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Let $\Gamma_p : \mathbb{Z}_p \to \mathbb{Z}_p^{\times}$ be the $p$-adic gamma function. I thought that I had successfully calculated $\Gamma_p(1 - 1/4)$, but sage is telling me I'm wrong (this is more than a sign error, which I have probably made along the way). Here is what I've done:

Let $p > 2$ be a prime such that $p \equiv 1 \bmod 4$, and $q = p^2$. We will assume that all roots of unity lie in $\mathbb{C}_p^{\times}$. Let $\omega_q : \mathbb{F}_q^{\times} \to \mu_{q-1}\subseteq \mathbb{Z}_p[\mu_q]$ be the Teichmüller character and $\zeta$ a primitive $p$-th root of unity. We denote $\chi = \omega_q^{(q-1)/4}$. Let $\operatorname{Tr} : \mathbb{F}_q \to \mathbb{F}_p$ and $N : \mathbb{F}_q \to \mathbb{F}_p$ be the trace and norm respectively. For a finite field $F$ and multiplicative character $\psi$ on $F$, the Gauss sum for $\psi$ is defined to be \begin{align*} G(\psi) = \sum_{c \in F} \psi(c) \zeta^{\operatorname{Tr}_{F/\mathbb{F}_p}(c)} \end{align*} and the Jacobi sum for $\psi$ is defined \begin{align*} J(\psi) = \sum_{c \in F} \psi(c) \psi(1-c). \end{align*} Let $\pi$ be the maximal prime in $\mathbb{Z}_p(\zeta)$ such that $\pi \equiv 1-\zeta \mod{\pi^2}$.

It follows from the Gross–Koblitz formula that \begin{align*} \Gamma_p\left(1 - \frac{1}{4}\right) ^2= -\frac{\pi^{(p-1)/2}G(\chi)}{q}. \end{align*}

Also, if $\omega = \omega_q \rvert_{\mathbb{F}_p} : \mathbb{F}_p \to \mu_{p-1}$ and $\psi = \omega^{(p-1)/4}$, then since \begin{align*} \chi(c) = \omega_q^{(p-1)/4}(c^{p+1}) = \omega_q^{(p-1)/4}(N(c)) = \omega^{(p-1)/4}(N(c)) \end{align*} $G(\chi) = G^2(\psi)$ by the Hasse–Davenport lifting relation.

Now, $\psi$ is a quartic character on $\mathbb{F}_p$, which means $\psi^2$ a quadratic character. So, from the well known value
\begin{align*} G(\psi^2) = i\pi^{(p-1)/2} \end{align*} and from the relation $J(\psi) = G^2(\psi)/G(\psi^2)$, we have \begin{align*} G^2(\psi) = i\pi^{(p-1)/2} J(\psi). \end{align*}

Putting the pieces together, we now have \begin{align*} \Gamma_p\left(1 - \frac{1}{4}\right) ^2= \frac{iJ(\psi)}{p}. \end{align*}

Again, this all seems correct to me, but sage is telling me I'm wrong. I've been looking at this for a while, and I can't seem to figure out where the mistake is. My first guess is that I messed up the Hasse–Davenport part, but I can't quite see how.

Edits: The value for $G(\psi^2)$ is really $i\pi^{(p-1)/2}$ (see comments below). However, this did not seem to fix the issue in sage.

To check my answer, I calculate $iJ(\psi)/p$ and $\Gamma_p(1-1/4)$ mod $p^2$ with $p = 13$ and compare. I know that $J(\psi) \equiv \sum_{c = 1}^{p-1}c^{p(p-1)/4}(1-c)^{p(p-1)/4}$ mod $p^2$, and that $i$ is either congruent to $70$ or $99$ mod $p^2$.

Input:

p = 13
m = 4
n = (p-1)/4
(Mod(99*sum((i^(p*n))*((1-i)^(p*n))for i in (1..p-1))/p,p^2),Mod(prod(i^2 for i in(1..(p^2-1)/4) if gcd(i,p)==1),p^2))

Output: (140, 114)

Input:

p = 13
m = 4
n = (p-1)/4
(Mod(70*sum((i^(p*n))*((1-i)^(p*n))for i in (1..p-1))/p,p^2),Mod(prod(i^2 for i in(1..(p^2-1)/4) if gcd(i,p)==1),p^2))

Output: (29, 114)

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    $\begingroup$ @KConrad's answer to your earlier question seems to say that $G(\psi^2)$ equals $\pi^{(p - 1)/2}\Gamma_p(1/2)$. Why do you say that it equals $\pi^{(p - 1)/2}$? $\endgroup$
    – LSpice
    Mar 9 at 18:04
  • $\begingroup$ @LSpice I thought that since $p \equiv 1 \mod 4$ the quadratic Gauss sum is $\sqrt{p}$. $\endgroup$ Mar 9 at 18:19
  • $\begingroup$ It's certainly $\pm\sqrt p$, and I think it can be either. I don't know enough about the Gamma function to know whether that agrees with @KConrad's answer. $\endgroup$
    – LSpice
    Mar 9 at 18:44
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    $\begingroup$ I'm seeing now that I made the mistake of thinking $\pi^{(p-1)/2} = \sqrt{p}$, when really $\pi^{(p-1)/2} = \sqrt{-p}$. As @KConrad says, $\Gamma_p(1/2) = \sqrt{-\left(\frac{-1}{p} \right)}$, which is $\sqrt{-1}$ if $p \equiv 1 \mod 4$. So my answer should be $\sqrt{-1}J(\psi)/p$. I'll see if sage agrees. $\endgroup$ Mar 9 at 18:58
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    $\begingroup$ Even classically (in $\mathbf C$), to say the Gauss sum of the Legendre symbol is the positive real $\sqrt{p}$ requires a convention, such as using powers of $e^{2\pi i/p}$ in the definition of the Gauss sum. Over the $p$-adics, there is no canonical choice of square root of $p$. Perhaps you can link the choice of $\sqrt{p}$ as the value of a Gauss sum with the choice of nontrivial $p$th root of unity used in the Gauss sum, but this may be delicate. You definitely need to be careful about how you translate Gauss sum formulas in the $p$-adics from knowing them in the complex numbers. $\endgroup$
    – KConrad
    Mar 11 at 8:09

1 Answer 1

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As usual, this was a goofy mistake on my part. Along with the mistake pointed out by LSpice in the comments, I realized that I had been calculating the wrong thing in Sage. With $p = 13$, calculating $iJ(\psi)/p$ mod $p^2$ requires me to look at $J(\psi)$ mod $p^3$, (since there is a $p$ in the denominator). So, once I input $\sum_{c = 1}^{p-1}c^{p^2(p-1)/4}(1-c)^{p^2(p-1)/4}$ and $i \equiv 99 \mod p^2$, everything matches up and I can finally stop pulling out my hair.

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