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

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

*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
$$ (g_L,g_R)~~\sim~~(-g_L,-g_R), \qquad g_L,g_R~\in~SL(2,\mathbb{R})~:=~\{g\in {\rm Mat}_{2\times 2}(\mathbb{R}) \mid \det g~=~1\}.\qquad\tag{4}$$
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.

*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:


*

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

