# Confusion about homology with local coefficients - where is my mistake?

In the Chapter Local Coefficients of his book Algebraic Topology Hatcher writes the following, where $$X$$ is a path-connected space having a universal cover $$\tilde{X}$$ and fundamental group $$\pi$$.

For a $$\mathbb{Z}[\pi]$$-module $$M$$ let $$\pi´$$ be the kernel of the homomorphism $$\pi \to Aut(M)$$ given by the group action $$\pi \circlearrowright M$$. If $$X´\to X$$ denotes the cover corresponding to the normal subgroup $$\pi´\subset \pi$$, then $$C_n(\tilde{X})\otimes_{\mathbb{Z}[\pi]}M\cong C_n(X´)\otimes_{\mathbb{Z}[\pi]}M\cong C_n(X´)\otimes_{\mathbb{Z}[\pi/\pi´]}M.$$

Let us denote $$G:=\pi/\pi´$$. Let us for simplicity assume that $$X$$ is a triangulated closed manifold (seen as a CW-complex) and we consider the cellular chain complex. Then $$C_n(X´)$$ is a free $$\mathbb{Z}[G]$$-module with basis the $$n$$-cells $$\sigma:\Delta^n\to X$$ of the base space $$X$$, i.e. $$C_n(X´)\cong \bigoplus_{\sigma}\mathbb{Z}[G].$$ Therefore $$C_n(\tilde{X})\otimes_{\mathbb{Z}[\pi]}M\cong C_n(X´)\otimes_{\mathbb{Z}[G]}M \cong \bigoplus_{\sigma}\mathbb{Z}[G]\otimes_{\mathbb{Z}[G]} M\cong \bigoplus_\sigma M =C_n(X;M).$$ If we furthermore assume that the triangulation of $$X$$ is fine enough, s.t. $$G$$ respects the boundary homorphisms, then the entire chain complex $$\big(C_n(\tilde{X})\otimes_{\mathbb{Z}[\pi]}M\big)_{n\geq 0}$$ is the same as the usual chain complex with coefficients in $$M$$. But then the homology with local coefficients wouldn´t be distinct from the usual homology, i.e. $$H_n^\Pi(X;M)\cong H_n(X;M).$$ But this is certainly wrong.

I would be glad for explanations where and why I go wrong.

Thank you in advance.

## 1 Answer

You are almost correct, the problem is in your assumption that a fine enough triangulation will yield that $$G$$ respects the boundary morphisms.

In fact we don't need to work cellularly, this works just as well with the singular chain complex (because $$\Delta^n$$ is contractible, so up to choosing a basepoint we have unique liftings $$\Delta^n\to X$$ to $$\Delta^n\to X'$$), and I'll rather use this to show how the differentials differ (it seems clearer to me that way)

Also, for simplicity, I'll take $$\pi' = 0$$ so $$X' = \tilde X$$.

Now we need to analyze the map $$C_n(\tilde X)\otimes_{\mathbb Z[G]}\otimes M \to C_n(X;M)$$. For each $$\sigma : \Delta^n\to X$$, fix one $$\tilde \sigma : \Delta^n\to \tilde X$$ that lifts it. By the usual lifting property, this is equivalent to choosing a lift $$p_{\sigma}$$ of $$\sigma(e_0)$$ for each $$\sigma$$, and this will come in handy.

This choice is what gives you the isomorphism $$C_n(\tilde X) = \bigoplus_{\sigma : \Delta^n\to X}\mathbb Z[G]$$

Now your map takes $$\tau\otimes m$$, writes it as $$(g\cdot \tilde\sigma) \otimes m$$ for some $$g\in \pi, \sigma : \Delta^n \to X$$, and then sends this to $$\sigma \otimes g\cdot m$$ (where I see $$C_n(X;M) = C_n(X;\mathbb Z)\otimes M$$). Note that $$\sigma = p\circ \tau$$ so it's almost $$p\circ \sigma \otimes m$$, but it's twister by the difference

Now look at $$\partial (\tau\otimes m) = \partial \tau \otimes m = \sum_{i=0}^n (-1)^i d_i\tau \otimes m$$, write $$d_i\tau$$ as $$g_i\cdot (\widetilde{p\circ d_i\tau})$$, and this is sent to $$\sum_{i=0}^n (-1)^i p\circ d_i\tau \otimes g_i\cdot m$$

Compare to what happens if you first send it over to $$p\circ \tau\otimes g\cdot m$$ ($$g$$ such that $$g\cdot \widetilde{p\circ\tau} = \tau$$), and then take the boundary : although clearly $$d_i(p\circ \tau) = p\circ d_i\tau$$, you have $$g\cdot m$$ instead of the varying $$g_i$$ : you get $$\sum_{i=0}^n (-1)^i p\circ d_i\tau \otimes g\cdot m$$

So a priori there's no reason for the two differentials to agree, they're somehow "twisted" by $$\pi$$, and that's precisely what local coefficients do : they twist the differential.

In fact, what you can do is, instead of fixing $$p_\sigma$$ for each $$\sigma$$, fix an antecedent $$p_x$$ for each $$x\in X$$. Then we define $$\tilde\sigma$$ to be the unique lift of $$\sigma$$ that sends $$e_0$$ to $$p_{\sigma(e_0)}$$ : this gives something more uniform, and you can check that in this situation, the differentials are almost the same : for $$i\neq 0$$, $$g_i = g$$, but for $$i=0$$, there's a problem in that $$d_0\tau$$ doesn't have the same evaluation in $$e_0$$ as $$\tau$$.

More precisely, let $$\tau = \tilde\sigma$$ (so that there's no need for a $$g$$). Then for each $$i$$, $$d_i\tau$$ is a lift of $$d_i\sigma$$, and for $$i\neq 0$$, $$d_i\tau (e_0) = \tau(e_0) = p_{\sigma(e_0)}$$, so that $$d_i\tau = \widetilde{d_i\sigma}$$ : there is no $$g_i$$. But for $$i=0$$, $$d_0\tau(e_0) = \tau(e_1)$$ which is not necessarily $$p_{\sigma(e_1)}$$ : it is related to it by some $$g\in \pi$$ (because it's also a lift of $$\sigma(e_1)$$), so that one of the differentials looks like $$p\circ d_0\tau \otimes g\cdot m+ \sum_{i=1}^n (-1)^i p\circ d_i\tau \otimes m$$, and the other is $$\sum_{i=0}^n (-1)^i p\circ d_i\tau \otimes m$$.

So they're almost the same except for a twist on the $$0$$th face - this is the same twist as the one in something called the standard resolution, when computing group (co)homology, if you know about that.

Now you can clearly (hopefully) see that there's no reasonable condition to impose on the triangulation or the cell-structure or whatever that will make sure that this twist disappears and that the two differentials actually agree.

• Thanks a lot. Do you know a book/article/... with some discussed examples that can give one a better geometric understanding? Or some examples that could be good to try on my own? [We didn‘t have any examples in our lecture] – Frieder Jäckel Nov 22 '19 at 19:57
• Unfortunately I don't know the standard references. You can check this document (homepages.math.uic.edu/~mholmb2/serre.pdf) from page 12 it discusses local coefficients in a slightly different way (I think) and it can be interesting. For examples you could try on your own : you could try to look at $X= S^1$ with any abelian group $M$ and automorphism $f: M\to M$, or $\mathbb RP^2$ with an abelian group together with an involution $s:M\to M$ (e.g. multiplication by $-1$ on $\mathbb Z$) – Maxime Ramzi Nov 22 '19 at 20:04