Skew symmetry of indices in cocycles of Cech cohomology

Question

In Cech cohomology, the coboundary operator $$\delta:C^p(\underline U, \mathcal F)\to C^{p+1}(\underline U, \mathcal F) $$ is defined by the formula $$ (\delta \sigma)_{i_0,\dots, i_{p+1}} = \sum_{j=0}^{p+1}(-1)^j \sigma_{i_0,\dots, \hat {i_j},\dots i_{p+1}}{\Huge_|}_{U_{i_0}\cap\dots\cap U_{i_{p+1}}}. $$ In the book by Griffiths and Harris, the authors claim that $\delta\sigma=0$ implies the skew-symmetry condition $$ \sigma_{i_0,\dots,i_p} = -\sigma_{i_0,\dots, i_{q-1},i_{q+1},i_q,i_{q+2},\dots,i_p}. $$ How can I see this?

EDIT: Or is it possible that the skew symmetry is merely an assumption for all cochains?

Answer 1

There are two reasonable ways to define Cech cohomology, and they give the same answer in end. The exact answer to your question depends on which route you take.

Let $X$ be a topological space. Let $U_i$ be an open cover, with the index $i$ running through some set $I$. For $i_0$, $i_1$, ..., $i_p \in I$, write $U_{i_0 i_1 \cdots i_p}$ for $U_{i_0} \cap \cdots \cap U_{i_p}$. $\def\cE{\mathcal{E}}$Let $\cE$ be a sheaf of abelian groups on $X$, we will write the group operation additively.

Without skew symmetry $\def\cC{\mathcal{C}}$Define $\cC^p$ to be $\prod_{(i_0, \ldots, i_p) \in I^{p+1}} \cE(U_{i_0 \cdots i_p})$. Define a boundary map $\cC^p \to \cC^{p+1}$ by the usual formula. Then the Cech cohomology (definition 1) is the cohomology of the complex $\cC^{\bullet}$. We use the terms Cech cochain, Cech cocycle and Cech coboundary in the standard ways. If $c \in \cC^p$, I'll write $c(i_0, \ldots, i_p)$ for the contribution to $c$ from $\cE(U_{i_0 \cdots i_p})$

Note that, a priori, we do not impose any relation between $c(i_0, \ldots, i_p)$ and in $c(i_{\sigma(0)}, \ldots, i_{\sigma(p)})$. We also do not impose that $i_0$, $i_1$, ..., $i_p$ are distinct.

When $p=1$, this definition forces $p$-cocycles to be skew symmetric. Proof: Let $c$ be a $1$-cocycle. Then $(dc)(i,i,i) = c(i,i) - c(i,i) + c(i,i) = 0$ so $c(i,i)=0$. Also, $(dc)(i,j,i) = c(i,j) - c(i,i) + c(j,i) = 0$ so $c(i,j) = - c(j,i)$. However, this is not the case for general $p$. For example, if we only have one open set $U_i$, then any element of $\cC^2$ is a cocycle. More generally, if you choose a $(p-1)$-cochain $c$ at random (for $p \geq 2$), then $dc$ will probably not be skew symmetric.

With skew symmetry Define $\cC_a^p$ to be the subgroup of $\cC^p$ consisting of skew symmetric cochains. Then $\cC_a^{\bullet}$ is a subcomplex of $\cC^{\bullet}$; we write $\iota$ for this inclusion. We define Cech cohomology (definition 2) to be the cohomology of $\cC_a^{\bullet}$. So we obtain a map $\iota_{\ast}$ from $H^{\bullet}(\cC_a^{\bullet})$ (defn 2) to $H^{\bullet}(\cC^{\bullet})$ (defn 1).

Moreover, fix a total order on $I$. Define an inverse map $\alpha: \cC^p \to \cC^p_{a}$ by defining $\alpha(c)(i_0, \ldots, i_p)$ to be $0$ if $(i_0, \ldots, i_p)$ have a repeated element, and setting $\alpha(c)(i_0, \ldots, i_p) = (-1)^{\sigma} c(i_{\sigma(0)}, \ldots, i_{\sigma(p)})$ if $\sigma$ is a permutation such that $i_{\sigma(0)} < \cdots < i_{\sigma(p)}$. Then $\alpha \circ \iota = \mathrm{Id}$. So we get a map $\alpha_{\ast}$ from $H^{\bullet}(\cC^{\bullet})$ (defn 1) to $H^{\bullet}(\cC_a^{\bullet})$ (defn 2). Since $\alpha \circ \iota = \mathrm{Id}$, we have $\alpha_{\ast} \circ \iota_{\ast} = \mathrm{Id}$.

In fact, $\iota \circ \alpha$ is chain homotopic to the identity, so we also have $\iota_{\ast} \circ \alpha_{\ast} = \mathrm{Id}$. See Brian Conrad's notes for a proof.

So, if you use definition 2, then cochains are antisymmetric by definition. And, if you are using definition 1, then you know that any cocycle $c$ is cohomologous to its skew symmetrization $\alpha(c)$.

As a final remark, in more general settings, definition 1 is the right one to generalize, not definition 2. However, definition 2 is certainly much easier for computation.

Answer 2

Presumably, you mean $\sigma_{i_0,\ldots,i_p} = -\sigma_{i_0,\ldots,i_{q-1},i_{q+1},i_q,i_{q+2},\ldots,i_p}$. Let us consider the case $p=1$. Then we want to show that $\sigma_{i_0,i_1}=-\sigma_{i_1,i_0}$ for any $(i_0,i_1)$. But $\delta\sigma = 0$ means that for all $i$, we have $$\sigma_{i_1,i}|_{V_i}-\sigma_{i_0,i}|_{V_i}+\sigma_{i_0,i_1}|_{V_i} = 0,$$ where $V_i := U_{i_0} \cap U_{i_1} \cap U_i$. On the other hand, applied to $(i_1,i_0,i)$ instead, it says $$\sigma_{i_0,i}|_{V_i}-\sigma_{i_1,i}|_{V_i}+\sigma_{i_1,i_0}|_{V_i} = 0.$$ Adding these equations yields that $$\sigma_{i_0,i_1}|_{V_i} + \sigma_{i_1,i_0}|_{V_i} = 0.$$ But the $\{V_i\}$ are a covering of $U_{i_0} \cap U_{i_1}$, and hence indeed $\sigma_{i_0,i_1} = -\sigma_{i_1,i_0}$. The general case is the same proof, but with more indices hanging around on either side. [Edit: It doesn't actually appear to be that obvious.]