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There are various ways to construct $KU$ as an $E_\infty$ ring spectrum; I will take that as given. Using this, we can make $KU\otimes G_+$ into an $E_\infty$ ring for any commutative topological group $G$, and we can make $F(X_+,KU)$ into an $E_\infty$ ring for any space $X$. In particular, we can take $G=X=S^1$ and get $E_\infty$ rings $KU\otimes S^1_+$ and $F(S^1_+,KU)$. In the category of naive ring spectra these are just the square-zero extensions $KU\oplus\Sigma KU$ and $KU\oplus\Sigma^{-1}KU$ respectively, but $\Sigma KU\simeq\Sigma^{-1}KU$ by Bott periodicity, so $KU\otimes S^1_+\simeq F(S^1_+,KU)$ as naive ring spectra. Can this be improved to an equivalence of $E_\infty$ algebras over $KU$?

As a first step, one could check that the action of power operations on homotopy groups is consistent. That sounds like it might be hard work, so I am checking whether anyone else has done it already.

One reason I ask is as follows. The spectrum $R=THH(F(S^1_+,S^0))$ plays an important role in the disproof by Burklund, Hahn, Levy and Schlank of Ravenel's Telescope Conjecture, so I would like to understand it better. (They actually use a $p$-adic version, but I am using the integral one for the moment.) One can calculate that $\pi_0(R)$ is the ring of numerical polynomials, which is the same as $KU_0(BS^1)$. This suggests that there might be an interesting map $R\to KU\otimes BS^1_+$ or $R\to KU\otimes S^1_+\otimes BS^1_+$. If we had an $E_\infty$ isomorphism $KU\otimes S^1_+\simeq F(S^1_+,KU)$ then we could apply $THH^{KU}(-)$ to it and that would do the job (using some auxiliary arguments that I will not spell out here).

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    $\begingroup$ A small remark is that the $p$-adic case is at least somewhat related to the chromatic Fourier transform, as in Hopkins-Lurie Ambidexterity Corollary 5.3.26 or Barthel-Carmeli-Schlank-Yanovski Theorem A/Theorem D. I didn't think this through and don't see immediately how to connect them. $\endgroup$ Commented Nov 12, 2023 at 15:36
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    $\begingroup$ The computation of strict units of E in the Nullstellensatz paper also suggests that there is some interesting commutative algebra map $KU[S^1] \to KU^{S^1}$, at least up to p-completion and going to $KU^{nr}$ $\endgroup$ Commented Nov 12, 2023 at 18:42
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    $\begingroup$ @MaximeRamzi Yes, I think my question is at least closely related to the question of whether the Bott element is a strict unit in $KU$. I guess you are referring to Theorem H of "The chromatic nullstellensatz" which implies that the strict unit spectrum of a suitable algebraic extension of $KU^\wedge_p$ is $\Sigma^2\mathbb{Z}_p\oplus\overline{\mathbb{F}_p}^\times$. The question is whether there is a corresponding $\Sigma^2\mathbb{Z}$ in the strict unit spectrum of $KU$ itself. $\endgroup$ Commented Nov 12, 2023 at 19:08
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    $\begingroup$ Snaith's theorem implies that the Bott element is a strict element, right ? I mean it's much more elementary than Snaith I guess, it's essentially the inclusion from Picard into the semiring of finite dimensional C-vector spaces (topologized). $\endgroup$ Commented Nov 12, 2023 at 19:47
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    $\begingroup$ Oh, right, shouldn't that answer the question completely ? Namely the equivalence of KU-modules between the two is classified by the Bott element, and this lifts to a strct element by ""Snaith", so the equivalence lifts to an E_oo-equivalence ? $\endgroup$ Commented Nov 12, 2023 at 19:51

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This seems to be an answer, based on discussion with Maxime Ramzi in the comments.

The space of $E_\infty$ $KU$-algebra maps from $KU\otimes S^1_+\to F(S^1_+,KU)$ is the same as the space of $E_\infty$ maps from $\Sigma^\infty_+S^1=\Sigma^\infty_+\Omega^\infty\Sigma H$ to $F(S^1_+,KU)$, or equivalently maps of spectra from $\Sigma H$ to $gl_1(F(S^1_+,KU))=F(S^1_+,gl_1(KU))=gl_1(KU)\times \Omega gl_1(KU)$, or equivalently maps from $\Sigma^2 H\to\Sigma gl_1(KU)\times gl_1(KU)$. Ignoring the first factor, maps of spectra from $\Sigma^2 H$ to $gl_1(KU)$ correspond to ring maps from $\Sigma^\infty_+\Omega^\infty\Sigma^2H=\Sigma^\infty_+BU(1)$ to $KU$.

Let $\mathcal{L}$ be the symmetric bimonoidal topological category of finite sets equipped with a complex line bundle, and let $\mathcal{V}$ be the symmetric bimonoidal topological category of finite-dimensional complex vector spaces. We have a symmetric bimonoidal functor $\mathcal{L}\to\mathcal{V}$ given by $(X,L)\mapsto\bigoplus_xL_x$, and this gives an $E_\infty$ map $K(\mathcal{L})\to K(\mathcal{V})$. It is standard to identify $K(\mathcal{V})$ with $kU$ and $K(\mathcal{L})$ with $\Sigma^\infty_+BU(1)=\Sigma^\infty_+\Omega^\infty\Sigma^2H$, so we have an $E_\infty$ map $\Sigma^\infty_+\Omega^\infty\Sigma^2H\to kU\to KU$, and thus a map $KU\otimes S^1_+\to F(S^1_+,KU)$ as in the previous paragraph. There are still some details to check, but I think that this should be the required equivalence.

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    $\begingroup$ Slight rewrite: There are E_\infty maps BU(1) \to BU_\otimes \to GL_1(KU) inducing isos on \pi_2. Hence there is an E_\infty map from S^1 to the fiber of GL_1(F(S^1_+, KU)) = F(S^1_+, GL_1(KU)) --> GL_1(KU) that induces iso on \pi_1. The adjoint E_\infty ring map S[S^1] --> F(S^1_+, KU) over KU is thus surjective on \pi_1, and its extension KU[S^1] --> F(S^1_+, KU) to an E_\infty KU-algebra map over KU is a \pi_1-iso, and therefore an equivalence. $\endgroup$ Commented Nov 12, 2023 at 21:54

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