Timeline for Link surgery on $S^2\times S^1$
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Nov 21, 2013 at 18:14 | comment | added | Steven Sivek | For n=2 you do have a product link. You can see it by writing the pages $(S^1\times I)$ of the open book collectively as $(S^1 \times I) \times S^1$, and then gluing in two solid tori $S^1\times D^2$ by using the disks $\{*\} \times D^2$ to cap off the annuli $(\{*\} \times I) \times S^1$ (which consist of one arc from each page). Thus in the surgered manifold the annuli $(\{*\}\times I)\times S^1$ can be closed up to form spheres $\{*\}\times S^2$, one for each point of $S^1$, and the link $L_2$ contributes a pair of points to each of those spheres. | |
| Nov 21, 2013 at 17:13 | comment | added | domenico fiorenza | Completely clear now, thanks. A last question: is there a way I can visualize the image of the link $L_n$ after the surgery? For instance, for $n=2$ one has a product $\{p_1,p_2\}\times S^1$ link in $S^2\times S^1$ before the surgery; is it a product link after the surgery, too? What does happen for $n=3$ and for higher $n$'s? (if that can be described easily) | |
| Nov 21, 2013 at 15:56 | vote | accept | domenico fiorenza | ||
| Nov 21, 2013 at 15:55 | comment | added | Steven Sivek | The argument only works for $n>0$ because when $n=0$ it isn't really an open book decomposition anymore, and then you really do have $S^2\times S^1$. You can check the n=2 case directly by Kirby calculus, though: it's 0-surgery on an unknot and on two of its meridians, and if you blow up one of those meridians you get a chain of unknots with surgery coefficients -1,-1,0,0. Blow down the second to get a chain with coefficients 0,1,0, and then the middle one, and then either of the remaining ones and you're left with a single 0-framed unknot, which gives $S^1\times S^2$. | |
| Nov 21, 2013 at 15:48 | comment | added | domenico fiorenza | Very informative, thanks. I'm a bit lost in the counting: I'd expect that with zero marked points I have the trivial surgery and so (the connected sum of) one copy of $S^2\times S^1$. Is maybe $n-1$ in the answer actually an $n+1$ or is the argument failing for $n=0$? | |
| Nov 21, 2013 at 15:26 | history | answered | Steven Sivek | CC BY-SA 3.0 |