Representation of a group is a homomorphism from the group to the $GL(n,R)$. But I don't quite understand what is the charge of representation. Could someone explain?
Asked
Active
Viewed 232 times
1 Answers
1
I'm not sure how deep your question is. Since you are employing highly ambiguous language, I'll assume the most simple-minded language that physicists talk to each other in.
Typically, a "charge" is a Lie algebra generator, for the Lie group involved, in your case GL(n, R). For instance, if we took the SU(2) subgroup of your GL(2,R), you'd have three charges, the three $\mathfrak{su}(2)$ algebra elements/generators.
In a given representation, e.g. the doublet representation, they'd be, e.g., the three Pauli matrices, suitably normalized. Their eigenvalues on a standardized doublet eigenvector are then called the respective charges of the doublet, e.g. 1 or -1 for $T_3$.
Is this what you are asking about?
Response to comment
You’ll have to choose among these suggestions…
The states of the theory are complex vectors of a given dimension characteristic of the representation; they are operated upon by the representation matrices of the group elements outlined. Such states thus connected are equivalent by the symmetry described by the group discussed… So, if, e.g., the hamiltonian of the system is symmetric under the group, the states thus connected have the same properties conferred by the hamiltonian, such as mass, and they are dubbed "degenerate"... Angular momentum is the archetypal paradigm of such symmetries (rotation group).
Cosmas Zachos
- 67,623