I) Introduction
I.1)
The mathematical structure is quite clear: given a spacetime $M$, and a Lie group $G$ (the gauge group), we can construct the Principal bundle $P^{G}_{M}$. The connection $1$-form in $P^{G}_{M}$ introduces the gauge $\mathcal{A}$ field in physics.
Given the spacetime and the principal bundle, matter fields are introduced in another manifold called the associated bundle $A_{P}$. Inside this space, we craft the gauge covariant derivative:
$$D_{\mu} \phi = \partial_{\mu} \phi+\rho(\mathcal{A}_{s})\phi, \tag{1}$$
where, $\rho$ is the representation $\rho: \mathfrak{g} \to GL(V)$ of a given Lie group $G$; $\mathfrak{g}$ is the lie algebra.
Now, the mathematics is imprecise, for example the $\mathcal{A}$ inside $\rho(\mathcal{A})$ in $(1)$, isn't just the connection, but rather the local connection $1$-form $\mathcal{A}_{s}$. But my point is: given the principal bundle and the associated bundle the connection is "just one".
I.2)
In Weinberg-Salam model, the gauge group is $SU(2)_{L} \times U(1)_{Y}$. This group have this form because we didn't "measure" the right electron neutrino $\nu_{{e}_{R}}$. So, we need a space to acomodate a doublet:
$$\psi = \begin{pmatrix} \nu_{e_{L}} \\ e_{L} \\ \end{pmatrix},$$
together with a singlet:
$$\phi = e_{R}.$$
So, the Weinberg-Salam-Glashow model deals with "triplets":
$$\xi = \begin{pmatrix} \nu_{e_{L}} \\ e_{L} \\ e_{R} \end{pmatrix} \implies \Psi = \begin{pmatrix} \psi \\ \phi \end{pmatrix} \equiv \begin{pmatrix} L \\ R \end{pmatrix}$$
I.3)
The covariant derivative for Weinberg-Salam-Glashow model is:
$$D_{\mu} = \partial_{\mu} + \rho(\mathcal{A}) = \partial_{\mu} - \frac{ig}{2}W^{a}_{\mu}\sigma_{a} - \frac{ig'}{2}B_{\mu} \tag{2}. $$
II) Question
Most texts books says that when you act the covariant derivative in the triplet, $D_{\mu}\Psi$, some internal calculation "breaks" the single $D_{\mu}$ in $(2)$ into two different ones:
$$D_{\mu}\Psi =\Big(\partial_{\mu} - \frac{ig}{2}W^{a}_{\mu}\sigma_{a} - \frac{ig'}{2}B_{\mu}\Big)\Psi \implies \tag{3}$$
$$ D_{\mu}L = \Big(\partial_{\mu} - \frac{ig}{2}W^{a}_{\mu}\sigma_{a} - \frac{ig'}{2}B_{\mu}\Big)L \tag{4}$$
$$ D_{\mu}R = \Big(\partial_{\mu} - \frac{ig'}{2}B_{\mu}\Big)R \tag{5}$$
My question is: what is the calculation that breakes $(3)$ into $(4)$ and $(5)$?
III) My attempt
$$D_{\mu}\Psi = \Biggr(\partial_{\mu} \otimes \mathbb{I}_{3\times3} - \frac{ig}{2}\begin{pmatrix} W^{3}_{\mu} & W^{1}_{\mu} - iW^{2}_{\mu} & 0\\ W^{1}_{\mu} + iW^{2}_{\mu} & W^{3}_{\mu} & 0 \\ 0 & 0 & 0 \end{pmatrix} - \frac{ig'}{2}\begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & 0 \\ 0 & 0 & B_{\mu} \end{pmatrix}\Biggr)\Psi $$
this matrix expression breaks the $D_{\mu}$ into different action for left particles and right particles, but is clear that won't give us the right set of equations.
So, I shall ask again: what is the calculation that breakes $(3)$ into $(4)$ and $(5)$? Please, consider to write a step-by-step calculation answer.
