3-manifolds and their invariants

수학노트
http://bomber0.myid.net/ (토론)님의 2010년 3월 23일 (화) 21:10 판
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introduction
  • volume of figure 8 knot complement
    • 2.02988321281930725
      \(V=\frac{9\sqrt{3}}{\pi^2}\zeta_{\mathbb{Q}(\sqrt{-3})}(2)=3D(e^{\frac{2i\pi}{3}})=2D(e^{\frac{i\pi}{3}})=2.029883212819\cdots\)
  • what is \(\zeta_{\mathbb{Q}(\sqrt{-3})}(2)\)? numrically 1.285190955484149
  1. L[x_] := Im[PolyLog[2, x]] + 1/2 Log[Abs[x]] Arg[1 - x]
    f[x_, y_] :=
     L[x] + L[1 - x*y] + L[y] + L[(1 - y)/(1 - x*y)] + L[(1 - x)/(1 - x*y)]
    Print["five term relation"]
    Table[f[i, j], {i, 0.1, 0.9, 0.1}, {j, 0.1, 0.9, 0.1}] // TableForm
    N[3 L[Exp[2 I*Pi/3]], 20]
    N[2 L[Exp[I*Pi/3]], 20]
    N[3 (L[Exp[2 I*Pi/3]] - L[Exp[4 I*Pi/3]])/2, 20]
    N[Pi^2*L[Exp[2 I*Pi/3]]/(3 Sqrt[3]), 20]

 

 

 

복소이차수체의 데데킨트 제타함수

\(\zeta_{K}(2)=\frac{\pi^2}{6\sqrt{|d_K|}}\sum_{(a,d_k)=1} (\frac{d_K}{a})D(e^{2\pi ia/|d_k|})\)

\(\zeta_{\mathbb{Q}\sqrt{-3}}(2)=\frac{\pi^2}{6\sqrt{3}}(D(e^{2\pi i/3})-D(e^{4\pi i/3}))\)

\(\zeta_{\mathbb{Q}\sqrt{-7}}(2)=\frac{\pi^2}{3\sqrt{7}}(D(e^{2\pi i/7})+D(e^{4\pi i/7})-D(e^{6\pi i/7}))\)

  1. L[x_] := Im[PolyLog[2, x]] + 1/2 Log[Abs[x]] Arg[1 - x]
    N[Sum[JacobiSymbol[a, 7]*L[Exp[2 I*Pi*a/7]], {a, 1, 6}], 20]
    N[L[Exp[2 I*Pi/7]] + L[Exp[4 I*Pi/7]] - L[Exp[6 I*Pi/7]], 20]

 

 

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