@AaronMazel-Gee That localization is also just $-∧Σ^∞\widetilde{EP}$. I have been writing a short note on how you do the standard G-spectra constructions in spectral Mackey functors. I might put it online at some point in the near future

Correct me if I am wrong: In G-spectra, there are two fibre sequences, one is called Tate diagram, which we use EG+ -> S^0 and X -> F(EG+, X), another one is called isotropy separation sequence, using EP+ -> S^0, where EP is the classifying space of all proper subgroups of G. The later one is related to geometric fixed point. When G is cyclic of order p, we get the same thing, but for other G, these two sequences shall behave very very differently right?

Those are only two of a family of sequences

Every time you have a family of subgroups F closed under conjugation and subsets you can form a G-space EF such that (EF)^H is contractible if H∊F and empty otherwise

Then you can form the cofiber $EF_+→S^0→\widetilde{EF}$

Yeah of course

Then $-\wedge \Sigma^∞\widetilde{EF}$ is a smashing localization onto the G-spectra that are supported away from F

(that is those G-spectra such that E^H=0 if H∊F)

This recovers both of your cases although if G is not cyclic of order p the two localizations are different

(You also get F(EF_+,-) that is the complementary completion although I think it is mostly useful in the case F={e})

For example, if H is a normal subgroup you can take F to be the family of subgroups not containing H. Then the category of spectra supported away from F is equivalent to G/H-spectra and the localization is just geometric fixed point at H

(I chose H to be normal for simplicity, there is also a similar answer for H non normal but the category where the "geometric fixed point" have value is not G/H-spectra of course)

i need to think about it more, but the smashing localization description seems quite useful

Thank you!

And another question, what's the reason behind the name "Tate spectrum"? Is it something like, homotopy groups of Tate spectrum of HZ, is Tate cohomology of G?

exactly - if you have a G-module M, then HM^{tG} is equivalent to the classical Tate cohomology of G with coefficients in M

It might help to notice that (-)^{tG} is a lax symmetric monoidal functor, so if M is a R[G]-module, (HM)^{tG} is a HR-module (and as Saul says it is precisely the one corresponding to the chain complex computing Tate cohomology)

To the Barwick clan: what's the universal property of A^eff(C)? I'm hoping for something along the lines of Lurie's HA.5.2.1.18 for the twisted arrow category

If I have a G-spectrum, how can I test whether it's compact? The two naive guesses (compact on all fixed pts or compact on all geometric fixed pts) both fail, I think.

that's a good question

in the case of G = C_p, it's necessary and sufficient for the genuine e-fixed points and the C_p-geometric fixed points to be compact

oh, that's the same as all geometric fixed points being compact - just woke up over here

are you sure it's not equivalent to all geometric fixed points being compact?

So $G$-spectra is rigidly-compactly generated, so compact is the same as dualizable

Unfortunately it is not true that if all geometric fixed points are dualizable, the original G-spectrum is dualizable

(This is because the map $F(X,Y)^{\phi H}→F(X^{\phi H},Y^{\phi H})$ is not an equivalence)

Denis, do you have a counterexample to hand?

Σ^∞_+EG should do the trick

ah yes

Basically you want some additional condition on the map $E^{\phi C_p}→E^{tC_p}$ (to get invertible instead of dualizable it is some kind of "non-degeneracy" condition, I expect it to be pretty much the same in the dualizable case)

right, I remember thinking about this at AIM

I wonder if there's something more checkable

This is a question that needs to be answered. Somehow I was unaware that I didn't know the answer to this question

I guess it's true that for a finite C_p-spectrum E, $E^{\phi C_p} \to E^{tC_p}$ is p-completion

but it's not clear to me that that's sufficient, and it's a pretty unsatisfying condition anyway

Yeah I was hoping to avoid the Segal conjecture :) and \tilde{EC_p} has compact geometric fixed points for all subgroups but isn't compact. (Since EC_{p+} isn't compact on trivial geometric fixed points, as a local system!, that one doesn't provide a counterexample.)

what do you mean by as a local system?

Well: the value of 'trivial geometric fixed points' lives in Fun(BC_p, Sp) most naturally. I could further forget down to spectra, but compact means different things in both places, I think. In particular EC_{p+}, viewed as the constant local system with value S^0, isn't compact in Fun(BC_{p+}, Sp) is it? Whereas the 'underlying spectrum' is S^0 and that's compact.

@Dylan That is true, but it is still dualizable which I believe is the easier condition to check

fair

wait, Denis, responding to an earlier comment: if I have a dualizable object, then it should be invertible off it is so on geometric fixed points, because I just need to check evaluation and coevaluation are equivalences and geometric fixed points are monoidal

Oh right, that works

(off= iff, above)

so what's the non-degeneracy condition you mentioned?

If I recall correctly, it should be that E^{phi C_p}→E^{tC_p} is an invertible element of π_*(E^{tC_p})

I'll see if I can find the computations I did back when I was thinking about this

but doesn't EC_{p+} have e-geometric fixed points zero?

No, it has e-geometric fixed point S^0 (the unit in local systems) that happens not to be compact

oh right, wrong way round

its C_p-geometric fixed points are zero

Yes, I think I can prove that the condition is $E^{phi C_p}→E^{tC_p}$ is an invertible element of π_*(E^{tC_p})

I was thinking of E~C_p

how do we regard that map as an element of pi_*?

E^{phi C_p} = S^n for some n

(more generally you might want the Picard-graded homotopy groups)

ah, this is the invertible case

The proof goes like this: suppose E is invertible with inverse F. Then E∧F=S^0. Without loss of generality let us assume that E and F have underlying spectra S^0 (it is just a suspension). Then the gluing map in E∧F is E^{phi C_p}∧F^{phi C_p}→E^{tC_p}∧F^{tC_p}→(E∧F)^{tC_p}=(S^0)^{tC_p}, that is the product of the gluing maps of E and F. Since the gluing map of S^0 is the 1 we are done.

Viceversa, if E^{phi C_p}→(S^0)^{tC_p} is invertible pick the element of π_*((S^0)^{tC_p}) corresponding to the inverse and use it as a gluing map for $(E^{phi C_p})^\vee→(S^0)^{tC_p}$. Then the previous computation yields that the resulting spectrum is an inverse for E

I gues this tells us what to do next: take a duality datum in C_p spectra and apply geometric and Tate constructions to all the maps in the duality datum to see what condition it says on the gluing map

What flavor of monoidal is the tate construction?

It is lax symmetric monoidal

One of the reasons I am skeptical that there is going to be a nice answer for a general G, is that it would allow us immediately to compute Pic(Sp^G), which is known but the description is not that nice

well then one part of your argument is suspicious: how do you get the gluing map of a smash product to factor thru the smash product of tate spectra?

What do you mean "immediately"? Checking that some element of the Tate construction is invertible doesn't seem so "immediate" to me... you'd have to compute the gluing map as an element in the homotopy, which sounds annoying

in practice

If E, F are C_p-spectra then the gluing datum of E∧F is the composition $(E∧F)^{\phi C_p} = E^{\phi C_p}∧F^{\phi C_p}→E^{tC_p}∧F^{tC_p}→(E∧F)^{tC_p}$. And ok, maybe I overstated the "immediately" but it still seems too easy

By the way, one way of seeing the factorization I asserted above is to check that the map Sp^G→Sp^{hG} sending every G-spectrum to the underlying spectrum is symmetric monoidal. Exactly how to prove this now depends on your model for G-spectra, but it is true

Suppose $A\to B$ is a map of commutative ring spectra such that (i) $B\otimes_A B\to B$ is an equivalence, and (ii) $B$ is dualizable as an $A$-module. Does it follow that $B$ is a retract of $A$ as an $A$-module?

@CharlesRezk Unless I'm mistaken (i) implies that $-⊗_AB$ is a smashing localization on $A$-modules and (ii) implies that it is conservative . Doesn't this imply that $A→B$ is an equivalence?

Hi all! Is there a way to "detect" existence of simply-connected 4-mflds with a (free) action of a discrete group on them?

Isn't the 0-module dualizable?

For ordinary rings, dualizable modules are those which are finitely presented flat (=finitely generated projective).

Hmm.. the 0 module is definitely dualizable, I don't know where I got that impression, that is clearly bogus. For example the inclusion of a summand in the product of two rings is an easy counterexample to $A\to B$ being an equivalence...

In this ordinary case, the local dimension function $d:Spec A\to Z_{\geq0}$ of $B$ will be continuous, taking values either $0$ or $1$. So I think $B$ turns out to be equal the localization of $A$ to the open set on which $B$ is supported. So you actually get $A=B\times B'$ as rings.

@LuigiM What do you mean by "detect"?

@CharlesRezk prove that they exists/ cannot exists. My line o thoughts was something like: ok by Freedman theorem we can build simply-connected 4-manifolds with given intersection form, if we choose carefully the int. form we can even claim that the mfld is smooth and spin for instance. The classification i up to homeo so I was curious whether we can detect something more, i.e. the presence of a free action of a group G on them

Are you asking whether the classification of simply connected 4-manifolds extends to an equivariant classification?

Since you're demanding the action be free, you're just asking about the classification of 4-manifolds with given fundamental group, which can be done for certain finite groups

@MikeMiller and then we take the universal cover. I see. we should even be able to control spin-behaviour

maybe I was a little bit bold about control spin-behaviour, but for sure it is a starting point

@DenisNardin So if the map of commutative ring spectra $A\to B$ is as I assert, then $B\otimes_A-$ is not merely colimit preserving, but also limit preserving (on $A$-modules). So if $\mathcal{C}$ be the class of $A$-modules $M$ such that $B\otimes_A M\approx 0$, then $\mathcal{C}$ is a full stable subcategory closed under limits ...

So you expect a left adjoint to the inclusion of $\mathcal{C}$ into $A$-modules, which presumably is another smashing localization $A\to B'$. One might hope to show that $A\to B\times B'$ is an equivalence ...

Well, I don't know why it would be a smashing localization ...

Duh, $\mathcal{C}$ is also closed under colimits, so it would be smashing.

@CharlesRezk I don't know if it helps, but you could apply 2.30 of arxiv.org/pdf/1507.06869.pdf, which (I think?) implies that B \simeq L_BA, the latter term being the Bousfield localization of A at B (in the category of A-modules), if I understand correctly

I'm aware that $B=L_B(A)$ (a fact which probably goes back to Bousfield's original paper).

Ah OK. What is a less hi-tech way to see that?

We can think of a fiber bundle F->E->B like a short exact sequence, and in this analogy Spin(7) has "composition factors" S^k for various k in more than one way, but they reduce to the same thing if we break apart S^7 and S^3 as Hopf bundles. Is there any kind of theory of "composition factors" like this?

On MSE I asked if there is something analogous to Jordan-Holder when a bundle has spheres as factors. I don't have the background to say if my question is silly or not.

@DenisNardin right, thanks

so in the notation $triv = i : Fin \rightleftarrows GFin : j = (-)^G$ i used above, taking Burnside and then Mackey gives $j^* : Sp \rightleftarrows Sp^G : i^* = (-)^G$ -- the right adjoint is genuine fixedpoints. can this also be computed as the right Kan extension $j_*$, or do i need to worry about additivity?

i'm confused, because the composite $Sp^{hG} := Fun(BG,Sp) \xrightarrow{Borel} Sp^G \xrightarrow{(-)^G} Sp$ should just be $(-)^{hG}$. but this source can be identified as Mackey functors on Burnside of the full subcategory $GFin^{free} \subset GFin$ on those of the form $\coprod_i G/e$, and its left adjoint ("forget the genuine structure") is just restriction along this inclusion. so the functor "Borel" is (additive?) RKE along this inclusion

so i've identified both arrows as RKE's, meaning i can compute their composite as a single RKE. but then the composite $GFin^{free} \hookrightarrow GFin \xrightarrow{(-)^G} Fin$ is constant at the empty set! so on Burnsides, the functor is constant at the zero object of $Burn(Fin)$

...meaning that the RKE is constant, at the limit of a certain functor $Burn(GFin^{free}) \to Spectra$. but that source has a zero object too, so the limit is the value there, namely the spectrum 0.

so......something must be wrong here.