Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,851 answers , 20,616 comments
1,470 users with positive rep
501 active unimported users
More ...

Where does in GUT symmetry breaking $U(1)$ come from?

+ 4 like - 0 dislike
22 views

In GUTs one starts with some larger group, like $SU(5)$, which is then broken into smaller groups, for example

$$SU(5) ~\longrightarrow~ SU(3) \times SU(2) \times U(1)$$

This can be seen, for example, by looking at the Dynkin diagram for $SU(5)$: Removing one node leaves us with the Dynkin diagrams for $SU(3)$ and $SU(2)$.

My problem is understanding where $U(1)$ comes from. I've read several statements about that, but couldn't fit the puzzle pieces together. Dynkin diagrams are one the level of Lie algebras. Removing one node means we remove one generator. For example, in this 2010 handout from the the course Symmetries in Physics by Michael Flohr:

By removing a node, the rank of the subalgebra is reduced by one, and the simple roots are a subset of the original simple roots. On the level of the groups, we thus find, $$G=G_1\times G_2 \times U(1),$$ where the additional $U(1)$ factor comes from the left out Cartan generator. For example, $SU(n+m)$ can be reduced in this way into $$SU(n)\times SU(m)\times U(1).$$ This is the classical ansatz for a GUT: $SU(5)$ gets broken into $$SU(3)\times SU(2)\times U(1).$$

  • In this paper by John Baez it is claimed that $SU(3) \times SU(2) \times U(1)$ is no subgroup of $SU(5)$ and some homomorphism is used the justify why $U(1)$ appears.

EDIT:

In Lie Algebras In Particle Physics: from Isospin To Unified Theories Georgi writes:

"The Cartan generator that have been left out generate U(1) factors"

and my problem is understanding why this is the case.

Any ideas to clarify this would be awesome!

This post imported from StackExchange Physics at 2015-08-13 21:35 (UTC), posted by SE-user Tim
asked Apr 16, 2015 in Theoretical Physics by Tim (20 points) [ no revision ]

1 Answer

+ 0 like - 0 dislike
  1. Ref. 1 does not seem to mention a symmetry-breaking $U(1)$, which must belong to the part of $SU(5)$ which is not in the standard model. In this answer, we will assume that OP is really asking about the weak hypercharge $U(1)$ gauge factor of the standard model.

  2. At the Lie algebra level, recall that the Lie algebra $su(n)$ consists of Hermitian traceless $n\times n$ matrices and has rank $n-1$. In the inclusion $$\underbrace{su(n)}_{\text{rank } n-1} \oplus \underbrace{su(m)}_{\text{rank } m-1} \oplus u(1)~\subset~ \underbrace{su(n+m)}_{\text{rank } n+m-1},$$ the maximal Abelian subalgebra [of diagonal Hermitian traceless $(n+m)\times(n+m)$ matrices] is the same on both sides. The Lie algebra on the left-hand (right-hand) side is reductive (simple), respectively. A traceless Cartan generator for the above $u(1)$ is
    $$\begin{pmatrix} -m{\bf 1}_{n\times n} & 0 \cr 0 & n{\bf 1}_{m\times m} \end{pmatrix}~\in~su(n+m).$$

  3. Ref. 1 makes the point that the Lie group homomorphism $$G~:=~SU(3) \times SU(2) \times U(1)~\ni~ (g,h,\alpha)~~\stackrel{\Phi}{\mapsto}~~ \begin{pmatrix} \alpha^{-2}g & 0 \cr 0 & \alpha^3 h \end{pmatrix}~\in~ SU(5) $$ is not injective. Here the $U(1)$ factor is the weak hypercharge $U(1)$. In fact, the kernel $${\rm Ker}(\Phi) ~\cong~\mathbb{Z}_6,$$ is generated by the element $$\left(e^{i2\pi/3}{\bf 1}_{3\times 3},~-{\bf 1}_{2\times 2},~e^{i\pi/3}\right)~\in~G.$$ The Lie group $G$ is not a subgroup of $SU(5)$, but the quotient group $G/{\rm Ker}(\Phi)$ is a subgroup of $SU(5)$.

References:

  1. J.C. Baez, Calabi-Yau Manifolds and the Standard Model, arXiv:hep-th/0511086.
This post imported from StackExchange Physics at 2015-08-13 21:35 (UTC), posted by SE-user Qmechanic
answered Apr 16, 2015 by Qmechanic (2,790 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOverf$\varnothing$ow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...