12:05 AM
What is $k[x]_{f}$ here? Do you mean $k[x, x^{-1}]$?

@PiotrPstrągowski Yeah I think you're right, that's a typo from when I wasn't writing the base change, yep thanks

The key here is that a base-change of a flat morphism is flat, and $k[x, x^{-1}]$ is flat over $k[x]$. In this case you can see it explicitly - $k[x, x^{-1}]$ is a filtered colimit of $k[x]$ along the diagram of multiplications by powers of $x$. Since (even derived) tensor products commute with filtered colimits, you see that your localization is a filtered colimit of $R$ along maps induced by multiplication by powers of $f$.
Taking homotopy groups also commutes with filtered colimits, so you see that the homotopy groups of the localization will be the localization of homotopy groups (in the sense of classical algebra).

17 hours later…
5:23 PM
X is a spectrum with action of a loop space G. What conditions are needed for a statement like, "The G-action on X is trivial if and only if X^BG+ \to X_{hG} is an equivalence"?

5:48 PM
@JohnBerman can you say something about the source to the question? I can not imagine a single case where these properties are correlated.
But the problem might be with my imagination of course

Suppose the G-action is trivial, then X^{BG} \simeq X^{hG}, so this seems to be asking also that the Tate construction vanishes? (up to some shift, I guess)
(I guess using X^{BG} and X^{hG} is a bad choice of notation, but I think you understand)

Oh, I see the confusion. I meant X\otimes\Sigma^\infty BG_{+}=X[BG].
Certainly if the G-action is trivial, then X[BG]\to X_{hG} is an equivalence. To what extent is the converse true?
(I am getting used to trying to appease more classical algebraic topologists -- probably not necessary here.)

@JohnBerman I'm confused. What's the map X[BG] -> X_hG?

X -> X_hG is compatible with trivial G-actions, so it induces X[BG] -> X_hG

Ah, I see.
Actually, I'm still confused. When G acts trivially, won't this give you the composite X[BG] -transfer> X -inclusion> X[BG]?

6:07 PM
@WilliamBalderrama yes its not an iso for the trivial action usually I think.

Okay now you guys are making me nervous. Let's look at X=*. The map * -> BG exhibits BG as a colimit of G acting on . That must mean that the induced map BG=_hG -> BG is an equivalence, right?
oops * became italics there. I meant BG=pt_hG -> BG

Okay so.... just to clarify notation, in your first comment you're writing $X\wedge BG_+\simeq X_{hG}$ where the second is the homotopy orbits?

Yes, in the first comment I was writing a smash product, but in hindsight it was confusing.

And certainly if the action is trivial then this is true. That's an interesting question though, when is the converse true...

6:23 PM
I think the issue is in saying "the map". A map pt -> BG in (say) Fun(BG,Gpd_infty) is by adjunction the same as a map BG -> BG in Gpd_infty. There are at least two maps: the equivalence and the composite BG -> pt -> BG. I don't see how to define a map X[BG] -> X_hG in general that picks out the equivalence when G acts trivially.

@WilliamBalderrama the map pt.→BG is the map of constant diagrams?
I think I'm confused about that statement. I'm used to thinking about the colimit/diagonal adjunction between the functor category and $Gpd_\infty$. Is that the adjunction you're referring to?

@JonathanBeardsley Yes, we're referring to the same thing.

I might be mixing something up here, but is it not true that a map of constant diagrams $p^\ast(pt.)\to p^\ast BG\in Fun(BG,Gpd_\infty)$ corresponds to a map $BG\to BG\times BG$ under that adjunction?
Where I'm using $p^\ast$ to denote the diagonal functor.
Or... no perhaps I'm screwing that up.

Doesn't p^\ast(pt) -> p^\ast(BG) corresponds to BG = p_! p^\ast(pt) -> BG?

Right sorry, I think I was just applying $p_!$ to that map.
Okay so I think we can say really precisely what that map is right?
Like the adjoint is pretty explicit.
Like it's something like $p_!p^\ast(pt.)\to p_!p^\ast(BG)\to BG$, where that second map is the counit?
Again, very possible I'm messing this up.
So I think it should be some map $BG\to BG\times BG\to BG$?

6:35 PM
Yeah, in this case there is a map p^\ast(pt) -> p^\ast(BG) which is adjoint to an equivalence BG -> BG. My issue is that I don't see how to construct a map X \times BG -> X_hG in general such that: (1) this map is natural in G-spaces X (2) when G acts trivially on X, this map is an equivalence.

Why do we expect there to be such a map?

I don't expect there to be such a map.

Ah ok. Me neither.

That's why I don't get the original question.

I see, but this is implicitly assumed by John's question.
I guess we could still ask the question: is it possible for $X_{hG}$ to be equivalent to $X\wedge BG_+$ when the $G$ action is non-trivial, right?
(I feel like.... no?)

6:39 PM
So there are two constructions that are relevant here:
1. A functor BG->C extends to a colimit diagram colim(F):BG^\triangleright -> C, which looks like X\to X_hG.
2. Given any map f:X -> Y, there is a composite BG^\triangleright -> \Delta^1 -> C which exhibits the compatibility of the map with the trivial G-action on X.

My claim is that construction (1) applied to X with trivial G-action produces the same functor as construction (2) applied to X->X_hG

I see. So the latter functor takes all of BG to {0} and the cone point to {1}?

Proof: The functor BG^\triangleright -> C of construction (1) factors through \Delta^1 -> C since BG->C factors through pt->C. So both (1) and (2) are of the form BG^\triangleright -> \Delta^1 -> C, where \Delta^1 -> C is the same map X -> X_hG in either case.
@JonathanBeardsley That's right. It's also the functor induced by BG->pt by taking cones on both.

Right, so the map $f:X\to Y$ actually produces a map of cones
A natural transformation in $Fun(BG^{\triangleright},Sp)$.

I think @WilliamBalderrama is saying I am wrong that the colimit diagram BG^\triangleright -> C factors through \Delta^1 = pt^\triangleright. Let me think about that...

I think I'm with William here at the moment: for Y with trivial G-action, the construction taking a map X --> Y to a map X_{hG} --> Y is natural in Y. In particular, if X has trivial action, all such maps factor through the one obtained by considering the identity X --> X, which is the collapse X[BG] --> X.

6:48 PM
I think I'm just a bit dense, but the statement "exhibits the compatibility of the map with the trivial G-action on X" is kind of opaque to me.

That's an excellent point @Arpon ... I see you all are right

Re the original question, in case it's relevant: I think that if G is connected and Y,Z are bounded below spectra with G-action, then a G-equivariant map Y --> Z is an equivalence if the induced map on orbits is.

@Arpon Right, but I don't know any map whose induced map on orbits relates X_hG and X[BG].

yep agreed :/

Is it all relevant to look at what ABGHR do to prove that $X_{hg}$ is equivalent to $X[BG]$?

6:57 PM
So here's the actual situation: I have an HZ-module X with S^1-action. I know that X\otimes_Z Q=0, and the S^1-action on X\otimes_HZ HFp is trivial for each p. I would like to conclude that the S^1-action on X is trivial, maybe with a mild condition on X.

Oh yeah... this makes me think of that Blumberg stuff we chatted about a while back... But if the answer's not in there...
:.(
I don't really know how to make a crying emoticon.

Lol, angry squinty person.

7:10 PM
@RuneHaugseng thanks Rune and @BryanShih for these suggestions. These are also the two books I was considering (though I had not heard of Leinster's book until Chris Rogers mentioned it to me). I think I will probably end up going with Leinster's book because (1) it is more basic than Riehl's book, and I think that will be better for my students, and (2) I haven't really found much else which is accessible and readable.

2 hours later…
9:34 PM
@JonathanBeardsley In the "applied CT" direction there's also this book by Fong and Spivak arxiv.org/abs/1803.05316 which starts pretty basic
(Though I feel like there's so much pressure these days for everything we do at universities to be "applied" in some sense that I'm not sure it's something we should encourage further...)

@RuneHaugseng heh, yeah i'd be pretty nervous about going in the applied direction
Especially since I know next to nothing about it.
I think the direction we're going to end up going, to make it at least a little relevant to the low-dimensional topologists' students who might take the course, is a primer in very basic category theory capped off by a discussion of (1+1)-TQFTs and their characterization by Frobenius algebras.
(i.e. the goal would be to get to monoidal categories and then apply them)
Speaking of which, anyone have a good elementary reference for the TQFT/Frobenius algebra stuff?

I guess the standard answer is Joachim Kock's book

9:52 PM
@RuneHaugseng oh right.... thanks. forgot about that. that's another one Chris Rogers suggested to me.

10:18 PM
@JohnBerman what about the following example? we have the map Z/p^2-p->Z/p^2 which we may think of as a circle action on Z/p^2 (+) Z/p^2[1]. Isn't it becoming trivial after tensoring with F_p? (I think it should, because the map p become 0 after tensoring with F_p, but I didn't check very carefully that the next cell, witnessing d^2=0, is not destroyed. But I think it shouldn't).

1 hour later…
11:44 PM
the group algebra Hℤ[S^1] is equivalent to H(ℤ[σ]/σ^2), and so you can definitely get examples by looking at differential graded modules over the latter ring. e.g. ℤ ⊕ ℤ[1] with σ acting by p is an example that base extends to the trivial circle action over 𝔽_p, but isn't rationally trivial; but you can tensor it with ℤ/p^2 and get @S.carmeli's construction, and this shows that it does become the trivial module after base extension
(because if 𝔽_p ⊗ A has trivial action, then so does 𝔽_p ⊗ A ⊗ ℤ/p^2)

Good point! Can we check that the circle action is trivial by tensoring up to Z/p^n for all p?

All p, or all n?

Oh I meant all p and all n
So I guess if the circle action is trivial on \otimes_Z Z/p^n for all n, does that imply it is trivial on \otimes_Z Z_p?

Well, a circle action on M is a map of algebras ℤ[S^1] → End(M). let's say M is dualizable? then End(M) has an arithmetic square; and if M is rationally trivial, then End(M) is the product of End(M^_p)
Let's see. If you're dualizable, then End(M^_p) is the limit of ℤ/p^n ⊗ End(M) = End_{ℤ/p^n}(M/p^n).
The algebra ℤ[σ]/(σ^2) is almost finitely presented and End(M) will be trivial above some finite stage, so yeah, I think that works.

Right, this should work out... In my case, it is sort of easy to see that the S^1-action is trivial on Z/p and also true but not as clear on Z/p^n, so I have been trying to cut corners! Apparently the corners are not cuttable.
But thanks!

11:55 PM
For a general M it's a little scarier because you might run into lim^1 terms but a specific case might be analyzable