1
$\begingroup$

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?

Improve this question
$\endgroup$
1
  • $\begingroup$ If the field lives in the fundamental representation, it is a column vector, not a matrix. $\endgroup$
    – AccidentalFourierTransform
    Commented Jun 17, 2018 at 17:02

1 Answer 1

Reset to default
2
$\begingroup$

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 $\psi \rightarrow U\psi$.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thank you, now it's clearer. What about the field in the adjoint representation? $\endgroup$
    – ablagi
    Commented Jun 17, 2018 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
    Commented Jun 17, 2018 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
    Commented Jun 17, 2018 at 17:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.