@loch I finally figured out something that had been causing me a lot of trouble

Ultimately it stemmed from a simple mistake

If $M$ is a right $A$-comodule, it is naturally a left $A^\vee$-module: apply the comodule structure and then contract. But I had the handedness wrong for weeks, since c'mon, it pops out an A on the right side!

yeah noncommutative stuff always confuses me

@loch when I complained to other people they thought it was obvious so rip me :p

if it makes you feel better i had to google the defn of comodule so it wasn't obvious to me :)

@loch They show up more often in algebraic topology

any examples?

@loch 1) the usual diagram defining the product on cohomology of a space instead defines a coproduct on homology (with field coefficients, to apply Kunneth)

2) the Steenrod algebra, naturally isomorphic to the set of homotopy classes of maps (of spectra) between the Eilenberg MacLane spaces, $[H\Bbb F_2, H \Bbb F_2]$, has a composition product, and a coproduct coming from the product $H\Bbb F_2 \wedge H \Bbb F_2 \to H \Bbb F_2$ (apply this in the first factor). Thus it is naturally a bialgebra (even a Hopf algebra), and it has both kinds of actions on $H^*(X;\Bbb Z/2)$

One (well, Milnor) takes the dual of this sometimes, as the coproduct is much more manageable than the product, and proves that the dual is a polynomial algebra (nice! easy) with an interesting coproduct, encoding the relations between the steenrod operations

(from the alg geo point of view, the dual of the steenrod algebra is the ring of functions on some algebraic group, and this drives a certain perspective; eric petersen wrote a book on formal geometry in homotopy theory)

relevant to me, but still formal: if $A$ is an augmented dg-algebra over a ring $R$, its bar construction $B(M, A, N)$ ($M$ a right module and $N$ a left module) is $$\bigoplus M \otimes A^{\otimes i} \otimes N$$ with a differential encoding the differentials on $M$, $A$, and $N$, as well as the product structures. there is a natural map $B(M, A, N) \to B(M, A, R) \otimes_R B(R, A, N)$ given by splitting up the tensor factors in all the possible ways.

in particular, this makes $BA = B(R, A, R)$ into a coalgebra and $B(M, A, R)$ into a right $BA$-comodule

if $A = C_* G$ and $M = C_* X$, where $X$ is a G-space, this recovers the fact that $BG$ is a coalgebra (via the diagonal map, aka the map induced by the inclusion into the product $BG \to B(G \times G)$), and that $(X \times_G EG)$ has a natural map to $X \times_G (EG \times_G EG) \cong X \times_G EG \times BG$; this gives $X_{hG}$ a comodule structure

more familiarly this gives the ring structure on $H^*_G$ and the module structure on $H^*_G(X)$

for my purposes, basically i was using it as a way to dualize and find a module structure

oh wait

now that I think of it $BG$ as a simplicial set also looks like $G^n \rightarrow G^{n-1} \rightarrow \ldots $

Yeah, exactly

The algebraic model just presents chains on BG

(BA is just notation inspired by that)

Though I think perhaps what you are describing sounds more like Hochschild homology - I am worried about the attention paid to A o A^op

oh yeah that's where the context where i learnt bar complex i think

hmm

so i shouldn't have a $A \otimes A^{op}$ here

but rather here im just taking products of $G$?

It's the totalization of the resolution -> A o A -> A -> R

Where the last map is some augmentation

Well, depending what you parse that as

I mean to include R in my totalization

If you truncate there and start with A you get EA = B(R,A,A) or something

But the model for BG I have in mind is "Milnor's infinite join construction"

@loch in a day or so I can send you my complete appendix where I do this and related constructions in painful detail :p

especially now that the handedness thing is sorted

I would call that EA or B(R,A,A) - it's the resolution of R as an A-module. Then BA is what you get when you quotient by A on the right

The handedness always confuses me somewhat

The resolution of R as an A-bimodule is B(A, A, A) - your 'lowest term' is a copy of A otimes A

And then you'll tensor a bimodule with this over (A otimes A^op)

Hm I think I should just pick up a reference and go through this more carefully

@loch perfect, I almost have one ready for you :p

You night like the discussion of bar constructions in Emily Riehl's book "categorical homotopy theory"

@MikeMiller let me see

The reference I use is Gugenheim and May, differential torsion products (a book from the 60s) and the updated version of it Barthels-May-Riehl

I occasionally used the Hochschild literature but May's sign conventions were easiest to use

i think whenever the word simplicial shows up it scares me a little bit

Me too, but I think that's mostly because I never really learned a philosophy to any of it