# Subgroup of Lorentz Group Generated by Boosts

It is common knowledge that a composition of boosts is not a boost, but involves a rotation. Further, in discussions of Thomas precession, it is often stated that boosting in $x$, then $y$, then back in $x$, then back in $y$ gives a rotation. If this is true, then spatial isotropy would suggest that any rotation can be written as a composition of boosts, and that therefore the Lorentz group is generated, as a group, by boosts. Is this true?

In most cases, the word "generator" is used in this context to mean "infinitesimal generator," but that is not what I am interested in. I mean group generators, which are often called topological generators in Lie theory. There are many examples to show that the number of topological generators of a Lie group can be less than the manifold dimension. Here are my specific questions:

1. Do boosts generate the Lorentz group as a group?
2. If not, what is the subgroup of the Lorentz group generated by boosts (i.e., the smallest subgroup containing all boosts?)
3. What is the minimal number of topological generators of the Lorentz group?
4. Where are these things written down?

EDIT: I mean the component of the Lorentz group containing the identity; i.e., the proper orthochronous Lorentz group or restricted Lorentz group.

• By "Where are these things written down?" are you looking for a particular reference (e.g., a book) in which these items could be found? Commented Jan 25, 2015 at 0:54
• Hi Kyle. Books, research papers, or whatever. I'm surprised I can't find this easily on google. After all, it's not too obvious to mention, since everyone is always introduced to the six standard generators. On the other hand, it's inconceivable that it wouldn't be well known.
– ben
Commented Jan 25, 2015 at 1:59
• Comment to the question (v3): Did you mean the restricted Lorentz group rather than the Lorentz group everywhere? Please then edit the question accordingly. Commented Feb 27, 2015 at 12:01
• Related: physics.stackexchange.com/q/525974/226902 (NOT about topological generators, but the "usual" notion of infinitesimal generators). Also: physics.stackexchange.com/q/480463/226902 and physics.stackexchange.com/q/643484/226902 Commented Oct 7, 2022 at 10:58

First Some Definitions Before I answer, let's first make a subtle nomenclature issue clear to check that I'm understanding your definitions: a set $$\mathcal{T}=\{X_1,\,X_2,\,\cdots,\,X_M\}$$ of Lie algebra vectors is called a set of topological generators for a group $$\mathfrak{G}$$ iff the closure of the smallest group containing elements of the form $$\exp(\tau\,X_j)$$ for $$\tau\in\mathbb{R}$$ and $$X_j\in\mathcal{T}$$ is the group $$\mathfrak{G}$$, i.e.

$$\mathfrak{G} = \overline{\left<\{e^{\tau\,X}|\,\tau\in\mathbb{R};,\,X\in\mathcal{T}\}\right>}$$

where, of course, there is implicitly some supergroup either equal to or containing $$\mathfrak{G}$$ whose group topology we use to define our notion of closure.

By this definition, the Lie algebra generator $$(i,\,i\alpha)$$ of the one-parameter irrational slope subgroup $$\{(e^{i\,\phi},\,e^{i,\,\alpha\,\phi})|\,\phi\in\mathbb{R}\}$$ of the torus $$\mathbb{T}^2=\{(e^{i\,\phi},\,e^{i,\,\theta})|\,\phi,\,\theta\in\mathbb{R}\}$$ when $$\alpha$$ is irrational is itself a topological generator of the whole torus.

Another definition which I think will be useful in this context is where $$\mathfrak{G}$$ is the smallest group containing the exponential of the smallest Lie algebra $$\mathfrak{g}$$ containing $$\mathcal{T}$$. This is a stronger concept that topological generation, and I haven't seen a name for it. Let us call it simply "connected immersed Lie subgroup generation" (CILSG) or "connected analytic subgroup generation" (there are some authors who do not call a Lie group that is immersed without being topologically embedded in a supergroup a "Lie subgroup" but rather an "analytic subgroup"). In this case $$\mathfrak{G}$$ is the group of all finite products of elements of the form $$\exp(\tau\,X_j)$$ for $$\tau\in\mathbb{R}$$ and $$X_j\in\mathcal{T}$$ is the group $$\mathfrak{G}$$:

$$\mathfrak{G} = \left\{\left.\prod\limits_{j=1}^M\,e^{\tau_j\,X_j}\right|\;X_j\in\mathcal{T};\;\tau_j\in\mathbb{R};\;M\in\mathbb{N}\right\}=\bigcup\limits_{k=1}^\infty\,\exp(\mathfrak{g}_s(\mathcal{T}))^k$$

where $$\mathfrak{g}_s(\cdot)$$ means "smallest Lie algebra containing the argument set of vectors".

By this definition, the CILSG of the the Lie algebra generator $$(i,\,i\alpha)$$ is simply the one parameter subgroup $$\{(e^{i\,\phi},\,e^{i,\,\alpha\,\phi})|\,\phi\in\mathbb{R}\}$$. We don't form its closure.

I shall only address connected immersed Lie subgroup generation (CILSG) in the following will not address strict topological generation further because the question of whether a particular subgroup is closed or not gets a bit messy and is highly particular to a given example. Basically what you need to do is test whether all the possible one parameter subgroups of a given group are topological embeddings. Then the Lie group in question is an embedding (this is much harder to apply than its sounds). See section 2.3 "Virtual Lie Subgroups" as well as Theorem 5.4 in:

V. V. Gorbatsevich, A. L. Onishchik and E. B. Vinberg, "Foundations of Lie theory and Lie transformation groups", Springer-Verlag, 1997

Instead, my CILSG definition is the image of the smallest Lie algebra $$\mathfrak{h}$$ containing the generators in question under what Rossmann calls the functor $$\Gamma$$ inverting $$\mathrm{Lie}$$ in the Lie Correspondence (the one to one correspondence between Lie subalgebras $$\mathfrak{h}\subset\mathfrak{g}$$ of the Lie algebra $$\mathfrak{g}$$ of a Lie group $$\mathfrak{G}$$ and its analytic (immersed Lie) subgroups. See Section 2.5 of

Wulf Rossmann, "Lie Groups: An Introduction Through Linear Groups", Oxford Graduate Texts in Mathematics

With these definitions, we have the following:

1. The smallest Lie algebra containing three linearly independent "infinitessimal" boosts is the whole of the Lie algebra $$\mathfrak{so}(1,\,3)$$. Accordingly, the group comprising finite products of members of the one parameter subgroups generated by three boosts in linearly independent directions is the whole of the identity connected component $$SO(1,\,3)$$ of the Lorentz group. In other words, it is the proper, orthochronous Lorentz group. In this case, the generators are also topological generators, because the closure of $$SO(1,\,3)$$ in $$O(1,\,3)$$ is $$SO(1,\,3)$$ itself.

2. The smallest Lie algebra containing two linearly independent "infinitessimal" boosts is the Lie algebra spanned by these two boosts together with an infinitessimal rotation within the plane spanned by the two directions (i.e. about an axis normal to this plane). Accordingly, the group comprising finite products of members of the one parameter subgroups generated by two boosts in linearly independent directions is an isomorphic copy of the connected component of the group $$SO(1,2)$$.

I cannot answer whether the closure of the group in 2. can ever be the whole of $$SO(1,\,3)$$; I suspect that it cannot be.

Where Are These Things Written Down?

Whereas it seems to be fairly widely known that the set of all finite products of exponentials of a given set of Lie algebra members is precisely the connected Lie group with the smallest Lie algebra containing the set under the Lie correspondence, this is a surprisingly hard piece of information to find written down: I too have had your experience.

I have written up a proof in section 2 of my paper below on ArXiv, but it has not yet been peer reviewed. I hope it shall be soon in the form of a book that I am publishing. But until that time, this is the best I can offer. It is not a hard or subtle proof, so I have high confidence in it and hopefully you'll be able to see what's going on too:

R. W. C. Vance, "Fine Entanglement and State Manipulation of Two Spin Coupled Qubits: A Lie Theoretic Overview" arXiv:1502.05200 [quant-ph]

If I am not mistaken, boosts only generate the connected part of identity of the Lorentz group (the restricted Lorentz group).

• Thanks akhmeteli... that's what I meant (now edited), but where can I find the original proof of this?
– ben
Commented Jan 25, 2015 at 2:08

Let us consider a closed subgroup of the (proper orthochronous) Lorentz group which includes all boost transformations. (We are actually interested in the smallest such subgroup, but what follows does not need this assumption.) As it is a closed subgroup, due to Cartan’s theorem, it is an embedded Lie subgroup of the Lorentz group. Hence its Lie algebra is a sub algebra of the Lorentz group Lie algebra which includes the three generators of the boosts and their commutators: the rotations around the three spatial axes. In summary, it also contains the generators of the rotations. The associated Lie subgroup is therefore the whole Lorentz group. This reasoning shows that the three families of boost transformations topologically generate the whole proper orthochronous Lorentz group.

The families of two linearly independent boosts generate a subgroup of transformations which only acts in the spatial plane spanned by the two generators (it being the identity transformation in the spatially orthogonal direction). This includes the rotations around the normal vector to that plane which pop out due to the polar decomposition theorem. Taking limits of sequences of these transformations, the action along the orthogonal direction remains the identity.

In summary, two families of boosts along independent spatial directions are not enough to topologically generate the whole Lorentz group.

As a consequence the minimal number of families of boost transformations necessary to reconstruct the whole Lorentz group is three: along three linearly independent spatial directions.