Timeline for Finding a certain value of $\Gamma_p$
Current License: CC BY-SA 4.0
12 events
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| Apr 8, 2022 at 21:06 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Mar 11, 2022 at 8:09 | comment | added | KConrad | 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. | |
| Mar 9, 2022 at 20:29 | answer | added | matt stokes | timeline score: 1 | |
| Mar 9, 2022 at 19:53 | history | edited | matt stokes | CC BY-SA 4.0 |
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| Mar 9, 2022 at 19:45 | history | edited | matt stokes | CC BY-SA 4.0 |
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| Mar 9, 2022 at 19:33 | history | edited | matt stokes | CC BY-SA 4.0 |
added 146 characters in body
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| Mar 9, 2022 at 18:58 | comment | added | matt stokes | 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. | |
| Mar 9, 2022 at 18:44 | comment | added | LSpice | 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. | |
| Mar 9, 2022 at 18:19 | comment | added | matt stokes | @LSpice I thought that since $p \equiv 1 \mod 4$ the quadratic Gauss sum is $\sqrt{p}$. | |
| Mar 9, 2022 at 18:06 | history | edited | LSpice | CC BY-SA 4.0 |
Teichmuller -> Teichmüller, and other minor changes
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| Mar 9, 2022 at 18:04 | comment | added | LSpice | @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 $\pi^{(p - 1)/2}$? | |
| Mar 9, 2022 at 17:59 | history | asked | matt stokes | CC BY-SA 4.0 |