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I am reading Serre's Faisceaux Algébriques Cohérents (Henceforth FAC) and he uses some terminology I have not seen. I have searched around a bit but can't get a clean and clear definition.

Question: For $\mathfrak{U}$ an open cover of a topological space, what is the definition of $\mathrm{dim}(\mathfrak{U})$?

The proof in which the terminology appears does seem to provide some clues:

Clue / Context: Here is the proposition and proof where the notion is used.

Corollary. $H^{q}(\mathfrak{U},\mathscr{F}) = 0$ for $q > \mathrm{dim}(\mathfrak{U})$.

By the definition of $\mathrm{dim}(\mathfrak{U})$ we have $U_{i_0} \cap \cdots \cap U_{i_q} = \emptyset$ for $q > \mathrm{dim}(\mathfrak{U})$, if the indices $i_0,\dots,i_q$ are distinct; hence $C^{'q}(\mathfrak{U},\mathscr{F})=0$, which shows that $$H^{q}(\mathfrak{U},\mathscr{F}) = H^{'q}(\mathfrak{U},\mathscr{F}) = 0.$$

It appears to be the largest number of overlapping sets?

The above selection is from this English translation, and can be found on page 25.

Many of my tags are related to the context of the document my question comes from, not the question itself. Feel free to edit this if you find it inappropriate.

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    $\begingroup$ That makes sense. A covering has a nerve. This is a simplicial complex whose vertices correspond members of a covering, and one gets a simplex when the intersection of the open sets corresponding to its vertices are non-empty. It makes sense to define the dimension of the covering as the dimension (as a simplicial complex) of its nerve. $\endgroup$ – Lord Shark the Unknown Jan 8 at 20:30
  • $\begingroup$ Thanks Lord Shark. My simplicial homology is not strong, and I actually don't see entirely why it makes sense. If you care to elaborate in an answer below I would gladly up-vote. When you say 'this is a simplicial complex' are you referring to the complex formed by the $K_{p}(I)$'s or the corresponding $C^{q}(\mathfrak{U},\mathscr{F})'s?$ $\endgroup$ – Prince M Jan 8 at 21:25
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You are right, Serre uses the phrase "By the definition of $\mathrm{dim}(\mathfrak{U})$" but actually nowhere defines $\mathrm{dim}(\mathfrak{U})$ in FAC.

So we should take the explanation after the above phrase as the definition. That is, $\mathrm{dim}(\mathfrak{U})$ is the unique number in $\mathbb{N} \cup \{ \infty \}$ with the property that for all $q \in \mathbb{N}$ we have $q > \mathrm{dim}(\mathfrak{U})$ if and only if $U_{i_0} \cap \cdots \cap U_{i_q} = \emptyset$ for any choice of distinct indices $i_0,\dots,i_q$. Equivalently, we have $q \le \mathrm{dim}(\mathfrak{U})$ if and only if there exist distinct indices $i_0,\dots,i_q$ such that $U_{i_0} \cap \cdots \cap U_{i_q} \ne \emptyset$.

This shows that $$\mathrm{dim}(\mathfrak{U}) = \\ \sup \{q \in \mathbb{N} \mid \text{There exist distinct indices } i_0,\dots,i_q \text{ suich that } U_{i_0} \cap \cdots \cap U_{i_q} \ne \emptyset\} .$$

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  • $\begingroup$ So there could be weird cases where we have a topological space with $n$ disconnected components, and each component is open. Then taking each component as an open cover, the dim in this case would be 0? $\endgroup$ – Prince M Jan 9 at 22:56
  • $\begingroup$ It also seems like weird things could happen with open covers being concentrated over one area, like many overlapping sets in a small area and then little to no overlaps elsewhere. Are my thoughts trivial, does this just mean you chose a bad open cover? Irrelevant in the algebro-geometric case? $\endgroup$ – Prince M Jan 9 at 22:58
  • $\begingroup$ The dimension of a single cover may be rather erratic as your examples show. In fact, a single cover does not contain much Information about the space $X$. It is the collection of all open covers of $X$ which provides sufficient information. See Section 22 of FAC. $\endgroup$ – Paul Frost Jan 10 at 0:45
  • $\begingroup$ Yes, I was aware this theory intends to take the direct limit I just want to sort out what can and cannot happen for one stationary open cover in the system. Thank you! You and Lord Shark have been very helpful. $\endgroup$ – Prince M Jan 10 at 16:41
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    $\begingroup$ I add another answer because comments do not have enough space. $\endgroup$ – Paul Frost Jan 10 at 18:01
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You have to be cautious. Serre takes the limit of a direct system of abelian groups indexed by open covers $\mathfrak{U}$. There is no corresponding concept of a direct system of open covers.

Serre introduces a pre-ordering on the set $\mathcal{C}$ of open covers by defining $\mathfrak{U} \le \mathfrak{V}$ if $\mathfrak{U}$ is finer than $\mathfrak{V}$, i.e. if each $U \in \mathfrak{U}$ is contained in some $V \in \mathfrak{V}$. For example, if $\mathfrak{U} \subset \mathfrak{V}$, then $\mathfrak{U} \le \mathfrak{V}$. But there are no reasonable "maps" between open covers, so we cannot form something like a limit of the system of open covers.

Intituively, a cover $\mathfrak{V}$ is refined by adding to $\mathfrak{V}$ "small" open sets and by throwing away "big" open sets from $\mathfrak{V}$.

Moreover, the topology $\mathfrak{T}$ of $X$ belongs to $\mathcal{C}$, however it is not the finest open cover, but the coarsest. In fact, each $\mathfrak{U} \subset \mathfrak{T}$, hence $\mathfrak{U}$ is finer than $\mathfrak{T}$, and the covers $\mathfrak{U}$ such that $\mathfrak{T}$ is finer than $\mathfrak{U}$ are precisely those with $X \in \mathfrak{U}$. So we do not approach $\mathfrak{T}$ by taking finer and finer covers.

As Lord Shark the Unknown explained, one can associate to each open cover $\mathfrak{U}$ a simplicial complex $K(\mathfrak{U})$. The collection of all these complexes forms a system whose "maps" are contiguity classes of certain ordinary simplicial maps $K(\mathfrak{U}) \to K(\mathfrak{V})$ which arise for $\mathfrak{U} \le \mathfrak{V}$. This is used to define the Cech homology groups and Cech cohomology groups of the space $X$.

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  • $\begingroup$ Good answer Paul. $\endgroup$ – Prince M Jan 16 at 18:54
  • $\begingroup$ edit: answer(s)* $\endgroup$ – Prince M Jan 16 at 18:55

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