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Let $M$ and $N$ be $3$-manifolds obtained by zero-surgery on (left-handed) trefoil and figure-eight knot respectively.

What is the easy way to prove that $M$ and $N$ are not homeomorphic?

Note: When they are knot homology spheres (they are both homology $S^1 \times S^2$'s), I cannot use the classical invariants.

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    $\begingroup$ Heegaard Floer homology (as well as Khovanov homology) detects several knots. For your question, see for example: arxiv.org/pdf/math/0604079.pdf $\endgroup$ – Oğuz Şavk Oct 4 '20 at 21:22
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If you're happy bringing in heavy machinery then you could compute some sort of Floer homology, like the 'hat' version of Heegaard Floer homology: this has rank 2 for $S^3_0(3_1)$ and rank 4 for $S^3_0(4_1)$, so they're different.

On the other hand, in cases like this where you have a very specific pair of 3-manifolds in mind, it's often enough to distinguish their fundamental groups by counting covers of some finite order. Here you could work this out by hand starting from Wirtinger presentations of the respective knot groups, or you could just ask SnapPy to do it:

In[1]: len(Manifold('3_1(0,1)').covers(9))
Out[1]: 4


In[2]: len(Manifold('4_1(0,1)').covers(9))
Out[2]: 2

This counts 9-fold covers of each 0-surgery, and since the numbers are different, they must have different fundamental groups.

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They can also be distinguished geometrically. Both knots are genus one fibered knots, so both $M$ and $N$ are torus bundles over the circle.

The complement of the figure eight is hyperbolic, so the monodromy of the fibration is pseudo-Anosov. Hence the monodromy on $N$ is Anosov, and $N$ has Sol geometry.

The complement of the trefoil is Seifert fibered with base orbifold a disk with cone points of order 2 and 3. The longitude intersects an ordinary circle fiber on the boundary 6 times, so $M$ is Seifert fibered over the sphere with cone points of order 2, 3, and 6.

Since $M$ and $N$ are closed manifolds admitting distinct geometric structures, they are not homeomorphic.

Furthermore, it is possible to check that the monodromy of $M$ as a torus bundle has order 6, which is finite, so $M$ in fact admits a Euclidean structure.

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As long as we're collectively throwing the kitchen sink at this, note that the Alexander polynomial of the knot is an invariant of the 0-surgered manifold. So since the figure eight and trefoil knots have different polynomials, the 0-surgeries are not homeomorphic.

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This is a topological counterpart to Josh's geometric argument.

Since $H_1(M) \cong H_1(N) \cong \mathbb{Z}$, there is only one 6-fold cyclic cover of either of them; call them $M'$, $N'$. Both $M'$ and $N'$ are both torus bundles over the circle, and their monodromies are the sixth powers of the monodromies of $M$ and $N$. Since the monodromy of $M$ is a sixth root of the identity but the monodromy of $N$ isn't, $M'$ is $T^3$, while $N'$ is not.

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This is slightly simpler than the arguments above. The double covers of (0,1) surgeries for these two manifolds have different homologies.

The double cover of the (0,1) surgery on the figure 8 knot complement has homology Z/5Z + Z. The double cover of the (0,1) surgery on the trefoil knot complement has homology Z/3Z + Z.

It is unclear how well distinguishing manifolds by the homologies of their covers works and it is known to fail for certain Sol torus bundles. However, when it succeeds, it provides a concrete invariant for distinguishing manifolds.

Also, the double cover of any two bridge knot complement S^3\TBL(p,q) is the complement of a null-homologous knot complement in L(p,q). Hence it has homology Z/pZ + Z as does the double cover of (0,1) surgery (I am implicitly using that the (0,1) curve lifts in cyclic covers). In this case, the trefoil is the (3,2) two bridge knot and the figure eight knot is the (5,2) two bridge knot. So this is an effective technique for distinguishing many pairs of (0,1) surgeries on two-bridge knot complements.

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