special relativity - How to prove that orthochronous Lorentz transformations $O^+(1,3)$ form a group?

Orthochronous Lorentz transform are Lorentz transforms that satisfy the conditions (sign convention of Minkowskian metric $+---$) $$ \Lambda^0{}_0 \geq +1.$$ How to prove they form a subgroup of Lorentz group? All books I read only give this result, but no derivation.

Why is this condition $ \Lambda^0{}_0 \geq +1$ enough for a Lorentz transform to be orthochronous?

The temporal component of a transformed vector is $$x'^0=\Lambda^0{}_0 x^0+\Lambda^0{}_1 x^1+\Lambda^0{}_2 x^2+\Lambda^0{}_3 x^3,$$ the positivity of $\Lambda^0{}_0$ alone does not seem at first glance sufficient for the preservation of the sign of temporal component.

And how to prove that all Lorentz transform satisfying such simple conditions can be generated from $J_i,\ K_i$?

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For those who think that closure and invertibility are obvious, keep in mind that $$\left(\bar{\Lambda}\Lambda \right)^0{}_0\neq \bar{\Lambda}^0{}_0\Lambda^0{}_0,$$ but instead $$\left(\bar{\Lambda}\Lambda \right)^0{}_0= \bar{\Lambda}^0{}_0\Lambda^0{}_0+\bar{\Lambda}^0{}_1\Lambda^1{}_0+\bar{\Lambda}^0{}_2\Lambda^2{}_0+\bar{\Lambda}^0{}_3\Lambda^3{}_0.$$

And I'm looking for a rigorous proof, not physical "intuition".

Answer

Let the Minkowski metric $\eta_{\mu\nu}$ in $d+1$ space-time dimensions be

$$\eta_{\mu\nu}~=~{\rm diag}(1, -1, \ldots,-1).\tag{1}$$

Let the Lie group of Lorentz transformations be denoted as $O(1,d;\mathbb{R})=O(d,1;\mathbb{R})$. A Lorentz matrix $\Lambda$ satisfies (in matrix notation)

$$ \Lambda^t \eta \Lambda~=~ \eta. \tag{2}$$

Here the superscript "$t$" denotes matrix transposition. Note that the eq. (2) does not depend on whether we use east-coast or west-coast convention for the metric $\eta_{\mu\nu}$.

Let us decompose a Lorentz matrix $\Lambda$ into 4 blocks

$$ \Lambda ~=~ \left[\begin{array}{cc}a & b^t \cr c &R \end{array} \right],\tag{3}$$

where $a=\Lambda^0{}_0$ is a real number; $b$ and $c$ are real $d\times 1$ column vectors; and $R$ is a real $d\times d$ matrix.

Now define the set of orthochronous Lorentz transformations as

$$ O^{+}(1,d;\mathbb{R})~:=~\{\Lambda\in O(1,d;\mathbb{R}) | \Lambda^0{}_0 > 0 \}.\tag{4}$$

The proof that this is a subgroup can be deduced from the following string of exercises.

Exercise 1: Prove that

$$ |c|^2~:= ~c^t c~ = ~a^2 -1.\tag{5}$$

Exercise 2: Deduce that

$$ |a|~\geq~ 1.\tag{6}$$

Exercise 3: Use eq. (2) to prove that

$$ \Lambda \eta^{-1} \Lambda^t~=~ \eta^{-1}. \tag{7}$$

Exercise 4: Prove that

$$ |b|^2~:= ~b^t b~ = ~a^2 -1.\tag{8}$$

Next let us consider a product

$$ \Lambda_3~:=~\Lambda_1\Lambda_2\tag{9}$$

of two Lorentz matrices $\Lambda_1$ and $\Lambda_2$.

Exercise 5: Show that

$$ b_1\cdot c_2~:=~b_1^t c_2~=~a_3-a_1a_2.\tag{10}$$

Exercise 6: Prove the double inequality

$$ -\sqrt{a_1^2-1}\sqrt{a_2^2-1} ~\leq~ a_3-a_1a_2~\leq~ \sqrt{a_1^2-1}\sqrt{a_2^2-1},\tag{11}$$

which may compactly be written as $$| a_3-a_1a_2|~\leq~\sqrt{a_1^2-1}\sqrt{a_2^2-1}.\tag{12}$$

Exercise 7: Deduce from the double inequality (11) that

$$ a_1\neq 0 ~\text{and}~ a_2\neq 0~\text{have same signs} \quad\Rightarrow\quad a_3>0. \tag{13}$$ $$ a_1 \neq 0~\text{and}~ a_2\neq 0~\text{have opposite signs} \quad\Rightarrow\quad a_3<0. \tag{14}$$

Exercise 8: Use eq. (13) to prove that $O^{+}(1,d;\mathbb{R})$ is stabile/closed under the multiplication map.

Exercise 9: Use eq. (14) to prove that $O^{+}(1,d;\mathbb{R})$ is stabile/closed under the inversion map.

The Exercises 1-9 show that the set $O^{+}(1,d;\mathbb{R})$ of orthochronous Lorentz transformations form a subgroup.$^{\dagger}$

References:

1. S. Weinberg, Quantum Theory of Fields, Vol. 1, 1995; p. 57-58.

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$^{\dagger}$A mathematician would probably say that eqs. (13) and (14) show that the map

$$O(1,d;\mathbb{R})\quad \stackrel{\Phi}{\longrightarrow}\quad \{\pm 1\}~\cong~\mathbb{Z}_2\tag{15}$$

given by

$$\Phi(\Lambda)~:=~{\rm sgn}(\Lambda^0{}_0)\tag{16}$$

is a group homomorphism between the Lorentz group $O(1,d;\mathbb{R})$ and the cyclic group $\mathbb{Z}_2$, and a kernel

$$ {\rm ker}(\Phi)~:=~\Phi^{-1}(1)~=~O^{+}(1,d;\mathbb{R}) \tag{17}$$

is always a normal subgroup.

For a generalization to indefinite orthogonal groups $O(p,q;\mathbb{R})$, see this Phys.SE post.