Symmetry breaking of $SO(10)$ or $Spin(10)$

Question

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?

This post imported from StackExchange Physics at 2020-11-08 17:31 (UTC), posted by SE-user annie marie heart

Comments

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.

This post imported from StackExchange Physics at 2020-11-08 17:31 (UTC), posted by SE-user user178876

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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?

This post imported from StackExchange Physics at 2020-11-08 17:31 (UTC), posted by SE-user Cosmas Zachos

Answer 1

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.

This post imported from StackExchange Physics at 2020-11-08 17:31 (UTC), posted by SE-user Qmechanic

Comments

You may need to explain the cup expression to us... thanks...

This post imported from StackExchange Physics at 2020-11-08 17:31 (UTC), posted by SE-user annie marie heart

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Is that ant symmetric tensor wedge product?

This post imported from StackExchange Physics at 2020-11-08 17:31 (UTC), posted by SE-user annie marie heart

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"The trivial representation 1 comes from contraction with the metric." which metric?

This post imported from StackExchange Physics at 2020-11-08 17:31 (UTC), posted by SE-user annie marie heart

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I also hope to understand this ${\rm Sym}^2V={\bf 1}\oplus{\bf 54}$ and what is the 54 better?

This post imported from StackExchange Physics at 2020-11-08 17:31 (UTC), posted by SE-user annie marie heart

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Thanks, can you explain that "The symmetric tensor product Sym$^2V$?" How do you define Sym$^2V$?

This post imported from StackExchange Physics at 2020-11-08 17:31 (UTC), posted by SE-user annie marie heart

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I accept your answer first, but please clarfy

This post imported from StackExchange Physics at 2020-11-08 17:31 (UTC), posted by SE-user annie marie heart

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I updated the answer.

This post imported from StackExchange Physics at 2020-11-08 17:31 (UTC), posted by SE-user Qmechanic