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Let $k$ be a field of characteristic $p$ and $D^b(k)$ be the infinity (equivalently, DG) category of perfect complexes over $k$. Let $C_p(=\mathbb{Z}/p)$ be the cyclic group on $p$ elements. For a $C_p$ equivariant complex $V\in D^b(k)^{C_p}$, define the "Tate invariants functor", $T(V):= \mathrm{Cone}(N:V_{hC_p}\to V^{hC_p})$. This will be a perfect module over the coefficient ring $T(k)$.

Define the "universal power operation functor" to be the functor $\hat{T}:D^b(k)\to D^b(T(k))$, with $V\mapsto T(V^{\otimes p})$. This functor is famously exact (in the infinity-categorical sense), hence given by smashing by a $k-T(k)$ bimodule spectrum, $\hat{T}(k)$, whose bimodule structure involves all the Steenrod operations (See Lurie's "Rational and p-adic Homotopy Theory" 2.2, though I hear the construction is much more classical).

Any $\sigma\in C_p$ induces an automorphism of $T^{\otimes p}$ by permuting the tensor components, and this descends to a $C_p$-action on the functor $\hat{T}$ by natural transformations, which by Yoneda's lemma must be given an action on the bimodule spectrum $\hat{T}(k)$. My question: is this action trivial?

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  • $\begingroup$ Hi Dmitry. What does the symbol "D^b(T(k))" mean? $\endgroup$ Commented Jan 12, 2015 at 1:41
  • $\begingroup$ Hi David, I mean the category of perfect modules over $T(k)$, which I think it's enough to consider as a DG algebra over $k$. (Correct me if this doesn't make sense). $\endgroup$ Commented Jan 12, 2015 at 2:19

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