How do I get the expression for the Killing form of $\mathfrak{u}(2)$?

Question

I saw in the book of Mark Hamilton "Mathematical of gauge theory" in the pag. 126 that the killing form to lie algebra $\mathfrak{u}(2)$ is

$$B(X, Y) = 4 Tr(XY) - 2 Tr(X)Tr(Y).$$ I want prove this for this case (not the general case $\mathfrak{gl}(2, \mathbb{C})$). I am trying take the basis $(T^0, T^j) := (i Id, i\sigma_j)$, where $\sigma_j$ are Pauli's matrices, then, by one hand, a matrix $A \in \mathfrak{u}(2)$ can be write as

$$A = \frac{Tr(A)}{2i} T^0 + a_j T^j.$$ So,

$$ad(A) = \sum_{j=1}^{3} a_j ad(T^j),$$ because $ad(T^0)=0$. But, when I use the definition

$$B(A,B) := Tr(ad(A)ad(B)) = \sum_{j=1}^{3} \sum_{l=1}^{3} a_j b_l Tr(ad(T^j)ad(T^l)),$$ I don't get progress, and seem to me that the $Tr(A)$ won't appear because the $T^0$ componente don't appear. Can anyone help me?

Appreciate.

Answer 1

Denote with $\{t_1,t_2,t_3\}$ a basis for $\mathfrak{su}(2)$ such that $[t_i,t_j]=\epsilon_{ijk}t_k$ (implicit sums over repeated indices will be assumed whenever suitable; I'll also not distinguish between lower and upper indices). Note that, in terms of the Pauli matrices, these are $t_i=-\frac{i}{2}\sigma_i$.

1. (First method) The Killing form is defined as $B(X,Y)={\rm Tr}({\rm ad}(X)\circ{\rm ad}(Y))$, for $X,Y\in\mathfrak{su}(2)$ with decompositions $X=X_i t_i,Y=Y_i t_i$.

   We have $$({\rm ad}(t_i)\circ{\rm ad}(t_j))(t_k) = [ t_i, [t_j, t_k]] = \sum_{\ell m}\epsilon_{jk\ell} \epsilon_{i\ell m} t_m = \sum_{\ell m}\epsilon_{jk\ell} \epsilon_{m i\ell} t_m. $$ We can now use the identity $\epsilon_{jk\ell} \epsilon_{m i\ell} = \delta_{jm}\delta_{jk} - \delta_{ji}\delta_{km}$, to obtain $$({\rm ad}(t_i)\circ{\rm ad}(t_j))(t_k) = \delta_{ki} t_j - \delta_{ji} t_k. $$ For the trace, you need to take the $t_k$ term in this expression, and then sum over $k$: $${\rm Tr}({\rm ad}(t_i)\circ{\rm ad}(t_j)) = \sum_k (\delta_{ki}\delta_{jk} - \delta_{ji}) = -2 \delta_{ij},$$ and thus $$B(X,Y)\equiv {\rm Tr}({\rm ad}(X)\circ{\rm ad}(Y)) = \sum_{ij} X_i Y_j {\rm Tr}({\rm ad}(t_i)\circ{\rm ad}(t_j)) = -2\sum_{i} X_i Y_i. $$ If you represent $X,Y$ as matrices in the standard way, then $${\rm Tr}(XY) = \sum_{ij} X_i Y_j \left(-\frac{1}{4}{\rm Tr}(\sigma_i \sigma_j)\right) = -\frac{1}{2}\sum_{i} X_i Y_i. $$ We conclude that $$B(X,Y) = 4{\rm Tr}(XY).$$

2. (Another method) We can also leverage the general expression of the Killing form in terms of the structure coefficients. If $[t_i,t_j]=\sum_k {c_{ij}}^k t_k$, then $B(t_i,t_j)=\sum_{mn} {c_{im}}^n {c_{jn}}^m$. In the case of ${c_{ij}}^k=\epsilon_{ijk}$, as is the case for $\mathfrak{su}(2)$, we get $$B(t_i,t_j) = \sum_{mn} \epsilon_{imn}\epsilon_{jnm} = - \sum_{mn} \epsilon_{imn}\epsilon_{jmn} = -2\delta_{ij},$$ and the rest follows as before.

A more general discussion about deriving the Killing forms for the classical Lie algebra is found in How to derive the general formula for the Killing form for classical Lie algebras?.