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Given $n$ points $p_1,\dots,p_n$ in $S^2$ one gets a product link $L_n=\{p_1,\dots,p_n\}\times S^1$ inside the closed 3-manifold $S^2\times S^1$, which can be looked at as a trivially framed link (by picking a tangent vector $v_i$ at each $p_i$ and dragging it along the factor $S^1$ to get the framing). I'm interested in the result of the surgery of $S^2\times S^1$ along $L_n$. Is it diffeomorphic to some ``space with a name'' (e.g., $S^3$, $S^1\times S^1\times S^1$, a lens space, etc?)

I'm interested in the answer for an explict computation of a Reshetikhin-Turaev invariant I'm faced with. Yet, the problem is purely topological and I expect there is well known answer, but I haven't been able to find it or work it out myself so far.

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The manifold you're describing has a very natural open book decomposition (see e.g. John Etnyre's lecture notes) in which the pages are the $n$-punctured spheres $(S^2 \backslash \cup_i N(p_i)) \times \{*\}$ and the monodromy is trivial. We can view this as a Murasugi sum of n-1 copies of the abstract open book $(S^1\times I, \mathrm{Id})$, which describes $S^1\times S^2$, and since Murasugi sums of open books correspond to connected sums of 3-manifolds, the manifold you want is $\#^{n-1}(S^1\times S^2)$.

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  • $\begingroup$ 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 $n=0$? $\endgroup$ – domenico fiorenza Nov 21 '13 at 15:48
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    $\begingroup$ 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 $S^1\times S^2$. $\endgroup$ – Steven Sivek Nov 21 '13 at 15:55
  • $\begingroup$ 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) $\endgroup$ – domenico fiorenza Nov 21 '13 at 17:13
  • $\begingroup$ 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. $\endgroup$ – Steven Sivek Nov 21 '13 at 18:14

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