The $(1,0)$ representation of $\text{SL}(2,\mathbb{C})$ is realized on two indexed symmetric spinors $\psi^{ab}$ transforming like $$D^{(1,0)}(A)\psi^{ab}=\sum_{c,d=1}^2A^a_cA^b_d\psi^{cd}$$ for all $A\in\text{SL}(2,\mathbb{C})$. Notice that this is a four dimensional representation the indices range in $a,b\in\{1,2\}$. Ramond's book claims this representation is equivalent to that on tensors $B_{\mu\nu}$ which are antisymmetric $B_{\mu\nu}=-B_{\nu\mu}$ and self-dual $B_{\mu\nu}=\frac{1}{2}g_{\mu\lambda}g_{\nu\kappa}\epsilon^{\lambda\kappa\rho\sigma}B_{\rho\sigma}$. These of course transform like $$\tilde{D}^{(1,0)}(A)B_{\mu\nu}=\Lambda(A)^{\rho}_{\mu}\Lambda(A)^\sigma_\nu B_{\rho\sigma}$$ where $\Lambda:\text{SL}(2,\mathbb{C})\rightarrow L^\uparrow_+$ is the covering map onto the orthocronous proper Lorentz group.

I have two problems with this. On the one hand, I don't see how non-trivial antisymmetric tensors exist. Indeed $$B_{oi}=\frac{1}{2}g_{0\lambda}g_{i\kappa}\epsilon^{\lambda\kappa\rho\sigma}B_{\rho\sigma}=\frac{1}{2}(1)(-1)\epsilon^{0i\rho\sigma}B_{\rho\sigma}=-\frac{1}{2}\epsilon^{0ijk}B_{jk}.$$ Then, with the convention $\epsilon^{0123}=1$ we obtain $B_{03}=-B_{12}$. On the other hand $$B_{ij}=\frac{1}{2}g_{i\lambda}g_{j\kappa}\epsilon^{\lambda\kappa\rho\sigma}B_{\rho\sigma}=\frac{1}{2}(-1)(-1)\epsilon^{ij\rho\sigma}B_{\rho\sigma}=\frac{1}{2}\epsilon^{ij0k}B_{0k}+\frac{1}{2}\epsilon^{ijk0}B_{k0}=\epsilon^{ij0k}B_{0k}=\epsilon^{0ijk}B_{0k}.$$ We thus obtain $B_{12}=B_{03}$. I am surely making a very stupid mistake with my index management but I would be very thankful if anyone could point it out.

My second problem is that I don't know how to correctly map self-dual antisymmetric tensors into the spinors to show the equivalence of the representations. If one of the above equations is correct while the other one being trivial, the space of such tensors is three dimensional. This coincides with the dimension of the symmetric spinors with two indices. Can someone help me find the correct mapping? I am guessing it should be similar to the map which shows that the $(1/2,1/2)$ representation is just the vector representation.

  • $\begingroup$ Notice that the index positions on the left-hand and right-hand sides of your last equation do not match, and raising and lowering indices can yield signs. $\endgroup$ – user178876 Jan 25 '19 at 16:25
  • $\begingroup$ I believe that they do not match in the same way that indices here en.wikipedia.org/wiki/… don't. I don't think this is the problem. $\endgroup$ – Iván Mauricio Burbano Jan 25 '19 at 17:48

The definition of self-duality depends on the signature. For the Lorentz group you need to put a factor of $i$. The reason is that taking the double dual should give the identity. Indeed if we define $$ (B_{\mu\nu})^\star = \frac{i}{2}\varepsilon_{\mu\nu\rho\lambda}g^{\rho\rho'}g^{\lambda\lambda'}B_{\rho'\lambda'}\,, $$ this is a good duality because $$ \begin{aligned} (B_{\mu\nu})^{\star\star} &= -\frac{1}{4}\varepsilon_{\mu\nu\rho\lambda}\varepsilon_{\rho'\lambda'\kappa'\sigma'}g^{\rho\rho'}g^{\lambda\lambda'}g^{\kappa\kappa'}g^{\sigma\sigma'}B_{\kappa\sigma} =\\&= -\frac{1}{4}\varepsilon_{\mu\nu\rho\lambda}\varepsilon^{\rho\lambda\kappa\sigma}B_{\kappa\sigma} = \\ &= -\frac{1}{4}\det(g)\,4 B_{\mu\nu} = B_{\mu\nu}\,. \end{aligned} $$ Regardless of the properties of $B_{\mu\nu}$ (except its antisymmetry). Self-duality states $B_{\mu\nu} = (B_{\mu\nu})^\star$ with the definition I gave above.

Now as for the mapping between tensors and spinors. I don't know if you are familiar with $\alpha\dot{\alpha}$ notation from Wess & Bagger's book of supersymmetry. In any case, one can define in four dimensions the Pauli matrices $$ \sigma^\mu_{\alpha\dot{\alpha}} = (-\mathbb{1}_{2\times2},\vec{\sigma}\,)\,,\qquad \bar{\sigma}^{\mu\dot{\alpha}\alpha} = (-\mathbb{1}_{2\times2},-\vec{\sigma}\,)\,. $$ Where $\vec{\sigma} = (\sigma^1,\sigma^2,\sigma^3)$ are the ordinary Pauli matrices and $\mathbb{1}_{2\times 2}$ is the identity matrix. With these you can define two objects $$ \sigma^{\mu\nu\phantom{\alpha}\beta}_{\phantom{\mu\nu}\alpha} = \frac{1}{4}\left(\sigma^\mu_{\alpha\dot{\alpha}}\bar{\sigma}^{\nu\dot{\alpha}\beta} - \sigma^\nu_{\alpha\dot{\alpha}}\bar{\sigma}^{\mu\dot{\alpha}\beta}\right)\,, $$ and $$ \bar{\sigma}^{\mu\nu\dot{\alpha}}_{\phantom{\mu\nu\alpha}\dot{\beta}} = \frac{1}{4}\left(\bar{\sigma}^{\mu\dot{\alpha}\beta}\sigma^\nu_{\beta\dot{\beta}} - \bar{\sigma}^{\nu\dot{\alpha}\beta}\sigma^\mu_{\beta\dot{\beta}} \right)\,. $$ The indices $\mu,\nu$ are Lorentz vector indices while the $\alpha$ and $\dot{\alpha}$ are $\mathrm{SL}(2)$ indices associated respectively to the fundamental Weyl left spinor and right spinor. The tensors I defined above are important for a number of reasons: first they are actually the generators in the left and right (respectively) spinor representation, second, they happen to have (anti)-self-duality properties. Namely $$ (\sigma^{\mu\nu\phantom{\alpha}\beta}_{\phantom{\mu\nu}\alpha})^\star = \sigma^{\mu\nu\phantom{\alpha}\beta}_{\phantom{\mu\nu}\alpha} \,,\qquad (\bar{\sigma}^{\mu\nu\dot{\alpha}}_{\phantom{\mu\nu\alpha}\dot{\beta}})^\star = -\bar{\sigma}^{\mu\nu\dot{\alpha}}_{\phantom{\mu\nu\alpha}\dot{\beta}}\,. $$ So you can define $$ B^{\mu\nu} = \psi_\beta^{\phantom{\beta}\alpha}\sigma^{\mu\nu\phantom{\alpha}\beta}_{\phantom{\mu\nu}\alpha}\,. $$ If you are confused by the position of the indices, the rule is that indices $\alpha$ and $\dot{\alpha}$ are raised and lowered by the two dimensional epsilon: $\epsilon^{\alpha\beta},\,\epsilon^{\dot{\alpha}\dot{\beta}}$. So you can also write $$ B^{\mu\nu} = \psi^{\alpha\gamma}\sigma^{\mu\nu\phantom{\alpha}\beta}_{\phantom{\mu\nu}\alpha}\epsilon_{\beta\gamma}\,. $$ Moreover there is a symmetry property $\sigma^{\mu\nu\phantom{\alpha}\beta}_{\phantom{\mu\nu}\alpha}\epsilon_{\beta\gamma} = \sigma^{\mu\nu\phantom{\alpha}\beta}_{\phantom{\mu\nu}\gamma}\epsilon_{\beta\alpha}$, which is consistent with the fact that $\psi^{\alpha\gamma}$ is symmetric in its indices (as you remarked). Similar considerations apply for the conjugated representation of course.

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    $\begingroup$ Excellent answer. My word. I love it. :-) $\endgroup$ – DanielC Jan 25 '19 at 22:03
  • $\begingroup$ Thank you for the complete answer! However, I don't see how multiplying by an i solves the consistency issues in my calculation above. Can you help me with this? $\endgroup$ – Iván Mauricio Burbano Jan 26 '19 at 1:44
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    $\begingroup$ I think if I follow the same steps with an $i$ I get for the first $B_{03} = -i B_{12}$ and for the second $B_{12} = i B_{03} = i(-i)B_{12}$ which is consistent. $\endgroup$ – MannyC Jan 26 '19 at 1:52

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