Intuition of non-free homology groups?

Solution 1:

I think people should ignore all this nonsense about counting holes and just look at what actually happens in more examples.

In particular, your intuition about top homology having something to do with enclosing volumes is not quite correct. I interpret this to mean that you have in mind a manifold which is the boundary of another manifold (the same way that the sphere $S^n$ is the boundary of the disk $D^{n+1}$), and it's not true that a manifold has to be a boundary in order to have nonvanishing top homology. The simplest closed counterexample is $4$-dimensional: the complex projective plane $\mathbb{CP}^2$ is a $4$-manifold with $H_4 = \mathbb{Z}$, but it's known not to be the boundary of a $5$-manifold.

Whether top homology vanishes or not instead has to do with orientability.

It's possible to directly visualize the case of $\mathbb{RP}^2$, so let's do it. In this case $\pi_1 \cong H_1 \cong \mathbb{Z}_2$, so the goal is to visualize why there's some loop which isn't trivial but such that twice that loop is trivial. Visualize $\mathbb{RP}^2$ as a disk $D^2$, but where antipodal points on the boundary have been identified. We'll consider loops starting and ending at the origin.

I claim that a representative of a generator of $\pi_1 \cong H_1$ is given by the loop which starts at the origin, goes up to the boundary, gets identified with the opposite point, and goes up back to the origin. Try nudging this loop around for a bit so you believe that it's not nullhomotopic: the point is that you can't nudge it away from the boundary because the two (antipodal, hence identified) points it's intersecting the boundary at can never annihilate.

Now we want to visualize why twice this loop is nullhomotopic. It will be convenient to nudge the loop so that it hits the boundary at four points, which come in two antipodal pairs $A, A', B, B'$, so that the loop hits them in that order before returning to the origin. At this point it would be helpful to draw a diagram if you haven't already; the disk, and the two loops in it, should look a bit like a tennis ball in profile. Now, nudge the loop so that $A, B'$ get closer together, and hence, since they're constrained to be antipodes, $A', B$ also get closer together. Eventually you'll have nudged them enough that you'll see that you can finally pull the curve away from the boundary: as you do so, $A', B$ annihilate each other, and then $A, B'$ annihilate each other.

A similar visualization works for the Klein bottle, visualized as a square with its sides identified appropriately.

Solution 2:

I'm not sure how to describe this precisely, but i had this question myself and this is a qualitative answer that I found intuitive:

In these spaces you can place a loop which is not trivial, but if you trace the loop twice, the result can be continuously deformed within the space to become trivial. That is the loop is it's own inverse.

In the $\mathbb{RP}^2$ case, consider $\mathbb{RP}^2$ as the quotient space of the square in the standard way, draw a line from a corner to the opposite corner. You can convince yourself that this is a non trivial loop, but by continuously rotating the line about $180^\circ$, the opposite ends are always identified under the quotient map so we are continuously moving the loop, but we reverse the orientation, so the loop is it's own inverse.

We can do a similar thing with the Klein bottle. Consider the Klein bottle as the cylinder, with top and bottom circles identified with opposite orientation. The loop around the top circle is then the same as the loop around the bottom circle with opposite orientation. Mapping it continuously to itself by moving it down the cylinder shows then that it is in some sense it's own inverse.