How are the following compatible: 1.) The $K(n)$-local sphere is bounded below. 2.) $K(n)$ localization is smashing. 3.) For any spectrum $X$ and any $k \geq 0$, the map $X\langle k \rangle \to X$ is a $K(n)$-local equivalence. It seems to me that if $d$ is the connectivity of $S_{K(n)}$, then the connectivity of $S_{K(n)} \wedge X\langle k \rangle$ has to be at least $d+k$, which grows to $\infty$ as $k$ increases, so that $S_{K(n)} \wedge X$ is infinitely connective and hence zero.
here $X\langle k \rangle$ is the $k$-connective cover of $X$.
Is $S_{K(n)}$ not bounded below?
Of course not, I'm being stupid.
I wonder what I was thinking of... Is $S_{E(n)}$ bounded below?
@TimCampion nope, it's not. for example, the E(1)-local sphere differs from the K(1)-local sphere only in degree 0,-1,-2 (think about the chromatic pullback square) so the rest is just as unbounded as L_{K(1)}S
Maybe while I'm airing my misconceptions, I'll mention another thing I'm confused about. The telescope conjecture is that $\langle T(0) \vee \dots \vee T(n) \rangle = \langle K(0) \vee \dots \vee K(n)\rangle$ where brackets denote Bousfield classes and $T(h) = v_h^{-1} X$ where $X$ is a finite spectrum with type $h$ self map $v_h$.
But there's no conjecture saying that $\langle T(n) \rangle = \langle K(n) \rangle$, is there?
There shouldn't be I suppose -- $L_{T(n)}$ is finite and (hence?) smashing while $L_{K(n)}$ is not
I don't think L_{tel(n)} is automatically finite (this is a special property of L_{tel(0)\vee ... \vee tel(n)}), and, indeed, I think the telescope conjecture is equivalent to the assertion that L_{tel(n)} is the same as L_{K(n)}. (but I could be wrong- I get lost in the sea of things equivalent to the telescope conjecture)
Regarding the earlier discussion, any $E(n)$ (or $L_n^f$) local spectrum is unbounded. I'm very embarrased to admit that I didn't notice this fact until very recently.
Goal: To gain a better understanding of Euclidean space, $\Bbb R^3,$ conformally compactified into a unit ball.
Question: How can I visualise and mathematically describe Euclidean space, $\Bbb R^3,$ conformally compactified into a unit ball? Is my model of it correct?
My attempt:
Start w...
Is the concept of simplicial ideals homtopy-invariant?
Or could we describe it in terms of the language where the category simplicial commutative rings is defined to be the non-abelian derived category of that of polynomials?
Suppose that C is an infinity-category with a Grothendieck topology (I'm even interested in the case when C is just a 1-category). Suppose that D is a full subcategory with the property that every object in C is covered by an object in D. Is it true that restriction Sh(C) -> Sh(D) is an equivalence? Is the same true if I replace "sheaves" with "hypercomplete sheaves"?
I don't think so. Consider the étale site of Spec(R), where R is the field of real numbers. The subcategory generated by Spec(C) satisfies your assumption but no form of your desired conclusion, I think.
@PaulVanKoughnett It's true for hypercomplete sheaves if C has finite limits and D is closed under them. That's because a map of presheaves X→Y is ∞-connected in the topos iff X→X^{S^n}×_{Y^{S^n}}Y is a covering morphism for every n
It's false for ordinary sheaves, and very false in general (the ∞-topoi associated with the atomic topology tend to be interesting in general)
@PaulVanKoughnett Ok, I'm trying to reconstruct the proof I remembered. The point is the following: restriction gives a functor Sh(C)^→Sh(D)^ which has a left adjoint by the adjoint functor theorem, we need to prove that unit and counit are equivalences, The idea is to use the criterion I gave to prove they are ∞-connected
It's sort of a modification of the proof for topoi in SGA
Step 1 should be to show that the restriction functor Sh(C)^→Sh(D)^ is conservative, by using that every object in C has a hypercover by (disjoint unions of) objects in D
This allows you to check only the counit, which should be easy-ish
If you need a written reference, 2.22 here has a somewhat weaker result that might be enough for your purposes
If you're familiar with the proof in SGA, it helps to recognize that the condition of being "couvrant et bicouvrant" is exactly the criterion I gave above for ∞-connectedness
Okay, interesting. What I would naively want to do is say that, given F in Sh(D), you can extend F to x in C by taking a cover y -> x with y in D and defining F(x) as the limit of F applied to the Cech nerve of the cover. Then one has to check that this is independent of the cover and defines a sheaf. I guess in general, one doesn't know that the other objects in the Cech nerve are in D, only that they're covered by something in D, so one only ends up with a hypercover.
But that raises another question -- suppose that C has finite limits, D is closed under finite limits in C, and also, if x is in C and y is in D, the pullback y \times_x y is also in D. Does the claim then hold for non-hypercomplete sheaves?
@PaulVanKoughnett I don't think so, but it's hard to think of an example
The problem is that it's hard to talk about non hypercomplete topoi in term of topologies. On some level you are leaving out some information (not literally, ofc) because you cannot give an "intrinsic" characterization of which maps are sent to equivalences by the sheafification
Or, at least, if you can I don't know it (and I'd love to hear about it)
Goal: To gain a better understanding of Euclidean space, $\Bbb R^3,$ conformally compactified into a unit ball.
Question: How can I visualise and mathematically describe Euclidean space, $\Bbb R^3,$ conformally compactified into a unit ball? Is my model of it correct?
My attempt:
Start w...
