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Stream: theory: algebraic topology

Topic: relaxed good covers in Cech cohomology

Naso (Jul 26 2022 at 10:30):

On https://en.wikipedia.org/wiki/Good_cover_(algebraic_topology) it says

"...for the purposes of computing the Čech cohomology it suffices to have a more relaxed definition of a good cover in which all intersections of finitely many open sets have contractible connected components."

I tried to compute the cohomology of the circle using two open arcs that intersect in a disjoint union of two contractible sets, but I got 0 for $H^1$. Did I make a mistake or does the definition of Cech cohomology for a cover need to be modified for this relaxed definition of good cover?

Matteo Capucci (he/him) (Jul 26 2022 at 12:10):

Uhm that definition looks subtly wrong. I think they meant to write 'contractible and connected'

Naso (Jul 26 2022 at 12:16):

isn't that just a usual 'good open cover' then?

Naso (Jul 26 2022 at 12:18):

I thought they may be referring to a generalisation of the Cech cohomology like the groupoid van Kampen theorem where you have many basepoints https://en.wikipedia.org/wiki/Seifert–Van_Kampen_theorem#Van_Kampen's_theorem_for_fundamental_groupoids

John Baez (Jul 26 2022 at 18:29):

Reading Wikipedia it looks like they really wanted the finite intersections of open sets to have "contractible connected components" in their "relaxed" definition of good cover. Unfortunately, they don't say how this works. I don't think it will be a big problem except for computing $H^n$ for low $n$, like $n = 0$ and $n = 1$, which are probably the cases @naso was looking at.

Their original "non-relaxed" definition was the usual thing: the finite intersections of open sets are "contractible (and therefore connected)".

Naso (Jul 27 2022 at 02:23):

Thanks John. Why do you think there's less of a problem for $n > 1$ ?

John Baez (Jul 27 2022 at 02:36):

It's just a gut feeling, but you'll notice that the only difference homologically between a single contractible set and a disjoint union of several contractible sets is in the $H^0$, not the $H^n$ for $n > 0$. I was expecting this would only propagate up to affect $H^1$ of the answer.

I'm having trouble quickly proving this, so my hunch could be overoptimistic. But as a slight pointer in this direction, note that Mayer-Vietoris says that if $X = U \cup V$, if you change the $H^0$ of $U$ or $V$ while keeping $U \cap V$ contractible it can't change the $H^n$ of $X$ for $n > 1$.

John Baez (Jul 27 2022 at 02:39):

Anyway, if I were you I would avoid these "relaxed good covers" until you figure this stuff out! They're probably not worth the trouble.

Peter Arndt (Jul 27 2022 at 07:00):

For a constant sheaf with value $\mathbb{Z}$ I do actually get the right $H^1$ and $H^0$. The complex is $0 \to \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z} \times \mathbb{Z} \to 0$ where the middle map is $(a,b) \mapsto (a-b,b-a)$. The kernel of this map is the diagonal, hence $\cong \mathbb{Z}$ and the image is the subgroup generated by $(1,-1)$. Therefore the quotient is $H^1\cong \mathbb{Z}$.

Naso (Jul 27 2022 at 12:28):

Is the second $\mathbb{Z} \times \mathbb{Z}$ because $U \cap V$ has two connected components? So the complex is $0 \to F(U) \times F(V) \to F(U \cap V) \times F(U \cap V) \to 0$ ? I know there are different 'conventions' for the Cech complex, whether you allow arbitrary indices or only strictly increasing ... I was using the strictly increasing one so I just had a single $F (U \cap V) = F(U_{12})$.

Edit: oh, I think I see the problem.. I was looking at the constant PREsheaf and not the constant sheaf

Naso (Jul 27 2022 at 12:31):

Thanks for the nice intuition and advice, John

Peter Arndt (Jul 28 2022 at 11:38):

Exactly, the constant presheaf would have just on $\mathbb{Z}$ at level 1, but the constant sheaf has two copies, because of the two connected components.