"Non-unitary c(2,2k+1) minimal models"의 두 판 사이의 차이

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26번째 줄: 26번째 줄:
 
Let's specify p=2, p'=2k+1, r=1, s=i
 
Let's specify p=2, p'=2k+1, r=1, s=i
  
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<math>\sum_{n=-\infty}^{\infty}(q^{2(2k+1)n^2+(2k+1-2i)n}-q^{(2n+1)((2k+1)n+i)})=\sum_{n=-\infty}^{\infty}(-1)^{n}q^{\frac{(2k+1)n^2}{2}}q^{\frac{n(2k-2i+1)}{2}}</math>
+
<math>\sum_{n=-\infty}^{\infty}(q^{2(2k+1)n^2+(2k+1-2i)n}-q^{(2n+1)((2k+1)n+i)})=\sum_{n=-\infty}^{\infty}(q^{2(2k+1)n^2+(2k+1-2i)n}-\sum_{n=-\infty}^{\infty}q^{(2n+1)((2k+1)n+i)})</math>
  
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<math>=\sum_{n=-\infty}^{\infty}(q^{2(2k+1)n^2+(2k+1-2i)n}-\sum_{n=-\infty}^{\infty}q^{(2(-n)+1)(-(2k+1)(-n)+i)})</math>
 +
 
 +
<math>=\sum_{n=-\infty}^{\infty}(q^{(2n)((2k+1)(2n)+(2k+1-2i))}-\sum_{n=-\infty}^{\infty}q^{(2n-1)((2k+1)(2n-1)+2k-2i+1)/2}</math>
 +
 
 +
<math>=\sum_{n=-\infty}^{\infty}(-1)^{n}q^{\frac{(2k+1)n^2}{2}}q^{\frac{n(2k-2i+1)}{2}}</math>
  
 
<math>=\sum_{n=-\infty}^{\infty}(-1)^n(q^{\frac{(2k-2i+1)}{2}})^{n}(q^{\frac{(2k+1)}{2}})^{n^2}=\prod_{m=1}^{\infty}(1-q^{(2k+1)m})(1-q^{\frac{2k-2i+1}{2}}q^{\frac{2k+1}{2}(2m-1)})(1-q^{-\frac{2k-2i+1}{2}}q^{\frac{2k+1}{2}(2m-1)})</math>
 
<math>=\sum_{n=-\infty}^{\infty}(-1)^n(q^{\frac{(2k-2i+1)}{2}})^{n}(q^{\frac{(2k+1)}{2}})^{n^2}=\prod_{m=1}^{\infty}(1-q^{(2k+1)m})(1-q^{\frac{2k-2i+1}{2}}q^{\frac{2k+1}{2}(2m-1)})(1-q^{-\frac{2k-2i+1}{2}}q^{\frac{2k+1}{2}(2m-1)})</math>

2010년 10월 9일 (토) 05:11 판

introduction
  •  

 

 

central charge and conformal dimensions
  • central charge
    \(c(2,2k+1)=1-\frac{3(2k-1)^2}{2k+1}\)
  • primary fields have conformal dimensions
    \(h_j=-\frac{j(2k-1-j)}{2(2k+1)}\), \(j\in \{0,1,\cdots,k-1\}\) or by setting i=j+1
    \(h_i=-\frac{(i-1)(2k-i)}{2(2k+1)}\) \(i\in \{1,2, \cdots,k\}\) (this is An's notation in his paper)
  • effective central charge
    \(c_{eff}=c-24h_{min}\)
    \(c_{eff}=\frac{2k-2}{2k+1}\)

 

 

character formula and Andrew-Gordon identity

Let's specify p=2, p'=2k+1, r=1, s=i

\(\sum_{n=-\infty}^{\infty}(q^{2(2k+1)n^2+(2k+1-2i)n}-q^{(2n+1)((2k+1)n+i)})=\sum_{n=-\infty}^{\infty}(q^{2(2k+1)n^2+(2k+1-2i)n}-\sum_{n=-\infty}^{\infty}q^{(2n+1)((2k+1)n+i)})\)

\(=\sum_{n=-\infty}^{\infty}(q^{2(2k+1)n^2+(2k+1-2i)n}-\sum_{n=-\infty}^{\infty}q^{(2(-n)+1)(-(2k+1)(-n)+i)})\)

\(=\sum_{n=-\infty}^{\infty}(q^{(2n)((2k+1)(2n)+(2k+1-2i))}-\sum_{n=-\infty}^{\infty}q^{(2n-1)((2k+1)(2n-1)+2k-2i+1)/2}\)

\(=\sum_{n=-\infty}^{\infty}(-1)^{n}q^{\frac{(2k+1)n^2}{2}}q^{\frac{n(2k-2i+1)}{2}}\)

\(=\sum_{n=-\infty}^{\infty}(-1)^n(q^{\frac{(2k-2i+1)}{2}})^{n}(q^{\frac{(2k+1)}{2}})^{n^2}=\prod_{m=1}^{\infty}(1-q^{(2k+1)m})(1-q^{\frac{2k-2i+1}{2}}q^{\frac{2k+1}{2}(2m-1)})(1-q^{-\frac{2k-2i+1}{2}}q^{\frac{2k+1}{2}(2m-1)})\)

\(\prod_{m=1}^{\infty}(1-q^{(2k+1)m})(1-q^{(2k+1)m-i})(1-q^{(2k+1)m-(2k-i+1)})=\prod_{n\neq 0,\pm i\pmod {2k+1}}(1-q^n)^{-1}\)

 

 

different expressions for central charge
  • from above
    \(h_i-c(2,2k+1)/24\)
    \(c(2,2k+1)=1-\frac{3(2k-1)^2}{2k+1}\)
    \(h_i=-\frac{(i-1)(2k-i)}{2(2k+1)}\), \(i\in \{1,2, \cdots,k\}\)
  • L-values
    \(\frac{k}{12}+\frac{2k+1}{12}-\frac{j(N-j)}{2N}\)

 

 

Dirichlet L-function

\(L(-1, \chi) = \frac{1}{2f}\sum_{n=1}^{\infty}{\chi(n)}{n}\)

\(n\geq 1\) 이라 하자. 일반적으로 \(\chi\neq 1\)인 primitive 준동형사상 \(\chi \colon(\mathbb{Z}/f\mathbb{Z})^\times \to \mathbb C^{*}\)에 대하여 \(L(1-n,\chi)\)의 값은 다음과 같이 주어진다

\(L(1-n,\chi)=-\frac{f^{n-1}}{n}\sum_{(a,f)=1}}\chi(a)B_n(\frac{a}{f})\)

\(L(-1,\chi)=L(1-2,\chi)=-\frac{f}{2}\sum_{(a,f)=1}}\chi(a)B_2(\frac{a}{f})\)

여기서 \(B_n(x)\) 는 베르누이 다항식(\(B_0(x)=1\), \(B_1(x)=x-1/2\), \(B_2(x)=x^2-x+1/6\), \(\cdots\))

 

Let N=2k+1

\(\omega=\exp \frac{2\pi i}{2k+1}\)

G: group of Dirichlet characters of conductor N which maps -1 to 1

G has order k and cyclic generated by \(\chi\)

\(c_i=\frac{1}{2k}\sum_{s=1}^{k}\omega^{is}L(-1,\chi^s)\)

Then, 

\(c_i=-\frac{2k+1}{12}+\frac{j(N-j)}{2N}\)

where j satisfies \(\chi(j)=\omega^{k-i}\)

Vacuum energy is given by

\(d_i=\frac{1}{2}L(-1,\chi^{k})-c_i\)

 

Since

\(L(-1,\chi^{k})=\frac{N-1}{12}=\frac{k}{6}\),  the vacuum energy 

\(d_i=\frac{1}{2}L(-1,\chi^{k})-c_i=\frac{k}{12}+\frac{2k+1}{12}-\frac{j(N-j)}{2N}=\frac{j(2k+1-j)}{2(2k+1)}-\frac{k+1}{12}\).

These are equal to \({h_i-c/24}\)

 

 

  1. k := 5
    f[k_, j_] := (2 k)/
       24 + ((2 k + 1)/12 - (j (2 k + 1 - j))/(2 (2 k + 1)))
    Table[{j, f[k, j]}, {j, 1, 2 k}] // TableForm
    Table[{j, -24*f[k, j]}, {j, 1, 2 k}] // TableForm
    d[k_, j_] := (2 (k - j) + 1)^2/(8 (2 k + 1)) - 1/24
    Table[{j, d[k, j]}, {j, 1, 2 k}] // TableForm
    Table[{j, -24*d[k, j]}, {j, 1, 2 k}] // TableForm
    cef[k_, j_] := -((j (2 k - 1 - j))/(2 (2 k +
             1))) - (1 - (3 (2 k - 1)^2)/(2 k + 1))/24
    Table[{j, cef[k, j]}, {j, 0, 2 k - 1}] // TableForm
    Table[{j, -24*cef[k, j]}, {j, 0, 2 k - 1}] // TableForm

 

 

  1. w := Exp[2 Pi*I*1/k]
    L[j_] := -(2 k + 1)/2*
      Sum[DirichletCharacter[2 k + 1, j, a]*
        BernoulliB[2, a/(2 k + 1)], {a, 1, 2 k}]
    c[k_, i_] := 1/(2 k) Sum[w^(i*s)*L[Mod[3*s, 2 k]], {s, 1, k}]
    Table[DiscretePlot[{Re[DirichletCharacter[2 k + 1, j, a]],
       Im[DirichletCharacter[2 k + 1, j, a]]}, {a, 0, 2 k + 1},
      PlotLabel -> j], {j, 1, EulerPhi[2 k + 1]}]
    Table[c[i], {i, 1, 2 k}]

 

 

 

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