Okay here's another one: when computing Cech cohomology, we're supposed to take the (co?)limit over all covers. Is there some way of guaranteeing that one cover is good enough to get all the information?
If the cover is, say, faithfully flat, can I just computer Cech cohomology for THAT cover and be done with it?
What are the methods people use to compute this business?
Oh, I suppose I need to find an acyclic resolution, is the point.
I.e. a cover such that the sheaf is trivial on intersections.
@Jon You have basically figured it out on your own but 1) Yes you always want the first of the two conditions you've written. Sometimes $G$ is the pullback to B of something global, in which case you get the second condition by transitivity of the pullback
2) You need that your cover is acyclic for your sheaf, i.e. that the sheaf has no higher cohomology on the finite intersections (this is just to ensure the degeneracy of the Mayer-Vietoris spectral sequence)
(the most common application in algebraic geometry is when you have a quasi-coherent sheaf on a separated scheme. Then any cover by affines does the trick)
But..... hmmmm, if I have a good enough cover.... isn't it enough just to know that it converges for that cover?
I mean, if I know that the sheaf descends along the cover of interest?
Sorry, this is getting a little far afield, the idea is that for some fixed ring spectrum I'd like to compute Map(Spec(R),BG) where by G I really mean Spec(H) for H some spectral bialgebra.
Now, I of course don't have any good notion of Spec
Yeah, I really only have a theoretical notion of it as the thing classifying principal G-bundles.
Ah yes. Okay. I think I see what you mean.
Hm...
I think I'll need to use some kind of intuition to guess what the right space of morphisms should be, and then prove a categorical equivalence between that and some $\infty$-category of "torsors"
My favourite definition of torsor is a sheaf in whatever topology your considering now togeter with an action of $G$ (the sheaf corepresented by H) that is simply transitive
This should work everytime and be computed exactly by the Cech complex you described above
Hm, reading Milne's notes he claims that a connected diagonalizable algebraic group is always isomorphic to a finite product of copies of G_m : this can't be true, right? If he said smooth algebraic diagonalizable algebraic groups then I agree, but not otherwise. What about say the pth roots of unity?
@Adeel: what are you looking for in a reference about vector bundles? Is it some correspondence between vector bundles on X and some locally trivial geometric objects over X? (because I've seen people take Spec Sym (perfect complex) plenty of times over derived schemes... I think.
@user101036 Is this a problem with the characteristic of the ground field? Aren't all groups smooth? (take any point in the smooth locus and use left translations to prove that all other points are smooth)
@bananastack so I think I've basically worked out the right definition of Spec Sym of a perfect complex, but still it would be nice if someone had written it down already so i could just reference it
@bananastack It fails in characteristic p since not all group schemes are reduced. It actually fails already for perfect fields of characteristic p. Take for example k[x,x^{-1}]/((x-1)^p) (pth roots of unity and comultiplication taking x --> x \otimes x) .
@user101036 @DenisNardin yes, I didn't remember what the assumptions were to have your group be automatically reduced. Is Milne working in char 0? Or maybe he just forgot an assumption.
there's a certain quote I like regarding making mistakes which cheers me up sometimes. Maybe my sense of what is encouraging is a bit off, but nonetheless
Basic question: so, the steenrod algebra is of course non-commutative so we can't study it as an algebraic group directly. However, it is a cocommutative hopf algebra, so dualizing we should get something . Is it profiteable to view the Steenrod algebra under this dualization?
so from the chromatic viewpoint this is what we would want to be true. If we take some complex-oriented cohomology theory and consider the algebra of cohomology operations, is this always automorphisms of the underlying formal group?
the first is that there can be extra stuff: e.g. [HZ, HZ] has lots of of stuff in it that comes in as "derived" stuff rather than coming from the formal group
if your cohomology theory is Landweber exact, this problem goes away
(in some other cases it is not too bad, e.g. the Morava K-theories K(n) have some stuff coming from the formal group and a little exterior contribution)
however, the second reason is that cohomology operations aren't really "multiplicative" enough in general.
e.g. with K-theory, K_* K is the coordinate ring of Aut(G_m)
but you can take those elements and add them, for example.
the interpretation that I've heard, and I think I first saw in Ando-Hopkins-Rezk, is that the algebra of cohomology operations is usually closer to something like a ring of measures on automorphisms of the formal group
where "measure" is just something to make you more comfortable with the dualization (a measure on X is something that takes a function on X and returns a number, so it's basically something in the dual)
the multiplication on cohomology operations then corresponds to the convolution product for measures
you could also instead try to hope that it was the "group algebra" of automorphisms of the formal group. (for finite discrete groups, these two processes agree with each other)
that doesn't usually work as well when your groups are not discrete, and so this perspective could be viewed as a slight correction on trying to take the group algebra
@Espen: the answer is no at the infinity-categorical level. suppose f : spaces -> spaces is an (infinity-)functor that preserves homotopy colimits. then f is completely determined by what it does to a point: in fact it must have the form X x (-) for some uniquely determined space X (its value at a point), and in fact this is an equivalence of infinity-categories. this is a version of the eilenberg-watts theorem for spaces
now it's not hard to see that the only invertible such functor is the identity
since the only space which is invertible under product is the point
so in fact the infinity-group of automorphisms of Spaces not only has trivial pi_0 but is contractible
@QiaochuYuan Thanks, that's interesting. Can we also say this on the 1-categorical level if the equivalence comes from a zig-zag of Quillen equivalences?
anyway, if you know the autoequivalence comes from Quillen equivalences of some decent model categories then you have an autoequivalence at the level of (∞, 1)-categories so it is equivalent to the identity
but I don't see any reason to believe that every autoequivalence of the (evil) homotopy category comes from such
in Haus there is a very easy way of recognising closed subsets
I don't know how to do that in Top
hmmm... actually, maybe I do. We can recognise the Sierpiński space as the unique two-pointed space that is neither discrete nor indiscrete (these concepts being invariant under autoequivalence), so it's just a matter of distinguishing the closed point from the open point
but the closed point has this distinguishing property: the class of all pullbacks of the inclusion of the closed point is closed under arbitrary intersection
okay, let me see if I follow you. it suffices, more or less, to show that an autoequivalence preserves the point and the sierpinski space. i think it preserves the point because the point is the unique tiny object (object X such that Hom(X, -) preserves all colimits)
hence we can talk about the sierpinski space, and now the last step is to distinguish the closed and the open point
this is precisely the step that fails in the finite case, so this is presumably the hardest step
ah, but now you can use the fact that closed sets are closed under arbitrary intersection but open sets aren't, great
and we can talk about those because we can talk about homs into the sierpinski space and then talk about pullbacks along the inclusion of either the closed or the open point
