Simplicity of O(p,q) Let $\operatorname{O}(p,q)$ be the indefinite orthogonal group of signature $(p,q)$ (over  $\mathbb{R}$) and $\operatorname{O}_0(p,q)$ its identity component.
After doing a bit of research, I am given to understand that:

*

*As a Lie group, $\operatorname{O}(p,q)$ (equivalently $\operatorname{O}_0(p,q)$) is simple unless:

*

*if $p + q \leqslant 2$ because $\operatorname{O}_0(p,q)$ is abelian.

*$p=4$ and $q=0$ because $\operatorname{PSO}(4, \mathbb{R}) \approx \operatorname{SO}(3, \mathbb{R}) \times \operatorname{SO}(3, \mathbb{R})$.

*$p=q=2$ because $\operatorname{O}_0(2,2) \approx \operatorname{PSL}(2, \mathbb{R}) \times \operatorname{PSL}(2, \mathbb{R})$.



*It happens to be true that $\operatorname{O}_0(p,q)$ is simple as an abstract group if and only if it is simple as a Lie group (for any $p,q$). (In fact, when $p,q > 0$ and $p+q \geqslant 3$, there are analogous results for quadratic forms over any field.)

Did I get this right? Do you have a favorite reference for these facts?
 A: Yes this is right and there are a few ways to see this. Perhaps most simply we can appeal to the classification of simple Lie groups. The real classification is a little complicated so let's just assume we know the complex one. This tells us that $O(n,\mathbb{C})$ is simple for $n=3$ and $n>4$. As you say, $O(2,\mathbb{C})$ is abelian and $O(1,\mathbb{C})$ is discrete so neither are even semisimple and the same is true of their real forms $O(1,1)$, $O(2)$, $O(1)$.
For any complex simple Lie group, it's real forms are also simple (each $O(p,q)$ is a real form of $O(p+q,\mathbb{C})$). Therefore, $O(p,q)$ is simple if $p+q = 3$ or $p+q>4$. So we only need to worry about the last remaining case: the real forms of $O(4,\mathbb{C})$.
There are 3 of these (ignoring swapping p and q around which gives isomorphic groups) $O(4)$, $O(3,1)$ and $O(2,2)$. Now, $O(4,\mathbb{C})$ is semisimple so these are all semisimple but they could be simple as well. As you have said the first and the last of these are not simple but the middle one is. On a Lie algebra level:
$$ \mathfrak{so}(4) = \mathfrak{so}(3) \oplus \mathfrak{so}(3) = \mathfrak{su}(2) \oplus \mathfrak{su}(2) $$
$$ \mathfrak{so}(2,2) = \mathfrak{so}(2,1) \oplus \mathfrak{so}(2,1) = \mathfrak{sl}(2,\mathbb{R}) \oplus \mathfrak{sl}(2,\mathbb{R}) $$
These are the compact and split real forms respectively and correspond to taking the comapct and split real froms of each summand in $\mathfrak{so}(4,\mathbb{C}) = \mathfrak{so}(3,\mathbb{C}) \oplus \mathfrak{so}(3,\mathbb{C})$.
On the other hand:
$$ \mathfrak{so}(3,1) = \mathfrak{sl}(2,\mathbb{C}) $$
Where we view the right hand side as a Lie algebra over the reals.
Edit: One thing I should mention as well is that there is another real form of $O(4,\mathbb{C})$ not of the form $O(p,q)$. This is often denoted $O^*(4)$ and corresponds to the subgroup of $O(4)$ preserving a compatible quaternioic structure on $\mathbb{C}^4$. I believe it is not simple but I'm not sure.
