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A version of local Tate duality stated: Let $K$ be a finite extension of $\mathbb Q_p$, $A$ be a finite $G_K=Gal(\overline K/K)$ module. Then for $0\le i\le 2$, the cup product induces a perfect pairing $H^i(G_K,A)\times H^{2-i}(G_K,A')\to \mathbb Q/\mathbb Z$, where $A'=Hom(A,\overline K^\times)$. In particular, there is a canonical isomorphism $$H^i(G_K,A)\cong H^{2-i}(G_K,A')^\vee,$$ where $^\vee=Hom(\cdot,\mathbb Q/\mathbb Z).$

However, another version is used when studying the global/local deformation ring: Let $K$ be a finite extension of $\mathbb Q_p$, $\mathbb F$ be its residue field, and $V$ be a finite dimensional $\mathbb F$-vector space with a continuous $G_K$ action. Let $V^*$ be the dual representation. Then for $0\le i\le 2$, there is a canonical isomorphism $$H^i(G_K,V) \cong H^{2-i}(G_K,V^*(1))^*. $$

Can I ask whether it is possible to derive the second version from the first one? If not, is there a reference for the proof of the second version?

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  • $\begingroup$ You have explained the notation $A'$ and $(\ )^\vee$, but not $V^*(1)$ and $(\ )^*$. Please kindly explain! $\endgroup$
    – Mikhail Borovoi
    Commented Apr 22, 2023 at 12:34
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    $\begingroup$ Probably, when expalining, you will see yourself that (2) follows from (1)... $\endgroup$
    – Mikhail Borovoi
    Commented Apr 22, 2023 at 12:40
  • $\begingroup$ $()^*=Hom_{\mathbb F}(\cdot,\mathbb F)$ is the $\mathbb F$-linear dual and $V^*(1)=V^*\otimes_{\mathbb Z_p}\mathbb Z_p(1)$ means the Tate twist of $V^*$, where $\mathbb Z_p(1)$ is the p-adic cyclotomic character of $G_K$. $\endgroup$
    – user14411
    Commented Apr 22, 2023 at 14:13
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    $\begingroup$ Does it help to note that if $\mathbb{L} / \mathbb{F}$ is an extension of finite fields, then the functors on $\mathbb{L}$-vector spaces given by $Hom_{\mathbb{L}}(-, \mathbb{L})$ and $Hom_{\mathbb{F}}(-, \mathbb{F})$ are canonically the same? $\endgroup$
    – David Loeffler
    Commented Apr 22, 2023 at 21:31
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    $\begingroup$ @MikhailBorovoi The natural transformation $Hom_L(-, L) \to Hom_F(-, F)$ is composing with trace. The transformation $Hom_F(-, F) \to Hom_L(-, L)$ is the composite of base-extension $Hom_F(V, F) \to Hom_L(V \otimes_F L, L)$ with the embedding $V \into V \otimes_F L$. $\endgroup$
    – David Loeffler
    Commented May 4, 2023 at 18:31

1 Answer 1

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EDIT: I treat the general case. Write ${\Bbb F}={\Bbb F}_q$ where $q=p^l$ for some natural $l$. Then ${{\Bbb F}}_q\supseteq {{\Bbb F}}_p={\Bbb Z}/p{\Bbb Z}$.

Using a comment of @DavidLoeffler, we obtain that for an ${\Bbb F}$-vector space $W$, we can identify \begin{multline*} W^*={\rm Hom}_{{\Bbb F}}(W, {\Bbb F})\cong {\rm Hom}_{{{\Bbb F}_p}}(W, {\Bbb F}_p) ={\rm Hom}(W, {\Bbb Z}/p{\Bbb Z})\\ \cong{\rm Hom}(W, \tfrac1p{\Bbb Z}/{\Bbb Z}) ={\rm Hom}(W, {\Bbb Q}/{\Bbb Z})=W^\vee. \end{multline*}

By definition, our $V$ is a finite $G_K$-module and $$ V^*(1)\cong{\rm Hom}_{{\Bbb F}}(V,{{\Bbb F}})\otimes_{{{{\Bbb F}_p}}}\mu_p$$ where $\mu_p$ denotes the group of roots of unity of degree dividing $p$ in $\overline K^\times$. We can identify \begin{multline*} V^*(1)\cong{\rm Hom}_{{\Bbb F}}(V,{{\Bbb F}})\otimes_{{{{\Bbb F}_p}}}\mu_p\cong {\rm Hom}(V,{\Bbb Z}/p{\Bbb Z})\otimes_{{{\Bbb F}_p}} \mu_p\\ \cong {\rm Hom}(V,\mu_p)={\rm Hom}(V, \overline K^\times)=V'. \end{multline*}

We conclude that $$H^{2-i}(G_K,V^*(1))^*\cong H^{2-i}(G_K, V')^\vee. $$ Now we see that the second assertion of the question is a special case of the first one.

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  • $\begingroup$ I cannot treat the case of $q=p^l$ when $l>1$ because I don't understand your definition of $V^*(1)$. Try to unravle the definition and to construct a $G_K$-equivariant bilinear pairing $$V\times V^*(1) \to \mu_q. $$ $\endgroup$
    – Mikhail Borovoi
    Commented Apr 22, 2023 at 17:33
  • $\begingroup$ Thanks for your answer. In the definition of $V^*(1)$, we view $V^*=Hom_{\mathbb F}(V,\mathbb F)$ as a $\mathbb Z_p$-module by letting $\mathbb Z_p$ acts on $\mathbb F$ via $\mathbb Z_p\to \mathbb F_p\to \mathbb F$. In case it helps, this statement of local Tate duality appears on p.32 of link, though his notation is slightly different from mine. $\endgroup$
    – user14411
    Commented Apr 22, 2023 at 21:04

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