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@QiuyuRen There's no need to invoke Floer homology here. It helps to label the two copies of $Y$: you start with $Y_1$ and attach a 2-handle and a 3-handle to get a cobordism $W: Y_1 \to Y_2$. Reversing the cobordism (but not its orientation) gives us $W^\dagger: -Y_2 \to -Y_1$, and the theorem says that inclusion induces an isomorphism $\pi_1(-Y_1) \to \pi_1(W^\dagger)$. If you reverse $W^\dagger$ again then the map $\pi_1(Y_1) \to \pi_1(W)$ is still the same isomorphism, because it's insensitive to orientation or to the direction of the cobordism.
@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.
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.
The condition is $\frac{1}{n} \geq 2g-1$, so you still have to deal with 1-surgery on the right-handed trefoil (i.e., the Poincaré homology sphere). But this is known to have $I(Y) \neq 0$ anyway, so it's fine. In general, the difference between $I(Y)$ and the reduced instanton homology $\hat{I}(Y)$ (which relates to $I^\#(Y)$ by the cited theorem 1.3) is measured by the Frøyshov invariant $h(Y)$, so in particular $I(Y)$ vanishes iff $\dim I^\#(Y)=1$ and $h(Y)=0$.
For (1), you could just iterate through NonalternatingKnotExteriors() in SnapPy. Volume computations are fast, so for every knot K whose volume was within 0.001 of M.volume(), I used M.is_isometric_to(K) to compare them. It found 16n847920 in just over 45 minutes, which is fine for isolated examples but I'd also like to know if there's a faster way.
I think SnapPy's identify method doesn't search through 16-crossing knots because there are too many of them, but if you set "M=Manifold('DT:[(-8,16,28,14,-2,6,-20,10,24,-32,-12,-30,18,-22,4,-26)]')" then "Manifold('16n847920').is_isometric_to(M)" should still return True.
@HJRW: for strongly invertible knots I imagine the involution $f$ as a 180-degree rotation about an axis which meets the knot twice. This is an oriented map $(S^3,K) \to (S^3,-K)$: it reverses the string orientation of $K$ but doesn't send it to the mirror $(-S^3,\pm K) \cong (S^3,\pm m(K))$, which is why $K$ need not be amphichiral. The quotient of the knot complement is obtained from $S^3/f \cong S^3$ by removing $N(K)/f \cong B^3$, so it's $B^3$.
I don't claim that one always exists, but that this construction applies when it does. E.g. if you splice together two copies of the same knot complement $E_K$ in this way, then it suffices to find an irreducible representation $\rho: \pi_1(E_K) \to SU(2)$ with $\rho(\mu)=\rho(\lambda)$ and use this $\rho$ on each copy of $E_K$. But this is the same as an irreducible representation of the (-1)-surgery group $\pi_1(S^3_{-1}(K)) \cong \pi_1(E_K)/\langle\mu\lambda^{-1}\rangle$, and Kronheimer and Mrowka's proof of property P showed that for $\pm1$-surgery, such representations always exist.
Rigidity can fail even for an irreducible homology sphere $Y$. Say $Y$ splits along an incompressible torus $T$ as $Y=Y_1 \cup_T Y_2$ (example: glue two nontrivial knot complements, meridian to longitude and vice versa), and $\rho:\pi_1(Y)\to SU(2)$ is irreducible on each $Y_i$ separately. Then $\rho$ is reducible on $T$, with image in a $U(1)$ subgroup, and you can "bend" it by the same trick as for connected sums: keep $\rho|_{\pi_1(Y_1)}$ fixed and replace $\rho|_{\pi_1(Y_2)}$ with $g\rho g^{-1}$ for elements $g$ in that subgroup. This gives a $U(1)$ family of non-conjugate representations.
