Is the gauge group of the Standard Model an invariant subgroup of $SU(5)$?

Question

It is well known that the Standard Model (SM) gauge group is a subgroup of $SU(5)$:

$$\begin{equation} SU(3) \times SU(2)\times U(1) ~\subset~SU(5) \end{equation}$$ This can be easily checked using the method of Dynkin diagrams. Is this subgroup an invariant subgroup such that, $$\begin{equation} g _{SU(5)} g _{SM} g _{SU(5)} = g _{SM} ' \,, \end{equation}$$ where $g_{SU(5)}$ ($g_{SM}$) is an element of $SU(5)$ ($SM$)?

Background: The reason I'm interested in this is because then its necessarily true that the non-SM gauge group generators of $SU(5)$ can be written as solely off-diagonal matrices and the SM as solely diagonal, which simplifies calculations.

This post imported from StackExchange Physics at 2014-11-27 10:39 (UTC), posted by SE-user JeffDror

Answer 1

The answer is no.

The easiest way to see this I think it to look at the concrete embedding of $SU(3)\times SU(2)\times U(1)$ in $SU(5)$.  That embedding is given by

$$(A,B,\mathrm{e}^{\mathrm{i}\theta})\mapsto \begin{pmatrix}A\mathrm{e}^{\mathrm{i}\theta} & 0 \\ 0 & B\mathrm{e}^{-\mathrm{i}\theta}\end{pmatrix}$$ .

Given this, one can simply just compute to see that the conjugate of something of this form is not necessarily of this form.  For example, the conjugate of this by

$$-\frac{1}{\sqrt{2}}\begin{pmatrix}1 & 0 & \mathrm{i} & 0 & 0 \\ \mathrm{i} & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & \mathrm{i} & 0 \\ 0 & \mathrm{i} & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & \sqrt{2}\end{pmatrix}=:\begin{pmatrix}X & Y \\ U & V\end{pmatrix}$$

is

$$\begin{pmatrix}XAX^*\mathrm{e}^{\mathrm{i}\theta}+YBY^*\mathrm{e}^{-\mathrm{i}\theta} & XAU^*\mathrm{e}^{\mathrm{i}\theta}+YBV^*\mathrm{e}^{-\mathrm{i}\theta} \\ UAX^*\mathrm{e}^{\mathrm{i}\theta}+VBY^*\mathrm{e}^{-\mathrm{i}\theta} & UAU^*\mathrm{e}^{\mathrm{i}\theta}+VBV^*\mathrm{e}^{-\mathrm{i}\theta}\end{pmatrix}$$ .

Taking

$$\theta :=0,\qquad A:=\frac{1}{\sqrt{2}}\begin{pmatrix}1 & \mathrm{i} & 0 \\ \mathrm{i} & 1 & 0 \\ 0 & 0 & \sqrt{2}\end{pmatrix},\qquad B:=\frac{1}{\sqrt{2}}\begin{pmatrix}1 & \mathrm{i} \\ \mathrm{i} & 1\end{pmatrix}$$ ,

a computation shows that this conjugate is not in the sub-group of $SU(5)$ listed above isomorphic to $SU(3)\times SU(2)\times U(1)$.

Answer 2

The more commonly used notion of "invariant subgroup" is "normal subgroup": we say $H \subseteq G$ is a normal subgroup of $G$ if $h \in H$ implies $g h g^{-1} \in H$ for all $g \in G$.  I have never heard anyone consider subgroups $H$ obeying your condition: $h \in H$ implies $g h g \in H$ for all $g \in G$.  This implies, for example, that the square of every element of $g$ lies in $H$.

$\mathrm{SU}(3) \times \mathrm{SU}(2) \times \mathrm{U}(1)$ is actually not a subgroup of $\mathrm{SU}(5)$.  $(\mathrm{SU}(3) \times \mathrm{SU}(2) \times \mathrm{U}(1))/\mathbb{Z}_6$ is a subgroup of $\mathrm{SU}(5)$, and this fact explains many interesting patterns in the Standard Model.

Jonathan Gleason's answer actually shows that $(\mathrm{SU}(3) \times \mathrm{SU}(2) \times \mathrm{U}(1))/\mathbb{Z}_6$ is not a normal subgroup of $\mathrm{SU}(5)$.  While this is not the question you asked, this is probably the right answer.

One can also take a more abstract approach.  $\mathrm{SU}(n)$ is a simple Lie group.  For any simple Lie group $G$, all normal subgroups are either subgroups of the center of $G$ (which is a discrete group), or the whole group $G$.   In the case $G = \mathrm{SU}(n)$ the center consists of multiples of the identity matrix, and it's isomorphic to $\mathbb{Z}/n$.    This does the job.