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

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imported>Pythagoras0
imported>Pythagoras0
41번째 줄: 41번째 줄:
  
 
* '''[GW1998]'''
 
* '''[GW1998]'''
* dimension of symmetric algebra and exterior algebra of V_k
+
* dimension of symmetric algebra and exterior algebra of $V_k$
  
 
 
 
 
49번째 줄: 49번째 줄:
 
===symmetric power of sl(2) representations===
 
===symmetric power of sl(2) representations===
  
* 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>
+
* q-binomial type formula (Heine formula,[[useful techniques in q-series]])
* 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>
+
:<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
 +
:<math>\begin{bmatrix} k+j\\ k\end{bmatrix}_{q}</math>
 +
where the q-analogue of the natural number is defined as <math>[n]_{q}=\frac{q^n-q^{-n}}{q-q^{-1}}</math>
  
 
;proof
 
;proof
57번째 줄: 59번째 줄:
 
Fix a k throughout the argument.
 
Fix a k throughout the argument.
  
Let <math>F_j(q)</math> be the character of j-th symmetric power of V_k.
+
Let <math>F_j(q)</math> be the character of j-th symmetric power of $V_k$.
 
+
:<math>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}</math>
<math>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}</math>
 
  
 
where <math>m_0+m_1+\cdots+m_k=j</math>
 
where <math>m_0+m_1+\cdots+m_k=j</math>
  
 
Now consider the generating function
 
Now consider the generating function
 
+
:<math>F(z,q)=\sum_{j=0}^{\infty}F_j(q)z^j</math>
<math>F(z,q)=\sum_{j=0}^{\infty}F_j(q)z^j</math>
 
  
 
I claim that
 
I claim that
 +
:<math>F(z,q)=\sum_{j=0}^{\infty}F_j(q)z^j=\prod_{j=0}^{k}(1-zq^{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
 
+
:<math>(1-zq^{k-2j})^{-1}=\sum_{m=0}^{\infty}z^mq^{m(k-2j)}</math>
To prove that see the power series expansion of a factor:
+
Therefore
 
+
:<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>
<math>(1-zq^{k-2j})^{-1}=\sum_{m=0}^{\infty}z^mq^{m(k-2j)}</math>. Therefore
 
 
 
<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-zq^{k-2j})^{-1}=\sum_{j=0}^{\infty}z^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>■
 
  
 
 
 
 
88번째 줄: 85번째 줄:
  
 
* q-binomial type formula (Gauss formula,[[useful techniques in q-series]][[q-analogue of summation formulas|q-analogue of summation formulas]])
 
* q-binomial type formula (Gauss formula,[[useful techniques in q-series]][[q-analogue of summation formulas|q-analogue of summation formulas]])
:<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>
+
:<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>
* the character of j-th exterior algebra of V_k is :<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  
 +
:<math>\begin{bmatrix} k+1 \\ j\end{bmatrix}_{q}q^{j(j-1)/2}</math>
  
  
95번째 줄: 93번째 줄:
  
 
analogous to the above. ■
 
analogous to the above. ■
 
 
  
 
==Clebsch-Gordan coefficients==
 
==Clebsch-Gordan coefficients==

2013년 12월 14일 (토) 15:27 판

introduction

 

 

specialization

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

 

representation theory

  • integrable weights and Weyl vector

\[\omega=\frac{1}{2}\alpha, \rho=\omega\]

\[\operatorname{ch}L(k\omega)=\frac{e^{(k+1)\omega}-e^{-(k+1)\omega}}{e^{\omega}-e^{-\omega}}=e^{k\omega}+e^{(k-2)\omega}+\cdots+e^{-k\omega}\]


character formula and Chebyshev polynomial of the 2nd kind

  • \(U_{n+1}(x) = 2xU_n(x) - U_{n-1}(x)\)
  • character evaluated at an element of SU(2) with the eigenvalues $e^{i\theta}, e^{-i\theta}$ is given by the Chebyshev polynomials

\[U_k(\cos\theta)= \frac{\sin (k+1)\theta}{\sin \theta}\]

  • \(w=e^{i\theta}\),\(z=w+w^{-1}=2\cos\theta\)
  • \(p_k(z)=\frac{w^{k+1}-w^{-k-1}}{w-w^{-1}}\)
  • \(p_k(z)^2=1+p_{k-1}(z)p_{k+1}(z)\)

 

 

Hermite reciprocity

  • [GW1998]
  • dimension of symmetric algebra and exterior algebra of $V_k$

 

 

symmetric power of sl(2) representations

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

  • 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

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

  • the character of j-th exterior algebra of $V_k$ is

\[\begin{bmatrix} k+1 \\ j\end{bmatrix}_{q}q^{j(j-1)/2}\]


proof

analogous to the above. ■

Clebsch-Gordan coefficients

 

 

Catalan numbers

  1. f[n_] := Integrate[(2 Cos[Pi*x])^n*2 (Sin[Pi*x])^2, {x, 0, 1}]
    Table[Simplify[f[2 k]], {k, 1, 10}]
    Table[CatalanNumber[n], {n, 1, 10}]

 

 

 

history

 

 

related items

 

 

encyclopedia


 

 

books

  • [GW1998]Goodman and Wallach,Representations and invariants of the classical groups


articles

  • Bacry, Henri. 1987. “SL(2,C), SU(2), and Chebyshev Polynomials.” Journal of Mathematical Physics 28 (10) (October 1): 2259–2267. doi:10.1063/1.527759.