using, e.g., the Hopf Fibration), and we consider the tubular neighborhood U of the solid torus $D^2\times S^1$, given by:
U=$D^2\times S^1 \times S^1$ of the solid torus.
The complement $S^4 -U$:= $E^4$ in $S^4$ of this tubular neighborhood U is a topological $D^2\times S^1\times S^1$ , and the boundary $\partial E^4 $ of this $E^4$ is itself a 3-torus, i.e. $\partial E^4:= E^4 = S^1\times S^1\times S^1$.
We then use a result by Montesinos, whereby maps defined on the bounding T3$\partial E^4$, which induce certain maps in homology, extend to the whole complement $\partial E^4$, E^4$, and then we construct ,
I think, a map whose restriction to $S_3$ induce a map of the Montesinos type.
But I cannot make sense of why/how this argument shows that the given map extends.
Any Ideas?
Thanks in Advance.
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edited Jul 25 2011 at 8:57
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Hi, Again:
I am trying to understand an argument I wrote down a long time ago, to show that a given
element of $M_3$ (mapping-class group of a) genus-3-surface, defined on a trivially-embedded copy of $S_3$ in $S^4$ extends (into the ambient $S^4$); by trivially embedded we mean that this $S_3$ bounds a topological $H_3$, i.e., a 3-handlebody; a 3-ball with 3 handles attached.
So consider $S_3$, and two basis curves $c_3$ and $c_4$, which meet at exactly one point, e.g., a meridian and a parallel in the same sub-torus of $S_3$, and let the map be given by
$C_4C_3C_4^{-1}$, where $C_i$ is the Dehn twist about $c_i$.
The argument had to see with a sort of surgery on $S^4$ , which we get by joining (identifying) a $B^3$x$S^1$ B^3 \times S^1$ and an $S^2$x$D^2$ S^2\times D^2$ along their common boundary $S^2\times S^1$ (no problem so far).
NOw######:
We then somehow consider a solid torus trivially-embedded in $B^3$ --which means that its
complement in $S^3$ is also a solid torus (we can show this last using, e.g., the Hopf Fibration),
and we consider the tubular neighborhood:
U=$D^2$x$S^1$x$S^1$
U=$D^2\times S^1 \times S^1$ of the solid torus.
The complement $S^4 -U$:= $E^4$ in $S^4$ of this tubular neighborhood U is a $D^2\timesS^1\timesS^1$ D^2\times S^1\times S^1$ , and the boundary of this $E^4$ is itself a 3-torus $\partial E^4$:= $S^1\times\timesS^1\timesS^1$.E^4:= S^1\times S^1\times S^1$.
We then use a result by Montesinos, whereby maps defined on the bounding T3, which induce certain maps in homology, extend to the whole complement $\partial E^4$, and then we construct ,
I think, a map whose restriction to $S_3$ induce a map of the Montesinos type.
But I cannot make sense of why/how this argument shows that the given map extends.
Any Ideas?
Thanks in Advance.
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edited Jul 25 2011 at 8:52
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Hi, Again:
I am trying to understand an argument I wrote down a long time ago, to show that a given
element of $M_3$ (mapping-class group of a) genus-3-surface, defined on a trivially-embedded copy of $S_3$ in $S^4$ extends (into the ambient $S^4$); by trivially embedded we mean that this $S_3$ bounds a topological $H_3$, i.e., a 3-handlebody; a 3-ball with 3 handles attached.
So consider $S_3$, and two basis curves $c_3$ and $c_4$, which meet at exactly one point, e.g., a meridian and a parallel in the same sub-torus of $S_3$, and let the map be given by
$C_4C_3C_4^{-1}$, where $C_i$ is the Dehn twist about $c_i$.
The argument had to see with a sort of surgery on $S^4$ , which we get by joining (identifying) a $B^3$x$S^1$ and an $S^2$x$D^2$ along their common boundary $S^2$x$S^1$ S^2\times S^1$ (no problem so far).
NOw######:
We then somehow consider a solid torus trivially-embedded in $B^3$ --which means that its
complement in $S^3$ is also a solid torus (we can show this last using, e.g., the Hopf Fibration),
and we consider the tubular neighborhood:
U=$D^2$x$S^1$x$S^1$ of the solid torus.
The complement $S^4 -U$:= $E^4$ in $S^4$ of this tubular neighborhood U is a $D^2$x$S^1$x$S^1$ D^2\timesS^1\timesS^1$ , and the boundary of this $E^4$ is itself a 3-torus $\delE^4$:= \partial E^4$:= $S^1$x$S^1$x$S^1$.S^1\times\timesS^1\timesS^1$.
We then use a result by MontesionsMontesinos, whereby maps defined on the bounding T3, which induce certain maps in homology, extend to the whole complement $\del \partial E^4$, and then we construct ,
I think, a map whose restriction to $S_3$ induce a map of the Montesinos type.
But I cannot make sense of why/how this argument shows that the given map extends.
Any Ideas?
Thanks in Advance.
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asked Jul 25 2011 at 4:54
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Show a Map Defined on $S_3$ (trivially-embedded) in S^4 extends.
Hi, Again:
I am trying to understand an argument I wrote down a long time ago, to show that a given
element of $M_3$ (mapping-class group of a) genus-3-surface, defined on a trivially-embedded copy of $S_3$ in $S^4$ extends (into the ambient $S^4$); by trivially embedded we mean that this $S_3$ bounds a topological $H_3$, i.e., a 3-handlebody; a 3-ball with 3 handles attached.
So consider $S_3$, and two basis curves $c_3$ and $c_4$, which meet at exactly one point, e.g., a meridian and a parallel in the same sub-torus of $S_3$, and let the map be given by
$C_4C_3C_4^{-1}$, where $C_i$ is the Dehn twist about $c_i$.
The argument had to see with a sort of surgery on $S^4$ , which we get by joining (identifying) a $B^3$x$S^1$ and an $S^2$x$D^2$ along their common boundary $S^2$x$S^1$ (no problem so far).
NOw######:
We then somehow consider a solid torus trivially-embedded in $B^3$ --which means that its
complement in $S^3$ is also a solid torus (we can show this last using, e.g., the Hopf Fibration),
and we consider the tubular neighborhood:
U=$D^2$x$S^1$x$S^1$ of the solid torus.
The complement $S^4 -U$:= $E^4$ in $S^4$ of this tubular neighborhood U is a $D^2$x$S^1$x$S^1$ , and the boundary of this $E^4$ is itself a 3-torus $\delE^4$:= $S^1$x$S^1$x$S^1$.
We then use a result by Montesions, whereby maps defined on the bounding T3, which induce certain maps in homology, extend to the whole complement $\del E^4$, and then we construct ,
I think, a map whose restriction to $S_3$ induce a map of the Montesinos type.
But I cannot make sense of why/how this argument shows that the given map extends.
Any Ideas?
Thanks in Advance.
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