Is the exponential map to the indefinite special orthogonal groups $SO^+(p,q)$ surjective?

Question

Is the exponential map to the identity component of the special indefinite orthogonal groups $$ \mathrm{exp} \colon \mathfrak{so}(p,q) \to SO^+(p,q)$$ surjective?

Answer 1

1. For the special orthogonal group $SO(d)$ and the restricted Lorentz group $SO^+(d,1)\cong SO^+(1,d)$, the exponential maps $$\begin{align}\exp: so(d)~~\longrightarrow~~& SO(d), \cr \exp: so(d,1)~~\longrightarrow~~& SO^+(d,1),\end{align}\tag{1} $$ are surjective. See also this related Phys.SE post and links therein.

2. Simplest counterexample. The exponential map $$\exp: sl(2,\mathbb{R})\oplus sl(2,\mathbb{R})~~\longrightarrow~~ SL(2,\mathbb{R})\times SL(2,\mathbb{R})\tag{2} $$ for the split group $$SO^+(2,2)~\cong~[SL(2,\mathbb{R})\times SL(2,\mathbb{R})]/\mathbb{Z}_2 \tag{3}$$ is not surjective. Here the $\mathbb{Z}_2$-action identifies $$\begin{align} (g_L,g_R)~~\sim~~&(-g_L,-g_R), \cr g_L,g_R~\in~& SL(2,\mathbb{R})~:=~\{g\in {\rm Mat}_{2\times 2}(\mathbb{R}) \mid \det g~=~1\}.\end{align}\tag{4}$$ One may show that a pair $$(g_L,g_R)~\in~SL(2,\mathbb{R})\times SL(2,\mathbb{R})\tag{5}$$ with $${\rm tr}(g_L)<-2\quad\text{and}\quad{\rm tr}(g_R)>2\tag{6}$$ (or vice-versa $L\leftrightarrow R$) is not in the image of the exponential map, even after $\Bbb{Z}_2$-modding.

3. More generally, one may prove for the indefinite orthogonal groups $SO^+(p,q)$, where $p,q\geq 2$, that the exponential map $$\exp: so(p,q)~~\longrightarrow~~ SO^+(p,q)\tag{7} $$ is not surjective, cf. e.g. Ref. 1.

References:

1. D.Z. Dokovic & K.H. Hofmann, Journal of Lie Theory 7 (1997) 171. The pdf file is available here.