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Let $F$ be an unramified extension of $\mathbf{Q}_p$ of degree $n$, and let $K = F(\alpha)$ where $\alpha$ satisfies $\alpha^{p^n - 1} = -p$.

I'm interested in the local $\varepsilon$-factors attached to finite-order characters of the Galois group of $G_k = \operatorname{Gal}(K(\mu_{p^k}) / F)$, for $k \gg 0$. Specifically, I'd like to know if there's any relation between these $\varepsilon$-factors and the quantities $$ \tau(u, \omega) = \sum_{g \in G_k} \omega(g)^{-1} g(u) $$ for suitable $u \in K(\mu_{p^k})$. If $n = 1$, so $F = \mathbf{Q}_p$ and $K=\mathbf{Q}_p(\mu_p)$, then $\tau(u, \omega)$ for $u = \zeta_{p^k}$ is just a Gauss sum, and it's well known that these are the epsilon-factors. Can one say anything useful for $n \ge 2$?

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  • $\begingroup$ is the $g$ in the sum supposed to be the same as the summation index $\gamma$ ? $\endgroup$
    – Chandan Singh Dalawat
    Commented Dec 6, 2013 at 13:09
  • $\begingroup$ Yes, it was, sorry! $\endgroup$
    – David Loeffler
    Commented Dec 6, 2013 at 17:40
  • $\begingroup$ when $n=1$, you have $F=\mathbf{Q}_p$ of course, but why do you have $K=F$ ? Your definition gives $K=F(\root{p-1}\of{-p})$, so $[K:F]=p-1$, and in fact $K=\mathbf{Q}_p(\zeta_p)$. $\endgroup$
    – Chandan Singh Dalawat
    Commented Dec 8, 2013 at 3:43
  • $\begingroup$ You are, of course, right; $K$ is not $F$ (but $K(\mu_{p^k})$ is $F(\mu_{p^k})$ for $k \ge 1$) $\endgroup$
    – David Loeffler
    Commented Dec 8, 2013 at 10:31

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