Let $X$ be a smooth, projective variety and $F$ a coherent sheaf on $X$. Let $\{U_i\}_{i \in I}$ be an open affine covering of $X$ and $\{f_{ij}\}_{i<j}$ with $f_{ij} \in \Gamma(U_{ij},F)$ satisfying the cocycle condition. Denote by $\alpha \in H^1(X,F)$ corresponding to the collection $\{f_{ij}\}_{i<j}$.

Choose $i_0 \in I$ and a proper affine open subset $V \subset U_{i_0}$. Consider the new open affine covering $\mathcal{V}$ consisting of $V$ and $U_i$ for $i \in I$. We know that $H^1(X,F)$ does not depend on the choice of the covering. What is the Cech cocyle corresponding to $\alpha$ under the new open covering $\mathcal{V}$?

Any hint/reference will be most welcome.

  • 1
    The general discussion is Tag 09UY. You should be able to work it out in this simple case. – R. van Dobben de Bruyn Oct 11 at 19:42
  • @R.vanDobbendeBruyn I tried that but the notation is confusing in the reference. In this case 2 indices in $I$ maps to the same element in $J$. Could you please write down a small sketch in the case cardinality of $I$ is $3$. – Jana Oct 11 at 19:51
up vote 2 down vote accepted

Let $U = \coprod_i U_i$ and $U'=V \amalg \coprod_i U_i$. Since there is $V \to U_{i_0}$, there is a map $U' \to U$ over $X$, and hence a map $U'\times_X U'\to U\times_X U$ (also over $X$). The original cocycle is a map $U\times_X U\to F$, and then the new cocycle is the induced map $U'\times_X U' \to U\times_X U \to F$.

To unpack that, it means that on $V\cap U_j$ the new cocycle is defined as the restriction of $f_{i_0j}$ along $V\cap U_j \hookrightarrow U_{i_0}\cap U_j$, and on $U_i\cap V$ it is defined as the restriction of $f_{ii_0}$ along $U_i\cap V \hookrightarrow U_i\cap U_{i_0}$

  • My confusion is that if $i_0=j$ then there is no $f_{i_0,j}$ or am I supposed to take such $f_{i_0,j}$ to be zero? – Jana Oct 12 at 7:15
  • You can define $f_{ji}=-f_{ij}$ and $f_{ii}=0$ and then ignore the ordering, and then do what I did in my answer. – David Roberts Oct 12 at 9:58
  • @Jana did that help? I can add more detail if needed. – David Roberts Oct 18 at 11:44
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    Thanks a lot. That helped. No need for further details. – Jana Oct 19 at 9:38

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