# Are cyclic codes bounded by a continuous function?

In coding theory, we know that if you take the function

$$$$\alpha_q(\delta) := \limsup_{n \rightarrow \infty} \ \max \{ R(C) \mid \delta(C) \ge \delta \mid C \subseteq \mathbb{F}_q^n\}.$$$$

then the function is continuous. Here $$R(C)$$ is the rate of a code $$C$$ and $$\delta(C)$$ the relative minimum distance. This follows by spoiling a code appropriately to have codes in the vicinity of any point in the graph. A proof is in Tsfasman and Vladut's "Algebraic-geometric codes" book, for example.

I think the original proof is by Manin, but I am not sure (would like to know!)

Now if I restrict only to cyclic codes and define a similar function as the following:

$$$$\alpha_q^{\text{cyclic}}(\delta) := \limsup_{n \rightarrow \infty} \ \max \{ R(C) \mid \delta(C) \ge \delta \mid C \text{ is cyclic} \}. \label{eq:defi_of_Alp_cyc}$$$$

Is it known that this function is also continuous? The usual proof for the general case cannot be extended (as far as I could tell).

EDIT: This question is an open problem, because it proves an open problem.

However, I am now changing my demand. Instead of 'cyclic', I can choose other class of codes like 'Goppa' codes or 'transitive' codes and make similar constructions $$\alpha_q^{\text{Goppa}}$$ or $$\alpha_q^{\text{trans}}$$ for instance. Are any such functions known to be continuous?

• On a second thought, I think proving it to be continuous will solve the long-standing "good cyclic codes" conjecture. – Breakfastisready Jan 10 at 21:54

To answer the question on the origin of $$\alpha_q$$. The original proof goes back to Manin's article "What is the maximum number of points on a curve over $$\mathbb F_2$$", J. Fac. Sci. Univ. Tokyo, 1981. Manin and Marcolli have some more recent articles ([1], [2]) studying $$\alpha_q$$ from a sort of a statistical mechanical perspective.
If we prove $$\alpha_q^{\text{cyclic}}$$ to be continuous, we solve the long-standing problem of "are there good cyclic codes?".
This is because $$\alpha_q^{\text{cyclic}}(0) = 1$$ trivially and by continuity we would then have $$\alpha_q^{\text{cyclic}}(\delta)> 0$$ for some $$0 < \delta < 1$$. This proves good cyclic codes.