6
$\begingroup$

I am currently studying the proper orthochronous Lorentz group $\text{SO}^+(1,3)$ and I have run into some confusion. I started by defining the Poincare group to be the set of all transformations satisfying $$ g = \Lambda^T g \Lambda $$ and defined its Lie algebra to be the set of all matrices $X$ such that $e^{tX} \in \text{O}(1,3)$ for all $t \in \mathbb{R}$. Plugging in $\Lambda = e^{tX}$ into the above equation, differentiating with respect to $t$ and taking $t$ to zero, it is easy to show that the following must be true $$ X^Tg + g X = 0 $$ This relation shows that $X$ has six independent components, seeming to prove that $\mathfrak{o}(1,3)$ has dimension equal to six. But it is well known that the dimension of the Poincare algebra is ten. Where are the other four elements of the Lie algebra? This proof seems to have generated the Lie algebra $\mathfrak{so}(1,3)$, but I have never used the fact that $\text{det} \, \Lambda = 1$ or $\Lambda^0 \,_0 \geq 1$. My thoughts are that this has something to do with the surjectivity of the exponential map (or lack thereof). I know that the exponential map is surjective onto $\text{SO}^+(1,3)$, but what about $\text{O}(1,3)$? Furthermore, I have seen it claimed before that the lie algebras $\mathfrak{so}^+(1,3)$ and $\mathfrak{o}(1,3)$ are identical (An Introduction to Tensors and Group Theory for Physicists by Jeevanjee) How can this be when they clearly have different dimension?

$\endgroup$
6
$\begingroup$

Unfortunately you can't just define the Poincare group to be that, because in the standard treatment it is defined a little bit differently. What you defined is actually the Lorentz group. The Poincare group contains translations as well.

The Lorentz group $O(1,3)$ is the group of all $\Lambda \in GL(4, \mathbb{R})$ such that

$$\Lambda^T \eta \Lambda = \eta,$$

with $\eta = \operatorname{diag}(-1,1,1,1)$. This can be seen as the group of all "changes of orthonormal frames in spacetime".

Remember that one orthonormal reference frame is a set of vectors $\{e_\mu\}$ such that $g(e_\mu,e_\nu)=\eta_{\mu\nu}$. In that sense, given two such frames, the change of frame that takes components in one of them to components in the other is given by these elements.

With this, the proper Lorentz group is $SO(1,3)$ which basically means that you pick all elements of $O(1,3)$ with determinant $+1$. The orthocronous part just means that if $\Lambda \in O(1,3)$ has $\Lambda^0_0 > 0$ then it preserves the sense of time-like vectors.

Some authors seems to include "by default" the orthocronous requirement in the group $SO(1,3)$ (see for example Analysis, Manifolds and Physics by Choquet-Bruhat, vol. 1, page 290). Others leave it separately, so that you end up with a group $SO^+(1,3)$, but this is a question of convention.

Now the Poincare group $P(1,3)$ (which I don't know any standard notation for) is the group of all Lorentz transformations together with all spacetime translations. In other words, we have:

$$P(1,3)= \{(a,\Lambda) : a\in \mathbb{R}^4, \Lambda \in SO(1,3)\}$$

together with the multiplication defined by

$$(a_1,\Lambda_1)\cdot (a_2,\Lambda_2)=(a_1+\Lambda_1 a_2, \Lambda_1\Lambda_2)$$

Think like that: while elements of $SO(1,3)$ relates orthonormal frames with coincident origins, elements of $P(1,3)$ also allows for the shift of origin as well.

The action of $SO(1,3)$ in the Minkowski vector space $\mathbb{R}^{1,3}$ (not to be confused with flat spacetime - this is actually the "model" for spacetime's tangent spaces, which just happens to be possible to identify with spacetime itself in the flat case) is given by usual matrix multiplication, i.e., given $ \Lambda \in SO(1,3)$ you have:

$$\Lambda \cdot v = \Lambda v, \quad \forall v\in \mathbb{R}^{1,3}.$$

On the other hand, the action of $P(1,3)$ on the Minkowski vector space $\mathbb{R}^{1,3}$ is characterized by the fact that given $(a,\Lambda)\in P(1,3)$ you have:

$$(a,\Lambda)\cdot v = a + \Lambda v, \quad \forall v\in \mathbb{R}^{1,3}.$$

I don't know if it helps, but people like to compare this to the case in $\mathbb{R}^3$ where you have the rotation group $SO(3)$ and the Euclidean group $E(3)$ comprising rotations in $SO(3)$ with translations in $\mathbb{R}^3$ thus forming the group of rigid motions. This could be seen as the analogous construction in spacetime.

EDIT: Regarding the semi-direct product construction mentioned in comments, recall that given groups $N,H$ with $\varphi : H\to \operatorname{Aut}(N)$ a homomorphism into the group of automorphisms of $N$, we can build the semi-direct product as the set $N\times H$ with the product:

$$(a,b)\cdot (c,d)=(a\varphi(b)(c), bd)$$

the resulting group is denoted $N\rtimes H$. In the particular case it is clear that we have this construction with $N = \mathbb{R}^{1,3}$, $H = SO(1,3)$ and $\varphi : SO(1,3)\to \operatorname{Aut}(\mathbb{R}^{1,3})$ given by $\varphi(\Lambda)(v) = \Lambda v$. Thus

$$P(1,3)=\mathbb{R}^{1,3}\rtimes SO(1,3)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.