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Let $X$ be a topological space with and open cover $\{U_i\}$ and let $\mathcal F$ be a sheaf of abelian groups on $X$. A $n$-cochain is a section $f_{i_0,\ldots,i_n}\in U_{i_0,\ldots,i_n}:= U_{i_0}\cap\ldots\cap U_{i_n}$; we can costruct the following abelian group (written in additive form):

$$ \check C^n(\mathcal U,\mathcal F):=\!\!\prod_{(i_o,\ldots,i_n)}\!\!\mathcal F(U_{i_0,\ldots,i_n}) $$

Now my question is the following:

we consider oredered sequences $(i_o,\ldots,i_n)$ ? Because in this case in the direct product we have each group repetead $(n+1)!$ times, that is the number of permutations of the set $\{i_o,\ldots,i_n\}$.

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I think the answer is yes, for when you define the differential, the order of the indices matters. – Olivier Bégassat Jun 13 '12 at 16:39
You can consider ordered or unordered sequences, with repeated indexes or not. The cochains complexes are different, but cohomologies are the same. See page 180 of Liu's Algebraic geometry. – Andrea Jun 13 '12 at 18:24
up vote 9 down vote accepted

Have a look at :
There are three complexes which are homotopic, and so, induce the Čech Cohomology :
1) Without ordering the open sets : the Čech complex of singular cochains : $C^n(\mathcal U,\mathcal F)=\displaystyle\prod_{i_0,\ldots,i_n}\mathcal F(U_{i_0,\ldots,i_n})$.
2) Considering alternating open sets : the Čech complex of alternating cochains : $C^n_{\mathrm{alt}}(\mathcal U,\mathcal F)$ where $\omega_{\varphi(i_0,\ldots,i_n)}=\varepsilon(\varphi)\omega_{i_0,\ldots,i_n}$ where $\varphi$ where $\varphi$ is in the symmetric group $\mathfrak S_n$ and where $\varepsilon(\varphi)$ is the sign of $\varphi$.
3) Taking care of the order : the Čech complex of ordered cochains : $C^n_<(\mathcal U,\mathcal F)=\displaystyle\prod_{i_0<\ldots<i_n}\mathcal F(U_{i_0,\ldots,i_n})$, for a total order $<$ on $I$ where $\{U_i\}_{i\in I}$.

As shown in this link, these complexes induce the same cohomology which is the usual Čech cohomology.

The advantage of the complex of alternating cochains over singular cochains is that we can easily use refinements for $\mathcal U$ (for the inductive limit) because we doesn't need the order.
The advantage of the complex of singular cochains over alternating cochains is that we can use non-injective refinements.

An other inconvenient of the ordored cochains is that we need a total order $<$.

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which version do you prefer? (The question is extended also to @Georges Elencwajg) The ordered version requires the axiom of choice! – Dubious Jun 13 '12 at 17:24
I prefer the singular cochains which are more adapted to calculate inductive limits. However, I may have not enough perspective to have a relevant opinion. By the way, hello M. Elencwajg, you were my professor last year, you should recognize me by my initials. – JBC Jun 13 '12 at 17:42
Yes I do, Mr. C...o ! And I'm very proud of your perfect answer: +1 (I hadn't seen it when I started writing mine: I would have abstained if I had). I'm amazed at the maturity of your mathematical style and also at your mastery of English. I'm looking forward to many fine posts from you on this site. Meanwhile, a very warm welcome! – Georges Elencwajg Jun 13 '12 at 18:20

You can choose: both versions exist!

a) You can take the product $\check C^n(\mathcal U,\mathcal F)=\!\!\prod_{(i_o,\ldots,i_n)}\!\!\mathcal F(U_{i_0,\ldots,i_n})$ over all $n+1$-tuples, so that indeed there will be much redundancy in your groups i.e. they will be repeated.

b) Or you can put some total order on $I$ and consider the complex $$ \check C'^n(\mathcal U,\mathcal F):=\!\!\prod_{i_o\lt\ldots\lt i_n}\!\!\mathcal F(U_{i_0,\ldots,i_n}) $$
which is clearly more economical.

b') There is a variant where you use the subcomplex $\check C_{alt}^\bullet(\mathcal U,\mathcal F)\subset \check C^n(\mathcal U,\mathcal F)$ of the complex in a) consisting of alternate families: for example if $n=1$ you require that $s_{ij}=-s_{ji}\in \mathcal F(U_i\cap U_j) \;\text {for } \;i\neq j$ and $s_{ii}=0.$

These complexes give the same cohomology groups: it is pretty clear for b) and b') and it requires a calculation to show that the inclusion of complexes $$\check C_{alt}^\bullet(\mathcal U,\mathcal F)\hookrightarrow \check C^\bullet(\mathcal U,\mathcal F)$$ yields isomorphisms at the level of cohomology groups $$ \check H_{alt }^n(\mathcal U,\mathcal F)\hookrightarrow \check H^n(\mathcal U,\mathcal F) $$

In that generality I must admit that all this is a bit boring.
For familiarizing oneself with effective calculations in low degrees, I recommend §12 of Forster's Lectures on Riemann Surfaces in which he very explicitly computes, for example, the first Čech cohomology group $\check H^1$ of some sheaves on Riemann surfaces.

Let me say, as an answer to Galoisfan's question, that version b) is much more powerful.
Here is an example:

Let $X$ be a Riemann surface and $\mathcal F$ a coherent sheaf.
If $\mathcal U=\lbrace U_0,U_1\rbrace$ is a covering of $X$ consisting of two open sets, then obviously $\check C'^n(\mathcal U,\mathcal F)=0$ for $n\geq 2$ since you cannot extract a sequence ot three strictly increasing numbers from $\lbrace 0,1\rbrace$ !
But if $U_0,U_1\subsetneq X$ are strict open subsets, they are Stein and Leray's theorem says that $ \check H^n(\mathcal U,\mathcal F) = \check H^n(X,\mathcal F) $, the genuine cohomology of $\mathcal F$ (i.e. the inductive limit over the coverings of $X$).
So version b) lets you prove that all genuine cohomology groups $\check H^n(X,\mathcal F) \; (n\geq 2)$ are zero: quite a remarkable theorem!

In the same vein, for every algebraic variety $X$ that can be covered by $n+1$ open affine subsets ($\mathbb P^n$ for example ) and every coherent algebraic sheaf $\mathcal F$ on $X$, the totally ordered version of Čech cohomology shows that $\check H^k(X,\mathcal F) =0$ for $k\gt n$.

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