Riemannian and Weyl tensors as spinor representation

There is the way of converting vector indices to spinor indices, for example, Maxwell stress tensor $$F_{[\mu\nu]}$$ can be decomposed to $$(1,0) \oplus (0,1)$$ irreducible representations of $$\mathfrak{su}(2)_L\times \mathfrak{su}(2)_R$$:

$$$$F_{[\mu\nu]} \sim (\sigma_{[\mu\nu]})^{\alpha\beta} F_{(\alpha\beta)} + (\sigma_{[\mu\nu]})^{\dot{\alpha}\dot{\beta}} F_{(\dot{\alpha}\dot{\beta})}$$$$

Let's consider the Riemannian tensor $$R_{[\mu\nu]\vert[\rho\sigma]}$$:

$$$$R_{[\mu\nu][\rho\sigma]} \sim (\sigma_{[\mu\nu]})^{\alpha\beta} (\sigma_{[\rho\sigma]})^{\gamma\delta} R_{(\alpha\beta)(\gamma\delta)} + (\sigma_{[\mu\nu]})^{\dot{\alpha}\dot{\beta}} (\sigma_{[\rho\sigma]})^{\dot{\gamma}\dot{\delta}} R_{(\dot{\alpha}\dot{\beta})(\dot{\gamma}\dot{\delta})}$$$$

But objects $$R_{(\alpha\beta)(\gamma\delta)}$$ and $$R_{(\dot{\alpha}\dot{\beta})(\dot{\gamma}\dot{\delta})}$$ are reducible representations of $$\mathfrak{su}(2)_L\times \mathfrak{su}(2)_R$$. So we need decompose such objects further.

How do you decompose $$R_{(\alpha\beta)(\gamma\delta)}$$? What is meaning of the obtained objects ( objects like $$W_{(\alpha\beta\gamma\delta)}$$, $$R_{(\alpha\beta)}$$, $$R$$)?

• There are two books by Moshe Carmeli on this... May 10 at 17:11
• @DanielC could you please provide concrete reference? May 10 at 17:28
• Try M.Carmeli - „Group Theory and General Relativity”, Chapter 8. May 10 at 23:16

The most useful reference is ch. 13 of GR by Wald. But in book there are some mistakes in numerical coefficients.

Expansion of Rimannian thensor:

$$$$R_{\alpha\dot{\alpha}\beta\dot{\beta}\gamma\dot{\gamma}\delta\dot{\delta}} = \Psi_{( \alpha\beta\gamma\delta)} \epsilon_{\dot{\alpha}\dot{\beta}} \epsilon_{\dot{\gamma}\dot{\delta}} + \Phi_{(\dot{\alpha}\dot{\beta})(\gamma\delta)} \epsilon_{\alpha\beta} \epsilon_{\dot{\gamma}\dot{\delta}} + \Lambda (\epsilon_{\alpha\gamma}\epsilon_{\beta\delta} + \epsilon_{\beta\gamma}\epsilon_{\alpha\delta}) \epsilon_{\dot{\alpha}\dot{\beta}} \epsilon_{\dot{\gamma}\dot{\delta}} + c.c.$$$$

$$$$16 \, R_{abcd} = - \Psi_{( \alpha\beta\gamma\delta)} (\sigma_{ab})^{\alpha\beta} (\sigma_{cd})^{\gamma\delta} + \Phi_{(\dot{\alpha}\dot{\beta})(\gamma\delta)} (\sigma_{ab})^{\alpha\beta} (\tilde{\sigma}_{cd})^{\dot{\gamma}\dot{\delta}} -4 \Lambda (\eta_{a d} \eta_{bc} - \eta_{ac} \eta_{bd} - i \epsilon_{abcd}) + c.c.$$$$

$$16 R = -4 \Lambda g^{ac}g^{bd} (\eta_{a d} \eta_{bc} - \eta_{ac} \eta_{bd}) = -4 \Lambda (4-16) = 48 \Lambda \;\;\; \Rightarrow \;\;\; \Lambda = \frac{R}{3}$$

$$$$\sigma^{ab}_{\alpha\beta} \sigma^{cd}_{\gamma\delta} R_{abcd} = - \Psi_{( \alpha\beta\gamma\delta)} + 4 \Lambda \epsilon_{(\alpha\gamma} \epsilon_{\beta)\delta} + c.c.$$$$

• You can also find a detailed derivation of this decomposition in "Spinors and Space-time", Volume-I , by Roger Penrose, Wolfgang Rindler.
– KP99
Jul 3 at 10:24
• Btw, isn't $\Lambda=R/24$ ?
– KP99
Jul 3 at 10:25
• @KP99 I edited answer. I don't know, how Penrose obtain $R/24$. Maybe he used nonstandart sigma-matrices. Jul 3 at 11:00
• OK, I will redo that calculation, but since this is just a relation between scalars $\Lambda$ and R, I don't think it should depend on choice of sigma matrix. Also I found this same relation R/24 in other literatures. Anyway, let's see
– KP99
Jul 3 at 11:55
• Please check your calculation in the second line: You shouldn't get $16R_{abcd}=.....-4\Lambda(...)$, instead it should be $R_{abcd}=...-2\Lambda(...)$. For confirmation: see the tensor decomposition of $R_{abcd}$ in 4 dimension, it will have a term containing $R/12g_{abcd}$, but your calculation shows $R/3g_{abcd}$
– KP99
Jul 5 at 9:45

Calculation for Ricci scalar R in terms of $$\Lambda$$ (notations and identities borrowed from R.Penrose (1984)):

The Infeld - van der Waerden symbols are defined as: $${g_a}^{AA'}={g_a}^{\textbf{a}}{\epsilon_{\textbf{A}}}^{A}{\epsilon_{\textbf{A}'}}^{A'}$$ and it satisfies the following identities:

(1)$$g_{ab}=\epsilon_{AB}\epsilon_{A'B'}{g_a}^{AA'}{g_b}^{BB'}$$

(2i)$${g_a}^{AA'}{g_{AA'}}^b={g_a}^b$$ and (2ii) $${g_{AA'}}^a{g_a}^{BB'}={\epsilon_A}^B{\epsilon_{A'}}^{B'}$$

(3)The "Clifford relation" :$$2{{g_{(a|}}^A}_{A'}{g_{|b)B}}^{B'}=-{\epsilon_B}^Ag_{ab}$$

And finally the transformation rule for spinor to spacetime indices given by: $${\chi_{a...c}}^{d...f}={\chi_{AA'...CC'}}^{DD'...FF'}{g_a}^{AA'}...{g_c}^{CC'}{g_{DD'}}^d...{g_{FF'}}^f$$

Now refer to the relation (4.6.20) from the reference: $$R_{ABA'B'}=6\Lambda\epsilon_{AB}\epsilon_{A'B'}-2\Phi_{ABA'B'}$$ We have Ricci tensor given by $$R_{cd}=R_{(cd)}=R_{ABA'B'}{g_{(c}}^{AA'}{g_{d)}}^{BB'}$$ From relation (1) I can write $$R_{ABA'B'}=6\Lambda g_{ab}{g_{AA'}}^a{g_{BB'}}^b-2\Phi_{ABA'B'}$$, So

$$R_{cd}=6\Lambda g_{ab}{g_{(c}}^{AA'}{g_{d)}}^{BB'}{g_{AA'}}^a{g_{BB'}}^b-2\Phi_{ABA'B'}{g_{(c}}^{AA'}{g_{d)}}^{BB'}$$

By applying (2i) we get $$R_{cd}=6\Lambda g_{ab}{g_{(c}}^a{g_{d)}}^b-2\Phi_{(cd)}=6\Lambda g_{cd}-2\Phi_{cd}$$. Now contract with $$g^{cd}$$ on both sides. Since $${\Phi_a}^a=0$$ we get $$R=24\Lambda$$

EDIT(1):Given the identity $${g^a}_{AA'}g^{bA'B}=g^{ab}{\delta_A}^B-i{{\sigma^{ab}}_A}^B$$ verify that $${\sigma^{[ab]}}_{AB}=i{g^{[a}}_{AA'}{g^{b]A'}}_B$$. Now $$R_{abcd}=R_{[ab][cd]}=R_{AA'BB'CC'DD'}{g_{[a}}^{AA'}{g_{b]}}^{BB'}{g_{[c}}^{CC'}{g_{d]}}^{DD'}$$ Expanding Riemann curvature spinor in terms of Weyl and Ricci Spinor and $$\Lambda$$ we will get the following expression: $$R_{abcd}=-\Psi_{ABCD}{\sigma_{[ab]}}^{AB}{\sigma_{[cd]}}^{CD}-\Phi_{ABC'D'}{\sigma_{[ab]}}^{AB}{\bar{\sigma}_{[cd]}}^{C'D'}+c.c.+2\Lambda (g_{ac}g_{bd}-g_{ad}g_{bc})$$ Note the anti-symmetrization in space-time indices in sigma matrix. If we focus on the $$\Lambda$$ part, we see that on contraction with $$g^{bd}$$ we get back the familiar $$6\Lambda g_{ac}$$ term as in (4.6.20)

• Where are mistakes in my answer? Jul 3 at 17:33
• I think that there are differences in defenitions 1,2,3 Jul 3 at 17:35
• @Nikita I just copied those expressions (1,2,3) from the textbook "Spinors and Space-time" Volume I. Can you re-send your calculation for R and $\Lambda$? The mistake isn't actually clear from your previous answer :( Ok , just tell what is the definition for sigma matrix you used in your calculation? I will compare those with Infeld van der Waerden symbols. That would help at least.
– KP99
Jul 3 at 17:54