# how to split indices for a Spinor object?

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: $$\{S_{01},S_{02},S_{03}\}=-i\sigma_{y}\{\sigma_{x},\sigma_{y},\sigma_{z}\}$$ $$i\{S_{01},S_{02},S_{03}\}=\{S_{23},S_{31},S_{12}\}$$

• 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?
– bmf
Mar 18, 2022 at 19:06
• 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.
– Ali
Mar 19, 2022 at 8:51
• What is the exact relation between the subscripts of $S$ and those of $\sigma$s? Mar 19, 2022 at 8:59
• 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. Mar 19, 2022 at 9:12
• I have edited all the typos and tried to clarify
– Ali
Mar 19, 2022 at 9:24

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_] :=
TensorContract[
TensorProduct[S[μ, ν], S[λ, ρ], tenϕ],
{{1, 5}, {2, 6}, {3, 7}, {4, 8}}
]


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

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