# Why is the Symmetry Group for the Electroweak force SU(2)xU(1) and not U(2)

Let me first say that I'm a layman who's trying to understand group theory and gauge theory, so excuse me if my question doesn't make sense.

Before symmetry breaking, the Electroweak force has 4 degrees of freedom (B0, W1, W2, and W3 right?) and after symmetry breaking, we are left with the weak SU(2) bosons (W+, W-, Z) and the photon. Why, then, is the symmetry group before Electroweak symmetry breaking SU(2)xU(1) and not just U(2) (Since a Unitary group of n dimension contains n^2 degrees of freedom?) What am I missing?

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–  Qmechanic May 14 at 18:40

The correct electroweak gauge group is $SU(2)_L \times U(1)_Y$ where $Y$ denotes the weak hypercharge. After the Higgs field spontaneously breaks this exact symmetry, third generator of $SU(2)_L$ (weak isospin) and weak hypercharge combine to give the remaining unbroken $U(1)_{em}$.

Gauge bosons and fermions fall under different representations of this gauge group and transform non-trivially under the action of the group. For example $W_{1,2,3}$ transforms as a triplet under $SU(2)_L$. Left handed fermions transform as $SU(2)_L$ doublets. This simply means that under the action of a symmetry transformation, components of each representation are mixed in a special way depending on the representation of the gauge group. So quantum numbers of our fields determine how they transform and interact with each other. For the electroweak sector of the Standard Model, the correct properties are obtained with the gauge group $SU(2)_L \times U(1)_Y$ and not with $U(2)$.

Having said that in grand unified theories one starts with a larger gauge group that contains the SM gauge group and try to break it down to the SM gauge group.

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Nice question! The short answer is that the group is not $SU(2)\times U(1)$, it is $SU(2)_L \times U(1)_{em}$. In other words the two groups act on different standard model particles differently. For example the left handed neutrino does interact weakly and so transforms under the $SU(2)_L$, but is electrically neutral so it doesn't transform under the $U(1)_{em}$. So you can't combine the two groups the way you would think you could naively.

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Not that this answer is bad in any way, but if you have time, I think a more detailed demonstration of this (i.e. actually showing how the use of U(2) would result in a different Lagrangian) would be very helpful. –  David Z Mar 30 at 17:45
Thanks for the comment! I will try to add more later. Indeed there's a bit more to it than what I said. For example if I had a pure yang mills theory based on $U(N)$, I would still write the group as $SU(N) \times U(1)$ bc the generator associated with the $U(1)$ commutes with everything else so the associated gauge field decouples. I don't have the time right now to say this with equations though. –  Andrew Mar 30 at 17:56
@Andrew also, this is incorrect, as HenryMcFly points out below. The GWS gauge group is $\mbox{SU}(2)_L\times\mbox{U}(1)_Y$, which is sopntaneously broken down to $\mbox{U}(1)_{EM}$. They have the same group structure, but the charges are different. Just to avoid confusing the questioner over this. Otherwise this answer is perfect. –  Flint72 Apr 16 at 12:46

Actually we have the following Lie algebra isomorphism $$u(1)\oplus su(2)\cong u(2),$$ and there exists the following Lie group isomorphism

$$[U(1)\times SU(2)]/\mathbb{Z}_2 ~\cong~ U(2).$$

In other words, there is a two-to-one map between $U(1)\times SU(2)$ and $U(2)$. So in that sense the Glashow-Salam-Weinberg $U(1)\times SU(2)$ model already contains a $U(2)$ gauge group.

Of course, the various matter and gauge fields transform naturally in the weak hypercharge and weak isospin language, i.e., under $U(1)\times SU(2)$ .

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The group isomorphism is $U(2)\ni g\to (\sqrt{\det g}, g/\sqrt{\det g}) \sim (-\sqrt{\det g}, -g/\sqrt{\det g})\in [U(1)\times SU(2)]/\mathbb{Z}_2$. If we try to endow the GSW electroweak model with the $\mathbb{Z}_2$-reduced gauge group $U(2)$ [instead of the $U(1)\times SU(2)$ gauge group], then a matter field transforming in a (hypercharge,isospin) irrep $(Y,I)$ may becomes multi-valued, depending on the fractional value of the hypercharge $Y$. –  Qmechanic May 11 at 16:15