Suppose I have an $(\infty,2)$-category modeled by, say, a fibrant scaled simplicial set. View that as a stratified set: is there hope of describing a fibrant replacement for this in the model structure for saturated complicial sets (i.e. $(\infty,\infty)$-categories)?

Or, say, starting from a 2-trivial saturated complicial set. Ideally, I'd want the underlying simplicial set not to be modified. I think I may have a candidate for this but I'm up for suggestions. It's possible that having a replacement for $(\infty,3)$ is also enough for what I want to do.

What's an example of two finite spectra with the same $\mathbb{F}_p$ homology as comodules over the steenrod algebra but of different chromatic type?

@SaalHardali the type 2 spectrum V(p, v1^j) will do, which has cohomology in degrees 0, 1, j |v1| + 1, and j |v1| + 2. taking j ≠ 1, there’s no room for indecomposables other than beta to act, so gives a dual steenrod comodule indiscernible from M(p) v Susp^(j |v1| + 1) M(p)

i’m worried i may have some off-by-one errors, but also confident that there are too few places to make such errors to prevent me from picking an adversarial j no matter how many i were to make

@EdoardoLanari A 2-trivial saturated complicial set is already fibrant in the model structure for n-trivial saturated complicial sets, for all 2 <= n <= infinity.

So to replace a fibrant scaled simplicial set by a fibrant saturated complicial set, you only have to mark the equivalence 1-simplices and all simplices in dimension 3 and above.

@EricPeterson @SaalHardali I think what your argument can be rephrased as saying that the map $v_{1}^{j}: M(p) \rightarrow M(p)$ is of Adams filtration at least two for $j \geq 2$ (when it exists). In classical language, it means that it induces the zero map on mod $p$ homology, and the homology of the cofibre is the trivial extension of the homologies of $M(p)$.

Then, by taking appropriate cofibres of $v_{i}^{j}$ for larger $i$ you can make examples of arbitrary type > 0.

@AlexanderCampbell I'm not sure this answers my question. Look at a scaled simplicial set X as a stratified set whose marking outside of dimension 2 is the minimal one (only degenerate simplices). Why is the one obtained by marking 1-equivalences and all n-simplices for n>2 weakly equivalent to the initial one? It is not entirely clear to me how to see in X that ideally those simplices want to be marked. Of course I agree if we look at n-trivial fibrant replacement for some n.

@EdoardoLanari Ah, I see, you’re right.

And yet I feel like my answer should still be correct. The fact of X being a fibrant scaled simplicial set should mean that for any n-simplex x with n>2, there should be (n+1)-simplifies witnessing that x is an equivalence. So at least for m-trivial saturated complicial sets, my answer should be right.

Ah that’s what you said in your last sentence. I have no idea for the infinite dimensional case.

Of course, that's what I meant with my last comment. If from some dimension onward everything is marked, then clearly we can "detect" markings by saturation. But I really want $(\infty,\infty)$ here.

ops! sorry I haven't read your last comment :)

Given that there exist Kan complexes with non-maximal markings which are fibrant saturated complicial sets, what I said is false in general for the (infinity,infinity) case.

Probably...

@SaalHardali Another example along the same lines would be the mod $p^n$ Moore spectrum for $n>1$ and a wedge of two spheres.

@JohnPalmieri @PiotrPstrągowski @EricPeterson Thanks for the examples. Apparently it's a lot more commonplace than I imagined initially.

@PiotrPstrągowski I like this perspective! So roughly speaking $HF_p$ has a chance to detect chromatic type of spectra whose BP-homology has a "shallow" landweber filtration sort of like being "irreducible"... I have a new appreciation for the existence of strongly type n complexes.

@SaalHardali another way to say this is that each Milnor operation Q_i has an infinite family of higher order variants (all of its Massey powers in the ‘spectral Steenrod algebra’), just like there are higher Bocksteins. A type n complex built using a high power of v_i will have a nontrivial higher order Q_i operating it. Since “being detected by higher order operations” is code for “detected in higher Adams filtration” this is another way of saying Piotr’s remark

@DylanWilson That's another cool perspective. Thanks!

Related: the classical Adams spectral sequence for a type $n$ spectrum will have a vanishing line of the appropriate slope (parallel to the action of $v_n$ in Ext), but not necessarily at $E_2$.