# A simple question about matrix product with spinor indices

I have a big problem with dotted and undotted spinor indices.

For example, suppose we have two convolutions: $$\sigma^{\dot {a} a}F_{ab}, \quad \sigma^{\dot {a} a}F_{\dot {a} \dot {b}}, \quad F_{ab} = F_{ba}, \quad F_{\dot {a} \dot {b}} = F_{\dot {b} \dot {a}}.$$ Is it right that without index notation they are reduced to $$\sigma^{\dot {a} a}F_{ab} = \hat {\sigma}\hat {F}, \quad \sigma^{\dot {a} a}F_{\dot {a} \dot {b}} = \hat {F}\hat {\sigma}?$$ Is it right that I can change the indices' order, $F_{\dot {b} \dot {a}} = F_{\dot {a} \dot {b}}$, so I will get $F_{\dot {b}\dot {a}}\sigma^{\dot {a} a} = \hat {F}\hat {\sigma}$?

What if $F_{ \dot {a} \dot {b}} \neq F_{\dot {b} \dot {a}}$? Should I write something like $\hat {F}^{T}\hat {\sigma}$?

Let the undotted index $a$ correspond to the irrep $(\frac12,0)$ of the Lorentz group. Then, $F_{ab}$ corresponds to the tensor product $(\frac12,0)\otimes (\frac12,0)$ which decomposes into the two irreps $(1,0)$ and $(0,0)$ of the Lorentz group. $(1,0)$ is given by the symmetric part of $F_{ab}$ while the antisymmetric part gives the scalar $\epsilon^{ab}F_{ab}$. So if one is working with irreps of the Lorentz group, then the symmetry of $F_{ab}$ is known at the start. If you start out without any symmetry, it makes sense to break $F_{ab}$ into the two irreps before working with it.