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In the description of Wikipedia of flipped SO(10) model, it says that:

  1. In flipped $SO(10)$ models, however, the gauge group is $[SO(10)_F × U(1)_B]/Z_4$.

  2. If we suppose $[SU (5) × U(1)_χ ]/ Z_5$ is a subgroup of $SO(10)_F$, then we have the intermediate scale symmetry breaking $[SO(10)_F × U(1)_B]/Z_5 → [ SU(5) × U(1)_χ]/Z_5$ where $\chi=-{A\over 4}+{5B\over 4}$.

My question is that

  • should $[SO(10)_F × U(1)_B]/Z_4$ and $[SO(10)_F × U(1)_B]/Z_5$ mentioned above, they are the same group or not? Is that one of them has a typo and indeed are the same as $[SO(10)_F × U(1)_B]/Z_4$?

  • What exactly are the $A$ and $B$ in $\chi=-{A\over 4}+{5B\over 4}$? I think that $B$ stands for a $U(1)$, and $B$ does not imply baryon, correct? But $A$ does not show up anywhere on the Wikipedia page.

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    $\begingroup$ $SO(10)$ does not even have a normal $Z_5$ subgroup so the second option makes little sense. $\endgroup$ Mar 7, 2021 at 22:59

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  1. Wiki "In flipped $SO(10)$ models, however, the gauge group is not just $SO(10)$ but $SO(10)_F \times U(1)_B$ or $[SO(10)_F \times U(1)_B]/\mathbf Z_4$." $\to$

My suggestion on the correction to a gauge group $Spin(10)$ but $Spin(10)_F \times U(1)_B$ or $[Spin(10)_F \times U(1)_B]/\mathbf Z_4$. This is because the $Spin(10)$ has a $\mathbf Z_4$ center.

  1. Wiki "If we suppose $[SU(5) \times U(1)_A]/\mathbf Z_5$ is a subgroup of $SO(10)_F$, then we have the intermediate scale symmetry breaking $[SO(10)_F \times U(1)_B]/\mathbf Z_4 \to [SU(5) \times U(1)_\chi]/\mathbf Z_5$"

This statement seems wrong. My suggestion on the correction to

$U(5) = [SU(5) \times U(1)_A]/\mathbf Z_5$ or $U(5)' = [SU(5) \times U(1)_\chi]/\mathbf Z_5$. $U(5)$ is a subgroup of $SO(10)$. But $U(5)$ is not a subgroup of $Spin(10)$. Because we cannot lift the inclusion of $U(5)$ to $SO(10)$ to its double cover $Spin(10)$. So we cannot have $[SU(5) × U(1)_A]/\mathbf Z_5$ as a subgroup of $Spin(10)_F$. I think that $[Spin(10)_F × U(1)_B]/\mathbf Z_4$ does not contain U(5) = $[SU(5) \times U(1)_A]/\mathbf Z_5$ nor contain $U(5)' = [SU(5) \times U(1)_\chi]/\mathbf Z_5$ either as a subgroup. So the whole thing there does not work in this part.

Maybe other experts can comment.

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  • $\begingroup$ I did not give a complete understanding. There are some gaps. It will be nice someone can make comments about these... $\endgroup$ Aug 15, 2021 at 20:30
  • $\begingroup$ thanks so much + 1, I will be glad if you or someone can clarify more $\endgroup$ Aug 17, 2021 at 13:27

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