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

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!

1. Ref. 1 does not seem to mention a symmetry-breaking $$U(1)$$ that belongs to the part of $$SU(5)$$ that 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
$$\begin{pmatrix} -m\mathbb{1}_{n\times n} & \mathbb{0}_{n\times m} \cr \mathbb{0}_{m\times n} & n\mathbb{1}_{m\times m} \end{pmatrix}~\in~su(n+m).$$
3. Now let's consider at the group level. Define $$k~:=~{\rm gcd}(n,m), \quad n^{\prime}~:=~n/k \quad\text{and}\quad m^{\prime}~:=~m/k,$$ so that the integers $$n^{\prime}$$ and $$m^{\prime}$$ are co-prime. The Lie group homomorphism $$G~:=~SU(n) \times SU(m) \times U(1)~\ni~ (g,h,\alpha)~\stackrel{\Phi}{\mapsto}~\begin{pmatrix} \alpha^{-m^{\prime}}g & \mathbb{0}_{n\times m} \cr \mathbb{0}_{m\times n} & \alpha^{n^{\prime}} h \end{pmatrix}~\in~ SU(n+m)$$ is not injective. In fact, the kernel $${\rm Ker}(\Phi) ~\cong~\mathbb{Z}_{nm/{\rm gcd}(n,m)}$$ is generated by the elements $$\left(e^{i2\pi j/n}\mathbb{1}_{n\times n},~e^{-i2\pi j/m}\mathbb{1}_{m\times m},~e^{i2\pi jk/nm}\right)~\in~G, \quad j~\in~\{1,\ldots,k\}.$$ The Lie group $$G$$ is not a subgroup of $$SU(n+m)$$, but the quotient group $$G/{\rm Ker}(\Phi) ~\cong~G/\mathbb{Z}_{nm/{\rm gcd}(n,m)}$$ is a subgroup of $$SU(n+m)$$.