Is the adjoint representation of $SU(2)$ its triplet representation? Is the triplet representation of $SU(2)$ the same as its adjoint representation? Where the convention for the adjoint representation used is the one used in particle physics, where the structure constants are real and antisymmetric:
$$ \mathrm{ad}(t^b_G)_{ac} = i f^{abc} $$
I was under the impression that is was, but I see two different forms of the generators in the triplet representations used, one being just the real skew symmetric generators of the $SO(3)$ rotation group, which agrees with the adjoint representation, and the other being:
$$ T^1 = \frac{1}{\sqrt{2}} \left(\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{matrix}\right) \quad T^2 = \frac{1}{\sqrt{2}} \left(\begin{matrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0\end{matrix}\right) \quad T^3= \left(\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1\end{matrix}\right)$$
These two representations do not agree, I assume that my idea about the adjoint reperesentation of $SU(2)$ being its triplet representation is wrong, but why?
 A: Your issue is excessive contrast between the spherical basis and the Cartesian basis.
The hermitian generators (spin-1 matrices of quantum mechanics) in the spherical tensor basis
$$ T^1 = \frac{1}{\sqrt{2}} \left(\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{matrix}\right) \quad T^2 = \frac{1}{\sqrt{2}} \left(\begin{matrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0\end{matrix}\right) \quad T^3= \left(\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1\end{matrix}\right)$$
may be unitarily transformed to the hermitian generators $UT^jU^\dagger=L~^j$ in the Cartesian basis. These, then, satisfy the 
very same Lie algebra. 
The equivalence transformation is
$$
U=  \frac{1}{\sqrt{2}} \left(\begin{matrix} -1 & 0 & 1 \\ i & 0 & i \\ 0 & \sqrt{2} & 0\end{matrix}\right) ~,
$$
so that 
$$ L_{1} = i\left(\begin{matrix}0&0&0\\0&0&-1\\0&1&0\end{matrix}\right) , \quad
 L_{2} = i\left(\begin{matrix}0&0&1\\0&0&0\\-1&0&0\end{matrix}\right) , \quad
 L_{3} = i\left(\begin{matrix}0&-1&0\\1&0&0\\0&0&0\end{matrix}\right),
$$
that is, i times the real antisymmetric basis of classical angular momentum.


*

*Observe how U sends the eigenvectors of $L_3$ to the evident eigenvectors of $T_3$. You may find this conversion in good angular momentum books, such as M E Rose's.

