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The exponential map for the restricted Lorentz group is surjective. An outline of why is shown on the wiki page Representation Theory of the Lorentz Group.

Is there a more general theorem that states that for some class of Lie groups or Riemannian manifolds (which includes the restricted Lorentz group), the exponential map is surjective?

There is a theorem stating that compact, connected Lie groups have surjective exponential maps. But as the restricted Lorentz group is not compact, this isn't applicable.

Qmechanic
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I Dunno
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  • Great question. I have wondered this myself many times. A while back I stumbled across these notes http://www.cis.upenn.edu/~cis610/cis61005sl8.pdf, and from the discussion there, it seems to me implied that the answer to the question is (at least currently) no, since otherwise I would think the author of the notes would have commented on such a theorem. – joshphysics Jan 23 '15 at 06:37
  • @Qmechanic, Folland's Quantum Field Theory book says that the restricted and proper Lorentz groups are the same thing. – I Dunno Jan 28 '15 at 22:02
  • Hm, you are right: on p. 9 Folland does say that. Note however that e.g. Wikipedia and Goldstein call $SO(1,d)$ proper and $SO^+(1,d)$ restricted. – Qmechanic Jan 28 '15 at 22:12

2 Answers2

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  1. The consensus in the literature seems to be that the surjectivity of the exponential map $$\tag{1}\exp: so(1,d;\mathbb{R}) \to SO^+(1,d;\mathbb{R})$$ for the restricted Lorentz group for general spacetime dimensions $D=d+1$ does not have a short proof.

  • The case $d=1$ is trivial.

  • The case $d=2$ can be proved via the isomorphism $SO^+(1,2;\mathbb{R})\cong SL(2,\mathbb{R})/\mathbb{Z}_2$, cf. e.g. this Phys.SE post.

  • The case $d=3$ can be proved via the isomorphism $SO^+(1,3;\mathbb{R})\cong SL(2,\mathbb{C})/\mathbb{Z}_2$, cf. e.g. Wikipedia and this Phys.SE post.

  1. Already the exponential map $\exp: sl(2,\mathbb{R}) \to SL(2,\mathbb{R})$ is not surjective, cf. e.g. this MO.SE answer and this Phys.SE post. Note that the Lie algebras $$\tag{2}so(1,2;\mathbb{R}) ~\cong~ sl(2,\mathbb{R}) $$ are isomorphic, but only the Lie group $SO^+(1,2;\mathbb{R})$ for the lefthand-side of the isomorphism (2) has a surjective exponential map; not the Lie group $SL(2,\mathbb{R})$ for the righthand-side. A counterexample such as (2) undoubtedly makes it delicate to try to formulate a generalization of (1) beyond the restricted Lorentz groups $SO^+(1,d;\mathbb{R})$ and case-by-case-proofs. See also this Math.SE post.
Qmechanic
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    Potentially interesting reference: 1. Andrew Baker, Matrix Groups: An Introduction to Lie Group Theory; Chapter 6. – Qmechanic Dec 14 '15 at 16:27
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@Qmechanic: I believe there are problems with Baker's discussion of the surjectivity in "Matrix groups". Here is a quote from Jean Gallier and Jocelyn Quaintance's lecture notes at U Penn: (http://www.seas.upenn.edu/~jean/diffgeom.pdf)

"We warn our readers about Chapter 6 of Baker’s book [16]. Indeed, this chapter is seriously flawed.The main two Theorems (Theorem 6.9 and Theorem 6.10) are false, and as consequence, the proof of Theorem 6.11 is wrong too. Theorem 6.11 states that the exponential map exp: so(n,1) → SO0(n,1) is surjective, which is correct, but known proofs are nontrivial and quite lengthy (see Section 6.2). The proof of Theorem 6.12 is also false, although the theorem itself is correct (this is our Theorem 6.17, see Section 6.2). The main problem with Theorem 6.9 (in Baker) is that the existence of the normal form for matrices in SO0(n,1) claimed by this theorem is unfortunately false on several accounts. Firstly, it would imply that every matrix in SO0(n, 1) can be diagonalized, but this is false for n ≥ 2. Secondly, even if a matrix A ∈ SO0(n,1) is diagonalizable as $A = PDP^{−1}$, Theorem 6.9 (and Theorem 6.10) miss some possible eigenvalues and the matrix P is not necessarily in SO0(n,1) (as the case n = 1 already shows). For a thorough analysis of the eigenvalues of Lorentz isometries (and much more), one should consult Riesz [146] (Chapter III)".

Frederic Thomas
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mike stone
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