# the index in representation theory

I'm studying quantum field theory. In my text book, generators of compact groups are normalized by $\rm{Tr}$$(T^aT^b)=\frac{1}{2}\delta^{ab}. However, the index T(R) is defined by \mathrm{Tr}$$(T^a_RT^b_R)=T(R)\delta^{ab}$, where the $R$ is the label of representations and the $T^a_R$ is a representation matrix. This definition seems inconsistent with normalization of generators unless $T(R)=\frac{1}{2}$ in any representations. Whats happened?

-
I am not sure, but it seems to me that the normalization takes place inside either the matrix group itself or in the adjoint rep (the latter being a more generic choice). The images of the generators in the representation space are not constrained by that normalization. In short, $T^a$ and $T_R^a$ are different matrices. The former being a special case of the latter (some "natural" representation is set aside for the purposes of normalization). IIRC a similar thing is done with Casimir elements. – Jyrki Lahtonen Jan 25 '13 at 8:42
I don't understand the notations, but I can see no inconsistency. Generators are normalised one way, and an index is defined another way, where's the conflict? – Marc van Leeuwen Jan 25 '13 at 8:45
@Jyrki Lahtonen $T^a$ is not a matrix but a operator.Can I think the factor of $\frac{1}{2}$ is the normalization of operators and $T(R)$ is the normalization of matrices, so the two normalization factors is chosen independently since representations, which are maps from operators to matrices, just have to keep the algebras of generators? – user59608 Jan 25 '13 at 9:37
@Marc van Leeuwen I thought $\mathrm{Tr}$$(T^aT^b)=\frac{1}{2}\delta^{ab} must be true in any representations. Because I have not chosen a representation yet. Can I normalize again or independently after fixing a representation? – user59608 Jan 25 '13 at 9:43 How do you define the trace of an operator without reference to the space the operator is acting on, i.e. without turning it into a matrix? Anyway, the representation must respect the relations between the normalized generators, and this forces their traces to change according to the space they act on. For a simple example, let the the operator x\frac{\partial}{\partial x} act on the span of \{1,x,x^2,\ldots,x^n\} and see its trace change as n grows. – Jyrki Lahtonen Jan 25 '13 at 9:48 ## 1 Answer I don't see why the index should be the same in all representations. The equations you quote merely imply that T(D)=\frac12, where D is the defining representation to which the generators T^a belong. - You mean the equation \rm{Tr}$$(T^aT^a)=\frac{1}{2}\delta^{ab}$ maybe only valid in the defining representation in this text? But under eq.(70.10), it is described the SO(N) is not the case. – user59608 Jan 25 '13 at 11:57
@user59608: No, that's not what I mean. First, there's a typo there; it should be $\operatorname{Tr}(T^aT^b)$. Further, that sentence makes no sense, since the $T^a$ are by definition the matrices in the defining representation, so it's unclear what it would mean for this equation to be valid in another representation. – joriki Jan 25 '13 at 12:22
@user59608: About $SO(N)$: This is a new point you raise that wasn't in the question. My interpretation of the text would be that $(69.8)$ is introduced explicitly for $SU(N)$ (see the bottom of page $407$); $(70.6)$ then appears to be used more generally, but without actually making use of the factor (since it's only used to show that something is zero). So I'd count this as an oversight and regard $(69.8)$ as defining only the standard normalization for $SU(N)$, and not the standard normalization for $SO(N)$ that the text underneath $(70.10)$ refers to. – joriki Jan 25 '13 at 12:24
$T^a$ implies the defining representation? Then (70.6) can be replaced by $\mathrm{Tr}$$(T^aT^b)\propto\delta^{ab}$? – user59608 Jan 25 '13 at 14:00
@user59608: Sorry, it seems we're talking past each other; I can never really make any sense of your questions. I'm not sure what you mean by "implies" here. The $T^a$ are introduced even before the concept of representations and thus cannot refer to anything other than the defining representation. This is reemphasized on p. $412$ underneath equation $(70.1)$. So the use of the notation $T^a$ implies a reference to the defining representation; perhaps this is what you meant? – joriki Jan 25 '13 at 14:52