"Theta functions in affine Kac-Moody algebras"의 두 판 사이의 차이
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imported>Pythagoras0 |
imported>Pythagoras0 |
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| 11번째 줄: | 11번째 줄: | ||
\begin{align} | \begin{align} | ||
\Theta_{k,\lambda} &=e^{-\frac{|\bar{\lambda}|^2\delta}{2k}}\sum_{\gamma \in M}e^{t_{\gamma}(\lambda)} \\ | \Theta_{k,\lambda} &=e^{-\frac{|\bar{\lambda}|^2\delta}{2k}}\sum_{\gamma \in M}e^{t_{\gamma}(\lambda)} \\ | ||
| + | &=e^{-\frac{|\bar{\lambda}|^2\delta}{2k}}e^{\lambda}\sum_{\gamma \in M}e^{k\gamma}q^{\frac{k(\gamma,\gamma)}{2}+(\gamma,\lambda)} \\ | ||
&=e^{k\Lambda_0}\sum_{\mu\in Q^{\vee}+\frac{\bar{\lambda}}{k}}e^{k\mu-\frac{1}{2}k( \mu,\mu ) \delta}\\ | &=e^{k\Lambda_0}\sum_{\mu\in Q^{\vee}+\frac{\bar{\lambda}}{k}}e^{k\mu-\frac{1}{2}k( \mu,\mu ) \delta}\\ | ||
&=e^{k\Lambda_0}\sum_{\mu\in Q^{\vee}+\frac{\bar{\lambda}}{k}}e^{k\mu}q^{\frac{k}{2}( \mu,\mu )} | &=e^{k\Lambda_0}\sum_{\mu\in Q^{\vee}+\frac{\bar{\lambda}}{k}}e^{k\mu}q^{\frac{k}{2}( \mu,\mu )} | ||
2014년 12월 15일 (월) 02:00 판
introduction
notation
- let $k\in \mathbb{Z}_{\geq 1}$ be the level
- Let $M=Q^{\vee}$. This is also the $\mathbb{Z}$-span of $W\theta$ where $\theta$ is the highest root
- for $\gamma\in M$, define $t_{\gamma} : \mathfrak{h}^{*}\to \mathfrak{h}^{*}$ by
$$t_{\gamma}(\lambda)=\lambda+\lambda(c)\gamma-\left(\frac{1}{2}\lambda(c)|\gamma|^2+(\gamma,\lambda)\right)\delta $$
- note $\lambda=\bar{\lambda}+k\Lambda_0$
- note that $|\lambda|^2=|\bar{\lambda}|^2$
- definition
$$ \begin{align} \Theta_{k,\lambda} &=e^{-\frac{|\bar{\lambda}|^2\delta}{2k}}\sum_{\gamma \in M}e^{t_{\gamma}(\lambda)} \\ &=e^{-\frac{|\bar{\lambda}|^2\delta}{2k}}e^{\lambda}\sum_{\gamma \in M}e^{k\gamma}q^{\frac{k(\gamma,\gamma)}{2}+(\gamma,\lambda)} \\ &=e^{k\Lambda_0}\sum_{\mu\in Q^{\vee}+\frac{\bar{\lambda}}{k}}e^{k\mu-\frac{1}{2}k( \mu,\mu ) \delta}\\ &=e^{k\Lambda_0}\sum_{\mu\in Q^{\vee}+\frac{\bar{\lambda}}{k}}e^{k\mu}q^{\frac{k}{2}( \mu,\mu )} \end{align} $$
$A_1$ example
- level k=1, $\lambda=0$
- let $z=e^{-\alpha_1}$
$$ \Theta_{1,0}=1 + q (1/z + z) + q^4 (1/z^2 + z^2) + q^9 (1/z^3 + z^3)+\cdots $$