Generators of Lie groups in physics I asked this question in physics SE here. But I was not satisfied.
As we solving something related to symmetry transformations, we need Lie groups. Also Lie algebra is very important to generate those transformations.
E.g. the generators of $SO(3)$:
$$\begin{pmatrix}0&i&0\\-i&0&0\\0&0&0\end{pmatrix},
\begin{pmatrix}0&0&i\\0&0&0\\-i&0&0\end{pmatrix},
\begin{pmatrix}0&0&0\\0&0&i\\0&-i&0\end{pmatrix}.
$$
Why there is an i in the front.. The thing it is not so wrong if we just consider the Lie algebra as a vector space. But when we use the Lie bracket we will also get those real ones.. But the Lie algebra is three real dimensional.. Now, it is 6.
 A: Just a reminder of facts concerning physics usage. 
Because quantum physics relies on unitary operators, the symmetry group elements are unitary matrices, so then exponentials of iJ where the Js are Hermitean operators, the standard convention for operators in physics. Commutators of such Hermitean Js are antihermitean, so not in the Lie algebra: in physics structure constants are normally pure imaginary, so they can multiply hermitean generators. 
This is all there is to it, as @Dietrich Burde instructs you. The highly unconventional adjoint rep generators J you wrote down conform to this convention, but, of course, as you might be familiar, physics normally uses an equivalent representation for them, this one. Multiplying them by i and exponentiating yields a unitary group element in both cases but real orthogonal group elements only for your pure imaginary (Hermitean) basis. I gather you took the real antisymmetric generators of classical rotations and physics and multiplied them by i to make them Hermitean, assuming somehow physicsists use those, which they rarely would.
