Skip to main content
minor edit: added the word "perfect" in a couple places
Source Link

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?

Let $k$ be a field of characteristic $p$ and $D^b(k)$ be the infinity (equivalently, DG) category of 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 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?

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?

edited typo
Source Link

Let $k$ be a field of characteristic $p$ and $D^b(k)$ be the infinity (equivalently, DG) category of 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(\mathbb{F}_p)^{C_p}$$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 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?

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

Let $k$ be a field of characteristic $p$ and $D^b(k)$ be the infinity (equivalently, DG) category of 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 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?

Source Link

Twisting of the power functor

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