# 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... Commented May 10, 2021 at 17:11
• @DanielC could you please provide concrete reference? Commented May 10, 2021 at 17:28
• Try M.Carmeli - „Group Theory and General Relativity”, Chapter 8. Commented May 10, 2021 at 23:16

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? Commented Jul 3, 2021 at 17:33
• I think that there are differences in defenitions 1,2,3 Commented Jul 3, 2021 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. Commented Jul 3, 2021 at 17:54

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. Commented Jul 3, 2021 at 10:24
• Btw, isn't $\Lambda=R/24$ ? Commented Jul 3, 2021 at 10:25
• For $\Lambda$ part note that you can use the identity (I am using e for $\epsilon$): $e_{A'B'}e_{C'D'}+e_{A'D'}e_{B'C'}-e_{A'C'}e_{B'D'}=0$ to convert $\Lambda(e_{AC}e_{BD}+e_{AD}e_{BC})e_{A'B'}e_{C'D'}+cc$ to $2\Lambda(e_{AC}e_{BD}e_{A'C'}e_{B'D'}-e_{AD}e_{BC}e_{A'D'}e_{B'C'})$. Now you have to contract with the g matrices $g^{AA'}_{[a}g^{BB'}_{b]}g^{CC'}_{[c}g^{DD'}_{d]}$, ok? Commented Jul 9, 2021 at 12:38
• You should get $2\Lambda(g_{[a}^{AA'}g_{b]DD'}g_{AA'[c}g_{d]}^{DD'}-g_{D[a}^{A'}g_{b]}^{B'}_Cg_{B'[b}^Cg_{d]}^D_{A'})$ .... You can now expand the anti-symmetrisation [...] in a,b,c,d indices in these g matrices and then make use of identities (1),(2i),(2ii) mentioned in my answer to get a simpler expression in metric only: $2\Lambda(g_{ac}g_{bd}-g_{ad}g_{bc})$.. this part you have to try it yourself Commented Jul 9, 2021 at 12:46
• For some reason I can't edit the above long equation Commented Jul 9, 2021 at 12:53