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I apologize for using German language in the title, but this question came to my mind after watching the French movie "le théorème de Marguerite" in which the protagonist gets an insight about Goldbach's conjecture after seeing the "Goldbach's pyramid" upside down ("à l'envers").

It seems a recent preprint by Matomäki and Merikoski suggests that the existence of a Siegel zero would violate Goldbach's conjecture. On the other hand, I asked in a previous question whether it would entail the existence of complex zeros of Zeta off the critical line.

I also edited recently an answer to a question of mine about the isomorphy of the $\Omega$-motive of a mirror pair of Calabi-Yau varieties giving rise to a unique L-function as a consequence of the mirror symmetry.

As I am an inveterate pun lover, I wanted to coin the term "Spiegel Vermutung" which translates as "mirror conjecture" because of the phonologic proximity between "Spiegel" and "Siegel", stating that:

Spiegel Vermutung/Specular conjecture/conjecture spéculaire :

The Landau-Siegel zero conjecture holds if and only if the Grand Riemann Hypothesis is equivalent to the quantitative form of Goldbach's conjecture given by Hardy-Littlewood.

Do we now, as of November 10th 2023, have compelling evidence for this conjecture I just stated?

Edit November 11th 2023: another reason to believe in this conjecture is the formal analogy between GRH which can be reformulated as $\bar{s}=1-s$ where $s$ stands for a non trivial zero of an L-function and the symmetric roles played by $p$ and $q$ in $p=2N-q\Longleftrightarrow\{N\pm r\}=\{p,q\}$. This analogy appears to be almost obvious if we replace formally $N$ with $\frac{1}{2}$ and $r$ with $it$.

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    $\begingroup$ I've never understood what exactly is meant by claims such as "GRH $\iff$ GB". For example, is the fundamental theorem of calculus equivalent to Fermat's last theorem? I suppose the answer would have to be yes since both are true. $\endgroup$
    – Christian Remling
    Commented Nov 10, 2023 at 18:23
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    $\begingroup$ @ChristianRemling it probably means either both are true or both are false? $\endgroup$
    – Kenta Suzuki
    Commented Nov 10, 2023 at 18:26
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    $\begingroup$ @ChristianRemling It means simply that we can prove this equivalence. For proven statements this is a mostly empty statement but for conjectures it tells us something about two statements given we don't know whether they are true. $\endgroup$
    – Wojowu
    Commented Nov 10, 2023 at 18:32
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    $\begingroup$ There can't be a good answer to this question since "compelling evidence" is vague. The truth is that likely most people won't be interested in investigating this. $\endgroup$
    – Stanley Yao Xiao
    Commented Nov 10, 2023 at 18:58
  • $\begingroup$ @wojowu even for proven statements there can sometimes be meaningful content if the equivalence proof is “easier” than establishing the truth of either side. For example, in CS “fine-grained complexity theory” studies equivalences between problems where each direction of the equivalence is computable by an extremely efficient Turing machine. Of course formalizing that the proof of equivalence is “easier” than directly establishing the validity of either proposition sounds like mostly a pain for normal math, but the intuition is still there. $\endgroup$
    – Mark Schultz-Wu
    Commented Nov 10, 2023 at 18:59

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