@JonBeardsley well, thanks for the honesty! :). But I thought I've seen you discussing descent for much more complicated gadgets around these forums. Maybe I'm misremembering.
Well sure, I do discuss descent abstractly all the time. You're not misremembering.
I still haven't the slightest idea how to answer your question.
I have no idea how to think about P1, or even really what the "standard open cover" of P1 is, haha. I in fact know very very little.
That's like... actual "algebraic geometry."
Also I'm miserably embroiled in trying to work something out for my thesis that I haven't been able to get anywhere with for at least a month, so I'm very cranky and don't have a lot of brain power to spare.
well, hmm, should we take the sphere S^2, an open cover made up of two open discs and take sheaves of Q-vector spaces, and consider a map Q --> Q[2] representing some non trivial cohomology class in H^2(S^2)?
I also just don't really know about this statement "quasicategories help with gluing." I think algebraic geometers mean something specific by this, but I have no idea what it is. For me, quasicategories just let me get away with saying fewer confusing words.
I think I've heard statements to the effect that you don't have to worry about transversality of submanifolds when looking at their intersections, and so forth.
Or whatever it is that algebraic geometers call submanifolds. Codimension 1 squiggles.
And this seems to come from the fact that you're not taking hard products anymore, you're taking derived products (or homotopy products, or homotopy limits, or whatever), so you're not forgetting any of the higher coherence data.
The one thing I can say is that people were trying to construct moduli spaces, for example moduli spaces of vector bundles. They realized they needed stacks. Sure. But when they tired to construct gadgets parameterizing chain complexes up to quasi-isomorphism they realized they just could not do it with normal stacks. That's where you need higher stacks. There is a relation between morphisms E --> F[-k] and gluing that I don't quite remember how it worked.
the thing about transversality is more about the derived structure, rather than the "higher" one.
Yeah, I suppose that's another distinction I don't really understand. It's sometimes hard for me to see how one can "derive" things without essentially including some higher homotopy coherence data.
Or, I guess that just means like, strictly invert quasi-isomorphisms or whatever.
I'm just trying to understand how having a descent theorem (in the oo-sense) helps with the fact that, no matter how I look at it, a map like Q --> Q[2] vanishes if I pass to a contractible open.
yeah, no worries, my original question was about P1
I'll say it again
take P1
as a variety and consider O and O(-n), two line bundles, I want to understand how oo-descent helps with the fact that I have global maps O-->O(-n)[1] which are non-trivial, but locally trivial.
take the standard open cover of P1 (so two open affines), with intersection A
the only "gluing" condition seems to be a map O --> O(-n) on A
no, I don't think so. also, this is a Zariski open cover (so an honest cover), nothing interesting happens to the n-uple covers, they just stay A
(if it were an etale cover then multiple interesections become interesting) (had it been, I don't know, some weird cover with proper maps then I agree one should derive everything)
I think here triple intersections are irrelevant. Let me find the thing I was quoting earlier.
Prop 4.3 of hamilton.tcd.ie/events/swisk.pdf. You only need double intersections when you have a situation as simple as this one. But this is good, it means one should be able to actually see everything explicitly.
So, we want to cook up a global map O-->O(-n)[1] on P1, with standard opens U and V, with intersection A.
On U and V there is nothing to say, but on A, we need to specify a "homotopy" (ie a map O|A-->O(-n)|A) witnessing the fact that the "zeroable maps" on U and V agreed.
as we said before, one could certainly pick the zero map, but one does not have to.
ok, both O|A and O(-n)|A become k[t,t^-1] =: R, the functions on A.
and there are a gazillion maps I could choose (all of R)
now, the fact that I started with O(-n) and not O(666) should make a difference at this point, having to do with the transition functions defining O(-n)
ah! will I finally be using the degeneracy maps of the Cech diagram? I've never used them in my life...
so, I mean, that's some cocycle data, and then you need to figure out when two cocycles are cohomologous; that's presumably where the transition functions come in, and presumably how things get cut down to size
@SeanTilson Yeah, it's the paper from which I learned all the theory. The thing is that in Corollary 7.6, they jump from "the realizations of such X's is B Aut(T,k)" to "In particular E(T,k) has a unique E_oo structure".
The thing is that the obstruction groups are zero pretty badly so maybe it gives it somehow
Actually fun thing : In their bigger paper "Moduli Problems for structured ring spectra", in the introduction they only claim to have shown what the moduli space of such X's are, and not that the Lubin Tate theories are E-oo
@AaronMazel-Gee Thanks, I was thinking that something like this might do it. But as Tyler said, I'm not sure there is even a map a priori. I'm perfectly happy for E being a Lubin-Tate theory. Do you know in that case how to prove that they are E-oo ? It's corollary 7.6 in their paper. They say it so chill, it seems like it should follow straight from the cotangent complex being contractible
@TylerLawson Thanks, it's cute when Adams-Novikov pops up. It might be doable because of the coinduced thing that simplifies, but I wanted to ask you to make sure : the point in computing [X,E] is to hope for the degree 0 part to be a field and thus everything is invertible ?
I think I can cheat here and call the result saying that the Goerss-Hopkins obstruction groups are the same as Robinson's, and Robinson really builds an E-oo structure on the desired spectrum
@TylerLawson @Bogdan @AaronMazel-Gee Maybe what Aaron was pointing out is that you don't need a map because you can just use $X$ as it has the same homotopy etc as Tyler points out. Could that be it?
@Bogdan The point was hoping that the isomorphism E_* X -> E_* E, viewed as an element in Hom_{comod} (E_* X, E_* E) = Ext^0(E_* X, E_* E), survives the spectral sequence and so is the map in E-homology X -> E (which then would be an equivalence in E-homology)
@SeanTilson Right, well, that depends on whether your goal was to work with the given spectrum or not; in the Lubin-Tate case you could say that the result is an essentially unique Lubin-Tate spectrum for each formal group law in question, but sometimes you might really have a spectrum you're interested in as a start
Hey, all (and I guess all = Tyler right now): what's your favorite way to compute Ext groups for wildly noncommutative rings? I'm looking for new tricks.
Hey guys... why is it that the ANSS for MO converges to the 2-completion of the sphere? (At least... I think that's true). Like, is it some basic arithmetic fact about $\pi_\ast(MO)$?
What I'm really trying to think about I guess is whether or not a Thom spectrum is HZ-orientable, and whether or not this can be determined arithmetically, or even by looking at just pi_0 of the Thom spectrum.
But, here's how bad it is: Take a free k-module V of finite rank, find an automorphism of f V \otimes V which satisfies the braid relation on V \otimes V \otimes V, and then let A be the tensor algebra on V, not with the multiplication in the tensor algebra, but rather via a shuffle product based upon f
If f just switches the two factors, the result is literally the shuffle product on the tensor algebra
But unfortunately that's the uninteresting case of what I have in mind.
Well, in the case that V=k has rank one, and the braiding is "multiply by -1," I can show that the Ext groups of this algebra are the homology of \coprod_n Conf_n(C). You can see the classical computation of the homology of \Omega^2 S^2 falling out of this like magic.
Anyone know of a resource that describes an equivalence (in any situation) between modules over an algebra and comodules over its Koszual dual coalgebra?
Or, here's a related question: suppose I have two spaces $X$ and $Y$ over a common space $Z$. Is there any relationship between the mapping space $map_Z(X,Y)$ and the cotensor product of $X$ and $Y$ over $Z$? Where this is the equalizer of the diagram $X\times Y\rightrightarrows X\times Z\times Y$ where the coactions are given by postcomposing the map to $Z$ with the diagonal on $X$ and $Y$ respectively?
Hrm... maybe that's just the fibered product over $Z$?
W/r/t my first question above, the paper of Hess and Shipley on K-theory via comodules describes an equivalence between X-comodules and \Omega X-modules, with suitable finiteness conditions.
