While learning general representation theory of Lie algebra, I am having hard times to compute myself the value of the quadratic Casimir operator for some basic examples.
To fix the notation, according to Ref. [1], the quadratic Casimir operator is defined by
$$ C = g^{\mu\nu} T_\mu T_\nu
\tag{1}\label{def}
$$
where $T_\mu$ are the generators of the algebra, satisfying
$$ [T_\mu, T_\nu] = C^\rho_{\mu\nu}T_\rho $$
and the metric $g_{\mu\nu}$ is defined as
$$ g_{\mu\nu}\equiv C^\alpha_{\mu\beta}C^\beta_{\nu\alpha}$$.
$g^{\mu\nu}$ is the inverse of $g_{\mu\nu}$, i.e., $g^{\mu\nu}g_{\nu\rho}=\delta^{\mu}_\rho$.
Ref. [1] reports the following Weyl's formula for the eigenvalues $C(\Lambda)$ of the representation of $C$
$$C(\Lambda) = \langle \Lambda |\Lambda + \delta\rangle
\tag{2}\label{weyl}
$$
where $\Lambda$ is the maximum weight of the representation and $\delta$
$$\delta\equiv \frac{1}{2} \sum_{\alpha>0} \alpha$$
is half of the sum of the positive roots.
(In other words, $C$ is represented by $C(\Lambda)I$, $I$ the identity matrix.)
Now let me take as a basic example the defining representation of $su(4)$. With the "physicists" convention, the algebra is the one of hermitian traceless matrix, and is spanned by 3 commuting elements generating the Cartan subalgebra
$$(T_\alpha)_{ij}=\delta_{ij}(\delta_{\alpha,i}-\delta_{\alpha+1,i})\qquad \alpha=1,2,3
\tag{3}\label{cartan}
$$
and the remaining $4^2-1-3=12$ generators can be chosen to be matrices like these
$$\begin{pmatrix}0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix},\qquad
\begin{pmatrix}0 & 0 & i & 0\\ 0 & 0 & 0 & 0\\ -i & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix},\quad \text{etc.}
$$
Computing by brute force the sum of Eq. (\ref{def}) I get $C=15/32$ for the defining representation.
On the other hand, the maximum weight of the defining representation is $\Lambda=(1,0,0)$ according to Ref. [1] and also to a direct evaluation from Eq. (\ref{cartan}). Also, the positive roots are $(1,-1,0)$, $(1,0,-1)$, $(0,1,-1)$ from which I get $\delta=(1,0,-1)$. This also agrees with a formula reported in Ref. [1]
$$
\delta_\rho = n/2-\rho,\qquad su(n)
\tag{4}\label{delta}
$$
With Eq. (\ref{delta}), and using Eq. (\ref{weyl}) one gets $C(\Lambda)=2$, different than the explicit computation.
More generally, for the defining representation of $su(n)$, which has $\Lambda=(1,0,\ldots,0)$ [1], I would get $C(\Lambda)=\frac{n^2-1}{2n^2}$ by summing explicitly Eq. (\ref{def}), whereas apparently with Eq. (\ref{weyl}) I would get $C(\Lambda)=n/2$.
I also noticed that Ref.[2] cites a slightly different Weyl formula, perhaps a different normalization:
$$
C(\Lambda) = \langle \Lambda + \delta|\Lambda + \delta\rangle-\langle \delta| \delta\rangle = \langle \Lambda |\Lambda + 2\delta\rangle
\tag{5}\label{weyl2}
$$
-
What is the source of the discrepancy between $C$ as obtained with an explicit summation, and the (apparent) result of Eq. (\ref{weyl}) or Eq. (\ref{weyl2})?
-
What is the correct way of using Eq. (\ref{weyl}) (or perhaps Eq. (\ref{weyl2})) to compute $C(\Lambda)$, for a generic representation with maximum weight $\Lambda$?
[1]
J.D. Vergados, "Group and Representation Theory", World Scientific
[2]
B. C. Hall, "Lie Groups, Lie Algebras and Representations", Springer
Best Answer
$\DeclareMathOperator{\Tr}{Tr}\DeclareMathOperator{\ad}{ad}$ 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).