# Matrix Representations of the Poincare Group

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## Main Question or Discussion Point

I'm trying to 'see' what the generators of the Poincare Group are. From what I understand, it has 10 generators. 6 are the Lorentz generators for rotations/boosts, and 4 correspond to translations in ℝ1,3 since PoincareGroup = ℝ1,3 ⋊ SO(1,3).

The 6 Lorentz generators are easy enough to find in the literature. They are:

I cannot find the ℝ1,3 generators explicitly stated anywhere. My naive guess is that since the other four generators correspond to translations in ℝ1,3, we get the other 4 generators by exponentiating the 4 translation matrices for ℝ1,3.

Is that correct?

## Answers and Replies

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A. Neumaier
2019 Award
To get the Poincare group you need to take a semidirect product with the translation group. This means that you get a 5-dimensional representation by 5x5 matrices ##\pmatrix{A & t \cr 0 & 1}##, where ##A## is a Lorentz transformation and ##t## a translation vector. The relevant orbit of this matrix action is the set of 5D vectors
##\pmatrix{x \cr 1}##, where ##x## is a point in Minkowski space.

PeroK, dextercioby, vanhees71 and 2 others
Thanks Prof. Neumaier, so would the generators then be the six rotations/boosts

$$R_x = \begin{pmatrix} 1 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0\\ 0 & 0 & cos \theta & -sin \theta & 0\\ 0 & 0 & sin \theta & cos \theta & 0\\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} ~ R_y = \begin{pmatrix} 1 & 0 & 0 & 0 & 0\\ 0 & cos\theta & 0 & sin\theta & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & -sin\theta & 0 & cos\theta & 0\\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} ~ R_z = \begin{pmatrix} 1 & 0 & 0 & 0 & 0\\ 0 & cos\theta & -sin\theta & 0 & 0\\ 0 & sin\theta & cos\theta & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1 \end{pmatrix}$$
$$B_x = \begin{pmatrix} cosh \theta & sinh \theta & 0 & 0 & 0\\ sinh \theta & cosh \theta & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} ~ B_y = \begin{pmatrix} cosh \theta & 0 & sinh \theta & 0 & 0\\ 0 & 1 & 0 & 0 & 0\\ sinh \theta & 0 & cosh \theta & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} ~ B_z = \begin{pmatrix} cosh \theta & 0 & 0 & sinh \theta & 0\\ 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ sinh \theta & 0 & 0 & cosh \theta & 0\\ 0 & 0 & 0 & 0 & 1 \end{pmatrix}$$

Plus the four the translations represented like this?

$$T_t = \begin{pmatrix}1 & 0 & 0 & 0 & -c t\\ 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\end{pmatrix} T_x = \begin{pmatrix} 1 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & x\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} ~ T_y = \begin{pmatrix} 1 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & y\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} ~ T_z = \begin{pmatrix} 1 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & z\\ 0 & 0 & 0 & 0 & 1 \end{pmatrix}$$

Last edited:
A. Neumaier
2019 Award
would the generators then be the six rotations/boosts
Plus the four the translations represented like this?
Generators have no free parameters left but each of your matrices contains such a parameter.

What you wrote down is not describing generators but the 1-parameter groups they are generating. Taking the derivatives at zero gives the generators.

vanhees71 and cuallito
Thanks professor, like a true scientist, I'm always happy to know when I'm wrong