Matrix Representation of Lorentz Group Generators Let $\Lambda^{\alpha}{}_{\beta}$ denote a generic Lorentz transformation.
Then, an infinitesimal transformation can be written like
$$\Lambda^{\mu}{}_{\nu}  = \delta^{\mu}{}_{\nu} + \omega^{\mu}{}_{\nu} $$
where
$$\omega^{ij} = \epsilon^{ijk}\theta_k$$
$$\omega^{i0} = - \omega^{0i} = \delta^i$$
where $i,j,k$ run from 1 to 3 and $\delta^i$ is a parametre related with boosts. Then, an infinitesimal transformation has a matrix representation
\begin{pmatrix}
1 & -\delta_1 & -\delta_2 & -\delta_3\\
-\delta_1 & 1 & \theta_3 & -\theta_2\\
-\delta_2 & -\theta_3 & 1 & \theta_1\\
-\delta_3 & \theta_2 & -\theta_1 & 1
\end{pmatrix}
However, we can also write
$$\Lambda^{\mu}{}_{\nu}  = \delta^{\mu}{}_{\nu} + i\frac{\omega^{\alpha \beta}}{2}\left(J_{\alpha \beta} \right)^{\mu}{}_{\nu} $$
where $J_{\alpha \beta}$ are the generators of the group. I want to prove that $J_{01}$ is of the form
\begin{pmatrix}
0 & -i & 0 & 0 \\
-i  & 0  & 0  & 0 \\
0  & 0  & 0  & 0 \\
0  & 0  & 0  & 0 
\end{pmatrix}
My problem is in understanding the notation in
$$\Lambda^{\mu}{}_{\nu}  = \delta^{\mu}{}_{\nu} + i\frac{\omega^{\alpha \beta}}{2}\left(J_{\alpha \beta} \right)^{\mu}{}_{\nu} $$
For example, I tried to compute $\left(J_{01} \right)^{0}{}_{1}$ by doing
$$\Lambda^{0}{}_{1}  = \delta^{0}{}_{1} + i\frac{\omega^{01}}{2}\left(J_{01} \right)^{0}{}_{1} $$
$$\Leftrightarrow - \delta_1 = 0 -i \frac{\delta_1}{2}\left(J_{01} \right)^{0}{}_{1}$$
which yields $\left(J_{01} \right)^{0}{}_{1} = -2i$, which is not correct. What am I doing wrong?
 A: In the following text, we use $\eta_{\mu\nu} = \text{diag}(-1,+1,+1,+1)$.
For an infinitesimal homogeneous Lorentz transformation, we have
$$ {\omega^\mu}_\nu = \begin{pmatrix} 0 & \zeta_1 & \zeta_2 & \zeta_3 \\ \zeta_1 & 0 & -\theta_3 & \theta_2 \\ \zeta_2 & \theta_3 & 0 & -\theta_1 \\ \zeta_3 & -\theta_2 & \theta_1 & 0  \end{pmatrix}, $$
and
$$\begin{aligned}
{\Lambda^\mu}_\nu &= {\delta^\mu}_\nu + {\omega^\mu}_\nu \\
&= {\delta^\mu}_\nu + \eta^{\rho\mu}\omega_{\rho\nu} \\
&= {\delta^\mu}_\nu + \eta^{\rho\mu}\omega_{\rho\sigma}{\delta^\sigma}_\nu \\
&= {\delta^\mu}_\nu + \frac{1}{2}\omega_{\rho\sigma}(\eta^{\rho\mu}{\delta^\sigma}_\nu - \eta^{\sigma\mu}{\delta^\rho}_\nu) \\
&= {\delta^\mu}_\nu + \frac{i}{2}\omega_{\rho\sigma}{(S_V^{\rho\sigma})^\mu}_\nu \\
\end{aligned}$$
where the vector representation of the generators are defined as
$$ {(S_V^{\rho\sigma})^\mu}_\nu \equiv -i({\eta}^{\rho\mu}{\delta^\sigma}_\nu - {\eta}^{\sigma\mu}{\delta^\rho}_\nu). $$
Note that $\omega_{\rho\sigma}$ and $S_V^{\rho\sigma}$ are antisymmetric in the indices $(\rho\sigma)$, and
$$ {(S_V^{0i})}^\dagger = - S_V^{0i}, \quad {(S_V^{ij})}^\dagger = S_V^{ij}. $$

If we define
$$ \boldsymbol\theta \equiv (\theta_1, \theta_2, \theta_3), \quad \boldsymbol\zeta \equiv (\zeta_1, \zeta_2, \zeta_3), $$
$$ \mathbf{J} \equiv (S_V^{23},S_V^{31},S_V^{12}),\quad \mathbf{K} \equiv (S_V^{01},S_V^{02},S_V^{03}), $$
then
$$ {\Lambda^\mu}_\nu = {\delta^\mu}_\nu - i{{(\boldsymbol\theta \cdot \mathbf{J} + \boldsymbol\zeta \cdot \mathbf{K})}^\mu}_\nu. $$
The explicit expressions of the matrices $J_i$ and $K_i$ are
$$
J_1 = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -i \\ 0 & 0 & i & 0 \end{pmatrix}, \quad
J_2 = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & i \\ 0 & 0 & 0 & 0 \\ 0 & -i & 0 & 0 \end{pmatrix}, \quad
J_3 = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -i & 0 \\ 0 & i & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}
$$
$$
K_1 = \begin{pmatrix} 0 & i & 0 & 0 \\ i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}, \quad
K_2 = \begin{pmatrix} 0 & 0 & i & 0 \\ 0 & 0 & 0 & 0 \\ i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}, \quad
K_3 = \begin{pmatrix} 0 & 0 & 0 & i \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ i & 0 & 0 & 0 \end{pmatrix}
$$
A: Thanks to @Charlie and @Cosmas Zachos I was able to find the correct answer.
It simply suffices to develop the sum
$$\frac{\omega^{\alpha \beta}}{2}\left(J_{\alpha \beta} \right)^{\mu}{}_{\nu} = -\delta_1 \left(J_{01} \right)^{\mu}{}_{\nu}  - \delta_2 \left(J_{02} \right)^{\mu}{}_{\nu}  - \delta_3 \left(J_{03} \right)^{\mu}{}_{\nu} + \theta_3\left(J_{12} \right)^{\mu}{}_{\nu} - \theta_2 \left(J_{13} \right)^{\mu}{}_{\nu}  + \theta_1\left(J_{23} \right)^{\mu}{}_{\nu} $$
where I used
$$\left(J_{\alpha \beta} \right)^{\mu}{}_{\nu} = - \left(J_{\beta \alpha }  \right)^{\mu}{}_{\nu} $$
and the other properties mentioned above.
