Timeline for Estimates of product of eigenvalues gaps for Wigner matrices
Current License: CC BY-SA 4.0
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| Jan 17, 2022 at 16:57 | comment | added | Carlo Beenakker | (i) the eigenvalues of $W_n$ cover an interval of length $\simeq n^{1/2}$ on the real axis; the scaling with $n^{1/2}$ ensures that the eigenvalue density $\rho(\lambda)$ approaches an $n$-independent limit (the semicircle) in the large-$n$ limit (after the rescaling they cover an interval of length $2\sqrt 2$); (ii) the convergence $\sum_i u(w_i)$ to an integral refers to the expectation value of the sum, with sample-to-sample fluctuations vanishing as $n^{-1/2}$. | |
| Jan 17, 2022 at 16:49 | comment | added | Ludwig | @CarloBeenakker: I finally went through your answer. I have a few questions: (i) since I assume that the Wigner matrix is normalized, why do you need to introduce the $n^{1/2}$ scaling in the first part of the proof? (ii) in which sense the summation converges to the integral in the large $n$ limit? | |
| Jan 16, 2022 at 21:49 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Jan 16, 2022 at 21:37 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Jan 16, 2022 at 21:08 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Jan 16, 2022 at 20:57 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Jan 16, 2022 at 20:45 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Jan 16, 2022 at 20:39 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Jan 16, 2022 at 18:29 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Jan 16, 2022 at 17:32 | comment | added | Ludwig | @CarloBeenakker: Okay, thanks. I edited my question to clarify notation. | |
| Jan 16, 2022 at 16:53 | comment | added | Terry Tao | The rigorous proof of convergence here (say, in the almost sure sense) is slightly tricky due to the (mild) singularity of $\log |\lambda-\mu|$ on the diagonal $\lambda=\mu$. For instance, one has to ensure that the spectrum is a.s. simple, which was only established in 2017. But the lower bound on the spectral gap in mathscinet.ams.org/mathscinet-getitem?mr=3627428 should suffice to deal with this issue (at least when the diagonal has bounded variance). | |
| Jan 16, 2022 at 16:45 | comment | added | Ludwig | Interesting approach, thanks. I'm missing something though: (i) in my question I assume that $i$ is fixed, e.g. $i=1$, and only $j$ varies, (ii) in your (double) sum to (double) integral transformation, isn't a factor $n$ missing? | |
| Jan 16, 2022 at 16:16 | history | answered | Carlo Beenakker | CC BY-SA 4.0 |