Possibly stupid question that I don't really want to ask on the main site: Has anyone conjectured that some of the sporadic groups are the F_1 rational points of the groups of Lie type?
Following a computation of mapping class groups in 5 dimensions, I'm trying to calculate $\Omega_6^{spin}(\text{CP}^\infty)$. The obvious approach is to run AHSS, which doesn't have any differentials in the appropriate range, and I get that the answer is either $\Bbb Z^2$ or $\Bbb Z^2 \oplus \Bbb Z/2$, depending on extension problems.
Is it clear to anyone how to resolve these extension problems, or if there is an alternate method of calculation?
@MikeMiller I don't know how to resolve the extension problem, but here are a few ideas.
You could just look for two linearly independent manifolds in that cobordism group. Let $M$ be a spin 6-manifold with a map to $\mathbb{CP}^\infty$, which is classified by a $B\in H^2(M)$. Then $[M] \frown B^3$ and $[M]\frown p_1(M) B$ are both cobordism invariants, and perhaps you can find two spin manifolds on which they're (0, 1) and (1, 0) or something like that
wait, I misread your original question. That won't actually help; sorry
in that case, here's a better way to calculate it. If you set up the Adams SS with p = 2, you can drastically simplify the $E_2$ page because $\mathit{MSpin}\wedge\mathbb{CP}^\infty_+\to\mathit{ko}\wedge\mathbb{CP}^\infty_+$ is an isomorphism up to degree 7.
Well, I like the idea of picking 3 elements that project to generators of the corresponding subquotients, and then trying to understand what relations exist between them. I am having some trouble coming up with them, unfortunately. (I feel like how to come up with these should be "reasonably clear" if I really understood AHSS.)
then, because $H^*(\mathit{ko}; \mathbb Z/2) = A\otimes_{A(1)} \mathbb Z/2$, the $E_2$ page simplifies via a change-of-rings theorem to $E_2^{s,t} = \mathrm{Ext}_{A(1)}^{s,t}(H^*(\mathbb{CP}^\infty, \mathbb Z/2), \mathbb Z/2)$
and because this is over A(1) and not A, it's considerably more tractable
unfortunately, you could still have extension problems at the end. But maybe this computation would tell you something new
a similar ASS approach, using sciencedirect.com/science/article/pii/0040938368900268 : in Table 6.2 it looks like both of those classes are in the image of the map Omega^fr –> Omega^spin, and then according to Table 7.2 all extensions in that degree of Omega^fr are detected by Sq^1, no funny business
I'm rather ignorant of Adams, but this seems like fine motivation to learn. I have a silly question: what, precisely, is this converging to? The group I want, tensored with Z/2? (that sounds like too much to ask for.)
@skd yep, that's the place I learned this approach from! There's also a followup paper by Beaudry-Campbell with more pictures and more computations: arxiv.org/abs/1801.07530
(if anyone cares, the point is that this is known to be isomorphic to the mapping class group of diffeomorphisms $S^2 \times S^3$ which are trivial on homology; see the brief discussion here)
do you really need to use A(1), though? you're trying to understand ko^*(CP^oo) up to degree 7, but this can be done by the atiyah-hirzebruch spectral sequence
as an A(1)-module, H^*(CP^oo) is a sum of a bunch of H^*(CP^2)'s, so the E_2-page of the ASS is just a sum of a bunch of Ext(H^*(CP^2))'s
this has classes only in even degrees, so it collapses
now you get a nontrivial class in pi_{4n+2}(bo smash CP^oo) by a generator in filtration zero of degree 4n+2. moreover, eta-multiplication kills this class (because the spectral sequence is concentrated in even degrees)
in particular, since S^{-2} CP^2 is the cone on eta, you can extend this element S^{4n+2} -> bo smash CP^oo to a map S^{4n} CP^2 -> bo smash CP^oo
@Arun Diarmuid has at least not published anything on the subject since then, but I don't know. (I am sort of just fiddling, I don't think about this much.) Note that in this particular case the conjecture is covered by older work of Fang.
actually, i'm not really sure about the difference between bu/ku and bo/ko. i've used them synonymously, but it seems like older papers use them differently. are they actually different?
(I think I broadly follow but need to spend tonight learning how people do ASS computations :) I really appreciate the care you've taken in this response)
In particular the answer to the original question way back was that Omega_6^spin(CP^infty) = Z oplus Z. It would be interesting to have explicit generators as Arun suggested; CP^3 is one clear one, giving us one summand
user131753
Does there exist any fully faithful functor from $\mathbf{Top}$ to $\mathbf{Grp}$ or vice versa @skd? (The only functor so far I know from $\mathbf{Top}$ to $\mathbf{Grp}$ is the fundamental group functor which is not fully faithful).
you're basically asking: can i construct an invariant F(-) of topological spaces such that a map f: X -> Y is a homeomorphism/homotopy equivalence if and only if the induced map F(X) -> F(Y) of groups/rings/whatever is an isomorphism
user131753
Actually I was trying to find some functor that is fully faithful so that it would reflect isomorphism.
You would be able to tell whether spaces are homeomorphic using your output groups and that sounds very hard to believe - even for spaces with only finitely many points how could you imagine constructing this?
the best invariants Top -> Groups that we know of are (co)homology, K-theory, and homotopy groups (in increasing order of strength) --- and none of these detect homeomorphisms/htpy equivalences
These capture the homotopy type fully in the case that a CW complex is simply connected. The algebraic datum are the homotopy groups and some cohomology operations.
@Arun If you evaluate p1 cup B for K3 x CP^1 (the map is projection to the second factor and then inclusion), we get 48. I wonder if one can actually do better than this, or if this represents a generator. I might expect not, since cupping with B is like saying "take a Poincare dual 4-manifold and calculate p_1(M) over it"; if that reduced to p_1 of the 4mfd itself, and as that 4mfd is spin, it would be divisible by 48
Hm, I am not sure why I think that 4 manifold is spin. Its normal bundle is the pullback of the tautological line bundle, which would change w2 if it was non-trivial.
user131753
@MikeMiller Given any group $(G,\ast)$, I viewed the binary operation $\ast$ as a family of functions $\mathscr{F}_X:=\{f_x:X\to X\mid \forall x\in X\}$ which for a given $x\in X$ is defined as $f_x(y)=\ast(x,y)$ for all $(x,y)\in X\times X$. Define a new category $\mathbf{GrpFunc}$ by taking the objects as the pairs $(X,\mathscr{F}_X)$ and the "obvious" homomorphisms between them.
@MikeMiller indeed, CP2 inside CP3 isn't spin, but is the Poincaré dual to the choice of B you used for the first generator iirc
user131753
Anyway what I was trying to do was to associate to each group $(X,\ast)$ the family $\mathscr{F}$ and on it the subspace topology induced by $X^X$ (which is given the product topology), say $\tau_{\mathscr{F}_X}$. Thus I was trying to associate $(X,\ast)$ with $(\mathscr{F}_X,\tau_{\mathscr{F}_X})$.
OK, $S^2 \times S^2 \times S^2$, with B the sum of the generators of the three copies $H^2(S^2)$, is linearly independent from CP^3 inside the bordism group. But these two are probably not a generating set.
user131753
However if $\varphi$ is a group homomoprhism between $G$ and $H$ then I couldn't figure out to which continuous map between $\mathscr{F}_G$ and $\mathscr{F}_H$ I would associate.
Here's a silly reason why there is no fully faithful functor from Top to Grp: any fully faithful functor induces isomorphisms of automorphism groups for every object. It's known that every group is the automorphism group of some topological space (see eudml.org/doc/160702), but no group has an automorphism group that is non-trivial cyclic of infinite or odd order
So in Lurie's construction of the straightening and unstraightening functors (over an ordinary category $C$), he produces a Quillen equivalence between $(sSet^+)/N(C)$ and $(sSet^+)^C$, where the second category is the category of simplicial functors from $C$ (with the trivial simplicial structure) into the simplicially enriched category of marked simplicial sets. This second category he equips with the "projective model structure." Am I to understand that the model structure on $sSet^+$...
then is the one determined by thinking of the model structure on $(sSet^+)/\ast$ first, and that is what is meant by assigning the projective model structure to the category of simplicial functors from $C$ to (the presumed model category) $sSet^+$?
In other words, Lurie never describes a model structure on $sSet^+$, but he does describe one on $sSet^+/S$ for any simplicial set $S$, and I'm guessing he just wants me to use the one where $S=\ast$.
Lord... okay actually... I suspect I'm supposed to be using the model structure on $sSet^+$ which has coCartesian equivalences, rather than Cartesian equivalences.
Ah, but because $\ast$ is a Kan complex it doesn't matter.
