In ref to the paper "Non-Abelian Gauge Lepton Symmetry as the Gateway to Dark Matter", A bi doublet has been defined in eq. (2) for the fermion fields $\nu ,e, N,E$ as $\begin{pmatrix}\nu & N\\ e & E\end{pmatrix} $ . I tried writing down the lagrangian kinetic term for this bi doublet with the covariant derivative terms containing the vector bosons related to both the $SU(2)$. But i am getting a $2x2$ matrix for the kinetic term which is $$\begin{pmatrix} \bar{\nu} D_{\mu} \nu + \bar{e} D_{\mu}e & \bar{\nu} D_{\mu} N + \bar{e} D_{\mu} E\\ \bar{N} D_{\mu} \nu + \bar{E} D_{\mu} e & \bar{N} D_{\mu}N + \bar{E} D_{\mu}E\end{pmatrix}.$$ I couldn't diagonalise it . How to get the correct kinetic terms and interaction terms from this matrix? is the matrix correct? I see that diagonal terms look like the kinetic terms and the off diagonal terms being interaction terms. Howto handle such bi doublet fields?
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$\begingroup$ Just write the Lagrangian for the first doublet and add it to the Lagrangian of the second doublet. The fact that they've been packaged into a matrix is just abuse of notation. $\endgroup$– Connor BehanJun 20, 2021 at 11:08
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$\begingroup$ Do you mean to take basically both columns as independent and write like a usual doublet? in that case, v and N fields don't interact, similarly, e and E don't interact. $\endgroup$– MadScientistJun 20, 2021 at 12:19
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$\begingroup$ Adding the two free Lagrangians is how you get the kinetic terms. You indeed need to add all the interactions allowed by symmetry after that. But it looks like this has already been done in equation (6) which has $\nu \phi N$ and $e \phi E$ terms. $\endgroup$– Connor BehanJun 20, 2021 at 22:14
1 Answer
It appears that you are misunderstanding the notation. Before descending into covariant derivatives, let's focus on the left-handed leptons, i.e., those with the left-chiral projector on the left, henceforth suppressed, $$M= \begin{pmatrix}\nu & N\\ e & E\end{pmatrix} , $$ which Ernie tells you precisely transform by conventional SM $SU(2)_L$ on the left, and conjectural $SU(2)_N$ on the right.
How does this field matrix transform? Obviously, $$ M'= e^{i\vec\theta_L\cdot \vec \tau} M e^{-i\vec\theta_N\cdot \vec \tau}. $$ This is the identical formalism you learned about the custodial symmetry of the SM from.
The left SU(2) scrambles the column components as in the SM, and the conjectural SU(2) on the right scrambles the row components, i.e. mixes the known leptons with the conjectural ones denoted by capital letters. That's all.
It is then evident that $$ \operatorname{Tr} {M^\dagger} \gamma_0 (i{\partial\!\!\!\big /}) M=\operatorname{Tr} \begin{pmatrix}\bar\nu & \bar e \\ \bar N & \bar E\end{pmatrix} (i{\partial\!\!\!\big /}) \begin{pmatrix}\nu & N\\ e & E\end{pmatrix} \\ = i ( \bar\nu {\partial\!\!\!\big /}\nu + \bar e {\partial\!\!\!\big /} e+ \bar N{\partial\!\!\!\big /} N + \bar E {\partial\!\!\!\big /} E ) $$ is a singlet under the action of both SU(2)s, on the left and on the right. Observe how the exponentials cancel when you substitute M' for M.
You are now ready to insert the left- and right-acting gauge field (W and X) completions into the covariant derivatives (B Doesn't matter: it's ambidextrous!): invent your own notation; and handle the other multiplets, left-doublet/right-singlet, left-singlet/right-doublet, and singlet.
As Ernie stresses, everything you write should be SU(2)×SU(2) invariant. But, without the above trace, you simply don't have $SU(2)_N$ invariance, only covariance.
