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Let $Y$ be a closed oriented $3$-manifold and $K$ be a knot in $Y$. We say $K$ is the unknot if $K$ is contained in a local $3$-ball in $Y$ and is unknotted therein.

Generalized Property R:
If a Dehn surgery on $K$ yields $Y\#S^1\times S^2$, must $K$ be the unknot? When $Y=S^3$, this is the property R conjecture solved in the affirmative by Gabai.

Generalized Property P:
If an integral Dehn surgery on $K$ yields $Y$, must $K$ be the unknot? When $Y=S^3$, this is the property P conjecture solved in the affirmative by Kronheimer-Mrowka.

I am less optimistic about the generalized Property P question. In this case, the restriction to integral surgeries is necessary: lens spaces give many counterexamples if one were to allow all rational Dehn surgeries.

If these questions are out of reach in the full generality, can one prove positive results for special families of $3$-manifolds? For example, can one answer the generalized property R question for $Y=\#^n(S^1\times S^2)$ in the affirmative?

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    $\begingroup$ The generalized property P is sometimes called cosmetic surgery conjecture. arxiv.org/abs/math/9911247 $\endgroup$
    – Ian Agol
    Commented Jul 5 at 23:35

2 Answers 2

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The generalized Property R conjecture stated above is known for nullhomologous knots $K$ in a rational homology 3-sphere $Y$. The only surgery that can produce $Y \# (S^1\times S^2)$ is the zero-surgery $Y_0(K)$, and then Hom and Lidman (arXiv:2006.11249) proved that $K$ must be unknotted.

The result is also known if $K$ is a nullhomotopic knot in an arbitrary 3-manifold $Y$. Here we might as well assume that the exterior $Y \setminus N(K)$ is prime, and then it's known more generally that $Y_0(K)$ doesn't have any $S^1\times S^2$ summands. Hom and Lidman (same article) established the case where $Y$ is prime and a rational homology 3-sphere, and then Ni (arXiv:2201.10613) proved it in full generality.

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    $\begingroup$ The second condition should hold for Y aspherical. Such a surgery gives a ribbon homology cobordism from Y to Y which is then a homotopy equivalence by arxiv.org/abs/2204.10730. Then the 2-handle must be attached along a homotopically trivial curve. $\endgroup$
    – Ian Agol
    Commented Jul 6 at 16:26
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    $\begingroup$ Actually, is asphericity needed? We take a 2-handle cobordism from $Y$ to $Y_0(K) \cong Y \# (S^1 \times S^2)$, attach a 3-handle to cancel the $S^1\times S^2$, and call the composite cobordism $W: Y \to Y$. Turning $W$ upside down gives a ribbon homology cobordism $W^\dagger: -Y \to -Y$. Theorem 1.5 of the linked paper says that the inclusion of the outgoing $-Y$ into $W^\dagger$ -- equivalently, the incoming $Y$ into $W$ -- induces an isomorphism on $\pi_1$. This isomorphism $\pi_1(Y) \to \pi_1(W)$ is the quotient by the attaching curve in $\pi_1(Y)$, which must then be trivial. $\endgroup$
    – Steven Sivek
    Commented Jul 7 at 6:52
  • $\begingroup$ @StevenSivek: It seems the isomorphism $\pi_1(Y)\to\pi_1(W)$ is still from the outgoing $Y$ (it's not the way how Floer homology works, unfortunately). $\endgroup$
    – Qiuyu Ren
    Commented Jul 7 at 9:12
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    $\begingroup$ @IanAgol If K is rationally nullhomologous then it's a $\mathbb{Q}$-homology cobordism and that theorem applies. If it's $\mathbb{Q}$-homologically essential then any Dehn surgery on K should decrease $b_1$ by 1, so it can't produce $Y\#(S^1\times S^2)$ anyway. $\endgroup$
    – Steven Sivek
    Commented Jul 7 at 14:26
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    $\begingroup$ @StevenSivek: Okay, I think there’s still an issue with my claim about being a ribbon homology cobordism if $b_1(Y)>0$. In W† if the 1-handle attached to Y_2 is not homologically killed by the 2-handle, then one cannot conclude it is a homology cobordism. $\endgroup$
    – Ian Agol
    Commented Jul 7 at 18:10
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To answer the last question of myself: The generalized property R question is also true for $Y=\#^n(S^1\times S^2)$. This follows from the main Theorem of Scharlemann's Producing reducible 3-manifolds by surgery on a knot (beware of the addendum at the end of the paper). To apply the theorem, one should first remove redundant $S^1\times S^2$ prime factors in $Y\backslash K$.

In fact, Scharlemann's result together with Steven's answer and Ian's comment there reduces the proof of the generalized property R question to the case when $Y=Y'\#L$ for some irreducible $Y'$ with $b_1(Y')>0$ and $L$ a (possibly empty) connected sum of lens spaces ($S^1\times S^2$ excluded).

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