What is the QCD gluon field (mathematically)?

In QCD, the gluon field is described as $$A^a_\mu$$. In the covariant derivative for the Lagrangian, it is multiplied by the Gell-Mann $$SU(3)$$ generator matrices $$\lambda_a$$ ($$a=1..8$$) as $$\lambda_aA^a_\mu$$. My question is, what is the mathematical structure of $$A_\mu$$ (for a given a)? Since $$\mu=0..3$$, it must have 4 components (I'm assuming $$\mu$$ does not mean $$A(x_\mu)$$), but the $$\lambda$$'s are $$3\times 3$$ matrices. Since $$\lambda$$ is a 3x3 matrix, and $$A_\mu$$ is a 1x4 vector, how can the two be "multiplied" together?

Is each element of $$A_\mu$$ a $$3\times 3$$ matrix? If not, then what? Or does the notation $$\lambda_aA^a_\mu$$ mean something other than normal matrix multiplication is going on in this case?

What is the mathematical structure of $$\lambda_aA^a_\mu$$ (like, a 1x4 vector, or 3x3 matrix, or a 1x4 vector of 3x3 matrices)?

What does an element of $$A_\mu$$ represent physically?

$$A$$ is a $$4$$-vector taking values in $$3 \times 3$$ matrices. More precisely, it's a dual $$4$$-vector taking values in the Lie algebras $$su(3)$$, which is an 8-dimensional subspace of the 9-dimensional space of $$3\times 3$$ matrices.

If you pick coordinates $$x^\mu$$ on spacetime, you get a basis $$dx^\mu$$ for the dual vectors and you can expand $$A = \sum_\mu A_\mu \otimes dx^\mu$$. Each component $$A_\mu$$ lives in $$su(3)$$, i.e., is a $$3\times 3$$ matrix.

The symbol $$\otimes$$ is the tensor product, which is a kind of external multiplication, not the same thing as the usual internal multiplication. In matrix terms, it means that each component of the 4-vector $$A$$ is a $$3\times 3$$ matrix. (Equivalently, you can think that $$A$$ is a $$3\times 3$$ matrix, each entry of which is a $$4$$-vector.)

If you pick a basis $$\lambda_a$$ for $$su(3)$$, you can further expand each $$A_\mu = \sum_a A^a_\mu \lambda_a$$. Now each component $$A^a_\mu$$ is just a real-valued function. Combining these two, you can write $$A = \sum_\mu \sum_a A^a_\mu (\lambda_a \otimes dx^\mu).$$

• Thank you very much! I see now that I was confusing $A_\mu$ with $A$. Commented May 31, 2020 at 12:47
• Is that also true of the SU(2) symmetry of the electroweak sector? Are the 4-vector in SU(2) taking $2 \times 2$ matrix as values? Commented Jun 29, 2020 at 16:21

For a specified $$a$$, $$A^\mu$$ is a four-vector. Four-vectors are defined to be object that "transforms like four-vectors", i.e. with Lorentz transformations $$A'^{\mu} = \Lambda^\mu_\nu A^\nu.$$

In a suitable basis, you can write it out as a column vector: $$\left ( \begin{array}{c}A_0 \\ A_1 \\ A_2 \\ A_3 \end{array} \right ).$$

Different $$a$$'s imply different values for the entries of the four-vector.

I should also say that it is a vector field meaning that there will be a different numerical value of the same $$A^\mu$$ at each point in space (and time).

The $$\lambda$$s are just the generators of the $$SU(3)$$ group, so they are just the $$3\times 3$$ Gell-Mann matrices.

• It is also a real field since the gluons do not possess an electromagnetic charge. Commented May 29, 2020 at 17:33
• SuperCiocia - I already understand everything you said, but my point is that $\lambda$ is a 3x3 matrix, and $A_\mu$ is a 1x4 vector, so how can the two be "multiplied" together? Does the notation $\lambda_aA^a_\mu$ mean that something other than standard matrix multiplication is going on in this case? P.S. - edited my original post for clarity. Commented May 30, 2020 at 12:37
• At this point the better answer that more directly answers your question is @user1504. He explains what object is it. Commented May 30, 2020 at 21:51