Cohomology with U(1) Coefficients

Question

I am trying to work through this Karoubi paper on twisted bundles for research purposes.

I understand most of the construction, however, one point is still confusing me. The paper seems to insist on the following isomorphism:

$H^2 (X, U(1)) \simeq H^3 (X, Z)$

or sometimes:

$H^2 (X, C^{\times})$ instead of U(1).

Using the Universal Coefficient Theorem, it seems like:

$H^2 (X, U(1)) \simeq (U(1))^{\beta_2} \times T_2$

but we know that:

$H^3 (X, Z) \simeq Z^{\beta_3} \times T_2$ from the UCT for Z into Z.

So the torsion part is correct, but the Betti number part seems like it's off by one. Am I missing something obvious?

PS: I used the following facts for the UCT:

$\forall G, Ext_Z (Z, G) \simeq 0$

$Ext_Z (Z/n, U(1)) \simeq U(1)/nU(1) \simeq 0$

$\forall G, Hom_Z (Z, G) \simeq G$

$Hom_Z (Z/n, U(1)) \simeq \{x \in U(1) | nx = 0\} \simeq Z/n$

Answer 1

The notation $H^2(X, U(1))$ is somewhat ambiguous. It could mean one of two possible things depending on whether or not $U(1)$ is taken with the discrete topology or with the usual topology. With the usual topology, we have $H^2(X, U(1)) \cong H^3(X, \mathbb{Z})$. With the discrete topology, we have a short exact sequence

$$0 \to \mathbb{Z} \to \mathbb{R} \to U(1) \to 0$$

inducing a long exact sequence

$$\dots \to H^2(X, \mathbb{Z}) \to H^2(X, \mathbb{R}) \to H^2(X, U(1)) \to H^3(X, \mathbb{Z}) \to H^3(X, \mathbb{R}) \to \dots$$

so things are more complicated in this case.

Answer 2

You are missing the definition of $H^2(X, U(1))$ that Karoubi uses: it is defined via continuous cochains, rather than discrete ones that are required for the UCT to work. The shortest way to define continous cohomology for the topological group $U(1)$ is to say that it is the cohomology of the sheaf of continuous functions from $X$ to $U(1)$. Note that we have an exponential exact sequence of topological groups $$0 \to \mathbb Z \to \mathbb C \xrightarrow{exp} \mathbb C^\times \to 0$$ $\mathbb C$ is the sheaf of continuous $\mathbb C$-valued functions and admits a partition of unity, thus its higher cohomologies vanish. From the exact sequence in cohomology we thus get $$H^2(X, \mathbb C^\times)=H^3(X, \mathbb Z)$$ A homotopy equivalence of topological abelian groups induces an isomorphism in cohomology (at least for CW $X$), thus the statement.

If we had instead considered the discrete group $\mathbb C^\times$, then we would have a similar exact sequence of sheaves, but now $\mathbb C$ would be a constant sheaf associated with abelian group $\mathbb C$. Thus $H^i(X,\mathbb C)$ is torsion-free and nontrivial (singular cohomology with $\mathbb C$ coefficients), and the map $H^i(X, \mathbb Z) \to H^i(X, \mathbb C)$ has at least the torsion of $H^i(X, \mathbb Z)$ as its kernel. The exact sequence in cohomology would thus imply that $H^2(X, \mathbb C^\times)$ has the same torsion as $H^3(X, \mathbb Z)$, as implied by UCT, but the free part can be complicated. Note that we can no longer conclude that $H^2(X, \mathbb C^\times) = H^2(X, U(1))$. Since $\mathbb C^\times = U(1) \times \mathbb R$, we can say that the torsion in $H^2(X, U(1))$ is the same, but the free part will be even more complicated.

Also note that in your computation via UCT you have implicitly assumed all cohomology to be finitely generated. This is obviously true for finite CW complexes, but false in general. Any abelian group can be realised as the specified cohomology of some space. The calculations shown above still stand in this case.

Also note that the free part of $H^i(X,\mathbb Z)$ and $H^i(X,\mathbb C)$ are also different in general. For example, consider $X = \mathbb{RP}^\infty$. Is integer cohomology is free in even degrees and zero in odd ones, while its complex comohology vanish in positive degrees.

The most topological approach to the continuous cohomology used above relies on the notion of the classifying space. For any abelian group $A$ we have a natural isomorphism $H^i(X, A) = HoTop(X, B^i A)$ where $B^i A$ is a space with all homotopy groups zero except for $\pi_i B^i A = A$. The set on the right is the set of homotopy classes of maps. With this notion it is easy to prove that $U(1) = B\mathbb Z$, that a homotopy equivalence of groups implies isomorphic cohomology and that $H^i(X, B^k A) = H^{i+k}(X, A)$.