# A doubt in Yang-Mills procedure

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

Thanks

• 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. Jul 14, 2018 at 14:39
• 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. Jul 14, 2018 at 16:04

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.