Levi-Civita symbol in 2-spinor notation I'm reading An Introduction to Twistor Theory, by Huggett and Tod, and I don’t get the result we're being given page 17: the 2-spinor form of the 4 dimensional Levi-Civita symbol.
\begin{equation}
\epsilon_{abcd} = i (\epsilon_{AC}\epsilon_{BD}\epsilon_{A'D'}\epsilon_{B'C'}-\epsilon_{AD}\epsilon_{BC}\epsilon_{A'C'}\epsilon_{B'D'}) 
\end{equation}
To get to this expression, we are supposed to write $\epsilon_{ABCDA'B'C'D'}$ as a sum of symmetric and antisymmetric spinors. The explanation is to just apply repeatedly the procedure shown on the previous page on the spinor equivalent of a valence 2 tensor. But I don't understand how to write this sum. I tried two different things :

*

*(anti)Symmetrising on only two indices at a time :

$\epsilon_{ABCDA'B'C'D'} = \epsilon_{(AB)CDA'B'C'D'} + \epsilon_{[AB]CDA'B'C'D'} + \epsilon_{AB(CD)A'B'C'D'} + \epsilon_{AB[CD]A'B'C'D'} + \epsilon_{A(BC)DA'B'C'D'}  +\epsilon_{A[BC]DA'B'C'D'} + \epsilon_{(DA)BCA'B'C'D'} + \epsilon_{[DA]BCA'B'C'D'}$
And then following on the rest. I already see two problems : there's going to be a term in $\epsilon_{AB}$, which there isn't in the final expression. Secondly, I should probably add a factor looking like $\frac{1}{n}$ (no idea what $n$ should be equal to though)


*(anti)Symmetrising over every (un)primed at indices at once:

$\epsilon_{ABCDA'B'C'D'} = \epsilon_{(ABCD)A'B'C'D'} + \epsilon_{[ABCDA]'B'C'D'} $
This isn't even correct, I detailed everything and didn't find $\epsilon_{ABCDA'B'C'D'}$ again.
So here are my questions: a) what is exactly meant by applying the procedure repeatedly?
b) Where does the $i$ come from ?
 A: I'll be using the two component spinor conventions of this book, since I'm not familiar with your reference. There are two methods I can think of that should work, then you can adapt them to your notation.
We want to simplify the following expression as much as possible
$$\epsilon_{\alpha\dot{\alpha}\beta\dot{\beta}\gamma\dot{\gamma}\delta\dot{\delta}}:= (\sigma^a)_{\alpha\dot{\alpha}}(\sigma^b)_{\beta\dot{\beta}}(\sigma^c)_{\gamma\dot{\gamma}}(\sigma^d)_{\delta\dot{\delta}}\epsilon_{abcd} \tag{1}$$
Method one
As a first step rewrite it as follows
$$ \epsilon_{\alpha\dot{\alpha}\beta\dot{\beta}\gamma\dot{\gamma}\delta\dot{\delta}}=-\text{i}(\sigma^a)_{\alpha\dot{\alpha}}(\sigma^b)_{\beta\dot{\beta}}(\sigma^c)_{\gamma\dot{\gamma}}\big(\sigma_a\tilde{\sigma}_b\sigma_c-\eta_{ac}\sigma_{b}+\eta_{bc}\sigma_{a}+\eta_{ab}\sigma_{c}\big)_{\delta\dot{\delta}} \tag{2}$$
where I have made use of the identity
$$\sigma_a\tilde{\sigma}_b\sigma_c=\eta_{ac}\sigma_{b}-\eta_{bc}\sigma_{a}-\eta_{ab}\sigma_{c}+\text{i}\epsilon_{abcd}\sigma^d~.$$
Then just use the identity
$$(\sigma^a)_{\alpha\dot{\alpha}}(\sigma_a)_{\beta\dot{\beta}}=-2\varepsilon_{\alpha\beta}\varepsilon_{\dot{\alpha}\dot{\beta}}$$
repetitively to simplify the right hand side of (2).
Method two
First split the sigma matrices in (1) into two sets of pairs and try to convert each into double sigma matrices $$\sigma^{ab}:=-\frac{1}{4}\big(\sigma^a\tilde{\sigma}^b-\sigma^b\tilde{\sigma}^c\big)~,\qquad \tilde{\sigma}^{ab}:=-\frac{1}{4}\big(\tilde{\sigma}^a\sigma^b-\tilde{\sigma}^b\sigma^c\big)~.$$
I'll do it explicitly for one pair (I wont write the other two sigma's for ease of notation):
\begin{align}
(\sigma^a)_{\alpha\dot{\alpha}}(\sigma^b)_{\beta\dot{\beta}}\epsilon_{abcd}&=\big((\sigma^a)_{(\alpha|\dot{\alpha}}(\sigma^b)_{|\beta)\dot{\beta}}+(\sigma^a)_{[\alpha|\dot{\alpha}}(\sigma^b)_{|\beta]\dot{\beta}}\big)\epsilon_{abcd} \tag{3}
\end{align}
Term one: Due to the antisymmetry in $(a,b)$ and the symmetry in $(\alpha,\beta)$, the two indices $(\dot{\alpha},\dot{\beta})$ are antisymmetric. Any rank-2 spinor $F_{\alpha\beta}$ which is antisymmetric $F_{\alpha\beta}=-F_{\beta\alpha}$, is proportional to the spinor metric with the half the trace as proportionality constant: $F_{\alpha\beta}=\frac{1}{2}\varepsilon_{\alpha\beta}F^{\delta}{}_{\delta}$. Implementing  this observation on (3) we get
\begin{align}
(\sigma^a)_{\alpha\dot{\alpha}}(\sigma^b)_{\beta\dot{\beta}}\epsilon_{abcd}
&=\frac{1}{2}\Big(\varepsilon_{\dot{\alpha}\dot{\beta}}(\sigma^a)_{(\alpha}{}^{\dot{\delta}}(\sigma^b)_{\beta)\dot{\delta}}+\varepsilon_{\alpha\beta}(\sigma^a)^{\delta}{}_{(\dot{\alpha}|}(\sigma^b)_{\delta|\dot{\beta})}\Big)\epsilon_{abcd} \\
&=\Big(\varepsilon_{\dot{\alpha}\dot{\beta}}\big(\sigma^{ab}\big)_{\alpha\beta}-\varepsilon_{\alpha\beta}\big(\tilde{\sigma}^{ab}\big)_{\dot{\alpha}\dot{\beta}}\Big)\epsilon_{abcd}
\end{align}
Then do the same thing for the other two sigma matrices in (1). Next we use the (anti-)self duality properties of the double sigma matrices:
$$ 
\frac{1}{2}\big(\sigma^{ab}\big)\epsilon_{abcd}=-\text{i}\sigma_{cd}~, \qquad \frac{1}{2}\big(\tilde{\sigma}^{ab}\big)\epsilon_{abcd}=\text{i}\tilde{\sigma}_{cd}~. 
$$
To finish it off, you will need the following identities
$$
\big(\sigma^{ab}\big)_{\alpha\beta}\big(\sigma_{ab}\big)_{\gamma\delta}=-2\varepsilon_{\alpha(\gamma}\varepsilon_{\delta)\beta}~,\qquad \big(\sigma^{ab}\big)_{\alpha\beta}\big(\tilde{\sigma}_{ab}\big)_{\dot{\alpha}\dot{\beta}}=0~,\qquad \big(\tilde{\sigma}^{ab}\big)_{\dot{\alpha}\dot{\beta}}\big(\tilde{\sigma}_{ab}\big)_{\dot{\gamma}\dot{\delta}}=2\varepsilon_{\dot{\alpha}(\dot{\gamma}}\varepsilon_{\dot{\delta})\dot{\beta}}~.
$$
