$mathcal{L}_{I}^{C C} $ global $U(1)$ invariance

It is well known that the SM lagrangian has the $U(1)$ symmetry. Considering for the moment the global symmetry $U(1)$, I am having some trouble seeing it in the following terms:

$$mathcal{L}_{I}^{C C}=-frac{g}{sqrt{2}} sum_{ell=e, mu, tau}left(W_{mu}^{+} bar{nu}_{ell L} gamma^{mu} ell_{L}+W_{mu}^{-} bar{ell}_{L} gamma^{mu} nu_{ell L}right)$$

after the transformation I should obtain $mathcal{L}^{‘}=mathcal{L}$:

$$mathcal{L}_{I}^{C C}=-frac{g}{sqrt{2}} sum_{ell=e, mu, tau}left(W_{mu}^{+} bar{nu}_{ell L} e^{i alpha_l}gamma^{mu} ell_{L}e^{-i beta_l}+W_{mu}^{-} bar{ell}_{L} e^{i beta_l}gamma^{mu} nu_{ell L}e^{-i alpha_l}right).$$

This should implies that $alpha_l=beta_l$ which is not true, because we know that lepton are charged and neutrinos no.

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In case of interaction lagrangian of QED as far as I know, you can see that it will conserve the electric charge doing:
$$mathcal{L}_{I}^{‘,QED}= – q_e bar{psi}e^{iq_e}{A!!!/}e^{-iq_e}psi=- q_e bar{psi}{A!!!/}psi =mathcal{L}_{I}^{QED}$$

Honestly I have never transformed a gauge boson under a global symmetry, however, the only way in which the previous Lagrangian is invariant, according to a transformation that is connected to the electric charge, is :
$$mathcal{L}_{I}^{‘,C C}=-frac{g}{sqrt{2}} sum_{ell=e, mu, tau}left(W_{mu}^{+} e^{-iq_W}bar{nu}_{ell L} e^{iq_{nu}}gamma^{mu} ell_{L}e^{-iq_{ell}}+W_{mu}^{-}e^{iq_W} bar{ell}_{L} e^{iq_{l}}gamma^{mu} nu_{ell L}e^{-iq_{nu}}right)= mathcal{L}_{I}^{C C}$$
with $q_w=e,q_{ell}=-e,q_{nu}=0$

maybe is correct in this way, but I’m not sure…