Following this question: [Can we get that $ P(N^{2/3}(\lambda\_N-\lambda\_{N-1})\le c)\ge 1-\epsilon$?](https://mathoverflow.net/questions/436173/can-we-get-that-pn2-3-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
$$
P(N^{2/3}(\lambda\_N-\lambda\_{N-1})\ge c)\ge 1-\epsilon?
$$



Does there exist a constant $c>0$ such that $$ P(N^{2/3}(\lambda_N-\lambda_{N-1})\ge c)\ge 1-\epsilon? $$
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 $N$.


