# Exact Gauge Symmetry of the Standard Model

by sweetser
Tags: gauge symmetry, standard model
 P: 361 Hello: The gauge symmetry of the standard model is written in authoritative places like wikipedia :-) as U(1)xSU(2)xSU(3). This would have 12 elements in its Lie algebra corresponding to one photon, W+, W- and W0 or Z, and the 8 gluons. I recall reading discussions that such a characterization is not exactly right. U(1) is a subgroup of SU(2), and SU(2) is a subgroup of SU(3). My vague memory is recalling something like a "flag" must be added to U(1) so that it is nicely separated from SU(2), which also needs one of these flags or markers. If someone in the know could fill me in, it will be appreciated as an hour of Googling did not turn up this detail of the standard model. Thanks, Doug
 P: 986 U(1), SU(2), and SU(3) in this notation are all distinct groups. 8 generators of SU(3) are gluons. W+, W-, Z and gamma are linear combinations of generators of SU(2) and U(1).
P: 361
 Quote by hamster143 U(1), SU(2), and SU(3) in this notation are all distinct groups.
Agreed. My question is how is that shown in math notation. If one takes an element of math symmetry U(1) and multiplies by an element of SU(2), the result belongs in the group SU(2). In the physics context, the electroweak symmetry U(1)xSU(2) is a different gauge symmetry than weak symmetry of SU(2). I recall some sort of notation to indicate that [perhaps $$U(1)_{\gamma}$$, but I don't know what to call the gamma, and if correct, the subscripts for SU(2) and SU(3)]

Thanks,
Doug

P: 527
Exact Gauge Symmetry of the Standard Model

 Quote by sweetser ... If one takes an element of math symmetry U(1) and multiplies by an element of SU(2), the result belongs in the group SU(2)...
At the level of the abstract group itself, you dont multiply elements of U(1) with the elements of SU(2). The direct product group is constructed through elements $$(a,b)$$ with $$a$$ in U(1) and $$b$$ in SU(2). Multiplication is defined as: $$(a,b)*(c,d) = (ac,bd)$$.

Now, this gauge group acts on an internal space carried by the particles. This internal space is a representation of the direct product group $$U_Y(1)xSU(2)$$ where the Y denotes the weak hypercharge. These representations can be constructed by taking the direct product of representations of the original groups, i.e. $$R_{U_Y(1)xSU(2)} = R_{U_Y(1)}\otimes R_{SU(2)}$$.

For the force carriers we use the adjoint representation of the gauge group. The adjoint representation of the group U(1) is fairly simply - it's just an expontential, and hence the space it acts on is one-dimensional (complex). The adjoint representation of the SU(2) part is a 3-dimensional one. So the total representation space is 4-dimensional. So the 'flag' that you are talking about is really a reference to the representation that assigned to particular particle (i.e gluons don't carry U(1)xSU(2) etc).

The same goes for SU(3). It contains an SU(2) subgroup (it even contains a SU(2)xU(1) subgroup), but you dont identify this group with the electroweak part. It's 'just' some result from group theory. So you have to be careful with what symmetry structure is assigned to what force carrier (this basically comes down to a choice of what particles carry which representation).

In that train of thought, there have been GUT proposals that make use of for instance SU(5) or SO(10), since these groups contain the SU(3)xSU(2)xU(1) as a subrgoup.

By introducing a coupling with a Higgs field and demanding that the Higgs boson acquires some vacuum expectation value, the gauge symmetry is no longer exact and said to be broken. It does not vanish completely though, as the symmetry group is said to break down to a subgroup. In the case of electroweak theory this subgroup is the $$U_em(1)$$ group identified with electromagnetism. The label is there because the elements of this subgroup can, naturally, be identified with elements of the larger group $$U_Y(1)xSU(2)$$ (it's a subgroup after all!). And as it turns out, these elements do not correspond with the group $$U_Y(1)$$ (which is also a subgroup if you consider the elements (a,e) - with e the identity element of $$SU(2)$$ and a elements of $$U_Y(1)$$. This is why the distinction has to be made.

HW Helper
P: 3,018
 Quote by sweetser Agreed. My question is how is that shown in math notation. If one takes an element of math symmetry U(1) and multiplies by an element of SU(2), the result belongs in the group SU(2). In the physics context, the electroweak symmetry U(1)xSU(2) is a different gauge symmetry than weak symmetry of SU(2). I recall some sort of notation to indicate that [perhaps $$U(1)_{\gamma}$$, but I don't know what to call the gamma, and if correct, the subscripts for SU(2) and SU(3)] Thanks, Doug

People write $$U(1)_Y \otimes SU(2)_L \otimes SU(3)_c$$ where the U(1) factor is the hypercharge abelian group. Is this what you were looking for
P: 136
 Quote by xepma In the case of electroweak theory this subgroup is the $$U_em(1)$$ group identified with electromagnetism. The label is there because the elements of this subgroup can, naturally, be identified with elements of the larger group $$U_Y(1)xSU(2)$$ (it's a subgroup after all!). And as it turns out, these elements do not correspond with the group $$U_Y(1)$$ (which is also a subgroup if you consider the elements (a,e) - with e the identity element of $$SU(2)$$ and a elements of $$U_Y(1)$$. This is why the distinction has to be made.
I think we should emphasize that part: The U(1) of electromagnetism is not the U(1) factor appearing in SU(2)xU(1) of the electroweak group. It comes from a "mixture" of the two factors. Otherwise, e.g., the massive weak bosons appearing after the breaking wouldn't be charged under the U(1) of EM! Hence the "Y" label on the electroweak group's U(1) factor.

Also, from the original poster:
 The electroweak symmetry U(1)xSU(2) is a different gauge symmetry than weak symmetry of SU(2)
Well, the SU(2) factor in the SU(2)xU(1) electroweak symmetry group is itself a symmetry (subgroup). After the electroweak group is broken to U(1)' (the electromagnetic group), there is no longer an SU(2) *symmetry*. I think what you may have been getting at, though, is that just as the EM U(1) comes from a "mixture" of the original SU(2)xU(1), the three massive bosons W+,W-,Z of the weak interaction are associated with an SU(2) "mixture" of the original SU(2)xU(1). That "weak SU(2)" are broken symmetries.

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