On the $C_{p}$-equivariant dual Steenrod algebra

The map

$$\mathbb{P}(\mathbb{C}[z])\to\mathrm{SP}^{\infty}(S^{\lambda})$$

to the infinite symmetric product, which sends a polynomial $f(z)$ to its set of roots (with multiplicity), is an equivariant homeomorphism. The group-completion of the latter is a model for $K(\underline{\mathbb{Z}},\lambda)$ by the equivariant Dold-Thom theorem. But $\mathrm{SP}^{\infty}(S^{\lambda})$ is already group-complete: the monoid of connected components of the fixed points is $\mathbb{N}/p=\mathbb{Z}/p$. ∎

This follows from a more general observation. If $L$ is a one-dimensional complex representation, and $V$ is an arbitrary complex representation, then the function assigning a linear map to its graph,

$$\mathrm{Hom}_{\mathbb{C}}(L,V)\longrightarrow\mathbb{P}(V\oplus L)-\mathbb{P}(V),$$

is an equivariant homeomorphism. So it induces an equivalence on one-point compactifications

$$S^{L^{\vee}\otimes V}\cong\mathbb{P}(V\oplus L)/\mathbb{P}(V).$$

∎

Using the model for $\mathrm{B}_{C_{p}}S^{1}$ given by $\mathbb{P}(\mathbb{C}[z])$, we see that, in fact, $\mathbb{P}(\mathbb{C}[z])$ has the structure of a filtered monoid. It follows that the product in homology respects the filtration by the classes $\{e_{i}\}$. Thus, for $i\geq 0$, we have:

$$e_{p^{i}}^{p}=\sum_{j\leq p^{i+1}}c_{i,j}e_{j}$$

where the coefficients lie in $\pi_{\star}\underline{\mathbb{Z}}$. When $j<p^{i+1}$ we see that the virtual representations $|c_{i,j}|$ have positive virtual dimension and their fixed points also have positive virtual dimension. The homotopy of $\underline{\mathbb{Z}}$ vanishes in these degrees (see, e.g., [FL04, Theorem 8.1(iv)]), so we must have

$$e^{p}_{p^{i}}=c_{i,p^{i+1}}e_{p^{i+1}}$$

where $|c_{0,p}|=\lambda-2$ and $|c_{i,p^{i+1}}|=0$ when $i\geq 1$. In both cases, the restriction map on $\pi_{\star}\underline{\mathbb{Z}}$ is injective in this degree, so the result follows from the nonequivariant calculation. ∎

Since $p=\mathrm{tr}(1)$, the class $pN(t_{i})$ is the transfer of the class $\mathrm{res}(t_{i})^{p}$ into degree $(2p^{i}-2)\rho_{C_{p}}$. Notice that $(2p^{i}-2)\rho_{C_{p}}-|t_{i}^{p}|=\lambda-2$ (after identifying the $\lambda^{k}$ suspensions with $\lambda$ for $(k,p)=1$), and the transfer of $1$ into this degree is $\theta$, so we have

$$pN(t_{i})=\theta t_{i}^{p}=0.$$

∎

By Lemma 2.6, we have $e_{1}^{p}\doteq\theta e_{p}$ so that $\theta t_{1}=\sigma(\theta e_{p})=0$, since $\sigma$ annihilates decomposables. For the remaining classes, consider the commutative diagram

where $[\theta]=\Omega^{\infty}(\theta)$. Thus, to show that $\theta t_{i}=0$ for $i\geq 2$, it is enough to show that $[\theta]_{*}e_{p^{i}}$ is decomposable in $\pi_{\star}(\underline{\mathbb{Z}}_{(p)}\otimes K(\underline{\mathbb{Z}},2)_{+})$ for $i\geq 2$.

Write

$$\underline{\mathbb{Z}}_{(p)}\otimes K(\underline{\mathbb{Z}},2)_{+}=\underline{\mathbb{Z}}_{(p)}\{\gamma_{i}(\beta_{1})\}$$

where the elements $\gamma_{i}(\beta_{1})$ are the standard module generators of $H_{*}(\mathbb{C}P^{\infty};\mathbb{Z})$, and write $\beta_{(i)}=\gamma_{p^{i}}\beta_{1}$. To show that $[\theta]_{*}(e_{p^{i}})$ is decomposable for $i\geq 2$, it is enough to establish the following two claims:

1. (a)

   $[\theta]_{*}(e_{p^{i}})\doteq\frac{p^{i-1}\theta}{u_{\lambda}^{p^{i}(p-1)-1}}\beta_{(i)}$, and

2. (b)

   $\beta_{(i-1)}^{p}\doteq p\beta_{(i)}$.

Claim (b) is just the classical computation of the product in homology for $H_{*}(\mathbb{C}P^{\infty},\mathbb{Z})$. For claim (a), let $\iota_{\lambda}$ denote the fundamental class in cohomology for $K(\underline{\mathbb{Z}},\lambda)$ and $\iota_{2}$ the same for $K(\underline{\mathbb{Z}},2)$. Then we have $[\theta]^{*}(\iota_{2})=\theta\iota_{\lambda}$ by design, and hence

$$[\theta]^{*}(\iota_{2}^{j})=\theta^{j}\iota_{\lambda}^{j}.$$

The map on homology is now determined by the relation

$$\langle[\theta]_{*}e_{p^{i}},\iota_{2}^{j}\rangle=\theta^{j}\langle e_{p^{i}},\iota^{j}_{\lambda}\rangle\in\pi_{\star}\underline{\mathbb{Z}}_{(p)}.$$

Since $\theta^{j}$ is a transferred class, the value above is also a transfer, and hence determined by its restriction to an underlying class. But $\mathrm{res}([\theta])=[p]$ and we clearly have $[p]_{*}(\mathrm{res}(e_{p^{i}}))=p^{i}\beta_{(i)}$, which agrees with the restriction of $\frac{p^{i-1}\theta}{u_{\lambda}^{p^{i}(p-1)-1}}\beta_{(i)}$. This completes the proof. ∎

By assumption (i), it is enough to check that $f^{\Phi C_{p}}$ is an equivalence; by assumption (ii), it is enough to check that $\pi_{*}(f^{\Phi C_{p}})$ is an injection; and by assumption (iii) it is enough to check that $\pi_{*}(f^{\Phi C_{p}})[b^{-1}]$ is an injection.

Again by (i), the map $f^{tC_{p}}$ is an equivalence. So, from the diagram

we see that it is enough to check that the vertical maps are injective on homotopy. More generally, we show that whenever $X$ is a bounded below $\underline{\mathbb{Z}}$-module, the map

$$\pi_{*}X^{\Phi C_{p}}[b^{-1}]\to\pi_{*}X^{tC_{p}}$$

is injective. Indeed, by Proposition 5.2 and the fact that the Tate construction commutes with limits of Postnikov towers (see, e.g., [NS18, I.2.6]), we have

$$\lim_{n}\left((\tau_{\leq n}X)^{\Phi C_{p}}[b^{-1}]\right)\stackrel{{\scriptstyle\simeq}}{{\to}}\lim_{n}(\tau_{\leq n}X)^{tC_{p}}\simeq X^{tC_{p}}.$$

Therefore, we need only check that

$$\pi_{*}X^{\Phi C_{p}}[b^{-1}]\to\pi_{*}\lim_{n}\left((\tau_{\leq n}X)^{\Phi C_{p}}[b^{-1}]\right)$$

is injective. Since the maps $X^{\Phi C_{p}}\to(\tau_{\leq n}X)^{\Phi C_{p}}$ have increasingly connective fibers, we can replace the left hand side by $(\lim_{n}\pi_{*}(\tau_{\leq n}X)^{\Phi C_{p}})[b^{-1}]$ and reduce to showing that

$$(\lim_{n}\pi_{*}(\tau_{\leq n}X)^{\Phi C_{p}})[b^{-1}]\to\lim_{n}\pi_{*}\left((\tau_{\leq n}X)^{\Phi C_{p}}[b^{-1}]\right)$$

is injective. Finally, this reduces to showing that the kernel of

$$\lim_{n}\pi_{*}(\tau_{\leq n}X)^{\Phi C_{p}}\to\lim_{n}\pi_{*}\left((\tau_{\leq n}X)^{\Phi C_{p}}[b^{-1}]\right)$$

consists of elements annihilated by a power of $b$. This is clear because, for each $j$, the system $\{\pi_{j}(\tau_{\leq n}X)^{\Phi C_{p}}\}_{n}$ is eventually constant. ∎

Let $\mathcal{E}$ denote the full subcategory of $\underline{\mathbb{Z}}$-modules $M$ for which

$$M^{\Phi C_{p}}[b^{-1}]\to M^{tC_{p}}$$

is an equivalence. Then $\mathcal{E}$ is stable, closed under retracts, and closed under suspending by representation spheres.

The map $M^{\Phi C_{p}}[b^{-1}]\to M^{tC_{p}}$ is one of $\underline{\mathbb{Z}}^{\Phi C_{p}}=\mathbb{F}_{p}[b]$-modules, and hence one of $\mathbb{F}_{p}$-modules, so it must be a retract of

$$(M/p)^{\Phi C_{p}}[b^{-1}]=M^{\Phi C_{p}}[b^{-1}]/p\to M^{tC_{p}}/p=(M/p)^{tC_{p}}.$$

Thus $M/p\in\mathcal{E}$ if and only if $M\in\mathcal{E}$. So, by replacing $M$ with $M/p$ and considering the Postnikov tower, we are reduced to proving the proposition in the case where $M\in\mathsf{Mod}^{\heartsuit}_{\underline{\mathbb{Z}}}$ is a Mackey functor which is a module over $\underline{\mathbb{F}}_{p}$.

In particular, $M^{e}$ is an $\mathbb{F}_{p}[C_{p}]$-module. Let $\gamma$ denote the generator of $C_{p}$ so that $\mathbb{F}_{p}[C_{p}]=\mathbb{F}_{p}[\gamma]/(1-\gamma)^{p}$. Let $F_{j}M\subseteq M$ be the sub-Mackey functor generated by $(1-\gamma)^{j}M^{e}\subseteq M^{e}$. This is a finite filtration with associated graded pieces given by Mackey functors with trivial underlying action. So, since $\mathcal{E}$ is a thick subcategory, we are reduced to the case when $M$ is a discrete $\underline{\mathbb{F}}_{p}$-module with trivial underlying action.

For the next reduction we recall some notation. If $N$ is any Mackey functor, denote by $N_{C_{p}}$ the Mackey functor $N\otimes C_{p+}$ and, if $A$ is an abelian group, denote by $\underline{A}_{\mathrm{tr}}$ the Mackey functor whose transfer map is the identity on $A$ and whose restriction map is multiplication by $p$. We also recall that the transfer extends to a map of Mackey functors $\mathrm{tr}:N_{C_{p}}\to N$.

Now consider the two exact sequences

$$0\to\mathrm{im}(\mathrm{tr})\to M\to M/\mathrm{im}(tr)\to 0$$

$$0\to\mathrm{ker}(\mathrm{tr})\to\underline{M}^{e}_{\mathrm{tr}}\to\mathrm{im}(\mathrm{tr})\to 0$$

If $N$ is any Mackey functor with $N^{e}=0$, then $N\in\mathcal{E}$ since then $N=N^{\Phi C_{p}}$ is bounded above and hence $N^{\Phi C_{p}}[b^{-1}]=0$. Thus, from the exact sequences above, we are reduced to the case where $M$ is of the form $\underline{V}_{\mathrm{tr}}$ for an $\mathbb{F}_{p}$-vector space $V$ (with trivial action). Now recall that $(\underline{\mathbb{F}}_{p})_{\mathrm{tr}}=\Sigma^{2-\lambda}\underline{\mathbb{F}}_{p}$ and hence $\underline{V}_{\mathrm{tr}}=\Sigma^{2-\lambda}\underline{V}$. So we are reduced to showing that the constant Mackey functor $\underline{V}$ lies in $\mathcal{E}$, where $V$ is an $\mathbb{F}_{p}$-vector space with trivial action. This certainly holds for $V=\mathbb{F}_{p}$, and in general we have

$$\underline{V}^{\Phi C_{p}}\simeq\underline{\mathbb{F}}^{\Phi C_{p}}_{p}\otimes_{\mathbb{F}_{p}}V,$$

since geometric fixed points commutes with colimits, and

$$V^{tC_{p}}\simeq\mathbb{F}_{p}^{tC_{p}}\otimes_{\mathbb{F}_{p}}V$$

by direct calculation. (Notice this holds even when $V$ is infinite-dimensional). This completes the proof. ∎

Combine Proposition 5.1 with the two lemmas below. ∎

First observe that, by our construction in the proof of Lemma 4.4, the map $\widehat{N(t_{i})}$ restricts to the map $t_{i}^{p-1}\hat{t}_{i}$, since the nullhomotopy witnessing $pN(t_{i})=0$ was chosen to restrict to the nullhomotopy chosen for $pt_{i}^{p}$ that came from the already chosen nullhomotopy of $pt_{i}$. The upshot is that the map

$$S^{0}\oplus S^{0}[Nt_{i}]\otimes T_{\theta}(t_{i})\to\underline{\mathbb{Z}}\otimes\underline{\mathbb{Z}}$$

restricts on underlying spectra to the map

$$S^{0}[t_{i}]/(pt_{i})\to\mathbb{Z}\otimes\mathbb{Z}$$

obtained just from the relation $pt_{i}=0$ and extended via the multiplicative structure.

In particular, on mod $p$ homology $f^{e}$ induces a ring map

$$\mathbb{F}_{p}[t_{i}]\otimes\Lambda(x_{i})\to\mathbb{F}_{p}[\xi_{i}]\otimes\Lambda(\tau_{i}).$$

We know that $t_{i}$ maps to $\xi_{i}$ and that $\beta x_{i}=t_{i}$, so that $\beta(f_{*}^{e}(x_{i}))=\xi_{i}$. Modulo decomposables, $\tau_{i}$ is the only element whose Bockstein is $\xi_{i}$. So $x_{i}$ must map to $\tau_{i}$, mod decomposables. It follows that $f^{e}$ is a mod $p$ equivalence, and hence an equivalence. ∎