Poincare transformations and "three kinds of infinitesimal variations"

  1. The "three kinds of infinitesimal variations" are defined as follows. $$\begin{align} \Delta u(x) &~:=~ u^{\prime}(x^{\prime})-u(x) \qquad\text{ total infinitesimal variation}, \tag{2.2}\cr \delta u(x) &~:=~ u^{\prime}(x)-u(x) \qquad \text{ local/vertical infinitesimal variation},\tag{2.3}\cr \mathrm{d}u(x)&~:=~ u(x^{\prime})- u(x) \qquad\text{ differential/horizontal infinitesimal variation}.\tag{2.4} \end{align}$$ Here the words horizontal and vertical spaces refer to spacetime and $u$-target space, respectively. While the terminology/names & notation vary from author to author, these three infinitesimal variations are often introduced in physics field theory textbooks.

  2. The statement $[\delta,\partial_\mu]=0$ means that vertical infinitesimal variations $\delta$ commute with spacetime-derivatives $\partial_{\mu}\equiv\frac{\partial}{\partial x^{\mu}}$. Be aware that total infinitesimal variations $\Delta$ do not necessarily commute with spacetime-derivatives $\partial_{\mu}$.

$\delta x^\mu$ has one upper index, so you just write down all possible terms which give some object with an upper index. Also $\delta x^\mu$ should only depend on $x^\mu$ and constants, because what else should it depend on? So you could come up with the idea to classify the contributions to $\delta x^\mu$ with respect to the order in $x$.

Zeroth order in $x$ is just a constant. Remember, it should carry an upper index, so it is a constant vector. $\delta x^\mu$ is infinitesimal, so each contribution should be infinitesimal. So you end up with an infinitesimal constant vector $$(\delta x^\mu)^{(0)}=\delta \omega^\mu$$ to zeroth order in $x$.

First order in $x$ is something of the form "constant times x". Since $x$ is a vector, the most general form would be contracting it with a constant tensor. The tensor should be infinitesimal (since $x$ itself is not infinitesimal). To get back an object with one upper index, the index structure should be as follows

$$(\delta x^\mu)^{(1)}=\delta\omega^\mu_\lambda x^\lambda$$

We can lift one of the indices using the metric tensor: $\delta\omega^\mu_\lambda=\delta\omega^{\mu\nu}g_{\nu\lambda}$ and will end up with

$$(\delta x^\mu)^{(1)}=\delta\omega^{\mu\nu}g_{\nu\lambda} x^\lambda$$

In total, up to first order in $x$ we get exactly (2.12).