Quotient of cohomology groups with different coefficents

Question

Let $M$ be a smooth manifold, (if necessary I'm ok with assuming that $M$ is four dimensional, orientable, and closed). I wish to understand the quotient: $$Q=H^1(M;\mathbb{R})/H^1(M;\mathbb{Z})$$ I have seen this pop up in a paper I am reading on Seiberg-Witten theory but I unfortunately don't really understand how to make sense of this space. Particularly, I am not sure I fully understand how to realize $H^1(M;\mathbb{Z})$ as a subgroup of $H^1(M;\mathbb{R})$, since I am unsure of how to relate their two constructions (i.e. as homomorphism groups from chains to the reals/integers). I believe that I am perhaps missing something naive, as my algebraic topology is definitely lack luster. If anyone could provide any insight I would be most appreciative.

Answer 1

Universal coefficients gives, for any abelian group $A$ and any $X$, an isomorphism

$$H^1(X, A) \cong \text{Hom}(H_1(X), A)$$

(since $H_0(X)$ is free, so the Ext term vanishes). If in addition $H_1(X)$ is finitely generated (which is always true if $X$ is a closed manifold), write $H_1(X) \cong \mathbb{Z}^r \oplus T$ where $T$ is the torsion subgroup. Since $\mathbb{R}$ and $\mathbb{Z}$ are both torsion-free we have

$$H^1(X, \mathbb{Z}) \cong \text{Hom}(H_1(X), \mathbb{Z}) \cong \text{Hom}(\mathbb{Z}^r, \mathbb{Z}) \cong \mathbb{Z}^r$$

and similarly

$$H^1(X, \mathbb{R}) \cong \text{Hom}(H_1(X), \mathbb{R}) \cong \text{Hom}(\mathbb{Z}^r, \mathbb{R}) \cong \mathbb{R}^r.$$

Moreover the universal coefficients isomorphism is natural in $A$, which implies that the coefficient map $H^1(X, \mathbb{Z}) \hookrightarrow H^1(X, \mathbb{R})$ is the obvious inclusion $\mathbb{Z}^r \hookrightarrow \mathbb{R}^r$. So their quotient is a torus $(\mathbb{R}/\mathbb{Z})^r$.