When is $SO(m,n)$ simple as a Lie group? What are the Zariski and Euclidean components?

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Let $SO(m,n)=\operatorname{SO}(m,n)(\mathbb{R})$ denote the real $(m+n) \times (m+n)$ matrices, with determinant $1$, which preserve the quadratic form $x_1 + \cdots + x_m - x_{m+1} \cdots - x_{m+n}$ on $\mathbb{R}^{m+n}$. I am only interested in the case where $m+n \geq 3$ and $m,n \geq 1$.

On page 430 in Introduction to Arithmetic Groups Dave Witte Morris (free copy at http://deductivepress.ca/) it says that (its identity component) is simple if $m+n \neq 4$.

I was wondering whether it is actually simple (as a Lie group) when $m+n=4$ (with $SO(4)$ being the only case where it is not?).

Also, is the identity component actually always index $2$ in my case (and the only exceptions being $SO(n)$). The book only says it is at most $2$? If so, is it true that this index-$2$ component is not Zariski closed (i.e $SO(m,n)$ is always Zariski connected)?

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For Connectedness issues: See Chapter 1 section 3 problems 9 and 10 of the Onischchik-Vinberg book, available here http://people.maths.ox.ac.uk/drutu/tcc2/onishchik-vinberg.pdf

For simplicity: It seems that they are all simple except for $SO(2,2)$, see Chapter 5 Section 1.6 of the Onischchik-Vinberg book for the classification of Real simple Lie algebras.