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?


2 Answers 2


$A$ 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 $$ A = \sum_\mu \sum_a A^a_\mu (\lambda_a \otimes dx^\mu). $$

  • $\begingroup$ Thank you very much! I see now that I was confusing $A_\mu$ with $A$. $\endgroup$
    – ZenFox42
    Commented May 31, 2020 at 12:47
  • $\begingroup$ 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? $\endgroup$
    – Anon21
    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.

  • $\begingroup$ It is also a real field since the gluons do not possess an electromagnetic charge. $\endgroup$
    – proton
    Commented May 29, 2020 at 17:33
  • $\begingroup$ 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. $\endgroup$
    – ZenFox42
    Commented May 30, 2020 at 12:37
  • $\begingroup$ At this point the better answer that more directly answers your question is @user1504. He explains what object is it. $\endgroup$
    – SuperCiocia
    Commented May 30, 2020 at 21:51

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