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 $\mathfrak{su}(2)_L\times \mathfrak{su}(2)_R$:

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

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

\begin{equation} 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})} \end{equation}

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 $W_{(\alpha\beta\gamma\delta)}$, $R_{(\alpha\beta)}$, $R$)?

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

\begin{equation} 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. \end{equation}

\begin{equation} 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. \end{equation}

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

\begin{equation} \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. \end{equation}

  • $\begingroup$ You can also find a detailed derivation of this decomposition in "Spinors and Space-time", Volume-I , by Roger Penrose, Wolfgang Rindler. $\endgroup$
    – KP99
    Jul 3 at 10:24
  • $\begingroup$ Btw, isn't $\Lambda=R/24$ ? $\endgroup$
    – KP99
    Jul 3 at 10:25
  • $\begingroup$ @KP99 I edited answer. I don't know, how Penrose obtain $R/24$. Maybe he used nonstandart sigma-matrices. $\endgroup$
    – Nikita
    Jul 3 at 11:00
  • $\begingroup$ 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 $\endgroup$
    – KP99
    Jul 3 at 11:55
  • $\begingroup$ 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 $R/3g_{abcd}$ $\endgroup$
    – 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:


(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)

  • $\begingroup$ Where are mistakes in my answer? $\endgroup$
    – Nikita
    Jul 3 at 17:33
  • $\begingroup$ I think that there are differences in defenitions 1,2,3 $\endgroup$
    – Nikita
    Jul 3 at 17:35
  • $\begingroup$ @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. $\endgroup$
    – KP99
    Jul 3 at 17:54

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