"Finite dimensional representations of Sl(2)"의 두 판 사이의 차이

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34번째 줄: 34번째 줄:
 
* <math>U_{n+1}(x) & = 2xU_n(x) - U_{n-1}(x)</math><br> U_0[x]=1<br> U_1[x]=2 x<br> U_2[x]=-1+4 x^2<br> U_3[x]=-4 x+8 x^3<br>
 
* <math>U_{n+1}(x) & = 2xU_n(x) - U_{n-1}(x)</math><br> U_0[x]=1<br> U_1[x]=2 x<br> U_2[x]=-1+4 x^2<br> U_3[x]=-4 x+8 x^3<br>
 
*  character evaluated at an element of SU(2) with the eigenvalues e^{i\theta}, e^{-i\theta} is given by the Chebyshev polynomials<br><math>U_k(\cos\theta)= \frac{\sin (k+1)\theta}{\sin \theta}</math><br>
 
*  character evaluated at an element of SU(2) with the eigenvalues e^{i\theta}, e^{-i\theta} is given by the Chebyshev polynomials<br><math>U_k(\cos\theta)= \frac{\sin (k+1)\theta}{\sin \theta}</math><br>
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* <math>w=e^{i\theta}</math>, <math>z=w+w^{-1}=2\cos\theta</math><br><math>p_k(z)=\frac{w^{k+1}-w^{-k-1}}{w-w^{-1}}</math><br><math>p_{0}(z)=1</math><br><math>p_{1}(z)=z</math><br><math>p_{2}(z)=z^2-1</math><br><math>p_{3}(z)=z^3-2z</math><br><math>p_k(z)^2=1+p_{k-1}(z)p_{k+1}(z)</math><br>
+
* <math>w=e^{i\theta}</math>, <math>z=w+w^{-1}=2\cos\theta</math><br><math>\operatorkame{ch}L(k\omega)=p_k(z)=\frac{w^{k+1}-w^{-k-1}}{w-w^{-1}}</math><br><math>p_{0}(z)=1</math><br><math>p_{1}(z)=z</math><br><math>p_{2}(z)=z^2-1</math><br><math>p_{3}(z)=z^3-2z</math><br><math>p_k(z)^2=1+p_{k-1}(z)p_{k+1}(z)</math><br>
  
 
 
 
 
51번째 줄: 51번째 줄:
 
<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">symmetric power of sl(2) representations</h5>
 
<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">symmetric power of sl(2) representations</h5>
  
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*  q-binomial type formula<br><math>\prod_{j=0}^{k}(1-zq^{k-2j})^{-1}=\sum_{j=0}^{\infty}z^j\begin{bmatrix} k+j\\ k\end{bmatrix}_{q}</math><br>
+
*  q-binomial type formula (Heine formula, [[useful techniques in q-series]])<br><math>\prod_{j=0}^{k}(1-zq^{k-2j})^{-1}=\sum_{j=0}^{\infty}z^j\begin{bmatrix} k+j\\ k\end{bmatrix}_{q}</math><br>
 
*  the character of j-th symmetric power of V_k is<br><math>\begin{bmatrix} k+j\\ k\end{bmatrix}_{q}</math><br> where the q-analogue of the natural number is defined as <br><math>[n]_{q}=\frac{q^n-q^{-n}}{q-q^{-1}}</math><br>
 
*  the character of j-th symmetric power of V_k is<br><math>\begin{bmatrix} k+j\\ k\end{bmatrix}_{q}</math><br> where the q-analogue of the natural number is defined as <br><math>[n]_{q}=\frac{q^n-q^{-n}}{q-q^{-1}}</math><br>
  
72번째 줄: 72번째 줄:
 
I claim that
 
I claim that
  
−
<math>F(t,q)=\sum_{j=0}^{\infty}F_j(q)t^j=\prod_{j=0}^{k}(1-tq^{k-2j})^{-1}</math>. 
+
<math>F(z,q)=\sum_{j=0}^{\infty}F_j(q)z^j=\prod_{j=0}^{k}(1-zq^{k-2j})^{-1}</math>. 
  
 
To prove that see the power series expansion of a factor:
 
To prove that see the power series expansion of a factor:
  
−
<math>(1-tq^{k-2j})^{-1}=\sum_{m=0}^{\infty}t^mq^{m(k-2j)}</math>. Therefore
+
<math>(1-zq^{k-2j})^{-1}=\sum_{m=0}^{\infty}z^mq^{m(k-2j)}</math>. Therefore
  
−
<math>\prod_{j=0}^{k}(1-tq^{k-2j})^{-1}=\sum_{m_0,\cdots,m_k}t^{m_0+\cdots+m_k}q^{(k-0)m_0+(k-2)m_1+\cdots+(2-k)m_{k-1}+(0-k)m_k}</math>
+
<math>\prod_{j=0}^{k}(1-zq^{k-2j})^{-1}=\sum_{m_0,\cdots,m_k}z^{m_0+\cdots+m_k}q^{(k-0)m_0+(k-2)m_1+\cdots+(2-k)m_{k-1}+(0-k)m_k}</math>
  
 
Now we can easily check
 
Now we can easily check
  
−
<math>\prod_{j=0}^{k}(1-tq^{k-2j})^{-1}=\sum_{j=0}^{\infty}t^j\begin{bmatrix} k+j\\ k\end{bmatrix}_{q}</math>■
+
<math>\prod_{j=0}^{k}(1-zq^{k-2j})^{-1}=\sum_{j=0}^{\infty}z^j\begin{bmatrix} k+j\\ k\end{bmatrix}_{q}</math>■
  
 
 
 
 
90번째 줄: 90번째 줄:
 
<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">exterior algebra of sl(2) representations</h5>
 
<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">exterior algebra of sl(2) representations</h5>
  
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*  q-binomial type formula (Gauss formula)<br><math>\prod_{j=0}^{k}(1+tq^{k-2j})}=\sum_{j=0}^{k+1}\begin{bmatrix} k+1 \\ j\end{bmatrix}_{q}q^{j(j-1)/2}t^j</math><br>
+
*  q-binomial type formula (Gauss formula, [[useful techniques in q-series]][[q-analogue of summation formulas|]])<br><math>\prod_{j=0}^{k}(1+zq^{k-2j})}=\sum_{j=0}^{k+1}\begin{bmatrix} k+1 \\ j\end{bmatrix}_{q}q^{j(j-1)/2}z^j</math><br>
 
*  the character of j-th exterior algebra of V_k is<br><math>\begin{bmatrix} k+1 \\ j\end{bmatrix}_{q}q^{j(j-1)/2}</math><br>
 
*  the character of j-th exterior algebra of V_k is<br><math>\begin{bmatrix} k+1 \\ j\end{bmatrix}_{q}q^{j(j-1)/2}</math><br>
  

2010년 11월 22일 (월) 21:18 판

introduction

 

 

specialization
  • Cartan matrix
    \(\mathbf{A} = \begin{pmatrix} 2 \end{pmatrix}\)
  • root system
    \(\Phi=\{\alpha,-\alpha\}\)

 

 

representation theory
  • integrable weights and Weyl vector
    \(\omega=\frac{1}{2}\alpha\)
    \(\rho=\omega\)
  • there is a unique k+1 dimensional irreducible module \(V_k\) with the highest integrable weight \(\lambda=k\omega\)
  • Weyl-Kac formula
    \(\operatorkame{ch}L(k\omega)=\frac{e^{(k+1)\omega}-e^{-(k+1)\omega}}{e^{\omega}-e^{-\omega}}=e^{k\omega}+e^{(k-2)\omega}+\cdots+e^{-k\omega}\)

 

 

character formula and Chebyshev polynomial of the 2nd kind
  • \(U_{n+1}(x) & = 2xU_n(x) - U_{n-1}(x)\)
    U_0[x]=1
    U_1[x]=2 x
    U_2[x]=-1+4 x^2
    U_3[x]=-4 x+8 x^3
  • character evaluated at an element of SU(2) with the eigenvalues e^{i\theta}, e^{-i\theta} is given by the Chebyshev polynomials
    \(U_k(\cos\theta)= \frac{\sin (k+1)\theta}{\sin \theta}\)
  • \(w=e^{i\theta}\), \(z=w+w^{-1}=2\cos\theta\)
    \(\operatorkame{ch}L(k\omega)=p_k(z)=\frac{w^{k+1}-w^{-k-1}}{w-w^{-1}}\)
    \(p_{0}(z)=1\)
    \(p_{1}(z)=z\)
    \(p_{2}(z)=z^2-1\)
    \(p_{3}(z)=z^3-2z\)
    \(p_k(z)^2=1+p_{k-1}(z)p_{k+1}(z)\)

 

 

Hermite reciprocity
  • [GW1998]
  • dimension of symmetric algebra and exterior algebra of V_k

 

 

symmetric power of sl(2) representations
  • q-binomial type formula (Heine formula, useful techniques in q-series)
    \(\prod_{j=0}^{k}(1-zq^{k-2j})^{-1}=\sum_{j=0}^{\infty}z^j\begin{bmatrix} k+j\\ k\end{bmatrix}_{q}\)
  • the character of j-th symmetric power of V_k is
    \(\begin{bmatrix} k+j\\ k\end{bmatrix}_{q}\)
    where the q-analogue of the natural number is defined as 
    \([n]_{q}=\frac{q^n-q^{-n}}{q-q^{-1}}\)

 

(proof)

Fix a k throughout the argument.

Let \(F_j(q)\) be the character of j-th symmetric power of V_k.

\(F_j(q)=\sum_{m_0,\cdots,m_k}q^{(k-0)m_0+(k-2)m_1+\cdots+(2-k)m_{k-1}+(0-k)m_k}\)

where \(m_0+m_1+\cdots+m_k=j\)

Now consider the generating function

\(F(z,q)=\sum_{j=0}^{\infty}F_j(q)z^j\)

I claim that

\(F(z,q)=\sum_{j=0}^{\infty}F_j(q)z^j=\prod_{j=0}^{k}(1-zq^{k-2j})^{-1}\). 

To prove that see the power series expansion of a factor\[(1-zq^{k-2j})^{-1}=\sum_{m=0}^{\infty}z^mq^{m(k-2j)}\]. Therefore

\(\prod_{j=0}^{k}(1-zq^{k-2j})^{-1}=\sum_{m_0,\cdots,m_k}z^{m_0+\cdots+m_k}q^{(k-0)m_0+(k-2)m_1+\cdots+(2-k)m_{k-1}+(0-k)m_k}\)

Now we can easily check

\(\prod_{j=0}^{k}(1-zq^{k-2j})^{-1}=\sum_{j=0}^{\infty}z^j\begin{bmatrix} k+j\\ k\end{bmatrix}_{q}\)■

 

 

exterior algebra of sl(2) representations

 

 

 

history

 

 

related items
  • [[affine sl(2) $A^{(1)} 1$]]

 

 

encyclopedia

 

 

books

 

 

articles

[[2010년 books and articles|]]

 

 

question and answers(Math Overflow)

 

 

blogs

 

 

experts on the field

 

 

links
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