# Skew symmetry of indices in cocycles of Cech cohomology

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?

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.

• Thanks for this great answer! So this actually seems to be a false claim in Griffiths & Harris, p. 39. – Klaas May 14 '15 at 21:40

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.]

• You are right, I forgot the $-$ sign. But I am not sure how your method would work in the general case. For $p=2$ I would write $$0 = (\delta\sigma)_{0,1,2,3} = \sigma_{1,2,3} - \sigma_{0,2,3} + \sigma_{0,1,3} - \sigma_{0,1,2}$$ $$0 = (\delta\sigma)_{1,0,2,3} = \sigma_{0,2,3} - \sigma_{1,2,3} + \sigma_{1,0,3} - \sigma_{1,0,2}$$ But on adding these equations, only two pairs (not three) cancel. – Klaas May 6 '15 at 15:37
• Oh, sorry, I can't see it either spontaneously. My claim was misguided. – Thomas Poguntke May 6 '15 at 18:36
• hm... can anybody else give me a hint? – Klaas May 7 '15 at 10:43