I am studying special relativity. I am trying to understand how to define the group elements of the Lorentz group, $\text{O}(1,3)$. I understand from [1] that the Lorentz group is has (at least 3) subgroups. These are $\text{O}^+(1,3)$, $\text{SO}(1,3)$ and $\text{SO}^+(1,3)$. From [1],

Every element in $\text{O}(1,3)$ can be written as the semidirect product of a proper, orthochronous transformation [i.e., an element of $\text{SO}^+(1,3)$] and an element of the discrete group $\left\{1, P, T, PT\right\}$, where P and T are the parity and time reversal operators [i.e., $P = \textrm{diag}(1, −1, −1, −1)$ and $T = \textrm{diag}(−1, 1, 1, 1)$].

This quote makes clear how $\text{O}(1,3)$ and $\text{SO}^+(1,3)$ are related to each other. This fact is referred to by @arctic tern in his/her answer in [2]. However, this quote does not help me define $\text{O}(1,3)$.

To be clear what I am looking for is something like a set theoretic definition of $\text{O}(1,3)$. Something like what I see in the example of Lie groups found in [3].

For example, I believe that $\text{O}(1,3)\subset \text{GL}(4, \mathbb{R})$. Therefore, I should be able to define the Lorentz group with some additional predicates.

$$\text{O}(1,3) \equiv \left\{M\in \text{GL}(4, \mathbb{R}) \mid \operatorname{Predicate 1}, \operatorname{Predicate 2}, \ldots \operatorname{Predicate n}\right\}.$$

The Lorentz group is given as an orthogonal group. However, though the inverse of the Lorentz boosts are equal to the transpose of the Lorentz boosts, it does not appear that the the inverse of the Lorentz rotations are equal to the transpose of the Lorentz rotations.


How is the Lorentz group, $\text{O}(1,3)$, defined using set theoretic notation?


[1] https://en.wikipedia.org/wiki/Lorentz_transformation#boost

[2] Difference between the Lorentz group and the restricted Lorentz group

[3] https://en.wikipedia.org/wiki/Lie_group


1 Answer 1


Write $\eta = \left[ \begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right]$. Then

$$O(1, 3) = \{ M \in GL_4(\mathbb{R}) : M\,\eta\,M^T = \eta \}.$$

This condition is equivalent to the condition that $M$ preserves the Minkowski / Lorentzian inner product on 4-vector $v$ and 4-vector $w$ as

$$(v, w) = -v_0 \,w_0 + v_1\, w_1 + v_2 \,w_2 + v_3\, w_3.$$

In other words, $(M\,v, M\,w) = (v, w)$, which is the condition used by the Wikipedia article. This is because $(v, w)$ can be written $v \cdot \eta\,w$ where $\cdot$ is the usual dot product. Thus, $$(Mv, Mw) = Mv \cdot \eta\,M\,w = v \cdot M^T \,\eta\,M\,w.$$ Setting this equal to $(v, w) = v \cdot \eta\,w$ for all $v, w$ gives the desired result (after taking one last transpose).

This is why you don't get that the inverse is the transpose (which would define the ordinary orthogonal group $O(4)$). Instead, we have $M^{-1} = \eta\,M^T \,\eta^{-1}$. Said another way, there is a Lorentzian transpose we should be taking here instead.

One last note. There is a sign ambiguity. Specifically, we could replace $\eta$ with $-\eta$ in the definition of $\eta$ and still get the same Lorentz group.

  • $\begingroup$ ok. I worked through everything. I have one last comment. Would it not be better to add another predicate to your list. Namely, that $\left[ \det(M) = +1 \lor \det(M) = -1\right] ?$ $\endgroup$ Jul 25, 2022 at 21:59
  • 1
    $\begingroup$ @Michael: that's automatic. Taking the determinant of $M \eta M^T = \eta$ gives $\det(M)^2 = 1$. $\endgroup$ Jul 25, 2022 at 22:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .