Let $O(k,l)$ be the orthogonal group associated to the quadratic form $q$ on $\mathbb{R}^{k+l}$ with signature $(k,l)$. Let $O^+_+(k,l)$ be the connected component of the identity, i.e. the connected components of the elements in $O(k,l)$ which maintein the orientation on each maximal positive subspace $W\subset \mathbb{R}^{k+l}$ and $W^\perp$. I want to prove that $O^+_+(k,l)$ is connected.

I've read the references of the wiki page http://en.wikipedia.org/wiki/Indefinite_orthogonal_group but no one seems to really prove it.

So i thought that maybe we can apply the same argument of the proof of the connectedness of $SO(k)$. I consider the hyperboloid $q(x)=1$ for $x\in \mathbb{R}^{k+l}$ and I have to prove that $O^+_+(k,l)$ acts transitively on one connected component of the hyperboloid (this seems legit to me, although I'm not sure about how many connected components the hyperboloid has). Then the stabilizer of a point is of the form \begin{pmatrix} 1 & 0 & \cdots & 0 \\ 0 & a_{1,1} & \cdots & a_{1,k+l-1} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & a_{k+l-1,1} & \cdots & a_{k+l-1,k+l-1} \end{pmatrix} where \begin{pmatrix} a_{1,1} & \cdots & a_{1,k+l-1} \\ \vdots & \ddots & \vdots \\ a_{k+l-1,1} & \cdots & a_{k+l-1,k+l-1} \end{pmatrix} is in $O^+_+(k-1,l)$ which is connected by induction hypothesis. And so also $O^+_+(k,l)$ is connected.

But i don't know how to find out how many connected components the hyperboloid has and how to prove that $O^+_+$ acts transitively on one of them.. Do you have any ideas?

  • $\begingroup$ In the case of $SO(k)$, the conic $q(x)=1$ is a sphere. $\endgroup$
    – cactus314
    Feb 6, 2013 at 17:06
  • $\begingroup$ yeah, i know that, but this is not what i need. $\endgroup$
    – ciccio
    Feb 6, 2013 at 17:19

1 Answer 1


Do you recall how the proof that $SO(k)$ is connected goes? There's a few. One goes like this. $SO(k)$ fibers over $S^{k-1}$ with fiber $SO(k-1)$. The base is connected, so by induction connectedness boils-down to arguing that $SO(1)$ is connected, but this is a point. Similarly, you can argue that $O(k)$ has precisely two components by this induction argument.

There are arguments that use elementary matrix operations as well, but in my opinion they work better for proving $GL_n^+$ is connected.

Let's adapt the first argument to your case. We'll show $O(k,l)$ has precisely four components provided $k,l \geq 1$, which is equivalent to your problem.

Given a "timelike" vector in $\mathbb R^{k+l}$, $O(k,l)$ preserves the fact that it's timelike (by design), and preserves the length of the vector. So $O(k,l)$ acts on the unit timelike vectors. Moreover, this action is transitive. The proof that the action is transitive boils down to arguing that any unit timelike vector can be complemented to an orthonormal basis of $\mathbb R^{k+l}$, consisting of $k$ unit timelike and $l$ unit spacelike vectors, respectively. So that gives a bundle $O(k,l) \to H(k,l)$ where $H(k,l) \subset \mathbb R^{k+l}$ is the unit timelike vector subspace. This is a hyperboloid of some kind, and it is connected precisely when $k \geq 2$ and $l \geq 1$. Moreover, the fiber of this fibration is $O(k-1,l)$. Since we already know $O(0,l)$ has precisely two components $O(0,l) = O(l)$, by induction we're done.


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