Does anyone know where I can find a table of the homology and cohomology groups, with different coefficients, of standard spaces – $S^1\times S^1$, Klein bottle, projective space, etc.?
[Math] A table of homology and cohomology groups
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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.
Suppose that $X$ is a space such that all of its homology groups are finitely generated. This holds, for example, if $X$ is a "levelwise finite" CW complex (finitely many cells in each dimension), which is a very broad class of spaces. Then universal coefficients implies that all of the cohomology groups are also finitely generated. Moreover, the isomorphism class of each homology and cohomology group is determined by rank and torsion subgroup, and universal coefficients implies that
- $H^k(X, \mathbb{Z})$ and $H_k(X, \mathbb{Z})$ have the same rank, and
- The torsion subgroup of $H^k(X, \mathbb{Z})$ is isomorphic to the torsion subgroup of $H_{k-1}(X, \mathbb{Z})$.
Hence the two sequences of isomorphism classes of groups determine each other in this case. They are nearly the same, except that the torsion subgroups are shifted one degree.
Without the finitely generated hypothesis this is no longer true. For every abelian group $A$ and positive integer $n \ge 1$ it is possible to construct a Moore space $X = M(A, n)$, which is a space whose homology vanishes except in degree $n$, where it is isomorphic to $A$, and in degree $0$, where it is $\mathbb{Z}$. By universal coefficients, the cohomology of a Moore space is
- $H^0(X, \mathbb{Z}) \cong \mathbb{Z}$
- $H^n(X, \mathbb{Z}) \cong \text{Hom}(A, \mathbb{Z})$
- $H^{n+1}(X, \mathbb{Z}) \cong \text{Ext}^1(A, \mathbb{Z})$
and all other cohomology vanishes. So the question is whether an abelian group $A$ is determined up to isomorphism by the isomorphism class of $\text{Hom}(A, \mathbb{Z})$ and $\text{Ext}^1(A, \mathbb{Z})$, and the answer is no. Counterexamples cannot be finitely generated: the first one that comes to mind is the following. We have
$$\text{Hom}(\mathbb{Q}, \mathbb{Z}) = 0$$
(straightforward) and
$$\text{Ext}^1(\mathbb{Q}, \mathbb{Z}) \cong \mathbb{R}$$
(nontrivial), and $\text{Ext}^1(-, -)$ preserves direct sums in the first factor, so it follows that the groups $\mathbb{Q}$ and $\mathbb{Q} \oplus \mathbb{Q}$ cannot be distinguished this way, since the two groups are not isomorphic, but $\mathbb{R} \cong \mathbb{R} \oplus \mathbb{R}$ (as abelian groups).
Best Answer
Here are some (integral) homology computations.
I've made this answer "community wiki", so feel free to add anything I've missed.