# Counterexample: Homology with Local Coefficients versus Homology with Module Coefficients

Definitions Let $X$ be a nice space with universal covering $\widetilde{X}$.

• Homology $H_*(X,R)$ with Ring Coefficients $R$ is the homology of \begin{align*} C_n(X;R) :=&\; \text{ free left $R$-module with basis $S_n(X)$}, \\ \cong&\; R \otimes_{\mathbb{Z}} C_n(X) \end{align*} where $C_*(X)$ is the singular chain complex.
• Homology $H_*(X;M)$ with Module Coefficients $M$, where $M$ is a left $R$-module, is the homology of $$C_n(X;M) := M \otimes_R C_n(X;R)$$

• Homology $H_*(X;A)$ with Local Coefficients in $A$, where $A$ is a left $R[\pi_1X]$-module, is the homology of $$C_n(\widetilde{X};R) \otimes_{R[\pi_1X]} A$$ where we view $C_n(\widetilde{X};R)$ as a right $R[\pi_1X]$-module using the monodromy action on $\widetilde{X}$.

Question: Is the local homology with local coefficients given by a left $R[\pi_1X]$-module $A$ (def. 3) the same as the homology with module coefficients in $A$, viewed as a $R[\pi_X]$-module (def. 2)? In other words, is the expression $$H_n(X;A)$$ even well-defined? If not, what is a counter example?

They are distinguished by their $\pi_1X$ action. Your definition of coefficients in a module, is the so-called untwisted homology by:

$$C_n(X;M) := M \otimes_R C_n(X;R) = M \otimes_R R^{|n|} = M^{|n|},$$ where $|n|$ denotes the number of $n$-cells of $X$ with no $\pi_1$ action on it. (3) in contrast is the twisted homology. As suggested by the name, (2) is the "regular" case, and (3) generalizes it by a non-trivial $\pi_1X$ action.

So if you want, define $M$ as a $R[\pi_1X]$-module by the trivial action, which then gives module homology as a special case of local coefficients. You can read about this in Hatcher's Algebraic Topology, more specifically you are going to learn if the action on $A$ is given by $a:\pi_1 X\to Aut(A)$, it suffices to look at a cover corresponding to $ker(a)$ while tensoring over $R[\pi_1(X)/ker(a)]$ the group ring over the deck transformations. In the trivial action case, that gives you the equivalence (up to the left- and right-stuff which you have to be careful with).

As you noticed correctly, this potentially leads to confusion. When being sloppy we get weird looking things like

$$H_1(S^1;\mathbb Z/3) = \mathbb Z/3 \neq 0 = H_1(S^1;\mathbb Z/3),$$ as computed here Homology with twisted coefficients of the circle . The difference is that on the left hand side I interpreted $\mathbb Z/3$ as a $\mathbb Z$-module, while on the right hand side I mean a module with a non-trivial fundamental group action, that is a $\mathbb Z[\mathbb Z]$-module with non-trivial $\mathbb Z$-action.

Therefore it is common practice to potentially think of every homology coefficients $A$ as $\mathbb [\pi_1X]$-modules via $a: \pi_1X \to Aut(A)$ and write $$H_i(X;A),$$ if $a=1$ and otherwise specify $$H_i(X;a),$$ by explicitely mentioning the twisting. However if the twisting is clear from context, often the former is (ab)used anyway.

• Thank you, this clears up my understanding. But still the thing is, given a ring $R$ and a $R[\pi_1X]$-module $A$, there are two ways to arrive at $H_*(X;A)$. First use the definition of twisted homology. Second, consider $S = R[\pi_1X]$ as the new base ring (forget that its a group ring!), and view $A$ as a usual module over $S$, and we get $H_*(X;A)$ as the untwisted homology with coefficients in the $S$-module $A$. Are they the same? – xxpauly Jul 29 '16 at 7:55
• In other words, is twisting again just a special case of taking the untwisted homology with coefficients in a $R[\pi_1X]$-module? I don't think it is, because the former involves $\widetilde{X}$, while the second one uses only $X$ itself. – xxpauly Jul 29 '16 at 8:25