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LSpice
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Are tamely ramified representations $\widehat{K^\text{tr}}$-admissible?

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 $$ H^1 (G_K/I_K^{(p)}, \operatorname{GL}_d(\widehat{\mathbb{Q}_p^\text{tr}}))=0 ?$$

Konstantin
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