Let us consider a gauge group, e.g. $SU(N)$. One usually says that a fermionic field $\psi$ belongs to the fundamental representation of the group.

As far as I understand, the fundamental representation is made of matrices that belong to $SU(N)$. Then why the field, being a matrix, transforms as $$\psi \mapsto U\psi$$ and not as $$\psi \mapsto U\psi U^\dagger?$$

The adjoint representation instead should be made of matrices belonging to the Lie algebra $\mathfrak{su}(N)$. What is the physical meaning of a field in the adjoint representation and how does it transform?

  • $\begingroup$ If the field lives in the fundamental representation, it is a column vector, not a matrix. $\endgroup$ – AccidentalFourierTransform Jun 17 '18 at 17:02

A representation of a group can refer to both the group homomorphism, i.e. M: $SU(N) \rightarrow GL(N)$ and the vector space on which the representation acts. In this case a field transforming in the fundamental representation means that it lives in the vector space on which the fundamental representation of SU(N) acts, hence it is a column vector which transforms as $\psi \rightarrow U\psi$.

  • $\begingroup$ Thank you, now it's clearer. What about the field in the adjoint representation? $\endgroup$ – ablagi Jun 17 '18 at 17:21
  • $\begingroup$ The adjoint is a little more tricky because then the representation and the vector space on which the representation acts are the same. $\endgroup$ – CStarAlgebra Jun 17 '18 at 17:24
  • $\begingroup$ Ah ok! So it is the vector space of the $N^2-1 \times N^2-1$ matrices that acts on itself. How is it useful in this context? $\endgroup$ – ablagi Jun 17 '18 at 17:29

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