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I have been introduced to electroweak theory in lectures and I wanted to check I understand the notation for the doublets, triplets etc.

Take the first generation lepton left handed doublet $L$ and we have the kinetic term $$i\bar{L} \cancel{D} L, \tag{1}$$ where $L = (\nu_L, e_L)^{T}$.

The covariant derivative is a 2 by 2 complex matrix with the connection corresponding to the gauge group $SU(2)\times U(1)$. This part of the kinetic term acts on the doublet.

But then it is contracted with the gamma matrices (from the slash). The notation is compact but somewhat confusing to me, it is not clear what the gamma matrix is acting on. Initially I interpreted the components of the doublet as 2 Weyl spinors since we are considering just the left-handed component of the spinor, of course however that is only separable in the Weyl representation. Once I realised that by $\nu_L$ we really mean more generally (in any gamma representation) $\nu_L = \frac{1}{2}(1+\gamma^5)\nu$ then the components of the doublet is a Dirac bispinor each. I have assumed that the gamma matrices, contracted with the covariant derivative, acts component-wise on the doublet.

Perhaps a more illustrative but ugly notation to show what contracts with what would be

$$i (\bar{\nu}_L \gamma^{\mu}, \bar{e}_L \gamma^{\mu} ) D_{\mu} (\nu_L, e_L)^{T}. \tag{2}$$

Is this the correct interpretation of this notation?

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    $\begingroup$ Yes, nobody can stop you from using your ugly notation; you already used it before, as the bar entails $\gamma_0$. In this pure left-handed term all spinors are left-handers, including the conjugate which should have been written as $\overline{\nu_L}$ instead, etc, so they and their gammas factor out of the problem, the 2x2 structure you are focussing on. It is a tensor product. Personally, I would have slipped the Ls to the Right of the gammas, for clarity. $\endgroup$ Mar 7, 2021 at 17:17

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TL:DR: Well, it should be stressed that the gauge-covariant derivative $D_{\mu}$ is a $2\times 2$ matrix wrt. the doublet-indices in OP's eq. (2).

It is probably most clarifying/pedagogical to simply write out Lorentz indices $\mu\in\{0,1,2,3\}$, Dirac indices $\alpha,\beta\in\{1,2,3,4\}$, $SU(2)$-indices $a\in\{1,2,3\}$, and doublet indices $j,k\in\{1,2\}$, although admittedly not very elegant. Then the kinetic term for the left-handed fermions reads

$$i\bar{L} \cancel{D} L~=~i\bar{L} \gamma^{\mu}D_{\mu} L~=~i\bar{L}^j \gamma^{\mu} D_{\mu}^{jk} L^k~=~i\bar{L}^j_{\alpha} (\gamma^{\mu})_{\alpha\beta} D_{\mu}^{jk} L^k_{\beta},\tag{A}$$

where the electroweak gauge-covariant derivative is $$ D_{\mu}^{jk}~=~\delta^{jk}\partial_{\mu}-ig W_{\mu}^a (\tau_a)^{jk}-ig^{\prime} B_{\mu}\delta^{jk}Y. \tag{B}$$ (In the last formula, be aware of conventional factors of $\frac{1}{2}$ depending on which author you ask.)

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