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I came across this paper by V.I.Yorgov named "Binary self-dual codes with automorphisms of odd order". (Actually I was first reading another paper by Borello that cited this paper. It later become important that I actually understand the proof of this theorem.) Unfortunately I don't know any Russian, so I couldn't read it. I wonder if anyone could help me with this? Or anyone knows how to prove the theorem?

The paper is here:

http://www.mathnet.ru/links/64c42446ed8bceff5260b4ea78ce9c1f/ppi1198.pdf

A brief explanation of the theorem Borello is citing:

Let $\sigma$ be the automorphism of order $p$. Assume the length of the code is $n$, here $n$ is a multiple of $p$. Consider the ambient space $ V=\mathbb{F}_2^n$, then $V$ can be decomposed into $V(\sigma) \oplus V(\sigma)^\bot$, where $V(\sigma)$ is the space fixed by $\sigma$ and $V(\sigma)^\bot$ is its dual.

And $C$ can be decomposed into $C(\sigma)\oplus C(\sigma)^\bot$. Here $C(\sigma)=C \cap V(\sigma)$ and $C(\sigma)^\bot=C \cap V(\sigma)^\bot$.

On the other hand, $C$ can be divide into several orbits of $\sigma$, each one of them isomorphic to $ \mathbb{F}_2[p] \cong \mathbb{F}_2[x]/(x^p + 1) \cong \mathbb{F}_2[x]/(x+1) \oplus \mathbb{F}_2[x]/(1 + x + \cdots + x^{p-1})\cong \mathbb{F}_2 \oplus \mathbb{F}_{2^{p-1}}$. It's easy to see that $V(\sigma)$ is several copies of $ \mathbb{F}_2[x]/(x+1)$ is fixed by $\sigma$, and $V(\sigma)^\bot$ is several copies of $ \mathbb{F}_2[x]/(1 + x + \cdots + x^{p-1})$.

On $\mathbb{F}_{2^{p-1}}$, there is an involution defined by the Frobenius map $ \tau: \alpha \mapsto \alpha^{2^\frac{p-1}{2}}$. Therefore we can defined a Hermitian Conjugate on $C(\sigma)^\bot$ using this map. It turns out to be inverse the last $p-1$ bits in each orbit.

And the theorem says that if $C$ is self-dual if and only if $C(\sigma)$ is self-dual and $C(\sigma)^\bot$ is Hermitian self-dual. I assume Hermitian self-dual means its dual is its Hermitian conjugate.

I wonder if anyone knows how to prove this?

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2 Answers 2

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According to MathSciNet, a translated version of the paper is available in the journal Problems of Information Transmission ISSN: 0032-9460. Your local librarian should be able to help you track down a paper copy of that journal (as far as Google can tell me, there are no digital copies from prior to Springer's acquisition of it in the 2000s).

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    $\begingroup$ Indeed, the page linked by OP includes the statement "Англоязычная версия: Problems of Information Transmission, 1983, 19:4, 260–270", which Google Translate informs me means "English version: Problems view the Transmission of Information Part, 1983, 19 : 4," $\endgroup$ Commented Jul 26, 2016 at 22:18
  • $\begingroup$ Obviously machine translation isn't perfect, but I saw this no sooner than following OP's link, which Google immediately offered to translate. (Perhaps this is a Chrome feature that other web browsers don't share.) In any case, your answer is absolutely the best one, and let me add that academic librarians are generally eager to be asked to track down papers like this. $\endgroup$ Commented Jul 26, 2016 at 22:32
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    $\begingroup$ @TheoJohnson-Freyd: giving the pointer to the English version only in Russian seems like a bad expository choice in the paper… $\endgroup$ Commented Jul 27, 2016 at 2:08
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    $\begingroup$ @PeterLeFanuLumsdaine It is a Russian website! (The mention of the English version is on the front page that links to the PDF.) In any case, Willie Wong's answer is the obviously correct one: (1) look up the article in various archives, e.g. mathscinet, and hope that one lists an English translation (2) ask your academic librarian to get you that paper. This is a good approach for many papers. But let me repeat that machine translation is more than good enough to handle the phrase "English version", and so it popped out for me when I followed OP's link. $\endgroup$ Commented Jul 27, 2016 at 12:18
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Cutting and pasting from the linked PDF into Google Translate provides an acceptable first pass. Mathematical notation is generally lost, but can be restored by going back and forth to the original document.

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