# Direct sum of the spinors and EM field tensor

EM field tensor refer to the direct sum of $(1, 0), (0, 1)$ spinor representation of the Lorentz group. How to show it?

Each of these spinor representations corresponds to the symmetrical spinor tensor rank 2. So first I got the relation between an arbitrary antisymmetrical tensor $M_{\mu \nu}$ and spinor tensor $h_{ab \dot {a}\dot {b}}$:

$$M_{\mu \nu} \to h_{ab \dot {a}\dot {b}} = \epsilon_{a b}h_{(\dot {a}\dot {b})} + \epsilon_{\dot {a} \dot {b}}h_{(ab)},$$ where $h_{(ab)}, h_{(\dot {a}\dot {b})}$ are symmetrical spinor tensors, but I don't understand how to identity $h_{ab \dot {a}\dot {b}}$ with $(1, 0) + (0, 1)$.

Maybe I need to build the transformation law of $h_{ab \dot {a}\dot {b}}$, then the transformation law of "bispinor" $(1, 0) + (0, 1)$, and then to see, that they are identically?

• If you agree that $M_{\mu\nu}$ is equivalent to $h_{ab|\dot{a}\dot{b}}$ then you just need to use that $M_{\mu\nu}=-M_{\nu\mu}$ and remember that $\mu$ and $\nu$ go to $a\dot{a}$ and $b\dot{b}$. You also have to enjoy a nice fact that an anti-symmetric bispinor is always proportional to $\epsilon_{ab}$ or $\epsilon_{\dot{a}\dot{b}}$. So you get the decomposition you wrote. – John Sep 23 '13 at 19:47
• @ John. I know the expressions for $h_{(ab)}, h_{(\dot {a} \dot {b})}$. But it's not obvious for me that $h_{ab \dot {a}\dot {b}}$ is the direct sum $(0, 1) + (1, 0)$. – user8817 Sep 23 '13 at 20:03
• This is what I explained in the comment that $h_{ab|\dot{a}\dot{b}}$ decomposes as you wrote as a consequence of $M_{\mu\nu}=-M_{\nu\mu}$. So it proves that it is $(1,0)+(0,1)$ since it is written as a direct sum (note that $\epsilon$ cannot mix with $h_{ab}$ under Lorentz transformations. – John Sep 23 '13 at 20:31
• @John . The direct sum of (1,0),(0,1) can be represented as the vector with components Aab,B~a˙b˙. But in this case it doesn't. Why? – user8817 Sep 24 '13 at 16:24
• Your first sentence is meaningless. $\epsilon_{ab}$ is an invariant tensor, it behaves as a singlet $(0,0)$. Take some $(1,0)$, say $V_{ab}$, then $\epsilon_{ab}V_{cd}$ is $(1,0)$ too, and $\epsilon_{uv}\epsilon_{ab}V_{cd}$ etc., simplfy because $\epsilon$ is invariant. Then, following the definition of a direct sum, you can check that the two terms on the rhs of your equation do not mix, they transform independently, so it is a direct sum. The number of $\epsilon$ factors is irrelevant. – John Sep 24 '13 at 16:31