$ \def\Ext{\mathrm{Ext}}$ I have the following situation: I have a complex algebraic variety $ X$ . I want to compute the weight filtration on various abelian covers $ Y \to X$ . As a concrete example, take $ X = \mathbb{A}^2 \setminus \{ (x_1, x_2) : x_1 x_2 = 1 \}$ and let $ Y_{\ell}$ be the cover where we adjoin an $ \ell$ -th root of $ x_1 x_2 – 1$ . Topologically, $ X$ retracts onto the locus $ \{ |x_1| = |x_2|,\ |x_1 x_2-1| = 1 \}$ , which is a pinched $ 2$ -sphere, and $ Y_{\ell}$ retracts onto a “necklace” of $ \ell$ $ 2$ -spheres. One can compute that $ Y_{\ell}$ has betti numbers $ (1,1,\ell)$ with $ H^1$ in weight $ 2$ , and $ H^2$ splitting up into one piece in weight $ 4$ and the rest in weight $ 2$ . See Section 7.1 of this paper for the details.

It is natural to organize our data in the following way. Let $ A = H_1(X, \mathbb{Z})$ , so abelian covers of $ X$ come with an action of $ A$ . For any character $ \chi$ of $ A$ , let $ H^{p, \chi}(X)$ be the $ \chi$ component of $ H^{\ast}(Y, \mathbb{C})$ where $ Y$ is any abelian cover of $ X$ such that the image of $ H_1(Y)$ in $ H_1(X)$ is contained in $ \mathrm{Ker}(A)$ . This is unaltered by passing to a larger cover $ Z$ since, for any finite $ G$ -cover $ Z \to Y$ , we have $ H^{\ast}(Y, \mathbb{C}) \cong H^{\ast}(Z, \mathbb{C})^G$ . Moreover, it is well defined to speak of the weight filtration on $ H^{\ast, \chi}(X, \mathbb{C})$ . In the above example, $ H^{0,0} = H^{1,0} = \mathbb{C}$ , in weights $ 0$ and $ 2$ ; $ H^{2,\chi}$ is one dimensional for all $ \chi$ , with weight $ 4$ if $ \chi$ is trivial and weight $ 2$ otherwise.

I was talking with a topologist, and he told me I should organize the data as follows (errors in this are my own). Let $ R$ be the group algebra $ \mathbb{C}[A]$ . Let $ X^{ab}$ be the universal abelian cover of $ X$ and let $ M_p = H_p(X^{ab}, \mathbb{C})$ . Then $ M_p$ is an $ R$ -module called the Alexander module, and it is finitely generated. Let $ L_{\chi}$ be $ \mathbb{C}$ considered as an $ R$ -module where $ A$ acts by $ \chi$ . Then there is a spectral sequence relating $ H^{\ast,\chi}$ and $ \Ext^{\ast}_R(M_{\ast}, L_{\chi})$ . (This is basically the universal coefficient theorem.)

In the example I gave, $ A = \mathbb{Z}$ so $ R$ is the Laurent polynomial ring $ \mathbb{C}[z]$ . The abelian cover $ X^{ab}$ is an infinite chain of spheres so $ H_0$ is one dimensional, $ H_1$ vanishes and $ H_2 \cong \mathbb{C}^{\infty}$ . As $ R$ -modules, we have $ M_0 \cong L_0 \cong R/z$ , $ M_1 = 0$ and $ M_2 \cong R$ . We have $ $ \Ext^0(M_0, L_{\chi}) = \Ext^1(M_0, L_{\chi}) = \begin{cases} \mathbb{C} & \chi = 1 \ 0 & \chi \neq 1 \end{cases}$ $ and $ $ \Ext^0(M_2, L_{\chi}) \cong \mathbb{C}$ $ for all $ \chi$ . So this predicts the dimensions exactly.

But how do I see that $ \Ext^0(M_2, L_1)$ is in weight $ 4$ , while $ \Ext^0(M_2, L_{\chi})$ is in weight $ 2$ for all other $ \chi$ ?