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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,

Hunter
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    Comment to the question (v1): Georgi is saying that $M=(M_{ab}){1\leq a,b\leq n}$ is a real symmetric matrix with matrix elements $M{ab}:=\mathrm{Tr} (T_a T_b)\in \mathbb{R}$. – Qmechanic Oct 23 '13 at 00:29
  • Ah ok thank you! I cannot see how you get this from what he writes, but that makes sense. – Hunter Oct 23 '13 at 00:53

1 Answers1

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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.

Selene Routley
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  • Thank you for your reply. I have another question. Do you know why we can only define the inner product (i.e. Killing form) for generators in the adjoint representation? It seems to me that the trace of the generators satisfies the conditions of the inner product in any arbitrary representation of the generators. – Hunter Oct 23 '13 at 16:18
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    @Hunter Good question. Let me think about this one. I think you can define an inner product as you say in any representation (e.g. for unitary groups), it's simply that unless it's in the adjoint representation, properties of this bilinear form don't have special significance like the theorem in Helgason. Another one is this: the group is semisimple iff the Killing form is nondegenerate. I'm pretty sure this is the reason why one doesn't talk about Killing forms in other reps, but there may still be a reason to define an inner product (to define a metric, for example) there. – Selene Routley Oct 24 '13 at 01:46
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    @Hunter Of course - silly me (I was disarmed by the clarity of your thinking sword! - ouch!) - there is the other reason that one also needs the Lie algebra representation to be a faithful one to convey the greatest possible information about the underlying Lie group. My prejudice in thinking is of using the algebra to study the Lie group - so I have only thought of the Killing form in the adjoint rep's image - but of course one generally has more general reasons to use other, nonfaithful, reps. Moreover, not even the adjoint rep is faithful to the group if the latter has a nondiscrete centre – Selene Routley Oct 25 '13 at 00:07
  • Thank you for your reply. I will have to think more about your answers. I think your level of understanding is much better than mine, so I'm not sure if I fully understand you. Either way, it is always good to be pushed to better understand such an important topic ;). If I have more questions, I will be sure to ask them here on this forum. – Hunter Oct 26 '13 at 10:56
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    @Hunter I'm sorry you don't fully understand me, but I think I know what you mean - feel free to ask any questions here after having mulled over this topic (you'll have to post the question in a comment here to contact me, or send me an email at the address on my user page and we can chat). A very simple application of the Killing form in the adjoint rep of $SU(2)$ can be found here - the original asker of the question seemed to lose interest in his question but maybe you can gather something from it! – Selene Routley Oct 27 '13 at 23:57
  • I have asked a question about an isomorphism relation in group theory on Math SE, and unfortunately people on there are not reacting to my question. If you have some spare time, I would be very grateful if you could have a look at it and see if you agree/disagree with my proof. – Hunter Mar 26 '14 at 14:42
  • Dear @Hunter Sorry for my lack of response. YOur proof looks basically sound although I haven't got time right at the moment to go through with a fine tooth come- clearly you are reasoning in fundamental and sound terms. Manolito Pérez's answer is the one I like - it's the most concise and really all you need to worry about. In general, though, you have to be careful of quotients and products: in this case you're beginning with a direct product $G\times H$ so it all works, but in general $G = K/H$ does not imply $K = G\times H$ but rather the ... – Selene Routley Mar 27 '14 at 21:33
  • ... semidirect product is relevant (the main difference being that the semidirect product between two groups is not unique. – Selene Routley Mar 27 '14 at 21:36
  • No problem. Do you have any resources/books where they explicitly discuss this stuff in great detail? Most books (I know) briefly discuss the direct product and the quotient group, but they never go in much detail (and they don't discuss the relation between them). This seems odd to me, because it is so important in theoretical physics. Also, thanks for your comment about the fact that $G=K/H$ does not imply $K=G \times H$. I really want to learn more about. Maybe I should ask it on physics SE, but I don't think such a question will be allowed. – Hunter Mar 27 '14 at 22:02