@SaalHardali All of the $v_{n}$-maps are zero in $H\mathbb{F}_{p}$-homology, so that all of the inclusions $X_{n} \hookrightarrow X_{n+1}$ as in Sanath's answer are $H\mathbb{F}_{p}$-injective.

It is not immediate to me that $\varinjlim X_{n}$ is not in the subcategory generated by colimits by $H \mathbb{F}_{p}$, but I agree that this is the first one to check.

@PiotrPstrągowski Yes, thank you. What if we suppose there's a non-trivial map to this guy from $HF_p$ then look at the map on homology. But what we're mapping from is the dual steenrod algebra which is a countable union of projective objects in the the category of comodules so it must have projective dimension at most 1. So in the adams spectral sequence for the homotopy groups of the mapping space we get only 2 lines. So it's a short exact sequence.

Maybe if we do this carefully we can get that there are no maps and then its obvious that this object can't be in the category

Its probably not that easy and I'm missing something.

@SaalHardali What about the Brown-Comenetz dualizing spectrum $I = I_{\mathbb Q / \mathbb Z}$? At least $p$-locally, I've been told that $I \wedge I = 0$. By the Kunneth formula, this implies that $K(n)_\ast(I) = 0$ for any $n$, including $0$ and $\infty$. So $I$ is dissonant and has vanishing homology.

Wait -- something is wrong -- $L_{K(n)} I$ is supposed to be an invertible object by Gross Hopkins duality, right?

Or maybe it's a slightly different version of $I$ that's invertible...

@TimCampion But why does $I$ being $HF_p$-acyclic mean it's not in the category generated by it? I mean it's also $I$-acylic like you said and it's still in the category generated by $I$. Maybe you had a different argument in mind?

Also Brown-Comenetz recieves many non-trivial maps from every p-finite spectrum so it seems like a rather shaky candidate for this sort of thing. It might still work though...

Actually I hope it works because it's already a counter example to a quadrillion other things so why not.

@SaalHardali Oh wow -- now I'm thinking that every bounded-above spectrum is in the colimit closure of $H\mathbb Z$. So in particular, $I$ is in the colimit closure of $H\mathbb Z$.

Is that really true? Gosh...

@TimCampion Good catch.

I have a feeling I once knew this and forgot.

@TimCampion am I misunderstanding what 'colimit closure' means? Because the smallest subcategory of spectra containing HZ and closed under colimits would, in particular, consist entirely of connective spectra (since those are closed under colimits). So most bounded above spectra would be excluded

er -- I suppose I should say "colimit closure of shifts of $H\mathbb Z$"

@DylanWilson good catch

Ah yes, I was assuming he meant shifts too. Thanks for the correction.

It's easy to get all bounded spectra, and then by taking the colimit of the Whitehead tower you get any bounded-above spectrum

It's basically the same as bousfield localization w.r.t. the bounded spectra.

meaning these are the acyclics.

So I'm asking if we know someone in the image of this localization which is not harmonic.

I may have done too many contrapositives. The question is whether the (p-complete) Dissonant spectra are strictly contained in the colimit closure of suspensions of $H\mathbb{F}_p$

i vote "probably not"

One containment is true: everything in the colimit closure of desuspensions of $H\mathbb Z$ is dissonant.

oh! you're going to hate this

here: let's take a finite type 2 spectrum X with a v_2-self-map.

I'm excited!

we get an induced map Tel(V) -> L_{K(2)}(V) with fiber F.

I think F is K(n)-acyclic for all n and has no maps from HF_p because HF_p is both K(n)-locally and telescopically trivial.

I regret ever asking this question.

so F would be dissonant and have no maps from HF_p

so a positive solution to your question would imply that the telescope conjecture is true

So do we have an example whose non-triviality doesn't depend on the telescopic conjecture?

:D

Wait it also works in the other direction?

Ah yes!

Nevermind, It's the best answer because I'm a believer!

Ah wait no. I was too quick to celebrate. I thought for a second Telescopic implies my thing but it's the other way round so I'm sad again,

:/

@TylerLawson Thanks for this example though!

I agree that an example that doesn't depend on falsifying the telescope conjecture would be a little more fun

I still don't understand why something in the colimit closure of shifts of $HF_p$ must admit a map from $HF_p$?

Oh -- maybe it needn't, but it must admit a map from itself, and it itself is both telescopically and $K(n)$-locally trivial.

@TimCampion given X, the collection of objects {Y | [Sigma^n Y, X] is trivial for all n} is closed under hocolims (it implies that all the mapping spaces are contractible)

@TylerLawson oh -- I guess so. I was worried about a $lim^1$ term but I guess I don't have to worry since by assumption all the mapping spaces in the $holim$ are zero.

Anyway, if Saal's conjecture holds, and the dissonant spectra are exactly those in the colimit closure of shifts of $H\mathbb Z$, then the harmonic spectra would be exactly those spectra $X$ such that $F(H\mathbb Z, X) = 0$.

As it is, we know that being harmonic implies this condition.

Which is already striking -- it hadn't occurred to me before that there can be no nonzero map $H\mathbb Z \to \Sigma^\infty Y$ for a space $Y$, for instance.

I'm putting my money on this colimit of the generalized moore spectra.

Ah, so I shouldn't say "Saal's conjecture" then :)

You can call it this once we know it's false.

That would be a fitting name

:)

@TimCampion Also about the thing you said. It took me some time to realize this but anything harmonic is basically immune to arbitrary fiddling with it's lower homotopy groups. Because it can be essentially recovered from any connective cover of itself (barring the difference between chromatically complete and harmonic).

So in particular you can't map bounded above stuff into it.

That's my intuition for this anyways.

I think this is an elementary question, but given two Cartesian fibrations $E\to X$ and $E'\to X$, how can I classify the "functors of Cartesian fibrations" from $E\to E'$ relative to all possible simplicial sets maps over $X$?

Oh or here's maybe a simpler question: how much can I test Cartesianness on fibers? Like suppose I have a triangle X<-p-E-->E'-q->X, where p and q are Cartesian. If I know that, when I restrict to x\in X, the induced map E_x-->E'_x is Cartesian, do I get that E-->E' is Cartesian?

@JonathanBeardsley No. For example, let p be the identity on Delta[1] and let q be sigma^0 : Delta[2] --> Delta[1]. Then p and q are cartesian fibrations, and delta^1 : Delta[1] --> Delta[2] is a functor from p to q over Delta[1] which is cartesian on fibres, but not cartesian.

The Steenrod algebra is sometimes defined as the algebra of all stable cohomology operations. Then again, it's sometimes defined as the homotopy of the mapping spectrum $F(HF_p, HF_p)$. How does one show these are equivalent? That is, how does one show that there are no phantom maps $HF_p \to \Sigma^n HF_p$ for any $n$?

@JonathanBeardsley A morphism in a fibre of a cartesian fibration is cartesian if and only if it is invertible, so given two cartesian fibrations p : E--> B and q : F --> B, every functor f : E --> F such that qf = p is cartesian on fibres.

@JonathanBeardsley (My second answer supersedes the first, but I cannae delete it.)