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At some point of the demonstration of Yang Mills theory we assume an ansatz that $A_{\mu}=t^a A_{\mu}{}^a$ where $a=1, \ldots,n^2-1$ and the $t^a$ are the generators of the $SU(n)$ symmetry in order to have a hermitian $n \times n A_{\mu}$.

I don't understand exactly what are the $A_{\mu}{}^a$ are they a constant in that space? Are they a matrix? Like a vector field times the identity? Is the definition of $A_{\mu}$ a matrix product?

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  • $\begingroup$ The $A^a_\mu(x)$ are the gauge fields. You have 8 of them for SU(3) (QCD), as the index $a=1,\cdots,8$ dictates and that makes sense because we have 8 gluons. The object $A_\mu:=t_aA^a_\mu(x)$ is indeed a matrix. $\endgroup$ – Thomas Wening Jan 14 at 17:31
  • $\begingroup$ Or, for a hopefully more familiar example, consider the U(1) symmetry of electromagnetism. There you have A_mu, a four-vector. $\endgroup$ – puppetsock Jan 14 at 17:58
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If the structure group of the theory is $G$, with Lie algebra $\mathfrak g$, then the (local) Yang-Mills fields, are locally defined differential $1$-forms that take value in the Lie algebra $\mathfrak g$.

If the local gauge neighborhood is $U$, and we suppose that it is also a coordinate neighborhood, then we may write $$ A(x)=A_\mu^a(x)\mathrm dx^\mu\otimes t_a. $$

If spacetime is $n$ dimensional, and $G$ is $N$ dimensional, then the $A^a_\mu(x)$ are $n\times N$ local functions on the gauge domain $U$ that constitute the component functions of the local Yang-Mills field $A$.

For each value of the indices $\mu$ and $a$, $A^a_\mu$ is an ordinary real-valued function. The expression $A_\mu=A^a_\mu t_a$ (for a fixed value of $\mu$) is a $\mathfrak g$-valued local function. Usually in physics, $G$ is a matrix group, and $\mathfrak g$ is a matrix Lie algebra, in this case the generators $t_a$ are matrices, so $A_\mu^a t_a$ is (also) a matrix-valued function.

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