Meaning of “transforms like the adjoint” in the context of Yang Mills Theory and connection to Lie Algebra

Question

In Srednicki Chapter 69, we say something transforms like the adjoint if its transformation under the $SU(N)$ group action is $$W\rightarrow UWU^\dagger$$ The Field strength and the covariant derivative both transform like this. What to me seems to be going on (based on Chapter 70) is that (field strength)$F$ has an index in the conjugate of the fundamental and an index in the fundamental representation of $SU(N)$ and a those transform like $\psi\rightarrow U\psi$ and $\psi^\dagger \rightarrow \psi^\dagger U^\dagger $. Furthermore, we know that $F$ is traceless hermitian so it can be expressed as a linear combination of the generators so it is a vector sized $N^2-1$ in the adjoint representation. Srednicki also mentions $N\otimes\bar{N}=1\oplus A$ so it seems that objects that have an index in each of this reps (e.g. $F^i_j)$ are just objects that are vectors in the adjoint (i.e. $F^a(T^a)$). So my questions are:

1. Is this the reason this transformation is called “transforms like the adjoint” even though this is also an $N$ by $N$ matrix in the fundamental.
2. If we choose to represent $F$ as a matrix in the adjoint rather than a state, what is its transformation rule? Is it still $UFU^\dagger$?Its transformation rule as a vector is given in the solution to 69.1 (at least partially). Infinitesimally it transforms like $$F^a \rightarrow F^b +iF^cf^{cba}\theta^a$$ but what if I write it as $F^{bc}\rightarrow?$
3. The way something like the gauge field matrix in Yang Mills is introduced in Srednicki is an $N$ by $N$ matrix of fields that is definitionally traceless and hermitian. Then it is presented as a consequence that it is spanned by the generators of the algebra. Implicit (at least to me) is that this matrix was built with a conjugate fundamental and a fundamental index which is precisely why it is a vector in the adjoint. Is it a contradiction to define an object with these two indices and have it be not traceless or not hermitian?
4. How does the transformation connect to the Lie Algebra given that that is the vector space spanned by the traceless, hermitian generators.

Answer 1

I'm not sure which subquestions are still obscure. Using a,b,... for the $N^2-1$ adjoint indices and i,j,... for the fundamental ones, and summation over repeated indices, you have $$ F^a \qquad \leftrightarrow \qquad (F^a T^a)_{ij}, $$ so the $N\times N$ traceless hermitean matrices on the right correspond isomorphically to the ($N^2-1$) - dim "isovectors" on the left; they span the su(N) Lie algebra.

They transform under the group action as $$ R_{ab} F^b \qquad \leftrightarrow \qquad U(F^a T^a) U^\dagger ,\\ R_{ab}=2 \operatorname{Tr}(T^aUT^b U^\dagger) \qquad \leftrightarrow \qquad U=e^{-i\theta^c T^c} , $$ and expanding in θ may serve to verify all statements on the $(N^2-1)\times (N^2-1)$ generators of R, so
$$ R_{ab}=\delta_{ab}-f_{abc} \theta^c +..., \qquad U={\mathbb I} -i\theta^c T^c +... . $$ (Recall for N=2, R is real orthogonal.) These R are the adjoint representation group matrices, and the -if s are the Lie algebra matrices in said representation.

Concerning your modified 3., there is no "contradiction" in defining an object outside the Lie algebra, but it would have nonsensical transformation properties, and would not satisfy the Lie Algebra, which is used all over. The "magic" of R relies on the completeness and trace-orthogonality of the generators of the Lie Algebra.

Answer 2

"The adjoint representation" of a Lie group $G$ is in the end only a name for the Lie algebra $\mathfrak{g}$ of that group. The $W\mapsto UWU^\dagger$ transformation rule comes from the fact that when we consider $G$ as a matrix group (e.g. the $\mathrm{SU}(N)$ matrices inside of all $N$-by-$N$ matrices), then the natural action on the Lie algebra $\mathfrak{g}$ as a matrix algebra is by conjugation.

That is, the Lie algebra has a basis $T^a$, and elements of the Lie algebra $A$ can be expanded as $A = A^a T^a$. But simultaneously you could think about them as being N-by-N matrices ${A^i}_j$, and the $W\mapsto UWU^\dagger$ notation uses that interpretation: ${F^i}_j\mapsto {U^i}_k {F^k}_l {{U^\dagger}^l}_j$. Simultaneously, in the "adjoint notation" in terms of the basis of the Lie algebra, this is $F^a \mapsto \mathrm{ad}(U)^a_b F^b$, where $\mathrm{ad}(U)$ is the adjoint representation of $U$, itself a $\mathrm{dim}(\mathfrak{g})$-matrix. It's important not to confuse these two kinds of matrices of different dimensions - the N-by-N matrices $U$ acting by conjugation and the $\mathrm{dim}(\mathfrak{g})$-by-$\mathrm{dim}(\mathfrak{g})$ matrices $\mathrm{ad}(U)$ acting by normal matrix-vector multiplication are different objects.

Now, when we come to the spacetime-dependent gauge transformations, then those are no longer linear transformations on the space of gauge fields $A$ (see also this answer of mine) - the term with the derivative in it spoils the linear transformation behaviour. However, for $F$, it is restored. The transformation the question writes as $$ F^a \mapsto F^a + \mathrm{i}F^c f^{abc} \theta^b$$ in the "adjoint notation" is just $F\mapsto F + \mathrm{i}[F,\theta]$ in terms of N-by-N matrices, since the structure constants of the adjoint encode the commutator that comes from the infinitesimal form of the conjugation action.