Let us consider the Minkowski space $(\mathbb{R}^{4},\eta)$ and the mass shell $H\_{m}$, $m\ge 0$, given by:
\begin{eqnarray}
H\_{m}:=\{x=(x\_{0},x\_{1},x\_{2},x\_{3}) \in \mathbb{R}^{4}: \hspace{0.1cm} x\cdot \tilde{x} = m^{2}, x\_{0}>0\} \tag{1}\label{1}
\end{eqnarray}
where $\tilde{x}$ is defined by $x=(x\_{0},x\_{1},x\_{2},x\_{3}) \mapsto \tilde{x} := (x\_{0},-x\_{1},-x\_{2},-x\_{3})$ and $x\cdot \tilde{x}$ is the usual Lorentz inner product:
$$x \cdot \tilde{x} = \eta(x,x).$$
Further, define $j\_{m}: \mathbb{R}^{4}\to \mathbb{R}^{3}$ (if $m=0$, take $\mathbb{R}^{3}\setminus \{0\}$ instead) as the homeomorphism $x=(x\_{0},x\_{1},x\_{2},x\_{3}) \mapsto (x\_{1},x\_{2},x\_{3})$. The set function:
\begin{eqnarray}
\mu\_{m}(E) := \int\_{j\_{m}(E)} \frac{d^{3}x}{\sqrt{m^{2}+|x|^{2}}} \tag{2}\label{2}
\end{eqnarray}
defines a measure on $H\_{m}$.


In the literature, $\mu\_{m}$ is called Lorentz invariant measure because it is said to be invariant under Lorentz transformations on $\mathcal{L}\_{+}^{\uparrow}$.


I have some really basic questions about this settings, but which are not explicitly stated anywhere I know.


**Q1:** The "Lorentz inner product" $x\cdot\tilde{x}$ might be misleading since $x\cdot \tilde{x}$ is not positive-definite, so it does not define an inner product in mathematical terms. Does it play any role in defining the topology on $\mathbb{R}^{4}$, or is $\mathbb{R}^{4}$ equipped with its usual Euclidean topology? Note that the topology is important because $j\_{m}$ is assumed to be an homeomorphism and, also, because $\mu\_{m}$ is a measure on $H\_{m}$ which, I assume, becomes a measure space when equipped with the Borel $\sigma$-algebra inherited from $\mathbb{R}^{4}$.


**Q2:** I know that $H\_{m}$ is invariant under $\mathcal{L}\_{+}^{\uparrow}$ because, if $\Lambda \in \mathcal{L}\_{+}^{\uparrow}$ we have: (a) $(\Lambda x)\_{0} > 0$ whenever $x\_{0}>0$, and (b) $\Lambda x \cdot \tilde{\Lambda x} = x\cdot \tilde{x} = m^{2}$, since it is a Lorentz transformation. But what is the precise meaning of $\mu\_{m}$ being Lorentz invariant? Is it as usually stated in ergodic theory, i.e. $\mu\_{m}(\Lambda^{-1}E) = \mu\_{m}(E)$ for every measurable $E$?



Topology on Minkowski space $\mathbb{R}^{4}$ and Lorentz invariant measure
**Q1** The topology on $\mathbb{R}^4$ is the usual one. This is the general case for Lorentzian geometry: the topology is the one defined by the charts in your atlas.


**Q2** Given a fixed Lorentz transformation $\Lambda$, it sends $H\_m\to H\_m$ as you observed. It is a linear transformation of $\mathbb{R}^4$, and hence restricts to a diffeomorphism of $H\_m$ to itself. Invariance of $\mu\_m$ is simply that it agrees with its pushforward by $\Lambda$, which is the same as what you wrote.


**Remark**: when $m > 0$, your $\mu\_m$ is the volume measure associated to the induced Riemannian metric on $H\_m$, and so invariance can also be seen by the fact that $\Lambda$ is an isometry of $\mathbb{R}^4$ and hence restricts to an isometry of $H\_m$.


