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Bounty Ended with 150 reputation awarded by Gil Kalai
The original answer was completly wrong in scale. The editted post reflects a corrested (log, not loglog) scale
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ofer zeitouni
ofer zeitouni
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My initial answer was wrong. Instead of completely deleting it, I left it at the bottom of the post. I use notation now that slightly differ from my original post.

As before, I give only a partial answer related to the max eigenvalue question and the limsup. Recall that the tail estimates for the max eigenvalue is $$(P(\lambda_1(n) >2\sqrt{n}+tn^{-1/6})\sim e^{-Ct^{3/2}}$$ (see e.g. Ledoux's reviews or http://math.univ-lyon1.fr/~aubrun/recherche/smalldeviations/smalldeviations.pdf).

Now, the question is what is the scale in which the recentered and rescaled maximal eigenvalues start behaving like independent random variables. In my original post I essentially claimed this starts happening at scale roughly $n$. This is wrong: the correct scale is given by the so-called GUE minor process (described by Baryshnikov in 2001). Relevant to Gil's problem is the scaling limit computed by Forrester and Nagao: http://arxiv.org/pdf/0801.0100.pdf. They show that the correct scaling in $n^{2/3}$, that is that the (centered and rescaled) maximal eigenvalues of the GUE(n) and GUE(m) decorelate if $m-n>>n^{2/3}$ and are strongly correlated in $m-n<<n^{2/3}$. Using that, and Borel Cantelli, one gets the correct scale. The limsup can be taken over the sequence $a_n=n^3$ instead of $n$. If you go over that sequence, you get from Borel-Cantelli that you cannot exceed $t_{a_n}a_n^{-1/6}$ infinitely often as long as $$\sum_n e^{-C t_{a_n}^{3/2}}<\infty.$$ This will happen as soon as $t_{a_n}^{3/2}>C'\log n$, i.e. $t_{a_n}\sim C'' (\log n)^{2/3}$. Unravelling the definitions, this will give you an upper bound on the limsup that is $t_n\sim (\log n)^{2/3}$; that is, this will show that the function $F(n)$ is at most $C (\log n)^{2/3}/n^{1/6}$. The lower bound on the limsup will follows a similar pattern, using the asymptotic independence stated above, and will give the same function $F(n)$. One can also compute constants (for the a.s. convergence) in the same manner.

Filling in the details in the above is beyond a mathoverflow answer, I may return to it in detail later.

I speculate that the liminf will use the lower tail estimates, which are of the form $e^{-Ct^{3})}$. Working with the same argument would then give $(\log n)^{1/3}/n^{1/6}$ as the correct scaling.

OLD (WRONG) POST:

The following is a partial answer with some hints on how to complete it; I may revisit it later with more details. The answer below refers to the max eigenvalue question, to the limsup, and does not attempt to get the constants, only the scale.

The following is a partial answer with some hints on how to complete it; I may revisit it later with more details. The answer below refers to the max eigenvalue question, to the limsup, and does not attempt to get the constants, only the scale.

My initial answer was wrong. Instead of completely deleting it, I left it at the bottom of the post. I use notation now that slightly differ from my original post.

As before, I give only a partial answer related to the max eigenvalue question and the limsup. Recall that the tail estimates for the max eigenvalue is $$(P(\lambda_1(n) >2\sqrt{n}+tn^{-1/6})\sim e^{-Ct^{3/2}}$$ (see e.g. Ledoux's reviews or http://math.univ-lyon1.fr/~aubrun/recherche/smalldeviations/smalldeviations.pdf).

Now, the question is what is the scale in which the recentered and rescaled maximal eigenvalues start behaving like independent random variables. In my original post I essentially claimed this starts happening at scale roughly $n$. This is wrong: the correct scale is given by the so-called GUE minor process (described by Baryshnikov in 2001). Relevant to Gil's problem is the scaling limit computed by Forrester and Nagao: http://arxiv.org/pdf/0801.0100.pdf. They show that the correct scaling in $n^{2/3}$, that is that the (centered and rescaled) maximal eigenvalues of the GUE(n) and GUE(m) decorelate if $m-n>>n^{2/3}$ and are strongly correlated in $m-n<<n^{2/3}$. Using that, and Borel Cantelli, one gets the correct scale. The limsup can be taken over the sequence $a_n=n^3$ instead of $n$. If you go over that sequence, you get from Borel-Cantelli that you cannot exceed $t_{a_n}a_n^{-1/6}$ infinitely often as long as $$\sum_n e^{-C t_{a_n}^{3/2}}<\infty.$$ This will happen as soon as $t_{a_n}^{3/2}>C'\log n$, i.e. $t_{a_n}\sim C'' (\log n)^{2/3}$. Unravelling the definitions, this will give you an upper bound on the limsup that is $t_n\sim (\log n)^{2/3}$; that is, this will show that the function $F(n)$ is at most $C (\log n)^{2/3}/n^{1/6}$. The lower bound on the limsup will follows a similar pattern, using the asymptotic independence stated above, and will give the same function $F(n)$. One can also compute constants (for the a.s. convergence) in the same manner.

Filling in the details in the above is beyond a mathoverflow answer, I may return to it in detail later.

I speculate that the liminf will use the lower tail estimates, which are of the form $e^{-Ct^{3})}$. Working with the same argument would then give $(\log n)^{1/3}/n^{1/6}$ as the correct scaling.

OLD (WRONG) POST:

The following is a partial answer with some hints on how to complete it; I may revisit it later with more details. The answer below refers to the max eigenvalue question, to the limsup, and does not attempt to get the constants, only the scale.

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ofer zeitouni
ofer zeitouni
  • 7.5k
  • 1
  • 22
  • 38

The following is a partial answer with some hints on how to complete it; I may revisit it later with more details. The answer below refers to the max eigenvalue question, to the limsup, and does not attempt to get the constants, only the scale.

First, the upper bound: recall that the tail estimates for the max eigenvalue is $$(P(\lambda_1(n) >2\sqrt{n}+t\sqrt{n})\sim e^{-Cnt^{3/2}}$$ (see e.g. Ledoux's reviews or http://math.univ-lyon1.fr/~aubrun/recherche/smalldeviations/smalldeviations.pdf). Also recall (a kindly neighbor next door, A. Dembo, pointed out the usefulness of this observation in the current context) that the sequence $\lambda_1(n)$ is monotone in $n$. So, the limsup can be taken over a sequence $a_n=(1+\epsilon)^n$ instead of $n$. If you go over that sequence, you get from Borel-Cantelli that you cannot exceed $t_{a_n}\sqrt{a_n}$ infinitely often as long as $$\sum_n e^{-C t_{a_n}^{3/2} a_n}<\infty.$$ This will happen as soon as $a_nt_{a_n}^{3/2}>C'\log n$, i.e. $t_{a_n}\sim C'' (\log n/a_n)^{3/2}$. Unravelling the definitions, this will give you an upper bound on the limsup that is $t_n\sim (\log \log n /n)^{2/3}$; that is, this will show that the function $F(n)$ is at most $C (\log \log n)^{2/3}/n^{1/6}$.

(Remark to @Carlo Beenakker: the difference with TSAW is that the tail estimates there are proportional to $t^3$, not $t^{3/2}$; This will be in line with the liminf computation in the RM case, in which case I expect the same discrepency in exponents of the $\log \log n$ as you noted, see below.)

For the lower bound (on the limsup), you need to take a sequence that is sparse enough so that you have approximate independence. Here is a leap of faith, that would require a separate proof: if you take an $N\times N$ GUE and resample the top $\epsilon N\times \epsilon N$ entries, the maximum eigenvalue does not change much if $\epsilon$ is small. I believe this follows from delocalization results for the eigenvectors of GUE, but have not checked details. If that is true, then for an exponentially growing sequence $a_n=C^n$, the sequence $\lambda_1(a_n)$ will behave like an i.i.d. sequence, to which you now can again apply the independent Borel-Cantelli to match the previous upper bound.

I speculate that the liminf will use the lower tail estimates, which are of the form $e^{-C(nt^{3/2})^2}$. Working with the same argument would then give $(\log \log n/n^2)^{1/3}$ as the correct scaling.