Given a presehaf of abelian groups $\mathcal{F}$ and an open cover $\mathcal{U}:=\{U_i\}_{i=0}^n$, I can define the Cech cohomology groups $\check{H}^q(\mathcal{U},\mathcal{F})$.

It s well known that elements of $\check{H}^0(\mathcal{U},\mathcal{F})$ (i.e. 0-cocycles) satisfy the compatibility condition

$$ z_i\mid_{U_i\cap U_j}=z_j\mid_{U_i\cap U_j}~\forall i,j $$

and thus can be interpreted as compatibile families.

Can we interpret the elements of the first cohomology group $\check{H}^1(\mathcal{U},\mathcal{F})$ in a similiar way? I'm trying to achieve a concrete definition of this group.

  • $\begingroup$ Isn't the $1$-th Cech cohomology the group of families satisfying the compatibility condition, while the $0$-th Cech cohomology is the group of global sections? $\endgroup$ – darij grinberg Jul 9 '15 at 12:29
  • $\begingroup$ $\check{H}^0(\mathcal{U},\mathcal{F})\cong Z^0(\mathcal{U}, \mathcal{F})$, thus elements of $\check{H}^0$ are $0$-cocycles. The general formula for the coboundary map $\delta^q$ is $\delta^q(c)_{(i_0,\dots, i_{q+1})}=\sum_{j=0}^{q+1}(-1)^jc_{(i_0,\dots,\hat{i}_j, \dots i_{q+1})}\mid_{U_{i_0\dots i_{q+1}}}$. For the case $q=0$ we have $\delta^0(z)_{ij}=z_j\mid_{U_i\cap U_j}-z_i\mid_{U_i\cap U_j}$. Thus $z\in Z^0$ iff it satisfy the compatibility condtion. Notice that each compatible family induces a unique golbal section. $\endgroup$ – John Jul 9 '15 at 16:27
  • $\begingroup$ I think you're right. It's been a while since I've read anything about Cech cohomology... $\endgroup$ – darij grinberg Jul 9 '15 at 18:18

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