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Pauli matrices satisfy following relation $$[\sigma_i,\sigma_j]=2i\epsilon_{ijk}\sigma_k$$ While looking through models of noncommutative geometry of spacetime I have seen people defining following commutation relation to model the noncommutative nature of spacetime $$[x_{\mu},x_{\nu}]=i\theta_{\mu\nu}$$ I am thinking to have commutation relation defined as following $$\color{blue}{[x_{\mu},x_{\nu}]=i\tilde{\epsilon}_{\mu\nu\rho\sigma}x^{\rho}x^{\sigma}}$$ where $\tilde{\epsilon}_{\mu\nu\rho\sigma}$ is the tensor constructed by multiplying Levi-Civita and appropriate factor of determinant of metric.

Is there any model of noncommutative spacetime based on the above commutation relation?

Can we infer something about this commutation relation using the algebra of pauli matrix which satisfy algebra of $SO(3)$ group?

Edit: As Cosmas Zachos pointed out it would be simply zero but since I meant operator $\hat{x}$ in the above equation so it doesn't imply $\hat{x}^{0}\hat{x}^{1}$ = $\hat{x}^{1}\hat{x}^{0}$ so it won't be zero as an explicit example $$\hat{x}_0\hat{x}_1=i\sqrt{-g}\hspace{2pt}\epsilon_{01\rho\sigma}\hat{x}^{\rho}\hat{x}^{\sigma}$$ $$=i\sqrt{-g}\hspace{2pt}(\hat{x}^2\hat{x}^3-\hat{x}^3\hat{x}^2)$$ $$\neq0$$

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  • $\begingroup$ ? The right hand side vanishes from the antisymmetry of $\tilde \epsilon$, and the symmetry of the xs. $\endgroup$ Commented Jun 26, 2020 at 13:17
  • $\begingroup$ @CosmasZachos Oh! I missed it. But wouldn't this property of antisymmetric and symmetric tensors contracted to give zero suggest that we cannot define similar kind of operator equation on spacetime operator (for $D\geq$1+3) as of the Pauli matrices? $\endgroup$ Commented Jun 26, 2020 at 14:42
  • $\begingroup$ Pauli matrices are very different objects. You have to be more explicit than "suggests"... $\endgroup$ Commented Jun 26, 2020 at 14:48
  • $\begingroup$ @CosmasZachos I just did an explicit calculation and it's not zero because here $x^{\alpha}$ is an operator($\hat{x}^{\alpha}$) so we can't take it for granted that $\hat{x}^{\alpha}\hat{x}^{\beta}=\hat{x}^{\beta}\hat{x}^{\alpha}$. $\endgroup$ Commented Jun 26, 2020 at 15:01
  • $\begingroup$ Have you checked the Jacobi identity? $\endgroup$ Commented Jun 26, 2020 at 15:32

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