# 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).