How does one rigorously lift a map $Sp \rightarrow Sp$ of spectra to equivariant spectra?

Question

This is in part motivated from my attempt to understand tate diagonal in III.1 of Thomas Nikolaus, Peter Scholze, On topological cyclic homology, arXiv:1707.01799. I just want to make my understanding precise.

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Particular goal:

How the map (III.1) $T_p: Sp \rightarrow Sp$ $$ X \mapsto (X \otimes \cdots \otimes X)^{tC_p}$$ is defined rigorously. What I could define: I could define a map $$ Sp \rightarrow Sp^{\times n} \rightarrow Sp$$ $$ X \mapsto (X,\ldots, X) \mapsto X \otimes \cdots \otimes X $$ using monoidal structure of $Sp^\otimes$ of spectra.

Question How do I rigorously lift this to map $Sp^{BC_p}$? ( allowing me to apply the Tate functor $(-)^{tC_p} : Sp^{BC_p} \rightarrow Sp$. )

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EDIT: Most of my question of this goal have been resolved in the replies below (of which all are nice answers). I still have the following confusion

how does one prove formulas fo the underlying(under the notation of Maxime) the adjunction: $$ U:Sp^{BG} \rightarrow Sp:Ind, CoInd $$ of "forgetful"/"inclusion"? where Ind and CoInd are left and right adjoint respectively. i.e. It seems that $$ \bigoplus_g X \simeq UInd X $$ $$ U CoInd X \simeq ?? $$

In particular I am confused about the computation $CoInd(Sp) \simeq Sp^{\times n}$.

Answer 1

Let $C$ be a complete $\infty$-category.

Let $U:Fun(BC_n,C)\to C$ denote the forgetful functor, $\mathrm{CoInd}$ its right adjoint, and $(-)^{triv}$ the functor given by precomposition along $BC_n\to *$.

Then we have a canonical equivalence $U(X^{triv})\to X$ which yields, by adjunction, a map $X^{triv}\to \mathrm{CoInd}(X)$ which is $C_n$-equivariant.

Apply this to $C= Cat_\infty$ and $X=Sp$ yields a $C_n$-equivariant map $Sp \to \mathrm{CoInd}(Sp)$. Now $\mathrm{CoInd}(Sp) = Sp^{\times n}$ with the permutation action.

Now $Sp$ can be canonically seen as a commutative monoid in $Cat_\infty$, that is, a certain type of functor $Fin_*\to Cat_\infty$, which we can then obviously restrict to $Fin$ to get $Fin\to Cat_\infty$, informally given by $n\mapsto Sp^{\times n}$.

In particular, we get a $\Sigma_n$-equivariant map $Sp^{\times n}\to Sp$ corresponding to the smash product, and the action of $\Sigma_n$ on $Sp^{\times n}$ restricts to the permutation action of $C_n$

You can prove this by dealing with the universal case.

Another way to do that, which certainly agrees, is to note that in $CAlg(Cat_\infty)$, products and coproducts agree, i.e. it is preadditive and so induction and co-induction agree. In particular you get for free a $C_n$-equivariant map $\mathrm{CoInd}(Sp)\to Sp$ (from $Sp\to U(Sp^{triv})$) which is also given by smash product.

Anyways, it follows that both $Sp\to Sp^{\times n}$ and $Sp^{\times n}\to Sp$ are $C_n$-equivariant

Your left Kan extension construction will not work. Left Kan extending along $*\to BG$ is left adjoint to the forgetful functor, i.e. it is induction - when composed with the forgetful map, this looks like $\bigoplus_{g\in G}$, so if you left Kan extend $X\mapsto X\otimes ... \otimes X$, you will get $\bigoplus_{g\in C_p}X\otimes... \otimes X$, and no permutation action.

As Harry already pointed out, the answer to your side question is "yes", the inclusion has both a left and a right adjoint, in particular it preserves limits and colimits.

Answer 2

I actually worked this out a few months ago (with a hint from Denis Nardin) and wrote this in a message to a friend of mine:

Consider SymmMonCat as a symmetric monoidal category with the cocartesian monoidal structure. I filled in the details for my own sake: SymmMonCat itself has a symmetric monoidal structure, the cocartesian symmetric monoidal structure. Every symmetric monoidal category is canonically an algebra in this symmetric monoidal category wrt the coproduct. Choose our object C as a functor Δ^0→SymmMonCat. Since SymmMonCat has finite coproducts, the functor extends to a functor FinSet→SymmMonCat . Let BΣ_n×Δ^1→FinSet be the map from yesterday. Then composing BΣ_n×Δ^1→FinSet→SymmMonCat, we get an arrow Δ^1→SymmMonCat^{BΣ_n} classifying the equivariant fold map. Finally, to obtain the arrow we wanted, take the composite with the limit functor SymmMonCat^BΣ_n→SymmMonCat. This gives a map Δ^1→SymmMonCat classifying the map C→C^{BΣ_n} that does what we wanted. That's the excruciating detail. The short of it is consider the fold map for an object C in SymmMonCat, which is BΣ_n-equivariant. Then take homotopy fixed points levelwise.

Getting from here to the cyclic version is the obvious thing (consider the canonical permutation representation to get your map by restriction). I don't mind elaborating more, but I don't have time right this second to flesh this out.

Edit: Oh, also the mysterious map $BΣ_n\times \Delta^1\to FinSet$ is the map corresponding to the map sending $0$ to $\langle n \rangle$ with the obvious $\Sigma_n$-action and 1 to $\langle 1 \rangle$ with the trivial $\Sigma_n$-action. It's the map that canonically factors through the cone $B\Sigma_n \star \Delta^0\to FinSet$.

Also, yes, the inclusion of ∞-groupoids in ∞-categories has both a left and right adjoint (the left adjoint is 'invert everything' and the right adjoint is taking the core ∞-groupoid).

Answer 3

There are several ways to do this, depending on how much technology one is interested in using.

One way to do it is to use the fact that the $\infty$-category of commutative monoid objects in a category with finite products, $\mathsf{CMon}(\mathcal{C})$, can be computed as $\mathsf{Fun}^{\times}(\mathsf{Span}(\mathrm{Fin})^{op}, \mathcal{C})$- i.e. the $\infty$-category of product-preserving presheaves on the $(2,1)$-category of spans of finite sets, with values in $\mathcal{C}$. (See, e.g., Theorem 6.5 in the paper of Nardin https://arxiv.org/pdf/1608.07704.pdf for a proof in a more general context; essentially the proof is by right Kan extending from the restriction to finite pointed sets.) Now, $\mathrm{Map}_{\mathsf{Span}(\mathrm{Fin})}(\bullet, \bullet)$ is then the groupoid of finite sets, and in particular receives a map $\mathrm{B}\Sigma_n \to \mathrm{Map}_{\mathsf{Span}(\mathrm{Fin})}(\bullet, \bullet)$. Now take $\mathcal{C}=\mathsf{Cat}_{\infty}$ and consider the functor $J \mapsto \mathsf{Sp}^{\times J}$ given by the symmetric monoidal structure on $\mathsf{Sp}$. Composing we get: $\mathrm{B}\Sigma_n \to \mathrm{Map}_{\mathsf{Span}(\mathrm{Fin})}(\bullet, \bullet) \to \mathrm{Map}_{\mathsf{Cat}_{\infty}}(\mathsf{Sp}, \mathsf{Sp})$. That's the same as a functor $\mathsf{Sp} \to \mathsf{Sp}^{\mathrm{B}\Sigma_n} = \mathsf{Sp}^{h\Sigma_n}$, which is what you're after (you can take $n=p$ and restrict to $C_p$ if you want).

But maybe you don't want to use that fact about spans. That's fine. You can follow the approach from the beginning of section 2.2 in DAGXIII (https://www.math.ias.edu/~lurie/papers/DAG-XIII.pdf). The point is this: if $\mathcal{D}$ is a symmetric monoidal $\infty$-category, i.e. a commutative monoid/algebra object in $\mathsf{Cat}_{\infty}$, then we automatically get a functor $\mathrm{Sym}(\mathcal{D}) \to \mathcal{D}$ from the free commutative algebra object. The free commutative algebra object is computed as $\coprod \mathcal{D}^{\times n}_{h\Sigma_n}$. Restricting to the $n$th summand gives a map $\mathcal{D}^{\times n}_{h\Sigma_n} \to \mathcal{D}$ which refines the functor $(X_1, ..., X_n) \mapsto X_1\otimes \cdots \otimes X_n$. We may view it just as well as a functor $\mathcal{D}^{\times n} \to \mathcal{D}$ in $\mathsf{Fun}(\mathrm{B}\Sigma_n, \mathsf{Cat}_{\infty})$ (since taking colimits is adjoint to the constant diagram functor). On the other hand, we also have a diagonal map $\mathcal{D} \to \mathcal{D}^{\times n}$ which is obtained by applying $\mathrm{Fun}(-, \mathcal{D})$ to the $\Sigma_n$-equivariant map of sets $\{1, ..., n\} \to \bullet$, and hence is $\Sigma_n$-equivariant. Composing gives $\mathcal{D} \to \mathcal{D}^{\times n} \to \mathcal{D}$ with a $\Sigma_n$-equivariant structure that does what you want.

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Also:

• This functor $X \mapsto (X^{\otimes p})^{tC_p}$ is not 'the Tate diagonal'. The Tate diagonal is a natural transformation (the unique lax symmetric monoidal natural transformation) $X \to (X^{\otimes p})^{tC_p}$.