12:54 AM
given a triangulated category, a sequence of maps $X \to Y \to Z \xrightarrow{f} X[1]$ is called "antidistinguished" if $X\to Y \to Z \xrightarrow{-f} X[1]$ is distinguished. this defines a new triangulated category structure on the same category, with the same shift functor. what's up with this construction? i feel like it must be obvious, but i don't see how to lift this to an involution on stable $\infty$-categories.

1 hour later…
2:20 AM
@AaronMazel-Gee I don't think the antidistinguished triangles satisfy TR 2. So there's no involution on triangulated categories, even. I think what your observation says is that you could give a variant of the triangulated category axioms fomulated in terms of antidistinguished triangles rather than distinguished ones. Maybe there would be "fewer weird signs" in the axioms.

2 hours later…
4:38 AM
a while ago i got confused about something related to the "ASS-flat" discussion that i think i never really sorted out. i'm not sure exactly how to state even my confusion:
for an algebra map R –> S you can form the associated cobar complex, which has levelwise objects (S, S (x)_R S, S (x)_R S (x)_R S, …) as well as R-module maps between them. there's also an R-module iso (S (x)_R S) (x)_S (S (x)_R S) –> S (x)_R S (x)_R S which lets you write the levels above level 0 as tensor powers of a fixed S module, and this map is often important for making 1-truncation-type observations
but the maps in the cosimplicial object don't become maps of S-modules after making this identification. i feel at a loss as to why, morally, S-modules breeze in to the picture and then breeze back out
so uh what's up with that, why am i pulling an S-module trick when little else (nothing?) in the picture respects S-module structure

1 hour later…
5:55 AM
maybe what's happening is that $C = S \otimes_R S$ is a a module in two ways (left and right). They are isomorphic but not canonically. when you take the tensor product $C \otimes_S C$ you are using both the left and the right $S$-module structures. On a second thought maybe this doesn't help at all here

2 hours later…
7:39 AM
@TimCampion hmm, i think i agree. i just took this on faith from something i was reading...so i guess they were maybe just wrong then. thanks! i was feeling like it was supposed to be some sort of "twist" of the spectral enrichment by an involution of the category of spectra, or something...

9 hours later…
4:19 PM
@AaronMazel-Gee yeah, that's what I was thinking initially, but it was pretty hard to square with the fact that being tensored over finite spectra (or being stable) is a property of an $\infty$-category, just like being an additive (or abelian) category is a property of a category -- the enrichment is unique if it exists, so apparently it can't be twisted.
@EricPeterson I think this makes sense to me when I dualize everything. If I have a map $X \leftarrow Y$, then I get a simplicial object $(Y, Y \times_X Y, Y \times_X Y \times_X Y, \dots)$ -- for example if $X \leftarrow Y$ is an open cover, then this is the Cech complex. Then the isomorphism $(Y \times_X Y) \times_Y (Y \times_X Y) \leftarrow Y \times_X Y \times_X Y$ is the Segal condition corresponding to $\Lambda^1[2] \to \Delta[2]$.
Note that the inner simplicial structure maps are $S$-module maps, even though the outer ones aren't.
And tensoring over $S$ corresponds to the fact that the Segal maps being pullbacks over higher parts of the simplicial object than just the 0th one.
This is okay, because they're pullbacks over inner face maps.

4:38 PM
i see: 'little else' is a serious overstatement, for one
i'll have to think about it later, but this probably calms me down. ty

@TimCampion oh, certainly, i had that same confusion. i was figuring it would be some operation on stable (unenriched) $\infty$-categories which induces such a twist on their canonical spectral enrichments.