Question: 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.

Answer: 
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.
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.