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In Naor and Neiman's Assouad embedding theorem - "Assouad’s theorem with dimension independent of the snowflaking" Revisita Mathematica, the authors derive quantitative estimates on the distortion required to embed a doubling metric space into Euclidean space. However, these estimates are up to an absolute constant ($c$ in the paper's notation and hidden by the $\lesssim$ notation).

What is this constant ($c$) and, if it is not given explicitly, are there upper (maybe even lower) bounds for it available?

To what I can gather from the proof of Lemma 3.3 and from equation (3.1), it seems that: for any $\alpha :=1-\epsilon\in (\frac1{2},1)$ any $\theta \in (0,1)$, and any doubling metric space with doubling constant $K$; we have that $$ \Big\lceil \frac{c\log(K)}{\theta} \Big\rceil =: N \leq \frac{ -2\log \Big( e \big( 1 + K^{O(\log\log(K) - \log(1-\alpha))} \big) \Big) }{ \theta\, \log\big( \frac{\epsilon}{\log(K)} \big) }. $$ Therefore, "c large enough" means that $$ c := \frac{ -2 \, \log \Big( e \big( 1 + K^{O(\log\log(K) - \log(1-\alpha))} \big) \Big) }{ \log(K)\big(\log(\epsilon)-\log(K)\big) }. $$ Still, the expression is not entirely clear since what is O hiding? Also I don't see how c is "universal" in the sense that it clearly depends on $K$ and on $\epsilon$...

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  • $\begingroup$ The link doesn't work. Give a full reference? $\endgroup$
    – Aryeh Kontorovich
    Commented Aug 25, 2022 at 9:15
  • $\begingroup$ @AryehKontorovich done and also fixed the wierd link $\endgroup$
    – ABIM
    Commented Aug 25, 2022 at 11:27
  • $\begingroup$ Should be Naor and Neiman (Ofer is the first name) $\endgroup$
    – Aryeh Kontorovich
    Commented Aug 25, 2022 at 15:06
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    $\begingroup$ @AIM Yeah I see you mean the formulation from Theorem 1.2. But I still think that the part about $N$ could be relevant. At least that what I remember from a more classical Assouad. I mean that you can take a smallish constant in ≲ for $N$ but than you get a big one in ≲ for $D$ and vice versa. $\endgroup$
    – Vladimir Zolotov
    Commented Aug 27, 2022 at 15:04
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    $\begingroup$ @VladimirZolotov Both of these comments are rather obvious and the last bit is not very helpful. I'm asking in case someone has computed or knows good estimates on c. $\endgroup$
    – ABIM
    Commented Aug 28, 2022 at 1:42

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