Symmetric powers, localisation and Frobenius

Question

I am trying to understand the proof of lemma 2.2.18 in Lucas Mann's thesis. Its statement is surprising for me, because it talks about general rings which are not necessarily characteristic $p$, and out of nowhere Frobenius appears in the statement. the statement is this:

Let $A$ be a ring and $C\subset D_{\ge0}(A)$ a subcategory which is stable under cofibers. Let $L:D_{\ge0}(A)\to C$ be left adjoint to the forgetful functor, and satisfying these conditions:

1. if $L(Q)=0$ then $L(Q\otimes M)=0$ for every $M$.
2. if $L(Q)=0$, then for every prime $m$ and every integer $m$, $L(\phi_{p^m}^{*}Q)=0$, where $\phi_{p^m}\colon A\to A/p$ is the Frobenius $x\mapsto x^{p^m}$.

Then the map from $\operatorname{Ring}_C\to\operatorname{Ring}_A$ admits a left adjoint ($\operatorname{Ring}_C$ is the category of $A$-algebras whose underlying "module" lies in $C$).

The proof is based on the fact that if $L(Q)=0$ then $L(\operatorname{Sym}(Q))=0$. For the proof of this fact, it gives a reference to Proposition 12.26 in Scholze's Lectures on Analytic Geometry, where they only give a sketch of the proof and I can't understand it. Can someone explain the proof of this fact in more detail or give a reference?

Answer 1

I want to come back to this, and provide a sketch of the Dold-Puppe argument so people who don't speak German (such as myself) don't have to try and sift through the linked paper to try and find the argument. I'll try and add relevant parts of it on where to fill in details I leave out. If $SP^i$ is the ith derived symmetric power on $\mathcal{D}_{\geq 0}(\mathbb{Z})$, then the key points are as follows:

• We have natural multiplication maps $\alpha_{i,n-i}:SP^i(X)\otimes SP^{n-i}(X)\to SP^n(X)$ coming from multiplication on the symmetric algebra.
• There is also a natural map $\beta_{i,n-i}:SP^n(X) \to SP^{i}\otimes SP^{n-i}(X)$ coming from the transfer homomorphism.
• When we compose $\alpha_{i,n-i}\circ \beta_{i,n-i}$, we get multiplication by the binomial coefficient ${n \choose i}$. (this construction is section 10 of Dold-Puppe)
• By Spectral Algebraic Geometry 25.2.4.1, we have that if $X$ is $m$-connected, for $m>1$, then $SP^n(X)$ is $m+2n-2$-connected.

From this so far, one can already deduce that the Goodwillie derivative of $SP^n$ vanishes for $n>1$, if $n$ is not a prime power. For this, note that we have the greatest common divisor of ${n\choose i}$ for $1\leq i\leq n-1$ is 1 for $n$ not a power of a prime. For each $i$, we can look at ${n\choose i}=:r_i$, and then if we localize to $\mathbb{Z}[\frac{1}{r_i}]$, then $\alpha_{i,n-i}\circ \beta_{i,n-i}$ becomes an equivalence. In particular, when localized here, $SP^n(X)\simeq SP^{i}(X)\otimes SP^{n-i}(X)$, which is $(m+2i-2)+(m+2(n-i)-2)$-connected, so in particular, $\Omega^{m}SP^n(\Sigma^m(\mathbb{Z}))$ is at least $m+2n-4$-connected for $m>1$, at least after inverting $r_i$, but since the $r_i$ for $1\leq i\leq n-1$ have gcd 1, then we must have that $\Omega^{m}SP^n(\Sigma^m(\mathbb{Z}))$ is at least $m+2n-4$-connected for $m>1$ already, and this shows the vanishing of the Goodwillie derivative.

What goes wrong when $n$ is a prime power, $p^j$, is that the gcd of the $r_i$ is now equal to $p$, so by the same proof as above, the Goodwillie derivative must be p-power torsion. It remains to show that the Goodwillie derivative is suitably finite. For this, we can use techniques i.e., from Lemma 2.9 of this paper, note (in Raskin's notation) that $B_{SP^n}(X,X)=\oplus_{i=1}^{n-1}SP^i(X)\otimes SP^{n-i}(X)$, so that we have an exact triangle $B_{SP^n}(X,X)[1]\to SP^n(X)[1]\to fil_2SP^n(\Sigma X)$. It is not hard to show that in fact, $gr_k(\Sigma X)=\bigoplus_{i_1,...,i_k>0, i_1+...+i_k=n}SP^{i_1}(X)\otimes\ldots \otimes SP^{i_k}(X)[k]$. In particular, if $X$ is $m$-connected $m>1$, the $k$th graded piece is $km+2n-2k+k$-connected, so that, $fil_2 SP^{n}(\Sigma X)\to SP^n(\Sigma X)$ induces an equivalence on homology in degrees $\leq 3m-3$ (one can do better but this is all we need). So, we have exact sequences $H_i(B_{SP^n}(X))\to H_i(SP^n(X))\to H_{i+1}(SP^n(\Sigma X)) \to H_{i-1}(B_{SP^n}(X))$ (see also Dold-Puppe 12.6), for $i\leq 3m-4$. Finally, we use that $B_{SP^n}(X)$ is at least $2m+2n-4$-connected, to see that (for $m>>0$), $H_i(SP^n(X))\to H_{i+1}(SP^{n}(\Sigma X))$ is an isomorphism for $i\leq 2m-4$, so that in particular, for $m>>0$, $H_i(\Omega^{m}SP^{n}(\Sigma^m\mathbb{Z}))\to H_i(\Omega^{m+1}SP^{n}(\Sigma^{m+1}\mathbb{Z}))$ is an isomorphism for $i\leq m-4$, and it is not hard to show the desired finite generation of these homology groups (by looking at iterative bar constructions), so when we pass to the sequential colimit, since the homology groups stabilize at a finite level, they must be finitely generated.