# Invariant symbol, group representation

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.

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).

• One tensor contracted with another transforms as a singlet. The invariant symbol is what lets you contract. Nov 25 at 20:53

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.