Why are discontinuous Lorentz transformations excluded from the Poincare group?

[redtree]

TL;DR Summary

The restricted Lorentz group

The full Lorentz group includes discontinuous transformations, i.e., time inversion and space inversion, which characterize the non-orthochronous and improper Lorentz groups, respectively. However, these groups are excluded from the Poincare group, in which only the proper, orthochronous Lorentz group is included. From my understanding, the exclusion of the discontinuous Lorentz transformations is necessary for the Poincare group to be considered a Lie Group with a Lie algebra as represented by the generators.

My question is two-fold: 1) Is my understanding correct that the exclusion of the discontinuous Lorentz transformations from the Poincare group is done so that the Poincare group can be represented as generators of a Lie algebra? 2) Do other distinct reasons exist for excluding the discontinuous Lorentz transformations from the Poincare group?

 

[vanhees71]

That's correct. Only the part of the Poicare group that is smoothly connected with the unit element forms a Lie group.

This proper orthochronous Poincare group is the symmetry group of special-relativstic spacetime, and all special-relativistic dynamical laws must be Poincare invariant.

Indeed the other disconnected pieces of the full Poincare group are not symmetries of Nature since the weak interaction breaks space reflection P an time-reversal invariance T (and the charge-conjugation symmetry, which is added to the discrete trafos within relativistic quantum field theory). Today it has been experimentally demonstrated that the weak interaction violates all the discrete symmetries: P, T, C, CP, CT, and PT. Only the combination CPT is a symmetry as predicted by quantum field theory.

 

[redtree]

Your response leads to a related question that I have about the Poincare group. The question involves distinguishing between a basis transformation and the evolution of a system. The Poincare group restricts transformations to those that are continuous and preserve the norm of the 4-vector. For basis (or frame) transformations, it seems clear that all transformations should preserve the norm. However, the generators of the Lie Algebra of the Poincare group are often used to describe how particular particles or fields evolve, thereby restricting the evolution of state configurations to those that preserve the norm.

Clearly, the evolution of states might preserve the norm, but why would one assume that the evolution of states necessarily does so. In other words, it seems that a frame transformation and the evolution of a system are being treated as if they are equivalent.

For example, in quantum mechanical terms, all frame transformations should obviously preserve the norm of the 4-vector. Thus, for example, frame transformations of $E_0$ and $\vec{p}$ should preserve the norm of the 4-momentum. But why should the evolution of quantum mechanical system, for example in the case of the Schrödinger equation for a single nonrelativistic particle in one dimension, limit values for 3-momentum $\vec{p}=\hbar \vec{k}$ for a given energy $E_0$, such that the norm 4-momentum is invariant? In position terms, one can also consider a similar restriction to preserve 4-vector invariance. Given an invariant interval between events in position space, every time interval would limit the possible positions of the system to those that preserve the 4-space interval. That type of determinism seems inconsistent with the probabilistic nature of quantum mechanics.

Thanks!

 

[vanhees71]

First of all one should be aware that the only consistent relativistic quantum theory is quantum field theory, because when dealing with collisions with relativistic energy exchange there's always some probability to destroy and create particles, i.e., you need a formalism which describes such processes, and for the quantum field theory is the most convenient way.

Further there is a clear particle interpretation only for asymptotic free states, i.e., you describe a scattering process as starting with usually two particles prepared far away from each other such that interactions between them can first be neglected. Then they scatter and interact and after this collision process there fly out other particles which then are also far from each other such that their interaction can be neglected.

To describe free, non-interacting particles you look for the unitary representations of the proper orthochronous Poincare group (or since we deal with quantum theory with its covering group, where the Lorentz subgroup is substituted with its covering group $\mathrm{SL}(2,\mathbb{C})$, leading to the posibility of particles with half-integer spin).

The useful irreducible representations together with the assumption of the existence of a stable ground state and "microcausality", which ensures that there cannot be faster-than-light causality violating influences, then lead to either massive representations, where the energy-momentum eigenvalues fulfill the "onshell condition" $(E^2/c^2-\vec{p}^2=m^2 c^2$, describing massive particles or massless representations where $E^2/c^2-\vec{p}^2=0$, describing massless field quanta (like photons, which are the quanta of the electromagnetic field).

The representations are formulated in terms of field operators, which fulfill relativistic field equations like the Klein-Gordon equation (for particles with spin 0), the Dirac equation (for particles with spin 1/2), and vector fields which in the standard model are the gauge fields describing the strong and the electro-weak interactions. For free particles all these field equations imply the on-shell condition for energy-momentum eigensolutions. The field operators transform under Poincare transformations as the analogous classical fields transform, i.e., "locally". E.g. a vector Field transforms as
$$A^{\prime \mu}(x')={\Lambda^{\mu}}_{\nu} A^{\nu}(x)={\Lambda^{\mu}}_{\nu} A^{\nu}[\Lambda^{-1}(x-a)].$$
This, together with the microcausality conditions and that the energy spectrum should be bounded from below implies the existence of antiparticles (which may be identical with the particles if one has strictly neutral particles), the spin-statistics relation (integer-spin particles are bosons, half-integer-spin particles are fermions), and the symmetry under CPT transformations.

For more on the approach to the derivation of the field equations for relativistic QFT, see

https://itp.uni-frankfurt.de/~hees/publ/lect.pdf

 

[redtree]

To describe free, non-interacting particles you look for the unitary representations of the proper orthochronous Poincare group (or since we deal with quantum theory with its covering group, where the Lorentz subgroup is substituted with its covering group $\mathrm{SL}(2,\mathbb{C})$, leading to the posibility of particles with half-integer spin).

Thanks for the response. As you note, the Poincare group models the evolution of certain free, non-interacting particles, such that, in the absence of external influences and given a stable wavefunction, there is no difference between a basis transformation and the evolution of the particle. Of course, this implies that the evolution of the particle respects the invariance of the 4-vector (or the rest mass) and that this 4-vector invariance holds in both position and momentum space. In position space, 4-position invariance means that observation of the particle at a given time interval restricts its 3-position such that the 4-position remains invariant. In other words, the spacetime distance between events for such an evolving particle remains invariant.

 

[vanhees71]

Formally the Poincare invariance means that the field operators transform as the corresponding classical fields to under (proper orthochronous) Poincare transformations. Another constraint is microcausality, i.e., local operators, that represent observables, should commute with the Hamilton density at space-like separated spacetime-arguments. Then the Poincare invariance of S-matrix elements is guaranteed.

 

[George Jones]

Summary:: The restricted Lorentz group

The full Lorentz group includes discontinuous transformations, i.e., time inversion and space inversion, which characterize the non-orthochronous and improper Lorentz groups, respectively. However, these groups are excluded from the Poincare group, in which only the proper, orthochronous Lorentz group is included. From my understanding, the exclusion of the discontinuous Lorentz transformations is necessary for the Poincare group to be considered a Lie Group with a Lie algebra as represented by the generators.

My question is two-fold: 1) Is my understanding correct that the exclusion of the discontinuous Lorentz transformations from the Poincare group is done so that the Poincare group can be represented as generators of a Lie algebra? 2) Do other distinct reasons exist for excluding the discontinuous Lorentz transformations from the Poincare group?

I am having a little trouble parsing "necessary for the Poincare group to be considered a Lie Group with a Lie algebra as represented by the generators"

The full Poincare group is a semi-direct product of the full Lorentz group with spacetime translations. There is no requirement that a Lie group needs to be connected, and both the full Lorentz group and and full Poincare group are Lie groups. The Lie algebra of a Lie group can be identified with the tangent space of the identity, which is in one of the connected components of the full Poincare Lie group. Parity, which is not in this connected component, acts on Dirac spinors.

 

[vanhees71]

The reason, why we deal more frequently with Dirac spinors than Weyl spinors indeed is that the strong and the electromagnetic interaction are space-reflection invariant and to describe space reflections you need both Weyl-spinor representations $(1/2,0) \oplus (0,1/2)$, which then leads to the Dirac field.

 

[redtree]

Both the full Lorentz group and and full Poincare group are Lie groups.

Time and spatial inversion are discontinuous transformations. What is the tangent of a discontinuity?

 

[George Jones]

I think you are a bit confused with respect to continuity and Lie groups. I understand your confusion, but I do not know if I can explain it very well. Also, what do you mean by\

Time and spatial inversion are discontinuous transformations.

Here goes.

Let $G$ be the full Lorentz group considered as a group of $4\times4$ matrices. Each element of $G$ is a mapping from $\mathbb{R}^4$ to $\mathbb{R}^4$.

Let $P$ be the element of $G$ that has $\left(1,-1-,1,-1\right)$ on the diagonal, and zeros elsewhere. $I$ and $P$ are in different connected components of $G$. As a mapping that is a linear transformation, $P : \mathbb{R}^4 \rightarrow \mathbb{R}^4$ is continuous, but this is irrelevant for the Lie group structure of $G$.

In order for $G$ to be a Lie group, $G$ must be a topological space, and it is group multiplication and inversion that must be continuous, not the group elements. (also calculus must be possible).

Group multiplication is a mapping $G \times G \rightarrow G$. It is this mapping that must be continuous, not the elements of $G$. This mapping is continuous iff the inverse image of every open set in $G$ is an open set in $G \times G$.

For example, consider an open set $H$ in $G$ that contains the element $P$. In turns out that (though I will not show), with respect to the group multiplication map $G \times G \rightarrow G$, the inverse image of $H$ is an open subset of $G \times G$. There no problem with discontinuity of group multip;lication, even when the element $P$ is involved.

 

[dextercioby]

I do not understand where questioning the topology of the Lorentz of the Poincaré group comes from. The full Lorentz group is $\text{O}(1,3)$ and is obviously a Lie group. Then its semidirect product with $(\mathbb{R}^4, +)$ is obviously a Lie group.

 

[George Jones]

Time and spatial inversion are discontinuous transformations.

Okay, I think that this is what you mean; following on from my previous post #10.

The full Lorentz group (and thus the full Poincare group) is not path connected. In particular, there does not exist a continuous map $\gamma$ from the real interval $\left[0,1\right]$ in into $G$ such that $\gamma\left(0\right) = I$ and $\gamma\left(1\right) = P$, but this has no bearing on the continuity of group multiplication and inversion, which is what is necessary for $G$ to be a topological group.

 

[vanhees71]

I do not understand where questioning the topology of the Lorentz of the Poincaré group comes from. The full Lorentz group is $\text{O}(1,3)$ and is obviously a Lie group. Then its semidirect product with $(\mathbb{R}^4, +)$ is obviously a Lie group.

But the space-reflection matrix is not even continuously connected to the identity. I also always thought that only $\mathrm{SO}(1,3)^{\uparrow}$ is a Lie group. On the other hand, if you are only interested on the structure as a differentiable manifold in a local sense, then the entire Lorentz group can be seen as a Lie group, and you can define the tangent space and Lie algebra also around the space-reflection element.

 

[George Jones]

But the space-reflection matrix is not even continuously connected to the identity. I also always thought that only $\mathrm{SO}(1,3)^{\uparrow}$ is a Lie group.

Then, e.g., $O\left(3\right)$ and $GL\left(n,\mathbb{R}\right)$ also would also not be Lie groups, as none of these are connected.

A topological (Lie) group is a topological space (differentiable manifold) such that group multiplication and inversion are continuous (smooth). Neither connectedness nor Lie algebras play direct roles in this definition. Intuitively, continuity and smoothness are local properties.

In particular, it is quite possible for the group operations to be continuous (smooth) for disconnected spaces., i,e., a Lie group can be a topological space that is not connected. See, for example, exercise 1 on page 147 of Sexl and Urbantke. Also, from page 157 of the second edition of Modern Differential Geometry for Physicists by Chris Isham: "is a Lie group, and is denoted $O\left(p,q;\mathbb{R}\right)$".

On the other hand, if you are only interested on the structure as a differentiable manifold in a local sense, then the entire Lorentz group can be seen as a Lie group, and you can define the tangent space and Lie algebra also around the space-reflection element.

Yes, but why is this necessary?

 

[vanhees71]

I'm not a mathematician. So maybe I misunderstood the strict definition of a Lie group. I always thought each group element must be smoothly connected with the identity. I guess I remember something wrong.

 

[samalkhaiat]

There are three (real) Lorentz (Lie) groups all having the same Lie algebra:

1) The full Lorentz group $O(1,3)$. It is the group of all invertible linear maps of $\mathbb{R}^{4}$ that preserves the Minkowski metric: $$O(1,3) = \Big\{ \Lambda \in GL(4 , \mathbb{R}) : \Lambda^{T}\eta \Lambda = \eta \Big\}.$$ As a manifold, $O(1,3)$ is not connected because there are two branches corresponding to $\mbox{det}(\Lambda) = \pm 1$. Spatial orientation is preserved only on the $\mbox{det}(\Lambda) = + 1$ branch.

2) The proper Lorentz group $$SO(1,3) = \Big\{ \Lambda \in O(1,3): \mbox{det}(\Lambda) = +1 \Big\} .$$

3) The proper orthochroneous Lorentz group $$SO^{\uparrow} (1,3) = \Big\{ \Lambda \in SO(1,3): \Lambda^{0}{}_{0} \geq +1 \Big\} .$$

Note that $\Lambda^{T}\eta \Lambda = \eta$ also implies that $|\Lambda^{0}{}_{0}| \geq 1$. This means that the full Lorentz group $O(1,3)$ consists of 4 disconnected pieces:

i) The proper orthochroneous component $\mathscr{L}_{+}^{\uparrow}$ which consists of matrices $\{ \Lambda \}$ satisfying $\mbox{det}(\Lambda) = +1, \ \Lambda^{0}{}_{0} \geq +1$. In fact, the component $\mathscr{L}_{+}^{\uparrow} = SO^{\uparrow} (1,3)$ is the only subgroup of $O(1,3)$, the identity component.

ii) The improper orthochroneous component $$\mathscr{L}_{-}^{\uparrow} = \big\{ \Lambda : \mbox{det}(\Lambda) = -1, \ \Lambda^{0}{}_{0} \geq +1 \big\} ,$$ which consists of matrices of the form $\Lambda_{P} SO^{\uparrow} (1,3)$.

iii) The improper non-orthochroneous component $$\mathscr{L}_{-}^{\downarrow} = \big\{ \Lambda : \mbox{det}(\Lambda) = -1, \ \Lambda^{0}{}_{0} \leq -1 \big\} ,$$ consisting of the matrices of the form $\Lambda_{T} SO^{\uparrow} (1,3)$.

iv) The proper non-orthochroneous component $$\mathscr{L}_{+}^{\downarrow} = \big\{ \Lambda : \mbox{det}(\Lambda) = +1, \ \Lambda^{0}{}_{0} \leq -1 \big\} ,$$ with matrices of the form $\Lambda_{PT} SO^{\uparrow} (1,3)$, where $\Lambda_{PT} = \Lambda_{P}\Lambda_{T}$ takes $x^{\mu} \to - x^{\mu}$. And we have the following disjoint unions $$O(1,3) = SO^{\uparrow} (1,3) \cup \Lambda_{P} SO^{\uparrow} (1,3) \cup \Lambda_{T} SO^{\uparrow} (1,3) \cup \Lambda_{PT} SO^{\uparrow} (1,3) ,$$ $$SO(1,3) = SO^{\uparrow} (1,3) \cup \Lambda_{PT} SO^{\uparrow} (1,3).$$

One can show that $SO^{\uparrow} (1,3)$ is a principal $SO(3)$-bundle over the space $H_{3}^{+}(1)$ which consists of one sheet of the hyperbolic paraboloid $H_{3}(1)$. That is, as topological spaces, we have $$SO^{\uparrow} (1,3) / SO(3) \cong H^{+}_{3}(1).$$ Since $H^{+}_{3}(1)$ is non-compact, it follows that $SO^{\uparrow} (1,3)$ is also non-compact. But, $H^{+}_{3}(1)$ is connected, so the connectivity of $SO^{\uparrow} (1,3)$ follows from the connectivity of $SO(3)$. Indeed, as a manifold $SO^{\uparrow} (1,3)$ is connected but not simply-connected.

 

[PeterDonis]

@redtree, why did you post this question in the High Energy, Nuclear, and Particle Physics forum? Normally we would expect a thread like this to be in the relativity forum. Is there some particular particle physics issue that is driving your question?

 

[PeterDonis]

Moderator's note: A subthread that was shifting topic to derivations of the Lorentz transformation has been moved to a separate thread here:

https://www.physicsforums.com/threa...ity-in-1-postulate-derivations-of-sr.1011358/

 

[redtree]

I'm not a mathematician. So maybe I misunderstood the strict definition of a Lie group. I always thought each group element must be smoothly connected with the identity. I guess I remember something wrong.

This was my understanding as well.

See: https://encyclopediaofmath.org/wiki/Lie_group

 

[redtree]

@redtree, why did you post this question in the High Energy, Nuclear, and Particle Physics forum? Normally we would expect a thread like this to be in the relativity forum. Is there some particular particle physics issue that is driving your question?

I am interested in understanding when the Poincare group is used not only to model a basis transformation but also the evolution of a system such as in the case of free non-interacting particles, see post #5.

Which actually raises a related question (which I was holding either for another thread or until the present question was fully elucidated): why is there no (statistical) entropy effect on the wavefunction for an evolving system? As a system evolves, entropy should increase and a well-formulated probability function should reflect this change in entropy.

See for example: https://www3.nd.edu/~lent/pdf/nd/Quantum_Operator_Entropies__PRE_final.pdf

When utilizing the Poincare group as a model of an evolving system, there is an assumption that the probability amplitude function remains the same, and it is this constancy (among other things) that allows in special cases, such as the evolution of non-interacting particles, a basis transformation and a system's evolution to be treated as equivalent.

 

[George Jones]

This was my understanding as well.

See: https://encyclopediaofmath.org/wiki/Lie_group

The link that you give says the opposite of this.

 

[redtree]

Per post #16, perhaps my confusion was that for every Lie group, I thought there exists a single set of generators that represents the Lie algebra for the entire group. However, groups with disconnected sub groups, such as $O(1,3)$ with subgroups classified by $det \pm 1$ and $\Lambda_0^0 > 1 \lor <1$, cannot have a single set of generators spanning the entire group.

 

[George Jones]

Per post #16, perhaps my confusion was that for every Lie group, I thought there exists a single set of generators that represents the Lie algebra for the entire group. However, groups with disconnected sub groups, such as $O(1,3)$ with subgroups classified by $det \pm 1$ and $\Lambda_0^0 > 1 \lor <1$, cannot have a single set of generators spanning the entire group.

There are Lie groups that are vectors spaces, e.g., $\mathbb{R}^n$, but most Lie groups are not vector spaces, e.g., $O\left(n\right)$. What, then, does "spanning the entire group" mean? In any case, the generators are not even elements of the Lie group.

 

[redtree]

What, then, does "spanning the entire group" mean? In any case, the generators are not even elements of the Lie group.

Given a Lie algebra, one can use it as infinitesimal generators to construct group elements near the identity, then by repeated multiplication push the group elements further and further away from the identity, and eventually integrate it out to cover the whole group.

If subgroups of a larger group are disconnected, can the same generators be used to construct all the elements of all subgroups?

 

[springbottom]

Given a Lie algebra, one can use it as infinitesimal generators to construct group elements near the identity, then by repeated multiplication push the group elements further and further away from the identity, and eventually integrate it out to cover the whole group.

If subgroups of a larger group are disconnected, can the same generators be used to construct all the elements of all subgroups?

Specifically, the construction is via the exponential map (taking Lie Algebra ~ tangent space at identity and exponentiating to get elements in the Lie Group). It isn't necessary that this exponential map is surjective even in the connected case (https://math.stackexchange.com/questions/301504/on-surjectivity-of-exponential-map-for-lie-groups)

But you are right, that in the case the Lie Group is disconnected (e.g. the 'full' Poincare group), the exponential map will only ever get you elements connected to the identity. If you include "P" and "T" transformations (these are discrete), then these can map you to the other components of the Poincare Group.

 

[George Jones]

Specifically, the construction is via the exponential map (taking Lie Algebra ~ tangent space at identity and exponentiating to get elements in the Lie Group). It isn't necessary that this exponential map is surjective even in the connected case (https://math.stackexchange.com/questions/301504/on-surjectivity-of-exponential-map-for-lie-groups)

Some further results:

Any element of a connected matrix Lie group can be written as a finite product of exponentials of Lie algebra elements,

For any connected compact Lie group the exponential map is surjective, but not injective,

 

[samalkhaiat]

Per post #16,... such as $O(1,3)$ with subgroups classified by $det \pm 1$ and $\Lambda_0^0 > 1 \lor <1$, cannot have a single set of generators spanning the entire group.

1) There are no subgroups. I have already said in #16, $O(1,3)$ has one, and only one, Lie subgroup, and that is $SO^{\uparrow}(1,3)$. This is because $SO^{\uparrow}(1,3)$ is the only component of $O(1,3)$ that contains the group identity element $e$ (i.e., the identity matrix $e = I_{4}$). The identity element $I_{4}$ is not contained in anyone of the other components $\Lambda_{P}SO^{\uparrow}(1,3)$, $\Lambda_{T}SO^{\uparrow}(1,3)$ and $\Lambda_{PT}SO^{\uparrow}(1,3)$. So these components are not groups, let alone Lie subgroups. In general, if $G$ is a real matrix Lie group, then only a closed subgroup of $G$ is a smooth manifold. This is the reason why $O(1,3)$ and $SO^{\uparrow}(1,3)$ share the same Lie algebra $\mathfrak{o}(1,3) = \mathfrak{so}(1,3)$: any smooth (time-like) curve in $O(1,3)$ passing through the identity element must lie entirely in $SO^{\uparrow}(1,3)$ because the determinant (which is continuous in $\mathbb{R}^{4 \times 4}$) cannot jump from $+1$ to $-1$. This means that $SO^{\uparrow}(1,3)$ and $O(1,3)$ share the same tangent space at the identity, $T_{e}\left(SO^{\uparrow}(1,3)\right) = T_{e}\left(O(1,3) \right)$, i.e., the same Lie algebra $\mathfrak{o}(1,3) = \mathfrak{so}(1,3)$.

2) Of course, the exponential map $\mbox{exp}: \mathfrak{o}(1,3) \to O(1,3)$ is well-defined but it is not surjective because there are matrices in $O(1,3)$ with determinant $-1$, whereas (symbolically) $$\mbox{det}\left( e^{\mathfrak{o}(1,3)}\right) \equiv \mbox{det}\left( e^{\mathfrak{so}(1,3)}\right) = e^{\mbox{Tr}\left( \mathfrak{so}(1,3)\right)} = 1.$$ So, exponentiation (of the algebra $\mathfrak{o}(1,3) = \mathfrak{so}(1,3)$) leads to the group $SO^{\uparrow}(1,3)$ instead of $O(1,3)$. Of course, that is not a problem because the other 3 components of $O(1,3)$ can be obtained by multiplying $SO^{\uparrow}(1,3)$ with $\Lambda_{P}, \ \Lambda_{T}$ and $\Lambda_{PT}$.

3) For every Lie group $G$, the definition $\mathfrak{g} = T_{e}(G)$ gives uniquely a Lie algebra. Given a Lie algebra, however, there may be more than one Lie group for which it is the Lie algebra. And, in infinite dimensions, the group may not exist. For example, it is known that no Lie group corresponds to the complexified Virasoro algebra (without central extension). Groups with the same Lie algebra are said to be locally isomorphic. Examples of locally isomorphic groups : $SO(3)$ and $SU(2)$; $SO(4)$ and $SU(2) \times SU(2)$; $SO^{\uparrow}(1,3)$ and $SL(2 , \mathbb{C})$. All locally isomorphic groups can be obtained by factoring the simply connected covering group by various discrete normal subgroups of its fundamental group. For example $SO^{\uparrow}(1,3) \cong SL(2, \mathbb{C}) / \mathbb{Z}_{2}$.

4) Finally, I leave you to think about the following

Exercise (i). In the textbooks, you often read the following sentence: “The full Lorentz group $O(1,3)$ is a real Lie group consisting of four disconnected components (or pieces)”. What does the word “disconnected” mean in that sentence, when we know for sure that the component $SO^{\uparrow}(1,3)$ is a real and connected Lie group?

Exercise (ii). Show that the exponential map $\mbox{exp} : \mathfrak{so}(1,3) \to SO^{\uparrow}(1,3)$ is well-defined and “surjective”. Or, if you don’t understand the mathematical jargons, show that any Lorentz matrix $\Lambda \in SO^{\uparrow}(1,3)$ can be written as $$\Lambda = \mbox{exp} \left( \frac{1}{2}\omega^{\rho \sigma}M_{\rho \sigma} \right),$$ where $\omega^{\mu\nu} = - \omega^{\nu\mu}$ are the real parameters, and the matrices $M_{\rho \sigma}$, whose elements are given by $$(M_{\rho \sigma})^{\mu}{}_{\nu} = \delta^{\mu}_{\rho} \ \eta_{\sigma \nu} - \delta^{\mu}_{\sigma} \ \eta_{\rho \nu} ,$$ form a basis of the Lie algebra $\mathfrak{so}(1,3)$ of the Lorentz group $SO^{\uparrow}(1,3)$, i.e., $$\big[M_{\mu\nu} , M_{\rho \sigma} \big] = \eta_{\mu \sigma} M_{\nu\rho} - \eta_{\mu\rho} M_{\nu\sigma} + \eta_{\nu\rho} M_{\mu\sigma} - \eta_{\nu\sigma} M_{\mu\rho} .$$

 

[vanhees71]

But $\mathrm{O}(1,3)$ as $\mathrm{SO}(1,3)$ (the proper Lorentz group including the "grand reflection" $PT$, which is not in the part connected to the identity and $\mathrm{SO}(1,3)^{\uparrow}$ (the proper orthochronous Lorentz group, and that's the symmetry group of Nature).