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Peter Nelson
Peter Nelson
3:53 AM
hey @AaronMazel-Gee you don't happen to have a good reference for Motivic Morava E-theories do you?
 
Aaron Royer
Aaron Royer
Hi @Sven. Sorry I missed you this afternoon, hopefully you'll be back. I don't know about a good theory in general, but I've come across certain ideals of equivariant K-theory which arise by restricting the sort of representation types which can occur at fixed-points.
 
Aaron Mazel-Gee
Aaron Mazel-Gee
well, the main thing is here: arxiv.org/abs/0806.0274
 
Aaron Royer
Aaron Royer
In general these are equivariant spectra, but not genuine equivariant spectra. Is this the sort of thing you have in mind?
 
Aaron Mazel-Gee
Aaron Mazel-Gee
i don't know how much they've been studied, beyond just having been constructed there
 
Peter Nelson
Peter Nelson
ooh a character map too!
 
Aaron Mazel-Gee
Aaron Mazel-Gee
3:54 AM
our result is related to this one, but distinct: arxiv.org/abs/1010.3944
(besides being at all heights, and p-complete instead of integral)
there is this funny distinction between two different things that one might mean by "an essentially unique E_\infty structure", which is the source of the non-comparability
 
Qiaochu Yuan
Qiaochu Yuan
4:42 AM
@Zhen: yes, that's precisely what stability means. in the context of spaces the closest analogous result is that if F -> E -> B is a fiber sequence, then the homotopy fiber of F -> E is Omega B; this isn't quite the same result because B is not the homotopy cofiber, but B can be compared to the cofiber via the blakers-massey theorem
once you believe that long fiber sequences are a natural thing to look at and that stability means fiber and cofiber sequences agree then i think that's where all the magic is
 
 
3 hours later…
Zhen Lin
Zhen Lin
8:09 AM
indeed
something I'm having trouble with right now, though, is convincing myself that the isomorphism between the fibre and the loop space of the cofibre induces isomorphisms in homology that fit into the long exact sequence
 
 
4 hours later…
user101036
user101036
12:02 PM
is it possible to develop k-theory for simplicial sets, internal for simpliial sets? Like, without using realization of any sort. Has anyone done this?
 
 
1 hour later…
Saul Glasman
Saul Glasman
1:12 PM
what kind of k-theory?
 
user101036
user101036
Well, some k-theory that corresponded nicely to topological k-theory, i.e the one with vector bundles and bott periodicity
I'm perfectly happy if one restricts to kan complexes
This is not what the homotopy hypothesis say I suppose, but if \infinity-groupoids are spaces up to weak homotopy equivalence, does this say anything about the kind of invariants we can define internally in each model? So to be more precise with what I mean:
Suppose we have some homotopy invariant that is constructed "internal" to topological spaces. Can we construct this homotopy invariant internally in any other model of infinity-groupoids?
 
Saul Glasman
Saul Glasman
that's a good question
 
 
2 hours later…
Josh Shadlen
Josh Shadlen
2:50 PM
@user101036 I'm not sure this is a satisfying answer but a theorem of Snaith says that you can make complex K-theory by taking the free E-infinity ring on CP^\infty and inverting the Bott element. Take this for your definition and model CP^\infty as B^2\mathbb{Z}. Now it's all combinatorics! You can find a write-up of the proof in these notes of Akhil Mathew: http://math.berkeley.edu/~amathew/snaith.pdf

I don't think this will give you a very satisfying description of the K-theory of a given simplicial set, and I'm also not sure whether one can get KO this way (though there's an obvious gu…
 
Jon Beardsley
Jon Beardsley
3:11 PM
@user101036 can you say more about what you mean by "internally"? do you mean that it's representable?
 
user101036
user101036
I'm not sure how to specify it, I mean somehow that you shouldn't pass to topological spaces with some realization. So, for example just taking the geometric realization of a simplicial set and developing K-theory for the realization would not be "internal" to the category.
For vector bundles over a topological space X, the data required to specify it doesn't draw from any other categories than topological spaces and vector spaces if I recall correctly. We can then form the grothendieck group on vector bundles and get a version of k-theory. So can something analogous work for kan complexes …
 
Jon Beardsley
Jon Beardsley
Well, yeah okay. I mean... hmmm. I'm thinking of Thomason's creating topological K-theory by inverting the bott element in alg. k-theory. that might not really be relevant here though.
 
user101036
user101036
of course, the same construction doesn't fall naively through if I'm not wrong, but can we define a cohomology theory on kan complexes, with it being "internal" in the vague sense above, that is naturally isomorphic to the K-theory of the geometric realization as topological spaces
And in more generality:
If we have a cohomology theory on topological spaces, can we form it as well in other models of infinity-groupoids? I have no idea, but my guess is no
 
Adeel
Adeel
well, don't you need a notion of vector bundles on a Kan complex in order to talk about its K-theory?
 
Jon Beardsley
Jon Beardsley
I mean, I'm also thinking about locally constant sheaves of Kan complexes. But again, maybe this isn't exactly... "internal"
 
user101036
user101036
3:20 PM
I think locally constant sheaves of Kan complexes is something I would say is internal in the sense I'm thinking of
 
Jon Beardsley
Jon Beardsley
Couldn't you also do something with starting with a Kan complex whose geometric realization was equivalent to BUxZ?
But maybe that's kind of... ad-hoc and not really internal in the sense you're describing.
 
user101036
user101036
I don't really know what I'm describing to be honest.
 
Jon Beardsley
Jon Beardsley
Yeah, I think I sort of get it though. Like, if we started out living in Kan complexes and had never heard of "topological spaces" could we get an invariant that was equivalent to K-theory.
 
Adeel
Adeel
isn't there a simplicial construction of the spectrum representing K-theory though?
 
user101036
user101036
Eactly
*exactly
 
Adeel
Adeel
3:23 PM
oh, that's your question i guess
 
user101036
user101036
So i think Jon's question is the best interpretation of internal
 
Jon Beardsley
Jon Beardsley
I don't know. I mean... what IS, e.g. real K-theory? It's coherent bundles of the 1-simplices, I feel like...
Or, sorry.
I guess that would be 1-d KO
I mean, the point is, you want something like... I dunno, locally constant bundles of simplices of fixed dimension..
And it seems like one could do that.
But, I feel like one might have to go through some unnatural feeling contortions to get, I dunno, decent O-actions and U-actions in there.
But maybe you could arrange it in such a way that they just sort of, are there, at the end.
On might also have to bring in some kind of technology like building an $E_\infty$-monoid object in Kan-complexes and then applying an infinite delooping functor, which also seems doable.
 
Josh Shadlen
Josh Shadlen
The cohomology theory and the spectrum you can get without touching the category of spaces by this Snaith construction I mentioned, I think the thing is that it's not at all clear from that description that K-theory is made from vector bundles. Any old cohomology theory does not have a geometric object attached to it.
 
Jon Beardsley
Jon Beardsley
Right.
 
Josh Shadlen
Josh Shadlen
(which is sad)
 
Jon Beardsley
Jon Beardsley
3:34 PM
Haha. Yeah, it'd be nice if we could go the other way.
 
Bogdan
Bogdan
4:06 PM
@user101036 I don't know how relevant that is and how much he published, but Marc Stephan gr-he.epfl.ch/page-77164-en.html has a combinatorial model for K-theory, combinatorial not in sSet but in semi-simplicial spectra. Maybe his technique works for sSet ?
 
user101036
user101036
that is very interesting
 
Bogdan
Bogdan
Yeah it seems interesting, I googled for a minute and didn't find anything written though. Maybe he'll give a talk on it at the YTM there this summer!
 
Tyler Lawson
Tyler Lawson
4:22 PM
@user101036 i believe that the "combinatorial Grassmannians" / "matroid Grassmannians" are meant to be objects that try to model classifying spaces without passing through geometric realization
 
Aaron Royer
Aaron Royer
I feel like "construct KU-theory using vector bundles" and "construct KU-theory without knowing about topological spaces" are mutually exclusive goals, for the simple reason that you need to know something about the topological structure of the complex numbers to talk sensibly about vector bundles.
If you want KU without using topological spaces, Snaith's construction does it, or if you're willing to work p-completely I don't see any reason topological spaces necessarily need to be involved in constructing Morava E-theory
 
Peter Nelson
Peter Nelson
well, you need MU to build Morava E-theory...
 
Jon Beardsley
Jon Beardsley
do you?
oh yeah.
LEFT
 
Peter Nelson
Peter Nelson
right.
 
Jon Beardsley
Jon Beardsley
hehe
 
Peter Nelson
Peter Nelson
4:34 PM
:)
 
Aaron Royer
Aaron Royer
I'm not convinced that MU has to be constructed using manifolds
 
Jon Beardsley
Jon Beardsley
Me neither.
Haha. I'm so not convinced that I'm trying to prove it.
 
Aaron Royer
Aaron Royer
Any way I can imagine doing it is totally post-hoc
and I don't see how it would have arisen naturally
 
Tyler Lawson
Tyler Lawson
@NatStapleton what kind of relationship are you interested in? (their E-cohomologies are still different, and in the colimit one becomes Q(CP^infty) and one becomes BU)
 
 
4 hours later…
CPM
CPM
8:30 PM
the ring R=\oplus Rep(S_n) is a hopf algebra (I think) under the maps induction and restriction and the dual is the hopf algebra of symmetric functions. Are they dual under the above maps?
Also, it is usual in topology to think about Symm as H*(BU), if the above is correct can I think about H_*(BU) as being iso to R?
On this page ncatlab.org/nlab/show/Schur+functor it says Symm is the decategorification of the category of Schur functors but I would have thought it more natural to say R, does it just not make a difference as Symm is self dual?
 
Jon Beardsley
Jon Beardsley
Question for the room: is there something specific I can check to see if two functors in the $\infty$-category of functors between $\infty$-categories, $Fun(C,D)$ are equivalent? In particular I'm interested in cosimplicial objects.
 
Denis Nardin
Denis Nardin
@JonBeardsley Well equivalent as functors means just that there is a natural equivalence between them, so I suppose you could construct a map $C\times\Delta^1\to D$ implementing it. I'm not sure what else can be said in this generality (i.e. without hypothesis on $D$)
 
Jon Beardsley
Jon Beardsley
Ah, well in particular I'm interested in equivalent cosimplicial functors into spectra.
I mean, I've got two functors, and I know that there's a levelwise equivalence between them that is a cosimplicial equivalence (in the sense that it agrees with the edges) but can I run into problems with higher morphisms?
 
Adeel
Adeel
8:47 PM
it's enough to have a morphism which is an objectwise equivalence
 
Jon Beardsley
Jon Beardsley
Okay, I was actually just thinking that that statement might be in HTT somewhere...
 
Denis Nardin
Denis Nardin
So you are asking whether an arrow $C\times \Delta^1\to D$ is an equivalence in $Fun(C,D)$ iff it is a pointwise equialence?
 
Jon Beardsley
Jon Beardsley
Yeah I guess so.
Sorry, I'm thinking this might be kind of trivial.
 
Denis Nardin
Denis Nardin
Then this is certainly true, I'd use it as obvious and you could probably cook a proof with horn-filling conditions if you feel like that
 
Jon Beardsley
Jon Beardsley
For some reason it feels like I'm getting something for nothing.
 
Denis Nardin
Denis Nardin
8:51 PM
I mean the statement you want is that $C\times \Lambda^n_0 \to C\times \Delta^n$ is a left anodyne morphism
 
Jon Beardsley
Jon Beardsley
Perhaps it is non-trivial to show that, in the context of $\infty$-categories, such an edge actually exists.
 
Adeel
Adeel
in general, i think it is reasonable to assume the obvious analogue of any 1-categorical result is true for infinity-categories, and to just be aware of the exceptions :)
 
Jon Beardsley
Jon Beardsley
Maybe, but in my case, it's not really 1-categorical result so much as a model-category theoretic result. In the case of model categories this equivalence really only holds after totalization if you assume fibrancy.
And that's what I'm really after.
So I need the equivalence to be good up to all higher homotopies. And I guess that's sort of what gave me pause.
 
Denis Nardin
Denis Nardin
Well in the $\infty$-category world every object is always fibrant and cofibrant
 
Jon Beardsley
Jon Beardsley
Right.
So, this is why it felt like I was getting something for nothing.
 
Adeel
Adeel
8:54 PM
i find working with infinity-categories is closer to working with 1-categories than with model categories
anyway, if you want a reference, you can look at Joyal's notes: mat.uab.cat/~kock/crm/hocat/advanced-course/Quadern45-2.pdf
Theorem 5.14
probably in HTT somewhere as well
 
Denis Nardin
Denis Nardin
If you have a functor $\Delta\times\Delta^1\to Spectra$ which is an objectwise equivalence you're good to go
 
 
2 hours later…
Craig Westerland
Craig Westerland
10:54 PM
@NatStapleton Let me write N = \Sigma_n \wr U(1) for the maximal torus normalizer in U(n); then there's an obvious map BN --> BU(n). The homotopy fibre is U(n)/N, whose Euler characteristic is 1; thus the Becker-Gottlieb transfer \Sigma^\infty BU(n) --> \Sigma^\infty BN splits the suspension spectrum of BU(n) off of that of BN
So the same is certainly true for the plus-construction on BN, since the suspension spectrum kills the difference. But I suspect that I'm missing your point. In any case, it's certainly true that BN^+ --> BU(n) is split surjective in any homology theory that you like.
 
Jon Beardsley
Jon Beardsley
11:22 PM
Sigh. Nonabelian cohomology gives me the vapors.
 
Zhen Lin
Zhen Lin
11:47 PM
@DenisNardin If I recall correctly the proof is actually a bit elaborate, unless you develop some machinery/theory first
 

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