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

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.

• What is the definition of “orthochronous” when there are multiple temporal dimensions? – G. Smith Jul 30 '19 at 2:20
• @G.Smith $\det(\Lambda_a)>0$ (and in fact $\ge 1$). – Abhimanyu Pallavi Sudhir Jul 30 '19 at 5:17

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 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 $$P_{\tau\sigma}~=~{\rm diag}(\tau,\underbrace{1, \ldots, 1}_{n-2\text{ elements}},\sigma) ~=~\begin{bmatrix} \tau && \cr &\mathbb{1}_{(n-2)\times(n-2)} & \cr && \sigma \end{bmatrix}_{n \times n},$$ $$\qquad \tau,\sigma ~\in~\mathbb{Z}_2~:=~\{\pm 1\}, \tag{4}$$ 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. $$\Box$$

--

$$^1$$ 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.