Am I too old to go?

@AaronRoyer I think that, for example over F2, all the Squares Sq^i act, even on just the E_2 structure. But you don't have the Adem relations between them, as well as the Cartan formula, so maybe the whole thing that acts is the free associative algebra on the Sq^i, with no relations and no coalgebra structure. I say that because of Lurie's notes where all he uses is the E_2 structure (lecture 2 in the notes on Sullivan conjecture)

I think you need E-4 for Adem, and don't remember for Cartan formula

Not sure if that was obvious to you or if it helps a bit!

@JonBeardsley nope! We skew towards younger grad students but we also have older grad students and postdocs every year

Oh good. :-)

Topic and mentors look righteous.

@TylerLawson @CraigWesterland Thank you for your replies! I think that what I wanted to know (and wasn't specific about) is how close the canonical map B(\Sigma_n \wr S^1)_+ -> BU(n) is to being an equivalence. I know that it isn't an equivalence, but I thought people might know how to "fix" that.

@NatStapleton so it's a rational equivalence, and if n < p it's a p-local equivalence (but you knew those). this approximates U(n) by what you can see from U(1) -- my understanding is that if you "fix" it by approximating U(n) by what you can see from U(1), U(p), ... U(p^k), it becomes a p-local equivalence for n < p^{k+1} and a K(m)-local equivalence for m <= k. there's a couple of papers by Arone-Lesh about this

huh, thanks!

@AaronRoyer For maps of E_n-algebras f: X --> Y, f_*: H_*(X, F_p) --> H_*(Y, F_p) preserves the (lower indexing) Dyer-Lashof operations Q_0, ..., Q_{n-1}, as these are the only ones which are defined for E-n-algebras in full generality. It will of course preserve the ring structure, and the Browder operation (Lie bracket), but for X which is E_k for k>n, that bracket is 0, so that might not be too interesting in your case.

The translation from Dyer-Lashof-operations Q_i to Steenrod operations P^j when you're applying this to cochains on something is a complicated formula that I can't remember exactly, but I'm pretty sure can be extracted from May's "A general algebraic approach to Steenrod operations."

@CraigWesterland Alright cool, I'll check it out. It turns out I really only care about the cochains on S^1 and maybe BS^1, and Mandell's lecture that I cited handles S^1, at least. Namely, P^0 acts on the fundamental class of S^1 even in the E_2-structure

Also, thanks!

what is an example of a fibration with a section that is non-trivial? Maybe there aren't any, but they must be quite pathological right?

maybe I'm misinterpreting, but this is very common and not pathological

for example, consider the Möbius strip fibered over S^1

or actually any nontrivial vector bundle

I guess these are only nontrivial as vector bundles, not as topological fibrations, but examples there shouldn't be much harder to come by

eg how about the Klein bottle regarded as a fibration over S^1 - that definitely has a section but it's not trivial

if G -> H is a split surjective group homomorphism with kernel K then BG -> BH is a fibration with a section, but it's trivial iff G ~= H x K in a manner respecting the projection

oh, sorry, I meant something more along the lines if there were any principal fibrations which were non-trivial but had sections

I guess otherwise any example could be , just any orientable manifold and the orientation cover

Surely that has a section

for a principal G-bundle, giving a section is equivalent to giving a trivialization

@user101036 yes but that's kind of a silly example because it's just two copies of the manifold

right, we just take s(b)g , where g is some element of the group and s the section

thanks1

*thanks

Has anybody internalized what's going on here yet? math.harvard.edu/~lurie/papers/Waldhausen.pdf

oh dude I didn't see that show up

that's going to give me something to do

Yeah it's pretty baller, I've been looking at it for a couple of days

I kind of played with this structure at MSRI and didn't get anywhere, now I know why

oh wow, i hadn't seen that either

whoa, yeah this looks awesome