Timeline for Known results on cyclic difference sets
Current License: CC BY-SA 3.0
14 events
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| Jan 2, 2014 at 4:18 | comment | added | Yuichiro Fujiwara | I don't think so unless I'm missing something trivial. The former question requires you to list up all possible nonisomorphic ones for each $v$, which is quite a difficult problem even for a fixed $v$ if it's large. In fact, it's already very difficult to enumerate all of them (i.e., compute the number of nonisomorphic ones) for fixed $v$ and $n$, let alone to construct all of them... | |
| Jan 2, 2014 at 3:45 | comment | added | Binzhou Xia | @YuichiroFujiwara To my knowledge, the answer to the latter question is not known yet. For the former question, is the answer "yes" up to isomorphisms of designs? | |
| Jan 1, 2014 at 15:24 | comment | added | Yuichiro Fujiwara | As for your questions, the answer to the first one seems very likely to be "no" because we need to know all inequivalent cyclic difference sets that belong to $C_v$ for infinitely many $v$. You can check all relevant papers published in these several years; older results should be covered by the linked article, CRC Handbook, and reference therein. In principle, you should be able to tell if the answer to the latter question is known on your own this way as well, although it might require a lot of reading... I'm guessing it's not known yet if such $N$ exists. | |
| Jan 1, 2014 at 15:00 | comment | added | Yuichiro Fujiwara | A typical definition of equivalence between $D_0$ over group $G_0$ and $D_1$ over $G_1$ is that they're equivalent if there is a group isomorphism $\pi$ between $G_0$ and $G_1$ such that $\{\pi(d) \mid d \in D_0\} = D_1 +g$ for some $g$. In your case, $G_0$ and $G_1$ are both cyclic. So, we only need to consider the group automorphisms, i.e., the unit group. So your definition of equivalence is pretty much the same. Equivalent difference sets are all isomorphic. But the converse may not be true, although in the cyclic case, as far as I know, all known isomorphic difference sets are equivalent. | |
| Jan 1, 2014 at 14:59 | comment | added | Yuichiro Fujiwara | A common definition (given in CRC Handbook of Combinatorial Designs) is that two difference sets $D_0$ and $D_1$ are isomorphic if the designs $\operatorname{dev}(D_0)$ and $\operatorname{dev}(D_1)$ are isomorphic. As you probably already know, by developing a difference set, you get a sharply point-transitive design. So, if the corresponding designs are isomorphic in the usual sense in design theory, your difference sets are isomorphic. This notion is different from the equivalence you wrote there. (cont.) | |
| Jan 1, 2014 at 14:09 | history | edited | Binzhou Xia | CC BY-SA 3.0 |
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| Jan 1, 2014 at 14:08 | comment | added | Binzhou Xia | @YuichiroFujiwara Thanks for your comment which helps me to edit my question. (hope it is clearer now!) What do you mean by 'isomorphic' of cyclic difference set? If $D'=cD+d$ for some $c,d\in\mathbb{Z}/v\mathbb{Z}$ and $c$ coprime to $v$ then I say $D'$ is equivalent to $D$. Do you mean just this equivalence or any more? | |
| Jan 1, 2014 at 13:58 | history | edited | Binzhou Xia | CC BY-SA 3.0 |
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| Jan 1, 2014 at 12:15 | comment | added | Yuichiro Fujiwara | I guess you identify two isomorphic cyclic difference sets when defining the set $C_v$...? If that's the case, are $p$ and $f$ fixed, too? For example, is $\vert C_v \vert$ the number of all nonisomorphic ones of the same parameters? Or do you mean $C_v = \bigcup_{p, f} C_{v,p,f}$, where $C_{v,p,f}$ is the set of nonisomorphic difference sets of order $v$ with $k-\lambda = p^f$ and $\operatorname{gcd}(k-\lambda, v)=1$? In any case, when you say, "$C_v$ is explicitly known," do you mean a complete characterization of cyclic difference sets (up to isomorphism, I guess) is known for some cases? | |
| Jan 1, 2014 at 11:23 | history | edited | Binzhou Xia | CC BY-SA 3.0 |
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| Dec 31, 2013 at 9:44 | history | edited | Binzhou Xia | CC BY-SA 3.0 |
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| Dec 31, 2013 at 9:19 | history | edited | Binzhou Xia | CC BY-SA 3.0 |
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| Dec 31, 2013 at 4:49 | answer | added | Yuichiro Fujiwara | timeline score: 3 | |
| Dec 31, 2013 at 3:06 | history | asked | Binzhou Xia | CC BY-SA 3.0 |