Lorentz Group Generators: Two Methods Members of the Lorentz group obey $$\eta=\Lambda^{T}\eta\Lambda$$ where $\eta=\textrm{diag}(1,-1,-1,-1)$ is the Minkowski metric.
First, in matrix form write $$\Lambda=I+T$$ where $T$ is an infinitesimal generator. The above condition implies to first order that $$T=-\eta T^{T}\eta$$ and so $T$ is anti-symmetric in the space-space and time-time components, and symmetric in the space-time components.
Next, in index form write $$\Lambda^{\mu}_{\,\,\nu}=\delta^{\mu}_{\,\,\nu}+\omega^{\mu}_{\,\,\nu}$$ The component version of the defining equation then implies that, to first order, $$\omega^{\mu}_{\,\,\nu}=-\omega^{\,\,\mu}_{\nu}$$
and so the generators are fully anti-symmetric.
As far as I can see, $\omega^{\mu}_{\,\,\nu}$ is just the index form of $T$. Therefore, I'm a little confused as to why they seem to obey different conditions. Could anybody please help?
 A: I actually got confused by the third comment... The first equation in component form reads
$$ \Lambda^{\mu}{}_{\alpha}\, \eta_{\mu\nu}\, \Lambda^{\nu}{}_{\beta} = \eta_{\alpha\beta} $$
so that with the expansion near the identity:
$$ (\delta^{\mu}{}_{\alpha} + \omega^{\mu}{}_{\alpha})\, \eta_{\mu\nu}\, (\delta^{\nu}{}_{\beta} + \omega^{\nu}{}_{\beta}) + o(\omega) = \eta_{\alpha\beta} + \omega^{\mu}{}_{\alpha}\, \eta_{\mu\beta} +  \eta_{\alpha\nu}\, \omega^{\nu}{}_{\beta} + o(\omega) = \eta_{\alpha\beta} $$
$$ \Longrightarrow\quad  \omega^{\mu}{}_{\alpha}\, \eta_{\mu\beta} = -  \eta_{\alpha\nu}\, \omega^{\nu}{}_{\beta} \quad \Longleftrightarrow\quad  \omega_{\beta\alpha} = - \omega_{\alpha\beta}$$
or also
 $$\omega^{\gamma}{}_{\alpha} = -  \eta_{\alpha\nu}\, \omega^{\nu}{}_{\beta}\, \eta^{\beta\gamma} =: -\omega_{\alpha}{}^{\gamma} \qquad (Eq)$$

Let $M$ be a matrix with component $M^i{}_j$ at line $i$ and column $j$.
 $$\textbf{Inconsistent notation}\qquad (M^T)^i{}_j := M^j{}_i$$
One has to understand the transposition as a "dual" map: $M: E\rightarrow F$ then $M^T: F^* \rightarrow E^*,\ \lambda \mapsto \lambda \circ M $. 
Now let $(\mathbf{e}_1,\cdots, \mathbf{e}_n)$ be a basis of $E$ and $(\boldsymbol{\epsilon}^1,\cdots, \boldsymbol{\epsilon}^n)$ its dual in $E^*$. A vector $\mathbf{x}\in E$, resp. $\boldsymbol{\varphi}\in E^*$ decomposes as
$$\mathbf{x}= x^{\mu}\, \mathbf{e}_{\mu}\quad \text{and}\quad \boldsymbol{\varphi} = \varphi_{\nu}\, \boldsymbol{\epsilon}^{\nu}$$
This motivates
$$ (M^T)_i{}^j := M^j{}_i$$
this is the coefficient of the line $i$ column $j$ of $M^T$ matrix of $M^T: F^* \rightarrow E^*$ w.r.t. a basis of the domain and the target.
It thus seems that what you wrote in comment is true as an equality of maps from $F^* $ to $ E^*$ (I distinguish the domain and target on purpose because if one says $\mathbf{R}^4$ then... Map defined by (Eq))
Other thing to understand: $\eta_{\mu\nu}$ as the matrix component of a map $\eta: E \rightarrow E^*$ w.r.t. a base in the domain and its dual in the target space.
