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Henry
Henry
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It turned out that it was a very simple calculation to compute $\langle \square,\square+2\rho\rangle$ for $sl(n)$ using Cartan matrices. For $sl(n)$, $\langle \square,\square+2\rho\rangle = 2-\frac{1}{N}$$\langle \square,\square+2\rho\rangle = 2-\frac{1}{n}$, and hence $$t_{\square\square} = q^{2-\frac{1}{N}}.$$$$t_{\square\square} = q^{2-\frac{1}{n}}.$$ Note, by the way, that the convention used here is $q=e^{\frac{\pi i}{k+h^\vee}}$, where $k$ is the bare level, and $h^\vee$ is the dual Coxeter number, which is just $n$ for $sl(n)$.

It turned out that it was a very simple calculation to compute $\langle \square,\square+2\rho\rangle$ for $sl(n)$ using Cartan matrices. For $sl(n)$, $\langle \square,\square+2\rho\rangle = 2-\frac{1}{N}$, and hence $$t_{\square\square} = q^{2-\frac{1}{N}}.$$ Note, by the way, that the convention used here is $q=e^{\frac{\pi i}{k+h^\vee}}$, where $k$ is the bare level, and $h^\vee$ is the dual Coxeter number, which is just $n$ for $sl(n)$.

It turned out that it was a very simple calculation to compute $\langle \square,\square+2\rho\rangle$ for $sl(n)$ using Cartan matrices. For $sl(n)$, $\langle \square,\square+2\rho\rangle = 2-\frac{1}{n}$, and hence $$t_{\square\square} = q^{2-\frac{1}{n}}.$$ Note, by the way, that the convention used here is $q=e^{\frac{\pi i}{k+h^\vee}}$, where $k$ is the bare level, and $h^\vee$ is the dual Coxeter number, which is just $n$ for $sl(n)$.

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Henry
Henry
  • 1.4k
  • 1
  • 8
  • 20

It turned out that it was a very simple calculation to compute $\langle \square,\square+2\rho\rangle$ for $sl(n)$ using Cartan matrices. For $sl(n)$, $\langle \square,\square+2\rho\rangle = 2-\frac{1}{N}$, and hence $$t_{\square\square} = q^{2-\frac{1}{N}}.$$ Note, by the way, that the convention used here is $q=e^{\frac{\pi i}{k+h^\vee}}$, where $k$ is the bare level, and $h^\vee$ is the dual Coxeter number, which is just $n$ for $sl(n)$.