My question is this:

I saw the next relation in a Yang Mills theory paper:

$$L_{i}P^{\mu \nu}_{i}=P^{\mu \nu}$$

With $L_{i}$ a generator of su(2) and for any $P^{\mu \nu}_{i}$.

But I can't understand then what is the function of these generators. Do they multiplied by another matrix give a new generator?

This is because these relations in Yang Mills procesure are confusing for me:

$$F^{\mu \nu}_{i}= \partial^{\mu} A^{\nu} _{i}- \partial^{\nu} A^{\mu}_{i} - g\epsilon_{ijk}A_{j}^{\mu} A_{k}^{\nu}$$

$$F^{\mu \nu}_{i}L_{i}= (\partial^{\mu} A^{\nu} _{i}- \partial^{\nu} A^{\mu}_{i} - g\epsilon_{ijk}A_{j}^{\mu} A_{k}^{\nu})L_{i}$$

$$F^{\mu \nu}= \partial^{\mu} A^{\nu}- \partial^{\nu} A^{\mu}+igA_{j}^{\mu} A_{k}^{\nu}[L_{j},L_{k}]$$

$$F^{\mu \nu}= \partial^{\mu} A^{\nu}- \partial^{\nu} A^{\mu}+ig[A^{\mu}, A^{\nu}]$$

So, does a generator transform a matrix to an another generator? A group element? Or what is the meaning of this equality $L_{i}P^{\mu \nu}_{i}=P^{\mu \nu}$.


  • $\begingroup$ Li are matrices, while Pi are scalar functions. In more familiar terminology (space vectors), Li are the aces unit vectors (i,j,k) and Pi are the coordinates (x,y,z). The P (without the "i" index) is the total vector, like r (with arrow above) is r=x * i + y * j + z * k. $\endgroup$
    – Panos C.
    Jul 14, 2018 at 14:39
  • $\begingroup$ Once you appreciated the basic spinor map from vectors to matrices you can move to more elaborate entities such as Y-M. But it is pointless to do so until you are completely comfortable with Pauli vectors. $\endgroup$ Jul 14, 2018 at 16:04

1 Answer 1


It is quite confusing. The $L_i$ are fixed matrices that do not depend on spacetime, and the $P_i^{\mu\nu}$ is a tensor field that does depend on spacetime but is not a matrix (for fixed $\mu$ and $\nu$). If we were to explicitly include the matrix indices $a$ and $b$ and the spacetime dependence, we would have $$P^{\mu\nu}_{ab}(x) = \sum_{i=1}^d P_i^{\mu \nu}(x)\, L^i_{ab}, \qquad a,b \in \{1, \dots, N\},$$ where $d$ is the dimensionality of the gauge group (and its Lie algebra) and $N$ is the dimension of the particular gauge group representation of the matter fields on which this operator acts. As you can see, there's quite a proliferation of indices, which is why the matrix indices are usually suppressed.

The point is that the set of linear operators over an $N$-dimensional vector space itself forms an $N^2$-dimensional vector space, because you can take linear combinations of linear operators to get new ones. In particular, you can fix a set of basis operators/matrices and then write an arbitrary operator as a linear combination of these basis operators. (In our case, we're not considering the set of all linear operators, but only a subspace, which is the relevant representation of the relevant Lie algebra of the gauge group.) Think of the $L_i$ as fixed basis elements, and the $P_i^{\mu \nu}(x)$ (with $\mu$ and $\nu$ held fixed) as the vector of coefficients of the linear combination, which varies continuously over spacetime.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.