The Geometric Proof of the Hecke Conjecture

Kaida Shi

(Department of , Zhejiang Ocean University,

Zhoushan City, 316004, Zhejiang Province, China)

Abstract Beginning from the resolution of Dirichlet L function L(s, χ) , using the inner product formula of two infinite-dimensional vectors in the complex space, the author proved the baffling problem---- Hecke conjecture. Keywords: Dirichlet L function L(s, χ) , Hecke conjecture, non-trivial zeroes, solid of rotation, axis-

cross section, bary-centric coordinate, inner product, infinite-dimensional vectors. MR(2000): 11M

1 Introduction

For investigating the distribution problem of prime numbers within the arithmetic series, first, L. Dirichlet imported the Dirichlet L function L(s, χ) . Although its property and effect is similar with Riemann Zeta functionζ (s) , but the different matter is, the research about the distribution of the real zeros of the Dirichlet L function L(s, χ) which correspond real property is a very difficult thing. But because of this, it has very important meaning. For solving above-mentioned problem, Hecke put forward a famous proposition as follows: Hecke Conjecture: Suppose that χ is a real property, then we have

L(s, χ) ≠ 0, 0 < s < 1. Now, let’s prove the proposition.

2 The resolution of Dirichlet L function L(s, χ)

Considering the geometric meaning of the Dirichlet L function

∞ χ(n) χ(1) χ(2) χ(3) χ(n) χ = = + + + + + , (1) L(s, ) ∑ s s s s L s L n=1 n 1 2 3 n we have the following figure:

1

w

C(1) A1

A 2 χ (1) B1 χ(2) B2 A 3 1s 2 s χ(3) B3 3s

O D1(1) D2(2) D3(3) … z Fig. 2

In this figure, the areas of the rectangles ADOCADDB11,,, 2 211 ADDB 3322L are respectively:

χ(1) χ(2) χ(3) χ(n) ⋅ ⋅ ⋅ ⋅ s 1, s 1, s 1, L, s 1, L 1 2 3 n therefore, the geometric meaning of the Dirichlet L function L(s, χ) is the sum of the areas of a series of rectangles within the complex space s.

Using the inner product formula between two infinite-dimensional vectors, the Dirichlet L function L(s, χ) equation

χ(1) χ(2) χ(3) χ(n) χ = + + + + + = =+σ L(s, ) s s s L s L 0, ()s it 1 2 3 n can be resolved as

χ(1) χ(2) χ(3) χ(n) 1 1 1 1 χ = ⋅ = (2) L(s, ) ( σ , σ , σ ,L, σ ,L) ( it , it , it ,L, it ,L) 0. 1 2 3 n 1 2 3 n or

1 1 1 1 χ(1) χ(2) χ(3) χ(n) χ = ⋅ = . (3) L(s, ) ( σ , σ , σ ,L, σ ,L) ( it , it , it ,L, it ,L) 0 1 2 3 n 1 2 3 n

From the expressions (2) and (3), we obtain:

2 ∞ χ 2 (n) ∞ 1 → → ⋅ ⋅ cos(N , N ) = 0 ; (4) ∑ 2σ ∑ 2it 1 2 n=1 n n=1 n and

∞ 1 ∞ χ 2 (n) → → ⋅ ⋅ cos(N , N ) = 0 . (5) ∑ 2σ ∑ 2it 3 4 n=1 n n=1 n

But in the complex space, if the inner product between two vectors equals to zero, then these two vectors are

r r π r r π perpendicular, namely, (N , N ) = , (N , N ) = , therefore, we have 1 2 2 3 4 2

r r = cos(N1, N 2 ) 0, and

r r = cos(N3 , N 4 ) 0.

3 The relationship between the volume of the rotation solid and the area of its axis-cross section within the complex space

1 From the expression (1), we can know that when Re(s ) = , the series (harmonic series) within the 2

r r = first radical expression of left side is divergent. Because cos(NN12 , )0 , therefore the situation of the complex series within the second radical expression of left side will change unable to research. Hence, we must transform the Riemann Zeta functionζ ()s equation. For this aim, let’s derive the relationship between the volume of the rotation solid and the area of its axis-cross section within the complex space.

We call the cross section which pass through the axis z and intersects the rotation solid as axis-cross section. According to the barycentric formula of the complex plane lamina:

3 b  ∫ zf[() z− gzdz ()] ξ = a ,  b [()fz− gzdz ()]  ∫a

 b 1 22−  2 [()()]fz gzdz η = ∫a b .  [()fz− gzdz ()]  ∫a we have

b π [()()]fz22− gzdz πη = ∫a 2 b . (6) [()fz− gzdz ()] ∫a

The numerator of the fraction of right side of the formula (**) is the volume of the rotation solid, and the denominator is the area of the axis-cross section. Theη of left side of the formula (6) is the longitudinal coordinate of the barycenter of axis-cross section. Its geometric explanation is: taking the axis-cross section around the axis z to rotate the angle of 2π , we will obtain the volume of the rotation solid.

4 The proof of the Hecke conjecture

Because we have the volume formula of the rotation solid formed by rotating the rectangle

ADDnnn−−11 B naround the axis z:

2 2 2 n πχ πχ πχ = (n) = (n) n = (n) Vn dz z n−1 , ∫n−1 n 2s n 2s n 2s therefore, the sum of the volumes of a series of the cylinders formed by rotating a series of the rectangles ADOCADDB11,,, 2 211 ADDB 3322Laround the axis z is:

∞ χ 2 (1) χ 2 (2) χ 2 (3) χ 2 (n) = = π + + + + + . V ∑ Vn ( 2s 2s 2s L 2s L) n=1 1 2 3 n

But the sum of the areas of a series of the rectangles ADOCADDB11,,, 2 211 ADDB 3322Lis:

4 χ(1) χ(2) χ(3) χ(n) + + + + + . s s s L s L 1 2 3 n

By the formula (6), we have

χ 2 (1) χ 2 (2) χ 2 (3) χ 2 (n) π + + + + + ( 2s 2s 2s L 2s L) 1 2 3 n

χ(1) χ(2) χ(3) χ(n) = πη + + + + + (7) 2 ( s s s L s L). 1 2 3 n

Substituting the Dirichlet L function L(s, χ) equation

χ(1) χ(2) χ(3) χ(n) χ = + + + + + = L(s, ) s s s L s L 0 1 2 3 n into (7), we can obtain the transformed equation:

χ 2 (1) χ 2 (2) χ 2 (3) χ 2 (n) + + + + + = . 2s 2s 2s L 2s L 0 1 2 3 n

This can be explained geometrically as: if the area of the axis-cross section equals to zero, then it corresponds to the volume of the rotation solid also equals to zero.

Because s = σ + 0i , therefore, above expression becomes

χ 2 (1) χ 2 (2) χ 2 (3) χ 2 (n) + + + + + = . 2σ 2σ 2σ L 2σ L 0 1 2 3 n

According to the inner product formula between two infinite-dimensional vectors, above expression can be written as:

χ 4 (1) χ 4 (2) χ 4 (3) χ 4 (n) 1 1 1 1 + + + + + ⋅ + + + + + ⋅ = 4σ 4σ 4σ L 4σ L 4ti 4ti 4ti L 4ti L cos(N1 , N 2 ) 0, 1 2 3 n 1 2 3 n

In the expression, when t = 0 , although the series

∞ ∑sin(0 ⋅ln n) = 0 + 0 + 0 + L+ 0 +L = 0, n=1 but because the series

5 ∞ ∑cos(0 ⋅ ln n) = 1+1+1+ L+1+ L = ∞, n=1

therefore, the complex series

1 + 1 + 1 + + 1 + 0i 0i 0i L 0i L 1 2 3 n ∞ = ∑[cos(0ln n) − isin(0ln n)] n=1 = (1+1+1+L+1+L) − − i ⋅ (0 + 0 + 0 +L+ 0 + L) = ∞.

Because “infinite” multiplied by zero doesn’t equal to zero, therefore

∞ χ 4 ∞ (n) ⋅ 1 ⋅ ∑ 4σ ∑ 0i 0 n=1 n n=1 n

∞ χ 4 = (n) ⋅ ∞ ⋅ ≠ ∑ 4σ 0 0. n=1 n

On the other hand, from the expression (3), we have:

χ 4 χ 4 χ 4 χ 4 1 + 1 + 1 + + 1 + ⋅ (1) + (2) + (3) + + (n) + ⋅ = 4σ 4σ 4σ L 4σ L 4it 4it 4it L 4it L cos(N 3 , N 4 ) 0, 1 2 3 n 1 2 3 n

As is known to all, among the series:

l, l + q, l + 2q, LL (l, q) = 1. we can know χ(n; q) = 1. According to the character of the property, we have

χ(n; q) = χ(n + q; q) , also, we have

χ(n + q; q) = χ(n + 2q; q) = χ(n + 3q; q) = LL = 1, therefore, the real properties are infinite. Hence, we can obtain:

6 ∞ χ 4 (n) ∞ = χ 4 (n)(cos(0 ⋅ log n) − isin(0 ⋅ log n)) ∑ 0i ∑ n=1 n n=1 = 1+ L+1+ L+1+L +1+L = ∞. therefore, we have

∞ 1 ∞ χ 4 (n) ⋅ ⋅ 0 ∑ 4σ ∑ 0i n=1 n n=1 n

∞ 1 = ⋅ ∞ ⋅0 ≠ 0. ∑ 4σ n=1 n

So, we have proved that when the χ is real property, then we have

L(s, χ) ≠ 0, 0 < s < 1.

References

1. Pan C. D., Pan C. B., Basic of Analytic , Beijing: Science Press, 1991. pp.79-109; pp.229-267.

2. Lou S. T., Wu D. H., Riemann hypothesis, Shenyang: Liaoning Education Press, 1987. pp.152-154.

3. J. R., On the representation of a large even integer as the sum of a prime and the product of at most two primes,

Science in China, 16 (1973), No.2, pp. 111-128..

4. Pan C. D., Pan C. B., Goldbach hypothesis, Beijing: Science Press, 1981. pp.1-18; pp.119-147.

5. Hua L. G., A Guide to the Number Theory, Beijing: Science Press, 1957. pp.234-263.

6. Chen J. R., Shao P. C., Goldbach hypothesis, Shenyang: Liaoning Education Press, 1987. pp.77-122; pp.171-205.

7. Chen J. R., The Selected Papers of Chen Jingrun, Nanchang: Jiangxi Education Press, 1998. pp.145-172.

8. Lehman R. S., On the difference π(x)-lix, Acta Arith., 11(1966). pp.397～410.

9. Hardy, G. H., Littlewood, J. E., Some problems of “patitio numerorum” III: On the expression of a number as a sum of primes, Acta. Math., 44 (1923). pp.1-70.

10. Hardy, G. H., Ramanujan, S., Asymptotic formula in combinatory analysis, Proc. London Math. Soc., (2) 17 (1918). pp.

75-115.

11. Riemann, B., Ueber die Anzahl der Primzahlen unter einer gegebenen Große, Ges. Math. Werke und

Wissenschaftlicher Nachlaß , 2, Aufl, 1859, pp 145-155.

7 12. E. C. Titchmarsh, The Theory of the Riemann Zeta Function, Oxford University Press, New York, 1951.

13. A. Selberg, The Zeta and the Riemann Hypothesis, Skandinaviske Mathematiker Kongres, 10 (1946).

Appendix

The inner product formula between two vectors in real

space can be extended formally to complex space

Dear Mr. Referee,

Thank you for your review, please observe following examples.

Example 1 A{1+ i, 3}, B{−i, 2i}, C{1, −i}.

AB = {−1− 2i, −3 + 2i}, AC = {−i, −3 − i}, BC = {1+ i, −3i}. AB = (−1− 2i)2 + (−3 + 2i) 2 = 2 − 8i,

AC = (−i) 2 + (−3 − i)2 = 7 + 6i,

BC = (1+ i) 2 + (−3i) 2 = − 9 + 2i. According to Cosine theorem

2 2 2 AB + AC − BC = 2 AB AC cos(AB, AC) , we have

2 2 2 + − AB AC BC (2 − 8i) + (7 + 6i) − (−9 + 2i) = = 9 − 2i. 2 2 On the other hand, we have

AB ⋅ AC = AB AC cos(AB, AC) = (−1− 2i)(−i) + (−3 + 2i)(−3 − i) = 9 − 2i.

8 According to the Cosine theorem

2 2 2 AC + BC − AB = 2 AC BC cos(AC, BC)

2 2 2 + − AC BC AB (7 + 6i) + (−9 + 2i) − (2 − 8i) = = −2 + 8i. 2 2

On the other hand, we have

AC ⋅ BC = AB AC cos(AB, AC) = (−i)(1+ i) + (−3 − i)(−3i) = −2 + 8i.

− − = 1 θ = 1 − ⋅ + ⋅ 15 8i = 1 − − S ∆ABC AB AC sin 1 2 8i 7 6i 15 8i, 2 2 62 − 44i 2

− − = 1 θ = 1 + ⋅ − + ⋅ 15 8i = 1 − − S ∆ACB AC BC sin 21 7 6i 9 2i 15 8i. 2 2 − 75 − 40i 2

Example 2 A{1+ i, 1− i, 2i}, B{1− i, 1+ i, −2i}, C{1, 0, i}.

AB = {−2i, 2i, −4i}, AC = {−i, −1+ i, −i}, BC = {i, −1− i, 3i}.

AB = (−2i) 2 + (2i)2 + (−4i) 2 = − 24,

AC = (−i) 2 + (−1+ i)2 + (−i) 2 = − 2 − 2i,

BC = (i)2 + (−1− i) 2 + (3i)2 = −10 + 2i. According to the Cosine theorem

2 2 2 AB + AC − BC = 2 AB AC cos(AB, AC) , we have

2 2 2 + − AB AC BC (−24) + (−2 − 2i) − (−10 + 2i) = = −8 − 2i. 2 2 On the other hand, we have

AB ⋅ AC = AB AC cos(AB, AC) = (−2i)(−i) + (2i)(−1+ i) + (−4i)(−i) = −8 − 2i.

According to the Cosine theorem

9 2 2 2 AC + BC − AC = 2 AC BC cos(AC, BC) , we have

2 2 2 + − AC BC AB (−2 − 2i) + (−10 + 2i) − (−24) = = 6. 2 2

On the other hand, we have

AC ⋅ BC = AC BC cos(AC, BC) = (−i)(i) + (−1+ i)(−1− i) + (−i)(3i) = 6.

− + = 1 θ = 1 − ⋅ − − ⋅ 12 16i = − + S ∆ABC AB AC sin 1 24 2 2i 3 4i, 2 2 48 + 48i

− + = 1 θ = 1 − − ⋅ − + ⋅ 12 16i = − + S ∆ACB AC BC sin 21 2 2i 10 2i 3 4i. 2 2 24 +16i

Example 3 A{8i, 14, 8 − i, 1}, B{6, 15i, 17, −8}, C{3 − i, 10 + 7i, 11, 3i}.

AB = {6 − 8i, −14 +15i, 9 + i, −9}, AC = {3 − 9i, −4 + 7i, 3 + i, −1+ 3i},

BC = {−3 − i, 10 − 8i, −6, 8 + 3i}.

AB = (6 − 8i) 2 + (−14 + 5i)2 + (9 + i) 2 + (−9)2 = 104 − 498i,

AC = (3 − 9i) 2 + (−4 + 7i) 2 + (3 + i) 2 + (−1+ 3i) 2 = −105 −110i,

BC = (−3 − i) 2 + (10 − 8i) 2 + (−6)2 + (8 + 3i) 2 = 135 −106i. According to the Cosine theorem

2 2 2 AB + AC − BC = 2 AB AC cos(AB, AC) , we have

2 2 2 + − AB AC BC (104 − 498i) + (−105 −110i) − (−135 −106i) = = −68 − 251i. 2 2 On the other hand, we have

10 AB ⋅ AC = AB AC cos(AB, AC) =

= (6 − 8i)(3 − 9i) + (−14 +15i)(−4 + 7i) + (9 + i)(3 + i) + (−9)(−1+ 3i) = −68 − 251i.

According to the Cosine theorem

2 2 2 AC + BC − AC = 2 AC BC cos(AC, BC) , we have

2 2 2 + − AC BC AB (−105 −110i) + (135 −106i) − (104 − 498i) = = −37 +141i. 2 2

On the other hand, we hand

AC ⋅ BC = AC BC cos(AC, BC) =

= (3 − 9i)(−3 − i) + (−4 + 7i)(10 − 8i) + (−6)(3 + i) + (−1+ 3i)(8 + 3i) = −37 +141i.

1 1 − 7323 + 6714i S ∆ = AB AC sinθ = 104 − 498i ⋅ −105 −110i ⋅ ABC 2 1 2 − + 65700 40850i 1 = − 7323 + 6714i, 2

1 1 − 7323 + 6714i S ∆ = AC BC sinθ = −105 −110i ⋅ 135 −106i ⋅ ACB 2 21 2 − − 25835 3720i 1 = − 7323 + 6714i. 2

11