Extreme gap problems in random matrix theory
Renjie Feng
BIMCR, Peking University
Renjie Feng (BICMR) 1 / 21
Extreme gap problems in random matrix theory Renjie Feng BIMCR, - - PowerPoint PPT Presentation
Extreme gap problems in random matrix theory Renjie Feng BIMCR, Peking University Renjie Feng (BICMR) 1 / 21 Joint density of eigenvalues of G E G E: Given n point 1 , , n ( > 0) with the joint density n 1 e n
BIMCR, Peking University
Renjie Feng (BICMR) 1 / 21
n
4 λ2 k
i<j
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(Γ(1+β/2))n , β = 1 is corresponding to COE, β = 2 for CUE,
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n
I
1 < tn 2 · · · < tn k be the first k smallest eigenangles gaps, denote
k = (72π)−1/3tn k , then as a consequence,
n→+∞ P(τ n k ∈ [x, x + dx]) =
Renjie Feng (BICMR) 4 / 21
n
I
1 < tn 2 · · · < tn k be the first k smallest eigenangles gaps, denote
k = (72π)−1/3tn k , then as a consequence,
n→+∞ P(τ n k ∈ [x, x + dx]) =
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n
(n
β+2 β+1 (θi+1−θi),θi)
Γ(3β/2+1)Γ(β+1) . In particular, the result holds for
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n
(n
β+2 β+1 (θi+1−θi),θi)
Γ(3β/2+1)Γ(β+1) . In particular, the result holds for
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k as the k-th smallest gap, and
k = n(β+2)/(β+1) × (Aβ/(β + 1))1/(β+1)tn k ,
n→+∞ P(τ n k ∈ [x, x + dx]) =
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k as the k-th smallest gap, and
k = n(β+2)/(β+1) × (Aβ/(β + 1))1/(β+1)tn k ,
n→+∞ P(τ n k ∈ [x, x + dx]) =
Renjie Feng (BICMR) 6 / 21
k as the k-th smallest gap, and
k = n(β+2)/(β+1) × (Aβ/(β + 1))1/(β+1)tn k ,
n→+∞ P(τ n k ∈ [x, x + dx]) =
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β+1
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n
β+2 β+1 (θi2−θi1),··· ,n β+2 β+1 (θi2k −θi2k−1).
β =
n→+∞
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n
β+2 β+1 (θi2−θi1),··· ,n β+2 β+1 (θi2k −θi2k−1).
β =
n→+∞
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Renjie Feng (BICMR) 9 / 21
n
4 3 (λi+1−λi),λi)1|λi|<2−η
k = (
k has the limiting
3 (k−1)!x3k−1e−x3, same as CUE.
Renjie Feng (BICMR) 10 / 21
n
4 3 (λi+1−λi),λi)1|λi|<2−η
k = (
k has the limiting
3 (k−1)!x3k−1e−x3, same as CUE.
Renjie Feng (BICMR) 10 / 21
n
4 3 (λi+1−λi),λi)1|λi|<2−η
k = (
k has the limiting
3 (k−1)!x3k−1e−x3, same as CUE.
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n−1
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k = cβn(β+2)/(β+1)tk has the limiting density
−nβ
n
V (λi)
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k = cβn(β+2)/(β+1)tk has the limiting density
−nβ
n
V (λi)
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Lp
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1≤j,k≤n
α
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1≤j,k≤n
α
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1≤j,k≤n
α
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1 2 (nmk − (32 ln n) 1 2 )/4 − (3/8) ln(2 ln n),
k } will tend to a Poisson distribution and we have the limit of the
n→+∞ P(τk ∈ I) =
2 . In particular, the limiting density for the largest gap
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Figure: Density of eigenvalues of GUE
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k = (2 ln n)
1 2 (nS(I)mk − (32 ln n) 1 2 )/4 + (5/8) ln(2 ln n),
k } will tend to a Poisson distribution and we have the limit of the
n→+∞ P(τ ∗ k ∈ I1) =
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β ln n/n for CβE/GβE, but how to prove?
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β ln n/n for CβE/GβE, but how to prove?
Renjie Feng (BICMR) 18 / 21
β ln n/n for CβE/GβE, but how to prove?
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