Question about the Killing form in Georgi

Question

I'm trying to understand the Killing form described on page 49 in the book by Howard Georgi. He starts by saying that one defines the inner product between two generators $T_a$ and $T_b$ in the adjoint representation as follows: \begin{equation} \mathrm{Tr} (T_a T_b) \end{equation} and then subsequently he says that this is a real symmetric matrix. I'm not sure why this is the case, because the trace would just be a number? First, I thought this might just be a small mistake.

However, he derives that a linear transformation on the generators $X_a$ (in an arbitrary representation): \begin{equation} X_a \rightarrow X_a' = L_{ab}X_b \end{equation} results in the following transformation: \begin{equation} \mathrm{Tr} (T_a T_b) \rightarrow \mathrm{Tr} (T_a' T_b') = L_{ac}L_{bd} \mathrm{Tr} (T_c T_d) \end{equation} And then he states we can diagonalize the trace by choosing an appropriate $L$ such that we can write (after dropping the primes): \begin{equation} \mathrm{Tr} (T_a T_b) = k^a \delta_{a b} \end{equation} I really don't understand where this equation comes from. I've never heard of diagonalizing the trace (because this is a number, not a matrix) and I couldn't find anything useful on Google. Any help with my problem would be much appreciated.

Best regards,

Answer 1

We're simply treating the Lie algebra of the relevant Lie group here purely as a vector space and making linear transformations on that linear space. Since $\operatorname{Tr}(X\,Y) = \operatorname{Tr}(Y\,X)$ is generally true, the matrix of the trace is symmetric. The $L_{a,b}$ are like generalized rotations and, as long as they have nonsingular matrices, keep all the information of the Lie algebra.

Some of Georgi's comments I don't think are general. The form he is defining is the Killing form in the adjoint representation and it is not always an inner product. He is assuming that the group concerned (i) has a finite centre and (ii) is compact, for we have the following remarkable theorem:

Given that a Lie group has a finite centre, the Killing form on a group's Lie algebra is negative definite if and only if the group is compact.

A good reference for this is: S. Helgason "Differential geometry Lie groups and symmetric spaces" Chap. II, section 6, prop. 6.6.

I love this theorem - it's really quite spesh when you think about it - telling us as it does something about the group's global properties from information encoded locally (in the Lie algebra).

So the Killing form is the negative of an inner product for compact groups with finite centres. Once we have an inner product, we can of course define orthogonality, orthonormality and unitary transformations of the Lie algebra. Although I've not seen this before, this is going to be how the diagonalisation you speak of can be done. Once you have an inner product, the Gramm-Schmidt procedure can be worked through, and this is how your $L_{a,b}$ are going to be derived.

For the unitary groups (which I suspect Georgi is dealing with - we are talking about Prof SU(5) / SO(10) here!), the negative of the Killing form is even more readily seen to be an inner product:

$$\left<X,\,Y\right> = \operatorname{Tr}(X^\dagger\, Y) = -\operatorname{Tr}(X\, Y)$$

because of course the Lie algebra members are skew-Hermitian.