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Consider the theory of a single generation of $e, \nu_L$ matter content. The initial lagrangian is

$$ \mathcal{L} = i\bar{\ell}\not \partial\ell + i\bar{e}_R\not \partial e_R \tag{1} $$

where

$$ \ell = {\nu_L \choose e_L} $$

is an $SU(2)_L$ doublet.

It is often stated that this theory has the global symmetry $SU(2)_L \times U(1)_Y$. I see the $SU(2)_L$ symmetry, however I see two independent $U(1)$ symmetries, which leads to my following question.

Question: Why is isn't the global symmetry group of (1) given by

$$ SU(2)_L \times U(1)_L \times U(1)_R. $$

I understand that we can choose $\ell$ and $e_R$ to have the same phase, but it seems like when doing so we are losing a degree of freedom, so what exactly happens to this degree of freedom?

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  • $\begingroup$ There is indeed another $U(1)$ in the global symmetry and people do talk about it -- it's just called lepton number! Presumably this isn't mentioned explicitly because people want to focus on the global symmetries that are gauged. $\endgroup$
    – knzhou
    Commented Feb 20, 2019 at 2:01
  • $\begingroup$ I see.. So there actually is an additional one as well right? Since if you expand out the lagrangian in terms of the $\nu, e_L, e_R$ it's clear that there are 3 independent U(1)s $\endgroup$
    – InertialObserver
    Commented Feb 20, 2019 at 2:03

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