Orthochronous indefinite orthogonal group $O^+(m, n)$ forms a group

Question

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.

Answer 1

1. Let us repeat that the orthochronous indefinite orthogonal group is defined as $$ O^+(p,q;\mathbb{R})~:=~\left\{\begin{bmatrix} a & b \cr c& d\end{bmatrix}\in O(p,q;\mathbb{R})\mid \det(a)>0\right\}.\tag{1}$$ OP wants to check that this is indeed a subgroup. i.e. that it is closed/stabile under the multiplication & inversion maps.

2. OP is right: A generalization of the algebraic proof in my Phys.SE answer here does not seem feasible. Instead let us (as OP already suggested) use the non-trivial$^1$ topological fact that the indefinite orthogonal group $$O(p,q;\mathbb{R})~=~\sqcup_{\tau,\sigma \in \mathbb{Z}_2} C_{\tau\sigma}, \qquad n~:= p+q, \qquad p,q ~\geq ~1,\tag{2} $$ has 4 connected components $$ \begin{align}C_{\tau\sigma} &~=~P_{\tau\sigma} \cdot SO^+(p,q;\mathbb{R}) \cr &~=~\left\{\begin{bmatrix} a & b \cr c& d\end{bmatrix}\in O(p,q;\mathbb{R})\mid {\rm sgn}\det(a)=\tau ~\wedge~{\rm sgn}\det(d)=\sigma \right\},\end{align}\tag{3} $$ which are labelled by $2\times 2=4$ elements $$\begin{align} P_{\tau\sigma}~=~{\rm diag}(\tau,\underbrace{1, \ldots, 1}_{n-2\text{ elements}},\sigma) ~=~&\begin{bmatrix} \tau && \cr &\mathbb{1}_{(n-2)\times(n-2)} & \cr && \sigma \end{bmatrix}_{n \times n}, \cr \tau,\sigma ~\in~&\mathbb{Z}_2~:=~\{\pm 1\}, \end{align}\tag{4}$$ of the Klein 4-group $\mathbb{Z}_2\times\mathbb{Z}_2$.

3. We are here using the non-trivial$^2$ fact that the restricted indefinite orthogonal group $SO^+(p,q;\mathbb{R})$ is path-connected, i.e. any restricted element can be continuously connected to the identity element $\mathbb{1}_{n\times n}$. Since multiplication (and taking inverses) are continuous operations, there must be a group homomorphism $$O(p,q;\mathbb{R})\qquad \stackrel{\Phi}{\longrightarrow}\qquad \mathbb{Z}_2\times\mathbb{Z}_2.\tag{5}$$ In other words, the multiplication table of connected components is dictated by the Klein 4-group.

4. We conclude that the orthochronous indefinite orthogonal group $$ O^+(p,q;\mathbb{R})~=~C_{++}\sqcup C_{+-} \tag{6}$$ corresponds to the subgroup $\{1\}\times \mathbb{Z}_2$ of the Klein 4-group, and is hence itself a subgroup. This answers OP's title question. $\Box$

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$^1$ The fact that it has at least 4 connected components is trivial, since $$a^ta-\mathbb{1}_{p\times p}~=~c^tc~\geq~ 0\qquad\text{and}\qquad d^td-\mathbb{1}_{q\times q}~=~b^tb~\geq~ 0\tag{7}$$ are semi-positive matrices, so that $$|\det(a)|\geq 1\qquad\text{and}\qquad|\det(d)|\geq 1.\tag{8}$$

$^2$ Note for starters that the exponential map $\exp: so(p,q;\mathbb{R})\to SO^+(p,q;\mathbb{R})$ is not surjective if $p,q\geq 2$, cf. e.g. this Math.SE post.