Transfer operator for the Gauss' continued fraction map. Structure of ...

Preliminaries, previous results Arithmetic of decomposition formulas. Examples Solution

Transfer operator for the Gauss’ continued fraction map. Structure of the eigenvalues and trace formulas. Giedrius Alkauskas, Vilnius University, Lithuania

End of February 2015, TU Clausthal

Giedrius Alkauskas, Vilnius University, Lithuania

Structure of the eigenvalues

Preliminaries, previous results Arithmetic of decomposition formulas. Examples Solution

Contents 1

Preliminaries, previous results Transfer operator Trace formulas Conjectures and results

2

Arithmetic of decomposition formulas. Examples Arithmetic of decomposition formulas Direct connections

3

Solution Plans for the attack Solution

Giedrius Alkauskas, Vilnius University, Lithuania

Structure of the eigenvalues

Preliminaries, previous results Arithmetic of decomposition formulas. Examples Solution

Transfer operator Trace formulas Conjectures and results

Definition 1 Warning In order to simplify a bit, this talk, though directly and intricately related to the theory of continued fractions, dynamical systems and modular forms, will not mention these topics. Also, there will be almost no functional analysis! Many mathematicians contributed to this topic: C.F. Gauss, R.O. Kuzmin, P. L´evy, A.Y. Khinchin, E. Wirsing, K.I. Babenko, S.P. Jur’ev, D. Knuth, D. Mayer, Ph. Flajolet, B. Vall´ee, D. Zagier, D. Hensley, L. Vepˇstas, A.J. MacLeod, P. Sebah, K. Briggs. In the talk, we fix the notation √ 1+ 5 φ= . 2 Giedrius Alkauskas, Vilnius University, Lithuania

Structure of the eigenvalues

Preliminaries, previous results Arithmetic of decomposition formulas. Examples Solution

Transfer operator Trace formulas Conjectures and results

Definition 2 Let D be the disc {z ∈ C : |z − 1| < 32 }. Let V be the Banach space of functions which are analytic in D and are continuous in its closure, with the supremum norm. The Perron-Frobenius, or the transfer operator for the Gauss’ continued fraction map, (also called the Gauss-Kuzmin-Wirsing operator), is defined for functions f ∈ V by L[f ](z) =

∞ X m=1

 1  1 f . (z + m)2 z + m

Our chief interest is the point spectrum of this operator. As was shown by Babenko, this operator is of trace class and is nuclear of ∞ P order 0: |λn |ε < +∞. n=1

Throughout the presentation, “n” is reserved for the index of eigenvalue (canonical or not)? Giedrius Alkauskas, Vilnius University, Lithuania

Structure of the eigenvalues

Preliminaries, previous results Arithmetic of decomposition formulas. Examples Solution

Transfer operator Trace formulas Conjectures and results

Numerical values

n 1 2 3 4 5 6 7 8

(−1)n+1 λn 1.0000000000000000 0.3036630028987326 0.1008845092931040 0.03549615902165984 0.01284379036244026 0.004717777511571031 0.001748675124305511 0.0006520208583205029

n 9 10 11 12 13 14 15 16

(−1)n+1 λn 0.0002441314655245158 0.000091689083768593376 0.000034516546163854253 0.000013017697877023030 0.0000049167823024644912 0.0000018593073515090423 0.00000070381134308703980 0.00000026664134344795640

Table : The eigevalues λn for 1 ≤ n ≤ 16.

Giedrius Alkauskas, Vilnius University, Lithuania

Structure of the eigenvalues

Preliminaries, previous results Arithmetic of decomposition formulas. Examples Solution

Transfer operator Trace formulas Conjectures and results

The nature of the eigenvalues λn for n ≥ 2 is unknown. It is widely believed that these constants are unrelated to other most important constants in mathematics - but note the appearance of extended ring of periods (Zagier-Kontesevich) further in our work! Now, more that 480 digits of λ2 have been calculated by K. Briggs, but one can get rigorous certificates only for the several first few digits of λ2 and λ3 . On the other hand, the trace of the operator L can be given explicitly. As was shown by D. Mayer and K.I. Babenko (also, Ph. Flajolet and B. Vall´ ee), we have Trace formulas

Tr(L)

=

∞ X

Z∞ λn =

n=1

=

Tr(L2 )

=

∞ 2k  1 1 1X − √ + (−1)k−1 (ζ(2k) − 1), 2 2 k=1 k 2 5 ∞ X n=1

ξ`

=

0

∞

X J1 (2x) 1 dx = x e −1 ξ −2 + 1 `=1 `

λ2n =

Z∞Z∞ (e x 0

0

1 1 1 ··· , `+ `+ `+

√ ∞ X J1 (2 xy )2 1 dx dy = ; y − 1)(e − 1) (ξ ξ )−2 − 1 i,j j,i i,j=1

ξi,j =

Giedrius Alkauskas, Vilnius University, Lithuania

1 1 1 1 ··· , i+ j+ i+ j+

`, i, j ∈ N.

Structure of the eigenvalues

Preliminaries, previous results Arithmetic of decomposition formulas. Examples Solution

Transfer operator Trace formulas Conjectures and results

Proposition. Trace of the square of the operator We have an identity: 1 4 3 2 √ + √ + √ + √ 5 + 3 5 3 + 2 3 21 + 5 21 16 + 12 2   ∞ X 2 3 2 k 2k − 2 + (−1) ζ 2 (k) − 1 − k − k − k k −2 2 3 4

Tr(L2 ) =

k=2

= 1.103839653617+ . The kth term of this series is asymptotically 12 ( 54 )k (πk)−1/2 . The formulas for trace of the square of the operator(s) are crucial in our proof: we construct eigenvalues explicitly by a special method (without relying on functional analysis), and show that the given collection amounts to all eigenvalues, not a single is missed, since that would invalidate the trace formulas for the square. Giedrius Alkauskas, Vilnius University, Lithuania

Structure of the eigenvalues

Preliminaries, previous results Arithmetic of decomposition formulas. Examples Solution

Transfer operator Trace formulas Conjectures and results

Conjectures (1988-1995) The following three statements are true: i) Simplicity. The eigenvalues are simple, |λn | strictly decreases. ii) Sign. The eigenvalues have alternating sign: (−1)n+1 λn > 0. λn n→∞ λn+1

iii) Ratio. There exists lim

√

= − 3+2

5

= −φ2 .

The first two were raised by Mayer and Roepstorff (1988), reiterated by MacLeod. The last was raised by MacLeod (1992) (with a constant ≈ −2.6), seconded by Flajolet and Vall´ee (1995) (with the constant −φ2 ). The ratio conjecture is very cautious, since nobody posed a question on true asymptotics of λn . Motivation:

L0 [f ](z) =

 1  1 f . (z + 1)2 z + 1

Giedrius Alkauskas, Vilnius University, Lithuania

Structure of the eigenvalues

Preliminaries, previous results Arithmetic of decomposition formulas. Examples Solution

Transfer operator Trace formulas Conjectures and results

These are true, and much more! Theorem 1 [GA] (Asymptotics) We have the formula φ−2n φ−2n (−1)n+1 λn = φ−2n + C · √ + d(n) · , n n √ 4 5 · ζ(3/2) √ where the constant C = = 1.1019785625880999+ ; 2 π here ζ(?) is the Riemann zeta function, and the function d(n) is bounded. Based on high precision numerical computations of P. Sebah, who computed λn for 1 ≤ n ≤ 150, we have: d(1) = 0.5160, d(2) = 0.6042, d(3) = 0.5222, d(4) = 0.4662, d(5) = 0.4343, d(10) = 0.3850, d(20) = 0.3688, d(30) = 0.3646, d(40) = 0.3626, d(50) = 0.3615, d(70) = 0.3604, d(100) = 0.3595, d(130) = 0.3590, d(148) = 0.35887, d(149) = 0.35886, d(150) = 0.35885. Giedrius Alkauskas, Vilnius University, Lithuania

Structure of the eigenvalues

Preliminaries, previous results Arithmetic of decomposition formulas. Examples Solution

Transfer operator Trace formulas Conjectures and results

Theorem 2 [GA] (Arithmetic and decomposition of trace formulas) (i) There exist positive functions Wv (n), v ≥ 0, n ≥ 1, defined by W0 (n) = 1, (0,1) W1 (n) = 54 · φ−2n Pn−1 (3/2) (Jacobi polynomials), and then by a certain explicit recurrence, such that for n ≥ 1, (−1)n+1 λn = φ−2n

∞ X

W` (n),

`=0

√ 4 B 5 W` (n) = √ √ + 3/2 for `, n ≥ 1, 2 π · `3/2 n ` n

for a bounded function B = B(`, n). (ii) This decomposition is compatible and gives the decomposition of trace formulas for the powers of L: for the first, the second, and the third powers, we have, respectively, ∞ X

X

n=1 ∞ X

(−1)n+1 φ−2n W`−1 (n)

=

φ−4n Wi−1 (n)Wj−1 (n)

=

i+j=` n=1

X

∞ X

Giedrius Alkauskas, Vilnius University, Lithuania

X i+j=`

(−1)n+1 φ−6n Wi−1 (n)Wj−1 (n)Wk−1 (n)

i+j+k=` n=1

1 ξ`−2 + 1

=

,

1 , (ξi,j ξj,i )−2 − 1

X i+j+k=`

Structure of the eigenvalues

1 . (ξi,j,k · · · )−2 + 1

Preliminaries, previous results Arithmetic of decomposition formulas. Examples Solution

Transfer operator Trace formulas Conjectures and results

Decomposition of trace formulas Recall once more that Tr(L) =

∞ X

λn =

n=1

∞ X

1 . +1

ξ −2 `=1 `

So, in the trace formulas now we are able to crystallize the contribution of each individual eigenvalue. Thus, this defines an infinite matrix  ∞ (−1)n+1 φ−2n W`−1 (n) , n,`=1

whose elements in rows add up to eigenvalues, elements in columns add up to (ξ`−2 + 1)−1 , and the sum of all real numbers in the matrix is equal to Tr(L). This is canonical, and similar decomposition holds for Tr(Lk ), k ∈ N also (as given above). Giedrius Alkauskas, Vilnius University, Lithuania

Structure of the eigenvalues

Preliminaries, previous results Arithmetic of decomposition formulas. Examples Solution

Arithmetic of decomposition formulas Direct connections

Continuation of this work Transfer operator for the Gauss’ continued fraction map. II. Fine arithmetic of the decomposition formulas (in preparation). Theorem 2 (continued) (iii)? Let a = {`1 , `2 , . . . , `k }, `i ∈ N. Let us define Ωa (w ) =

∞ Y k X n=1

 W`i (n) w n ,

|w | < 1.

i=1

Then all Ωa (w ) = Ω`1 ,`2 ,...,`k (w ) are “arithmetic” functions. For example, Ω∅ (w ) =

w , 1−w

φ2 + w

Ω1 (w ) =

Giedrius Alkauskas, Vilnius University, Lithuania

2

(φ4

1
1/2 − 2 . − w )(1 − w )

Structure of the eigenvalues

Preliminaries, previous results Arithmetic of decomposition formulas. Examples Solution

Arithmetic of decomposition formulas Direct connections

Examples 1 Identities involving W0 (n) = 1, and W1 (n): ∞ X n=1 ∞ X

(−1)n+1 φ−2n =

1 φ2 + 1

=

1 ξ1−2 + 1

,

(−1)n+1 φ−4n Pn−1 (3/2)

=

1 1 √ = −2 , 2 2+4 ξ2 + 1

∞ 5 X −6n (0,1) φ Pn−1 (3/2) 4 n=1

=

1 1 √ = , (ξ1,2 ξ2,1 )−2 − 1 4 3+6

∞ 5X (0,1) (−1)n+1 φ−8n Pn−1 (3/2) 4 n=1

=

1 1 √ . = (ξ1,1,2 ξ1,2,1 ξ1,1,2 )−2 + 1 6 10 + 20

5 4

(0,1)

n=1

Identities involving W2 (n). Though W2 (n) is given by a series, not in a closed form, we have: ∞ X

(−1)n+1 W2 (n)φ−2n = 0.0839748528310781+ =

n=1

1 2 √ . = −2 13 + 3 13 ξ3 + 1

This can be double-checked by computer, so the above Theorem 2 must hold true! We are left only with filling all the needed technical details ,. Giedrius Alkauskas, Vilnius University, Lithuania

Structure of the eigenvalues

Preliminaries, previous results Arithmetic of decomposition formulas. Examples Solution

Arithmetic of decomposition formulas Direct connections

Examples 2 A still more interesting example occurs for higher powers. Numerically, we have: ∞ h1i X A = W2 (n)φ−4n = 0.0428848639793538+ ∈ P , π n=1

B =

∞ X

W12 (n)φ−4n = 0.0356498091111648+ ∈ P

n=1

h1i π

,

1 2 √ = = 0.0455447255899809+ , −2 − 1 (ξ ξ 21 + 5 21 1,3 3,1 ) 1 1 √ = D = = 0.0303300858899106+ . (ξ2,2 ξ2,2 )−2 − 1 16 + 12 2 And thus, we get the correct identity C

=

2A + B = 2C + D = 0.1214195370698725+ . Giedrius Alkauskas, Vilnius University, Lithuania

Structure of the eigenvalues

Preliminaries, previous results Arithmetic of decomposition formulas. Examples Solution

Arithmetic of decomposition formulas Direct connections

Direct relations This is not an exotic topic! Trace formulas of D. Mayer were crucial in proving his fundamental result that for a certain generalization of L, call it Ls , to a complex parameter s, one has: Z (s) = det(1 −

L2s )

=

∞ Y

(1 − λ2n (s)),

n=1

where Z (s) is the Selberg zeta function for PSL2 (Z). Maass forms, period functions of modular forms, the work by J. Lewis & D. Zagier, Ann. Math. (2), 2001, 191-258. Every factor 1 − λn (s) vanishes at (halfs) of some Riemann zeros and some spectral parameters associated with even Maass forms. Every factor 1 + λn (s) vanishes at some set of zeros of Z (s), spectral parameters of odd Maass forms. Giedrius Alkauskas, Vilnius University, Lithuania

Structure of the eigenvalues

Preliminaries, previous results Arithmetic of decomposition formulas. Examples Solution

Plans for the attack Solution

The main functional equation Let us now forget everything said, and focus on the following Question. How can one solve the functional equation? λf (z) =

 1  1 + λf (z + 1), f (z + 1)2 z + 1

λ ∈ R, f (z) is defined in the cut plane C \ (−∞, −1), is not identically 0, with the regularity condition sup

|(z + 1)f (z)| < +∞.