Difference betwenn rings used in calculating homologies

Question

I need to know where the rings changes the calculus of singular homology. After i mount the matrix of operator i have to obtain dimension of image and kernel. If i have, for example the matrix: \begin{pmatrix} 1 & 1 \\ -1 & -1 \\ 1 & -1 \end{pmatrix} what will be the difference calculate $H_{n}(X, \mathbb{R})$ and $H_{n}(X,\mathbb{Z})$ where X is a topologic space ? And in which cases will be different ? Thanks !

Answer 1

With respect to "direct computations" via matrices etc, be careful with torsion for example. As an illustration, consider the maps $$\mathbb{Z} \stackrel{2 i_1}\to \mathbb{Z}\oplus \mathbb{Z} \stackrel{ p_2}\to \mathbb{Z},$$ where $2i_1: x \mapsto (2x,0)$ and $p_2:(x,y) \mapsto y$. We then have that the homology on the middle is $\mathbb{Z}/2\mathbb{Z}$. Now do the calculation with the same complex but $\mathbb{R}$'s instead.

Now, one good way to answer your question generally is via the universal coefficient theorem. It states the existence of a split exact sequence

$$0 \to H_i(X;\mathbb{Z}) \otimes G\to H_i(X; G) \to \mathrm{Tor}( H_{i-1}(X;G), G) \to 0. $$

That is, we have the following isomorphism:

$$H_i(X; G) \cong (H_i(X;\mathbb{Z}) \otimes G) \oplus \mathrm{Tor}(H_{i-1}(X;G),G). $$ This says that, if you calculate the homology (or at least the $i$-th and $(i-1)$-th) singular homology with $\mathbb{Z}$ coefficients, you know the homology with any coefficient (provided, of course, you know what $\mathrm{Tor}$ and the tensorization are).

Now, if $G=\mathbb{R}$, we have a flat abelian group, therefore $\mathrm{Tor}$ vanishes. If the homology is free, then it is simply a case of changing $\mathbb{Z}$ to $\mathbb{R}$, since $$H_i(X;G) \simeq H_i(X;\mathbb{Z}) \otimes \mathbb{R} \cong \mathbb{Z}^n\otimes \mathbb{R}\cong \mathbb{R}^n.$$ If the homology is finitely generated, it is almost as easy, since we will have $$H_i(X;G) \simeq H_i(X;\mathbb{Z}) \otimes \mathbb{R} \cong (\mathbb{Z}^n\oplus \text{the part with torsion})\otimes \mathbb{R}\cong \mathbb{R}^n,$$ and the algorithm is essentially "erasing the torsion, and changing $\mathbb{Z}$'s to $\mathbb{R}$'s".

Of course, the converse is not so simple: you can't, from the $\mathbb{R}$-coefficients homology go to the $\mathbb{Z}$-coefficients only via the homology groups themselves, since for instance $H_*(\mathbb{R}P^3;\mathbb{R}) \cong H_*(S^3,\mathbb{R})$, which can be seen from the computations above, but $H_*(\mathbb{R}P^3; \mathbb{Z})$ is not isomorphic to $H_*(S^3,\mathbb{Z})$ as is well-known.