# Two definitions of power operations — how do they relate?

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 $$A = Map(\Sigma^{\infty}_+X,R)$$.

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.
• 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. – Mr. Palomar Jun 25 '20 at 8:05