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@%\input commons.tex
%\author{Anton Mellit}
%\title{Necessary geometric constructions}
%\date{\today}
%\begin{document}
%\bibliographystyle{alpha}
%\maketitle

\section{Hypercovers}\label{hypercovers}
\subsection{Abstract cell complex}
\begin{defn}
An \emph{abstract cell complex} is a graded set
\[
\sigma = \bigcup_{i\geq 0} \sigma_i
\]
together with homomorphisms $d_i:\Z[\sigma_i]\To\Z[\sigma_{i-1}]$ such that $d_i\circ d_{i+1}=0$ and $\veps\circ d_0=0$, where $\veps$ is the \emph{augmentation map} $\Z[\sigma_0]\To\Z$. The elements of the set $\sigma_i$ are called \emph{cells} of dimension $i$ and the homomorphisms $d_i$ are called \emph{boundary maps}.
\end{defn}
\[
\begin{CD}
\cdots @@>d>>\Z[\sigma_2] @@>d>> \Z[\sigma_1] @@>d>> \Z[\sigma_0] \\
& & & & & & @@V{\veps}VV\\
& & & & & & \Z
\end{CD}
\]

If an abstract cell complex is given we denote it usually by $\sigma$ and encode the boundary maps by the coefficients $D_{ab}$:
\[
d a = \sum_{b\in\sigma_{i-1}} D_{ab} b, \;a\in\sigma_i.
\]


The basic example is the standard simplex of dimension $n$, $\Delta_n$. In this case $\sigma_i$ is the set of all subsets of $\{0,1,\ldots,n\}$ of size $i+1$ and, writing subsets as increasing sequences, 
\[
d_i(j_0,\ldots,j_i) = \sum_{k=0}^i (-1)^k (j_0,\ldots,\hat{j}_k,\ldots,j_i).
\]

Given two abstract cell complexes $\sigma$, $\sigma'$ the \emph{product} 
\[
\sigma'' = \sigma\times \sigma'
\]
is again an abstract cell complex in the following way:
\[
\sigma''_i = \bigcup_{j=0}^i \sigma_j\times\sigma'_{i-j},
\]
\[
d_i(a\times b) = (d_j a) \times b + (-1)^j a\times d_{i-j} b \qquad \text{for $a\in\sigma_j$, $b\in\sigma'_{i-j}$.}
\]
For example, the cube is defined as
\[
\square_n = \Delta_1\times\cdots\times\Delta_1,\qquad\text{the product has $n$ terms.}
\]

\subsection{Hypercover}
Let $X$ be a topological space and $\sigma$ an abstract cell complex. 

\begin{defn}
A \emph{hypercover} of $X$ (indexed by $\sigma$) is a system of open subsets $\U=(U_a)$ indexed by elements $a\in\sigma$ such that:

\begin{enumerate}
\item Whenever a cell $b$ belongs to the boundary of a cell $a$, $U_a\subset U_b$.
\item For any point $x\in X$ consider all cells $a$ such that $x\in U_a$. These cells form a subcomplex of $\sigma$ by the first property, call it $\sigma_x$. We require the complex of abelian groups coming from $\sigma_x$ to be a resolution of $\Z$ with the augmentation map induced by $\varepsilon$.
\end{enumerate}
\end{defn}

\begin{example}
Put $\sigma=\Delta_n$ and let $X$ be a space covered by $n+1$ open subsets $U_0,\dots U_n$. For any sequence $(j_0,\dots,j_i)\in\sigma_i$ we put 
\[
U_{(j_0,\dots,j_i)}= U_{j_0}\cap\dots\cap U_{j_i}.
\]
This clearly gives a hypercover which is called the \v{C}ech hypercover.
\end{example}

If $\sigma$ and $\sigma'$ are two abstract chain complexes, $(U_a)$ and $(U'_{a'})$ are hypercovers of spaces $X$, $X'$ indexed by $\sigma$ and $\sigma'$ correspondingly, by making all products $(U_a\times U'_{a'})$ one gets a hypercover of $X\times X'$ indexed by $\sigma\times\sigma'$, the \emph{product} hypercover.

\subsection{Hypersection}
Let $X$ be a topological space, $\sigma$ an abstract chain complex and $\U=(U_a)$ a hypercover.

Let $\Z_a$ denote the constant sheaf with fiber $\Z$ on $U_a$ extended by zero to $X$. Let 
\[
\Z_{\U i} = \bigoplus_{a\in \sigma_i} \Z_a.
\]
If $U_a\subset U_b$, there is a canonical morphism $r_{ab}:\Z_a \rightarrow \Z_b$. We define $d_i:\Z_{\U i} \rightarrow \Z_{\U i-1}$. If $a\in\sigma_i$ then the morphism from $\Z_a$ to $\Z_{\U i-1}$ is 
\[
\sum_{b\in\sigma_{i-1}}  D_{ab} r_{a b},\qquad \text{where}
\]
\[
d a = \sum_{b\in\sigma_{i-1}} D_{ab} b.
\]
Denote the corresponding sequence of sheaves by $\Z_\U$. The augmentation map $\veps:\Z_{\U 0} \rightarrow \Z_X$ is defined as the sum of the canonical morphisms $\veps_a:\Z_a\rightarrow\Z_X$ for $a\in\sigma_0$.
\[
\begin{CD}
\cdots @@>d>> \Z_{\U 2} @@>d>> \Z_{\U 1} @@>d>> \Z_{\U 0} \\
& & & & & & @@V{\veps}VV \\
& & & & & & \Z_X
\end{CD}
\]

\begin{prop}
The sequence $\Z_\U$ is a resolution of $\Z_X$.
\end{prop}
\begin{proof}
For any point $x\in X$ the stalk of $\Z_\U$ at $x$ is simply the complex of abelian groups corresponding to $\sigma_x$. So the statement follows from the second condition of hypercover.
\end{proof}

In other words, we have obtained a \emph{quasi-isomorphism}
\[
\veps:\Z_{\U} \rightarrow \Z_X.
\]
For any sheaf or complex of sheaves $\F$ on $X$ this gives a morphism of complexes
\[
\veps^*: \HHom(\Z_X, \F) \rightarrow \HHom(\Z_\U, \F).
\]
Note that $\HHom(\Z_X, \F) = \Gamma(X, \F)$ and for any $a\in\sigma$ $\HHom(\Z_a, \F) = \Gamma(U_a, \F)$. We denote the complex $\HHom(\Z_\U, \F)$ by $\Gamma(\U, \F)$ and call elements of this complex \emph{hypersections}. 

The complex of hypersections can be defined more precisely as follows. Let $\F$ be a complex of sheaves on $X$ written as
\[
\F^0\rightarrow\F^1\rightarrow\dots.
\]
Then we put
\[
\Gamma(\U,\F)^i = \prod_{j\geq0,a\in \sigma_j} \Gamma(U_a, \F^{i-j}).
\]
The coboundary of a hypersection $s=(s_a)\in \Gamma(\U,\F)^i$ is a hypersection $ds=(s'_a)\in \Gamma(\U,\F)^{i+1}$, where
\begin{equation}\label{dsa}
s'_a = d s_a + (-1)^{i-j+1} \sum_{b\in \sigma_{j-1}} D_{ab} s_b|_{U_a},\qquad a\in\sigma_j.
\end{equation}
One can check directly that $d^2=0$. Let $d^2s=(s''_a)$, $a\in\sigma_j$.
\begin{multline*}
s''_a = d s'_a + (-1)^{i-j} \sum_{b\in \sigma_{j-1}} D_{ab} s'_b|_{U_a}
=
(-1)^{i-j+1} \sum_{b\in\sigma_{j-1}} D_{ab} d s_b|_{U_a} +\\ (-1)^{i-j} \sum_{b\in \sigma_{j-1}} D_{ab} d s_b|_{U_a} + \sum_{b\in \sigma_{j-1}} \sum_{c\in \sigma_{j-2}} D_{ab} D_{bc} s_c|_{U_a}=0.
\end{multline*}

The augmentation morphism $\veps^*:\Gamma(X,\F)\rightarrow \Gamma(\U,\F)$ is defined by sending a section $s\in\Gamma(X,\F^i)$ to the hyperchain $(s'_a)$ where $s'_a$ is not zero only for $a\in \sigma_0$ and equals to the restriction of $s$ to $U_a$.

\begin{rem}
In the case of the \v{C}ech hypercover (corresponding to an open cover) we obtain the \v{C}ech complex.
\end{rem}
\begin{rem}
If I took the definition of $\HHom$ from \cite{harts:rd}, p. 64 the sign in (\ref{dsa}) would be $(-1)^{i+1}$. I assume that switching from a chain complex $\Z_\U$ to the corresponding cochain complex adds the multiplier $(-1)^j$.  My choice of sign is made in order to get the sign  correct in the case of the \v{C}ech hypercover.
\end{rem}

\subsection{Hyperchain}
We assume some class of topological chains on $X$ is given, i.e. semialgebraic chains. For any open set $U\subset X$ we denote by $C_i(U)$ the group of chains with support on $U$ of dimension $i$, which is the free abelian group generated by maps of the chosen class from the standard simplex of dimension $i$ to $U$. Suppose an abstract cell complex $\sigma$ and a hypercover $(U_a)$ is given. 

The complex of \emph{hyperchains} is defined similarly to the complex of hypersections, but with a change of sign. We put
\[
C_i(\U) = \bigoplus_{j\geq 0} C_{i-j}(\U, j), \qquad\text{where}
\]
\[
C_{i}(\U,j) = \bigoplus_{a\in \sigma_j} C_{i}(U_a).
\]
Given a hyperchain $\xi\in C_{i-j}(U_a)\subset C_i(\U)$ for $a\in\sigma_j$ its boundary is 
\[
\partial_h \xi:= \partial\xi + (-1)^{i-j} \sum_{b\in\sigma_{j-1}} D_{a b} j_b(\xi).
\]
Here $\partial\xi\in C_{i-1-j}(U_a)$ is the ordinary boundary of $\xi$ and $j_b(\xi)$ denotes the same chain as $\xi$, but considered as an element of $C_{i-j}(U_b)$ if $U_a\subset U_b$. The definition of the boundary map is then extended to $C_i(\U)$ by linearity. 

The augmentation morphism $\veps:C(\U)\rightarrow C(X)$ sends all chains in $C_{i-j}(U_a)$ with $j>0$ to $0$ and the ones with $j=0$ to themselves.

We have the following lifting property for lifting chains to hyperchains. Here the word \emph{subdivision} of a chain or of a hyperchain means some iterated barycentric subdivision of all of its simplices. It is clear that the operation of taking subdivision commutes with the boundary operation.
\begin{lem}
Let $\xi$ be a chain on $X$, $\eta=\partial \xi$, $\bar\eta$ a hyperchain such that $\veps\bar\eta=\eta$ and $\partial_h \bar\eta=0$. Then, after possibly replacing $\eta$, $\bar\eta$, $\xi$ with a subdivision, there exists a hyperchain $\bar\xi$ such that $\partial_h\bar\xi=\bar\eta$ and $\veps \bar\xi=\xi$.
\end{lem}
\[
\begin{CD}
\bar\xi\in C_i(\U) @@>{\partial_h}>> \bar\eta\in C_{i-1}(\U) \\
@@V{\veps}VV @@V{\veps}VV \\
\xi\in C_i(X) @@>{\partial}>> \eta\in C_{i-1}(X)
\end{CD}
\]
\begin{proof}
The complex $C_\bullet(\U)$ is the total complex of the bicomplex $C_\bullet(\U,\bullet)$ with the horizontal differential induced by the boundary map of chains in space $X$ and vertical one induced by the boundary map of the complex $\sigma$:
\[
\begin{CD}
& & C_{i-1}(\U, 1) @@>>> C_{i-2}(\U,1) \\
& & @@VVV @@VVV\\
C_i(\U,0) @@>>> C_{i-1}(\U,0) @@>>> C_{i-2}(\U,0)\\
@@V{\veps}VV @@V{\veps}VV\\
C_i(X) @@>>> C_{i-1}(X)
\end{CD}
\]
Clearly, it is enough to prove that the vertical complexes are exact (up to a subdivision). That is, we need to prove that if 
\[
\gamma=\sum_{a\in\sigma_{j}} \gamma_a \in C_i(\U,j) \;\text{is such that}
\]
\[
\sum_{a\in\sigma_{j}} D_{ab} \gamma_a=0 \; \text{for any $b\in\sigma_{j-1}$,}
\]
then there exists 
\[
\bar\gamma=\sum_{c\in\sigma_{j+1}} \bar\gamma_c \in C_i(\U,j+1) \;\text{such that}
\]
\[
\sum_{c\in\sigma_{j+1}} D_{ca} \bar\gamma_c = \gamma'_a \; \text{for any $a\in\sigma_{j}$,}
\]
where $\gamma'_a$ is a subdivision of $\gamma_a$.

It is enough to prove the statement for multiples of a single simplex. Let $s$ be a simplex in $X$ which enters $\gamma_a$ with coefficient $s_a\in\Z$. Then 
\[
\sum_{a\in\sigma_{j}} D_{ab} s_a=0 \; \text{for any $b\in\sigma_{j-1}$.}
\]
Therefore the cycle $\sum_a s_a a$ of $\sigma$ is closed. Since the simplex $s$ must belong to $U_a$ for all $a$ for which $s_a\neq 0$, the mentioned cycle is a closed cycle of $\sigma_x$ for all $x$ in the closure of $s$. For each $x$ one can therefore represent it as a boundary of some cycle of $\sigma_x$, say $t_x$,
\[
t_x = \sum_{c\in\sigma_{j+1}} t_{x c} c.
\]
Here $t_{x c}$ is zero unless $x\in U_c$. Let $V_x$ be the open set which is the intersection of all $U_c$ for which $t_{xc}$ is not zero. Then $V_x$ form a cover of the support of $s$, so there exists a finite subcover. Let it be $V_1=V_{x_1}$, $V_2=V_{x_2}$, \dots, $V_k=V_{x_k}$. One can subdivide the simplex $s$ so that each simplex of the subdivision belongs to one of the chosen open sets. Let us do this and denote the subdivision by $s'$ so that
\[
s' = \sum_{l=1}^k s'_l, \; |s'_l| \subset V_k.
\]
Now it is clear that one may put
\[
\bar\gamma_l = \sum_{c\in\sigma_{j+1}} t_{x_l c} j_c(s'_l)
\]
because $s'_l$ belongs to $U_c$ for every $c$ for which $t_{x_l c}$ is not zero. Then 
\[
\sum_{c\in\sigma_{j+1}} D_{ca} t_{x_l c} s'_l = s_a s'_l \; \text{for any $a\in\sigma_{j}$,}
\]
therefore $\bar\gamma:=\sum_{l=1}^k \bar\gamma_l$ satisfies the required condition.
\end{proof}

This immediately implies
\begin{cor}
If $\xi$ is a closed chain on $X$ then there exists a closed hyperchain $\bar\xi$ whose augmentation is a subdivision of $\xi$. The hyperchain $\bar\xi$ is unique up to a boundary of a hyperchain whose augmentation is $0$.
\end{cor}

\begin{rem}
One extends the construction of hyperchains to the cases of chains on an open subset of $X$ and to the relative chains. It is clear that the boundary maps commute with the natural projections from chains to relative chains.
\end{rem}

\subsection{Integration}
We again fix a space $X$, an abstract cell complex $\sigma$ and a hypercover $(U_a)$. Suppose some class of chains and some class of differential forms are given such that one can integrate form along a chain. We have the de Rham complex of sheaves of differential forms $\Omega^\bullet$:
\[
\Omega^0\ra\Omega^1\ra\dots.
\]
Also we have the complex of chains on $X$:
\[
\dots\ra C_1(X)\ra C_0(X).
\]

The hypersections of $\Omega^\bullet$ will be called the \emph{hyperforms}. Given a hyperform $\omega$ of degree $d$ and a hyperchain $\xi$ of degree $d$ we define the \emph{integral}:
\[
\int_\xi \omega = \sum_{a\in\sigma} \int_{\xi_a}\omega_a.
\]
Here $\omega_a \in\Gamma(U_a, \Omega^{d-\dim a})$ and $\xi_a\in C_{d-\dim a}(U_a)$ are the components of $\omega$ and $\xi$.

The main property is the Stokes theorem:
\begin{prop}
For a hyperform $\omega$ of degree $d$ and a hyperchain $\xi$ of degree $d+1$ one has
\[
\int_{\partial_h \xi} \omega = \int_{\xi} d\omega.
\]
\end{prop}
\begin{proof}
By the definition
\[
\int_{\xi} d\omega = \sum_{a\in\sigma} \int_{\xi_a} (d \omega_a + (-1)^{d-\dim a+1}\sum_{b\in\sigma_{\dim a - 1}} D_{ab} \omega_b).
\]
Applying the Stokes formula this is further equal to 
\[
\sum_{a\in\sigma}(\int_{\partial\xi_a}\omega_a + (-1)^{d-\dim a+1}\sum_{b\in\sigma_{\dim a - 1}} D_{ab} \int_{\xi_a} \omega_b) = \sum_{a\in\sigma} \int_{\partial \xi_a} \omega = \int_{\partial_h\xi} \omega.
\]
\end{proof}

\subsection{Products}\label{products}
For two spaces $X$, $X'$, two abstract complexes $\sigma$, $\sigma'$, two hypercovers $(U_a)$, $U_{a'}'$ we have the product hypercover. There is exterior product on chains and on sheaves. For example, consider the exterior product on chains. For two chains $c\in C_i(X)$, $c'\in C_{i'}(X)$ we obtain the product chain $c\times c'\in C_{i+i'}(X\times X')$. This operation satisfies the property
\[
\partial(c\times c')=(\partial c) \times c' + (-1)^i c\times \partial c'.
\]

Introduce an exterior product on hyperchains. Let $c\in C_{i-j}(U_a)\subset C_i(\U)$, $c'\in C_{i'-j'}(U'_{a'})\subset C_{i'}(\U')$. Here $\dim a = j$, $\dim a'=j'$. Then put
\[
c\times_h c':=(-1)^{j(i'-j')} c\times c'\in C_{i+i'-j-j'}(U_a\times U'_{a'}) \subset C_{i+i'}(\U\times\U').
\]
The superscript $h$ stands for "hyper".

One can check that this satisfies
\begin{prop}
\[
\partial_h(c\times_h c') = (\partial_h c) \times_h c' + (-1)^i c\times_h \partial_h c'.
\]
\end{prop}

On hypersections the exterior product is defined in a similar way. If $s=(s_a)\in \Gamma(\U, \F^\bullet)^i$ and $s'=(s'_{a'})\in \Gamma(\U', \F'_\bullet)^{i'}$, then
\[
(s\times s')_{a\times a'} = (-1)^{\dim a (i'-\dim a')} s_a\times s'_{a'}.
\]
One has a similar formula
\begin{prop}
\[
d(s\times s') = (ds) \times s' + (-1)^i s\times ds'.
\]
\end{prop}

\subsection{Residues}
Let $X$ be a projective algebraic variety over $\C$, $\sigma$ an abstract cell complex and $\U=(U_a)$ a hypercover of $X$ in the Zariski topology. 

\begin{defn}
By a \emph{refinement} of $\U$ we understand any hypercover $\U'=(U_a')$ of $X$ in the analytic topology such that for any $a\in \sigma$ $U_a'\subset U_a$.
\end{defn}

Suppose a finite family $M$ of irreducible subvarieties of $X$ is given. These subvarieties will be called \emph{special}.

\begin{defn}
A refinement $\U'=(U_a')$ is called \emph{nice} if for any cell $a\in \sigma$ and any special subvariety $Z$ such that $\dim Z < \dim a$ one has $U_a'\cap Z = \varnothing$. 
\end{defn}

If we have a nice refinement $\U'=(U_a')$ we can make the following construction. Fix a special subvariety $Z$. Take the fundamental class of $Z$, represent it by a chain (of dimension $2\dim Z$) and lift it to a closed hyperchain $(c_a)$ with respect to the hypercover $(U_a' \cap Z)$. Note that for $a\in\sigma_{\dim Z+1}$ the set $U_a'\cap Z$ is empty. Therefore for every $a\in \sigma_{\dim Z}$ the chain $c_a$ is a closed chain of dimension $\dim Z$. Moreover, if we choose a different representation of the fundamental class or a different lift, the hyperchain will differ by a boundary. This implies that the chain $c_a$ for $a\in\sigma_{\dim Z}$ will change by a boundary. Thus the class of $c_a$ in the homology of $Z\cap U_a'$ does not depend on the choices made. Denote this class by $h_a(Z, \U')\in H_{\dim Z}(Z\cap U_a')$. 

\begin{defn}
For any meromorphic differential form $\omega$ on $Z\cap U_a$ of degree $\dim Z$ which is regular outside special subvarieties of $Z$ we define its residue as
\[
\res_{a,\U'}^{\Int} \omega = \int_{h_a(Z, \U')} \omega.
\]
\end{defn}

\subsection{Construction of nice refinements}\label{construction_refinements}
We will construct some nice refinements in certain special situation and relate the corresponding residues with the ordinary residues.

Let $X$ be a product of smooth projective curves, $X=X_1\times\cdots\times X_n$. Suppose on each curve a finite open cover (in the Zariski topology) is given. As we have seen before, this gives a hypercover of $X_k$ indexed by the standard simplex of dimension, say $m_k$, the \v{C}ech hypercover. By taking products we obtain an abstract cell complex
\[
\sigma = \Delta_{m_1}\times \Delta_{m_2}\times\cdots\times\Delta_{m_n}
\]
and the product hypercover $\U$ of $X$ indexed by $\sigma$. For any finite family $M$ of irreducible subvarieties of $X$ we construct a nice refinement of $\U$. Here we describe the construction.

Without loss of generality we may assume that the set of special subvarieties $M$ satisfies the following conditions:
\begin{enumerate}
\item For any open set $U_a$ belonging to the hypercover the irreducible components of its complement are in $M$.
\item For any two sets in $M$ the irreducible components of their intersection is also in $M$.
\item If $M$ contains a set $Z$, then it also contains the singular locus of $Z$.
\item If $M$ contains a set $Z$, then for any subset $L\subset\{1,\ldots,n\}$ $M$ contains the irreducible components of the set where the projection from $Z$ to the product $\times_{k\in L} X_k$ is not \'etale (this may be the whole set $Z$).
\end{enumerate}

Let $X_k$ be one of the curves above. Choose a Riemannian metric on $X_k$. Let $n(S,\veps)$ denote the $\veps$-neighbourhood of a set $S$. For any $k$ we give a refinement of the hypercover of the curve $X_k$. This depends on two real numbers $\veps>\veps'>0$. 

Suppose $X_k$ is covered by the open sets $U_{k,0},\ldots,U_{k,m_k}$. Denote the complement $X_k\setminus U_{k,0}$ by $S_k$. The set $S_k$ is a finite set of points $p_1,p_2,\ldots,p_r$. 

Consider those points among $p_1,p_2,\ldots,p_r$ which are covered by $U_{k,1}$. Suppose they are $p_1,p_2,\ldots,p_{r_1}$. Then consider those points among the remaining ones which are covered by $U_{k,2}$. Suppose they are $p_{r_1+1},\ldots,p_{r_2}$, etc. In this way we obtain a decomposition of the set $S_k$ into $m_k$ subsets, some of them empty:
\[
S_k = \bigcup_{i=1}^{m_k} S_{k,i}.
\]
Let $\wt{S_k} = S_k\cup \{\eta_k\}$, where $\eta_k$ is the generic point of $X_k$. Let $S_{k,0} = \{\eta_k\}$.
For $p\in\wt{S_k}$ put 
\[
U^p_k(\veps, \veps') = \begin{cases} n(p, \veps) & \text{if $p$ is a closed point, $p\in S_k$,}\\ X_k\setminus \overline{n(S_k, \veps')} & \text{if $p$ is the generic point.}\end{cases}
\]
For $p\in S_k$ put
\[
R^p_k(\veps, \veps') = U^p_k(\veps, \veps') \cap U^{\eta_k}_k(\veps, \veps'),\; R_k(\veps, \veps') = \bigcup_{p\in S_k} R^p_k(\veps,\veps'),\; U^S_k(\veps,\veps')=\bigcup_{p\in S_k} U^p_k(\veps,\veps').
\]

We define the refinement in the following way. Put 
\[
U_{k,i}'(\veps, \veps') = \bigcup_{p\in S_{k,i}} U^p_k(\veps, \veps'),\;i=1,\ldots,m_k.
\]
This is a cover of $X_k$. We obtain $\U_k'(\veps,\veps')$ as the \v{C}ech hypercover associated to this cover. If $\veps$ is small enough, this is a refinement of the original cover, i.e. $U_{k,i}'\subset U_{k,i}$. Moreover, if $\veps$ is small enough, the open sets $U^p_k$ for $p\in S_k$ are non-intersecting disks. Let $q>0$ be a number such that as soon as $\veps<q$ these two conditions are satisfied.

For $p\in \wt{S_k}$ we define by $a_k(p)$ the $0$-cell $a$ such that $p\in S_{k,a}$. For $p\in S_k$ we define by $a_k(\eta_k,p)$ the $1$-cell $a$ which joins $a_k(\eta_k)$ and $a_k(p)$. Then for any cell $a$ of the standard simplex $\Delta_{m_k}$ the elements of the hypercover are given as follows:
\[
U_{k,a}'(\veps, \veps') = \begin{cases} 
  \bigcup_{p\in \wt{S_k},a_k(p)=a} U^p_k(\veps, \veps') & \text{if $\dim a=0$,}\\
  \bigcup_{p\in S_k, a_k(p, \eta_k) =a} R^p_k(\veps, \veps') & \text{if $\dim a=1$,}\\
  \varnothing & \text{otherwise.}
\end{cases}
\]

We consider vectors of real numbers $\vec\veps=(\veps_1,\ldots,\veps_n, \veps_1',\ldots,\veps_n')$ such that for each $k$, $q>\veps_k>\veps_k'>0$. Denote
\[
\U(\vec\veps) = \times_{k=1}^n \U_k(\veps_k,\veps_k').
\]
This is a refinement of $\U$. We denote for any $L\subset\{1,\dots,n\}$ 
\[
R_L(\vec\veps) = \times_{k\in L} R_k(\veps_k,\veps_k'),\; U^S_L(\vec\veps) = \times_{k\in L} U^S_k(\veps_k,\veps_k'),\; S_L=\times_{k\in L} S_k.
\]
For any $p\in S_L$ we denote $p_k=\pi_k p$ and put
\[
R_L^p(\vec\veps) = \times_{k\in L} R_k^{p_k}(\veps_k, \veps_k').
\]
We will simply write $R_L$, $U_L^S$, $R_L^p$ when there is no confusion.
We put 
\[
X_L=\times_{k\in L} X_k,\; \pi_L:X\ra X_L\;\text{the projection.}
\]

\begin{defn}
We say that the choice of real numbers $\veps_m,\ldots,\veps_n, \veps_m',\ldots,\veps_n'$ is \emph{good} if $q>\veps_k>\veps_k'>0$ for $k=m,\ldots,n$ and there is an increasing sequence of positive real numbers 
\[
\delta_m, \delta_m',\delta_{m+1},\delta_{m+1}',\ldots,\delta_n,\delta_n'
\]
such that the following conditions are satisfied for any special set $Z$, an index $k$ ($m\leq k\leq n$) and $p\in S_k$:
\begin{enumerate}
\item If $x\in Z$ is such that $\dist(\pi_k(x),p)\leq\veps_k$, then $\dist(x, Z')<\delta_k$.
\item If $x\in X$ is such that $\dist(x, Z)<\delta_{k-1}'$ and $\dist(\pi_k(x), p)<\veps_k$, then $\dist(x, Z')<\delta_k$.
\item If $x\in X$ is such that $\dist(x, Z')<\delta_{k-1}'$, then $\dist(\pi_k(x), p)\leq\veps_k'$.
\end{enumerate}
And the following condition is satisfied for any two special sets $Z_1$, $Z_2$ and an index $k$, $m\leq k\leq n$: 
\begin{enumerate}
\item[(iv)] If $x\in X$ is such that $\dist(x, Z_1)<\delta_k$ and $\dist(x, Z_2)<\delta_k$, then $\dist(x, Z_1\cap Z_2)<\delta_k'$.
\end{enumerate}
We have denoted by $\pi_k$ the projection $X\rightarrow X_k$ and by $Z'$ the intersection $Z\cap\pi_k^{-1} p$. The second and the third conditions are required only for $k>m$.
\end{defn}

Next we prove that good choices indeed exist. For this it is enough to show
\begin{lem}\label{lem:good_choices}
For every good choice of real numbers $\veps_{m+1},\ldots,\veps_n, \veps_{m+1}',\ldots,\veps_n'$ there exists a number $t>0$ such that for any $\veps_m, \veps_m'$ satisfying $t>\veps_m>\veps_m'>0$ the choice $\veps_{m},\ldots,\veps_n$, $\veps_{m}',\ldots,\veps_n'$ is good. Therefore good sequences exist.
\end{lem}
\begin{proof}
Suppose there is a sequence $\delta_{m+1},\ldots,\delta_n'$ as in the definition above. Consider $Z\in M$ and $Z'=Z\cap \pi_{m+1}^{-1} S_{m+1}$. We first show that we can choose $\delta_m'$ such that the second and the third conditions are satisfied for $k=m+1$.

Consider the set 
\[
V=\{x\in X: \dist(x, Z')<\delta_k\; \text{or}\; \dist(\pi_{m+1}(x), S_{m+1})> \veps_{m+1}\}.
\]
This is an open neighbourhood of $Z$ by the first condition for $k=m+1$. Therefore if $\delta_m'>0$ is small enough, then the $\delta_m'$-neighbourhood of $Z$ is also contained in $V$. Therefore the second condition is satisfied for $k=m+1$.

Since $Z'$ is a compact set and $\pi_{m+1}(Z')\subset S_{m+1}$, if $\delta_m'>0$ is small enough, the third condition is satisfied.

Therefore we can choose $\delta_m'>0$ such that $\delta_m'<\delta_{m+1}$ and both the second and the third conditions are satisfied for all $Z$ and $k=m+1$.

Next we choose $\delta_m$ such that the fourth condition is satisfied. To see that this can be done for $Z_1$, $Z_2$ we consider the compact set $X\setminus n(Z_1\cap Z_2, \delta_m')$. The two continuous functions $\dist(\bullet, Z_1)$ and $\dist(\bullet, Z_2)$ do not attain simultaneously value zero on this set. Therefore if $\delta_m$ is small enough, these functions cannot attain simultaneously value less than $\delta_m$. This is equivalent to the fourth condition.

Next consider $Z\in M$ and $Z'=Z\cap \pi_m^{-1} S_m$. The set
\[
\{x\in Z: \dist(x,Z')\geq \delta_m\} 
\]
is compact, hence the continuous function $d(x)=\dist(\pi_m(x), S_m)$ attains its minimum. Suppose $\epsilon_m$ is less than the minimal value of this function. It follows that if $x\in Z$ and $d(x)\leq \epsilon_m$, then $x$ cannot belong to the set above. Therefore $\dist(x,Z')< \delta_m$ and the first condition is satisfied. This implies existence of $t>0$ with the required properties.
\end{proof}

\subsection{Flags of subvarieties}
We consider flags of subvarieties of $X$. A flag of length $m$ is a sequence of irreducible subvarieties $Z_\bullet=(Z_0\supset Z_1 \supset \cdots \supset Z_m)$. We say that a flag $Z_\bullet$ \emph{starts} with $Z_0$ and \emph{ends} with $Z_m$. We require $Z_m$ to be not empty.

\begin{defn}
Let $L\subset\{1,\ldots,n\}$, $L=\{k_1,\ldots,k_l\}$. Let $p\in S_L$. We say that a flag $Z_\bullet=Z_0\supset\cdots\supset Z_l$ is $L,p$\emph{-special} if
\begin{enumerate}
\item $Z_i$ is special for $0\leq i\leq l$,
\item $Z_i$ is an irreducible component of $Z_{i-1}\cap \pi_{k_i}^{-1} p_{k_i}$ for $1\leq i\leq l$.
\end{enumerate}
\end{defn}

\begin{defn}
We say that a flag $Z_\bullet=Z_0\supset\cdots\supset Z_l$ is \emph{strict at index $i$} if $Z_i\neq Z_{i-1}$. We say that a flag is \emph{strict} if it is strict at all indices.
\end{defn}

Suppose $\vec\veps$ is good for a sequence of numbers $\vec\delta=(\delta_1,\delta_1',\ldots,\delta_n,\delta_n')$. The \emph{$\vec\delta$-neighbourhood} of an $L$-special flag $Z_\bullet$ is defined to be the set
\[
n(Z_\bullet, \vec\delta) = \{x\in X: \dist(x, Z_i)<\delta_{k_i},\;i\geq 1\}.
\]

\begin{prop}\label{prop:sf0}
Suppose $Z\subset X$ is special, $L\subset\{1,\ldots,n\}$, $x$ is a point for which either $x\in Z$ or $\dist(x, Z)<\delta_{\min L -1}'$ (if $\min L>1$). Suppose $p\in S_L$ and $\pi_L(x)\in U_L^p(\vec\veps)$. Then there exists an $L,p$-special flag $Z_\bullet$ starting with $Z$ such that $x$ belongs to its $\vec\delta$-neighbourhood.
\end{prop}
\begin{proof}
We construct the flag step by step. If 
\[
\dist(x,Z_{i-1})<\delta_{k_{i-1}}'\leq\delta_{k_i-1}',\; \dist(\pi_{k_{i}} x, p_{k_i})<\veps_{k_i},
\]
by the second property of good sequences we obtain 
\[
\dist(x,Z_{i-1}\cap \pi_{k_i}^{-1} p_{k_i})<\delta_{k_i}.
\]
Therefore on each step we can choose $Z_i$ as an irreducible component of $Z_{i-1}\cap \pi_{k_i}^{-1} p_{k_i}$ so that $\dist(x,Z_i)<\delta_{k_i}$.
\end{proof}

\begin{prop}\label{prop:sf}
Let $x$, $Z$, $L$ and $p$ be as in the proposition above. Take an $L,p$-special flag $Z_\bullet$ such that $x$ belongs to the $\vec\delta$-neighbourhood of $Z_\bullet$. If $L'\subset L$ is such that $\pi_{L'}(x)\in R_{L'}(\vec\veps)$, then $Z_\bullet$ is strict at all indices $i$ for which $k_i\in L'$.
\end{prop}
\begin{proof}
If $k_i\in L'$, then $\dist(\pi_{k_i}, p_{k_i})>\veps_{k_i}'$. By the third property of good sequences we obtain $\dist(x, Z_{k_i})\geq \delta_{k_i}'>\delta_{k_{i-1}}'$. Since $\dist(x, Z_{k_{i-1}})<\delta_{k_{i-1}}$, $Z_{k_{i-1}}\neq Z_{k_i}$.
\end{proof}

\begin{cor}
If $\vec\veps$ is a good choice of numbers, then for any special subvariety $Z$ of dimension less than $d$ and $L\subset\{1,\ldots,n\}$ of size $d$ the intersection $Z\cap \pi_L^{-1} R_L(\vec\veps)$ is empty. Therefore the refinement $\U'(\vec\veps)$ is nice.
\end{cor}
\begin{proof}
If the intersection was not empty, by Proposition \ref{prop:sf} there would exist a strict $L,p$-special flag which starts with $Z$. Therefore the dimension of $Z$ would be at least $d$. 
\end{proof}

\subsection{Decomposition of the residue according to flags}
Let $X$, $\sigma$, $\U$, $M$ be as in the previous section. Let $Z$ be a subvariety of $X$, $Z\in M$. Let $d=\dim Z$ and $\omega$ be a meromorphic differential form on $Z$ of degree $d$ which is regular outside the special subvarieties of $Z$. Let $a\in\sigma_d$. Let $\U'=\U(\vec\veps)$ be the nice refinement of $\U$ corresponding to a good choice of numbers $\vec\veps$. We will show how to compute
the residue $\res_{a,\U'} \omega$.

Recall that the residue was defined as the integral of $\omega$ along the class $h_a(Z, \U')\in H_d(Z\cap U_a')$. Since 
\[
\sigma=\prod_{k=1}^n \Delta_{m_k},
\]
the cell $a$ is given by a sequence of cells $a_1\in\Delta_{m_1},\ldots,a_n\in\Delta_{m_n}$. The open set $U_a'$ is the product of sets $U_{k,a_k}(\veps_k, \veps_k')$ for $k=1,\ldots,n$. Note that for $k=1,\ldots,n$ the set $U_{k,a_k}(\veps_k, \veps_k')$ is not empty only if either $a_k$ is a point or $a_k$ is the edge joining the $0$ vertex and some other vertex. Therefore we get a non-zero residue only if there is a set $L\subset\{1,\ldots,n\}$ of size $d$ and 
\[
\dim a_k= \begin{cases} 1 & \text{for $k\in L$,}\\ 0 & \text{otherwise.}\end{cases}
\]
Moreover in this case we have $U_a'\subset \pi_L^{-1} R_L$. 

Consider the set of strict special $L,p$-flags for all $p\in S_L$, starting with $Z$. Denote this set by $\Fl_L(Z)$. This is a finite set. We have seen that any point of $Z\cap \pi_L^{-1} R_L$ belongs to the $\vec\delta$-neighbourhood of a flag from $\Fl_L(Z)$. In fact we have
\begin{prop}\label{prop:unfl}
Any point $x\in Z\cap \pi_L^{-1} R_L$ belongs to the $\vec\delta$-neighbourhood of a unique flag from $\Fl_L(Z)$. 
\end{prop}
\begin{proof}
Suppose $x$ belongs to the $\vec\delta$-neighbourhood of two flags $Z_\bullet$ and $Z_\bullet'$. Let $i$ be the minimal index for which $Z_i\neq Z_i'$. Since $\dist(x, Z_i)<\delta_{k_i}$ and $\dist(x, Z_i')<\delta_{k_i}$, by the fourth property of good sequences we have $\dist(x, Z_i\cap Z_i')<\delta_{k_i}'$. Let $Y$ be an irreducible component of $Z_i\cap Z_i'$ such that $\dist(x, Y)<\delta_{k_i}'$. Let $K$ be the subset of elements of $L$ which are greater than $k_i$. By Proposition \ref{prop:sf} one can construct a strict $K$-special flag starting from $Y$ whose $\vec\delta$-neighbourhood contains $x$. But one can see that the length of the flag equals to the dimension of $Z_i$ and is at least one more than the dimension of $Y$. Hence such flag does not exist and we obtain a contradiction.
\end{proof}

This gives us a possibility to decompose $Z\cap \pi_L^{-1} R_L$ according to the set $\Fl_L(Z)$.
\begin{cor}\label{cor:unfl}
The set $Z\cap \pi_L^{-1} R_L$ is the union of non-intersecting open sets $R_{L,Z_\bullet} = Z\cap \pi_L^{-1} R_L \cap n(Z_\bullet, \vec\delta)$, one for each $Z_\bullet\in \Fl_L(Z)$. Correspondingly we obtain the decomposition
\[
\res_{a,\U'}^{\Int}\omega = \sum_{Z_\bullet \in \Fl_L(Z)} \res_{a,\U',L,Z_\bullet}^{\Int}\omega,\;\text{where}\;
\res_{a,\U',L,Z_\bullet}^{\Int}\omega = \int_{h_a(Z,\U')\cap R_{L,Z_\bullet}}\omega.
\]
\end{cor}

Next we show, for a given flag $Z_\bullet\in \Fl_L(Z)$, how to determine the cell $a$ for which $R_{L,Z_\bullet}$ is a union of several connected components of $U_a'$. We want this cell to be the product of cells $\times_{k=1}^n a_k$ with $\dim a_k=1$ for $k\in L$ and $\dim a_k=0$ for $k\notin L$. We will see that such a cell will be unique. 

The last variety in the flag, $Z_d$, is a point from $S_L$. 
\begin{prop}
Suppose $x\in R_L$. If $x\in R_{L,Z_\bullet}$ and $\pi_L(x)\in R_L^p$, then $p=Z_d$.
\end{prop}
\begin{proof}
Since $x\in Z\cap\pi_L^{-1}R_L^p$, one has an $L,p$-flag whose $\vec\delta$-neighbourhood contains $x$. By Proposition \ref{prop:unfl} this flag must be $Z_\bullet$. Hence $Z_\bullet$ is $L,p$-special. Thus $Z_d=p$.
\end{proof}

For an index $k\notin L$ and a flag $Z_\bullet\in \Fl_L(Z)$ we define an element $p_k$ of $\wt{S_k}$ as follows. Let $i$ be the maximal index for which $k_i<k$ or $0$ if $k<k_1$. If $\pi_k Z_i$ is a point in $S_k$ we let $p_k$ be this point. Otherwise we put $p_k=\eta_k$, the generic point.
This gives an element $p(Z_\bullet)\in\wt{S_{L^c}}$, where $L^c=\{1,\ldots,n\}\setminus L$. 

On the other hand if $x\in Z\cap R_L$ then for any $k\in L^c$ $\pi_k(x)$ belongs to $U_k^{p_k'}(\veps,\veps')$ for exactly one $p_k'\in \wt{S_k}$. If it was not true, then $x$ would belong to $R_{L\cup\{k\}}$, which is a contradiction. This defines another point $p(x)\in\wt{S_{L^c}}$.
\begin{prop}
For any $x\in Z\cap R_L$ if $x\in R_{L,Z_\bullet}$, then $p(Z_\bullet)=p(x)$.
\end{prop}
\begin{proof}
Let $K$ be the set of such $k$ that either $k\in L$ or $p_k'\in S_k$ (i.e. $p_k$ is not the generic point). For $k\in L$ let $p_k'$ be such that $\dist(\pi_k(x), p_k')<\veps_k$. This defines a point $p'\in S_K$. Then for every $k\in K$ $\dist(\pi_k(x), p_k')<\veps_k$. Thus $\pi_K(x)\in U_L^{p'}$. 

By Proposition \ref{prop:sf0} there exists a $K, p'$-special flag $Z'$ starting from $Z$ whose $\vec\delta$-neighbourhood contains $x$. This flag must be strict for all indices which correspond to elements of $L$. Since the number of this indices equals the dimension of $Z$ the flag must be not strict at all other indices. Therefore this flag defines a $L,p'$-special flag, which must be $Z$ by the Proposition \ref{prop:unfl}. At the same time this shows that $p_k=p_k'$ for all $k\in K$. 

Let $k\notin K$. Then $\dist(\pi_k(x), S_k)>\veps_k'$. Suppose $i$ is the maximal index for which $k_i<k$ or $0$ if $k<k_1$. If $i\neq 0$, by the third property of good sequences 
\[
\dist(x, Z_{k_i}\cap\pi_k^{-1} S_k)\geq \delta_{k-1}'\geq\delta_{k_i}\geq\dist(x, Z_{k_i}).
\]
Therefore $Z_{k_i}\neq Z_{k_i}\cap\pi_k^{-1} S_k$ which means that $p_k=\eta_k$. The case $i=0$ is obvious.
\end{proof}

We see that 
\begin{cor}
For every flag $Z_\bullet\in \Fl_L(Z)$ there is a canonically defined cell $a_L(Z_\bullet)$ of dimension $d$. For any cell $a$ of dimension $d$ the set $Z\cap U_a'$ is the disjoint union of open sets $R_{L,Z_\bullet}$ where $Z_\bullet$ runs over all $Z_\bullet\in \Fl_L(Z)$ with $a_L(Z_\bullet)=a$.
\end{cor}
\begin{rem}
Therefore we may omit $a$ in the notation $\res_{a,\U',L,Z_\bullet}^{\Int}$ and write $\res_{\U',L,Z_\bullet}^{\Int}$ instead.
\end{rem}

The set $a_L(Z_\bullet)$ can be reconstructed from the points $\pi_L(Z_d)\in S_L$ and $p(Z_\bullet)\in \wt{S_{L^c}}$. In fact
\[
a_L(Z_\bullet) = \times_{k\in L} a_k(\eta_k, \pi_k(Z_d)) \times \times_{k\in L^c} a_k(p_k(Z_\bullet)).
\]

\subsection{Computation of the residue using iterated residues}
Note that $Z$ and $X_L$ have the same dimension. 
\begin{prop}
If the restriction of $\pi_L$ to $Z$ is not surjective, then the set $\Fl_L(Z)$ is empty.
\end{prop}
\begin{proof}
If not, then $\dim \pi_L(Z)<d$. Let $L=\{k_1,\ldots,k_d\}$. Let $Z_\bullet\in \Fl_L(Z)$. Consider the corresponding flag of irreducible subvarieties of $X_L$:
\[
\pi_L(Z)=\pi_L(Z_0)\supset \pi_L(Z_{1}) \supset \cdots \supset \pi_L(Z_{d}).
\]
Because of the dimension reasoning there must be an index $i$ with $\pi_L(Z_i)=\pi_L(Z_{i-1})$. This implies
\[
\pi_{k_i}(Z_{i-1}) = \pi_{k_i}(Z_i) \subset S_{k_i}.
\]
Therefore $Z_i=Z_{i-1}$, so the flag is not strict, which is a contradiction.
\end{proof}

We may therefore suppose without loss of generality that $\pi_L:Z\ra X_L$ is surjective. Hence the set of points on $Z$ where this map is not \'etale is a proper closed subvariety. The irreducible components of this subvariety have dimension $d-1$ or less and are special. Therefore $\pi_L^{-1} R_L\cap Z$ belongs to its complement. This means the following is true.
\begin{prop}
The projection
\[
\pi_L:\pi_L^{-1} R_L\cap Z\ra R_L
\]
is an unramified covering.
\end{prop}

Let $p=\pi_L Z_d\in S_L$.
Since for a flag $Z_\bullet\in \Fl_L(Z)$ the set $R_{L,Z_\bullet}$ is open and closed in $\pi_L^{-1} R_L\cap Z$ we get
\begin{prop}
The projection
\[
\pi_L:R_{L,Z_\bullet} \ra R_L^p
\]
is an unramified covering.
\end{prop}

\begin{rem}
By the proposition we see that since $R_L^p$ is a product of annuli, $R_{L,Z_\bullet}$ is a disjoint union of products of annuli.
\end{rem}

We are going to determine the cycle $h_a(Z,\U')\cap R_{L,Z_\bullet}\in H_d(R_{L,Z_\bullet})$. The $d$-th homology group of a product of $d$ annuli is $\Z$. Hence there is a canonical generator $h_c$ of $H_d(R_L^p)$. In fact $h_c$ can be defined as the product $c_1\times\cdots\times c_d$, where $c_i$ is the circle in $R_{k_i}^{p_{k_i}}$ going around $p_{k_i}$ counterclockwise. 

\begin{prop}
The cycle $h_a(Z,\U')\cap R_{L,Z_\bullet}$ is the pullback of $(-1)^{\frac{d(d-1)}2}h_c$ via the projection $\pi_L:R_{L,Z_\bullet} \ra R_L^p$.
\end{prop}
\begin{proof}
\begin{rem}
It is clear that we can decrease numbers $\veps_k$ and increase numbers $\veps_k'$. The sequence obtained in this way will be also good. Moreover the open subsets of the new hypercover are contained in the corresponding open subsets of the old one. Therefore the residues computed with respect to these hypercovers are equal. Therefore one can assume that the projection $\pi_{L}:R_{L,Z_\bullet} \ra R_{L}^{p}$
extends to an unramified covering for the closures of $R_{L,Z_\bullet}$ in $Z$, $R_{L}^{p}$ in $X_L$.
\end{rem}

Consider the commutative diagram:
\[
\begin{CD}
H_{2d}(X) @@>>> H_{2d}(\ol{R_{L,Z_\bullet}}, \partial R_{L,Z_\bullet}) @@<{\pi_L^*}<< H_{2d}(\ol{R_L^p},\partial R_L^p) \\
@@V{h_a}VV @@V{h_a}VV @@V{h_a}VV\\
H_d(U_a') @@>>> H_d(R_{L,Z_\bullet}) @@<{\pi_L^*}<< H_d(R_L^p)
\end{CD}
\]
We see that it is enough to prove that the image of the fundamental class of $H_{2d}(\ol{R_L^p},\partial R_L^p)$ by the map $h_a$ in $H_d(R_L^p)$ is $(-1)^{\frac{d(d-1)}2}h_c$. There is a direct product decomposition $\ol{R_L^p}=\times_{k\in L} \ol{R_k^{p_k}}$. The hypercover on $\ol{R_L^p}$ is the product hypercover. For $k\in L$ let $\phi_k$ be the fundamental class of $H_2(\ol{R_L^p}, \partial R_L^p)$. Let us lift it to a hyperchain $\wt{\phi_k}$. 

The hypercover of $\ol{R_L^p}$ is the \v{C}ech hypercover associated to the cover with two open sets:
\begin{align*}
V_0 &=\ol{n(p_k, \veps_k)}\setminus \ol{n(p_k, \veps_k)}\;& \text{corresponding to the cell}\;& a(\eta_k),\\
V_1 &=n(p_k, \veps_k)\setminus n(p_k, \veps_k)\;& \text{corresponding to the cell}\;& a(p_k),\\
V_0\cap V_1 &=n(p_k, \veps_k)\setminus\ol{n(p_k, \veps_k)}\;& \text{corresponding to the cell}\;& a(\eta_k,p_k).
\end{align*}

Let $\veps'<r<\veps$. Consider topological chains
\begin{multline*}
c^k_0 = \ol{n(p_k, \veps_k)}\setminus n(p_k,r)\in C_2(V_0),\;
c^k_1 = \ol{n(p_k, r)}\setminus n(p_k,\veps_k')\in C_2(V_1),\\
c^k_{01} = \partial n(p_k,r)\in C_1(V_0\cap V_1).
\end{multline*}
They define a hyperchain $c^k$. We have $c_0+c_1=\phi_k$. The hyperchain is closed, therefore $h_{01}(\phi_k)=c^k_{01}$, which is the canonical generator of $H_1(R_k^{p_k})$.

Since the product of $\phi_k$ is the fundamental class of $H_{2d}(\ol{R_L^p},\partial R_L^p)$, the product of the hyperchains $c=\times_{k\in L} c^k$ lifts the fundamental class. The term of $c$ at $\times_{k\in L}(\eta_k, p_k)$ is, by the definition of the product for hyperchains, $(-1)^{\frac{d(d-1)}2} \times_{k\in L} c^k_{01}$. This is exactly $(-1)^{\frac{d(d-1)}2} h_c$.
\end{proof}

Let $Z_\bullet$ be a strict $L,p$-special flag, $L=\{k_1,\ldots,k_d\}$, $p=(p_{k_1},\ldots,p_{k_d})$. Let $Z'=Z_1$, $k=k_1$, $L'=\{k_2,\ldots,k_d\}$. Let $Z_\bullet'$ be the flag $Z_1\supset\cdots\supset Z_d$, $Z_\bullet'\in \Fl_{L'}(Z')$. Let $t_i$ be a local parameter on $X_{k_i}$ at the point $p_{k_i}=\pi_{k_i} Z_d$. Let $p'=\pi_{L'} p$. For $Z'$ we have the projection 
\[
\pi_{L'}:R_{L',Z_\bullet'} \ra R_{L'}^{p'},
\]
which is also an unramified covering. Let 
\[
\wt Z = Z'\times_{X_{L'}} Z = (X_{k_1}\times Z') \times_{X_L} Z. 
\]
We obtain the canonical diagrams
\[
\begin{CD}
\wt Z @@>{\rho}>> Z  &\qquad & \wt Z @@>{\rho}>> Z\\
@@V{\tau}VV @@VVV     @@V{\wt\tau}VV @@VVV     \\
Z' @@>>> X_{L'}      & & (X_{k_1}\times Z') @@>>> X_L
\end{CD}
\]
The diagonal embedding $Z'\ra Z'\times_{X_{L'}} Z'$ induces a morphism $\Delta:Z'\ra \wt Z$, which is a section to the natural projection $\tau:\wt Z \ra Z'$. Let 
\[
R_Z=Z\cap\pi_L^{-1}(U_k^{p_k}\times R_{L'}^{p'}) \cap n(Z_\bullet, \vec\delta).
\]
Consider the fiber products over $U_k^{p_k}\times R_{L'}^{p'}$:
\[
\begin{CD}
\wt R @@>{\rho}>> R_Z \\
@@V{\wt\tau}VV @@V{\pi_L}VV     \\
U_k^{p_k}\times R_{L',Z_\bullet'} @@>>> U_k^{p_k}\times R_{L'}^{p'}
\end{CD}
\]
Again, we have a section $\Delta:R_{L',Z_\bullet'}\ra \wt R$. Let $\wt R'$ be the union of connected components of $\wt R$ which intersect the image of $\Delta$. Let $\rho'$ be the restriction of $\rho$ to $\wt R'$.
\begin{rem}
The set $R_Z$ is open and closed in $Z\cap\pi_L^{-1}(U_k^{p_k}\times R_{L'}^{p'})$ because Proposition \ref{prop:unfl} and the first sentence of Corollary \ref{cor:unfl} work if we replace $R_L$ by $U_k^{p_k}\times R_{L'}^{p'}$.
\end{rem}
\begin{rem}\label{rem:RZ-nice}
All special subsets of $Z$ which intersect $R_Z$ are contained in $Z'$. Therefore the map $\pi_L$ on the diagram is an unramified covering outside $\{p_k\}\times R_{L'}^{p'}$ and $\pi_k^{-1} p_k \cap R_Z = R_{L',Z'_\bullet}$.
\end{rem}

\begin{prop}
The map $\rho'$ is an analytic isomorphism.
\end{prop}
\begin{proof}
Since $\rho'$ is a base change of the unramified covering $\pi_{L'}:R_{L',Z'_\bullet}\ra R_{L'}^{p'}$, it is an unramified covering itself. Therefore it is enough to construct a continuous section $s':R_Z\ra \wt R'$ to $\rho'$ which extends the diagonal map. This is equivalent to constructing a retraction $s:R_Z\ra R_{L',Z'_\bullet}$ which respects the projection $\pi_{L'}$.

Take a compact connected set $A\subset R_{L'}^{p'}$ such that $\pi_{L'}^{-1}(A)\cap R_{L',Z'_\bullet}=A_1\cup\cdots\cup A_m$ is a disjoint union of spaces isomorphic to $A$.

One can choose disjoint open subsets $V_1,\ldots, V_m$ in $\pi_{L'}^{-1}(A)\cap R_Z$ such that $A_i\subset V_i$. The set $C=\pi_{L'}^{-1}(A)\cap R_Z\setminus(V_1\cup\cdots\cup V_m)$ is closed. Therefore $\pi_L(C)$ is closed. Take $\alpha>0$ such that $n(p_k,\alpha)\times A$ does not intersect $\pi_{L}(C)$. This means that the open set $\pi_L^{-1}(n(p_k,\alpha)\times A) \cap R_Z$ is contained in the union of the sets $V_i$.

Let $V_i'= \pi_L^{-1}(n(p_k,\alpha)\times A) \cap V_i$. Let us prove that $V_i'$ is connected for each $i$. If not, then $V_i'=B_1\sqcup B_2$ with $B_j$ open, closed and nonempty. Since $A_i$ is connected, for some $j$ $B_j$ does not intersect $A_i$. Therefore $\pi_L B_j$ is open, closed and nonempty. Hence it must be the whole $U_k^{p_k}\times A$. This contradicts the assumption that $B_j$ does not intersect $A_i$.

One can constract a deformation retract retracting $\pi_{L'}^{-1}(A)\cap R_Z$ inside $\pi_L^{-1}(n(p_k,\alpha)\times A)\cap R_Z$. Therefore there are exactly $m$ connected components of $\pi_{L'}^{-1}(A)\cap R_Z$, each containing exactly one $A_i$. Therefore there is a unique map $s_A:\pi_{L'}^{-1}(A)\cap R_Z\ra \pi_{L'}^{-1}(A)\cap R_{L',Z'_\bullet}$ which is identity on $\pi_{L'}^{-1}(A)\cap R_{L',Z'_\bullet}$ and makes the diagram below commutative.
\[
\begin{CD}
\pi_{L'}^{-1}(A)\cap R_Z  @@>>> U_k^{p_k}\times A\\
@@V{s_A}VV @@VVV\\
\pi_{L'}^{-1}(A)\cap R_{L',Z'_\bullet} @@>>> A
\end{CD}
\]
Patching these $s_A$ together gives $s$ as required proving the first statement of the proposition.
\end{proof}

Take a meromorphic form $\omega$ on $Z$ which is holomorphic outside the special subvarieties of $Z$. The form $\omega$ can be written as
\[
\omega = f d t_1 \wedge d t_2 \wedge \cdots \wedge d t_n,
\]
where $f$ is a rational function on $Z$. Let $K$ be the field of fractions of $Z'$. Then $Z\times \spec K$ is a curve and $\Delta(\spec K)$ is a point. Therefore the algebraic residue $\res_{\Delta(\spec K)} \rho^* f d t_1$ is defined. We have
\begin{prop}
Put
\[
\omega' = (\res_{\Delta(\spec K)} \rho^* f d t_1) d t_2 \wedge \cdots \wedge d t_n.
\]
Then
\[
\res_{\U',L,Z_\bullet}^{\Int} \omega = (-1)^{d-1} 2\pi\I \res_{\U',L',Z_\bullet'}^{\Int} \omega'.
\]
\end{prop}
\begin{proof}
By the definition
\[
\res_{\U,L,Z_\bullet} \omega = \int_{h_a\cap R_{L,Z_\bullet}} \omega.
\]
Since $\rho'$ is an isomorphism,
\[
\int_{h_a\cap R_{L,Z_\bullet}} \omega = \int_{\rho'^*(h_a \cap R_{L,Z_\bullet})} \rho^* \omega.
\]
We have 
\[
\rho'^* (h_a \cap R_{L,Z_\bullet}) = (-1)^{\frac{d(d-1)}2}\wt R'\cap \rho^* \pi_L^* h_c = (-1)^{\frac{d(d-1)}2}\wt R'\cap \wt{\tau^*}(h_{c k_1}\times (R_{L',Z'_\bullet}\cap\pi_{L'}^* h_c')),
\]
where $h_{c k_1}$ is the circle in $R_{k_1}^{p_{k_1}}$ and $h_c'$ is the product of circles in $R_{L'}^{p'}$.
By Fubini's theorem
\[
\int_{\wt R'\cap \wt{\tau^*}(h_{c k_1}\times (R_{L',Z'_\bullet}\cap\pi_{L'}^* h_c'))} \rho^* \omega = \int_{R_{L',Z'_\bullet}\cap\pi_{L'}^* h_c'} g(z') d t_2\wedge\cdots\wedge d t_d,
\]
where for $z'\in R_{L',Z'_\bullet}$
\[
g(z') = \int_{\wt R'\cap \tau^{-1} z' \cap \pi_{k_1}^{-1} h_{c k_1}} \rho^* f d t_1.
\]
The last integral is nothing else than 
\[
2\pi \I \res_{\Delta(z')} \rho^* f d t_1.
\]
Therefore
\[
g(z') d t_2\wedge\cdots\wedge d t_d = 2\pi\I\omega'
\]
Taking into account that
\[
R_{L',Z'_\bullet}\cap\pi_{L'}^* h_c' = (-1)^{\frac{(d-1)(d-2)}2}R_{L',Z'_\bullet}\cap h_{a',Z'},
\]
where $a'$ is the cell obtained from $a$ by replacing the component $a(\eta_{k_1},p_{k_1})$ with the component $a(p_{k_1})$, we obtain the statement.
\end{proof}

Let us denote by $\res_{L,Z_\bullet} \omega$ the iterated algebraic residue of $\omega$ with respect to the flag $Z_\bullet$. This is defined by induction on the dimension of $Z$ by the formula
\[
\res_{L,Z_\bullet}f d t_1\wedge\cdots\wedge d t_d = \res_{L',Z_\bullet'} (\res_{\Delta(\spec K)} \rho^* f d t_1) d t_2 \wedge \cdots \wedge d t_d.
\]
Thus we obtain a formula for our residue.
\begin{cor}\label{cor_3_1_22}
\[
\res_{\U',L,Z_\bullet}^{\Int} \omega = (-1)^{\frac{d(d-1)}2}(2\pi\I)^d \res_{L,Z_\bullet} \omega.
\]
\end{cor}

\subsection{Gauss-Manin}
Suppose we have a morphism of algebraic varieties $X\ra S$. Let $\sigma$ be an abstract cell complex and $\U=(U_a)$ a hypercover on $X$ with respect to Zariski's topology. We assume that $S=\spec R$ is affine and all the open sets of the hypercover are affine. 

Associated to the de Rham complex on $X$ we have the complex of hypersections
\[
\Omega_X^0(\U)\ra\Omega_X^1(\U)\ra\cdots.
\]
Consider the complex of hypersections corresponding to the relative de Rham complex.
\[
\Omega_{X/S}^0(\U)\ra \Omega_{X/S}^1(\U)\ra\cdots.
\]
These are both complexes of $R$-modules.

\begin{prop}
The following natural sequence is exact for all $k\geq 0$:
\[
\Omega_S^1(S)\otimes_R\Omega_X^{k-1}(\U)\ra \Omega_X^k(\U)\ra \Omega_{X/S}^k(\U)\ra 0.
\]
\end{prop}
\begin{proof}
Since the corresponding sequence of sheaves is exact, it induces an exact sequence over any affine set.
\end{proof}

We define Gauss-Manin connection as follows. Let $\omega\in\Omega_{X/S}^k(\U)$ be closed. Lift it to $\ol\omega\in\Omega_{X}^k$. Then $d \ol\omega\in \kernel(\Omega_X^{k+1}(\U)\ra\Omega_{X/S}^{k+1}(\U))$. Choose $\ol\eta\in \Omega_S^1(S)\otimes_R\Omega_X^{k}(\U)$ which maps to $d \ol\omega$. Let $\eta$ be the projection of $\ol\eta$ in $\Omega_{X/S}^{k}(\U)\otimes_R\Omega_S^1(S)$. We say that $\eta$ is a Gauss-Manin derivative of $\omega$. Of course, this construction depends on several choices. Although $\eta$ is not well-defined, by abuse of notation we will write
\[
\eta=\nabla\omega
\]
if $\eta$ is a Gauss-Manin derivative of $\omega$.

Suppose we have a family of hyperchains $c_s\in C_k(\U) $, $s\in S$. This means that $c_s$ is a linear combination of simplices $c_s^i$ with each $c_s^i$ being a map from $\Delta\times S$ to $X$ which composed with $X\ra S$ gives the projection $\Delta\times S\ra S$. Here $\Delta$ denotes a standard simplex. We require the maps $c_s^i$ to be of the same type as we require for simplices, i.e. semi-algebraic. For $\omega\in\Omega_{X/S}^k(\U)$ consider the integral
\[
f(s)=\int_{c_s} \omega, \; s\in S.
\]
For a path $s_0 s_1$ in $S$ let $c_{s_0 s_1}\in C_{k+1}(\U)$ denote the corresponding hyperchain over this path. It provides a homotopy:
\[
\partial_h c_{s_0 s_1} + (\partial_h c)_{s_0 s_1}= c_{s_1}-c_{s_0}.
\]
\begin{lem}
Let $\omega\in\Omega_{X/S}^k(\U)$ be a closed relative hyperform and $c=(c_s)_{s\in S}$ be a family of hyperchains, $c_s\in C_k(\U)$.
If $\eta=\nabla\omega$ with $\ol\eta$ and $\ol\omega$ as in the definition of $\nabla$, then we have
\[
d \int_{c_s} \omega = \int_{c_s} \nabla\omega + R(\partial_h c, \ol\omega).
\]
Here $R$ is the bilinear operator which is constructed as follows. The value of $R$ on a vector in $S$ represented by a path $s_t$ is
\[
\langle [s_t], R(\partial_h c, \ol\omega) \rangle = \frac{\partial}{\partial t}\bigg\vert_{t=0} \int_{(\partial_h c)_{s_0 s_t}} \ol\omega.
\]
\end{lem}
\begin{proof}
For a fixed $s_0\in S$ let $s_t$ be a path in $S$ starting from $s_0$. We have
\[
f(s_t)-f(s_0)=\int_{\partial_h c_{s_0 s_t}+(\partial_h c)_{s_0 s_t}} \ol\omega.
\]
The first summand can be transformed as
\[
\int_{\partial_h c_{s_0 s_t}} \ol\omega = \int_{c_{s_0 s_t}} \ol\eta.
\]
Over the path we are considering $\Omega_S^1$ is generated by $dt$. Therefore $\ol\eta=dt\wedge \ol\eta_0$. We obtain
\[
\int_{c_{s_0 s_t}} \ol\eta=\int_{c_{s_0 s_t}} dt \wedge \ol\eta_0=
\int_{c_{s_0 s_t}} (d(t\ol\eta_0) - t d\ol\eta_0)=\int_{\partial_h c_{s_0 s_t}} t\ol\eta_0 - \int_{c_{s_0 s_t}} t d\ol\eta_0.
\]
Examining the second term we see
\[
{\frac{\partial}{\partial t}}\bigg\vert_{t=0} \int_{c_{s_0 s_t}} t d\ol\eta_0 =0.
\]
The first one transforms to
\[
\int_{\partial_h c_{s_0 s_t}} t\ol\eta_0 = \int_{c_{s_t}-c_{s_0}} t\ol\eta_0 -\int_{(\partial_h c)_{s_0 s_t}} t\ol\eta_0.
\]
The second term gives
\[
{\frac{\partial}{\partial t}}\bigg\vert_{t=0} \int_{(\partial_h c)_{s_0 s_t}} t\ol\eta_0=0.
\]
The first one gives
\[
{\frac{\partial}{\partial t}}\bigg\vert_{t=0} \int_{c_{s_t}-c_{s_0}} t\ol\eta_0 = \lim_{t\ra 0} \int_{c_{s_t}} \ol\eta_0=\int_{c_{s_0}} \ol\eta_0.
\]
\end{proof}

For example, if $c_s$ is closed:
\begin{cor}
If $c_s$ is a closed family of hyperchains and $\omega$ is a closed relative hyperform, then
\[
d \int_{c_s} \omega = \int_{c_s} \nabla\omega.
\]
\end{cor}
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