@HarryGindi The BV tensor does not arise from a Day convolution - the BV-tensor of two free operads is not free, so there's no monoidal structure on Omega that induces it.

@RuneHaugseng There's no pro-monoidal structure that induces it either

in general, Day convolutions arise from something more general than a monoidal structure

and pro-monoidal structures can sometimes be traced back along dense full embeddings

that's the content of the question

Professor Heuts e-mailed me back and he's looking into it

the question here is why the trace of Operads on Ω does not exist

There's actually a lot of neat stuff in Day's original paper (the generalized Day convolution for pro-monoidal structures)

does anyone know anything about the homotopy theory of differential (not-graded) algebras? (theres still a notion of quasi-iso, so they form a relative category)

@AaronMazel-Gee How do you even define a quasi-iso in that situation, is it a ring R with a Z-linear derivation R->R? How do you get the (co)homology groups to define quasi-isos?

When looking at the $p$-primary components of the stable homotopy ring, Hatcher says that it inherits the ring structure. I am a bit confused about this however as all the degree $0$ elements have been taken out so where exactly is the identity?

I suppose in a way I was hoping for ${}_p \pi^s_*$ to be a $\Bbb Z_p$-algebra but it seems just out of reach.

@AliCaglayan Usually people say "the p-primary part of π_*X", they mean the p-completion of π_*X (more accurately: the homotopy of the p-completion of X, but there's no difference when X is the sphere spectrum), and it is clear that that inherits the ring structure

In particular, π_0S^_p = Z_p

As in p-adic completion??

yes

ok I need to do some more reading, thanks anyway :)

As I said, the p-completion is better-behaved if done directly at the level of spectra, but if you're not familiar with spectra you can take the p-completion of the homotopy groups as a first approximation (they very often coincide)

I have looked at spectra a bit. What kind is good for this, for the sphere anyway? I guess it doesn't matter

Well, the trick answer is that we're interested in the p-completion because that's what the Adams spectral sequence computes, and the ASS is our best computational tool

(also, if a group is a finitely generated abelian group, all that the p-completion does is discard the torsion that is not p-power torsion and replace the Z's by Z_p's so it's not that destructive)

@HarryGindi yes, just kernel/image as always

Ok good so it is a $\Bbb Z_p$-algebra. That's good because I was hoping to maybe do some rep theory on it

All positive degree elements are nilpotent, and it's awfully non-noetherian

I am aware that it is slightly difficult to put your hands on. But I was interested in seeing where exactly rep theory techniques break (and possibly how to fix it) Thank you for the discussion as always Denis :)

Question for the room: does anyone know a description of the set of (homotopy classes of) E_1-maps from C_2 to GL_1S? (alternatively maps of pointed spaces BC_2→BGL_1S). I think it must have a C_2 summand, but I wonder if there's anything else

Re:

@AaronMazel-Gee so what's interesting here is that the natural functor sending a differential algebra (A, d) to the DGA (A <-d- A <-d- ...) is faithful but not full? It preserves quasi-isos but doesn't reflect them. I guess that kills any hope of viewing the homotopy theory of differential algebras as a sub-theory of that of DGAs?

Oof, I kinda hate that I ended statements with question marks, but I wanted to sound less sure of myself so my poor punctuation still stands

@HarryGindi In fact, the category of trees Ω cannot be endowed with a promonoidal category structure with respect to the BV-tensor product. Indeed, such a promonoidal category would be a 6-tuple $(Ω, P, J, \lambda, \rho, \alpha)$, ...

where for any trees $Sm T$ and $U$, $P(S, T, U) = dSet(\Omega^S, \Omega^T\otimes \Omega^U)$, $J= dSet(-, \Delta^0)$, $\lambda$ and $\rho$ are isomorphisms for the tensor product of a representable presheaf $\Omega^T$ of a tree $T$ with the representable unit $\Delta^0$ and (if one compute the associated coend) $\alpha$ requires the canonical morphism of denroidal sets $(\Omega^S\otimes \Omega^T)\otimes \Omega^U \to \Omega^S\otimes (\Omega^T \otimes \Omega^U)$ to be an isomorphism

Now, we know by the paper of Heuts, Hinich and Moerdijk that $\alpha$ is not an isomorphism in general. (Note that the target of that morphism is just the dendroidal nerve of the symmetric operad $S\otimes T \otimes U$, while the source is not, in general).

@AndreaGagna I know, that's why I'm trying to find out when the trace exists!

maybe there is some condition stronger than density of the embedding that allows us to trace back

it's a sort of inverse to Day's reflection theorem

but I'm definitely aware that in the dendroidal case it doesn't exist

Andrea helped me figure out where I went wrong, I misread a certain thing in Day's paper, and it's not stating the existence of cotensors, it's stating that colimits are computed pointwise

an insanely strong condition

A fascinating blog post : noncommutativeanalysis.wordpress.com/2017/12/05/the-nightmare

@DenisNardin this is really bugging me. Are we talking about p-adic integers or cyclic groups here?

@ReubenStern oh, no this is great! there's a Z-action on Ch by translation, and DiffAlg is its fixedpoints (strict/strong...maybe also homotopy?)

@ReubenStern haha it just so happens that i'm reading a book right now that uses this all the time, so i'm actually pretty used to it

"inherent vice", by thomas pynchon. it's really great. i think one might appreciate it more after reading other pynchons, but i'm sure it's good standalone too.

usually he's super-literary with all kinds of references to anything from chemistry to ancient greek philosophy to higher math to whatever... but this is, like, a pulpy detective thriller.

it's really of amusing, the diction he gives his various beach bums / stoners / deadbeats (it's set in L.A.) -- even when they're talking about acid trips and whatnot, they sometimes display these little flourishes of erudition. i get the sense that pynchon just couldn't help but make his prose beautiful.

it's free to use for your first two semesters. after that, you can either pay by the class, or get your department/institution to spring for a membership