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I have a question regarding the following passage from Srednicki's QFT book (p. 415) (https://web.physics.ucsb.edu/~mark/qft.html).

Notations are $R$ = some representation of a lie group, $\bar{R}$ is the complex conjugate representation. I assume $R$ is a real representation so that the generators $T^a$ are hermitian.

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I'm not clear what the argument he is making. Why does an "invariant symbol" imply that there is a 1 dimensional subspace of the tensor product of the 2 representation that does not transform (transforms as a singlet).

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  • $\begingroup$ One tensor contracted with another transforms as a singlet. The invariant symbol is what lets you contract. $\endgroup$ Nov 25 at 20:53
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The author finds out how that symbol transforms under group action: it stays the same! (~is invariant). Specifically, since he only works things out to first order in θ, he finds vanishing O(θ) transform pieces to be added to it.

This is a rare occurrence: other symbols have all sort of pieces of O(θ) added to them upon transformation. These pieces, if these symbols are in a representation r of the group, are $i\theta^a T_r^a$ acting on those symbols.

But for this invariant symbol, he finds $T_r^a=0 $, the trivial map, for all a. The symbol just goes to itself, as the group action on it is just the identity operator. So the representation space spanned is one-dimensional: a point. This is called a singlet (trivial) representation, the smallest possible one.

Not worth reading that much into it.

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