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  $\mathcal N=2$ Weyl multiplet and chiral superfields

+ 5 like - 0 dislike

It is more or less known that a given antisymmetric tensor $F$ in two indices can be written in terms of spinorial indices, splitting into self-dual and anti-self-dual parts
$$ F_{\mu\nu} =
\varepsilon_{\dot\alpha\dot\beta} +
\varepsilon_{\alpha\beta} .$$

My question is the following: start from equations 5.3 and 5.4 of [1]
$$ W_{\mu\nu}^{ij} = T_{\mu\nu}^{ij} -
R_{\mu\nu\lambda\rho}\theta^i\sigma_{\lambda\rho}\theta^j + \cdots $$
which is self-dual in Lorentz indices $\mu\nu$ and antisymmetric in $\mathrm{SU}(2)$ indices $ij$.
Then it is squared to
$$W^2= \varepsilon_{ij}\varepsilon_{kl} W_{\mu\nu}^{ij}W_{\mu\nu}^{kl}.$$
How do I see (possibly using the first equation above) that this is the same as starting from a tensor (see formula 4.1 and below of [2]) $W_{\alpha\beta}$ with self-dual $T$ as top component, where $\alpha$, $\beta$ denote symmetric spinor indices and
\varepsilon^{\alpha\alpha'}\varepsilon^{\beta\beta'} ?$$

  [1]: http://arxiv.org/abs/hep-th/9307158
  [2]: http://arxiv.org/abs/hep-th/9912123

asked May 19, 2014 in Theoretical Physics by jj_p (150 points) [ revision history ]

1 Answer

+ 3 like - 0 dislike

I guess that in $W_{\alpha \beta}$ the indices $i$ and $j$ are here but not written and are contracted in $W^2$ in the same way that in the first $W^{2}$. So I can forget these indices in what follows.

I will show that the two $W^{2}$ are the same if we have the relation  

$W_{\mu\nu} =\frac{1}{2}W_{\alpha\beta}(\sigma_\mu)^{\alpha\dot\alpha}(\sigma_\nu)^{\beta\dot\beta}\varepsilon_{\dot\alpha\dot\beta}$

which is the same thing suggested by the first equation of the question up to the factor 1/2. I don't know if this factor is relevant. Given this relation, we have

$W_{\mu\nu}W^{\mu\nu} = \frac{1}{4}W_{\alpha\beta}W_{\alpha'\beta'}(\sigma_\mu)^{\alpha\dot\alpha}(\sigma^\mu)^{\alpha'\dot\alpha'} (\sigma_\nu)^{\beta\dot\beta} (\sigma^\nu)^{\beta'\dot\beta'} \varepsilon_{\dot\alpha\dot\beta}\varepsilon_{\dot\alpha'\dot\beta'}$.

Now the key point is the identity (*)  $(\sigma_\mu)^{\alpha\dot\alpha}(\sigma^\mu)^{\alpha'\dot\alpha'} = 2 \varepsilon^{\alpha \alpha'}\varepsilon_{\dot\alpha \dot\alpha'}$. Using (*) and the analoguous relation with the $\alpha$'s replaced by the $\beta$'s, and simplifying the $\epsilon$'s (we have $\varepsilon^{\dot\alpha\dot\alpha'}\varepsilon_{\dot\alpha\dot\beta}\varepsilon^{\dot\beta\dot\beta'}  \varepsilon_{\dot\alpha'\dot\beta'}=1$), we obtain

$W_{\mu\nu}W^{\mu\nu}= W_{\alpha\beta}W_{\alpha'\beta'} \varepsilon^{\alpha\alpha'}\varepsilon^{\beta\beta'}$.

Proof of (*): the identity (*) is a part of a large class of identities generally referred to as Fierz identities. These identities are all consequences of the fact that the four matrices $\sigma^\mu$ are a basis of the four dimensional vector space of two by two complex matrices. See for example this question:


Using $\sigma^0 = 1$, $Tr(\sigma^i)=0$, and $Tr(\sigma^i \sigma^j)=2 \delta^{ij}$ for $i,j=1,2,3$, we find that any two by two complex matrix $M$ can be written $M = \frac{1}{2} Tr(M) 1 + \frac{1}{2}Tr(M \sigma^i) \sigma^i$. Applying this  to $M$ being the matrix $M_{\alpha \dot\alpha}(\beta \dot\beta)=\delta_{\alpha \beta} \delta_{\dot \alpha \dot\beta}$ (the matrices $M(\beta \dot\beta)$ form a natural basis of the space of two by two complex matrice) gives

$\delta_{\alpha \beta} \delta_{\dot\alpha \dot\beta}= \frac{1}{2} \delta_{\beta \dot\beta} \delta_{\alpha \dot\alpha} + \frac{1}{2}(\sigma^i)_{\dot\beta \beta}(\sigma^i)_{\alpha \dot\alpha}$. Therefore,

$-(\sigma^i)_{\dot\beta \beta}(\sigma^i)_{\alpha \dot\alpha}=\delta_{\beta \dot\beta} \delta_{\alpha \dot\alpha}  -2\delta_{\alpha \beta} \delta_{\dot\alpha \dot\beta} $.

In signature (+---), we have $(\sigma^\mu)_{\alpha \dot\alpha} (\sigma^\mu)_{\dot\beta \beta}=\delta_{\beta \dot\beta} \delta_{\alpha \dot\alpha} -(\sigma^i)_{\beta \dot\beta}(\sigma^i)_{\alpha \dot\alpha}$. Combining with the previous equality gives

$(\sigma^\mu)_{\alpha \dot\alpha} (\sigma^\mu)_{ \dot\beta \beta}= 2 (\delta_{\beta \dot\beta} \delta_{\alpha \dot\alpha}-\delta_{\alpha \beta} \delta_{\dot\alpha \dot\beta})=2 \varepsilon_{\alpha \dot\beta}\varepsilon_{\dot\alpha \beta}$, hence (*).

answered May 20, 2014 by 40227 (5,140 points) [ revision history ]
edited May 20, 2014 by 40227

thanks for the answer; I'm not sure I can either upvote or choose it as the preferred answer, basing on my current reputation, but that's what I was asking for

Hi @jj_p welcome to PhysicsOverflow :-)

for voting on answers you need 50 rep, and the SE feature of accepting answers is disabled as many people considered it to be not too useful  for high-level physics discussions.

BTW if there are SE questions (your own would be highly welcome here too) you would like to see on PhysicsOverflow, you can suggest them to be imported here

Thanks for the welcome and info @Dilaton

I'll check whether there's something worth being imported among my questions :)

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