# A $p$-adic “Jacobi sum” with an unramified character

Working over a $p$-adic field with absolute value $|\cdot|$, let $\chi$ be a character on ${\mathfrak o}^\times$ with conductor $n\ge 1$, meaning that $n$ is the smallest integer such that $\chi$ is trivial on $U_n=1+{\mathfrak p}^n$. I'm trying to calculate the integral, with ${\rm Re}(s)\gg 1$, $$J(\chi,|\cdot|^s)=\int_{\mathfrak o^\times}\chi(x)|x-1|^s\ dx$$ Note that if we replace $|\cdot|^s$ with a character $\chi'$ with conductor $n$, this would reduce to the typical Jacobi sum, and we'd have the formula $$J(\chi,\chi')={G(\chi,\psi)G(\chi',\psi)\over G(\chi\chi',\psi)}$$ where $\psi$ is an additive character with conductor $n$ and $G(\chi,\psi)$ is the Gauss sum $$G(\chi,\psi)=\int_{\mathfrak o^\times}\chi(x)\psi(x)\ dx$$ Of course, this formula wouldn't work for $\chi'$ unramified.

Is anyone aware of how to calculate this sort of Jacobi sum?

I think I've figured out how to do it. It's essentially the same argument as in the "classical" case, just interpreted slightly differently. Here's the sketch, which may lose some change-of-measure constants. $$J(\chi,|\cdot|^s)=\int_{\mathfrak o^\times}\chi(x)|x-1|^s\ dx$$ Extend $\chi$ by zero to all of $k$ and apply Fourier inversion to get $$=\int_{\mathfrak o^\times}\int_k |x-1|^s\psi(-ax)\int_{\mathfrak o^\times}\chi(y)\psi(ay)\ dy\ da\ dx$$ where $\psi$ has conductor $0$. The integral over $y$ vanishes unless ${\rm ord}(a)=-n$, so the integral becomes $$=\int_{\mathfrak o^\times}\int_{\mathfrak o^\times}|x-1|^s\psi(-\varpi^{-n}ux)\int_{\mathfrak o^\times}\chi(y)\psi(\varpi^{-n}uy)\ dy\ du\ dx$$ $$=\int_{\mathfrak o^\times}\int_{\mathfrak o^\times}|x-1|^s\psi(-\varpi^{-n}ux)\bar\chi(u)\int_{\mathfrak o^\times}\chi(y)\psi(\varpi^{-n}y)\ dy\ du\ dx$$ $$=G(\chi,\psi_{\varpi^{-n}})\int_{k}\int_{\mathfrak o^\times}{\bf 1}_{\mathfrak o^\times}(x)|x-1|^s\psi(-\varpi^{-n}ux)\bar\chi(u)\ du\ dx$$ $$=G(\chi,\psi_{\varpi^{-n}})\int_{k}\int_{\mathfrak o^\times}{\bf 1}_{\mathfrak o^\times}(1+x)|x|^s\psi(-\varpi^{-n}ux)\psi(-\varpi^{-n}u)\bar\chi(u)\ du\ dx$$ $$=q^{-ns-1}G(\chi,\psi_{\varpi^{-n}})\int_{k}\int_{\mathfrak o^\times}{\bf 1}_{\mathfrak o^\times}(1-\varpi^{n}u^{-1}x)|x|^s\psi(x)\bar\psi(\varpi^{-n}u)\bar\chi(u)\ du\ dx$$ Now, as a function of $x$, the characteristic function ${\bf 1}_{\mathfrak o^\times}(1-\varpi^{n}u^{-1}x)$ differs from ${\bf 1}_{\mathfrak o}(\varpi^{n}x)$ only on a set of measure zero, so we have $$J(\chi,|\cdot|^s)=q^{-ns-1}\big|G(\chi,\psi_{\varpi^{-n}})\big|^2\int_{\varpi^{-n}\mathfrak o}|x|^s\psi(x)\ dx$$ This final integral can be calculated using the well-known formula for $\int_{|x|=p^i}\psi(x)\ dx$, giving (if I did everything correctly), $$J(\chi,|\cdot|^s)=q^{-ns-1}\big|G(\chi,\psi_{\varpi^{-n}})\big|^2{2-q^s\over 1-q^{-s}}$$
• And, of course, the norm-square of the Gauss sum is $q^{-n}$. – B R Dec 12 '12 at 22:52