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Is there any infinite family of $v$ for which all the $(v,k,\lambda)$-cyclic difference sets with $k-\lambda$ a prime power coprime to $v$ have been determined?

A subset $D=\{a_1,\ldots,a_k\}$ of $\mathbb{Z}/v\mathbb{Z}$ is said to be a $(v,k,\lambda)$-cycic difference set if for each nonzero $b\in\mathbb{Z}/v\mathbb{Z}$, there are exactly $\lambda$ ordered pairs $(a_s,a_t)\in D^2$ such that $a_s-a_t=b$. For a $(v,k,\lambda)$-difference set $D$, $k-\lambda$ is called the order.

Let $C_{v,n}$ be the set consisting of all cyclic difference sets of $\mathbb{Z}/v\mathbb{Z}$ with order $n$, and $$ C_v=\bigcup\limits_{\text{$n>1$ is a prime power coprime to $v$}}C_{v,n}. $$ For a fixed $v$, $C_v$ can be explicitly written down if $v$ is not too large. My question actually is: have we already know $C_v$ for infinitely many $v$s'?

I would pose another question related to this: does there exist an $N$ such that for all $v>N$, $|C_v|>0$?

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  • $\begingroup$ 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? $\endgroup$
    – Yuichiro Fujiwara
    Commented Jan 1, 2014 at 12:15
  • $\begingroup$ @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? $\endgroup$
    – Binzhou Xia
    Commented Jan 1, 2014 at 14:08
  • $\begingroup$ 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.) $\endgroup$
    – Yuichiro Fujiwara
    Commented Jan 1, 2014 at 14:59
  • $\begingroup$ 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. $\endgroup$
    – Yuichiro Fujiwara
    Commented Jan 1, 2014 at 15:00
  • $\begingroup$ 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. $\endgroup$
    – Yuichiro Fujiwara
    Commented Jan 1, 2014 at 15:24

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I'm not sure exactly what you mean because if you prove that there exists a cylcic $(v, k, \lambda)$-difference set for all $v$ except those that are excluded by known nonexistence results, you actually solved the existence problem entirely. So you should impose some condition(s) on parameters. But it can easily become trivial (e.g., cyclic difference sets with the classical parameters). So, it requires some careful and sensible choice of additional condition for your question to make sense and become interesting.

Whatever restriction you choose, here's a recent survey:

Q. Xiang, Recent progress in algebraic design theory, Finite Fields Appl. 11, (2005) 622–653.

And its preprint is available for free from the author's website:

http://titan.math.udel.edu/~xiang/surveyffa.pdf

There have been progress both on their existence and nonexistence. For example, as Theorem 3.4 of the paper says, fairly recently two of the major nonexistence conjectures were proved under the condition that $k-\lambda$ is a prime power greater than $3$. But if you ask if a cyclic $(v,k,\lambda)$-difference set exists for all $v$ that is not excluded by this recent nonexistence result (and the trivial conditions such as meeting the Bruck–Ryser–Chowla condition), then the answer is no, and there are open cases.

Small open parameters can be found in the following table:

http://www.ccrwest.org/diffsets.html

In general, we still have many major open problems when it comes to difference sets. The prime power conjecture (i.e., if $\lambda = 1$, then $n$ is a prime power) is open, too.

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  • $\begingroup$ Thanks! I've edited my question to make it more precise. $\endgroup$
    – Binzhou Xia
    Commented Dec 31, 2013 at 9:45

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