Timeline for Is there a non-degenerate quadratic form on every finite abelian group?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 25, 2020 at 6:19 | vote | accept | Sebastien Palcoux | ||
| Oct 14, 2020 at 23:46 | comment | added | Konstantinos Kanakoglou | @Sebastien Palcoux: essentially you are asking whether any fin Abelian group bicharacter can be written in terms of a non-degenerate quadratic form ? | |
| Oct 14, 2020 at 23:36 | answer | added | Qiaochu Yuan | timeline score: 6 | |
| Oct 14, 2020 at 23:28 | comment | added | Jack L. | @Sebastien Palcoux: The old idea works! I just observed that a parity tweak was all that was needed. | |
| Oct 14, 2020 at 15:25 | comment | added | Sebastien Palcoux | @JackL. For the cyclic group $C_n$, it should be $q: e^{\frac{2\pi i k}{n}} \mapsto e^{\frac{2\pi i k^2}{n}}$, with $k$ integer. Let $A$ be an abelian group, it is of the form $\prod_i C_{n_i}$. Let $N$ be $lcm_i(n_i)$, then $C_{n_i}$ is a subgroup of $C_N$. By combining that with the idea in your answer, something natural can be written; but is it a non-degenerate quadratic form? If $N$ is odd, it should be ok, but if $N$ is even, it is degenerate because of the multiplicative constant $2$. Do you see a way to fix the even case? | |
| Oct 14, 2020 at 13:32 | comment | added | Jack L. | Thanks. Then I’ve considered shelving my posted answer and have a rethink of it. | |
| Oct 14, 2020 at 13:25 | comment | added | Sebastien Palcoux | There is a problem with my first comment because in general $e^{i(\theta+2\pi)^2} \neq e^{i\theta^2}$. | |
| Oct 14, 2020 at 12:45 | comment | added | Sebastien Palcoux | @JackL. A bicharacter $B$ is non-degenerate if for all non-identity $g ∈ G$ there exists some elements $h,h’ ∈ G$ such that $B(g, h), B(h’,g) \neq 1$. | |
| Oct 14, 2020 at 12:02 | answer | added | Jack L. | timeline score: 4 | |
| Oct 14, 2020 at 11:41 | comment | added | Jack L. | What do you mean when you say the bi-character is non-degenerate (do you mean non-trivial, as in it becomes a constant function?). Sorry, couldn’t figure it out from the statement (unless it’s obvious in a way I can’t seem to see). | |
| Oct 14, 2020 at 10:39 | comment | added | Sebastien Palcoux | @PedroTamaroff: Yes, I guess you mean the quadratic form $q: e^{i\theta} \mapsto e^{i\theta^2}$. Then $b(e^{i \theta_1},e^{i \theta_2}) = e^{2i \theta_1\theta_2}$ is a non-degenerate bicharacter. So it is ok for the cyclic groups. | |
| Oct 14, 2020 at 9:40 | comment | added | Pedro | You can realize any cyclic group as a subgroup of $S^1\subseteq \mathbb C^\times$. Isn't there any way to get going from there, using the bona-fide quadratic form $z\mapsto z^2$ in the circle? Do you have any ideas? | |
| Oct 14, 2020 at 9:22 | history | edited | YCor | CC BY-SA 4.0 |
fixed typo
|
| Oct 14, 2020 at 9:02 | history | asked | Sebastien Palcoux | CC BY-SA 4.0 |