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Is there an exhaustive list of conditions satisfied by rational surgery coefficients assigned to the components of the Hopf link in $S^3$ such that the resulting 3-manifold by Dehn surgery acting on $S^3$ along this Hopf link is again $S^3$?

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If you're doing surgery with coefficients $p/q$ and $p'/q'$, the condition is that ${pp}'-{qq}' = \pm 1$.

In order to see this, let's think of what we're doing: we're drilling out of $S^3$ two solid torus neighbourhoods of the two components, obtaining $M$, and then we're gluing in solid tori, $T$ and $T'$. Since $M$ is a thickened torus (because this is the Hopf link complement), if we glue $T'$ first, we get $M'$ which is again a solid torus, and then when we glue $T$ we get a lens space $L$. Therefore, we only need to figure out when $H_1(L)$ is trivial.

Since I'm a bit lazy, I'll do it with Mayer–Vietoris: we need to look at subspace generated by the two boundary slopes $s$ and $s'$ in $H_1(M) = H_1(T^2) = \mathbb{Z}^2$. We pick coordinates $\lambda, \mu$ on $H_1(M)$ given by Seifert longitude and meridian of one of the components. Since it's the Hopf link, these are also the meridian and Seifert longitude of the other component (note that I've swapped the order of the two, but that I haven't changed any orientation!). In this basis, the surgery slopes are $(p,q)$ and $(q',p')$, so $H_1(L)$ (which is the quotient $H_1(M)/\langle s, s'\rangle$, by Mayer–Vietoris) has order $\det \left(\begin{array}{cc} p & q'\\ q & p'\end{array}\right) = pp'-qq'$, so $H_1(L)$ is trivial if and only if $pp'-qq' = \pm 1$, as claimed.

With a little bit more care, one can pin down which lens space one gets by doing any rational surgery on the Hopf link, but right now I'm a bit lazy...

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    $\begingroup$ Thanks!! @Marco Golla $\endgroup$ – shashank markande Aug 26 '20 at 1:33

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