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On this slide

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it just says that $\mathcal{P}$ and $\mathcal{G}$ are the Poincoré and Galilei groups, but I do not understand what they are made of.

What does $\mathbb{R}^{1,3}$ mean?

Why does $\mathcal{G}$ have the extra $(\mathbb{R}^1 \times \mathbb{R}^3)$?

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$\mathbf{R}^{1,3}$ is Minkowski spacetime. It's a 4-dimensional real vector space with inner product \begin{align}x\cdot y = - x_0 y_0 + x_1 y_1 + x_2 y_2 + x_3 y_3.\end{align}

Note that in each case the decomposition is given in the form \begin{align} \text{Group = (Boosts and Rotations) } \times \text{(Spacetime translations)}. \end{align} (Strictly speaking, this should be a semidirect product rather than a direct product, but feel free to ignore this detail.)

For the case of the Poincaré group $\mathcal{P}$, the Lorentz subgroup $SO(1,3)$ includes both rotations and boosts — boosts by themselves aren't a subgroup. But for the Galilei group, boosts (parameterized by a velocity vector in $\mathbf{R}^3$) and rotations ($SO(3)$) are independent.

The "extra" $(\mathbf{R}^1 \times \mathbf{R}^3)$ appearing in $\mathcal{G}$ is the group of non-relativistic spacetime translations, the analogue of $\mathbf{R}^{1,3}$ for $\mathcal{P}$. Again, for $\mathcal{G}$, space and time translations decompose.

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  • $\begingroup$ so what does $\mathbb{R}^{1,3}$ mean? Translations in space(3)-time(1)? How is that different, in its represenatoin, from $(\mathbf{R}^1 \times \mathbf{R}^3)$? $\endgroup$ – SuperCiocia Dec 15 '18 at 12:38
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    $\begingroup$ $\mathbb{R}^{1,3}$ is the vector space endowed with the Minkowski inner product that d_b wrote out; if you evaluate it for $x = y$, you simply get the Minkowski metric. Acting on it is the group SO(1,3) which preserves the Minkowski inner product — it is the analog of SO(3) which preserves the usual cartesian inner product (as well as the orientation). $\endgroup$ – Max Lein Dec 19 '18 at 1:12

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