Timeline for Sums of quadratic forms over finite abelian groups
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jul 17 at 7:12 | comment | added | Dave Benson | You might also be interested in Ed Brown's paper, "Generalizations of the Kervaire invariant", especially the discussion of quadratic forms in Section 3. It's not quite what you're asking, but it's related. | |
| Jul 17 at 2:13 | history | edited | LSpice | CC BY-SA 4.0 |
Capitalise title, while this is on the front page
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| Jul 17 at 2:05 | answer | added | Eric S. | timeline score: 2 | |
| Jul 13, 2017 at 13:07 | comment | added | user85913 | @EhudMeir The reduction to the cyclic case is not at all obvious (to me). But Chapter 5 of Scharlau's book Quadratic and Hermitian Forms appears to be very relevant. | |
| Jul 13, 2017 at 11:42 | comment | added | Ehud Meir | @t.c. I can see how to reduce to the case where $|A|$ is a prime power. However: how do we reduce the non-cyclic case to the cyclic case? | |
| Jul 13, 2017 at 8:07 | comment | added | user85913 | @LSpice, I'm not sure whether calling the sum a ``quadratic Gauss sum'' agrees with standard usage when $|A|$ is not prime, but I think at least that the computation is still ok: if $A=\mathbb Z/n$, then $q$ has the form $q(a,b)=\exp(2\pi i abc /n)$ for some $c\in (\mathbb Z/n)^*$, and the sum is $\sum_{a\in \mathbb Z/n} \exp(2\pi i a^2 c/n)$, which I believe was computed by Gauss. | |
| Jul 12, 2017 at 16:49 | comment | added | LSpice | @t.c., I think that further $\lvert A\rvert$ should be prime, no? | |
| Jul 12, 2017 at 14:44 | comment | added | user85913 | If $A$ is cyclic then this is an example of a quadratic Gauss sum, which have been computed by Gauss. | |
| Jul 12, 2017 at 13:45 | history | asked | Ehud Meir | CC BY-SA 3.0 |