Lorentz boost as a conformal transformation

Question

The conformal group is the set of transformation that preserve angles. With this idea, then a conformal transformation is such that $x\rightarrow x^\prime$ and $$ g^\prime_{\mu\nu}(x^\prime) = \Omega(x)g_{\mu\nu}(x),\quad \Omega: \mathbb{R}^d\mapsto\mathbb{R} $$ with $d$ the spacetime dimension.

It can be proven that the whole set of conformal transformations is the sum of dilations, translations, the so-called special conformal transformations, and Lorentz transformations. These last ones comprise rotations and boosts.

Let's say our metric $g_{\mu\nu}$ is Minkowski's. We know that a boost produces the contraction of lengths in the direction of the boost. Let's consider a couple of vectors with same origin in the $yz$-plane and a boost in $y$-axis. Then, we would see a reduction on the $y$-components of these vectors while the $z$-components are preserved. This clearly modifies the angle between the vectors.

My question is: how is then the whole Lorentz group (not just the subgroup of rotations) part of the conformal group?

Answer 1

A four vector which lies in the $yz$ plane as you describe has $t$ and $x$ components of zero before you act with the boost. If you take two such vectors $\textbf{u}$ and $\textbf{v}$, the verbose way of writing the angle between them is \begin{equation} \theta = \mathrm{arccos} \left [ \frac{u^\mu g_{\mu\nu} v^\nu}{\sqrt{(u^\alpha g_{\alpha\beta} u^\beta)(v^\gamma g_{\gamma\delta} v^\delta)}} \right ]. \end{equation} This only involves Lorentz invariant index contractions so it must be preserved. In fact the numerator and denominator are preserved separately making it trivial that the Lorentz group is contained in the conformal group.

When you talk about the angle changing, you are talking about the angle between three vectors. I.e. a spatial angle computed with respect to \begin{equation} \begin{pmatrix} 0 & 0 \\ 0 & \delta_{ij} \end{pmatrix}, \end{equation} not the Minkowski metric used to define Lorentz transformations.