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

+ 3 like - 0 dislike
216 views

It is well known that the Standard Model (SM) gauge group is a subgroup of $SU(5)$: $$SU(3) \times SU(2)\times U(1) ~\subset~SU(5)$$ This can be easily checked using the method of Dynkin diagrams. Is this subgroup an invariant subgroup such that, $$g _{SU(5)} g _{SM} g _{SU(5)} = g _{SM} ' \,,$$ 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

edited Mar 3, 2015

+ 5 like - 0 dislike

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 $SU(3)\times SU(2)\times U(1)$.

answered Nov 27, 2014 by (265 points)

 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): Email me at this address if my answer is selected or commented on: 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$\hbar$ysicsOverfl$\varnothing$wThen 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.