The Stacks project

20.23 The alternating Čech complex

This section compares the Čech complex with the alternating Čech complex and some related complexes.

Let $X$ be a topological space. Let $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ be an open covering. For $p \geq 0$ set

\[ \check{\mathcal{C}}_{alt}^ p(\mathcal{U}, \mathcal{F}) = \left\{ \begin{matrix} s \in \check{\mathcal{C}}^ p(\mathcal{U}, \mathcal{F}) \text{ such that } s_{i_0 \ldots i_ p} = 0 \text{ if } i_ n = i_ m \text{ for some } n \not= m \\ \text{ and } s_{i_0\ldots i_ n \ldots i_ m \ldots i_ p} = -s_{i_0\ldots i_ m \ldots i_ n \ldots i_ p} \text{ in any case.} \end{matrix} \right\} \]

We omit the verification that the differential $d$ of Equation (20.9.0.1) maps $\check{\mathcal{C}}^ p_{alt}(\mathcal{U}, \mathcal{F})$ into $\check{\mathcal{C}}^{p + 1}_{alt}(\mathcal{U}, \mathcal{F})$.

Definition 20.23.1. Let $X$ be a topological space. Let $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ be an open covering. Let $\mathcal{F}$ be an abelian presheaf on $X$. The complex $\check{\mathcal{C}}_{alt}^\bullet (\mathcal{U}, \mathcal{F})$ is the alternating Čech complex associated to $\mathcal{F}$ and the open covering $\mathcal{U}$.

Hence there is a canonical morphism of complexes

\[ \check{\mathcal{C}}_{alt}^\bullet (\mathcal{U}, \mathcal{F}) \longrightarrow \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F}) \]

namely the inclusion of the alternating Čech complex into the usual Čech complex.

Suppose our covering $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ comes equipped with a total ordering $<$ on $I$. In this case, set

\[ \check{\mathcal{C}}_{ord}^ p(\mathcal{U}, \mathcal{F}) = \prod \nolimits _{(i_0, \ldots , i_ p) \in I^{p + 1}, i_0 < \ldots < i_ p} \mathcal{F}(U_{i_0\ldots i_ p}). \]

This is an abelian group. For $s \in \check{\mathcal{C}}_{ord}^ p(\mathcal{U}, \mathcal{F})$ we denote $s_{i_0\ldots i_ p}$ its value in $\mathcal{F}(U_{i_0\ldots i_ p})$. We define

\[ d : \check{\mathcal{C}}_{ord}^ p(\mathcal{U}, \mathcal{F}) \longrightarrow \check{\mathcal{C}}_{ord}^{p + 1}(\mathcal{U}, \mathcal{F}) \]

by the formula

\[ d(s)_{i_0\ldots i_{p + 1}} = \sum \nolimits _{j = 0}^{p + 1} (-1)^ j s_{i_0\ldots \hat i_ j \ldots i_{p + 1}}|_{U_{i_0\ldots i_{p + 1}}} \]

for any $i_0 < \ldots < i_{p + 1}$. Note that this formula is identical to Equation (20.9.0.1). It is straightforward to see that $d \circ d = 0$. In other words $\check{\mathcal{C}}_{ord}^\bullet (\mathcal{U}, \mathcal{F})$ is a complex.

Definition 20.23.2. Let $X$ be a topological space. Let $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ be an open covering. Assume given a total ordering on $I$. Let $\mathcal{F}$ be an abelian presheaf on $X$. The complex $\check{\mathcal{C}}_{ord}^\bullet (\mathcal{U}, \mathcal{F})$ is the ordered Čech complex associated to $\mathcal{F}$, the open covering $\mathcal{U}$ and the given total ordering on $I$.

This complex is sometimes called the alternating Čech complex. The reason is that there is an obvious comparison map between the ordered Čech complex and the alternating Čech complex. Namely, consider the map

\[ c : \check{\mathcal{C}}_{ord}^\bullet (\mathcal{U}, \mathcal{F}) \longrightarrow \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F}) \]

given by the rule

\[ c(s)_{i_0\ldots i_ p} = \left\{ \begin{matrix} 0 & \text{if} & i_ n = i_ m \text{ for some } n \not= m \\ \text{sgn}(\sigma ) s_{i_{\sigma (0)}\ldots i_{\sigma (p)}} & \text{if} & i_{\sigma (0)} < i_{\sigma (1)} < \ldots < i_{\sigma (p)} \end{matrix} \right. \]

Here $\sigma $ denotes a permutation of $\{ 0, \ldots , p\} $ and $\text{sgn}(\sigma )$ denotes its sign. The alternating and ordered Čech complexes are often identified in the literature via the map $c$. Namely we have the following easy lemma.

Lemma 20.23.3. Let $X$ be a topological space. Let $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ be an open covering. Assume $I$ comes equipped with a total ordering. The map $c$ is a morphism of complexes. In fact it induces an isomorphism

\[ c : \check{\mathcal{C}}_{ord}^\bullet (\mathcal{U}, \mathcal{F}) \to \check{\mathcal{C}}_{alt}^\bullet (\mathcal{U}, \mathcal{F}) \]

of complexes.

Proof. Omitted. $\square$

There is also a map

\[ \pi : \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F}) \longrightarrow \check{\mathcal{C}}_{ord}^\bullet (\mathcal{U}, \mathcal{F}) \]

which is described by the rule

\[ \pi (s)_{i_0\ldots i_ p} = s_{i_0\ldots i_ p} \]

whenever $i_0 < i_1 < \ldots < i_ p$.

Lemma 20.23.4. Let $X$ be a topological space. Let $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ be an open covering. Assume $I$ comes equipped with a total ordering. The map $\pi : \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F}) \to \check{\mathcal{C}}_{ord}^\bullet (\mathcal{U}, \mathcal{F})$ is a morphism of complexes. It induces an isomorphism

\[ \pi : \check{\mathcal{C}}_{alt}^\bullet (\mathcal{U}, \mathcal{F}) \to \check{\mathcal{C}}_{ord}^\bullet (\mathcal{U}, \mathcal{F}) \]

of complexes which is a left inverse to the morphism $c$.

Proof. Omitted. $\square$

Remark 20.23.5. This means that if we have two total orderings $<_1$ and $<_2$ on the index set $I$, then we get an isomorphism of complexes $\tau = \pi _2 \circ c_1 : \check{\mathcal{C}}_{ord\text{-}1}(\mathcal{U}, \mathcal{F}) \to \check{\mathcal{C}}_{ord\text{-}2}(\mathcal{U}, \mathcal{F})$. It is clear that

\[ \tau (s)_{i_0 \ldots i_ p} = \text{sign}(\sigma ) s_{i_{\sigma (0)} \ldots i_{\sigma (p)}} \]

where $i_0 <_1 i_1 <_1 \ldots <_1 i_ p$ and $i_{\sigma (0)} <_2 i_{\sigma (1)} <_2 \ldots <_2 i_{\sigma (p)}$. This is the sense in which the ordered Čech complex is independent of the chosen total ordering.

Lemma 20.23.6. Let $X$ be a topological space. Let $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ be an open covering. Assume $I$ comes equipped with a total ordering. The map $c \circ \pi $ is homotopic to the identity on $\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F})$. In particular the inclusion map $\check{\mathcal{C}}_{alt}^\bullet (\mathcal{U}, \mathcal{F}) \to \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F})$ is a homotopy equivalence.

Proof. For any multi-index $(i_0, \ldots , i_ p) \in I^{p + 1}$ there exists a unique permutation $\sigma : \{ 0, \ldots , p\} \to \{ 0, \ldots , p\} $ such that

\[ i_{\sigma (0)} \leq i_{\sigma (1)} \leq \ldots \leq i_{\sigma (p)} \quad \text{and} \quad \sigma (j) < \sigma (j + 1) \quad \text{if} \quad i_{\sigma (j)} = i_{\sigma (j + 1)}. \]

We denote this permutation $\sigma = \sigma ^{i_0 \ldots i_ p}$.

For any permutation $\sigma : \{ 0, \ldots , p\} \to \{ 0, \ldots , p\} $ and any $a$, $0 \leq a \leq p$ we denote $\sigma _ a$ the unique permutation of $\{ 0, \ldots , p\} $ such that $\sigma _ a(j) = \sigma (j)$ for $0 \leq j < a$ and such that $\sigma _ a(a) < \sigma _ a(a + 1) < \ldots < \sigma _ a(p)$. So if $p = 3$ and $\sigma $, $\tau $ are given by

\[ \begin{matrix} \text{id} & 0 & 1 & 2 & 3 \\ \sigma & 3 & 2 & 1 & 0 \end{matrix} \quad \text{and} \quad \begin{matrix} \text{id} & 0 & 1 & 2 & 3 \\ \tau & 3 & 0 & 2 & 1 \end{matrix} \]

then we have

\[ \begin{matrix} \text{id} & 0 & 1 & 2 & 3 \\ \sigma _0 & 0 & 1 & 2 & 3 \\ \sigma _1 & 3 & 0 & 1 & 2 \\ \sigma _2 & 3 & 2 & 0 & 1 \\ \sigma _3 & 3 & 2 & 1 & 0 \\ \end{matrix} \quad \text{and} \quad \begin{matrix} \text{id} & 0 & 1 & 2 & 3 \\ \tau _0 & 0 & 1 & 2 & 3 \\ \tau _1 & 3 & 0 & 1 & 2 \\ \tau _2 & 3 & 0 & 1 & 2 \\ \tau _3 & 3 & 0 & 2 & 1 \\ \end{matrix} \]

It is clear that always $\sigma _0 = \text{id}$ and $\sigma _ p = \sigma $.

Having introduced this notation we define for $s \in \check{\mathcal{C}}^{p + 1}(\mathcal{U}, \mathcal{F})$ the element $h(s) \in \check{\mathcal{C}}^ p(\mathcal{U}, \mathcal{F})$ to be the element with components

20.23.6.1
\begin{equation} \label{cohomology-equation-first-homotopy} h(s)_{i_0\ldots i_ p} = \sum \nolimits _{0 \leq a \leq p} (-1)^ a \text{sign}(\sigma _ a) s_{i_{\sigma (0)} \ldots i_{\sigma (a)} i_{\sigma _ a(a)} \ldots i_{\sigma _ a(p)}} \end{equation}

where $\sigma = \sigma ^{i_0 \ldots i_ p}$. The index $i_{\sigma (a)}$ occurs twice in $i_{\sigma (0)} \ldots i_{\sigma (a)} i_{\sigma _ a(a)} \ldots i_{\sigma _ a(p)}$ once in the first group of $a + 1$ indices and once in the second group of $p - a + 1$ indices since $\sigma _ a(j) = \sigma (a)$ for some $j \geq a$ by definition of $\sigma _ a$. Hence the sum makes sense since each of the elements $s_{i_{\sigma (0)} \ldots i_{\sigma (a)} i_{\sigma _ a(a)} \ldots i_{\sigma _ a(p)}}$ is defined over the open $U_{i_0 \ldots i_ p}$. Note also that for $a = 0$ we get $s_{i_0 \ldots i_ p}$ and for $a = p$ we get $(-1)^ p \text{sign}(\sigma ) s_{i_{\sigma (0)} \ldots i_{\sigma (p)}}$.

We claim that

\[ (dh + hd)(s)_{i_0 \ldots i_ p} = s_{i_0 \ldots i_ p} - \text{sign}(\sigma ) s_{i_{\sigma (0)} \ldots i_{\sigma (p)}} \]

where $\sigma = \sigma ^{i_0 \ldots i_ p}$. We omit the verification of this claim. (There is a PARI/gp script called first-homotopy.gp in the stacks-project subdirectory scripts which can be used to check finitely many instances of this claim. We wrote this script to make sure the signs are correct.) Write

\[ \kappa : \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F}) \longrightarrow \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F}) \]

for the operator given by the rule

\[ \kappa (s)_{i_0 \ldots i_ p} = \text{sign}(\sigma ^{i_0 \ldots i_ p}) s_{i_{\sigma (0)} \ldots i_{\sigma (p)}}. \]

The claim above implies that $\kappa $ is a morphism of complexes and that $\kappa $ is homotopic to the identity map of the Čech complex. This does not immediately imply the lemma since the image of the operator $\kappa $ is not the alternating subcomplex. Namely, the image of $\kappa $ is the “semi-alternating” complex $\check{\mathcal{C}}_{semi\text{-}alt}^ p(\mathcal{U}, \mathcal{F})$ where $s$ is a $p$-cochain of this complex if and only if

\[ s_{i_0 \ldots i_ p} = \text{sign}(\sigma ) s_{i_{\sigma (0)} \ldots i_{\sigma (p)}} \]

for any $(i_0, \ldots , i_ p) \in I^{p + 1}$ with $\sigma = \sigma ^{i_0 \ldots i_ p}$. We introduce yet another variant Čech complex, namely the semi-ordered Čech complex defined by

\[ \check{\mathcal{C}}_{semi\text{-}ord}^ p(\mathcal{U}, \mathcal{F}) = \prod \nolimits _{i_0 \leq i_1 \leq \ldots \leq i_ p} \mathcal{F}(U_{i_0 \ldots i_ p}) \]

It is easy to see that Equation (20.9.0.1) also defines a differential and hence that we get a complex. It is also clear (analogous to Lemma 20.23.4) that the projection map

\[ \check{\mathcal{C}}_{semi\text{-}alt}^\bullet (\mathcal{U}, \mathcal{F}) \longrightarrow \check{\mathcal{C}}_{semi\text{-}ord}^\bullet (\mathcal{U}, \mathcal{F}) \]

is an isomorphism of complexes.

Hence the Lemma follows if we can show that the obvious inclusion map

\[ \check{\mathcal{C}}_{ord}^ p(\mathcal{U}, \mathcal{F}) \longrightarrow \check{\mathcal{C}}_{semi\text{-}ord}^ p(\mathcal{U}, \mathcal{F}) \]

is a homotopy equivalence. To see this we use the homotopy

20.23.6.2
\begin{equation} \label{cohomology-equation-second-homotopy} h(s)_{i_0 \ldots i_ p} = \left\{ \begin{matrix} 0 & \text{if} & i_0 < i_1 < \ldots < i_ p \\ (-1)^ a s_{i_0 \ldots i_{a - 1} i_ a i_ a i_{a + 1} \ldots i_ p} & \text{if} & i_0 < i_1 < \ldots < i_{a - 1} < i_ a = i_{a + 1} \end{matrix} \right. \end{equation}

We claim that

\[ (dh + hd)(s)_{i_0 \ldots i_ p} = \left\{ \begin{matrix} 0 & \text{if} & i_0 < i_1 < \ldots < i_ p \\ s_{i_0 \ldots i_ p} & \text{else} & \end{matrix} \right. \]

We omit the verification. (There is a PARI/gp script called second-homotopy.gp in the stacks-project subdirectory scripts which can be used to check finitely many instances of this claim. We wrote this script to make sure the signs are correct.) The claim clearly shows that the composition

\[ \check{\mathcal{C}}_{semi\text{-}ord}^\bullet (\mathcal{U}, \mathcal{F}) \longrightarrow \check{\mathcal{C}}_{ord}^\bullet (\mathcal{U}, \mathcal{F}) \longrightarrow \check{\mathcal{C}}_{semi\text{-}ord}^\bullet (\mathcal{U}, \mathcal{F}) \]

of the projection with the natural inclusion is homotopic to the identity map as desired. $\square$

Lemma 20.23.7. Let $X$ be a topological space. Let $\mathcal{F}$ be an abelian presheaf on $X$. Let $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ be an open covering. If $U_ i = U$ for some $i \in I$, then the extended alternating Čech complex

\[ \mathcal{F}(U) \to \check{\mathcal{C}}_{alt}^\bullet (\mathcal{U}, \mathcal{F}) \]

obtained by putting $\mathcal{F}(U)$ in degree $-1$ with differential given by the canonical map of $\mathcal{F}(U)$ into $\check{\mathcal{C}}^0(\mathcal{U}, \mathcal{F})$ is homotopy equivalent to $0$. Similarly, for any total ordering on $I$ the extended ordered Čech complex

\[ \mathcal{F}(U) \to \check{\mathcal{C}}_{ord}^\bullet (\mathcal{U}, \mathcal{F}) \]

is homotopy equivalent to $0$.

Second proof. Since the alternating and ordered Čech complexes are isomorphic it suffices to prove this for the ordered one. We will use standard notation: a cochain $s$ of degree $p$ in the extended ordered Čech complex has the form $s = (s_{i_0 \ldots i_ p})$ where $s_{i_0 \ldots i_ p}$ is in $\mathcal{F}(U_{i_0 \ldots i_ p})$ and $i_0 < \ldots < i_ p$. With this notation we have

\[ d(x)_{i_0 \ldots i_{p + 1}} = \sum \nolimits _ j (-1)^ j x_{i_0 \ldots \hat i_ j \ldots i_ p} \]

Fix an index $i \in I$ with $U = U_ i$. As homotopy we use the maps

\[ h : \text{cochains of degree }p + 1 \to \text{cochains of degree }p \]

given by the rule

\[ h(s)_{i_0 \ldots i_ p} = 0 \text{ if } i \in \{ i_0, \ldots , i_ p\} \text{ and } h(s)_{i_0 \ldots i_ p} = (-1)^ j s_{i_0 \ldots i_ j i i_{j + 1} \ldots i_ p} \text{ if not} \]

Here $j$ is the unique index such that $i_ j < i < i_{j + 1}$ in the second case; also, since $U = U_ i$ we have the equality

\[ \mathcal{F}(U_{i_0 \ldots i_ p}) = \mathcal{F}(U_{i_0 \ldots i_ j i i_{j + 1} \ldots i_ p}) \]

which we can use to make sense of thinking of $(-1)^ j s_{i_0 \ldots i_ j i i_{j + 1} \ldots i_ p}$ as an element of $\mathcal{F}(U_{i_0 \ldots i_ p})$. We will show by a computation that $d h + h d$ equals the negative of the identity map which finishes the proof. To do this fix $s$ a cochain of degree $p$ and let $i_0 < \ldots < i_ p$ be elements of $I$.

Case I: $i \in \{ i_0, \ldots , i_ p\} $. Say $i = i_ t$. Then we have $h(d(s))_{i_0 \ldots i_ p} = 0$. On the other hand we have

\[ d(h(s))_{i_0 \ldots i_ p} = \sum (-1)^ j h(s)_{i_0 \ldots \hat i_ j \ldots i_ p} = (-1)^ t h(s)_{i_0 \ldots \hat i \ldots i_ p} = (-1)^ t (-1)^{t - 1} s_{i_0 \ldots i_ p} \]

Thus $(dh + hd)(s)_{i_0 \ldots i_ p} = -s_{i_0 \ldots i_ p}$ as desired.

Case II: $i \not\in \{ i_0, \ldots , i_ p\} $. Let $j$ be such that $i_ j < i < i_{j + 1}$. Then we see that

\begin{align*} h(d(s))_{i_0 \ldots i_ p} & = (-1)^ j d(s)_{i_0 \ldots i_ j i i_{j + 1} \ldots i_ p} \\ & = \sum \nolimits _{j' \leq j} (-1)^{j + j'} s_{i_0 \ldots \hat i_{j'} \ldots i_ j i i_{j + 1} \ldots i_ p} - s_{i_0 \ldots i_ p} \\ & + \sum \nolimits _{j' > j} (-1)^{j + j' + 1} s_{i_0 \ldots i_ j i i_{j + 1} \ldots \hat i_{j'} \ldots i_ p} \end{align*}

On the other hand we have

\begin{align*} d(h(s))_{i_0 \ldots i_ p} & = \sum \nolimits _{j'} (-1)^{j'} h(s)_{i_0 \ldots \hat i_{j'} \ldots i_ p} \\ & = \sum \nolimits _{j' \leq j} (-1)^{j' + j - 1} s_{i_0 \ldots \hat i_{j'} \ldots i_ j i i_{j + 1} \ldots i_ p} \\ & + \sum \nolimits _{j' > j} (-1)^{j' + j} s_{i_0 \ldots i_ j i i_{j + 1} \ldots \hat i_{j'} \ldots i_ p} \end{align*}

Adding these up we obtain $(dh + hd)(s)_{i_0 \ldots i_ p} = - s_{i_0 \ldots i_ p}$ as desired. $\square$

Comments (2)

Comment #3950 by Nicolas Müller on

I think in the definition of d, below definition 01FH the index of on the RHS should end with , not .

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