[Math] How to use Weyl dimension formula

lie-algebras

Let $V(\lambda)$ be a highest weight module of a semi-simple Lie algebra with highest weight $\lambda$. The Weyl dimension formula is $\dim V(\lambda) = \frac{\prod_{\alpha>0} (\lambda+\rho, \alpha)}{\prod_{\alpha>0}(\rho, \alpha)}$. By multiplying $\frac{2}{(\alpha, \alpha)}$ for each $\alpha$, we have $\dim V(\lambda) = \frac{\prod_{\alpha>0} \langle \lambda+\rho, \alpha \rangle}{\prod_{\alpha>0} \langle \rho, \alpha \rangle }$. Here $\langle \cdot, \cdot \rangle$ the form such that $\langle \lambda_i, \alpha_j \rangle = \delta_{ij}$ where $\lambda_j$ are fundamental weights and $\alpha_j$ are simple roots. Let $\Delta^{\vee}$ be a base of $\Phi^{\vee}$ (dual root system). Then $\alpha^{\vee} = \sum_{i} k_i \alpha_i^{\vee} $. How to compute the coefficients $k_i$? If we can compute $k_i$, then we know how to use the Weyl dimension formula.

For type $G_2$, I computed that $\alpha_1=2\lambda_1-\lambda_2$, $\alpha_2=-3\lambda_1+2\lambda_2$, $\alpha_1+\alpha_2=-\lambda_1+\lambda_2$, $2\alpha_1+\alpha_2=\lambda_1$, $3\alpha_1+\alpha_2=3\lambda_1-\lambda_2$, $3\alpha_1+2\alpha_2=\lambda_2$. These are all positive roots. $\rho=\lambda_1+\lambda_2$.

Let $\lambda=m_1\lambda_1+m_2\lambda_2$. How can we obtain the formula $$\dim V(\lambda) = 1/120 (m_1+1)(m_2+1)(m_1+m_2+2)(m_1+2m_2+3)(m_1+3m_2+4)(2m_1+3m_2+5)?$$

Best Answer

If I well remember the notation in rooth systems, I think the following equality will be useful for you.

Let $\alpha$ be $\sum_{i=1}^lc_i\alpha_i$. We have

$$ \langle\lambda,\alpha\rangle = (\lambda, \alpha^{\vee})=\frac{2(\lambda,\alpha)}{(\alpha,\alpha)}=\frac{2(\lambda,\sum_{i=1}^lc_i\alpha_i)}{(\alpha,\alpha)}= \sum_{i=1}^lc_i\frac{2(\lambda,\alpha_i)}{(\alpha,\alpha)}=\sum_{i=1}^lc_i\frac{(\alpha_i,\alpha_i)}{(\alpha,\alpha)}\frac{2(\lambda,\alpha_i)}{(\alpha_i,\alpha_i)}=\sum_{i=1}^lc_i\frac{(\alpha_i,\alpha_i)}{(\alpha,\alpha)}(\lambda,\alpha_i^{\vee})=\sum_{i=1}^lc_i\frac{(\alpha_i,\alpha_i)}{(\alpha,\alpha)}\langle\lambda,\alpha_i\rangle$$

So in your notation $$k_i= c_i\frac{(\alpha_i,\alpha_i)}{(\alpha,\alpha)}$$ Now I think you could easy check the formula for $G_2$ starting from its roots system.

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Compute Casimir operator for $su(n)$ representations via the Weyl formula

$\DeclareMathOperator{\Tr}{Tr}\DeclareMathOperator{\ad}{ad}$ Before doing the calculations for $\mathfrak{su}(4)$, here are the corresponding ones for $\mathfrak{su}(2)$ first.

Let $x = -\frac{i}{2} \sigma_1$, $y = -\frac{i}{2} \sigma_2$ and $z = -\frac{i}{2} \sigma_3$, where the $\sigma_i$ are the Pauli matrices. Besides using Pauli matrices, I am using mathematicians' notations, so that elements in $\mathfrak{su}(2)$ are skew-hermitian, rather than hermitian as in physics texts. We then have

$$ [x,y] = z \text{ and cyclic}.$$

Let us denote the Killing form by $B(-,-)$, so that

$$ B(x_1,x_2) = \Tr(\ad_{x_1} \circ \ad_{x_2}), $$

where $x_1, x_2 \in \mathfrak{su}(2)$. We then have $B(x,x) = B(y,y) = B(z,z) = -2$ and $x$, $y$ and $z$ are $B$-orthogonal.

Let $\Lambda$ be the highest weight for the fundamental representation of $\mathfrak{su}(2)$. In that representation, we then have

$$ x^2 = y^2 = z^2 = -\frac{1}{4} I.$$

So the Casimir element is therefore

$$ C = -\frac{1}{2}\left( - \frac{1}{4}I - \frac{1}{4}I - \frac{1}{4}I\right) = \frac{3}{8}I.$$

Let $e_1$ and $e_2$ denote the following diagonal matrices

$$ e_1 = \operatorname{diag}(i,0) \text{ and } e_2 = \operatorname{diag}(0,i).$$

Let $h = e_1 - e_2$, which spans a Cartan subalgebra of $\mathfrak{su}(2)$. Note that $B(h,h) = -8$, by a straight-forward calculation, so that $B = -4 g_{Euc}$, where $g_{Euc}$ is the Euclidean inner product induced on the real span of $h$ from the Euclidean inner product on the real span of $e_1$ and $e_2$, so that for instance $g_{Euc}(e_1-e_2,e_1-e_2) = 2$.

If $D$ is the real span of $e_1$ and $e_2$, so that $D$ consists of pure imaginary diagonal $2$ by $2$ matrices, let $D^*$ be its dual. Moreover, let $e^1, e^2$ be the dual basis of $e_1$, $e_2$, so that we have

$$ (e_i, e^j) = \delta_i^j$$

where $(-,-)$ here denotes the natural pairing between $D$ and its dual space $D^*$, and $\delta_i^j$ is the Kronecker delta.

We then have $\delta = \Lambda = \frac{1}{2}(e^1-e^2)$. We now have to be careful and use $B^{-1}$ on the real span of $e^1-e^2$, rather than $B$, as we are now looking at elements in the dual of the span of $h$, i.e. in the dual of the chosen Cartan subalgebra. Most authors just write $B$, rather than $B^{-1}$, as it is implicitly understood that one should use the inverse inner product, but I will write it as $B^{-1}$ to stress this point. Moreover, it is customary to introduce a minus sign and consider minus $B^{-1}$ rather than $B^{-1}$, in order to get a positive-definite inner product, rather than a negative-definite one.

Then

$$-B^{-1}(\Lambda+2\delta, \Lambda) = \frac{1}{4}g^{-1}_{Euc}(\frac{3}{2}(e^1-e^2),\frac{1}{2}(e^1-e^2)) = \frac{1}{4} \times \frac{3}{4} \times 2 = \frac{3}{8}.$$

Note that if we had used instead $\Lambda + \delta$ rather than $\Lambda + 2 \delta$, we would have gotten the incorrect value $1/4$, so the correct formula is the one which has $\Lambda + 2 \delta$.

(Note that I did not go over the references themselves, so there is a chance that their notation may be a little different).

Edit 1: I then did the computation for $\mathfrak{su}(4)$ and I obtained that, for the fundamental representation,

$$C = \frac{15}{32}I$$

from the definition of $C$. On the other hand, $$\delta = \frac{3}{2}e^1 +\frac{1}{2}e^2 -\frac{1}{2}e^3 -\frac{3}{2}e^4$$ and $$\Lambda = \frac{1}{4}\left( 3e^1-e^2-e^3-e^4 \right)$$ so that $$\Lambda + 2 \delta = \frac{1}{4}\left( 15e^1 + 3e^2 - 5e^3 - 13e^4 \right).$$

Moreover, the Killing form $B$ is $$B = -8 g_{Euc},$$ so that $$-B^{-1}(\Lambda, \Lambda+2\delta) = \frac{15}{32}$$ which is the correct answer indeed.

Remark: the calculation of the Killing form and its inverse were kind of long, but luckily, there were many zeros. It would take a long time to type everything, but let me know if something is not clear.

Edit 1: the positive roots of $\mathfrak{su}(4)$ with respect to our choice of Cartan subalgebra and simple roots, is $e^i - e^j$ for $1 \leq i<j \leq 4$.

If we let $T_i = e_i-e_{i+1}$, for $i = 1, \ldots, 3$, then the $T_i$ form a basis of the chosen Cartan subalgebra of $\mathfrak{su}(4)$. If we adjoin to them $T_4 = (1/4)(e_1+e_2+e_3+e_4)$, then the $T_i$, for $1 \leq i \leq 4$, form a basis for the diagonal matrices in $\mathfrak{u}(4)$, and their dual basis, is given by

$$T^1 = \frac{1}{4}(3e^1-e^2-e^3-e^4)$$ $$T^2 = \frac{1}{2}(e^1+e^2-e^3-e^4)$$ $$T^3 = \frac{1}{4}(e^1+e^2+e^3-3e^4)$$ and $$T^4 = \frac{1}{2}(e^1+e^2+e^3+e^4).$$

Note that $T^i$ for $i = 1, \ldots, 3$ form a basis dual to the $T_i$ for $i = 1, \ldots, 3$. One can then check, that with respect to that basis, one has

$$e^1 - e^2 = (2,-1,0)$$ $$e^2 - e^3 = (-1,2,-1)$$ $$e^3 - e^4 = (0,-1,2)$$ and so on.

One can thus check that $\delta = (1,1,1)$ and $\Lambda = (1,0,0)$. Therefore $\Lambda + 2\delta = (3,2,2)$.

Moreover, with respect to the $T_i$ basis ($1 \leq i \leq 3$), one has $$ g_{Euc} = \left( \begin{array}{ccc} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{array} \right).$$ Inverting it gives $$ g_{Euc}^{-1} = \frac{1}{4} \left( \begin{array}{ccc} 3 & 2 & 1 \\ 2 & 4 & 2 \\ 1 & 2 & 3 \end{array} \right).$$

We then deduce that $$\frac{1}{8} g_{Euc}^{-1}(\Lambda+2\delta, \Lambda) = \frac{15}{32},$$ as required.

Edit 2: a reference I personally find very helpful, though it is perhaps best as a second book on the topic, is Knapp's "Lie groups beyond an introduction". In the appendices at the end of the book, if I remember well, you have various tables for various complex semisimple Lie algebras, their root systems, their $\delta$ etc. I am personally mostly self-taught on this beautiful topic, and I find that Knapp's book is great for looking up various notions and theorems regarding the structure and representation theory of Lie groups/Lie algebras (especially the complex semisimple ones).

Deriving Jacobi triple from the Weyl-Kac denominator formula of affine $\mathfrak{sl}_2$

Let $(\delta, \varpi_1, \Lambda_0)$ be the basis of $\mathfrak{h}^*$ which is dual to the basis $(d, \alpha_1^\vee, c)$ of $\mathfrak{h}$, where $c$ is the canonical central element and $d$ is the derivation. The element $\varpi_1$ is a lift of the fundamental weight of $\mathfrak{sl}_2$ to the affine algebra, but is not a fundamental weight of the affine algebra: instead the element $\Lambda_1 = \varpi_1 + \Lambda_0$ is. In this basis the missing roots and coroots are $$ \alpha_1 = 2 \varpi_1, \quad \alpha_0 = \delta - 2 \varpi_1, \quad \alpha_0^\vee = c - \alpha_1^\vee.$$ The reason we use this basis is because it shows the structure of the Weyl group a bit better: abstractly, the affine Weyl group (for untwisted affine Lie algebras) is the semidirect product of the finite coroot lattice and the finite Weyl group. In terms of the basis $(\delta, \varpi_1, \Lambda_0)$ of $\mathfrak{h}^*$, the matrices (with zeros omitted) of the simple reflections are $$ s_0 = \begin{pmatrix} 1 & 1 & -1 \\ & -1 & 2 \\ & & 1 \end{pmatrix}, \quad s_1 = \begin{pmatrix} 1 & & \\ & -1 & \\ & & 1\end{pmatrix}. $$ We can see the semidirect product structure by defining $t = s_0 s_1$, then we have $s_1 t^k = t^{-k} s_1$, and the Weyl group is the semidirect product of $\mathbb{Z}$ (generated by $t$) and $\mathbb{Z}_2$ (generated by $s_1$). In terms of matrices, we have $$ t^k = \begin{pmatrix} 1 & -k & -k^2 \\ & 1 & 2k \\ & & 1 \end{pmatrix}.$$

A little computation shows that $$ t^k \rho - \rho = (-k - 2k^2)\delta + 4k \varpi_1, \quad s_1 t^k \rho - \rho = (-k -2k^2) \delta + (-4k - 2) \varpi_1.$$ Therefore letting $q = e^{-\delta}$ and $r = e^{\alpha_1} = e^{2 \varpi_1}$ we get $$ \begin{aligned} \sum_{w \in W} (-1)^w e^{w \rho - \rho} &= \sum_{k \in \mathbb{Z}}(e^{t^k \rho - \rho} - e^{s_1 t^k \rho - \rho}) \\ &= \sum_{k \in \mathbb{Z}}q^{k + 2k^2}r^{2k} - \sum_{k \in \mathbb{Z}}q^{k + 2k^2}r^{-2k-1} \\ &= \sum_{l \in 2\mathbb{Z}} q^{(l^2 + l)/2} r^l - \sum_{l \in 2\mathbb{Z} + 1}q^{(l^2 - l)/2} r^{-l} \\ &= \sum_{l \in 2\mathbb{Z}} q^{(l^2 + l)/2} r^l - \sum_{l \in 2\mathbb{Z} + 1}q^{(l^2 + l)/2} r^{l} \\ &= \sum_{l \in \mathbb{Z}} (-1)^l q^{(l^2 + l)/2} r^l. \end{aligned}$$

The first equality is using the semidirect product fact, the third equality is reindexing the first sum over even integers $l = 2k$ and the second sum over odd integers $l = 2k + 1$, and the fourth equality is using the $l \mapsto -l$ symmetry in the second sum. The final answer is equivalent to yours because I used $r = e^{\alpha_1}$ rather than $r = e^{- \alpha_1}$.

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Compute Casimir operator for $su(n)$ representations via the Weyl formula

Deriving Jacobi triple from the Weyl-Kac denominator formula of affine $\mathfrak{sl}_2$

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