Timeline for Finite field "contour" sum
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Mar 4, 2021 at 19:43 | vote | accept | Jeanne Scott | ||
| Jun 21, 2016 at 13:22 | history | edited | Dan Petersen | CC BY-SA 3.0 |
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| Apr 25, 2015 at 22:18 | history | edited | Will Sawin | CC BY-SA 3.0 |
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| Apr 25, 2015 at 22:15 | comment | added | Will Sawin | @GHfromMO yes, I think I miswrote both those things. I'll fix it. | |
| Apr 25, 2015 at 21:53 | history | edited | GH from MO | CC BY-SA 3.0 |
fixed typos
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| Apr 25, 2015 at 21:47 | comment | added | GH from MO | In your 5th display, I think $\overline{\chi}^* \left(a -b \sigma + (r+1) \sigma \right)$ should be $\overline{\chi}^* \left(a -b \sigma - (r+1) \sigma \right)$. Also, I could not understand what you meant by "where you are taking $a = \sqrt{\delta} b$". | |
| Apr 25, 2015 at 21:36 | history | edited | GH from MO | CC BY-SA 3.0 |
fixed typos
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| Apr 25, 2015 at 21:07 | history | edited | Will Sawin | CC BY-SA 3.0 |
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| Apr 22, 2015 at 17:27 | comment | added | Will Sawin | @GHfromMO I will soon, and also edit my answer to include a monodromy computation. | |
| Apr 22, 2015 at 15:50 | comment | added | GH from MO | What I meant is: I understood why the second display equals the third display. I also understand now that the trace of the Frobenius of $\mathbb{F}_{q^2}$ on $\mathcal{L}_\chi$ at $a+b\sigma$ equals $\chi(a+b\sigma)\chi(a-b\sigma)$. Correct me if I am wrong. At any rate, if you have a chance, include further details for the sake of non-expert readers like me. | |
| Apr 22, 2015 at 15:31 | comment | added | Will Sawin | @GHfromMO Really I should replace the $\chi$s with $\mathcal L_{\chi}$s to clarify what I am doing. | |
| Apr 22, 2015 at 15:30 | comment | added | Will Sawin | No. Already in the first display I am writing an element of $\mathbb F_q^2$ as a sum of two elements. This corresponds to a sheaf on a curve. I am then taking the same sheaf on the same curve and looking at it over $\mathbb F_{q^2}$, so $a$ and $b$ are now elements of $\mathbb F_{q^2}$. What happens to a multiplicative character as you go to a larger field is you compose with the norm map, which corresponds to this doubling. | |
| Apr 22, 2015 at 14:41 | comment | added | GH from MO | $\chi$ is a multiplicative character, not an additive one, so I still don't see how $\chi$ is doubled. I understood the second point, each $\alpha\in\mathbb{F}^\times_{q^2}$ corresponds to a unique pair $(a,b)\in\mathbb{F}_{q^2}$ in the second display. | |
| Apr 22, 2015 at 14:17 | comment | added | Will Sawin | @GHfromMO They are not equal. Rather, they are related by the theory of exponential sums over finite fields. The second sum is the generalization of the first sum from $\mathbb F_q$ to $\mathbb F_{q^2}$. $\chi$ is doubled because we are replacing the trace of $Frob_q$ with the trace of $Frob_{q^2} Frob_{q}^2$. In the second display I do not mean to restrict to a special sort of $\alpha$ - I am taking the equation $a^2- \delta b^2=(\delta/4)(4-r^2)$ and looking at it over a larger field. | |
| Apr 22, 2015 at 14:14 | comment | added | GH from MO | I am sorry, but there are several things I don't understand. Why is the first display equal to the second display? The first display involves two values of $\chi$, the second display involves four values of $\chi$. Also, why is the second display equal to the third display? With the notation $\alpha=a+\sigma b$, the second display is restricted to $\alpha$ such that $\alpha^{1+q}=\frac{\delta}{4}(4-r^2)$, and that information is missing in the third display. | |
| Apr 22, 2015 at 2:45 | history | answered | Will Sawin | CC BY-SA 3.0 |