Universal Cover of the Lorentz Group in $n$ Dimensions I'm interested in learning some Lorentzian geometry and Spin geometry. From what I've heard, the idea is to lift the psuedo-orthonormal frame bundle, an $O(1,n)$-bundle, to a double cover, a principal bundle with group that perhaps we'll call $Spin(1,n).$ 
Now, in the Riemannian case, this double cover $Spin(n)$ is simply connected. Does this hold for $Spin(1,n),$ and while we're at it, more general $Spin(k,l)$? 
All I can seem to find on the internet is $n = 3$ stuff, in which case $Spin(1,3) \simeq SL(3, \mathbb{C}).$ 
Idea: $Spin(k,l)$ can be constructed inside a Clifford algebra, just like $Spin(n).$ I think it will be simply connected. But how to show this...?
Many thanks in advance.
 A: Unlike the definite case, the indefinite special orthogonal group $SO(k, l)$ is neither connected nor compact. It has two connected components; the connected component of the identity is denoted $SO^+(k, l)$. The maximal compact subgroup of $SO^+(k, l)$ is $SO(k)\times SO(l)$, so $\pi_1(SO^+(k, l)) \cong \pi_1(SO(p))\times\pi_1(SO(q))$.
Theorem $2.10$ of Lawson & Michelsohn's Spin Geometry states that there is an exact sequence
$$0 \to \mathbb{Z}_2 \to \operatorname{Spin}(k, l) \to SO(k, l) \to 0.$$
Furthermore, if $(k, l) \neq (1, 1)$, the two-sheeted coverings are non-trivial over each connected component of $SO(k, l)$.
It follows that $\operatorname{Spin}(k, l)$ is disconnected. One often takes connectedness as part of the definition of a simply connected space, but lets suppose we just want to know whether each connected component is simply connected. As the connected components of a topological group are homeomorphic, it is enough to check whether one is simply connected
First, lets deal with the case $(k, l) = (1, 1)$. Here $\operatorname{Spin}(1, 1) \cong SO(1, 1)\times\mathbb{Z}_2$. The connected component of the identity is $SO^+(1, 1)\times\{0\}$ which deformation retracts onto $SO(1)\times SO(1)\times\{0\} \cong *$ which is simply connected.
For $(k, l) \neq (1, 1)$, the topological group $\operatorname{Spin}(k, l)$ has two connected components; let $\operatorname{Spin}^+(k, l)$ denote the connected component of $\operatorname{Spin}(k, l)$ mapping onto $SO^+(k, l)$. As the two-sheeting covering $\operatorname{Spin}^+(k, l) \to SO^+(k, l)$ is non-trivial, $\operatorname{Spin}^+(k, l)$ is simply connected if and only if the fundamental group of $SO^+(k, l)$ is $\mathbb{Z}_2$ which is the case if and only if $k = 1$ and $l \geq 3$, or $k \geq 3$ and $l = 1$.
In conclusion, the indefinite spin groups $\operatorname{Spin}(k, l)$ are always disconnected, but the connected components are simply connected if and only if $(k, l) = (1, 1)$, $k = 1$ and $l \geq 3$, or $k \geq 3$ and $l = 1$.
