Gauge SU(2) with real triplet I have come across a model of gauge $SU(2)$ with a real triplet. 
The covariant derivative for $SU(2)$ complex doublet is written as $$D_\mu=\partial_\mu-igT^aA^a_\mu$$ where $T^a$ are generators of the group in fundamental representation. Which intuitively tell me for a triplet the covariant derivative should be the same except $T^a$ now are the generators in adjoint representation i.e. a 3x3 matrix. But i have come acros this model of gauge $SU(2)$ with a real triplet which writes kinetic part of the lagrangian as $$(\partial_\mu\phi-gA_\mu \times\phi)^2$$ where phi is a real triplet. Please some one explain me this derivation\origin new covariant derivative(any reference will also be helpful).
 A: In general, the covariant derivative acting upon a field in a representation $\rho : G \to V_\rho$, is given by
$$ D = \mathrm{d} - \mathrm{i} g \rho(A) \wedge $$
or, in coordinates/generators, indeed
$$ D_\mu = \partial_\mu - \mathrm{i} g A_\mu^a \rho(T^a)$$
where $\rho(T^a)$ are the generators of $G$ in the chosen representation.
Now, for the triplet of $\mathrm{SU}(2)$ (the rep labeled by $j = 1$), we are in the fundamental representation of of $\mathrm{SO}(3)$, since $\mathrm{SU}(2)$ is its double cover. The representation of the Lie algebra consists of all skew-symmetric $3\times 3$ matrices (since these are the generators of ordinary rotations). Let ($\rho$ suppressed in the following)
$$ T^1 = \left(\begin{matrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0\end{matrix}\right) \quad T^2 = \left(\begin{matrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0\end{matrix}\right) \quad T^3= \left(\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0\end{matrix}\right)$$
and just expand $\phi = \phi^i e_i$ with $e_b$ as the standard basis of $\mathbb{R}^3$ to see how $A_\mu^a T^a\phi$ becomes the cross product through the action of the $T^a_{ij}$ on the $e_i$.

Another way to see this without the tedious matrix computations is recalling that the vector product is defined by
$$ (a \times b)^i = \epsilon^{ijk} a_j b_k$$
and that the adjoint representation of any Lie group $G$ upon its Lie algebra is induced by1
$$ \mathrm{ad}(T^a)_{ij} = f^{aij}$$
As it happens, the triplet of $\mathrm{SU}(2)$ is its adjoint (since there are three generators), and the structure constants of $\mathrm{SU}(2)$ are just the Levi-Civita symbols $\epsilon^{ijk}$. Plugging this into $D_\mu \phi$ yields the vector product $(A_\mu \times \phi)$-term.

1This is a consequence of the $T^a$ themselves being naturally elements of the Lie algebra, the structure constants being given by $[T^a,T^b] = f^{abc}T^c$ and the Baker-Campbell-Hausdorff formula applied to the adjoint action of $g \in G$ on $T^a$: $\mathrm{Ad}(g)(T^a) = gT^ag^{-1}$. 
