Finding the form of an Infinitesimal Lorentz matrix In the context of Lie groups, when looking for the form of the Lorentz generators we expand a general Lorentz matrix using some infinitesimal parameter $\epsilon$ such that $\Lambda = \mathbb{1} + \epsilon X$.
The form of $X$ is restricted by the metric preserving equation ie $\Lambda^T \eta \Lambda = \eta$. We obtain that $$\mathbb{1} + \epsilon ( X^T + X ) + \epsilon^2 X^TX= \mathbb{1}.$$ To leading over we want to infinitseimal contriburion to vanish so we have $X=-X^T$.
The bit I don't understand is that once we impose this condition the leading infinitesimal contribution is now $\epsilon^2 X^TX$. Why do we not require $X^TX=0$?
 A: $\Lambda=1+\epsilon X$, with $X$ a generator, is only correct up to order $\epsilon$. You need a slightly more rigorous definition of the Lorentz generators to completely justify this. Suppose $q(t)$ is a smooth path through the space of Lorentz transformations that passes through $\mathbb{1}$. Specifically, for each $t\in\mathbb{R}$, $q(t)$ is a Lorentz transformation, and $q(0)=\mathbb{1}$. Then $q'(0)$ is a tangent vector to the identity $\mathbb{1}$. These tangent vectors are essentially "infinitesimal" Lorentz transformations, and any generator can be written in the form $q'(0)$ for some function $q$.
I'm going to use explicit index notation to avoid ambiguities (statements like $\Lambda^T\eta\Lambda=\eta$ or $X=-X^T$ depend on whether indices are up or down). Now, since $q(t)$ is a Lorentz transformation, we know that
$$\eta_{\mu\nu} \big[q(t)\big]^{\mu\rho}\big[q(t)\big]^{\nu\sigma}=\eta^{\rho\sigma}$$
$$\implies 0=\frac{d}{dt}\big(\eta_{\mu\nu} q^{\mu\rho}q^{\nu\sigma}\big)=\eta_{\mu\nu}q^{\nu\sigma}\frac{d}{dt}\big(q^{\mu\rho}\big)+\eta_{\mu\nu} q^{\mu\rho} \frac{d}{dt}\big(q^{\nu\sigma}\big).$$
For $t=0$, using $q(0)=1$, we get
$$0=\eta_{\mu\nu}\delta^{\nu\sigma}\frac{d}{dt}\big(q^{\mu\rho}\big)|_{t=0}+\eta_{\mu\nu} \delta^{\mu\rho} \frac{d}{dt}\big(q^{\nu\sigma}\big)|_{t=0}={\delta_\mu}^\sigma\frac{d}{dt}\big(q^{\mu\rho}\big)|_{t=0}+{\delta_\nu}^\rho \frac{d}{dt}\big(q^{\nu\sigma}\big)|_{t=0}$$
$$=\frac{d}{dt}\big(q^{\sigma\rho}\big)|_{t=0}+\frac{d}{dt}\big(q^{\rho\sigma}\big)|_{t=0}.$$
Let $X^{\mu\nu}=\frac{d}{dt}\big(q^{\mu\nu}\big)|_{t=0}$. Then $X$ is a tangent vector to the identity, and a generator of the Lorentz group by definition. We have shown
$$X^{\sigma\rho}=-X^{\rho\sigma}.$$
Edit Alternatively, if you're willing to accept that $e^{\epsilon X}$ is a Lorentz transformation for any generator $X$, then you can expand
$$e^{\epsilon X}=1+\epsilon X+\frac{\epsilon^2}{2}X^2+\frac{\epsilon^3}{3!}X^3+...$$
and solve $(e^{\epsilon X})^T \eta e^{\epsilon X}=\eta$ and match the coefficients of $\epsilon^n$ on each side (all of the equations give the same requirement: $X=-X^T$).
