My question is based on Qmechanic's answer here which proves that $O^+(m, 1)$ forms a group -- that if two Lorentz transformations have positive time-time co-ordinate, so does their product. The key is that with the Lorentz transformation written in the form:

$$\Lambda = \left[\begin{array}{cc}\Lambda_a & \Lambda_b^t \cr \Lambda_c &\Lambda_R \end{array} \right].$$

We can show that $|(\Lambda\tilde{\Lambda})_a-\Lambda_a\tilde{\Lambda}_a|\le \sqrt{(\Lambda_a^2-1)(\tilde{\Lambda}_a^2-1)}$ which implies that positive $\Lambda_a,\tilde{\Lambda_a}$ imply positive $(\Lambda\tilde{\Lambda})_a$.

Well, the trouble is that this uses the Cauchy-Schwarz inequality in Step 6, and therefore doesn't work for the general case of $O^+(m, n)$. How would one generalise the proof to prove the orthochronous indefinite orthogonal group $O^+(m, n)$ is a group?

Here's what I've tried so far: defining $O^{+}(m,n)$ as the subset of $O(m,n)$ with elements $\Lambda$ which satisfy $\det(\Lambda_a)>0$ (and in fact $\ge 1$),

1. As before, $(\Lambda\tilde{\Lambda})_a=\Lambda_a\tilde{\Lambda}_a+\Lambda_b^T\tilde{\Lambda}_c$.

2. From multiplying out $\Lambda^T\eta \Lambda=\eta$ and $\Lambda\eta \Lambda^T=\eta$, we see that $\Lambda_a^2-\Lambda_c^T\Lambda_c=\Lambda_a^2-\Lambda_b^T\Lambda_b=I$ and analogous for $\tilde{\Lambda}$.

3. So $\det\left((\Lambda\tilde{\Lambda})_a-\Lambda_a\tilde{\Lambda}_a\right)=\det\left(\Lambda_b^T\tilde{\Lambda}_c^T\right)=\sqrt{\det\left(\Lambda_a^2-I\right)\det\left(\tilde{\Lambda}_a^2-I\right)}$.

Well, I'm not sure how to proceed at this point. Does $\det(X-PQ)=\det((P^2-I)(Q^2-I))^{1/2}$ imply that $\det P\ge 1\land\det Q\ge 1\Rightarrow \det X>0$ in general?

The "topological proof" from Ron Maimon does not work either, as the orbit of the unit time vector is connected when $n>1$. I suspect that a more powerful technique than looking at the orbit of the unit time vector would be to look at the topology of the Lie group itself -- but I'm not that familiar with this stuff.