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Let $K$ be a finite field extension of $\mathbb{Q}_p$. Let $G_K$ denote the absolute Galois group of $K$, $I_K$ the inertia subgroup and $I_K^{(p)}$ the $p$-Sylow subgroup of $I_K$, i.e. the wild ramification group. Let us denote the maximal unramified extension by $K^\text{nr}$ and the maximal tamely ramified extension by $K^{tr}$. It is true that we have $$ H^1 (G_K/I_K, \operatorname{GL}_d(\widehat{\mathbb{Q}_p^\text{nr}}))=0. $$ Does there exist a similar result for $I^{(p)}_K$? I.e. do we have $$ H^1 (G_K/I_K^{(p)}, \operatorname{GL}_d(\widehat{\mathbb{Q}_p^\text{tr}}))=0 ?$$

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    $\begingroup$ What does the hat indicate? $\endgroup$
    – LSpice
    Commented Jan 26, 2022 at 19:07
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    $\begingroup$ @LSpice The $p$-adic completion of $\mathbb{Q}_p^{\mathrm{nr}}$ resp. $\mathbb{Q}_p^{\mathrm{tr}}$ $\endgroup$
    – Konstantin
    Commented Jan 26, 2022 at 19:08

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