4
$\begingroup$

The Spin(10) grand unification has a symmetry breaking of SO(10), or Spin(10).

In Wikipedia, it says,

"The symmetry breaking of SO(10) is usually done with a combination of (( a $45_H$ OR a $54_H$) AND ((a $16_H$ AND a $\overline{16}_H$) OR (a $126_H$ AND a $\overline{126}_H$)) )."

I suppose that 16 has something to do with the 16 spinor representation of SO(10), and 45 has something to do with ${10 \choose 2} = 45$, while 126 has something to do with $\frac{1}{2}{10 \choose 5} = 126$.

  • What does 54 stands for in the representation theory?

  • So what are so special about these number: 16,45,54, and 126 in these models? And their roles in the representation theory?

Improve this question
$\endgroup$
2
  • $\begingroup$ 45 is just the number of independent entries in an antisymmetric $10\times10$ matrix and there is nothing special about the numbers. Some people like to use an $\overline{126}$-plet because it allows one to break the $B-L$ symmetry to matter parity, others don't like any of the large representations because it seems hard to impossible to get them out of string theory. $\endgroup$
    – user178876
    Jul 30, 2018 at 21:11
  • 2
    $\begingroup$ Table 41 in your Slansky. the 45 is (01000) in dynkin index notation, the 54 the (20000) and the 16 and 126 spinors. write down the SO(10) invariants of them and the SU(5) invariance of the SSB vacua. Now look in the next table at 16x16bar and 126x126bar to see what higgs reps you need to saturate with the fermion bilinears.... what do you see? $\endgroup$
    – Cosmas Zachos
    Jul 30, 2018 at 22:22

1 Answer 1

Reset to default
3
$\begingroup$

Let us call the defining representation of $SO(10)$ for $V={\bf 10}$. The vector space $V$ is endowed with an invariant metric form: $V\times V\to \mathbb{R}$ (of positive/Euclidean signature). Then we have:

  1. The antisymmetric tensor product $\bigwedge{}^2V\equiv V\wedge V={\bf 45}$.

  2. The totally antisymmetric tensor product $\bigwedge{}^3V={\bf 120}$.

  3. The symmetric tensor product ${\rm Sym}^2V\equiv V\odot V={\bf 1}\oplus{\bf 54}$. The trivial representation ${\bf 1}$ comes from contraction with the metric. The ${\bf 54}$ may be thought of as the traceless part of ${\rm Sym}^2V$.

  4. $\bigwedge{}^5V={\bf 126}^+\oplus{\bf 126}^-$, corresponding to selfdual and anti-selfdual 5-forms. (The Hodge star operator is defined via the metric.)

  5. ${\bf 16}_{L/R}$ are the left/right Weyl spinors in 10D.

See also this related Phys.SE post.

Improve this answer
$\endgroup$
7
  • 1
    $\begingroup$ You may need to explain the cup expression to us... thanks... $\endgroup$
    – ann marie cœur
    Jul 31, 2018 at 12:57
  • $\begingroup$ Is that ant symmetric tensor wedge product? $\endgroup$
    – ann marie cœur
    Jul 31, 2018 at 12:58
  • $\begingroup$ "The trivial representation 1 comes from contraction with the metric." which metric? $\endgroup$
    – ann marie cœur
    Jul 31, 2018 at 12:58
  • $\begingroup$ I also hope to understand this ${\rm Sym}^2V={\bf 1}\oplus{\bf 54}$ and what is the 54 better? $\endgroup$
    – ann marie cœur
    Jul 31, 2018 at 12:59
  • $\begingroup$ Thanks, can you explain that "The symmetric tensor product Sym$^2V$?" How do you define Sym$^2V$? $\endgroup$
    – ann marie cœur
    Aug 1, 2018 at 20:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.