Show a Map Defined on $S_3$ (trivially-embedded) in S^4 extends. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T14:28:03Z http://mathoverflow.net/feeds/question/71184 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71184/show-a-map-defined-on-s-3-trivially-embedded-in-s4-extends Show a Map Defined on $S_3$ (trivially-embedded) in S^4 extends. Larry 2011-07-25T04:54:57Z 2011-07-25T22:50:15Z <p>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. </p> <p>EDIT: My previous layout may have been unclear:</p> <p>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 $C_4C_3C_4^{-1}$,</p> <p>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 number is 1), e.g., a meridian and a parallel in the same sub-torus of $S_3$.</p> <p>Then the map h defined on (the embedded copy of) $S_3$ which we want to have extended into $S^4$ is defined by :</p> <p>h:=$C_4C_3C_4^{-1}$, where $C_i$ is the Dehn twist about $c_i$.</p> <p>The argument showing that h extends had to see with a sort of surgery on $S^4$ , where we decompose $S^4$ by joining (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).</p> <p>NOw######, this is how the original argument went:</p> <p>We somehow consider a solid torus $D^2\times S^1$ trivially-embedded in the 3-ball $B^3$ --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, 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:</p> <p>U=$D^2\times S^1 \times S^1$ of the solid torus.</p> <p>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. $\partial E^4 = S^1\times S^1\times S^1$.</p> <p>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 $\partial E^4$ induce a map of the Montesinos type.</p> <p>But I cannot make sense of why/how this argument shows that the given map extends.</p> <p>Any Ideas?</p> <p>Thanks in Advance.</p>