I am wondering if anybody can help me with my question. My main relation is from gravity and is known as self-dual Riemann. When it is written in the Spinorial form it is like: $$G_{\mu \nu \alpha\beta}= S_{\mu \nu}^{A B} S_{\alpha \beta}^{C D} \phi_{A B C D}$$. We also know that S can be read from Pauli matrices, for example: $$ S_{01}=-i\sigma_{y}\sigma_{x}= \left[\begin{array}{1}-1&0\\0&1\end{array}\right]$$, so its elements could be read as follow: $S_{01}^{00}=-1,S_{01}^{01}=0,... $. I am familiar with xAct to some extent, but I do not have any idea how I can introduce all of the elements of G (which has both spinorial and space time indices)for Mathematica neatly. Do you have any idea? Using the Spinors package can not be helpful because S is a special matrix and is not soldering and I must have its form not just compute it abstractly. The exact relation between $S$ with $\sigma$ is: $$\{S_{01},S_{02},S_{03}\}=-i\sigma_{y}\{\sigma_{x},\sigma_{y},\sigma_{z}\}$$ $$i\{S_{01},S_{02},S_{03}\}=\{S_{23},S_{31},S_{12}\}$$

  • $\begingroup$ Hi, just a clarification: when you write I do not have any idea how I can introduce all of these elements for Mathematica neatly which ones do you mean exactly? $\endgroup$
    – bmf
    Mar 18, 2022 at 19:06
  • $\begingroup$ Hi, I have edited my question. I want to compute G, so I have to enter S and \phi which have both spinorial and space time indices. $\endgroup$
    – Ali
    Mar 19, 2022 at 8:51
  • $\begingroup$ What is the exact relation between the subscripts of $ S $ and those of $ \sigma $s? $\endgroup$ Mar 19, 2022 at 8:59
  • $\begingroup$ BTW, the OP's defining expression for $ G $ might contain typos, because the indices are different on the two sides of the equation. So please make checks for the whole post. $\endgroup$ Mar 19, 2022 at 9:12
  • $\begingroup$ I have edited all the typos and tried to clarify $\endgroup$
    – Ali
    Mar 19, 2022 at 9:24

1 Answer 1


The OP hid the key information that $ S $ and $ \phi $ are symmetric, so I had to dig it out in the literature.

Clear[S, tenϕ, tenG]

tenϕ = Array[Subscript[ϕ, Sequence @@ Sort[{##} - 1]] &, {2, 2, 2, 2}];

S[0, j_] := -I Dot @@ PauliMatrix[{2, j}] /; j > 0
S[i_, i_] := ConstantArray[0, {2, 2}]
{S[2, 3], S[3, 1], S[1, 2]} = I {S[0, 1], S[0, 2], S[0, 3]};
S[i_, j_] := S[j, i];

tenG[μ_, ν_, λ_, ρ_] := 
  TensorProduct[S[μ, ν], S[λ, ρ], tenϕ], 
  {{1, 5}, {2, 6}, {3, 7}, {4, 8}}

enter image description here

So tenG gives the same results as in Eq. (B11) in the paper added by the OP.

  • $\begingroup$ wow, thank you very much $\endgroup$
    – Ali
    Mar 20, 2022 at 12:28
  • $\begingroup$ Hi @Ali, if my answer helps you truly, please consider voting it up and accepting it. $\endgroup$ Mar 22, 2022 at 7:57
  • $\begingroup$ @ Αλέξανδρος Ζεγγ, I have learned your answer. But this way,I do not have each S with four index although tried. $\endgroup$
    – Ali
    Apr 12, 2022 at 8:57

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