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0.1 ζ                                                                                                          1


0.1          ζ
0.1.1

                       ζ(s)
                   s                                                                            ζ(s)
                                                                               s=2

                                      ∞
                                                        1   1   1         π2
                         ζ(2) =           (2) = 1 +       + 2 + 2 + ··· =                                  (1)
                                                        22 3   4          6
                                    n=1



                                                            π2
                                                              π2
                                                                                     1735              Euler


                                                                                                  1


                                                                                            1




                                ∞
                                                   1 1 1
                       ζ(1) =         (1) = 1 +     + + + ···
                                                   2 3 4
                                n=1
                                 ∞
                                                               1        1        1
                           =        (1 − p−1 )−1 =                                 1 ···                   (2)
                                p
                                                            (1 − 2 ) (1 − 3 ) (1 − 5 )
                                                                 1        1


                                                                           =

                                             ∞                  ∞
                                                       −s
                                    ζ(s) =         n        =       (1 − p−s )−1                           (3)
                                             n=1                p
   1
                 (1)
       Leibniz
                                           1  1 1 1  1          π
                                    1−       + − + −    + ··· =
                                           3  5 7 9  11         4
                                           1    1   1   1         π3
                                      1−      + 2 − 2 + 2 − ··· =
                                           32  5   7   9          32
 Dirichlet
                              1  1 1 1  1    1             1        √
                        1−      − + + −    −    + · · · = √ log (1 + 2)
                              3  5 7 9  11   13             2
       log
0.1 ζ                                                                                                             2



          Euler                            2m                                                             (1)

                                                              B2m 2m
                        ζ(2m) = (−1)m+1 22m−1                      π ,               m ∈ 1, 2, 3, . . .         (4)
                                                             (2m)!
                                                    Bn
                                                                     ∞
                                                  t                             tn
                                                 t−1
                                                     =                     Bn                                   (4)
                                                e                               n!
                                                                     n=1




          2m                                                          × π 2m
                                                                                                      2m        B2m
               B2m                  ζ(2m)



                  2m     B2m                  B2m                          ζ(2m)
                    2                 1                     6              ζ(2) = 1 π 2 = 2×3 π 2
                                                                                   6
                                                                                             1

                    4                 1                    30              ζ(4) = 90 π = 2×312 ×5 π 4
                                                                                   1 4
                                                                                      1
                    6                 1                    42              ζ(6) = 33 ×5×7 π 6
                                                                                        1
                    8                 1                    30              ζ(8) = 2×33 ×52 ×7 π 8
                                                                                        1
                   10                 5                    66              ζ(10) = 35 ×5×7×11 π 10
                                                                                           691
                   12               691                  2730              ζ(12) = 36 ×53 ×72 ×11×13 π 12
                    .
                    .                 .
                                      .                     .
                                                            .              .
                                                                           .
                    .                 .                     .              .

               (1)                                                                           2



                                                                            B2m
                                                ζ(1 − 2m) = −
                                                                            2m


          (π                )

      2
                          Re(s) > 1
                                           „ «                     „     «
                                            s                        1−s
                                π −2/3 Γ       ζ(s) = π −(1−s)/2 Γ        ζ(1 − s)
                                            2                         2

                                                         Z       ∞
                                                Γ(s) =               e−x xs−1 dx
                                                             0



(4)         2m ≥ 2
                                                                           B2m
                                                 ζ(1 − 2m) = −
                                                                           2m
0.1 ζ                                                                                      3

                                               虚数軸
                                                       ●


                                                                     非自明なゼロ点は実部1/2
                        オイラーの発見した                      ●             の直線上にあるであろう
                        自明なゼロ点                                       リーマン予想




                 ○          ○
                 -4         -2            -1       0       1/2   1      2
                                                                                実軸

                                                                     実部が1より大きい
                                                                     領域にはゼロ点はない
                                                       ●




                                                       ●




                            Figure 1:



                      2m         ζ(1 − 2m)
                        2        ζ(−1) = − 12 = − 221
                                             1
                                                     ×3
                                           1       1
                        4        ζ(−3) = 120 = 23 ×3×5
                        6        ζ(−5) = − 252 = − 22 ×32 ×7
                                             1          1
                                           1       1
                        8        ζ(−7) = 240 = 24 ×3×5
                       10        ζ(−9) = − 132 = − 22 ×3×11
                                             1          1
                                           691           691
                       12        ζ(−11) = 32760 = 23 ×32 ×5×7×11
                        .
                        .        .
                                 .
                        .        .

                                               ζ(1 − 2m) = − B2m
                                                             2m             2m = n

                                                 ζ(−2) = 0, ζ(−4) = 0, . . .
        ζ(0) =   −1
                  2                                                                  s>1



   •                                (3)

   •                         f (s) = f (1 − s)




   •                                                   1/2

                                                           1/2              (         )
0.1 ζ                                                                                          4


0.1.2

                                (4)3




      • 1735                                                                      ζ(2m + 1)
                           ζ(3)            sin

                                           4


      • 1882                        π


      • 1979                 ζ(3)

      • 1884   -1885


      • 2000                                                                      (ζ(s)
                                         ζ(3), ζ(5), ζ(7), . . .
                                  ζ(3), ζ(5), . . . , ζ(a)                            δ(a)
        a→∞
                                                 log a
                                    δ(a) ≥               (1 + O(1))
                                               1 + log 2



      • 2004                                      ζ(5), ζ(7), ζ(9), ζ(11)



                                                         5

  3
                                  W.Zudilin                       (4)
                            4

                            ∞
                           X 1        (2πi)2k B2k
                       2           =−             ,      k = 1, 2, 3, . . .
                           n=1
                               n2k       (2k)!

  (4)
  4
  5
                                                                                          K3
  Picard-Fuchs                                                          Landau-Ginzburg

          Sato-Tate                     Calabi-Yau
           K3                          Calabi-Yau                                    Calabi-Yau
                                                                              1
0.1 ζ                                                                                                         5



      Ramanujan                                                                       2006

                                                       ∞
                                    7π 3                             1
                             ζ(3) =      −2
                                    180                        n3 (e2πn   − 1)
                                                      n=1


                                                           ∞
                                        19π 7                         1
                          ζ(7) =              −2
                                        56700                   n7 (e2πn   − 1)
                                                       n=1

           4m − 1
                     ∞                                                      2m
                              1                      1                                      B2j B4m−2j
 ζ(4m − 1) = −2                         2πn
                                       (e      − 1) − (2π)4m−1                    (−1)j
                          n4m−1                      2                                    (2j)!(4m − 2j)!
                    n=1                                                     j=0




                                  21



                                                                                                       2003
                                                   6 sin        Taylor
                              ∞
                                     (−1)m 2m+1     x3 x5
                  sin x =                    x  =x−    +    ···
                                   (2m + 1)!        3!   5!
                             m=0

        sin x                                  Z
                                   ∞
                                                x2
                    sin x = x           (1 −          )
                                               n2 π 2
                                  n=1
                                x2         x2           x2
                          = x(1 −  )(1 − 2 2 )(1 − 2 2 ) × · · ·
                                π2        2 π          3 π
                                   ∞
                              1        1
                          =x− 2             x3 + · · ·
                             π         n2
                                             n=1

          x3                                Taylor                                               1/6
                                                                  1
                                              x3                  π2
                                                                     ζ(2)

                                                           π2
                                               ζ(2) =
                                                           6
  6
                         9    20                            I

                                                           I       Euler                   (4)

                                                           2003
0.1 ζ                                                                                            6


                                                 n = 2, 4, 6, 8, . . .



                                            ζ(5), ζ(7), ζ(9), ζ(11), . . .

                                                                              7




0.1.3

                                                                     1

      •                  s                                               ζ(s) =(         )×π s

      •                  s

      •                  s                                               ζ(s) = 0

      •                  s                                               ζ(s) =(         )




               Nr , Dr


  1.                                             2/m                          2m    4
                              2/m                        2m      4


  2.                                   Dr                                 p         Dr
                  p−1         2m

                                   1

      ζ(−1) < 0, ζ(−3) > 0, ζ(−5) < 0, ζ(−7) > 0, ζ(−9) < 0, ζ(−11) > 0, . . .


                     s → −∞


           2
                                            2m + 1
  7
      Zagier         ζk (s)
                    s=2                 ζk (2)   3
0.1 ζ                                                                         7




                   ζ(−11)                 691



    2m       Nr                    Dr       ζ(1 − 2m)
                                            p =?      p Dr
                                            p − 1 2m
        2     1                 22 · 3      ζ(−1) = − 12 = − 221
                                                         1
                                                                ×3
                                            p=2
                                            p=3
                                            p=5
        4     1              23 · 3 · 5                1      1
                                            ζ(−3) = 120 = 23 ×3×5
                                            p=2
                                            p=3
                                            p=5
                                            p=7
        6     1             22 · 32 · 7     ζ(−5) = − 252 = − 22 ×32 ×7
                                                         1         1

                                            p=2
                                            p=3
                                            p=5
                                            p=7
                                            p = 11
        8     1              24 · 3 · 5                1      1
                                            ζ(−7) = 240 = 24 ×3×5
                                            p=2
                                            p=3
                                            p=5
                                            p=7
                                            p = 11
        10    1             22 · 3 · 11     ζ(−9) = − 132 = − 22 ×3×11
                                                         1         1

                                            p=2
                                            p=3
                                            p=5
                                            p=7
                                            p = 11
                                            p = 13
        12   691   23 · 32 · 5 · 7 · 11                691          691
                                            ζ(−11) = 32760 = 23 ×32 ×5×7×11
                                            p=2
                                            p=3
                                            p=5
                                            p=7
                                            p = 11
                                            p = 13
         .
         .     .
               .                      .
                                      .     .
                                            .
         .     .                      .     .
0.1 ζ                                                                                      8


0.1.4    691

     Fermat       350
                                                1994     Wiles
                                  3                      a
         x=y=z=0

                                      xa + y a = z a

19                                              Kummer


                                                                            Q




     • Dirichlet                            k      ε              ε−1               Ok
                                                                            r = r1 +r2 −1
                                   r1                             r2


                                  Ok ∼ Zr ⊕(
                                   ×
                                     =                        )



     •
                    Ik
                                   Pk                                   Cl(k) = Ik /Pk



                             19                                                        1

                                                                √
     •        1                                          k = Q( −m)           9
                                        1967                 Baker Stark
                                                         m = 1, 2, 3, 7, 11, 19, 43, 67, 163
                         1              m
         Siegel
                                                             √
     •                                                 k = Q( m)                  1
                                                          Gauss
0.1 ζ                                                                                                             9



                                                Kummer                 Fermat
                                             Kummer                       Q 1            n        ξ
              Q(ξ)   (ξ n   = 1)                                                             Q(e2πi/p )           p
                                                   p    Q(e2πi/p )
                                                           1850

      p                            h            Q(e2πi/p )                       h
          p                     8              p
                                 B2m , m = 1, 2, . . . , (p − 3)/2                 p
                                p                                  xp + y p = z p x = y = z = 0
                                                   Kummer



      •       1 2 ≤ 2m ≤ p − 3                                        2m
                                             p

      •       2 Q(e2πi/p )                   p

                                       p                              Fermat                  Kum-
mer                                              691                        ζ(12)(          ζ(−11))
                                                   Kummer                        x691 + y 691 = z 691
                                                                          691
                                                                                     691


                Kummer                        Herbrand                       Ribet
                 Kummer             2                                           Galois
                                             ∼
                   ω : σ ∈ Gal(Q(e2πi/p )/Q − → (Z/p)×
                                             −

   •    2’                                 Cl(Q(e2πi/p ))                p           x                    σ   ∈
Gal(Q(e2πi/p )/Q


                                             σ(x) = ω(σ)(1−r) x



          2’            1          1930           Herbrand                                                    1
           2’         Ribet            1976                                          Galois
                                                                                                    Ribet
                691                                                                           Ramanujan
∆                                                                        Ribet               1980
Frey                            Fermat
                                                                                             Fermat

  8
      h                     p                               37, 59, 67, 101, . . .
0.1 ζ                                                                                            10




0.1.5

                                                       1 1
                                       ζ(1) = 1 +       + + ···
                                                       2 3
                                                           Riemann                           1
                      Re s > 1



                                          lim (s − 1)ζ(s) = 1
                                        s→1+0

                                               ζ(s)     1                            s   1
                                                                                         9


                             Q                                                               F
                                                             Dirichlet

                                                        2r1 (2π)r2 hF
                             lim (s − 1)ζF (s) =                      RF
                             s→1                            |DF | wF

                                             ζF (s)            hF
                                       lim  r1 +r2 −1
                                                        =−        RF
                                       s→0 s                   wF
  9
                         9       20                                 I
                                                   1  1 1 1
                                  log(2) = 1 −       + − + − ...
                                                   2  3 4 5
                                                            Re s > 1
                                                      1    1   1   1
                         (1 − 21−s )ζ(s) = 1 −           + s − s + s − ···
                                                      2s  3   4   5


                                        lim (s − 1)ζ(s)
                                       s→1+0

                                                (s − 1)
                                   = lim                · (1 − 21−s )ζ(s)
                                       s→1+0   1 − 21−s
                                      1
                                   =      · log 2
                                    log 2
                                   =1

                                                                               s=1
                                                              log
      π                                      log
                                                                                         π
          log
                log                                                      log
                                                                                                 π
0.1 ζ                                                                                                            11



                   r1                                                                                   r2
                                                               hF        F                wF        F             1
                        DF       F                                  RF       F

                                    α∈F
                                        log |α(i)|   (i = 1, . . . , r1 ),
                        l(i) α =
                                        2 log |α(j)| (j = r1 + 1, . . . , r1 + r2 )
                                                                             η1 , . . . , ηr
                                                     l(1) η1    l(1) η2      · · · l(1) ηr
                                                     l(2) η1    l(2) η2      · · · l(2) ηr
                             R[η1 , . . . , ηr ] =
                                                      ···           ···      ··· ···
                                                     l(r) η1    l(r) η2      · · · l(r) ηr
  (η1 , . . . , ηr )




      • Beilinson
      • Birch-Swinnerton-Dyer



      Ramanujan                      Selberg                    Langlands




0.1.6

      2005                           Witten                                           Langlands
                                                                                                             6
Witten                              6                                                                        4
                        4
                               S-
Schimmrigk                                   q-
                                        -           Langlands
                Witten                        Arthur 1989
                                                    Langlands
                                                                                               10




 10

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Zetavalue

  • 1. 0.1 ζ 1 0.1 ζ 0.1.1 ζ(s) s ζ(s) s=2 ∞ 1 1 1 π2 ζ(2) = (2) = 1 + + 2 + 2 + ··· = (1) 22 3 4 6 n=1 π2 π2 1735 Euler 1 1 ∞ 1 1 1 ζ(1) = (1) = 1 + + + + ··· 2 3 4 n=1 ∞ 1 1 1 = (1 − p−1 )−1 = 1 ··· (2) p (1 − 2 ) (1 − 3 ) (1 − 5 ) 1 1 = ∞ ∞ −s ζ(s) = n = (1 − p−s )−1 (3) n=1 p 1 (1) Leibniz 1 1 1 1 1 π 1− + − + − + ··· = 3 5 7 9 11 4 1 1 1 1 π3 1− + 2 − 2 + 2 − ··· = 32 5 7 9 32 Dirichlet 1 1 1 1 1 1 1 √ 1− − + + − − + · · · = √ log (1 + 2) 3 5 7 9 11 13 2 log
  • 2. 0.1 ζ 2 Euler 2m (1) B2m 2m ζ(2m) = (−1)m+1 22m−1 π , m ∈ 1, 2, 3, . . . (4) (2m)! Bn ∞ t tn t−1 = Bn (4) e n! n=1 2m × π 2m 2m B2m B2m ζ(2m) 2m B2m B2m ζ(2m) 2 1 6 ζ(2) = 1 π 2 = 2×3 π 2 6 1 4 1 30 ζ(4) = 90 π = 2×312 ×5 π 4 1 4 1 6 1 42 ζ(6) = 33 ×5×7 π 6 1 8 1 30 ζ(8) = 2×33 ×52 ×7 π 8 1 10 5 66 ζ(10) = 35 ×5×7×11 π 10 691 12 691 2730 ζ(12) = 36 ×53 ×72 ×11×13 π 12 . . . . . . . . . . . . (1) 2 B2m ζ(1 − 2m) = − 2m (π ) 2 Re(s) > 1 „ « „ « s 1−s π −2/3 Γ ζ(s) = π −(1−s)/2 Γ ζ(1 − s) 2 2 Z ∞ Γ(s) = e−x xs−1 dx 0 (4) 2m ≥ 2 B2m ζ(1 − 2m) = − 2m
  • 3. 0.1 ζ 3 虚数軸 ● 非自明なゼロ点は実部1/2 オイラーの発見した ● の直線上にあるであろう 自明なゼロ点 リーマン予想 ○ ○ -4 -2 -1 0 1/2 1 2 実軸 実部が1より大きい 領域にはゼロ点はない ● ● Figure 1: 2m ζ(1 − 2m) 2 ζ(−1) = − 12 = − 221 1 ×3 1 1 4 ζ(−3) = 120 = 23 ×3×5 6 ζ(−5) = − 252 = − 22 ×32 ×7 1 1 1 1 8 ζ(−7) = 240 = 24 ×3×5 10 ζ(−9) = − 132 = − 22 ×3×11 1 1 691 691 12 ζ(−11) = 32760 = 23 ×32 ×5×7×11 . . . . . . ζ(1 − 2m) = − B2m 2m 2m = n ζ(−2) = 0, ζ(−4) = 0, . . . ζ(0) = −1 2 s>1 • (3) • f (s) = f (1 − s) • 1/2 1/2 ( )
  • 4. 0.1 ζ 4 0.1.2 (4)3 • 1735 ζ(2m + 1) ζ(3) sin 4 • 1882 π • 1979 ζ(3) • 1884 -1885 • 2000 (ζ(s) ζ(3), ζ(5), ζ(7), . . . ζ(3), ζ(5), . . . , ζ(a) δ(a) a→∞ log a δ(a) ≥ (1 + O(1)) 1 + log 2 • 2004 ζ(5), ζ(7), ζ(9), ζ(11) 5 3 W.Zudilin (4) 4 ∞ X 1 (2πi)2k B2k 2 =− , k = 1, 2, 3, . . . n=1 n2k (2k)! (4) 4 5 K3 Picard-Fuchs Landau-Ginzburg Sato-Tate Calabi-Yau K3 Calabi-Yau Calabi-Yau 1
  • 5. 0.1 ζ 5 Ramanujan 2006 ∞ 7π 3 1 ζ(3) = −2 180 n3 (e2πn − 1) n=1 ∞ 19π 7 1 ζ(7) = −2 56700 n7 (e2πn − 1) n=1 4m − 1 ∞ 2m 1 1 B2j B4m−2j ζ(4m − 1) = −2 2πn (e − 1) − (2π)4m−1 (−1)j n4m−1 2 (2j)!(4m − 2j)! n=1 j=0 21 2003 6 sin Taylor ∞ (−1)m 2m+1 x3 x5 sin x = x =x− + ··· (2m + 1)! 3! 5! m=0 sin x Z ∞ x2 sin x = x (1 − ) n2 π 2 n=1 x2 x2 x2 = x(1 − )(1 − 2 2 )(1 − 2 2 ) × · · · π2 2 π 3 π ∞ 1 1 =x− 2 x3 + · · · π n2 n=1 x3 Taylor 1/6 1 x3 π2 ζ(2) π2 ζ(2) = 6 6 9 20 I I Euler (4) 2003
  • 6. 0.1 ζ 6 n = 2, 4, 6, 8, . . . ζ(5), ζ(7), ζ(9), ζ(11), . . . 7 0.1.3 1 • s ζ(s) =( )×π s • s • s ζ(s) = 0 • s ζ(s) =( ) Nr , Dr 1. 2/m 2m 4 2/m 2m 4 2. Dr p Dr p−1 2m 1 ζ(−1) < 0, ζ(−3) > 0, ζ(−5) < 0, ζ(−7) > 0, ζ(−9) < 0, ζ(−11) > 0, . . . s → −∞ 2 2m + 1 7 Zagier ζk (s) s=2 ζk (2) 3
  • 7. 0.1 ζ 7 ζ(−11) 691 2m Nr Dr ζ(1 − 2m) p =? p Dr p − 1 2m 2 1 22 · 3 ζ(−1) = − 12 = − 221 1 ×3 p=2 p=3 p=5 4 1 23 · 3 · 5 1 1 ζ(−3) = 120 = 23 ×3×5 p=2 p=3 p=5 p=7 6 1 22 · 32 · 7 ζ(−5) = − 252 = − 22 ×32 ×7 1 1 p=2 p=3 p=5 p=7 p = 11 8 1 24 · 3 · 5 1 1 ζ(−7) = 240 = 24 ×3×5 p=2 p=3 p=5 p=7 p = 11 10 1 22 · 3 · 11 ζ(−9) = − 132 = − 22 ×3×11 1 1 p=2 p=3 p=5 p=7 p = 11 p = 13 12 691 23 · 32 · 5 · 7 · 11 691 691 ζ(−11) = 32760 = 23 ×32 ×5×7×11 p=2 p=3 p=5 p=7 p = 11 p = 13 . . . . . . . . . . . .
  • 8. 0.1 ζ 8 0.1.4 691 Fermat 350 1994 Wiles 3 a x=y=z=0 xa + y a = z a 19 Kummer Q • Dirichlet k ε ε−1 Ok r = r1 +r2 −1 r1 r2 Ok ∼ Zr ⊕( × = ) • Ik Pk Cl(k) = Ik /Pk 19 1 √ • 1 k = Q( −m) 9 1967 Baker Stark m = 1, 2, 3, 7, 11, 19, 43, 67, 163 1 m Siegel √ • k = Q( m) 1 Gauss
  • 9. 0.1 ζ 9 Kummer Fermat Kummer Q 1 n ξ Q(ξ) (ξ n = 1) Q(e2πi/p ) p p Q(e2πi/p ) 1850 p h Q(e2πi/p ) h p 8 p B2m , m = 1, 2, . . . , (p − 3)/2 p p xp + y p = z p x = y = z = 0 Kummer • 1 2 ≤ 2m ≤ p − 3 2m p • 2 Q(e2πi/p ) p p Fermat Kum- mer 691 ζ(12)( ζ(−11)) Kummer x691 + y 691 = z 691 691 691 Kummer Herbrand Ribet Kummer 2 Galois ∼ ω : σ ∈ Gal(Q(e2πi/p )/Q − → (Z/p)× − • 2’ Cl(Q(e2πi/p )) p x σ ∈ Gal(Q(e2πi/p )/Q σ(x) = ω(σ)(1−r) x 2’ 1 1930 Herbrand 1 2’ Ribet 1976 Galois Ribet 691 Ramanujan ∆ Ribet 1980 Frey Fermat Fermat 8 h p 37, 59, 67, 101, . . .
  • 10. 0.1 ζ 10 0.1.5 1 1 ζ(1) = 1 + + + ··· 2 3 Riemann 1 Re s > 1 lim (s − 1)ζ(s) = 1 s→1+0 ζ(s) 1 s 1 9 Q F Dirichlet 2r1 (2π)r2 hF lim (s − 1)ζF (s) = RF s→1 |DF | wF ζF (s) hF lim r1 +r2 −1 =− RF s→0 s wF 9 9 20 I 1 1 1 1 log(2) = 1 − + − + − ... 2 3 4 5 Re s > 1 1 1 1 1 (1 − 21−s )ζ(s) = 1 − + s − s + s − ··· 2s 3 4 5 lim (s − 1)ζ(s) s→1+0 (s − 1) = lim · (1 − 21−s )ζ(s) s→1+0 1 − 21−s 1 = · log 2 log 2 =1 s=1 log π log π log log log π
  • 11. 0.1 ζ 11 r1 r2 hF F wF F 1 DF F RF F α∈F log |α(i)| (i = 1, . . . , r1 ), l(i) α = 2 log |α(j)| (j = r1 + 1, . . . , r1 + r2 ) η1 , . . . , ηr l(1) η1 l(1) η2 · · · l(1) ηr l(2) η1 l(2) η2 · · · l(2) ηr R[η1 , . . . , ηr ] = ··· ··· ··· ··· l(r) η1 l(r) η2 · · · l(r) ηr (η1 , . . . , ηr ) • Beilinson • Birch-Swinnerton-Dyer Ramanujan Selberg Langlands 0.1.6 2005 Witten Langlands 6 Witten 6 4 4 S- Schimmrigk q- - Langlands Witten Arthur 1989 Langlands 10 10