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David Loeffler
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No. The Gauss sum is not a $p$-adic measure. One cheap way to see this is as follows: if $\chi$ has conductor $p^n$, the $p$-adic valuation of $G(\chi)$ is $n/2$, where $n$ is the conductor of $\chi$. But if $\mu$ is a mearuemeasure, the asymptotics of $\int \chi \mathrm{d}\mu$ for $\chi$ of increasing $p$-power conductor, are governed by the $\lambda$ and $\mu$ invariants of $\mu$, and in particular the valuations of these numbers must tend to a limit -- they cannot tend to $\infty$.

No. The Gauss sum is not a $p$-adic measure. One cheap way to see this is as follows: the $p$-adic valuation of $G(\chi)$ is $n/2$, where $n$ is the conductor of $\chi$. But if $\mu$ is a mearue, the asymptotics of $\int \chi \mathrm{d}\mu$ for $\chi$ of increasing $p$-power conductor, are governed by the $\lambda$ and $\mu$ invariants of $\mu$, and in particular must tend to a limit -- they cannot tend to $\infty$.

No. The Gauss sum is not a $p$-adic measure. One cheap way to see this is as follows: if $\chi$ has conductor $p^n$, the $p$-adic valuation of $G(\chi)$ is $n/2$. But if $\mu$ is a measure, the asymptotics of $\int \chi \mathrm{d}\mu$ for $\chi$ of increasing $p$-power conductor, are governed by the $\lambda$ and $\mu$ invariants of $\mu$, and in particular the valuations of these numbers must tend to a limit -- they cannot tend to $\infty$.

Source Link
David Loeffler
  • 37k
  • 3
  • 89
  • 194

No. The Gauss sum is not a $p$-adic measure. One cheap way to see this is as follows: the $p$-adic valuation of $G(\chi)$ is $n/2$, where $n$ is the conductor of $\chi$. But if $\mu$ is a mearue, the asymptotics of $\int \chi \mathrm{d}\mu$ for $\chi$ of increasing $p$-power conductor, are governed by the $\lambda$ and $\mu$ invariants of $\mu$, and in particular must tend to a limit -- they cannot tend to $\infty$.