Construction 3.3. We consider the category consisting of pairs $(B,\unicode[STIX]{x1D6E4})$ where $B$ is a commutative ring and $\unicode[STIX]{x1D6E4}$ is a formal group over  $B$ . A morphism $(B,\unicode[STIX]{x1D6E4})\rightarrow (B^{\prime },\unicode[STIX]{x1D6E4}^{\prime })$ consists of a morphism of commutative rings $f:B\rightarrow B^{\prime }$ and an isomorphism of formal groups $f_{\ast }\unicode[STIX]{x1D6E4}\simeq \unicode[STIX]{x1D6E4}^{\prime }$ . This category is cofibered in groupoids over the category of commutative rings, and is encoded in the moduli stack of formal groups  $M_{FG}$ : that is, it is (by definition) the category of maps $\operatorname{Spec}B\rightarrow M_{FG}$ from affine schemes to $M_{FG}$ .

Proof. Let $B^{\prime }$ be the $B$ -algebra which classifies coordinates modulo degrees at least $n+1$ on the formal group  $\unicode[STIX]{x1D6E4}$ . Abstractly, $B^{\prime }\simeq B[x_{1}^{\pm 1},x_{2},\ldots ,x_{n}]$ . The $G$ -action on $(B,\unicode[STIX]{x1D6E4})$ induces a $G$ -action on $(B^{\prime },\unicode[STIX]{x1D6E4}^{\prime })$ , where $\unicode[STIX]{x1D6E4}^{\prime }$ is the extension of scalars of $\unicode[STIX]{x1D6E4}$ to  $B^{\prime }$ .

We need to show that $G$ acts freely on $\operatorname{Hom}(B^{\prime },k)$ for any field  $k$ . Let $\unicode[STIX]{x1D70E}\in G\setminus \{1\}$ fix an element of $\operatorname{Hom}(B^{\prime },k)$ given by a map $\unicode[STIX]{x1D719}:B^{\prime }\rightarrow k$ . We get that $\unicode[STIX]{x1D70E}$ fixes the restriction $\unicode[STIX]{x1D719}|_{B}:B\rightarrow k$ , which means that the prime ideal $\mathfrak{p}=\text{ker}(\unicode[STIX]{x1D719}|_{B})$ is fixed by $\unicode[STIX]{x1D70E}$ , and that $\unicode[STIX]{x1D70E}$ acts trivially on the residue field $k(\mathfrak{p})$ , so that $\unicode[STIX]{x1D70E}\in I_{\mathfrak{p}}$ . Note that the fiber above $\unicode[STIX]{x1D719}|_{B}$ of the map (which is preserved by  $\unicode[STIX]{x1D70E}$ )

(3.1) $$\begin{eqnarray}\operatorname{Hom}(B^{\prime },k)\rightarrow \operatorname{Hom}(B,k)\end{eqnarray}$$

is given by the set of coordinates to degree $n$ on the formal group $(\unicode[STIX]{x1D719}|_{B})_{\ast }(\unicode[STIX]{x1D6E4})$ , and the action of $\unicode[STIX]{x1D70E}$ on this set of coordinates is given by composing with the image of  $f_{\unicode[STIX]{x1D70E}}$ . However, since the image of $f_{\unicode[STIX]{x1D70E}}\in k[\![X]\!]$ is not the identity, it acts freely on the above set of coordinates up to degree  $n$ . In particular, the action of $\unicode[STIX]{x1D70E}$ on the fiber of (3.1) above $\unicode[STIX]{x1D719}|_{B}$ has no fixed points, which is a contradiction and which completes the proof.◻

Proof. We will apply Proposition 3.4. Fix a prime ideal $\mathfrak{p}\subset R$ and $\unicode[STIX]{x1D70E}\in I_{\mathfrak{p}}$ . The first claim is that $\unicode[STIX]{x1D70E}\in S$ . In fact, this follows from the surjection $R/\mathfrak{p}\rightarrow R/\mathfrak{m}=k$ . Since $\unicode[STIX]{x1D70E}$ acts trivially on $R/\mathfrak{p}$ , it must act trivially on  $k$ .

In applying Proposition 3.4, it suffices to consider the inertia subgroups  $I_{\mathfrak{p}}$ ; since these are contained in  $S$ , we conclude that it suffices to consider the case when $G$ itself acts trivially on the residue field. Choose $n$ large enough such that each power series $f_{\unicode[STIX]{x1D70E}}(X)\in R[\![X]\!]$ for $\unicode[STIX]{x1D70E}\neq 1$ has a coefficient in degrees at most $n$ different modulo $\mathfrak{m}$ from  $X$ . The claim is that $(\operatorname{Spec}R)/G\times _{M_{FG}}M_{FG}^{(n)}$ is representable by an affine scheme. This now follows from Proposition 3.4 because each $f_{\unicode[STIX]{x1D70E}}(X)-X$ for $\unicode[STIX]{x1D70E}\neq 1$ has a coefficient which is a unit in some degree at most $n$ .◻

Proof. By [Reference Mathew and MeierMM15, Theorem 5.8], it suffices to show that $(\operatorname{Spec}\unicode[STIX]{x1D70B}_{0}E_{n})/G\rightarrow M_{FG}$ is an affine morphism of stacks. However, this follows from Lemma 3.5 (with $S=G_{\mathbb{S}}=G\cap \mathbb{S}$ ) because $\mathbb{S}_{n}$ is defined as a subgroup of automorphisms of the Honda formal group over $\mathbb{F}_{p^{n}}$ . The above reasoning shows that there exists $m$ such that $(\operatorname{Spec}\unicode[STIX]{x1D70B}_{0}E_{n})/G\times _{M_{FG}}M_{FG}^{(m)}$ is affine.◻

Proof. It is a straightforward calculation (cf. [Reference Mathew, Naumann and NoelMNN15, Proposition 3.26]) that $f\otimes _{\mathbb{Z}_{p}}\mathbb{F}_{p}$ is also a uniform ${\mathcal{F}}_{p}$ -isomorphism, so that we may assume that $A_{0}$ and $A_{0}^{\prime }$ are $\mathbb{F}_{p}$ -algebras by the universal property of the Witt vectors. It suffices to show that $\operatorname{Hom}(k,A_{0})\simeq \operatorname{Hom}(k,A_{0}^{\prime })$ .

Recall that an $\mathbb{F}_{p}$ -algebra $C_{0}$ is said to be perfect if the Frobenius $F:C_{0}\rightarrow C_{0}$ is an isomorphism. The inclusion of perfect $\mathbb{F}_{p}$ -algebras into all $\mathbb{F}_{p}$ -algebras has a right adjoint, called the inverse limit perfection, which sends an $\mathbb{F}_{p}$ -algebra $C_{0}$ to the inverse limit of the Frobenius system $\cdots \rightarrow C_{0}\stackrel{F}{\rightarrow }C_{0}$ . Since $k$ is perfect, and since the uniform ${\mathcal{F}}_{p}$ -isomorphism condition guarantees that $A_{0}\rightarrow A_{0}^{\prime }$ induces an isomorphism on inverse limit perfections, an adjunction argument now gives the desired claim.◻

Proof. By the previous lemma, it suffices to show that the reduction map $\unicode[STIX]{x1D70B}_{0}(R^{h\tilde{G}})\rightarrow \unicode[STIX]{x1D70B}_{0}(R)^{\tilde{G}}$ is a uniform ${\mathcal{F}}_{p}$ -isomorphism. This follows as in [Reference MathewMat15, § 4] from the fact that the HFPSS for the $\tilde{G}$ -action on $R$ has a horizontal vanishing line at a finite stage by the theory of ‘descendability’ (cf. [Reference MathewMat16, §§ 3–4]) and the fact that everything in positive filtration at $E_{2}$ is annihilated by the $p$ -part of $|\tilde{G}|$ .◻

Proof. Let $\unicode[STIX]{x1D6E4}=G/\tilde{G}$ be the quotient group, so that by the Galois correspondence, the morphism of $\mathbf{E}_{\infty }$ -rings $A=R^{hG}\rightarrow B=R^{h\tilde{G}}$ is a faithful $\unicode[STIX]{x1D6E4}$ -Galois extension. We claim first that it is actually a Galois extension on homotopy groups, i.e.  $A\rightarrow B$ is étale, in the sense that $\unicode[STIX]{x1D70B}_{0}(A)\rightarrow \unicode[STIX]{x1D70B}_{0}(B)$ is étale and the natural map $\unicode[STIX]{x1D70B}_{\ast }A\otimes _{\unicode[STIX]{x1D70B}_{0}A}\unicode[STIX]{x1D70B}_{0}B\rightarrow \unicode[STIX]{x1D70B}_{\ast }B$ is an isomorphism; this will enable us to make the desired conclusion. We refer to [Reference Baker and RichterBR07] for the relationship between Galois extensions on homotopy and Galois extensions in the sense of Rognes [Reference RognesRog08], and to [Reference LurieLur11, § 4] as a convenient reference for some of the facts we need.

In particular, by [Reference LurieLur11, Corollary 4.15], we need to show that $\unicode[STIX]{x1D6E4}$ acts freely on $\unicode[STIX]{x1D70B}_{0}(B)$ . We will do this by functoriality. By the universal property of the ( $p$ -typical) Witt vectors, we obtain a $G$ -equivariant map

$$\begin{eqnarray}\mathbb{W}(k)\rightarrow \unicode[STIX]{x1D70B}_{0}(R),\end{eqnarray}$$

where $G$ acts on $\mathbb{W}(k)$ via the induced action on $k$ . Of course, the $G$ -action on $\mathbb{W}(k)$ factors through $\unicode[STIX]{x1D6E4}=G/\tilde{G}$ , and we claim that this lifts to a $\unicode[STIX]{x1D6E4}$ -equivariant map

$$\begin{eqnarray}\mathbb{W}(k)\rightarrow \unicode[STIX]{x1D70B}_{0}(B)=\unicode[STIX]{x1D70B}_{0}(R^{h\tilde{G}}).\end{eqnarray}$$

The equivariance alone gives the $\unicode[STIX]{x1D6E4}$ -equivariant lift $\mathbb{W}(k)\rightarrow \unicode[STIX]{x1D70B}_{0}(R)^{\tilde{G}}$ , which of course receives a map from $\unicode[STIX]{x1D70B}_{0}(R^{h\tilde{G}})$ . Using Lemma 3.9, we can upgrade it to the claimed $\unicode[STIX]{x1D6E4}$ -equivariant map $\mathbb{W}(k)\rightarrow \unicode[STIX]{x1D70B}_{0}(B)$ .

The $\unicode[STIX]{x1D6E4}$ -action on $W(k)$ is free (Example 3.2). It follows that the $\unicode[STIX]{x1D6E4}$ -action on $\unicode[STIX]{x1D70B}_{0}(B)$ is free. Applying [Reference LurieLur11, Corollary 4.15], we conclude that $A\rightarrow B$ is an algebraic $\unicode[STIX]{x1D6E4}$ -Galois extension in that $\unicode[STIX]{x1D70B}_{0}(A)\rightarrow \unicode[STIX]{x1D70B}_{0}(B)$ is one, and $\unicode[STIX]{x1D70B}_{\ast }(A)\otimes _{\unicode[STIX]{x1D70B}_{0}(A)}\unicode[STIX]{x1D70B}_{0}(B)\rightarrow \unicode[STIX]{x1D70B}_{\ast }(B)$ is an isomorphism.

Finally, we prove the result about the Picard group. Let $M$ be an invertible $A$ -module. By assumption, $M\otimes _{A}B$ is a suspension of a free $B$ -module of rank one; by desuspending $M$ appropriately, we may assume that $M\otimes _{A}B\simeq B$ . We would like to see that $M\simeq A$ . Since $B$ is faithfully flat over  $A$ , we can conclude that $\unicode[STIX]{x1D70B}_{0}(M)$ is an invertible $\unicode[STIX]{x1D70B}_{0}(A)$ -module and that the map $\unicode[STIX]{x1D70B}_{0}(M)\otimes _{\unicode[STIX]{x1D70B}_{0}(A)}\unicode[STIX]{x1D70B}_{\ast }(A)\rightarrow \unicode[STIX]{x1D70B}_{\ast }(M)$ is an isomorphism (because this happens after base-change to $B$ ). In particular, $M$  simply arises from an invertible module over  $\unicode[STIX]{x1D70B}_{0}(A)$ .

Therefore, it suffices to show that the Picard group of the (ordinary) commutative ring $\unicode[STIX]{x1D70B}_{0}(A)$ is trivial. This will follow if we can show that $\unicode[STIX]{x1D70B}_{0}(A)$ is a (possibly non-noetherian) local ring. Since $\unicode[STIX]{x1D70B}_{0}(A)\rightarrow \unicode[STIX]{x1D70B}_{0}(B)$ is finite étale, it suffices to show that $\unicode[STIX]{x1D70B}_{0}(B)$ is a local ring, i.e. that its spectrum has a unique closed point. However, $\unicode[STIX]{x1D70B}_{0}(B)\rightarrow \unicode[STIX]{x1D70B}_{0}(R)^{\widetilde{G}}$ is an ${\mathcal{F}}_{p}$ -isomorphism as above, so that it induces an isomorphism on Zariski spectra. Therefore, the fact that $\unicode[STIX]{x1D70B}_{0}(R)^{\widetilde{G}}$ is a local ring forces $\unicode[STIX]{x1D70B}_{0}(B)$ to be one, and completes the proof.◻

Proof. Observe first that $\operatorname{Pic}(E_{n}^{hG})=\unicode[STIX]{x1D70B}_{0}(\mathfrak{pic}(E_{n})^{hG})$ is a torsion group (cf. [Reference Mathew and StojanoskaMS16, § 5.3]), so it suffices to show that the suspension generates the $q$ -part $\operatorname{Pic}(E_{n}^{hG})_{q}$ of $\operatorname{Pic}(E_{n}^{hG})$ for any prime number  $q$ . We consider two cases.

(i) $q=p$ . In this case, the map

$$\begin{eqnarray}\unicode[STIX]{x1D719}_{p}:\operatorname{Pic}(E_{n}^{hG})_{p}\rightarrow \operatorname{Pic}(E_{n}^{hG_{p}})_{p}\end{eqnarray}$$

(induced by base-change along $E_{n}^{hG}\rightarrow E_{n}^{hG_{p}}$ ) is injective thanks to the transfer map

$$\begin{eqnarray}\operatorname{Pic}(E_{n}^{hG_{p}})_{p}\simeq \unicode[STIX]{x1D70B}_{0}\,\mathfrak{pic}(E_{n})^{hG_{p}}\rightarrow \operatorname{Pic}(E_{n}^{hG})_{p}\simeq \unicode[STIX]{x1D70B}_{0}\,\mathfrak{pic}(E_{n})^{hG}.\end{eqnarray}$$

Namely, recall that if $X$ is a $p$ -local spectrum with a $G$ -action, the composite map $X^{hG}\rightarrow X^{hG_{p}}\stackrel{\text{tr}}{\rightarrow }X^{hG}$ is an equivalence. Since the map $\unicode[STIX]{x1D719}_{p}$ carries the suspension of the unit to itself, the fact that this class generates the target implies that it must generate the source.

(ii) $q\neq p$ . Let $G_{q}\subseteq G$ be a $q$ -Sylow subgroup. As above, we observe that the map $\unicode[STIX]{x1D719}_{q}$ defined via

$$\begin{eqnarray}\unicode[STIX]{x1D719}_{q}:\operatorname{Pic}(E_{n}^{hG})_{q}\rightarrow \operatorname{Pic}(E_{n}^{hG_{q}})_{q}\end{eqnarray}$$

is injective. So, it suffices to consider the case where $G$ is itself a $q$ -group.

We consider the HFPSS converging to $\operatorname{Pic}(E_{n}^{hG})=\unicode[STIX]{x1D70B}_{0}\,\mathfrak{pic}(E_{n})^{hG}$ . In this case, since $G$ is a $q$ -group, we observe that the only contributions come from $H^{0}(G,\mathbb{Z}/2)$ (which is $\mathbb{Z}/2$ , generated by the suspension) and $H^{1}(G,(E_{n})_{0}^{\times })\simeq H^{1}(G,(\mathbb{F}_{p^{n}})^{\times })$ via the Teichmüller lift. In other words, we need to show (to prove cyclicity of the Picard group in this case) that the group $H^{1}(G,(\mathbb{F}_{p^{n}})^{\times })=\operatorname{Hom}(G,\mathbb{F}_{p^{n}}^{\times })$ is generated by the class $\unicode[STIX]{x1D713}$ corresponding to the $G$ -action on $\unicode[STIX]{x1D70B}_{-2}(E_{n})\otimes _{(E_{n})_{0}}\mathbb{F}_{p^{n}}$ . (This will give the order of the Picard group as well as solve the extension problem.) However, we know (by general facts about even-periodic ring spectra) that this is simply the action on the Lie algebra, i.e. the homomorphism of (2.1). However, as we remarked before, the kernel of (2.1) is pro- $p$ , so its restriction to $G$ must be injective. Hence, $G$ injects into $\mathbb{F}_{p^{n}}^{\times }$ , which forces $H^{1}(G,\mathbb{F}_{p^{n}}^{\times })$ to be generated by the class of the injection.◻

Proof. In the additive spectral sequence, by Lemma 2.8, the classes $e_{c}$ are never the source of a $d_{2p-1}$ differential, but can be the target. In particular, there are additive differentials

$$\begin{eqnarray}d_{2p-1}(\unicode[STIX]{x1D6FD}^{a}\unicode[STIX]{x1D6FF}^{b})\stackrel{\cdot }{=}b\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}^{a+p-1}\unicode[STIX]{x1D6FF}^{b-n^{2}}.\end{eqnarray}$$

For this image to be (a multiple of) $e_{c}$ , we must have $b=n^{2}+n-1-c(pn-1)\equiv c-1~\text{mod}~p$ .

1. – If $c\not \equiv 1~\text{mod}~p$ , then $e_{c}$ is the target of $d_{2p-1}$ on an element with $t=2p(c-1)$ . Since such $c\geqslant 2$ , we have that $t\geqslant 2p>2p-1$ , so that Theorem 1.4 implies that this differential can be imported in the Picard spectral sequence. Hence, $e_{c}$ itself is killed.

2. – If $c\equiv 1~\text{mod}~p$ , this differential is zero (as then $b\equiv 0~\text{mod}~p$ ), so $e_{c}$ survives the $d_{2p-1}$ .

The next and only possible additive differential is a $d_{2n^{2}+1}$ . We have

$$\begin{eqnarray}d_{2n^{2}+1}(\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FF}^{n^{3}})\stackrel{\cdot }{=}\unicode[STIX]{x1D6FD}^{n^{2}+1},\end{eqnarray}$$

which determines the rest. Namely, we have, for all $a,b\in \mathbb{Z}$ , $a\geqslant 0$ ,

$$\begin{eqnarray}d_{2n^{2}+1}(\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}^{a}\unicode[STIX]{x1D6FF}^{bp-1})\stackrel{\cdot }{=}\unicode[STIX]{x1D6FD}^{a+n^{2}+1}\unicode[STIX]{x1D6FF}^{pb-n^{3}-1}.\end{eqnarray}$$

When $c\equiv 1~\text{mod}~(p)$ , we have that $n-1-c(pn-1)\equiv -1~\text{mod}~p$ , hence

$$\begin{eqnarray}d_{2n^{2}+1}(e_{c})\stackrel{\cdot }{=}\unicode[STIX]{x1D6FD}^{pc+n^{2}}\unicode[STIX]{x1D6FF}^{c(1-pn)+n-n^{3}-1}.\end{eqnarray}$$

To import this differential to the Picard spectral sequence, we need to have that

$$\begin{eqnarray}2n^{2}+1\leqslant t(e_{c})=2pc-2.\end{eqnarray}$$

This holds whenever

$$\begin{eqnarray}p-2+\frac{5}{2p}\leqslant c\quad \Leftrightarrow \quad c\geqslant p-1.\end{eqnarray}$$

Since $c\equiv 1~\text{mod}~(p)$ , we see we can import all differentials except that involving $e_{1}$ , and the result follows.◻

Proof of Theorem 4.1.

We have seen that in the range $t\geqslant 2$ of the HFPSS for $\unicode[STIX]{x1D70B}_{t-s}\,\mathfrak{pic}(E^{hC_{p}})$ , we only have a single possible contribution to the 0-stem, namely a group of order at most  $p^{n}$ , arising from the class  $e_{1}$ . This class has bidegree $(2p-1,2p-1)$ , and it follows from Theorem 7.1 that there is a differential

(4.2) $$\begin{eqnarray}d_{2n^{2}+1,\times }(e_{1})\stackrel{\cdot }{=}d_{2n^{2}+1}(e_{1})+\unicode[STIX]{x1D701}\unicode[STIX]{x1D6FD}{\mathcal{P}}^{n/2}(e_{1}),\end{eqnarray}$$

for $\unicode[STIX]{x1D701}\in \mathbb{F}_{p}^{\times }$ . Let $f_{t}=\unicode[STIX]{x1D6FD}^{n^{2}+n+1}\unicode[STIX]{x1D6FF}^{-n^{2}(n+1)}$ so that we have (in the additive spectral sequence)

$$\begin{eqnarray}d_{2n^{2}+1}(e_{1})\stackrel{\cdot }{=}f_{t}.\end{eqnarray}$$

While it is possible to work out exactly what $\unicode[STIX]{x1D6FD}{\mathcal{P}}^{n/2}(e_{1})$ is, we will not need to be so precise. Instead we claim that the group $\mathbb{F}_{p^{n}}e_{1}$ on the $E_{2}$ -page can contribute at most an element of order $p$ to Picard group.

Choose $\unicode[STIX]{x1D709}^{\prime }\in \mathbb{F}_{p^{n}}$ such that $\unicode[STIX]{x1D6FD}{\mathcal{P}}^{n/2}(e_{1})=\unicode[STIX]{x1D709}^{\prime }f_{t}$ . Then we have

$$\begin{eqnarray}d_{2n^{2}+1}(e_{1})=\unicode[STIX]{x1D709}f_{t}\quad \text{and}\quad \unicode[STIX]{x1D6FD}{\mathcal{P}}^{n/2}(e_{1})=\unicode[STIX]{x1D709}^{\prime }f_{t},\end{eqnarray}$$

for $\unicode[STIX]{x1D709},\unicode[STIX]{x1D709}^{\prime }\in \mathbb{F}_{p^{n}}$ and $\unicode[STIX]{x1D709}\neq 0$ . It follows from (4.2) that, for any $a\in \mathbb{F}_{p^{n}}$ , we have

$$\begin{eqnarray}\displaystyle d_{2n^{2}+1,\times }(ae_{1}) & = & \displaystyle d_{2n^{2}+1}(ae_{1})+\unicode[STIX]{x1D701}\unicode[STIX]{x1D6FD}{\mathcal{P}}^{n/2}(ae_{1})\nonumber\\ \displaystyle & = & \displaystyle a\unicode[STIX]{x1D709}f_{t}+a^{p}\unicode[STIX]{x1D701}\unicode[STIX]{x1D6FD}{\mathcal{P}}^{n/2}(e_{1})\nonumber\\ \displaystyle & = & \displaystyle a\unicode[STIX]{x1D709}f_{t}+a^{p}\unicode[STIX]{x1D701}\unicode[STIX]{x1D709}^{\prime }f_{t}\nonumber\\ \displaystyle & = & \displaystyle (a\unicode[STIX]{x1D709}+a^{p}\unicode[STIX]{x1D701}\unicode[STIX]{x1D709}^{\prime })f_{t}.\nonumber\end{eqnarray}$$

Since $\unicode[STIX]{x1D709}\neq 0$ , there are at most $p$ choices of $a$ that make the above vanish, and so we do indeed have an upper bound of $\mathbb{Z}/p$ for the contribution of $e_{1}$ in $\operatorname{Pic}(E^{hC_{p}})$ .

Finally, we know from Proposition 6.5 that $H^{1}(C_{p},\unicode[STIX]{x1D70B}_{1}\,\mathfrak{pic}(E))\simeq \mathbb{Z}/p$ , while as usual it follows from [Reference Baker and RichterBR05] that the invariants $H^{0}(C_{p},\unicode[STIX]{x1D70B}_{0}\,\mathfrak{pic}(E))$ are $\mathbb{Z}/2$ . Thus, we have the following in the 0-stem of the Picard spectral sequence:

1. – a group of order at most $2$ in $t=s=0$ ;

2. – a group of order at most $p$ in $t=s=1$ ; and

3. – a group of order at most $p$ in $t=s=2p-1$ .

This gives an upper bound of $2p^{2}$ for the order of the Picard group, but as noted previously this is also a lower bound, and the result follows. A picture of the Picard spectral sequence for $n=2$ , $p=3$ and $G=C_{3}$ is shown in Figure 2.◻

Figure 2.

The HFPSS for $\unicode[STIX]{x1D70B}_{\ast }\,\mathfrak{pic}\,E_{p-1}^{hC_{p}}$ in the case $n=2$ , $p=3$ . The differential out of $(0,5)$ has kernel $\mathbb{Z}/3$ . (Here $\bullet$ denotes $\mathbb{Z}/3$ , 

 denotes $\mathbb{F}_{9}$ , and $\times$ denotes $\mathbb{Z}/2$ .)

Proof. This follows by a straightforward computation. ◻

Proof. This follows from (5.8) taking $x_{i}=(-1)^{i}y_{i}$ . We multiply it by $\prod _{i~\text{odd}}(\exp _{p}(-y_{i})/\exp _{p}(y_{i})^{-1})$ and invoke (5.9).◻

Construction 5.8. Suppose $\mathfrak{m}^{\bullet }$ is a nonunital cosimplicial commutative $\mathbb{Z}[1/(p-1)!]$ -algebra such that $\mathfrak{m}^{\bullet ,p+1}=0$ .Footnote ² We have a descending multiplicative filtration by powers

$$\begin{eqnarray}\mathfrak{m}^{\bullet }\supset \mathfrak{m}^{\bullet ,2}\supset \cdots .\end{eqnarray}$$

Consider also the cosimplicial abelian group $1+\mathfrak{m}^{\bullet }$ .Footnote ³ We also have a filtration of cosimplicial abelian groups

$$\begin{eqnarray}1+\mathfrak{m}^{\bullet }\supset 1+\mathfrak{m}^{\bullet ,2}\supset \cdots ,\end{eqnarray}$$

and the successive quotients are isomorphic to those of the additive one.

Both lead to spectral sequences starting from the associated graded modules $\mathfrak{m}^{\bullet ,j}/\mathfrak{m}^{\bullet ,j+1}$ (for $1\leqslant j\leqslant p$ ) converging to the cohomology of $\mathfrak{m}^{\bullet }$ (respectively $1+\mathfrak{m}^{\bullet }$ ), i.e. the $E_{1}$ -pages are identified. We have spectral sequences

(5.12) $$\begin{eqnarray}\displaystyle E_{1}^{i,j} & = & \displaystyle H^{i}(\mathfrak{m}^{\bullet ,j}/\mathfrak{m}^{\bullet ,j+1})\;\Longrightarrow \;H^{i}(\mathfrak{m}^{\bullet })\end{eqnarray}$$

(5.13) $$\begin{eqnarray}\displaystyle E_{1}^{i,j,\times } & = & \displaystyle H^{i}(\mathfrak{m}^{\bullet ,j}/\mathfrak{m}^{\bullet ,j+1})\;\Longrightarrow \;H^{i}(1+\mathfrak{m}^{\bullet }).\end{eqnarray}$$

The $E_{1}$ -terms are identified. We will denote the differentials at the $k$ th page by $d_{k}$ and $d_{k}^{\times }$ , respectively.

Using the truncated exponential, we obtain isomorphisms of filtered cosimplicial abelian groups

$$\begin{eqnarray}\mathfrak{m}^{\bullet }/\mathfrak{m}^{\bullet ,p}\simeq (1+\mathfrak{m}^{\bullet })/(1+\mathfrak{m}^{\bullet ,p}),\quad \mathfrak{m}^{\bullet ,2}\simeq 1+\mathfrak{m}^{\bullet ,2}\end{eqnarray}$$

inducing the above isomorphism on the associated graded modules.

Construction 5.10. Let $x\in P^{i}$ . We define

$$\begin{eqnarray}\unicode[STIX]{x1D719}(x)\stackrel{\text{def}}{=}\unicode[STIX]{x1D707}_{p}(\unicode[STIX]{x2202}^{0}x,\unicode[STIX]{x2202}^{1}x,\unicode[STIX]{x2202}^{2}x,\ldots ,\unicode[STIX]{x2202}^{i}x)\in \operatorname{Sym}^{p}P^{i+1}.\end{eqnarray}$$

Here the $\unicode[STIX]{x2202}^{k}$ are the cosimplicial operators. We claim that:

1. (i) if $x$ is a cocycle in $P^{i}$ , then $\unicode[STIX]{x1D719}(x)$ is a cocycle in $\operatorname{Sym}^{p}P^{i+1}$ ;

2. (ii) if $x,y\in P^{i}$ are cohomologous cocycles, then $\unicode[STIX]{x1D719}(x)$ and $\unicode[STIX]{x1D719}(y)$ are cohomologous in $\operatorname{Sym}^{p}P^{i+1}$ .

The proofs of both these claims will be contained in Proposition 5.11 below. As a result, we will also abuse notation and use $\unicode[STIX]{x1D719}$ to denote the induced map in cohomology.

Proof. We unwind the definitions. A class $x\in H^{i}(\mathfrak{m}^{\bullet }/\mathfrak{m}^{\bullet ,2})$ that has survived to the $E_{p-1}$ stage of the spectral sequence, by definition, lifts to a class in $H^{i}(\mathfrak{m}^{\bullet }/\mathfrak{m}^{\bullet ,p})$ , which we represent by a degree $i$ cycle  $\widetilde{x}$ . The short exact sequence of cosimplicial abelian groups

$$\begin{eqnarray}0\rightarrow \mathfrak{m}^{\bullet ,p}\rightarrow \mathfrak{m}^{\bullet }\rightarrow \mathfrak{m}^{\bullet }/\mathfrak{m}^{\bullet ,p}\rightarrow 0,\end{eqnarray}$$

leads to a connecting homomorphism

$$\begin{eqnarray}\unicode[STIX]{x1D6FF}:H^{i}(\mathfrak{m}^{\bullet }/\mathfrak{m}^{\bullet ,p})\rightarrow H^{i+1}(\mathfrak{m}^{\bullet ,p}).\end{eqnarray}$$

The differential $d_{p-1}(x)$ is given by the image of $\unicode[STIX]{x1D6FF}(\widetilde{x})$ .

In the multiplicative setting, we know that the cycle $\exp _{p}(\widetilde{x})\in (1+\mathfrak{m}^{\bullet })/(1+\mathfrak{m}^{\bullet ,p})$ lifts the image of $x$ (or rather, $1+x$ ) in the $E_{1}$ -page of the multiplicative spectral sequence. Now $d_{p-1,\times }(x)$ is given as follows: one considers the short exact sequence of cosimplicial abelian groups

$$\begin{eqnarray}0\rightarrow (1+\mathfrak{m}^{\bullet ,p})\rightarrow (1+\mathfrak{m}^{\bullet })\rightarrow (1+\mathfrak{m}^{\bullet })/(1+\mathfrak{m}^{\bullet ,p})\rightarrow 0,\end{eqnarray}$$

and the induced connecting homomorphism

$$\begin{eqnarray}\unicode[STIX]{x1D6FF}_{\times }:H^{i}((1+\mathfrak{m}^{\bullet })/(1+\mathfrak{m}^{\bullet ,p}))\rightarrow H^{i+1}(1+\mathfrak{m}^{\bullet ,p}).\end{eqnarray}$$

Unwinding the definitions, the multiplicative differential $d_{p-1,\times }(x)$ is given by

$$\begin{eqnarray}d_{p-1,\times }(x)=\unicode[STIX]{x1D6FF}_{\times }(\exp _{p}(\widetilde{x}))\in H^{i+1}(1+\mathfrak{m}^{\bullet ,p})\simeq H^{i+1}(\mathfrak{m}^{\bullet ,p}),\end{eqnarray}$$

or rather the image in the quotient of this at the $E_{p-1}$ -page.

It follows now that we need to compare the additive and the multiplicative boundary maps $\unicode[STIX]{x1D6FF},\unicode[STIX]{x1D6FF}^{\times }$ . Given a cycle $\widetilde{x}\in \mathfrak{m}^{\bullet }/\mathfrak{m}^{\bullet ,p}$ in degree  $i$ , projecting to a cycle and induced cohomology class $x\in \mathfrak{m}^{\bullet }/\mathfrak{m}^{\bullet ,2}$ , we claim that

(5.14) $$\begin{eqnarray}\unicode[STIX]{x1D6FF}_{\times }(\exp _{p}(\widetilde{x}))=1+\unicode[STIX]{x1D6FF}(\widetilde{x})+\unicode[STIX]{x1D719}(\widetilde{x})\in H^{i+1}(1+\mathfrak{m}^{\bullet ,p})\simeq H^{i+1}(\mathfrak{m}^{\bullet ,p}).\end{eqnarray}$$

To prove this, we unwind the definitions again. Lift $\widetilde{x}$ to a chain $x^{\prime }\in \mathfrak{m}^{\bullet }$ (in degree  $i$ ). The additive connecting homomorphism $\unicode[STIX]{x1D6FF}(\widetilde{x})$ is obtained by considering the image in cohomology of the alternating sum $y^{\prime }=\sum _{k}(-1)^{k}\unicode[STIX]{x2202}^{k}x^{\prime }$ , which is an $(i+1)$ -cocycle in $\mathfrak{m}^{\bullet ,p}$ . For the multiplicative version, we have to consider the cocycle

$$\begin{eqnarray}\unicode[STIX]{x1D6FF}_{\times }(\exp _{p}(\widetilde{x}))=\mathop{\prod }_{k}(\unicode[STIX]{x2202}^{k}\exp _{p}(x^{\prime }))^{(-1)^{k}}=\mathop{\prod }_{k}\exp _{p}(\unicode[STIX]{x2202}^{k}x^{\prime })^{(-1)^{k}}\in 1+\mathfrak{m}^{\bullet ,p+1}.\end{eqnarray}$$

Now we use (5.10), which gives us

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6FF}_{\times }(\exp _{p}(\widetilde{x})) & = & \displaystyle \exp _{p}\Bigl(\sum (-1)^{k}\unicode[STIX]{x2202}^{k}x^{\prime }\Bigr)+\unicode[STIX]{x1D707}_{p}(\unicode[STIX]{x2202}^{0}x^{\prime },\unicode[STIX]{x2202}^{1}x^{\prime },\ldots ,\unicode[STIX]{x2202}^{i}x^{\prime })\nonumber\\ \displaystyle & = & \displaystyle \exp _{p}(y^{\prime })+\unicode[STIX]{x1D707}_{p}(\unicode[STIX]{x2202}^{0}x^{\prime },\unicode[STIX]{x2202}^{1}x^{\prime },\ldots ,\unicode[STIX]{x2202}^{i}x^{\prime })\nonumber\\ \displaystyle & = & \displaystyle 1+y^{\prime }+\unicode[STIX]{x1D707}_{p}(\unicode[STIX]{x2202}^{0}x^{\prime },\unicode[STIX]{x2202}^{1}x^{\prime },\ldots ,\unicode[STIX]{x2202}^{i}x^{\prime }).\nonumber\end{eqnarray}$$

In cohomology, this is the desired claim, provided we can justify the claims about $\unicode[STIX]{x1D719}$ made in Construction 5.10.

The justification of Construction 5.10 now follows from reversing the above argument. Keep the notation from there. We consider the commutative nonunital cosimplicial algebra

$$\begin{eqnarray}A^{\bullet }=\bigoplus _{0<k\leqslant p}\operatorname{Sym}^{k}P^{\bullet }\end{eqnarray}$$

and its quotient $\overline{A}^{\bullet }=\bigoplus _{0<k<p}\operatorname{Sym}^{k}P^{\bullet }$ . Choose a cycle $x\in P^{\bullet }\subset \overline{A}^{\bullet }$ , which clearly extends to a cycle in $A^{\bullet }$ . Then $\exp _{p}(x)$ is a cycle in $1+\overline{A}^{\bullet }$ . The coboundary of this class in $\operatorname{Sym}^{p}P^{\bullet }\simeq 1+\operatorname{Sym}^{p}P^{\bullet }\subset 1+A^{\bullet }$ is given by $\unicode[STIX]{x1D719}(x)$ by the above analysis. This shows that $\unicode[STIX]{x1D719}(x)$ is a cycle and that its cohomology class depends only on that of  $x$ .◻

Proof of Proposition 5.12.

We will analyze the last example more carefully to determine the precise multiple.Footnote ⁵ Note that $\overline{\mathfrak{m}}/\overline{\mathfrak{m}}^{2}\simeq \mathbb{F}_{p}$ (generated by the image of $1-\unicode[STIX]{x1D701}_{p}$ ) and $\overline{\mathfrak{m}}^{p}\simeq \mathbb{F}_{p}$ (generated by the image of $(1-\unicode[STIX]{x1D701}_{p})^{p}$ ). The (additive) differential $d_{p-1}$ comes from the short exact sequence

(5.17) $$\begin{eqnarray}0\rightarrow \overline{\mathfrak{m}}^{p}\rightarrow \overline{\mathfrak{m}}\rightarrow \overline{\mathfrak{m}}/\overline{\mathfrak{m}}^{p}\rightarrow 0,\end{eqnarray}$$

and we have the following commutative diagram of short exact sequences.

(5.18)

Here the map $\mathbb{Z}/p^{2}\rightarrow \overline{\mathfrak{m}}$ carries $1\mapsto 1-\unicode[STIX]{x1D701}_{p}$ . Note that $p(1-\unicode[STIX]{x1D701}_{p})\equiv (1-\unicode[STIX]{x1D701}_{p})^{p}$ in $\overline{\mathfrak{m}}$ as one sees from the expression $\prod _{i=1}^{p-1}(T-\unicode[STIX]{x1D701}_{p}^{i})=T^{p-1}+T^{p-2}+\cdots +1$ with $T=1$ .

Take $X_{\bullet }=K(\mathbb{Z}/p,n)$ , so that $H^{n}(X_{\bullet },\mathbb{F}_{p})=H^{n+1}(X_{\bullet },\mathbb{F}_{p})=\mathbb{F}_{p}$ . We let $\unicode[STIX]{x1D704}$ be a generator of the former, so that $\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D704}$ is a generator of the latter for $\unicode[STIX]{x1D6FD}$ the Bockstein. Again, this analysis is based on the cosimplicial commutative ring $\overline{\mathfrak{m}}^{X_{\bullet }}$ .

1. (i) Note that $H^{n}(X_{\bullet },\overline{\mathfrak{m}}/\overline{\mathfrak{m}}^{p})\simeq H^{n}(X_{\bullet },\mathbb{F}_{p})\otimes _{\mathbb{F}_{p}}\overline{\mathfrak{m}}/\overline{\mathfrak{m}}^{p}\simeq \overline{\mathfrak{m}}/\overline{\mathfrak{m}}^{p}\unicode[STIX]{x1D704}$ and similarly for $H^{n+1}(X_{\bullet },\overline{\mathfrak{m}}^{p})$ . The differential in question comes from the connecting homomorphism in the short exact sequence (5.17), so that by naturality of (5.18), it carries $(1-\unicode[STIX]{x1D701}_{p})\unicode[STIX]{x1D704}$ to $(1-\unicode[STIX]{x1D701}_{p})^{p}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D704}\in H^{n+1}(X_{\bullet },\overline{\mathfrak{m}^{p}})$ .

2. (ii) The associated graded module of $\overline{\mathfrak{m}}$ comes from the reduction of a $\mathbb{Z}$ -algebra, so $\unicode[STIX]{x1D6FD}{\mathcal{P}}^{0}$ is the composite of the Bockstein and ${\mathcal{P}}^{0}$ . Now ${\mathcal{P}}^{0}((1-\unicode[STIX]{x1D701}_{p})\unicode[STIX]{x1D704})=(1-\unicode[STIX]{x1D701}_{p})^{p}\unicode[STIX]{x1D704}\in H^{n}(X,\overline{\mathfrak{m}}^{p})$ since ${\mathcal{P}}^{0}$ is the Frobenius on underlying rings. Applying $\unicode[STIX]{x1D6FD}$ to that, we obtain $(1-\unicode[STIX]{x1D701}_{p})^{p}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D704}$ .

Putting these together, it follows that if the class corresponding to $(1-\unicode[STIX]{x1D701}_{p})\unicode[STIX]{x1D704}$ is to survive in the multiplicative spectral sequence (as it must), the formula must be as desired.◻