Timeline for Looking for a reference: double orthogonal complement in $(\mathbb{Z}/q\mathbb{Z})^n$
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 7, 2013 at 21:28 | vote | accept | Michael Blondin | ||
| Nov 29, 2013 at 13:08 | comment | added | KConrad | Set $\langle v,u\rangle = e^{2\pi i(v \cdot u)/q}$. Then $v \cdot u = 0 \bmod q$ if and only if $\langle v,u\rangle = 1$. Each character of the finite abelian group $({\mathbf Z}/q{\mathbf Z})^n$ has the form $v \mapsto \langle v,u\rangle$ for a unique $u \in ({\mathbf Z}/q{\mathbf Z})^n$. (The term used here is that the group $({\mathbf Z}/q{\mathbf Z})^n$ is "self-dual": it's isomorphic to its own character group.) In any finite abelian group $G$ we have the property $(H^{\perp})^{\perp} = H$ for all subgroups $H$ of $G$. Try Terras, "Fourier Analysis on Finite Groups and Applications". | |
| Nov 29, 2013 at 12:07 | answer | added | David Wilding | timeline score: 2 | |
| Sep 30, 2013 at 8:16 | answer | added | Marc Palm | timeline score: 2 | |
| Sep 29, 2013 at 20:34 | review | First posts | |||
| Sep 29, 2013 at 20:36 | |||||
| Sep 29, 2013 at 20:19 | history | asked | Michael Blondin | CC BY-SA 3.0 |