Are there two different adjoint representations used in the SM?

Question

We know from Lie representation theory that the Lie algebra is a vector space. Therefore a representation of the Lie group can be transformation of this vector space itself which we call the Adjoint Representation. An element of this vector space, is itself represented by a matrix. For example, in the case of $SU(3)$, to show the adjoint representation, we write $V\psi(x)V^{-1}$, where $V$ is a $3*3$ special unitary matrix, $\psi(x)$ is also a $3*3$ (traceless) matrix, i.e. an element in the Lie algebra vector space. So we are multiplying three $3*3$ matrices. A new vector, or $3*3$ matrix, is produced, which belongs to or is an element in the Lie algebra vector space.

But in a SM textbook I see another definition/use of the Adjoint Representation. This time we consider the (abstract) commutation relations among the generators of the $SU(3)$ and interpret them as a vector acts on the another to produce a new vector. In other words, use the structure constants as elements to build a $8*8$ matrix to represent the Lie algebra vectors: $L_{1},...L_{8}$. Now if we consider an octet, or an 8 component column object, $\psi(x)=(\psi_{1}, ..., \psi_{8})^T$ each of $\psi_{i}$ a fermion or Dirac spinor, with four complex components, i.e. spinor indices suppressed, we can write for example $L_{1}\psi(x)$ to make or show a transformation of the fermion fields.

Now my question is that in the example of $SU(3)$ are these two different Adjoint Representations related somehow, as one of them involves the operation of $3*3$ matrices and the another one involves the operation of $8*8$ matrices?

My first impression is that in the first case, the 3 dimensional one, we are using the adjoint representation of the group using the Lie algebra as a vector space whereas in the second case, the 8 dimensional one, we are using the adjoint representation of the Lie algebra using the Lie algebra itself as a vector space. And both of them are useful in writing the Lagrangian density of the SM. Is this correct?

Answer 1

It's actually self-explanatory, but you have made a hash of terminology and illustrations to concoct a mystery. You are talking about the same object, really.

Your unitary V s are group elements, exponentials of elements L in the Lie algebra (in which you also take ψ to be: in your setup, it is traceless hermitian). The dimensionality of the matrices you represent these two sets with is 3×3, for simplicity, but it need not be! The combinatorics is identical for any dimensionality. The vector space of the L s is 8-dimensional, closed under commutation, and, again, these may be n×n, not necessarily 3×3 matrices.

Thus, your conjugation map $V\psi V^{-1}$ amounts to $$e^L \psi e^{-L}= \operatorname{Ad}_{L} \psi= \exp( \operatorname{ad }_L ) ~~\psi\\ =\psi +[L,\psi]+ [L,[L,\psi]]~/2!+[L, [L,[L,\psi]]]~/3!+... $$ in the Lie algebra, for all L in the Lie algebra.

This is a linear map of all matrices ψ, which you may also parameterize as 8-vectors, to transformed ψ s. At the group level, the map is conjugation, but at the algebra level it is commutation by the L s. In any case, a linear map of 8-vectors can be summarized as an 8×8 matrix, and all 8 of these matrices are built up of the structure constants, and satisfy the algebra (by the Jacobi identity), so they are also a representation: the adjoint representation.

Before you tie yourself up in knots, recall the completely computable case of SU(2), su(2), where you may compute all exponentials of Pauli matrices explicitly, and you may see how Pauli 3-vectors of 2×2 matrices may be characterized in these two alternate manners.