20
$\begingroup$

A few weeks ago Jean-Louis Verger-Gaugry announced a proof of Lehmer's conjecture, see https://arxiv.org/pdf/1709.03771.pdf. The key result (Theorem 5.28, p. 122) is a Dobrowolski type minoration of the Mahler Measure $M(\beta)$, namely \begin{align}\label{eq:1} M(\beta) \geq \Lambda_r\mu_r-\frac{\Lambda_r\mu_r \arcsin(\kappa/2)}{\pi}\frac{1}{\log(n)},\end{align} where $\Lambda_r\mu_r = 1.15411\ldots$ and $\kappa=0.171573\ldots$ are some constants and $n=\mathrm{dyg}(\beta)$ is some function of $\beta$ assumed to be at least 260.

This result should follow directly from the asymptotic expansion in Theorem 5.27 given by $$\log M_r(\beta) = \log \Lambda_r \mu_r + \frac{\mathcal{R}}{\log(n)} + O\left(\left(\frac{\log\log n}{\log n}\right)^2\right),$$ where $M(\beta)\geq M_r(\beta)$ and $\mathcal{R}$ depends on $\beta$ and $n$ satisfying $$|\mathcal{R}| < \frac{\arcsin(\kappa/2)}{\pi}.$$

Indeed, if we would now that the error term is positive we obtain the desired lower bound for $M(\beta)$ by taking the exponential of $\log M_r(\beta)$ and using the series expansion of the exponential. However, it is not shown that this error is positive and I don't see how one can show this fact (if it indeed turns out to be true). So, my question is:

What happened with the error term occurring in the expansion of $\log M_r(\beta)$, but no longer occurring in the lower bound for $M(\beta)$?

$\endgroup$
7
$\begingroup$

There are other strange assertions in the paper. For instance p. 131. The author wants to prove (Theorem 7.3) that that a certain meromorphic function $P/f$ has no pole, where the holomorphic function $f$ has (at least) a simple zero at $\omega\in \mathbb C$, and $P$ is a polynomial. The proof is very odd:First of all, it is never proved that $U=P/f$ has no pole. Secondly, $U$ is treated... like it was already proved that it has... no pole. More precisely, the author considers $U$ as a formal series in $X$ (like $1/X= 1/z$ is a Laurent series). Then he derives $P=Uf$ and gets $P'=U'f +Uf'$. But now he writes "specializing the formal variable $X$ to the complex variable $z$" and gets $$P'(z) = U'(z)f(z) +U(z)f'(z),$$ forgetting that this formal hocus-pocus does not make $U(z)$ (and $U'(z)$) a defined complex number! He then makes $z=\omega$ and obtains $P'(\omega) = U(\omega) f'(\omega)$, forgetting that $U$ is still a formal series. In fact this argument holds for $P=1$ and $f=z$. It would then be proved that $U(z) =1/z$... is holomorphic at $0$!

Added: there are (at least) two other places where the same argument is used : p110 and p125. The 'fracturability' defined p. 4 is this kind of strange factorization of polynomials, like $1$ is factorized as $1=(1/z)z$, and $1/z$ becomes a holomorphic function...

$\endgroup$
5
  • 4
    $\begingroup$ I think such an attitude is not appropriate for this site. Maybe you could rewrite your answer using less harsh language? $\endgroup$ Jul 10 '18 at 7:50
  • 1
    $\begingroup$ Ok, sorry, I changed the style. $\endgroup$
    – Kimjungun
    Jul 10 '18 at 9:33
  • 2
    $\begingroup$ Yes, it is better. Downvoters might consider removing their votes or commenting on the actual content of the answer. $\endgroup$ Jul 10 '18 at 9:39
  • 2
    $\begingroup$ It is announced that the proof contains at least one fatal error, using the same argument as described above. See arxiv.org/pdf/1809.10600.pdf $\endgroup$ Nov 22 '18 at 13:17
  • $\begingroup$ Hi. A different version of the paper has been anounced here hal.archives-ouvertes.fr/hal-02322497 I am not a specialist in this topic, but I'd like to see your opinion. $\endgroup$ Oct 29 '19 at 19:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.