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

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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

asked Nov 26, 2014 in Theoretical Physics by JeffDror (650 points) [ revision history ]
edited Mar 3, 2015 by Arnold Neumaier

1 Answer

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

answered Nov 27, 2014 by Jonathan Gleason (255 points) [ revision history ]

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