21
$\begingroup$

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 $M(\beta)$?

$\endgroup$

1 Answer 1

8
$\begingroup$

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? $\endgroup$ Commented Jul 10, 2018 at 7:50
  • 1
    $\begingroup$ Ok, sorry, I changed the style. $\endgroup$
    – Kimjungun
    Commented Jul 10, 2018 at 9:33
  • 2
    $\begingroup$ Yes, it is better. Downvoters might consider removing their votes or commenting on the actual content of the answer. $\endgroup$ Commented Jul 10, 2018 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 $\endgroup$ Commented Nov 22, 2018 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. $\endgroup$ Commented Oct 29, 2019 at 19:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .