To prove $\frac{\mathrm{d}^3\vec{p}}{E}$ is Lorentz invariant is to prove
$$\frac{\mathrm{d}^3\vec{p}}{E} = \frac{\mathrm{d}^3\vec{p}'}{E'} \qquad(\mathrm{d}^3\vec{p} := \mathrm{d}p_x \mathrm{d}p_y \mathrm{d}p_z). \tag{1}$$
Suppose reference system $\Sigma'$ is moving at speed $v=\beta$ ($c=1$)...
I have no idea what the expression $\mathrm{d}p^\mu\mathrm{d}p_\mu$ is supposed to mean, but $\mathrm{d}^4p$ is invariant simply because it's a volume element and so it transforms under linear transformations $p\mapsto \Lambda p$ by a factor of $\mathrm{det}(\Lambda)$, and the determinant of Lorentz transformations connected to the identity is 1.
@JakeRose A Lorentz transformation connected to the identity cannot change the branch of solutions to $p^2 = m^2$ a vector lies on, because the condition $p^2 = m^2$ is invariant and so every transformation of $p$ still needs to lie on one of the branches
that is, a vector and its transformation need to be continuously connected via a path that lies on the hyperboloids cut out by $p^2 = m^2$, and this is cannot be for vectors on different branches
According to the answer given here, the diagonal generalized Gell-Mann matrices are not unique. But what exactly is meant by this?
Are they just saying that we can multiply the diagonal matrices by a constant and still have a valid generator of the group since the matrices will still be orthogo...
To prove $\frac{\mathrm{d}^3\vec{p}}{E}$ is Lorentz invariant is to prove
$$\frac{\mathrm{d}^3\vec{p}}{E} = \frac{\mathrm{d}^3\vec{p}'}{E'} \qquad(\mathrm{d}^3\vec{p} := \mathrm{d}p_x \mathrm{d}p_y \mathrm{d}p_z). \tag{1}$$
Suppose reference system $\Sigma'$ is moving at speed $v=\beta$ ($c=1$)...
