Below are two different stories about power operations for $\mathbb{E}_\infty$-ring spectra, and I am struggling to see how they relate. In the following we let $R$ be an $\mathbb{E}_\infty$-ring spectrum.

  1. $R$ admits natural maps $R^{\wedge n}_{h\Sigma_n} \to R$. If $X$ is now a space, we may apply $(\,\cdot\,)^{\wedge n}_{h\Sigma_n}$ to a $\Sigma^\infty_+ X \to R$ representing an element of $R^0(X)$, which we then compose with the multiplications $R^{\wedge n}_{h\Sigma_n} \to R$ to obtain a map $$\mathbb{P}_n \colon R^0(X) \to R^0(X^{\times n}_{h\Sigma_n})$$ called the $n$-th total power operation of $R$ --- a multiplicative but non-additive map.
  2. There is a category $\mathsf{CAlg}(R)$ of $R$-algebras, and it admits a forgetful functor $U \colon \mathsf{CAlg}(R) \to \mathsf{Sp}$. One then defines a spectrum of power operations on $R$ to be the endomorphism spectrum $\operatorname{Map}(U,U)$.

Question. Are these two approaches in any way related?

At first glance it appears not so, but the forgetful functor $U$ admits a left adjoint $F$ sending a spectrum $Y$ to $R \wedge \bigoplus_n Y^{\wedge n}_{h\Sigma_n}$ --- a formula vaguely similar to what we see in the first definition. I guess if you apply the second definition to the $R$-algebra $\operatorname{Map}(\Sigma^\infty_+ X,R)$ and play around with the adjunction you can make the comparison precise, but I'm struggling with the details.


The first observation is that $U$ is representable in $CAlg(R)$ by $F(S)$ (with $F$ as in your question): $$Map_{CAlg(R)}(F(S),A) \simeq Map_{SMod}(S,A) \simeq U(A).$$

By some appropriately flowery version of Yoneda's Lemma, it follows that $$ Hom(U,U) \simeq Map_{CAlg(R)}(F(S),F(S)) \simeq Map_{SMod}(S,F(S)) \simeq R \wedge \bigvee_n B\Sigma_{n+}.$$ Applying $\pi_0$ to this, is easy to see that the operation corresponding to $1 \in R_0(B\Sigma_n)$ will be precisely the classic $n$th power operation, when applied to the $R$--algebra $A = Map(\Sigma^{\infty}_+X,R)$.

Added later by request ...

Suppose given $f \in \pi_0(F(R))$. Tracing through my equivalences, the associated operation $\theta_f: \pi_0(A) \rightarrow \pi_0(A)$, for $A$ a commutative $R$--algebra, is as follows.

Firstly $f$ can regarded as an $R$-module map $f:R \rightarrow F(R)$. Similarly, $x \in \pi_0(A)$ can be regarded as an $R$-module map $x:R \rightarrow A$. Then $\theta_f(x)$ is the composite $$ R \rightarrow F(R) \rightarrow F(A) \rightarrow A,$$ where the first map is $f$, the next is $F(x)$ and the the last is the structure map for the $R$--algebra: the wedge over $n$ of the maps $A^{\wedge n}_{h \Sigma_n} \rightarrow A$.

For the $n$th power operation, recall that $\displaystyle F(R) = R \wedge \bigvee_m B\Sigma_{m+}$, and let $f_n$ be the composite $$S^0 \rightarrow B\Sigma_{n+} \rightarrow R \wedge B\Sigma_{n+}\hookrightarrow F(R).$$ Then $\theta_{f_n}$ will be the $n$th power operation. You can now specialize to $A = Map(X,R)$ if you want.

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    $\begingroup$ Thanks. I'm struggling to follow the last sentence. Could you expand on that? $\endgroup$ – Mr. Palomar Jun 12 '20 at 8:47
  • $\begingroup$ Dear Nicholas, this is just a reminder that I've put up a bounty a few days ago in the hopes that someone could expand on the last sentence. As of yet I'm not at all following what is going on. I do not know what '$1$' means nor how any element of $\pi_0\Big(R \wedge \bigvee_n (B\Sigma_n)_+\Big)$ could ever give rise to a map of the form $R^0(X) \to R^0(X^{\times n}_{h\Sigma_n})$. I'd be grateful if you could help me out. $\endgroup$ – Mr. Palomar Jun 25 '20 at 8:05

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