- ig' U Y_{\mu}$$

Here, $W_{\mu}$ and $Y_{\mu}$ are the $SU(2)_L$ and the $U(1)_Y$ gauge fields given by:

$$W_{\mu} = \frac{{\vec{W}}_{\mu} {\vec{\tau}}}{2}$$

$$Y_{\mu} = \frac{B_{\mu} {\tau}_3}{2}$$

where $W_{\mu}=(W^1_{\mu},W^2_{\mu},W^3_{\mu})$ represents the triplet of $SU(2)_L$ gauge fields.

As usual, the covariant field strength tensors are defined as

$$\begin{aligned}
F_{\mu\nu}(x) & = & {\partial}_{\mu} W_{\nu}(x) - {\partial}_{\nu}
W_{\mu}(x) + ig [W_{\mu}(x),W_{\nu}(x)]  \\
B_{\mu\nu}(x) & = & {\partial}_{\mu} B_{\nu}(x) - {\partial}_{\nu}
B_{\mu}(x)\end{aligned}$$

The transformation properties under the $SU(2)_L \times U(1)_Y$ gauge transformations are

$$\begin{aligned}
ig W'_{\mu} & = & g_L (x) ig W_{\mu} g_L^{\dagger} (x) +  g_L (x)
{\partial}_{\mu} g_L^{\dagger} (x)  \\
ig' Y'_{\mu} & = & ig' Y_{\mu} + g_Y (x) {\partial}_{\mu}
g_Y^{\dagger} (x) \\
U' & = & g_L (x) U {g_Y}^{\dagger} (x)\end{aligned}$$

where

$$\begin{aligned}
g_L(x) & = & e^{i {\theta}_k (x) {\tau}^{k} /2}    \\
g_Y(x) & = & e^{i \beta(x).{\tau}^3 /2}\end{aligned}$$

Note that, as expected, the $GB$ fields, ${\pi}_i$, transform non-linearly .

In order to construct the chiral lagrangian to order $O(p^4)$ we define, following Longhitano in [@Long], the quantities $T$, $V_{\mu}$ and ${\cal D}_{\mu} O(x)$:

$$\begin{aligned}
T & = & U {\tau}^3 U^{\dagger}  \\
V_{\mu} & = & D_{\mu}U U^{\dagger}   \\
{\cal D}_{\mu} O(x) & = & {\partial}_{\mu} O(x) + 