Primes, partitions and permutations. Paul-Olivier Dehaye ETH Zürich, October 31 st

Transcription

1 Primes, Paul-Olivier Dehaye ETH Zürich, October 31 st

2 Outline Review of Bump & Gamburd s method A theorem of Moments of derivatives of characteristic polynomials Hypergeometric functions Markov chain on a Young graph Explain the title

3 Definitions Let Z U (θ) = N (1 ) e i(θ j θ) j=1 be the char. pol. of U U(N), which has eigenvalues { e iθ j }. A partition λ is a non-increasing sequence (λ 1,, λ l(λ), 0, 0, ) of integers trailing to 0s. We define its length l(λ) as the number of non-zero entries, and its weight λ as the total value λ i of the entries in the sequence. We always identify a partition with its Young diagram. The diagram for (6, 5, 3, 1), for instance, is.

4 Definitions (II) Schur polynomials are symmetric polynomials. They form compatible families, indexed by : s λ (x 1,, x N ) = s λ (x 1,, x N, 0). There is one reduction property: s λ (x 1,, x N ) = 0 if l(λ) > N. Excluding those, we have irreducible characters of U(N), and s λ (U)s µ (U) = U(N) { δλµ if N l(λ) 0 if l(λ)>n with s λ (U) := s λ (e iθ 1,, e iθ N ). For large N, the s λ are orthonormal over U(N).

5 The method of Bump and Gamburd Bump and Gamburd have a way to compute Z U (0) 2k U(N) which uses the dual Cauchy identity λ λ s λ (x 1, x 2,, x M )s λ t(y 1, y 2,, y N ) = s λ ( {1} 2k) s λ t (U) = det(id +U) 2k M,N (1+x m y n ). m,n = det(u) k det(id +U) 2k = s kn (U) det(id +U) 2k or (replacing U by U) Z U (0) 2k = ( 1) kn s kn (U) λ ( 1) λ s λ ({1} 2k) s λ t (U).

6 The method of Bump and Gamburd (II) Z U (0) 2k = ( 1) kn s kn (U) λ So Z U (0) 2k U(N) ( 1) λ s λ ({1} 2k) s λ t (U) ( = s Nk {1} 2k), which can be evaluated combinatorially, provides many alternate expressions and gives analytic continuations. This gives an interpreation of the Keating-Snaith geometric term as a dimension.

7 Evaluation There are many ways to evaluate a Schur function, based on (ratios of Vandermonde determinants) determinants (valid for any values of the variables) Jacobi-Trudi identity (2 expressions) Giambelli formula the hook-content formula (a dimension formula, only 1s) ( ) s λ {1} K = K + c( ) H( ) λ = K λ H(λ) The K λ notation is a generalization of the Pochhammer symbol, which provides analytic continuations in K.

8 Evaluation (II) ( ) s λ {1} K = K + c( ) H( ) λ = K λ H(λ) The way these expressions vanish when l(λ) > K is very important. Boxes can be grouped in many ways: along rows, columns or half-hooks [BHNY: columns] The numerators K λ can be obtained as exponentials of integrals against Russian diagrams of the logarithmic derivative of the Barnes G-function

9 Moments of derivatives We now consider the question of moments of derivatives of characteristic polynomials, i.e. we look at Z U (0) 2k Z U (0) 2h U(N) (remember: Z U (θ) = N j=1 ( 1 e i(θ j θ) ) ). We also look at V U (0) 2k V U (0) 2h U(N) with V U (θ) = e in(θ+π)/2 e i P N j=1 θ j /2 Z U (θ) The two are simple linear combinations obtained from the moments ( Z Z U (0) 2k ) U (0) r Z U (0) U(N)

10 A theorem of Okounkov and Generalized Binomial Theorem: If s µ stand for shifted Schur functions, then s λ (1 + a 1,, 1 + a n ) s λ ({1} n ) = µ l(µ) n (The case n = 1 is the binomial theorem). Also, s µ(λ 1,, λ n )s µ (a 1,, a n ). n µ s µ({n} k µ ( N µ)(k µ) ) = ( 1), H(µ) so s Nk (1 + a 1,, 1 + a n ) has a nice expression as a sum over. The Taylor expansion of a Schur function near the identity.

11 ( Z Z U (0) 2k ) U (0) r = ( 1) kn s (U) Z U (0) kn ( 1) λ s λ t (U) λ and hence 1 r s λ ({1} 2k r {1 ia 1,, 1 ia r }) aj =0 ( Z Z U (0) 2k ) U (0) r = Z U (0) U(N) ( 1 r s Nk {1} 2k r aj {1 ia 1,, 1 ia r }) =0.

12 Using, we get Proposition: When 0 r 2k, ( Z Z U (0) 2k U (0) Z U (0) Remarks: ) r U(N) = i r r! Z U (0) 2k U(N) µ r r has to be an integer At first k is an integer, and r 2k 1 ( N µ)(k µ) H(µ) 2 2k µ This can be extentded to a rational function of k The sum yields a polynomial in N of degree r, with coefficients even rational functions in k (Hughes, unpublished).

13 (I) r ( Z Z U (0) 2k ) U (0) r (iz) r Z U (0) U(N) r! = = Z U (0) 2k U(N) µ Z U (0) 2k U(N) µ ( k µ) (N µ)z µ 2k µ H(µ) 2 k µ s µ (z Id N N ) 2k µ H(µ)

14 (II) Proposition (Borodin, D.): r = ( Z Z U (0) 2k ) U (0) r (iz) r Z U (0) U(N) r! Z U (0) 2k U(N) µ = ( k µ) 1 (N µ)z µ ( 2k µ) H(µ) H(µ) Z U (0) 2k U(N) 1 F 1 ( k; 2k; z Id N N ) See work of Richards, Gross, Yan on hypergeometric functions of matrix arguments. At finite N, any problem of analytic continuation (in k, h) is not much harder than classical hypergeometric functions (cf. integral, differential equations, recurrence relations).

15 Hypergeometric functions have integral. When they have matrix arguments, Selberg integrals appear. As meromorphic functions of k, ( ) Z U (0) 2k ZU (0) r Z U (0) U(N) (iz) r r Z U (0) 2k = r! U(N) P 0 ez t i N i t k N i (1 t i ) k N (t i ) 2 dt 1 dt N 1 (1 t i ) k N (t i ) 2 dt 1 dt N 1 0 N 0 i t k N i

16 h V U (0) 2k V U (0) V U (0) V U (0) 2k U(N) 0 e 1 4 (2 P i t i N) 2 z N i 1 1 N 0 0 i t k N i = e 1 4 (2 P i t i +N) 2 z 0 0 2h U(N) z h h! = t k N i (1 t i ) k N (t i ) 2 dt 1 dt N (1 t i ) k N (t i ) 2 dt 1 dt N z N(N+2k) N j=1 Γ(2k + j)γ(j + 1) N ti k (1+t i ) k i 1 i<j N t i t j 2 dt 1 dt In the limit in N, I don t know. The existence of an analytic continuation in h in the limit in N is unproved and very possibly false.

17 Fact: dim λ := dim χ λ = λ! = #paths to λ in the Young graph. H(λ) Set (Poissonized Plancherel measure) which satisfies m(λ) = dim(λ)2, λ! λ s.t. λ =cst m(λ) = 1. The formula obtained before then takes the form r g(r) zr r! = λ f (λ)m(λ) z λ λ!, for f (λ) given by a product over the boxes of the partition λ.

18 (II) r ( Z Z U (0) 2k ) U (0) r (iz) r Z U (0) U(N) r! = Z U (0) 2k U(N) µ ( k µ)(n µ) m(µ) z µ ( 2k µ) µ! r g(r) zr r! = λ f (λ)m(λ) z λ λ!

19 If the RMT conjectures are to be believed, the are necessary to express the joint moments of the derivatives of ζ. In the limit in N, all λ appear, not just those with l(λ) N. Hypergeometric functions of operator arguments have not been studied, but the probabilistic interpretation is still present, and points to the study of the whole Young graph, not just restrictions of it.

20 Cauchy identities s λ (x 1, x 2,, x M )s λ (y 1, y 2,, y N ) = λ s λ (x 1, x 2,, x M )s λ t(y 1, y 2,, y N ) = λ M,N m,n M,N m,n 1 1 x m y n 1 + x m y n Theorem (Bourgade, D., Nikeghbali) For k N, when Rs 1, ζ(s) k = s λ ({1} k) s λ (p s ) λ and one can continue analytically s λ (p s ) to Rs > 0. This is merely a reorganization of an absolutely convergent sum. Note that when k or k N, this is not trivial. This preserves a symmetry of the Riemann zeta function (invariance under permutation of the primes).

21 s λ (x 1, x 2,, x M )s λ (y 1, y 2,, y N ) = λ s λ (x 1, x 2,, x M )s λ t(y 1, y 2,, y N ) = λ M,N m,n M,N m,n 1 1 x m y n 1 + x m y n Conjecture (D.) There exist functions f λ (z, s) such that s λ (f λ (γ i, s)) admits an analytic continuation to Rs > 0 and s λ (f λ (γ i, s)) = s λ t (p s ) Selecting N zeroes would then correspond to looking at λ with λ 1 N (cf. Keating-Snaith). The involution on is crucial (cf. graded Hopf algebra of symmetric functions has an antipode). By using we can index characters in a way that is more natural for the symmetric group, and less for the unitary

22 We then have 3 types of objects: primes zeroes eigenvalues of random matrices Question We know that lim N can we conjecture s λ (U), s µ (U) U(N) = δ λµ, lim T 1 T ( s λ t p 1/2+it) ( s ) µ t p 1/2+it dt = C λ t δ λµ T 0 and pass from proofs using orthonormality of characters of U(N) directly to conjectures using orthogonality?

23

24 Main computations Main computations Just as before, Z U (a 1 ) Z U (a r ) = λ ( 1) λ s λ t (U)s λ ( e ia 1,, e iar ). To the first order in small a, we have so e ia 1 ia, Z U (0)r = λ ( 1) λ s λ t (U) where j := aj. Also, 1 r s λ (1 ia 1,, 1 ia r ) a1 = =a r =0, Z U (0) k = ( 1) kn det U k Z U (0) k = ( 1) kn s kn (U)Z U(0) k.

25 Z U (0)r = λ ( 1) λ s λ t (U) Main computations and 1 r s λ (1 ia 1,, 1 ia r ) a1 = =a r =0 Z U (0) k = ( 1) kn s kn (U)Z U(0) k imply ( Z Z U (0) 2k ) U (0) r = ( 1) kn s (U) Z U (0) kn ( 1) λ s λ t (U) 1 r s λ ({1} 2k r {1 ia 1,, 1 ia r }) aj =0 and hence ( Z Z U (0) 2k ) U (0) r = Z U (0) U(N) ( 1 r s Nk {1} 2k r aj {1 ia 1,, 1 ia r }) =0. λ

26 Main computations Using, ( ) Z U (0) 2k Z U (0) r Z U (0) U(N) Z U (0) 2k U(N) = ( i) r µ r s µ = ( i) r µ r ( {N} k) 1 r s µ (a 1,, a r ) aj =0 s µ = ( i) r µ r 2k µ ( {N} k) sµ, p 1 r S r 2k µ ( {N} k) dim µ s µ = i r r! µ r with a condition that 0 r 2k. 2k µ 1 ( N µ)(k µ) H(µ) 2, 2k µ

27 Main computations

28 Main computations

29 Main computations

30 Main computations

31 Basics about slide transpose s k N s(u) = 0 when length rectangle < N k > Frobenius vect sort ones character symmetric group