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Following this question: Can we get that $ P(N^{2/3}(\lambda_N-\lambda_{N-1})\le c)\ge 1-\epsilon$?.

We know that for $\lambda_N\le \lambda_{N_1}\le \dots le\lambda_1$ (eigenvalues of GOE matrix)

$$ \lim_{N\to\infty}P(N^{2/3}(\lambda_N-2)\le s_1,\dotsc, N^{2/3}(\lambda_{N-k+1}-2)\le s_k)=F_{\beta, k}(s_1,\dotsc, s_k). $$

Can we say that for any $\epsilon>0$, there exists a constant $c>0$ such that $$ P(N^{2/3}(\lambda_N-\lambda_{N-1})\ge c)\ge 1-\epsilon? $$

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The probability density function of the spacing $\delta_N=\lambda_N-\lambda_{N-1}$ of the eigenvalues $\lambda_N$ and $\lambda_{N-1}$ at the edge of the spectrum decays linearly for $\delta_N\ll N^{-2/3}$, with a slope that is independent of $N$. So no matter how small $\epsilon$, you can always find a $c>0$ such that $P(N^{2/3}\delta_N\le c)\le \epsilon$, or equivalently, such that $P(N^{2/3}\delta_N\ge c)\ge 1-\epsilon$. The coefficient $c$ will depend on $\epsilon$ but it will not depend on $N$.

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  • $\begingroup$ Thank you! So this result does not contradict with mathoverflow.net/questions/436173/…? $\endgroup$
    – Hermi
    Commented Feb 24, 2023 at 20:11
  • $\begingroup$ no contradiction, the coefficients $c$ are different in the two inequalities, of course, here $c\ll 1$, in your earlier question $c\gg 1$. $\endgroup$
    – Carlo Beenakker
    Commented Feb 24, 2023 at 21:51

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